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http://mathoverflow.net/questions/807/describing-the-universal-covering-map-for-the-twice-punctured-complex-plane/118602
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## Describing the universal covering map for the twice punctured complex plane
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As is well known, the universal covering space of the punctured complex plane is the complex plane itself, and the cover is given by the exponential map.
In a sense, this shows that the logarithm has the worst monodromy possible, given that it has only one singularity in the complex plane. Hence we can easily visualise the covering map as given by the Riemann surface corresponding to log (given by analytic continuation, say).
Seeing how fundamental the exponential and logarithm are, I was wondering how come I don't know of anything about the case when two points are removed from the complex plane.
My main question is as follows: how can I find a function whose monodromy corresponds to the universal cover of the twice punctured complex plane (say ℂ∖{0,1}), in the same way as the monodromy of log corresponds to the universal cover of the punctured plane.
For example, one might want to try f(z) = log(z) + log(z-1) but the corresponding Riemann surface is easily seen to have an abelian group of deck transformations, when it should be F2.
The most help so far has been looking about the Riemann-Hilbert problem; it is possible to write down a linear ordinary differential equation of order 2 that has the required monodromy group.
Only trouble is that this does not show how to explicitly do it: I started with a faithful representation of the fundamental group (of the twice punctured complex plane) in GL(2,ℂ) (in fact corresponding matrices in SL(2,ℤ) are easy to produce), but the calculations quickly got out of hand.
My number one hope would be something involving the hypergeometric function 2F1 seeing as this solves in general second order linear differential equations with 3 regular singular points (for 2F1 the singular points are 0, 1 and ∞, but we can move this with Möbius transformations), but I was really hoping for something much more explicit, especially seeing as a lot of parameters seem to not produce the correct monodromy. Especially knowing that even though the differential equation has the correct monodromy, the solutions might not.
I'd be happy to hear about any information anyone has relating to analytical descriptions of this universal cover, I was quite surprised to see how little there is written about it.
Bonus points for anything that also works for more points removed, but seeing how complicated this seems to be for only two removed points, I'm not hoping much (knowing that starting with 3 singular points (+∞), many complicated phenomena appear).
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## 6 Answers
Others have already given a satisfactory qualitative description as a modular function under a suitable congruence group. Since the quotient in question is necessarily genus zero, there are explicit formulas for such functions.
I was mistaken in my comment to Tyler's answer. The function I provided there is invariant under a larger group than the one we want, and yields the universal cover for the complex plane with one and a half punctures.
The Dedekind eta product eta(z/2)^8/eta(2z)^8 is not only invariant under Gamma(2) (= F2), but it maps H/Gamma(2) bijectively to the twice punctured plane. You will have to post-compose with a suitable affine transformation to move the punctures to zero and one. An alternative description is: eta(z)^24/(eta(z/2)^8 eta(2z)^16).
This function arises in monstrous moonshine: eta(z)^8/eta(4z)^8 + 8 is the graded character of an element of order four, in conjugacy class 4C in the monster, acting on the monster vertex algebra (a graded vector space with some extra structure). It is invariant under Gamma0(4), which is what you get by conjugating Gamma(2) under the z -> 2z map.
Other elements of the monster yield functions that act as universal covers for planes with specific puncture placement and orbifold behavior. For the general case of more than 2 punctures, you have to use more geometric methods, due to nontrivial moduli. I think you idea of using hypergeometric functions is on the right track. I think Yoshida's book, Hypergeometric Functions, My Love has a few more cases worked out.
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
The function you want is, as David mentioned, a modular function. It appears in almost any proof of the Little Picard Theorem (I believe that Ahlfors' text on complex analysis has some discussion) and can be constructed using abstract nonsense by finding a conformal mapping of the domain
`1 > Re(z) > 0, |z-1/2| > 1/2`
onto the upper half-plane and then bouncing it around using Schwarz reflection. More concretely the group of Gamma(2) of 2x2 integer matrices that reduce to zero mod 2 acts freely (except for minus the identity matrix) on the upper half-plane by linear fractional transformations, and this gives an explicit action of the free group on two generators by deck transformations.
This leads into modular functions and forms which are, to put it mildly, well-studied.
Most introductory texts on modular forms usually have a thorough discussion and proof of this covering - for example, Goro Shimura's "Introduction to the Arithmetic Theory of Automorphic Functions".
EDIT: whoops, if I'm going to say "Schwarz reflection" I want half of the fundamental domain.
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Aha! That's what screwed me up. I had the mental image of half the fundamental domain in my head, and that lead me to the wrong group. By the way, if someone would vote Tyler's answer up, I'd appreciate it. He definitely deserves to be above me, since he got the details right and I didn't. – David Speyer Oct 17 2009 at 5:18
Very nice. I believe the correct modular function is called the Lambda modular function; it is indeed described in Ahlfors' book on complex analysis. Does this idea work for free groups on more generators? I don't know much about the algebraic properties of congruence subgroups, but it definitely seems plausible as the upper half plane is going to be the universal cover again. Although the fundamental domains are going to look pretty complicated. Is there a computer package or something that deals with this (algebra of congruence subgroups and plots of corresponding fundamental domains)? – Sam Derbyshire Oct 17 2009 at 15:39
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Incidentally if you use the z -> 2z map to replace \Gamma(2) with the isomorphic group \Gamma_0(4), the map is given by the trace of an element of order 4 (conjugacy class 4A) acting on the monster vertex algebra. It can be expressed (up to a constant shift of 24) as \Delta(2z)^2/(\Delta(z)\Delta(4z)). If you want covering maps for free groups with more generators, you will have to account for moduli, i.e., the maps aren't necessarily taken to each other by Mobius transformations. – S. Carnahan♦ Oct 17 2009 at 23:30
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I think MAGMA and SAGE can work with congruence subgroups. H. Verrill has written a java program that draws pictures of fundamental domains. – S. Carnahan♦ Oct 17 2009 at 23:35
The function you want is a modular function. The universal cover of C \ {0,1} is the upper half plane; a fundamental domain is { z : 0 < Re(z) < 1, |z-1/2| > 1/2 }. This is also a fundamental domain for the action of \Gamma_0(2) on the upper halfplane, where \Gamma_0(2) is the group of integer matrices whose lower left hand entry is even. There is a standard construction of a modular function which has this symmetry; but I'm forgetting the terminology. Scott will probably come along soon and fill in the details I'm missing.
A good reference for this sort of thing is Conformal Mapping, by Zeev Nehari.
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Isn't the fundamental domain -1 < Re(z) < 1, |z - 1/2| > 1/2, |z + 1/2| > 1/2, using Gamma(2) level structures (integer matrices which are the identity mod 2)? – Tyler Lawson Oct 17 2009 at 4:03
You are right, I am wrong. It is $\Gamma(2)$ which is isomorphic to the free group on 2 generators, not $\Gamma_0(2)$. If you want to write an answer explaining in more detail, I'll gladly vote it up. – David Speyer Oct 17 2009 at 4:12
Ok good, I was getting worried, but didn't dare to say anything. I started thinking of modular forms as soon as I realised the covering space was the upper half plane, but I didn't really see much point in it. I tried coming up with holomorphic modular functions never attaining 0 or 1, but that obviously failed. I'm very interested in hearing about how it goes in this case though; I love modular forms! – Sam Derbyshire Oct 17 2009 at 4:43
I think you're looking for polylogarithms. It's not just one function you're after; the fundamental group of the once-punctured plane is cyclic and doesn't have a ton of interesting representations, while the fundamental group of the twice-punctured plane is free of rank two and thus has a lot more beef.
Note, though, that the polylogarithms all have unipotent monodromy; you should think of them as seeing, not quite the full fundamental group of C - 0,1, but the pro-unipotent algebraic envelope of this. But as we learned from Deligne's monograph on the fundamental group of P^1 minus three points, there's a huge amount of content even here.
Not sure what reading to recommend for an introduction to this story, except for Deligne's paper itself, but if you search for things containing some combination of polylogarithm, multizeta, and iterated integral you'll find tons; among these, find something that suits your taste....
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Hey why not look at this paper?
http://www2.math.uu.se/research/pub/Jonzon1.pdf
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2
The paper is 58 pages long and does not answer the question, so there is no reason to look at it. – Misha Nov 14 at 11:07
I am sorry to revive such an old topic. But it really needs to be mentioned that this universal cover of the twice punctured plane was constructed by PICARD, and as one of the comments says, this was used in the proof of little Picard (and also big Picard). This is one of Picard's most famous constructions, so you will excuse my emphasizing it. This is explicitly constructed in Ahlfors' book on complex analysis.
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http://openwetware.org/index.php?title=User:Pranav_Rathi/Notebook/OT/2010/08/18/CrystaLaser_specifications&oldid=482001
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# User:Pranav Rathi/Notebook/OT/2010/08/18/CrystaLaser specifications
### From OpenWetWare
Revision as of 00:27, 22 December 2010 by Pranav Rathi (Talk | contribs)
## Specifications
We are expecting our laser any time. To know the laser more we are looking forward to investigate number of things. These specifications are already given by the maker, but we will verify them.
### Polarization
Laser is TM (transverse magnetic) or P or Horizontal linearly polarized. We investigated these two ways: 1) by putting a glass interface at Brewster’s angle and measured the reflected and transmitted power. At this angle all the light is transmitted because the laser is P-polarized, 2) by putting a polarizing beam splitter which uses birefringence to separate the two polarizations; P is reflected and S is transmitted, by measuring and comparing the powers, the desired polarizability is determined. We performed the experiment at 1.8 W where P is 1.77 W and S is less than .03 W*
### Beam waist at the output window
We used knife edge method (this method is used to determine the beam waist (not the beam diameter) directly); measure the input power of 1.86W at 86.5 and 13.5 % at the laser head (15mm). It gave us the beam waist (Wo) of .82mm (beam diameter =1.64mm).
### Possible power fluctuations if any
The power supply temperature is really critical. Laser starts at roughly 1.8 W but if the temperature of the power supply is controlled very well it reaches to 2 W in few minutes and stay there. It’s really stupid of manufacturer that they do not have any fans inside so we put two chopper fans on the top of it to cool it and keep it cool. If no fans are used then within an hour the power supply reaches above 50 degrees of Celsius and then, not only the laser output falls but also the power supply turns itself off after every few minutes.
### Mode Profile
Higher order modes had been a serious problem in our old laser, which compelled us to buy this one. So mode profiling is critical; we want our laser to be in TEM00. I am not going to discuss the technique of mode profiling; it can be learned from this link: [1] [2].
As a result it’s confirmed that this laser is TEM00 mode. Check out the pics:
## Specs by the Manufacturer
All the laser specs and the manual are in the document: [Specs[3]]
## Beam Profile
The original beam waist of the laser is .2mm, but since we requested the 4x beam expansion option, the resultant beam waist is .84 at the output aperture of the laser. As the nature of Gaussian beam it still converges in the far field. We do not know where? So there is a beam waist somewhere in the far field. There are two ways to solve the problem; by using Gaussian formal but, for that we need the beam parameters before expansion optics and information about the expansion optics, which we do not have. So the only way we have, is experimentally measure the beam waist along the z-axis at many points and verify its location for the minimum. Once this is found we put the AOM there. So the experimental data gives us the beam waist and its distance from the laser in the z-direction. We use scanning knife edge method to measure the beam waist.
### Method
• In this method we used a knife blade on a translation stage with 10 micron accuracy. The blade is moved transverse to the beam and the power of the uneclipsed portion is recorded with a power meter. The cross section of a Gaussian beam is given by:
$I(r)=I_0 exp(\frac {-2r^2}{w_L^2})$
Where I(r) is the Intensity as function of radius (distance in transverse direction), I0 is the input intensity at r = 0, and wL is the beam radius. Here the beam radius is defined as the radius where the intensity is reduced to 1/e2 of the value at r = 0. This can be seen by letting r = wL.
setup
Power Profile
The experiment data is obtained by gradually moving the blade across from point A to B, and recording the power. Without going into the math the intensity at the points can be obtained. For starting point A
$\mathbf{I_A(r=0)}=I_0 exp(-2)=I_0*.865$
For stopping point B
$\mathbf{I_B}=I_0 *(1-.865)$
By measuring this distance the beam waist can be measured and beam diameter is just twice of it:
$\mathbf{\omega_0}=r*.135-r*.865$
this is the method we used below.
• Beam waist can also be measured the same way in terms of the power. The power transmitted by a partially occluding knife edge:
$\mathbf {p(r)}=\frac{P_0}{\omega_0} \sqrt{\frac{2}{\pi}} \int\limits_r^\infty exp(-\frac{2r^2}{\omega^2}) dr$
After integrating for transmitted power:
$\mathbf {p(r)}=\frac{P_0}{2}{erfc}(2^{1/2}\frac{r}{\omega_0})$
Now the power of 10% and 90% is measured at two points and the value of the points substituted here:
$\mathbf{\omega_0}=.783(r*.1 - r.9)$
The difference between the methods is; the first method measures the value little higher than the second method (power), but the difference is still under 13%. So either method is GOOD.
#### Data
We measured the beam waist at every 12.5, 15 and 25mm, over a range of 2000mm from the output aperture of the laser head. The measurement is minimum at 612.5 mm from the laser, thus the beam waist is at 612.5±12.5mm from the laser. And it is to be 1.26±.1 mm.
#### Plot
Here the plot beam diameter Vs Z is presented. Experimental data is presented as blue and model is red. As it can be seen that model does not fit the data. Experimental beam expands much faster than the model; this proves that the beam waist before the expansion optics must be relatively smaller. To make a perfect model(using Gaussian optics formula) we would need the beam parameters before the expansion optics as i already mentioned.
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http://mathhelpforum.com/calculus/45338-continuity.html
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Thread:
1. continuity
Let $f: \bold{R} \to \bold{R}$ be the function $f(x) := \begin{cases} 1 \ \text{if} \ x \in \bold{Q} \\ 0 \ \text{if} \ x \not \in \bold{Q} \end{cases}$. Is this function continuous at any real number $x_0$?
It has no limit because $L = \lim_{n \to \infty} f(1/n) = \lim_{n \to \infty} 1 = 1$. Whereas $L = \lim_{n \to \infty} f(\sqrt{2}/n) = \lim_{n \to \infty} 0 = 0$. In other words, we can find both a rational and irrational sequence converging to $x_0$? Hence $f$ is not continuous at any real number $x_0$?
Is this correct?
2. No it's not correct. The function is not continuous at any point. (you were right on this).
And for
It has no limit because .
$\lim_{n\to \infty} f(\frac{1}{n})$ doesn't exist. It's equivalent to write $\lim_{x \to 0^+} f(x)$. When x tends to 0, f(x) doesn't tend to anything, it get the values 0 and 1 again and again and it doesn't tend to 1 as you said.
And about
In other words, we can find both a rational and irrational sequence converging to ? Hence is not continuous at any real number ?
I think the implication doesn't work. For example if you have the function $f(x)=x$, then you could find both a rational and irrational sequence converging to, say 3. But the functions is continuous at any point. So be careful.
3. Originally Posted by arbolis
No it's not correct. The function is not continuous at any point. (you were right on this).
And for $\lim_{n\to \infty} f(\frac{1}{n})$ doesn't exist. It's equivalent to write $\lim_{x \to 0^+} f(x)$. When x tends to 0, f(x) doesn't tend to anything, it get the values 0 and 1 again and again and it doesn't tend to 1 as you said.
And about I think the implication doesn't work. For example if you have the function $f(x)=x$, then you could find both a rational and irrational sequence converging to, say 3. But the functions is continuous at any point. So be careful.
It does exist (the limits of the two sequences) in the context of this question. $(1/n)_{n=0}^{\infty}$ is a sequence of rationals, thus $\lim_{n \to \infty} f(1/n) = 1$. Whereas, $(\sqrt{2}/n)_{n=0}^{\infty}$ is a sequence of irrationals. Thus $\lim_{n \to \infty} f(\sqrt{2}/n) = 0$. Both these sequences converge to $0$. But $1 \neq 0$, and so the limit at $0$ does not exist.
4. You are right, sorry. I didn't realize $\frac{1}{n}$ was rational.
So
In other words, we can find both a rational and irrational sequence converging to ? Hence is not continuous at any real number ?
should be correct in your case.
5. Originally Posted by particlejohn
$\lim_{n \to \infty} f(1/n) = 1$. Whereas, $(\sqrt{2}/n)_{n=0}^{\infty}$ is a sequence of irrationals. Thus $\lim_{n \to \infty} f(\sqrt{2}/n) = 0$. Both these sequences converge to $0$. But $1 \neq 0$, and so the limit at $0$ does not exist.
You have the right idea. Here is a way to sharpen your result.
If $r \in \mathbb{R}$ is any real number then $r$ is the limit of a sequence of rational numbers, $\left( {Q_n } \right) \to r$, and is also the limit of a sequence of irrational numbers, $\left( {I_n } \right) \to r$ now as you have noted $\left( {f\left( {I_n } \right)} \right) \to 0\,\& \,\left( {f\left( {Q_n } \right)} \right) \to 1$.
Since $r$ is either rational or irrational $f$ cannot be continuous at $r$.
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http://mathoverflow.net/questions/55006/n-th-roots-of-pythagorean-numbers/55056
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## n-th roots of Pythagorean numbers
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $F$ be the field ${\mathbb Q}(i)\subset \mathbb C$ and let $T\subset F$ be the set of all elements of complex absolute value 1. Let $n$ be a natural number $\ge 2$ and let $\mu_n(T)\subset\mathbb C$ be the set of all $n$-th roots of elements of $T$. Finally, let $E=F(\mu_n(T))$.
Question: Is the field extension $E/F$ finite or infinite?
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Since $1$ is in $T$, the n-th roots of unity $\mu_n(1)$ are in $E$. Thus $E$ contains the field of all roots of unity, hence is infinite. – Franz Lemmermeyer Feb 10 2011 at 12:13
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oops - n is supposed to be fixed. The answer is still that $E$ is infinite because the group of rational points on the unit circle is not finitely generated. I'll provide references later. – Franz Lemmermeyer Feb 10 2011 at 12:20
@Franz: If you could show that the group $T$ is an direct sum of infinitely many copies of $\mathbb Z$, that would prove that the extension is infinite. Otherwise it could be that $T$ is not finitely generated, but contains the $n$-th roots nevertheless, at least all but finitely many. – anton Feb 10 2011 at 12:59
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For every prime $p$ in $\mathbb{Z}$ that splits in $\mathbb{Z}[i]$ as $q\bar{q}$, $T$ contains $q/\bar{q}$. All such elements generate an infinite dimensional free subgroup in $T$. – Yaakov Baruch Feb 10 2011 at 14:28
## 2 Answers
Tan (The group of rational points on the unit circle, Math. Mag. 69 (1996), 163-171) proved that the group of rational points on the unit circle modulo torsion is isomorphic to infinitely many copies of $\mathbb Z$.
I have given a couple of references to related articles in Kreise und Quadrate modulo $p$, Math. Semesterber. 47 (2000), 51-73.
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@Franz : I guess you mean infinitely many copies of Z. Recently I came across the following unpublished note by Elkies : math.harvard.edu/~elkies/Misc/hilbert.pdf There he shows that the mentioned result is a nice application of Hilbert's Theorem 90. – François Brunault Feb 10 2011 at 20:58
@Francois: thanks. – Franz Lemmermeyer Feb 11 2011 at 6:58
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
(This may have errors - I'm not an algebraic number theorist.)
We have a complete description of the multiplicative structure of $F = \mathbb{Q}(i)$. It is: $$\mathbb{Q}(i)^\times \cong \left( \bigoplus_{p \cong 1 \mod 4} (\mathbb{Z} \oplus \mathbb{Z}) \right) \oplus \left( \bigoplus_{\text{other } p} \mathbb{Z} \right) \oplus \mathbb{Z}/4\mathbb{Z}.$$ Note that there are no $n$-divisible subgroups, except the 4th roots of unity (when $n$ is odd). This is a good sign of infinite degree.
Each prime $p$ congruent to 1 mod 4 can be written as product of primes $(a+ib)(a-ib)$, with $a$ and $b$ unique up to obvious symmetries. We find that $\frac{a+ib}{a-ib} \in T$, and is a primitive element in the copy of $\mathbb{Z} \oplus \mathbb{Z}$ in the big sum corresponding to $p$. In particular, it is not an $n$th power for $n \geq 2$.
We can now construct a sequence of fields $F=F_0 \subset F_1 \subset \dots$, where $F_k$ is given by starting with $F_{k-1}$, and adjoining an $n$th root of the number $\frac{a+ib}{a-ib}$ corresponding to some prime congruent to 1 mod 4 over which $F_{k-1}$ is unramified. Since finite extensions are ramified over finitely many primes, and adjoining the $n$th root creates ramification over $p$, we have strict containment at each step, and the chain does not terminate after finitely many steps. The union of the chain is an infinite degree extension that is contained in $E$, so $E$ has infinite degree over $F$.
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First error! I should have said that in realizing the description of $\mathbb{Q}(i)^\times$, we fixed a choice of generators for the splittings of the primes $p$ congruent to 1 mod 4. Otherwise, the number $\frac{a+ib}{a-ib}$ is only a well-defined primitive element in the copy of $\mathbb{Z} \oplus \mathbb{Z}$ after we quotient by the copy of $\mathbb{Z}/4\mathbb{Z}$ that makes the 4th roots of unity. – S. Carnahan♦ Feb 10 2011 at 18:31
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http://math.stackexchange.com/questions/16146/is-the-knaster-tarski-fixed-point-theorem-constructive
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Is the Knaster-Tarski Fixed Point Theorem constructive?
According to Tarski's Fixed Point Theorem, for a complete lattice $L$, and monotone function $f:L \rightarrow L$, the set of fixed points of $f$ forms complete lattice.
Definition of $lfp(f)$ and $gfp(f)$ is given in terms of set $\text{Red}(f)$ and $\text{Ext}(f)$. Where $\text{Red}(f)$ is set of all those points $l' \in L$, which satisfy $f(l') \le l'$. Similarly $\text{Ext}(f)$ is set of all those points $l'' \in L$, which satisfy $f(l'') \ge l''$.
$lfp(f) = \bigcap(\text{Red}(f))$ and $gfp(f) = \bigcup(\text{Ext}(f))$
I read somewhere that these definitions are not constructive in nature, so to reach least fixed point we try to find out least upper bound of the sequence $(f^n(\bot))_n$, i.e. $f(\bot) \sqcup f(f(\bot)) \sqcup f(f(f(\bot))) \dots$,
(Here $\bot$ is least element of lattice $L$ ).
Can someone please explain it further what do we mean by constructive definition here, and how these two definitions of $lfp(f)$ coincide?
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@user: I tried to format it. \meet seems to be giving error. Please replace with what you intended. – Aryabhata Jan 2 '11 at 19:24
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shouldn't the meet be a join? – Yuval Filmus Jan 2 '11 at 19:47
Yes you are right , I changed it to join operator. @Moron thanks for formatting the post. I will take care of it when posting next time. – chinu Jan 3 '11 at 5:05
2 Answers
The first thing to know about the word "constructive" is that it means too many things. There are so many sorts of "constructive mathematics" that significant context is needed to tell which one is meant.
That being said, the clearest sense in which the definitions you have above are nonconstructive is that they are "impredicative". This means that, when you look at the definition $\operatorname{lfp}(f) = \bigcap \operatorname{Red}(f)$, you see that the element $\operatorname{lfp}(f)$ is already a member of $\operatorname{Red}(f)$. In ordinary mathematics as it is practiced, this is just a curiosity, but for a time there was concern about the impredicative definitions.
The first motivation for this concern was Russell's paradox: if we seek to define "the set of all sets that do not contain themselves", we see that the set we are defining is already a member of the collection of "all sets" that we quantify over, and moreover this impredicative definition leads to a paradox. This led Russell to be interested in predicativism (the program of developing mathematics without impredicative definitions. Later Weyl in Das Kontinuum showed that is is possible to develop a significant amount of mathematics in a predicative way; this work has been further extended and clarified by the reverse mathematics program of Friedman and Simpson.
One way to conceptualize the difference between predicative and impredicative definitions (not formal, but informative) is that impredicative definitions often "identify" something that already exists, while predicative definitions directly "construct" it. For example, you could define the supremum of a nonempty bounded set of real numbers as the least element of the set of all upper bounds - this impredicative definitions identifies which upper bound is the supremum, but it doesn't help you actually construct it. Or you could directly construct a Cauchy sequence whose limit you can prove is both an upper bound for the original set and a lower bound for the set of all upper bounds. This would give you a predicative definition of the supremum.
The way to build the Cauchy sequence is as follows. Suppose $A$ is a nonempty set with an upper bound $x_0 \not \in A$. Pick $y_0 \in A$. Now, by induction, assume $y_i, x_i$ are given with $y_i \in A$, $x_i \not \in A$, $y_i < x_i$. Let $z = (x_i + y_i)/2$. If $z \in A$ let $y_{i+1} = z$ and $x_{i+1} = x_i$. Otherwise let $y_{i+1} = y_i$ and $x_{i+1} = z$. Then the sequence $(x_i)$ is Cauchy, and you can prove its limit is the least upper bound of $A$. You can see that this took a lot more work than the impredicative definition of the supremum, but it also gives a lot more information: if you knew enough about the set $A$ you could use this procedure to actually approximate the supremum to any precision you like.
Similarly, there are two proofs of the Knaster–Tarski theorem. One is impredicative but otherwise easy, while the other is predicative but uses transfinite induction. There are many other mathematical objects that have the same pattern of two definitions. Another common one is the algebra of Borel sets on a topological space.
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See this.
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Thanks for the link. The paper doesn't address the sense in which the original proof is not constructive. – Andres Caicedo Jan 2 '11 at 20:08
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http://mathoverflow.net/questions/43433/small-complete-categories-in-a-grothendieck-topos/50834
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## Small complete categories in a Grothendieck topos
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It is a classical theorem of Freyd that if a small category is complete (has all small limits—in fact, having small products suffices), then it is a preorder (has at most one morphism between any two objects). The proof of this theorem (which can be found here or in CWM) is non-constructive, i.e. it uses the Law of Excluded Middle. Therefore, it can potentially fail in the internal logic of an elementary topos. And in fact, it does fail in the effective topos, and more generally in realizability topoi, where there do exist small complete categories that are not preorders.
However, I have heard it said that Freyd's theorem cannot fail in a Grothendieck topos; i.e. that a small complete category in a Grothendieck topos must still be a preorder—despite the fact that the internal logic is still in general intuitionistic, so that Freyd's proof cannot work. Can someone explain why this is, or (even better) give a reference containing a proof?
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Good question. Since Grothendieck toposes are complete for higher-order intuitionistic logic we cannot hope to produce an internal argument. We have to use something about Grothendieck toposes, presumably cocompleteness. – Andrej Bauer Oct 24 2010 at 21:31
The question can be reduced easily to the following. In a Grothendieck topos, suppose we have a mono $m : A \to B$ such that, for every object $I$, there is a mono $A^I \to B$. What can $A$ be? In $\mathsf{Set}$ it has to be empty or a singleton. – Andrej Bauer Oct 25 2010 at 20:30
What do you mean formally for "internal complete category"? I find some different definitions about indexed categories (SKetches of a ELephant I, or Indexed Categories and their Applications, LNM 661). I formulate another one in terms of Kan extentions in the 2-cetegory of internal categories. How completeness is definible in the internal topos logic ?. this definition agree with the definitions of completeness of a indexed category? Excuse me for the mistakes and for the bad English, and thank for your time. – Buschi Sergio Nov 21 2010 at 17:20
@Buschi, the paper "The Discrete Objects in the Effective Topos" has a fairly exhaustive discussion of the relevant notions of "completeness" for internal categories, in the context of showing that such categories do exist in the effective topos. – Mike Shulman Nov 22 2010 at 0:30
## 1 Answer
Hi. I mentioned that I had thought about this on nForum a while back - sorry I didn't get back to you sooner. The following sketch of a proof is mainly due to Colin McLarty.
Two features which distinguish a Grothendieck topos from a more general topos are
1. That it has a geometric morphism to Sets, namely the global sections functor.
2. That it has an object of generators (i.e. there is an object G such that if $f,g: A \to B$ are not equal then there exists an arrow $h: G \to A$ with $fh \neq gh$)
Let $C$ be a small complete category object in a Grothendieck topos $T$ which is not a preorder. Then $C^G$ is also a small complete category in this topos essentially because exponentials commute. The global sections functor applied to $C^G$ gives a small complete category in the category of sets which is not a preorder (the property of being a small complete category is preserved by geometric morphisms, and the special property of G allows the property of being "not a preorder" to carry through), which is a contradiction.
It is a little easier to think about in the case of sheaves on some topological space. There a small complete category object which is not a preorder would have to fail to be a preorder on some open set, and the sections on that open set would be a small complete category which is not a preorder. $G$ takes the place of this open set above.
If you have any questions about this let me know. In particular I can write out all of the adjunctions showing various properties are preserved, but I don't want to get too nitty gritty if it isn't useful to you.
Kind regards, Steven Gubkin
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Excellent. Thank you, Steven! – Todd Trimble Dec 31 2010 at 22:21
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Thanks!! Does that proof really use boundedness (the existence of G)? If T is any Set-topos containing a small complete category C, then the adjunction argument seems to say that $\Gamma(C^G)$ is a small complete category in Set for any object G, hence a preorder; and thus the category T(G,C) is a preorder for any object G of T. Which should imply that C is a preorder in the internal logic of T, via Kripke-Joyal semantics. – Mike Shulman Dec 31 2010 at 22:34
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Also, it appears that the only property of Set used is that it is Boolean. If so, then something stronger is true: no topos which admits a (possibly bounded) geometric morphism to any Boolean topos can contain a small complete category. – Mike Shulman Dec 31 2010 at 22:36
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http://math.stackexchange.com/questions/231948/is-war-necessarily-finite
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# Is War necessarily finite?
War is an cardgame played by children and drunk college students which involves no strategic choices on either side. The outcome is determined by the dealing of the cards. These are the rules.
A standard $52$ card deck is shuffled and dealt evenly to two players face down. Every turn, both players lay down their top cards (a "battle"). The higher card wins, and the winning player gets both cards and places them at the bottom of his deck. If the battle is a tie, a "war" is declared, and each player lays down his next three cards. Of these six cards, the owner of the card with the highest value wins the turn, and all cards played that turn are placed at the bottom of his deck. If a player does not have $3$ cards to play in a war, he wins the turn by default. If the war is a tie, another war is held ("double war"), and so on until one of the players wins. The game ends when a player runs out of cards.
If the case occurs that the deals are entirely symmetric, e.g. both players decks are two two's, two threes, two fours, etc. in that order, we say that both players lose, as does everyone watching. (In other words, the game "fails.") This would be considered a finite game.
## Ordering Conventions.
To consider the game from a mathematical standpoint, the order that cards are placed into a player's hand after he wins a turn must be well-defined. There are many choices to be made here, especially considering that we can vary the conventions for double wars, triple wars, etc., which is where the most potential for variation lies. For the sake of simplicity, and because it coincides with how I pick the cards up when I am actually playing, I suggest that when a player wins a turn, he picks up all of the losing cards in the order that they were laid down (highest to lowest), followed all of the winning cards.
For example, say a turn begins with a battle in which players $\sf X$ and $\sf Y$ both lay down aces. Thus a war begins. Say that player $\sf X$ lays down a Jack, a King, and a $3$, in that order, and player $\sf Y$ lays down a Queen, a $4$, and a King, in that order. So a double war is held. Player $\sf X$ lays down $5$, $6$, $7$ and player $\sf Y$ lays down $8$, $9$, $10$. Now, player $\sf Y$ wins, so all of these cards are placed in his deck, ordered as such:
$$\sf (\ldots\text{previous deck}\ldots, \underbrace{\overbrace{\text{A},}^{\text{battle}}\overbrace{\text{J},\text{K},3,}^{\text{first war}}\overbrace{5,6,7,}^{\text{second war}}}_{\text{losing cards}}\underbrace{\overbrace{\text{A},}^{\text{battle}}\overbrace{\text{Q},4,\text{K},}^{\text{first war}}\overbrace{8,9,10}^{\text{second war}}}_{\text{winning cards}}).$$
A possible variation would be to pick up the cards in each war separately, in the order the wars were played, in which case the above example would yield
$$\sf (\ldots\text{previous deck}\ldots,\overbrace{\underbrace{A,}_{\text{losing}}\underbrace{A,}_{\text{winning}}}^{battle}\overbrace{\underbrace{J,K,3,}_{\text{losing}}\underbrace{Q,4,K,}_{\text{winning}}}^{\text{first war}}\overbrace{\underbrace{5,6,7,}_{\text{losing}}\underbrace{8,9,10}_{\text{winning}}}^{\text{second war}}).$$
## Questions.
First and foremost,
Is War necessarily finite?
Perhaps there exists some arrangement of starting decks which would produce an everlasting game, forever permuting cards between the two players. If this is not possible, can we prove it? Which ordering conventions always give rise to finite games (if any exist)?
and, for bonus points,
There are $\rm {52 \choose 26}(26!)^2\approx 10^{41}$ possible ways to deal the decks at the beginning of the game. Since suits are irrelevant, however, some arrangements will result in identical games. What is the expression for the number non-identical games? Thanks joriki.
A (substantially) tougher question: if we say that two games are similar if every turn has the same outcome, what is the expression for the number of non-similar games?
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– Gerry Myerson Nov 7 '12 at 5:53
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On the bonus question: $\binom{52}{26}(26!)^2=52!\approx8\cdot10^{67}$. The number of non-identical games is obtained by dividing through the permutations of the thirteen ranks: $52!/(4!)^{13}\approx9\cdot10^{49}$. – joriki Nov 7 '12 at 6:05
Thanks for the links @GerryMyerson. I should note that my suggested ordering convention is opposite what these other posts are doing (deliberately so, as I'd seen the $K,A$/$A,K$ example in the first link). In particular, I'm interested in knowing whether a periodic example like this can be constructed for any ordering convention. – Alexander Gruber Nov 7 '12 at 6:06
– B.D Nov 7 '12 at 8:52
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I have a puzzle book that have a variant of this game (2 suits of 5 cards, in a war you play one card instead of 3, you order the cards such that the war cards come first, then the winning card then the losing card). It asks the question "from how many moves can you be sure that the game will go on indefinitely ?", and its answer is 40, suggesting that infinite games DO arise (if they don't, while the question and answer could still be technically correct, that would be really shrewed) – mercio Nov 9 '12 at 15:20
show 3 more comments
## 1 Answer
A few years ago I studied this problem in the case in which there is only one suit in an $n$-card deck, so that the cards have a strict ranking from $1$ to $n$ and wars are not possible. I thought it would be easier to understand this simpler case before tackling decks with four suits in which wars are possible. As it turns out, I found even this case to be sufficiently difficult.
Under the convention that the losing card is picked up first, I found a way to construct cycles (so infinite games) for all values of $n$ except when $n = 2^k$ for some $k$. Under the convention that the winning card is picked up first, I found a way to construct cycles for all values of $n$ except when $n = 2^k$ or $n=3 \cdot 2^k$ for some $k$. Since $52 = 13 \cdot 2^2$, my constructions show how the 52-card deck with a strict ordering of the cards can exhibit cycling under either ordering convention.
Again, this is what I found for a simpler version of War than the OP is asking about, but perhaps this sheds some light on the more general case. Obviously the possibility of wars makes the problem much more difficult. I've wondered, though, if the method I found for generating cycles in the one-suit case could be modified to generate cycles when there is more than one suit. I worked on that question a little but didn't get very far.
(The paper is "Cycles in War," Integers 10: Article G2, 747-764, 2010, and can be found near the bottom of this page. The paper focuses on the "winning card is picked up first" convention, but I discuss the "losing card is picked up first" convention in the final section. I should also mention that I answered the question in the second of Gerry Myerson's links with a reference to this paper.)
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http://stats.stackexchange.com/questions/18754/sample-size-to-determine-if-success-proportion-is-zero
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# Sample size to determine if success proportion is zero
I have a lot of $N$ devices, and I'd like to know if any of them are faulty. What sample size do I need in order to make sure with $100(1-\alpha)\%$ confidence that none are faulty?
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## 1 Answer
You are asking for the number $m$ to sample so that when one or more devices are faulty, there will be at least a $100(1-\alpha)$% chance that at least one faulty device is in the sample. You have to handle the worst case of just one faulty device. Assuming the sample is random without replacement, we have (by definition) that there are $\binom{N}{m}$ such samples and $\binom{N-1}{m}$ of them do not include the faulty device. Therefore you need to find the smallest $m$ for which
$$\alpha \ge \binom{N-1}{m} / \binom{N}{m} = 1-\frac{m}{N}.$$
The solution is to take $m$ to be any whole number equal to or exceeding $(1-\alpha)N$. In brief, if you want to be (say) 95% confident there are no faulty devices, then you have to sample 95% of the lot.
You would be better off using a sequential sampling procedure if you can: sample the devices one at a time, stopping as soon as a faulty one is found or $100(1-\alpha)$% of them are sampled. For typical high confidence levels, the cost of sampling $100(1-\alpha)$% is so close to the cost of sampling $100$% that it's not worthwhile stopping there; plan to sample them all if necessary.
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http://mathoverflow.net/questions/122707?sort=oldest
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Which topological spaces are (topological) groups?
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General literature does not seem to offer a characterisation of topological groups among all topological spaces. Of course, being completely regular (uniform) is necessary, but separation properties, or indeed any sort of "niceness" like pseudo-metrisability are not sufficient, since topological groups, for instance, cannot have fixed point property.
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By "uniform" I think you mean that the space can be given a uniform structure, which is necessary as you say. Metrizability or any kind of separation axiom is not necessary; the indiscrete topology on any group gives a topological group. – Tom Goodwillie Feb 23 at 12:13
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Much easier is: which spaces are homeomorphic to a subset of a topological group. – Gerald Edgar Feb 23 at 14:39
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By the separation axiom complete regularity, I mean it in the weaker sense of Kelly and Willard - without the T0 axiom. It is this property which is equivalent to uniformisability (in the class of all topological spaces). This, I hope, every topological group (which is not assumed T0, I emphasise) must satisfy, as does the indiscrete topology. But then I regret having written metrisability and not pseudo-metrisability in my question. – N Unnikrishnan Feb 23 at 18:30
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@N Unnikrishnan: As for subsets, every $T_{3\frac12}$ space is homeomorphic to a subspace of $\mathbb R^\kappa$ (or $(S^1)^\kappa$ if you prefer a compact group) for some $\kappa$, no homogeneity is needed. I believe that likewise every completely regular space can be embedded into an appropriate abelian topological group whose Kolmogorov quotient is, say, $\mathbb R^\kappa$. – Emil Jeřábek Feb 23 at 18:42
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@Unnikrishnan: I think you're using the "accept" checkmark all wrong. You should reserve it till you have an answer you're really happy with, or until a long time goes by and it seems you're not getting any new answers. Don't just shift it to whatever the newest answer is – David White Feb 24 at 4:19
show 4 more comments
5 Answers
There is a homological criterion that is often helpful, to rule out the possibility for a topological space to admit a continuous group structure (even H-space structure):
The rational cohomology ring of a connected topological group (or H-space) $G$ is a connected graded-commutative Hopf algebra over $\mathbb{Q}$ and if $H^i(G;\mathbb{Q})$ is finite dimensional for all $i \ge 0$, then, by a theorem of Borel, $H^\ast(G;\mathbb{Q})$ is the tensor product of an exterior algebra on odd-dimensional generators and a polynomial algebra on even-dimensional generators.
For example $H^\ast(\mathbb{C}P^n;\mathbb{Q})=\mathbb{Q}[X]/(X^n),\; \deg x=2$ isn't of this form. Hence $\mathbb{C}P^n$ is no topological group.
For the theorem (and variations thereof) and further examples see Hatcher: Algebraic Topology, Section 3.C.
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Along the same lines, the fundamental group must be abelian. Note that these are restrictions on the (weak) homotopy type of the space. – Tom Goodwillie Feb 23 at 12:17
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Is this condition sufficient? Are there "fake groups" or "fake H-spaces" with the right structure on cohomology but no actual group structure? – mkreisel Feb 23 at 14:51
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@mkreisel: $S^n$ for $n\neq 1,3,7$ odd are counter-examples. For, $H^\ast(S^n)=\mathbb{Q}[X]/(X^n),$ $\deg x=n$ is exterior, but only $S^1,S^3,S^7$ admit a H-space structure (Hatcher, 3.C). – Ralph Feb 23 at 15:48
You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Every topological group is homogenous - this rules out spaces like $[0,1]$, $[0,\omega_1]$ or $\beta \mathbb{N}$.
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Among necessary conditions, I should have added homogeneity as well, as Cleft pointed out. But, is every topological group strongly homogeneous, i.e., can any two points be mapped onto each other by a self-homeomorphism? Or is it ever so? – N Unnikrishnan Feb 23 at 18:15
This is really a long comment on the answer by Cleft accepted by the OP.
Note that there are homogeneous spaces with the fixed point property (e.g. the Hilbert cube) and there are first countable non-metrizable homogeneous spaces (e.g. Alexandroff's double arrow space). Under additional axioms, there are also homogeneous spaces (even compact ones) which are hereditarily separable but not hereditarily Lindelof and the other way around: some that are hereditarily Lindelof but not hereditarily separable. And the list goes on.
So homogeneity is far from characterizing topological groups (even in the class of compact spaces). On the other hand, a (kind of vague) question due to Kunen (I think) is: can one say something interesting about compact right topological groups that cannot be said about compact homogeneous spaces? (besides things like "do not have the fixed point property" or "admits a group operation which is continuous in one variable").
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I did not in the least mean that cleft's answer gave a characterisation which we sought, when I accepted it. But the existence of homogeneous spaces with fixed point property was a really great reminder, in the least. But pardon me, how are first countability, hereditary separability and hereditary Lindeloffness related to being topological groups? Also, I shall be thankful if you tell me where to look for Kunen's question. – N Unnikrishnan Feb 23 at 18:52
A first countable topological group is metrizable. A compact group is hereditarily Lindelof iff it is hereditarily separable iff it is metrizable. As for the question, I remember reading about it in some paper by Jan van Mill, but can't remember which; anyway there wasn't anything else to read about it, it was just a comment towards the end of the paper (if I remember correctly). – Ramiro de la Vega Feb 23 at 23:08
@N Unnikrishnan: Let $G$ be a topological group, and $a\ b\in G$. Consider $h:G\rightarrow G$ defined by: $$\forall_{x\in G}\quad h(x) := a\cdot x^{-1}\cdot b$$
Then $h$ is a homeomorphism such that $h(a)=b$ and $h(b)=a$. This shows that (in your terminology above) every topological group is strongly homogeneous.
In general, the above homeomorphism $h_{a\ b} := h$ is not an involution. Indeed, in general, it is not its own inverse (with respect to composition)--actually, $h_{b\ a}$ is the inverse of $h_{a\ b}$ (it is an involution in the Abelian case though since then $h_{a\ b}=h_{b\ a}$).
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Dear Prof Holsztynski W, when I was student I studied your articles about shape theory (a categorical study) . By the way I remeber that for compact (T2 I seem) connected group the (weak) shape funcor is a equivalence. ams.org/journals/tran/1974-194-00/… – Buschi Sergio Feb 24 at 11:00
Hi Buschi, thank you for your kind words. I am rusty, but I think I can sketch a complete argument to support your statement about the equivalence. It would then work for all compact groups (not just for connected). K.Borsuk and I only started along this line. However, the general categorical approach gives a good foundation for this question. It was later James Keesling who seriously studied compact groups in the context of shape theory. (Also, another q. on MO--about Cech cohomology--is strongly (and unknowingly?) related to shape theory; I'll try to find it, to post an "answer"). – Wlodzimierz Holsztynski Feb 24 at 20:05
A necessary condition for a Hausdorff compact space to admit the structure of a topological group is the Suslin condition (I hope I am using proper terminology)
````every family of pair-wise disjoint open sets is countable.
````
This is so because Hausdorff compact topological groups admit Haar measure.
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http://physics.stackexchange.com/questions/39084/gauge-fixing-and-equations-of-motion/39096
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# Gauge fixing and equations of motion
Consider an action that is gauge invariant. Do we obtain the same information from the following:
1. Find the equations of motion, and then fix the gauge?
2. Fix the gauge in the action, and then find the equations of motion?
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## 2 Answers
I) Here we will assume that we ultimately want to consider the full quantum theory, usually written in terms of a gauge-fixed path integral
$$Z~=~\int \!{\cal D}\phi~ \exp\left(\frac{i}{\hbar}S_{\rm gf}[\phi]\right)$$
rather than just the classical action and the corresponding classical equations of motion (with or without gauge-fixing terms). If the gauge orbits have infinite volume (as is often the case), then we need to gauge-fix the path integral.
Of course, if we by brute force eliminate a field in the action via a gauge-fixing condition, then we can no longer carry out a variation of the action with respect to that field, and we would have lost information.
II) However, here we will only consider a 'softer' way to impose the gauge-fixing conditions via Lagrange multipliers, which may appear linearly or quadratically in the gauge-fixed action. The linear case leads directly to delta functions in the path integral, which impose the gauge-fixing conditions; while the quadratic case leads to Gaussian terms in the path integral, which suppress (but do not completely forbid) field configurations that violate the gauge-fixing condition. (Nevertheless, in a certain scaling limit, the Gaussian factors become delta functions.)
Together with the original fields, the (non-propagating, auxiliary) Lagrange multipliers are part of the fields $\phi$ in the path integral that is integrated over. In particular, the gauge-fixed action $S_{\rm gf}[\phi]$ can be varied with respect to these Lagrange multiplier fields as well.
These 'softly imposed' gauge-fixing conditions still do affect the variation of the action (as opposed to not imposing the gauge-fixing conditions). However, a more relevant question is:
Do the gauge-invariant physical observables of the theory depend on the specific gauge-fixing condition (e.g. Lorenz gauge, Coulomb gauge, etc)?
The answer is No, i.e. within the class of consistent gauge-fixing terms in the action, the specific form of the gauge-fixing terms in the corresponding equations of motion has no physical consequences.
III) For more general gauge theories, the equations of motion are e.g. not gauge-invariant, and it is better to encode the gauge symmetry via a (generalized) fermionic nilpotent BRST symmetry $\delta$ that squares to zero
$$\delta^2~=~0,$$
and preserves the original action
$$\delta S_0~=~0.$$
The gauge-fixed action $$S_{\rm gf}=S_0+\delta\psi$$ is the original action $S_0$ plus a BRST-exact term $\delta\psi$ that depends on the so-called gauge-fixing fermion $\psi$, which encodes the gauge-fixing condition.
A physical observable $F=F[\phi]$ in the theory is required to be BRST-closed
$$\delta F~=~0.$$
Formally, if the path integral measure is BRST-invariant, one may show that the correlation function for a physical observable
$$\langle F[\phi] \rangle ~=~ \frac{ \int \!{\cal D}\phi~ F[\phi]\exp\left(\frac{i}{\hbar}S_{\rm gf}[\phi]\right)} {\int \!{\cal D}\phi~ \exp\left(\frac{i}{\hbar}S_{\rm gf}[\phi]\right)}$$
is independent of the gauge-fixing fermion $\psi$. (Note however, that the gauge-fixing fermion $\psi$ is required to satisfy certain rank conditions, and it can e.g. not be chosen to be identically zero.)
IV) Finally, let us mention, that an even bigger class of Lagrangian gauge theories can be treated with the help of the Batalin-Vilkovisky (BV) formalism.
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No, it is not always consistent to first fix the gauge before deriving equations of motion. Consider electromagnetism coupled to matter. One can perform a gauge transformation to set $A_0=0$. However, if this is done in the action before deriving the equations of motion, then one will miss out on the $A_0$ equation of motion which guarantees the conservation of electric charge.
When attempting to derive sufficiently symmetric solutions, it is sometimes possible to begin by choosing an appropriate ansatz, even at the level of the action. But this is something of an art, and is not guaranteed to be consistent with the full equations of motion evaluated on that same ansatz.
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http://mathoverflow.net/questions/28712/on2-algorithm-to-approximate-the-sum-of-the-log-of-the-singular-values-of-a-ma
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## O(n^2) algorithm to approximate the sum of the log of the singular values of a matrix
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Given an $M \times N$ matrix of rank $N$ ($M \ge N$) with $i^{th}$ singular value $\sigma_i$, does their exist an $O(M^2)$ algorithm for approximating the sum $H =\sum_{i=1}^N \log(\sigma_i)$ with absolute error less than $M^{-\alpha}\log(\frac{\sigma_{max}}{\sigma_{min}})$ and $\alpha > 0$? Such an algorithm would be useful for fast computation of the entropy of normal distributions that arise in the probabilistic formulation of linear inverse problems.
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Isn't this equivalent to finding the determinant of $A^*A$? (which doesn't have, yet, an $O(M^2)$ algorithm) – Dror Speiser Jun 19 2010 at 7:33
Do you expect your matrices to be sparse? – S. Carnahan♦ Jun 19 2010 at 16:55
I wish I could make some assumptions about the matrix. In reality there is just such a large diversity of applications that result in different kinds of matrices. I suppose I should try to identify a few of my favorite problems and see what can be done (although in light of the response below it seems apparent that O(M^2) might be a bit too naive). – Gabriel Mitchell Jun 25 2010 at 16:46
## 1 Answer
$O(M^2)$ is barely enough time to read the entire matrix, let alone to do any meaningful computations. Suppose for instance that M=N was A was a permutation matrix except with the 1 coefficients replaced by coefficients between 1 and 2, so $\sigma_{max} / \sigma_{min} \sim 2$. In order to compute H to the desired accuracy, one has to locate every last non-trivial entry of the permutation matrix, which can barely be done in $O(M^2)$ operations (and in fact one has to lose a log or two to do the required high precision arithmetic).
If one is willing to have a little less accuracy (or perhaps if one places some "incoherence" assumptions on the matrix that spreads it out more than in the above permutation matrix-like example), then there is a chance that random sampling methods would work. Under reasonable hypotheses on the matrix, the singular values of a randomly selected set of rows or columns (or a random minor) are approximately proportional to the singular values of the whole. There is a certain amount of literature on this subject (see e.g. this paper).
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http://mathhelpforum.com/calculus/163968-cannot-find-my-own-mistake-derivation.html
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# Thread:
1. ## Cannot find my own mistake (derivation)
Hello.
So I have this:
$x^{y} = y^{x}$
which, in case $x$ and $y$ are positive, may be graphed like:
Assuming:
$x^{y} = y^{x} \Longrightarrow e^{y \cdot ln(x)} = e^{x \cdot ln(y)} \Longrightarrow y \cdot ln(x) = x \cdot ln(y) \Longrightarrow ln(x^{y}) = ln(y^{x})$
It seems plotting $x^{y} = y^{x}$ and $ln(x^{y}) = ln(y^{x})$ should be essentially the same.
However, my real problem is when I try to get $\frac{dy}{dx}$ out of those equations.
*Unfortunately first pictures showing my thorough work had to be removed because of some reader-disruptive flaws.*
Essence of derivation of equation #1:
$\frac{d}{dx}(x^{y} = y^{x})$
$y \cdot x^{y-1} \cdot \frac{dx}{dx} + x^{y} \cdot ln(x) \cdot \frac{dy}{dx} = x \cdot y^{x-1} \cdot \frac{dy}{dx} + y^{x} \cdot ln(y) \cdot \frac{dx}{dx}$
$\frac{dy}{dx} = \frac{y^{x} \cdot ln(y) - y \cdot x^{y-1}}{x^{y} \cdot ln(x) - x \cdot y^{x-1}}$
Essence of derivation of equation #2:
$\frac{d}{dx}(ln(x^{y}) = ln(y^{x}))$
$\frac{y}{x} \cdot \frac{dx}{dx} + ln(x) \cdot \frac{dy}{dx} = \frac{x}{y} \cdot \frac{dy}{dx} + ln(y) \cdot \frac{dx}{dx}$
$\frac{dy}{dx} = \frac{ln(y) - \frac{y}{x}}{ln(x) - \frac{x}{y}}$
Why is there a big difference?
2. Originally Posted by Pranas
Hello.
So I was looking for the derivative of this:
What is that very first line? This is false big time: the left hand is $yx^{y-1}$ , whereas the right one is $y^x\ln y$ ...
Tonio
But then I tried to put everything inside logarithm ln() , which, in case numbers are positive, I thought wouldn't change the answer. However it changed big time:
Can you explain, what is wrong with my thoughts?
.
3. tonio, could you please be at least a little more thorough?
$x^{y} = y^{x}$ is what i have.
$\frac{dy}{dx} = ?$ is what I need.
As long as I understand you saying $y \cdot x^{y-1} = y^{x} \cdot ln(y)$ that doesn't give me much.
4. Originally Posted by Pranas
tonio, could you please be at least a little more thorough?
$x^{y} = y^{x}$ is what i have.
$\frac{dy}{dx} = ?$ is what I need.
As long as I understand you saying $y \cdot x^{y-1} = y^{x} \cdot ln(y)$ that doesn't give me much.
Pranas, could you please be a little less sloppy? You did not say anything about having $x^y=y^x$ . It doesn't
appear anywhere in your message. It only appears the first equality $\frac{d}{dx}(x^y)=\frac{d}{x}(y^x)$ , which gives
you what I wrote you.
And don't bother in writing back asking for my help unless you first apologize.
Tonio
5. Originally Posted by tonio
Pranas, could you please be a little less sloppy? You did not say anything about having $x^y=y^x$ . It doesn't
appear anywhere in your message. It only appears the first equality $\frac{d}{dx}(x^y)=\frac{d}{x}(y^x)$ , which gives
you what I wrote you.
And don't bother in writing back asking for my help unless you first apologize.
Tonio
Yes, I apologize. I simply couldn't figure your post out at first, although it is very correct
Seems I should have written $\frac{d}{dx}(x^{y} = y^{x})$ ?
However $x^{y} = y^{x} \Longrightarrow e^{y \cdot ln(x)} = e^{x \cdot ln(y)} \Longrightarrow y \cdot ln(x) = x \cdot ln(y) \Longrightarrow ln(x^{y}) = ln(y^{x})$ so it seems like the disagreement I get
$\frac{d}{dx}(x^{y} = y^{x}) \neq \frac{d}{dx}(ln(x^{y}) = ln(y^{x}))$ prevails although it (I believe) really means the same.
6. Originally Posted by Pranas
Yes, I apologize. I simply couldn't figure your post out at first, although it is very correct
Seems I should have written $\frac{d}{dx}(x^{y} = y^{x})$ ?
However $x^{y} = y^{x} \Longrightarrow e^{y \cdot ln(x)} = e^{x \cdot ln(y)} \Longrightarrow y \cdot ln(x) = x \cdot ln(y) \Longrightarrow ln(x^{y}) = ln(y^{x})$ so it seems like the disagreement I get
$\frac{d}{dx}(x^{y} = y^{x}) \neq \frac{d}{dx}(ln(x^{y}) = ln(y^{x}))$ prevails although it (I believe) really means the same.
Well, that last statement is certainly true- $\frac{d(x^y= y^x)}{dx}\ne \frac{d ln(x^y)= ln(y^x)}{dx}$ and no one has said they are equal.
The problem goes back to your initial post: "I was looking for the derivative of this
$\frac{d x^y}{dx}= \frac{d y^x}{dx}$
which is very ambiguous: were you asking for the derivative of each side of that equation or were you asking how to arrive at that equation?
Now you say "Seems I should have written $\frac{d}{dx}(x^{y} = y^{x})$
If we let $z= x^y$, then $ln(z)= y ln(x)$ and so $\frac{1}{z}\frac{dz}{dx}= \frac{y}{x}+ ln(x)\frac{dy}{dx}$.
That is, $\frac{dz}{dx}= \frac{d x^y}{dx}= \left(\frac{y}{x}+ ln(x)\frac{dy}{dx}\right)z$
$\frac{dx^y}{dx}= yx^{y-1}+ y^x ln(x)\frac{dy}{dx}$.
Of course, that depends upon dy/dx. We cannot differentiate some function of x without knowing precisely what that function is- what y is as a function of x.
(Note that if y does NOT depend upon x, if y is a constant, then dy/dx= 0 and this just becomes the standard $\frac{dx^y}{dx}= yx^{y-1}$ from Calculus I.)
7. Indeed there were some obvious flaws in my "questionnaire", I am trying to correct most of it as the conversation evolves.
Originally Posted by HallsofIvy
Well, that last statement is certainly true- $\frac{d(x^y= y^x)}{dx}\ne \frac{d ln(x^y)= ln(y^x)}{dx}$ and no one has said they are equal...
For the sake of simplicity, let's say we're interested only in positive values of $x$ and $y$ (I've added a possible graphing below in this post).
Please be aware that I did not add $ln()$ just like that, I tried to rationally generate it as showed in this thread:
$x^{y} = y^{x} \Longrightarrow e^{y \cdot ln(x)} = e^{x \cdot ln(y)} \Longrightarrow y \cdot ln(x) = x \cdot ln(y) \Longrightarrow ln(x^{y}) = ln(y^{x})$
Maybe I am wrong, but at this moment I do not see how were the relations between $x$ and $y$ effected by that, therefore I assume I haven't changed the possible plot nor the value of $\frac{dy}{dx}$.
Originally Posted by HallsofIvy
...The problem goes back to your initial post: "I was looking for the derivative of this
$\frac{d x^y}{dx}= \frac{d y^x}{dx}$
which is very ambiguous: were you asking for the derivative of each side of that equation or were you asking how to arrive at that equation?...
I did not write what I really meant to at that point. Sorry.
Originally Posted by HallsofIvy
...Now you say "Seems I should have written $\frac{d}{dx}(x^{y} = y^{x})$
If we let $z= x^y$, then [tex]ln(z)= y ln(x) and so $\frac{1}{z}\frac{dz}{dx}= \frac{y}{x}+ ln(x)\frac{dy}{dx}$.
That is, $\frac{dz}{dx}= \frac{d x^y}{dx}= \left(\frac{y}{x}+ ln(x)\frac{dy}{dx}\right)z$
$\frac{dx^y}{dx}= yx^{y-1}+ y^x ln(x)\frac{dy}{dx}$...
Well, yes. I pretty much applied parallel method to yours on both $x^{y}$ and $y^{x}$. Then expressed $\frac{dy}{dx}$ (that would be my equation #1 in updated first message). As far as my mental calculation goes, answer seems to be identical as the one you're approaching in this part of the post.
Originally Posted by HallsofIvy
...Of course, that depends upon dy/dx. We cannot differentiate some function of x without knowing precisely what that function is- what y is as a function of x.
(Note that if y does NOT depend upon x, if y is a constant, then dy/dx= 0 and this just becomes the standard $\frac{dx^y}{dx}= yx^{y-1}$ from Calculus I.)
Indeed you're correct once more. That is a reasonable variation in a broad sense, although might be a little vulgar mapped on a plane, because I would imagine it as of having only a few points.
What I imagined as a relation between positive $x$ and $y$ in the function $x^{y} = y^{x}$ is like this:
In my language that is defined to be a simple function, only "unexpressed", because of not being represented by $y =$*operations involving only variable $x$ and constants*
P.S. We've had some confusion here, so I can politely remind, that what's still unclear for me is inequality $\frac{d(x^{y}= y^{x})}{dx}\ne \frac{d(ln(x^{y})= ln(y^{x}))}{dx}$.
Also I tried to do my best in editing the first post.
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http://physics.stackexchange.com/questions/30167/what-is-physical-meaning-of-this-partial-derivative/30174
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what is physical meaning of this partial derivative?
what is physical meaning of this partial derivative: $$\frac {\partial p_x}{\partial x}$$ i know how do i solve it when the case is just derivative but partial derivative is a bit Hectic!.
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1
This question is lacking some context. A partial derivative is nothing but a normal derivative with the other variables taken as constants. What the physical meaning is depends on the function $p_x$. – Alexander Jun 16 '12 at 0:31
If $p_x$ is momentum then the derivative is the spatial gradient along the `x`-axis. – ja72 Jun 16 '12 at 1:56
3 Answers
If $p_x$ is taken to mean momentum in the $x$ direction, $\frac{\partial p_x}{\partial x}$ simply means that momentum in the $x$ direction changes as you move along $x$. So for an object with a given mass, $\frac{\partial p_x}{\partial x}$ describes how its $x$ velocity changes as the object moves along $x$. In everyday life, that would just be called an acceleration or deceleration, such as braking a car that is moving east, if you call east $x$.
Where partials get more complicated is that you can have several such things going on at once. More specifically, you can have other independent accelerations or decelerations going on in $y$ and $z$ (think roller coasters). Finally, for a partial derivative that is expressed in terms of momentum instead of velocity, you cannot always assume the mass is constant, so you also cannot always assume that the effects are equivalent to velocity expressions.
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The partial derivative of a function $f(x_1,x_1,x_3,..)$ at a point $(k,x_2,x_3,...)$, e.g. with respect to $x_1$, is the slope of the tangent line to the curve obtained by the intersection of the plane $x_1=k$, with the plot of $f(x_1,x_2,...)$.
The partial derivative also gives the rate of change with respect to the quantity being changed. Your partial derivative would give the rate of change of the momentum in the $x$ direction with respect to change in $x$, it essentially is a measure of how the velocity along the $x$ axis, if mass is constant changes, as you move along $x$, i.e. the change in a quantity along the direction you are moving in.
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A partial derivative basically tells you how a function changes if I change just one of many variables it may depend on, while keeping all other variables constant.
On the other hand, a total derivative tells you the "TOTAL" information about the function. It describes how a function changes if all (or some) variables are changed together.
When a function depends only on one variable then ALL = one variable. In this case, a partial derivative is the same as the total derivative.
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http://mathoverflow.net/questions/101321/injective-von-neumann-algebra
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## Injective von Neumann algebra
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $G$ be a non-amenable countable discrete group. How can I show that the group von Neumann algebra $L(G)$ has no injective direct summand?
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I must be missing something. Why can't you take an abelian vN subalg of L(G) and then project onto it? – Yemon Choi Jul 4 at 17:08
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There is no reason that this subalgebra will be a direct summand – Owen Sizemore Jul 4 at 17:36
@Owen: right, for some reason I had "direct summand as a Banach space" in my mind – Yemon Choi Jul 4 at 23:52
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Why not accept Jesse Peterson's answer? – Yemon Choi Jul 9 at 23:32
For that matter, why haven't you accepted any answers? – S. Carnahan♦ Jul 11 at 4:49
show 1 more comment
## 1 Answer
Here is an adaptation of the standard proof that $G$ is amenable if $LG$ is injective. (I believe for instance that it is contained in the book of Brown and Ozawa).
Suppose $p \in LG$ is a non-zero central projection such that $p LG$ is injective. Thus, there exists a conditional expectation $E: \mathcal B(p \ell^2 G) \to p LG$. If we view $\ell^\infty G \subset \mathcal B(\ell^2 G)$ as diagonal multiplication operators (for $f \in \ell^\infty G$ and $\xi \in \ell^2 G$ we set $(M_f \xi)(\gamma) = f(\gamma) \xi(\gamma)$), and if we denote by $\tau$ a tracial state on $pLG$ then we can construct a state $\varphi$ on $\ell^\infty G$ by the formula $\varphi(f) = \tau \circ E(p M_f p)$. If $\gamma \in G$ then we have $$\varphi( f \circ \gamma) = \tau \circ E(p M_{f \circ \gamma} p)$$ $$= \tau \circ E(p \lambda_{\gamma^{-1}} M_f \lambda_{\gamma} p) = \tau( (p\lambda_{\gamma^{-1}}p) E(p M_f p) (p \lambda_{\gamma}p) ) = \varphi(f).$$ Thus $\varphi$ is an invariant mean for $G$ and so $G$ is amenable.
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thanks for the nice answer. – m07kl Jul 5 at 16:48
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http://mathoverflow.net/questions/49351/does-the-fact-that-this-vector-space-is-not-isomorphic-to-its-double-dual-require
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Does the fact that this vector space is not isomorphic to its double-dual require choice?
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $V$ denote the vector space of sequences of real numbers that are eventually 0, and let $W$ denote the vector space of sequences of real numbers. Given $w \in W$ and $v \in V$, we can take their "dot product" $w \cdot v$, and for any give $w \in W$, this defines a linear functional $V \to \mathbb{R}$. In fact, under this association, we see that $W \cong V^{\ast}$. Assuming choice, $W$ has a basis and has uncountable dimension, whereas $V$ has countable dimension. So $\mathrm{dim} (V) < \mathrm{dim} (V^{\ast}) \leq \dim(V^{\ast \ast})$ so $V$ is not isomorphic to its double-dual. In particular, the canonical map $\widehat{\cdot} : V \to V^{\ast \ast}$ defined by $\widehat{x} (f) = f(x)$ is not an isomorphism.
Question 1: Can you explicitly write down an element of $V^{\ast \ast}$ that isn't of the form $\widehat{x}$?
Question 2: What's the situation if we don't assume choice? (i.e. might the canonical map be an isomorphism, might $W$ not even have a basis, might $V$ be isomorphic to its double-dual but not via the canonical map, etc?)
In light of Daniel's and Yemon's comments, and Konrad's related question here, I'd like to reorganize my question(s). So consider the following statements:
1. The canonical map $\widehat{\cdot} : V \to V^{\ast \ast}$ is injective but not surjective.
2. The canonical map is not an isomorphism.
3. There is no isomorphism from $V$ to its double-dual.
4. The double-dual is non-trivial.
5. $V^{\ast}$ has a basis.
This is the question I'm really interested in, and which clarifies and makes more precise my original Question 1:
Revised Question 1: Are there models without choice where (1) holds? If so, can we find some $x \in V^{\ast \ast}$ witnessing non-surjectivity to describe a witness to non-surjectivity in choice models? Are there models of choice where (1) fails?
The following question is a more precise version of Question 2.
Revised Question 2: Under AC, all 5 statements above are true. There are $2^5 = 32$ ways to assign true-false values to the 5 sentences above that might hold in some model of ZF where choice fails. Not all 32 are legitimate possibilites, some of them are incompatible with ZF since, for instance, (1) implies (2), (5) implies (4), and the negation of (4) implies (1) through (3), etc. Which of the legitimate possibilites actually obtains in some model of ZF where choice fails?
This second question is rather broad, and breaks down into something on the order of 32 cases, so seeing an answer to any one of the legitimate cases would be cool.
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How explicit do you want? Can I for example define a functional on some subspace of $W$, and extend it to all of W using the axiom of choice, and then prove it isn't of the form $\hat x$? For example, if $w=(w_n)$ has a limit, let $f(w)=\lim w_n$; then extend this to a linear map on all of $W$. Such a map clearly doesn't come from $V$. – Daniel Litt Dec 14 2010 at 6:25
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A while ago, in a discussion on the secret blogging seminar, I raised the question of the "Boring Duals Hypothesis": Is it consistent with ZF for every vector space to be isomorphic to its double dual? I asked some logicians at the time, but never got an answer. It would be nice if that finally got resolved here. sbseminar.wordpress.com/2007/10/30/… – David Speyer Dec 14 2010 at 13:52
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I added a related question mathoverflow.net/questions/49388/… – Konrad Swanepoel Dec 14 2010 at 14:34
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@Konrad: It seems to me that the statement "Then $\{f_i: i\in \mathbb{N}\}$ generates $V^*$" seems to rely on choice--in particular, why is the restriction map $W^*\to V^*$ surjective? – Daniel Litt Dec 14 2010 at 20:36
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@Daniel: you're right. Surjectivity of the restriction $W^*\rightarrow V^*$ means exactly that any linear functional on $V$ can be extended to a linear functional on $V^*$, and I can't see how to do that without choice. So my argument doesn't work.... Thanks for pointing it out. – Konrad Swanepoel Dec 15 2010 at 11:58
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4 Answers
Let me give an alternative proof of the result in Ron Maimon's answer, that the canonical map `$V\to V^{**}$` can consistently, in the absence of the axiom of choice, be surjective. The only technical advantage of my proof is that it uses only the consistency of ZF plus countable choice plus "all sets (in Polish spaces) have the Baire property". That consistency was proved (by Shelah) relative to just ZF, whereas the consistency of "all sets are Lebesgue measurable" needs an inaccessible cardinal. I think my proof is also a bit simpler than the one using measure. (Also, I don't need to mention ultrafilters, which some people might consider an advantage.)
Let `$f:V^*\to\mathbb R$` be a linear map, and let me identify `$V^*$` with the space of infinite sequences of reals. Topologize `$V^*$` with the product topology, where each factor $\mathbb R$ has the usual topology of the reals. That makes `$V^*$` a Polish space, so I can use the assumption about Baire category. In particular, if I partition $\mathbb R$ into intervals of length 1, then the inverse images of these intervals under $f$ have the Baire property, and they can't all be meager, by the Baire category theorem. So at least one of them, call it `$f^{-1}(I)$`, differs by a meager set from a nonempty open set. Inside that nonempty open set, I can find a basic open set of the form `B=$\prod_iU_i$` where, for some $n$, the first $n$ of the factors `$U_i$` are intervals of some length $\delta$ and the later factors `$U_i$` are $\mathbb R$. For the first $n$ indices $i$, let `$U'_i$` be the interval with the same midpoint as $U_i$ but only half the length, and let $B'$ be the product that is like $B$ except using the `$U'_i$` instead of the `$U_i$` for the first $n$ factors. Consider an arbitrary `$z\in V^*$` whose first $n$ components are smaller than $\delta/2$ in absolute value. Then translation by $z$ in `$V^*$` maps $B'$ homeomorphically to another subset of $B$. So the two sets `$f^{-1}(I)\cap B'$` and `$\{x\in B':z+x\in f^{-1}(I)\}$` are both comeager in $B'$ and therefore must intersect. Let $x$ be in their intersection. Both $f(x)$ and $f(z+x)$ are in the interval $I$ of length 1. Subtracting (and remembering that $f$ is linear), we get that $|f(z)|\leq 2$.
Summarizing, we have that $|f(z)|\leq 2$ for all `$z\in V^*$` whose first $n$ components are smaller than $\delta/2$ in absolute value. By linearity, if the first $n$ components of $z$ are smaller than $\alpha\delta/2$ in absolute value, for some positive $\alpha$, then $|f(z)|\leq2\alpha$. In particular, if the first $n$ components of $z$ are zero, then so is $f(z)$. That is, the kernel of $f$ includes the subspace $N$ consisting of those `$z\in V^*$` whose first $n$ components vanish. So $f$ factors through the quotient `$V^*/N$`, which is finite-dimensional (in fact, $n$-dimensional). Knowing what linear functionals on finite-dimensional spaces look like, we immediately conclude that $f$ is given by inner product with a member of $V$ (having non-zero components in at most the first $n$ positions).
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An awesome answer. – Todd Trimble Jun 12 2011 at 14:55
You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
A simple uncountable linearly independent set in $V^*$ (V dual--- the space of infinite sequences of reals) is the collection of vectors:
$A^r_n = r^n$
For all nonzero real values r. There can be no linear relation between M of these, just restricting to the first M positions, because of the non-vanishing of the van-der-Monde determinant. This does nothing to answer the question.
Question 4 is simple： $V^*$ is the space of all infinite sequences, and the space of finite sequences dotted into these give nontrivial linear maps. The interesting question is whether these are all of the linear maps in the absence of choice.
For question 5: if probability arguments work (i.e. if every subset of R is measurable) then $V^*$ does not have a basis.
Preliminary comment: If m and n are two positive integers both distributed with probability distribution P, their sum m+n does not have the same distribtion P. To prove this, let N>0 be the first integer position where P(N) is nonzero. Then $m\ge N$, $n\ge N$, so $m+n\ge 2N$, so the probability that m+n is N is zero.
Suppose that there is a basis set B. Then any vector v in V^* has an integer degree n and n uniquely determined elements of B, $e_1,...,e_n$ such that v is a linear combination of the e's. Generate the sequence A_n by picking a gaussian random number with width 1 at each position n. Generate B_n in the same way. Then the degree of A is some integer n, with probability distribution P(n), and the degree of B is an integer m with a probability distribution P(m). The degree of $(A+B)/sqrt{2}$ is then m+n by adding the expansion of A and the expansion of B (there is no chance that A and B share basis elements, the probability of meeting any finite dimensional subspace is zero), and its distribution is the distribution of the sum of two random integers with distribution P. But $(A+B)/\sqrt{2}$ is identically distributed with A and B, therefore its degree must have distribution P, and there is no such distribution P on positive integers.
I'm repeating the argument in the form I came up with so as to make the rest of the answer more obvious. A closer inspection reveals that the nonmeasurable sets here are the sets of all vectors with degree n. These union up to the whole space, so they need to add to the full measure, but the larger n degrees necessarily swamp the lower n degrees in measure because of the properties of linear combination, forbidding a consistent assignment of measures.
For the main question, questions 1,2,3, I show that the existence of any map f in $V^{**}$, other than the ones given by dotting with elements of V, either leads to a probabilistic contradiction or a nonprincipal ultrafilter on the integers. Both of these cannot be "explicit", so there is no explicit linear map in V double-dual other than the canonical images of those in V.
Proposition: f of any infinite sequence of independent gaussian random numbers is gaussian distributed.
Proof: let two picks of the sequence be $S$ and $S'$, $(a S+ b S')\over \sqrt{a^2 + b^2}$ is identically distributed as S and S', so $(a f(S) + b f(S'))\over\sqrt{a^2 + b^2}$ is identically distributed as f(S). Note also that $f(-S)=-f(S)$, so that the probability distribution of $f$ is symmetric around zero. This implies that $f$ has the convolution properties of a Gaussian symmetric about zero, and is therefore a Gaussian with zero mean(or a delta function at zero, which I will call a Gaussian of zero width below).
This means that the linear function $f$ gives a map $\Sigma$ from every sequence of variances to a variance. The function $f$ has variance $\Sigma(\sigma_k)$ when evaluated on the sequence of gaussian random variables of variance $\sigma_k$
Proposition: If the variances $\sigma_k$ are each individually bigger than $\sigma'_k$, then $\Sigma(\sigma)>\Sigma(\sigma')$
Proof: Let $\alpha_k$ be such that $\alpha_k^2 + \sigma'_k^2 = \sigma_k^2$. Generate a sequence of gaussian random picks $A_k$ with variance $sigma'_k$, and a second gaussian sequence $B_k$ with variance $\alpha_k$. By construction, $A_k+B_k$ has variance sequence $\sigma_k$.
$f(A)$, $f(B)$, and $f(A+B)=f(A)+f(B)$ are each Gaussian distributed, and the variance of $f(A)$ squared plus the variance of $f(B)$ squared equals the variance of $f(A+B)$ squared. This proves the result.
Theorem: Define the sequence $A^N_n$ to be zero for $n\le N$, and a gaussian random number with unit variance for $n>N$. If probabilistic arguments work, there is an $N$ for which $f$ on $A^N$ is zero with probability 1.
Proof: If $f$ on $A^k$ is nonzero for all k, consider the infinite sum
$S = C_1 A^1 + C_2 A^2 + C_3 A^3 + C_4 A^4 ...$
since each $A^k$ is zero on the first $k$ positions, this just defines $S$ as a sequence of Gaussians of increasing variance (this is not really an infinite sum--- it is just a trick for writing an infinite sequence of variances in a more illuminating way). For any sequence $C_1, C_2, C_3 ...$, the function $f$ produces a Gaussian with a given width which is an non-decreasing function of the $C$'s.
By assumption, $f(A^k)$ has some nonzero variance $a_k$ for all $k$. Choose $C_k$ to be $1/a_k$. Then the width of $f$ on $S$ is greater than any integer $k$, a contradiction.
Therefore there is an integer $N$ such that $f$ has zero variance acting from some $A^N$ on, i.e. $f$ acting on a sequence which is zero at the first $N$ positions, and Gaussian random with unit variance on the remaining positions, gives 0 with certainty.
What remains is to deal with the case that f is nonzero on some specific vector v which has no chance of ever being generated by random Gaussian picking. To deal with this, you need the following:
Theorem: If $f$ is of zero width on all gaussian random variables, and $f$ is nonzero on some vector $v$ (and probability works), then there is a nonprincipal ultrafilter on the integers.
proof: Suppose $f(v)$ is nonzero for some infinite sequence $v$. By the zero width property, $f$ is zero on sequences which are nonzero only at finitely many places.
Define a set $S$ of integers to be "zeroing" iff: for any collection $g(S)$ of independent gaussian random variables defined on $S$, and a gaussian random variable $g$, $f$ acting on $g v_S + g(S)$ is certainly zero. $g v_S$ is $g$ times the restriction of $v$ to zero on the complement of S, while $g(S)$ is a sequence of random variables inside S, and zero outside S.
Any finite set is zeroing, while the full-set is nonzeroing since $f(gv+g(N))$ has the variance of $g$ times $f(v)$ (since $f$ is zero acting on the gaussian random sequence $g(N)$).
If $S$ is zeroing, $S$ complement is nonzeroing by linearity. (Note that the converse is not true--- S can be nonzeroing and also S complement nonzeroing--- this is not automatically an ultrafilter).
If $S$ is zeroing and $S'$ is a subset of $S$, then $S'$ is zeroing (by the positivity of adding gaussian widths).
If $S$ is zeroing and $S'$ is zeroing then $S$ union $S'$ is zeroing, since the sum of independent gaussian random variables on $S$ and $S'$ are again independent gaussian random variables on $S$ union $S'$, and any independent gaussian random variables on the union of $S$ and $S'$ can be decomposed in this way. Therefore the nonzeroing sets form a nonprincipal filter extending the finite-complement filter.
Using dependent choice, either you have an infinite sequence of disjoint restrictions of $v$ $S_1$, $S_2$, $S_3$, ... which are nonzeroing, or the restriction onto one of the sets makes an ultrafilter. An infinite sequence of disjoint nonzeroing restrictions leads to a probabilistic contradiction, as before, by considering
$A = \sum_k C_k g(S_k)$
with appropriate fast-growing choice of $C_k$, just like before. The terminating case gives an ultrafilter.
Putting the two theorems together, the function $f$ has to be certainly zero on Gaussian sequences from position $N$ onward, so $f$ is a linear function on the first $N$ values plus a residual. The residual is zero on the first N positions, and makes a linear function on the values past $N$, and the residual is zero on any gaussian random sequence.
If the residual function is nonvanishing on some vector v, then it either leads to a probalistic contradiction or it defines a nonprincipal ultrafilter on Z. So in a universe with every subset of R measurable, with dependent choice, and with no nonprincipal ultrafilter on the integers, V double dual is the canonical image of V, and questions 1,2,3 are answered.
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If every subset of R is measurable, then there are no nonprincipal ultrafilters. (By symmetry, an ultrafilter would have to have measure 1/2, but by this is impossible for a tail set.) – Goldstern Jun 6 2011 at 18:43
Thanks, sorry for the lapse. I must confess that it was initially surprising to me that V double dual is V (assuming probability is not contradictory, which I think one always should). The argument I found is too ad-hoc though--- there should be a simple non-measurable set. I will clean the answer up to remove the redundant assumption. – Ron Maimon Jun 8 2011 at 19:20
I have recently noticed$^1$ that in models of ZF+DC+"All sets of real numbers have the Baire property" (e.g. Solovay's model or Shelah's model mentioned by Andreas) there is a very interesting property for Banach spaces:
If $V$ is a Banach space, $W$ is a normed space and $T\colon V\to W$ is linear then $T$ is continuous.
In particular this means that if $V$ is a Banach space over $\mathbb R$ then every functional is continuous. This means that the algebraic dual of a Banach space is the same as the continuous dual of the space.
For example, $\ell_p$ for $p\in(1,\infty)$ have this property, using DC we can develop the basic tools of functional analysis just fine (except Hahn-Banach, though). This implies that $\ell_p$ is reflexive in the algebraic sense, not only in the topological sense.
We further remark$^2$ that the assertion $\ell_1\subsetneq\ell_\infty^\prime$ (where the $\prime$ denotes all the continuous functionals, but in our model these are all the functionals) implies the negation of "All sets of real numbers have the Baire property". The result is that $\ell_1$ is also reflexive in the algebraic sense and in the topological sense.
So with this in mind we have $V_p=\ell_p\oplus\ell_q$ (where $\frac1p+\frac1q=1$) is a self-dual as well algebraically reflexive space, and for distinct $p_1,p_2\in[1,2]$ we have $V_{p_1}\ncong V_{p_2}$ as well, so there are many non-isomorphic examples for this.
Notes:
1. After sitting to write all the details of the above I found out it is mentioned in Eric Schechter's book, Handbook of Analysis and its Foundations in the last section of chapter 27.
2. The above reference does not discuss the case of $p=1$, which is discussed later in the end of chapter 29.
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For the record, in 1973 the Belgian mathematician Henri Garnir, combining results of Schwartz on a measurable graph theorem and Solovay as mentioned here, that it is consistent with ZF without AC that every linear map from an ultrabornological space (in particular, Banach, Fréchet or an inductive limit of a sequence of Banach spaces which covers the spaces mentioned here) into any locally convex space is continuous. This in turn implies that the algebraic dual of such a space coincides with the topological dual and that the same holds for biduals for most spaces of interest. – jbc yesterday
The precise reference for Garnir's result is to be found in MR0477688 (and the word "proved" is missing in the previous comment). – jbc yesterday
jbc, but Garnir's work uses Lebesgue measurability of all sets, which itself requires the consistency strength of an inaccessible cardinal. On the other hand, the assumption that all sets of reals have the Baire property is equiconsistent with $\sf ZFC$. – Asaf Karagila 21 hours ago
edit I'm leaving the "answer" as is but now I realize that it doesn't get that far. I think it gives a subspace $W' \subset W=V^{\ast}$ with an uncountable basis. And $(W')^{\ast}$ includes but is much bigger than the set of all $\widehat{x}$. Maybe that is not that helpful.
No, it doesn't require the axiom of choice. It takes choice to show that $W$ has a basis but not to show that it has a subspace that has an uncountable basis. I remembered that I once saw a great construction due to Von Neumann so I Googled it and ended up at explicit big linearly independent sets. I suggest checking out the web site!
I think that should do it. Well actually Von Neumann's set is algebraically independent. Earlier in the answer it is pointed out that $T_r = \sum_{q_n < r} \frac{1}{n!}$ (where $q_0, q_1, \ldots$ is an enumeration $\mathbb{Q}$) is independent over $\mathbb{Q}$. So write each $T_r$ in binary and convert it into a $0,1$ element of $W$.
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@Andres I'm thinking that it does not require choice (as in the title and question 2). I might be off base though. – Aaron Meyerowitz Dec 14 2010 at 5:26
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@Aaron: it is not clear to me that, without choice, having a large subspace with an uncountable basis tells you anything about the size of the dual. – Qiaochu Yuan Dec 14 2010 at 5:30
Aaron, you're right, this answer doesn't seem to get us all that far. You're saying that without choice, we can still get $W$ to have an uncountable linearly independent subset. Does that guarantee us that the same is true of $W ^{\ast}$. If so, can we write down an element of $W ^{\ast}$ that isn't an $\widehat{x}$. Also, why is it that if the $T_r$ are linearly independent over $\mathbb{Q}$, then the sequences corresponding to their binary expansions are linearly independent over $\mathbb{R}$? – Amit Kumar Gupta Dec 14 2010 at 6:14
@Amit It probably isn't the best way to do it, but I think, independent of the other properties, that uncountable family of vectors is such that for any finite set , each one has positions where it has a $1$ and the others all have $0$'s. I might be wrong and there does not seem to be much point to finding out. – Aaron Meyerowitz Dec 14 2010 at 6:36
I don't think that works. Let $x^i$ be the sequence of 0's and 1's that has 0's in the locations congruent to $i (\mathrm{mod} 3)$ for $i = 0, 1, 2$. `$\{x^0, x^1, x^2\}$` is linearly independent but doesn't have the property you mention. – Amit Kumar Gupta Dec 14 2010 at 7:01
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http://physics.stackexchange.com/questions/48573/does-the-amount-of-gravitational-potential-energy-in-the-universe-increase-as-it/51735
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# Does the amount of gravitational potential energy in the universe increase as it expands?
It seems to me that extra gravitational potential energy is created as the universe expands and the distance between massive objects such as galaxy clusters increases; this implies that energy is not conserved in the universe. Is that right?
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– twistor59 Jan 8 at 9:06
I've awarded the bounty to @markovchain - best answer so far is "we don't know" which is perfectly honest if a bit unsatisfying! – roblev Jan 27 at 17:47
– markovchain Jan 28 at 1:08
## 5 Answers
That's actually a tricky question.
The short answer to the title question is yes, it does. But the answer to the follow up question about conservation is, it is still conserved.
In a much simpler universe, what hwlau said would be true - as the gravitational potential energy increases, the kinetic energy decreases. But we do know through the Hubble telescope that this is not the case. As the universe expands, the planets and dust and stars accelerate away from each other.
The answer is only hypothetical for now, which is dark energy. Here's the wikipedia page on it. The problem is, dark energy is very unknown. I'm not sure how this force will be able to counteract gravity. However, dark energy only manifests itself through gravity, and not through the other 3 fundamental forces, so there is definitely some relationship there. Dark energy can possibly negate the increased potential energy through some mechanism, but how it does that exactly is also unknown.
There's little I can give you other than the wikipedia link above, but notice when you open it, how the only explanations available are evidence arguing for dark energy's existence, what properties it should have, and other explanations for the expansion aside from dark energy (very interesting part). There is no explanation as to its mechanism, except that it causes things to accelerate against each other.
It's a good question. There is simply no good, accurate answer at the moment.
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hi, I see that dark energy may or may not exist, but I don't see how this is related to the question of whether the expanding universe increases the overall amount of gravitational potential energy. Your last sentence leaves a lot to the imagination! – roblev Jan 24 at 2:59
That is true, it does. But only because there really is little known about it. The simple answer is, yes, the amount of gravitational potential energy does increase as the universe expands. But as for the violation of conservation of energy in the universe, dark energy is a possible way to counteract this increase, so very likely, energy is conserved. I'll edit my answer to reflect this. :) – markovchain Jan 24 at 4:57
Here is an extract from this article on math.ucr.edu that explains why energy sometimes seems to be not "conserved".
I figured it would be better if I copy the entire paragraph and post it as answer (so that others can see) instead of providing the link as comment because you never know when the server might go down and the article would be lost.
Original by Michael Weiss and John Baez.
## Is Energy Conserved in General Relativity?
In special cases, yes. In general — it depends on what you mean by "energy", and what you mean by "conserved".
In flat spacetime (the backdrop for special relativity) you can phrase energy conservation in two ways: as a differential equation, or as an equation involving integrals (gory details below). The two formulations are mathematically equivalent. But when you try to generalize this to curved spacetimes (the arena for general relativity) this equivalence breaks down. The differential form extends with nary a hiccup; not so the integral form.
The differential form says, loosely speaking, that no energy is created in any infinitesimal piece of spacetime. The integral form says the same for a finite-sized piece. (This may remind you of the "divergence" and "flux" forms of Gauss's law in electrostatics, or the equation of continuity in fluid dynamics. Hold on to that thought!)
An infinitesimal piece of spacetime "looks flat", while the effects of curvature become evident in a finite piece. (The same holds for curved surfaces in space, of course). GR relates curvature to gravity. Now, even in newtonian physics, you must include gravitational potential energy to get energy conservation. And GR introduces the new phenomenon of gravitational waves; perhaps these carry energy as well? Perhaps we need to include gravitational energy in some fashion, to arrive at a law of energy conservation for finite pieces of spacetime?
Casting about for a mathematical expression of these ideas, physicists came up with something called an energy pseudo-tensor. (In fact, several of 'em!) Now, GR takes pride in treating all coordinate systems equally. Mathematicians invented tensors precisely to meet this sort of demand — if a tensor equation holds in one coordinate system, it holds in all. Pseudo-tensors are not tensors (surprise!), and this alone raises eyebrows in some circles. In GR, one must always guard against mistaking artifacts of a particular coordinate system for real physical effects. (See the FAQ entry on black holes for some examples.)
These pseudo-tensors have some rather strange properties. If you choose the "wrong" coordinates, they are non-zero even in flat empty spacetime. By another choice of coordinates, they can be made zero at any chosen point, even in a spacetime full of gravitational radiation. For these reasons, most physicists who work in general relativity do not believe the pseudo-tensors give a good local definition of energy density, although their integrals are sometimes useful as a measure of total energy.
One other complaint about the pseudo-tensors deserves mention. Einstein argued that all energy has mass, and all mass acts gravitationally. Does "gravitational energy" itself act as a source of gravity? Now, the Einstein field equations are
$$G_{mu,nu} = 8\pi T_{mu,nu}$$
Here $G_{mu,nu}$ is the Einstein curvature tensor, which encodes information about the curvature of spacetime, and $T_{mu,nu}$ is the so-called stress-energy tensor, which we will meet again below. $T_{mu,nu}$ represents the energy due to matter and electromagnetic fields, but includes NO contribution from "gravitational energy". So one can argue that "gravitational energy" does NOT act as a source of gravity. On the other hand, the Einstein field equations are non-linear; this implies that gravitational waves interact with each other (unlike light waves in Maxwell's (linear) theory). So one can argue that "gravitational energy" IS a source of gravity. In certain special cases, energy conservation works out with fewer caveats. The two main examples are static spacetimes and asymptotically flat spacetimes.
## Article references:
• Clifford Will, The renaissance of general relativity, in The New Physics (ed. Paul Davies) gives a semi-technical discussion of the controversy over gravitational radiation.
• Wheeler, A Journey into Gravity and Spacetime. Wheeler's try at a "pop-science" treatment of GR. Chapters 6 and 7 are a tour-de-force: Wheeler tries for a non-technical explanation of Cartan's formulation of Einstein's field equation. It might be easier just to read MTW!)
• Taylor and Wheeler, Spacetime Physics.
• Goldstein, Classical Mechanics.
• Arnold, Mathematical Methods in Classical Mechanics.
• Misner, Thorne, and Wheeler (MTW), Gravitation, chapters 7, 20, and 25
• Wald, General Relativity, Appendix E. This has the Hamiltonian formalism and a bit about deparametrizing, and chapter 11 discusses energy in asymptotically flat spacetimes.
• H. A. Buchdahl, Seventeen Simple Lectures on General Relativity Theory Lecture 15 derives the energy-loss formula for the binary star, and criticizes the derivation.
• Sachs and Wu, General Relativity for Mathematicians, chapter 3.
• John Stewart, Advanced General Relativity. Chapter 3 (Asymptopia) shows just how careful one has to be in asymptotically flat spacetimes to recover energy conservation. Stewart also discusses the Bondi-Sachs mass, another contender for "energy".
• Damour, in 300 Years of Gravitation (ed. Hawking and Israel). Damour heads the "Paris group", which has been active in the theory of gravitational radiation.
• Penrose and Rindler, Spinors and Spacetime, vol II, chapter 9. The Bondi-Sachs mass generalized.
• J. David Brown and James York Jr., Quasilocal energy in general relativity, in Mathematical Aspects of Classical Field Theory.
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I think your reasoning is partially correct.
Yes, the Universe is expanding. But, the energy is conserved. How?
From the second law of thermodynamics, $$dG = dH-TdS$$ Taking our universe as a closed system. And, our universe does not exchange heat with the surroundings (well, there are no one to exchange).
So, $$dH=0$$
As our universe is expanding, $dS$ increases. Therefore, $-TdS$ decreases. Hence, $dG$ decreases. The energy for the universe to do a work (can be seen as potential energy) is lost in expansion. As the potential energy of the universe is decreasing, it can be seen as Gravitational Potential Energy + Internal Energy (of Masses) + Electro Magnetic Radiation. The mass is conserved (for large objects (principle of momentum during collisions)), Radiation remains constant. The major change occurs in Gravitational Potential Energy. And, it decreases as $dG>0$
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OK but your argument just states that an expanding universe is a closed system... I can just as well state that it is not a closed system and this thermodynamics argument doesn't apply. – roblev Jan 24 at 3:09
And going back to basics (Newtonian mechanics) if the masses are unchanged, the distance between them increases, and the gravitational constant is unchanged then the total GPE increases. You are arguing total GPE decreases, how do you explain this basic break with Newtonian mechanics? – roblev Jan 24 at 3:15
Our universe is not in thermodynamical equilibrium, as a matter of fact. – Alexey Bobrick Jan 26 at 11:55
@roblev As the distance between masses increases, the potential energy between them decreases. E=Gmm/r – Fr34K Jan 28 at 14:10
– roblev Jan 29 at 1:53
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The problem with this question is that gravitational potential energy between massive objects is a Newtonian concept but the question of energy conservation in cosmology can only be discussed properly in terms in general relativity.
The general answer is that energy is always conserved if you take into account the energy in the gravitational field as well as the matter and radiation fields. Dark energy can also be accounted for.
The full treatment is long and gets technical so I refer to my article at http://vixra.org/abs/1305.0034
The answer in the Physics FAQ above is incorrect and the points raised are treated in my article
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quoting from your paper: "This means that as the universe expands new energy is being created out of no where. " This is/should be unacceptable in physics. Read this pdf A self-similar model of the Universe unveils the nature of dark energy. DE is an artifact in the process of measuring. – Helder Velez May 6 at 10:55
@Helder Valez You misunderstood the structure of the article. The numbered texts written in bold are paraphrasing incorrect assertions made by other people and are refuted by the text in italics. – Philip Gibbs May 6 at 11:58
my comment was addressed to all that in the last decades use free lunches in Physics. Since you depart from false premises then your article is also wrong. You will understand if you take some of your time to read the paper I've linked. – Helder Velez May 6 at 20:13
Your reasoning is not right. Energy can still conserved even expansion occurs.
Considering the simplest case that there are only two massive particles moving away from each other. As the distance increase, the potential energy increase while the kinetic energy decrease. Hence, the speed is slowing down, but they are still moving away from each other (expanding).
For large among of particles, you can also think of some kind of explosion driving them away from each other. The expansion is slowing down as time pass, but they are still expanding. There is no need for "creation of extra potential energy". To claim that the energy is not conserved, you need more evidences such as the speed distribution, etc.
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OK but why would the kinetic energy or speed of the two massive particles decrease just because the universe expanded, increasing the distance between them? The speed (measured in meters per second) shouldn't change because of expansion. – roblev Jan 8 at 13:07
The massive objects are still acting on each other gravitationally. – Deiwin Jan 9 at 11:14
I.e. the potential energy needs to come from somewhere and it comes out of the kinetic energy of those 2 objects. (I know this is a circular statement, as it relies on the conversation of energy, but it's simply a way to look at what's happening) – Deiwin Jan 9 at 11:19
yes the massive objects act on each other gravitationally, but somehow the universe expanding needs to "boost" the gravitational interaction to make the objects slow down more than if the universe was not expanding. Doesn't seem obvious to me. – roblev Jan 9 at 12:53
As bodies move away from each other due to the expansion of space, they are accelerating apart, not decelerating (that is, the more they move apart, the faster they move apart). Look up Hubble's law. – Kyle Jan 23 at 19:03
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http://math.stackexchange.com/questions/279014/results-that-came-out-of-nowhere
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# Results that came out of nowhere.
Most big results in mathematics are built on years and years of groundwork by the author and other mathematicians, such as Wiles' proof of FLT or the classification of finite simple groups. Nevertheless, there is this big romantic idea in pop culture where a young genius comes up with a brilliant proof of a major conjecture, based on some creative new idea which flies in the face of the mathematical community.
What are some examples where this has actually happened? That is, which results stem from independent work by a mathematician who came along "out of nowhere" and solved a huge problem by surprise through nonstandard techniques?
I read that the recent proposed proof of the ABC conjecture comes from years of autonomous theory by Prof Mochizuki, most of which ventures far outside of the current literature; this would seem to qualify as one example provided the proof turns out to be true.
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I have flagged this question to be made CW. – JavaMan Jan 15 at 2:55
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A totally personal remark: I don't like very much this kind of questions. That "romantic idea in pop culture" should be dispelled, not encouraged. – Giuseppe Negro Jan 15 at 3:17
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@GiuseppeNegro If it really happens, there is nothing to be dispelled. I think it is neat to hear about sudden historic advancements. But I do agree that the role of non-celebrity mathematicians should receive more attention in the media. – Alexander Gruber Jan 15 at 3:37
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The ABC conjecture has not been proved, Vesselin Dimitrov and Akshay Venkatesh found an error in the proof in October, 2012 and Mochizuki accepted the mistake. – dwarandae Jan 15 at 4:02
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@Jonathan, independently of what? You have to reconcile that view with Newton's remark that he stood "on the shoulder of giants." – alancalvitti Jan 15 at 4:04
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## 6 Answers
Marjorie Rice surprised Martin Gardner and most of the readers of Mathematical Games when she found new pentagon tilings in 1977.
With no formal training in mathematics beyond high school, she (Marjorie Rice) uncovered a tenth type of pentagon.... Her method of search was completely methodical, beginning with an analysis of what was already known.1
Several papers had supposedly proven that such tilings were not possible. Rice found three additional pentagon tilings in the years that followed.
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2
Can you point to someplace that describes these proofs and why they were wrong? The wikipedia page doesn't seem to mention them. – HappyEngineer Jan 30 at 20:38
Goedel's completeness and incompleteness theorems I believe qualify as results coming out of nowhere. Hilbert declared as one of the most important tasks for the 20th century a foundations for mathematics proved impossible by Goedel. So, this is not a case where a brilliant mathematician comes with a proof of a major conjecture but instead crumbles a major conjecture. As far as I know Goedel did not build on pervious work for his incompleteness theorems.
Grothendieck's work in algebraic geometry transformed the entire field and as far as I understand this was completely his doings.
Galois' work should perhaps be number one on the list. Certainly, nobody saw him coming and again as far as I know he developed almost all of his results on his own.
Then there is Ramanujan who proved in complete isolation an unbelievable range of results in number theory. He claimed that these results were given to him in his sleep by a goddess, so I guess that came out of nowhere.
Hamilton's creation of the quaternions can count as coming out of nowhere, at least considering the rudimentary development of abstract algebra at his time.
The discoveries of projective and hyperbolic models for planes that finally settled the quest for a proof of Euclid's proof might count as coming out of nowhere as no previous models were in existence (disregarding the fact that we all walk on a pretty good approximation of a sphere).
Cantor's creation of the heaven of set theory certainly fits the requirements (in particular, the uncountability of the real numbers --> indirect proof of existence of non-algebraic numbers).
The irrationality of the square root of 2 is a well known historical event but it is a bit hard to discren who precisely proved it so perhaps it did build on previous work, I do not know.
Since you mentioned popular movies depicting such acts of heroic mathematics, the development of game theory by Nash I believe fits the bill.
I'm sure there are more examples, but I'll stop here.
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The one I have to take objection to is Hamilton. A lot of people were thinking about numbers that encode three-dimensional rotations, and he just seems to have realized that the solution was in four dimensions before the next guy did. – Eric Stucky Jan 15 at 2:56
yes, I agree Eric. The reason I included it in the list is because of the famous story according to which the idea came out of nowhere to him, forcing him to stop along the way and write it on a bridge somewhere. – Ittay Weiss Jan 15 at 2:58
Alright, that's fair. – Eric Stucky Jan 15 at 3:04
Probably Ramanujan's work would be the easiest answer to come up with for your question.
However, I take slight issue with your classification of Mochizuki's work as being an example of "coming out of nowhere." It may be true that he worked for a while mostly on his own, but his results certainly did not come from nowhere. See this MO post and its various answers for more about the philosophy behind his work and the results that preceded it.
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I have read that post, actually. I was under the impression that the reason Mochizuki's work was still being verified is that it builds on a wealth of theory he developed essentially by himself for the last 20+ years. I don't mean to say that it does not build off of other results also. – Alexander Gruber Jan 15 at 2:43
1
– B.D Jan 15 at 2:48
I like that. +1 – Alexander Gruber Jan 15 at 2:58
– B.D Mar 20 at 10:54
Cantor's diagonal argument ${}{}$
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Shelah's proof that the Whitehead problem is independent of ZFC.
And while we're at it to some extent Cohen's technique of forcing. People already knew that the axiom of choice could be negated using atoms, and people knew how to have relative consistency and unprovability results, but Cohen's forcing was special in that it didn't generate some pathological model (like compactness arguments would) but rather take a nice model of SFC and produce another nice model of ZFC.
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RE: Cohen's forcing. See the comments in my answer. – B.D Jan 15 at 19:51
Shannon's 'Mathematical Theory of Communication' was a work that came out of 'nowhere' in that he defined a paradigm and then modeled it perfectly and completely... (IMO)
There were others interested in the field but no one had encompassed it or envisaged it correctly.
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http://en.wikipedia.org/wiki/Homeomorphic
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# Homeomorphism
(Redirected from Homeomorphic)
Not to be confused with homomorphism.
"Topological equivalence" redirects here. For topological equivalence in dynamical systems, see Topological conjugacy.
A continuous deformation between a coffee mug and a donut illustrating that they are homeomorphic. But there need not be a continuous deformation for two spaces to be homeomorphic — only a continuous mapping with a continuous inverse.
In the mathematical field of topology, a homeomorphism or topological isomorphism or bicontinuous function is a continuous function between topological spaces that has a continuous inverse function. Homeomorphisms are the isomorphisms in the category of topological spaces—that is, they are the mappings that preserve all the topological properties of a given space. Two spaces with a homeomorphism between them are called homeomorphic, and from a topological viewpoint they are the same. The word homeomorphism comes from the Greek words ὅμοιος (homoios) = similar and μορφή (morphē) = shape, form.
Roughly speaking, a topological space is a geometric object, and the homeomorphism is a continuous stretching and bending of the object into a new shape. Thus, a square and a circle are homeomorphic to each other, but a sphere and a donut are not. An often-repeated mathematical joke is that topologists can't tell their coffee cup from their donut,[1] since a sufficiently pliable donut could be reshaped to the form of a coffee cup by creating a dimple and progressively enlarging it, while preserving the donut hole in a cup's handle.
Topology is the study of those properties of objects that do not change when homeomorphisms are applied. As Henri Poincaré famously said, mathematics is not the study of objects, but instead, the relations (isomorphisms for instance) between them.[2]
## Definition
A function f: X → Y between two topological spaces (X, TX) and (Y, TY) is called a homeomorphism if it has the following properties:
• f is a bijection (one-to-one and onto),
• f is continuous,
• the inverse function f −1 is continuous (f is an open mapping).
A function with these three properties is sometimes called bicontinuous. If such a function exists, we say X and Y are homeomorphic. A self-homeomorphism is a homeomorphism of a topological space and itself. The homeomorphisms form an equivalence relation on the class of all topological spaces. The resulting equivalence classes are called homeomorphism classes.
## Examples
A trefoil knot is homeomorphic to a circle. Continuous mappings are not always realizable as deformations. Here the knot has been thickened to make the image understandable.
• The unit 2-disc D2 and the unit square in R2 are homeomorphic.
• The open interval (a, b) is homeomorphic to the real numbers R for any a < b. (In this case, a bicontinuous forward mapping is given by f = 1/(x − a) + 1/(x − b) while another such mapping is given by a scaled and translated version of the tan function).
• The product space S1 × S1 and the two-dimensional torus are homeomorphic.
• Every uniform isomorphism and isometric isomorphism is a homeomorphism.
• The 2-sphere with a single point removed is homeomorphic to the set of all points in R2 (a 2-dimensional plane).
• Let A be a commutative ring with unity and let S be a multiplicative subset of A. Then Spec(AS) is homeomorphic to {p ∈ Spec(A) : p ∩ S = ∅}.
• Rm and Rn are not homeomorphic for m ≠ n.
• The Euclidean real line is not homeomorphic to the unit circle as a subspace of R2 as the unit circle is compact as a subspace of Euclidean R2 but the real line is not compact.
## Notes
The third requirement, that f −1 be continuous, is essential. Consider for instance the function f: [0, 2π) → S1 defined by f(φ) = (cos(φ), sin(φ)). This function is bijective and continuous, but not a homeomorphism (S1 is compact but [0, 2π) is not).
Homeomorphisms are the isomorphisms in the category of topological spaces. As such, the composition of two homeomorphisms is again a homeomorphism, and the set of all self-homeomorphisms X → X forms a group, called the homeomorphism group of X, often denoted Homeo(X); this group can be given a topology, such as the compact-open topology, making it a topological group.
For some purposes, the homeomorphism group happens to be too big, but by means of the isotopy relation, one can reduce this group to the mapping class group.
Similarly, as usual in category theory, given two spaces that are homeomorphic, the space of homeomorphisms between them, Homeo(X, Y), is a torsor for the homeomorphism groups Homeo(X) and Homeo(Y), and given a specific homeomorphism between X and Y, all three sets are identified.
## Properties
• Two homeomorphic spaces share the same topological properties. For example, if one of them is compact, then the other is as well; if one of them is connected, then the other is as well; if one of them is Hausdorff, then the other is as well; their homotopy & homology groups will coincide. Note however that this does not extend to properties defined via a metric; there are metric spaces that are homeomorphic even though one of them is complete and the other is not.
• A homeomorphism is simultaneously an open mapping and a closed mapping; that is, it maps open sets to open sets and closed sets to closed sets.
• Every self-homeomorphism in $S^1$ can be extended to a self-homeomorphism of the whole disk $D^2$ (Alexander's trick).
## Informal discussion
The intuitive criterion of stretching, bending, cutting and gluing back together takes a certain amount of practice to apply correctly—it may not be obvious from the description above that deforming a line segment to a point is impermissible, for instance. It is thus important to realize that it is the formal definition given above that counts.
This characterization of a homeomorphism often leads to confusion with the concept of homotopy, which is actually defined as a continuous deformation, but from one function to another, rather than one space to another. In the case of a homeomorphism, envisioning a continuous deformation is a mental tool for keeping track of which points on space X correspond to which points on Y—one just follows them as X deforms. In the case of homotopy, the continuous deformation from one map to the other is of the essence, and it is also less restrictive, since none of the maps involved need to be one-to-one or onto. Homotopy does lead to a relation on spaces: homotopy equivalence.
There is a name for the kind of deformation involved in visualizing a homeomorphism. It is (except when cutting and regluing are required) an isotopy between the identity map on X and the homeomorphism from X to Y.
## See also
• Local homeomorphism
• Diffeomorphism
• Uniform isomorphism is an isomorphism between uniform spaces
• Isometric isomorphism is an isomorphism between metric spaces
• Dehn twist
• Homeomorphism (graph theory) (closely related to graph subdivision)
• Homotopy#Isotopy
• Mapping class group
• Poincaré conjecture
## References
1. Hubbard, John H.; West, Beverly H. (1995). Differential Equations: A Dynamical Systems Approach. Part II: Higher-Dimensional Systems. Texts in Applied Mathematics 18. Springer. p. 204. ISBN 978-0-387-94377-0.
2. Poincaré, Henri. "Chapter II: Mathematical Magnitude and Experiment". Science and Hypothesis.
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http://math.stackexchange.com/questions/210996/latin-squares-proving-the-unique-number-of-sudoku-that-can-be-generated?answertab=oldest
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# Latin Squares - Proving the Unique number of Sudoku that can be generated
I recently read that the Sudokus are just Latin Squares for $n = 9$. I know that proving the number of Latin Squares is considered difficult to generalize in terms of $n$. I would like to know if there is any method of proving the unique number of Sudokus taking a special case for $n = 9$.
The Wikipedia page on Latin Squares shows that for $n = 9 , number\ of\ possible\ Latin\ Squares = 5524751496156892842531225600$. I would like to know different possible methods used to arrive at the answer (e.g using Brute-force , or generating for the reduced Latin-Squares or any other combinatorial method).
PS: I have also read this question asked here and the paper referred by the answer. I am not able to understand why the method fails for $n>11$. Even the OEIS page does not have sequences for $n>11$
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## 1 Answer
Sudokus are a proper subset of the Latin squares of order $9$, as they have the added restriction of the $3 \times 3$ boxes. Thus, all sudokus are Latin squares, but not all Latin squares are sudokus. As with Latin squares, there is no computer-free proof that these numbers are correct. Typically, these are checked by performing independent computations (possibly by slightly different methods).
That being said, most of the searching can be eliminated by identifying symmetries. Both Latin squares, in general, and sudokus have "symmetries", which can be exploited to give a significant reduction in the search space. In a Latin square, for instance, we can permute the rows arbitrarily to give another Latin square (so it'd be a waste of time to count all of these separately). I'm not particularly familiar with the methods used in enumerating sudokus, but a website by Jarvis (and the linked papers) offers much detail into his enumeration method.
There are $R_9=377597570964258816$ reduced Latin squares of order 9. I don't think any computer ever has counted from 1 to $R_9$, let alone play around with Latin squares at each step. Thus, it's safe to say a brute force enumeration is completely out of the question.
To date, the easiest way to find $R_9$ is to use Sade's method which I mention in my answer to the linked question. Sade's method is the only feasible way for order $10$ or greater. Unfortunately, Sade's method was published only in a very obscure paper and is hard to obtain. But I describe it in great detail in my survey paper (here).
Roughly speaking, Sade's method saves an enormous amount of time by identifying Latin rectangles that have the same number of completions, and clumping them together. We then count the number of ways of extending each equivalence class by one row, then identify which of these extended Latin rectangles admit the same number of completions, and so on recursively.
(Note: In an earlier version of this post, I claimed it was the only way for $9 \times 9$ squares, but there is actually another way in this case.)
For $9 \times 9$ Latin squares, it's possible to iterate through representatives of the 19270853541 main classes of Latin squares on a computer, and at each step calculate the size of the autoparatopism group $\mathrm{Par}(L)$. The total number of $9 \times 9$ Latin squares is thus $$\sum_L \frac{6n!^3}{|\mathrm{Par}(L)|},$$ where the sum is over the 19270853541 representatives. Generating these representatives can be done via the "canonical construction path" described in:
B. D. McKay, Isomorph-free exhaustive generation, J Algorithms, 26 (1998), 306-324.
See this paper for more details (including the relevant definitions: "main class" and "autoparatopism"):
B. D. McKay, A. Meynert, W. Myrvold, Small Latin squares, quasigroups, and loops. J. Combin. Des. 15 (2007), 98-119.
Sade's method for $12 \times 12$ Latin squares would work fine, if we had a sufficiently powerful computer with enough memory (and the budget to use it, and know-how to program it efficiently). I think it's safe to say we (as a species) could find this number in the next 100 years or so (I'm hoping to see $R_{12}$ before I die).
Note that the number of Latin squares grows quite fast. Most people think $n!$ grows quickly. Well, Smetaniuk showed that $L_{n+1} \geq (n+1)!\ L_n$ (where $L_n$ counts non-reduced Latin squares). The sheer number of Latin squares is the obstacle here.
B. Smetaniuk, A new construction of Latin squares. II. The number of Latin squares is strictly increasing, Ars Combin., 14 (1982), pp. 131-145.
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– Nick Alger Oct 11 '12 at 11:05
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http://mathhelpforum.com/advanced-algebra/189839-deduce-lagranges-theorem-print.html
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# Deduce Lagrange's Theorem
Printable View
• October 8th 2011, 10:27 AM
dwsmith
Deduce Lagrange's Theorem
Let H be a group acting on a set A. Prove that the relation ~ on A defined by a~b iff $a=hb$ for some h in H is an equivalence relation.
I have already shown this.
Let H be a subgroup of the finite group G and let H act on G by left multiplication. Let x exist in G and let O be the orbit of x under the action of H. Prove that the map $H\to O$ defined by $h\mapsto hx$ is a bijection.
I have already shown this too.
From these two statements, deduce Lagrange's Theorem: if G is a finite group and $H\leq G$, then $|H|| |G|$.
I understand Lagrange's Theorem. I need an explanation on how I can deduce his theorem from the above.
• October 8th 2011, 10:31 AM
FernandoRevilla
Re: Deduce Lagrange's Theorem
Quote:
Originally Posted by dwsmith
I understand Lagrange's Theorem. I need an explanation on how I can deduce his theorem from the above.
Hint All the equivalence classes have the same cardinality.
• October 8th 2011, 10:41 AM
dwsmith
Re: Deduce Lagrange's Theorem
Quote:
Originally Posted by FernandoRevilla
Hint All the equivalence classes have the same cardinality.
I know that since each one is a coset but I still don't understand.
• October 8th 2011, 11:15 AM
Deveno
Re: Deduce Lagrange's Theorem
if every coset has the same size, then they all have the same size as the coset containing the identity, which is H.
so the coset containing the identity, H, has cardinality |H|, as does every coset Hg.
so, summing over our partition of G:
|G| = |H| + |Hg1| + |Hg2| +....+ |Hgk|, where k is the number of distinct cosets. and....?
• October 8th 2011, 11:17 AM
FernandoRevilla
Re: Deduce Lagrange's Theorem
Quote:
Originally Posted by dwsmith
I know that since each one is a coset but I still don't understand.
The set $G$ has the form $G=\cup_{i}C_i$ and the family of cosets $\mathcal{C}=\{C_i\}$ is pairwise disjoint with $|C_i|=|H|$. So, $|G|=|\mathcal{C}||H|$ .
• October 8th 2011, 11:33 AM
Deveno
Re: Deduce Lagrange's Theorem
and the pairwise-disjointness arises because being in the same coset is an equivalence relation (what you proved at the beginning), and equivalence classes partition the set they are on.
(equivalence classes are "quotient sets").
All times are GMT -8. The time now is 03:45 AM.
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http://electronics.stackexchange.com/questions/58603/nodal-analysis-for-current-controlled-current-source
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# Nodal analysis for current controlled current source
Okay here I go. I just started learning nodal analysis technique but I feel its difficult in solving the circuits consisting of dependent current and voltage sources.
Im unable to solve this circuit.
The current source is a dependent current source and whose value is 180 times Ix and Ix is the current flowing through the 0.7 v battery.
Generally to solve this type of circuits we find a relation to express Ix in terms of node voltages / resistance. example `Ix = (V2 - V1 ) / 500 ohms .` But here there is no resistance in the 0.7 v branch voltage. I am unable to find a solution to solve it. Could anyone kindly explain me the solution to this circuit.
NOTE I have found a solution by finding the thevinin's resistance across 0.7 v battery and replaced it with Rth in series with 0.7v and I have got the solution but I want to know, Can we get it using only KCL equations (nodal analysis) ?
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## 2 Answers
AFAIK, you can always solve any linear circuit the 'brute force' way using nodal analysis:
1. Write Kirchoff's Current Equations on all nodes except ground
2. For every circuit component, (i.e. resistors, capacitors etc.), write down their behaviour (for instance, ohm's law for a resistance, i = c dV/dt for a capacitance and so on)
3. At this point, we'll have a handful of equations with us. We can also try to eliminate as many equations from them as possible using any info we have; however in the end, we need to be left with N simultaneous equations in N unknowns. Solve them and we'll get all the node voltages and branch currents.
Coming to the circuit above, let's define the current through V2 as I2, and the ones through R_n as I_n. Let me also denote the node at the top as V_a, the node between the CCCS and R5 as V_c, the one between the R_7 and R_8 as V_b and the node in the middle as V_e. Now, writing KCL on these nodes will leave us with $$I_2 = I_7 + I_5\\ I_7 = I_8 + I_x\\ I_x + I_5 = I_6$$ respectively. Writing down the 'behaviour' of R6, R5, R7, R8, V2, V3 and the CCCS will respectively yield $$V_E = I_6 R_6 \\ V_A - V_C = I_5 R_5 \\ V_A - V_B = I_7 R_7 \\ V_B = I_8 R_8 \\ V_A = V_2 \\ V_B = V_E + 0.7\\ I_5 = 180 I_x$$
That's 10 linear equations in 10 unknowns. Solve them, and we'll find all I_x as 88.18uAmps...
Of course, 10 equations is a bit too much to solve by hand (I generally use Gauss-Jordan elimination to do this part), but as far as I've seen, this method works in situations where the usual 'text-book' approach using nodal and mesh analyses fail. Furthermore, we don't have to deal with the painful Thevenin equivalent/Super-mesh workarounds here...
On the downside, I'm not quite sure if this approach works with every possible circuit (so far I haven't seen any where it fails), so any negative feedback on this part is welcome :)
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Thank you so much.. That is the right answer :) but could you explain me the equation Ve = I5 * R6 ? how did we get . Sorry but I find it difficult myself to calculate voltages like Vb or Va. Could you tell me any simple way to understand those things or provide some good links where I can improve my skills kindly explain me this please. Thank you @nav – niko Feb 20 at 18:15
I guess Ve = I6 * R6 ? – niko Feb 20 at 18:18
Please make it complete – niko Feb 20 at 19:33
Sorry for that typo; It's V_E = I_6 * R_6 as you said. Fixed :) – nav Feb 21 at 9:49
@niko: Frankly speaking, analog circuits are real sources of pain. To solve them, we really don't need anything more than what our textbooks provide, but then, we'll always stumble upon some exceptional circuit topology where the 'regular' textbook methods fail; however AFAIK, they can always be solved using the same methods, only that we should not restrict ourselves to using a single method (nodal/mesh). If you can figure out how a "(resistor + current_source in series) parallel to (volt_source + resistor in parallel)" works, I think you'll be able to play with any circuit. Good luck:) – nav Feb 21 at 10:51
show 1 more comment
The problem with solving this circuit by nodal analysis is not the CCCS, it's the voltage source V3. In nodal analsyis, voltage sources can't be directly accomodated but have to be accounted for bye creating "supernodes". In this case you'd combine the nodes at the left and right of V3 to make a supernode.
But of course that means that Ix is no longer a variable in the analysis.
Luckily, the basics of nodal analysis give a clue how to solve this. We just need to apply KCL at one of the nodes connected to V3.
For example, if we define the currents through R7 and R8 as I7 and I8, both defined in the "top-to-bottom" direction as the circuit is drawn, then we can easily find
Ix = I7 - I8
Since I7 and I8 can be written in terms of the node voltages, we can now continue with the nodal analysis.
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http://math.stackexchange.com/questions/216926/is-normal-in-base
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# Is π normal in base π?
This question states that π is normal:
Does Pi contain all possible number combinations?
My understanding of this is that it means that the statistically, the distribution of every number is equal across the infinite range.
If the numbering system is base π, wouldn't the number just be 1, so not normal, or does the definition only mean the bases that the number would be infinately non-repeating?
e.g π, in base π is 1 and not infinately non-repeating π in base 10 is infinately non-repeating
Let me summerise, does π base π = 1 mean that π isn't normal, or is π base π excluded from the definition because it in not infinately non-repeating?
Cheers
Dave
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5
What does base $\pi$ mean after all? Usually one talks about bases $b$ only if $b$ is an integer $\ge 2$. – Hagen von Eitzen Oct 19 '12 at 14:58
1
You have to define what it means to be in a base that is not an integer. In particular, with integer base, "almost" every real number is uniquely represented in the base. I'm not sure this is true for non-integer bases. – Thomas Andrews Oct 19 '12 at 15:00
## 1 Answer
When we say that $x$ is normal, what we mean is that it's normal to base $b$ for every integer $b\ge2$. Base $\pi$ does not enter into the discussion.
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Both of you have mentioned that bases can only be integers... I can't see why this should be at all, surely what values are integers and what are not ate nearly a product of your choice of base. However, my original question was if pi as excluded from the rules for a normal number definition in base pi... which it seems is the case. In think I'll start a new question about integers being defined by your choose of base. Thanks all. – BanksySan Oct 21 '12 at 20:49
Having reconsidered... I see the argument. I was on the wrong track. Thanks all for answering though. – BanksySan Oct 21 '12 at 21:29
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Careful --- I didn't say "bases can only be integers" --- I said that when it comes to deciding whether something is normal bases can only be integers. The reason for this is clear; if you don't restrict the bases, there aren't any normal numbers, since no number $x$ is normal to base $x$. Bases don't have to be integers, and there are many papers devoted to the properties of expansions to non-integer bases. It's just in the context of normality that we can't get anywhere without the restriction. – Gerry Myerson Oct 21 '12 at 22:51
Thank for explaining it. – BanksySan Oct 22 '12 at 11:45
Kind of like how there would be no prime numbers without the restriction to integer divisors... – Alistair Buxton Nov 4 '12 at 10:54
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http://mathoverflow.net/questions/11354/extension-of-induced-reps-over-z-is-it-a-sum-of-induced-reps
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## Extension of induced reps over Z: is it a sum of induced reps?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $G$ be a finite group. If $L$ is a finite free $\mathbf{Z}$-module with an action of $G$, say $L$ is induced if it's isomorphic as a $G$-module to $Ind_H^G(\mathbf{Z})$ with $H$ a subgroup of $G$ and $H$ acting trivially on $\mathbf{Z}$. And, for want of better terminology, let's say $L$ is a sum-of-induceds if it's isomorphic to a direct sum of induced modules in the sense above. [EDIT: Ben Webster points out that "permutation representation" is a rather better name for this notion! It's just the $\mathbf{Z}$-module coming from the action of $G$ on a finite set.]
The question: Is a finite free $\mathbf{Z}$-module which is an extension of one sum-of-induceds by another, also a sum-of-induceds?
Over $\mathbf{Q}$ this is trivial because every short exact sequence splits. But this is not true over $\mathbf{Z}$. For example there are two actions of $\mathbf{Z}/2\mathbf{Z}$ on $\mathbf{Z}^2$ which become the group ring over $\mathbf{Q}$: one has the non-trivial element acting as $(1,0;0,-1)$ and the other has it acting as $(0,1;1,0)$. The latter is a non-split extension of the trivial 1-d representation by the non-trivial one (and also a non-split extension of the non-trivial one by the trivial one). Note that this non-split extension is induced.
There are mod $p$ extensions that don't split either. For example over $\mathbf{Z}/p\mathbf{Z}$ there is a non-trivial extension of the trivial representation by itself. But this extension does not lift to an extension of the trivial $\mathbf{Z}$-module by itself.
Why am I interested? For those that know what a $z$-extension of a connected reductive group over a number field is, my "real" question is: is a $z$-extension of a $z$-extension still a $z$-extension? I've checked the geometric issues here but the arithmetic one above is the one I haven't resolved. $G$ is a Galois group and the $\mathbf{Z}$-modules are the character groups of the central tori in question. If I've understood things correctly, a $z$-extension of a $z$-extension is a $z$-extension iff the question I ask above has a positive answer.
Note finally that applying the long exact sequence of cohomology, and using the fact that induced representations have no cohomology by Shapiro's Lemma, we see that the extension I'm interested in also has no cohomology (and furthermore its restriction to any subgroup has no cohomology either). Is this enough to show it's induced?
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2
I think "sum of induced" just means "permutation representation." Am I missing something? – Ben Webster♦ Jan 10 2010 at 18:41
I think you're right. – Kevin Buzzard Jan 10 2010 at 19:17
## 1 Answer
In my interpretation of your description, a sum-of-induceds is just the permutation representation of some (not necessarily connected) G-set. The representation $\mathrm{Hom}(V_1,V_2)$ for two permutation representations is itself permutation: it's the permutation action on product of the G-sets.
So, $\mathrm{Ext}^1(V_1,V_2)=H^1(G,\mathrm{Hom}(V_1,V_2))$ which you claim vanishes. So it sounds to me like every extension between permutation representations splits. [EDIT: I'm no longer nervous] I'm still a little nervous about this cohomology vanishing, but I don't know integral representation theory so well, so it's hard for me to say.
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I don't think I'm claiming that the ext group vanishes. I'm just saying that every element of the ext group gives rise to a sum-of-induceds representation, not that it's the sum of the two sum-of-induceds we're starting with. – Kevin Buzzard Jan 10 2010 at 19:12
Oh, but stop, the ext group clearly vanishes by Shapiro. So done! Thanks! – Kevin Buzzard Jan 10 2010 at 19:20
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For any finite group G, the first cohomology group H^1(G,Z) of the group G with constant integral coefficients Z vanishes, since there are no nonzero group homomorphisms G->Z. By the Shapiro Lemma, it follows that the first cohomology of G with coefficients in any integral permutational representation vanishes, too. Hence your argument is indeed correct and any extension of integral permutational representations splits. – Leonid Positselski Jan 10 2010 at 19:49
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http://math.stackexchange.com/questions/149647/maximal-subgroups-of-a-finite-p-group
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# Maximal subgroups of a finite p-group
I want to prove the following:
Let $G$ be a finite abelian $p$-group that is not cyclic. Let $L \ne {1}$ be a subgroup of $G$ and $U$ be a maximal subgroup of L then there exists a maximal subgroup $M$ of $G$ such that $U \leq M$ and $L \nleq M$.
Proof. If $L=G$ then we are done. Suppose $L \ne G$ . Let $|G|=p^{n}$ then $|L|=p^{n-i}$ and $|U|=p^{n-i-1}$ for some $0<i<n$. There is $x_{1} \in G$ such that $x_{1} \notin L$. Thus $|U\langle x_{1}\rangle|=p^{n-i}$ and does not contain L. There is $x_{2} \in G$ such that $x_{2} \notin L$ and $x_{2} \notin |U\langle x_{1}\rangle|$. Thus $|U\langle x_{1}\rangle\langle x_{2}\rangle|=p^{n-i+1}$. Continuing like this, we get $|U\langle x_{1}\rangle \langle x_{2}\rangle\cdots \langle x_{i}\rangle|=p^{n-1}$ is a maximal subgroup of $G$. The problem is, I am not sure that this subgroup does not contain $L$.
Thanks in advance.
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1
Let $G=C_4$, $L=\langle 2\rangle$, $U=\langle 0\rangle$. What maximal subgroup of $G$ will you pick that does not contain $L$? – Arturo Magidin May 25 '12 at 13:37
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It's false even with $G$ non-cyclic. Take $G = \langle x \rangle \oplus \langle y \rangle$ with $2x=4y=0$, $L=\langle x,2y \rangle$, $U = \langle x \rangle$. – Derek Holt May 25 '12 at 13:54
You need the condition $G/U$ not cyclic. – jug May 25 '12 at 13:58
## 1 Answer
Your proof fails if G is cyclic, by the uniqueness of sbgps. of a given divisor of the order of the group.
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assume G is not cyclic – user28083 May 25 '12 at 13:46
Try then a proof by some inductive argument on the number of different cyclic factors of G – DonAntonio May 25 '12 at 13:51
Thanks for everyone. It is not true. – user28083 May 25 '12 at 14:18
3
You're welcome. Perhaps you're interested in enhancing your acceptation rate? – DonAntonio May 25 '12 at 14:25
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http://math.stackexchange.com/questions/202199/is-this-log-form-simple-enough?answertab=active
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# Is this log form simple enough?
$$\frac{3}{\ln{2}-12}$$ Is this form simplified enough?
There is a number '$12$' below the fraction line, do i need to transform the $\log$ more to make it simpler?
I wrote that in a college math exam
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4
Simple enough for what? – Henning Makholm Sep 25 '12 at 14:01
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Yeah,it seems like we can't make it simplier – user42625 Sep 25 '12 at 14:02
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Whether it is simplified enough depends on what you are doing. If this is a homework answer, then it will depend on what your teacher expects, and what convensions have been set in class. If this is a part of your own research, then it depends on what additional manipulations you might need to make. It might be less ambiguous to ask if a simpler form can be found. – flodyninja Sep 25 '12 at 14:02
## 1 Answer
The only complaint I can see is that it obscures the fact that it is negative, so one might prefer $\frac {-3}{12-\ln 2}$ but otherwise I don't see a simpler form. I guess you could go to $$\frac 3{\ln \frac 2{e^{12}}}= \frac 1{\ln \sqrt [3]{\frac 2{e^{12}}}}=\frac 1{\ln \frac{\sqrt [3] 2}{e^4}}$$ but I don't think this is progress.
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http://gowers.wordpress.com/2009/01/27/
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# Gowers's Weblog
Mathematics related discussions
## Archive for January 27th, 2009
### Is massively collaborative mathematics possible?
January 27, 2009
Of course, one might say, there are certain kinds of problems that lend themselves to huge collaborations. One has only to think of the proof of the classification of finite simple groups, or of a rather different kind of example such as a search for a new largest prime carried out during the downtime of thousands of PCs around the world. But my question is a different one. What about the solving of a problem that does not naturally split up into a vast number of subtasks? Are such problems best tackled by $n$ people for some $n$ that belongs to the set $\{1,2,3\}$? (Examples of famous papers with four authors do not count as an interesting answer to this question.)
It seems to me that, at least in theory, a different model could work: different, that is, from the usual model of people working in isolation or collaborating with one or two others. Suppose one had a forum (in the non-technical sense, but quite possibly in the technical sense as well) for the online discussion of a particular problem. The idea would be that anybody who had anything whatsoever to say about the problem could chip in. And the ethos of the forum — in whatever form it took — would be that comments would mostly be kept short. In other words, what you would not tend to do, at least if you wanted to keep within the spirit of things, is spend a month thinking hard about the problem and then come back and write ten pages about it. Rather, you would contribute ideas even if they were undeveloped and/or likely to be wrong. (more…)
Posted in Mathematics on the internet, polymath1 | 188 Comments »
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http://mathhelpforum.com/algebra/4325-quadradic-equation.html
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# Thread:
1. ## Urgent!!!
Solve by using the quadratic formula $x^2-13x+30=0$
2. Surely, Brooke, you can use the quadratic formula to solve this yourself.
Please say you can. It's just a matter of plugging and chugging.
That's why the formula is so handy. Just be careful with your signs.
3. Originally Posted by Brooke
Solve by using the quadratic formula $x^2-13x+30=0$
Note,
$a=1,b=-13,c=30$
Thus,
$x=\frac{b\pm \sqrt{b^2-4ac} }{2a}$
Thus,
$x=\frac{-13\pm \sqrt{169-4(1)(30)}}{2(1)}$
Thus,
$x=\frac{-13\pm \sqrt{169-120}}{2}$
Thus,
$x=\frac{-13\pm 7}{2}$
Thus,
$x=\frac{-13+7}{2}=-3,x=\frac{-13-7}{2}=-10$
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http://physics.stackexchange.com/questions/46705/how-to-relate-speed-of-sound-with-relative-humidity/46707
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# How to relate speed of sound with relative humidity?
I am exploring the idea of measuring the humidity of a space using sound waves, however I am having trouble finding a mathematical relationship between the speed of sound and the humidity level.
$c_{air} = 331.3 \sqrt{1 + \frac{T}{273.15}}$ but this is for dry air (0%RH)
How can I factor the effects of humidity into this relationship?
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## 2 Answers
Speed of sound in a gas is given by the equation: $$c = \sqrt{\gamma R T}$$
where $\gamma = c_p/c_v$ ( $c_p$ and $c_v$ are specific heats), $R$ is the gas constant, and $T$ is temperature. The specific heat of a gas changes with humidity, so varying these will vary your calculated speed of sound.
This page has a calculator as well as a great explanation of how their formula works.
Hope this helps!
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The speed of sound in a gas is:
$$c = \sqrt{\gamma R T}$$
where $\gamma = c_p/c_v$ is the ratio of specific heats, $R$ is the specific gas constant and $T$ is temperature. Both $\gamma$ and $R$ depend on the composition of the gas, which includes humidity in air.
The specific heats are $c_p = 1.005+1.82H$ (see this answer) where $H$ is the absolute humidty and $c_v = c_p - R$. Finally, $R = R_{univ}/M_{gas}$ where $M_{gas}$ is the molecular weight of the gas (which depends on humidity).
To get it all in terms of relative humidity is just an exercise in unit conversion.
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http://mathoverflow.net/questions/71002?sort=newest
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## divisors on Abelian varieties
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $A$ be an Abelian variety. Let $L$ and $C$ be an effective divisor and a curve on $A$, respectively. If $C$ is not including in $L$, can we conclude that $L+C=A$? Thanks.
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## 2 Answers
Let us try to come up with a criterion for $L+C\ne A$. Let us assume both $L$ and $C$ are irreducible. (Otherwise consider separate irreducible components.) Let us also shift both $L$ and $C$ so that they pass through the origin. (I guess here I am implicitly assuming the field is algebraically closed, so $L$ and $C$ have points.)
Note that $L+C$ is an irreducible closed variety containing $L$. If $L+C\ne A$, then $L+C=L$. This implies $C+C+C+\dots+C\subset L$ for any number of summands. Clearly, the chain $C\subset C+C\subset\dots$ must stabilize; denote the limit by $B$. It is a semigroup: $B+B=B$. Therefore, it is an abelian variety. It has the following properties: $C\subset B$ and $B+L\subset L$. We thus arrive at the following criterion:
$L+C\ne A$ if and only if there is an abelian subvariety $B$ such that $C$ lies in a translate of $B$ and $L$ is the preimage of a divisor under $A\to A/B$.
P.S. Note that for such $L$ and $C$, we clearly have $L.C=0$, as suggested by Jack Huizenga in his answer.
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Not necessarily. For instance, if $L\subset A$ is a subgroup and $C$ is contained in a translate $L + x$ of $L$, then $L + C$ would be contained in $L+x$.
Additional hypotheses should make this work out, though. For instance, it should be true if $C$ intersects $L$ transversely at some point $x$. For then the differential of the map $L\times C \to A$ at the point $(x,x)$ will be surjective, and so the map is dominant, hence surjective in light of projectivity. I doubt the intersection actually has to be transverse (in particular, I believe $L.C >0$ should be enough), but somebody else can fill in the details.
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http://mathoverflow.net/questions/60775/projecting-the-unit-cube-onto-subspaces
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## Projecting the unit cube onto subspaces
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Given a set $S$ of non-zero vectors in $\mathbb{R}^n,$ and a subspace $L,$ consider $f(S,L)=\max_{s \in S}\frac{\|Ps\|_2}{\|s\|_2}$ where $Ps$ is the orthogonal projection of $s$ onto $L.$
Specifically, consider the set $X$ consisting of the $2^n-1$ (nonzero) vectors with all coordinates $0$ or $1$.
A recent question concerns criteria which might show, for a given subspace $L$, that $f(X,L)$ is small. Here I am concerned with choosing the subspace:
For each $n$ and `$d<n$`, what is the minimum over all $d$-dimensional subspaces of $f(X,L)?$
If that seems too broad, then the case $d=1$ would be of interest.
Summary There are several good answers here. It is a toss-up which one to choose. I'm going to repost the `$d>1$` case as a new question.
Here is a somewhat selective review, all for the case $d=1.$
1. Gerhard mentioned that $(1,-1,0,0,\cdots)$ attains $\sqrt{1/2}.$
2. Seva suggested that $(1,1/\sqrt{2},1/\sqrt{3},\cdots)$ is asymptotically optimal attaining $O(\sqrt{1/\log n}).$ This is indeed better but the smallest $n$ for which it beats $\sqrt{1/2}$ is (by my calculations) $n=1203.$ The nice optimality proof references $(1,\sqrt{2}-1,\sqrt{3}-\sqrt{2},\cdots)$ This is actually a better choice, it first beats $\sqrt{1/2}$ for $n=56.$ The two choices are comparable (up to scalar multiple) because $\sqrt{k+1}-\sqrt{k} \approx 1/2\sqrt{k}$. The approximation is pretty good, except for $k=0.$ As this is the largest entry, it takes longer to beat $\sqrt{1/2}$.
3. Denis essentially does mention $(1,\sqrt{2}-1,\sqrt{3}-\sqrt{2},\cdots)$ and, at least in one edit, suggests reflecting in the center: So for $n=6$ use $(a_1,a_2,a_3,-a_3,-a_2,-a_1)$ where $a_k=\sqrt{k}-\sqrt{k-1}.$ This is superior at $n=4$ and even at $n=3$ with the right rule for the center: It seems obvious that there should be symmetry and entries summing to $0$, but actually $(1,\sqrt{2}-1,-1)$ is the best choice for $n=3$.
4. Emil nicely lays this out in a comment. Here is my somewhat informal explanation (with some minor details skipped) , in case another perspective is useful. I suggest at the end that this should remind one of linear programming.:
There is no harm in assuming that the subspace is generated by a vector `$v=(v_1,v_2,\cdots,v_n)$` with `$v_1 > v_2 > \cdots > v_n$` (actually, only $\ge$ is clear, but I will ignore that in the interests of brevity.). By a mild abuse of notation, for `$j>0$ let $x_{j}$` be the vector whose first $j$ entries are $1$ and the rest $0$ while `$v_{-j}$` has first $n-j$ entries $0$ and the last $j$ equal to $1$.
Given $v$, Let $B \subset X$ be the set of vectors in $X$ which attain the maximum of $\frac{\|Px\|}{\|x\|}.$ I make two claims about $B$:
• Every vector in $B$ is $x_j$ for some $j$ with `$v_j > 0$` or else $x_{-j}$ for some $j$ with `$v_{n-j} < 0$` (think about why any other vector with $j$ entries equal to $1$ will be worse.)
• Unless $B$ is a basis of $\mathbb{R}^n$, There is a $v$ which gives a better ratio. (Because otherwise we should be able perturb the entries of $v$ in such a way as to lower $\frac{\|Px\|}{\|x\|}$ for the vectors already in $B$ (simultaneously keeping them equal and maximal) while raising that ratio slightly for some $x_j$ not yet in $B$.
Together these tell us that $B=\lbrace x_{1},x_{2},\cdots x_n\rbrace$ or else $B=\lbrace x_{1},x_{2},\cdots x_q;x_{q-n},\cdots,x_{-2},x_{-1}\rbrace$ where $v_q>0>v_{q+1}.$ Since the various ratios must all be the same, we deduce that `$v=(\alpha a_1,\alpha a_2, \cdots,\alpha a_q; -\beta a_{n-q},\cdots,-\beta a_2,-\beta a_1)$` for the `$a_k$` as above. Here $\alpha$ and $\beta$ are positive constants. A bit of reflection shows that they must be equal and hence can be taken to be $1$. Finally, $q$ should be $n/2$ (rounded if needed).
Comment: (disclaimer: I am an optimist but not an optimizer so the following may inexact.) Here we have the convex optimization problem of choosing $v$ so as to minimize the largest of $\frac{\|Px\|_2}{\|x\|_2}.$ Had it been $\frac{\|Px\|_1}{\|x\|_1}$ we would have been able to use linear programming. My argument above has the feel of linear programing, perhaps using duality. Perhaps a similar method could uncover optimal subspaces of dimension `$d \ge 2.$`
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I presume that the norm is the standard Euclidian and the projection is orthogonal. – Denis Serre Apr 6 2011 at 6:58
Yes, that was my intention. – Aaron Meyerowitz Apr 6 2011 at 7:01
If n=2, I imagine that the subspace (t, -t) gives the minimum value. For n > 2, and d=1, the subspace (t, -t, 0, ...,0) sets the limbo bar pretty low. For arbitrary d, looking at n = d+1 might help with determining the minimum, and will be something like the subspace orthogonal to (1,1,...,1) (d+1) ones, which you can generalize. This is not a proof so much as a goal. Gerhard "How Low Can You Go" Paseman, 2011.04.06 – Gerhard Paseman Apr 6 2011 at 7:08
@Aaron: there is a typo in the title of the question. @Everybody else: if you like the present question, you may wish to check the question it originated from (mathoverflow.net/questions/60604) and consider voting to re-open it. – Seva Apr 6 2011 at 17:51
## 2 Answers
My answer concerns with the case $d=1$ only. Without loss of generality, we can focus on the subspaces, generated by a vector with all coordinates non-negative. It is easy to verify that for the subspace $L$, generated by the vector `$(1,1/\sqrt{2},...,1/\sqrt{n})$`, the projection onto $L$ of any non-zero vector `$\epsilon\in\{0,1\}^n$` has lenght at most `$\frac{2}{\sqrt{\log n}}\,\|\epsilon\|$`. This is essentially the worst case as, on the other hand, for any non-zero vector $z\in R^n$ with non-negative coordinates there exists a non-zero vector `$\epsilon\in\{0,1\}^n$` such that $$\langle z,\epsilon \rangle \ge \frac{2}{\sqrt{\log n+4}}\,\|z\|\|\epsilon\|.$$
To see this, write $z=(z_1,...,z_n)$ and, without loss of generality, assume that $$z_1 \ge \dotsb \ge z_n \ge 0\quad \text{and}\quad \|z\|=1.$$ Let $\tau := 2/\sqrt{\log n+4}$. We will show that there exists $k\in[n]$ with $z_1+...+z_k\ge\tau\sqrt k$; choosing then $\epsilon$ to be the vector with the first $k$ coordinates equal to $1$ and the rest equal to $0$ completes the proof.
Suppose, for a contradiction, that $z_1+...+z_k<\tau\sqrt{k}$ for $k=1,...,n$. Multiplying this inequality by $z_k-z_{k+1}$ for each $k\in[n-1]$, and by $z_n$ for $k=n$, adding up the resulting estimates, and rearranging the terms, we obtain $$z_1^2+...+z_n^2 < \tau \big(z_1+(\sqrt2-1)z_2 +...+ (\sqrt n-\sqrt{n-1})z_n \big).$$ Using Cauchy-Schwarz now gives $$1 < \tau \Big( \sum_{k=1}^n \big(\sqrt k-\sqrt{k-1}\big)^2 \Big)^{1/2} \|z\| < \tau \sqrt{\log n +4}/2,$$ a contradiction.
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Assymptotically that might be optimal but if we want to achieve the absolute minimum then it seems better to have a mix of positive and negative coordinates. – Aaron Meyerowitz Apr 6 2011 at 11:43
@Aaron: absolutely. – Seva Apr 6 2011 at 12:34
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Edited. Conjecture for $d=1$: Define the sequence $v_1,\ldots,v_n$ by $v_1=1$ and $$v_{k+1}=\left(\sqrt{1+\frac1k}-1\right)(v_1+\cdots v_k).$$ Then the $\min\max$ equals $a$ where $$a^2\sum_1^nv_j^2=1.$$ It corresponds to the projection on the line spanned by $u:=(a_1,\ldots,a_n)$ where $a_j:=av_j$.
In this construction, the equality $\|Pe_I\|=\|e_I\|$ ($I$ a subset of indices) is achieved for every subset $I=(1,\ldots,p)$ with $1\le p\le n$.
We have $v_k=\sqrt{k}-\sqrt{k-1}\sim\frac{1}{2\sqrt k}$. Asymptotically, we have $a\sim\frac{2}{\sqrt{\log n}}$.
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At least for $n=3,$ Your example $(-a,0,a)$ is not quite as good as $(-5,2,5)$. – Aaron Meyerowitz Apr 6 2011 at 11:48
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@Aaron. See my edits. Now my $(a_1,a_2,a_3)$ is better than $(-5,2,5)$. – Denis Serre Apr 6 2011 at 13:22
@Dennis Aha, so using $\sqrt{2} \approx 7/5.$ in$(-1,\sqrt{2}-1,1).$ – Aaron Meyerowitz Apr 6 2011 at 14:46
@Aaron. One N in Denis in French. Even if there are two *A*'s in Aaron. – Denis Serre Apr 6 2011 at 15:14
What do you mean Aaron with this new vector $(-1,\sqrt2-1,1)$ ? – Denis Serre Apr 6 2011 at 15:20
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http://mathoverflow.net/questions/24958/showing-e-is-transcendental-using-its-continued-fraction-expansion
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Showing e is transcendental using its continued fraction expansion
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Can the transcendence of e be shown using its continued fraction expansion e = [2;1,2,1,1,4,1,1,6,...]?
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The following paper by Henry Cohn at least demonstrates some relation between the transcendence of $e$ and its continued fraction expansion: arxiv.org/PS_cache/math/pdf/0601/0601660v3.pdf – Pete L. Clark May 17 2010 at 1:38
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Thank you for the reference. Interesting read. – paarshad May 17 2010 at 2:42
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"Interesting read." Well, yes -- I said it was by Henry Cohn, so my statement implies yours. :) – Pete L. Clark May 17 2010 at 3:54
BTW, can anyone show a reference where I can find the proof that e=[2;1,2,11,4,11,6,1,1,...]? – TCL Mar 7 2011 at 17:35
@TCL: I hadn't seen this thread before, otherwise I would have answered. I discuss te continued fraction of $e$ at length in a blog post here: topologicalmusings.wordpress.com/2008/08/04/… Lots of references are given, and there is some useful discussion by Henry Cohn in the comments. – Todd Trimble Oct 19 2011 at 10:33
7 Answers
I have to join Gerry in his claim that $e$ is uniquely determined by its continued fraction, but transcendence proofs for the latter use the number $e$ instead. Without pretending to prove the transcendence of $e$ (proofs can be found in so many books and articles), I would like to mention that more general "pseudoperiodic" continued fractions of the form $\alpha=[b_0;b_1,\dots,b_s,(\overline{c_1+\lambda d_1,\dots,c_m+\lambda d_m})^\infty_{\lambda=0}]$ are known to be transcendental. This is because they can be expressed through the values of so-called Siegel's $E$-functions at rational points, and the transcendence result for the latter goes back to Siegel's 1929 paper.
The result where the use of continued fraction for $e$ is crucial is the following [C.S. Davis, Bull. Austral. Math. Soc. 20 (1979) 407--410]. For each $C<1/2$, the inequality $$\biggl|e-\frac pq\biggr|< C\frac{\log\log q}{q^2\log q}$$ has only finitely many solutions in rationals $p/q$, and at the same time there are infinitely many solutions of the inequality if one takes $C>1/2$. This means that the continued fraction gives us an efficient way to measure the "quality" of rational approximations to $e$, and this is exactly the thing continued fractions are designed for! This result was generalized to the general "pseudoperiodic" continued fractions in [B.G. Tasoev, Math. Notes 67 (2000) 786--791].
I know only one family of examples where continued fractions are used directly to prove the transcendence of the numbers they represent. This is related to Mahler's method and the numbers are Liouvillian numbers, so that they are "too good" approximated by rationals.
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In his article Transcendental continued fractions, Journal of Number Theory 13, November 1981, 456-462, Gideon Nettler shows that two numbers given by continued fractions $A = [a_0,a_1,a_2,...]$ and $B = [b_0, b_1, b_2, ...]$ have the property that $A$, $B$, $A \pm B$, $A/B$ and $AB$ are irrational if $\frac12 a_n > b_n > a_{n-1}^{5n}$ for sufficiently large $n$, and transcendental if $a_n > b_n > a_{n−1}^{(n−1)^2}$ for sufficiently large $n$. The growth of the $a_i$ in the continued fraction expansion of $e$ is so small that present methods seem useless for proving the transcendence of $e$ in this way.
Edit. Similarly, Alan Baker proved in Continued fractions of transcendental numbers (Mathematika 9 (1962), 1-8) that if $q_n$ denotes the denominator of the $n$-th convergent of a continued fraction $A$, and if $$\lim \sup \frac{(\log \log q_n)(\log n)^{1/2}}{n} = \infty,$$ then $A$ is transcendental.
Edit 2 I guess that the answer to your question should be a firm "yes" after all. In
• Über einige Anwendungen diophantischer Approximationen, Abh. Preu\ss. Akad. Wiss. 1929; Gesammelte Abhandlungen, vol I, p. 209-241
Siegel proved that all continued fractions $$\frac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + \ldots}}}$$ in which the $a_i$ form a nonconstant arithmetic sequence are transcendental. Applying this to the continued fraction expansion of $\frac{e-1}{e+1}$ gives the transcendence of $e$.
Siegel's proof uses, predictably, analytic machinery (solutions of Bessel and Riccati differential equations) going far beyond Liouville's theorem.
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Might it not be better to say that it has been proved that various Bessel functions take on transcendental values at integer or rational arguments, and that among these transcendental Bessel values are some which have partial quotients in arithmetic progression? – Gerry Myerson Oct 17 2010 at 10:03
I don't think so. Euler computed the value of continued fractions in which the $a_i$ form an arithmetic sequence in terms of solutions of the Riccati equation (E71 on the Euler archive). This means there's a direct way from the continued fraction to the differential equation. – Franz Lemmermeyer Oct 18 2010 at 7:58
Franz, commenting on your Edit 2: Siegel indeed settled the grounds of a new method in transcendental number theory (generalising Lindemann-Hermite-Weierstrass rather than Liouville). However, this method makes no use of continued fraction expansions. Quite opposite, the fact that "pseudo-periodic" CFs are transcendental followed from Siegel's general theorem on the transcendence of the values of Bessel functions at nonzero algebraic points. – Wadim Zudilin Dec 15 2010 at 11:08
@Wadim: yes. But you can rewrite Siegel's proof as follows: start with the periodic CFs, show that they satisfy the Bessel differential equation, and finally prove that the Bessel functions have transcendental values at nonzero algebraic points. – Franz Lemmermeyer Dec 15 2010 at 11:56
There is a not-very-helpful sense in which the answer is yes: the continued fraction expansion determines $e$, and therefore can be used to prove anything one can prove about $e$, including its transcendence.
We know so little about the continued fractions of real algebraic numbers of degree exceeding 2 that I would be surprised if there were any direct path from continued fraction to transcendence.
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@Gerry - Actually there has been quite a bit of work done on the transcendence of continued fractions. The first transcendence criteria for continued fractions were proved by E. Maillet, H. Davenport and K.F. Roth, A. Baker, and more recently improved by B. Adamczewski and Y. Bugeaud. For instance, see: math.univ-lyon1.fr/homes-www/adamczew/ABD.pdf and references therein. A particularly nice result is the following: if r is a real number with an infinite continued fraction expansion that begins with infinitely many palindromes, then r is either quadratic irrational or transcendental. – Amy Glen Jul 19 2010 at 20:21
To use the continued fraction for a number to prove it's transcendental, one usually shows that the rational approximations it affords are "too good" for an algebraic number. The traditional tool here is Liouville's theorem, but this has been improved to the more powerful Roth's theorem:
If $\alpha$ is algebraic, then for every $\varepsilon > 0$ there are only finitely many rational numbers $p/q$ satisfying $$\left|\alpha - \frac{p}{q}\right| < \frac{1}{q^{2+\epsilon}}.$$
Unfortunately, $e$ also satisfies this. Indeed, rational approximations to $e$ are uncommonly bad. As Wadim mentions, there are only finitely many $p/q$ satisfying the following inequality. $$\left|e - \frac{p}{q}\right| < \frac{\log \log q}{3 q^2 \log q}.$$
But Khinchin proved that, for almost all $\alpha$, $$\left|\alpha - \frac{p}{q}\right| < \frac{1}{q} \phi(q)$$ has infinitely many solutions if and only if $\sum_q \phi(q)$ diverges.
Additionally, Khinchin showed that the geometric means of the entries of the continued fraction expansion of a real number almost always converge to a universal constant. The geometric means of the entries of the continued fraction expansion of $e$ diverge.
If either of Khinchin's conditions hold for nonquadratic algebraic numbers, the transcendentality of $e$ would follow, but a proof is probably out of reach.
Finally, the continued fraction expansion of $e$ provides an immediate proof of its irrationality, because rational numbers have finite continued fraction expansions. We also have that $e$ is not a quadratic irrational, because those have (eventually) periodic continued fraction expansions.
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I thought that all the algebraic transcendentals had a repeated pattern in their continued fraction representation. I know this is true for all the quadratic surds.
The fact that the representation for e, as well as e^2^n don't truly repeat, in the conventional sense, would suggest that they are not algebraic, hence transcendental. Many 'non-algebraic' numbers, like e and pi, have simple representations as continued fractions.
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What is an "algebraic transcendental"? Do you mean "algebraic irrational"? Anyway, what you know about quadratic surds is essentially unique to them: an eventually periodic continued fraction is a quadratic surd. There is no known pattern to the entries in the continued fraction of the cube root of 2. – KConrad Feb 24 2011 at 5:04
In what way is the continued fraction representation of pi simple? – S. Carnahan♦ Aug 25 2011 at 4:15
@S. Carnahan: He might be referring not to the regular continued fraction for $\pi$, but others such as those given here: en.wikipedia.org/wiki/… – Todd Trimble Oct 19 2011 at 10:38
If there were to be a relation between the CF and the transcendence of e,one would feel that it would involve the pattern in the values of the CF. Considering also there appears to be no decipherable pattern to the CF of pi it may well be that such a path does not exist, as if it did then it seems likely that a similar relation would be present in pi.
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There is no decipherable pattern to the Simple CF of pi, but there are numerous Generalized CFs for pi, as [http://en.wikipedia.org/wiki/Continued_fraction#Generalized_continued_fraction] notes. One could, however, say that pi is even more transcendental than e (which does have a Simple CF pattern) is.
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http://mathhelpforum.com/geometry/202017-proving-angles-equal-if-bisectors-parallel.html
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Thread:
1. Proving angles are equal if bisectors are parallel
The question asks that if there's a parallelogram ABCD and bisectors of < ABC and < ADC are parallel then how can we prove that angles <DAB and <BCD are equal.
I have managed to work backwards and show that triangle DAE and FBC both have the same angles by working with the parallel lines. I have simply sketched in the angles congruent to <DAE and <BCF and gone from there. I let E be the point where the bisector from angle <DAB met the opposite side and F be the point at which the bisectors from <DCB touched the opposite side. How do I prove that it because the proof is dependent on the placement of E and F, That in all cases this will work?
2. Re: Proving angles are equal if bisectors are parallel
Hello, Idiotinabox!
Is there a typo in the problem?
$\text{Given: parallelogram }ABCD\text{ and bisectors of }\angle ABC\text{ and }\angle ADC\text{ are parallel.}$
$\text{Prove that: }\:\angle DAB \:=\: \angle BCD.$
In any parallelogram, opposite angles are always equal.
It has nothing to do with angle bisectors.
Code:
``` B C
* - - - - - - *
/ θ /
/ /
/ /
/ /
/ θ /
* - - - - - - *
A D```
3. Re: Proving angles are equal if bisectors are parallel
Problem makes sense if the opposite angle theorem of parallelograms is not allowed
4. Re: Proving angles are equal if bisectors are parallel
Sorry, it's not a parallelogram. My mistake
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http://mathoverflow.net/revisions/61226/list
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## Return to Question
Hi,
I've a question regarding the congruence subgroup.
We have the following congruence subgroup
```$\Gamma^* := \left\{ M \in \Gamma | XM \equiv X(mod \; \mathbb{Z}^2),\; m \cdot det|\begin
{smallmatrix} X \\ XM \end{smallmatrix}| \in \mathbb{Z} \right\}$```
of $\Gamma$.
Where
$\Gamma$ is a subgroup of $SL_2(\mathbb{Z})$,
`$M = \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \in \Gamma$`, `$X = \begin{pmatrix} \lambda & \mu \end{pmatrix}\in \mathbb{Q}^2$`, $m \in \mathbb{Z}$
This subgroup can also be written as
`$\Gamma^* = \left\{ \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \in \Gamma | (a-1)\lambda + c \mu ,\; b \lambda + (d-1)\mu,\; m(c \mu^2 + (d-a)\lambda\mu-b\lambda^2) \in \mathbb{Z} \right\}$`
and hence contains $\Gamma \cap \Gamma \left(\frac{N^2}{(N,m)}\right)$ if `$NX \in \mathbb{Z}^2$`.
My question is, why is $\Gamma^* \supset \Gamma \cap \Gamma \left(\frac{N^2}{(N,m)}\right)$ true? What does $\Gamma \left(\frac{N^2}{(N,m)}\right)$ mean exactly? Actually I only have found references to the main congruence subgroup $\Gamma(N)$ (but never anything to a more complex form).
Thanks, Jan
2 improved formatting
Hi,
I've a question regarding the congruence subgroup.
We have the following congruence subgroup
```$\Gamma^* := \left\{ M \in \Gamma | XM \equiv X(mod \; \mathbb{Z}^2),\; m \cdot det|\begin
{smallmatrix} X \\ XM \end{smallmatrix}| \in \mathbb{Z} \right\}$```
of $\Gamma$.
Where
$\Gamma$ is a subgroup of $SL_2(\mathbb{Z})$,
`$M = \{ \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \in \Gamma$`, `$X = \begin{pmatrix} \lambda & \mu \end{pmatrix}\in \mathbb{Q}^2$`
This subgroup can also be written as
`$\Gamma^* = \left\{ \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \in \Gamma | (a-1)\lambda + c \mu ,\; b \lambda + (d-1)\mu,\; m(c \mu^2 + (d-a)\lambda\mu-b\lambda^2) \in \mathbb{Z} \right\}$`
and hence contains $\Gamma \cap \Gamma \left(\frac{N^2}{(N,m)}\right)$ if `$NX \in \mathbb{Z}^2$`.
My question is, why is $\Gamma^* \supset \Gamma \cap \Gamma \left(\frac{N^2}{(N,m)}\right)$ true? What does $\Gamma \left(\frac{N^2}{(N,m)}\right)$ mean exactly? Actually I only have found references to the main congruence subgroup $\Gamma(N)$ (but never anything to a more complex form).
Thanks, Jan
1
# congruence subgroup
Hi,
I've a question regarding the congruence subgroup.
We have the following congruence subgroup
```$\Gamma^* := \left\{ M \in \Gamma | XM \equiv X(mod \; \mathbb{Z}^2),\; m \cdot det|\begin
{smallmatrix} X \\ XM \end{smallmatrix}| \in \mathbb{Z} \right\}$``` of $\Gamma$.
Where
$\Gamma$ is a subgroup of $SL_2(\mathbb{Z})$,
`$M = \{ \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \in \Gamma$`, `$X = \begin{pmatrix} \lambda & \mu \end{pmatrix}\in \mathbb{Q}^2$`
This subgroup can also be written as
`$\Gamma^* = \left\{ \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \in \Gamma | (a-1)\lambda + c \mu ,\; b \lambda + (d-1)\mu,\; m(c \mu^2 + (d-a)\lambda\mu-b\lambda^2) \in \mathbb{Z} \right\}$`
and hence contains $\Gamma \cap \Gamma \left(\frac{N^2}{(N,m)}\right)$ if `$NX \in \mathbb{Z}^2$`.
My question is, why is $\Gamma^* \supset \Gamma \cap \Gamma \left(\frac{N^2}{(N,m)}\right)$ true? What does $\Gamma \left(\frac{N^2}{(N,m)}\right)$ mean exactly? Actually I only have found references to the main congruence subgroup $\Gamma(N)$ (but never anything to a more complex form).
Thanks, Jan
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http://physics.stackexchange.com/questions/56443/franck-condon-principle-and-born-oppenheimer-approximation
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# Franck Condon Principle and Born Oppenheimer approximation
My question here is purely fundamental. I am confused with the concept in Franck Condon (FC) principle and Born Oppenheimer (BO) approximation. The FC principle is in accordance with the BO approximation or not? In FC there is a correlation between electronic states and the nuclear motion. So, BO approximation is broken. So, can we say FC is an example of breaking of BO approximation?
Secondly, in the case of mega electron volt ions, is BO approximation valid? In this case, the velocity of electrons are comparable to the velocity of the nucleus!
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I would suggest branching off your second point into a question of its own. What exactly do you mean by MeV ions? If that is the energy of the center-of-mass motion, then the BOA stays unchanged as a simple change of frame will put the ion to rest, and unless it collides with something else there is nothing that can couple to the COM motion. If, on the other hand, you have > 1 MeV in the internal nuclear motion of your molecule, it wil very quickly tear itself apart. – Emilio Pisanty May 10 at 23:22
## 3 Answers
The Franck-Condon principle is a direct consequence of the Born-Oppenheimer approximation stating that since nuclei are so much slower then electrons, they cannot move during the electronic excitation. There is no violation because the is no energy exchange between the electron and the nuclei - they both get necessary energy from the photon.
I'm not sure what you mean by megavolt ions. If the ion is just moving very fast, but nuclei and electron cloud travel together, then I don't see why the BO approximation should be violated.
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If we assume direct dissociation of molecules (when molecules are irradiated with laser) and pre dissociation, then why the predissociation is said to be non adiabatic (means Born Oppenheimer approximation is broken)? In both direct and predissociation, electrons are excited and cause the molecules to brake (electro-vibronic coupling). – albedo Mar 11 at 11:55
The Franc-Condon factor influences the process of electronic excitation, which is very vast. Predissociation happens much later, when nuclei move around sufficiently. Hence, FC factor has no direct relation to predissociation. – gigacyan Mar 12 at 7:37
First, one can derive the Franck-Condon principle from purely quantum mechanical considerations. In that case, the separation of the electronic and vibrational motion of the nucleus comes from the Born-Oppenheimer approximation which allows to treat separately the electronic and nuclear degrees of freedom.
Second, in the Born-Oppenheimer approximation the first step is to neglect the kinetic energy of the nucleus. This is justified by the assumption that the nucleus is heavy and moves slowly while the electrons move much faster so that we can consider that the adiabatic hypothesis applies. This will make sense if one assumes that the momentum of the nucleus and the electrons are of the same order of magnitude $p_N \simeq p_e$ so that the kinetic energy $E_N \ll E_e$ since we assume $m_N \gg m_e$. From this it follows that if the velocity of the nucleus and the electron are similar, the Born-Oppenheimer approximation still holds.
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The electronic transitions are non adiabatic. Electronic transitions are accompanied by vibrational transitions and hence electronic transition is coupled with nuclear motion. So, Born-Oppenheimer approximation is can not be valid! – albedo Mar 11 at 5:53
but if velocities of the nucleus and the electron are similar, then kinetic energy of the nucleus is three orders of magnitude larger than that of the electron! – gigacyan Mar 11 at 7:35
I think, it is not the kinetic energy but the velocity which is important here. The BO approximation is valid if the electrons can adjust quickly with the nuclear motion. In the normal cases, the electrons adjust very quickly with the nucleus (adiabatic process). But, for eg., if an electromagnetic wave interact with an ion, the ion oscillate faster, then the motion is non adiabatic. – albedo Mar 11 at 7:57
What the Franck-Condon principle is telling you is that since the electronic transitions are virtually instantaneous (since we assume the Born-Oppenheimer approximation) compared to the time scale of nuclear motion, this sets constraints to the vibrational transitions during the electronic transition since the new vibrational state should be instantaneously compatible with the nuclear positions and momentum in which the molecule was before. Born-Oppenheimer approximation does not decouple electronic and vibrational motion and indeed the energy of the electrons depend on the nuclear position. – DaniH Mar 11 at 17:56
@DaniH: I think you didn't get my point! I was saying about the non adiabatic case. The Born Oppenheimer approximation neglects the non adiabatic effects. For example, when there is a crossing between two Born Oppenheimer surfaces or conical intersections. What you said is true in the case of vertical transitions. Am I right? – albedo Mar 12 at 9:46
show 1 more comment
You are confused by a slightly misleading aspect of the usual presentation of the Franck-Condon principle.
The FCP does indeed rely on a separation of slow and fast timescales, but now the fast timescale is not that of the electronic motion but that of electronic transitions. The typical setting of single-photon transitions in a weak field is tricky to deal with in the time domain, but the take-home message from the first-order perturbation-theoretic analysis is that you can assume the transition to be instantaneous even if there is a (coherent) probability distribution for when that instant occurs.
Suppose, then, that you know that a transition has occurred. (You can do this by post-selecting the excited molecules, for example.) In that moment there is no correlation between the electronic and nuclear coordinates: wherever the nuclei were, they remain, and the electrons are upgraded to the BO excited state corresponding to those nuclear coordinates.
Right after the transition, then, the electronic potential energy surface changes to that of the excited state. The important thing, though, is that the nuclear wavepacket remains unchanged. It must, because the transition was instantaneous! What does happen, however, is that this wavepacket is no longer an eigenstate of the nuclear hamiltonian, and therefore it has to move. The nuclear wavepacket then begins to slosh around the excited-state potential well until otherwise disturbed.
(If the displacement of the minima is small, then the motion is harmonic and nothing very interesting happens. If the displacement is enough to let the wavepacket "see" the anharmonic edges of the well, on the other hand, then all sorts of interesting TDSE dynamics might happen, like spreading and re-interference.)
So what is all the hullabaloo about Franck-Condon factors/oscillations/so on? As in all TDSE evolutions, one can choose to decompose the initial wavepacket into a superposition of the eigenstates of the (new) potential well. The coefficients will probably oscillate with eigenstate number, but so far these oscillations are purely a mathematical artefact of how we're describing the evolution, and they are not physically measurable.
How then, do we measure the coefficients? Well, that task is really measuring the nuclear energy very precisely, i.e. to a precision greater than the spacing between the vibrational levels. Because of the Uncertainty Principle, this requires a measurement over a time that's longer than the period of the nuclear oscillations. (An example is electronic fluorescence, which happens on a long timescale.) This means you are making your system interact with some measuring device, such as the fluorescent EMR modes, over a long time, and the probability of interaction is a Fourier transform over all the system's degrees of freedom: in particular, the temporal motion of the nuclei gets Fourier transformed to the energy domain, and out you get (of course!) the FC factors.
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http://physics.stackexchange.com/questions/24425/how-does-f-frac-delta-mv-delta-t-equal-m-frac-delta-v-del?answertab=active
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How does $F = \frac{ \Delta (mv)}{ \Delta t}$ equal $( m \frac { \Delta v}{ \Delta t} ) + ( v \frac { \Delta m}{ \Delta t} )$?
That's how it's framed in my Physics school-book.
The question (or rather, the explanation) is that of the thrust of rockets and how the impulse is equal (with opposite signs) on the thrust-gases and the rocket itself.
$( m \frac { \Delta v}{ \Delta t} ) = -(v \frac {\Delta m}{ \Delta t} ) = F_i$
I suppose it's a problem with how I see the transfer of impulse and exactly which part of the equation relates to which part of the physical world (gases, rocket). So we can start from there.
Title equation:
$F = \frac{ \Delta (mv)}{ \Delta t} = \left ( m \frac { \Delta v}{ \Delta t} \right) + \left ( v \frac { \Delta m}{ \Delta t} \right)$
Grade: The equivalent of G-10 in the US.
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Hi M.Na'el, and welcome to Physics Stack Exchange! I took the "homework" tag off because it looks like you're really asking a conceptual question, not a homework question - but that's good! I rather like this question. – David Zaslavsky♦ Apr 26 '12 at 17:59
@DavidZ, Maybe, but I just wanted to be sure I don't get a final answer as my book gave. I wanted thorough steps hence the tag. In any case, you're the experts around here :) – Mussri Apr 28 '12 at 10:48
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Your curiosity is much appreciated :-) No need to worry, though; we always give explanations, not plain answers, regardless of what tags may be (or not be) on the question. – David Zaslavsky♦ Apr 28 '12 at 11:04
3 Answers
Here is a visualization:
Momentum is mass times velocity, so draw it as the area of a rectangle:
If we change the mass and velocity a little, we change the momentum:
The total change in the momentum is the sum of green, blue, and purple rectangles. Their sizes are just length times width, so overall we have
$\Delta p = m\Delta v + v\Delta m + \Delta v \Delta m$
This looks like the answer you're seeking except for the extra term at the end.
Suppose we cut $\Delta m$ and $\Delta v$ down to one tenth their current size. Then the first two terms become one tenth as large, but $\Delta m \Delta v$ becomes one hundredth as large. The purple box shrinks away much faster than the blue and green ones. Therefore, for very small changes, we can ignore the purple box and write
$\Delta p = \Delta(mv) = m\Delta v + v\Delta m$
we usually indicate this limiting procedure by changing the $\Delta$ to $\mathrm{d}$, so
$\mathrm{d} p = \mathrm{d}(mv) = m\mathrm{d} v + v\mathrm{d} m$
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nice diagrams, great answer. – tmac Apr 26 '12 at 21:54
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The idea about this school book derivation is that you can change moment by changing velocity (common case) or by changing mass.
Since you can change the momentum of the system (rocket plus gasses) only by the external force, and in case of the rocket there is no external force (neglect gravity for a moment), so the question is, why is rocket getting faster and faster?
$$F_\text{ext} = \frac{\text{d}p}{\text{d}t} = m \frac{\text{d}v}{\text{d}t} + v \frac{\text{d}m}{\text{d}t} = 0,$$
$$m \frac{\text{d}v}{\text{d}t} = - v \frac{\text{d}m}{\text{d}t}.$$
Important contribution comes from the fact that as rocket pushes gasses away it effectively decreases its mass (right side of the equation). If momentum is conserved, this means that velocity of the rocket increases (left side of equation).
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That's the way I saw at first but I still don't see how the non-changing $m$ and $v$ get into the same equation. The only explanation I had (which the book said nothing about) was that the equation measured momentum-change in very small time intervals hence why, for a brief moment, either $m$ or $v$ are constant-ized to measure the other. Doesn't this sound like the uncertainty principle? – Mussri Apr 26 '12 at 12:39
The idea is that this process happens in very short moment and changes are very small. You could imagine that you have $\bar{m}$ on the left side and $\bar{v}$ on the right. The very next moment you write the same equation but with a new mass and new velocity, so this is step-by-step process, analogous to integrating. – Pygmalion Apr 26 '12 at 13:45
It's only true when the changes $\Delta t$, $\Delta v$, $\Delta m$ are small and then it is known as the Leibniz rule, the rule for the derivative of a product, which Leibniz (but also Newton) discovered when they invented the calculus 3 centuries ago.
Just look at this proof: $$\frac{\Delta (mv)}{\Delta t} = \frac{(mv)_{new} - (mv)_{old}}{\Delta t} =\frac{(m_{old}+\Delta m)(v_{old}+\Delta v)-m_{old}v_{old}}{\Delta t} = \dots$$ Here I just used $x_{new}=x_{old}+\Delta x$ which holds for $X=m,v,t,mv$ or anything else. The new value is the old value plus the increment.
But now, expand the parentheses' products via the distribution law. You get; $$=\dots \frac{m_{old}\Delta v+ \Delta m v_{old}+\Delta m\,\Delta v}{\Delta t}$$ because the term $m_{old}v_{old}$ canceled (it was subtracted). Now, if each $\Delta X$ is significantly smaller than $X$, e.g. 100 times, then $\Delta m \,\Delta v$ is 10,000 times smaller and can be totally neglected. So you're only left with the first two terms in the numerator and they give you exactly the two terms you wanted to find.
So the increase of $mv$ is obtained either from an increase of $m$ or from an increase of $v$. The equation just encodes this simple observation quantitatively.
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So I should get it that the $\Delta t$ value is an infinitesimal and constantly changing along the time-line (but not in value) so that the change in $m$ and $v$ is also very small and can be neglected for a larger picture? I'm sorry but I only know the basics of Calculus from wikipedia; nothing more... – Mussri Apr 26 '12 at 12:41
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Yup, your comment sounds totally fine and is a way to learn Calculus in Newton's way. But the $\Delta$ symbols are meant for people who don't have to know Calculus and derivatives. You may still want to imagine that all these things are finite, just small, and the very small terms are neglected. But what they mean is really $dt$, $dm$, $dv$ etc. in the calculus-infinitesimal sense while these people also implicitly say "don't ask me about calculus, it is not my goal to explain it now, instead, I want to explain some physics you would quite properly formulate only if you knew calculus". – Luboš Motl Apr 26 '12 at 16:47
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– Mark Eichenlaub Apr 26 '12 at 20:50
@LubosM, I think it's now a problem of $\Delta$ and $d$... Can you provide any info-links on this point? I've only encountered cap-Delta in my studies and I thought $d$ was just a synonym used in wikipedia... – Mussri Apr 28 '12 at 11:08
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http://liviusmathblog.blogspot.com/2012/08/on-banchoff-cmutov-surfaces.html
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# Liviu's Math Blog
Hope to add bits of mathematics or mathematical news I find interesting.
## Tuesday, August 28, 2012
### On the Banchoff-Chmutov surfaces
This discusses a question posed on MathOverflow by Leon Lampret.
Denote by $T_n$ the Chebysev polynomial of the first kind and degree $n$ uniquely determined by the equality $\newcommand{\bR}{\mathbb{R}}$
$$T_n(\cos t)=\cos nt,\;\;\forall t\in\bR.$$
Denote by $U_n$ the Chebysev polynomial of degree $n$ and of the second kind uniquely determined by the equality
$$U_n(\cos t)=\frac{\sin (n+1) t}{\sin t},\;\;t\in\bR.$$
They are related by two equalities
$$T_n'= n U_{n-1}, \;\; T_n(x)^2 -(x^2-1)U_{n-1}(x)^2=1. \tag{1}$$
The polynomial $T_n$ has $n$ distinct real zeros located in $(-1,1)$ and thus, by Rolle's theorem, all its critical points are located in $(-1,1)$.
The polynomial $T_n$ is a solution of the second order linear differential equation
$$(1-x^2)y''-xy'+n^2 y=0,$$
which shows that all the critical points of $T_n$ are nondegenerate.
The Banchoff-Chmutov surface $Z_n$ is defined by
$$Z_n:=\Bigl\lbrace (x,y,z)\in\bR^3;\;\; \underbrace{T_n(x)+T_n(y)+ T_n(z)}_{=: f_n(x,y,z)} =0\;\bigr\rbrace.$$
Remark. (a) $Z_n$ is a smooth submanifold of $\bR^3$. To see this, we rely on the implicit function theorem. Observe that if $df_n(x_0,y_0,z_0)=0$, then
$$T_n'(x_0)=T_n'(y_0)=T_n'(z_0) =0.$$
In particular (1) implies
$$U_{n-1}(x_0)=U_{n-1}(y_0)=U_{n-1}(z_0)=0.$$
Invoking (1) again we deduce
$$T_n(x_0)=T_n(y_0)=T_n(z_0)=1.$$
This shows that $(x_0,y_0,z_0)\not\in Z_n$.
(b) If $n$ is even, then $Z_n$ is compact. Indeed, in this case $T_n$ is an even polynomial and
$$\lim_{|x|\to\infty} T_n(x)=\infty.$$
This implies that $Z_n$ is bounded, thus compact since it is obviously closed.
Assume that $n$ is even and consider the function
$$f: Z_n\to \bR,\;\; h(x,y,z)= z.$$
The critical points of $h$. A point $(x,y,z)$ on $Z_n$ is critical for $h$ iff
$$T'_n(x)= T_n'(y)=0, \;\; T_n(z)=-T_n(x)-T_n(y)$$
Now the critical points of $T_n$ are all located in the interval $[-1,1]$ and can be easily determined from the defining equality
$$T_n( \cos t) = \cos nt, \;\;t\in [0,\pi], \tag{A}$$
so that
$$T_n'(\cos t) = n\frac{\sin nt}{\sin t}$$
This nails the critical points of $T_n$ to
$$u_k = \cos \frac{k\pi}{n},\;\; k=1,\dotsc, n-1.$$
Note that
$$T_n(u_k)= \cos k\pi=(-1)^k$$
so that the critical points of $h$ on the surface $Z_n$ are
$$\bigl\lbrace (u_j,u_k,z);\;\; T_n(z)+(-1)^j+(-1)^k=0,\;\;j,k=1,\dotsc, n-1 \bigr\rbrace.$$
Now we need to count the solutions of the equations
$$T_n(x)=0,\;\pm 2.$$
The equation $T_n(x)=0$ has $n$ solutions, all situated in $[-1,1]$.
On the interval $[-1,1]$ we deduce from (A) that $|T_n|\leq 1$. The polynomial $T_n$ is even and is increasing on $[1,\infty)$. We conclude that the equation $T_n(x)=-2$ has no solutions, while the equality $T_n(x)=2$ has two solutions. Thus the critical set of $h$ splits into three parts
$$C_0= \lbrace (u_j,u_k,z);\;\;j+k\in 2\mathbb{Z}+1,\;\;T_n(z)=0\rbrace,$$
$$C_2^+= \lbrace (u_j,u_k,z);\;\;j,k\in 2\mathbb{Z}+1,\;\;T_n(z)=2, z>1\rbrace,$$
$$C_2^-= \lbrace (u_j,u_k,z);\;\;j,k\in 2\mathbb{Z}+1,\;\;T_n(z)=2, z<-1\rbrace.$$
From the above discussion we deduce that the points in $C_2^-$ are minima and the points in $C_2^+$ are maxima.
The function $h$ is a Morse function. Note that $h$ is defined implicitly, by solving for $z$ in the nonlinear equation
$$T_n(x)+ T_n(y) + T_n(z)=0. \tag{2}$$
Suppose that $(x_0,y_0,z_0)$ is a critical point of $h$, $x_0=u_j$, $y_0=u_k$. Differentiating (2) near this critical point we deduce $\newcommand{\pa}{\partial}$
$$\frac{\pa z}{\pa x}T_n'(z) +T_n'(x) =0,\;\; \frac{\pa z}{\pa y} T_n'(z) + T_n'(y)=0.$$
Differentiating the above agian we deduce that
$$\frac{\pa^2 z}{\pa x\pa y}|_{(u_j,u_k)}=0, \tag{3}$$
$$\frac{\pa^2 z}{\pa x^2}|_{(u_j,u_k)} T_n'(z_0) +T_n''(u_j)= 0,\;\; \frac{\pa^2 z}{\pa y^2}|_{(u_j,u_k)} T_n'(z_0) +T_n''(u_k)= 0.$$
Now let observe that $T_n(z_0)=0$ or $T_n(z_0)=2$. In the first case $T_n'(z_0)\neq 0$ bcauise $T_n$ has only simple zeros. In the second case $T_n'(z_0)\neq 0$ because in this case $|z_0|>1$ and $T_n$ has no critical points outside $(-1,1)$. Hence
$$\frac{\pa ^2 z}{\pa x^2}|_{(u_j,u_k)}= -\frac{T_n''(u_j)}{T_n'(z_0)},\;\;\frac{\pa ^2 z}{\pa y^2}|_{(u_j,u_k)}= -\frac{T_n''(u_k)}{T_n'(z_0)}. \tag{4}$$
The point $u_k$ is a local minimum for $T_n$ if $k$ is odd and a local maximum if $k$ is even. Moreover, these are nondegenerate critical points of $T_n$.
This proves that all the critical points of $h$ are nondegenerate. Moreover if $(u_j, u_k)\in C_0$ then the numbers $T_n''(u_j)$ and $T_n''(u_k)$ have opposite signs and invoking (3) and (4) we deduce that this point is a saddle point.
Thus the Euler characteristic of $Z_n$ is
$$\chi(Z_n)= {\rm card}\; C_2^+ +{\rm card}\; C_2^- -{\rm card}\; C_0.$$
Now observe that
$${\rm card}\; C_2^\pm = \Bigl(\;{\rm card}\; [1,n-1]\cap (2\mathbb{Z}+1) \;\Bigr)^2= \frac{n^2}{4},$$
$${\rm card} \; C_0 = n\Bigl( \frac{n(n-2)}{4}+\frac{n(n-2)}{4}\Bigr)=\frac{n^2(n-2)}{2}.$$
Thus the Euler characteristic of $Z_n$ is
$$\chi(Z_n)=\frac{n^2(3-n)}{2}. \tag{E}$$
## Here are images of $Z_2, Z_4, Z_6$, courtesy of StackExchange (hat tip to Igor Rivin)
Z_2
Z_6
Z_4
The above computations do not explain whether $Z_n$ is connected or not. To check that it suffices to look at the critical values of the above function corresponding to saddle points. These critical values are the zeros $\zeta_1<\dotsc <\zeta_n$ of $T_n$. The level set
$$Z_n\cap \lbrace z=\zeta_k\rbrace$$
is the algebraic curve
$$T_n(x)+T_n(y)=0. \tag{C}$$
This forces $|x|,|y|\leq 1$ because $T_n(x)> 1$ for $|x|> 1$ and $|T_n(x)\leq 1$ for $|x|\leq 1$. We can use the *homeomorphism*
$$[0,\pi]\ni t\mapsto x=\cos t\in [-1,1]$$
to give an alternate description to (C). It is the singular curve inside the square $[0,\pi]\times [0,\pi]$ with coordinates $(s,t)$ described by
$$\cos ns+ \cos nt =0.$$
This can be easily visualized as the intersection of the square with the grid
$$s\pm t\in (2\mathbb{Z}+1)\frac{\pi}{n}$$
which is connected. Now it is not very difficult to conclude using the Morse theoretic data on $h$ that $Z_n$ is connected.
Example. The equality (E) predicts that $\chi(Z_6)=-54$. Let us verify this directly. Here is a more detailed image of $Z_6$.
Z_6
We can give an alternate description of $Z_6$ as follows. Consider the $1$-dimensional simplicial complex $C\subset \bR^3$ depicted below
The surface $Z_6$ is homeomorphic to the boundary of a thin tubular neighborhoof $T$ of $C$ in $\bR^3$ and thus
$$\chi(Z_6)=\chi(\pa T)=2\chi(T)=2\chi(C).$$
Thus formula (E) predicts that
$$\chi(C)= -27.$$
Let us verify this directly. The vertex set of $C$ consists of
• 8 Green vertices of degree 3.
• 12 Red vertices of degree 4.
• 6 Blue vertices of degree 5.
• 1 Black vertex of degree 6.
The number $V$ of vertices of $C$ is thus
$$V= 8+12+6+1=27.$$
The total number $E$ of edges of $C$ is half the sum of degrees of vertices so that
$$E=\frac{1}{2}( 3\times 8+ 4\times 12+ 5\times 6+ 6\times 1)= \frac{1}{2} (24+48+30+6)=54.$$
Hence
$$\chi(C)= 27-54 =-27.$$
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http://mathhelpforum.com/advanced-algebra/185238-first-isomorphism-theorem-quotient-group-question-print.html
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# First Isomorphism Theorem/quotient group question
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• July 28th 2011, 03:55 AM
hmmmm
First Isomorphism Theorem/quotient group question
Let G be the group of invertible upper triangular 2x2 matrices over the real numbers. Determine if the following are normal subgroups and if they are use the frist isomorphism theorem to identify the quetient group G/H.
where H is
a) $a_{11}=1$
b) $a_{11}=a_{22}$
I have checked (hopefully correctly) that these are both normal subgroup (via $gHg^{-1}=H$) However I am unsure as to what the next part is actually asking me?
Thanks for any help
• July 28th 2011, 04:42 AM
Swlabr
Re: First Isomorphism Theorem/quotient group question
Quote:
Originally Posted by hmmmm
Let G be the group of invertible upper triangular 2x2 matrices. Determine if the following are normal subgroups and if they are use the frist isomorphism theorem to identify the quetient group H.
where H is
a) $a_{11}=1$
b) $a_{11}=a_{22}$
I have checked (hopefully correctly) that these are both normal subgroup (via $gHg^{-1}=H$) However I am unsure as to what the next part is actually asking me?
Thanks for any help
For a) the question is now wanting you to find a map from G to another group, H, such that the kernel is the set of all matrices such that $a_{11}=1$. It is analogous for b).
If you have any trouble finding these maps, just ask. People will happily surrender the answers, but you finding them yourself is much, much more useful!
Finally, what are your matrices over? $\mathbb{Z}$? $\mathbb{Q}$? $\mathbb{R}$? Some arbitrary field? (The question doesn't really make sense unless you know this...)
• July 28th 2011, 05:34 AM
hmmmm
Re: First Isomorphism Theorem/quotient group question
Sorry I should have said that it is over the real numbers I have edited it now.
Im a bit confused by your answer I want to find an isomorphism from G to H (H being a subgroup of G this isnt possible is it?) or do you just mean a different group H'
Thanks for the help sorry for my confusion
for a
So am I looking for a map $\phi:G\rightarrow G'$ such that $\phi(g)\rightarrow a_{11}$ and where $G'=(\mathbb{R^*},\times)$?
So the quotient group G/H is isomorphic to $\mathbb{R^*},\times)$
• July 28th 2011, 11:58 AM
Swlabr
Re: First Isomorphism Theorem/quotient group question
Quote:
Originally Posted by hmmmm
Sorry I should have said that it is over the real numbers I have edited it now.
Im a bit confused by your answer I want to find an isomorphism from G to H (H being a subgroup of G this isnt possible is it?) or do you just mean a different group H'
Thanks for the help sorry for my confusion
for a
So am I looking for a map $\phi:G\rightarrow G'$ such that $\phi(g)\rightarrow a_{11}$ and where $G'=(\mathbb{R^*},\times)$?
So the quotient group G/H is isomorphic to $\mathbb{R^*},\times)$
Yeah, sorry, I just meant a different group. The isomorphism looks correct, and would be my first guess, but you should check first that it is well-defined, surjective, and that the kernel is what you want it to be...
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http://stats.stackexchange.com/questions/46767/validation-error-less-than-training-error-implications
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# Validation error less than training error — implications?
I am running a neural net to predict used car prices, sample size is 800. Using both 10-fold cross validation (10 times) and 1/3 holdback (10 times), the $R^2$ for training is about 0.60 and for validation is about 0.68 for all 20 runs. The smallest difference in the 20 runs is training $R^2$ = 0.64 and validation $R^2$ = 0.68, so the training $R^2$ is always less than the validation $R^2$.
I am very used to seeing training $R^2$ bigger than validation $R^2$, which means overfitting. In the past when I have seen training $R^2$ less than validation $R^2$, it has been a transient phenomenon that disappeared when I re-ran the model. This is the first time that I have seen validation $R^2$ systematically larger than training $R^2$.
I have no idea what this means. Any thoughts?
-
2
A single large outlier in the training set (and not in any of the CV sets) might cause the larger error. Did you check the data distribution? – Robert Kubrick Dec 31 '12 at 16:34
1
Doing 10-fold cv for 10 reps is really not enough. You might want to do the 10-fold cv at least 30 times. Better yet, use bootstrap to get an estimate of your $R^2$. – user765195 Dec 31 '12 at 18:37
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http://math.stackexchange.com/questions/167529/is-this-polynomial-positive
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# Is this polynomial positive?
Let $p\geq 2$, and $p$ is not a half odd integer. $t\in R$.
Is the following polynomial positive:
$$T_k(t)=\left(\frac t2\right)^p\sum_{j=0}^k\frac{\left(-\frac{t^2}{4}\right)^j\Gamma(p+1)}{j!\Gamma(p+j+1)}.$$
Why the gamma notation? It's not a polynomial unless $p$ is a natural number. – John Bentin Jul 6 '12 at 16:02
Sorry, I did not mentioned that $p$ is not half odd integer. – Michael Jul 6 '12 at 17:45
$p$ is not a half-integer $\left(p \neq \frac{5}{2},\frac{7}{2},\frac{9}{2},\dots\right)$? – Vandermonde Jul 6 '12 at 18:07
Consider series $$T(t)=\left(\frac{t}{2}\right)^p\sum\limits_{j=0}^\infty\frac{\left(-\frac{t^2}{4}\right)^j\Gamma(p+1)}{j!\Gamma(j+p+1)}$$ This is related to the series representation of the Bessel function of order $p$ of the first kind. Indeed $$T(t)=\Gamma(p+1)\sum\limits_{j=0}^\infty\frac{(-1)^j}{j!\Gamma(j+p+1)}\left(\frac{t}{2}\right)^{2j+p}=\Gamma(p+1)J_p(t)$$ Since this series converges, then $$\lim\limits_{k\to\infty} T_k(t)=\Gamma(p+1)J_p(t)\tag{1}$$ It is known that Bessel functions of the first kind take negative and positive values infinitely many times on $(0,+\infty)$. Hence we may consider $t_0$ such that $\Gamma(p+1)J_p(t_0)<0$. From $(1)$ it follows that for some $k_0$ we would have $$T_k(t_0)<0\quad\text{ for all}\quad k>k_0.$$
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http://mathhelpforum.com/calculus/22661-sum-two-lipschitz-funtions.html
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# Thread:
1. ## sum of two lipschitz funtions
Let f : R ->R be a function.
We say that f is Lipschitz continuous if there is some
L > 0 such that |f(x) − f(y)| < L|x − y| for all x, y in R.
letting f, g : R -> R be lipschitz continious funtions,
(a) i want to show that f+g is lipschitz continious too
and also if f,g are bounded lip continuos funtions
(b) then i want to show that f.g is lipschitz continous too
does anyone know how to show these?
2. Originally Posted by dopi
Let f : R ->R be a function.
We say that f is Lipschitz continuous if there is some
L > 0 such that |f(x) − f(y)| < L|x − y| for all x, y in R.
letting f, g : R -> R be lipschitz continious funtions,
(a) i want to show that f+g is lipschitz continious too
Proof:
Since $f$ and $g$ are Lipschitz continuous, we have that $|f(x) - f(y)| < \frac {L_1}2|x - y|$ for $L_1 >0$ and $|g(x) - g(y)| < \frac {L_2}2|x - y|$ for $L_2 > 0$
Now let $L = \mbox{max} \{ L_1, L_2 \}$, and note that $(f + g)(x) = f(x) + g(x)$, then we have:
$|(f + g)(x) - (f + g)(y)| = |f(x) + g(x) - f(x) - g(x)| \le |f(x) - f(y)| + |g(x) - g(y)|$ $< \frac {L_1}2|x - y| + \frac {L_2}2|x - y| < \frac L2|x - y| + \frac L2|x - y| = L|x - y|$
QED
you do (b), it should be similar
3. Originally Posted by dopi
Let f : R ->R be a function.
and also if f,g are bounded lip continuos funtions
(b) then i want to show that f.g is lipschitz continous too
If $f$ is Lipschitz then $|f(x)-f(y)|\leq A|x-y|$ similarly $g$ so $|g(x)-g(y)|\leq A|x-y|$. We want to show $|f(x)g(x)-f(y)g(y)|\leq C|x-y|$ for some $C$. Note that $|f(x)g(x)-f(x)g(y)+f(x)g(y)-f(y)g(y)|\leq$ $|f(x)||g(x)-g(y)|+|g(y)||f(x)-f(y)| \leq M_1B|x-y|+M_2A|x-y| = C|x-y|$ where $C=M_1B+M_2A$. Q.E.D.
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http://mathhelpforum.com/calculus/36324-coordinate-systems.html
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# Thread:
1. ## Coordinate Systems
Question 1:
A line with gradient 4 touches the parabola with equation y^2 = 4x. Find the coordinates of the point of contact of this tangent with the parabola and an equation of the normal to the parabola at that point.
Question 2:
The line with equation y = m(x+a), where ‘m’ can vary but ‘a’ is constant, meets the parabola with equation y^2 = 4ax in two points P and Q. Find, in terms of a and m, the coordinates of the mid-point R of PQ. As m varies show that R lies on the curve with equation y^2 = 2a(x+a).
2. Originally Posted by geton
Question 1:
A line with gradient 4 touches the parabola with equation y^2 = 4x. Find the coordinates of the point of contact of this tangent with the parabola and an equation of the normal to the parabola at that point.
taking the derivative implicitly we get
$2y\frac{dy}{dx}=4$ but we know that $\frac{dy}{dx}=4$
so $2y(4)=4 \iff y=\frac{1}{2}$ so $x=\frac{1}{16}$
The normal line will have slope $m=-\frac{1}{4}$
$y-\frac{1}{2}=-\frac{1}{4}(x-\frac{1}{16})$
$y=-\frac{1}{4}x+\frac{33}{64}$
3. Thank you so much TheEmptySet.
4. I've done my second problem myself.
Thank you.
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http://physics.stackexchange.com/questions/tagged/electromagnetic-radiation
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I have seen it remarked in some problem sets that if you have an electromagnetic wave traveling in the $x$-direction with it's $y$-coordinate given as $y(x,t)=y_0\sin (\omega t +kx)$ and you want a ...
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http://math.stackexchange.com/questions/58677/limit-of-sin-circ-sin-circ-cdots-circ-sinx
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# Limit of $(\sin\circ\sin\circ\cdots\circ\sin)(x)$
I'm trying to find this limit:
$$\lim_{n \to \infty} \underbrace{\sin \sin \ldots \sin }_{\text{$n$ times}}x$$
Thank you
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2
Please use LaTeX in the future, just as Jonas did for you on this question. – Raphael Aug 20 '11 at 16:18
19
"Forgive me, father, for I have sinned. Many many times..." – Mark Schwarzmann Aug 20 '11 at 16:18
10
@user14829 $\Large \textrm{Thi}\int_{s}^{i} \LaTeX.$ – muntoo Aug 20 '11 at 18:04
6
$\top \mathbb{N} \chi$ – Jozef Aug 20 '11 at 20:26
3
This question is not a duplicate. In the other question, $n$ and $x$ are the same. – Qiaochu Yuan Aug 23 '11 at 1:19
show 4 more comments
## 2 Answers
First, you have to prove there is a limit. the first application of $\sin x$ will get us into $[-1,1]$ If $x \in [0,1]$ we have $0 \le \sin x \lt x$, so the sequence (after the first term, maybe) is monotonically decreasing and bounded below by $0$. Then, if there is a limit, you must have $\sin x=x$. Where is that true? The case below zero is for you.
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Here is a different approach that doesn't lead as directly to a rigorous proof as Ross Millikan's, but is more concrete and shows the rate of convergence. Let the $n$th member of the sequence be given by the function $x(n)$. Using the Taylor series of the sine function, and approximating the discrete function $x$ by a continuous one, we have $dx/dn\approx -(1/6)x^3$. Separation of variables and integration gives $x\approx \pm\sqrt{3/n}$ for large $n$.
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Thank you ben.. – Jozef Aug 20 '11 at 20:29
– Aryabhata Aug 23 '11 at 1:27
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http://mathhelpforum.com/differential-equations/167069-initial-value-problem.html
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# Thread:
1. ## initial value problem
Given that y''=y'+2y and that y(0)=0 and y(1)=1 Deduce that one of the initial value problems has a trivial solution.
Please tell me how can I get the trivial answer. If we set y(0)=0 and y'(0)=0 we get the correct answer, but the answer is saying that y(0)=0 and y'(0)=1 is also trivial(!) which doesn't make sense to me.
2. An IVP, if I am not mistaken, refers to an equation together with a SINGLE value. What you have there looks more like a boundary problem. In any case, it has two values, so it's not an IVP.
Perhaps they mean to give you two IVPs with the same equation... ? In that case y(0)=0 is obviously the one with the trivial solution.
3. Originally Posted by immortality
Given that y''=y'+2y and that y(0)=0 and y(1)=1 Deduce that one of the initial value problems has a trivial solution.
Please tell me how can I get the trivial answer. If we set y(0)=0 and y'(0)=0 we get the correct answer, but the answer is saying that y(0)=0 and y'(0)=1 is also trivial(!) which doesn't make sense to me.
$\displaystyle y(x)=\frac{e^{1-x}-e^{2x+1}}{1-e^3}$
$\displaystyle y'(x)=\frac{(2e^{3x}+1)e^{1-x}}{e^3-1}$
$\displaystyle y'(0)=\frac{(2e^{0}+1)e^{1}}{e^3-1}=\frac{3e}{e^3-1}\neq 1$
4. Originally Posted by immortality
Given that y''=y'+2y and that y(0)=0 and y(1)=1 Deduce that one of the initial value problems has a trivial solution.
You are given a boundary value problem. I don't know what "one of the initial value problems" means. One of what initial value problems? Certainly one possible initial value problem for this differential equation is "y''= y'+ 2y with y(0)= 0, y'(0)= 0". That uses the same equation and one of the given boundary values so perhaps that is what is meant.
Please tell me how can I get the trivial answer. If we set y(0)=0 and y'(0)=0 we get the correct answer, but the answer is saying that y(0)=0 and y'(0)=1 is also trivial(!) which doesn't make sense to me.
5. Originally Posted by hatsoff
An IVP, if I am not mistaken, refers to an equation together with a SINGLE value. What you have there looks more like a boundary problem. In any case, it has two values, so it's not an IVP.
An "initial value problem" for a second order equation would give a value of y and its derivative at a single value of the independent variable.
Perhaps they mean to give you two IVPs with the same equation... ? In that case y(0)=0 is obviously the one with the trivial solution.
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http://math.stackexchange.com/questions/43778/short-exact-sequence-of-modules-generated-by-a-set
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Short exact sequence of modules generated by a set
Let $0 \to A \stackrel{i}{\to} B \stackrel{p}{\to} C \to 0$ be a short exact sequence of $R$-modules.
Suppose that $A = \langle X \rangle$ and $C = \langle Y \rangle$
For each $y \in C$, choose $y' \in B$ such that $p(y')=y$. Prove that $$B = \langle i(x) \cup \{ y':y \in Y \} \rangle$$
I just can't seem to figure out how to get this to work - I don't think it should be hard!
Let $a \in A$, then $a = \sum r_i x_i$. Similarly $c = \sum r'_i y_i = \sum r'_i p(y'_i)$.
Obivously I should use exactness: $\operatorname{img} i = \operatorname{ker} p$
$b \in \operatorname{img} i \implies b = \sum r_i i(x_i)$
The $p(b) = 0$ gives that $pi(x_i)=0$ - but that is just clear from the definitions!
I am thinking that to be in the kernel of $p$, we must have $p(y'_i)=0$
Any hints to point me in the right direction?
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2 Answers
Let $b\in B$, write $p(b) = \sum_k r_k y_k$ (a minor remark: it is a bad idea to use the same letter for indices in a sum and for a homomorphism, like you did in your post). Now, wouldn't you just love to be able to claim that $b = \sum_k r_k y'_k$. Of course that's not true, but what do you know about the difference $b - \sum_k r_k y'_k$? You should be able to take it from here.
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thanks for that. (I hadn't even really noticed I was using $i$ for the homomorphism and the indicies!\$) – Juan S Jun 7 '11 at 3:07
You don't want to start with $a\in A$; rather, you should start with $b\in B$ (and try to show that $b$ can be expressed using the $\iota(x)$ and the $y'$).
To that end, note that you can certainly use the $y'$ to construct an element $b'$ of $B$ that has the same image under $p$ as $b$; this because the images of the $y'$ generate $C$.
But if $b$ and $b'$ have the same image under $p$, then $b-b'\in \mathrm{ker}(p) = \mathrm{Im}(\iota)$; so you can express $b-b'$ in terms of $\iota(X)$.
So: $b-b'$ can be expressed in terms of $\iota(X)$, and $b'$ can be expressed in terms of the $y'$. That means that $b$ can be expressed in terms of $\iota(X)$ and the $y'$, which proves that $B$ is contained in the submodule generated by the given set.
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http://physics.stackexchange.com/questions/tagged/coordinate-systems?sort=faq&pagesize=30
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# Tagged Questions
The coordinate-systems tag has no wiki summary.
4answers
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### Centrifugal Force and Polar Coordinates
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### coordinate change differential equation polar
I noticed that v [in step (2.5)] is not the same as the terms from the first formula, even if they are related.. I tried to understand how did he reach to this ...
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http://physics.stackexchange.com/questions/33365/whats-the-electric-field-intensity-inside-parallel-plate-capacitor-filled-with/33657
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# What's the electric field intensity inside parallel-plate capacitor filled with water?
```` _______________________
| |
--------------------------- |
copper plate |
--------------------------- |
PDMS plate |
--------------------------- =======
Microchannel filled === 500 DC V
with H2O E? |
--------------------------- |
PDMS plate |
--------------------------- |
copper plate |
--------------------------- |
|______________________|
Cross-section view
````
I'm sorry for your inconvenience to see the above diagram, since my reputation is too low to insert image file. >_<
Thinkness: copper=300um; PDMS(polymer)=300um; Microchannel depth=400um;
Dielectric constants of PDMS and H2O are 2.65 and 80, respectively.
My experimental purpose is to generate uniform electric field inside of water-filled microchannel by appling electric field externally. (“externally” here emphasizes no direct contact between metal electrodes and water; here PDMS, a dielectric polymer, serves as a barrier between them)
Question 1: If DC V = 500 V; then what is electric filed intensity inside of water-filled microchannel? As I know, in pure water, autoionization of water molecules can generate hydroxide and hydronium ions at a constant concentration, and these free ions can form Debye space charge layers that screen electric fields. Does it mean, no matter how high DC V is, the final electric field intensity inside water-filled microchannel would go to zero after instantaneous screening process?
Question 2: If alternating current voltage, like sinusoidal AC (100 kHz, peak voltages from 0 and 500 V, the bottom copper plate is grounded), is applied instead of DC, what would be electric field intensity inside microchannel? Taken electrostatic screening issue into account, what the minimal frequency of ac should I set?
Many thanks in advance for your attention to this matter.
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Could anyone help me out here??? >_< – lolol Aug 7 '12 at 4:24
Hi lolol - it's possible that nobody has answered your question because nobody has the required knowledge. In any case, I'll put a bounty on it and hopefully that will get it some attention. – David Zaslavsky♦ Aug 7 '12 at 20:15
@David Zaslavsky Even it's been a long time, I'd like to extend my heartful thanks to you, again. – lolol Oct 22 '12 at 8:16
## 3 Answers
About the autoionization of water ...
Wikipedia (http://en.wikipedia.org/wiki/Debye_length) gives a formula for water
$$\text{debye length in nm} = \frac{0.304}{\sqrt{I\text{ in molar}}}$$ where $I$ is ionic strength, which is 1E-7 for pure, pH-neutral water. That gives a screening length of 1$\mu$m.
So at DC, there will be an electric field in the bottom 1 micron and top 1 micron of the water, with no electric field in the central 99% of the water. The field will be canceled by the OH- ions spread near the top electrode and the H+ ions spread near the bottom electrode.
Despite what you say, this screening process is NOT instantaneous. The H+'s and OH-'s have to travel to the appropriate side. Therefore at AC the screening may be much less or completely negligible. To calculate the frequency required, you would do a calculation involving the electrical resistance of pure, pH-neutral water, the initial voltage pulling the charges, the quantity of charge that has to move from one side to the other, and the distance that it has to move. I don't have time for this part and don't have any intuition for what order of magnitude of frequency would be the cutoff between screening and non-screening behavior.
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Many thanks for your help~~ – lolol Oct 22 '12 at 8:18
1) First of all, screening is a response to an external field, and as such it can never fully counteract the effect, except in the case of a perfect conductor. As $\varepsilon_\text{water}$ is finite, we know we are not in the perfect conductor limit. If an external field is applied, charges will rearrange (via bulk motion of ions, re-orientation of polar molecules, and/or induced polarity in neutral atoms/molecules) to partially cancel the field. Gauss's Law, together with appropriate geometry considerations, tells us the electric field for fixed surface charge density $\sigma$ is $E = \sigma/\varepsilon$.
Of course, you aren't fixing charge density, but rather voltage. To find the charge density from the voltage, you need the capacitance of the system: $Q = CV$. This is a series of three capacitors, and so the capacitance of the system is the reciprocal of the sum of the reciprocals of the individual capacitances, which are $C_i = \varepsilon_i A/d_i$ for ideal parallel-plate capactors.
If there were only one material, you'd find the effect of $\varepsilon$ on capacitance is exactly balanced by its reduction of the electric field for a given charge density, and so $E = V/d$ no matter the material. Your problem is slightly more complicated, but in the end you should find a nonzero field strength in the water.
(Note: Everything here is in the limit of instantaneous transitions. Not being an experimentalist, I don't know off hand the length scale over which screening takes place. The problem is somewhat more complicated if that length is comparable to the thicknesses in your setup, since then step functions are replaced by exponential attenuation.)
2) You can use AC. Then you must take into account the frequency-dependence of $\varepsilon$ for each material. A very rough rule of thumb in most typical situations is that $\varepsilon$ is reduced at higher frequencies, since the charges have less time to rearrange by whatever mechanism. However, there are resonances that need to be taken into account.
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If I remember correctly from Electromagnetism, if you wait for the circuit to reach equilibrium with the capacitor (i.e. current = 0) doesn't the total voltage across the entire circuit have to be zero? So the if $\Delta V_{capacitor} \approx 500V$ and since in a parallel plate capacitor the electric field is relatively constant. So you have: $$\Delta V = EL = 500,$$ $$E = 500/L$$ where $E$ is the electric field and $L$ is the distance from one plate to the other?
If your hypothesis in Question 1 was correct and the electric field was zero, then there would never be any voltage going against the battery. Constant current = constant build-up of charge on the plates, I don't think that's likely.
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1
This is largely what I would expect as well. The one correction I would make is that the electric field won't be the same in the polymer layer as in the water, as they have different dielectric constants. But the field should be constant within a given layer. – Colin McFaul Aug 7 '12 at 22:23
Yup, when I said constant, I was only considering the water, but yes in a specific medium between parallel plates the electric field should be constant – Mike Flynn Aug 8 '12 at 12:58
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http://mathoverflow.net/questions/26374/for-which-classes-of-topological-spaces-euler-characteristics-is-defined
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## For which classes of topological spaces Euler characteristics is defined?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I would like to know something more than what is written on wikipedia http://en.wikipedia.org/wiki/Euler_characteristic
What would be some large (largest?) class of topological spaces for which $\chi$ is defined, so that all standard properties hold, for example that $\chi(X)=\chi(Y)+\chi(Z)$ if $X=Y \cup Z$, ($Y\cap Z=0$).
ADDED. The answer of Algori indicates that a reasonably large class of spaces for which Euler characteristics can be defined are locally compact spaces $X$, whose one point compactification $\bar X$ is a CW complex. Then we can define $\chi(X)=\chi(\bar X)-1$. For example, the Euler characteristics of an open interval according to this definition is $-1$. This definition rases a second (maybe obvious) question.
Question 2. Suppose $X$ is a locally compact space whose 1 point compactification is a $CW$ complex, and $Y$ is a subspace of $X$ such that both $Y$ and $X\setminus Y$ have this property. Is it ture that $\chi(X)=\chi(Y)+\chi(X\setminus Y)$?
Also, I was thinking, that Euler characteristics is more fundamental then homology.But can it be defined for spaces, where homology is not defined?
Finally, Quiaochu pointed out below that a very similar question was already discussed previously on mathoverflow.
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2
Much like divergent series, there is more than one extension of the Euler characteristic to spaces other than finite CW-complexes. E.g. there is the rational-cohomology Euler characteristic or the Morava K-theory Euler characteristic K(n), which assign 1 and p^n to the classifying space of a cyclic group of order p respectively, or one can apply divergent series techniques to form the alternating sum of dimensions. – Tyler Lawson May 29 2010 at 19:40
The open interval is a(n infinite) CW-complex. – algori May 29 2010 at 23:03
1
There is a notion of "combinatorial Euler characteristic," but it's not homotopy invariant; see mathoverflow.net/questions/1184/… . – Qiaochu Yuan May 30 2010 at 0:02
## 2 Answers
The answer to the question as it is stated is that there is probably no "largest" class of spaces for which the Euler characteristic makes sense.
The answer also depends on where you would like the Euler characteristic to take values. Here is the tautological answer (admittedly not a very exciting one): if you have a category $C$ of spaces closed under taking cones and cylinders, then there is the universal Euler characteristic for that category: just take the free abelian group $K(C)$ that has a generator $[X]$ for each $X\in C$ and quotient it by the span of $[X]+[Cone(f)]-[Y]$ for all $X,Y\in C$ and any morphism $f:X\to Y$ in $C$. The Euler characteristic of any $X$ in $C$ is set to be $[X]$. (There may be variations and/or generalizations of this approach.)
The group $K(C)$ is complicated in general but for some choices of $C$ it has interesting quotients. This can happen e.g. when $C$ admits a good "cohomology-like" functor. For example if $C$ is the category of spaces with finitely generated integral homology groups then $K(C)$ maps to $\mathbf{Z}$ and this gives the usual Euler characteristic. If one takes $C$ to be formed by spaces that admit a finite cover with finitely generated integral homology groups (typical examples are the classifying spaces of $SL_2(\mathbf{Z})$ and more generally of mapping class groups), then $K(C)$ does not map to $\mathbf{Z}$ any more, but it maps to $\mathbf{Q}$. This gives the rational Euler characteristic.
Finally, let me address the last remark by Dmitri. For some categories the group $K(C)$ maps to $\mathbf{Z}$ in several different ways. Let us take e.g. $C$ to be the category formed by spaces whose one-point compactification is a finite CW-complex (with proper maps as morphisms). Then there are (at least) two characteristics; one is obtained using the ordinary cohomology and another one comes from the Borel-Moore homology. On complex algebraic varieties both agree. But the Borel-Moore Euler characteristic of an open $n$-ball is $(-1)^n$.
Here is the answer to the second question: suppose $Y$ is a locally closed subspace of a locally compact space $X$ such that $X,Y,\bar Y,\bar Y\setminus Y, X\setminus\bar Y$ and $X\setminus Y$ are of the form "a finite CW-complex minus a point". Then $\chi(Y)+\chi(X\setminus Y)=\chi(X)$ where $\chi$ is the Euler characteristic computed using the Borel-Moore homology.
The case when $Y$ is closed follows from the Borel-Moore homology long exact sequence. In general we can write $\chi(X)=\chi(X\setminus\bar Y)+\chi(\bar Y)=\chi(X\setminus\bar Y)+\chi(\bar Y\setminus Y)+\chi(Y)$. In the last sum the sum of the first two terms gives $\chi(X\setminus Y)$ since $X\setminus\bar Y$ is open in $X\setminus Y$.
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Algori, thanks a lot! Are there some readable references on what you have stated (for example about K(C) and Borel-Moore)? – Dmitri May 29 2010 at 23:30
Dmitri -- unfortunately I don't have a reference. But to make up for it here are some comments: 1. the definition of $K(C)$ is a slight variation of the definition of the motivic measure, which in turn goes back to other similar definitions (e.g. the Grothendieck group for coherent sheaves etc) 2. the Borel-Moore homology of a space $X$ (with constant coefficients) is the homology of the one point compactification modulo the added point. – algori May 30 2010 at 1:32
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
The Euler characteristic is the alternate sum of the rank of the integer homology groups of the space.
So how do we end up with an integer ?
The homology groups should be finitely generated. This is the case for finite CW complexes, for spherical complexes (obtained by attaching finitely many cells to a finite set), for compact subspaces of euclidean spaces that are Absolute Neighborhood retracts (e.g. compact topological manifolds), and of course for any space homotopy equivalent to such spaces. Note that only finitely many homology groups are non-zeros for these spaces. The Euler characteristic is a well-defined integer and the standard properties hold.
Let me add a few remarks on other interesting kind of spaces. Some infinite dimensional manifolds have finitely generated homology groups, but infinitely many of them may be non-zero, e.g. $P^\infty C$. In order to extend the Euler characteristic to such spaces, one may want to allow $\infty$ as a possible value for the Euler characteristic (but beware of the $\infty - \infty$ problem), or try to use a normalisation procedure to get some real number (I heard a talk about that but unfortunately I can't recall any reference). I think that it makes sense to say, e.g., that the euler characteristic of $P^\infty C$ is infinite. If we restrict our attention to (inductive limits of) complex manifolds for example, all odd Betti numbers are zero, so one may try to define an Euler characteristic with values in $N\cup \lbrace \infty\rbrace$ and I would not be surprised if the standard properties hold for such an extension (any counterexamples welcome).
Finally, when leaving the category of finite CW-complexes, I think that the Euler characteristic actually depends on what homology theory you are using. I would be interested in some feedback here. Borel-Moore homology is said to be better behaved than singular homology in that respect.
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This is in no way a contradiction or criticism of your excellent answer, but I just want to point out that, w.r.t. "So how do we end up with an integer?"---we don't always end up with an integer. For example, orbifolds have (in general) fractional Euler characteristics. – Joseph O'Rourke May 29 2010 at 23:16
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Right; for example, for G a finite group the Eilenberg-Maclane space K(G, 1) should have Euler characteristic 1/|G|. The simplest example of this is when G = Z/2Z, for which K(Z/2Z, 1) is RP^{\infty}. The alternating sum of the Betti numbers is Grandi's series! – Qiaochu Yuan May 30 2010 at 0:27
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http://physics.stackexchange.com/questions/18390/focusing-laser-off-axis-illumination-diffraction-limit
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# Focusing laser & off-axis illumination & diffraction limit
I've thought I had a good understanding how resolution enhancement tricks works for projection lithography, until I tried to understand if it's possible to get sub-diffraction performance for focused laser beam:
Let's assume we have a laser with nice and shiny perfect aspherical focusing lens (with performance greatly exceeding diffraction limit for it's NA). This system moves around and draw stuff.
The question is - can we get enhanced resolution if we apply usual lithography trick of annular/quadruple illumination? Obviously, we won't reduce aberrations thanks to off-axis illumination, as there are none.
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– Antillar Maximus Dec 20 '11 at 15:30
## 1 Answer
Yes, in some cases, for some applications, depending on your definition of "beating the diffraction limit."
In the normal sense of imaging resolution, you can't beat the diffraction limit with a normal free-space optical system. But in lithography, what you need isn't really the same as resolution in the imaging sense. Because you are using the light to etch a material to a certain depth, all you need is a sufficient contrast between the bright part of your focal spot and the rest of it. If that contrast is sufficient, then your etched pattern may only become deep over a region smaller than the typical spot size estimate given by $2.44 \lambda \mathcal{f}/\#$.
To illustrate, think of the focal spot due to a perfect lens system with a circular aperture. It takes the form (ignoring scale factors and such) of: $$\left(J_1(r) \over r \right)^2$$
where $J_1$ is the Bessel function of the first kind. Optics people call this a "jinc" function, due to similarity to the sinc() function. It is squared to give intensity. An image of this spot looks like this: $$\left(J_1(2\pi r) \over r \right)^2$$
An annular aperture can be considered the difference of two circular apertures of slightly different size, so by superposition the field at the focus of a system with an annular aperture will be the difference of two jink functions with different sizes. Such a function looks like this: $$\left(\frac{J_1(2\pi r)}{r} - \frac{J_1(2.1\pi r)}{r}\right)^2$$
You can see that the spot is considerably more spread out in general, which would be a problem for an imaging system; but the central bright spot is significantly smaller. If the intensity is chosen properly, the outer rings will not affect the lithographic material, while the small central lobe will.
This is even more effective if the material is chosen such that it doesn't absorb light except for rare two-photon events. In this case the intensity dependance of the absorption goes like the square of the intensity, making the contrast between the central spot and the lobes even better. I'm not sure how common this is in industry yet.
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Thanks for great explanation, that makes sence now. There are chemically amplified resists commonly used nowadays which have this "square" sensitivity. – BarsMonster Dec 21 '11 at 4:13
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http://mathoverflow.net/questions/37273/products-of-linear-forms-in-3-variables/37279
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Products of linear forms in 3 variables
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
We say that a linear form $f=ax+by+cz$ of real coefficients is "irrational" if $a,b,c$ are linearly independent over the rationals. My question is: Are there 3 such "irrational" linear forms $f_1,f_2,f_3$ such that the product $f_1f_2f_3$ is of integral coefficients?
Note that for the "2-dimensional" analogue, we have $(x+\sqrt{2}y)(x-\sqrt{2}y)=x^2-2y^2$.
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2 Answers
This should be a comment to Robin's answer.
Take any irreducible polynomial $f \in \mathbb{Q}[x]$ of degree 3 with real roots, say $\alpha, \beta, \gamma$. Set $f_1 = x + \alpha y + \alpha^2 z$, $f_2 = x + \beta y + \beta^2 z$, $f_3 = x + \gamma y + \gamma^2 z$.
You can find plenty of polynomials here.
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You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
How about $x+2\cos(2\pi/7)y+4\cos^2(2\pi/7)z$, $x+2\cos(4\pi/7)y+4\cos^2(4\pi/7)z$, $x+2\cos(6\pi/7)y+4\cos^2(6\pi/7)z$ ?
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Sorry, but we are saying real linear forms... – lonekite Aug 31 2010 at 16:18
Just take a totally real cubic field instead. Davenport studied such products a lot. – Franz Lemmermeyer Aug 31 2010 at 16:27
Could you give an explicit example or a reference? Thanks. – lonekite Aug 31 2010 at 16:33
@lonekite, Robin's forms are linear. The cosine terms are numerical coefficients of the indeterminates $y$ and $z$. – Gerry Myerson Sep 1 2010 at 0:32
@Gerry: Yes, I understand. Robin first gave an example of complex linear forms with the same property, then revised it using the cosine function. Anyway, both Robin's example and felix' are very helpful to me. Thank you all. – lonekite Sep 1 2010 at 4:22
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http://physics.stackexchange.com/questions/11806/objects-in-physics-as-a-mathematician-would-see-them/11815
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# Objects in Physics as a mathematician would see them
I'm a mathematician with hardly any knowledge of physics. Before I start reading volumes of physics books, I have a few questions that have been bugging me and that will help me start reading physics.
Let's forget Quantum Mechanics for the purpose of the discussion, and focus on Relativity. I always fail to understand what the objects are and how to relate them intuitively to perceivable space and time. What are the objects?
Here's what I mean: we begin with spacetime being a pseudo-Riemannian manifold with some metric (is this metric assumed to be locally Lorentzian?). For each point in this manifold, there should be some parameters, right? Like whether there's a particle there, what particle is it (information that includes, for example, mass), is there a magnetic field there and so forth. Together, this pseudo-Riemannian manifold, with a pre-specified metric, and with a set of parameters for each point, is what described the universe in relativity, right? What precisely are those parameters in General Relativity?
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– genneth Jul 2 '11 at 18:51
## 6 Answers
General relativity is a mathematical theory that generalizes the special theory of relativity and mechanics. It turns out that one can treat the gravitational effects as non-inertial forces, thanks to one of the principles that governs this theory, that is also considered the most important one, namely the principle of equivalence postulated by Einstein. That the metric of space-time must be locally flat, i.e. Minkowskian, is a consequence of this principle. In fact, the principle of equivalence states that in a free falling frame of reference, i.e. a reference where there are no inertial forces, the laws of physics are those of special relativity. As you are a mathematician there is no need for me to recall that if a manifold has a null curvature tensor, then there are global coordinates on the manifold such that the components of the metric tensor, say $g$, are exactly those of the Minkowski metric $\eta = \operatorname{diag}(1,-1,-1,-1)$.
The metric $g$ on the manifold $M$ describing the space-time is not a-priori given, but it is determined by the distribution of matter. Now comes your question on the objects of general relativity. These objects are all sorts of geometrical object that you can construct on a (smooth) manifold. Thus you'll have tensors of any ranks, and even spinors of any rank. What makes them physical objects is just their interpretation. Einstein's field equations
$$\text{Ric}-\frac12\operatorname{Tr}(\text{Ric})g = \chi T,$$
where $\text{Ric}\in T^*M\otimes T^*M$ is the Ricci tensor, $g\in T^*M\otimes T^*M$ the metric on $M$, and $T\in T^*M\otimes T^*M$ the energy-stress tensor, gives you the metic $g$ in terms of the energy-matter distribution $T$ on $M$, that is just a tensor density on $M$.
The mathematical problem of studying the geodesics on a manifold $M$ described by a metric $g$ is the equivalent of the physical problem of determining the motion of a particle that is freely falling in the gravitational field generated by a matter distribution $T$ such that the resulting metric is $g$.
One of the most spectacular prediction of the general theory of relativity is the existence of so-called black holes. Again you find these object by studying the mathematical properties of the solutions of the Einstein's equations and then give them a physical interpretation. There are some tough situation in physics when you can't base your theories on observations and experiments. Therefore you must start with a mathematical theory and develop it as further as you can. Each result you obtain is then physically interpreted. This is quite the situation of general relativity (see, e.g. the problem of the detection of gravitaional waves, or the observation of naked singularities).
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This answer is very well suited for me. Thank you very much! Unfortunately, I can't upvote with my reputation... – Wesley Jul 2 '11 at 18:49
Those set of parameters are usually called matter fields. They contribute to GR on the "right hand side" of the equation
$$G_{\mu\nu} = T_{\mu\nu}$$
Since you mentioned that you are a mathematician, let me be slightly more technical and verbose. The space-time is given by a Lorentzian manifold $(M,g)$. The matter-fields should be considered to be a collection $\{\Phi_A\}_{A\in \mathcal{A}}$ where $\mathcal{A}$ is some indexing set, with each individual $\Phi_A$ being a section of some fibre bundle $(E_A,M,\pi_A)$ over $M$. (There are more general possibilities for matter fields, but let us stick to these now.) For example, the scalar field is given by a trivial complex line bundle over $M$, while the usual description of Maxwell's theory of electro-magnetism admits the formulation of the field (the vector potential) as a section of the cotangent bundle $T^*M$.
The dynamics of the fields are generally prescribed by some equations of motions, and their contribution to gravity is taken to be their contribution to the energy momentum tensor
$$T_{\mu\nu} = T_{\mu\nu}(\{ \Phi_A\}_{A\in \mathcal{A}})$$
in such a way that the condition
$$\nabla^\mu T_{\mu\nu} = 0$$
where $\nabla$ is the covariant derivative of the metric $g$ is satisfied. (This is due to that the identity must hold for the left hand side of Einstein's equation according to the contracted Bianchi identity.)
Note that general relativity by itself is a theory of gravity. It doesn't really specify what the matter fields are. It only requires that the matter fields, when they do exist, have dynamics that obey the conservation law given by the divergence condition on the energy-momentum tensor. To get an actual physical model of the world, you will have to come-up with some rules that govern the behaviour of the matter fields. In modern physics, this is generally through some sort of action principle, since the Euler-Lagrange equations will automatically be compatible with the divergence condition above, if you take $T_{\mu\nu}$ to be the Einstein-Hilbert stress-energy generated from the action principle.
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Thanks! I apologize for not being able to upvote, but this is also a very good answer for me! – Wesley Jul 2 '11 at 18:51
I would say, as far as general relativity is concerned, "object" is a fairly vacuous term. What one is really looking at is some perturbation in the stress-energy tensor. Einstein's Equations (without cosmological constant) are $G_{\mu\nu}= \frac{8 \pi G}{c^4} T_{\mu\nu}$, which defines a relation between Einstein curvature, and therefore the metric, and the stress-energy tensor at a given point. I think this tensor is what these "parameters" you are looking for are, as it defines the energy density, the energy flux, the shear stress, and the pressure.
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In GR, the "objects" (whatever they might be for your problem) need to be mapped to the stress-energy tensor, which the GR field equations relate to the curvature of the manifold. That mapping is not really part of GR though but it is surely part of the everyday application of GR, although it will differ depending on the problem and what level of description of the physical objects is used.
One introductory physical object which is often used as an example in introductory GR texts as far as I know is "dust", to quote Schutz: "'dust' is defined to be a collection of particles, all of which are at rest in some one Lorentz frame". You define a density of the particles in a volume, and the momentum flux component $\alpha$ of the particles across each perpendicular surface of constant $\beta$ in spacetime is one of the components of the stress-energy tensor $T^{\alpha\beta}$. So $T^{00}$ is the energy-components of the dust particles across a surface of constant time, that is, the energy density.
From there you can keep adding more complex objects like fluids or electromagnetic fields etc. as long as you map them to the stress-energy tensor.
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The 'objects' that are studies in relativistic physics are the same as what people are studying in non-relativistic physics. Consider a few examples:
• If you are concerned with relativistic mechanics one can consider extended objects like rods of length L and calculate moments of intertia. (Here the object is the rod)
• If you are studying quantum mechanics then you will study the Dirac equation rather than the Schrodenger equation (Here the objects being studies are electrons)
• If you are studying statistical mechanics you may study the equation of state using a gas of photons rather than some non-relativistic gas. (not if object here applies unless we're talking about the collection of photons themselves)
The basic things which are studied are still the same things you'd study in non-relativistic physics. Nothing really changes there. What does change is the geometry of the underlying manifolds where the objects of study live. This usually adds two things regardless of wether you're talking special or general relativity:
• Extra terms in the potentials used to describe interactions between the 'objects'
• Extra terms in the equations of motions which arise from the geometry of the space-time the objects live in.
General relativity is still physics so look to physics first to understand what you are trying to study. The questions being asked are still the same. I know sometimes it's hard to see that when you pull open a book that says general relativity on it and the discussion revolves around connections, curvature, exterior algebra, and so forth but the objects of study are still there as you correctly assert. :) My suggestion is to crack open a basic physics book and then try to find the relativistic analogs of what you see there in more advanced books like Misner, Thorne, and Wheeler's Gravitation or Robert Wald's General Relativity. They both have very physical approaches to GR and you may have an easier time getting at the 'objects' that way.
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Classical GR can only study objects which can be modelled by tensor fields (on the Lorentzian manifold of space-time). The metric, of course, is itelf a tensor field, and the Laws of Nature have to take the form of equating two tensor fields with the same covariance properties, i.e., of the same type. Now it would be pretty silly if the metric itself didn't get used in the equation, so this puts some limitations on one's search...
If matter is regarded as a kind of continuous distribution, like a density, it can be well modelled by the stress-energy tensor mentioned by the other posters. Hydrodynamics can be done pretty well too. Electromagnetism, less well, but something can be done. To answer your question very directly, then, the only parameters are the coordinates of space and time and the only objects are these tensor fields, and this limits GR so that it cannot do a good job considering quantum effects like wavicles or spin. Classical particles can be treated by letting the density of matter, considered as a kind of « fluid » (I think that is a better word than powder or dust) of mass-energy, have some singularities, they definitely do not get any kind of parameter of their very own. In this respect, it is very like Newtonian and Eulerian dynamics and hydrodynamics.
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http://math.stackexchange.com/questions/81354/linear-recursion-with-coefficients-depending-on-n
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# Linear recursion with coefficients depending on n
I am searching a way to solve a recursive definition for a sequence in the following form:
$a_n = c_{n-1}\cdot a_{n-1} + b_{n-1}$
So I'd like to have an explicit form of $a_n$ depending only on $c_i$ and $b_i$, but not on $a_n$. I know how to solve linear recursions, but the coefficients are not constant here. Or are there no well-known solutions for this case?
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## 1 Answer
As it is, the problem seems a bit too general (any sequence satisfies many such recursions). So while you can express $a_n$ is terms of the $b_i$ and $c_i$, it's not a simple expression in general. Indeed if you set $C_n = \prod_{i = 0}^{n} c_i$, then you can express $a_n$ as
$$a_{n+1} = C_n \, \left(a_0 + \sum_{k = 0}^n \frac{b_k}{C_k}\right)$$
But unless you have specific sequences $(b_n)_{n \ge 0}$ and $(c_n)_{n \ge 0}$, I doubt you can do much better than that.
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Well also if have specefic $(b_n)_{n≥0}$ and $(c_n)_{n≥0}$, it might be really difficult to find the solution if I dont' have the right idea. I was wondering if there is something like a characteristic polynomial for this case. – lumbric Nov 12 '11 at 15:10
My point is that since any sequence satisfies such a recursion, it would be hopeless to expect a general method to always give a simple expression (that would imply that any sequence has a simple expression). However, if $(b_n)$ and $(c_n)$ are such that you can find a simple expression for the $C_i$ and the series $\sum \frac{b_i}{C_i}$, then you get a simple expression for $a_n$. I don't know if that helps. – Joel Cohen Nov 12 '11 at 15:22
– Joel Cohen Nov 12 '11 at 15:36
– lumbric Nov 12 '11 at 15:36
You were some seconds faster... :) Since there is a genreal method for all linear recursions, I thought maybe this could be extended to this case somehow. But if you say that you don't think so it's also a very useful answere! – lumbric Nov 12 '11 at 15:39
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http://unapologetic.wordpress.com/2009/02/03/
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# The Unapologetic Mathematician
## The Determinant of an Upper-Triangular Matrix
We know that every linear endomorphism $T$ on a vector space $V$ over an algebraically-closed field $\mathbb{F}$ has a basis with respect to which its matrix is upper-triangular. That is, it looks like
$\displaystyle\begin{pmatrix}\lambda_1&&*\\&\ddots&\\{0}&&\lambda_d\end{pmatrix}$
So we’re interested in matrices of this form, whether the base field is algebraically-closed or not. One thing that’s nice about upper-triangular matrices is that their determinants are pretty simple.
Remember that we can calculate the determinant by summing one term for each permutation in the symmetric group $S_d$. Each term is the product of one entry from each row of the matrix. But we can see that in the last row, we have to pick the last entry, or the whole term will be zero. Then in the next to last row, we must pick either of the last two entries. But we can’t pick the last one, since that’s locked up for the last row, so we must again pick the entry on the diagonal.
As we work backwards up the matrix, we find that the only possible way of picking a nonzero term is to always pick the diagonal element. That is, we only need to consider the identity permutation $\pi(k)=k$. And then the determinant is simply the product of all these diagonal elements. That is,
$\displaystyle\det\begin{pmatrix}\lambda_1&&*\\&\ddots&\\{0}&&\lambda_d\end{pmatrix}=\prod\limits_{k=1}^d\lambda_k$
Notice that the entries above the diagonal don’t matter at all!
One way this comes in handy is in finding the eigenvalues of $T$. “But wait!” you cry, “Didn’t we find the $\lambda_i$ by looking for eigenvalues of $T$? Not quite. We found them by looking for eigenvalues of a sequence of restrictions of $T$ to smaller and smaller quotient spaces. We have no reason to believe (yet) that these actually correspond to eigenvalues of $T$.
But now we can easily find the matrix corresponding to $\lambda1_V-T$, since matrix of the identity transformation is the same in every basis. We find
$\displaystyle\begin{pmatrix}\lambda-\lambda_1&&*\\&\ddots&\\{0}&&\lambda-\lambda_d\end{pmatrix}$
This is again upper-triangular, so we can easily calculate the characteristic polynomial by taking its determinant. We find
$\displaystyle\prod\limits_{k=1}^d(\lambda-\lambda_k)$
Then it’s clear to see that this will be zero exactly when $\lambda=\lambda_k$. That is, the eigenvalues of $T$ are exactly the entries along the diagonal of an upper-triangular matrix for the transformation. Incidentally, this shows in passing that even though there may be many different upper-triangular matrices representing the same transformation (in different bases), they all have the same entries along the diagonal (possibly in different orders).
Posted by John Armstrong | Algebra, Linear Algebra | 4 Comments
## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://math.stackexchange.com/questions/257718/convergence-of-r-vs
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# Convergence of r.vs
I'm working on the question below and I appreciate if you can guide me on how I can solve it.
Here is the question:
Consider $X_j$'s j = 1, 2, ... as independent bernulli random variables. These random variables, although are independent, but are not identically distributed(so then I thought that I should use Lindeberg-Feller CLT). We knoe $P(X_j = 1) = P_j$ and we know that $P_j$'s are bounded between (0,1) {open interval}. I'm trying to show $(S_n - ES_n)/\sqrt{VarS_n}$ converges to N(0,1) in distribution.
Here is what I've done so far:
1) Consider $Y_j = (X_j - P_j)/(P_j*(1 - P_j))$ ==> EY_j = 0 {first condition of lindeberg-feller thm satisfied}
2) $\sum_{j = 11}^{n} E(Y_j^2) = \sum Var(Y_j^2)$ and is finite
3) I need to check the lindeberg-feller condition: $lim \sum_{j = 1}^{n} E(Y_j^2.1_{|Y_j| \gt \epsilon}) = 0$ ??? I don't know how to prove the third one.
Could you please guide me whether I'm on the right track to solve this question and also how I can show that lindeberg condition is satisfied.
Thanks a lot for your help.
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## 1 Answer
You'll have trouble if e.g. $P_j \to 0$ very rapidly. In fact if $\sum_{j=2}^\infty P_j < 1$, there is a nonzero probability that all $S_n = S_1$, and there is no hope of getting a normal limit. Or did you mean that there is some $\delta > 0$ such that all $P_j \in [\delta, 1-\delta]$?
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http://unapologetic.wordpress.com/2007/12/20/laws-of-limits/?like=1&source=post_flair&_wpnonce=9a71b83933
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# The Unapologetic Mathematician
## Laws of Limits
Okay, we know how to define the limit of a function at a point in the closure of its domain. But we don’t always want to invoke the whole machinery of all sequences converging to that point or that of neighborhoods with the $\epsilon$-$\delta$ definition. Luckily, we have some shortcuts.
First off, we know that the constant function $f(x)=1$ and the identity function $f(x)=x$ are continuous and defined everywhere, so we immediately see that $\lim\limits_{x->x_0}1=1$ and $\lim\limits_{x\rightarrow x_0}x=x_0$. Those are the basic functions we defined. We also defined some ways of putting functions together, and we’ll have a rule for each one telling us how to build limits for more complicated functions from limits for simpler ones.
We can multiply a function by a constant real number. If we have $\lim\limits_{x\rightarrow x_0}f(x)=L$ then we find $\lim\limits_{x\rightarrow x_0}\left[cf\right](x)=cL$. Let’s say we’re given an error bound $\epsilon$. Then we can consider $\frac{\epsilon}{|c|}$, and use the assumption about the limit of $f$ to find a $\delta$ so that $0<|x-x_0|<\delta$ implies that $|f(x)-L|<\frac{\epsilon}{c}$. This, in turn, implies that $|\left[cf\right](x)-cL|=|c||f(x)-L|<|c|\frac{\epsilon}{|c|}=\epsilon$, and so the assertion is proved.
Similarly, we can add functions. If $\lim\limits_{x\rightarrow x_0}f_1(x)=L_1$ and $\lim\limits_{x\rightarrow x_0}f_2(x)=L_2$, then we find $\lim\limits_{x\rightarrow x_0}\left[f_1+f_2\right](x)=L_1+L_2$. Here we start with an $\epsilon$ and find $\delta_1$ and $\delta_2$ so that $0<|x-x_0|<\delta_i$ implies $|f_i(x)-L_i|<\frac{\epsilon}{2}$ for $i=1,2$. Then if we set $\delta$ to be the smaller of $\delta_1$ and $\delta_2$, we see that $0<|x-x_0|<\delta$ implies $|\left[f_1+f_2\right](x)-L_1+L_2|<|f_1(x)-L_1|+|f_2(x)-L_2|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$.
From these two we can see that the process of taking a limit at a point is linear. In particular, we also see that $\lim\limits_{x\rightarrow x_0}\left[f_1-f_2\right](x)=\lim\limits_{x\rightarrow x_0}f_1(x)-\lim\limits_{x\rightarrow x_0}f_2(x)$ by combining the two rules above. Similarly we can show that $\lim\limits_{x\rightarrow x_0}\left[f_1f_2\right](x)=\lim\limits_{x\rightarrow x_0}f_1(x)\lim\limits_{x\rightarrow x_0}f_2(x)$, which I’ll leave to you to verify as we did the rule for addition above.
Another way to combine functions that I haven’t mentioned yet is composition. Let’s say we have functions $f_1:D_1\rightarrow\mathbb{R}$ and $f_2:D_2\rightarrow\mathbb{R}$. Then we can pick out those points $x\in D_1$ so that $f_1(x)\in D_2$ and call this collection $D$. Then we can apply the second function to get $f_1\circ f_2:D\rightarrow\mathbb{R}$, defined by $\left[f_1\circ f_2\right](x)=f_2(f_1(x))$. Our limit rule here is that if $f_2$ is continuous at $\lim\limits_{x\rightarrow x_0}f_1(x)$, then $\lim\limits_{x\rightarrow x_0}f_2(f_1(x))=f_2(\lim\limits_{x\rightarrow x_0}f_1(x))$. That is, we can pull limits past continuous functions. This is just a reflection of the fact that continuous functions are exactly those which preserve limits of sequences. In particular, a continuous function equals its own limit wherever it’s defined: $\lim\limits_{x\rightarrow x_0}f(x)=f(\lim\limits_{x\rightarrow x_0}x)=f(x_0)$.
As an application of this fact, we can check that $f(x)=\frac{1}{x}$ is continuous for all nonzero $x$. Then the limit rule tells us that as long as $\lim\limits_{x\rightarrow x_0}f(x)\neq0$, then $\lim\limits_{x\rightarrow x_0}\frac{1}{f(x)}=\frac{1}{\lim\limits_{x\rightarrow x_0}f(x)}$. Combining this with the rule for multiplication we see that as long as the limit of $g$ at $x_0$ is nonzero then $\lim\limits_{x\rightarrow x_0}\frac{f(x)}{g(x)}=\frac{\lim\limits_{x\rightarrow x_0}f(x)}{\lim\limits_{x\rightarrow x_0}g(x)}$.
Another thing that limits play well with is the order on the real numbers. If $f(x)\geq g(x)$ on their common domain $D$ then $\lim\limits_{x\rightarrow x_0}f(x)\geq\lim\limits_{x\rightarrow x_0}g(x)$ as long as both limits exist. Indeed, since both limits exist we can take any sequence converging to $x_0$. The image sequence under $f$ is always above the image sequence under $g$, and so the limits of the sequences are in the same order. Notice that we really just need $f(x)\geq g(x)$ to hold on some neighborhood of $x_0$, since we can then restrict to that neighborhood.
Similarly if we have three functions $f(x),$latex g(x)\$ and $h(x)$ with $f(x)\geq g(x)\geq h(x)$ on a common domain $D$ containing a neighborhood of $a$, and if $\lim\limits_{x\rightarrow a}f(x)=L=\lim\limits_{x\rightarrow a}h(x)$, then the limit of $g$ at $a$ exists and is also equal to $L$. Given any sequence $x_n\in D$ converging to $a$, our hypothesis tells us that $f(x_n)\geq g(x_n)\geq h(x_n)$. Given any neighborhood of $L$, $f(x_n)$ and $h(x_n)$ are both within the neighborhood for sufficiently large $n$, and then so will $g(x_n)$ be in the neighborhood. Thus the image of the sequence under $g$ is “squeezed” between the images under $f$ and $h$, and converges to $L$ as well.
These rules for limits suffice to calculate almost all the limits that we care about without having to mess around with the raw definitions. In fact, many calculus classes these days only skim the definition if they mention it at all. We can more or less get away with this while we’re only dealing with a single real variable, but later on the full power of the definition comes in handy.
There’s one more situation I should be a little more explicit about. If we are given a function $f$ on some domain $D$ and we want to find its limit at a border point $a$ (which includes the case of a single-point hole in the domain) and we can extend the function to a continuous function $\hat{f}$ on a larger domain $\hat{D}$ which contains a neighborhood of the point in question, then $\lim\limits_{x\rightarrow a}f(x)=\hat{f}(a)$. Indeed, given any sequence $x_n\in D$ converging to $a$ we have $f(x_n)=\hat{f}(x_n)$ (since they agree on $D$), and the limit of $\hat{f}$ is just its value at $a$. This extends what we did before to handle the case of $\frac{x}{x}$ at $x=0$, and similar situations will come up over and over in the future.
### Like this:
Posted by John Armstrong | Analysis, Calculus
## 4 Comments »
1. [...] Laws of Differentiation Just like we had the laws of limits we have a collection of rules to help us calculate derivatives. Let’s start with the most [...]
Pingback by | December 26, 2007 | Reply
2. [...] some cases establish the continuity of simple functions (like coordinate projections) and then use limit laws to build up a larger class. But this approach fails for functions superficially similar to the [...]
Pingback by | September 17, 2009 | Reply
3. i would luv 2 recieve questions
Comment by DANIEL DAVIES | October 4, 2009 | Reply
Comment by | October 4, 2009 | Reply
« Previous | Next »
## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://unapologetic.wordpress.com/2007/11/09/topologies-as-categories/?like=1&source=post_flair&_wpnonce=420e8d64d7
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# The Unapologetic Mathematician
## Topologies as Categories
Okay, so we’ve defined a topology on a set $X$. But we also love categories, so we want to see this in terms of categories. And, indeed, every topology is a category!
First, remember that the collection of subsets of $X$, like the collection of subobjects on an object in any category, is partially ordered by inclusion. And since every partially ordered set is a category, so is the collection of subsets of $X$.
In fact, it’s a lattice, since we can use union and intersection as our join and meet, respectively. When we say that a poset has pairwise least upper bounds it’s the same as saying when we consider it as a category it has finite coproducts, and similarly pairwise greatest lower bounds are the same as finite products. But here we can actually take the union or intersection of any collection of subsets and get a subset, so we have all products and coproducts. In the language of posets, we have a “complete lattice”.
So now we want to talk about topologies. A topology is just a collection of the subsets that’s closed under finite intersections and arbitrary unions. We can use the same order (inclusion of subsets) to make a topology into a partially-ordered set. In the language of posets, the requirements are that we have a sublattice (finite meets and joins, along with the same top and bottom element) with arbitrary meets — the topology contains the least upper bound of any collection of its elements.
And now we translate the partial order language into category theory. A topology is a subcategory of the category of subsets of $X$ with finite products and all coproducts. That is, we have an arrow from the object $U$ to the object $V$ if and only if $U\subseteq V$ as subsets of $X$. Given any finite collection $\{U_i\}_{i=1}^n$ of objects we have their product $\bigcap\limits_{i=1}^nU_i$, and given any collection $\{U_\alpha\}_{\alpha\in A}$ of objects we have their coproduct $\bigcup\limits_{\alpha\in A}U_\alpha$. In particular we have the empty product — the terminal object $X$ — and we have the empty coproduct — the initial object $\varnothing$. And all the arrows in our category just tell us how various open sets sit inside other open sets. Neat!
## 8 Comments »
1. So – I assume you’re doing this for some more intricate reason than just the sheer love for categories. This is the category we want functors from in order to define presheaves and sheaves, isn’t it?
Comment by | November 9, 2007 | Reply
2. There are sort of three answers here. On one hand, I’m doing it because I love the categories. On the other hand, this is the first step on the road to topoi. On the third hand, there should be some more-than-formal relation between limit (of nets) and limit (of functors). This I don’t really know, but it has to be there.
Comment by | November 9, 2007 | Reply
3. John –
Look into the theory of locales for the answer to the relationship of the limits. A good source is the text “Categorical Foundations” in the Encyclopedia of Mathematics series.
— Marc
Comment by Marceau | November 10, 2007 | Reply
4. John –
this being the first step on the road to topoi also means this is the first step on the road to sheaves? (just reraising that first question
Comment by | November 10, 2007 | Reply
5. Yes, I thought that what I’d said counted as an affirmative, if slightly veiled, answer to your question.
Comment by | November 10, 2007 | Reply
6. Thanks for letting me reaffirm my lack of knowledge in some of the cool parts of modern mathematics!
Comment by | November 10, 2007 | Reply
7. [...] in Category Theory, Topology at 2:54 am by saij This is just cool! Okay, so we’ve defined a topology on a set . But we also love categories, so we want to see this [...]
Pingback by | November 26, 2007 | Reply
8. [...] course we also want to consider the categorical perspective. What does a continuous map give us in terms of the topology? It’s a functor from the [...]
Pingback by | February 22, 2011 | Reply
« Previous | Next »
## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://mathoverflow.net/questions/35993/when-is-the-ring-of-invariants-of-a-finite-group-generated-by-symplectic-reflecti
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## When is the ring of invariants of a finite group generated by symplectic reflections a complete intersection ring?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let V be a finite dimensional symplectic vector space over $\mathbb{C}$. Let $G$ be a finite subgroup of the symplectic group $Sp(V),$ which is generated by symplectic reflections, i.e. by elements $g\in G,$ such that $rank(Id_V-g)=2.$ Then it is well-known that the ring of invariants $\mathbb{C}[V]^G$ is Gorenstein.
My question is assuming that V is an irreducible G-module and $dim V>2,$ when is $\mathbb{C}[V]^G$ a complete intersection ring? Of course when $dim V=2$ it is a complete intersection ring (Kleinian singularities), but I don't know other examples.
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http://en.wikipedia.org/wiki/Spacelike
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# Spacetime
(Redirected from Spacelike)
For other uses of this term, see Spacetime (disambiguation).
See also: Philosophy of space and time, Four-dimensionalism, and Arrow of time
Spacetime
Part of a series on
Special relativity General relativity
Types
Relation to gravity
In physics, spacetime (also space–time, space time or space–time continuum) is any mathematical model that combines space and time into a single continuum. Spacetime is usually interpreted with space as existing in three dimensions and time playing the role of a fourth dimension that is of a different sort from the spatial dimensions. From a Euclidean space perspective, the universe has three dimensions of space and one of time. By combining space and time into a single manifold, physicists have significantly simplified a large number of physical theories, as well as described in a more uniform way the workings of the universe at both the supergalactic and subatomic levels.
In non-relativistic classical mechanics, the use of Euclidean space instead of spacetime is appropriate, as time is treated as universal and constant, being independent of the state of motion of an observer. In relativistic contexts, time cannot be separated from the three dimensions of space, because the observed rate at which time passes for an object depends on the object's velocity relative to the observer and also on the strength of gravitational fields, which can slow the passage of time.
In cosmology, the concept of spacetime combines space and time to a single abstract universe. Mathematically it is a manifold consisting of "events" which are described by some type of coordinate system. Typically three spatial dimensions (length, width, height), and one temporal dimension (time) are required. Dimensions are independent components of a coordinate grid needed to locate a point in a certain defined "space". For example, on the globe the latitude and longitude are two independent coordinates which together uniquely determine a location. In spacetime, a coordinate grid that spans the 3+1 dimensions locates events (rather than just points in space), i.e. time is added as another dimension to the coordinate grid. This way the coordinates specify where and when events occur. However, the unified nature of spacetime and the freedom of coordinate choice it allows imply that to express the temporal coordinate in one coordinate system requires both temporal and spatial coordinates in another coordinate system. Unlike in normal spatial coordinates, there are still restrictions for how measurements can be made spatially and temporally (see Spacetime intervals). These restrictions correspond roughly to a particular mathematical model which differs from Euclidean space in its manifest symmetry.
Until the beginning of the 20th century, time was believed to be independent of motion, progressing at a fixed rate in all reference frames; however, later experiments revealed that time slows at higher speeds of the reference frame relative to another reference frame. Such slowing, called time dilation, is explained in special relativity theory. Many experiments have confirmed time dilation, such as the relativistic decay of muons from cosmic ray showers and the slowing of atomic clocks aboard a Space Shuttle relative to synchronized Earth-bound inertial clocks. The duration of time can therefore vary according to events and reference frames.
When dimensions are understood as mere components of the grid system, rather than physical attributes of space, it is easier to understand the alternate dimensional views as being simply the result of coordinate transformations.
The term spacetime has taken on a generalized meaning beyond treating spacetime events with the normal 3+1 dimensions. It is really the combination of space and time. Other proposed spacetime theories include additional dimensions—normally spatial but there exist some speculative theories that include additional temporal dimensions and even some that include dimensions that are neither temporal nor spatial (e.g. superspace). How many dimensions are needed to describe the universe is still an open question. Speculative theories such as string theory predict 10 or 26 dimensions (with M-theory predicting 11 dimensions: 10 spatial and 1 temporal), but the existence of more than four dimensions would only appear to make a difference at the subatomic level.[1]
## Spacetime in literature
Incas regarded space and time as a single concept, referred to as pacha (Quechua: pacha, Aymara: pacha).[2][3] The peoples of the Andes maintain a similar understanding.[4]
Arthur Schopenhauer wrote in §18 of On the Fourfold Root of the Principle of Sufficient Reason (1813): "...the representation of coexistence is impossible in Time alone; it depends, for its completion, upon the representation of Space; because, in mere Time, all things follow one another, and in mere Space all things are side by side; it is accordingly only by the combination of Time and Space that the representation of coexistence arises."
The idea of a unified spacetime is stated by Edgar Allan Poe in his essay on cosmology titled Eureka (1848) that "Space and duration are one." In 1895, in his novel The Time Machine, H.G. Wells wrote, "There is no difference between time and any of the three dimensions of space except that our consciousness moves along it", and that "any real body must have extension in four directions: it must have Length, Breadth, Thickness, and Duration."
Marcel Proust, in his novel Swann's Way (published 1913), describes the village church of his childhood's Combray as "... a building which occupied, so to speak, four dimensions of space—the name of the fourth being Time..."
### Mathematical concept
The first reference to spacetime as a mathematical concept was in 1754 by Jean le Rond d'Alembert in the article Dimension in Encyclopedie. Another early venture was by Joseph Louis Lagrange in his Theory of Analytic Functions (1797, 1813). He said, "One may view mechanics as a geometry of four dimensions, and mechanical analysis as an extension of geometric analysis".[5]
After discovering quaternions,[6] William Rowan Hamilton commented, "Time is said to have only one dimension, and space to have three dimensions. ... The mathematical quaternion partakes of both these elements; in technical language it may be said to be 'time plus space', or 'space plus time': and in this sense it has, or at least involves a reference to, four dimensions. And how the One of Time, of Space the Three, Might in the Chain of Symbols girdled be." Hamilton's biquaternions, which have algebraic properties sufficient to model spacetime and its symmetry, were in play for more than a half-century before formal relativity. For instance, William Kingdon Clifford noted their relevance.
Another important antecedent to spacetime was the work of James Clerk Maxwell as he used partial differential equations to develop electrodynamics with the four parameters. Lorentz discovered some invariances of Maxwell's equations late in the 19th century which were to become the basis of Einstein's theory of special relativity. Fiction authors were also involved, as mentioned above. It has always been the case that time and space are measured using real numbers, and the suggestion that the dimensions of space and time are comparable could have been raised by the first people to have formalized physics, but ultimately, the contradictions between Maxwell's laws and Galilean relativity had to come to a head with the realization of the import of finitude of the speed of light.
While spacetime can be viewed as a consequence of Albert Einstein's 1905 theory of special relativity, it was first explicitly proposed mathematically by one of his teachers, the mathematician Hermann Minkowski, in a 1908 essay[7] building on and extending Einstein's work. His concept of Minkowski space is the earliest treatment of space and time as two aspects of a unified whole, the essence of special relativity. (For an English translation of Minkowski's article, see Lorentz et al. 1952.) The 1926 thirteenth edition of the Encyclopædia Britannica included an article by Einstein titled "Space–Time".[8]) The idea of Minkowski space led to special relativity being viewed in a more geometrical way.
However, the most important contribution of Minkowski's geometric viewpoint of spacetime turned out to be in Einstein's later development of general relativity, since the correct description of the effect of gravitation on space and time was found to be most easily visualized as a "warp" or stretching in the geometrical fabric of space and time, in a smooth and continuous way that changed smoothly from point-to-point along the spacetime fabric.
## Basic concepts
Spacetimes are the arenas in which all physical events take place—an event is a point in spacetime specified by its time and place. For example, the motion of planets around the sun may be described in a particular type of spacetime, or the motion of light around a rotating star may be described in another type of spacetime. The basic elements of spacetime are events. In any given spacetime, an event is a unique position at a unique time. Because events are spacetime points, an example of an event in classical relativistic physics is $(x,y,z,t)$, the location of an elementary (point-like) particle at a particular time. A spacetime itself can be viewed as the union of all events in the same way that a line is the union of all of its points, formally organized into a manifold, a space which can be described at small scales using coordinates systems.
A spacetime is independent of any observer.[9] However, in describing physical phenomena (which occur at certain moments of time in a given region of space), each observer chooses a convenient metrical coordinate system. Events are specified by four real numbers in any such coordinate system. The trajectories of elementary (point-like) particles through space and time are thus a continuum of events called the world line of the particle. Extended or composite objects (consisting of many elementary particles) are thus a union of many world lines twisted together by virtue of their interactions through spacetime into a "world-braid".
However, in physics, it is common to treat an extended object as a "particle" or "field" with its own unique (e.g. center of mass) position at any given time, so that the world line of a particle or light beam is the path that this particle or beam takes in the spacetime and represents the history of the particle or beam. The world line of the orbit of the Earth (in such a description) is depicted in two spatial dimensions x and y (the plane of the Earth's orbit) and a time dimension orthogonal to x and y. The orbit of the Earth is an ellipse in space alone, but its world line is a helix in spacetime.[10]
The unification of space and time is exemplified by the common practice of selecting a metric (the measure that specifies the interval between two events in spacetime) such that all four dimensions are measured in terms of units of distance: representing an event as $(x_0,x_1,x_2,x_3) = (ct,x,y,z)$ (in the Lorentz metric) or $(x_1,x_2,x_3,x_4) = (x,y,z,ict)$ (in the original Minkowski metric)[11] where $c$ is the speed of light. The metrical descriptions of Minkowski Space and spacelike, lightlike, and timelike intervals given below follow this convention, as do the conventional formulations of the Lorentz transformation.
### Spacetime intervals
In a Euclidean space, the separation between two points is measured by the distance between the two points. The distance is purely spatial, and is always positive. In spacetime, the separation between two events is measured by the invariant interval between the two events, which takes into account not only the spatial separation between the events, but also their temporal separation. The interval, s2, between two events is defined as:
$s^2 = \Delta r^2 - c^2\Delta t^2 \,$ (spacetime interval),
where c is the speed of light, and Δr and Δt denote differences of the space and time coordinates, respectively, between the events. (Note that the choice of signs for $s^2$ above follows the space-like convention (−+++). Other treatments reverse the sign of $s^2$.)
Certain types of world lines (called geodesics of the spacetime) are the shortest paths between any two events, with distance being defined in terms of spacetime intervals. The concept of geodesics becomes critical in general relativity, since geodesic motion may be thought of as "pure motion" (inertial motion) in spacetime, that is, free from any external influences.
Spacetime intervals may be classified into three distinct types, based on whether the temporal separation ($c^2 \Delta t^2$) or the spatial separation ($\Delta r^2$) of the two events is greater.
#### Time-like interval
$\begin{align} \\ c^2\Delta t^2 &> \Delta r^2\\ s^2 &< 0 \\ \end{align}$
For two events separated by a time-like interval, enough time passes between them for there to be a cause–effect relationship between the two events. For a particle traveling through space at less than the speed of light, any two events which occur to or by the particle must be separated by a time-like interval. Event pairs with time-like separation define a negative squared spacetime interval ($s^2 < 0$) and may be said to occur in each other's future or past. There exists a reference frame such that the two events are observed to occur in the same spatial location, but there is no reference frame in which the two events can occur at the same time.
The measure of a time-like spacetime interval is described by the proper time, $\Delta\tau$:
$\Delta\tau = \sqrt{\Delta t^2 - \frac{\Delta r^2}{c^2}}$ (proper time).
The proper time interval would be measured by an observer with a clock traveling between the two events in an inertial reference frame, when the observer's path intersects each event as that event occurs. (The proper time defines a real number, since the interior of the square root is positive.)
#### Light-like interval
$\begin{align} c^2\Delta t^2 &= \Delta r^2 \\ s^2 &= 0 \\ \end{align}$
In a light-like interval, the spatial distance between two events is exactly balanced by the time between the two events. The events define a squared spacetime interval of zero ($s^2 = 0$). Light-like intervals are also known as "null" intervals.
Events which occur to or are initiated by a photon along its path (i.e., while traveling at $c$, the speed of light) all have light-like separation. Given one event, all those events which follow at light-like intervals define the propagation of a light cone, and all the events which preceded from a light-like interval define a second (graphically inverted, which is to say "pastward") light cone.
#### Space-like interval
$\begin{align} \\ c^2\Delta t^2 &< \Delta r^2 \\ s^2 &> 0 \\ \end{align}$
When a space-like interval separates two events, not enough time passes between their occurrences for there to exist a causal relationship crossing the spatial distance between the two events at the speed of light or slower. Generally, the events are considered not to occur in each other's future or past. There exists a reference frame such that the two events are observed to occur at the same time, but there is no reference frame in which the two events can occur in the same spatial location.
For these space-like event pairs with a positive squared spacetime interval ($s^2 > 0$), the measurement of space-like separation is the proper distance, $\Delta\sigma$:
$\Delta\sigma = \sqrt{s^2} = \sqrt{\Delta r^2 - c^2\Delta t^2}$ (proper distance).
Like the proper time of time-like intervals, the proper distance of space-like spacetime intervals is a real number value.
## Mathematics of spacetimes
For physical reasons, a spacetime continuum is mathematically defined as a four-dimensional, smooth, connected Lorentzian manifold $(M,g)$. This means the smooth Lorentz metric $g$ has signature $(3,1)$. The metric determines the geometry of spacetime, as well as determining the geodesics of particles and light beams. About each point (event) on this manifold, coordinate charts are used to represent observers in reference frames. Usually, Cartesian coordinates $(x, y, z, t)$ are used. Moreover, for simplicity's sake, the speed of light $c$ is usually assumed to be unity.
A reference frame (observer) can be identified with one of these coordinate charts; any such observer can describe any event $p$. Another reference frame may be identified by a second coordinate chart about $p$. Two observers (one in each reference frame) may describe the same event $p$ but obtain different descriptions.
Usually, many overlapping coordinate charts are needed to cover a manifold. Given two coordinate charts, one containing $p$ (representing an observer) and another containing $q$ (representing another observer), the intersection of the charts represents the region of spacetime in which both observers can measure physical quantities and hence compare results. The relation between the two sets of measurements is given by a non-singular coordinate transformation on this intersection. The idea of coordinate charts as local observers who can perform measurements in their vicinity also makes good physical sense, as this is how one actually collects physical data—locally.
For example, two observers, one of whom is on Earth, but the other one who is on a fast rocket to Jupiter, may observe a comet crashing into Jupiter (this is the event $p$). In general, they will disagree about the exact location and timing of this impact, i.e., they will have different 4-tuples $(x, y, z, t)$ (as they are using different coordinate systems). Although their kinematic descriptions will differ, dynamical (physical) laws, such as momentum conservation and the first law of thermodynamics, will still hold. In fact, relativity theory requires more than this in the sense that it stipulates these (and all other physical) laws must take the same form in all coordinate systems. This introduces tensors into relativity, by which all physical quantities are represented.
Geodesics are said to be time-like, null, or space-like if the tangent vector to one point of the geodesic is of this nature. Paths of particles and light beams in spacetime are represented by time-like and null (light-like) geodesics, respectively.
### Topology
Main article: Spacetime topology
The assumptions contained in the definition of a spacetime are usually justified by the following considerations.
The connectedness assumption serves two main purposes. First, different observers making measurements (represented by coordinate charts) should be able to compare their observations on the non-empty intersection of the charts. If the connectedness assumption were dropped, this would not be possible. Second, for a manifold, the properties of connectedness and path-connectedness are equivalent, and one requires the existence of paths (in particular, geodesics) in the spacetime to represent the motion of particles and radiation.
Every spacetime is paracompact. This property, allied with the smoothness of the spacetime, gives rise to a smooth linear connection, an important structure in general relativity. Some important theorems on constructing spacetimes from compact and non-compact manifolds include the following:[citation needed]
• A compact manifold can be turned into a spacetime if, and only if, its Euler characteristic is 0. (Proof idea: the existence of a Lorentzian metric is shown to be equivalent to the existence of a nonvanishing vector field.)
• Any non-compact 4-manifold can be turned into a spacetime.
### Spacetime symmetries
Main article: Spacetime symmetries
Often in relativity, spacetimes that have some form of symmetry are studied. As well as helping to classify spacetimes, these symmetries usually serve as a simplifying assumption in specialized work. Some of the most popular ones include:
• Axisymmetric spacetimes
• Spherically symmetric spacetimes
• Static spacetimes
• Stationary spacetimes.
### Causal structure
Main article: Causal structure
See also: Causality (physics) and Causality
The causal structure of a spacetime describes causal relationships between pairs of points in the spacetime based on the existence of certain types of curves joining the points.
## Spacetime in special relativity
Main article: Minkowski space
The geometry of spacetime in special relativity is described by the Minkowski metric on R4. This spacetime is called Minkowski space. The Minkowski metric is usually denoted by $\eta$ and can be written as a four-by-four matrix:
$\eta_{ab} \, = \operatorname{diag}(1, -1, -1, -1)$
where the Landau–Lifshitz space-like convention is being used. A basic assumption of relativity is that coordinate transformations must leave spacetime intervals invariant. Intervals are invariant under Lorentz transformations. This invariance property leads to the use of four-vectors (and other tensors) in describing physics.
Strictly speaking, one can also consider events in Newtonian physics as a single spacetime. This is Galilean–Newtonian relativity, and the coordinate systems are related by Galilean transformations. However, since these preserve spatial and temporal distances independently, such a spacetime can be decomposed into spatial coordinates plus temporal coordinates, which is not possible in the general case.
## Spacetime in general relativity
Main article: Spacetime in General relativity
General relativity
$G_{\mu \nu} + \Lambda g_{\mu \nu}= {8\pi G\over c^4} T_{\mu \nu}$
Introduction
Mathematical formulation
Resources · Tests
Fundamental concepts
Phenomena
Advanced theories
Scientists
Spacetime
In general relativity, it is assumed that spacetime is curved by the presence of matter (energy), this curvature being represented by the Riemann tensor. In special relativity, the Riemann tensor is identically zero, and so this concept of "non-curvedness" is sometimes expressed by the statement Minkowski spacetime is flat.
The earlier discussed notions of time-like, light-like and space-like intervals in special relativity can similarly be used to classify one-dimensional curves through curved spacetime. A time-like curve can be understood as one where the interval between any two infinitesimally close events on the curve is time-like, and likewise for light-like and space-like curves. Technically the three types of curves are usually defined in terms of whether the tangent vector at each point on the curve is time-like, light-like or space-like. The world line of a slower-than-light object will always be a time-like curve, the world line of a massless particle such as a photon will be a light-like curve, and a space-like curve could be the world line of a hypothetical tachyon. In the local neighborhood of any event, time-like curves that pass through the event will remain inside that event's past and future light cones, light-like curves that pass through the event will be on the surface of the light cones, and space-like curves that pass through the event will be outside the light cones. One can also define the notion of a 3-dimensional "spacelike hypersurface", a continuous 3-dimensional "slice" through the 4-dimensional property with the property that every curve that is contained entirely within this hypersurface is a space-like curve.[12]
Many spacetime continua have physical interpretations which most physicists would consider bizarre or unsettling. For example, a compact spacetime has closed timelike curves, which violate our usual ideas of causality (that is, future events could affect past ones). For this reason, mathematical physicists usually consider only restricted subsets of all the possible spacetimes. One way to do this is to study "realistic" solutions of the equations of general relativity. Another way is to add some additional "physically reasonable" but still fairly general geometric restrictions and try to prove interesting things about the resulting spacetimes. The latter approach has led to some important results, most notably the Penrose–Hawking singularity theorems.
## Quantized spacetime
Main article: Quantum spacetime
In general relativity, spacetime is assumed to be smooth and continuous—and not just in the mathematical sense. In the theory of quantum mechanics, there is an inherent discreteness present in physics. In attempting to reconcile these two theories, it is sometimes postulated that spacetime should be quantized at the very smallest scales. Current theory is focused on the nature of spacetime at the Planck scale. Causal sets, loop quantum gravity, string theory, and black hole thermodynamics all predict a quantized spacetime with agreement on the order of magnitude. Loop quantum gravity makes precise predictions about the geometry of spacetime at the Planck scale.
## Privileged character of 3+1 spacetime
See also: Kairos, Time perception, and Dreamtime
There are two kinds of dimensions, spatial (bidirectional) and temporal (unidirectional). Let the number of spatial dimensions be N and the number of temporal dimensions be T. That N = 3 and T = 1, setting aside the compactified dimensions invoked by string theory and undetectable to date, can be explained by appealing to the physical consequences of letting N differ from 3 and T differ from 1. The argument is often of an anthropic character.
The implicit notion that the dimensionality of the universe is special is first attributed to Gottfried Wilhelm Leibniz, who in the Discourse on Metaphysics suggested[13] that the world is "the one which is at the same time the simplest in hypothesis and the richest in phenomena." Immanuel Kant argued that 3-dimensional space was a consequence of the inverse square law of universal gravitation. While Kant's argument is historically important, John D. Barrow says that it "...gets the punch-line back to front: it is the three-dimensionality of space that explains why we see inverse-square force laws in Nature, not vice-versa." (Barrow 2002: 204). This is because the law of gravitation (or any other inverse-square law) follows from the concept of flux and the proportional relationship of flux density and the strength of field. If N = 3, then 3-dimensional solid objects have surface areas proportional to the square of their size in any selected spatial dimension. In particular, a sphere of radius r has area of 4πr ². More generally, in a space of N dimensions, the strength of the gravitational attraction between two bodies separated by a distance of r would be inversely proportional to rN−1.
In 1920, Paul Ehrenfest showed that if we fix T = 1 and let N > 3, the orbit of a planet about its sun cannot remain stable. The same is true of a star's orbit around the center of its galaxy.[14] Ehrenfest also showed that if N is even, then the different parts of a wave impulse will travel at different speeds. If N > 3 and odd, then wave impulses become distorted. Only when N = 3 or 1 are both problems avoided. In 1922, Hermann Weyl showed that Maxwell's theory of electromagnetism works only when N = 3 and T = 1, writing that this fact "...not only leads to a deeper understanding of Maxwell's theory, but also of the fact that the world is four dimensional, which has hitherto always been accepted as merely 'accidental,' become intelligible through it."[15] Finally, Tangherlini[16] showed in 1963 that when N > 3, electron orbitals around nuclei cannot be stable; electrons would either fall into the nucleus or disperse.
Properties of n+m-dimensional spacetimes
Max Tegmark[17] expands on the preceding argument in the following anthropic manner. If T differs from 1, the behavior of physical systems could not be predicted reliably from knowledge of the relevant partial differential equations. In such a universe, intelligent life capable of manipulating technology could not emerge. Moreover, if T > 1, Tegmark maintains that protons and electrons would be unstable and could decay into particles having greater mass than themselves. (This is not a problem if the particles have a sufficiently low temperature.) If N > 3, Ehrenfest's argument above holds; atoms as we know them (and probably more complex structures as well) could not exist. If N < 3, gravitation of any kind becomes problematic, and the universe is probably too simple to contain observers. For example, when N < 3, nerves cannot cross without intersecting.
In general, it is not clear how physical law could function if T differed from 1. If T > 1, subatomic particles which decay after a fixed period would not behave predictably, because time-like geodesics would not be necessarily maximal.[18] N = 1 and T = 3 has the peculiar property that the speed of light in a vacuum is a lower bound on the velocity of matter; all matter consists of tachyons.[17] However, signature (1,3) and (3,1) are physically equivalent. To call vectors with positive Minkowski "length" timelike is just a convention that depends on the convention for the sign of the metric tensor. Indeed, particle phyicists tend to use a metric with signature (+−−−) that results in positive Minkowski "length" for timelike intervals and energies while spatial separations have negative Minkowski "length". Relativists, however, tend to use the opposite convention (−+++) so that spatial separations have positive Minkowski length.
Hence anthropic and other arguments rule out all cases except N = 3 and T = 1 (or N = 1 and T = 3 in different conventions) — which happens to describe the world about us. Curiously, the cases N = 3 or 4 have the richest and most difficult geometry and topology. There are, for example, geometric statements whose truth or falsity is known for all N except one or both of 3 and 4.[citation needed] N = 3 was the last case of the Poincaré conjecture to be proved.
For an elementary treatment of the privileged status of N = 3 and T = 1, see chpt. 10 (esp. Fig. 10.12) of Barrow;[19] for deeper treatments, see §4.8 of Barrow and Tipler (1986) and Tegmark.[17] Barrow has repeatedly cited the work of Whitrow.[20]
String theory hypothesizes that matter and energy are composed of tiny vibrating strings of various types, most of which are embedded in dimensions that exist only on a scale no larger than the Planck length. Hence N = 3 and T = 1 do not characterize string theory, which embeds vibrating strings in coordinate grids having 10, or even 26, dimensions.
The Causal dynamical triangulation (CDT) theory is a background independent theory which derives the observed 3+1 spacetime from a minimal set of assumptions, and needs no adjusting factors. It does not assume any pre-existing arena (dimensional space), but rather attempts to show how the spacetime fabric itself evolves. It shows spacetime to be 2-d near the Planck scale, and reveals a fractal structure on slices of constant time, but spacetime becomes 3+1-d in scales significantly larger than Planck. So, CDT may become the first theory which doesn't postulate but really explains observed number of spacetime dimensions.[21]
## References
1. Kopeikin, Sergei; Efroimsky, Michael; Kaplan, George (2011). Relativistic Celestial Mechanics of the Solar System. John Wiley & Sons. p. 157. ISBN 3527634576. , Extract of page 157
2. Paul Richard Steele, Catherine J. Allen, Handbook of Inca mythology, p. 86, (ISBN 1-57607-354-8)
3. Shirley Ardener, University of Oxford, Women and space: ground rules and social maps, p. 36 (ISBN 0-85496-728-1)
4. R.C. Archibald (1914) Time as a fourth dimension Bulletin of the American Mathematical Society 20:409.
5. Gallier, Jean H. (2001). Geometric methods and applications: for computer science and engineering. Springer. p. 249. ISBN 0-387-95044-3. , Chapter 8, page 249
6. Minkowski, Hermann (1909), "Raum und Zeit", Physikalische Zeitschrift 10: 75–88
• Various English translations on Wikisource: Space and Time.
7. Einstein, Albert, 1926, "Space–Time", Encyclopædia Britannica, 13th ed.
8. Matolcsi, Tamás (1994). Spacetime Without Reference Frames. Budapest: Akadémiai Kiadó.
9. Ellis, G. F. R.; Williams, Ruth M. (2000). Flat and curved space–times (2nd ed.). Oxford University Press. p. 9. ISBN 0-19-850657-0.
10. Petkov, Vesselin (2010). Minkowski Spacetime: A Hundred Years Later. Springer. p. 70. ISBN 90-481-3474-9. , Section 3.4, p. 70
11. See "Quantum Spacetime and the Problem of Time in Quantum Gravity" by Leszek M. Sokolowski, where on this page he writes "Each of these hypersurfaces is spacelike, in the sense that every curve, which entirely lies on one of such hypersurfaces, is a spacelike curve." More commonly a space-like hypersurface is defined technically as a surface such that the normal vector at every point is time-like, but the definition above may be somewhat more intuitive.
12. Ehrenfest, Paul (1920). "How do the fundamental laws of physics make manifest that Space has 3 dimensions?". Annalen der Physik 61 (5): 440. Bibcode:1920AnP...366..440E. doi:10.1002/andp.19203660503. . Also see Ehrenfest, P. (1917) "In what way does it become manifest in the fundamental laws of physics that space has three dimensions?" Proceedings of the Amsterdam Academy 20: 200.
13. Weyl, H. (1922) Space, time, and matter. Dover reprint: 284.
14. Tangherlini, F. R. (1963). "Atoms in Higher Dimensions". Nuovo Cimento 14 (27): 636.
15. ^ a b c Tegmark, Max (April 1997). "On the dimensionality of spacetime". Classical and Quantum Gravity 14 (4): L69–L75. arXiv:gr-qc/9702052. Bibcode:1997CQGra..14L..69T. doi:10.1088/0264-9381/14/4/002. Retrieved 2006-12-16.
16. Dorling, J. (1970). "The Dimensionality of Time". American Journal of Physics 38 (4): 539–40. Bibcode:1970AmJPh..38..539D. doi:10.1119/1.1976386.
17. Barrow, J. D. (2002). The Constants of Nature. Pantheon Books. ISBN 0-375-42221-8.
18. Whitrow, G. J. (1955) " ," British Journal of the Philosophy of Science 6: 13. Also see his (1959) The Structure and Evolution of the Universe. London: Hutchinson.
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http://mathoverflow.net/questions/109190/indecomposable-modules-of-a-tensor-product-of-two-algebras
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Indecomposable modules of a tensor product of two algebras
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Hi,
I have two commutative finite-dimensional $k$-Algebras $A$ and $B$ ($k$ is a field).
I wonder, whether there is a way to get the finite-dimensional indecomposable modules of $A \otimes_k B$,
if I know the finite-dimensional indecomposable modules of $A$ and $B$.
For example, is there a way to do this, if $A=k[x]/(x^n)$ and $B=k[y]/(y^n)$?
Thank you very much.
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2 Answers
No, in general there isn't. In your particular case by accident for $n=1$ this is possible. But even for $n=2$, there are only $2$ indecomposable $A$-modules up to isomorphism. But Kronecker already clasified the $A\otimes_k B$-modules and there are infinitely many. For $n>2$ there are still only finitely many indecomposable $A$-modules (namely $n$) up to isomorphism, but the situation is in some sense worse. The indecomposable $A\otimes_k B$-modules are not even classifiable (one says $A\otimes B$ is wild).
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You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
There are a lot of indecomposable modules of $R=k[x]/x^n \otimes k[y]/y^n = k[x,y]/(x^n,y^n)$. If $k$ is infinite then there are infinitely many: $R/(x-ay)$ is a distinct indecomposable module for each $a\in k$. There is also $R$ mod any complicated but irreducible polynomial. I don't think you'll get a full classification that is nice in any way.
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http://unapologetic.wordpress.com/2007/06/11/products-and-coproducts/?like=1&source=post_flair&_wpnonce=fe7f791e1e
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# The Unapologetic Mathematician
## Products and Coproducts
Let’s consider the Cartesian product $X\times Y$ of two sets $X$ and $Y$. Classically we think of this as the set of all pairs $(x,y)$ with $x\in X$ and $y\in Y$. But we can also characterize it just in terms of functions.
Specifically, $X\times Y$ comes with two projection functions $\pi_X:X\times Y\rightarrow X$ and $\pi_Y:X\times Y\rightarrow Y$, defined by $\pi_X(x,y)=x$ and $\pi_Y(x,y)=y$. If we take any other set $S$ with functions $f_X:S\rightarrow X$ and $f_Y:S\rightarrow Y$ we can define the function $(f_X,f_Y):S\rightarrow X\times Y$ by $\left[(f_X,f_Y)\right](s)=(f_X(s),f_Y(s))$. Then we see that $\pi_X\circ(f_X,f_Y)=f_X$ and $\pi_Y\circ(f_X,f_Y)=f_Y$. Further, this function from $S$ to $X\times Y$ is the only such function.
Now let’s do away with those nasty elements altogether and draw this diagram:
What does this mean? Well, it’s like the diagram I drew for products of groups. The product of $X$ and $Y$ is a set $X\times Y$ with functions $\pi_X$ and $\pi_Y$ so that for any other set $S$ with functions to $X$ and $Y$ there exists a unique arrow to $X\times Y$ making the diagram commute. Since we’ve written this definition without ever really referring to elements we can just pick it up and drop it into any other category. Many descriptions of categorical products stop here, but let’s push a bit further.
Let’s consider a category $\mathcal{C}$ containing (among others) objects $X$ and $Y$. From this we’re going to build a new category. An object of our category will be a diagram that looks like $X\leftarrow^{f_X}S\rightarrow^{f_Y}Y$ in $\mathcal{C}$. A morphism from $X\leftarrow^{f_X}S\rightarrow^{f_Y}Y$ to $X\leftarrow^{f'_X}S'\rightarrow^{f'_Y}Y$ will be a morphism $S\rightarrow^{g}S'$ in $\mathcal{C}$ so that $f_X=f'_X\circ g$ and $f_Y=f'_Y\circ g$.
Now what’s a product in $\mathcal{C}$? It’s a terminal object of this category we’ve constructed! That is, it’s one of these diagrams so that every other diagram has a unique morphism (as defined above) to it. This definition makes sense in any category $\mathcal{C}$, though the category we build from a given pair of objects may not have a terminal object, so a given pair of objects of $\mathcal{C}$ may not have a product in $\mathcal{C}$. If every pair of objects of $\mathcal{C}$ has a product in $\mathcal{C}$, we say that $\mathcal{C}$ “has products”.
So, the existence of Cartesian products of sets shows that $\mathbf{Set}$ has products. Similarly, $\mathbf{Grp}$ has products, as do $\mathbf{Ring}$, $\mathbf{Gpd}$ (groupoids), $\mathbf{Cat}$ (small categories), and pretty much all our familiar categories from algebra.
What about something like a preordered set $(P,\preceq)$, considered as a category? What would “product” mean, when written in this language? Well, given elements $a$ and $b$ the product $a\times b$ will have arrows to $a$ and $b$, so $a\times b\preceq a$ and $a\times b\preceq b$. Also, for any other element $c$ with $c\preceq a$ and $c\preceq b$ we have $c\preceq a\times b$ — $c$ has an arrow to both $a$ and $b$, so it has an arrow to $a\times b$. That is, $a\times b$ is a greatest lower bound of $a$ and $b$, and the category has products if and only if every pair of elements has such a greatest lower bound.
And it gets better. If we consider a category $\mathcal{C}$ that has products, the product defines a functor $\times:\mathcal{C}\times\mathcal{C}\rightarrow\mathcal{C}$! If we have arrows $f_1:X_1\rightarrow Y_1$ and $f_2:X_2\rightarrow Y_2$ then I say we’ll have an arrow $f_1\times f_2:X_1\times X_2\rightarrow Y_1\times Y_2$. Indeed, if we consider $f_1\circ\pi_{X_1}:X_1\times X_2\rightarrow Y_1$ and $f_2\circ\pi_{X_2}:X_1\times X_2\rightarrow Y_2$ then we’ll get an arrow from $X_1\times X_2$ to $Y_1\times Y_2$. And this construction preserves compositions and identities. For compositions, start with this diagram:
and draw in the induced arrows $f_1\times f_2$, $g_1\times g_2$, and $(g_1\circ f_1)\times(g_2\circ f_2)$. Then use the uniqueness part of the universal property to show that the composite of the first two must be the same as the third. Do a similar thing to verify that identities are also preserved.
Finally, we can flip all the arrows in what we’ve said to get the dual notion: coproducts. Use this diagram:
and define the coproduct to be an initial object in a certain category of diagrams. Check that in $\mathbf{Set}$ this property is satisfied by disjoint unions. In $\mathbf{Grp}$ coproducts are free products. In a preorder, coproducts are least upper bounds. And, of course, the coproduct defines a functor from $\mathcal{C}\times\mathcal{C}$ to $\mathcal{C}$.
There’s a fair bit to digest here, but it’s worth it. The next few ideas are really very similar. Alternatively, you could take this to mean that if you don’t completely get it now there are a few more examples in the pipe that may help.
### Like this:
Posted by John Armstrong | Category theory
## 9 Comments »
1. [...] off is that and are both subsets of . That is, there are arrows and . So maybe the union is the coproduct of the two sets. Well, it’s a good guess, but there’s a problem. There may be some [...]
Pingback by | June 14, 2007 | Reply
2. [...] products, coproducts, equalizers, coequalizers, pullbacks, pushouts… We’ve got products and coproducts of two objects at a time, equalizers and coequalizers of two morphisms at a time, and pushouts and [...]
Pingback by | June 15, 2007 | Reply
3. [...] know that any functor that has a right adjoint preserves colimits! The disjoint union of sets is a coproduct, and the direct sum of vector spaces is a biproduct, which means it’s also a coproduct. Thus [...]
Pingback by | July 21, 2008 | Reply
4. Am I right in thinking that the category of diagrams you create in paragraph 5 only makes sense if the category C with which you start contains pull-backs? The reason I ask is that when trying to prove that you get a category, I got stuck on composing the morphisms. If you have pull-backs, I know what to do. Otherwise…
Comment by | September 19, 2008 | Reply
5. No, a morphism is just an arrow between the two middle terms that makes the triangles on the sides commute. Compose morphisms just by composing those arrows, and the larger triangles will automatically commute.
Comment by | September 19, 2008 | Reply
6. Ah…I see where I was confused now. The X and Y you chose are fixed and you are only letting the middle term change. I was letting the X and Y vary.
Comment by | September 19, 2008 | Reply
7. [...] We should also note that the category of root systems has binary (and thus finite) coproducts. They both start the same way: given root systems and in inner-product spaces and , we take the [...]
Pingback by | January 25, 2010 | Reply
8. [...] measurable spaces, but we have a broader perspective here. Indeed, we should be asking if this is a product object in the category of measurable spaces! That is, the underlying space comes equipped with projection [...]
Pingback by | July 15, 2010 | Reply
9. [...] we want to show that we have (finite) products in the category of manifolds. Specifically, if and are – and -dimensional smooth manifolds, [...]
Pingback by | March 7, 2011 | Reply
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://mathhelpforum.com/differential-geometry/194698-what-radius-convergence-power-series-x-print.html
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# what is the radius of convergence of the power series Σan xⁿ
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• December 27th 2011, 06:18 AM
sorv1986
what is the radius of convergence of the power series Σan xⁿ
suppose Σan xⁿ be a power series and an= the number of divisors of n⁵⁰.
Then what will be the radius of convergence of this power series?
Thanks in advance . regards.(Bow)
• December 27th 2011, 07:15 AM
FernandoRevilla
Re: what is the radius of convergence of the power series Σan xⁿ
Quote:
Originally Posted by sorv1986
suppose Σan xⁿ be a power series and an= the number of divisors of n⁵⁰. Then what will be the radius of convergence of this power series?
Hint $1\leq a_n\leq n^{50}\Rightarrow 1\leq a_n^{1/n}\leq (n^{1/n})^{50}$ . Now, apply the Sandwich Theorem.
All times are GMT -8. The time now is 03:54 PM.
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http://mathhelpforum.com/algebra/90357-simplifying-polynomial.html
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# Thread:
1. ## Simplifying this polynomial
I was reading this problem and read about how to solve for y:
$3x^2 + 2y^2 = 35$
It said the equation could be written like this:
I can't seem to figure out how they got that. Can someone explain?
2. Originally Posted by Phire
I was reading this problem and read about how to solve for y:
$3x^2 + 2y^2 = 35$
It said the equation could be written like this:
I can't seem to figure out how they got that. Can someone explain?
both sides have been divided by 35
could also look like $\frac{3x^2}{35} + \frac{2y^2}{35} = 1$
3. Originally Posted by pickslides
both sides have been divided by 35
could also look like $\frac{3x^2}{35} + \frac{2y^2}{35} = 1$
Yes, I knew at least that much, but how did they go further after that and get rid of the coefficients in the numerators? It couldn't have been just from dividing them out because the other terms of the equations don't look like they've been divided by them. The fractions in the denominators seem strange to me.
EDIT: Ah, nevermind, after re-writing the terms as division problems, it returns back to that form you wrote.
4. agreed it is a strange way to write this equation although if the question asked for the equation to be put in the form of
$\frac{x^2}{a}+\frac{y^2}{b}=1$
then it suits quite well..
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http://unapologetic.wordpress.com/2008/08/08/
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# The Unapologetic Mathematician
## Properties of Complex Numbers
Today I’ll collect a few basic properties of complex numbers.
First off, they form a vector space over the reals. We constructed them as an algebra — the quotient of the algebra of polynomials by a certain ideal — and every algebra is a vector space. So what can we say about them as a vector space? The easiest fact is that it’s two-dimensional, and it’s got a particularly useful basis.
To see this, remember that we have a basis for the algebra of polynomials, which is given by the powers of the variable. So here when we throw in the formal element $i$, its powers form a basis of the ring $\mathbb{R}[i]$. But we have a relation, and that cuts things down a bit. Specifically, the element $i^2$ is the same as the element $-1$.
Given a polynomial $p$ in the “variable” $i$, we can write it as
$c_ni^n+...+c_2i^2+c_1i+c_0$
We can peel off the constant and linear terms, and then pull out a factor of $i^2$:
$(c_ni^{n-2}+...+c_2)i^2+(c_1i+c_0)$
Now this factor of $i^2$ can be replaced by $-1$, which drops the overall degree. We can continue like this, eventually rewriting any term involving higher powers of $i$ using only constant and linear terms. That is, any complex number can be written as $c_0+c_1i$, where $c_0$ and $c_1$ are real constants. Further, this representation is unique. This establishes the set $\{1,i\}$ as a basis for $\mathbb{C}$ as a vector space over $\mathbb{R}$.
Now the additive parts of the field structure are clear from the vector space structure here. We can write the sum of two complex numbers $a_1+b_1i$ and $a_2+b_2i$ simply by adding the components: $(a_1+a_2)+(b_1+b_2)i$. We get the negative of a complex number by taking the negatives of the components.
We can also write out products pretty simply, since we know the product of pairs of basis elements. The only one that doesn’t involve the unit of the algebra is $i\otimes i\mapsto-1$. So in terms of components we can write out the product of the complex numbers above as $(a_1a_2-b_1b_2)+(a_1b_2+a_2b_1)i$.
Notice here that the field of real numbers sits inside that of complex numbers, using scalar multiples of the complex unit. This is characteristic of algebras, but it’s worth pointing out here. Any real number $a$ can be considered as the complex number $a+0i$. This preserves all the field structures, but it ignores the order on the real numbers. A small price to pay, but an important one in certain ways.
We also mentioned the symmetry between $i$ and $-i$. Either one is just as valid as a square root of $-1$ as the other is, so if we go through consistently replacing $i$ with $-i$, and $-i$ with $i$, we can’t tell the difference. This leads to an automorphism of fields called “complex conjugation”. It sends the complex number $a+bi$ to its “conjugate” $a-bi$. This preserves all the field structure — additive and multiplicative — and it fixes the real numbers sitting inside the complex numbers.
Studying this automorphism, and similar structures of other field extensions forms the core of what algebraists call “Galois theory”. I’m not going there now, but it’s a huge part of modern mathematics, and its study is ultimately the root of all of our abstract algebraic techniques. The first groups were automorphism groups shuffling around roots of polynomials.
Posted by John Armstrong | Fundamentals, Numbers | 20 Comments
## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://mathhelpforum.com/differential-equations/188786-finding-trial-solutions-2nd-order-homogeneous-de.html
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Thread:
1. Finding trial solutions of 2nd order homogeneous DE.
I am trying to find the trial solutions to the following differential equations. I would be able to solve them, but I am not sure how to go about simply determining trial solutions.
1) $xy''+y'+K^{2}xy=0$
2) $y''+3\sqrt{x}y=0$
2. Re: Finding trial solutions of 2nd order homogeneous DE.
What do you mean "trial solutions"?
3. Re: Finding trial solutions of 2nd order homogeneous DE.
That is what I don't understand. I guess no one will be able to assist on this. I thought trial solution was something specific for certain types of problems.
4. Re: Finding trial solutions of 2nd order homogeneous DE.
May we ask how you came upon the term "trial solution"?
5. Re: Finding trial solutions of 2nd order homogeneous DE.
Originally Posted by cheme
I am trying to find the trial solutions to the following differential equations. I would be able to solve them, but I am not sure how to go about simply determining trial solutions.
1) $xy''+y'+K^{2}xy=0$
2) $y''+3\sqrt{x}y=0$
I'll try to find a solution of 1)... but I don't know if it is or not the 'trial solution'...
Using t as independent variable the DE is...
$t\ y^{''} + y^{'} + K^{2}\ t\ y=0$ (1)
If $F(s)= \mathcal{L}\{f(t)\}$ is the Laplace Tranform of f(t), then a basic property is...
$\mathcal{L}\{t\ f(t)\}= -\frac{d}{d s} F(s)$ (2)
Now if we use the property (2) in (1) we obtain...
$-\frac{d}{ds}\{s^{2}\ Y(s) -s\ y(0)-y^{'}(0)\}+ s\ Y(s) -y(0) -K^{2}\ \frac{d}{ds} Y(s)=0$ (3)
... that with some steps becomes...
$(s^{2}+K^{2})\ Y^{'}(s) +s\ Y(s)=0$ (4)
... so that we have now a first order linear DE in s, of course more 'approachable' then (1). The (4) can be solved with 'standard procedure' obtaining...
$Y(s)= c\ e^{- \int \frac{s}{s^{2}+K^{2}}\ d s}= c\ e^{-\frac{1}{2}\ \ln (s^{2}+K^{2})}}= \frac{c}{\sqrt{s^{2}+K^{2}}}$ (5)
At this point if You have a look at the manual, You discover that is...
$y(t)= \mathcal{L}^{-1} \{\frac{c}{\sqrt{s^{2}+K^{2}}}\}= c\ J_{0}(K\ t)$ (6)
... where $J_{0}(*)$ is the Bessel function of first kind of order 0. Only one consideation: in (6) we have only one 'arbitrary constant' c even if the (1) is a linear ODE of order two. The [probable] reason is that the procedure finds only the solutions of (1) that are L-transformable, so that the other independent solution, probably with a singularity in t=0, is 'desaparecida'...
Kind regards
$\chi$ $\sigma$
6. Re: Finding trial solutions of 2nd order homogeneous DE.
Now we pass to...
$y^{''}+ \sqrt{x}\ y=0$ (1)
... which is a little modified respect to the ODE 2) in the original post [1 instead of 3...]. Here we only want to show how to proceed in cases like this...
In order to eliminate the 'unconfortable' term $\sqrt{x}$, let's suppose that is $y(x)= \sqrt{x}\ g(x)$. In thast case is...
$y^{'}= g^{'}\ \sqrt{x}+ \frac{g}{2\ \sqrt{x}}$ (2)
$y^{''}= g^{''}\ \sqrt{x} + \frac{g^{'}}{\sqrt{x}} - \frac{g}{4 x \sqrt{x}}$ (3)
... and, inserting (3) in (1), we have...
$x^{2}\ g^{''} + x\ g^{'} + (x^{2}-\frac{1}{4})\ g=0$ (4)
The (4) is identified as a Bessel differential equation of order $\nu= \frac{1}{2}$ and its solution is...
$g(x)= c_{1}\ J_{\frac{1}{2}} (x) + c_{2}\ J_{-\frac{1}{2}}(x)$ (5)
... so that the solution of (1) is...
$y(x)=\sqrt{x}\ \{ c_{1}\ J_{\frac{1}{2}} (x) + c_{2}\ J_{-\frac{1}{2}}(x)\}$ (6)
Kind regards
$\chi$ $\sigma$
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http://mathoverflow.net/questions/59996/sum-of-log-p-p-for-p-equivalent-to-l-mod-d
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## Sum of log p/p for p equivalent to l mod D
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It's fairly classical that for $D>1$ and $(D,l)=1$ one has $$\sum_{\stackrel{p\leq x}{p\equiv l\; (mod \; D)}}\frac{\log p}{p} = \frac{\log x}{\phi(D)} + \textrm{O}(1)$$ where if I understand correctly the dependence on $D$ in the $\textrm{O(1)}$ is captured by something like $$\frac{1}{\phi(D)}\sum_{\textrm{non-principal} \; \chi}\frac{-L'(1,\chi)}{L(1,\chi)}$$ which is $\ll_{\epsilon}D^{\epsilon}$ for any $\epsilon>0$. If this is correct, suppose I am interested in the sum $$\sum_{\stackrel{p\leq x}{\left(\frac{-D}{p}\right)=1}}\frac{\log p}{p}$$ where $\left(\frac{-D}{p}\right)$ is the Legendre symbol. There are $\phi(D)/2$ residue classes mod $D$ that $p$ can lie in, and so then this is just my previous sum $\phi(D)/2$ times, and I get that it's $1/2 \log x +\textrm{O}(1)$, where the dependence on $D$ in $\textrm{O}(1)$ is now something like $\textrm{O}(D^{1+\epsilon})$. Is this the best one can do? I was hoping I could get it to be $\textrm{O}(D^{\epsilon})$ but if that's not even correct perhaps I should stop trying.
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## 2 Answers
You're right that your proposed method for estimating the new sum, while correct, gives a worse error bound than we can obtain otherwise. Rather than quoting the classical result itself, I suggest going back to the proof of that classical result and modifying it to address the new sum.
Somewhat more precisely: the Legendre symbol $\big( \frac{-D}p \big)$, or rather its multiplicative generalization the Jacobi symbol $\big( \frac{-D}n \big)$, is a Dirichlet character (mod $D$)—call it $\chi_1(n)$. The sum you are interested in is just $$\sum_{p\le x} \frac{\log p}p \bigg( \frac{1+\chi_1(p)}2 \bigg).$$ (Not exactly, since this expression counts primes dividing $D$ with weight $1/2$, but that's easily dealt with.) Therefore you're left with needing to understand $\sum_{p\le x} (\log p)/p$ (no problem) and $\sum_{p\le x} (\chi_1(p)\log p)/p$. The latter sum is going to be $O(1)$ in the same way that the error term in the classical problem is $O(1)$; it is going to be related to $-L'(1,\chi_1)/L(1,\chi_1)$. So in fact you don't have all the different $-L'(1,\chi)/L(1,\chi)$ error terms to deal with, only the single one where $\chi=\chi_1$.
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1
Thanks, now that I see that it seems obvious. I had tried to work it out that way (since it did seem unnecessary to go through all of the characters to get information basically just about one) but for some reason I was failing. – Elena Mar 29 2011 at 18:51
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
That's a good question, as it illustrates that one should try to find the "correct" harmonic analysis for a given problem (of this type...). What you can do is represent the characteristic function of your set of interest (the $p$ where $(-D/p)=1$) as a combination of Dirichlet characters; by quadratic reciprocity, you only need the principal character and a real character. So you can get your error term as the sum of two error terms coming from $$\sum_{p\leq x}{\chi(p)\log(p)/p}=\delta(\chi)\log(x)+(Error).$$ This should give what you want.
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http://mathoverflow.net/questions/19152/why-is-a-topology-made-up-of-open-sets/49434
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Why is a topology made up of ‘open’ sets? [closed]
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I'm ashamed to admit it, but I don't think I've ever been able to genuinely motivate the definition of a topological space in an undergraduate course. Clearly, the definition distills the essence of many examples, but it's never been obvious to me how it came about, compared, for example, to the rather intuitive definition of a metric space. In some ways, the sparseness of the definition is startling as it tries to capture, apparently successfully, the barest notion of 'space' imaginable.
I can try to make this question more precise if necessary, but I'd prefer to leave it slightly vague, and hope that someone who has discussed this successfully in a first course, perhaps using a better understanding of history, might be able to help me out.
Added 24 March:
I'm grateful to everyone for their thoughtful answers so far. I'll have to think over them a bit before I can get a sense of the 'right' answer for myself. In the meanwhile, I thought I'd emphasize again the obvious fact that the standard concise definition has been tremendously successful. For example, when you classify two-manifolds with it, you get equivalence classes that agree exactly with intuition. Then in as divergent a direction as the study of equations over finite fields, there is the etale topology*, which explains very clearly surprising and intricate patterns in the behaviour of solution sets.
*If someone objects that the etale topology goes beyond the usual definition, I would argue that the logical essence is the same. It is notable that the standard definition admits such a generalization so naturally, whereas some of the others do not. (At least not in any obvious way.)
For those who haven't encountered one before, a Grothendieck topology just replaces subsets of a set $X$ by maps $$Y\rightarrow X.$$ The collection of maps that defines the topology on $X$ is required to satisfy some obvious axioms generalizing the usual ones.
Added 25 March:
I hope people aren't too annoyed if I admit I don't quite see a satisfactory answer yet. But thank you for all your efforts. Even though Sigfpe's answer is undoubtedly interesting, invoking the notion of measurment, even a fuzzy one, just doesn't seem to be the best approach. As Qiaochu has pointed out, a topological space is genuinely supposed to be more general than a metric space. If we leave aside the pedagogical issue for a moment and speak as working mathematicians, a general concept is most naturally justified in terms of its consequences. As pointed out earlier, topologies that have no trace of a metric interpretation have been consequential indeed.
When topologies were naturally generalized by Grothendieck, a good deal of emphasis was put on the notion of an open covering, and not just the open sets themselves. I wonder if this was true for Hausdorff as well. (Thanks for the historical information, Donu!) We can see the reason as we visualize a two-manifold. Any sufficiently fine open covering captures a combinatorial skeleton of the space by way of the intersections. Note that this is not true for a closed covering. In fact, I'm not sure what a sensible condition might be on a closed covering of a reasonable space that would allow us to compute homology with it. (Other than just saying they have to be the simplices of a triangulation. Which also reminds me to point out that homology can be computed for ordinary objects without any notion of topology.)
To summarize, a topology relates to analysis with its emphasis on functions and their continuity, and to metric geometry, with its measurements and distances. However, it also interpolates between these and something like combinatorial geometry, where continuous functions and measurements play very minor roles indeed.
For myself, I'm still confused.
Another afterthought: I see what I was trying to say above is that open sets in topology provide an abstract framework for describing local properties of functions. However, an open cover is also able to encode global properties of spaces. It seems the finite intersection property is important for this, but I'm not able to say for sure. And then, when I try to return to the pedagogical question with all this, I'm totally at a loss. There are very few basic concepts that trouble me as much in the classroom.
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8
Do you find the Kuratowski closure axioms intuitive? If so, then the proof of equivalence between the Kuratowski closure axioms and the standard axioms is not hard. – Qiaochu Yuan Mar 23 2010 at 23:14
3
Maybe "environments" is what some of us call "neighborhoods" (and others of us call "neighbourhoods"). – Gerry Myerson Mar 23 2010 at 23:36
5
filters and nets are equally powerful logically, nets are useful because you can use your intuition of sequences in metric spaces (more or less), filters are useful because the statements about convergence become much shorter and prettier.. – jef Mar 24 2010 at 3:08
4
Filters are equivalent to nets in a topological context, Andrew L. The difference is that filters also have application in logic and set theory as well. – Harry Gindi Mar 24 2010 at 13:16
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Regarding the effectiveness of the standard definition of "topology": I feel comfortable with its effectiveness in areas such as functional analysis and differential geometry. But I have never understood why the standard definitions of topology should be useful at all when working with finite fields and other discrete objects. Is there any way to motivate that for the non-expert? – Deane Yang Mar 24 2010 at 13:29
show 15 more comments
36 Answers
Topology is the art of reasoning about imprecise measurements, in a sense I'll try to make precise.
In a perfect world you could imagine rulers that measure lengths exactly. If you wanted to prove that an object had a length of $l$ you could grab your ruler marked $l$, hold it up next to the object, and demonstrate that they are the same length.
In an imperfect world however you have rulers with tolerance. Associated to any ruler is a set $U$ with the property that if your length $l$ lies in $U$, the ruler can tell you it does. Call such a ruler $R_U$.
Given two rulers $R_U$ and $R_V$ you can easily prove a length lies in $U\cup V$. You just hold both rulers up to the length and the length is in $U\cup V$ if one or the other ruler shows a positive match. You can think of $R_{U\cup V}$ as being a kind of virtual ruler.
Similarly you can easily prove that a point lies in $U\cap V$ using two rulers.
If you have an infinite family of rulers, $R_{U_i}$, then you can also prove that a length lies in $\bigcup_i U_i$. The length must lie in one of the $U_i$ and you simply exhibit the ruler $R_{U_i}$ matching for the appropriate $i$.
But you can't always do the same for $\bigcap_i U_i$. To do so might require an infinitely long proof showing that all of the $R_{U_i}$ match your length.
A topology is a (generalised) set of rulers that fits this description.
Your notion of 'measurement' in whatever problem you have might not match the notion that the above description tries to capture. But to the extent that it does, topology will work as a way to reason about your problem.
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Very nice explanation! I don't think I've ever seen anyone directly try to explain why open sets should be closed under arbitrary union but not arbitrary intersection. – Qiaochu Yuan Mar 23 2010 at 23:12
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Here's something that confuses me about this answer. I'm supposed to be thinking that your 'tolerances' are open conditions, giving open sets. Let's say instead that tolerances are closed conditions (which is to be honest how I always interpreted them: 1m+-5cm would lead me to believe that 95cm is a valid answer for something that was supposed to be 1m). If tolerances were closed then shouldn't your heuristic show that I can take arbitrary intersections but only finite unions? In short: I don't really understand the justification of the infinite unions v intersections bit. – Kevin Buzzard Mar 24 2010 at 13:30
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@gowers: "We could just as well have closed rulers...". This is the point I'm trying to make. If we had closed rulers then does the answer above then give a reasonable heuristic justification of the false statement that an arbitrary union of closed sets is closed? This is precisely what I'm trying to get to the bottom of. Yes I agree that the answer is lovely. But I am not yet convinced that it genuinely gives a conceptual explanation of the axiom that a union of open sets is open, in the sense that it seems to give equal credence to the false statement that a union of closed sets is closed. – Kevin Buzzard Mar 25 2010 at 17:23
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@Kevin: I think the point is that we should think about the rulers as open sets, because the property of belonging to an open set is more 'stable' than the property belonging to a closed set. For example, it is better to have a (95cm, 105cm) ruler rather than a [95cm, 105cm] ruler because if someone is exactly 95cm tall, then the [95cm, 105cm] ruler will answer yes, which is undesirable because she is actually on the boundary. On the other hand, if the (95cm, 105cm) ruler says yes, then we are more sure of the answer because our set is disjoint from its boundary (open). – Tony Huynh Mar 25 2010 at 18:14
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Just because open rules are "more desirable" has, from my point of view, nothing to do with it. Let me try and ask my question again! We have an answer above, and Qiaochu's comment directly below is "this gives a conceptual explanation of why an arbitrary union of open sets is open". I say "oh no it doesn't" and no-one has yet convinced me otherwise. Just because it works, and is "stable"/"desirable" does not convince me. My point is that if you change all the open sets to closed one the analogy seems just as good, but does not work. I'm just repeating myself now though so I'llnotpostanymore – Kevin Buzzard Mar 26 2010 at 7:33
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You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
The textbook presentation of a topology as a collection of open sets is primarily an artefact of the preference for minimalism in the standard foundations of the basic structures of mathematics. This minimalism is a good thing when it comes to analysing or creating such structures, but gets in the way of motivating the foundational definitions of such structures, and can also cause conceptual difficulties when trying to generalise these structures.
An analogy is with Riemannian geometry. The standard, minimalist definition of a Riemannian manifold is a manifold $M$ together with a symmetric positive definite bilinear form $g$ - the metric tensor. There are of course many other important foundational concepts in Riemannian geometry, such as length, angle, volume, distance, isometries, the Levi-Civita connection, and curvature - but it just so happens that they can all be described in terms of the metric tensor $g$, so we omit the other concepts from the standard minimalist definition, viewing them as derived concepts instead. But from a conceptual point of view, it may be better to think of a Riemannian manifold as being an entire package of a half-dozen closely inter-related geometric structures, with the metric tensor merely being a canonical generating element of the package.
Similarly, a topology is really a package of several different structures: the notion of openness, the notion of closedness, the notion of neighbourhoods, the notion of convergence, the notion of continuity, the notion of a homeomorphism, the notion of a homotopy, and so forth. They are all important, and it is somewhat artificial to try to designate one of them as being more "fundamental" than the other. But the notion of openness happens to generate all the other notions, and has a particularly elegant and simple axiomatisation, so we have elected to make it the basis for the standard minimalist definition of a topology. But it is important to realise that this is by no means the only way to define a topology, and adopting a more package-oriented point of view can be preferable in some cases (for instance, when generalising the notion of a topology to more abstract structures, such as topoi, in which open sets no longer are the most convenient foundation to begin with).
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The difference between the minimalist and 'packaged' version is exactly what Bill Farmer and I have called axiomatic theories and high-level theories (respectively). A minimalist 'basis' is useful when trying to connect theories (theory interpretations, aka homomorphisms) because less has to be proven, but that makes a really poor "working environment". Any mathematician using a theory (be it topology or Riemannian geometry) will automatically want to use the high-level version. When trying to mechanize mathematics, such issues are no longer philosophical! – Jacques Carette Jul 2 2010 at 12:38
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I really like this answer. Perhaps part of the so-called "mathematical maturity" is being able to recover the full package from the minimalist's definition. I'm not always able to do so. It took me a long time to realize the metric tensor really gives you a "package". I wish someone can tell me what the package the minimal definition of a scheme (as a ringed space $(X,\mathcal{O}_X)$ that is locally isomorphic to the spectrum of a ring) gives... Although I feel like I can handle it freely right now, I certainly don't know how to explain it to someone who is new to algebraic geometry. – temp May 8 2012 at 2:46
It may seem hard to add a new answer to all this, but here's mine. How to motivate the open set garbage of topological spaces:
Answer: Don't.
There are many ideas in mathematics that can be easily derived from some real situation, and I would count approximation (ie limits), metric spaces, and neighbourhoods as among these. I think that it is quite easy to motivate the neighbourhood definition of topological spaces, for example, by considering real world examples of needing approximations that can't be controlled by metrics (for example, if you always need your approximations to be greater than the true value).
But one can take this line too far and try to motivate everything in mathematics from real-world situations and this, I think, misses a great opportunity to teach something that all students of mathematics need to learn: that when something is presented to you in a particular way, you don't have to accept that viewpoint but can choose a different one more suited to what you want to do.
We try to teach them this with bases of vector spaces: don't use the basis given, use one that makes the matrix look nice (diagonal if possible!).
So here, we can present topological spaces as sets with lots of declared neighbouhoods satisfying certain simple, intuitive rules. But they are hard to work with so instead we work with open sets (sets which are neighbourhoods of all their points) because it makes life easier.
I should qualify the above a little. It's written as a counterpoint to all the previous replies which try to justify open sets based on some intuition. I'm not saying that those are wrong - far from it - just that with something like this, one should think carefully about the message one is sending to the students about mathematics. At some point, they have to learn that mathematics strives to be clear and elegant rather than intuitive and vague, and it's a good idea to do this with an example like topological spaces where we are still close to the intuition, rather than something like function spaces where intuition often takes a hike.
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+1000000 . – Mariano Suárez-Alvarez Mar 25 2010 at 1:58
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Yay! I got a downvote! Fantastic. With something like this, I don't care whether or not people agree with me. I care only that they have thought about it. I hope that those who think that this is a bad idea do so because they have a better one (if so, please tell me! I like hearing of how others explain stuff.). With teaching styles, everyone needs to develop one that "fits" them best. But often it's hard to do because we don't have many examples of what is possible. So I throw out things like this to help people start thinking about different ideas. – Andrew Stacey Mar 27 2010 at 12:54
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I think that this answer is one of the best on MO. – Harry Gindi Mar 27 2010 at 23:31
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I don't want to be sarcastic but surely a high school student can answer you back that $(x+y)^2=x^2+y^2$ makes life a lot easier for him also. This argument is really weak, philosophically. – alephomega Aug 21 2010 at 21:24
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There's easier, and then there's easier while remaining correct. There's a semi-fake Einstein quotation about this: ‘Everything should be made as simple as possible, but no simpler.’. – Toby Bartels Oct 9 2011 at 3:46
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Here's one of my favorite axiomatizations of topology!
To make a set $X$ into a topological space, you introduce a relation, "touches," between the elements of $X$ and the subsets of $X$. This relation must have the following properties:
1. No point touches the empty subset.
2. If $x$ is an element of $A$, then $x$ touches $A$.
3. If $x$ touches $A \cup B$, then $x$ touches $A$ or $x$ touches $B$.
4. If $x$ touches $A$, and every element of $A$ touches $B$, then $x$ touches $B$.
Here, $x$ is an arbitrary element of $X$, and $A$ and $B$ are arbitrary subsets of $X$.
The first three axioms agree very well with my intuitive concept of "touching," and I find the fourth one at least tolerable, if not totally self-evident. If you leave out the fourth axiom, you get the definition of a pretopological space (a set with a Čech closure operator).
In Joshi's Introduction to General Topology, and in most of the literature, this kind of relation is called a nearness relation (page 114). I think one of the first papers on these things was "Nearness--A Better Approach to Continuity and Limits," by P. Cameron, J. G. Hocking, and S. A. Naimpally, which talks about nearness relations on metric spaces.
The definition of continuity in terms of open sets really puzzled me at first. I think the definition in terms of the nearness relation is much clearer!
Let $X$ and $Y$ be topological spaces. A continuous map from $X$ to $Y$ is a map $f$ with the property that if $x$ touches $A$, then $f(x)$ touches $f(A)$.
The definitions of convergence and Hausdorffness are quite pretty, in my opinion, and the definition of connectedness is very intuitive. (WARNING: I'm kinda rusty at this, so these definitions may not be correct.)
The sequence $(x_n)$ converges to the point $x$ if $x$ touches every subsequence of $(x_n)$.
The topological space $X$ is Hausdorff if for any two distinct points $w, x \in X$, there is a subset $A$ of $X$ such that $w$ doesn't touch $A$ and $x$ doesn't touch the complement of $A$.
The topological space $X$ is disconnected if it has two subsets $A$ and $B$, with $A \cup B = X$, such that no point of $A$ touches $B$ and no point of $B$ touches $A$.
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this is just the same as kuratowski's closure operator. It is also confusing because there are also the Nearness spaces defined by Herrlich which use covers and are more general than topological spaces.. – jef Mar 24 2010 at 3:53
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@jef: Yes, it's the same as the closure operator, but I find it way easier to motivate. Maybe I'm just weird? Thanks for mentioning Herrlich's nearness spaces---I'd never heard of them before! If anyone is interested in the definition, here's a nice reference: "On Nearness Space," by Seung On Lee and Eun Ai Choi (ccms.or.kr/data/pdfpaper/jcms8_1/8_1_19.pdf). – Vectornaut Mar 24 2010 at 22:51
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@Vectornaut: Your definition of a continuous map is a nice formalization of informal “a map without breaks” or “a map which preserves infinitesimal distances”. a touches A ↔ a is infinitely close to A. Infinitesimal distance between points does not make sense, so we need to replace one of the points with a set. That was crucial. I went the same way, but directly from the Kuratowski's axioms. a touches A ↔ $a\in cl(A)$. Thank you very much for the references because I did not know even where to start my search or what name it is called. I wonder why point-set topology is not derived this way. – beroal Feb 19 2011 at 14:37
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@Vectornaut: There is a short answer by @Mike Benfield, which goes in parallel with mine. It seems that this point of view is not popular. If you have any further info, especially rigorously developed, I will be glad to hear that. – beroal Feb 19 2011 at 14:51
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To me, the concept of an open set is a distillation and abstraction of the properties of open intervals (on the real line) that are critical to defining and working with a continuous function. In my opinion students should never be introduced to the abstract concepts of topology (notably, the concepts of open sets and compactness) unless they have already mastered analysis of functions on the real line and finite-dimensional vector spaces and understand thoroughly the role of open sets and compactness in those settings.
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I am with you there Deane. – Charlie Frohman Mar 23 2010 at 23:17
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One of the comments on the original post said that you can define a topology in terms of neighbourhoods. I'd like to amplify on that comment because it's the answer I favour too, if you want to do things as intuitively as possible. In fact, you can do it with basic open neighbourhoods, which is often nicer, for reasons I'll come to in a moment.
The first step would be to axiomatize the notion of a basic open neighbourhood. So it would consist of properties like that if N is a b.o.n. of x then x is an element of N, that if y is also an element of N then there is a b.o.n. N' of y such that N' is a subset of N (and in many systems one would be able to take N'=N), that the intersection of two b.o.n.s of x is another one, and so on. Suppose we've got all that sorted out. Then the rest of the definitions are just like metric space definitions without the need to reformulate those definitions in terms of open sets. To give the most important example, a function $f:X\to Y$ is continuous at x if and only if the following condition holds: for every b.o.n. M of f(x) there exists a b.o.n. N of x such that $f(N)\subset M$. Of course, in a metric space the basic open neighbourhoods of x are the open balls $B_\epsilon(x)$.
In the usual definition of continuity for maps between topological spaces, one never talks about continuity at a point, but it is perfectly possible and natural to do so, as the above shows.
Here's another example: a set F is closed if and only if for every x not in F you can find a basic open neighbourhood N of x that is disjoint from F. Oh, and I should have said that a set U is open if and only if for every x in U you can find a basic open neighbourhood N of x such that $N\subset U$.
The one thing you can't do is reformulate these definitions in terms of sequences, for the simple reason that the sequence reformulations do not generalize to topological spaces (unless you replace them by nets).
Added later: I've just seen some more of the comments on the original post. Much of what I have said is implicit in those comments, but perhaps it is useful to have it spelt out.
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Here's an example. Suppose I define the product topology on X={0,1}^N. I won't tell you what an open set is. Instead I'll say that the b.o.n. B_n(x) is the set of all sequences that agree with x up to n. Then I can define the continuity at x of a map from X to X without ever having to say what an open set is. So basic open neighourhoods are playing a role similar to balls of radius epsilon in metric-space theory. To relate this definition to metric spaces I don't have to prove that a map between metric spaces is continuous if and only if the inverse image of an open set is open. – gowers Mar 26 2010 at 9:42
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This is an attempt to synthesize ideas that have appeared in other answers, for example sigfpe's and Tim Perutz's. Feel free to edit if you think the ideas can be better expressed.
The idea I want to back is that a topological space is an environment $X$ in which the notion of checking the truth of a statement locally makes sense. In the actual language of topological spaces, we want to be able to talk about statements which are true for a space $X$ if and only if they're true for every open set in an open cover of $X$, and the same should be true for every subspace of $X$. (For example, continuity and differentiability of a function both have this property.)
But whatever an open cover is, it should consist of elements chosen from a distinguished collection of subsets $\mathcal{P}$ of $X$ having certain properties. The empty set and $X$ should both be in $\mathcal{P}$ because checking a statement about $X$ is trivially equivalent to checking it on $X$ and on the empty set. $\mathcal{P}$ should be closed under arbitrary unions because a collection of open sets automatically forms an open cover of its union. $\mathcal{P}$ should be closed under binary intersections because one should be able to build an open cover of a subspace $S$ of $X$ by intersecting an open cover of $X$ with $S$, and if $S$ is itself open, an open cover of $S$ should be extendable to an open cover of $X$.
I don't think I've explained myself very well, though. I also wish I knew enough to say something about the relationship between topology and logic that the above seems to suggest. But one reason I like this perspective is that it suggests certain definitions naturally, such as the definition of compactness or of a manifold.
Some soapboxing: while I can see the pedagogical value of thinking about topological spaces as a natural generalization of metric spaces or even just of $\mathbb{R}$, I think the idea of a topological space is deeper than these roots suggest and I think Minhyong is looking for an answer that reflects this. In other words, I am of the opinion that the definition of a topological space is more natural than the definition of a metric space (or even of $\mathbb{R}$!), so one shouldn't use the latter to motivate the former. But this is just an opinion.
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Re soapboxing: I also think the idea of a topological space must be deeper than "just" a generalization of metric spaces or of $\mathbb{R}$; this is one of the reasons for this previous question of mine: mathoverflow.net/questions/14634/… – Kevin Lin Mar 24 2010 at 9:00
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I think I am missing here why the empty set should be included. It is certainly intuitive that a property that has been checked on $X$ is true on $X$; but, if I were a student presented with this definition, then I'd wonder, I think not unreasonably, why I should check it on $\emptyset$ as well. – L Spice Mar 24 2010 at 21:33
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I guess another answer is that "arbitrary union" includes the empty union. – Qiaochu Yuan Mar 24 2010 at 21:36
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After reading the comments on Sigfpe's answer I realized that it would be useful to make a rigorous argument to explain why "rulers" or as like to call them "observable properties" should be open sets. In the process I'd like to explain how we can view general topology as an idealized version of computation by interpreting topological spaces as data types and continuous maps as computable functions.
Computationally an observable property $P$ of a data type $A$ corresponds to a semi-decision procedure. In other words a computable function $\chi_P: A \to Unit$ which returns the unique value $()$ of type $Unit$ if $a \in A$ has the property $P$ and runs forever otherwise. We can interpret $P$ as a subset of $A$ and $\chi_P$ as it's characteristic function. Clearly observable properties pull back under computable functions since if $f:B \to A$ is a computable function $\chi_P \circ f$ is a semi-decision procedure.
Let's translate this into topological language. If we interpret $A$ and $B$ as topological spaces and $f:B \to A$ as a continuous map we have that $f^{-1}(P)$ is observable if $P$ is. Thus it makes sense to interpret observable properties as open sets. We can make this correspondence more precise if we notice that every open set $P$ in $A$ corresponds to a map $\chi_P: A \to \mathbb{S}$ where $\mathbb{S}$ is the Sierpinski space. Thus in our translation $Unit$ corresponds to $\mathbb{S}$, the open point of $\mathbb{S}$ corresponds to $()$, and the closed point $\bot$ of $\mathbb{S}$ corresponds to nontermination.
Now a question remains: why did we choose to represent observable properties by open sets instead of closed sets? The answer lies in the way observable properties behave under intersection and union. Let $P$ and $Q$ be observable properties. The intersection $P \cap Q$ is an observable property we can write a semi-decision procedure $\chi_{P \cap Q}$ by running $\chi_P$ and $\chi_Q$ in succession. Similarly notice that $P \cup Q$ is observable since we can write a semi-decision procedure $\chi_{P \cup Q}$ that runs $\chi_P$ and $\chi_Q$ in parallel and outputs $()$ if one of $\chi_P$ and $\chi_Q$ does. If you have an infinite number of computers it is clear that you can generalize this construction to an infinite union $\bigcup_{i \in I} P_i$ by running all the $\chi_{P_i}$ in parallel. However this will not work for an infinite intersection $\bigcap_{n \in \mathbb{N}} P_n$ because if $\chi_{P_n}$ takes $n$ seconds to terminate, then even running all the $\chi_{P_n}$ in parallel will not help.
I can't help but list out a few other things to ponder in light of this dictionary:
• A space $X$ is discrete if $=:X\times X \to \mathbb{S}$ is continuous
• A space $X$ is Hausdorff if $\neq: X\times X \to \mathbb{S}$ is continuous
• A space $X$ is compact if the map $\forall_X: (X \to \mathbb{S}) \to \mathbb{S}$ is continuous
• An observable property $P$ of $T$ is decidable if and only if $P$ is clopen
• On a sequential machine we can write a semi-decision procedure for a countable union of observable properties but not an uncountable union. Does this say anything about topology?
Here are some nice references:
• Alex Simpson - Topological Spaces from a Computational Perspective
• Martin Escardo - Synthetic Topology of Data Types and Classical Spaces
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Thanks, I was under the impression that "rulers" or "imprecise measurements" requires a geometrical intuition, but from this post I see that it does not. I am completely happy with sigfpe's answer now. This was something I was trying to get at as well, except for this is much clearer. – Tran Chieu Minh Mar 26 2010 at 4:06
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Not sure if anyone has mentioned this article, but Gregory Moore discussed the development of the notion of open sets vs other historical approaches, in the paper "The emergence of open sets, closed sets, and limit points in analysis and topology" in Historia Mathematica, no. 35, 2008, pages 220-241. Makes for an interesting read.
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Hello Minhyong. I think other people have already given several excellent answers about how to motivate topology, and I'm not sure I have much to add. But there are a couple of other parts to your question. I had the impression that the notion of a topological space was introduced by Hausdorff in his 1914 book on set theory (Mengenlehre). But my history is somewhat shaky, and it would be nice if someone else could confirm this. It certainly seems that by the early 20th century there was a radical shift in the way mathematics was done.
Jumping ahead to your added comment, I agree that a Grothendieck topology is a very natural extension of this idea. However, for Grothendieck this was secondary; the associated category of sheaves or topos was the important thing. Of course, you know that, but perhaps not everyone does.
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I risk reviving a settled matter.
I aim these sketchy remarks at the expert teacher - not at the neophyte student.
A motivation for open sets in topology might begin with a critique of measurement. Though we often think of measurement in continuous terms, practical measurement really always comes down essentially to answering Boolean questions. Thus the real-valued distance function $d(x,y)$ on a metric space carries the same information as a Boolean-valued function $D(x,y,r)$ where $D(x,y,r)=1$ iff $d(x,y)\leq r$. The Dedekind construction of the reals reflects the sort of divide-and-conquer process of actual measurement, the sort of process that takes us from $D$ to $d$, but not in finite terms.
Now in a non-metric topological space you just allow yourself a richer set of questions than you can index with a variable $r$ running over the reals.
Indeed mathematicians often identity sets with properties - having such and such a property means belonging to the (sub)set of all elements (of a given set) that have that property. Then we can measure the proximity of two things by which properties (that we care about) they share.
At a technical level, the previous paragraph has this reflection: just as a metric allows you to embed a metric space in a product of copies of the positive reals, a topology allows you to embed a general space in a product of copies of the $2$-element Sierpiński space.
Now a student might reasonably challenge the appearance of Sierpiński space in this fundamental role: why the asymmetry? why have just one open point? if binary decisions lie at the heart of the story, why not take as fundamental the $2$-element space discrete?
I say the choice of Sierpiński space mirrors an aspect of practical life. For certain questions, observation may, on the negative side, supply a full refutation, but on the positive side only at best lend support and never full confirmation. For example, we may discover after careful observation that two quantities are not equal, but often we can only amass evidence that they are equal pending more precise measurement. Another example, when we witness a demise we learn that something wasn't eternal, but observation can never confirm that something is eternal.
This concept of decidability motives the axiomatic closure properties of open sets. Intuitively, an open property admits confirmation by a finite amount of evidence. An arbitrary disjunction of open properties (a union) gets confirmed by confirming any one of them and thus also requires only a finite amount of evidence. But a conjunction of open properties requires confirming them all, so we must limit ourselves, at least a priori, to finite conjunctions (intersections).
In summary, the Sierpiński space may seem like a curiosity, if not a monstrosity, but it captures the essence of topology. We have open sets because we care about continuous maps to Sierpiński space, whether consciously or not. We care about continuous maps to Sierpiński space because we care about properties whether they are decidable or not (in the sense of the intuitionists, not in the sense of Turing), i.e., whether or not they disconnect the universe of possibilities. Accepting Sierpiński space commits you to accepting the subspaces of its self-products. The real conceptual hurdle for the topological neophyte lies in contemplating the ubiquity of undecidable properties.
Grothendieck topologies fit very nicely into this point of view (as it leads to topos theory). In essence, Grothendieck challenges the doctrine of identifying properties with subsets. For Grothendieck, a given property may only become visible if one breaks a symmetry or observes some distinction between objects that previously seemed identical. Thus singling out a property might require taking a cover rather than only passing to a subset.
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Great answer. I would only note that there is a large area of computer science known as Domain Theory that tries to make the notion of decidability a la Turing the same as the notion of decidability given by considering maps into the Sierpinski space. Here is a short primer on these ideas: homepages.inf.ed.ac.uk/als/Teaching/MSfS/l3.ps – Justin Hilburn Dec 20 2010 at 6:18
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I believe the correct embedding theorem into powers of the Sierpinski space is the following: a topological space $X$ embeds into a product of copies of the Sierpinski space iff it is Kolmogorov (a.k.a. $T_0$). So the Sierpinski space does not quite have the universal role that you suggest. – Pete L. Clark Dec 24 2010 at 9:30
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Two platitudes:
(1) On a metric space, $\mathbb{R}$-valued functions which are continuous in the ($\epsilon$-$\delta$)-sense are the same as those for which the preimage of an open is open. So one can achieve the aim of discussing continuity by using open sets.
(2) The standard open sets in a metric space satisfy the axioms for a topology.
However, the open sets in a metric space satisfy many other properties too (Hausdorff, etc.).
So - as a former colleague of mine pointed out - to motivate our definition we ought to say why we can't reasonably drop one of the axioms for a topology - the intersection axiom, say. After all, our examples will still satisfy the axioms, and we'll still be able to prove some standard lemmas about spaces and continuous functions.
The answer, I think, is that continuity really ought to be local: a function is continuous if it's continuous when restricted to each of the sets making up an open cover. In proving this, we use both the union and intersection axioms.
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The open sets determine and are determined by the closed sets, so one just has to make a choice about which to axiomatize. Axioms for one give complementary axioms for the other. My point was that the conventional axioms for open sets (or their complementary axioms for closed sets) appear to be minimal ones under which continuity is locally determined and constant functions are continuous. – Tim Perutz Jun 15 2010 at 22:36
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At least for me, the first time I learned about a metric space, was to discuss when sequences converge. I am not a math-history buff, but the concept of metric seems to stem from the need to formalize the concept of convergence. However, everything about of convergence depends only on the topology the metric generates. Hence one only needs to understand the open sets.
Topological spaces generalize metric spaces in the sense that every metric space gives rise to one and all concepts of convergence are captured by this topological space. Even more, the category of metric spaces and continuous maps sits inside the category of topological spaces as a full subcategory (this is just saying that a map between metric spaces is continuous if and only if the preimage of an open is open). However, you may object, and justifiably, that you can embed metric spaces into several categories, so, why is topological spaces the right generalization?
There are many "practical answers", for instance, the wide range of examples of abstract spaces which are not metric spaces, e.g., Spec(R) of a ring. However, the core of the matter is that topological spaces correctly axiomatize the notion of convergence. What I mean by this is, a topological space is completely determined by the convergence of the ultrafilters on its underlying set. This is seem most notably for compact Hausdorff spaces; a topological space is compact Hausdorff if and only iff every ultrafilter has a unique limit, and in fact, compact Hausdorff spaces are precisely the algebras of the ultrafilter monad. However, we are interested in non-compact examples, since we study unbounded metric spaces for example- but even here we can have ultrafilters with no limit- so, a complete generalization should take this into account. Furthermore, the space Spec(R) is very often non-Hasudorff, which means, ultrafilters which have a limit point, may have more than one. So, to understand convergence is to understand the set of limits of each ultrafilter. If X is a space and BX is its set of ultrafilters, we get a map BX->P(X) which sends each ultrafilter to its set of limit points (possibly empty). This corresponds to a relation $R \subset X \times BX$, which made be seen as a map BX->X in the bicategory of sets and relations. More precisely, we get a "relational algebra" for the ultrafilter monad. The converse is true as well: the category of topological spaces is equivalent to the category of relational algebras for the ultrafilter monad. This a theorem of Barr. The upshot is, there is a bijection between topologies on a set and "convergence systems for ultrafilters" on that set.
Anyhow, this probably goes way beyond what you can explain to most undergraduates.
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Topology can be defined directly, without open sets, as the study of "metric spaces without the metric", i.e., modulo metric deformations or homeomorphism. This matches reasonably well the intuition of a qualitative geometry, insensitive to stretching and bending.
One can then prove that the structure of open sets is a complete invariant (since we start from metrizable spaces), and one can observe, with some experience, that reasoning about the topology of metric spaces (i.e., proofs of properties that are invariant under deformation or homeomorphism) can be formulated directly in terms of this invariant. In other words, not only the topologically invariant maps but the constructions of those maps descend to the category of topological spaces in terms of open sets. This means that we can work natively in manifestly topologically invariant terms provided that the invariant thing --- the structure of open sets --- is taken as the object of study. This is a rare case of a total or near-total success of the Erlangen program, where thinking in terms of that which is invariant really suffices for the original purposes of the subject.
(I say near-total, but don't know of any example where topologically invariant properties of a metric spaces are most easily proved using one or more metrics.)
Once topology is set up in terms of open sets one can look at examples beyond the motivating intuition, such as Zariski topology, the long line or pathological spaces. As far as those extensions start to challenge the adequacy of the open-set formalism it is because they are based on phenomena different from the stretching and bending ideas abstracted from picturesque low-dimensional situations.
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But of course there are other structures on a metric space that are complete topological invariants, for example the convergence structure (specifying which sequences converge to which points). Abstracting this gives the category of [convergence spaces](ncatlab.org/nlab/show/convergence+space) (well, sequential convergence spaces without some motivation to generalise from sequences to nets), of which $Top$ is a full subcategory. So while this explanation is historically accurate, it doesn't explain why topological spaces precisely are so important (if they really are). – Toby Bartels Jan 14 2012 at 5:11
In this answer I will combine ideas of sigfpe's answer, sigfpe's blog, the book by Vickers, Kevin's questions and Neel's answers adding nothing really new until the last four paragraphs, in which I'll attempt to settle things about the open vs. closed ruler affair.
DISCLAIMER: I see that some of us are answering a question that is complementary of the original, since we are trying to motivate the structure of a topology, instead of adressing the question of which of the many equivalent ways to define a topology should be used, which is what the question literally asks for. In the topology course that I attended, it was given to us in the first class as an exercise to prove that a topology can be defined by its open sets, its neighbourhoods, its closure operator or its interior operator. We later saw that it can also be stated in terms of convergence of nets. Having made clear these equivalence of languages, its okay that anyone chooses for each exposition the language that seems more convinient without further discussion. However, I will mantain my non-answer since many readers have found the non-question interesting.
Imagine there's a set X of things that have certain properties. For each subset of S there is the property of belonging to S, and in fact each property is the property of belonging to an adequate S. Also, there are ways to prove that things have properties.
Let T be the family of properties with the following trait: whenever a thing has the property, you can prove it. Let's call this properties affirmative (following Vickers).
For example, if you are a merchant, your products may have many properties but you only want to advertise exactly those properties that you can show. Or if you are a physicist, you may want to talk about properties that you can make evident by experiment. Or if you predicate mathematical properties about abstract objects, you may want to talk about things that you can prove.
It is clear that if an arbitrary family of properties is affirmative, the property of having at least one of the properties (think about the disjunction of the properties, or the union of the sets that satisfy them) is affirmative: if a thing has at least one of the properties, you can prove that it has at least one of the properties by proving that property that it has.
It is also clear that if there is a finite family of affirmative properties, the property of having all of them is affirmative. If a thing has all the properties, you produce proofs for each, one after the other (assuming that a finite concatenation of proofs is a proof).
For example, if we sell batteries, the property P(x)="x is rechargeable" can be proved by putting x in a charger until it is recharged, but the property Q(x)="x is ever-lasting" can't be proved. It's easy to see that the negation of an affirmative property is not necessarily an affirmative property.
Let's say that the open sets are the sets whose characteristic property is affirmative. We see that the family T of open sets satisfies the axioms of a topology on X. Let's confuse each property with the set of things that satisfy it (and open with affirmative, union with disjunction, etc.).
Interior, neighbourhood and closure: If a property P is not affirmative, we can derive an affirmative property in a canonical way: let Q(x)="x certainly satisfies P". That is, a thing will have the property Q if it can be proved that it has the property P. It is clear that Q is affirmative and implies P. Also, Q is the union of the open sets contained in P. Then, it is the interior of P, which is the set of points for which Q is a neighbourhood. A neighbourhood of a point x is a set such that it can be proved that x belongs to it. The closure of P is the set of things that can't be proved not to satisfy P.
Axioms of separation: If T is not T0, there are x, y that can't be distinguished by proofs and if it is not T1, there are x, y such that x can't be distinguished from y (we can think that they are apparently identical batteries, but x is built in such a way that it will never overheat. So if it overheats, then it's y, but if it doesn't, you can't tell).
Base of a topology: Consider a family of experiments performable over a set X of objects. For each experiment E we know a set S of objects of X over which it yields a positive result (nothing is assumed about the outcome over objects that do not belong to S). If you consider the properties that can be proved by a finite sequence of experiments, the sets S are affirmative and the topology generated by them is the family of all the affirmative properties.
Compactness: I don't know how to interpret it, but I think that some people know, and it would be nice if they posted it. (Searchable spaces?)
Measurements: A measurement in a set X is an experiment that can be performed on each element of X returning a result from a finite set of possible ones. It may be a function or not (it is not a function if there is at least one element for which the result is variable). The experiment is rendered useful if we know for each possible result r a set T_r of elements for which the experiment certainly renders r and/or a set F_r for which it certainly doesn't, so let's add this information to the definition of measurement. An example is the measurement of a length with a ruler. If the length corresponds exactly with a mark on the ruler, the experimenter will see it and inform it. If the length fits almost exactly, the experimenter may think that it fits a mark or may see that it doesn't. If the length clearly doesn't fit any mark (because he can see that it lies between two marks, or because the length is out of range), he will inform it. It is sufficient to study measurements that have only a positive outcome and a negative outcome, a set T for which the outcome is certainly positive and a set F for which the outcome is certainly negative.
Imprecise measurements on a metric space: If X is a metric space, we say that a measurement in X is imprecise if there isn't a sequence x_n contained in F that converges to a point x contained in T. Suppose that there is a set of imprecise measurements available to be performed on the metric space. Suppose that, at least, for each x in X we have experiments that reveal its identity with arbitrary precision, that is, for each e>0 there is an experiment that, when applied to a point y, yields positive if y=x and doesn't yield positive if d(y,x)>e. Combining these experiments we are allowed to prove things. What are the affirmative sets generated by this method of proof? Let S be a subset of X. If x is in the (metric) interior of S, then there is a ball of some radius e>0 centered at x and contained in S. It is easy to find an experiment that proves that x belongs to S. If x is in S but not in the interior (i.e, it is in the boundary), we don't have a procedure to prove that x is in S, since it would involve precise measurement. Therefore, the affirmative sets are those that coincide with its metric interior. So, the imprecise measurements of arbitrary precision induce the metric topology.
Experimental sciences: In an experimental science, you make a model that consists of a set of things that could conceivably happen, and then make a theory that states that the things that actually happen are the ones that have certain properties. Not all statements of this kind are completely meaningful, but only the refutative ones, that is, those that can be proved wrong if they are wrong. A statements is refutative iff its negation is affirmative. By applying the closure operator to a non refutative statement we obtain a statement that retains the same meaning of the original, and doesn't make any unmeaningful claim.
An example from classical physics: Assume that the space-time W is the product of Euclidean space and an affine real line (time). It can be given the structure of a four-dimensional real normed space. Newton's first law of motion states that all the events of the trajectory of a free particle are collinear in space-time. To prove it false, we must find a free particle that incides in three non-collinear events. This is an open condition predicated over the space W^3 of 3-uples of events, since a small perturbation of a counterexample is also a counterexample. Assuming that imprecise measurements of arbitrary precision can be made, it is an affirmative property. I think that classical physicists, by assuming that these kind of measurements can be done, give exact laws like Newton's an affirmative set of situations in which the law is proved false. I also suspect (but this has more philosophical/physical than mathematical sense) that the mathematical properties of space-time (i.e. that it is a normed space over an Archimedean field) are deduced from the kind of experiments that can be done on it, so there could be a vicious circle in this explanation.
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This may be a naive answer, but for me the concept captured by a topological space is when a point is infinitely close to a set. This happens when the point is in the closure of the set (or, equivalently, when every neighborhood of the point intersects the set). The definition by open sets may obscure this.
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For the definition of "infinitely close" (or the way I would define "touches"), I prefer continuous functions: wherever the set goes, that point goes with it. (So a closed set is one that can go to a point without dragging more points with it--but that requires a tautological definition of continuity again.) – Elizabeth S. Q. Goodman Mar 25 2010 at 0:20
I would say that topology is defined in terms of open sets and closed sets. I think it is motivated by two theorems from the Calculus. The Bolzano-Weierstrass theorem and the intermediate value theorem.
In its simplest form, the Bolzano-Weierstrass theorem says that an infinite subset of a closed bounded interval $[a,b]$ of the real numbers has a limit point. You find that limit point as follows. As there are infinitely many points in the set, then there are infinitely many in the left half or the right half of $[a,b]$. Say the left half. Divide that interval in half and there are infinitely many in one half or the other. Proceed this way to produce a sequence of closed intervals $I_{n+1}\subset I_n$ with the length of the $n$th interval equal to $\frac{b-a}{2^n}$. By Cantor's theorem $\cap_n I_n$ is nonempty, and the point in there is your limit point.
It didn't really make a difference if you broke the interval into two pieces or 10 pieces. This leads to the notion of compactness, by saying that every open cover has a finite subcover. The fact that the intersection of the closed intervals $I_n$ is nonempty is the complementary notion that a collection of closed subsets with the finite intersection property has nonempty intersection. The proof of the Bolzano-Weierstrass theorem leads you to think of open and closed sets.
A similar analysis of the proof of the intermediate value theorem leads likewise to open sets and closed sets.
Really, the concept of a topology was an incredible creative leap, that allowed people to take ideas from the Calculus and apply them in other places. Similar leaps to me, are the notion of sigma algebra, distribution (in the PDE sense), and the construction of homological algebra. :)
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Without specifying a precise answer to the question, I am surprised that there has been so little emphasis on continuity as the motivating concept for topology - topological spaces seem to me to have been designed, so to speak, to capture the notion of continuity in as much generality as seemed possible at the time, and particularly in non-metric contexts - and incidentally clarifying some proofs by throwing away the metric structure.
What can we recover from the epsilon-delta formulation of continuity if we don't allow measurement? It is possible that this question is more readily answered by reference to closed sets than to open ones.
Clearly the concept then takes off in all sorts of directions, where intuitions motivated by metrics are confounded (as mine was initially with the Zariski Topology).
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There are several interpretations of the original question, but one is, why focus on open sets rather than closed sets? I have an unusual answer.
Suppose you want to do constructive mathematics. (Don't ask me why, you just do.) So you abstract the properties of open and closed subsets from the real line. Then you see that open subsets are closed under arbitrary union but only finitary intersection, OK. Dually, you see that closed sets are closed under arbitrary intersection but … not under finitary union! For example, the union of $[ 0 , 1 ]$ and $[ 1 , 2 ]$ cannot be proved to be closed. (The closure of the union is $[ 0 , 2 ]$, but to prove that the union itself is all of $[ 0 , 2 ]$ requires the lesser limited principle of omniscience. Or less formally, there is no definite method to decide whether a number is near $1$ is in $[ 0 , 1 ]$ or in $[ 1 , 2 ]$.) So open sets are better behaved and naturally you prefer to axiomatise them.
But as you continue with constructive topology, more advanced things fail, such as the Tychonoff Theorem (which implies the axiom of choice and thus excluded middle). Then you learn that this stuff works in locale theory, so you abandon traditional topological spaces for locales. And here the duality between open and closed is restored; a locale's frame of opens can just as well be interpreted as a coframe of closeds, and only tradition tells us to do the first.
So this answer only works in a very unusual frame of mind: setting off down an unusual path but not going all the way.
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It is indeed appropriate to ask "why open sets in topology ?" However, the answer is not so simple precisely since the concept of topology proves to be far more deep and complex than Hausdorff, Kuratowski or Bourbaki have ever imagined, and which may call the HKB topology. For starters, it turned out decades ago that the usual, open set based, that is, HKB topology leads to a category which is not Cartesian closed. And this creates serious difficulties when dealing with topologies on function spaces, and in particular, in duality theory for locally convex spaces. More simply, and without categories, there are most important topological type processes in mathematics which simply cannot be described by HKB topologies. Such are, for instance, in measure theory and partially ordered spaces. As a consequence, various more general concepts of pseudo-topologies have been suggested. What happened, however, is that the doors thus opened up proved to be too large for those who tried to pursue them, or would have liked to use them ... In other words, topologies beyond HKB are a far less cozy venture than usually customary in mathematics ... Some details about the above and relevant references may be found in
arXiv:1001.1866 [pdf, ps, other] Title: Beyond Topologies, Part I Authors: Elemer E Rosinger, Jan Harm van der Walt Subjects: General Mathematics (math.GM)
arXiv:1005.1243 [pdf, ps, other] Title: Rigid and Non-Rigid Mathematical Theories: the Ring $\mathbb{Z}$ Is Nearly Rigid Authors: Elemer E. Rosinger Subjects: General Mathematics (math.GM)
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First, note that a mapping between metric spaces is continuous if and only if the inverse image of an open set is always open. There are various concepts for metric spaces that you can likewise find equivalent formulations for in terms of open (and closed) sets, for example compactness. Convergence of a sequence to a point can be rephrased in terms of neighbourhoods of the point, with no reference to any ε. Then you could, for example, notice how you can talk about pointwise convergence of functions, but there is no corresponding metric. So you need a more general framework for talking about different kinds of convergence, and soon enough, topological spaces won't seem so strange anymore.
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Disclaimer: There are many topologists here and they may not like the philosophical flavour of my answer :-)
I think it all starts with the end... actually with the notion of "end". Take an open interval (a,b). It is bounded, yet you cannot reach its ends! At first this may look weird, but then one realizes this weirdness is to be attributed to the mathematically exact observation of this object (mathematicians can distinguish between so many things like point, set of points, boundedness, boundary etc.). Encountering such "weirdness" only shows the strong need for an exact and abstract formulation of "having no end". The lack of ends we then call "openness". If we are looking for the best generalization of this concept, the first approach would be of course a set-theoretical one. And in fact, in the case of intervals it turns out that the property "having no ends" is inhereted into arbitrary unions and finite intersections. Any attemps to expand this lead either to contradictions to our basic example or to unjustified reduction in generality.
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I have intentionally left out the postulate that the empty set and the whole set are also defined to be open. It seems to me rather a technical axiom and maybe there is a way to do most of the point-set-topology without it... For arbitrary unions, it seems intuitive to me that expanding an "endless set" with other "endless things" should be "endless" too (the key point is expanding). As for intersections, (my) intuition reaches only so far as to say that if A and B are "endless", then $A\cap B$ is "endless", in other words "endless" is a "shareable" property. – ex falso quodlibet Mar 24 2010 at 22:16
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I don't think that Grothendieck topologies should be viewed as analogous to ordinary topologies. It is true that a topology on a set induces various Grothendieck topologies on various categories, but so does every system of basic open neighborhoods. In my opinion, it is more fruitful to think of a Grothendieck topology as the analogue of a system of basic open neighborhoods and a topos as the analogue of a topological space.
Let me try to answer the following question, which may or may not be the question that is actually being asked: Why do we prefer topological spaces to systems of basic open neighborhoods, and topoi to Grothendieck topologies? I think that the answer has to do with morphisms.
To give a morphism of spaces with basic open neighborhoods, one must give a function that respects those neighborhoods. One can't, however, require that the pre-image of a basic open neighborhood be open. Instead, one has to require that each point contained in the pre-image of a basic open neighborhood have a basic open neighborhood inside the pre-image.
Not only is this definition more complicated than the one for topological spaces (and the extension to Grothendieck topologies is by no means obvious!), but there are multiple distinct but isomorphic systems of open neighborhoods on the same space. A topology is a maximal system of open neighborhoods in a given isomorphism class, which makes it a "best model" for a particular notion of nearness.
Another interpretation that makes this model appealing is that given a system of basic open neighborhoods, the open sets are the "local properties" (those that hold at a point if and only if they hold at all sufficiently nearby points). (If one believes a slogan like "there are two -1-categories: TRUE and FALSE" then open sets are "-1-sheaves"; this completes the formal analogy with Grothendieck topologies and their associated topoi, which are their associated categories of "0-sheaves".)
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Or, for that matter, why not study the higher topoi of stacks and higher stacks? If you're only interested in spaces, those theories add extra overhead without giving you anything new. If you're interested in something like the etale topology then you don't have the option of restricting your attention to topological spaces. – Jonathan Wise Jul 2 2010 at 0:39
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Think of the half-open interval $(0,1]$ with the usual open sets (e.g. $(1-\varepsilon,1]$ is an open neighborhood of 1. Then modify the collection of sets considered "open" so that every open neighborhood of 1 contains some set of the form $(1-\varepsilon,1] \cup (0,\varepsilon)$. See if students understand that this modification in which sets are considered open also modifies the way in which the space is connected together.
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Here's how I like to think about it.
We can all agree that a topological space should be a set $X$ together with some extra structure encoding how the points of $X$ fit together. It seems pretty reasonable ask that this structure is sophisticated enough to answer the following question whenever $x \in S \subset X$:
Given any choice of "direction" is there freedom to nudge $x$ some small "amount" in that direction without bumping into any points of $X \setminus S$?
We say that $x$ is an interior point of $S$ if the answer to the above question is "yes". I would say the following assertions about interior points are completely reasonable.
1. Any $x \in X$ is an interior point of $X$.
2. If $S \subset T \subset X$ and $x$ is an interior point of $S$, then $x$ is an interior point of $T$.
3. If $S,T \subset X$ and $x$ is an interior point of both $S$ and $T$, then $x$ is an interior point of $S \cap T$.
For instance, (1) holds because there are no points in $X \setminus X$ to concern ourselves about bumping into. (3) holds because, if I specify a direction, then I can move $x$ an amount $a_s$ (in this direction) without hitting points from $X \setminus S$ and an amount $a_t$ without hitting points from $X \setminus T$, so if I move $x$ the smaller of these two amounts, I won't hit anything in $X \setminus (S \cap T)$.
If we take the above as axioms for a machine that tells us which points are interior to which sets, and then define an open set to be a set each of whose points is an interior point, then it is simple to recover the standard axioms for open sets:
1. $\varnothing$ and $X$ are open.
2. The union of arbitrarily many open sets is open.
3. The intersection of two open sets is open.
The only issue I can see with this approach is that one might be able to convince oneself that interior points should satisfy more axioms. For instance, if $X = {0,1}$ and $1$ can be moved a little bit in any direction without bumping into $0$, then shouldn't it be possible to move $0$ a little bit in any direction without bumping into $1$? This would seem to preclude the existence of the Sierpiński topology ${\varnothing, {1} ,X}$. Or perhaps this is merely an invitation to be more imaginitive about the geometry of the situation? For instance, maybe there is a little round bowl with $1$ at the bottom and $0$ is sitting on the rim. If I give a $0$ a little push in the direction of $1$, no matter how small, $0$ will roll into the bowl and bump into $1$.
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Question for sigfpe (sorry, I don't have enough reputation to post comments, feel free to delete this). It can also take an infinite time to show that something belongs to a union of, i.e. one of, uncountably many sets, no?
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No: you just need to show the one member of the union to which it belongs. – JBL Mar 24 2010 at 0:40
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Don't think of it in terms of how long it takes to find a demonstration, but how long the demonstration would be once you've found it. If $l$ lies in $\bigcup_i U_i$ it might be hard to find a single $U_i$ it lies in. But once you have, the demonstration that $l\in\bigcup_i U_i$ reduce to showing that $l\in U_i$. Actually, with a little work you can tweak this into a question of how long it takes to compute, but that's another story. – Dan Piponi Mar 24 2010 at 0:43
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The complement of an open set isn't necessarily open and only open sets correspond to measuring devices. Thus Merlin can't tell you whether an element is in a uncountable intersection of open sets by looking at the uncountable union of their complements. This is supposed to mirror the behavior of semi-decidable propositions in computer science and logic. A good reference is the paper Synthetic Topology of Data Types and Synthetic Spaces by Martin Escardo. – Justin Hilburn Mar 24 2010 at 7:19
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My answer will not be of philosophical nature, neither historical, but perhaps pedagogical.
I find Munkress' Topology a great book. Among other merits, because its introduction, which I summarize as follows:
1. You recall what a metric space is. Define open balls and subsequently, open sets. Prove that, in a metric space:
1.1 The empty set and the total space are open sets.
1.2 The union of an arbitrary number of open sets is an open set.
1.3 The intersection of a finite number of open sets is an open set.
2. Recall what a continuous map between metric spaces is (the $\epsilon$-$\delta$ definition). Prove the theorem that says that a map between metric spaces $f: X \longrightarrow Y$ is continuous if and only if $f^{-1} (U) \subset X$ is an open set for every open set $U \subset Y$.
And you have a motivation for the definition of topological space and continuous map as well.
Of course this is not an historical explanation of how topological spaces arised, nor does it justify why you chose these properties of open sets in metric spaces and not others: "experience" has told us that these are the good ones. (For instance, if I'm not wrong, when Hausdorff first defined topological spaces included the property of being... Hausdorff among the axioms. "Experience" -and not an a priori argument- showed us that it could be interesting to work with non-Hausdorff topological spaces.)
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I have really quite enjoyed reading this thread, although I must confess I don't quite understand all of it. It has been many years since I have studied topology of any sort, and not only am I rusty, my mind is aging, and not as agile as it was in my youth. Anyway, for what it's worth, here is my take on basic definitions of topology. My first introduction to formal topology (rather than the specialized versions of it in real and complex analysis) was in simple homotopy theory, and the textbook I used gave the open sets axiomatically as a starting point. Although I saw (with some difficulty) that this was a generalization of the properties of the real vector spaces I was used to, and although it was in an algebraic style, I struggled with it's non-intuitive presentation. I believe (although I am hardly an expert) that "nearness" is a better approach to topology. It makes sense, even to a non-mathematician. For example, the statement of continuity in such a setting is simplicity itself: f:D->R is continuous at a point x in D, when: x is near a set A implies f(x) is near f(A). Furthermore, the axiomatic presentation of neighborhoods is simpler than the axioms for open sets in one important respect, you only need to consider the meet (for sets, intersection) of two neighborhoods, and the (possibly partial) order of the collection of the neighborhoods (for sets, the natural ordering on the power set, containment).
In a metric space, which automatically comes with a rich topological structure, you can easily resolve these notions to the traditional definitions. You lose the purely geometric flavor of topology in the process, though. Topology is concerned with matters in the small, and in the large, the nature of what are defined to be neighborhoods to a large part determines how intricate the spatial structure is.
As far as the aesthetic of why unions can be infinite, but intersections have to be finite, I have 2 thoughts: first, open and closed sets are somewhat dual notions, it is entirely possible to begin with the notion of a closed set, in which case you can allow only finite unions, but infinite intersections. In fact, this approach makes more sense for practical applications, since our physical tools for dealing with calculations (and our brains) are in fact finite. The second thought I have, is that when you take the union of two sets, you always get something "bigger", but when you take the intersection, you may get a null result. It seems natural to restrict the basic study to finite intersections, because infinite intersections can behave qualitatively different than infinte unions (similar to how, in whole numbers, substraction isn't always possible, but addition is).
Getting extra stuff is quite common in mathematics, you extend a field, or create a semi-direct product, or consider generated objects. But all this "extra stuff" is rather meaningless without some core thing that has some intrinsic behavior. In topology, I believe nearness should be that core thing. Much of topology's development was motivated by the idea of getting a handle on what a limit (and convergence) ought to mean, and these are notions which have their roots in approximation.
So, naively, one could argue, in real (one-dimensional) analysis we use open intervals as the basic building block, since we are often concerned about local behavior on very small intervals, and using that to extend to larger sets. extending this to a collection of open sets, or to a collection of neighborhoods can be shown to be equivalent constructions; however, using open sets focuses more on the "boundarylessness" of these sets (and thus emphasizes the density of the real numbers), whereas using neighborhoods focuses more on the "localness" of these intervals, lending itself more easily to abstraction while keeping some intuitive spatial idea.
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It might be a futile attempt to add anything worthwhile to this long list of interesting answers, but let me add my own pedestrian \$0.02:
One may think of topology as a set of rules about what's close to what. In other words, it tells me that if I pick a point in the space, then there are several rules (i.e., open sets) that tell me that with respect to some question this set of points is "close" to my chosen point. Considering many rules (i.e., intersecting the open sets) gives me better and better approximation of which points are "really close" to the chosen one. It seems clear that then the union and intersection of these rules would have to belong to the rules.
If we were in a Euclidean space, then we might agree that one way to measure what's close is to put a small (open) ball around a point. If we can't measure, we can't do this, so we need to do something more general and a single open set will not be enough (it's not enough even in a Euclidean space as the radius of the ball that defines "closeness" would certainly depend on the way we want to measure closeness).
So far both open and closed sets would qualify for these rules, but I feel that open sets work better: A rule of "closeness" should be independent of any single point. In other words, a rule should behave the same with respect to any point it applies to (i.e., any point contained in the corresponding set). This clearly picks open sets over closed sets.
I suppose one might say that none of this explains what happens with infinitely many rules/sets. I suppose we could say that if we take an infinite set of rules that define closeness, then on one hand we might still say that satisfying any one of the rules is still a reasonable rule while satisfying all the rules is a little bit too much to ask. If you feel this part of my argument is a little shaky, then we agree. I don't have a very good explanation for the behavior of infinite unions and intersections. If I was indeed trying to explain this to undergrads, then at this point I would probably switch over to see what happens in a Euclidean space with all this non-sense about rules of "closeness" and come to the conclusion that a good way to define rules is to say that their corresponding sets contain little balls around every point in them. Then deduce the axioms of open sets in a topology and then say that we should see what these give us if we forget that we were in a Euclidean space.
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I want to complete something said by Andrew Stacey above: like him I think that the only reason to motivate the use of the open sets it's because they are more easy to use. Topology is the study of property preserved by invertible continuous transformation (following the Erlangen program): this definition clearly need the notion of continuity, I've always thought of continuity as the relation of proximity of points, so the first thing to do topology is to define the notion of proximity and neighbourhood are most natural way to do so (at least for me). Anyway deal with neighbourhood is more complex than working with open set, for example the definition of topology with neighbourhood require five axioms while classical definition with open sets require just three axioms, so while it seems more natural the study of topology via neighbourhood it is more convenient dealing with open sets which allow to simplify proofing work.
I hope this answer my help.
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# Working with Data Files¶
So far, the data we have used in this book have all been either coded right into the program, or have been entered by the user. In real life data reside in files. For example the images we worked with in the image processing unit ultimately live in files on your hard drive. Web pages, and word processing documents, and music are other examples of data that live in files. In this short chapter we will introduce the Python concepts necessary to use data from files in our programs.
For our purposes, we will assume that our data files are text files–that is, files filled with characters. The Python programs that you write are stored as text files. We can create these files in any of a number of ways. For example, we could use a text editor to type in and save the data. We could also download the data from a website and then save it in a file. Regardless of how the file is created, Python will allow us to manipulate the contents.
In Python, we must open files before we can use them and close them when we are done with them. As you might expect, once a file is opened it becomes a Python object just like all other data. Table 1 shows the methods that can be used to open and close files.
Method Name Use Explanation
open open(filename,'r') Open a file called filename and use it for reading. This will return a reference to a file object.
open open(filename,'w') Open a file called filename and use it for writing. This will also return a reference to a file object.
close filevariable.close() File use is complete.
## Finding a File on your Disk¶
Opening a file requires that you, as a programmer, and Python agree about the location of the file on your disk. The way that files are located on disk is by their path You can think of the filename as the short name for a file, and the path as the full name. For example on a Mac if you save the file hello.txt in your home directory the path to that file is /Users/yourname/hello.txt On a Windows machine the path looks a bit different but the same principles are in use. For example on windows the path might be C:\Users\yourname\My Documents\hello.txt
You can access files in folders, also called directories, under your home directory by adding a slash and the name of the folder. For example we have been storing files to use with PyCharm in the CS150 folder inside the PyCharmProjects folder under your home directory. The full name for hello.py stored in the CS150 folder would be /Users/yourname/PyCharmProjects/CS150/hello.py
Here’s the important rule to remember: If your file and your Python program are in the same directory you can simply use the filename. open('myfile.txt','r') If your file and your Python program are in different directories then you should use the path to the file open(/Users/joebob01/myfile.txt).
## Reading a File¶
As an example, suppose we have a text file called qbdata.txt that contains the following data representing statistics about NFL quarterbacks. Although it would be possible to consider entering this data by hand each time it is used, you can imagine that it would be time-consuming and error-prone to do this. In addition, it is likely that there could be data from more quarterbacks and other years. The format of the data file is as follows
`First Name, Last Name, Position, Team, Completions, Attempts, Yards, TDs Ints, Comp%, Rating`
```Colt McCoy QB, CLE 135 222 1576 6 9 60.8% 74.5
Josh Freeman QB, TB 291 474 3451 25 6 61.4% 95.9
Michael Vick QB, PHI 233 372 3018 21 6 62.6% 100.2
Matt Schaub QB, HOU 365 574 4370 24 12 63.6% 92.0
Philip Rivers QB, SD 357 541 4710 30 13 66.0% 101.8
Matt Hasselbeck QB, SEA 266 444 3001 12 17 59.9% 73.2
Jimmy Clausen QB, CAR 157 299 1558 3 9 52.5% 58.4
Joe Flacco QB, BAL 306 489 3622 25 10 62.6% 93.6
Kyle Orton QB, DEN 293 498 3653 20 9 58.8% 87.5
Jason Campbell QB, OAK 194 329 2387 13 8 59.0% 84.5
Peyton Manning QB, IND 450 679 4700 33 17 66.3% 91.9
Drew Brees QB, NO 448 658 4620 33 22 68.1% 90.9
Matt Ryan QB, ATL 357 571 3705 28 9 62.5% 91.0
Matt Cassel QB, KC 262 450 3116 27 7 58.2% 93.0
Mark Sanchez QB, NYJ 278 507 3291 17 13 54.8% 75.3
Brett Favre QB, MIN 217 358 2509 11 19 60.6% 69.9
David Garrard QB, JAC 236 366 2734 23 15 64.5% 90.8
Eli Manning QB, NYG 339 539 4002 31 25 62.9% 85.3
Carson Palmer QB, CIN 362 586 3970 26 20 61.8% 82.4
Alex Smith QB, SF 204 342 2370 14 10 59.6% 82.1
Chad Henne QB, MIA 301 490 3301 15 19 61.4% 75.4
Tony Romo QB, DAL 148 213 1605 11 7 69.5% 94.9
Jay Cutler QB, CHI 261 432 3274 23 16 60.4% 86.3
Jon Kitna QB, DAL 209 318 2365 16 12 65.7% 88.9
Tom Brady QB, NE 324 492 3900 36 4 65.9% 111.0
Ben Roethlisberger QB, PIT 240 389 3200 17 5 61.7% 97.0
Kerry Collins QB, TEN 160 278 1823 14 8 57.6% 82.2
Derek Anderson QB, ARI 169 327 2065 7 10 51.7% 65.9
Ryan Fitzpatrick QB, BUF 255 441 3000 23 15 57.8% 81.8
Donovan McNabb QB, WAS 275 472 3377 14 15 58.3% 77.1
Kevin Kolb QB, PHI 115 189 1197 7 7 60.8% 76.1
Aaron Rodgers QB, GB 312 475 3922 28 11 65.7% 101.2
Sam Bradford QB, STL 354 590 3512 18 15 60.0% 76.5
Shaun Hill QB, DET 257 416 2686 16 12 61.8% 81.3
```
To open this file, we would call the open function. The variable, fileref, now holds a reference to the file object returned by open. When we are finished with the file, we can close it by using the close method. After the file is closed any further attempts to use fileref will result in an error.
```>>>fileref = open("qbdata.txt","r")
>>>
>>>fileref.close()
>>>
```
### Iterating over lines in a file¶
We will now use this file as input in a program that will do some data processing. In the program, we will read each line of the file and print it with some additional text. Because text files are sequences of lines of text, we can use the for loop to iterate through each line of the file.
A line of a file is defined to be a sequence of characters up to and including a special character called the newline character. If you evaluate a string that contains a newline character you will see the character represented as \n. If you print a string that contains a newline you will not see the \n, you will just see its effects. When you are typing a Python program and you press the enter or return key on your keyboard, the editor inserts a newline character into your text at that point.
As the for loop iterates through each line of the file the loop variable will contain the current line of the file as a string of characters. The general pattern for processing each line of a text file is as follows:
```for line in myFile:
statement1
statement2
...
```
To process all of our quarterback data, we will use a for loop to iterate over the lines of the file. Using the split method, we can break each line into a list containing all the fields of interest about the quarterback. We can then take the values corresponding to first name, lastname, and passer rating to construct a simple sentence as shown in Listing 1.
(files_for)
```
```
### Alternative File Reading Methods¶
In addition to the for loop, Python provides three methods to read data from the input file. The readline method reads one line from the file and returns it as a string. The string returned by readline will contain the newline character at the end. This method returns the empty string when it reaches the end of the file. The readlines method returns the contents of the entire file as a list of strings, where each item in the list represents one line of the file. It is also possible to read the entire file into a single string with read. Table 2 summarizes these methods and Session 2 shows them in action.
Note that we need to reopen the file before each read so that we start from the beginning. Each file has a marker that denotes the current read position in the file. Any time one of the read methods is called the marker is moved to the character immediately following the last character returned. In the case of readline this moves the marker to the first character of the next line in the file. In the case of read or readlines the marker is moved to the end of the file.
```>>> infile = open("qbdata.txt","r")
>>> aline = infile.readline()
>>> aline
'Colt McCoy QB, CLE\t135\t222\t1576\t6\t9\t60.8%\t74.5\n'
>>>
>>> infile = open("qbdata.txt","r")
>>> linelist = infile.readlines()
>>> print(len(linelist))
34
>>> print(linelist[0:4])
['Colt McCoy QB, CLE\t135\t222\t1576\t6\t9\t60.8%\t74.5\n',
'Josh Freeman QB, TB\t291\t474\t3451\t25\t6\t61.4%\t95.9\n',
'Michael Vick QB, PHI\t233\t372\t3018\t21\t6\t62.6%\t100.2\n',
'Matt Schaub QB, HOU\t365\t574\t4370\t24\t12\t63.6%\t92.0\n']
>>>
>>> infile = open("qbdata.txt","r")
>>> filestring = infile.read()
>>> print(len(filestring))
1708
>>> print(filestring[:256])
Colt McCoy QB, CLE 135 222 1576 6 9 60.8% 74.5
Josh Freeman QB, TB 291 474 3451 25 6 61.4% 95.9
Michael Vick QB, PHI 233 372 3018 21 6 62.6% 100.2
Matt Schaub QB, HOU 365 574 4370 24 12 63.6% 92.0
Philip Rivers QB, SD 357 541 4710 30 13 66.0% 101.8
Matt Ha
>>>
```
Method Name Use Explanation
write filevar.write(astring) Add astring to the end of the file. filevar must refer to a file that has been opened for writing.
read(n) filevar.read() Reads and returns a string of n characters, or the entire file as a single string if n is not provided.
readline(n) filevar.readline() Returns the next line of the file with all text up to and including the newline character. If n is provided as a parameter than only n characters will be returned if the line is longer than n.
readlines(n) filevar.readlines() Returns a list of strings, each representing a single line of the file. If n is not provided then all lines of the file are returned. If n is provided then n characters are read but n is rounded up so that an entire line is returned.
Now lets look at another method of reading our file using a while loop. This important because many other programming languages do not support the for loop style for reading file but they do support the pattern we’ll show you here.
(files_while)
```
```
The important thing to notice is that on line two we have the statement line = infile.readline() This is very important because the while condition needs to have a value for the line variable. We call this initial read the priming read.
### Glossary¶
open
You must open a file before you can read its contents.
close
When you are done with a file, you should close it.
read
Will read the entire contents of a file as a string. This is often used in an assignment statement so that a variable can reference the contents of the file.
readline
Will read a single line from the file, up to and including the first instance of the newline character.
readlines
Will read the entire contents of a file into a list where each line of the file is a string and is an element in the list.
### Exercises¶
The following sample file contains one line for each student in an imaginary class the students name is the first thing on each line, followed by some exam scores.
```joe 10 15 20 30 40
bill 23 16 19 22
sue 8 22 17 14 32 17 24 21 2 9 11 17
grace 12 28 21 45 26 10
john 14 32 25 16 89
```
1. Using the text file student_data.dat write a program that prints out the names of students that have more than six quiz scores.
2. Using the text file student_data.dat write a program that calculates the average grade for each student, and print out the student’s name along with their average grade.
3. Using the text file student_data.dat write a program that calculates the minimum and maximum grade grade for each student. Print out the students name along with their minimum and maximum scores.
Here is a file called lab_data.dat that contains some sample data from a lab experiment.
```44 71
79 37
78 24
41 76
19 12
19 32
28 36
22 58
89 92
91 6
53 7
27 80
14 34
8 81
80 19
46 72
83 96
88 18
96 48
77 67
```
1. Using the data file lab_data.data each line contains a an x,y coordinate pair.
Write a function called plotRegression that reads the data from this file and uses a turtle to plot those points and a best fit line according to the following formulas
$y = \bar{y} + m(x - \bar{x})$$m = \frac{\sum{x_iy_i - n\bar{x}\bar{y}}}{\sum{x_i^2}-n\bar{x}^2}$
Where $$\bar{x}$$ is the mean of the x-values, $$\bar{y}$$ is the mean of the y- values and $$n$$ is the number of points. If you are not familiar with the mathematical $$\sum$$ it is the sum operation. For example $$\sum{x_i}$$ means to add up all the x values.
Your program should analyze the points and correctly scale the window using setworldcoordinates so that that each point can be plotted. Then you should draw the best fit line, in a different color, through the points.
1. At the end of this chapter is a very long file called mystery.dat The lines of this file contain either the word UP or DOWN or a pair of numbers. UP and DOWN are instructions for a turtle to lift up or put down its tail. The pair of numbers are some x,y coordinates. Write a program that reads the file mystery.dat and uses the turtle to draw the picture described by the commands and the set of points.
```UP
-218 185
DOWN
-240 189
-246 188
-248 183
-246 178
-244 175
-240 170
-235 166
-229 163
-220 158
-208 156
-203 153
-194 148
-187 141
-179 133
-171 119
-166 106
-163 87
-161 66
-162 52
-164 44
-167 28
-171 6
-172 -15
-171 -30
-165 -46
-156 -60
-152 -67
-152 -68
UP
-134 -61
DOWN
-145 -66
-152 -78
-152 -94
-157 -109
-157 -118
-151 -128
-146 -135
-146 -136
UP
-97 -134
DOWN
-98 -138
-97 -143
-96 -157
-96 -169
-98 -183
-104 -194
-110 -203
-114 -211
-117 -220
-120 -233
-122 -243
-123 -247
-157 -248
-157 -240
-154 -234
-154 -230
-153 -229
-149 -226
-146 -223
-145 -219
-143 -214
-142 -210
-141 -203
-139 -199
-136 -192
-132 -184
-130 -179
-132 -171
-133 -162
-134 -153
-138 -145
-143 -137
-143 -132
-142 -124
-138 -112
-134 -104
-132 -102
UP
-97 -155
DOWN
-92 -151
-91 -147
-89 -142
-89 -135
-90 -129
-90 -128
UP
-94 -170
DOWN
-83 -171
-68 -174
-47 -177
-30 -172
-15 -171
-11 -170
UP
12 -96
DOWN
9 -109
9 -127
7 -140
5 -157
9 -164
22 -176
37 -204
40 -209
49 -220
55 -229
57 -235
57 -238
50 -239
49 -241
51 -248
53 -249
63 -245
70 -243
57 -249
62 -250
71 -250
75 -250
81 -250
86 -248
86 -242
84 -232
85 -226
81 -221
77 -211
73 -205
67 -196
62 -187
58 -180
51 -171
47 -164
46 -153
50 -141
53 -130
54 -124
57 -112
56 -102
55 -98
UP
48 -164
DOWN
54 -158
60 -146
64 -136
64 -131
UP
5 -152
DOWN
1 -150
-4 -145
-8 -138
-14 -128
-19 -119
-17 -124
UP
21 -177
DOWN
14 -176
7 -174
-6 -174
-14 -170
-19 -166
-20 -164
UP
-8 -173
DOWN
-8 -180
-5 -189
-4 -201
-2 -211
-1 -220
-2 -231
-5 -238
-8 -241
-9 -244
-7 -249
6 -247
9 -248
16 -247
21 -246
24 -241
27 -234
27 -226
27 -219
27 -209
27 -202
28 -193
28 -188
28 -184
UP
-60 -177
DOWN
-59 -186
-57 -199
-56 -211
-59 -225
-61 -233
-65 -243
-66 -245
-73 -246
-81 -246
-84 -246
-91 -245
-91 -244
-88 -231
-87 -225
-85 -218
-85 -211
-85 -203
-85 -193
-88 -185
-89 -180
-91 -175
-92 -172
-93 -170
UP
-154 -93
DOWN
-157 -87
-162 -74
-168 -66
-172 -57
-175 -49
-178 -38
-178 -26
-178 -12
-177 4
-175 17
-172 27
-168 36
-161 48
-161 50
UP
-217 178
DOWN
-217 178
-217 177
-215 176
-214 175
-220 177
-223 178
-223 178
-222 178
UP
-248 185
DOWN
-245 184
-240 182
-237 181
-234 179
-231 177
-229 176
-228 175
-226 174
-224 173
-223 173
-220 172
-217 172
-216 171
-214 170
-214 169
UP
-218 186
DOWN
-195 173
-183 165
-175 159
-164 151
-158 145
-152 139
-145 128
-143 122
-139 112
-138 105
-134 95
-131 88
-129 78
-126 67
-125 62
-125 54
-124 44
-125 38
-126 30
-125 27
-125 8
-126 5
-125 -9
-122 -15
-115 -25
-109 -32
-103 -39
-95 -42
-84 -45
-72 -47
-56 -48
-41 -47
-31 -46
-18 -45
-1 -44
9 -43
34 -45
50 -52
67 -61
83 -68
95 -80
112 -97
142 -115
180 -132
200 -146
227 -159
259 -175
289 -185
317 -189
349 -190
375 -191
385 -192
382 -196
366 -199
352 -204
343 -204
330 -205
315 -209
296 -212
276 -214
252 -208
237 -202
218 -197
202 -193
184 -187
164 -179
147 -173
128 -168
116 -164
102 -160
88 -158
78 -159
69 -162
57 -164
56 -165
51 -165
UP
68 -144
DOWN
83 -143
96 -141
109 -139
119 -146
141 -150
161 -155
181 -163
195 -169
208 -179
223 -187
241 -191
247 -193
249 -194
UP
-6 -141
DOWN
-15 -146
-29 -150
-42 -154
-51 -153
-60 -152
-60 -152
UP
-90 -134
DOWN
-85 -131
-79 -128
-78 -123
-80 -115
-82 -106
-80 -101
-76 -101
UP
-81 -132
DOWN
-76 -130
-71 -126
-72 -124
UP
43 -118
DOWN
44 -125
47 -135
41 -156
37 -160
40 -166
47 -171
47 -171
UP
-106 -153
DOWN
-107 -167
-106 -178
-109 -192
-114 -198
-116 -201
```
mystery.dat
|
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|
http://mathoverflow.net/questions/95475?sort=oldest
|
## When LCS is isomorphic to subspace of some function space?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Updated: Following Michael's suggestion, I rephrase the question slightly.
Given a locally convex (Hausdorff) topological vector space (LCTVS), when is it isomorphic to a subspace of some function space $Y^X$ (equipped with the product topology), where $Y$ is, say, some Banach space (if it helps simplify things, can assume $Y=\mathbb{C}$, the complex field), and $X$ is some set. We are free to choose X and Y.
If not all LCTVS have this property, then what kind of conditions do we need?
Any reference would be appreciated, thanks!
-
1
Now, perhaps you should tell us what you found when you looked in a textbook on locally convex spaces... – Gerald Edgar Apr 30 2012 at 3:13
I think I got the answer: It seems to be always true as long as the LCTVS is complete (which really is necessary for $Y^X$ is complete when $Y$ is so). It is in the book "Functional Analysis: Theory and Applications" by Robert E. Edwards, Ex. 6.17. – yaoliang Apr 30 2012 at 4:57
## 2 Answers
If I interpret the question correctly, Yaoliang would like to know which LCTVS are isomorphic to $\mathbb{C}^X$, where $X$ is a set (no topology), and $\mathbb{C}^X$ is given the product topology. If this is so, the answer is: very few. Actually, spaces of this form are fully determined by the cardinality of $X$.
Just to make an example: no infinite dimensional normed space can be isomorphic to a space of this form. Indeed, in this case the set $X$ would have to be infinite, but then every neighborhood of $0$ in $\mathbb{C}^X$ would contain a proper vector subspace, and this is never true for normed spaces.
-
Thanks Alberto! It is a nice counterexample. Can you tell me more why such spaces are fully determined by the cardinality of X? Let us assume the cardinality is infinite (otherwise not interesting). Note that we are free to choose X when realizing the LCTVS as a function space. I also don't mind replacing the complex field as some Banach space (or even more general, if it helps), i.e., the function space is some Banach space valued function over some set X. – yaoliang Apr 29 2012 at 17:18
1
A bijection between $X$ and $Y$ induces an isomorphism (of TVS) between $\mathbb{C}^X$ and $\mathbb{C}^Y$, so only the cardinality of $X$ matters. Notice also that in your class of spaces $\mathbb{C}^X$ there is only one infinite dimensional separable space, the one which you obtain for $X=\mathbb{N}$. The problem is that you are considering ALL functions from $X$ to $\mathbb{C}$ (and if you replace $\mathbb{C}$ by a larger vector spaces things get even worse). If you want to represent interesting TVS you should put more structure on $X$ and restrict the class of functions. – Alberto Abbondandolo Apr 29 2012 at 17:54
A potentially more interesting question might be which LCTVS are isomorphic to a subspace of $C^X$ for some X. – Michael Renardy Apr 29 2012 at 20:25
Yes, I thought subspaces of $C^X$ is also $C^{X'}$ for a different $X'$ :). I should ask differently. – yaoliang Apr 30 2012 at 2:21
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Yes, every LCTVS can be realized as a function space: every LCTVS is isomorphic to the dual of its dual space equipped with the weak* topology. This is a consequence of a version of the Hahn-Banach theorem which states that in a real LCTVS two disjoint closed convex sets, one of which is compact, can be strictly separated. See Section IV.3 of A Course in Functional Analysis by Conway.
-
... but why is the original LC topology equal to the product topology of the function space ??? – Gerald Edgar Apr 29 2012 at 12:35
You're right, I should have read the question more carefully. Not every LC topology is induced by a family of linear functionals --- for instance, if a topology is induced by a family of linear functionals then every open neighborhood of zero contains a finite codimension subspace. I guess the answer is that you can realize a LCTVS in this sense as a function space if and only if its original topology equals its weak topology. I'm not sure there's going to be any more concrete answer than that. – Nik Weaver Apr 29 2012 at 13:34
|
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|
http://math.stackexchange.com/questions/236455/find-all-complex-numbers-satisfying-the-equation?answertab=oldest
|
# Find all complex numbers satisfying the equation.
Find all complex numbers satisfying $\cos(z) = i$.
-
## 1 Answer
Hint Use the identity $\cos(z)= \frac{e^{iz}+e^{-iz}}{2}$ and then solve the resulting quadratic equation in $e^{iz}.$
-
|
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|
http://www.physicsforums.com/showthread.php?p=3806687
|
Physics Forums
## What is meant by that ??
This can considered as a logic question , when I say for example there exist x such that ... . Mustn't I define some set from which x belong .
In a book of set theory it defined a binary relation as following :
A set R is a binary relation if $(\forall x \in R)(\exists x)(\exists y)(z=(x,y))$
The way I understand $\exists x$ is as following , as he is referring to any set x which exist , So we must consider some Set containing all sets , Such set doesn't exist . So what set must be considered , how must I understand this I know that we didn't mention the universal set if it is clear from context . Here , there is no Universal set . we want x to be arbitrary .
Thanks
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Recognitions: Science Advisor It looks like the author is implicitly assuming x and y belong to some set (R?). Also the statement starts with for all x in R - shouldn't it be for all z in R?
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Quote by mahmoud2011 This can considered as a logic question , when I say for example there exist x such that ... . Mustn't I define some set from which x belong . In a book of set theory it defined a binary relation as following : A set R is a binary relation if $(\forall x \in R)(\exists x)(\exists y)(z=(x,y))$
One of those "x"s should be a "z".
The way I understand $\exists x$ is as following , as he is referring to any set x which exist , So we must consider some Set containing all sets , Such set doesn't exist . So what set must be considered , how must I understand this I know that we didn't mention the universal set if it is clear from context . Here , there is no Universal set . we want x to be arbitrary . Thanks
## What is meant by that ??
Quote by mahmoud2011 This can considered as a logic question , when I say for example there exist x such that ... . Mustn't I define some set from which x belong . In a book of set theory it defined a binary relation as following : A set R is a binary relation if $(\forall x \in R)(\exists x)(\exists y)(z=(x,y))$ The way I understand $\exists x$ is as following , as he is referring to any set x which exist , So we must consider some Set containing all sets
No. $\forall x$ and $\exists x$ are perfectly legitimate expressions as they are. Indeed, $\forall x \in R \varphi(x)$ is merely shorthand for $\forall x (x \in R \rightarrow \varphi(x))$.
You seem to be confusing this with the axiom of specification, which says that given a set A, you can define a set B as the set of all $x \in A$ such that $\varphi(x)$.
Quote by mahmoud2011 This can considered as a logic question , when I say for example there exist x such that ... . Mustn't I define some set from which x belong . In a book of set theory it defined a binary relation as following : A set R is a binary relation if $(\forall x \in R)(\exists x)(\exists y)(z=(x,y))$ The way I understand $\exists x$ is as following , as he is referring to any set x which exist , So we must consider some Set containing all sets , Such set doesn't exist . So what set must be considered , how must I understand this I know that we didn't mention the universal set if it is clear from context . Here , there is no Universal set . we want x to be arbitrary . Thanks
sorry it must be $(\forall z \in R)(\exists x)(\exists y)(z=(x,y))$
So how mus I understand it
Recognitions: Gold Member Science Advisor Staff Emeritus Every z in the relation consists of a pair of real numbers.
Quote by HallsofIvy Every z in the relation consists of a pair of real numbers.
But we are dealing wit general sets in set theory so we consider "any sets" also I don't know what must be meant for saying "any sets". When I say there exist x such that ... , I understand that as "x exists according to axioms of ZFC " , But When I was reading in Logic , and when defining quantifiers it was defined with a universal set . When we write for all x such that ... (Of course when is meant any set from the context) , I understand it as if for any x we can prove to exist according ZFC we have ... . And there exist x such that ... , I understand it as that we can prove the existence of some x according to ZFC such that ... . I am not sure if my way of thinking is formal and precise or not .
Quote by mahmoud2011 This can considered as a logic question , when I say for example there exist x such that ... . Mustn't I define some set from which x belong . In a book of set theory it defined a binary relation as following : A set R is a binary relation if $(\forall x \in R)(\exists x)(\exists y)(z=(x,y))$ The way I understand $\exists x$ is as following , as he is referring to any set x which exist , So we must consider some Set containing all sets , Such set doesn't exist . So what set must be considered , how must I understand this I know that we didn't mention the universal set if it is clear from context . Here , there is no Universal set . we want x to be arbitrary . Thanks
There is no set containing all sets, but there is a class containing all sets...Mathematicians say.
Quote by sigurdW There is no set containing all sets, but there is a class containing all sets...Mathematicians say.
That I am talking about , the book haven't considered classes yet ( I know some about them ) , but the question is when I say for all x such that P(x) , will it mean "if you have proved the existence of set x then P(x) holds" or what , and whwn I say there exists some x such that P(x) holds , will it mean " It can proven the existence of a set x in ZFC such that P(x)" , Here I consider the statements " It can be proven" and alike are informal . So Can any one explain to me What it is meant by them logically.
Quote by mahmoud2011 That I am talking about , the book haven't considered classes yet ( I know some about them ) , but the question is when I say for all x such that P(x) , will it mean "if you have proved the existence of set x then P(x) holds" or what , and when I say there exists some x such that P(x) holds , will it mean " It can proven the existence of a set x in ZFC such that P(x)" , Here I consider the statements " It can be proven" and alike are informal . So Can any one explain to me What it is meant by them logically.
1 When you say for all x such that P(x)...
It should mean that out of a certain set ordinarily called "the domain"
you have created a subset consisting of all objects ,x, satisfying the condition "P".
Note: You must distinguish between "x" as any object of a set and "x" considered as the set!
If x is any element of the set x then x contains itself as an element and there is a certain axiom in ZFC forbidding just that. (Since the axiom is independent of the other axioms, then you could use ZFC with the axiom replaced with its negation resulting in a set theory as consistent as ZFC is.)
2 When you say there exists some x such that P(x) holds, then you claim that your subset of the domain is not empty.
Quote by sigurdW 1 When you say for all x such that P(x)... It should mean that out of a certain set ordinarily called "the domain" you have created a subset consisting of all objects ,x, satisfying the condition "P". Note: You must distinguish between "x" as any object of a set and "x" considered as the set! If x is any element of the set x then x contains itself as an element and there is a certain axiom in ZFC forbidding just that. (Since the axiom is independent of the other axioms, then you could use ZFC with the axiom replaced with its negation resulting in a set theory as consistent as ZFC is.) 2 When you say there exists some x such that P(x) holds, then you claim that your subset of the domain is not empty.
I see you mean Axiom Schema of Comprehension right . So in the definition I have written at first such domain is not determined , I see that he wants to determine any set , and I can't take such domain to be the st of all sets because it doesn't exist . And I don't want to use notion of classes.
Quote by mahmoud2011 I see you mean Axiom Schema of Comprehension right . So in the definition I have written at first such domain is not determined , I see that he wants to determine any set , and I can't take such domain to be the st of all sets because it doesn't exist . And I don't want to use notion of classes.
Im not sure what you want to do, and how to advice you...An adventurous, perhaps stupid idea, is to take away one insignificant set from the "set of all sets" then you should have a set of almost all sets... right?
Quote by sigurdW Im not sure what you want to do, and how to advice you...An adventurous, perhaps stupid idea, is to take away one insignificant set from the "set of all sets" then you should have a set of almost all sets... right?
That I am talking about . My only question what 's the meaning of there exist x such that .... .
In the definition above it and and some of axioms of set theory for example Axiom of Extensionality we begin by saying for all x and y , .... . So what " for all " Exactly mean.
Quote by mahmoud2011 I see you mean Axiom Schema of Comprehension right . So in the definition I have written at first such domain is not determined , I see that he wants to determine any set , and I can't take such domain to be the st of all sets because it doesn't exist . And I don't want to use notion of classes.
Sorry , " Axiom of regularity " , I confused only their names .
Quote by mahmoud2011 That I am talking about . My only question what 's the meaning of there exist x such that .... . In the definition above it and and some of axioms of set theory for example Axiom of Extensionality we begin by saying for all x and y , .... . So what " for all " Exactly mean.
I dont remember the exact formulations of the set axioms...it was a long time since I looked at them. But it seems to me that you want to be sure that you understand the basic definitions in order to be sure that you understand the meaning of the axioms. Thats an honest strategy.
I dont mind looking at basic definitions... Do you have access to them? Some authors does not make the foundation of their theory clear and complete. Make a list of the statements you find unclear.
Are you really bothered about the meaning of "for all x..." and "there exists an x..."? The expressions are called "quantifiers" and belong to "Predicate Logic" they are perhaps assumed to already be defined and understood in the theory you are studying.
"for all x..." is called the universal quantifier and it affects a variable found in a sentence function: P(x) ... "x" is a variable so P(x) is a statement function not a statement...you get a statement if you replace x with an "individual constant" say ,a, ...then P(a) is (because of the quantifier) a true sentence.
An "interpretation" of the logical language used is some non empty set called "the Domain"
it contains all sentences all predicates and all constants:
"for all x..." now means: for each object x in the domain...
Having the universal we define the existential: "there is an x such that p(x)" means "not for all x : not p(x)"
(Note that the domain is assumed not exhibited! It might be infinite so we cant exhibit it.)
This is a very abbreviated view...does it make any sense to you?
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## Return to Answer
3 added more details regarding Figure 5 and low dimensions
The behavior for continuous charge distributions amounts to classical potential theory; for discrete charges, you get this behavior in the continuum limit.
It is true that the distribution is uniform for a sphere. On other manifolds things can be more complicated. See http://www.ams.org/notices/200410/fea-saff.pdf for a very nice exposition and further references. (In particularFor example, Figure 5 from that paper (included here thanks to Joseph O'Rourke) shows some remarkable behavior the limiting distribution for particles on a torus.) torus under an inverse $s$-th power law:
In this example, for $s \ge 2$ you get a uniform distribution, which is the default behavior when the energy for a continuous charge distribution diverges. For $s < 2$ the particles converge to the continuous distribution that minimizes energy. When $s<1$, this distribution is not even supported on the entire torus.
You don't see these phenomena for the sphere, because of its symmetry, but they are typical for less symmetric manifolds.
Incidentally, the behavior of 1 and 2 dimensions is not so strange. The charges do indeed end up on the boundary, if one uses a harmonic potential function (for example, a logarithmic potential in $\mathbb{R}^2$). The difficulty is that the Coulomb potential is not harmonic in $\mathbb{R}^1$ or $\mathbb{R}^2$. One way of thinking about it is that if you view the needle or disk as sitting inside $\mathbb{R}^3$, then the charges do all end up on the boundary, because the boundary in $\mathbb{R}^3$ is the entire set.
More generally, in $\mathbb{R}^n$, if you use an inverse $s$-th power law for the potential function, then all the charge will be on the boundary if $s \le n-2$ (because the potential function is superharmonic and therefore satisfies the minimum principle). When $s > n-2$, that does not happen.
2 Inserted the cited Fig. 5.
The behavior for continuous charge distributions amounts to classical potential theory; for discrete charges, you get this behavior in the continuum limit.
It is true that the distribution is uniform for a sphere. On other manifolds things can be more complicated. See http://www.ams.org/notices/200410/fea-saff.pdf for a very nice exposition and further references. (In particular, Figure 5 shows some remarkable behavior on a torus.)
Incidentally, the behavior of 1 and 2 dimensions is not so strange. The charges do indeed end up on the boundary, if one uses a harmonic potential function (for example, a logarithmic potential in $\mathbb{R}^2$). The difficulty is that the Coulomb potential is not harmonic in $\mathbb{R}^1$ or $\mathbb{R}^2$. One way of thinking about it is that if you view the needle or disk as sitting inside $\mathbb{R}^3$, then the charges do all end up on the boundary, because the boundary in $\mathbb{R}^3$ is the entire set.
1
The behavior for continuous charge distributions amounts to classical potential theory; for discrete charges, you get this behavior in the continuum limit.
It is true that the distribution is uniform for a sphere. On other manifolds things can be more complicated. See http://www.ams.org/notices/200410/fea-saff.pdf for a very nice exposition and further references. (In particular, Figure 5 shows some remarkable behavior on a torus.)
Incidentally, the behavior of 1 and 2 dimensions is not so strange. The charges do indeed end up on the boundary, if one uses a harmonic potential function (for example, a logarithmic potential in $\mathbb{R}^2$). The difficulty is that the Coulomb potential is not harmonic in $\mathbb{R}^1$ or $\mathbb{R}^2$. One way of thinking about it is that if you view the needle or disk as sitting inside $\mathbb{R}^3$, then the charges do all end up on the boundary, because the boundary in $\mathbb{R}^3$ is the entire set.
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http://terrytao.wordpress.com/2007/03/02/open-question-noncommutative-freiman-theorem/
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What’s new
Updates on my research and expository papers, discussion of open problems, and other maths-related topics. By Terence Tao
# Open question: noncommutative Freiman theorem
2 March, 2007 in math.CO, math.GR, question | Tags: additive combinatorics, Freiman's theorem, group theory
This is another one of my favourite open problems, falling under the heading of inverse theorems in arithmetic combinatorics. “Direct” theorems in arithmetic combinatorics take a finite set A in a group or ring and study things like the size of its sum set $A+A := \{ a+b: a,b \in A \}$ or product set $A \cdot A := \{ ab: a,b \in A \}$. For example, a typical result in this area is the sum-product theorem, which asserts that whenever $A \subset {\Bbb F}_p$ is a subset of a finite field of prime order with $1 \leq |A| \leq p^{1-\delta}$, then
$\max( |A+A|, |A \cdot A| ) \geq |A|^{1+\epsilon}$
for some $\epsilon = \epsilon(\delta) > 0$. (This particular theorem was first proven here, with an earlier partial result here; more recent and elementary proofs with civilised bounds can be found here, here or here. It has a number of applications.)
In contrast, inverse theorems in this subject start with a hypothesis that, say, the sum set A+A of an unknown set A is small, and try to deduce structural information about A. A typical goal is to completely classify all sets A for which A+A has comparable size with A. In the case of finite subsets of integers, this is Freiman’s theorem, which roughly speaking asserts that if $|A+A| = O(|A|)$, if and only if A is a dense subset of a generalised arithmetic progression P of rank O(1), where we say that A is a dense subset of B if $A \subset B$ and $|B| = O(|A|)$. (The “if and only if” has to be interpreted properly; in either the “if” or the “only if” direction, the implicit constants in the conclusion depends on the implicit constants in the hypothesis, but these dependencies are not inverses of each other.) In the case of finite subsets A of an arbitrary abelian group, we have the Freiman-Green-Ruzsa theorem, which asserts that $|A+A| = O(|A|)$ if and only if A is a dense subset of a sum P+H of a finite subgroup H and a generalised arithmetic progression P of rank O(1).
One can view these theorems as a “robust” or “rigid” analogue of the classification of finite abelian groups. It is well known that finite abelian groups are direct sums of cyclic groups; the above results basically assert that finite sets that are “nearly groups” in that their sum set is not much larger than the set itself, are (dense subsets of) the direct sums of cyclic groups and a handful of arithmetic progressions.
The open question is to formulate an analogous conjectural classification in the non-abelian setting, thus to conjecture a reasonable classification of finite sets A in a multiplicative group G for which $|A \cdot A| = O(|A|)$. Actually for technical reasons it may be better to use $|A \cdot A \cdot A| = O(|A|)$; I refer to this condition by saying that A has small tripling. (Note for instance that if H is a subgroup and x is not in the normaliser of H, then $H \cup \{x\}$ has small doubling but not small tripling. On the other hand, small tripling is known to imply small quadrupling, etc., see e.g. my book with Van Vu.) Note that I am not asking for a theorem here – just even stating the right conjecture would be major progress! An if and only if statement might be too ambitious initially: a first step would be to obtain a slightly looser equivalence, creating for each group G and fixed $\epsilon > 0$ a class ??? of sets for which the following two statements are true:
1. If A is a finite subset of G with small tripling, then A is a dense subset of $O(|A|^\epsilon)$ left- or right- translates of a set P of the form ???.
2. If P is a set of the form ???, then there exists a dense subset A of P with small tripling (possibly with a loss of $O(|A|^\epsilon)$ in the tripling constant).
An obvious candidate for ??? is the inverse image in N(H) of a generalised geometric progression of rank O(1) in an abelian subgroup of N(H)/H, where H is a finite subgroup of G and N(H) is the normaliser of H; note that property (2) is then easy to verify. Let us call this the standard candidate. I do not expect this standard candidate to fully suffice, though I do not know at present of a counterexample. (Update, Mar 5: Now I do – a discrete ball in a nilpotent group.) But it seems to be a reasonably good candidate nevertheless. In this direction, some partial results are known:
• For abelian groups G, from the Freiman-Green-Ruzsa theorem, we know that the standard candidate suffices.
• For $G = SL_2({\Bbb C})$, we know from work of Elekes and Király and Chang that the standard candidate suffices.
• For $G = SL_2({\Bbb F}_p)$, there is a partial result of Helfgott, which (roughly speaking) asserts that if A has small tripling, then either A is a dense subset of all of G, or is contained in a proper subgroup of G. It is likely that by pushing this analysis further one would obtain a candidate for ??? in this case.
• For $G = SL_3({\Bbb Z})$, there is a partial result of Chang, which asserts that if A has small tripling, then it is contained in a nilpotent subgroup of G.
• For the lamplighter group $G = {\Bbb Z}/2{\Bbb Z} \wr {\Bbb Z}$, there is a partial result of Lindenstrauss which (roughly speaking) asserts that if A has small tripling, then A cannot be nearly invariant under a small number of shifts. It is also likely that by pushing the analysis further here one would get a good candidate for ???.
• For a free non-abelian group, we know (since the free group embeds into $SL_2({\Bbb C})$) that the standard candidate suffices; a much stronger estimate in this direction was recently obtained by Razborov.
• For a Heisenberg group G of step 2, there is a partial result of myself, which shows that sets of small tripling also have small tripling in the abelianisation of G, and are also essentially closed under the antisymmetric form that defines G. This, in conjunction with the Freiman-Green-Ruzsa theorem, gives a characterisation, at least in principle, but it is not particularly explicit, and it may be of interest to work it out further.
• For G torsion-free, there is a partial result of Hamidoune, Lladó, and Serra, which asserts that $|A \cdot A| \geq 2|A|-1$, and that if $|A \cdot A| \leq 2|A|$ then A is a geometric progression with at most one element removed; in particular, the standard candidate suffices in this case.
These examples do not seem to conclusively suggest what the full classification should be. Based on analogy with the classification of finite simple groups, one might expect the full classification to be complicated, and enormously difficult to prove; on the other hand, the fact that we are in a setting where we are allowed to lose factors of O(1) may mean that the problem is in fact significantly less difficult than that classification. (For instance, all the sporadic simple groups have size O(1) and so even the monster group is “negligible”.) Nevertheless, it seems possible to make progress on explicit groups, in particular refining the partial results already obtained for the specific groups mentioned above. (Update, Mar 5: Via Akshay Venkatesh and Elon Lindenstrauss, I learnt that perhaps the closer analogy is with Gromov’s theorem, rather than the classification of finite simple groups. Which makes things slightly more hopeful… nevertheless, the question of revisiting the classification from a “O(1) perspective” may still be interesting.)
## 37 comments
Terry,
A very interesting post. Following some work I did with Tom Sanders, I’d like to suggest a possible splitting of non-commutative Freiman into two parts. The first is called “weak Freiman” and I can state it explicitly; the second is “strong Freiman” and I do not have any thoughts to add to those you have already put down in your post.
Weak Freiman is the statement that if A has small tripling then some large subset A’ of A is contained in a “Bourgain system”. A Bourgain system is a certain notion of what it means to be an approximate group. The idea comes from Bourgain’s 1998 paper on Roth’s theorem, hence the name.
In my paper with Sanders we only defined Bourgain systems in the abelian case, but I’d guess it extends to the non-abelian case as follows.
A Bourgain system in some group G is a collection (X_t)_{t = 0}^{10} of sets such that
(i) If g is in X_t then so is g^{-1}
(ii) X_t X_u and X_u X_t are both subsets of X_{t + u} when t,u > c_1(K)|A| such that A’ is contained inside a Bourgain system with |X_1| U_d(C), where d is not too big. Then the pull-back under pi of any ball about the identity matrix will be a Bourgain system of dimension about d (kind of a non-abelian Bohr set, by the way). What can one say about the structure of such sets? One would seem to need some kind of non-abelian geometry of numbers.
Anyway there are some great questions in this area waiting to be worked on…..
Ben
Terry, half my post got deleted. Any idea what happened?
Ben
3 March, 2007 at 2:50 am
Anonymous
Ben,
Most blogs limit the size of a “fast reply” to a few hundred characters. If you plan to post a long reply, it’s best to type it in notepad or some other text editor, and copy and paste it into the reply window. Then if it is truncated, simply copy and paste the rest in another reply.
(Another advantage of not typing long shpiels directly in the reply box is that you don’t risk losing the lot if you accidently hit the “back” button – Firefox is a bit better at saving input in forms, and I think the latest IE has improved, but it used to be maddening when this happened!)
Terence,
Have you ever considered the ABC Conjecture, or do you know if any progress has been made on it? Seems like a glass cliff face to me, but then I’d have said the same about primes in arbitrarily long arithmetic progressions and many of the other problems you have tackled.
Cheers
John R Ramsden
3 March, 2007 at 2:55 am
John R Ramsden
P.S. Oops – I should have addressed the ABC conjecture question to Terence _and_ Ben (and for that matter to anyone else who might be qualified to speculate).
Dear John,
Thanks for clearing up the truncation issue. I poked around but couldn’t find a way to modify the threshold for these “fast replies”. It seems that if one has a wordpress account then it is easier to make and then edit fancy comments, though.
Regarding the ABC conjecture, I tend to distinguish between two types of number theory problems: “many-solutions” problems and “no-solutions” problems. “many-solutions” problems ask for one or more solutions to some system of equations in integers or primes. Examples include the Goldbach or Waring problems, the twin prime conjecture, or finding long progressions in primes. “no-solutions” problems, on the other hand, ask that the number of solutions to a certain system in a certain range is in fact zero (or bounded independently of various parameters). Fermat’s last theorem is the most famous example of the latter; the ABC conjecture can also be rephrased in this form. (Specifically, if N_1, N_2, N_3 are large integers and A_i is the set of numbers of size at least (N_1 N_2 N_3)^{1+epsilon} with radical at most N_i, there are no solutions to a_1 + a_2 = a_3 with a_i in A_i.) Other examples of “no-solutions” problems include the finitude of Fermat primes and the (now solved) Catalan conjecture.
With “many-solutions” problems, one can try to solve the problem by getting a lower bound on the number of solutions; as long as the lower bound is non-trivial, one solves the problem. This suggests analytical methods, as one can tolerate error terms as long as they are dominated by the main term in the lower bound. But such techniques are much less likely to work for “no-solutions” problems; one has to get an upper bound which is less than 1, but all upper bounds are clearly at least zero, so one is aiming for an incredibly miniscule target and can basically afford no losses in error terms whatsoever. I believe that the best hope for such problems instead lies in methods from algebraic number theory.
Thinking about it a little more, I guess one of the indicators of difficulty of a number-theoretic problem is not so much whether it is asking for many solutions or for none, but rather just how large of a “conspiracy” amongst the natural numbers would be needed to disprove the conjecture.
For instance, take FLT. It would only take a single solution to a^n + b^n = c^n for some n > 2 to destroy the conjecture; this looks like a very small conspiracy, difficult to detect by most methods, and so the conjecture should be very difficult. But it turns out that such a conspirator must inevitably involve a very strange elliptic curve, which in turn drags in many other mathematical objects into the conspiracy, which makes it more feasible to detect and then destroy such a conspiracy.
On the other hand, a failure of the twin prime conjecture (say) would require an infinite number of conspiracies: past a certain point, every prime p is somehow forbidding p-2 and p+2 to be prime. We can’t yet say that all these conspiracies don’t happen at once, but it is plausible that one could set up some sort of computable statistic involving the primes which would detect and then exclude this scenario.
With arithmetic progressions, the conspiracy has to be even more “global”: if there are no prime progressions of length k, then whenever p, p+r, …, p+(k-2)r are large primes, then p+(k-1)r is mysteriously forced to be composite. Since p and r can be independently quite large, this gives a lot of structural information on the primality of p+(k-1)r which can hopefully be detected or exploited. (Indeed we sort of exploit this kind of information in our proof, via the machinery of “dual functions”. Rather amusingly, we don’t directly exclude the “conspiracy” of exotic structure in the primes, but instead we “co-opt” any such conspiracy, placing it into a homogenised model f_{U^\perp} of the primes which can then be forced (via Szemeredi’s theorem) to cough up progressions. In later papers we exclude these conspiracies directly.)
Ah, I got caught by the truncation issue also :-) . Actually I found the source of the problem: the less-than sign is interpreted as an HTML tag and so should be avoided in comments. I’m guessing that the portion of Ben’s post which was removed was the portion between a \lt sign and a \gt sign.
Returning back to the ABC conjecture, in order for this conjecture to be false one needs a sparse sequence of counterexamples, namely a sequence of solutions a+b=c with rad(abc) \leq c^{1+o(1)}. This sequence can be arbitrarily sparse and thus does not generate the type of aggregate bias which could be detectable by an analytical statistic (though this method might conceivably give a non-trivial upper bound to the number of ABC counterexamples). Instead, one would probably have to algebraically transform such a counterexample to another setting (possibly one from transcendence theory?) in which any conspiracy becomes large enough to be detectable. (cf. the “dispersion method” of Linnik.) Not that I have any idea how to achieve this…
Dear Terry,
Here is an attempt to revisit my post of yesterday without using less than/greater than symbols,
which HTML doesn’t seem to cope with.
Your post was very interesting. Following some work I did with Tom Sanders, I’d like to suggest a possible splitting of non-commutative Freiman into two parts. The first is called “weak Freiman” and I can state it explicitly; the second is “strong Freiman” and I do not have any thoughts to add to those you have already put down in your post.
Weak Freiman is the statement that if A has small tripling then some large subset A’ of A is contained in a “Bourgain system”. A Bourgain system is a certain notion of what it means to be an approximate group. The idea comes from Bourgain’s 1998 paper on Roth’s theorem, hence the name.
In my paper with Sanders we only defined Bourgain systems in the abelian case, but I’d guess it extends to the non-abelian case as follows.
A Bourgain system in some group G is a collection (X_t)_{t = 0}^{10} of sets such that
(i) If g is in X_t then so is g^{-1};
(ii) X_t X_u and X_u X_t are both subsets of X_{t + u} when t, u are at most 5;
(iii) 0 is in X_t for all t;
(iv) |X_{2t}| is no more than C|X_t| for some C and for all t at most 5.
log C may be interpreted as a kind of “dimension” of the Bourgain system.
What is the point of this definition? There are many places inside a Bourgain system where one can look and see
objects which are almost closed under addition. For example if t and u are “typical” values and if u is much less than
t then X_t + X_u will be almost the same as X_t. These properties are enough to enable Tom and I (in the abelian case)
to do a kind of “approximate harmonic analysis” on a Bourgain system. Actually all the ideas for doing that came from
Bourgain’s work and some unpublished notes of yours. For some applications (such as the one Tom and I had) this is all
one needs, and any more precise algebraic structural information about the Bourgain system is unnecessary. For example,
though the theorem you call the “Freiman-Green-Ruzsa Theorem” lets you relate an abelian Bourgain system to a coset
progression, we did not need this fact in our application.
Thus a Bourgain system (in the abelian case) qualifies as an approximate group in a quite strong sense.
I’m guessing that in the non-abelian case things are pretty similar, and one can do a kind of approximate representation
theory, at least of low-dimensional representations – I haven’t worked out the details yet.
So here’s my guess at a “weak Freiman”:
Suppose that |A+A| is at most K|A|. Then there is some subset A’ of A with size at least c_1(K)|A| such that
A’ is contained inside a Bourgain system of dimension c_2(K) with |X_1| having size at most c_3(K)|A|.
How might one prove such a weak Freiman theorem? Unfortunately one of the key steps in the proof of the Freiman-Green-Ruzsa theorem
is simply false for non-abelian groups. That is the existence of “good models”. Thus one can have a set A with *very*
small tripling (say about 3) with no large subset which is Freiman-isomorphic to a dense subset of a group.
This means that you can’t repeat our trick of transferring to a setting where the tools of (non-abelian) harmonic analysis
can be applied. I think there may be some hope of generalising a recent “energy-increment” argument of you and I; Tom
I can’t add much more to what you wrote concerning the correct statement of a “strong Freiman theorem”.
There is one example which I think ought to be instructive. Let G be a finite abelian group and suppose that pi from G to
U_d(C) is a unitary representation. Then the pull-back under pi of any ball about the identity matrix is
likely (assuming some amount of equidistribution of pi(G) in U_d(C)) to be a Bourgain system of dimension about d.
What can one say about the algebraic structure of such “unitary Bohr sets”? Is there a kind of “non-abelian geometry
of numbers” which allows one to find structure inside these sets? I think that the study of some
examples with small d (which I haven’t done yet) could be instructive here.
Anyway there are some great questions in this area waiting to be worked on…..
Ben
Ben please could you post some of these great questions in this area?
zerocold
5 March, 2007 at 6:53 am
akshay
Lindenstrauss has suggested that a non-commutative Freiman theorem should be connected to quantifications of Gromov’s theorem: a group of polynomial growth is virtually nilpotent. As far as I know, no quantitative form of that is known. Anyway, that may suggest candidates for ???
Thanks Akshay!
It does seem that virtual nilpotency is perhaps the key (the above-mentioned paper of Chang also makes this type of connection). Incidentally I realised that one can use a nilpotent group to concoct a set of small tripling for which my “standard candidate” doesn’t come close to approximating – take the Heisenberg group of upper-triangular matrices $\begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix}$ and consider the set A in which $a, b = O(N); c = O(N^2)$.
Perhaps the revised candidate for ??? would be the inverse image (under the projection N(H) -> N(H)/H by a finite group H) of a set B of small tripling contained inside a nilpotent group.
7 March, 2007 at 11:42 am
anonymous
You probably mean a,c = O(N) and c = O(N^2)
7 March, 2007 at 11:43 am
anonymous
You probably mean a,c = O(N) and b = O(N^2)
[...] an earlier blog Terry discussed Freiman’s theorem. The name of Freiman is attached to a growing body of [...]
I have a question about the case of a general Abelian group G. What is the best known result for the case G has *no* finite subgroups?
Dear HL,
In the torsion-free abelian case (i.e. G is abelian with no non-trivial finite subgroups), one can use the “Freiman isomorphism” trick to reduce the problem to the classical case $G={\Bbb Z}$ of the integers. (Quick sketch: since the set A one is studying is finite, one can reduce to the group generated by A, i.e. one can assume G is finitely generated. Since G is also abelian and torsion free, it is now isomorphic to a lattice ${\Bbb Z}^d$. Now, one cannot embed all of ${\Bbb Z}^d$ homomorphically inside the integers. But we are not really using all of ${\Bbb Z}^d$, since A is finite, we can work inside a big d-dimensional cube inside ${\Bbb Z}^d$, which can be mapped homomorphically and injectively into the integers in a variety of ways.)
In the integers, the classical theorem of Freiman applies: if $|A+A| \leq K|A|$, then A can be contained in a generalised arithmetic progression of rank at most F(K) and cardinality at most G(K) |A|. The best bounds for F and G are currently due to Chang, who obtained F(K) = K-1 and $G(K) = \exp( O( K^2 \log^3 K ) )$ for K large. There is also a variant due to Bilu and refined slightly by Ben Green and myself, which has the same hypothesis but concludes instead that A is contained inside $\exp( O_\epsilon( K^{O(1)} ) )$ translates of a generalised arithmetic progression of rank at most $\log_2 K + \epsilon$ and cardinality at most |A|, where $\epsilon > 0$ is arbitrary.
I’d like to throw out a specific recreational mathematics problem which I realized is on the same theme as non-commutative Freiman theorems.
It is known that the diameter of the Rubik’s cube group (in the half- and quarter-turn metric) is between 20 and 27. Meanwhile
here is a histogram of the word lengths of a random sample of 1,000 positions. You would guess just from looking at the histogram that the diameter is 20; it would be strange if it were more than 21.
Is there a way to prove that the diameter of the Rubik’s Cube group is 20, by combining Freiman-style results with simple-minded computer data? If not an outright proof, is there a probabilistic protocol, in the vein of Arthur-Merlin protocols or probabilistic primality?
For example, the histogram does immediately demostrate, if you are content with a probabilistic verification, that the diameter of the Rubik’s cube group is at most 36. That’s just by the pigeonhole principle: more than half of the positions can be solved in 18 moves. I have heard that for a time, this probabalistic verification was the best upper bound on the diameter of the cube group.
Actually, the pigeonhole principle even establishes a probabilistic verification that the diameter is at most 35, better than 36. More than a quarter of the positions can be solved in 17 or fewer moves, while only 3% require 19 or more moves. Therefore every ball of radius 17 intersects every ball of radius 18.
(Sorry, I made a mistake; this emendation could be attached to the previous post. )
2 April, 2007 at 5:02 pm
sum-product
two papers related to this problem came out
http://arxiv.org/abs/math/0703676
http://arxiv.org/abs/math/0703614
Dear Greg,
Your arguments seem to say that most pairs of points have distance at most 35 from each other. I never played Rubik, but I guess the underlying graph is vertex-transitive. In that case, the histogram actually would imply that most pair of points have distance at most 20 from each other, as the distribution of the distance between two random points is the same as the distribution of the distance between a fixed point and a random one. Best, Van.
5 April, 2007 at 10:48 pm
Anonymous
The Rubik’s cube graph is actually a Cayley graph of the Rubik’s cube group. So yes, it is vertex-transitive, in fact freely transitive. My question is not about most pairs of points, it’s about all pairs of points. (I.e., the graph diameter or worst case.) If it is true, as the histogram almost proves, that the distance between 97% of pairs is 18 or less and the distance between 25% of pairs is 17 or less, then the distance between all pairs of vertices is 35 or less. So my question is whether you can use approximate group theory somehow to improve this immediate but pretty inference.
5 April, 2007 at 10:55 pm
Greg Kuperberg
Oh oops, I accidentally signed that as “anonymous”!
I don’t understand your point. The histogram is about a set of random samples, how can it say something about all pairs ? In theory, the only thing a small set of random samples can say is that the number of bad pairs are small.
The point is that the histogram is a probabilistic verification of a mathematical assertion. If there existed even one pair of points in the cube whose distance is 36, then the probability of making such a histogram would be less than $10^{-100}$. Or if you are not satisfied with that histogram, you could make a bigger histogram whose odds of occurrence are less than $10^{-2000}$. Think about it.
Anyway, my question is whether one can improve such a probabilistic verification using better group theory ideas.
Dear Greg,
This is an interesting question. My guess is that one will have to understand the representation theory of the Rubik’s cube group in order to be able to convert these sorts of probabilistic statements into rigorous ones. There is a nascent theory of quasirandom groups, as pioneered by Gowers, which asserts roughly speaking that if a finite group G has no low-dimensional representations, then its multiplicative structure is highly expansive, thus we expect for instance A.A.A to be much larger than A if A is already rather large to begin with. This type of result probably can be used to show that large balls intersect each other rather often, which might indeed lead to some diameter bound. This type of thing has for instance been carried out for $SL_2(F_p)$ recently by Bourgain and Gamburd.
If there are small representations, there could be the very unlikely scenario that all the random elements you selected happen to lie in one corner of the representation – e.g. in a normal subgroup of small index. Then the data you have collected does not “span” enough of the group to definitively pin down the diameter.
First, my own perspective is that the probabilistic statements actually are rigorous, in a way: They are entirely rigorous probabilistic verifications rather than rigorous proofs. The situation is similar to that of numbers that are known as “probable primes”. It is not the same kind of lack of rigor as a physics discussion where many things have not been defined. So let us say that it would be good to change probabilistic verifications into deterministic proofs. This is true — although it is known using less elegant combinatorial methods that the cube group diameter is at most 27.
But I also think that it would be exciting to find a new probabilistic verification that the cube group diameter is 26 or better. In the competition between probabilistic verifications and deterministic proofs, the former can never do worse; it would be fun to have more examples where they can do better. Unfortunately the diameter < 35 result was a temporary victory.
Anyway, for either purpose, I agree that it would be nice to know the representation theory of the Rubik’s cube group. But that isn’t mysterious. The Rubik’s cube group is an index 12 subgroup of the disassembly group. That group is the Cartesian product of the corner group and the edge group. The corner group is the wreath product of S8 and C3, while the edge group is the wreath product of S12 and C2. The reason that the cube group has index 12 is that the corner rotations and edge flips satisfy congruences, while the parity of the edge permutation has to equal the parity of the corner permutation.
Note that the center of the cube group has order 2. The central involution is called the “superflip” move or position. It is known to have distance 20, and for all I know it is the unique position of distance 20. There may be a lesson in that too.
I realize that it is a bit dull to prove a theorem (or probabilistically verify a theorem) about one specific finite case instead of infinite sequences. But I think that this particular one is an interesting test case for new ideas. Among other things, it robs you of the license to prove estimates with bad constants.
28 August, 2007 at 9:47 am
Anonymous
Are there more continuous versions of the type of Freiman’s theorem? For instance could there be a result along the lines that if A is a subset of the reals of full Hausdorff Dimension, then A-A contains an interval (or if it doesn’t, then A has to have a certain form)?
Dear Anonymous,
There are some versions of Freiman’s theorem for measurable subsets E of Euclidean space in which E+E or E-E has comparable Lebesgue measure to E, see e.g. Proposition 7.1 in my paper
http://front.math.ucdavis.edu/math.CO/0601431
This result is in fact deduced rather easily from discrete versions of Freiman’s theorem.
The question of dimensional versions of Freiman’s theorem is a very interesting one: given a set A of reals for which A+A or A-A has the same Hausdorff dimension as A, what can one conclude about A? Note that in your above example it is certainly possible for A-A to not contain an interval, for instance A could be the union of a nested sequence G_n of subgroups of R of Hausdorff dimension 1-1/n (it is not too difficult to construct subgroups of any specified dimension, and with a little more work one can make them nested). Part of the problem here is that the quantitative bounds for discrete Freiman theorems are still too poor to say anything meaningful in this regime. Indeed, a toy discrete model of the statement “dim(A+A)=dim(A)” for some continuous set A would be the statement that $|A+A| \leq C_\epsilon |A|^{1+\epsilon}$ for some finite but large set A of integers, where $\epsilon > 0$ is small (more precisely, one could consider a sequence $A_n$ of such sets with cardinality going to infinity, and a sequence of $\epsilon_n$ going to zero slowly). At present we have no fully satisfactory Freiman-type characterisation of these sets. However we do have the Plunnecke-Ruzsa sumset theory which is effective in this regime. For instance, for a finite set A with the above property we can also bound quantities such as $|A-A|$ and $|A+A+A|$ by $O( |A|^{1+O(\epsilon)} )$. Similarly, in the continuous world, if A is a (say) compact set of reals with dim(A+A)=dim(A), then one can show that dim(A+A+A)=dim(A) and dim(A-A) = dim(A) as well. A deep theorem of Bourgain (part of what was once called the “Katz-Tao conjecture”) also asserts that dim(A.A) > dim(A) in this case.
[...] groups“, submitted to Contrib. Disc. Math.. This paper concerns the problem (discussed in this earlier blog post) of determining the correct analogue of Freiman’s theorem in a general non-abelian group . [...]
13 September, 2009 at 2:33 am
Ehud Hrushovski
Dear Terry,
approximate subgroups in particular. I have no precise answers to any of them, but it turns out that analogous issues have been studied in model theory for some time, with dimension theories replacing finite cardinalities or orders of magnitude. I tried to make this analogy precise in a paper I posted on ArXiv, “stable group theory and approximate subgroups”, arXiv:0909.2190. One of the statements made there shows a close connection between approximate subgroups and Lie groups, and has something of the flavor of Ben Green’s words on “approximate representations” on this page. Another consequence is that approximate subgroups generating \$G(F_q)\$ for a large prime power \$q\$ must contain a positive proportion of the elements of \$G(F_q)\$, where \$G\$ is a fixed simple algebraic group. -Udi
[...] theorem, Isaac Goldbring, type | by Terence Tao One of my favorite open problems, which I have blogged about in the past, is that of establishing (or even correctly formulating) a non-commutative analogue of [...]
[...] theorem | by Terence Tao Let be a finite subset of a non-commutative group . As mentioned previously on this blog (as well as in the current logic reading seminar), there is some interest in classifying those [...]
[...] exactly? There is not yet a full consensus on what the list of structured objects should be (see this earlier post for more discussion), but things are well understood in the abelian case, at least. Examples of [...]
[...] of stating and proving this statement is the noncommutative Freiman theorem problem, discussed in these earlier blog [...]
[...] new proof of a “noncommutative Freiman” type theorem in linear groups . As discussed in earlier blog posts, a general question in additive (or multiplicative) combinatorics is to understand the [...]
[...] free groups, nilpotent groups, solvable groups, or simple groups of Lie type) are also known; see these previous blog posts for a summary of several of these [...]
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http://stats.stackexchange.com/questions/19455/application-of-hidden-markov-model-to-crm
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Application of Hidden Markov Model to CRM
What are some of the applications of a HMM in the marketing field - specifically CRM and targeted marketing?
For example, is it mainly for predicting an outcome given a sequence? Such as if web site visitor visits {Page 2, Page 54, page 23} what is the probability they will $<next>$ visit page 22? Or visit page 22 in that session?
Is HMM a stand alone technique or normally coupled with other things?
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CRM? (Yes, I know what it is. But, it helps to define such things within the question.) – cardinal Dec 7 '11 at 3:10
2 Answers
It sounds like you're looking for an n-order Markov model and not a hidden Markov model. Typically, an HMM would be useful if you wanted to identify hidden states at each selection in a series of selections. An n-order Markov model would give you the conditional probability of choosing the n+1'st state given the previous n states, which would be useful in the example you just described.
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DaRob: Do you have any tutorials to point towards for this model? – B_Miner Dec 7 '11 at 0:25
A google search for 'hidden markov model marketing' turned up this paper: A Hidden Markov Model of Customer Relationship Dynamics. Google scholar and the references in the above paper may turn up a few more applications of HMM in marketing.
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http://cstheory.stackexchange.com/questions/14523/algebraic-or-numeric-invariants-of-complexity-classes
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# Algebraic (or numeric) invariants of complexity classes
I hope this question isn't too naive for this site.
In mathematics (topology, geometry, algebra) it is common for one to distinguish between two objects by coming up with an algebraic or numerical invariant, and proving the two objects have different values. I'm wondering to what extent this has been tried with complexity classes (or if it has, why I haven't ever heard about it). Algebraic structures show up a lot in theoretical computer science as a whole (cf Uses of algebraic structures in theoretical computer science), so why not in complexity theory?
In my naivete I can imagine a notion of equivalence of two languages: the existence of a polynomial-time reduction which is also reversible (or a bijection on the strings). I can also imagine that this notion is unsuitable: no finite languages of differing cardinality could be considered equivalent, even though we're more often interested in infinite languages.
Are there any other, weaker notions of isomorphism of languages that have yielded interesting results? Are there other kinds of numerically-flavored invariants that have been used to distinguish complexity classes?
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– Sasho Nikolov Nov 28 '12 at 20:06
While this is not entirely helpful, my (vague) understanding is that Mulmuley's geometric complexity program rests essentially on such an algebraic argument. More usefully, his P vs NC proof uses a counting characterization of problems computable in NC to separate it from P. – Suresh Venkat♦ Nov 29 '12 at 4:53
1
@SashoNikolov: one issue with resource-bounded measure is that, for classes closed under finite variations (as are essentially all complexity classes we ever consider), if they are measurable they have measure either 0 or 1. In that sense, as a numerical invariant resource-bounded measure only gives you three possibilities: measure 0, measure 1, or unmeasurable. Resource-bounded dimension can provide finer distinctions... – Joshua Grochow Nov 29 '12 at 16:40
## 1 Answer
First of all, the relation you defined is usually called polynomial-time isomorphism ($\cong^p$). Although isomorphism is an interesting notion that has been studied, the (weaker) relation that is more frequently of concern in complexity is polynomial-time equivalence: $A$ and $B$ are equivalent ($A \equiv_m^p B$) if there are polynomial-time many-one reductions (aka Karp reductions) from $A$ to $B$ and vice versa, but those reductions need not be inverses to one another, and need not even have polynomial-time inverses. Sometimes we also care about equivalence under polynomial-time Turing reductions rather than many-one ($\equiv_T^p$), aka Cook reductions. For example, either of these notions of equivalence is "good enough" for $P$ vs $NP$ (that is, you don't need to consider isomorphism classes).
From the perspective of polynomial-time equivalence, there is partial "good reason" that you don't hear about numerical invariants: they can't work in general. A theorem in Andrew Marks's thesis states that $\equiv_T^p$ is complete for countable Borel equivalence relations (the introduction of his thesis gives a good overview of Borel equivalence relations and their significance). In particular, this implies that there is no Borel function $f: 2^{\mathbb{N}} \to \mathbb{R}$ such that $A \equiv_T^p B$ iff $f(A) = f(B)$.
The reason I say this is only partially a good reason is that there still might be a Borel function $f:2^{\mathbb{N}} \to \mathbb{R}$ such that if $f(A) \neq f(B)$ then $A \not\equiv_T^p B$. Or there might be a non-Borel function that does the job, but if there is we're a little unlikely to find it...
But classifying all equivalence classes is also stronger than what we usually care about, since the number of equivalence classes that show up as natural complexity classes is comparatively small (despite the forbidding size of the complexity zoo). However, there are other "numerical" invariants we can associate to languages. One such is their density: the density of a language $A$ is the function $d_A(n) :=$ number of strings in $A$ of length $\leq n$. Note that density is preserved, up to a polynomial change, by poly-time isomorphisms, but not necessarily by polynomial-time equivalences (e.g. all languages in $P$ are polynomial-time equivalent, but they can have wildly different densities).
We know things like: if $A$ is polynomially sparse ($d_A(n) \leq poly(n)$) then $A$ cannot be $NP$-complete unless $P=NP$ (Mahaney's Theorem). There are lots of other results about sparse languages and their relation to complexity classes. For good surveys, see Cai and Ogihara "Sparse Sets versus Complexity Classes" in Complexity Theory Retrospective II (available online - just Google) and Hemaspaandra and Glaßer's pair of articles "A Moment of Perfect Clarity I,II" in SIGACT News.
As mentioned by @SureshVenkat, you can kind of view Geometric Complexity Theory in the light you're talking about. However, the algebraic objects used there - namely representations - are more akin to general properties of a language than to numerical properties per se, but at least they are properties of an algebraic flavor.
Finally, in algebraic complexity theory one numerical property that's worth mentioning, but probably won't work to resolve the big questions, is degree. (As in the degree of a polynomial.) Strassen's degree bound is still the only known super-linear lower bound on unrestricted algebraic circuits. Degree also gets used e.g. in Razborov-Smolensky and many other areas of low-level (Boolean) circuit complexity.
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## Is 'charged black hole' an oxymoron?
Quote by DaleSpam Hi Q-reeus, I would not over-interpret the M and the Q as representing some particular mass or charge. I would think of them simply as parameters of the metric. The M term can include rest mass, energy (including energy in EM fields), pressure, stress, etc. And Q could be an E-field boundary condition at the edge of the manifold rather than some charged particles actually located in the manifold.
Interesting perspective there DaleSpam. A bit out of my league and maybe Peter is best one to comment further there. As you will be aware, my real focus is on logical reason for any external E for a BH. Cheers.
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Quote by Q-reeus The local perspective is not the only legitimate one
We agree on the actual observable results, so I can't say that this view is "wrong". But it seems to me that taking this view often leads you into confusion, because it keeps you from applying common sense reasoning to disentangle causes. You have objects that "look different" whey they are far away than they do when they are close up, but instead of taking the obvious route of attributing the differences to the effect of the spacetime in between, you insist on saying that some "intrinsic" property of the objects has changed. And this prevents you from adopting a simple method of distinguishing the two possibilities: look at what local measurements say about the objects in their new location.
A simple analogy: two people, A and B, are standing next to a cube, and both of them agree that it looks white. Now the cube is moved to the other side of the room, and both of them agree that it now looks red. A says that the cube must have "changed color"; B says no, something about the space between must be altering the light reflected from the cube, changing it from white to red. They both agree on how the cube looks, but they disagree on why.
In one sense, the difference between A and B is just a "difference in pov"; after all, they both agree on all the experimental results. But suppose they now ask their friend C, who is standing on the other side of the room next to the cube, what color the cube looks to him. C answers that it looks white. B says, "You see? The cube is still white, but something about the space between us is making its apparent color change." How can A respond? If he tries to claim that the cube somehow "really has" changed color, even though it looks white to C, the one standing right next to it, won't he seem foolish? Wouldn't it be more reasonable for A (and B) to look for something in the middle of the room that could be changing the color of the light from the cube--a large red filter screen, perhaps? They could even shine some white light from their end of the room and ask C how it looks to him, and find that it looks red. In short, they could apply standard scientific techniques to figure out the causes of what they observe.
You can see the analogy, I hope. Consider the hydrogen atom that's slowly lowered into the gravity well. An observer, C, right next to it will not be able to ionize it with visible light; he will find its ionization energy to be exactly what it was when it was far away from all gravitating bodies and that energy was measured locally. Now observers A and B, at a much higher altitude, emit visible light and find that it ionizes the atom. A claims that the atom has somehow "weakened"; but B says no, it must be something about the spacetime between that is altering the light. After all, C finds the ionization energy to be the same as always. Furthermore, if they ask C how the "visible" light they are shining down looks to him, he will say it looks like gamma radiation; so obviously something about the spacetime between is changing the light. This is all standard scientific reasoning, but of course if you refuse to avail yourself of it, you will continue to be confused by these types of scenarios.
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Quote by Q-reeus Interesting perspective there DaleSpam. A bit out of my league and maybe Peter is best one to comment further there.
I would agree with what DaleSpam said. It is true that you can measure M and Q by looking at the behavior of test objects at very large radii, as I described in an earlier post; but doing that does not commit you to any particular belief about "where" the "mass" or "charge" being measured is located. It just means you've measured global properties of the spacetime.
Quote by jartsa Nobody seems to know how the total energy of a falling rock changes. Obviously that energy that is responsible of car crushing does increase.
What do you mean by that? Explanations come in layers--you explain phenomena at one level by showing how it works in terms of a more fundamental description. Sometimes that more fundamental description itself can be explained in terms of even more fundamental laws, but sometimes it can't.
We know how the energy of a rock changes as it falls, in the sense that it is perfectly described by General Relativity. We don't know, in any deep sense, why GR is true.
Light is more simple: When light enters a gravity field of a planet, the light slows down and the planet starts to move in the same direction as the light. So light loses energy when falling. Or as an extremely good aproximation the energy stays the same.
No, that's a very poor explanation, in my opinion. The General Relativity explanation, as I suggested, is in terms of parallel transport of vectors, and that applies both to a rock falling and a photon falling.
There was a long and technical discussion in this forum about the energy change of falling light, and those were the conclusion.
That's a very misleading conclusion.
So it must be the objects becoming weaker at lower altitude, which causes them to break more easily, when light from above hits them.
That's not what General Relativity says. What you're doing is taking the predictions of GR, and reinterpreting them according to a different model. There certainly can be two physical models that produce the same predictions (an example is Special Relativity and certain variants of aether theory), but talking in terms of forces getting weaker in a gravitational field is completely contrary to the General Relativity way of understanding gravity.
Curved spacetime can be thought of as taking a bunch of little regions of flat spacetime and "gluing" them together on the edges. Inside each little region, the laws of physics work almost exactly the same as they would in gravity-free space. The "curvature" comes in when you try to glue neighboring regions together. The vector corresponding to a slow-velocity rock or a low-energy photon in one region becomes a high-velocity rock or high-energy photon in another region.
It's exactly like trying to describe the surface of the Earth using flat maps. If each map only covers a small region of the Earth, say 100 x 100 kilometers, then you don't notice the curvature. But when you try to glue one map together with a second map, you will find that vectors don't precisely match up. A vector that is vertical on one map corresponds to a vector that is slightly tilted from parallel on the second map.
Hey I have one more scenario again: A charge in a gravity well is accelerated from 0 m/s to 100 m/s. Radiation energy is proportional to velocity change. As seen from higher altitude the velocity change was smaller than 100 m/s, and there is the reason why the radiation energy coming from the gravity well is smaller too.
I don't think that's a very good description at all. To go from "the light from the event is red-shifted" to "the velocity change must have been slow" is very dubious.
Imagine light from some event (say, a charged particle accelerating) comes to you from two different directions; for example, suppose there is a very massive star, or black hole between the event and you, and the light can go around in one direction, or the other. When the light gets to you, the two images can have different amounts of redshift. You can't explain that in terms of weaker or stronger electrical forces down in gravitational well. The way to explain it is to realize that light has to travel from the event to your eyes. Depending on the path it takes, the light is changed by its journey.
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Quote by Q-reeus A slab of dielectric lies immersed in an E field of a parallel plate capacitor. Both dielectric and capacitor are stationary in lab frame S', and E' there is below dielectric breakdown value. Now propel that slab to a relativistic velocity in S', normal to direction of E' such that in proper frame S of slab, E exceeds breakdown value and there is catastrophic failure and discharge. It is not legitimate to say that in S' the slab can be viewed as having 'weakened', since E has not changed in S'?
This is a different type of scenario: the large relative motion between capacitor and dielectric is a local effect, not an "at a distance" effect. There is no "spacetime in between"; everything happens locally. Even if you make the capacitor and the dielectric each a light-year long, the local field at any point in the dielectric, in the dielectric's rest frame, can still be attributed to the "local" part of the capacitor; it doesn't have to be transmitted over a distance.
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Quote by jartsa Nobody seems to know how the total energy of a falling rock changes.
Maybe you don't know, but that doesn't mean nobody knows. If the rock is freely falling, then its total energy is constant; as it falls, it gains kinetic energy and loses potential energy, but the sum of the two remains constant. (Strictly speaking, this is only true in certain spacetimes, such as Schwarzschild spacetime, but that's general enough to cover what we've been discussing.)
Quote by jartsa Obviously that energy that is responsible of car crushing does increase.
Yes, that's the kinetic energy. The potential energy decreases by the same amount.
Quote by jartsa Light is more simple: When light enters a gravity field of a planet, the light slows down and the planet starts to move in the same direction as the light.
If this happens with light, wouldn't it also happen with the falling rock? Why would the two be different?
One could adopt a viewpoint in which both the rock/light and the planet moved; this is just the center of mass frame. But the planet is so much more massive than either the rock or the light that the center of mass frame is not measurably different from the frame in which the planet is at rest. So that's the conventional approximation. The description I gave of the rock's motion above was in that approximation; and in that approximation, the light also "falls" towards the planet in such a way that its total energy, kinetic plus potential, remains constant. A light ray's kinetic energy is proportional to its frequency, so as the light falls, it blueshifts; or if it's rising, climbing out of a gravity well, it redshifts.
Quote by jartsa So light loses energy when falling. Or as an extremely good aproximation the energy stays the same.
The total energy, kinetic plus potential, stays the same. You have the kinetic energy backwards: it increases when falling and decreases when rising, just as a rock's does.
Quote by jartsa There was a long and technical discussion in this forum about the energy change of falling light, and those were the conclusion.
Do you have a reference? I think you must have misinterpreted something.
Quote by PeterDonis Do you have a reference? I think you must have misinterpreted something.
Here it is:
http://physicsforums.com/showthread.php?t=588937
Quote by Q-reeus I'm ignoring your earlier posting #37 and later one in #41 (but accept the nice sentiments in #39), as it all hinges on getting right what you say here. Maybe you have already changed again - I say 'again' because in #22 there is "I was mistaken in my first response--allowing a dropping mass to do work does change the total mass/energy."
I agree that if you get work out of lowering a mass toward a black, then the total energy of the system (the black hole + the mass you are lowering) will be less than if you don't extract work from it. It's not correct to describe this as "mass is reduced by the redshift formula"--it's that expression that seems completely wrong.
At an informal level, the total energy of a black hole is equal to the energy you dropped into it, minus the energy you pulled out of it. The total charge of a black hole is equal to the charge you dropped into it.
- which was a seeming backflip from your earlier position, and amazingly you seem to have done another 180 and it's back to the original stance.
I would say that it's a matter of trying to understand what you're saying. I would not say that dropping an electron into a black hole changes the mass of the electron. That's just a weird thing to say. Dropping an electron into a black hole changes the mass of the entire system, black hole + electron. Exactly how much the mass of the system is changed depends on how you lower the electron into the black hole. But I would not say that the mass of the electron changed. The definition of the mass of an electron is the total energy of the electron as measured in a local inertial frame in which the electron is at rest. That is not changed by lowering it into a black hole.
[Stop press: just now read your #46. Evidently then you believe the energy gain/loss is to be considered a wholly delocalized affair - shared amongst the entire system more or less equally?]
Yes, in general, energy is a function of an entire system. Mass is the energy as a measured in a frame in which the system is at rest (has zero total momentum).
Then consider the following: Suppose that mass/energy in a slowly lowered mass m is in the form of an unstable matter/antimatter doublet that self-annihilates and escapes entirely 'to infinity' as gamma rays. We surely agree that in escaping the gravitational potential well of central mass M, those rays - which carry all the energy tied up originally in m(r), are redshifted in coordinate measure.
Yes. The way I would say it is that the particle/antiparticle pair annihilate to produce a pair of photons with a characteristic frequency, as measured in the local rest frame of the pair. These photons then travel up out of their gravitational well and escape to infinite. Their frequencies are changed by their journey (according to the redshift formula).
Annihilate the same matter/antimatter doublet out there in distant space, and obviously the gamma rays are not redshifted at all.
Right.
This little experiment of the mind imo nicely indicates it is proper to consider the energy loss/gain in lowering/raising matter/energy of mass m(r) in a grav well as essentially confined to just that mass m(r) - provided m(r) << M.
No, I don't agree. When two particles annihilate, they produce two photons that go off in opposite directions. These two photons could have their paths warped by the gravity of a massive star or black hole, and then come back together. When they come back together, the two frequencies need not be the same. It doesn't make sense to attribute the difference in frequency to differences in the masses of the particles that produced the photons. Instead, the differences should be understood as a change in the momenta of the photons as they travel from where they are produced to where they are measured.
Unless that is one wishes to argue transporting matter/energy from a resting position at one potential to a resting position at another potential can be anything but a path independent process - assuming of course central mass M is taken as static.
Parallel transport of vectors is certainly path-dependent. In general, you can't compute redshift by noting where the photons started from, you have to take into account the path taken by the photons.
Re your's and to some extent PeterDonis's criticism of jartsa over his description of atoms being 'weaker' further down. The local perspective is not the only legitimate one and from a coordinate viewpoint I would agree with his thrust.
You can certainly describe things using whatever coordinates you like, but you have to be careful not to attribute physical effects to artifacts of your choice of coordinates.
There is something similar in SR. A slab of dielectric lies immersed in an E field of a parallel plate capacitor. Both dielectric and capacitor are stationary in lab frame S', and E' there is below dielectric breakdown value. Now propel that slab to a relativistic velocity in S', normal to direction of E' such that in proper frame S of slab, E exceeds breakdown value and there is catastrophic failure and discharge. It is not legitimate to say that in S' the slab can be viewed as having 'weakened', since E has not changed in S'?
The relevant value of E is E as measured in the frame in which the dielectric is at rest.
Notice too that in jartsa's scenario if an atom spontaneously decays radioactively releasing a gamma ray, it is redshifted as seen from outside of potential well. That can legitimately qualify it as a 'weaker' atom with weaker internal EM and nuclear fields in my book - it's all a matter of pov.
That's a very bad way of looking at it, in my opinion. The insight that Einstein formulated as the equivalence principle is that in any small region of spacetime, most phenomena--the ticking of clocks, the decay of particles, etc.--work exactly the same as they would in the absence of gravity. The complexity comes in when you try to relate phenomena in one region of spacetime to phenomena in another region. That's where the technical tool of "parallel transport" comes into play.
Quote by PeterDonis You can see the analogy, I hope. Consider the hydrogen atom that's slowly lowered into the gravity well. An observer, C, right next to it will not be able to ionize it with visible light; he will find its ionization energy to be exactly what it was when it was far away from all gravitating bodies and that energy was measured locally. Now observers A and B, at a much higher altitude, emit visible light and find that it ionizes the atom. A claims that the atom has somehow "weakened"; but B says no, it must be something about the spacetime between that is altering the light. After all, C finds the ionization energy to be the same as always. Furthermore, if they ask C how the "visible" light they are shining down looks to him, he will say it looks like gamma radiation; so obviously something about the spacetime between is changing the light. This is all standard scientific reasoning, but of course if you refuse to avail yourself of it, you will continue to be confused by these types of scenarios.
Are you picking on me now Peter? We both agreed earlier that there is legitimately a reduction in mass/energy when matter is lowered into a potential well. And it shows remotely - the gravitational contribution felt 'out there' is not that owing to plugging in the locally observed proper mass but the redshifted coordinate value. Agreed?
Quote by PeterDonis This is a different type of scenario: the large relative motion between capacitor and dielectric is a local effect, not an "at a distance" effect. There is no "spacetime in between"; everything happens locally. Even if you make the capacitor and the dielectric each a light-year long, the local field at any point in the dielectric, in the dielectric's rest frame, can still be attributed to the "local" part of the capacitor; it doesn't have to be transmitted over a distance.
Actually there are other factors to consider in that case and I jumped in too soon (dielectric polarization field transformation, plus effect of motion through the induced magnetic field in S'. There is cancellation but it gets messy). A cleaner example would be: same capacitor and field E' in S', but now there is a charge q lying at the end of a cantilever arm - oriented mutually orthogonal to both E' and later applied relative velocity v. We suppose initially the arm, at rest in S' barely resists the force on q from E'. Now set it in relative motion such that arm breaks under higher E in proper frame S of arm+charge. In S' we attribute no higher force on q owing to v. How to explain breakage? To say the material has relativistically weakened seems about ok to me. There is imo this much commonality with gravitational case - situation is viewed from differing proper frames in both cases.
Quote by Q-reeus Are you picking on me now Peter? We both agreed earlier that there is legitimately a reduction in mass/energy when matter is lowered into a potential well. And it shows remotely - the gravitational contribution felt 'out there' is not that owing to plugging in the locally observed proper mass but the redshifted coordinate value. Agreed?
I don't think it's appropriate to talk about mass being redshifted. You can talk about the frequency of a photon being redshifted, and maybe that's what you mean?
Quote by Q-reeus Actually there are other factors to consider in that case and I jumped in too soon (dielectric polarization field transformation, plus effect of motion through the induced magnetic field in S'. There is cancellation but it gets messy). A cleaner example would be: same capacitor and field E' in S', but now there is a charge q lying at the end of a cantilever arm - oriented mutually orthogonal to both E' and later applied relative velocity v. We suppose initially the arm, at rest in S' barely resists the force on q from E'. Now set it in relative motion such that arm breaks under higher E in proper frame S of arm+charge. In S' we attribute no higher force on q owing to v. How to explain breakage? To say the material has relativistically weakened seems about ok to me.
That seems like a very bad way of looking at it, in my opinion.
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Quote by jartsa
I've read that thread, and I don't see how you got the following out of it:
Quote by jartsa When light enters a gravity field of a planet, the light slows down and the planet starts to move in the same direction as the light. So light loses energy when falling. Or as an extremely good aproximation the energy stays the same.
There was no talk at all in that thread about the planet moving in the same direction as the light. You said the light losing energy when falling was your own idea, and indeed that's not in the thread either. Nor is the energy staying the same "to an extremely good approximation"; Jonathan Scott said that the frequency of the light stays the same, but he meant something different by "frequency", and he specified that that was relative to a particular coordinate system, the global Schwarzschild coordinates.
Let me try to summarize what the technical posts in that thread were actually saying. A photon emitted at a particular event in any spacetime will have a 4-momentum vector $k^{a}$ associated with it. Since photons travel on null geodesics, that 4-momentum vector will be parallel transported along the photon's worldline; this is the sense in which the photon "does not change" as it travels.
However, the observables associated with the photon are determined, not just by the photon's 4-momentum, but by geometric objects, vectors and tensors, associated with the observer. For example, the energy the photon is measured to have by that observer is the contraction of the photon's 4-momentum with the observer's 4-velocity $u^{b}$:
$$E = g_{ab} k^{a} u^{b}$$
So even if $k^{a}$ is unchanged as the photon travels, its observed energy can still change if either the metric $g_{ab}$ or the observers' 4-velocity $u^{b}$ changes. (We actually measure photon frequency, not energy, but the latter is just Planck's constant times the former.)
In the case of the standard Doppler shift, the measured energy (frequency) changes because the 4-velocity of the observer $u^{b}$ changes relative to that of the emitter, which determines the photon's 4-momentum $k^{a}$.
In the case of a photon falling into or climbing out of a gravity well, the energy (frequency) measured by static observers--observers who are "hovering" at a constant radius r--changes with r because the metric $g_{ab}$ changes. (The 4-velocity of "hovering" observers is the same at all r--all their 4-velocity vectors point "in the same direction".)
The "frequency staying the same" that Jonathan Scott was talking about was a different sense of "frequency": if I have a blinker, say, emitting flashes of light deep in a gravity well, such that it emits N flashes between Schwarzschild coordinate times t = 0 and t = 1, then an observer much higher up in the gravity well will also count N flashes between coordinate times t = 0 and t = 1. The two observers will differ in how much *proper* time they experience between those two coordinate times, so they will assign a different proper frequency (flashes per second of proper time) to the blinker; but the frequency relative to *coordinate* time is the same. This is all consistent with what I said above.
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Quote by Q-reeus We both agreed earlier that there is legitimately a reduction in mass/energy when matter is lowered into a potential well.
I agreed that there was a reduction in *energy at infinity*, because you are extracting work during the lowering process. But if you measure the rest mass of the object locally, you will get the same answer after it has been lowered as before.
Quote by Q-reeus And it shows remotely - the gravitational contribution felt 'out there' is not that owing to plugging in the locally observed proper mass but the redshifted coordinate value. Agreed?
Yes, but that doesn't mean what you think it means.
Quote by stevendaryl I don't think it's appropriate to talk about mass being redshifted. You can talk about the frequency of a photon being redshifted, and maybe that's what you mean?
Taking on board your comments in #59 and what I just read in #63, it seems we just have fundamentally different outlooks in matters discussed and best to just agree to disagree methinks. Have a nice day.
Quote by Q-reeus Taking on board your comments in #59 and what I just read in #63, it seems we just have fundamentally different outlooks in matters discussed and best to just agree to disagree methinks. Have a nice day.
Okay, but I think that you are deeply confused about these topics, and you are interpreting your confusion as a contradiction in General Relativity. The contradiction is in your own head.
Peter - where did your last post go - I so looked forward to saying 'gotcha'!
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http://cstheory.stackexchange.com/questions/6563/what-is-the-computational-complexity-of-solving-chess
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# What is the computational complexity of “solving” chess?
The basic idea of backwards induction is to start with all the possible final positions of a game in which player X wins. So for chess, look at all the ways White can checkmate Black. Now work backwards to all the possible moves/positions that would allow White to move in to one of those positions. If White ever found herself in such a position she could win by moving to the relevant checkmating move. Now we work backwards another step and so on. Eventually we get back to all the possible first moves White could make. The point is, once we've done this, we know that we have White's best response to any move Black makes.
Recently (last five years or so) Checkers was "solved" in this way. Obviously Noughts and Crosses (what the colonials might call "Tic-Tac-Toe") has been solved for ages. At the very least since this xkcd but presumably long before.
So the question is: what factors does this sort of procedure depend on? The number of possible legal positions, presumably. But also perhaps the number of legal moves at any given node... And given this, how complex is this sort of problem?
Bonus question: how long before a \$2000 PC can solve checkers in a day? Chess? Go? (Of course for this you also have to take into account increasing speed of home computers...)
I've added the graph-algorithms tag because you can represent these games as trees, but if I'm abusing the tag please add something more appropriate
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Questions about the complexity of chess have been rehashed in many places. The game tree has finite length, modulo the precise rules governing draws, but the game tree has very high branching factor so brute force solutions seem out of reach. To put it bluntly: the game tree has size something like $10^k$, for $k$ half-moves. I don't see a TCS question here? – András Salamon May 13 '11 at 18:14
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Is it cheeky to point out that the complexity is $O(1)$ by the fifty move rule? – Joe Fitzsimons May 13 '11 at 21:32
Seamus, cstheory's scope is research-level questions in tcs, please read the FAQ. You can use math.SE for non-research-level questions. – Kaveh♦ May 15 '11 at 7:19
– vzn May 9 at 21:47
## 4 Answers
As @Joe points out, chess is trivial to solve in $O(1)$ time using a lookup table. (An actual implementation of this algorithm would require a universe significantly larger than the one we live in, but this is a site for theoretical computer science. The size of the constant is irrelevant.)
There is obviously no canonical $n\times n$ generalization of chess, but several variants have been considered; their complexity depends on how the rules about moves without captures and repeating positions are generalized.
If a draw is declared after a polynomial number of capture-free moves, or after any position repeats a polynomial number of times, then any $n\times n$ chess game ends after a polynomial number of moves, so the problem is clearly in PSPACE. Storer proved that this variant is PSPACE-hard.
For the variant with no limits on repeated positions or capture-free moves, the number of legal $n\times n$ chess positions is exponential in $n$, so the problem is clearly in EXPTIME. Fraenkel and Lichtenstein proved that this variant is EXPTIME-hard.
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This probably isn't a terribly useful answer, but I think it is worth pointing out that chess has a maximum number of moves, and hence there is a finite number of possible games. The fifty move rule allows either player to claim a draw if 50 or more moves take place without movement of a pawn. We can reasonably assume that this is always used, since if there is any objective measure of the strength of each players positions then the weaker one will claim the draw. Further, the rules of chess require that whenever a pawn is moved it advances one square towards the opponents side of the board (whether moving directly forward, or taking diagonally), and hence each pawn can move at most 6 times. As there are 16 pawns in total, this puts the maximum number of moves at $50\times (16 \times6 + 1) + 1 = 4851$. In each move, the player moves one of at most 16 pieces. For a pawn there are at most 3 moves, 14 for a rook, 8 for a knight, 14 for a bishop, 28 for a queen and 8 for a king, for a total of 132 possible moves. This gives an upper bound of $132^{4851}$ on the total number of chess games. So, while this is a truely enormous number (approx $2^{34172}$), it does mean that the complexity is trivially $O(1)$. On the other hand, with such a naive approach would take approximately fifty thousand years for the problem to become tractable, assuming Moore's law continued indefinitely.
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Minor quibble: the maximum number of moves is more like 50 * (16 * 6 + 1) + 1. But then 16 * 6 of those are pawn moves, 16 of which are potentially drawn from 4 options, although one of those options cuts down the number of moves later... (Not to disagree with the basic point, which is sound. Curiously your estimate comes out at around 10^10^5, whereas Mathworld says that Hardy estimated it as 10^10^50. Wonder whether that's a bad transcription of his notes). – Peter Taylor May 15 '11 at 8:01
@Peter My mistake, sorry. I've corrected the answer now. I'm not sure whether this was how the estimate in Mathworld was arrived at. If it simply counted legal arrangements of the board (ignoring the fifty move rule), it may have been much bigger. – Joe Fitzsimons May 15 '11 at 19:25
Another quibble: you used an incorrect version of the fifty move rule. It should be "50 moves without a pawn move or a capture". The number of pieces on the board is of course monotonically decreasing during a game and the rule still allows a finite bound. Although the game is still finite without the fifty moves rule by the threefold repetition rule, which then yields a much larger bound. – Olaf May 9 at 6:52
Ah, yes. That increases the maximum number of moves by about a third. – Joe Fitzsimons May 10 at 6:58
You missed something about 50 move: after 50 move without pawn move and without any killed item, we have draw (the power of your calculation dramatically grows but still is constant). – Saeed 6 hours ago
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There are actually a couple of different questions here: (a) how much computing power does it take to do tree search for games, and (b) what's the computational complexity of these problems? The best all-purpose resource for this sort of thing is probably the Wikipedia page on Game Complexity, but to go into a bit more detail:
For (a), there are a lot of different practical algorithms that come into play, but they all boil down to some form of the tree search you noted; the biggest optimization that's generally used for the tree search itself is known as Alpha-Beta, which prunes branches of the tree once it's known that they can't be better than the best option already discovered. This is useful for evaluating positions 'on the fly' for chess (particularly with smart heuristics for ordering moves), because there are good estimates of the 'value' of a position; it generally gets a lot worse when having to compute a precise result for a position just because those heuristics don't hold. In general, if the tree has depth $d$ and a branching factor of $b$, then alpha-beta pruning cuts the number of nodes that need to be examined to roughly $b^{d/2}$ (from the naive value of $b^d$) - but even with this optimization, that's obviously a huge factor; consider that for the opening position of chess, $d$ is on the order of 60-100, and the branching factor $b$ is estimated to be in the range of 30-40.
In practice, pure tree search is supplemented by a bottom-up dictionary; for instance, the results of all 6-piece chess endgames are known, and many 7-piece endgames have been analyzed (see http://en.wikipedia.org/wiki/Endgame_tablebase), so the result of a game branch can be looked up in the 'dictionary' (a huge database of positions) once the position has been reduced to few enough pieces, shortcutting a lot of extra tree search that would otherwise be needed. This is what was done with checkers - databases were built up of all endgames with sufficiently few pieces, then extended to add more pieces and more, until the results of all 10-piece endgames were known; then tree search was used from the initial position, and essentially the two met in the middle.
Beyond these practical approaches, though, there's the (b) side of the question: what is the computational complexity of these sorts of problems? Abstractly, most problems of this ilk tend to fall into a couple of categories; they're either PSPACE-complete - which roughly means 'if you can solve this, you can solve any problem that takes polynomially much space' - or EXPTIME-complete (which roughly means 'if you can solve this, you can solve any problem that takes exponentially much time'), depending on how long the game can last; again, the Wikipedia page on EXPTIME-completeness has a pretty good discussion of the issues involved and what differentiates different games on this front.
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These estimates are way too high.
You're focused on branching based on legal moves. This makes lots of sense if you are trying to make a fast chess computer, but it's not how you would write a program to "solve" chess.
There <<<<< 13^64 possible game states in chess. Each square can only contain one of the chess pieces or nothing. You can iterate though them all and mark them as black win or white win in no more than 2^256 or so operations.
A more realistic guess of the number of reasonably achievable game states is around 2^100
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2
Because of the rule that repeating a board configuration three times is a draw, the state of the game does not include only the current positions of the pieces, but all past board configurations as well. But whatever; a constant is a constant. – JɛffE May 9 at 5:16
It also needs castling rights, but it only needs the past 100 board configurations. $\hspace{.2 in}$ – Ricky Demer May 10 at 1:30
You seem to be sweeping a lot of the complexity into "and mark them as black win or white win". – Joe Fitzsimons May 13 at 9:24
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http://mathoverflow.net/questions/2630/infinite-subgroups-with-finite-index/2637
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## Infinite subgroups with finite index
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Is there a general method to prove that an infinite subgroup of a group has finite index? Or, in other words, to prove that the quotient group is finite? I am particularly interested in classical groups, such as GL(n), SL(n), etc, over a nonarchimedean local field but I am looking for a general method, if there exists one.
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## 6 Answers
This is a somewhat tautological answer, but: if you can show that the subgroup contains the kernel of a finite representation (i.e. a homomorphism to a finite group), you're done. Intuitively: "I only need a finite number of things to go my way in order to belong to this subgroup."
If the group (or some representation of that group) is compact in some topology, and the subgroup contains the connected component of the identity (or an open neighbourhood of the identity), you're also done.
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
For this type of problem, "general methods" tend to be quite general indeed, but here are two ideas:
1) Find an action of $G$ on a finite set $X$ and an element $x \in X$ such that $H$ is the stabilizer of $x$. Then, by the Orbit-Stabilizer Theorem, $G/H$ is isomorphic to the orbit space $Gx$, so is finite.
2) Find a finite group $X$ and a homomorphism $f: G \rightarrow X$ such that $H$ contains the kernel of $f$. Then $f: G/\operatorname{ker}(f) \hookrightarrow X$, so $\operatorname{ker}(f)$ has finite index, so $H$, which contains $\operatorname{ker}(f)$, has finite index.
Note that both of these will, in principle, always work. In Case 1, take $X = G/H$. In Case 2, let $H' = \bigcap_{g \in G} gHg^{-1}$ be the normal core of $H$. It is easy to show that (since $H$ has finite index), $H'$ is a finite index normal subgroup of $G$ which is contained in $H$. Take $X = G/H'$ and $f$ to be the quotient map.
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I'm not sure if you're interested in classical groups over Q_p or over Z_p; in the latter case, one can often show that a closed subgroup H of G is in fact the whole group G once you know it projects surjectively onto some explicit finite quotient of G. See, for instance, [IV.3.4, Lemma 3] of Serre's "Abelian l-adic representations and elliptic curves." I explain how this works for GL_n (this is definitely not original to me, I just wanted the paper to be self-contained) in Lemma 3 of this paper about K3 surfaces.
Of course this won't work unless you know for some reason that H is closed; for instance, two random elements of GL_2(Z_p) will (I think) almost always generate a (discrete) free subgroup of GL_2(Z_p) which is dense, i.e. which projects surjectively onto every finite quotient, but which is obviously not finite-index.
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This may be tangentially related: a subgroup H of a nilpotent group G is in fact the whole group if it projects surjectively onto the abelianisation of G. – Terry Tao Oct 26 2009 at 16:49
Good point; similarly, if H and G are finitely generated pro-p groups, all you need is for H to surject onto the finite quotient G^{ab} / p G^{ab}. – JSE Oct 26 2009 at 17:05
From a computational perspective, you could try coset enumeration (or Knuth-Bendix) over the candidate subgroup (starting with generators given in terms of a known presentation of the overgroup). If it terminates, then the subgroup has finite index.
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I often try to show that a subgroup contains another subgroup known to be of finite index. There are various refinements. One technique is to prove that your subgroup contains an intersection of a finite number of subgroups, each of which is known to have finite index. For example, suppose you know that the central quotient G/Z(G) of a group G is finite. (One way to show this is to show that G is a union of finitely many abelian subgroups.) If it is too hard to show that your subgroup H contains the centre, but you can show that H contains the intersection of Z(G) with the derived subgroup [G,G] of G, then the index of H in G is finite.
Another idea is to try to show that (e.g.) every nilpotent quotient of your group is finite, and then show that your subgroup must contain some term of the lower central series. (You could replace nilpotent with soluble and the LCS with the derived series, and so on.)
This may not count as a general method; perhaps it is more of a "trick", but I've seen it used to good effect in proving some commutativity theorems for groups and rings. If you can show that your subgroup is a union of (at most) two subgroups known to have finite index, then you are done.
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You could try computing the virtual cohomological dimension of the ambient group, and then the subgroup. If they're torsion free, you only need to know the cohomological dimensions, as a theorem of Serre tells you that a torsion free group and its finite index subgroups have the same dimension. (You'd be surprised how often this works.)
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I guess that's more of a "show it has infinite index" test, but still useful. – Richard Kent Nov 4 2009 at 0:47
On the other hand, in a lot of settings, infinite index subgroups will have cd smaller than the ambient group. – Richard Kent Nov 4 2009 at 0:53
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http://gilkalai.wordpress.com/2008/06/12/billerafest/?like=1&source=post_flair&_wpnonce=966197e62c
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Gil Kalai’s blog
## Billerafest
Posted on June 12, 2008 by
I am unable to attend the conference taking place now at Cornell, but I send my warmest greetings to Lou from Jerusalem. The titles and abstracts of the lectures can be found here. Let me tell you about two theorems by Lou.
The first is the famous g-theorem: The g-theorem is a complete description of f-vectors (= vecors of face numbers) of simplicial d-polytopes. This characterization was proposed by Peter McMullen in 1970, and it was settled in two works. Billera and Carl Lee proved the sufficiency part of McMullen’s conjecture, namely for every sequence of numbers which satisfies McMullen’c conjecture they constructed a simplicial d-polytope P whose f-vector is the given sequence. Richard Stanley proved the necessity part based on the hard Lefschetz theorem in algebraic geometry. The assertion of the g-conjecture (the necessity part) for triangulations of spheres is open, and this is probably the one single problem I spent the most time on trying to solve.
The second theorem is a beautiful theorem by Margaret Bayer and Billera. Consider general d-polytopes. for a set $S \subset$ {0,1,2,…,d-1}, $S=${ $i_1,i_2,\dots,i_k$} , $i_1<i_2< \dots$, define the flag number $f_S$ as the number of chains of faces $F_1 \subset F_2 \subset \dots F_k$, where $\dim F_j=i_j$. Bayer and Billera proved that the affine dimension of flag numbers of d-polytopes is $c_d-1$ where $c_d$ is the dth Fibonacci number. ($c_1=1$, $c_2=2$, $c_3=3$, $c_4=5$, etc.) The harder part of this theorem was to construct $c_d$ d-polytopes whose sequences of flag numbers are affinely independent. The construction is simple: It is based on polytopes expressed by words of the form PBBPBPBBBPBP where you start with a point, and P stands for “take a pyramid” and B stands for “take a bipyramid.” And the word starts with a P (to the left) and has no two consecutive B’s.
Let’s practice the notions of f-vectors and flag vectors on the 24-cell . (The figure is a 3-dimensional projection into one of the facets of the polytope.)
This 4-polytope has 24 octahedral facets. It is self dual.
So $f_0=f_3=24$. And $f_1=f_2=96$. And $f_{02}=288$, $f_{03}=192$, etc.
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http://www.nag.com/numeric/CL/nagdoc_cl23/html/S/s17adc.html
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# NAG Library Function Documentnag_bessel_y1 (s17adc)
## 1 Purpose
nag_bessel_y1 (s17adc) returns the value of the Bessel function ${Y}_{1}\left(x\right)$.
## 2 Specification
#include <nag.h>
#include <nags.h>
double nag_bessel_y1 (double x, NagError *fail)
## 3 Description
nag_bessel_y1 (s17adc) evaluates the Bessel function of the second kind, ${Y}_{1}$, $x>0$.
The approximation is based on Chebyshev expansions.
For $x$ near zero, ${Y}_{1}\left(x\right)\simeq -2/\pi x$. This approximation is used when $x$ is sufficiently small for the result to be correct to machine precision. For extremely small $x$, there is a danger of overflow in calculating $-2/\pi x$ and for such arguments the function will fail.
For very large $x$, it becomes impossible to provide results with any reasonable accuracy (see Section 8), hence the function fails. Such arguments contain insufficient information to determine the phase of oscillation of ${Y}_{1}\left(x\right)$, only the amplitude, $\sqrt{2/\pi x}$, can be determined and this is returned. The range for which this occurs is roughly related to machine precision; the function will fail if $x\gtrsim 1$/ machine precision.
## 4 References
Abramowitz M and Stegun I A (1972) Handbook of Mathematical Functions (3rd Edition) Dover Publications
Clenshaw C W (1962) Chebyshev Series for Mathematical Functions Mathematical tables HMSO
## 5 Arguments
1: x – doubleInput
On entry: the argument $x$ of the function.
Constraint: ${\mathbf{x}}>0.0$.
2: fail – NagError *Input/Output
The NAG error argument (see Section 3.6 in the Essential Introduction).
## 6 Error Indicators and Warnings
NE_REAL_ARG_GT
On entry, ${\mathbf{x}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{x}}\le 〈\mathit{\text{value}}〉$.
x is too large, the function returns the amplitude of the ${Y}_{1}$ oscillation, $\sqrt{2/\pi x}$.
NE_REAL_ARG_LE
On entry, x must not be less than or equal to 0.0: ${\mathbf{x}}=〈\mathit{\text{value}}〉$.
${Y}_{1}$ is undefined, the function returns zero.
NE_REAL_ARG_TOO_SMALL
On entry, x must be greater than $〈\mathit{\text{value}}〉$: ${\mathbf{x}}=〈\mathit{\text{value}}〉$.
x is too close to zero, there is a danger of overflow, the function returns the value of ${Y}_{1}\left(x\right)$ at the smallest valid argument.
## 7 Accuracy
Let $\delta $ be the relative error in the argument and $E$ be the absolute error in the result. (Since ${Y}_{1}\left(x\right)$ oscillates about zero, absolute error and not relative error is significant, except for very small $x$.)
If $\delta $ is somewhat larger than the machine precision (e.g., if $\delta $ is due to data errors etc.), then $E$ and $\delta $ are approximately related by: $E\simeq \left|{xY}_{0}\left(x\right)-{Y}_{1}\left(x\right)\right|\delta $ (provided $E$ is also within machine bounds).
However, if $\delta $ is of the same order as machine precision, then rounding errors could make $E$ slightly larger than the above relation predicts.
For very small $x$, absolute error becomes large, but the relative error in the result is of the same order as $\delta $.
For very large $x$, the above relation ceases to apply. In this region, ${Y}_{1}\left(x\right)\simeq 2\mathrm{sin}\left(x-3\pi /4\right)/\pi x$. The amplitude $2/\pi x$ can be calculated with reasonable accuracy for all $x$, but $\mathrm{sin}\left(x-3\pi /4\right)$ cannot. If $x-3\pi /4$ is written as $2N\pi +\theta $ where $N$ is an integer and $0\le \theta <2\pi $, then $\mathrm{sin}\left(x-3\pi /4\right)$ is determined by $\theta $ only. If $x>{\delta }^{-1}$, $\theta $ cannot be determined with any accuracy at all. Thus if $x$ is greater than, or of the order of, the inverse of the machine precision, it is impossible to calculate the phase of ${Y}_{1}\left(x\right)$ and the function must fail.
None.
## 9 Example
The following program reads values of the argument $x$ from a file, evaluates the function at each value of $x$ and prints the results.
### 9.1 Program Text
Program Text (s17adce.c)
### 9.2 Program Data
Program Data (s17adce.d)
### 9.3 Program Results
Program Results (s17adce.r)
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http://math.stackexchange.com/questions/126963/about-a-sylow-subgroup-of-a-product
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About a Sylow subgroup of a product
Let $G$ be a finite group, $H$ and $K$ subgroups of $G$ such that $G=HK$. Show that there exists a $p$-Sylow subgroup $P$ of G such that $P=(P\cap H)(P\cap K)$.
I looked at the proof here, but I can't understand step "(3) It is clear in this situation that $P=(P\cap H)(P\cap K)$." Help.
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This should be a comment there, not a new question. – Steve D Apr 1 '12 at 18:41
I tried to put it as a comment there but I don't know how? – user28083 Apr 1 '12 at 18:46
@user28083: You don't have enough reputation yet to make a comment in someone else's question. – Arturo Magidin Apr 1 '12 at 20:44
2
@user28083: The comment by Jack Schmidt suggests one way of doing it: note that $|G|=|HK|=|H||K|/|H\cap K|$, and $|(P\cap H)(P\cap K)|=|P\cap H||P\cap K|/|P\cap H\cap K|$. Show that this equals the largest power of $p$ that divides $|G|$. – Arturo Magidin Apr 1 '12 at 20:52
|G|/ |(P∩H)(P∩K)|=(|H||K||P∩H∩K|)/(|P∩H||P∩K||H∩K|) is not divisible by p ==>|(P∩H)(P∩K)| is the largest p power that divides the order of G. Thank you very much. – user28083 Apr 1 '12 at 22:38
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http://stats.stackexchange.com/questions/tagged/skewness
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http://mathoverflow.net/questions/58826/how-many-semi-direct-products-are-there/58829
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## how many semi direct products are there?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
This question was initially proposed to me by two friends. Given an integer $n$, how many isomorphism classes are there for semidirect products $\mathbb{Z}/n\mathbb{Z}\rtimes\mathbb{Z}/2\mathbb{Z}$?
Maybe this is a really trivial question. I can tell that a semidirect product is the same as an integer $r\in\mathbb{Z}/n\mathbb{Z}$ with $r^2=1\mod[n]$, but are there isomorphisms between some of them? What happens for instance when n is squarefree, thus the product of fields.
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Such a group is called dihedral only when $r\equiv -1 \mod n$. – F. Ladisch Mar 18 2011 at 11:37
I wasn't aware of that. The question still stands. – Olivier Bégassat Mar 18 2011 at 11:40
I've corrected the wording. – Olivier Bégassat Mar 18 2011 at 11:56
## 2 Answers
Each Sylow $p$-subgroup of the copy of $\mathbb{Z}/n\mathbb{Z}$ will be normal (actually characteristic) in the semi-direct product. Moreover any element of the semi-direct product not in $\mathbb{Z}/n\mathbb{Z}$ will induce the same automorphism of each Sylow $p$-subgroup since $\mathbb{Z}/n\mathbb{Z}$ is abelian.
Thus two of your semi-direct products will be isomorphic precisely if each of the subproducts $\mathbb{Z}/p^r\mathbb{Z}$ by $\mathbb{Z}/2\mathbb{Z}$ are isomorphic where $p^r$ is the maximal power of $p$ dividing $n$.
In the square-free case this shows that there are $2^k$ non-isomorphic semi-direct products of the form you require, where $k$ denotes the number of odd prime factors of $n$.
In the general case this argument atill reduces you to the case $n$ is a prime power. I guess you'll still get two choices for each odd prime dividing $n$ and two choices for the prime $2$ if $4$ divides $n$ but only one otherwise.
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I suppose this means that I am claiming that if you have $r\neq r'$ in $\mathbb{Z}/n\mathbb{Z}$ there will be no isomorphisms between the corresponding semi-direct products. You can actually see this directly by observing that you can recover $r$ from the semi-direct product at least if $n$ is odd by realising that any element of order $2$ in the semi-direct product must act as multiplication by $r$ on any normal subgroup of order $n$. – Simon Wadsley Mar 18 2011 at 12:28
Actually, for the prime 2 you get four choices if 8 divides $n$, two choices if 4 divides $n$ but not 8, and only one choice (or rather: none :) if $n$ is odd. – Max Mar 18 2011 at 12:28
Right. Thanks. There are 4 roots of $-1$ mod $8$. – Simon Wadsley Mar 18 2011 at 12:31
There are 4 (square) roots of 1 mod 8. There are no (square) roots of -1 mod 8. – KConrad Mar 18 2011 at 16:34
Yes... I was clearly not thinking very clearly when I posted this. – Simon Wadsley Mar 18 2011 at 23:33
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
You are asking for isomorphism classes of split extensions $H$ of the module $N:=\mathbb{Z}/n\mathbb{Z}$ by the group $G:=\mathbb{Z}/2\mathbb{Z}$. This is a special case of a metacyclic group, by the way.
A first approximation is to determine all homomorphisms from $\mathbb{Z}/2\mathbb{Z}$ into the automorphism group $Aut(N)$ of $N$; each of these corresponds to a semidirect product (but we still might get some isomorphism classes multiple times). Now, it is well-known that $Aut(N)\cong \mathbb{Z}/\phi(n)\mathbb{Z}$, where $\phi$ denotes Euler's totient function.
Using the above (and the links I gave), it is not difficult to see that the 2-Sylow-subgroup $P$ of $Aut(N)$ is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^m \times Aut(\mathbb{Z}/2^k\mathbb{Z})$, where $m$ equals the number of distinct odd primes divisors of $n$, and $2^k$ is the largest power of $2$ dividing $n$. If $k=0$ or $k=1$, then $P\cong(\mathbb{Z}/2\mathbb{Z})^m$. If $k>1$, then $P\cong (\mathbb{Z}/2\mathbb{Z})^{m+1} \times \mathbb{Z}/2^{k-2}\mathbb{Z}$.
Counting the number of homomorphisms from $G$ into this, then for $k=0$ or $k=1$ we get $2^m$; for $k=2$ we get $2^{m+1}$ and for $k>2$ we get $2^{m+2}$.
With some more effort, one proceeds to verify that each of these homomorphisms leads to a unique isomorphism class; for that, you essentially have to verify that $G$ either acts trivially or non-trivially on each $p$-Sylow-subgroups; and that it really has four non-isomorphic actions on $\mathbb{Z}/2^k\mathbb{Z}$ if $k>2$ (once you get an action like in a dihedral group, once like in a semidihedral / quasidihedral group; once as in the direct product; and one more).
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Hi Max, I think your $2$ Sylow formula might be wrong. I agree with what you've said before that. Set $n=2^k\cdot\prod_{i=1}^r p_i^{m_i}$ the decomposition into prime factors. Then $\mathrm{Aut}(N)\simeq (\mathbb{Z}/n\mathbb{Z})^{\times}\simeq (\mathbb{Z}/2^k\mathbb{Z})^{\times}\times\Prod_i (\mathbb{Z}/p_i^{m_i}\mathbb{Z})^{\times}$. The factor corresponding to the prime $2$ depends on wether $k\leq 2$, but is a $2$ group. For $p_i$ odd prime however $(\mathbb{Z}/p_i^{m_i}\mathbb{Z})^{\times}\simeq \mathbb{Z}/(p_i-1)\mathbb{Z}\times\mathbb{Z}/p_i^{m_i-1}\mathbb{Z}$. – Olivier Bégassat Mar 18 2011 at 21:15
so, I think the formula for the $2$ Sylow should be $(\mathbb{Z}/2^k\mathbb{Z})^{\times}\times\prod_{i=1}^r \mathbb{Z}/2^{\nu_i}\mathbb{Z}$ where $\nu_i$ is the highest exponent of $2$ in $P_i-1$. Correct me if I'm wrong, but I think this is the correct formula. I just noticed that I changed your notation, what you call $m$ I call $r$. – Olivier Bégassat Mar 18 2011 at 21:21
For instance your formula yields that the $2$ Sylow of $\mathbb{Z}/17\mathbb{Z}$ should be $\mathbb{Z}/2\mathbb{Z}$ while it actually is $\mathbb{Z}/16\mathbb{Z}$. – Olivier Bégassat Mar 18 2011 at 21:24
Olivier, you are absolutely right, my formula for the 2-Sylow as given is incorrect. Luckily, the final result is still correct, as it only depends on the the subgroup of all elements of order 2. This still has the shape $\mathbb{Z}/2\mathbb{Z})^{m+a}$, where $a=0,1,2$ depending on whether $k=0,1$ or $k=2$ or $k>2$. – Max Mar 21 2011 at 13:03
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http://www.physicsforums.com/showthread.php?p=4176639
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Physics Forums
## My thought experiment on determining the moment of inertia of an irregular object
Hello everyone, (This might be a little long winded but I would very much appreciate some help on this!)
Today while in psychology class, I was thinking about a much more important subject, namely a project I want to start working on. Basically what I want to do is build a motor from the ground up. Like a real motor that would be in a machine. I want to start from complete theory and them move to actually building it once. (I am a senior undergraduate in Physics and I want to get into theoretical physics in my future.)
So my first thought would be, how would I determine how much current I need to run through each of the windings to get the motor to move? I then realized that I would need to know the torque required to move the armature, and this would be the LEAST amount of torque I would need. (I would obviously want more torque than this so the motor can actually drive things. And I would actually need a little more than this to get it moving due to friction effects.) Here is something I came up with to measure this torque.
Have an external force spin the armature to a constant frequency ω. When it reaches this point, cut the power to the armature (The force would be coming from some external motor) and see how long it takes it to stop spinning. The torque will be approximately equal to..
$$T≈\frac{ΔL}{Δt}=\frac{I(ω_2-ω_1)}{Δt}$$
So I would need to determine the rotational inertia of the armature first. It is spinning about one of its principle axis so it will not be a tensor. To determine the rotational inertia, the idea I came up with would be very similar to determining the torque. I would again drive the armature to spin at a constant frequency ω and then cut the power. But right when I cut the power, have a "negligible" string tied to one end of the armature. This string is attached to a block of known weight and will consequently do work on the block by moving it linearly. From the work-kinetic energy theorem we can find the kinetic energy.
$$W=ΔK$$
So here is the part I need help on. So for a rotating body the kinetic energy is..
$$K=\frac{1}{2}Iω^2$$
But ω would not be constant since the object is negatively accelerating (slowing down). So I would have..
$$dK=\frac{1}{2}I(dω)^2$$
Or..
$$∫dK=K=∫\frac{1}{2}I(dω)^2=\frac{1}{2}I∫(dω)^2$$
I have taken so many advanced physics and mathematics courses, I find it funny that I do not know how to integrate this. I have never worked with a squared differential before. How would I go about doing this?
Also, if anyone has any other ways about measuring the moment of inertia of something, or any comments about my way, feel free to let me know!
Thanks
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Quote by Xyius $$dK=\frac{1}{2}I(dω)^2$$
That does not work.
$$dK=\frac{1}{2}Id(\omega^2) = I\omega d\omega$$
This can be integrated, but where is the point in taking the derivative if you integrate afterwards?
Quote by mfb That does not work. $$dK=\frac{1}{2}Id(\omega^2) = I\omega d\omega$$ This can be integrated, but where is the point in taking the derivative if you integrate afterwards?
So that would mean that I can simply take the final and initial ω values. I will already have K from the work done by the block, and the values of ω, so the only unknown in the equation would be the moment of inertia.
Thanks! Hope this is a correct analysis. :]
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http://mathhelpforum.com/algebra/103889-solving-inequality.html
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# Thread:
1. ## Solving Inequality
$\frac {-4x^2} {2-x} < 2-4x$
im not sure about my result.
as far i can see i must look at 2 cases. 1st: x > 2 and 2nd x < 2
the first would give me the solution: $-\frac {2} {3} < x$
which leads me to the first set of solutions: x > 2
am i right here ?
the 2nd case would give me the same just with flipped inequality sign
so the overall solution i got is: all real numbers except the intervall $[-\frac {2}{3}, 2]$
somehow all my math programs give me the solution: -2/3 < x < 2
i cant see why that is
2. Originally Posted by coobe
$\frac {-4x^2} {2-x} < 2-4x$
im not sure about my result.
as far i can see i must look at 2 cases. 1st: x > 2 and 2nd x < 2
the first would give me the solution: $-\frac {2} {3} < x$
which leads me to the first set of solutions: x > 2
am i right here ?
the 2nd case would give me the same just with flipped inequality sign
so the overall solution i got is: all real numbers except the intervall $[-\frac {2}{3}, 2]$
somehow all my math programs give me the solution: -2/3 < x < 2
i cant see why that is
HI
Obviously , $x\neq 2$
$-\frac{4x^2}{2-x}<2-4x$
$0<2-4x+\frac{4x^2}{2-x}$
$\frac{8x^2-10x+4}{2-x}>0$
Since $8x^2-10x+4$ is complex , then $2-x>0\Rightarrow x<2$
3. Originally Posted by coobe
$\frac {-4x^2} {2-x} < 2-4x$
im not sure about my result.
as far i can see i must look at 2 cases. 1st: x > 2 and 2nd x < 2
the first would give me the solution: $-\frac {2} {3} < x$
which leads me to the first set of solutions: x > 2
am i right here ?
the 2nd case would give me the same just with flipped inequality sign
so the overall solution i got is: all real numbers except the intervall $[-\frac {2}{3}, 2]$
somehow all my math programs give me the solution: -2/3 < x < 2
i cant see why that is
Clearly $x \neq 2$.
Case 1: $2 - x < 0$
$\frac{-4x^2}{2 - x} < 2 - 4x$
$-4x^2 > (2 - 4x)(2 - x)$
$-4x^2 > 4 - 10x + 4x^2$
$8x^2 - 10x + 4 < 0$
$x^2 - \frac{5}{4}x + \frac{1}{2} < 0$
$x^2 - \frac{5}{4}x + \left(-\frac{5}{8}\right)^2 - \left(-\frac{5}{8}\right)^2 + \frac{1}{2} < 0$
$\left(x - \frac{5}{8}\right)^2 + \frac{7}{64} < 0$
This is clearly a contradiction.
Case 2: $2 - x > 0$.
$\frac{-4x^2}{2 - x} < 2 - 4x$
$-4x^2 < (2 - 4x)(2 - x)$
$-4x^2 < 4 - 10x + 4x^2$
$8x^2 - 10x + 4 > 0$
$\left(x - \frac{5}{8}\right)^2 + \frac{7}{64} > 0$
This is true for all $x$.
Thus, the solution to the inequality is $2 - x > 0$, or $x < 2$.
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http://www.physicsforums.com/showthread.php?t=139147
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Physics Forums
## Lagrangians and P-Conservation
I have a problem with a particle experiencing a central force towards some origin, as well as a gravitational force downwards. I've calculated the Lagrangian, and the equations of motion. Now I'm being asked to see if the system follows conservation of angular momentum. How do I do this? I know it has something to do with seeing if the system is invariant of rotation, but how do I check for that?
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If $$\frac{\partial L}{\partial q} =0$$ then the conjugate momentum $$\frac{\partial L}{\partial \dot{q}}$$ is a conserved quantity. If that doesn't clear things up, then post what you have for the Lagrangian.
But the conjugate momentum is the same as the angular momentum only in some cases. Compute H and chech if H commutes with J.
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http://mathoverflow.net/questions/106721/quantum-mechanics-basics/106734
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Quantum mechanics basics [closed]
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Hello. I'm thinking about where does the basic quantum mechanics things comes from. I mean the forms of operators and a Shroedinger equation. The more intuitive explanation is better.
To get forms of operators and Shroedinger eq., we can start from assumption that in our representation square of absolute value of wavefunction is probability density. Then it became clear that coordinate operator is just a multiplying by variable (obvious, looking for example to mean value expression). Next, Shroedinger equation anyway should be of the form $\frac{\partial \phi}{\partial t} = A \phi$ with some A. To see that $A$ is actually Hamiltonian multiplied by something, we can see that $\frac{\partial \phi}{\partial t}$ is an enegry multiplied by something. I have the only idea to explain this: assume additionally that in our representation defined momenta states are plane waves. Considering them, $i\frac{\partial \phi}{\partial t} = E$ is visible from De Broglie relation $E=\omega$ for plane waves (as well as $p=\frac{\partial \phi}{\partial x}$ visible from $p=k$). Then, the only thing i want explained deeper is De Broglie relations. Finally, now i also think that i want some explanation of introducing wavefunction as complex-valued.
Or maybe there is an other way? I guess, it would be more general to start from commutation relations, but i'm afraid it would be hard and abstract. But i appreciate if you try to explain where are they come from.
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2
I think you should try theoretical physics stackexchange. – Jon Bannon Sep 9 at 13:02
@Jon Bannon I think it is right way to ask on both forums and choose the answer which is more "intuitive" for OP, since it is often happens that physicists and mathematicians see the same subject in different ways. Any way I think that here are many people which might give there opinion on the question, so I do not think it reasonable to close. – Alexander Chervov Sep 9 at 13:09
There are some "related" questions: mathoverflow.net/questions/102313/… mathoverflow.net/questions/102415/… mathoverflow.net/questions/6200 To Ashley you can find more related questions if you press on "tags" under yours questions. – Alexander Chervov Sep 9 at 13:22
@Alexander: I agree. I suggested the physics site because of Ashley's response to your answer. – Jon Bannon Sep 9 at 13:50
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Re the de Broglie relation, there are a lot of different ways of getting at this, but one is to recognize that according to relativity, both $(t,\mathbb{x})$ and $(E,\mathbb{p})$ transform as vectors. If the E operator is going to be $\partial/\partial t$, then the $\mathbb{p}$ operator had better be $\nabla$. Note that this is more general than the Schrodinger equation, which is nonrelativistic. – Ben Crowell Sep 9 at 15:27
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2 Answers
Probably the explanation via Heisenberg picture is more intuitive (at least for mathematically minded person like me).
Equations of motion in classical mechanics can be described as Hamilton form as follows:
d\dt f = {H, f}
(here { } are Poisson brackets and these equations reduces to standard Hamilton equations of motions, if you take standard phase space R^2n p_i, q_i and standard Poisson bracket these equation will give d/dt p = -dH/dt ; d/dt q = dH/dt . However they make sense on arbitrary Poisson manifold.)
Quantization in Heisenberg picture turns these equations into
d/dt f = [H, f]
(I have omitted (i/h). )
That is more or less all, modula we need to explain what "[H, f]" is. It is commutator in non-commutative algebra which is deformation quantization of the Poisson algebra of classical observables. It should probably be pointed out here why deformation quantization might be considered as intuitively "clear". To explain this let us go in opposite direction: consider non-commutative algebras depending on parameter "h" such that for "h=0" algebra becomes commutative. The point is that Poisson bracket naturally arise in this step: define {f,g} = lim {h->0} (fg-gf) / h . Exercise - to check that such defined Poisson bracket satisfies the Jacobi identity (it follows from the associativity).
So the moral is that Poisson bracket is "shadow of non-commutativity", more precisely first order of the noncommutative product. The deformation quantization task is not construct the non-commutative algebra from the first this first order "shadow", it has been solved in certain generality by Kontsevich. Well, I am not sure how intuitive these reasons are, at least they are so for me.
Another important remark is to relate Heisenber picture to the Shrodinger one. This is quite easy linear algebra. So Heisenberg picture is evolution of "operators (=matrices)" "f" according to equation d/dt f = [H,f]. The linear statement is the following: such evolution on matrices equivalent to Schrodinger like evolution on VECTORS: d/dt v = H v. You can make this "equivalent" in precise statement in various way e.g. lemma: consider f(t) which satisfy the "Heisenberg" equation d/dt f=[H,f], then vector v(t)= f(t)v_0 will satisfy the "Schrodinger" equation d/dt v =Hv, for any vector v_0.
Another important remark is Stone von Neumann uniqueness theorem. It explains why usually operator corresponding to "p" is d/dq and corresponding to "q" is multiplication by "q".
Theorem says (up to details) that if you consider the algebra [p,q]=1 then it has the unique irreducible representation in the Hilbert space. The meaning of this theorem is the following: you can choose ANY other representation of operators corresponding to "p" and "q" ( not just "p"-> d/dq, q-> mult_q), but ALL these representation will be equivalent. So choose whatever you want and do not care.
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Thanks, that would be great if i could make that to look more intuitive for me. Currently, it's not. – Ashley Sep 9 at 12:03
I tried to extend the answer in more details. I was in a hurry to post the first draft version of the answer since I see votes to close yours question - if it will be close, no one can add answer, although it still be possible to edit already given answer :) – Alexander Chervov Sep 9 at 12:55
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Classical evolution from 0 to $t$ is given by symplectomorphisms of the phase space. For $t$ infinitesimal you get a symplectic vector field. When the vector field is actually Hamiltonian, its generating function is called energy/Hamiltonian. This is what the Hamilton equations tell you. Applying the art of quantization, you see that quantum evolution is given by an automorphism of the Hilbert space (evolution operator). Infinitesimal evolution is given by a certain endomorphism, which is now called the quantum Hamiltonian. – Pavel Safronov Sep 9 at 15:55
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I'll try to help with intuition on why wavefunctions are complex-valued.
First, it's not actually true that they have to be complex. When you quantize electromagnetic waves, the wavefunction is simply the electric and magnetic fields, which are real. The correct statement is that a spin-1/2 particle's wavefunction has to be complex.
As a simple example of why this is, consider the case of two planar sine waves that are moving in antiparallel directions and then merge and superpose.
If the wavefunction is a real scalar, then at a time when the two superposed waves are 180 degrees out of phase, their sum is identically zero. This violates conservation of energy and, perhaps more importantly, conservation of probability.
If the wavefunction is a complex scalar, then one can prove that solutions of the Schrodinger equation always conserve probability. This is easy to check in an example like the superposition of $e^{i(kx+\omega t)}$ with $e^{i(-kx+\omega t)}$.
To see that this argument doesn't prove that wavefunctions are always complex-valued, consider electromagnetic waves in the same situation of superposition of antiparallel plane waves. Because there is a right-handed relationship among $\mathbf{E}$, $\mathbf{B}$, and $\mathbf{k}$, you can't make both $\mathbf{E}$ and $\mathbf{B}$ cancel. For example, you could choose the polarization so that $\mathbf{E}$ would cancel, but then $\mathbf{B}$ wouldn't.
To see that the fundamental issue is conservation of probability, so that this is really something specific to quantum mechanics rather than classical physics, consider the case of sound waves, which can be represented as real scalar functions $f$ measuring the pressure. Rerunning the same argument about superposing antiparallel plane waves, we find that it's possible for $f$ to cancel, but that's OK, because $f$ doesn't have a probability interpretation. We also still have conservation of energy, because the energy depends not just on $f$ (potential) but also on $\partial f/\partial t$ (kinetic), so $f$ can vanish without making the energy vanish.
Complex numbers come up in a lot of places in quantum mechanics, not just in wavefunctions, and it's not always obvious when they're just a notational convenience. For example, Pauli basically reinvented the quaternions in 1924. His spin matrices $\sigma_1$, $\sigma_2$, and $\sigma_3$ are equivalent to the quaternions i, j, and k if you multiply them by i.
Operators can be complex-valued, but expectation values are always supposed to be real, since they correspond to measurable quantities.
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I don't completely agree. The Hilbert space of states of a quantum system is always a complex vector space. If one defines a wave function as coordinates of a state in a given basis, then it will be in general complex valued. This has no relation with the properties of the classical system that one quantizes. – unknown (google) Sep 9 at 16:56
@unknown (google): I agree with your view from 30,000 feet, but there really is a difference between how the complex number system relates to different fields. Under a gauge transformation, the electromagnetic field (expressed in terms of the 4-potential) transforms as $A_j\rightarrow A_j+\partial_j \phi$, while the wavefunction of an electron does $\Psi\rightarrow e^{i\phi}}\Psi$. One involves a complex phase, the other doesn't. Note also that electromagnetic fields are observable, while $\arg\Psi$ isn't. The complex phase ultimately arises from charge, which electrons have and photons don't. – Ben Crowell Sep 9 at 17:17
When you quantize the electromagnetic field the wave function is most definitely not the electric and magnetic fields. The fields are operators which act on a complex valued wave function, just as in the non-relativistic Schrodinger equation. You are confusing complex valued fields with the complex valued wave function of quantum mechanics. They are two different things. – Jeff Harvey Nov 26 at 22:09
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http://simple.wikipedia.org/wiki/Prime_counting_function
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# Prime counting function
The 60 first values of π(n)
In mathematics, the prime counting function is the function counting the number of prime numbers less than or equal to some real number x. It is written as $\pi(x)$, but it is not related to the number π.
This short article about mathematics can be made longer. You can help Wikipedia by .
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http://mathoverflow.net/questions/111473?sort=votes
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is the limit of ergodic functions still ergodic?
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under what conditions is the limit of a sequence of ergodic functions still ergodic? are there simple counter-examples to this general statement?
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1 Answer
A rotation $z\mapsto e^{2\pi i\alpha} z$ as a self-map of the unit circle is ergodic wrto the length measure iff $\alpha$ is irrational. So any sequence of irrational numbers converging to a rational number produces a counterexample.
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http://mathoverflow.net/questions/88026/pseudoisotopy-in-low-dimensions/88040
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## Pseudoisotopy in low dimensions
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Recall that the space $P(M)$ of (smooth) pseudoisotopies of the compact manifold $M$ is defined as the space of all diffeomorphisms $M\times I\to M\times I$ that fix every point in $(M\times 0)\cup (\partial M \times I)$. I have worked on these things in high-dimensional cases, but I realize that there are big gaps in my knowledge in low-dimensions. I am wondering how much of what I don't know is known.
If the dimension of $M$ is less than $2$ then $P(M)$ is contractible. According to the Smale Conjecture $P(D^2)$ (or equivalently $P(S^2)$) is contractible. I have the impression that Hatcher also proved that $P(M)$ is contractible when $M$ is a two-dimensional annulus. Is it perhaps known, or believed, that this holds for all compact $2$-manifolds? Or maybe it's not true. (I could have started by just asking Allen Hatcher or Ryan Budney, but I thought others might be interested in any answers.)
And what is known in the next dimension?
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## 1 Answer
When $M$ is a compact $2$-manifold, with or without boundary, $P(M)$ is known. When $M$ is a 3-manifold there's bits and pieces known, especially once you get to more fine detail like pseudo-isotopy embedding spaces. But at the level of $P(M)$ I don't think there's a complete description for a single $3$-manifold. For $4$-manifolds the situation is worse, but again there are some things known for pseudo-isotopy embedding spaces.
As you mention, $P(S^2)$ and $P(D^2)$ are contractible by the Smale Conjecture. I think $P(M)$ is contractible for an arbitrary $2$-manifold. The argument goes like this: You look at locally trivial fiber bundle $P(M) \to Diff(M)$. The fiber over the identity map is $Diff(M \times I)$, the group of diffeomorphisms fixing the entire boundary $\partial M \times I \cup M \times \partial I$. The other fibers are empty because diffeomorphisms are prescribed up to isotopy by their action on $\pi_1 M$. Except in a few cases, $Diff(M)$ has contractible components, so showing $P(M)$ is contractible is equivalent to showing $Diff(M \times I)$ is contractible. But $M \times I$ has incompressible annuli in it, corresponding to $C \times I$ where $C \subset M$ is a closed curve in $M$ that does not bound a disc in $M$. Hatcher's work on spaces of incompressible surfaces kicks in and tells you the space of these incompressible annuli is contractible. So you can reduce studying $Diff(M \times I)$ to things like $Diff(A \times I)$ where $A$ is an annulus or a pair of pants, but then you have vertical incompressible discs you can use and reduce to the point that $A$ is itself a disc.
The above argument works for any surface other than $\mathbb RP^2$, a torus or a klein bottle. But similar arguments cover these cases.
Here's the reference for the results of Hatcher's I'm using, it also defines the terminology like "incompressible annulus, disc" and so on. http://www.math.cornell.edu/~hatcher/Papers/emb.pdf
Allen had a student (Kiralis) who wrote some papers on pseudo-isotopy diffeomorphisms of 3-manifolds. That might be a place to look.
But the kind of things that I remember are mostly along the lines of pseudo-isotopy embeddings of knots and links in $D^3$ and $S^3$. I'm about to hop on an airplane. I'll edit this response sometime in the next few days and add some of these observations if they haven't already been made by someone else.
edit 1: Here are two unrelated observations.
(a) Let $N$ be a co-dimension zero solid torus in $M=\mathbb R^3$ or $M=S^3$. There's the pseudo-isotopy embedding fibration $P(N,M) \to Emb(N,M)$. If $N$ is an unknotted solid torus, the question of what the map the image of the map $\pi_0 P(N,M) \to \pi_0 Emb(N,M)$ is, this is a long-standing hard problem in knot theory. Another way to say it is `which knots in $S^3$ bound a disc in $D^4$?'. These knots are called slice knots. Ralph Fox has the Slice Ribbon Conjecture, which might be described as a hopeful combinatorial answer to the question. There are many useful tools for determining whether or not a given knot is slice,starting with the Alexander module and more recently tools from Heegaard Floer theory.
Two examples:
• Look at the class of knots that are a connect-sum of torus knots. Using the Alexander module Litherland proved such a knot bounds a disc in $D^4$ if and only if in the prime decomposition, the number of times a prime summand appears (like say a right-handed trefoil) is equal to the number of times its mirror-image appears (the left handed trefoil, in my example). Related previous MO thread
• Paolo Lisca has used Heegaard Floer theory to determine when a connect sum of 2-bridge knots in $S^3$ bounds a disc in $D^4$ REF, but in this case the answer is more elaborate.
(b) Just like how diffeomorphism groups $Diff(D^n)$ and spaces of long knots have actions of cubes operads, pseudo-isotopy embedding spaces and diffeomorphism groups also have such actions. I'm pretty sure there's a pseudo-isotopy splicing operad as well, but I haven't written out the details.
edit 2: Rick Litherland generalized a result of Zeeman (Deforming twist-spun knots TAMS 250 (1979) 311--331) showing that "Deform-spun knots" have complements that frequently fibre over $S^1$. This process called "Deform spinning" is just the boundary map in the pseudo-isotopy fibre sequence for spaces of knots. One of the nice things about this is Litherland gives a prescription for what the fibre is. In the case where you're looking at the pseudo-isotopy sequence for long knots in $\mathbb R^3$, it generates long embeddings of $\mathbb R^2$ in $\mathbb R^4$. So the fibre is a 3-manifold with boundary a sphere. This process produced some embeddings of 3-manifolds in the 4-sphere that nobody had known about at the time, like the once-punctured Poincare Dodecahedral Space (which without a puncture does not embed in $\mathbb R^4$, at least, not smoothly, it does admit a tame topological embedding). I got interested in this case largely because it represents sort of an extreme end of the terrain of your dissertation.
edit 3: I forgot to mention, I kept on pushing trying to understand why your dissertation broke down in co-dimension two. In some sense my paper "An obstruction to a knot being deform-spun via Alexander polynomials" is an answer. In a way it's not, since co-dimension two deform-spinning is more of a free loop space construction than a based loop space construction. But I found the exercise informative.
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http://mathoverflow.net/questions/21995/name-for-topology-making-group-action-continuous
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## Name for topology making group action continuous
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Fix a set $X$ with right $G$-action. Give $X$ a topology $\tau$ and make $G$ a topological group. (These topologies need not make the action continuous).
We can define another topology $\tau'$ on $X$ as the largest topology making the action $(X,\tau) \times G \to (X,\tau')$ continuous. (This is also called the quotient topology on $X$ with respect to the action $(X,\tau) \times G \to X$.)
Note that if the $G$-action is continuous for $\tau$ then $\tau'= \tau$.
For example, if $X = \mathbb{R}$, $\tau$ is the discrete topology and $G$ is $(\mathbb{R}, +)$ with the usual topology acting on $X$ by addition, then $G \times X / \sim = \mathbb{R}$ with the usual topology (unless I am much mistaken).
More interesting examples exist, e.g. the Skorokhod topology (again unless I am mistaken).
This construction feels useful enough that it must be well known and have a name. Can anyone provide me with more information?
[EDIT: actually I don't think it's necessary that $G$ is a topological group, just that it's a group with a topology. Although it is probably necessary for inversion to be continuous at the identity and for multiplication to be continuous on ${e} \times G$.]
[EDIT: made the presentation clearer to address the existing comments, changed title]
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In your example with $\mathbb R$, the action of $G$ is discontinuous. – Sergei Ivanov Apr 20 2010 at 21:53
Yes, it's supposed to be. If the action of G is continuous then you don't get a new topology. See Brad Hannigan-Daley's answer below. – Tom Ellis Apr 21 2010 at 6:14
## 1 Answer
If the action of $G$ on $X$ is continuous (i.e. the multiplication map $X\times G\to X$ is continuous) then the resulting topology is $\tau$:
Let $\tilde X$ denote $(X\times G)/\sim$, and let $\phi:X\to \tilde X:x\mapsto[x,e]$ be the identification you mentioned (with the factors $X,G$ reversed for convenience). Then $\phi$ is clearly continuous. Let $\psi:\tilde X\to X$ be the inverse of $\phi$, i.e. $\psi([x,g]) = xg$. For an open subset $U$ of $X$, $\psi^{-1}(U) = {(x,g):xg\in U}$. Pulling this back to $X\times G$ via the projection $X\times G\to\tilde X$ gives exactly the preimage of $U$ under the multiplication map $X\times G\to G$, which is open, and so $\psi^{-1}(U)$ is open in $\tilde X$. So $\phi$ is a homeomorphism.
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Yes I know. This is why I wrote "whose point set $X$" is acted on, rather than "with a right $G$ action". Perhaps I should have made it clearer. – Tom Ellis Apr 21 2010 at 6:09
I've clarified that now. – Tom Ellis Apr 21 2010 at 6:10
N.B. by the definition of quotient topology, $G \times X / \sim$ is the largest topology on $X$ such that the product $G \times \tau \to X$ is continuous. – Tom Ellis Apr 21 2010 at 6:18
@Brad: it's more conceptual to use the universal property of the quotient. – Martin Brandenburg Apr 21 2010 at 6:28
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http://www.scholarpedia.org/article/Control_of_partial_differential_equations
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Control of partial differential equations
From Scholarpedia
Jean-Michel Coron (2009), Scholarpedia, 4(11):6451.
Curator and Contributors
1.00 - Jean-Michel Coron
0.33 -
Leo Trottier
0.33 -
Eugene M. Izhikevich
0.17 -
Milos Rancic
0.17 -
Benjamin Bronner
A control system is a dynamical system on which one can act by using suitable controls. In this article, the dynamical model is modeled by partial differential equations of the following type $\tag{1} \dot y=f(y,u).$
The variable $$y$$ is the state and belongs to some space $$\mathcal{Y}\ .$$ The variable $$u$$ is the control and belongs to some space $$\mathcal{U}\ .$$ In this article, the space $$\mathcal{Y}$$ is of infinite dimension and the differential equation (1) is a partial differential equation.
There are a lot of problems that appear when studying a control system. But the most common one is the controllability problem, which is, roughly speaking, the following one. Let us give two states. Is it possible to steer the control system from the first one to the second one? In the framework of (1), this means that, given the state $$a\in \mathcal{Y}$$ and the state $$b\in \mathcal{Y}\ ,$$ does there exit a map $$u:[0,T]\rightarrow \mathcal{U}$$ such that the solution of the Cauchy problem $$\dot y=f(y,u(t)), \, y(0)=a,$$ satisfies $$y(T)=b\ ?$$ If the answer is yes whatever the given states are, the control system is said to be controllable. If $$T>0$$ can be arbitrary small one speaks of small-time controllability. If the two given states and the control are restricted to be close to an equilibrium one speaks of local controllability at this equilibrium. (An equilibrium of the control system is a point $$(y_e,u_e)\in \mathcal{Y}\times \mathcal{U}$$ such that $$f(y_e,u_e)=0$$). If, moreover, the time $$T$$ is small, one speaks of small-time local controllability.
A general framework for control systems modeled by linear PDE's
The framework
For two normed linear spaces $$H_1$$ and $$H_2 \ ,$$ we denote by $$\mathcal{L}(H_1;H_2)$$ the set of continuous linear maps from $$H_1$$ into $$H_2$$ and denote by $$\|\cdot\|_{\mathcal{L}(H_1;H_2)}$$ the usual norm in this space.
Let $$H$$ and $$U$$ be two Hilbert spaces. Just to simplify the notations, these Hilbert spaces are assumed to be real Hilbert spaces (the case of complex Hilbert spaces follows directly from the case of real Hilbert spaces). The space $$H$$ is the state space and the space $$U$$is the control space. We denote by $$(\cdot, \cdot)_H$$ the scalar product in $$H\ ,$$ by $$(\cdot,\cdot)_U$$ the scalar product in $$U\ ,$$ by $$\|\cdot\|_H$$ the norm in $$H$$ and by$$\|\cdot\|_U$$ the norm in $$U\ .$$
Let $$S(t),\, t\in[0,+\infty)\ ,$$ be a strongly continuous semigroup of continuous linear operators on$$H\ .$$ Let $$A$$ be the infinitesimal generator of the semigroup $$S(t), \, t\in[0,+\infty)\ .$$ As usual, we denote by $$S(t)^*$$ the adjoint of $$S(t)\ .$$ Then $$S(t)^*, \ t\in[0,+\infty),$$ is a strongly continuous semigroup of continuous linear operators and the infinitesimal generator of this semigroup is the adjoint $$A^*$$ of $$A\ .$$ The domain $$D(A^*)$$ is equipped with the usual graph norm $$\|\cdot\|_{D(A^*)}$$ of the unbounded operator $$A^*\ :$$
$$\|z\|_{D(A^*)}:=\|z\|_{H}+\|A^*z\|_{H}, \, \forall z\in D(A^*).$$
This norm is associated to the scalar product in $$D(A^*)$$ defined by
$$(z_1,z_2)_{D(A^*)}:=(z_1,z_2)_{H}+(A^*z_1,A^*z_2)_H, \, \forall (z_1,z_2) \in D(A^*)^2.$$
With this scalar product, $$D(A^*)$$ is a Hilbert space. Let $$D(A^*)'$$ be the dual of $$D(A^*)$$ with respect to the pivot space $$H\ .$$ In particular,
$$D(A^*)\subset H\subset D(A^*)'.$$
Let
$\tag{2} B\in \mathcal{L}(U,D(A^*)').$
In other words, $$B$$ is a linear map from $$U$$ into the set of linear functions from $$D(A^*)$$ into $$\mathbb{R}$$ such that, for some $$C>0\ ,$$
$$|(Bu)z|\leqslant C \|u\|_{U}\|z\|_{D(A^*)},\, \forall u \in U, \, \forall z\in D(A^*).$$
We also assume the following regularity property (also called admissibility condition):
$\tag{3} \forall T>0, \exists C_T>0 \text{ such that } \int_0^T\|B^*S(t)^* z\|_{U}^2dt \leqslant C_T \|z\|^2_H, \, \forall z\in D(A^*).$
In (3) and in the following, $$B^*\in \mathcal{L}(D(A^*);U)$$ is the adjoint of $$B\ .$$ It follows from (3) that the operators
$$(z\in D(A^*)) \mapsto ((t\mapsto B^*S(t)^* z)\in C^0([0,T];U)),$$
$$(z\in D(A^*)) \mapsto ((t\mapsto B^*S(T-t)^* z)\in C^0([0,T];U))$$
can be extended in a unique way as continuous linear maps from $$H$$ into $$L^2((0,T);U)\ .$$ We use the same symbols to denote these extensions.
Note that, using the fact that $$S(t)^*\ ,$$ $$t\in[0,+\infty)\ ,$$ is a strongly continuous semigroup of continuous linear operators on$$H\ ,$$ it is not hard to check that (3) is equivalent to
$$\exists T>0, \exists C_T>0 \text{ such that } \int_0^T\|B^*S(t)^* z\|_{U}^2dt \leqslant C_T \|z\|^2_H, \, \forall z\in D(A^*).$$
The control system we consider here is
$\tag{4} \dot y =Ay +Bu, \, t\in(0,T),$
where, at time $$t\ ,$$ the control is $$u(t)\in U$$ and the state is$$y(t)\in H\ .$$
Let $$T>0\ ,$$ $$y^0\in H$$ and $$u\in L^2((0,T);U)\ .$$ We are interested in the Cauchy problem
$\tag{5} \dot y =Ay +Bu(t) , \, t\in (0,T),$
$\tag{6} y(0)=y^0.$
We first give the definition of a solution to (5)-(6). Let us first motivate our definition. Let $$\tau \in [0,T]$$ and $$\varphi :[0,\tau]\rightarrow H\ .$$ We take the scalar product in $$H$$ of (5) with $$\varphi$$ and integrate on$$[0,\tau]\ .$$ At least formally, we get, using an integration by parts together with (6),
$$(y(\tau),\varphi(\tau))_{H}-(y^0,\varphi(0))_{H}-\int_0^\tau (y(t),\dot \varphi (t) +A^*\varphi (t))_{H}dt=\int_0^\tau (u(t),B^*\varphi (t))_U dt.$$
Taking $$\varphi(t)=S(\tau-t)^*z^\tau\ ,$$ for every given $$z^\tau \in H\ ,$$ we have formally $$\dot \varphi (t) +A^*\varphi (t)=0\ ,$$ which leads to the following definition.
Definition (solution of the Cauchy problem)
Let $$T>0\ ,$$ $$y^0\in H$$ and $$u\in L^2((0,T);U)\ .$$ A solution of the Cauchy problem (5)-(6) is a function $$y\in C^0([0,T];H)$$ such that $\tag{7} (y(\tau),z^\tau)_H-(y^0,S(\tau)^*z^\tau)_H = \int_0^\tau (u(t),B^*S(\tau-t)^*z^\tau)_U dt, \, \forall \tau \in[0,T], \, \forall z^\tau \in H.$
Note that, by the regularity property (3), the right hand side of (7) is well defined.
With this definition one has the following theorem.
Theorem 1 (well posedness of the Cauchy problem)
Let $$T>0\ .$$ Then, for every $$y^0\in H$$ and for every $$u\in L^2((0,T);U)\ ,$$ the Cauchy problem (5)-(6) has a unique solution $$y\ .$$ Moreover, there exists $$C=C(T)>0\ ,$$ independent of $$y^0\in H$$ and $$u\in L^2((0,T);U)\ ,$$ such that
$\tag{8} \|y(\tau)\|_{H}\leqslant C (\|y^0\|_{H}+\|u\|_{L^2((0,T);U)}), \, \forall \tau \in [0,T].$
For a proof of this theorem, see, for example, (Jean-Michel Coron, 2007, pages 53-54).
Controllability of linear control systems
Different types of controllability
In this section we are interested in the controllability of the control system (4). In contrast to the case of linear finite-dimensional control systems, many types of controllability are possible and interesting. We define here three types of controllability.
Definition (exact controllability)
Let$$T>0\ .$$ The control system (4) is exactly controllable in time $$T$$ if, for every $$y^0\in H$$ and for every $$y^1\in H\ ,$$ there exists $$u \in L^2((0,T);U)$$ such that the solution $$y$$ of the Cauchy problem
$\tag{9} \dot y =Ay + Bu(t), \, y(0)=y^0,$
satisfies$$y(T)=y^1\ .$$
Definition (null controllability)
Let$$T>0\ .$$ The control system (4) is null controllable in time $$T$$ if, for every $$y^0\in H$$ and for every $$\tilde y^0\in H\ ,$$ there exists $$u \in L^2((0,T);U)$$ such that the solution of the Cauchy problem (8) satisfies $$y(T)=S(T)\tilde y^0\ .$$
Let us point out that, by linearity, we get an equivalent definition of "null controllable in time $$T$$" if, in the definition above, one assumes that $$\tilde y^0=0\ .$$ This explains the usual terminology "null controllability".
Definition (approximate controllability)
Let$$T>0\ .$$ The control system (4) is approximately controllable in time $$T$$ if, for every $$y^0\in H\ ,$$ for every $$y^1\in H\ ,$$ and for every $$\varepsilon>0\ ,$$ there exists $$u \in L^2((0,T);U)$$ such that the solution $$y$$ of the Cauchy problem (8) satisfies $$\|y(T)-y^1\|_H\leqslant \varepsilon\ .$$
Clearly
(exact controllability) $$\Rightarrow$$ (null controllability and approximate controllability).
The converse is false in general (see, for example, the heat control equation at this link). However, the converse holds if $$S$$ is a strongly continuous group of linear operators. More precisely, one has the following theorem.
Theorem 2 (null controllability/exact controllability)
Assume that $$S(t)\ ,$$ $$t\in \mathbb{R}\ ,$$ is a strongly continuous group of linear operators. Let $$T>0\ .$$ Assume that the control system (4) is null controllable in time $$T\ .$$ Then the control system (4) is exactly controllable in time $$T\ .$$
Proof of Theorem. Let $$y^0\in H$$ and $$y^1\in H\ .$$ From the null controllability assumption applied to the initial data $$y^0-S(-T)y^1\ ,$$ there exists $$u\in L^2((0,T);U)$$ such that the solution $$\tilde y$$ of the Cauchy problem
$$\dot {\tilde y}=A\tilde y +Bu(t), \,\tilde y (0)=y^0-S(-T)y^1,$$
satisfies
$\tag{10} \tilde y(T)=0.$
One easily sees that the solution $$y$$ of the Cauchy problem
$$\dot y=A y +Bu(t), \, y (0)=y^0,$$
is given by
$\tag{11} y(t)=\tilde y(t)+S(t-T)y^1, \, \forall t \in [0,T].$
In particular, from (10) and (11),
$$y(T)=y^1.$$
This concludes the proof of the theorem.
Methods to study controllability
Rouhgly speaking there are essentially two types of methods to study the controllability of linear PDE, namely direct methods and duality methods.
Direct methods
Among these methods, let us mention in particular
• The extension method. See, for example, (David L. Russell, 1974), the proof of Theorem 5.3, pages 688-690, in (David L. Russell, 1978), (Walter Littman, 1978) and Section 2.1.2.2, pages 30-34, in (Jean-Michel Coron, 2007).
• Moment theory. See, for example, (Werner Krabs, 1992), (Vilmos Komornik and Paola Loreti, 2005), (Sergei Avdonin and Sergei Ivanov, 1995), and Section 2.6, pages 95-99, in (Jean-Michel Coron, 2007). For an example of an application of the moment theory for a Schrödinger equation, see at this link.
• Flatness. This approach has been initiated in the framework of control theory in finite dimension in (Michel Fliess, Jean Lévine, Philippe Martin and Pierre Rouchon, 1995). For applications of this method to the control of linear PDE, see, in particular, (Hugues Mounier, Joachim Rudolph, Michel Fliess and Pierre Rouchon, 1998), (Béatrice Laroche, Philippe Martin and Pierre Rouchon, 2000), (Nicolas Petit and Pierre Rouchon, 2001), as well as this article by Pierre Rouchon in Scholarpedia.
Duality methods
Let us now introduce some "optimal control maps". Let us first deal with the case where the control system (4) is exactly controllable in time $$T\ .$$ Then, for every $$y^1\ ,$$ the set $$U^T(y^1)$$ of $$u\in L^2((0,T);U)$$ such that
$$(\dot y=Ay+Bu(t), \, y(0)=0)\Rightarrow (y(T)=y^1)$$
is nonempty. Clearly the set $$U^T(y^1)$$ is a closed affine subspace of $$L^2((0,T);U)\ .$$ Let us denote by $$\mathcal{U}^T(y^1)$$ the projection of $$0$$ on this closed affine subspace, i.e., the element of $$U^T(y^1)$$ of the smallest $$L^2((0,T);U)$$-norm. Then it is not hard to see that the map
$$\begin{array}{rrcl} \mathcal{U}^T:&H&\rightarrow&L^2((0,T);U) \\ &y^1&\mapsto&\mathcal{U}^T(y^1) \end{array}$$
is a linear map. Moreover, using the closed graph theorem (see, for example, Theorem 2.15 on page 50 in (Rudin, 1973)) one readily checks that this linear map is continuous.
Let us now deal with the case where the control system (4) is null controllable in time $$T\ .$$ Then, for every $$y^0\ ,$$ the set $$U_T(y^0)$$ of $$u\in L^2((0,T);U)$$ such that
$$(\dot y=Ay+Bu(t), \, y(0)=y^0)\Rightarrow (y(T)=0)$$
is nonempty. Clearly the set $$U_T(y^0)$$ is a closed affine subspace of $$L^2((0,T);U)\ .$$ Let us denote by $$\mathcal{U}_T(y^0)$$ the projection of $$0$$ on this closed affine subspace, i.e., the element of $$U_T(y^0)$$ of the smallest $$L^2((0,T);U)$$-norm. Then, again, it is not hard to see that the map
$$\begin{array}{rrcl} \mathcal{U}_T:&H&\rightarrow&L^2((0,T);U) \\ &y^0&\mapsto&\mathcal{U}_T(y^0) \end{array}$$
is a continuous linear map.
The main results of this section are the following ones.
Theorem 3 (exact controllability)
Let $$T>0\ .$$ The control system (4) is exactly controllable in time $$T$$ if and only if there exists $$c>0$$ such that
$\tag{12} \int_0^T\|B^*S(t)^* z\|_U^2dt \geqslant c \|z\|_{H}^2, \, \forall z \in D(A^*).$
Moreover, if such a $$c>0$$ exists and if $$c^T$$ is the maximum of the set of $$c>0$$ such that (12) holds, one has
$\tag{13} \left\|\mathcal{U}^T\right\|_{\mathcal{L}(H;L^2((0,T);U))}=\frac{1}{\displaystyle \sqrt{c^T}}.$
Theorem 4 (approximate controllability)
The control system (4) is approximately controllable in time $$T$$ if and only if, for every $$z\in H\ ,$$
$\tag{14} (B^*S(\cdot)^*z =0 \text{ in }L^2((0,T);U))\Rightarrow (z=0).$
Theorem 5 (null controllability)
Let$$T>0\ .$$ The control system (4) is null controllable in time $$T$$ if and only if there exists $$c>0$$ such that $\tag{15} \int_0^T\|B^*S(t)^* z\|_U^2dt \geqslant c \|S(T)^*z\|^2_{H}, \, \forall z \in D(A^*).$
Moreover, if such a $$c>0$$ exists and if $$c_T$$ is the maximum of the set of $$c>0$$ such that (15) holds, then $\tag{16} \left\|\mathcal{U}_T\right\|_{\mathcal{L}(H;L^2((0,T);U))}=\frac{1}{\displaystyle \sqrt{c_T}}.$
Theorem 6 (null controllability/approximate controllability)
Assume that, for every $$T>0\ ,$$ the control system (4) is null controllable in time $$T\ .$$ Then, for every $$T>0\ ,$$ the control system (4) is approximately controllable in time $$T\ .$$
For a proof of these theorems, see, for example Section 2.3.2 in (Jean-Michel Coron, 2007). Inequalities (12) and (15) are usually called observability inequalities for the abstract linear control system $$\dot y =Ay +Bu\ .$$ The difficulty is to prove them! For this purpose, there are many methods available (but still many open problems). Among these methods, let us mention in particular
• Multiplier methods. See in particular, (Jacques-Louis Lions, 1988), (Vilmos Komornik, 1994) and (Enrique Zuazua, 2006). For a simple example where this method is used, see at this link.
• Microlocal analysis. See in particular (Claude Bardos, Gilles Lebeau and Jeffrey Rauch, 1992) and the appendix 2 of the book (Jacques-Louis Lions, 1988).
Remark. In contrast with Theorem 6, note that, for a given$$T>0\ ,$$ the null controllability in time $$T$$ does not imply the approximate controllability in time $$T\ .$$ For example, let $$L>0$$ and let us take $$H:=L^2(0,L)$$ and $$U:=\{0\}\ .$$ We consider the linear control system
$\tag{17} y_t+y_x=0,\, t\in (0,T),\, x\in (0,L),$
$\tag{18} y(t,0)=u(t)=0,\, t\in (0,T).$
Through examples in the next section, we shall see how to put this control system in the abstract framework $$\dot y =Ay +Bu\ .$$ As one can see in the section at this link, whatever $$y^0\in L^2(0,L)$$ is, the solution to the Cauchy problem
$$y_t+y_x=0,\, t\in (0,T),\, x\in (0,L),$$
$$y(t,0)=u(t)=0,\, t\in (0,T),$$
$$y(0,x)=y^0(x),\, x\in (0,L),$$
satisfies
$$y(T,\cdot)=0, \text{ if } T\geqslant L.$$
In particular, if $$T\geqslant L\ ,$$ the linear control system (17)-(18) is null controllable but is not approximately controllable.
Again, there are two possibilities to study numerically the controllability of a linear control system: direct methods, duality methods. The most popular ones use duality methods and in particular the Hilbert Uniqueness Method (HUM) introduced in (Jacques-Louis Lions, 1988). For the numerical approximation, one uses often discretization by finite difference methods. However a new problem appear: the control for the discretized model does not necessarily lead to a good approximation to the control for the original continuous problem. In particular, the classical convergence requirements, namely stability and consistency, of the numerical scheme used does not suffice to guarantee good approximations to the controls that one wants to compute. Observability/controllability may be lost under numerical discretization as the mesh size tends to zero. To overcome this problem, several remedies have been used, in particular, filtering, Tychonoff regularization, multigrid methods, and mixed finite element methods. For precise informations and references, we refer to the survey papers (Enrique Zuazua, 2005; 2006).
Complements
On the controllability of linear PDE, we have already given references to books and papers. But there are of course many other references which must also be mentioned. If one restricts to books or surveys we would like to add in particular (but this is a very incomplete list):
• The survey (Fatiha Alabau-Boussouira and Piermarco Cannarsa, 2008) which deals, in particular, with abstract evolution equations, wave equations, heat equations and quadratic optimal control for linear PDE.
• The book (Alain Bensoussan, 1992) on stochastic control.
• The books (Alain Bensoussan, Giuseppe Da Prato, Michel Delfour and Sanjoy Mitter, 1992; 1993) which deal, in particular, with differential control systems with delays and partial differential control systems with specific emphasis on controllability, stabilizability and the Riccati equations.
• The book (Ruth Curtain and Hans Zwart, 1995) which deals with general infinite-dimensional linear control systems theory. It includes the usual classical topics in linear control theory such as controllability, observability, stabilizability, and the linear-quadratic optimal problem. For a more advanced level on this general approach, one can look at the book (Olof Staffans, 2005).
• The book (René Dáger and Enrique Zuazua, 2006) on partial differential equations on planar graphs modeling networked flexible mechanical structures (with extensions to the heat, beam and Schrödinger equations on planar graphs).
• The book (Abdelhaq El Jaï and Anthony Pritchard, 1988) on the input-output map and the importance of the location of the actuators/sensors for a better controllability/observability.
• The books (Hector Fattorini, 1999; 2005) on optimal control for infinite-dimensional control problems (linear or nonlinear, including partial differential equations).
• The book (Andrei Fursikov, 2000) on the study of optimal control problems for infinite-dimensional control systems with many examples coming from physical systems governed by partial differential equations (including the Navier-Stokes equations).
• The book (Vilmos Komornik and Paola Loreti, 2005) on harmonic (and nonharmonic) analysis methods with many applications to the controllability of various time-reversible systems.
• The book (John Lagnese and Günter Leugering, 2004) on optimal control on networked domains for elliptic and hyperbolic equations, with a special emphasis on domain decomposition methods.
• The books (Irena Lasiecka and Roberto Triggiani, 2000a; 2000b) which deal with finite horizon quadratic regulator problems and related differential Riccati equations for general parabolic and hyperbolic equations with numerous important specific examples.
• The survey (David Russell, 1978) which deals with the hyperbolic and parabolic equations, quadratic optimal control for linear PDE, moments and duality methods, controllability and stabilizability.
• The book (Marius Tucsnak and George Weiss, 2006) on passive and conservative linear systems, with a detailed chapter on the controllability of these systems.
• The survey (Enrique Zuazua, 2006) on recent results on the controllability of linear partial differential equations. It includes the study of the controllability of wave equations, heat equations, in particular with low regularity coefficients, which is important to treat semi-linear equations, fluid-structure interaction models.
There are many important problems which are not discussed in this paper. Perhaps the more fundamental ones are optimal control theory and the stabilization problem. For the optimal control theory, see references already mentioned above. The stabilization problem is the following one. We have an equilibrium which is unstable (or not enough stable) without the use of the control. Let us give a concrete example. One has a stick that is placed vertically on one of his fingers. In principle, if the stick is exactly vertical with a speed exactly equal to 0, it should remain vertical. But, due to various small errors (the stick is not exactly vertical, for example), in practice, the stick falls down. In order to avoid this, one moves the finger in a suitable way, depending on the position and speed of the stick; one uses a feedback law (or closed-loop control) which stabilizes the equilibrium. The problem of the stabilization is the existence and construction of such stabilizing feedback laws for a given control system. More precisely, let us consider the control system (1) and let us assume that $$f(0,0)=0\ .$$ The stabilization problem is to find a feedback law $$y\rightarrow u(y)$$ such that 0 is asymptotically stable for the closed loop system $$\dot y = f(y,u(y))\ .$$
Again, as for the controllability, the first step to study the stabilization problem is to look at the linearized control system at the equilibrium. Roughly speaking one expects that a linear feedback which stabilizes (exponentially) the linearized control system stabilizes (locally) the nonlinear control system. This is indeed the case in many important situations. For example, for the Navier control system mentioned in the section Controllability of nonlinear control systems, see in particular (Viorel Barbu, 2003), (Viorel Barbu and Roberto Triggiani, 2004), (Viorel Barbu, Irena Lasiecka and Roberto Triggiani, 2006), (Andrei Fursikov, 2004), (Jean-Pierre Raymond, 2006; 2007) and (Rafael Vázquez, Emmanuel Trélat and Jean-Michel Coron, 2008).
When the linearized control system cannot be stabilized it still may happen that the nonlinearity helps. This is for example the case for the Euler control system described in the section at this link. See (Jean-Michel Coron, 1999) and (Olivier Glass, 2005). (See also at this link for the controllability.)
The most popular approach to construct stabilizing feedbacks relies on Lyapunov functions. See Chapter 12 in (Jean-Michel Coron, 2007) for various methods to design Lyapunov functions.
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Sponsored by: Jean-Jacques Slotine, Nonlinear Systems Lab, MIT, Cambridge, MA
Reviewed by: Anonymous
Accepted on: 2009-11-04 15:22:43 GMT
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http://mathhelpforum.com/calculus/167085-derivate.html
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# Thread:
1. ## Derivate
Hello everyone!
I have to prove that the weak derivate of function $h(x)=sinx$ on interval $(0,2\pi)$ is function $g(x)=cosx$.
Anybody know any hint how to do this?
Thank you in advance!
2. Have a go at
$\displaystyle \lim_{h\to 0 }\dfrac{\sin (x+h)- \sin x}{h}$
3. Ok so we get: $\displaystyle \lim_{h\to 0 }\dfrac{\sin (x+h)- \sin x}{h}=cosx$. So the interval is not important here and weak derivate we prove at the same way like "normal" derivate?
4. Originally Posted by cantona
Ok so we get: $\displaystyle \lim_{h\to 0 }\dfrac{\sin (x+h)- \sin x}{h}=cosx$. So the interval is not important here and weak derivate we prove at the same way like "normal" derivate?
Use the fact that $\sin(A+B)=\sin(A)\cos(B)+\sin(B)\cos(A)$. This will allow you to simplify the left hand side and then cancel "h".
5. Originally Posted by cantona
Ok so we get: $\displaystyle \lim_{h\to 0 }\dfrac{\sin (x+h)- \sin x}{h}=cosx$. So the interval is not important here and weak derivate we prove at the same way like "normal" derivate?
If a function, h(x), is differentiable in the conventional sense, then it's weak derivative is equivalent to it's conventional derivative, g(x). That is to say any function, f(x), which which is equal to the its conventional derivative, g(x), except over some set of Lebesgue measure zero, is also a weak derivative of the function, h(x).
I'm not sure the responses reflect that this refers to Lebesgue integration.
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http://nrich.maths.org/4926/solution?nomenu=1
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## 'Magic Potting Sheds' printed from http://nrich.maths.org/
### Show menu
We received lots of good solutions to this problem - well done everyone! Many of you spotted that Mr McGregor should put $7$ plants in his potting shed at the beginning, and put $8$ plants in each garden. Well done to Henry from Finton House School, Ruth from Manchester High School for Girls, Liam from Wilbarston School, Mel from Christ Church Grammar School, Rachel from Beecroft Public School in Australia, Yanqing from Devenport High School for Girls and Daniel from Junction City High School for their detailed explanations of how they arrived at the answer.
Henry from Finton House School wrote:
"$1 \times2 \times2 \times2 = 8$.
8 therefore seems likely to be the number in the garden. Let's try it.
The number in the shed at the end must be the number in the garden.
Now what number do we double to get to $8$? It must be $4$.
$4 + 8 = 12$. $12$ divided by $2 = 6$.
$6 + 8 = 14$. $14$ divided by $2 = 7$."
Liam used similar logic:
"Just work backwards from the last garden. Imagine there to be $8$ plants in each garden. (You can't have odd numbers in a garden as the last garden must be double the whole number of plants left after the 2nd garden was planted. I chose $8$ because it's a conveniently sized power of 2.) There must have been $4$ plants left after the 2nd garden was planted so before it was planted there must have been $12$ which is double $6$. $6+8=14$. So Mr McGregor needs to put $14/2$ or $7$ plants in his magic potting shed at the beginning!"
Yanqing and Rachel used algebra. Here is Yanqing's solution:
"First, we make the number of plants put in the shed $n$, and the number planted each night $x$. So by the first morning, the number has doubled to $2n$ in the shed. We plant $x$ of them, leaving $2n-x$ in the shed overnight. By the second morning, we have $2(2n-x)=4n-2x$ in the shed. Planting $x$ of them, we are left with $4n-2x-x=4n-3x$ in the shed. By the third morning, there should be $2(4n-3x)=8n-6x$ plants in the shed. There need to be $x$ plants in the shed, as we need to plant all of them, so $8n-6x=x$ and $8n=7x$.
We can now say that the ratio of $n$ to $x$ is $7:8$, so the smallest values for $n$ and $x$, where they are both positive whole numbers, are obviously $7$ and $8$.
Other numbers which will work are all multiples of $7$."
Rachel also found that $8n= 7x$ and concluded that:
"Now you can see that $8n$ or $7x$ could equal $56$, which makes $n = 7$ and $x = 8$.
This works when you try it out, and if you multiply both numbers by another number, those new numbers work too."
Daniel concluded that:
"If you want to have the same amount of plants in each garden you must start with a multiple of $7$ plants in the shed and each day plant the same multiple of $8$ plants in the garden."
James from C.G.S.B found such a solution:
"Start with $35$ ... then put $40$ in each garden"
And so did Mel from Christ Church Grammar School!
"You start off with $301$ plants in the shed. You put $344$ in each garden."
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http://aimath.org/textbooks/beezer/Qarchetype.html
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Archetype Q
Archetype Q Summary: Linear transformation with equal-sized domain and codomain, so it has the potential to be invertible, but in this case is not. Neither injective nor surjective. Diagonalizable, though.
◊ A linear transformation: (Definition LT)
\begin{equation*} \ltdefn{T}{\complex{5}}{\complex{5}}, \lt{T}{\colvector{x_1\\x_2\\x_3\\x_4\\x_5}}= \colvector{-2 x_1 + 3 x_2 + 3 x_3 - 6 x_4 + 3 x_5\\ -16 x_1 + 9 x_2 + 12 x_3 - 28 x_4 + 28 x_5\\ -19 x_1 + 7 x_2 + 14 x_3 - 32 x_4 + 37 x_5\\ -21 x_1 + 9 x_2 + 15 x_3 - 35 x_4 + 39 x_5\\ -9 x_1 + 5 x_2 + 7 x_3 - 16 x_4 + 16 x_5} \end{equation*}
◊ A basis for the null space of the linear transformation: (Definition KLT)
\begin{equation*} \set{\colvector{3\\4\\1\\3\\3}} \end{equation*}
◊ Injective: No. (Definition ILT)
Since the kernel is nontrivial Theorem KILT tells us that the linear transformation is not injective. Also, since the rank can not exceed 3, we are guaranteed to have a nullity of at least 2, just from checking dimensions of the domain and the codomain. In particular, verify that
\begin{align*} \lt{T}{\colvector{1\\3\\-1\\2\\4}}&=\colvector{4\\55\\72\\77\\31}& \lt{T}{\colvector{4\\7\\0\\5\\7}}&=\colvector{4\\55\\72\\77\\31} \end{align*}
This demonstration that $T$ is not injective is constructed with the observation that
\begin{align*} \colvector{4\\7\\0\\5\\7}&=\colvector{1\\3\\-1\\2\\4}+\colvector{3\\4\\1\\3\\3} \end{align*} and \begin{align*} \vect{z}&=\colvector{3\\4\\1\\3\\3}\in\krn{T} \end{align*}
so the vector $\vect{z}$ effectively "does nothing" in the evaluation of $T$.
◊ A basis for the range of the linear transformation: (Definition RLT)
Evaluate the linear transformation on a standard basis to get a spanning set for the range (Theorem SSRLT):
\begin{equation*} \set{ \colvector{-2\\-16\\-19\\-21\\-9},\,\colvector{3\\9\\7\\9\\5},\, \colvector{3\\12\\14\\15\\7},\,\colvector{-6\\-28\\-32\\-35\\-16},\, \colvector{3\\28\\37\\39\\16} } \end{equation*}
If the linear transformation is injective, then the set above is guaranteed to be linearly independent (Theorem ILTLI). This spanning set may be converted to a nice'' basis, by making the vectors the rows of a matrix (perhaps after using a vector reperesentation), row-reducing, and retaining the nonzero rows (Theorem BRS), and perhaps un-coordinatizing. A basis for the range is:
\begin{equation*} \set{ \colvector{1\\0\\0\\0\\1},\,\colvector{0\\1\\0\\0\\-1},\, \colvector{0\\0\\1\\0\\-1},\,\colvector{0\\0\\0\\1\\2} } \end{equation*}
◊ Surjective: No. (Definition SLT)
The dimension of the range is 4, and the codomain ($\complex{5}$) has dimension 5. So $\rng{T}\neq\complex{5}$ and by Theorem RSLT the transformation is not surjective.
To be more precise, verify that $\colvector{-1\\2\\3\\-1\\4}\not\in\rng{T}$, by setting the output equal to this vector and seeing that the resulting system of linear equations has no solution, i.e. is inconsistent. So the preimage, $\preimage{T}{\colvector{-1\\2\\3\\-1\\4}}$, is empty. This alone is sufficient to see that the linear transformation is not onto.
◊ Subspace dimensions associated with the linear transformation. Examine parallels with earlier results for matrices. Verify Theorem RPNDD. \begin{align*} \text{Domain dimension: }5&& \text{Rank: }4&& \text{Nullity: }1 \end{align*}
◊ Invertible: No.
Neither injective nor surjective. Notice that since the domain and codomain have the same dimension, either the transformation is both onto and one-to-one (making it invertible) or else it is both not onto and not one-to-one (as in this case) by Theorem RPNDD.
◊ Matrix representation (Theorem MLTCV):
\begin{equation*} \ltdefn{T}{\complex{5}}{\complex{5}}, \lt{T}{\vect{x}}=A\vect{x}, A= \begin{bmatrix} -2&3&3&-6&3\\ -16&9&12&-28&28\\ -19&7&14&-32&37\\ -21&9&15&-35&39\\ -9&5&7&-16&16 \end{bmatrix} \end{equation*}
◊ Eigenvalues and eigenvectors (Definition EELT, Theorem EER):
\begin{align*} \eigensystem{T}{-1}{\colvector{0\\2\\3\\3\\1}}\\ \eigensystem{T}{0}{\colvector{3\\4\\1\\3\\3}}\\ \eigensystem{T}{1}{\colvector{5\\3\\0\\0\\2},\,\colvector{-3\\1\\0\\2\\0},\,\colvector{1\\-1\\2\\0\\0}} \end{align*}
◊ A diagonal matrix representation relative to a basis of eigenvectors, $B$. \begin{align*} B&= \set{ \colvector{0\\2\\3\\3\\1},\,\colvector{3\\4\\1\\3\\3},\, \colvector{5\\3\\0\\0\\2},\,\colvector{-3\\1\\0\\2\\0},\, \colvector{1\\-1\\2\\0\\0}} \\ &\\ \matrixrep{T}{B}{B}&=\begin{bmatrix} -1&0&0&0&0\\ 0&0&0&0&0\\ 0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1 \end{bmatrix} \end{align*}
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http://quant.stackexchange.com/questions/254/black-equivalent-volatility
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# Black-Equivalent Volatility
Can Someone Explain to me what this term means, and how it's used?
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1
Depending on the context, I think this might mean the same thing as "Black Implied Volatility." The Black Model is to FX Options, Caplets, etc as Black Scholes is to Equity Options. So Black-Equivalent Volatility is likely the volatility that, if using the closed form Black model pricing formula, gives the same price as the market. – phlsmk Feb 8 '11 at 13:08
I think what you said makes sense. I came to the same realization. – dragunov Feb 9 '11 at 10:51
## 1 Answer
Quite a lot of options on asset $S(t) > 0$ have a payoff at tinme $T$ equal (at least approximately -- it's a bit more complicated in the case of e.g. credit index options)
$(S(T) - K)^+$
You can always find a number $\sigma$ such that, when plugged into Black formula together with strike $K$, spot price $S(t)$, interest rate $r$ and time to expiry $T-t$, you will recover the market price of the option $V(t)$. This number is called the Black implied volatility of the option. Basically, it's a quoting convention for the option prices. Traders use it because:
• it makes it easier for them to compare prices of options on different days, with different strikes
• Black vols tend to be similar across strikes and expiries (not always!)
• it is better (in the sense: you're less likely to suffer lots of arbitrage) to interpolate market prices in the $\sigma$ space then directly; that is, if prices for strikes $K_1$ and $K_2$ are quoted, it's better to use some interpolation method on their Black vols $\sigma_1$ and $\sigma_2$ than on their prices $V_1$ and $V_2$
• it fits their intution better (a paramount argument)
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Thanks, very informative post :-) – dragunov Feb 9 '11 at 10:52
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http://quant.stackexchange.com/questions/1356/diversification-rebalancing-and-different-means
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# Diversification, Rebalancing and Different Means
I have found many financial authors making generalizations about GM and AM but they are wrong in certain circumstances. Could someone explain their reasoning?
My fact why they are wrong is based Jensen inequality:
$\sum^{n}_{i=1} p_{i} f(x_{i}) \geq f( \sum_{i=1}^{n} p_{i} x_{i})$
for concave up functions, verified here. Now the special case is:
$\sqrt[n]{x_{1}x_{2}...x_{n}} \leq \frac{x_{1}+x_{2}+...+x_{n}}{n}$
For concave down functions, the corresponding result can be obtained by reversing the inequality, more here.
Author 1
Rebalancing and diversification go hand in hand. There is no diversification benefit without rebalancing – otherwise the total return will simply be the weighted average of the long-term geometric returns. If you don’t rebalance to asset types, you will get no diversification benefit. If you can’t rebalance to an asset type, you cannot get diversification benefits.
Rebalancing benefits increase as volatility rises, and decreases in less volatile times. The benefit of rebalancing after a 10% movement is more than 10 times the benefit after a 1% movement, and the benefit from rebalancing after a 50% move is more than 5 times the benefit after a 10% move. The greatest benefit comes in times, like 2008-2009, when there are wild movements in portfolios. -- diversification does not assure profit or protect against loss in a declining that affects numerous asset types. Source.
Suppose we have a concave-down environment so
$\sqrt[n]{x_{1}x_{2}...x_{n}} \geq \frac{x_{1}+x_{2}+...+x_{n}}{n}$.
1. Now following the logic in the paragraph the total return is simply the weighted average of the long-term geometric returs. We know from the latter result that it is greater than or equal to the arithmetic mean. Is there a diversification benefit with rebalancing?
2. Then the numbers thrown in the next bolded sentence are a bit odd. Why would there be such a benefit if we just noticed that it is not necessarily so if our environment is concave-down environment (valuations going down)?
3. But the last sentence saves this writer! No error here.
Author 2.
William Bernstein here goes one step further, ignoring the Jensen:
The effective (geometric mean) return of a periodically rebalanced portfolio always exceeds the weighted sum of the component geometric means.
The implicit premise behind such statements probably is that the market in the long-run is rising, very well right to some extent. But with that assumption, the problem of asset allocation simplifies to the Jensen -- an even with such premise, it should be noticed to the reader (or such sentences are wrong).
Author 3.
Many works rely on the ambiguous assertions such as Ilmanen's Expected Return -book, page 485:
the arithmetic mean of a series is always higher than the geometric mean (AM > GM) except when there is a zero volatility (AM = GM). A simple Taylor series expansion show a good approximate is $GM \approx AM - \frac{Variance}{2}.$
...but the bolded sentence is wrong. You can make a function with concave values and $AM < GM$.
Questions
1. What are the authors meaning here?
2. Are they wrong or do they have some some hidden premises that I am missing?
3. Why are they stating such issues about AM and GM as they cannot always be true?
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I have no idea what "rebalancing" is. Judging from the amount of activity on this question, I'm not the only one. – Gerry Myerson Jun 26 '11 at 0:57
@Gerry Suppose that you have a stock portfolio built out of many different stocks of differing expected returns and volatility (roughly, standard deviation of rate of return). When you design the portfolio, you choose the stocks to be in particular ratios to minimize variance for a particular rate of return. However, if you then leave the portfolio to its own devices, the differences in returns will change all of the ratios, making things less efficient. Rebalancing is the act of recalculating the ratios you want and buying/selling shares so the portfolio matches the desired ratios. – Aaron Jun 26 '11 at 5:26
@Aaron, thanks. I hope that with this additional information someone takes an interest in your question. If not, maybe there's another forum more attuned to investment problems. – Gerry Myerson Jun 26 '11 at 6:21
@Ben: yes you are actually right, it requires understanding of math and also some finance to really understand the question. How to move there? I have flagged it for moderation attention to move it. – hhh Jun 26 '11 at 16:15
The main question to answer before we can address the posting question is: how are we measuring the benefit of rebalancing? Is it strictly a matter of volatility? Strictly expected returns? Some combination of the two? It doesn't make any sense to compare the benefits in the way they do without a particular metric in mind. Does the paper list one? – Aaron Jun 29 '11 at 1:56
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## 1 Answer
I believe the assumption here in the first quote is that returns are either strictly positive or strictly negative and the authors are comparing the effect of volatility on geometric returns to arithmetic. The issue of diversification benefits have little to do with this difference as opposed to a time varying covariance matrix. The benefits of diversification in volatile times assumes low correlation and the assumption that asset correlations do not have a strong positive correlation with volatility. This is a common simplification for teaching basic finance. The second quote will be true when the ratio of return to risk is approximately equal across assets and assets are not perfectly positively correlated. The third quote is true on a long time horizon because real interest rates are positive. This follows from the fundamental assumption that a dollar today is worth more than a dollar tomorrow.
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http://crypto.stackexchange.com/questions/3442/aes-encryption-the-relevance-of-static-matrix-in-mixcolumns-operation?answertab=oldest
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# Aes encryption -The relevance of static matrix in mixcolumns operation
Can someone explain to me the relevance of the static matrix used for the mixcolumns operation in aes encryption.i.e the relevance of why the byte is multiplied by 2 + next byte multiplied by 3 + next byte +next byte
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Not a programming problem. Flagged to move to crypto. – erickson Aug 2 '12 at 17:16
Welcome to Cryptography Stack Exchange. Your question was migrated here because of being not directly related to software development (the topic of Stack Overflow), and being fully on-topic here. Please register your account here, too, to be able to comment and accept an answer. – Paŭlo Ebermann♦ Aug 7 '12 at 17:16
## 1 Answer
Well, how that matrix works is important for the security properties of AES. To see why it is important, we need to consider state differences, that is, if we consider two different inputs to AES and how they both flow through the cipher, how does those two differ in the internal state of the cipher at various places.
The important property of the mixcolumn operation is that it is a Maximum Distance Separable (or MDS) operation. That is, if we consider two distinct inputs to a single column of the mixcolumn operation, and the two inputs differ in $A$ bytes, and if the two corresponding outputs differ in $B$ bytes, then $A+B \ge 5$.
This implies that if two consider two inputs to a single column of the mixcolumn, if those two inputs differ in a single byte (that is, $A=1$), then the two outputs will always have different values for each byte (that is, $B=4$).
If you go through how this applies to AES, that means that if round $I$, two encryptions has an internal state that differs in a single byte, then at round $I+2$, the two encryptions will have states that differs in all 16 bytes (and hence AES has wonderful avalanche).
More generally, this behavior allows us to prove strength against differential and linear attacks; in that any differential or linear characteristic will necessarily go through a large number of sboxes; the resulting probability that the differential/linear characteristic will survive all those sboxes is negligible.
All this may not be true (or, at least, would be considerably harder to prove), if we replace the mixcolumn operation with an operation that is not MDS.
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http://4dpiecharts.com/2011/01/16/when-1-x-x/
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# 4D Pie Charts
Scientific computing, data viz and general geekery, with examples in R and MATLAB.
Home > MATLAB, R > When 1 * x != x
## When 1 * x != x
16th January, 2011
Trying to dimly recall things from my maths degree, it seems that in most contexts the whole point of the number one is that it is a multiplicative identity. That is, for any x in your set, 1 * x is equal to x. It turns out that when you move to floating point numbers, in some programming lanugages, this isn’t always true.
In R, try the following
```x <- Inf + 0i
1 * x
[1] Inf + NaNi
```
Something odd is happening, and my first reaction was that this is a bug. In fact, I reported it as such (in a slightly different form) but the terrifyingly wise Brian Ripley came up with the cause.
When you multiply two complex numbers, this is what happens
$(a + bi)(c + di) = (ac - bd) + (ad + bc)i$
In this instance
$(Inf + 0i)(1 + 0i) = (Inf * 1 - 0 * 0) + (Inf * 0 + 0 * 1)i = Inf + NaNi$
So, there is a reasonable argument that R is doing the right thing.
MATLAB works a little differently since all it’s numbers are implicitly complex. The inner workings of MATLAB are somewhat opaque for commercial reasons, but in order for compiled MATLAB code to work, I think it’s reasonable to assume that MATLAB stores its variables in a class that is similar to the MWArray class used by compiled code. As far as I can tell, each `double` matrix contains two matrices for storing real and imaginary components, and has an `IsComplex` property that is true whenever the complex part has any nonzero values. If `IsComplex` is false, only the real matrix is used for arithmetic.
Thus in MATLAB, when you define `x = Inf + 0i` it immediately resolves itself to simply `Inf`, and we don’t get the same weirdness.
```x = Inf + 0i
1 * x
Inf
```
Other languages are divided on the subject: python and ruby agree with R but Mathematica (or at least Wolfram Alpha) agrees with MATLAB.
$\lim_{n \to \infty} (n + 0i)(1 + 0i) = \lim_{n \to \infty} (n * 1 - 0 * 0) + (n * 0 + 0 * 1)i = \lim_{n \to \infty} n = Inf$
This concurs with the MATLAB answer and I’m inclined to agree that it makes more sense, but the issue isn’t entirely clear cut. Changing the way complex arithmetic works for a single edge case is likely to introduce more confusion than clarity. It is perhaps sufficient for you to remember that complex infinity is a bit pathological with arithemtic in some languages, and add checks in your code if you think that that will cause a problem.
### Like this:
Tags: complex numbers, matlab, r
1. 17th January, 2011 at 6:38 am | #1
A more rigorous mathematical proof would be to look at the norm. There is only one “infinity” in the complex plane. And R knows it:
> is.finite(Inf+3i)
[1] FALSE
even though its maths are wrong:
> Inf*1i
[1] NaN+Infi
(same reason as yours!)
2. 17th January, 2011 at 19:29 pm | #2
Another way of understanding the two different results comes from thinking about limits. I’ve discussed your example in that context in this post.
1. 17th January, 2011 at 22:57 pm | #1
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http://physics.stackexchange.com/questions/47288/are-carnot-engine-efficieny-and-fourier-heat-trasmission-law-related?answertab=oldest
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# Are Carnot engine efficieny and Fourier heat trasmission law related?
It just occured to me that the efficiency of Carnot cycles is $\eta= \frac{T_1 - T_2}{T_1}$, that is, the efficiency decreases as the difference between reservoir temperatures decreases. On the other side, Fourier's law states that the dissipation of heat is proportional to the temperature gradient, that is, to the temperature difference.
My question, then, is: are these two results related? Do they both have a common cause?
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2
It's a good question (+1), but to the best of my knowledge the answer is no. – Nathaniel Dec 21 '12 at 3:23
Fourier heat transmission law is like Hooke's law. They are empirical approximation. – C.R. Dec 21 '12 at 5:14
## 1 Answer
No, they are not directly related.
The law of heat conduction you cite can be thought of as one example of a general diffusion process, and involves retaining the first terms in a Taylor expansion of fluxes through zone boundaries. (see, for example, these MIT Introduction to Solid State Chemistry lecture notes on diffusion). On the other hand, the Carnot efficiency is an exact expression for the work that an engine undergoing a single Carnot cycle performs divided by the total energy flowing into the engine as heat during that cycle.
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I understand. However, they must both have a "cause" that explains them, don't them? Perhaps using statistical mechanics? My interest was if the causes or both phenomena could be traced back to a common one. – carllacan Dec 21 '12 at 14:50
2
I do not think it is useful to look for a common cause connecting these formulae in this case. They are both expressed in terms of temperatures, and so they manifestly demonstrate their accordance with the second law of thermodynamics, but then again all laws of nature possess this feature. I do not think there is a connection that runs deeper than that. Something else to notice is that the formula for Carnot efficiency depends on the ratio of the two temperatures ($1 - T_2/T_1$), whereas a nondimensional formulation for heat conduction still depends on the temperature difference. – kleingordon Dec 22 '12 at 21:22
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http://math.stackexchange.com/questions/256181/lifting-a-continuous-linear-functional
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# Lifting a continuous linear functional
A homework problem:
Let $\phi:X \twoheadrightarrow Y$ be a continuous surjection between compact Hausdorff spaces. Let $\lambda$ be a continuous linear functional on $C(Y)$.
Show that there is a continuous linear functional $\theta$ on $C(X)$ such that:
1. $\|\theta\|=\|\lambda\|$
2. $\theta(f \circ \phi)=\lambda(f)$ and any $f \in C(Y)$
My attempt: I am stuck about in the beginning. I read about the Hahn-Banach theorems in Rudin's functional analysis. Those theorems seem relevant because they talk about existence of functionals, but there are several versions and I'm confused about which of them to use, if any (I am allowed to quote this book). I think maybe the second condition can tell me how to define a functional on some subspae of $C(X)$ and then use an "extension theorem" (that's how some of those theorems are refered to in the book), but that's just general intuition.
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## 1 Answer
Hints:
1) Show that $E=\{f\circ \varphi:f\in C(Y)\}$ is linear subsapce of $C(Y)$.
2) Show that $f_1\circ\varphi=f_2\circ\varphi\implies f_1=f_2$
3) Using 2) show that you have well defined functional $\theta_0:E\to\mathbb{C}:g\mapsto \lambda(f)$ where $g=f\circ\varphi$.
4) Show that $\Vert f\circ\varphi\Vert=\Vert f\Vert$
5) Using 4) prove that $\Vert\theta_0\Vert=\Vert\lambda\Vert$
6) Using Hahn-Banach get norm preserving extenesion $\theta$ of $\theta_0$. This will be the desired functional
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