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http://mathoverflow.net/questions/77014/recovering-the-alexander-polynomial-from-ocneanus-homflypt/77023
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## Recovering the Alexander Polynomial from Ocneanu’s HOMFLYPT
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Let $H_q(n)$ denote the Hecke algebra associated to the symmetric group $S(n)$: this is the $\mathbb{Z}[q^{\pm 1/2}]$ algebra generated by $T_1, \ldots, T_{n-1}$ satisfying the braid relations along with $T_i^2=(q-1)T_i + q$. Ocneanu's trace $\tau_z (T_{\omega_{\mu}}) = z^{l(\omega_{\mu})}$ defined on fundamental elements is the unique normalized trace on our Hecke algebra that can be jiggled to yield invariants of oriented links: this gives the two-parameter HOMFLYPT polynomials $P_L(q,z)$. These polynomials satisfy the skein relation
$$\Big ( \frac{z}{z-q+1} \Big )^{1/2} P_{L_+} (q,z) - \Big ( \frac{z}{z-q+1} \Big )^{-1/2} P_{L_-}(q,z) = (q^{1/2} - q^{-1/2}) P_{L_0} (q,z).$$
which motivates the common change of variables $x = \sqrt{\frac{z}{z-q+1}}$, $y = q^{1/2} - q^{-1/2}$. Note that the target ring of this trace has to be at least $\mathbb{Z}[q^{\pm 1/2}, z^{\pm 1/2}, (z-q+1)^{ \pm 1/2}]$ if we want to write down the associated invariant $P_L(q,z)$.
Many people take the HOMFLYPT polynomials to be those obtained after the specialization $z=q^{N}/[N]$, which seems to be equivalent to $x=\sqrt{\frac{z}{z-q+1}} = q^{N/2}$. Setting $N=2$ recovers the Jones polynomial, and setting $N=0$ is supposed to recover the Alexander polynomial.
• How am I supposed to correctly obtain the Alexander polynomial in terms of the original $q$ and $z$?
It seems that $x=q^{0/2}=1$ gives the correct specialization. Still, doesn't $\sqrt{\frac{z}{z-q+1}}=1$ force $q=1$, leaving $z$ free? How does the right-hand side of the skein relation survive then?
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## 1 Answer
This confusion arises because you have defined the HOMFLY polynomial over the ring $\mathbb{Z}(z,1/z,q,1/q)$. The correct ring is the ring generated by $z$,$1/z$, $q$, $1/q$, $\delta$ subject to the relation $z-1/z=\delta(q-1/q)$. This avoids denominators and can be specialised. This is a blow-up of the Laurent polynomial ring.
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I've edited the statement of the problem above. In this deduction, I'm not sure how to use this $\delta$ of yours. Thanks for your help! - Alex – Alexander Moll Oct 3 2011 at 16:47
Your question is hard to follow. If you want to see the conventions I use see arxiv.org/abs/1007.2579. In particular $\delta$ is the value of a single closed loop. I am worried about your $[0]=1/1-q$ as $[0]=0$. I would normally put $z=q^n$. Putting $n=0$ then means $z-1/z=0$ so $\delta=0$ or $q=\pm 1$. – Bruce Westbury Oct 3 2011 at 17:32
I've used the $q$ and $z$ from D. Goldschmidt's "Group Characters, Symmetric Functions, and Hecke Algebras" above. I'm hesitant to accept that $\delta$ is the value of a single closed loop, since HOMFLYPT returns 1 on the unknot katlas.org/wiki/Image:0_1.gif – Alexander Moll Oct 4 2011 at 3:15
You're right that $[0] = \frac{1}{1-q}$ can't be the right (unbalanced) $q$-analog of $0$ - I've edited the question accordingly – Alexander Moll Oct 4 2011 at 3:42
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http://math.stackexchange.com/questions/234562/domain-codomain-and-range/234565
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# Domain, codomain, and range
This question isn't typically associated with the level of math that I'm about to talk about, but I'm asking it because I'm also doing a separate math class where these terms are relevant. I just want to make sure I understand them because I think I may end up getting answers wrong when I'm over thinking things.
In my first level calculus class, we're now talking about critical values and monotonic functions. In one example, the prof showed us how to find the critical values of a function $$f(x)=\frac{x^2}{x-1}$$ He said we have to find the values where $f' (x)=0$ and where $f'(x)$ is undefined.$$f'(x)=\frac{x^2-2x}{(x-1)^2}$$
Clearly, $f'(x)$ is undefined at $x=1$, but he says that $x=1$ is not in the domain of $f(x)$, so therefore $x=1$ is not a critical value. Here's where my question comes in:
Isn't the "domain" of $f(x)$ $\mathbb{R}$, or $(-\infty,\infty)$? If my understanding of Domain, Codomain, and Range is correct, then wouldn't it be the "range" that excludes $x=1$?
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## 3 Answers
If $f:X\to Y$ is a function from $X$ to $Y$, we usually call $X$ the domain, $Y$ the codomain, and $f(X)$ the range. Some authors call $Y$ the range, in which case $f(X)$ is called the image.
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I'm still confused - does this mean then that it's correct to say that x=1 is not in the domain, as opposed to not in the range? My mental image of domain and range has been totally messed up thanks to my introduction to the formal definition of functions. – agent154 Nov 11 '12 at 1:04
Oh, ok.. I think I may understand now. If we want to get fussy, then $1/x$ is not really a function, because it is undefined at some points in the domain... thus if we exclude 0 from the domain, $1/x$ becomes a function. For this reason it is implicitly understood that 0 is not in the domain? Otherwise, would this be called a partial function? – agent154 Nov 11 '12 at 1:07
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@agent154: Yes, the function $f(x)=\frac1x$ is a partial function from $\Bbb R$ to $\Bbb R$ and a function from $\Bbb R\setminus\{0\}$ to $\Bbb R$. In the context of your calculus class it is indeed understood that $0$ is excluded from the domain. – Brian M. Scott Nov 11 '12 at 1:13
Strictly speaking, a function is a subset $f$ of $(\hbox{domain}) \times (\hbox{codomain})$ such that if $(a, b), (a, c) \in f$, then $b = c$. That is, a specification of the domain and codomain are part of the definition of a function. But this is usually considered too technical for a calculus course, which is what the original poster was talking about, so ...
... in the typical (American) calculus course, you consider real-valued functions defined on a subset of the reals, where a "function" is given by an equation like "$f(x) = \dfrac{x^2}{x - 1}$'', as in the original post. Then the convention is that "domain" means "the largest subset of the reals on which the expression in the defining equation is defined", unless there is a statement to the contrary. (I've heard this referred to as the "natural domain".) Thus, for this $f$, the (natural) domain by this convention is $\{x \mid x \ne 1\}$. And so $x = 1$ is not a critical point, since it's not in the "domain".
If for some reason I want a smaller subset, I must say so explicitly. For example, I might say "the function $f(x) = x^2$ defined on $[0, \infty)$".
Thus, in (calculus, precalculus, college algebra) textbooks you will see questions like: "What is the domain of $f(x) = \dfrac{1}{x}$?" By the strict definition of a function, the question makes no sense, since you haven't defined the function unless you've stated what the domain is up front. But by the convention described above, the domain is the nonzero reals.
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$x=1$ is not in the domain because when $x=1$, $f(x)$ is undefined. And by definition, strictly speaking, a function defined on a domain $X$ maps every element in the domain to one and only element in the codomain.
The domain and codomain of a function depend upon the set on which $f$ is defined and the set to which elements of the domain are being mapped; both are usually made explicit by including the notation $f: X \to Y$, e.g. along with defining $f(x)$ for $x\in X$.
$X$ is then taken to be the domain of $f$, and $Y$ the codomain of $f$, though you'll find that some people interchange the terms "codomain" and "range". So "range" is a bit ambiguous, depending on the text used and how it is defined, because "range" is sometimes defined to be the set of all values $y$ such that there is some $x \in X$ for which it is true that $f(x) = y$, i.e. $f[X]$.
One way to circumvent any ambiguity related to use of "range" to refer to $f[X]$ is to note that many prefer to define $f[X]$ to be the "image" of $X$ under $f$, often denoted by $\text{Im}f(x)$, with the understanding that $f[X] = \text{Im}f(x) \subseteq Y.\;\; f[X]=\text{Im}f(x) = Y$ when $f$ is onto $Y$.
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http://mathoverflow.net/questions/100676/uniformity-on-a-convex-subspace-of-mathbbrn
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Uniformity on a convex subspace of $\mathbb{R}^n$
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Let's consider a set of points (in cartesian coordinates) in a convex subspace of $\mathbb{R}^n$ and more precisely in a subspace of the standard (n-1)-simplex ($x_i \in[0,1]$ s.t. $\displaystyle \sum_{i=1}^{n} x_i = 1$ ).
A typical subspace of mine would be the $x_i \in[a_i,b_i]$ s.t $\displaystyle \sum_{i=1}^{n} x_i = 1$ where $a_i,b_i \in(0,1)$
How can you test (or measure how) those points are uniformly distributed over this subspace ?
For the moment the only hint I had looking at articles is the distance-to-boundary test but it seems hard to evaluate the minimum distance of a given point to the simplex (or to a simplex-subspace) "boundaries" with a general formula.
The aim is to implement this test in R.
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http://mathoverflow.net/questions/112079?sort=votes
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## What is Gödel’s pairing function on ordinals?
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I find many references to Gödel's pairing function on ordinals but I have not found a definition. What is it?
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## 3 Answers
Define an order on pairs of ordinals $(\alpha,\beta)$ by ordering first by maximum, then by first coordinate, then by second coordinate. That is, one pair preceeds another if the maximum is smaller, or they have the same maximum and the first coordinate is smaller, or they have the same maximum and first coordinate, but the second coordinate is smaller. This is clearly a linear order, and it is a well-order, since none of these three quantities can descend infinitely. Furthermore, every proper initial segment of the order is a set, consisting of pairs with the same or smaller maximum (and indeed, the reason for using the order-by-maximum part of the definition is precisely to ensure that the order is set-like; the lexical order itself is not set-like on Ord).
Thus, every pair $(\alpha,\beta)$ is the $\xi$ th element in this order for some unique $\xi$, we may view $\xi$ as the code of $(\alpha,\beta)$. Every pair has a unique code and every ordinal is a code.
This pairing function is highly robust and absolute, since the definition of the order is absolute to any model of even very weak set theories that contain those ordinals. Another attractive feature is that whenever $\kappa$ is an infinite cardinal (or even merely a sufficiently indecomposable ordinal), then $\kappa$ is closed under pairing, in the sense that any pair of ordinals below $\kappa$ is coded by an ordinal below $\kappa$. This is how we know $\kappa^2=\kappa$ for well-ordered cardinals. In particular, this method of coding also works on natural numbers.
But I'm unsure if this coding is due to Gödel or was known earlier (perhaps even Cantor?); so it may not be the answer you seek.
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Thanks. The absoluteness is just the kind of thing I wanted to check from the definition. – Colin McLarty Nov 11 at 16:15
I think that this coding is how Zermelo proved that $\aleph_\alpha\times\aleph_\alpha=\aleph_\alpha$. I'm not sure whether or not it was his discovery or someone else's and it can probably be checked in his 1904 paper. – Asaf Karagila Nov 11 at 16:26
Maybe the ordinal pairing functions are called Gödel coding not because Gödel invented this particular ordinal pairing function, but rather just because it is analogous to Gödel coding of sequences? – Joel David Hamkins Nov 11 at 16:32
Okay, according to Jech Set Theory historical notes the ordering is due to Hessenberg (from his book - which I couldn't find - "Grundbegriffe der Mengenlehre", 1906). – Asaf Karagila Nov 11 at 16:56
I have not checked the original sources, but I guess that Godel's pairing function is the inverse of this function described by Joel Hamkins. This inverse have a direct description in Shoenfield's Mathematical Logic, page 251. This (inverse) function is used by Shoenfield in the definition of the constructible model. One place to look is Godel's book on constructible sets and the consistency of GCH. – Rodrigo Freire Nov 11 at 22:07
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According to this .pdf file the definition is this:
Consider the canonical ordering on $\mathsf{Ord\times Ord}$: $$(\alpha,\beta)\prec(\gamma,\delta)\iff\begin{cases} \max\lbrace\alpha,\beta\rbrace\lt\max\lbrace\gamma,\delta\rbrace & \lor \\ \max\lbrace\alpha,\beta\rbrace=\max\lbrace\gamma,\delta\rbrace\land\alpha\lt\gamma&\lor\\ \max\lbrace\alpha,\beta\rbrace=\max\lbrace\gamma,\delta\rbrace\land\alpha=\gamma\land\beta\lt\gamma \end{cases}$$
The pairing function, if so, $G(\alpha,\beta)=\operatorname{otp}\lbrace(\gamma,\delta)\in\mathsf{Ord\times Ord}\mid(\gamma,\delta)\prec(\alpha,\beta)\rbrace$.
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In case it isn't clear: this is exactly the same order and coding as in my answer. – Joel David Hamkins Nov 11 at 17:57
Yes, I only saw Joel's answer after posting my own. It is not hard to see that we describe the same order. – Asaf Karagila Nov 11 at 18:16
Asaf and Joel have answered the question. Let me add a remark that expands the fact that it helps us prove that $\kappa\times$ and $\kappa$ have the same size. All the claims here can be verified rather easily. Multiplication and exponentiation are in the ordinal sense.
It is customary to write $\Gamma(\alpha,\beta)$ for the order type of the predecessors of $(\alpha,\beta)$ under the order than Asaf denotes $\prec$. For example, $\Gamma(\omega,\omega\cdot2)=\omega^2+\omega$.
An ordinal $\alpha$ is (additively) indecomposable iff $\alpha\gt 0$ and whenever $\beta,\gamma\lt\alpha$, then $\beta+\gamma\lt \alpha$. One can easily check that the indecomposable $\alpha$ are precisely those of the form $\omega^\beta$. Say that $\alpha$ is multiplicatively indecomposable iff $\alpha>0$ and $\beta\gamma\lt \alpha$ whenever $\beta,\gamma\lt\alpha$. Then $\alpha$ is multiplicatively indecomposable iff it is $1$ or has the form $\omega^{\omega^\beta}$.
Ok, we can now state the remark; unfortunately I would not know who to credit for this observation, I think of it as folklore: An ordinal $\alpha$ is multiplicatively indecomposable iff it is closed under Gödel pairing, that is, $\Gamma(\beta,\gamma)\lt\alpha$ whenever $\beta,\gamma\lt\alpha$. In particular, $\Gamma(\kappa,\kappa)=\kappa$ for any infinite cardinal $\kappa$, which of course implies that $\kappa\times\kappa$ and $\kappa$ have the same size. Also, if $\kappa$ is uncountable, then there are $\kappa$ ordinals $\alpha$ below $\kappa$ such that $\Gamma(\alpha,\alpha)=\alpha$. Of course, all of this works well in $\mathsf{ZF}$ and all the definitions involved are absolute.
I prefer a different approach when verifying that $\kappa\times\kappa$ and $\kappa$ have the same size, one that (again) is absolute and goes through in $\mathsf{ZF}$, but only requires the use of additively indecomposable ordinals: One first checks that there is a (recursive) bijection $h:\omega\times\omega\to\omega$ with $h(0,0)=0$. Then, given ordinals $\alpha,\beta$, use their Cantor's normal form to write them as $$\alpha= \omega^{\alpha_1}n_1 + \omega^{\alpha_2}n_2 + \dots + \omega^{\alpha_k}n_k$$ and $$\beta= \omega^{\alpha_1}n'_1 + \omega^{\alpha_2}n'_2 + \dots + \omega^{\alpha_k}n'_k$$ where $\alpha_1 \gt \alpha_2 \gt \dots \gt \alpha_k$ are ordinals, and $n_1,\dots,n_k, n'_1,\dots,n'_k$ are natural numbers. (Note that these representations are not unique, but at least one of $n_i$ and $n_i'$ is non-zero iff $\alpha_i$ appears as an exponent in the canonical form of $\alpha$ or $\beta$).
Now set $$H(\alpha,\beta)=\omega^{\alpha_1}h(n_1,n'_1)+\omega^{\alpha_2}h(n_2,n'_2)+\dots+ \omega^{\alpha_k}h(n_k,n'_k).$$ Then $H$ is a bijection between $\alpha\times\alpha$ and $\alpha$ whenever $\alpha$ is indecomposable. And an easy inductive argument, appealing to the explicit proof of Schröder-Bernstein, allows us to use $H$ to argue that there is, provably in $\mathsf{ZF}$, a class function that assigns to each infinite ordinal $\alpha$ a bijection between $\alpha\times\alpha$ and $\alpha$.
(Of course, the existence of this class function can also be argued from $\Gamma$, using that there are $\kappa$ ordinals $\alpha$ below $\kappa$ with $\Gamma(\alpha,\alpha)=\alpha$, but this second approach is somewhat easier.)
I found this argument a while ago, but then saw that Levy gives essentially the same approach in his textbook on set theory. Again, I am not sure who to credit for this construction, it seems to go back to Gerhard Hessenberg's 1906 book, "Grundbegriffe der Mengenlehre".
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In the comments to Joel's answer I wrote that Jech attributes this proof to Hessenberg. – Asaf Karagila Nov 11 at 17:02
Thanks, Asaf. I'll look it up. – Andres Caicedo Nov 11 at 17:59
It is basically the same idea as the Hessenberg (commutative) addition operation on ordinals. For that, you sort the two Cantor normal forms to have the same terms, as here, and just add coordinate-wise. The twist for coding is not to just add the similar terms, but also to apply a natural number pairing function also. – Joel David Hamkins Nov 11 at 18:09
Hmm... the attribution seems right. I have not seen Hessenberg's book, but Oliver Deiser's "Einführung in die Mengenlehre" describes Hessenberg's argument in page 301, and it is reasonably close to the one above. – Andres Caicedo Nov 11 at 18:09
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Thanks Asaf and Joel! I didn't expect this argument to go back this far. – Andres Caicedo Nov 11 at 18:10
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http://math.stackexchange.com/questions/101676/proof-of-no-simple-group-of-order-992
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# Proof of no simple group of order 992
Prove there are no simple groups of order $992$.
Factorise it. $31 \times 2^5$ so you have $|G|=31 \times 2^5 \geq n_{31}(31-1)+ n_{2}(2^5-1)+1$
Putting it in Sylow theorem. So how do you get the contradiction? Or is this totally wrong.
I need to use Sylow theorem to prove this. Hmm, can someone describe how you prove this. I know the Sylow the theorems well the proof of them.
It seems from the notes you have to start with $n_{2}>1$ and then $n_2=1(mod2)$. However, I don't understand this at all.
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The $2$-Sylow subgroups could have non-trivial intersection, since a group of order $32$ need not be cyclic, so your inequality is not correct. – Zev Chonoles♦ Jan 23 '12 at 15:57
@ZevChonoles What inequality do you use? I got given two ways, one is a formula like that and the other is a method I don't understand. – simplicity Jan 23 '12 at 16:03
How many possible Sylow 31-groups can there be? Remember that groups of order 31 are cyclic, so you can be assured the different Sylows have trivial intersection. – Steve D Jan 23 '12 at 16:15
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I like this particular number, since you can have 31 cyclic Sylow 2-subgroups, but instead of 31*(32-1)+1 = 962 elements of order a power of 2, you can get only 512 elements of order a power of 2. The Sylows overlap quite a bit. – Jack Schmidt Jan 23 '12 at 16:29
## 1 Answer
What you write is not correct: the number of elements of order a power of $2$ is not necessarily equal to $2^5-1$ times the number of $2$-Sylow subgroups because, as Zev Chonoles points out, you do not have that two distinct $2$-Sylow subgroups must intersect trivially (which is what goes behind that particualr inequality).
The number of $31$-Sylow subgroups must divide $31\times 32$ and be congruent to $1$ modulo $31$; so either there is a single $31$-Sylow subgroup (in which case the group is not simple), or there are thirty two $31$-Sylow subgroups.
If there are thirty two $31$-Sylow subgroups, then since any two distinct ones must intersect trivially (the groups are cyclic of prime order, so the only proper subgroup is trivial), they account for $32(31-1) + 1$ elements of $G$.
That means that there are $32\times 31 - 32\times 30 = 32$ elements whose order is not $31$. Since a $2$-Sylow subgroup must contain $32$ elements, there are only enough elements left over for a single $2$-Sylow subgroup, which must therefore be normal.
So $G$ will have either a single $31$-Sylow subgroup, or a single $2$-Sylow subgroup. Either way, it is not simple.
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How do you know there are 32 subgroups. This confused me more. What you using to conclude number of 31-sylow subgroup divides $31\times32$? – simplicity Jan 23 '12 at 16:28
@simplicity: Sylow's Third Theorem states: "If $G$ is a finite group and $p$ is a prime, then the number of $p$-Sylow subgroups of $G$ divides $|G|$ and is congruent to $1$ modulo $p$." Since $|G|=992 = 2^5\times 31=32\times 31$, and the only divisors of $|G|$ that are congruent to $1$ modulo $31$ are $1$ and $32$, then the number of $31$-Sylow subgroups of $G$ is either $1$ or $32$. – Arturo Magidin Jan 23 '12 at 16:31
Oh I feel so silly now. I thought you was doing (31+1), but yeah I get that now. Oh, yes that makes perfect sense now. The proof makes sense. – simplicity Jan 23 '12 at 16:34
Cam you give motivation for $32(31-1)+1$ comment. – simplicity Jan 23 '12 at 16:42
You could immediately say that the number of $31$-Sylows divides $32$, right? It's the index of the normalizer. This is minor, since the right numbers pop out quickly. [Also, did you learn from Lang's book? It's the only one I've seen that uses "$p$-Sylow" :)] – Dylan Moreland Jan 23 '12 at 16:48
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http://maherararblog.com/
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LTE, WIMAX, MIMO, MATLAB
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It is ultimately the user experience that will make or break mass adoption of 4G technologies such as LTE and WiMax. The end user does not really care what kind of PHY layer, MAC layer or network protocols these technologies use. The end user needs to be able to surf the Internet reliably, to...
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• ### Useful tips for completing your Ph. D. on time
Many people I know found it surprising that I was able to finish my Ph. D. within the normal four-year period. I do not blame them for that. After all the Arar inquiry was still going on. Back then I still had all the extreme pressure that came with the media attention not to mention [...]
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## Motorola’s 4TXx8RX LTE eNB May be the Solution
Nov 27th
Posted by Maher Arar in LTE
No comments
Since my last post I have been surfing the Internet looking for a commercial LTE basestation that would satisfy the bandwidth-hungry applications such as HD video conferencing and HD video streaming. Motorola seems to be the only company that has gone beyond the minimum requirements defined by 3GPP release 8. It is indeed a smart move.
More specifically they are now offering a basestation that can transmit with up to 4 antennas and receive on 8 antennas. While this falls short of what is needed Motorola seems to be on right track. I would not be surprised if we soon see an offering with up to 12 or more antennas on the receive side.
Bravo to Motorola for this kind of forward thinking
## 64% of Mobile Internet traffic will consist of video
May 21st
Posted by Maher Arar in Business
No comments
As a follow up on my two previous posts I would like to share this important information: 64% of Mobile Internet traffic will consist of video. This was mentioned in a report produced by the Institute of the Future.
Despite the fact that the report was commissioned by Skype, a major player in the personal video conferencing market, the arguments and numbers presented in the report make perfect sense. What backs this claim is the forcast made by Cisco that video will account for 91% of Internet traffic by the year 2013. No wonder why Cisco recently acquired Tandem for \$3 billion.
Because HQ and HD video traffic require a consistent high bandwidth (> 500 kbps), wireless operators will be faced with a dilemma: loose money or sacrifice QoS.
The ONLY way for wireless operators to avoid this dilemma is to increase the cell capacity beyond what is minimally required by 4G standards. The use of MU-MIMO (multi-user MIMO) will be a must. What this means is that operators have to seriously think about installing basestations with high number of antennas. As of today’s date there is no commercial basestation with more than 4 antennas.
This represents a business opportunity for a young wireless start-up. The key to success is to target very high number of antennas so that a sufficient increase in cell capacity is achieved. A BS with 16 antennas will provide a 4x increase in cell capacity compared to the one that has only 2 antennas (i.e. simultaneously serving 8 users versus serving only 2 users, respectively, assuming that each user has two anntenas).
## Welcome to the real world!
Jan 22nd
Posted by Maher Arar in Business
No comments
I have just read two recent articles that every operator should pay attention to. The first one discusses the lower-than-expected download speed that a typical user terminal experienced, i.e. 5Mbps versus the 40Mbps advertised speed. This is not surprising given that the speed depends very much on many factors including distance from the basestation, the number of active users in the cell, intra-cell interference, etc.
The second article discusses the fact that the lack of a proper scheduling algorithm that is suitable for 4G may hinder the expected performance and consequently user experience. This demonstrates that we still have a long way to go before we can have a reliable 4G network. My fear is that in their rush to compete for market share operators may end up doing more harm than good when it comes to promoting the use of their newly installed 4G mobile network. If the end user has a bad taste for 4G repairing the damage may not be possible.
## User experience will ultimately decide the fate of 4G
Jan 4th
Posted by Maher Arar in Business
4 comments
It is ultimately the user experience that will make or break mass adoption of 4G technologies such as LTE and WiMax. The end user does not really care what kind of PHY layer, MAC layer or network protocols these technologies use.
The end user needs to be able to surf the Internet reliably, to stream audio and video efficiently while paying the lowest price possible. The promotional headlines highlighting the fact that 4G will provide 100Mbps data rates is of no meaning to most consumers.
In my opinion, providing a reliable network access for HD video streaming, HD live TV and two-way HD video conferencing will be the main challenge for operators as the demand for these services is on the rise while compression technologies have reached their theoretical limits.
For 4G to be a success operators have to meet the above consumers’ expectations while managing to make an acceptable ROI. Despite the bandwidth efficiency of OFDMA, as is used in LTE and WiMax, making money will be a huge challenge if QoS needs to be maintained at an acceptable level. This is because a substantial percentage of subscribers in a typical cell will either be streaming HD video or will be having a two-way HD videoconferencing session.
For the technical folks out there, let us remember that the advertised 100Mbps rate is a peak over-the-air data rate. What that means is that the average user will experience much less effective rates. This is fine if the majority of users are only surfing or talking, both of which are either bursty in nature (i.e. surfing) or requires very low data rate (i.e. voice) .
Will beyond-4G technologies such as LTE-advanced will be the answer? Only time will tell.
## Useful tips for completing your Ph. D. on time
Dec 31st
Posted by Maher Arar in Personal
1 comment
Many people I know found it surprising that I was able to finish my Ph. D. within the normal four-year period. I do not blame them for that. After all the Arar inquiry was still going on. Back then I still had all the extreme pressure that came with the media attention not to mention that my post traumatic stress level was at its peak. Coordinating my supporters’ and lawyers’ efforts was another huge task.
While luck may have played a role (like some people would like to believe) I am convinced that the following tips have allowed me, and I am sure will allow any future Ph. D. candidate, to finish on time:
## 1. Choose a research subject that you are passionate about
Before I even applied for admission I knew exactly what research area I was interested in. Choosing a topic that you are passionate about is crucial as it will help you keep the motivation levels high despite all the obstacles that you may face.
In my case, I found that the use of MIMO (i.e. multiple antennas at both the transmit and receive ends) to boost data rates without the need to use extra spectrum was an intriguing subject that was worth investigating.
## 2. Make sure your research topic has practical applications if not now but will in few years from now
This is important in the sense that your research should not be purely theoretical. This in turn will help you in your mission and keep you focused on your goal. Why? because you can feel the importance of what you are doing by simply reading business publications and by surfing the internet.
In my case I was able to easily find out that MIMO was going to be hot in WLAN and future 3G/4G as spectrum scarcity was on the rise and its acquisition cost was becoming more and more prohibitive.
## 3. Make sure you have a plan
Every Ph. D. candidate seems to agree that this is the most important thing they should be doing and yet I was amazed how many Ph.D. candidates I met did not have any plan at all. Knowing my research subject before I even took courses not only helped me choose the right ones but it also helped me choose the right topics for my term papers.
I was able to use not only the knowledge gained from writing these term papers but also many of the actual sections, figures, illustrations, etc. This will save you alot of time later when it is time to write your thesis. Never underestimate the time it takes to a produce a quality thesis.
## 4. Set achievable quarterly milestones
When it comes to measuring your productivity do it in quarters. With the help of your supervisor implement a milestone system by which you can either celebrate the outcome (and hence boost your motivation levels) or hold your self accountable for what went wrong. Make sure these milestones are reasonably achievable.
## 5. Build relationship of trust with your supervisor
You have to keep your supervisor as happy as you can and never play tricks with him/her. Try as much as you can to meet the deadlines for the milestones you and him/her have agreed to.
In my case, I was able to meet most of the deadlines to the point where I was reminding him about them. That helped both of us build the required trust that normally lacks in many student/supervisor relationships. If you do this you will find that your supervisor will be your best ally.
Voila I have just shared with you my secret of how I was able to finish my Ph. D. in a record time given all the circumstances mentioned above. Now let me go and celebrate this milestone of being able to write a blog post and share useful information with the rest of the world.
And the by the way, happy new year!!!!
## Ph. D. presentation slides
Dec 17th
Posted by Maher Arar in MIMO
No comments
In case you are interested:
Presentation Thesis
## Today I successfully defended my Ph. D. thesis
Dec 16th
Posted by Maher Arar in MIMO
9 comments
Congrats to me! I can finally say that I am a Dr. or may be Dr.2 to be more precise (because of my honorary Ph.D. degree from Nipissing University). It was an extremely tough year. I had to work more than full time (you figure out what that means) to be able to finish by the end of the year.
My supervisor asked me what is next and I told him that I have a lot of thoughts (not plans!) about what to do next. I do not want to think about it now. It is time to enjoy and celebrate this achievement. I will certainly make up my mind by early 2010.
If you ever think of embarking on a Ph. D. adventure at the University of Ottawa you may find the following presentation inspiring enough! Enjoy watching.
## Why did I decide to start blogging?
Dec 14th
Posted by Maher Arar in Personal
1 comment
I have been wondering for a while whether I should jump on the Web 2.0 wagon and join the rest of the world. It was not an easy decision to make. After all I have always valued my privacy and have always preferred to keep my opinions to myself.
I have finally surrendered as I have become convinced that my life will never be private again. Also, we live in a constant surveillance society any way and the BIG BROTHER is watching 24/7. Time will tell if the decision I took this week is the correct one.
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http://www.nag.com/numeric/cl/nagdoc_cl23/html/G01/g01intro.html
|
# NAG Library Chapter Introductiong01 – Simple Calculations on Statistical Data
## 1 Scope of the Chapter
This chapter covers three topics:
• summary statistics
• statistical distribution functions and their inverses;
• testing for Normality and other distributions.
## 2 Background to the Problems
### 2.1 Summary Statistics
The summary statistics consist of two groups. The first group are those based on moments; for example mean, standard deviation, coefficient of skewness, and coefficient of kurtosis (sometimes called the ‘excess of kurtosis’, which has the value $0$ for the Normal distribution). These statistics may be sensitive to extreme observations and some robust versions are available in Chapter g07. The second group of summary statistics are based on the order statistics, where the $i$th order statistic in a sample is the $i$th smallest observation in that sample. Examples of such statistics are minimum, maximum, median, hinges and quantiles.
### 2.2 Statistical Distribution Functions and Their Inverses
Statistical distributions are commonly used in three problems:
• evaluation of probabilities and expected frequencies for a distribution model;
• testing of hypotheses about the variables being observed;
• evaluation of confidence limits for arguments of fitted model, for example the mean of a Normal distribution.
Random variables can be either discrete (i.e., they can take only a limited number of values) or continuous (i.e., can take any value in a given range). However, for a large sample from a discrete distribution an approximation by a continuous distribution, usually the Normal distribution, can be used. Distributions commonly used as a model for discrete random variables are the binomial, hypergeometric, and Poisson distributions. The binomial distribution arises when there is a fixed probability of a selected outcome as in sampling with replacement, the hypergeometric distribution is used in sampling from a finite population without replacement, and the Poisson distribution is often used to model counts.
Distributions commonly used as a model for continuous random variables are the Normal, gamma, and beta distributions. The Normal is a symmetric distribution whereas the gamma is skewed and only appropriate for non-negative values. The beta is for variables in the range $\left[0,1\right]$ and may take many different shapes. For circular data, the ‘equivalent’ to the Normal distribution is the von Mises distribution. The assumption of the Normal distribution leads to procedures for testing and interval estimation based on the ${\chi }^{2}$, $F$ (variance ratio), and Student's $t$-distributions.
In the hypothesis testing situation, a statistic $X$ with known distribution under the null hypothesis is evaluated, and the probability $\alpha $ of observing such a value or one more ‘extreme’ value is found. This probability (the significance) is usually then compared with a preassigned value (the significance level of the test), to decide whether the null hypothesis can be rejected in favour of an alternate hypothesis on the basis of the sample values. Many tests make use of those distributions derived from the Normal distribution as listed above, but for some tests specific distributions such as the Studentized range distribution and the distribution of the Durbin–Watson test have been derived. Nonparametric tests as given in Chapter g08, such as the Kolmogorov–Smirnov test, often use statistics with distributions specific to the test. The probability that the null hypothesis will be rejected when the simple alternate hypothesis is true (the power of the test) can be found from the noncentral distribution.
The confidence interval problem requires the inverse calculation. In other words, given a probability $\alpha $, the value $x$ is to be found, such that the probability that a value not exceeding $x$ is observed is equal to $\alpha $. A confidence interval of size $1-2\alpha $, for the quantity of interest, can then be computed as a function of $x$ and the sample values.
The required statistics for either testing hypotheses or constructing confidence intervals can be computed with the aid of functions in this chapter, and Chapter g02 (for regression), Chapter g04 (for analysis of designed experiments), Chapter g13 (for time eries), and Chapter e04 (for nonlinear least squares problems).
Pseudorandom numbers from many statistical distributions can be generated by functions in Chapter g05.
### 2.3 Testing for Normality and Other Distributions
Methods of checking that observations (or residuals from a model) come from a specified distribution, for example, the Normal distribution, are often based on order statistics. Graphical methods include the use of probability plots. These can be either $P-P$ plots (probability–probability plots), in which the empirical probabilities are plotted against the theoretical probabilities for the distribution, or $Q-Q$ plots (quantile–quantile plots), in which the sample points are plotted against the theoretical quantiles. $Q-Q$ plots are more common, partly because they are invariant to differences in scale and location. In either case if the observations come from the specified distribution then the plotted points should roughly lie on a straight line.
If ${y}_{i}$ is the $i$th smallest observation from a sample of size $n$ (i.e., the $i$th order statistic) then in a $Q-Q$ plot for a distribution with cumulative distribution function $F$, the value ${y}_{i}$ is plotted against ${x}_{i}$, where $F\left({x}_{i}\right)=\left(i-\alpha \right)/\left(n-2\alpha +1\right)$, a common value of $\alpha $ being $\frac{1}{2}$. For the Normal distribution, the $Q-Q$ plot is known as a Normal probability plot.
The values ${x}_{i}$ used in $Q-Q$ plots can be regarded as approximations to the expected values of the order statistics. For a sample from a Normal distribution the expected values of the order statistics are known as Normal scores and for an exponential distribution they are known as Savage scores.
An alternative approach to probability plots are the more formal tests. A test for Normality is the Shapiro and Wilk's $W$ Test, which uses Normal scores. Other tests are the ${\chi }^{2}$ goodness-of-fit test and the Kolmogorov–Smirnov test; both can be found in Chapter g08.
### 2.4 Distribution of Quadratic Forms
Many test statistics for Normally distributed data lead to quadratic forms in Normal variables. If $X$ is a $n$-dimensional Normal variable with mean $\mu $ and variance-covariance matrix $\Sigma $ then for an $n$ by $n$ matrix $A$ the quadratic form is
$Q=XTAX.$
The distribution of $Q$ depends on the relationship between $A$ and $\Sigma $: if $A\Sigma $ is idempotent then the distribution of $Q$ will be central or noncentral ${\chi }^{2}$ depending on whether $\mu $ is zero.
The distribution of other statistics may be derived as the distribution of linear combinations of quadratic forms, for example the Durbin–Watson test statistic, or as ratios of quadratic forms. In some cases rather than the distribution of these functions of quadratic forms the values of the moments may be all that is required.
### 2.5 Energy Loss Distributions
An application of distributions in the field of high-energy physics where there is a requirement to model fluctuations in energy loss experienced by a particle passing through a layer of material. Three models are commonly used:
(i) Gaussian (Normal) distribution;
(ii) the Landau distribution;
(iii) the Vavilov distribution.
Both the Landau and the Vavilov density functions can be defined in terms of a complex integral. The Vavilov distribution is the more general energy loss distribution with the Landau and Gaussian being suitable when the Vavilov argument $\kappa $ is less than $0.01$ and greater than $10.0$ respectively.
### 2.6 Vectorized Functions
A number of vectorized functions are included in this chapter. Unlike their scalar counterparts, which take a single set of parameters and perform a single function evaluation, these functions take vectors of parameters and perform multiple function evaluations in a single call. The input arrays to these vectorized functions are designed to allow maximum flexibility in the supply of the parameters by reusing, in a cyclic manner, elements of any arrays that are shorter than the number of functions to be evaluated, where the total number of functions evaluated is the size of the largest array.
To illustrate this we will consider nag_prob_gamma_vector (g01sfc), a vectorized version of nag_gamma_dist (g01efc), which calculates the probabilities for a gamma distribution. The gamma distribution has two parameters $\alpha $ and $\beta $ therefore nag_prob_gamma_vector (g01sfc) has four input arrays, one indicating the tail required (tail), one giving the value of the gamma variate, $g$, whose probability is required (g), one for $\alpha $ (a) and one for $\beta $ (b). The lengths of these arrays are ltail, lg, la and lb respectively.
For sake of argument, lets assume that ${\mathbf{ltail}}=1$, ${\mathbf{lg}}=2$, ${\mathbf{la}}=3$ and ${\mathbf{lb}}=4$, then $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left({\mathbf{ltail}},{\mathbf{lg}},{\mathbf{la}},{\mathbf{lb}}\right)=4$ values will be returned. These four probabilities would be calculated using the following parameters:
| | | | | |
|-----------------------------------------------------------------------------|-----------------------------------------------------------------------------|-----------------------------------------------------------------------------|-----------------------------------------------------------------------------|-----------------------------------------------------------------------------|
| $i$ | Tail | $g$ | $\alpha $ | $\beta $ |
| $1$ | ${\mathbf{tail}}\left[1\right]$ | ${\mathbf{g}}\left[1\right]$ | ${\mathbf{a}}\left[1\right]$ | ${\mathbf{b}}\left[1\right]$ |
| $2$ | ${\mathbf{tail}}\left[1\right]$ | ${\mathbf{g}}\left[2\right]$ | ${\mathbf{a}}\left[2\right]$ | ${\mathbf{b}}\left[2\right]$ |
| $3$ | ${\mathbf{tail}}\left[1\right]$ | ${\mathbf{g}}\left[1\right]$ | ${\mathbf{a}}\left[3\right]$ | ${\mathbf{b}}\left[3\right]$ |
| $4$ | ${\mathbf{tail}}\left[1\right]$ | ${\mathbf{g}}\left[2\right]$ | ${\mathbf{a}}\left[1\right]$ | ${\mathbf{b}}\left[4\right]$ |
## 3 Recommendations on Choice and Use of Available Functions
Descriptive statistics / Exploratory analysis:
summaries:
frequency / contingency table,
one variable nag_frequency_table (g01aec)
mean, variance, skewness, kurtosis (one variable),
from frequency table nag_summary_stats_freq (g01adc)
from raw data nag_summary_stats_1var (g01aac)
median, hinges / quartiles, minimum, maximum nag_5pt_summary_stats (g01alc)
quantiles:
approximate:
large data stream of fixed size nag_approx_quantiles_fixed (g01anc)
large data stream of unknown size nag_approx_quantiles_arbitrary (g01apc)
unordered vector nag_double_quantiles (g01amc)
Distributions:
Beta:
central:
deviates nag_deviates_beta (g01fec)
probabilities and probability density function nag_prob_beta_dist (g01eec)
vectorized deviates nag_deviates_beta_vector (g01tec)
vectorized probabilities nag_prob_beta_vector (g01sec)
non-central:
probabilities nag_prob_non_central_beta_dist (g01gec)
binomial:
distribution function nag_binomial_dist (g01bjc)
vectorized distribution function nag_prob_binomial_vector (g01sjc)
Durbin–Watson statistic:
probabilities nag_prob_durbin_watson (g01epc)
Energy loss distributions:
Landau:
density nag_prob_density_landau (g01mtc)
derivative of density nag_prob_der_landau (g01rtc)
distribution nag_prob_landau (g01etc)
first moment nag_moment_1_landau (g01ptc)
inverse distribution nag_deviates_landau (g01ftc)
second moment nag_moment_2_landau (g01qtc)
Vavilov:
distribution nag_prob_vavilov (g01euc)
initialization nag_init_vavilov (g01zuc)
F:
central:
deviates nag_deviates_f_dist (g01fdc)
probabilities nag_prob_f_dist (g01edc)
vectorized deviates nag_deviates_f_vector (g01tdc)
vectorized probabilities nag_prob_f_vector (g01sdc)
non-central:
probabilities nag_prob_non_central_f_dist (g01gdc)
gamma:
deviates nag_deviates_gamma_dist (g01ffc)
probabilities nag_gamma_dist (g01efc)
probability density function nag_gamma_pdf (g01kfc)
vectorized deviates nag_deviates_gamma_vector (g01tfc)
vectorized probabilities nag_prob_gamma_vector (g01sfc)
vectorized probability density function nag_gamma_pdf_vector (g01kkc)
Hypergeometeric:
distribution function nag_hypergeom_dist (g01blc)
vectorized distribution function nag_prob_hypergeom_vector (g01slc)
Kolomogorov–Smirnov:
probabilities:
one-sample nag_prob_1_sample_ks (g01eyc)
two-sample nag_prob_2_sample_ks (g01ezc)
Normal:
bivariate:
probabilities nag_bivariate_normal_dist (g01hac)
multivariate:
probabilities nag_multi_normal (g01hbc)
quadratic forms:
cumulants and moments nag_moments_quad_form (g01nac)
moments of ratios nag_moments_ratio_quad_forms (g01nbc)
univariate:
deviates nag_deviates_normal (g01fac)
probabilities nag_prob_normal (g01eac)
probability density function
scalar nag_normal_pdf (g01kac)
vectorized nag_normal_pdf_vector (g01kqc)
reciprocal of Mill's Ratio nag_mills_ratio (g01mbc)
Shapiro and Wilk's test for Normality nag_shapiro_wilk_test (g01ddc)
vectorized deviates nag_deviates_normal_vector (g01tac)
vectorized probabilities nag_prob_normal_vector (g01sac)
Poisson:
distribution function nag_poisson_dist (g01bkc)
vectorized distribution function nag_prob_poisson_vector (g01skc)
Student's t:
central:
bivariate:
probabilities nag_bivariate_students_t (g01hcc)
univariate:
deviates nag_deviates_students_t (g01fbc)
probabilities nag_prob_students_t (g01ebc)
vectorized deviates nag_deviates_students_t_vector (g01tbc)
vectorized probabilities nag_prob_students_t_vector (g01sbc)
non-central:
probabilities nag_prob_non_central_students_t (g01gbc)
Studentized range statistic:
probabilities nag_prob_studentized_range (g01emc)
von Mises:
probabilities nag_prob_von_mises (g01erc)
χ 2:
central:
deviates nag_deviates_chi_sq (g01fcc)
probabilities nag_prob_chi_sq (g01ecc)
probability of linear combination nag_prob_lin_chi_sq (g01jdc)
non-central:
probabilities nag_prob_non_central_chi_sq (g01gcc)
probability of linear combination nag_prob_lin_non_central_chi_sq (g01jcc)
vectorized deviates nag_deviates_chi_sq_vector (g01tcc)
vectorized probabilities nag_prob_chi_sq_vector (g01scc)
Scores:
Normal scores, ranks or exponential (Savage) scores nag_ranks_and_scores (g01dhc)
Normal scores:
accurate nag_normal_scores_exact (g01dac)
variance-covariance matrix nag_normal_scores_var (g01dcc)
Note: the Student's $t$, ${\chi }^{2}$, and $F$ functions do not aim to achieve a high degree of accuracy, only about four or five significant figures, but this should be quite sufficient for hypothesis testing. However, both the Student's $t$ and the $F$-distributions can be transformed to a beta distribution and the ${\chi }^{2}$-distribution can be transformed to a gamma distribution, so a higher accuracy can be obtained by calls to the gamma or beta functions.
Note: nag_ranks_and_scores (g01dhc) computes either ranks, approximations to the Normal scores, Normal, or Savage scores for a given sample. nag_ranks_and_scores (g01dhc) also gives you control over how it handles tied observations. nag_normal_scores_exact (g01dac) computes the Normal scores for a given sample size to a requested accuracy; the scores are returned in ascending order. nag_normal_scores_exact (g01dac) can be used if either high accuracy is required or if Normal scores are required for many samples of the same size, in which case you will have to sort the data or scores.
## 4 Functions Withdrawn or Scheduled for Withdrawal
| | | |
|-----------------------------------|-------------------|------------------------------|
| WithdrawnFunction | Mark ofWithdrawal | Replacement Function(s) |
| nag_deviates_normal_dist (g01cec) | 24 | nag_deviates_normal (g01fac) |
## 5 References
Hastings N A J and Peacock J B (1975) Statistical Distributions Butterworth
Kendall M G and Stuart A (1969) The Advanced Theory of Statistics (Volume 1) (3rd Edition) Griffin
Tukey J W (1977) Exploratory Data Analysis Addison–Wesley
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http://math.stackexchange.com/questions/76820/why-t-mathbbs2-otimes-t-mathbbs2-is-trivial/76825
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# Why $T\mathbb{S}^{2}\otimes T\mathbb{S}^{2}$ is trivial?
I encountered this problem in an homework problem set of algebraic topology. Naturally I thought about Bott Periodicity which implies $K(\mathbb{S}^{2})\cong \mathbb{Z}$. But I am beware that we are working in $KO$ theory instead of $K$ theory, so this is not applicable. After checking online I found $KO(\mathbb{S}^{2})\cong \mathbb{Z}_{2}$, however I do not know a simple proof of it. My thoughts are:
1) Would it be appropriate to work this via Chern class of $\mathbb{S}^{2}$? Since we have ch($\mathbb{S}^{2}\otimes \mathbb{S}^{2}$)=ch($\mathbb{S}^{2}$)+ch($\mathbb{S}^{2}$), I only need to show $ch(\mathbb{S}^{2})$ has characteristic 2. So we have $ch(\mathbb{S}^{2})=ch(\mathbb{C}\mathbb{P}^{1})$. Geometrically it is "clear" that the tautologous bundle should have order 2 because of the half-twisting, but I encountered the same problem in here: namely the complex algebra can be different from the real algebra. Since Chern class is defined over hermitian vector bundles I do not know how to carry out this any further.
2) On the other hand it should be simple to solve this problem without using characteristic classes, as the author did not introduce the concept in that section. But I do not know how to work out the trivilization nicely. To prove it indeed trivialize I would need four independent sections; I do not know how to find them.
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## 1 Answer
What about the fact that vector bundles over $S^2$ are classified by clutching construction, and the map classifying $TS^2$ is twice the generator of $\pi_1(SO(2))$. If you think about this as a path of matrices $A(t)$, the tensor square bundle is classified by the path $A(t) \otimes A(t)$ in $SO(4)$, which is then trivial in $\pi_1 SO(4)$ (being divisible by 2, for example).
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Could you elaborate how the map classifying $TS^{2}$ be twice the generator of $\pi_{1}(SO(2))$? Also the step from $SO(2)\otimes SO(2)$ to $SO(4)$ seems unclear to me (do you just put them on diagonal blocks?) Anyway this is very helpful. – Changwei Zhou Oct 29 '11 at 0:29
– Max Oct 29 '11 at 0:40
Thanks for clarifying detail. I heard of Kronecker product but never imaged it to be useful. Thanks you - I will check for Hatcher. – Changwei Zhou Oct 29 '11 at 0:48
I am trying to type this in latex. Do you know where can I find information about the fundamental group of the special orthogonal groups? You used it explicitly in here but I feel it would be hard to write your argument as explicitly in local coordinates (consider the Kronecker product, for example). I feel it is related to the spin-representation. – Changwei Zhou Oct 29 '11 at 1:52
1
The computation follows from the fact that $SO(3)$ is $RP3$ (en.wikipedia.org/wiki/SO%283%29#Topology) and the fundamental group is stable after that (en.wikipedia.org/wiki/Orthogonal_group#Homotopy_groups, look at the fiber bundle and use the homotopy long exact sequence). The generator in $\pi_1(SO(n))$ is hence induced (under inclusion) from the generator in $RP3=SO3$ which is a circle of rotations around the same axis. – Max Oct 29 '11 at 15:08
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http://physics.stackexchange.com/questions/29673/modeling-stochastic-process-with-frequency-dependent-power-spectrum
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# Modeling stochastic process with frequency-dependent power spectrum
I'm trying to model of Johnson-Nyquist noise propagation in a nonlinear circuit. An ideal (linear) resistor can be modeled very nicely by the Fokker-Planck equation (equivalently, the drift-diffusion equation), where charge $V/R$ flows through the resistor on average, but there's also random flow of charge either way across the resistor characterized by "diffusion coefficient" $k_BT/R$. I get a nice differential equation describing how charge probabilistically flows through my circuit. Everything is good.
Then stage 2 is to have a resistor with frequency-dependent resistance (like all resistors in the real world). Here I get stuck...
A time-domain-based analytical solution seems impossible because---with frequency-dependent resistance---it would seem that charge transport across the resistor right now depends on the entire history of charge transport in the past.
A time-domain numerical (monte carlo) solution seems impossible because the relevant frequencies vary over many orders of magnitude I don't know how to construct a time-domain stochastic model with a predetermined power spectrum.
Any kind of frequency-domain solution seems impossible because other parts of the circuit are extremely nonlinear and therefore mix different frequencies together.
Any advice? Am I missing some trick?
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I'm out of my depth here, but something like a Drude model (ref wikipedia) for the resistor could be implemented by adding a small series inductance. – Art Brown Jun 13 '12 at 17:15
## 2 Answers
Just taking a stab in the dark here, but perhaps it would be possible to take your white noise signal and feed it through a parallel bank of digital time-domain bandpass filters, chosen so that their combined spectra approximate the desired spectrum?
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You get the bounty ... that helped get me thinking although I ended up doing something a bit different. – Steve B Jun 15 '12 at 14:58
OK, a little bit like zephyr's answer, I started with white noise, took the FFT, and then scaled each frequency component up or down according to the square-root of the power at that frequency, then did inverse-FFT to get the time-domain signal.
An equivalent approach would have been to generate the FFT directly by giving each component the appropriate magnitude and a phase randomly chosen between 0 and 2pi. (Obviously the positive and negative frequencies need to have opposite phases to keep the signal real.) Then, again, inverse-FFT back.
It kind of makes sense to me that every frequency component should have a predetermined magnitude and a random phase ... although I'm not 100% confident. Anyway so far the results all seem to be making sense.
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http://mathhelpforum.com/calculus/127377-prove-definition-e.html
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# Thread:
1. ## Prove the definition of e
Prove that
$lim_{h-->0}$ $(1+h)^{1/h}$= e
given that: If g is continuous at c and f is continuous a g(c), then the composition f o g is continuous at c
2. Originally Posted by wopashui
Prove that
$lim_{h-->0}$ $(1+h)^{1/h}$= e
given that: If g is continuous at c and f is continuous a g(c), then the composition f o g is continuous at c
By definition, we want $e^x$ to be the function that is its own derivative. We want to try to find what value $e$ is.
So let $f(x) = e^x$
$f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$.
So $f'(x) = \lim_{h \to 0}\frac{e^{x + h} - e^x}{h}$
$= \lim_{h \to 0}\frac{e^xe^h - e^x}{h}$
$= \lim_{h \to 0}\frac{e^x(e^h - 1)}{h}$
$= e^x \lim_{h \to 0}\frac{e^h - 1}{h}$.
So for $f(x)$ to be equal to $f'(x)$, we would require
$\lim_{h \to 0}\frac{e^h - 1}{h} = 1$
$\lim_{h \to 0}(e^h - 1) = \lim_{h \to 0}h$
$\lim_{h \to 0}e^h - \lim_{h \to 0}1 = \lim_{h \to 0}h$
$\lim_{h \to 0}e^h = \lim_{h \to 0}1 + \lim_{h \to 0}h$
$\lim_{h \to 0}e^h = \lim_{h \to 0}(1 + h)$
$\lim_{h \to 0}(e^h)^{\frac{1}{h}} = \lim_{h \to 0}(1 + h)^{\frac{1}{h}}$
$\lim_{h \to 0}e = \lim_{h \to 0}(1 + h)^{\frac{1}{h}}$
$e = \lim_{h \to 0}(1 + h)^{\frac{1}{h}}$.
3. Originally Posted by wopashui
Prove that
$lim_{h-->0}$ $(1+h)^{1/h}$= e
given that: If g is continuous at c and f is continuous a g(c), then the composition f o g is continuous at c
I find that definition rather bazaar. By that I mean beginning students do not have the tools to deal with that abstraction.
My all time favorite calculus textbook is the first edition of Gillman & McDowell, 1972.
It is normal is size( not the dictionary size), does integral as betweeness, and has this definition of e.
The number $e$ is defined as the number such that $\int_1^e {\frac{{dx}}{x}} = 1$
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http://math.stackexchange.com/questions/8157/if-a-set-contain-2n1-elements-and-if-the-number-of-subsets-which-contain-at
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# If a set contain $(2n+1)$ elements and if the number of subsets which contain at most n elements is 4096, then what is the value of $n$?
If a set contain $(2n+1)$ elements and if the number of subsets which contain at most n elements is 4096, then what is the value of $n$?
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I recently found a nice proof for the same. :-) – Quixotic Dec 23 '10 at 14:22
## 2 Answers
There are $2^{2n + 1}$ subsets in total. Now note that the function which takes a subset to its complement is a bijection, and that a subset has $n$ or fewer elements if, and only if, its complement has more than $n$ elements. Thus the number of subsets with $n$ or fewer elements is equal to the number with more than $n$; thus the number of each is $2^{2n + 1}/2 = 2^{2n}$. Now solve for $n$.
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Use the identity $(1+1)^n= n_{C_0}+n_{C_1}+\dots+n_{C_n}$ and the fact that $n_{C_r}=n_{C_{n-r}}$.
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http://mathoverflow.net/questions/7914?sort=oldest
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## Torsion line bundles with non-vanishing cohomology on smooth ACM surfaces
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I am looking for an example of a smooth surface $X$ with a fixed very ample $\mathcal O_X(1)$ such that $H^1(\mathcal O(k))=0$ for all $k$ (such thing is called an ACM surface, I think) and a globally generated line bundle $L$ such that $L$ is torsion in $Pic(X)$ and $H^1(L) \neq 0$.
Does such surface exist? How can I construct one if it does exist? What if one ask for even nicer surface, such as arithmetically Gorenstein? If not, then I am willing to drop smooth or globally generated, but would like to keep the torsion condition.
More motivations(thanks Andrew): Such a line bundle would give a cyclic cover of $X$ which is not ACM, which would be of interest to me. I suppose one can think of this as a special counter example to a weaker (CM) version of purity of branch locus.
To the best of my knowledge this is not a homework question (: But I do not know much geometry, so may be some one can tell me where to find an answer. Thanks.
EDIT: Removed the global generation condition, by Dmitri's answer. I realized I did not really need it that much.
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Some (hopefully) positive criticism: You should add (edit) some discussion to motivate your question and show some thoughts/progress you have about it, even if it is very little. This serves many purposes: 1) it gets people interested, 2) it shows you care about the problem (which I believe you do), and 3) it makes it easier for others to start working on it. – Andrew Critch Dec 6 2009 at 0:27
Thanks! See above. – Hailong Dao Dec 6 2009 at 0:43
I wonder what is so interesting about ACM surfaces (I'm sure there is something if you're asking)? – Ilya Nikokoshev Dec 6 2009 at 0:56
In general, ACM varieties have all the intermediate cohomolgy vanish. They correspond to Cohen-Macaulay rings, which have maximal depth, hence the name. From both algebraic (maximal depth) and geometric (no cohomology) I think they are nice. There are no wiki entry for them though (: – Hailong Dao Dec 6 2009 at 1:10
What is the ground field that you consider in this question? If this is over $C$ and the surface is smooth then a globally generated torsion line bundle is trivial. – Dmitri Dec 9 2009 at 0:33
show 4 more comments
## 1 Answer
Let us show that a globaly generated torsion line bundle $L$ on a (compact) complex surface is trivial. Ideed, a globally generated line bundle has at least one section, say $s$. Let us take it. If $s$ has no zeros, then $L$ is trivial. But if $s$ vanishes somewhere then any positive power $L^n$ has a section $s^n$ that vanishes at the same points. So any power of $L$ is not trivial, i.e. $L$ is not a torsion bundle, contradiction.
Notice that we did not use the fact that the surface is smooth. And we also did not use the fact that we work with a surface...
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Thanks! What about without the global generation condition? I realized I did not need it that much. – Hailong Dao Dec 9 2009 at 15:58
Dear Hailong, it will be very helpfull if you specify what you mean by $H^1(\mathcal O(k))=0$. For example is $k$ an integer number? In this case, can it be $0$? If it $k=0$, does this mean for you that $O(0)$ is just the structure sheaf of $X$? In this case it follows that your surface has $H^1(X)=0$. Also do I understand correctly that for you $mathcal O(1))$ is ANY ample line bundle? – Dmitri Dec 9 2009 at 17:00
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Dear Dmitri, I did mean all $k$(here I fixed a very ample line bundle $mathcal O_X(1)$). This is quite restrictive, I am aware of that. – Hailong Dao Dec 9 2009 at 18:06
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http://mathoverflow.net/questions/31271/approximation-with-continuous-functions/31366
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## Approximation with continuous functions
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Is it true that for every function $\mathbb{R} \to \mathbb{R}$ there exists a sequence of continuous functions $f_n(x): \mathbb{R} \to \mathbb{R}$ such that for any $x \in \mathbb{R}$ $f_n(x)$ converges to $f(x)$?
I started with characteristic function of rationals and tried to find corresponding sequence and got stuck. So additional question if this statement is true for this function.
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12
See "Baire function" - en.wikipedia.org/wiki/Baire_function – Yemon Choi Jul 10 2010 at 6:02
Is $f$ at least measurable? and are you asking about convergence everywhere, or almost everywhere? – Piero D'Ancona Jul 11 2010 at 7:17
## 5 Answers
Quick answer since it is late:
What you want to do is look up Baire function'' (in Wikipedia, for example).
Here is a simple way of seeing that the answer is negative: Any continuous function $f:{\mathbb R}\to{\mathbb R}$ is determined by its value on the rationals, so by an easy counting argument, there are only as many continuous functions as there are reals. Since a sequence of reals can be easily coded by a single real, there are only $|{\mathbb R}|$-many functions that are limit of sequences of continuous functions (you could replace "pointwise limit" with just about anything you want as long as the countable sequence suffices to describe the new function). But there are $2^{|{\mathbb R}|}$ many functions from ${\mathbb R}$ to itself.
This argument shows that even if you iterate the process (the Wikipedia entry talks about class $n$ Baire functions for all $n\in{\mathbb N}$. You need to go on much longer through a transfinite process), you have to iterate it for a very long time if you hope to capture all functions this way.
Let me add something of an advertisement, now that I have some time. Pete Clark's comments in another answer show that $\chi_{\mathbb Q}$ is not the pointwise limit of continuous functions. For this, he described a characterization of the Baire class 1 functions that clearly $\chi_{\mathbb Q}$ does not satisfy.
The argument above, on the other hand, only refers to cardinality considerations, so it does not apply to specific examples.
One can refine the argument (essentially, by a sophisticated use of Cantor's diagonalization) by appealing to techniques of descriptive set theory. Here, one studies definable'' classes of functions $f:{\mathbb R}\to{\mathbb R}$ or, more generally, of subsets of ${\mathbb R}^n$, and it is therefore the right setting for this type of problems.
The simplest kind of definability a function my have is that its graph is Borel (this is the case if the function is continuous, for example). From here, a very large hierarchy of levels of complexity of subsets of ${\mathbb R}^m$ is defined, starting by taking projections of Borel subsets of ${\mathbb R}^{m+1}$, and complements, and then iterating this procedure.
The fact that we can actually iterate the procedure, i.e., that the hierarchy does not collapse, is where Cantor's diagonalization appears. Anyway, any class of functions with a simple description is easily seen to belong to a (tipically, very short) initial segment of this hierarchy, and so we know it cannot capture the class of all functions. Many variants of your question are seen immediately to have negative answers through this procedure, which has the advantage of separating levels of complexity in a more refined way than mere cardinality.
An excellent reference you may want to look at is Alekos Kechris's book.
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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.
Well, if you look at the Dirichlet function, which is the characteristic function of the rationals in $[0,1]$ it can be written as the double pointwise limit: $$f(x)= \lim_{k\to\infty}\left(\lim_{j\to\infty}\left(\cos(k!\pi x)^{2j}\right)\right)$$ This definition holds for any $x \in \mathbb{R}$ as well.
(Taken from Wikipedia, Nowhere continuous function)
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Yep. So it's of Baire class 2, not of Baire class 1 (which, as already mentioned, is the set described by the asker). – Nate Eldredge Jul 12 2010 at 15:23
I quote a theorem due to Lusin:
Let $X$ be a locally compact Hausdorff space and let $\mu$ be a regular Borel measure on $X$ such that $\mu(K)<\infty$ for every compact $K\subseteq X$. Suppose $f$ is a complex measurable function on $X$, $\mu(A)<\infty$, $f(x)=0$ if $x\in X\setminus A$, and $\epsilon>0$. Then there exists a continuous complex function $g$ on $X$ with compact support such that
$\mu({x:f(x)\neq g(x)})<\epsilon$.
Furthermore, the function $g$ can be chosen such that
\$sup_{x\in X}|g(x)|\leq sup_{x\in X}|f(x)|.
As an immediate corollary (which is more relevant to your question), observe that:
If the hypotheses of Lusin's theorem are satisfied and if $|f|\leq 1$, then there is a sequence ${g_n}$ of continuous complex functions with compact support such that $|g_n|\leq 1$ for all $n$ and
$f(x)=\lim_{n \to \infty}g_n(x)$
almost everywhere with respect to $\mu$.
Note that the proof of Lusin's Theorem requires Urysohn's lemma for locally compact Hausdorff spaces, or at least a variation of it. (A very similar argument to that used to prove Urysohn's lemma for Normal Hausdorff spaces establishes the fact that any locally compact Hausdorff space is completely regular.) For more details on these results and their proofs, see Chapter 2 of the second edition of Walter Rudin's Real and Complex Analysis. (The results can be more precisely located on pages 56 and 57.)
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Quick, but less sharp, answer: The bounded Borel measurable functions are closed under bounded pointwise convergence. So any bounded non-measurable function is not the pointwise limit of continuous functions.
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The question is ill-posed. A function continuous in one topology might not be continuous in another topology. The same goes for convergence. When you precisely formulate your question, I am sure you will be able to come with the answer.
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Huh? I think it's clear that the OP is talking about the classical topology on R and pointwise convergence... – Andy Putman Jul 10 2010 at 6:17
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This is an unhelpful answer. There is a standard topology on $\mathbb{R}$, and surely that is the one the OP intends. Just asking the question precisely (as the OP has done) does not suggest the answer. Rather, as Yemon Choi points out, this question is the beginning of theory of Baire classes: a function is in Baire class one if it is a pointwise limit of continuous functions. Such functions have to have a dense subset of points of continuity, ruling out the characteristic function of $\mathbb{Q}$. – Pete L. Clark Jul 10 2010 at 6:20
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Slight correction: a real function $f$ is in Baire class one iff: for every nonempty closed subset $P \subset \mathbb{R}$ without isolated points, there exists $x \in P$ such that the restriction of $f$ to $P$ is continuous at $x$. Taking $P$ to be a closed interval $[a,b]$, this shows that there is a dense subset of points at which $f$ is either left- or right- continuous. This is not the case for $\chi_{\mathbb{Q}}$. – Pete L. Clark Jul 10 2010 at 6:28
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http://quant.stackexchange.com/tags/binary-options/hot
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# Tag Info
## Hot answers tagged binary-options
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### Do binary options make any sense?
The text of your question doesn't actually match the question title. The answer to your title is of course yes binary options make sense. And as others have pointed out with binary options your reward is limited, and conversely the risk involved in writing them is less. To answer your additional question you can replicate a binary option with a tight call ...
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### How does volatility affect the price of binary options?
The price of a binary option, ignoring interest rates, is basically the same as the CDF $\phi(S)$ (or $1-\phi(S)$ ) of the terminal probability distribution. Generally that terminal distribution will be lognormal from the Black-Scholes model, or close to it. Option price is $$C = e^{-rT} \int_K^\infty \psi(S_T) dS_T$$ for calls and P = e^{-rT} ...
5
### Do binary options make any sense?
The key difference besides the cap is that there is nothing in between: its 100 or nothing (binary!) - with traditional options you have S-K as long as S>K (for the call). You can find out more here: http://en.wikipedia.org/wiki/Binary_option
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### Do binary options make any sense?
They would make sense in certain narrow applications; one can perhaps think about scenarios where binary option might be the most efficient, quickest or easiest way to either benefit from a particular insight OR to hedge against some sort of event ... the real question is whether the volume in the markets for binary options will continue to sufficient to ...
2
### Creating a doubling and halving position
First of all, don't forget that there are two different probability measures at play here: the frequentist market measure that reflects actual observation of the market "in the long run" and the market-neutral martingale measure which is pertinent for pricing options. More or less, we can take the frequentist measure, and "back out" the effects of market ...
2
### Do binary options make any sense?
For a long or short position in a vanilla put, the maximum risk/reward is known and capped. For a long position in a vanilla call, the maximum risk is also always known and capped. The maximum reward is therefore known and capped for a short vanilla call position. However, the reward is unbounded for a long vanilla call position, and therefore the risk is ...
Only top voted, non community-wiki answers of a minimum length are eligible
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http://en.wikipedia.org/wiki/Continuum_hypothesis
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# Continuum hypothesis
In mathematics, the continuum hypothesis (abbreviated CH) is a hypothesis, advanced by Georg Cantor in 1878, about the possible sizes of infinite sets. It states:
There is no set whose cardinality is strictly between that of the integers and that of the real numbers.
Establishing the truth or falsehood of the continuum hypothesis is the first of Hilbert's 23 problems presented in the year 1900. The contributions of Kurt Gödel in 1940 and Paul Cohen in 1963 showed that the hypothesis can neither be disproved nor be proved using the axioms of Zermelo–Fraenkel set theory, the standard foundation of modern mathematics, provided ZF set theory is consistent.
The name of the hypothesis comes from the term the continuum for the real numbers.
## Cardinality of infinite sets
Main article: Cardinal number
Two sets are said to have the same cardinality or cardinal number if there exists a bijection (a one-to-one correspondence) between them. Intuitively, for two sets S and T to have the same cardinality means that it is possible to "pair off" elements of S with elements of T in such a fashion that every element of S is paired off with exactly one element of T and vice versa. Hence, the set {banana, apple, pear} has the same cardinality as {yellow, red, green}.
With infinite sets such as the set of integers or rational numbers, this becomes more complicated to demonstrate. The rational numbers seemingly form a counterexample to the continuum hypothesis: the integers form a proper subset of the rationals, which themselves form a proper subset of the reals, so intuitively, there are more rational numbers than integers, and more real numbers than rational numbers. However, this intuitive analysis does not take account of the fact that all three sets are infinite. It turns out the rational numbers can actually be placed in one-to-one correspondence with the integers, and therefore the set of rational numbers is the same size (cardinality) as the set of integers: they are both countable sets.
Cantor gave two proofs that the cardinality of the set of integers is strictly smaller than that of the set of real numbers (see Cantor's first uncountability proof and Cantor's diagonal argument). His proofs, however, give no indication of the extent to which the cardinality of the integers is less than that of the real numbers. Cantor proposed the continuum hypothesis as a possible solution to this question.
The hypothesis states that the set of real numbers has minimal possible cardinality which is greater than the cardinality of the set of integers. Equivalently, as the cardinality of the integers is $\aleph_0$ ("aleph-naught") and the cardinality of the real numbers is $2^{\aleph_0}$, the continuum hypothesis says that there is no set $S$ for which
$\aleph_0 < |S| < 2^{\aleph_0}. \,$
Assuming the axiom of choice, there is a smallest cardinal number $\aleph_1$ greater than $\aleph_0$, and the continuum hypothesis is in turn equivalent to the equality
$2^{\aleph_0} = \aleph_1. \,$
There is also a generalization of the continuum hypothesis called the generalized continuum hypothesis (GCH) which says that for all ordinals $\alpha\,$
$2^{\aleph_\alpha} = \aleph_{\alpha+1}.$
A consequence of the hypothesis is that every infinite subset of the real numbers either has the same cardinality as the integers or the same cardinality as the entire set of the reals.
## Impossibility of proof and disproof in ZFC
Cantor believed the continuum hypothesis to be true and tried for many years to prove it, in vain. It became the first on David Hilbert's list of important open questions that was presented at the International Congress of Mathematicians in the year 1900 in Paris. Axiomatic set theory was at that point not yet formulated.
Kurt Gödel showed in 1940 that the continuum hypothesis (CH for short) cannot be disproved from the standard Zermelo–Fraenkel set theory (ZF), even if the axiom of choice is adopted (ZFC). Paul Cohen showed in 1963 that CH cannot be proven from those same axioms either. Hence, CH is independent of ZFC. Both of these results assume that the Zermelo–Fraenkel axioms themselves do not contain a contradiction; this assumption is widely believed to be true.
The continuum hypothesis was not the first statement shown to be independent of ZFC. An immediate consequence of Gödel's incompleteness theorem, which was published in 1931, is that there is a formal statement (one for each appropriate Gödel numbering scheme) expressing the consistency of ZFC that is independent of ZFC. The continuum hypothesis and the axiom of choice were among the first mathematical statements shown to be independent of ZF set theory. These independence proofs were not completed until Paul Cohen developed forcing in the 1960s.
The continuum hypothesis is closely related to many statements in analysis, point set topology and measure theory. As a result of its independence, many substantial conjectures in those fields have subsequently been shown to be independent as well.
So far, CH appears to be independent of all known large cardinal axioms in the context of ZFC.
Gödel and Cohen's negative results are not universally accepted as disposing of the hypothesis, and Hilbert's problem remains an active topic of contemporary research (see Woodin 2001a). Koellner (2011a) has also written an overview of the status of current research into CH.
## Arguments for and against CH
Gödel believed that CH is false and that his proof that CH is consistent only shows that the Zermelo–Fraenkel axioms do not adequately describe the universe of sets. Gödel was a platonist and therefore had no problems with asserting the truth and falsehood of statements independent of their provability. Cohen, though a formalist[citation needed], also tended towards rejecting CH.
Historically, mathematicians who favored a "rich" and "large" universe of sets were against CH, while those favoring a "neat" and "controllable" universe favored CH. Parallel arguments were made for and against the axiom of constructibility, which implies CH. More recently, Matthew Foreman has pointed out that ontological maximalism can actually be used to argue in favor of CH, because among models that have the same reals, models with "more" sets of reals have a better chance of satisfying CH (Maddy 1988, p. 500).
Another viewpoint is that the conception of set is not specific enough to determine whether CH is true or false. This viewpoint was advanced as early as 1923 by Skolem, even before Gödel's first incompleteness theorem. Skolem argued on the basis of what is now known as Skolem's paradox, and it was later supported by the independence of CH from the axioms of ZFC, since these axioms are enough to establish the elementary properties of sets and cardinalities. In order to argue against this viewpoint, it would be sufficient to demonstrate new axioms that are supported by intuition and resolve CH in one direction or another. Although the axiom of constructibility does resolve CH, it is not generally considered to be intuitively true any more than CH is generally considered to be false (Kunen 1980, p. 171).
At least two other axioms have been proposed that have implications for the continuum hypothesis, although these axioms have not currently found wide acceptance in the mathematical community. In 1986, Chris Freiling presented an argument against CH by showing that the negation of CH is equivalent to Freiling's axiom of symmetry, a statement about probabilities. Freiling believes this axiom is "intuitively true" but others have disagreed. A difficult argument against CH developed by W. Hugh Woodin has attracted considerable attention since the year 2000 (Woodin 2001a, 2001b). Foreman (2003) does not reject Woodin's argument outright but urges caution.
Solomon Feferman (2011) has made a complex philosophical argument that CH is not a definite mathematical problem. He proposes a theory of "definiteness" using a semi-intuitionistic subsystem of ZF that accepts classical logic for bounded quantifiers but uses intuitionistic logic for unbounded ones, and suggests that a proposition $\phi$ is mathematically "definite" if the semi-intuitionistic theory can prove $(\phi \or \neg\phi)$. He conjectures that CH is not definite according to this notion, and proposes that CH should therefore be considered not to have a truth value. Peter Koellner (2011b) wrote a critical commentary on Feferman's article.
## The generalized continuum hypothesis
The generalized continuum hypothesis (GCH) states that if an infinite set's cardinality lies between that of an infinite set S and that of the power set of S, then it either has the same cardinality as the set S or the same cardinality as the power set of S. That is, for any infinite cardinal $\lambda\,$ there is no cardinal $\kappa\,$ such that $\lambda <\kappa <2^{\lambda}.\,$ An equivalent condition is that $\aleph_{\alpha+1}=2^{\aleph_\alpha}$ for every ordinal $\alpha.\,$ The beth numbers provide an alternate notation for this condition: $\aleph_\alpha=\beth_\alpha$ for every ordinal $\alpha.\,$
This is a generalization of the continuum hypothesis since the continuum has the same cardinality as the power set of the integers.
Like CH, GCH is also independent of ZFC, but Sierpiński proved that ZF + GCH implies the axiom of choice (AC), so choice and GCH are not independent in ZF; there are no models of ZF in which GCH holds and AC fails. To prove this, Sierpiński showed GCH implies that every cardinality n is smaller than some Aleph number, and thus can be ordered. This is done by showing that n is smaller than $2^{\aleph_0+n}\,$ which is smaller than its own Hartogs number (this uses the equality $2^{\aleph_0+n}\, = \,2\cdot\,2^{\aleph_0+n}$; for the full proof, see Gilllman (2002).
Kurt Gödel showed that GCH is a consequence of ZF + V=L (the axiom that every set is constructible relative to the ordinals), and is consistent with ZFC. As GCH implies CH, Cohen's model in which CH fails is a model in which GCH fails, and thus GCH is not provable from ZFC. W. B. Easton used the method of forcing developed by Cohen to prove Easton's theorem, which shows it is consistent with ZFC for arbitrarily large cardinals $\aleph_\alpha$ to fail to satisfy $2^{\aleph_\alpha} = \aleph_{\alpha + 1}.$ Much later, Foreman and Woodin proved that (assuming the consistency of very large cardinals) it is consistent that $2^\kappa>\kappa^+\,$ holds for every infinite cardinal $\kappa.\,$ Later Woodin extended this by showing the consistency of $2^\kappa=\kappa^{++}\,$ for every $\kappa\,$. A recent result of Carmi Merimovich shows that, for each n≥1, it is consistent with ZFC that for each κ, 2κ is the nth successor of κ. On the other hand, Laszlo Patai proved, that if γ is an ordinal and for each infinite cardinal κ, 2κ is the γth successor of κ, then γ is finite.[citation needed]
For any infinite sets A and B, if there is an injection from A to B then there is an injection from subsets of A to subsets of B. Thus for any infinite cardinals A and B,
$A < B \to 2^A \le 2^B.$
If A and B are finite, the stronger inequality
$A < B \to 2^A < 2^B \!$
holds. GCH implies that this strict, stronger inequality holds for infinite cardinals as well as finite cardinals.
### Implications of GCH for cardinal exponentiation
Although the Generalized Continuum Hypothesis refers directly only to cardinal exponentiation with 2 as the base, one can deduce from it the values of cardinal exponentiation in all cases. It implies that $\aleph_{\alpha}^{\aleph_{\beta}}$ is (see: Hayden & Kennison (1968), page 147, exercise 76):
$\aleph_{\beta+1}$ when α ≤ β+1;
$\aleph_{\alpha}$ when β+1 < α and $\aleph_{\beta} < \operatorname{cf} (\aleph_{\alpha})$ where cf is the cofinality operation; and
$\aleph_{\alpha+1}$ when β+1 < α and $\aleph_{\beta} \ge \operatorname{cf} (\aleph_{\alpha})$.
## References
• Cohen, P. J. (1966). Set Theory and the Continuum Hypothesis. W. A. Benjamin.
• Cohen, Paul J. (December 15, 1963). "The Independence of the Continuum Hypothesis". Proceedings of the National Academy of Sciences of the United States of America 50 (6): 1143–1148. doi:10.1073/pnas.50.6.1143. JSTOR 71858. PMC 221287. PMID 16578557.
• Cohen, Paul J. (January 15, 1964). "The Independence of the Continuum Hypothesis, II". Proceedings of the National Academy of Sciences of the United States of America 51 (1): 105–110. doi:10.1073/pnas.51.1.105. JSTOR 72252. PMC 300611. PMID 16591132.
• Dales, H. G.; W. H. Woodin (1987). An Introduction to Independence for Analysts. Cambridge.
• Enderton, Herbert (1977). Elements of Set Theory. Academic Press.
• (lecture slides)
• Foreman, Matt (2003). "Has the Continuum Hypothesis been Settled?" (PDF). Retrieved February 25, 2006.
• Freiling, Chris (1986). "Axioms of Symmetry: Throwing Darts at the Real Number Line". Journal of Symbolic Logic (Association for Symbolic Logic) 51 (1): 190–200. doi:10.2307/2273955. JSTOR 2273955.
• Gödel, K. (1940). The Consistency of the Continuum-Hypothesis. Princeton University Press.
• Gillman, Leonard (2002). "Two Classical Surprises Concerning the Axiom of Choice and the Continuum Hypothesis". American Mathematical Monthly 109.
• Gödel, K.: What is Cantor's Continuum Problem?, reprinted in Benacerraf and Putnam's collection Philosophy of Mathematics, 2nd ed., Cambridge University Press, 1983. An outline of Gödel's arguments against CH.
• Seymour Hayden and John F. Kennison: "Zermelo–Fraenkel Set Theory" (1968), Charles E. Merrill Publishing Company, Columbus, Ohio.
• Koellner, Peter (2011a). "The Continuum Hypothesis". Exploring the Frontiers of Independence (Harvard lecture series).
•
• Kunen, Kenneth (1980). . Amsterdam: North-Holland. ISBN 978-0-444-85401-8.
• Maddy, Penelope (June 1988). "Believing the Axioms, I". Journal of Symbolic Logic (Association for Symbolic Logic) 53 (2): 481–511. doi:10.2307/2274520. JSTOR 2274520.
• Martin, D. (1976). "Hilbert's first problem: the continuum hypothesis," in Mathematical Developments Arising from Hilbert's Problems, Proceedings of Symposia in Pure Mathematics XXVIII, F. Browder, editor. American Mathematical Society, 1976, pp. 81–92. ISBN 0-8218-1428-1
•
• Merimovich, Carmi (2007). "A power function with a fixed finite gap everywhere". Journal of Symbolic Logic 72 (2): 361–417. doi:10.2178/jsl/1185803615.
• Woodin, W. Hugh (2001a). "The Continuum Hypothesis, Part I" (PDF). Notices of the AMS 48 (6): 567–576.
• Woodin, W. Hugh (2001b). "The Continuum Hypothesis, Part II" (PDF). Notices of the AMS 48 (7): 681–690.
Primary literature in German
• Cantor, Georg (1878), "Ein Beitrag zur Mannigfaltigkeitslehre", 84: 242–258 .
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http://physics.stackexchange.com/questions/24295/relationship-between-the-angle-of-the-floor-and-the-angular-velocity-in-a-bank?answertab=oldest
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# Relationship between the “angle of the floor” and the angular velocity in a banked turn?
Wel, imagine that you're in a carousel, and the floor is, let's say at $\theta=0$ so it's totally horizontal, if $\theta=90$ the floor would be vertically.
The object put above the floordoesn't move until the carousel starts.
My question is:
With what angular velocity($\omega$) an object would have to go to stay where it is knowing the angle of the floor($\alpha$), the radius of the carousel($r$) and the gravity($g$)?
What i'm asking is when you combine this two types of movements.
Something like this:
Based on the Newton's second law of motion($F=ma$)
I finally get this equation:
$$\omega = \sqrt{-\dfrac{g\tan{\theta}}{r\cos{\theta}}}$$
Being:
$g=\text{gravity}=-9,8^m/_{s^2}$
$r=\text{radius of the carousel}=1m$
$\theta=\text{angle of the floor}=x\text{ axis in the graph below this}$
$\omega=\text{angular velocity}=y\text{ axis in the graph below this}$
The graph of this function is
And if I'm right,
Is it true that this type of motion doesn't depend on his mass?
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I think the problem is that nobody understands the question. Neither do I understand the question. – Pygmalion Apr 24 '12 at 10:30
## 1 Answer
I almost agree with your result.
For an object in a circular path of radius $r$ moving with angular velocity $\omega$ the acceleration towards the center is $r\omega^2$. This acceleration is horizontal in your top diagram, so the component of the acceleration acting along the slope (the purple arrow in your diagram) is $rw^2cos(\theta)$.
You've correctly drawn the acceleration of the object down the slope, F$_{||}$ as $g.sin(\theta)$, so just set them equal and you get:
$$rw^2cos(\theta) = g.sin(\theta)$$
$$\omega = \sqrt{\frac{g}{r} \frac{sin(\theta)}{cos(\theta)}}$$
We just disagree in that I've got $sin$ where you've got $tan$.
Note that I've just equated the accelerations in my treatment above. You could equate forces and include the mass is you want to, but the mass is the same on both sides of the equation and will cancel out, so you're correct that in this case the mass doesn't affect the result.
I suspect you've put off people from looking at your question by drawing so many diagrams, hence the lack of response. Only the top diagram is needed. Drawing a good clear diagram is often the most important part of attacking a problem like this.
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Yeah. I was wrong in the calculations. Nos I know where I have failed. Thanks! and $v$ can be written as $v=\sqrt{gr\tan{\theta}}$ and $\omega$ as $\omega=\sqrt{\frac{g\tan{\theta}}{r}}$. But this is obvious. – Garmen1778 Apr 24 '12 at 19:27
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http://m-phi.blogspot.com/2011/08/every-thing-is-what-it-is-and-not.html
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# M-Phi
A blog dedicated to mathematical philosophy.
## Saturday, 13 August 2011
### "Every thing is what it is, and not another thing"
In general, I'm not keen on reductionism. Bad for the health. That isn't to say we don't know how to reduce $A$s to $B$s (or, reduce theory $T$ in $L$ to theory $T^{\prime}$ in $L^{\prime}$, or, to interpret one structure $\mathcal{A}$ in another $\mathcal{B}$).
One can reduce natural numbers to finite sets (usually, reduce $\mathbb{N}$ to $\omega$.) One can reduce integers to equivalence classes of pairs of natural numbers. For example, identify $\mathbb{Z}$ with the set $\{[(n, m)]: n, m \in \mathbb{N}\}$, where $[(n, m)]$ is the equivalence class of pairs $(k, p)$ such that $n+p = k+m$. The intuitive idea is that the ordered pair $(n,m)$ represents the integer $n-m$. But there are infinitely many such; so we take the equivalence class of all. By a similar method (this time with $\times$ instead of $+$), we get the rationals, $\mathbb{Q}$. One can reduce real numbers to equivalence classes of Cauchy sequences of rationals, or to Dedekind cuts of rationals. One can reduce the ordered pair $(x,y)$ to some set, say $\{\{x\}, \{x, y\}\}$. One can reduce ordered pairs of numbers (and, more generally, finite sequences of numbers) to numbers. See what I did? - I conflated ordered $n$-tuples with finite sequences. No matter: we can reduce either to the other. And relations can be reduced to sets of ordered $n$-tuples and functions to special relations satisfying a uniqueness condition on one argument place. One can reduce ordinals to transitive $\epsilon$-well-founded sets and reduce cardinals to ordinals.
Even so, I don't think numbers are sets; I don't think real numbers are equivalence classes of Cauchy sequences; etc. These reductions are examples of interpretations or encodings - successful ones, but generally non-unique. This is the message of Paul Benacerraf's famous 1965 article, "What Numbers Could Not Be" (Philosophical Review 74). Given the non-uniqueness of these reductions, one can perhaps try and argue that one is special or distinguished, but this is quite implausible (despite the convenience of identifying $\mathbb{N}$ with $\omega$.) Or one can argue that there are no numbers, integers, ordered pairs, etc., and all there are are sets or classes (and so we consider mathematical theories to be convenient definitional extensions of standard set theory, while agreeing that these extensions are not unique). That seems to be Quine's view. Or one can think of these special mathematical entities as sui generis. Or one can think of them structurally (e.g., to be a real number is to be a node in the abstract real number structure). This is Shapiro's and Resnik's view. Or one can be a nominalist (there are no natural numbers, rationals, reals, sets, functions, etc.). This is Field's view.
(Constructivism, as an ontological claim (e.g., Heyting), seems indefensible: if numbers are "mental constructions", then either there are infinitely many mental constructions (one for $0$, one for $1$, and so on) or only finite many numbers. The former contradicts empirical fact and the second contradicts mathematical fact. Perhaps numbers are "possible" mental constructions, in the mind of God? Well, then one might as well just admit that they are not constructions after all. Constructivism, as an epistemological claim, concerning which proofs are legitimate, is not affected by this argument.)
The views are: sets are basic and there's a special reduction of mathematical entities to sets (implausible, I think); there are only sets and classes and all other mathematical entities are introduced by convenient but non-unique definitions (Quine); different kinds of mathematical entities are sui generis; mathematical entities are nodes/places/positions in abstract structures (Shapiro); mathematical entities are mental constructions (Heyting); there aren't any mathematical entities (Goodman, Field).
Of these, I think my preferred view is the sui generis one. Nominalism and constructivism are inconsistent with science. Quine's view is too austere. Structuralism is a bit too weird (the "nodes" in the domain of an abstract structure have an odd status, but maybe structuralism is sui generis-ism in disguise).
So, the natural numbers are just what they are, not another thing, which brings me to Bishop Butler:
If the observation be true, it follows, that self-love and benevolence, virtue and interest, are not to be opposed, but only to be distinguished from each other; in the same way as virtue and any other particular affection, love of arts, suppose, are to be distinguished. Every thing is what it is, and not another thing. (Bishop Joseph Butler 1726, Fifteen Sermons Preached at the Rolls Chapel. Preface.)
Ok, so that's not mathematics, I admit; but, as Tom Lehrer once joked, the idea's the important thing.
#### 15 comments:
1. Anonymous
I'm not a constructivist, but 'indefensible' presents a challenge that can't be refused.
Firstly, just because there are infinitely many numbers doesn't mean a constructivist has to propose infinitely many actions by which they are constructed - perhaps the whole structure of the natural numbers is constructed in one act (or a finite sequence of acts), but the individual numbers are parts of the structure in such a way that they can all be said to have been constructed together through this process? This might make the hypothetical constructivist in question to be a structuralist as well (constructuralist?), but they’ll be one with a distinctive position (a distinctively constructivist one - their constructivism won‘t be drowned in their structuralism).
Secondly, even if the acts/constructions in question needn’t be actually completed ones (in the way that painting the Eiffel Tower blue is an action that could be done, but not the outcome of anyone’s real action to date), and the constructivist must give an account of possible actions which will look like some other theory of possible things (probably Platonism, but maybe even something like fictionalism), the fact that these ideal or fictional things are ideal or fictional constructions, rather than any other kind of ideal or fictional thing, is still a substantive claim. Wouldn’t a second hypothetical constructivist who said e.g. ’Yes, numbers are fictions, but specifically fictional constructions, and here’s how the fiction of constructions works: ...’ be saying something interesting? It seems to me that if the fictional things we’re studying are constructions and fictionalists spend all their time talking about some guy called Holmes or whatever, they might be missing important details, because Holmes isn’t a construction. (Well, the character is a construction, but in the fiction, no-one built Holmes.)
OK, I seem to have made my hypothetical constructivists into hybrids with other positions. Although, the second one need not take any position on those possible constructions - they could just say ’Whatever nature they have as merely possible things, they nonetheless have the nature of constructions’, and perhaps also give details as to what this means for Platonists, fictionalist, etc. in detail. I also haven’t really defended constructivism so much as hypothesised people who could, and I suspect if a real constructivist were to turn up, they’d just attack your argument in some existing constructivist way. Oh well, I’ve had fun at least.
2. Anonymous
Sorry, same anon. Since the above isn’t really directed at your own position (though I don’t think it’s irrelevant, or I wouldn’t leave it here), I’ll try a question: are the non-negative integers identical with the natural numbers? And secondly, do you think the question makes sense? I think it does for at least e.g. structuralists - they can ask whether the structure of the integers contains that of the natural numbers as a part or if it is indivisible, and if only the non-negative integers are that part, or whether e.g. all the integers bigger than 6, or even all the integers less than 6, or even all the odd numbers less than 6, are also the natural numbers (since there are such natural isomorphisms with the natural numbers, though here structuralism and reductionism can blur into each other a bit).
If the natural numbers’ being ‘sui generis’ is meant to rule out their being the same kind of thing even as otherwise similar mathematical objects like the integers (as opposed to being a different kind of thing from sets), that sounds a bit off to me. Surely they’re in some sense the same kind of object as the integers. And if they are, then I think it’s in that sense that philosophers ask what kind of thing they are. Absolutely everything is of its own kind in the sense of being itself and nothing else, that’s just what your quote says, but I think to philosophers who are asking what kind of thing numbers could possibly be, ‘numbers, of course’ isn’t going to be a satisfying answer (even if it turns out to be the only one possible).
3. Thanks, anon! (Glad you couldn't resist; it's Saturday and not much on the telly.)
The argument I gave is basically an argument that having each number $n$ as a particular mental construction requires a supertask, while also insisting that our minds can't do those. Maybe it would require infinite amount of energy (this would require supposing there is a finite lower bound on the energy content of a mental construction: which seems right).
The first suggestion you consider is to suppose that the whole structure, (N, 0, S), say, is "constructed" tout court by some mental act. (David Miller once made a similar suggestion, comparing it to giving birth). Yes, that becomes some sort of structuralist. But it does leaves inner working of the infinite construction rather mysterious, unless it simply means define (e.g., either explicitly, or implicitly, by some categorical axiom system).
(In mathematical parlance, one does often say "construct" just to mean "define" or "prove to exist" - but these definitions and proofs, understood as being tokened in certain human actions, are small and finite.)
The second idea involves going either modal or fictional. But I'd protest that neither fictionalism nor modal nominalism is constructivism. The fictionalist has their fiction - usual some specific (finite) piece of text. Say, some specific Conan Doyle story. Since the fictionalist applies an error theory to this story, there simply aren't any referents to account for. So, it isn't a view according to which there are numbers, and they're mental constructions. Rather it's a view that there aren't numbers, there is only talk of numbers. One could go mentally modal - talking of "possible mental constructions", but I think this requires postulating supertasks again.
4. Thanks for the question. The short answer is. Not sure.
I'd like to have inclusions:
$\mathbb{N} \subset \mathbb{Z} \subset \dots \subset \mathbb{C}$.
Then, the natural number 0 is the integer 0.
(The Benacerraf argument focuses on non-unique identification of numbers with sets of various kinds.) But this would, perhaps, violate the spirit of sui generis-ism.
On the other hand, it seems that structuralism has to reject the identification of the natural number 0 and the integer 0: for the former is a node in the natural number structure while the latter is a node in a non-isomorphic structure, and presumably domains of distinct abstract structures have to be disjoint.
5. Anonymous
If constructing the numbers as individual objects is something like drawing more and more tally marks or counting on a very large number of fingers (and I can’t think of any other way of doing it, though I wonder how you’re meant to ‘construct’ 0 using tally marks or finger counting), then I agree that constructing all of them would require completing a supertask. I would, like you, be surprised if humans could do this. I’m not sure what you mean when you say that postulating an infinity of possible mental constructions involves also postulating supertasks, though. Each of those possible mental constructions is finitely completable (though perhaps some are very long indeed, which is probably just as bad for human calculators as if they were infinite - I don’t know how we could determine how long they would have to take, as it doesn‘t seem a determinate enough question to answer mathematically).
As to constructing the whole structure at once, I don’t know how to do it. I’m not a constructivist, after all. I’m not sure what to start with (surely there must be something you construct the natural numbers from - if they come out of nowhere, that’s not a construction). When reductionists start with sets, they usually build the naturals first and work from there. Could it ever be OK to start with the real numbers and construct the naturals from them? I’m not sure how to do that in a way not involving supertasks; you can maybe invoke the integers as the group the acts on something real in some particular way - could that constitute constructing them all at once? - then pick one element to be 0, define arithmetic and isolate the natural numbers.
On fictionalism, I was proposing, perhaps stupidly, that whether the thing that is claimed to not really exist is a non-existent construction as opposed to a non-existent something else might make a difference. It’s not quite the same as pointing out that even though Holmes is fictional, it’s important that he is a fictional detective rather than a fictional bus driver, because if the corresponding fiction in our case is the theory of the natural numbers, the constructivistic fictionalist doesn’t want to say that it is a truth in that theory that the numbers are constructions. It’s more like someone being a fictionalist about deities, so they think there aren’t any but that some claims about them are especially appropriate (in a way that mimics being true), and being specifically a monotheist fictionalist rather than a polytheist one. Even though they think there aren’t any, they think there isn’t one rather than that there aren’t 7. (OK, obviously if there isn’t one, there aren’t 7, but the claim that there are 7 isn’t even relevant, or at least isn’t appropriate, whereas the claim that there is one is relevant, appropriate, and false.) But in our case, what are appropriate or not to fictionalists are claims of arithmetic, not philosophical claims like ‘numbers are constructions’, so perhaps I’m getting my levels muddled.
6. Hi - many interesting points there. On getting the natural numbers from the reals, there is a difficulty based on a famous technical result (due to Tarski), concerning the first-order theory of real numbers (technIcally, RCF, for "real-closed field").
RCF is a complete theory. (Every sentence in the language of RCF is either a theorem or refutable.)
This means that Peano arithmetic, which is essentially undecidable, is not interpretable in RCF. One cannot therefore (at least not in the first-order theory) define a special subset of reals and prove that they satisfy the desired properties of successor, addition, multiplication and induction. One can define 0, 1, 2, etc., "one-by-one", as: 0, (0+1), (0+1+1), etc. But one can't get "x is an ancestor of 0 under successor". To overcome this, one needs to quantify over sets of reals (adding a comprehension principle). This will give a theory which is incomplete, and can define "x is a natural number". So, if one wants to start with the (first-order theory of the) reals, one needs to add sets of reals in order to get the natural numbers.
7. I think the difficulty of combining (ontological) constructivism and fictionalism is that they disagree about not only what the numbers, etc., are, but crucially about whether there even are such things.
(i) Constructivism: there are numbers, but they're "mental constructions" in some sense. (A view like this is articulated by Brouwer's student Arend Heyting, for example.)
but
(ii) Fictionalism: there are no numbers, but there is talk about numbers. (The sort of fictionalist view I am thinking of is Field's Science Without Numbers, and more recently, Mary Leng's Mathematics and Reality.)
These are claims in the metatheory, of course, and not the object theory. Both would like to satisfy something like the assertibility of ground level arithmetic claims - but it's important they have very different semantic metatheories. The fictionalist's semantic metatheory is, usually, a standard Tarskian semantics (or maybe deflationary) along with an error theory (existential claims of arithmetic are false). While the ontological constructivist will almost certainly defend some sort of assertibility semantics for arithmetic (thus joining with epistemological constructivism - resulting in the rejection of certain instances of LEM).
8. Anonymous
I see the problem with starting from the reals. I wasn’t sure how much structure to suggest starting with. Actually, I was thinking more about starting from Euclidean geometry (where there is a notion of ‘construct’ slightly different from just ‘define’, which I hoped would be closer to the kind of construction that constructivists might accept). E.g. if an equilateral triangle is constructed, maybe a constructivist could say that by inspection of it and its rotations, we come to construct arithmetic mod 3 all at once (so each rotation is associated with, though not identified with, a number, and addition with concatenation). That seems adequately finite to me, and reassuringly traditionalist. But is there any finite constructible structure that will give us an infinite arithmetic (other than the real numbers themselves) in the same sort of fashion?
The plan was for a constructivist fictionalist to be able to say ‘There are no numbers, there is acceptable talk of numbers as constructions, any talk of things other than constructions isn’t even arithmetic.’ I’m sure constructivism does posit numbers, but I don’t see it has to. If constructivists really are committed to those semantics, then it does look as if there is a problem, though, because the conditions for asserting that a+b=c or something like that are probably going to require the person making the assertion to have a construction, or possible construction, to hand, and the theory is that there are no such things to be provided. Here is where the level confusion was, I think: epistemological constructivists can’t be error theorists about arithmetical statements themselves, because they don’t think about (some of? - isn’t it OK to talk about the strictly finite statements as true?) them in terms of truth, just acceptability; it’s the acceptability conditions where constructions come in, and I think they could be error theorists here - it’s acceptable to claim a+b=c if you have a supporting claim about a corresponding construction, and the latter claim is truth evaluable, and strictly false (because of fictionalism about possible constructions), but still assertible. This doesn’t help me with respect to ontological constructivists, though, and they’re the ones we’re really interested in. I’m tempted to say that their claims about the constructive nature of numbers are at the same level as the epistemological constructivist’s claims about acceptability, and so can be both truth evaluable and false, but I’m not really sure.
Also, on structuralism, I suppose I was suggesting that there could be a debate over whether distinct structures have to be disjoint. A structuralist who thinks that structures actually contain objects, but that it makes no sense to ask which thing an object is outside of a particular structure would probably have to say the do have to be disjoint. That sounds like Benacerraf’s view, in that one paper at least (and maybe Shapiro’s? - I’m really not sure there), but I don’t think it’s essential to structuralism. Though now I’ve thought about it some more, it does sound a bit unstructuralist to say that the things called ‘0’ in the naturals and the reals are the same object (if they are objects at all).
9. Anonymous
I thought I posted this earlier, but it's not there, so I'll try again. Hopefully, it won't now appear twice.
I see the problem with starting from the reals. I wasn’t sure how much structure to suggest starting with. Actually, I was thinking more about starting from Euclidean geometry (where there is a notion of ‘construct’ slightly different from just ‘define’, which I hoped would be closer to the kind of construction that constructivists might accept). E.g. if an equilateral triangle is constructed, maybe a constructivist could say that by inspection of it and its rotations, we come to construct arithmetic mod 3 all at once (so each rotation is associated with, though not identified with, a number, and addition with concatenation). That seems adequately finite to me, and reassuringly traditionalist. But is there any finite constructible structure that will give us an infinite arithmetic (other than the real numbers themselves) in the same sort of fashion?
The plan was for a constructivist fictionalist to be able to say ‘There are no numbers, there is acceptable talk of numbers as constructions, any talk of things other than constructions isn’t even arithmetic.’ I’m sure constructivism does posit numbers, but I don’t see it has to. If constructivists really are committed to those semantics, then it does look as if there is a problem, though, because the conditions for asserting that a+b=c or something like that are probably going to require the person making the assertion to have a construction, or possible construction, to hand, and the theory is that there are no such things to be provided. Here is where the level confusion was, I think: epistemological constructivists can’t be error theorists about arithmetical statements themselves, because they don’t think about (some of? - isn’t it OK to talk about the strictly finite statements as true?) them in terms of truth, just acceptability; it’s the acceptability conditions where constructions come in, and I think they could be error theorists here - it’s acceptable to claim a+b=c if you have a supporting claim about a corresponding construction, and the latter claim is truth evaluable, and strictly false (because of fictionalism about possible constructions), but still assertible. This doesn’t help me with respect to ontological constructivists, though, and they’re the ones we’re really interested in. I’m tempted to say that their claims about the constructive nature of numbers are at the same level as the epistemological constructivist’s claims about acceptability, and so can be both truth evaluable and false, but I’m not really sure.
Also, on structuralism, I suppose I was suggesting that there could be a debate over whether distinct structures have to be disjoint. A structuralist who thinks that structures actually contain objects, but that it makes no sense to ask which thing an object is outside of a particular structure would probably have to say the do have to be disjoint. That sounds like Benacerraf’s view, in that one paper at least (and maybe Shapiro’s? - I’m really not sure there), but I don’t think it’s essential to structuralism. Though now I’ve thought about it some more, it does sound a bit un-structuralist to say that the things called ‘0’ in the naturals and the reals are the same object (if they are objects at all).
10. Hi - the disappearance is caused because by comments being "anonymous", which get caught by Blogger's spam filter. So, when I notice something has been caught, I have to "release" it.
The situation with RCF is the same as with first-order geometry in n dimensions. Call it $EG_1(n)$.
The result is due to Tarski.
$EG_1(n)$ is complete (and decidable).
Any sentence $\phi$ in the language of $EG_1(n)$ is either a theorem or refutable. So, we can't get arithmetic going in this theory. Tarski also gives a representation theorem for this theory:
$\mathcal{A} \models EG_1(n)$
iff
$\mathcal{A}$ has the form
$(\mathcal{F}^n, Bet(\mathcal{F}^n), Cong(\mathcal{F}^n))$,
where $\mathcal{F}$ is a real-closed field. Model-theoretically, first-order geometry is intimately related to the first-order theory of real numbers.
http://en.wikipedia.org/wiki/Tarski's_axioms
11. The question about the domain identity for structuralism is the really central problem that arises, I think.
If N is the abstract natural number structure and Z is the abstract integer structure, then it looks like the relation of the domain of former to the domain of the latter cannot be inclusion, for 0 in N seems to be what it is in virtue of the whole surrounding structure - N. And 0 in Z seems to be what it is in virtue of the whole surrounding structure - Z. So it seems that, because these are distinct structures, the 0 in N has to be distinct from the 0 in Z.
This is all rather speculative, though. I think it's probably a consequence of Shapiro's structuralism.
Hopefully, the sui generis approach doesn't face this problem.
12. Campbell Brown
Everything is what it is: $\forall x (x = x)$.
Nothing is another thing: $\forall x \forall y (x \neq y \rightarrow x \neq y)$.
The quote from Bishop Butler -- 'Everything is what it is, and not another thing' -- seems to say merely that identity is reflexive. Reductionists needn't deny that. If numbers are sets, then saying that numbers are sets is just saying that they are what they are, not that they are another thing.
13. Hi Campbell, yes, you're right. Literally read, it just means what you say. Still, Bishop Butler manages to get his point across ...
It's quite hard to articulate this exactly. I think it's some sort of meta-claim of non-implication in disguise, maybe: "From the possibility of reducing $A$s to $B$s (or the structural similarity of $A$s and $B$s), it doesn't follow that As are Bs".
14. Hi Jeff and Campbell,
Like Jeff, I'm sympathetic to the sui generis view (which I used to call 'unruffled Butlerism').
Campbell's point is a good one. A suggestion: we obviously have some working criteria for deciding whether a purported reduction of mathematical objects is genuinely *reductive* (ignoring the question of adequacy or truth), or just trivial in some sense.
I think what the Butlerian-Benaceraffian really wants to do is to take those reductivity-criteria and re-employ them as falsity-criteria.
15. To clarify, by 'purported reduction' above I meant reduction of the sort which does entail identity - the sorts of claims discussed in Benacerraf's article. (Reductions in a weaker sense can be true on the Butler-Benacerraf view.)
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http://stats.stackexchange.com/questions/7542/how-to-perform-goodness-of-fit-test-and-how-to-assign-probability-with-uniform/7547
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# How to perform goodness of fit test and how to assign probability with uniform distribution?
I have to demonstrate that a generator of VoIP calls generates calls uniformly distributed between callers. In particular the distribution is the uniform (min, max) one where the volume per caller distribution is uniformly distributed between a minimum and maximum. So by running a test with 10000 users and a min value equal to 30 calls per week and a max value equal to 90 calls/week i obtain that not all the users respect this limits.
The situation is depicted in figure:
The few users that generate <30 or >90 calls spoil the chi-square goodness of fit test and I don't know how can i proceed with a goodness of fit test. Most of the values is within the interval (and from these we obtain low chi-square value), but the few out-of-range values spoil the final chi-square calculation.
In your opinion what is the best way to operate? What should I do with the "out-of-range"values? Thank you.
PS: the chi-square goodness-of-fit test performed is reported in the following figure:
where still I don't know what to do with the out-of-range values.
UPDATE
after talking with people linked with the project we concluded that the generator does not satisfy the uniform distribution. We have to do a theoretic analysis of what we expect really to be the distribution at the end of the generation based on the inputs.
This means that I have to do it! More details: The generator assigns a "probability" between 0 and 1 to the callers (with a particular method, that probably is the problem). Then it generates a random value from 0 to 1 and it finds the associated user and assigns the call to him.
The generator generates calls for a week with the constant rate equal to 1 call per second, this means that it generates ca. 604800 total calls.
My goal is to distribute the callers between the min and max number of calls in a week. For example if I have 10000 users and the min limit is equal to 30 calls per week and max = 90 calls per week I should obtain something about:
30 calls : 163 users.
31 calls : 163 users.
....
90 calls : 163 users.
So 163 users generate 30 calls in a week..etc, etc and finally 163 users generate 90 calls in a week. How should I assign the probability to callers in order that the generator distributes the callers uniformly between the range 30-90?
-
11
You have demonstrated the calls are not uniformly distributed. The $\chi^2$ calculation is not "spoiled": it worked! – whuber♦ Feb 23 '11 at 16:23
ok, mmm..maybe is there a "conceptual" error during the implementation of the generator? because for my thesis, my supervisor asked me to demonstrate formally that the generator (which is the result of months&months of study and they use for a long time) really follows the uniform distribution. Maybe can I use a sort of "approximation"? – Maurizio Feb 23 '11 at 16:35
5
What would you like to approximate? Obviously you cannot validly demonstrate this generator produces uniform results: the data flatly contradict that. The next step depends on what actions you contemplate. Could the data be wrong? Could they inaccurately reflect what the generator is doing? Could they indicate an unexpected phenomenon? Could the design of the generator be wrong? You need to raise (and eventually address) questions like these. The one question that is definitively settled, though, is the one you originally asked: these data are not uniform! – whuber♦ Feb 23 '11 at 16:45
thanks, i think it is better to speak with who has implemented the generator before reaching hasty conclusions. – Maurizio Feb 23 '11 at 16:52
1
Then the answer would be to apply a goodness of fit test such as the chi square. What people are doing fitting to the data generated interval where data falls into cells outside the range doesn't make sense to me. If you are strict about the interval once you find a single observation outside the interval you can unequivocally reject the hypothesis of uniformity on the interval without any goodness of fit test. – Michael Chernick May 3 '12 at 20:03
show 7 more comments
## 2 Answers
Perhaps you should randomly permute the callers, then, per your example, claim that the first 163 made 30 calls, the next 163 made 31 calls, etc. By 'permutation', I mean sort them in a random order. The simple way to do this is to use your random number generator to assign each caller a number, then sort them by that number.
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What you are describing resembles a "continuous uniform distribution",
http://mathworld.wolfram.com/UniformDistribution.html
-Ralph Winters
-
7
@Ralph The whole point of the comments following the question is that although the distribution of the data might "resemble" a uniform distribution, it is extremely unlikely that the data were truly obtained from this distribution. That's why we use quantitative methods in statistics: they keep us from fooling ourselves (and others) with suggestive patterns or resemblances that are not justifiable. – whuber♦ Feb 23 '11 at 20:46
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http://mathhelpforum.com/algebra/197463-finding-polynomials-lowest-possible-degree-satisfy-given-condition-print.html
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# Finding polynomials of the lowest possible degree that satisfy a given condition
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• April 17th 2012, 03:23 PM
LJoe21
Finding polynomials of the lowest possible degree that satisfy a given condition
Hello there,
I need to find the polynomials $f(x)$ and $g(x)$, which are of the lowest possible degree and satisfy the given condition:
$(x^4-2x^3-4x^2+6x+1) \cdot f(x) + (x^3-5x-3) \cdot g(x) = x^4$
All I want to know is how to find the lowest possible degree that the polynomials should be of. I can solve the rest using the method of undetermined coefficients—but if there's another way (preferably easier), then feel free to share.
This is my first post here, so please bear with me if I did anything wrong.
Thanks in advance.
• April 17th 2012, 05:09 PM
skeeter
Re: Finding polynomials of the lowest possible degree that satisfy a given condition
Quote:
Originally Posted by LJoe21
Hello there,
I need to find the polynomials $f(x)$ and $g(x)$, which are of the lowest possible degree and satisfy the given condition:
$(x^4-2x^3-4x^2+6x+1) \cdot f(x) + (x^3-5x-3) \cdot g(x) = x^4$
All I want to know is how to find the lowest possible degree that the polynomials should be of. I can solve the rest using the method of undetermined coefficients—but if there's another way (preferably easier), then feel free to share.
This is my first post here, so please bear with me if I did anything wrong.
Thanks in advance.
note that the right side of the equation is degree 4. since the two products are summed, that means both products on the left side would have to be 4th degree, also.
f(x) is multiplying a 4th degree poly ... so, wouldn't f(x) have to have degree 0 (i.e. , a constant) ?
so, what degree would g(x) have to be to form a product that is also 4th degree?
• April 17th 2012, 11:56 PM
LJoe21
Re: Finding polynomials of the lowest possible degree that satisfy a given condition
Well, the solutions are (according to my workbook):
$f(x) = 9x^2 - 26x - 21$
$g(x) = -9x^3 + 44x^2 - 39x - 7$
and I've absolutely no idea how to determine that $f(x)$ should be of second degree and $g(x)$ of third.
• April 18th 2012, 06:32 AM
HallsofIvy
Re: Finding polynomials of the lowest possible degree that satisfy a given condition
This has nothing to do with your original question. Your original question asked for the lowest degree of polynomials that would satisfy the given equation. The answer given in your second post are NOT of lowest degree.
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http://mathhelpforum.com/differential-geometry/130569-interior-compact-subet-c-0-empty.html
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# Thread:
1. ## Is the interior of a compact subet of C^0 empty?
This question is very similar to one I asked here: http://www.mathhelpforum.com/math-he...ons-empty.html
Let C0 be the set of continuous functions (from [a,b] to R) and a subspace of Cb, the set of bounded functions (from [a,b] to R), defined with the supnorm metric.
Let A be a compact subset of C0.
Is the Interior of A empty?
My first thought was that the proof is the same as in the link above (that is, yes it is empty since we can easily find a function g in any open ball around f in A where g is not continuous), but I'm worried this doesn't apply here because the subset A may only contain functions with certain properties, and so we don't know if g is in A.
Am I making any sense here? I think I successfully confused myself.
2. Originally Posted by southprkfan1
This question is very similar to one I asked here: http://www.mathhelpforum.com/math-he...ons-empty.html
Let C0 be the set of continuous functions (from [a,b] to R) and a subspace of Cb, the set of bounded functions (from [a,b] to R), defined with the supnorm metric.
Let A be a compact subset of C0.
Is the Interior of A empty?
My first thought was that the proof is the same as in the link above (that is, yes it is empty since we can easily find a function g in any open ball around f in A where g is not continuous), but I'm worried this doesn't apply here because the subset A may only contain functions with certain properties, and so we don't know if g is in A.
Am I making any sense here? I think I successfully confused myself.
As stated here, the answer is exactly the same as in the previous question: you can easily find a discontinuous function in any neighbourhood of a given continuous function.
However, this problem, unlike the previous one, makes sense (and is much more interesting) if you work entirely within C[a,b], the space of continuous functions on [a,b]. The reason is that this time the set A is not the whole space but a compact subset of it. So it makes sense to ask if A can contain an open subset of C[a,b]. I think that the answer is no, and that the reason is the Arzelà–Ascoli theorem, which characterises the compact subsets of C[a,b]. I don't have time to think it through carefully just now, but I think that it is not possible for an open subset of C[a,b] to be equicontinuous. It follows from the A–A theorem that such a set can never be compact in C[a,b].
3. Originally Posted by Opalg
As stated here, the answer is exactly the same as in the previous question: you can easily find a discontinuous function in any neighbourhood of a given continuous function.
However, this problem, unlike the previous one, makes sense (and is much more interesting) if you work entirely within C[a,b], the space of continuous functions on [a,b]. The reason is that this time the set A is not the whole space but a compact subset of it. So it makes sense to ask if A can contain an open subset of C[a,b]. I think that the answer is no, and that the reason is the Arzelà–Ascoli theorem, which characterises the compact subsets of C[a,b]. I don't have time to think it through carefully just now, but I think that it is not possible for an open subset of C[a,b] to be equicontinuous. It follows from the A–A theorem that such a set can never be compact in C[a,b].
I agree with Opalg and will work on it as well. I think I've done a problem before about open subsets of $\mathcal{C}\left[X,\mathbb{R}$ for compact metric spaces $X$. Let me look for it.
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http://physics.stackexchange.com/questions/24492/upper-critical-dimension-in-field-theory
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# upper critical dimension in field theory
Is there field theory which describe a second order phase transition without upper critical dimension ? Mermin-Wagner says something about lower critical dimension but nothing about upper dimension.
-
You need a way to declare that two field theories in different dimensions are "the same". There are natural ways to do this for simple examples, but it is easy to make up a model where there is no upper critical dimension because you extrapolate wrong. – Ron Maimon Apr 28 '12 at 5:32
2
– Manishearth♦ May 7 '12 at 6:09
Thanks @Manishearth for pointing that out to PanAkry, I was just about to leave the same 'citizenship patrol' comment ;) – NewAlexandria Dec 29 '12 at 23:54
## 1 Answer
The upper critical dimension is the dimension where the statistical field theory is well described by a mean field theory. It is also the dimension where the fluctuation theory turns into a free field theory. You can avoid having an upper critical dimension by tuning the kinetic terms properly:
Consider the Euclidean action:
$$S= \int |q|^{2n} |\phi|^2 + \lambda \phi^4 d^n x$$
This field theory never has an upper critical dimension. But this is because the dimensional extrapolation is wrong. For any fixed power of q, there is an upper critical dimension.
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Does this action correspond to some physical model? It is not clear : "This field theory never has an upper critical dimension... there is an upper critical dimension"! – PanAkry Apr 28 '12 at 12:25
@PanAkry: Not really physical. The point of this example is that I continued the theory wrong into higher dimensions. The right continuation holds the power of q fixed (the exponent doesn't change with n), and this continuation does have an upper critical d. – Ron Maimon Apr 28 '12 at 14:26
Why the downvote? It answers the question--- this is a silly mathematical continuation without upper critical dimension. – Ron Maimon May 9 '12 at 22:12
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http://mathhelpforum.com/business-math/177340-sinking-fund.html
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# Thread:
1. ## Sinking fund
Hi,
I’m trying to solve this problem;
A bank has three types of account in which the interest rate depends on the amount invested. The ordinary account offers a return of 6% and is available to any customer. Extra account offers 7%, only available to customers with 5,000\$ or more to invest, and super extra account offers 8% and is available only to customers with 20,000\$ or more to invest. In each case interest is compounded annually and is added to the investment at the end of the year. A person saves 4,000\$ at the beginning of the year for 25 years. Calculate the total amount saved assuming the money is transferred to a higher-interest account at the earliest opportunity.
Any help would be much appreciated.
Thanks
2. Here is an Excel table that I made: fund.xls.
Code:
```Year Balance Interest
1 $4,000.00 0.06
2 $8,240.00 0.07
3 $12,816.80 0.07
4 $17,713.98 0.07
5 $22,953.95 0.08
6 $28,790.27 0.08
7 $35,093.49 0.08
8 $41,900.97 0.08
9 $49,253.05 0.08
10 $57,193.29 0.08
11 $65,768.76 0.08
12 $75,030.26 0.08
13 $85,032.68 0.08
14 $95,835.29 0.08
15 $107,502.12 0.08
16 $120,102.28 0.08
17 $133,710.47 0.08
18 $148,407.31 0.08
19 $164,279.89 0.08
20 $181,422.28 0.08
21 $199,936.06 0.08
22 $219,930.95 0.08
23 $241,525.42 0.08
24 $264,847.46 0.08
25 $290,035.25 0.08
26 $313,238.07 0.08```
The second column contains the balance $b_n$ at the beginning of year $n$. The third column contains the interest rate $r_n$ used at the end of year $n$. Then $b_1=4,000$ and
$b_{n+1}=<br /> \begin{cases}<br /> b_n(1+r_n)+4,000, & n< 25\\<br /> b_{25}(1+r_{25}), & n=25<br /> \end{cases}<br />$ and $r_{n}=<br /> \begin{cases}<br /> 8\%, & b_n\ge 20,000\\<br /> 7\%, & 5,000\le b_n<20,000\\<br /> 6\%, & \mbox{otherwise}<br /> \end{cases}<br />$
3. Thanks for the reply. The answer you got is the same as in the textbook. However this question is in preparation for an exam and I will not be able to use EXCEL. If anyone is able to produce the an answer without using excel I would be very grateful.
Cheers
4. Well, you probably need a calculator unless you are amazing at calculations, rounding and estimates. In fact, you can do all 25 iterations in reasonable time. It is also possible to calculate $b_5$, after which the interest rate stabilizes. Let $r=1+r_5=\dots=1+r_{25}=1.08$. Then
$b_6=r b_5+4000$
$b_7=r(r b_5+4000)+4000=r^2b_5+4000(r+1)$
$b_8=r(r(r b_5+4000)+4000)+4000=r^3b_5+4000(r^2+r+1)$
Following the pattern and using the formula for the sum of geometric progression,
$b_{5+k}=r^kb_5+4000(r^{k-1}+\dots+r+1)=r^k+4000(r^k-1)/(r-1)$
Thus,
$b_{25}=b_{5+20}=r^{20}b_5+4000(r^{20}-1)/(r-1)$
and $b_{26}=rb_{25}$.
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http://mathoverflow.net/questions/110636/books-on-analytic-functions-on-banach-spaces-over-a-non-archimedean-field/110674
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## Books on analytic functions on Banach spaces over a non-Archimedean field
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I'm looking for good textbooks on analytic functions on Banach spaces over a non-Archimedean field. If you know one(s), please let me know.
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## 3 Answers
One beautiful book is Peter Schneider's Nonarchimedean Functional Analysis, appeared in the Springer Monographs in Mathematics in 2006. A more analytic one (with less emphasis on Functional Analysis and more on Calculus) in Alain Robert's A course in $p$-adic Analysis, GTM 198. But I still think the bible is S. Bosch, U. Güntzer and R. Remmert's Non-Archimedean Analysis, appeared in the Grundlehren der mathematischen Wissenschaften, 261.
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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.
I am not sure why the question is about "analytic functions on Banach spaces". This subject from infinite-dimensional analysis seems never studied in the non-Archimedean setting. Already the p-adic analogs of analytic functions behave quite differently from their classical counterparts. General monographs on non-Archimedean analysis contain only minimal information on p-adic analytic functions. There are good introductions in the books by Robert and Koblitz. However there are some books devoted specifically to this subject:
Hu, Pei-Chu; Yang, Chung-Chun. Meromorphic functions over non-Archimedean fields. Dordrecht: Kluwer Academic Publishers. 2000;
A. Escassut, Analytic elements in $p$-adic analysis. Singapore: World Scientific, 1995.
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There is a Bourbaki's volume of the resume of the theory of (possibly infinite dimensional) differential and analytic manifolds over Archimedean(i.e. real and complex) and non-Archimedean fields. It contains definitions and results but no proofs on the title subject. – Makoto Kato Oct 25 at 19:01
Although they are a little older and Anatoly's remark on general monographs might apply here, maybe A. C. M. van Rooij: Non-archimedean functional analysis and W. H. Schikhof: Ultrametric Calculus still contain helpful information.
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http://mathoverflow.net/questions/69179/integration-by-parts-for-a-general-negative-definite-self-adjoint-operator/69648
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Integration by parts for a general negative-definite self-adjoint operator.
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I suspect I am asking a very stupid question.
Suppose you have self-adjoint negative-definite operator $L$ densely defined on a space $L^2(\pi)$, with $Lf = \nabla \cdot ( A(x)\nabla f)$, for some symm. pos. def matrix A. Here assume that differentiation $\nabla = (D_i)_{i=1,..,n}$ is a skew-adjoint operator densely defined on $L^2(\pi)$, that is, $\mathcal{D}(D_i) = \mathcal{D}(D_i^*)$ and $D_i^* = -D_i$, for $i=1,\ldots n$.
Now, I'm trying to make sense of the statement that $f \in \mathcal{D}((-L)^\frac{1}{2})$. This would imply that $((-L)f, f) < \infty$, but I'm not sure if we can ALWAYS write this as:
$$((-L)f, f) = \int_\Omega \nabla f\cdot A(x) \nabla f \space \pi(dx)$$
I'm sorry if this is a stupid question but for some reason I can't convince myself of this fact.
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Am I right that you want to study this problem in the abstract (i.e. $\pi$ is a measure on some unspecified measurable space etc.)? Then your assumption that the operators $D_i$ are only skew-Hermitian seems too weak to me. A more natural assumption would be that the $D_i$ are skew-adjoint, that is, the domain of the adjoint $D_i^*$ equals that of $D_i$ (in addition to the skew-symmetry). – Florian Jun 30 2011 at 15:11
Yes, actually I should have mentioned that assumption also. Thanks for pointing this out. – RadonNikodym Jun 30 2011 at 15:14
@Florian. Yes I would to know of results where $\pi$ is as general as possible (for example the underlying $\sigma$-algebra\$ not necessarily countably generated, etc) – RadonNikodym Jun 30 2011 at 15:19
What is the question? I just read it three times, and am not sure what the question is. My best guess is: "What does $f\in D((-L)^(1/2))$ mean?" – Helge Jul 7 2011 at 3:46
The question is: Given f with the properties specified can I write $(-Lf,f)$ as $\int \nabla f \cdot A(x) \nabla f \pi(dx)$ – RadonNikodym Jul 7 2011 at 14:13
show 1 more comment
3 Answers
It is possible to make sense of $T^{1/2}$ without some of the particulars mentioned, when $T$ is a positive self-adjoint (densely-defined) operator on a Hilbert space. Namely, Friedrichs' argument (as in Riesz-Nagy, for example) shows that the resolvent $(T-\lambda)^{-1}$ exists and is a bounded operator for $\lambda$ not positive real. In particular, $T^{-1}$ is a bounded operator. It is also positive, so by standard (bounded-operator) spectral calculus admits a positive square root, whose inverse is the desired $T^{1/2}$.
Edit: after seeing some reactions, it is worth clarifying, as follows. Again, the square root of a positive unbounded (densely defined) _truly_self-adjoint_ operator exists without necessarily expressing the square root in terms of differential operators. The domain $D(\sqrt{T})$ is the same as the domains $D(\sqrt{T-\lambda})$ for $\lambda$ not in $[0,\infty)$.
There is the subordinate issue of whether one knows that the operator is genuinely self-adjoint, or only "symmetric". In the latter case, the question of self-adjoint extensions is non-trivial, depending on things abstracting "boundary conditions", tho' there is always the Friedrichs "minimal" extension.
In a similar vein, if we truly know a-priori that the operators $D_i$ are well-behaved in the sense that their domains and their adjoints' domains agree (the skew-adjoint version of self-adjoint), and assuming the domain(s) of the various expressions genuinely agree with that of the original operator, the seemingly formal computations are (by fiat) correct.
In practice, yes, there would be a non-trivial issue of specifying a common domain for the $D_i$ so that they are "genuinely" skew-adjoint, and so that the implied domain of the symmetric second-order operator is equal to that of its adjoint (so it is truly self-adjoint).
G. Grubb's book "Distributions and Operators" discusses many concrete examples of such things.
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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.
In short, "yes, probably, but you should be careful about boundary conditions."
The long version:
First a cautionary result. Let the Hilbert space be $L^2(0,1)$ and let $Lf =f''$ on the domain of functions $f\in L^2(0,1)$ with two derivatives in $L^2$ and such that $f(0)=-f'(1)$ and $f(1)=f'(0)$. This is a self-adjoint operator and by integration by parts
$$(f,(-L)f)= 2 \mathrm{Re}(\overline{f(1)}f(0))+ \int_0^1 |f'(x)|^2 dx.$$
However, this counter example is crazy, since it is not possible to interpret $L$ as $- D^\dagger D$ for an operator $D$ which takes one derivative. Also the boundary conditions I used would be very unlikely to arise in practice. Nevertheless it shows that some caution is needed.
So, let me interpret your question as follows:
"Let $L$ be the (unique!) operator defined on the domain $\mathcal{D}(L)\subset \mathcal{D}(\nabla)$ of functions $f$ such that $A(x)\nabla f \in \mathcal{D}(\nabla^\dagger)$, with $Lf=-\nabla^\dagger A(x) \nabla f$. Does the identity $$((-L)f,f)=\int \nabla f \cdot A(x)\nabla f d\pi$$ hold for $f\in \mathcal{D}(L)$?"
To begin with it may not be obvious that such an operator exists, or that it is self-adjoint, however this is the case and further more your integration by parts identity always holds. In fact, under the standard construction -- originally due to Friedrichs I think -- the answer is trivially yes since integration by parts is essentially the definition of $L$!
The Friedrichs construction is based on a theorem from functional analysis that says that any closed, positive quadratic form $q$ on a Hilbert space is the quadratic form of a positive self-adjoint operator. (See, for example, Thm. VIII.15 in Reed and Simon Vol. I.) In the present case we would define the quadratic form
$$q(g,f)= \int \overline{\nabla g(x)} \cdot A(x) \nabla f(x) d\pi(x)$$
which is easily shown to be closed and positive so long as $\nabla$ is a closed operator and $A(x)$ is symmetric positive definite as you assume. The proof of the theorem goes by showing that the domain $\mathcal{D}$ of functions $f$ such that $|q(g,f)| \le C \|g\|$ is dense so that it makes sense (by the Riesz thm. on linear functionals) to define an operator $L$ on this domain by the identity
$$q(g,f)= (g,(-L)f).$$
Note that the identity you want is a special case of this defintion!
It is easy to see now that the domain of $L$ consists of all functions $f$ such that $A(x)\nabla f \in \mathcal{D}(\nabla^\dagger)$ and that the quadratic form domain agrees with the domain of $\sqrt{-L}$ so that we have
$$\|\sqrt{-L}f\|^2 =q(f,f).$$
Note that, none of what was done above relied on the derivatives being implemented as anti-self-adjoint operators as you asked for. Returning to the one-dimensional case with Hilbert space $L^2(0,1)$ and $A=1$, first let $\nabla = d/dx$ on the domain of functions in $L^2(0,1)$ with one derivative in $L^2$. The resulting operator $L_N$ is the Neumann second derivative defined on the domain $\mathcal{D}(L_N)$ of twice differentiable functions with derivatives that vanish at $0$ and $1$ and the identity holds, however $\nabla$ is not anti-self-adjoint.
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The BC described at the beginning is not crazy. It represents an operator interpolating between periodic and anti-periodic boundary conditions. – Helge Jul 6 2011 at 18:09
Some comments:
• Self-adjointness of $L$ imposes a non-trivial condition on the measure $\pi$. I believe it has to be essentially Lebesgue. For example try $A = 1$, $\pi = 1 + x$, and $M = [0,1]$. Then $L$ is not self-adjoint.
• You should ask if for $f \in \mathcal{D}( (-L)^{\frac{1}{2}})$, one has $$( (-L)^{\frac{1}{2}} f, (-L)^{\frac{1}{2}} f) = \int \nabla f \cdot A \nabla f \pi(dx),$$ since $-L f$ is only defined for $f \in \mathcal{D}(L)$.
Is this an accurate interpretation of your question?
• The formulation above is non-trivial, since one doesn't have that $$(-\Delta)^{\frac{1}{2}} = - i \nabla.$$ It is given by multiplication by $|k|$ in the fourier basis. But my best guess is that for the usual cases the formula you stated is still correct... But I am also not completely sure how to check if for non-constant $A$.
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http://stats.stackexchange.com/questions/10592/how-to-calculate-tridiagonal-approximate-covariance-matrix-for-fast-decorrelati/10638
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# How to calculate tridiagonal approximate covariance matrix, for fast decorrelation?
Given a data matrix $X$ of say 1000000 observations $\times$ 100 features, is there a fast way to build a tridiagonal approximation $A \approx cov(X)$ ?
Then one could factor $A = L L^T$, $L$ all 0 except $L_{i\ i-1}$ and $L_{i i}$, and do fast decorrelation (whitening) by solving $L x = x_{white}$. (By "fast" I mean $O( size\ X )$.)
(Added, trying to clarify): I'm looking for a quick and dirty whitener which is faster than full $cov(X)$ but better than diagonal. Say that $X$ is $N$ data points $\times Nf$ features, e.g. 1000000$\times$ 100, with features 0-mean.
1) build $Fullcov = X^T X$, Cholesky factor it as $L L^T$, solve $L x = x_{white}$ to whiten new $x$ s. This is quadratic in the number of features.
2) diagonal: $x_{white} = x / \sigma(x)$ ignores cross-correlations completely.
One could get a tridiagonal matrix from $Fullcov$ just by zeroing all entries outside the tridiagonal, or not accumulating them in the first place. And here I start sinking: there must be a better approximation, perhaps hierarchical, block diagonal → tridiagonal ?
(Added 11 May): Let me split the question in two:
1) is there a fast approximate $cov(X)$ ?
No (whuber), one must look at all ${N \choose 2}$ pairs (or have structure, or sample).
2) given a $cov(X)$, how fast can one whiten new $x$ s ?
Well, factoring $cov = L L^T$, $L$ lower triangular, once, then solving $L x = x_{white}$ is pretty fast; scipy.linalg.solve_triangular, for example, uses Lapack.
I was looking for a yet faster whiten(), still looking.
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Do the columns have a natural ordering to them? Or do you want to find a tridiagonal approximation under some ("optimal") permutation of the columns? I'm assuming that when you say $A = \mathrm{Cov}(X)$ you're speaking of the covariance structure of the features. Can you confirm this? – cardinal May 10 '11 at 12:01
No, there's no natural ordering, and yes, covariance of the 100 features. Methods that add up a full covariance matrix, then approximate it, would be >> O(size X); I'm looking for a fast simple approximation, which will necessarily be crude. – Denis May 10 '11 at 12:07
So, you want a tridiagonal approximation under some (to be determined by the data) permutation, yes? – cardinal May 10 '11 at 12:09
added, tried to clarify. If a good (satisficing) permutation could be found in O(Nfeatures), yes, that would do. – Denis May 10 '11 at 14:09
There are approximations when the variables have additional structure, such as when they form a time series or realizations of a spatial stochastic process at various locations. These effectively rely on assumptions that let us relate the covariance between one pair of variables to that between other pairs of variables, such as between pairs separated by the same time lags. Calculations can be $O(Nf \log(Nf)$ in such cases. Absent such a model, I don't see how you can avoid computing all pairwise covariances. – whuber♦ May 10 '11 at 16:10
## 2 Answers
Merely computing the covariance matrix--which you're going to need to get started in any event--is $O((Nf)^2)$ so, asymptotically in $N$, nothing is gained by choosing a $O(Nf)$ algorithm for the whitening.
There are approximations when the variables have additional structure, such as when they form a time series or realizations of a spatial stochastic process at various locations. These effectively rely on assumptions that let us relate the covariance between one pair of variables to that between other pairs of variables, such as between pairs separated by the same time lags. This is the conventional reason for assuming a process is stationary or intrinsically stationary, for instance. Calculations can be $O(Nflog(Nf)$ in such cases (e.g., using the Fast Fourier Transform as in Yao & Journel 1998). Absent such a model, I don't see how you can avoid computing all pairwise covariances.
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On a whim, I decided to try computing (in R) the covariance matrix for a dataset of about the size mentioned in the OP:
````z <- rnorm(1e8)
dim(z) <- c(1e6, 100)
vcv <- cov(z)
````
This took less than a minute in total, on a fairly generic laptop running Windows XP 32-bit. It probably took longer to generate `z` in the first place than to compute the matrix `vcv`. And R isn't particularly optimised for matrix operations out of the box.
Given this result, is speed that important? If N >> p, the time taken to compute your approximation is probably not going to be much less than to get the actual covariance matrix.
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http://math.stackexchange.com/questions/21459/bourbaki-exercise-on-connected-sets
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# Bourbaki exercise on connected sets
This is Exercise I.11.4 of Bourbaki's General Topology.
Let $X$ be a connected space.
a) Let $A$ be a connected subset of $X$, $B$ a subset of $\complement A$ which is both open and closed in $\complement A$. Show that $A\cup B$ is connected.
b) Let $A$ be a connected subset of $X$ and $B$ a component of the set $\complement A$. Show that $\complement B$ is connected (use a)).
I have managed to show a).
My attempts to show b): Let $C$ be a nonempty clopen subset of $\complement B$. By a), $B\cup C$ is connected. Since $B$ is a component of $\complement A$, $B\cup C$ cannot be a subset of $\complement A$. So $C$ has to contain an element of $A$. Since $A$ is connected, $A\subset C$. But I'm stuck here. I can't see how this implies $C=\complement B$.
Can someone point me in the right direction?
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## 1 Answer
Suppose $B^{c} = U \cup V$ with $U,V$ disjoint and open. Then $U$ and $V$ are also closed and if they are non-empty, they both must contain $A$ by your argument, so they can't be disjoint.
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Thank you. This argument is pretty obvious, but somehow it was totally out of my reach. – Stefan Walter Feb 11 '11 at 10:58
@Stefan: This happens to me all the time... Cheer up, you managed the important and more difficult part :) – t.b. Feb 11 '11 at 11:05
The part b) it's very difficult! I have no idea what can I do My try: Let´s suppose that A,B form a separation of M−C i.e M−C=A∪B and also holds that A ¯ ¯ ¯ ∩B=A∩B ¯ ¯ ¯ =∅ . Clearly M=C∪A∪B is a disjoint union. And since C it´s a connected component, C it's closed. If I prove that A∪B it's closed, then I'm ready. How can I prove that no limit points of A or of B , lies in C ? Please help me! And clearly I can suppose that X lies in A without loss of generality. But also considering that I don't know how to do it. – Arkj May 2 '12 at 7:16
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http://mathhelpforum.com/discrete-math/173846-set-theory-show-these-equivalent-print.html
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# Set theory: Show these are equivalent.
Printable View
• March 8th 2011, 07:34 AM
Showcase_22
Set theory: Show these are equivalent.
Quote:
Show that the following are equivalent:
a). $A$ is finite
b). $A$ admits a well order <whose inverse> is also a well-order.
c). The structure $(\mathcal{P}(A), \subset_{\mathcal{P}(A)})$ is well-founded: that is every $S \subset \mathcal{P}(A)$ has a $\subset$-minimal element.
I decided to go for $a) \Rightarrow b) \Rightarrow c) \Rightarrow a)$.
For $a) \Rightarrow b)$:
Assign each element of $A$ an ordinal number. Since $A$ is finite, and there are an infinite number of ordinals, we will always be able to do this. We can then order the ordinals (and the elements of $A$ they represent) in the following way:
ie. $\phi$, $\{ \phi \}$, $\{\phi, \{ \phi \} \} \ \ldots$
or: $\phi, \ \mathcal{P}(\phi), \ \mathcal{P}\mathcal{P} (\phi ), \ \ldots$
This is a partial order and since every element contains $\phi$, every non-empty subset contains a least element.
For $b) \Rightarrow c)$
If we take any $S \subset \mathcal{P}(A)$, $S$ contains sets of elements of $A$.
Since $A$ is well-ordered, each of the subsets of $S$ has a least element.
We can take all these least elements and make a new set.
We can repeat this argument until we get a singleton set (the number of steps required to do this is equal to $|S|$).
This singleton set is the $\subset$-minimal element.
For $c) \Rightarrow a)$ I run into some trouble.
I figured the best way was to use a proof by contradiction. So suppose $A$ is infinite.
Let $S$ be any subset of $\mathcal{P}(A)$ and let $B \subset \mathcal{P}(A)$ be the $\subset$-minimal element of $S$.
I think I need to show that $\mathcal{P}(A)$ is finite which is equivalent to a finite number of $B \subset \mathcal{P}(A)$. Are there any tips for this?
• March 9th 2011, 01:16 PM
MoeBlee
The problem depends on what your previous definition of 'finite' is.
And it would help to know what previous theorems have been established.
What book are you working from? Suppes maybe? What exercise number?
• March 10th 2011, 08:34 AM
DrSteve
The second way you are claiming to order the ordinals is incorrect. for example, the power set of 2 is not 3 - it is a set with 4 elements.
All times are GMT -8. The time now is 11:31 AM.
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http://mathhelpforum.com/algebra/38002-expanding-binomials-print.html
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# expanding binomials
Printable View
• May 11th 2008, 01:52 PM
Nick87
expanding binomials
Promise no more questions today! ^_^
The reason I'm stuck on these is because I've completely forgotten what im doing, after returning to these from months ago.
I need to:
1) Express $x^2 - 8x + 15$ in the form $(x + a )^2 + b$ where "a" and "b" are the numbers to be found.
This is what I've got thus far:
$y = x^2 - 8x + 15$
$= (x - 4 )^2 - (4)^2 + 15 = 0$
$(x - 4)^2 - 16 + 15$
Then I need to plot for the original equation. Not alot (Thinking)
2) Expand the following using Pascal's Triangle
$(2 + x)^3$
I could manage the $(x+1)^n$ expansion but not sure how to go about this one.
Thanks
• May 11th 2008, 01:59 PM
Moo
Hello,
Quote:
Originally Posted by Nick87
Promise no more questions today! ^_^
The reason I'm stuck on these is because I've completely forgotten what im doing, after returning to these from months ago.
I need to:
1) Express $x^2 - 8x + 15$ in the form $(x + a )^2 + b$ where "a" and "b" are the numbers to be found.
This is what I've got thus far:
$y = x^2 - 8x + 15$
$= (x - 4 )^2 - (4)^2 + 15 = 0$
$(x - 4)^2 - 16 + 15$
Then I need to plot for the original equation. Not alot (Thinking)
Sorry if I misunderstood what you've said, but you got it :D
Quote:
2) Expand the following using Pascal's Triangle
$(2 + x)^3$
I could manage the $(x+1)^n$ expansion but not sure how to go about this one.
Thanks
So you know the formula for $(x+1)^n$
It's the same for $(2+x)^3$
$=C(3,0) 2^0 x^3+C(3,1)2^1 x^2+C(3,2) 2^2 x^1+C(3,3) 2^3 x^0$
Where $C(n,p)={p \choose n}$
• May 11th 2008, 02:07 PM
Nick87
So for the first one $a=4$ and $b=-16$?
It says to use my answer to sketch the graph of $y = x^2 - 8x + 15$ so what does that mean? Can't I just draw the graph like usual?
• May 11th 2008, 02:11 PM
Moo
Quote:
Originally Posted by Nick87
So for the first one $a=4$ and $b=-16$?
It says to use my answer to sketch the graph of $y = x^2 - 8x + 15$ so what does that mean? Can't I just draw the graph like usual?
Erm...
$y=(x-4)^2-16+15=(x-4)^2-1$
--> $a=-4$ and $b=-1$ (Wink)
To graph the thing, represent yourself :
- you know how to graph y=x²
- imagine that if you add a certain number to x, it's like translating to the right or the left the graph
- adding a constant (here, b) is like increasing or decreasing the curve, that is to say "go up or go down", because you add b to the ordinate of each point
• May 11th 2008, 02:18 PM
Nick87
Quote:
Originally Posted by Moo
Erm...
$y=(x-4)^2-16+15=(x-4)^2-1$
--> $a=-4$ and $b=-1$ (Wink)
*Slaps self* of course its -1!
So what I need to do now is plot the orignal graph but with the translation ${a \choose b}$?
• May 11th 2008, 02:23 PM
Moo
Quote:
Originally Posted by Nick87
*Slaps self* of course its -1!
So what I need to do now is plot the orignal graph but with the translation ${a \choose b}$?
Yep !
I have an hesitation, perhaps it's -a. Check on the graph which one it is :D
• May 11th 2008, 02:24 PM
Nick87
Thank you so much for all your help!
All times are GMT -8. The time now is 12:14 PM.
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http://physics.stackexchange.com/questions/11904/why-arent-there-compression-waves-in-electromagnetic-fields
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# Why aren't there compression waves in electromagnetic fields?
I just started learning about optics, and in the book I'm reading they explain how the electrical field caused by a single charged particle could be described by a series of field lines, and compare them to ropes, to provide an intuition of the concept.
Then they say that and that if we wiggle the particle up and down, that would produce transversal waves in the horizontal field lines, but no waves in the vertical lines. I know that the physical analogy is not to be taken literally, but I don't understand why wouldn't that cause compression waves in the vertical lines.
I mean, even though the direction of the field in the points directly above and below the particle doesn't change, the intensity does. And I assume it wouldn't instantly. So what am I missing?
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3
It follows from Maxwell's equations in vacuum, that the $\vec{E}$ and $\vec{B}$ fields are divergenceless, i.e., incompressible. By a Fourier transform one sees that $\vec{E}$ and $\vec{B}$ must propagate as transversal degrees of freedom. – Qmechanic♦ Jul 5 '11 at 21:15
To add to what Qmechanic wrote: the divergenceless property of the electric and magnetic fields are general properties of electromagnetic theories. The same expression also holds for non-linear theories of electromagnetism, as well as propagation in homogeneous (but not necessarily isotropic) media (for example, crystal optics). And in those cases you also have only transversal degrees of freedom. – Willie Wong Jul 5 '11 at 21:21
2
Thanks, but as I said, I am (1) just starting to learn about the subject (so I haven't yet reached Maxwell's equations), and (2) looking for an intuitive way to understand why this happens -- i.e., I'm not doubting that it does. – Waldir Jul 5 '11 at 21:26
## 5 Answers
If you are going to pursue this physical analogy, which can be useful at times, then you must consider the electric field lines to be of constant tension. That is, the tension of these lines is a constant no matter how much you stretch them. This is different from ordinary ropes or strings or whatever, where the more you stretch them, the higher the tension.
More technically, if you examine the Maxwell stress tensor for a pure electric field, you will find a tension term along the direction of the field and a pressure term transverse to the field. So you in the static case, you can think of the electric field lines as being in balance between tension along field lines and a pressure pushing different lines apart.
For an ordinary stretched string or something like that, if you move the end of the string longitudinally then the stretching or compression changes the tension and the difference in tension will propagate along the string, producing a longitudinal wave. In the case of electric field lines, there is no change in tension to propagate along the line, since the tension is a fixed constant. I hope this helps with your intuition.
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Thank you. I am afraid I am not yet prepared to understand all the details of your answer, but it gives me a general idea about how this differs from day-to-day physics and why the analogy fails in that particular aspect. – Waldir Jul 6 '11 at 21:53
It depends on what you mean by "compression wave". When we typically think of compression wave, we think of sound waves, where the air (the medium) has a pressure differential between the peak and trough of the wave.
In Electromagnetism, the wave is not a change in the medium^, it is a change in the electromagnetic field.* Because of this, we have to ask, what "compresses" in the compression wave? One possible answer is that the "EM Field" gets more dense, or more strongly positive, at which point we are back where we started: the analogy gets us nowhere, it is neither wrong, nor more insightful. We also find that it starts to break down (what about "strongly negative" E-Field, this doesn't really work in a pressure analogy).
So, the E-Field doesn't have compression waves because it doesn't modify the medium in which it is traveling.
^In this case, medium is understood to be the vacuum, or space-time, not the macroscopic medium (or dielectric). In a dielectric, it is kind of possible for EM waves to PRODUCE compression waves (waves of varying density of the medium), but they cannot fundamentally BE compression waves.
*For a long time, this wasn't well understood, which is why (pre-Einstein), the dominant belief in physics was in a "luminiferous aether" as the medium in which EM waves traveled. Michelson and Morley actually "disproved" this in 1887 with their seminal experiment (though I believe Michelson spent the rest of his career trying to improve upon his initial measurement and find the aether). Combined with their null result, and Einstein's Theory of Special Relativity which came out 34 years later, the idea of a "medium" in which EM waves propagate is largely considered false.
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You're right, "compression" was the incorrect word to use -- what I meant is a longitudinal wave. But since I am aware that the whole analogy wasn't literal, I didn't think too hard about using the scientifically correct words, but rather about posing the problem in a way I could visualize it more easily. In this case, I assumed that moving the charged particle up would mean that a point directly above it would be under a stronger influence of its field, since it's closer, and that this increase would not happen instantaneously, but propagate at the speed of light -- hence a longitudinal wave. – Waldir Jul 6 '11 at 21:48
Something to keep in mind is that longitudinal waves are actually possible in EM, however they require a medium in which to propagate (you can't do this in a vacuum). Typically they show up in waveguides in a dielectric. – Andrew Spott Jul 6 '11 at 23:09
Let us consider a horizontally pointing laser pointer that is being wiggled up and down. At some distance away the laser beam makes a bright spot on a wall. The spot moves up and down, say 10 meters, when the pointer device is moving up and down two millimeters. This is caused by relativistic aberration: http://en.wikipedia.org/wiki/Relativistic_aberration
Okay, now we wiggle the laser pointer horizontally and in the beam's direction. Now the bright spot on the wall does not move at all, just the size of the spot changes a tiny amount.
Now Quantum Electro Dynamics tells us that electric field consists of stream of virtual photons, but I'm not saying that aberration of virtual photons is an explanation of anything at all.
I'm just saying that transverse wiggle and longitudinal wiggle are very different, because of relativistic effects, that are different in transverse and longitudinal directions.
Oh yes I must add that a sensitive instrument can detect a movement of a charge from any direction, and the information travels at the speed of light.
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Well, is not 100% percent accurate to say that there aren't longitudinal EM waves. In a waveguide there are allowed propagation modes that have non-zero electric and magnetic components in the direction of propagation:
http://en.wikipedia.org/wiki/Transverse_mode
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It would be better to answer this question in another way, rather than by trying to think about moving field lines, which themselves are a pure analogy to explain field strength. If one has a charged particle lets say an electron, which itself has mass and therefore will not move at light speed. If this electron is impressed with an oscillatory energy such that it is forced to move in a cyclic manner the field around it will also oscillate with the electron and as such also produce a quadrature phase component which is the magnetic field. This proves by observation that a moving electron charge has some form of relationship with free space. Also remember that free space passes through every molecular system and thus these characteristics are also available in all material bodies. It has been formulated by Maxwell and others that free space has two characteristics which can be measured by extrapolation. These are called Permittivity and Permeability. Permittivity is the characteristic of space which supports electric fields and permeability is the characteristic which supports magnetic fields. Both these characteristics take a finite time to set up and collapse. If one produces an instant velocity change to an electric field the associated magnetic field will take a short time to be set up in free space. If one gives enough energy to the electron by increasing its frequency there will come a time where the electric field which is produced by the moving electron cannot collapse fast enough in free space before the next oscillatory part of the field comes behind it. This then forces away the previous cycle with its quadrature magnetic component. This is the point where one has an EM oscillatory field travelling away from the source. Also each oscillatory wavelet of electric and magnetic fields can be considered a single photon which is a self contained energy source without mass. This then has to move away at light speed because this is the speed at which the changing electric field takes to set up the quadrature magnetic field. This is a free space characteristic and is why photons or oscillatory EM wavelets have to travel only at that speed. This was found well before Einstein used the concept and is the manner in which electromagnetic communication operates. Stan
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This answer would be more readable if it was structured into paragraphs. – Claudius Nov 15 '12 at 20:13
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http://math.stackexchange.com/questions/74298/different-results-for-row-reduction-in-matlab?answertab=votes
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# Different results for row reduction in Matlab
Consider the following matrix: $\left[ \begin{array}{ccc} -0.05 & 0.45 & 0 \\ 0.05 & -0.45 & 0 \end{array} \right]$
Row reducing the above matrix in Matlab using the rref() function produces what I would expect (just adding top row to bottom row and scaling top row): $\left[ \begin{array}{ccc} 1.0 & -9.0 & 0 \\ 0 & 0 & 0 \end{array} \right]$
But if I remove the last column of just zeros, and row reduce that matrix, I get a 2x2 identity matrix: $\left[ \begin{array}{ccc} -0.05 & 0.45 \\ 0.05 & -0.45 \end{array} \right] \sim \left[ \begin{array}{ccc} 1 & 0 \\ 0 & 1 \end{array} \right]$
I can't see how removing the last column changes anything; adding the top row to the bottom row will still produce the result above, just without the last column of $0$'s. But I'm quite sure Matlab is right and I'm not, so what am I missing here?
Edit: I have managed to reproduce the above and I believe it's all due to rounding errors. If you input $M = \left[ \begin{array}{ccc} 0.95 & 0.45 \\ 0.05 & .55 \end{array} \right]$ and then do $A = M - eye(2)$. rref(A) will now give the $2 \times 2$ identity matrix. If I enter the result of $M-eye(2)$ directly, that is $B= \left[ \begin{array}{ccc} -0.05 & 0.45 \\ 0.05 & -0.45 \end{array} \right]$, then rref(B) returns the expected $\left[ \begin{array}{ccc} 1 & -9 \\ 0 & 0 \end{array} \right]$ .
Here's a screenshot as an example:
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Are you sure you're getting an identity matrix for the $2\times 2$ case? The matrix you have there is singular, after all. – J. M. Oct 20 '11 at 13:35
1
I've just introduced your matrix on my Matlab and said to row reduce it with rref(A) and the result is the correct one: your second matrix, without the third column of zeros. – Agustí Roig Oct 20 '11 at 13:48
@J.M. and Agusti, I think I might have narrowed it down to a rounding error, causing the top and bottom row to be slightly different (but not shown due to lack of decimals shown in Matlab). As you are both more experienced here than me, do you think I should answer my own post or delete it? Thank you! – Jodles Oct 20 '11 at 15:40
What version of MATLAB are you using? No, don't delete; this might be a nice cautionary tale... (You might want to post a screenshot of what you gave MATLAB as input.) – J. M. Oct 20 '11 at 15:48
1
It's a cautionary example for using rank functions. For the example above rank(A) returns 1, but rref(A) thinks it's 2. You can tweak it though >> rref(A) ans = 1 0 0 1 >> rref(A,1E-15) ans = 1.0000 -9.0000 0 0 >> I'm surprised that the default eps value ends up with wrong result as well. – karakfa Oct 20 '11 at 17:00
show 1 more comment
## 1 Answer
This is already documented in MATLAB.
Roundoff errors may cause this algorithm to compute a different value for the rank than `rank`, `orth` and `null`.
You need to type `format long e`. After that you can see the difference if you execute `N-eye(2)` resulting with
````N-eye(2)
ans =
-5.000000000000004e-002 4.500000000000000e-001
5.000000000000000e-002 -4.500000000000000e-001
````
Here, the trouble is already visible in the (1,1) element of the matrix. But also
```` [1 1]*(N-eye(2))
ans =
-4.163336342344337e-017 5.551115123125783e-017
````
gives you the error between seemingly identical elements.
The reason why you get correct results with an additional zero column is (my personal view) due to the term in the tolerance computation for `rref` given by `(max(size(A))*eps *norm(A,inf))`. Here, the first term is 3 instead of 2 and that should make the small difference between selecting a rank 1 and rank 2 matrix.
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http://physics.stackexchange.com/questions/32871/simple-pde-vs-string-theory?answertab=oldest
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# Simple PDE vs String Theory
For the sake of simplicity, I’d like to believe that there is one master non-linear partial differential equation governing physics. In particular, consider a Klein-Gordon form:
$$\frac{\partial^2 f}{\partial t^2}=v^2 \left (\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}+\frac{\partial^2 f}{\partial y^2} \right ) - m^2f$$
The state function $f$ (with $\partial f/\partial t$) here is naturally a function of space $xyz$ and time $t$. But, let $v$ and $m$ depend on the local $f$ and $\partial f/\partial t$ (so that the equation is non-linear and probably impossible to solve analytically). It seems $f$ needs to be a two-number vector (to allow two light polarizations, or two electron spins), but I'd like to leave that unspecified for now.
Then, is there an argument why nature can't have this simple non-linear form (where quarks and other standard model elements appear as something like solitons)? I don't understand why most physicists are focusing on more-complicated 12D string theories.
(To simplify even further, it would be nice if $m$=0, but I believe Einstein and others sought such an "electromagnetic-only theory for the electron" without success, even after adding two more dimensions! If there's no general argument for my question above, I wonder if there is a specific argument for why nature can't have the $m$=0 form.)
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Should the third term in the sum in parentheses be $\partial^2 f/\partial z^2$? – William Feb 15 at 21:45
yep, that was a typo – bobuhito Feb 20 at 4:20
## 2 Answers
The idea that nature is described by a nonlinear system of equations was the idea that Einstein had in the 1920s, and motivated his search for a unified field theory. It doesn't work, and it's philosophically less worthwhile than current theories anyway, so even if it did work, it wouldn't be simpler than string theory, or as elegant.
The idea that you can describe what's going on with local equations is false, as is demonstrated conclusively by Bell's inequality violations. The Bell inequality tells you that you can send electrons to far-away locations with spins that can be measured in 3 directions, A,B,C. The spin of the two electrons in each direction are 100% correlated (it's actually anti-correlated, but same difference for the argument), so if you measure the spin in direction A, and one electron is up the other is 100% certain to be up. Same for direction B and C, the two electrons always report the same spin in any of the three directions.
The spin in directions A and B are 99% correlated, meaning if you measure A on one of the electrons is up, then B on the other electron is up 99% of the time, and B is down 1% of the time. The spin in directions B and C are 99% correlated, so if you measure B is down on one electron, C is up on the other electron 1% of the time.
From the 100% correlation of the electrons, you conclude that the nonlocal field state (hidden variable) on one electron has the property that
A and B are 99% the same, 1% different B and C are 99% the same, 1% different
From this you deduce that
A and C must be at least 98% the same
meaning that whatever field configuration is happening to make A, the field configuration for C can only give different results 2% of the time, the sum of those times when it gives different answers than B plus the times B gives different answers than A.
This bound is called Bell's inequality, and it is violated by quantum mechanics. A and C are different 96% of the time.
This means any type of local-in-space description, linear, nonlinear, complicated, simple, whatever, will never ever work to describe nature. Your description is either nonlocal in the sense of faster-than-light communication, or nonlocal in the sense of having a global notion of state which is entangled nonlocally by measurements. This is why nobody looks for nonlinear field equations to describe nature anymore. It can't possibly work.
But the main ideas of Einstein's nonlinear field theories have survived to inspire developments in later physics.
• The pions are excitations of a sigma-model, which is a type of nonlinear field theory. They are small oscillations of the quark condensate in the vacuum.
• The proton can be thought of as the topological soliton of the sigma model. In quantum mechanics, it can still be a fermion even though the sigma model has no fundamental fermionic variables.
• The field equations of 11-dimensional supergravity, which are a central part of string theory, generalize General Relativity in pretty much the only nontrivial ways known--- they give the biggest extension of spacetime symmetry possible, and they include a new field, constrained by the supersymmetry.
So these ideas are not a dead end, but they cannot work without quantum mechanics by themselves. If you want to understand quantum behavior emerging from some sort of nonlinear dynamics underneath, this dynamics can't be local.
-
Good point! But, let me be clear that I was envisioning a quantum interpretation of $f$ (please change $f$ to $\psi$ if it changes your interpretation). To handle Bell's 2-particle spin correlations, I think I could use a “four-number vector” for $f$ (instead of a “two-number vector”, as suggested in my original question but really “unspecified”, so I consider the question still open). – bobuhito Jul 26 '12 at 6:11
In general, N-particle spin correlation seems to need a "$2^N$-number vector" which, I admit, does seem like a bad kluge...but, doesn’t string theory similarly blow up with multiple particles (e.g., one particle uses 12D, but two particles use 24D)? – bobuhito Jul 26 '12 at 6:11
@bobuhito: Bell's inequality closes the debate on local realism, there is no local realism. All QM theories have this exponential blowup when you simulate them, in string theory it's a little different because the wavefunctions are defined asymptotically, so it's not 10d 20d 30d wavefunctions, but more like 9d 18d to describe the "in" state of scattering. Anyway, this is nature's blowup, and unless we find that highly entangled quantum systems don't exist in nature, so that quantum computers fail at a few hundred or few thousand qubits, it is not reasonable to give non-quantum descriptions. – Ron Maimon Jul 27 '12 at 2:28
The theory you've written down is linear in $f$. If $f$ and $g$ are solutions, then so is $f+g$. This means that your field theory has no interactions between the different modes of $f$.
-
it's non-linear - please read everything – bobuhito Jul 26 '12 at 3:59
My mistake. A new complaint, however: the equation fails to be Lorentz-invariant if $v$ depends on $f$. You're back to free theory if you want $m=0$ + Lorentz invariance. – user1504 Jul 26 '12 at 4:11
It's possible that the speed of light is different in a high-field background...we've probably only tested the weak-field limit. – bobuhito Jul 26 '12 at 5:35
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http://mathhelpforum.com/number-theory/184931-another-quadratic-reciprocity-problem-print.html
|
# Another quadratic reciprocity problem
Printable View
• July 21st 2011, 04:02 PM
alexmahone
Another quadratic reciprocity problem
For which positive integers n do there exist integers x and y such with (x, n) = 1, (y, n) = 1, such that $x^2+y^2 \equiv 0\ (mod\ n)$?
• July 21st 2011, 05:04 PM
alexmahone
Re: Another quadratic reciprocity problem
It is easy to see that n = 2 satifies the given congruence. If n is odd, every prime p dividing n should satisfy $\left(\frac{-y^2}{p}\right)=1 \implies \left(\frac{-1}{p}\right)=1 \implies p \equiv 1\ (mod\ 4)$.
The answer given in the book is: $n=2^{\alpha}\prod p^{\beta}$ where $\alpha$ = 0 or 1, the primes p in the product are all $\equiv 1\ (mod\ 4)$, and $\beta=\beta(p)=0,1,2,...$.
How do they get the $2^\alpha$ in the product?
All times are GMT -8. The time now is 01:20 PM.
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http://math.stackexchange.com/questions/217022/how-to-tackle-markov-chains-with-transition-cost
|
How to tackle markov chains with transition cost?
A simple example would be:
I have 3 states, A,B,C
The transition matrix for the chain is:
```` A B C
A 0.1 0.3 0.6
B 0.4 0.2 0.4
C 0.5 0.2 0.3
````
and the transition cost matrix is:
```` A B C
A 3 5 1
B 6 8 3
C 9 2 3
````
so, for each step, say, from A to C, the chance is 0.6, and the cost is 1
If the chain starts from A, and goes on 10 steps (so there are 11 states including A and 10 transitions), how to get the expected sum of the cost?
What's is this kind of problem called? I am able to get the expected number of visited states, but feel a bit confused as the cost is associated with transition but the states.
-
1 Answer
The distribution at step $n$ is $\pi_n=\pi_0T^n$, where $\pi_0$ is the initial distribution, in your case $(1,0,0)$, and $T$ is the transition matrix. The expected cost of the transition from $n$ to $n+1$ is $\pi_nKe$, where $K$ is the componentwise product of the transition matrix and the transition cost matrix and $e$ is a column vector with all entries $1$. Thus the expected sum of costs of the first $m$ transitions is
$$\sum_{n=0}^{m-1}\pi_0T^nKe=\pi_0\left(\sum_{n=0}^{m-1}T^n\right)Ke=(1,0,0)\left(\sum_{n=0}^{m-1}T^n\right)\pmatrix{2.4\\5.2\\5.8}\;.$$
Diagonalizing $T$ allows you to express the sum over its powers as $m$ for the eigenvalue $1$ and $(1-\lambda^m)/(1-\lambda)$ for the other two; in the long run, the eigenvalue $1$ dominates and the cost is approximately
$$m\pi Ke=m\pi\pmatrix{2.4\\5.2\\5.8}\;.$$
The equilibrium distribution is $\pi=\frac1{47}(16,11,20)$ (computation), so the long-term average cost per step is $1058/235\approx4.5$.
-
perfect. btw i don't get what it is following the computation link to wolfram – colinfang Oct 19 '12 at 20:03
1
@colinfang: The equilibrium distribution is the left eigenvector for eigenvalue $1$. So I subtracted $1$ on the diagonal, took the transpose, computed the right eigenvectors, got $(4/5,11/20,1)$ from Wolfram|Alpha and normalized it to sum to $1$. – joriki Oct 19 '12 at 20:32
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http://xianblog.wordpress.com/2012/09/13/abc-as-knn/
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# Xi'an's Og
an attempt at bloggin, from scratch…
## ABC as knn…
Gérard Biau, Frédéric Cérou, and Arnaud Guyader recently posted an arXiv paper on the foundations of ABC, entitled “New insights into Approximate Bayesian Computation“. They also submitted it to several statistics journals, with no success so far, and I find this rather surprising. Indeed, the paper analyses the ABC algorithm the way it is truly implemented (as in DIYABC for instance), i.e. with a tolerance bound ε that is determined as a quantile of the simulated distances, say the 10% or the 1% quantile. This means in particular that the interpretation of ε as a non-parametric bandwidth, while interesting and prevalent in the literature (see, e.g., Fearnhead and Prangle’s discussion paper), is only an approximation of the actual practice.
The authors of this new paper focus on the mathematical foundations of this practice, by (re)analysing ABC as a k-nearest neighbour (knn) method. Using generic knn results, they thus derive a consistency property for the ABC algorithm by imposing some constraints upon the rate of decrease of the quantile as a function of n. (The setting is restricted to the use of sufficient statistics or, equivalently, to a distance over the whole sample. The issue of summary statistics is not addressed by the paper.) The paper also contains a perfectly rigorous proof (the first one?) of the convergence of ABC when the tolerance ε goes to zero. The mean integrated square error consistency of the conditional kernel density estimate is established for a generic kernel (under usual assumptions). Further assumptions (on the target and on the kernel) allow the authors to obtain precise convergence rates (as a power of the sample size), derived from classical k-nearest neighbour regression, like
$k_N \approx N^{(p+4)/(m+p+4)}$
in dimensions m larger than 4…. The paper is completely theoretical and highly mathematical (with 25 pages of proofs!), which may explain why it did not meet with success with editors and/or referees, however I definitely think (an abridged version of) this work clearly deserves publication in a top statistics journal as a reference for the justification of ABC! The authors also mention future work in that direction: I would strongly suggest they consider the case of the insufficient summary statistics from this knn perspective.
### Share:
This entry was posted on September 13, 2012 at 12:12 am and is filed under Statistics with tags ABC, convergence, DIYBAC, economics, k-nearest neighbours. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.
### One Response to “ABC as knn…”
1. Keith O'Rourke Says:
September 14, 2012 at 9:31 pm
Neat, if you can efficiently search Andrew G’s blog, I proposed Bayes as inherently being a k-nearest neighbour procedure in 2005/6. I had, in a pinch, come up with (what Don Rubin used in his 1984 paper) a two stage sampling model in order to explain Bayes to Epi students.
Not that I did anything with it (or could I with computational resources back then), but I remember the clear lack of enthusiam to see it that way.
It just seemed wrong or worse too simple!
Cancel
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http://mathoverflow.net/revisions/36268/list
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## Return to Answer
2 fix typo
(this is a variation on Darij's answer)
Note that "holomic" holonomic" generating functions $f(x)$, i.e. generating functions satisfying a linear differential equation with polynomial coefficients (in $x$) enjoy rich closure properties.
For a (very brief) summary, see Table 1 of a paper on guessing (sorry, I couldn't find a better reference, but there should be one...). Moreover, all these closure properties can be made "effective": if we know the order and the degree of the coefficients of the diffential equation for $f(x)$ and for $g(x)$, we can (easily) compute bounds for order and coefficient degree of the differential equation for $f+g$, $f\cdot g$, etc.
Moreover, it turns out that the (Taylor) coefficient sequence of a holonomic generating function satisfies a linear recurrence with polynomial coefficients, i.e., is P-recursive, and the P-recursive sequences are precisely the coefficient sequences of holonomic generating functions.
Thus, to find such a recurrence (or differential equation), it is often easiest to use gfun for Maple by Bruno Salvy and Paul Zimmermann, or the Mathematica equivalent by Mallinger (I think). Unfortunately, the FriCAS package described in 1 only does the guessing part, i.e., you would need to compute the bounds yourself and then check. In the case at hand the results are below.
However, you also asked, whether the solution is hypergeometric. To check this, you simply feed the recurrence you found below into Petkovsek's program Hyper, it will tell you whether there is a hypergeometric solution. (most probably no, because if the degree of the coefficient polynomials of the hypergeometric solution are not huge, guessPRec and guessHolo would have found it.)
In case you are dealing with orthogonal polynomials, the Askey Wilson scheme by Koekoek and Swarttouw is a great reference to find the basic information.
```(1) -> guessHolo(cons(x, [(legendreP(n, x)-(1-1/n)*legendreP(n-2, x)) for n in 2..30]), variableName==t)
(1)
[
[
n
[t ]f(t):
4 2 2 5 3 4 2 ,,
((- 2t - 2t )x + (t + 6t + t)x - 2t - 2t )f (t)
+
3 2 4 2 3 ,
((- 3t - 7t)x + (3t + 15t + 2)x - 8t - 2t)f (t)
+
2 2 3 2
((t - 3)x + (t + 3t)x - 4t + 2)f(t)
=
0
,
2 2 3 2 3 4 3 2 3
3t x - 2t 15t x - 13t x 35t x - 39t x + 6t 4
f(t)= x + ---------- + -------------- + --------------------- + O(t )]
2 6 8
]
Type: List(Expression(Integer))
(2) -> guessPRec cons(x, [(legendreP(n, x)-(1-1/n)*legendreP(n-2, x)) for n in 2..30])
(2)
[
[
f(n):
4 3 2 2 4 3 2
((4n + 28n + 67n + 63n + 18)x - 4n - 28n - 68n - 68n - 24)
*
f(n + 2)
+
4 3 2 3
(- 8n - 52n - 118n - 107n - 30)x
+
4 3 2
(8n + 52n + 120n + 114n + 36)x
*
f(n + 1)
+
4 3 2 2 4 3 2
((4n + 24n + 51n + 46n + 15)x - 4n - 24n - 52n - 48n - 16)f(n)
=
0
,
2
3x - 2
f(0)= x, f(1)= -------]
2
]
Type: List(Expression(Integer))
```
1
(this is a variation on Darij's answer)
Note that "holomic" generating functions $f(x)$, i.e. generating functions satisfying a linear differential equation with polynomial coefficients (in $x$) enjoy rich closure properties.
For a (very brief) summary, see Table 1 of a paper on guessing (sorry, I couldn't find a better reference, but there should be one...). Moreover, all these closure properties can be made "effective": if we know the order and the degree of the coefficients of the diffential equation for $f(x)$ and for $g(x)$, we can (easily) compute bounds for order and coefficient degree of the differential equation for $f+g$, $f\cdot g$, etc.
Moreover, it turns out that the (Taylor) coefficient sequence of a holonomic generating function satisfies a linear recurrence with polynomial coefficients, i.e., is P-recursive, and the P-recursive sequences are precisely the coefficient sequences of holonomic generating functions.
Thus, to find such a recurrence (or differential equation), it is often easiest to use gfun for Maple by Bruno Salvy and Paul Zimmermann, or the Mathematica equivalent by Mallinger (I think). Unfortunately, the FriCAS package described in 1 only does the guessing part, i.e., you would need to compute the bounds yourself and then check. In the case at hand the results are below.
However, you also asked, whether the solution is hypergeometric. To check this, you simply feed the recurrence you found below into Petkovsek's program Hyper, it will tell you whether there is a hypergeometric solution. (most probably no, because if the degree of the coefficient polynomials of the hypergeometric solution are not huge, guessPRec and guessHolo would have found it.)
In case you are dealing with orthogonal polynomials, the Askey Wilson scheme by Koekoek and Swarttouw is a great reference to find the basic information.
```(1) -> guessHolo(cons(x, [(legendreP(n, x)-(1-1/n)*legendreP(n-2, x)) for n in 2..30]), variableName==t)
(1)
[
[
n
[t ]f(t):
4 2 2 5 3 4 2 ,,
((- 2t - 2t )x + (t + 6t + t)x - 2t - 2t )f (t)
+
3 2 4 2 3 ,
((- 3t - 7t)x + (3t + 15t + 2)x - 8t - 2t)f (t)
+
2 2 3 2
((t - 3)x + (t + 3t)x - 4t + 2)f(t)
=
0
,
2 2 3 2 3 4 3 2 3
3t x - 2t 15t x - 13t x 35t x - 39t x + 6t 4
f(t)= x + ---------- + -------------- + --------------------- + O(t )]
2 6 8
]
Type: List(Expression(Integer))
(2) -> guessPRec cons(x, [(legendreP(n, x)-(1-1/n)*legendreP(n-2, x)) for n in 2..30])
(2)
[
[
f(n):
4 3 2 2 4 3 2
((4n + 28n + 67n + 63n + 18)x - 4n - 28n - 68n - 68n - 24)
*
f(n + 2)
+
4 3 2 3
(- 8n - 52n - 118n - 107n - 30)x
+
4 3 2
(8n + 52n + 120n + 114n + 36)x
*
f(n + 1)
+
4 3 2 2 4 3 2
((4n + 24n + 51n + 46n + 15)x - 4n - 24n - 52n - 48n - 16)f(n)
=
0
,
2
3x - 2
f(0)= x, f(1)= -------]
2
]
Type: List(Expression(Integer))
```
|
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|
http://math.stackexchange.com/questions/161479/name-of-proof-that-area-of-squarearea-of-rectangle-of-the-same-perimeter/161639
|
Name of proof that area of square>area of rectangle of the same perimeter
What is the proof called for the fact that the area of a square is always greater than the area of a non-square rectangle of the same perimeter?
-
1
How about this: "Proof of the fact that area of a square>area of a rectangle of the same perimeter" – user31029 Jun 22 '12 at 2:16
2
This is a consequence what is commonly called the arithmetic-geometric mean inequality. – Tim Duff Jun 22 '12 at 2:23
This is a consequence of $(a+b)(a-b)=a^2-b^2$. – Jonas Meyer Jun 22 '12 at 2:55
4
Facts of this kind are sometimes called isoperimetric inequalities, of which the best-known is the fact that the area of a circle is always greater than that of any other region with the same perimeter. – Nate Eldredge Jun 22 '12 at 3:49
– math-visitor Jun 22 '12 at 4:08
1 Answer
I don't know what the name of the proof is (if there is a standard known name), but as stated in one of the comments above it is a consequence of the Arithmetic-Geometric (AM-GM) Mean Inequality which states that for non-negative real numbers $x_i$ the following inequality holds: $$\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}\ge \sqrt[n]{x_1 x_2\cdots x_n}$$ With equality if, and only if $x_1=x_2=\cdots=x_n$
So suppose we fix the perimeter of a rectangle as $2(a+b)$ with side lengths $a$ and $b$. Hence, a square with the same perimeter has a side length of $\displaystyle\frac{a+b}{2}$. So by applying the AM-GM inequality with $n=2$ it follows that $\displaystyle \left(\frac{a+b}{2}\right)^2 \ge ab$. So we see that the area of the rectangle doesn't exceed the area of the square, and the areas are equal if and only if $a=b$. So in other words, for a fixed perimeter the maximum area of a rectangle is when it is in fact a square.
Alternatively, one can apply this to a fixed area and deduce that the perimeter of a rectangle is always greater than or equal to the perimeter of a square. This time suppose that $x$ and $y$ are the lengths of a rectangle with a fixed area $xy$. Hence, a square with the same area has side-lengths $\sqrt{ab}$. So, by AM-GM we obtain $2(x+y)\ge 4\sqrt{xy}$. Now we see that the for a fixed area $xy$, the rectangle with the smallest perimeter is a square.
This can be easily extended and generalized into $n$ dimensions and we can say that an $n$-cube has the smallest perimeter among all $n$-dimensional boxes with the same volume and that an $n$-cube has the largest volume among all $n$-dimensional boxes with the same perimeter.
-
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http://mathoverflow.net/questions/24859/finding-the-union-of-n-random-circles-arbitrarily-or-conspiratorially-placed-on/96364
|
## Finding the union of N random circles arbitrarily (or conspiratorially) placed on a two-dimensional surface
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Please consider a two-dimensional surface populated with a set of Cartesian coordinates $(x_i, y_i)$ for $N$ circles with individual radii $r_i$, where $r_{min} < r_i < r_{max}$.
Here, the number of circles, $N$, may be large - ranging from hundreds to tens of thousands. The circles may sparsely populate the plane in some places, and in others, be 'conspiratorially' packed together. Furthermore, $r_{min}$ / $r_{max}$ are not necessary defined in such a way to allow for accurate convex hull, spline interpolation, etc.
While we can always perform a monte carlo sampling of coordinates on the plane (or over some defined lattice), is there an efficient deterministic method of calculating the exact area given by the union of all $N$ circles?
Update - After a more extensive literature search (and thanks to "jc" for mentioning Edelsbrunner!), I was able to find a few relevant papers in the literature. First, the problem was of finding the union of 'N' discs was first proposed by M. I. Shamos in his 1978 thesis:
Shamos, M. I. “Computational Geometry” Ph.D. thesis, Yale Univ., New Haven, CT 1978.
In 1985 Micha Sharir presented an O(n $log^2n$) time and O(n) space deterministic algorithm for the disc intersection/union problem (based on modified Voronoi diagrams): Sharir, M. Intersection and closest-pair problems for a set of planar discs. SIAM .J Comput. 14 (1985), pp. 448-468.
In 1988, Franz Aurenhammer presented another, more efficient O(n log n) time and O(n) space algorithm for circle (disc) intersection/union using power diagrams (generalizations of Voronoi diagrams): Aurenhammer, F. Improved algorithms for discs and balls using power diagrams. Journal of Algorithms 9 (1985), pp. 151-161.
It would be really neat if anyone could be point me to an implementation of one of the two determistic algorithms above, perhaps in a computational geometry package (neither appear trivial to put into practice)...
-
## 5 Answers
CGAL has a general voronoi diagram module that's quite customizable. While I've never used it to build power diagrams, it should not be hard to add in the right kind of distance function to generate the diagrams you need:
http://www.cgal.org/Manual/3.5/doc_html/cgal_manual/Voronoi_diagram_2/Chapter_main.html
-
### 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.
I believe Herbert Edelsbrunner has written a few papers on the topic (for balls in $\mathbb{R}^d$), though his main interest was in d=3. You might try starting with "The union of balls and its dual shape", which discusses some structure useful in all dimensions.
-
I would perform DFS on a quadtree to the desired level of depth. Can't prove it is optimal, but it is very space efficient and can be done in parallel.
````At each level you need to determine::
If one of the circles does not touch this box then I return empty.
If all remaining circles cover this box then I return the size of the box.
Any circles that cover this box I ignore them on future recursions.
````
-
Dear Chad, Thanks, I really appreciate you taking the time to answer my question! However, I'm looking for an exact/analytical solution to the problem... – Rob Grey Jul 6 2010 at 3:12
The problem was solved completely by the following papers:
A New Algorithm for Dynamic Computing the Perimeter of Union of Circular Arcs， CHEN Jian-xun ZHAO Hui Dept. of Computer Science & technology, Wuhan University of Science & Technology, Wuhan Hubei 430081; Journal of Computer-Aided Design & Computer Graphics，1998-03
-
Do you happen to have a version of this last paper in English?
-
"this last paper": Which paper? "Last" in this context does not pick out a paper unambiguously. – Joseph O'Rourke May 8 2012 at 19:27
Sorry. I meant the paper "A New Algorithm for Dynamic Computing the Perimeter of Union of Circular Arcs", by CHEN Jian-xun and ZHAO Hui. The only related reference I found is en.cnki.com.cn/Article_en/… but that is another paper (for union instead of perimeter) by CHEN Jian-xun and MA Heng Tai. I would like to see either of these papers by CHEN Jian-xun. Thanks! – Cristina Fernandes May 8 2012 at 20:10
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http://en.wikipedia.org/wiki/Canonical_ensemble
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# Canonical ensemble
Statistical mechanics
• NVE Microcanonical
• NVT Canonical
• µVT Grand canonical
• NPH Isoenthalpic–isobaric
• NPT Isothermal–isobaric
• µVT Open statistical
Models
Scientists
This article . Relevant discussion may be found on Talk:Canonical ensemble. Please improve it by verifying the claims made and adding inline citations. Statements consisting only of original research may be removed. (November 2012)
The canonical ensemble in statistical mechanics is a statistical ensemble representing a probability distribution of microscopic states of the system. For a system taking only discrete values of energy, the probability distribution is characterized by the probability $p_i$ of finding the system in a particular microscopic state $i$ with energy level $E_i$, conditioned on the prior knowledge that the total energy of the system and reservoir combined remains constant. This is given by the Boltzmann distribution,
$p_i = \tfrac{1}{Z}e^{-\frac{E_i}{kT}} = e^{-\frac{E_i -A}{kT}}$
where
$Z=e^{-\frac{A}{kT}}$
is the normalizing constant explained below (A is the Helmholtz free energy function). The Boltzmann distribution describes a system that can exchange energy with a heat bath (or alternatively with a large number of similar systems) so that its temperature remains constant. Equivalently, it is the distribution which has maximum entropy for a given average energy $\langle E \rangle$.
It is also referred to as the NVT ensemble: the number of particles $(N)$ and the volume $(V)$ of each system in the ensemble are constant and the ensemble has a well-defined temperature $(T)$, given by the temperature of the heat bath with which it would be in equilibrium.
The quantity $k$ is the Boltzmann constant, which relates the units of temperature to units of energy. It may be suppressed by expressing the absolute temperature using thermodynamic beta,
$\beta = \frac{1}{kT}$.
The quantities $A$ and $Z$ are constants for a particular ensemble, which ensure that $\Sigma p_i$ is normalised to 1. $Z$ is therefore given by
$Z = \sum_{i} e^{-\frac{E_i}{kT}} = \sum_{i} e^{-\beta E_i}$.
This is called the partition function of the canonical ensemble. Specifying this dependence of $Z$ on the energies $E_i$ conveys the same mathematical information as specifying the form of $p_i$ above.
The canonical ensemble (and its partition function) is widely used as a tool to calculate thermodynamic quantities of a system under a fixed temperature. This article derives some basic elements of the canonical ensemble. Other related thermodynamic formulas are given in the partition function article. When viewed in a more general setting, the canonical ensemble is known as the Gibbs measure, where, because it has the Markov property of statistical independence, it occurs in many settings outside of the field of physics.
## Deriving the Boltzmann factor from ensemble theory
Let $E_i\,$ be the energy of the microstate $i\,$ and suppose there are $n_i\,$ members of the ensemble residing in this state. Further we assume the total number of members in the ensemble, $\mathcal{N}\,$, and the total energy of all systems of the ensemble, $\mathcal{E}\,$, are fixed, i.e.,
$\mathcal{N}= \sum_i n_i , \,$
$\mathcal{E}= \sum_i n_i E_i \,.$
Since systems in the ensemble are indistinguishable with respect to a macrostate, for each set $\{n_i\} \,$, the number of ways of shuffling systems is equal to
$W (\{n_i\}) = \frac{\mathcal{N}!}{ \prod_{i} n_i!} .$
So for a given $\{n_i\}\,$, there are $W(\{n_i\})\,$ rearrangements that specify the same state of the ensemble.
The most probable distribution is the one that maximizes $W (\{n_i\})\,$. The probability for any other distribution to occur is extremely small in the limit $\mathcal{N} \rightarrow \infty \,$. To determine this distribution, one should maximize $W (\{n_i\})\,$ with respect to the $n_i\,$'s, under two constraints specified above. This can be done by using two Lagrange multipliers $\alpha \,$ and $\beta\,$. (The assumption that $\mathcal{N} \rightarrow \infty \,$ would be invoked in such calculation, which allows one to apply Stirling's approximation.) The result is
$n_i = e^{-\alpha -\beta E_i} \,$.
This distribution is called the canonical distribution. To determine $\alpha \,$ and $\beta\,$, it is useful to introduce the partition function as a sum over microscopic states
$Z(\beta) = \sum_j e^{-\beta E_j} .\,$
Comparing with thermodynamic formulae, it can be shown that $\beta\,$, is related to the absolute temperature $T\,$ as, $\beta=1/k_B T\,$. Moreover the expression
$F=- \frac{\ln Z(\beta)}{\beta}$
is identified as the Helmholtz free energy $F$. A derivation is given here. Consequently, from the partition function we can obtain the average thermodynamic quantities for the ensemble. For example, the average energy among members of the ensemble is
$\langle E \rangle = \frac{ \mathcal{E}}{ \mathcal{N} } = - \frac{\partial}{\partial \beta } \ln Z(\beta) \,$.
This relation can be used to determine $\beta\,$. $\alpha\,$ is determined from
$e^{\alpha} = \frac{ Z(\beta)}{ \mathcal{N}}$.
## A derivation from heat-bath viewpoint
Illustration of a system of interest suspended in a heat bath. The system of interest is taken to be small compared to the heat bath.
Define the following:
• S - the system of interest
• S′ - the heat reservoir in which S resides; S is small compared to S′
• S* - the system consisting of S and S′ combined together
• m - an indexing variable which labels all the available energy states of the system S
• Em - the energy of the state corresponding to the index m for the system S
• E′ - the energy associated with the heat bath
• E* - the energy associated with S*
• Ω′(E) - denotes the number of microstates available at a particular energy E for the heat reservoir.
It is assumed that the system S and the reservoir S′ are in thermal equilibrium. The objective is to calculate the set of probabilities pm that S is in a particular energy state Em.
Suppose S is in a microstate indexed by m. From the above definitions, the total energy of the system S* is given by
$E^\ast = E' + E_m \,$
Notice E* is constant, since the combined system S* is taken to be isolated.
Now, arguably the key step in the derivation is that the probability of S being in the m-th state, $\; p_m$, is proportional to the corresponding number of microstates available to the reservoir when S is in the m-th state. Therefore,
$p_m = C'\Omega'(E') \,$
for some constant $\; C'$. Taking the logarithm gives
$\ln p_m = \ln C' + \ln \Omega' (E') = \ln C' + \ln \Omega' (E^* - E_m) \,$
Since Em is small compared to E*, a Taylor series expansion can be performed on the latter logarithm around the energy E*. A good approximation can be obtained by keeping the first two terms of the Taylor series expansion:
$\ln \Omega'(E') = \sum_{k=0}^\infty \frac{(E' - E^\ast )^k }{k!} \frac{d^k \ln \Omega' (E^\ast)}{dE'^k} \approx \ln \Omega'(E^\ast) - \frac{d}{dE'} \ln \Omega'(E^\ast) E_m$
The following quantity is a constant which is traditionally denoted by β, known as the thermodynamic beta.
$\beta = \frac{d}{dE'} \ln \Omega'(E^\ast) = \left . \frac{d}{dE'} \ln \Omega'(E') \right |_{E'=E^\ast}$
Finally,
$\ln p_m = \ln C' + \ln \Omega'(E^\ast) - \beta E_m \,$
Exponentiating this expression gives
$p_m = C' \Omega'(E^\ast) e^{-\beta E_m}$
The factor in front of the exponential can be treated as a normalization constant C, where
$C = C' \Omega'(E^\ast) \,$
From this
$p_m = C e^{-\beta E_m} \,$
### Normalization to recover the partition function
Since probabilities must sum to 1, it must be the case that
$\sum_m p_m = 1 = \sum_m C e^{-\beta E_m} = C \sum_m e^{-\beta E_m} \iff C = \frac{1}{\sum_m e^{-\beta E_m}} \equiv \frac{1}{Z(\beta)}$
where $Z$ is known as the Partition function for the canonical ensemble.
### Note on derivation
As mentioned above, the derivation hinges on recognizing that the probability of the system being in a particular state is proportional to the corresponding multiplicities of the reservoir (the same can be said for the grand canonical ensemble). As long as one makes that observation, it is flexible as how one might proceed. In the derivation given, the logarithm is taken, then a linear approximation based on physical arguments is used. Alternatively, one can apply the thermodynamic identity for differential entropy:
$d S = {1 \over T} (d U + P d V - \mu d N)$
and obtain the same result. See the article on Maxwell-Boltzmann statistics where this approach is employed.
The canonical ensemble is also called the Gibbs ensemble, in honor of J.W. Gibbs, widely regarded with Boltzmann as being one of the two fathers of statistical mechanics. In his definitive 1901 book "Elementary Principles in Statistical Mechanics", Gibbs viewed an ensemble as a list of the allowed states of the system (each state appearing once and only once in the list) and the associated statistical weights. The states do not interact with each other, or with a reservoir, until Gibbs treats what happens when two complete ensembles at two different temperatures are allowed to interact weakly (Gibbs, pp 160). Gibbs writes that "...the distribution in phase..." (the phase space density in modern language) "...[is] called canonical...[if] the index of probability" (the logarithm of the statistical weight of the phase space density) "...is a linear function of the energy..." (Gibbs, Ch. 4). In Gibbs' formulation, this requirement (his equation 91), in modern notation
$P = e^{\frac{E-A}{kT} } \,$
is taken to define the canonical ensemble and to be the fundamental postulate. Gibbs does show that a large collection of interacting microcanonical systems approaches the canonical ensemble, but this is part of his demonstration (Gibbs, pp 169–183) that the principle of equal a priori probabilities, therefore the microcanonical ensemble, are inferior to the canonical ensemble as an axiomatization of statistical mechanics, at every point where the two treatments differ.
Gibbs original formulation is still standard in modern mathematically rigorous treatments of statistical mechanics, where the canonical ensemble is defined as the probability measure
$e^{ {E - A \over kT} } dp \, dq$
with p and q being the canonical coordinates.
### Characteristic state function
The characteristic state function of the canonical ensemble is the Helmholtz free energy function, as the following relationship holds:
$Z(T,V,N) = e^{- \beta A} \,\;$
## Quantum mechanical systems
By applying the canonical partition function, one can easily obtain the corresponding results for a canonical ensemble of quantum mechanical systems. A quantum mechanical ensemble in general is described by a density matrix. Suppose the Hamiltonian H of interest is a self adjoint operator with only discrete spectrum. The energy levels $\{ E_n \}$ are then the eigenvalues of H, corresponding to eigenvector $| \psi _n \rangle$. From the same considerations as in the classical case, the probability that a system from the ensemble will be in state $| \psi _n \rangle$ is $p_n = C e^{- \beta E_n}$, for some constant $C$. So the ensemble is described by the density matrix
$\rho = \sum p_n | \psi _n \rangle \langle \psi_n | = \sum C e^{- \beta E_n} | \psi _n \rangle \langle \psi_n|$
(Technical note: a density matrix must be trace-class, therefore we have also assumed that the sequence of energy eigenvalues diverges sufficiently fast.) A density operator is assumed to have trace 1, so
$\operatorname{Tr} (\rho) = C \underbrace{\sum e^{- \beta E_n}}_Q = 1,$
which means
$C = \frac{1}{\sum e^{- \beta E_n} } = \frac{1}{Q}.$
Q is the quantum-mechanical version of the canonical partition function. Putting C back into the equation for ρ gives
$\rho = \frac{1}{\sum e^{- \beta E_n}} \sum e^{- \beta E_n} | \psi _n \rangle \langle \psi_n| = \frac{1}{ \operatorname{Tr}( e^{- \beta H} ) } e^{- \beta H} .$
By the assumption that the energy eigenvalues diverge, the Hamiltonian H is an unbounded operator, therefore we have invoked the Borel functional calculus to exponentiate the Hamiltonian H. Alternatively, in non-rigorous fashion, one can consider that to be the exponential power series.
Notice the quantity
$\operatorname{Tr}( e^{- \beta H} )$
is the quantum mechanical counterpart of the canonical partition function, being the normalization factor for the mixed state of interest.
The density operator ρ obtained above therefore describes the (mixed) state of a canonical ensemble of quantum mechanical systems. As with any density operator, if A is a physical observable, then its expected value is
$\langle A \rangle = \operatorname{Tr}( \rho A ).$
## Issues in the traditional models of the derivation of the canonical distribution
Understanding and clear presentation of the derivation of the canonical distribution are difficult both for students and for teachers. The difficulty is caused by the complexity of the subject, but also by the circumstance that mathematical schemes allowing to receive a desirable result are confused with physical models.
These schemes consider only the eigenstates of the system to be system states. However, a great number of quantum superpositions of the system eigenstates corresponds to the same value of the energy of the system.
In one of the schemes the system S is considered to be a part of a huge system U, usually called “Universe”. The system environment W (or addition to system U) is often called “the thermostat”. The “Universe” is described by the microcanonical ensemble. It means that the "Universe" is in equilibrium, the energy of the "Universe" lies in a very small interval, only the eigenstates of the "Universe" are possible and all eigenstates are equiprobable.
Another scheme - the method of the most probable distribution - assumes that the "Universe" consists of a very great number of systems identical to the system under consideration. In both schemes the system interaction with its environment is considered extremely weak – to make it possible to talk about certain quantum state of the system S. At the same time, the transition of the system from one eigenstate to another is considered to be caused by its energy exchange with the environment. It is obvious that the canonical distribution can be used to calculate the observed quantities only if during the measurement the system has time to visit all states of the spectrum repeatedly. However, the aforementioned schemes do not correlate the values of the contact with the environment, spectral diapason of the system energy and measurement time.
The schemes used for the derivation of canonical distribution do not call into question the absolute accuracy of quantum mechanics. However, one must remember that both classical and quantum mechanics have resulted from the observation of systems with small number of objects. If the number of objects (e.g. particles) in a system is small and calculations are possible, the mechanics show amazing accuracy. One might assume that in systems with macroscopically great number of particles quantum mechanics would also be absolutely exact. However, this assumption contradicts the irreversibility of evolution of the macrosystems, the second law of thermodynamics and the experimental data received on concrete physical objects.
The experimental evidence of the existence of the probabilistic processes which are not described by the standard quantum formalism allows to consider the canonical distribution as a result of the averaging on various system states with the same energy of the squared modules of the coefficients of expansion of the system state function on the eigenfunctions. But the question arises of what exactly are the system states in light of the existence of the probabilistic processes which are not considered by the standard formalism of the quantum mechanics (in particular, recording the state function as a superposition of eigenfunctions of the whole macrosystem may not be quite adequate).
## Relations with other ensembles
A generalization of this is the grand canonical ensemble, in which the systems may share particles as well as energy. By contrast, in the microcanonical ensemble, the energy of each individual system is fixed.
## See also
Statistical mechanics
• NVE Microcanonical
• NVT Canonical
• µVT Grand canonical
• NPH Isoenthalpic–isobaric
• NPT Isothermal–isobaric
• µVT Open statistical
Models
Scientists
## References
• L. D. Landau and E. M. Lifshitz, "Statistical Physics, 3rd Edition Part 1", Butterworth-Heinemann, Oxford, 1996.
• J. A. White and S. Velasco, "The Ornstein-Zernike equation in the canonical ensemble" (2001) Europhys. Lett. 54 pp. 475–481. (requires subscription)
• Harvey Gould and Jan Tobochnik, Draft Chapters of Thermal and Statistical Physics Textbook, in preparation for Princeton University Press, 2010
• Juha Javanainen, PHYS-5500, Statistical Mechanics, 2008
• R. Balescu, “Equilibrium and Nonequilibrium Statistical Mechanics”, A Wiley-Interscience Publication, New York, 1975.
• V.A. Skrebnev, R.N. Zaripov, “Investigation of the equilibrium establishment in the spin system by using dipole magic echo”, Appl. Magn. Reson. 16, 1-17, 1999.
• V.A. Skrebnev, Canonical distribution and incompleteness of quantum mechanics, arXiv:1201.5078v1 [physics.gen-ph][original research?].
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http://mathoverflow.net/questions/68647/understanding-the-wiki-page-on-verdier-duality/94478
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## Understanding (the wiki page on) Verdier duality
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My familiarity with concepts related to derived categories is only tangential, and little by little I intend to get more comfortable with them. I was playing around with Caldararu's introduction to the topic, and looking up various things on the web.
Here is my question (that I have every confidence is trivial for experts):
On the wiki page on Verdier duality http://en.wikipedia.org/wiki/Verdier_duality it says the following: Let $F$ be a field, and $X$ a finite dimensional (dimension is defined here cohomologically, but for our purposes a finite dimensional manifold will do) locally compact space.
In the part about Poincare duality, it says: $H^k(X,F)=[F,X[k]]$. What is the interpretation of this notation? As I see it, $[F,X[k]]$ means $Hom(F,X[k])$ in the derived category. But this means that $X$ is seen as a complex. How? And why would $Hom(F,X[k])$ equal $H^k(X,F)$?
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I guess that there is a typo and that $[F,X[k]]$ should be understood as $[F,F[k]]$. – DamienC Jun 23 2011 at 20:27
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Why is $[F,F[k]]=H^k(X,F)$? – James D. Taylor Jun 23 2011 at 20:30
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Very shortly, for any sheaf $S$ (i.e. a complex of sheaves concentrated in degree $0$), $[F[-k],S]=H^k(X,S)$. – DamienC Jun 23 2011 at 20:37
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Thanks! You mean $[S[-k],S]$, right? Because $H^k(X,S)$ doesn't depend on $F$. – James D. Taylor Jun 23 2011 at 20:40
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I do really mean $F$, because I ma working with the derived category of $F$-modules (as in the wiki). But you can just take $F=\mathbb{Z}$ if you like. – DamienC Jun 23 2011 at 20:54
show 1 more comment
## 2 Answers
I think here is how one should understand the last paragraph of the wiki.
Consider $f:X\to pt$. We have (all functors are derived and my $Hom$ are sheaf $Hom$) $$Hom_{pt}(f_!F,F)=f_*Hom_X(F,f^!F)$$
$f_!F=\Gamma_c(X,F)$, so the l.h.s. computes the dual of the cohomology with compact support of the constant sheaf $F$, i.e. the dual of the cohomology with compact support of $X$.
$Hom_X(F,f^!F)= \Gamma(f^!F)$ and thus the r.h.s. is $f_*(\Gamma(f^!F))=\Gamma(X,f^!F)$, the homology of $X$.
EDIT : to answer precisely the question I include a summary of the comments.
1. there is a typo in the wiki: $[F,X[k]]$ should be understood as $[F,F[k]]$.
2. for any sheaf of $F$-modules $S$ (concentrated in degree 0), $[F[−k],S]=H^k(X,S)$.
3. Contrary to what is claimed in the wiki, there is no duality between $H^k(X,F)=[F[−k],F]$ and $H_k(X,F)=[F[−k],D_X]$ (where $D_X:=f^!F$ for $f:X\to pt$). The duality is, either between $H^k_c(X,F)$ and $H_k(X,F)$, or between $H^k(X,F)$ and $H_k^{BM}(X,F)$. And as far as I understand $H^*(X,S)$ is dual to $H_c^*(X,S^\vee)$.
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I'm still a little confused. Under some conditions, in Poincare duality we have $D_X=F[-n]$, right? So $H_k(X,F)=Hom(F[-k],F[-n])=Hom(F,F[k-n])=H^{k-n}(X,F)=0$. I'm doing something wrong. – James D. Taylor Jun 23 2011 at 21:11
Oh, I guess you're saying that $Hom(F,F[a])\neq H^a(X,F)$, but rather $H^a(X,F)=Hom(O_X,F[a])$. But still, how does one get from $Hom(F,F[k-n])$ to the desired $H^{n-k}(X,F)$ dual? I guess $Hom(F,F[k-n])=Hom(O_X,F[k-n]$ dual $)$. Now what? – James D. Taylor Jun 23 2011 at 21:17
Aha! Here's what's missing. For some reason: $Hom(O_X,C)\cong$ to the dual of $H^(X,$ dual of $C)$. If I can only figure out why that's true, I'll be satisfied. – James D. Taylor Jun 23 2011 at 21:19
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Sorry. The point is that there is another mistake in the wiki: Poincaré duality is - either between $H^*$ and $H^{BM}_*$ (Borel-Moore homology) - either between $H_c^*$ and $H_*$ - not between $H^*$ and $H_*$. Then one has $H^k(X,F)=[F[-k],F]$, $H_k(X,F)=[F[-k],D_X]$, and $H^k_c(X,F)$ is given by the formula in you comment. – DamienC Jun 23 2011 at 22:22
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@euklid345: nobody's complaining (IMO spotting a typo and/or a mistake is not complaining). – DamienC Jan 11 2012 at 8:02
show 5 more comments
### 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.
I just revamped what was written. Perhaps now it is more understandable:New Wiki Entry on Verdier duality.
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http://mathoverflow.net/revisions/59180/list
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## Return to Answer
2 added 2 characters in body; deleted 6 characters in body
This is of the "to puzzle graduate students" variety, but I was taken enough with it to write it up in a short note:
Let $V$ be a vector space over a field $F$, of dimension at least $2$, and consider coverings $\mathcal{C} = \{W_i\}$(W_i)$of$V$by proper subspaces. Does there exist a countable covering$\mathcal{C}$? (It depends on$\dim V\$ and `$\# F$`, but perhaps not exactly as you expect!)
1 [made Community Wiki]
This is of the "to puzzle graduate students" variety, but I was taken enough with it to write it up in a short note:
Let $V$ be a vector space over a field $F$, of dimension at least $2$, and consider coverings `$\mathcal{C} = \{W_i\}$` of $V$ by proper subspaces. Does there exist a countable covering $\mathcal{C}$? (It depends on $\dim V$ and `$\# F$`, but perhaps not exactly as you expect!)
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http://mathoverflow.net/questions/63278?sort=newest
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## Complexity of computing matrix rank over integers
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Does computing the rank of an integer matrix have complexity polynomial in the size of the input?
The Gaussian elimination algorithm is polynomial in the number of elementary operations (addition and multiplication), but the intermediate size of of the matrix entries may go up exponentially. Are there other algorithms with better complexity? Can anyone give a reference?
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## 3 Answers
Gaussian elimination is a polynomial-time algorithm. While it may not be obvious on the first sight, it can be implemented so that the intermediate entries have only polynomial size (bit length), because they happen to be equal to determinants of certain submatrices of the original matrix (or ratios thereof, depending on the version). See e.g. Edmonds and Bareiss.
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Right, using Bareiss' algorithm space needed is at most 0.5 n lg n + O(n) due to a theorem of Hadamard. This means that each operation takes time $O(n(\log n)^{2+\varepsilon})$ giving overall time complexity $O(n^4(\log n)^{2+\varepsilon}).$ – Charles Apr 28 2011 at 15:36
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I asked more-or-less the same question 12 years ago on sci.math.num-analysis, but for the case when the integer entries are small. Thom Mulders replied describing this wonderful modular method.
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The answer to your question is yes. Note that you can bound determinant(in fact you need to bound the size of lattice spanned by rows of the matrix) of the matrix with integer of size polynomial in the length input. Let $p$ be a prime with is large than this bound then the rank of the integer matrix will be equal to the rank of the matrix mod $p$.
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http://mathhelpforum.com/advanced-algebra/2466-when-lr-not-complete-space.html
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# Thread:
1. ## When is lR NOT a complete space?
Hey all!
I was wondering... Everywhere I see examples of completes spaces, I also see that $\mathbb{R}$ is complete with the usual metric. That's fine with me, but I was wondering why I see no example of a metric for which $\mathbb{R}$ is not complete.
For those wondering - I was thinking about all this when trying to prove (or disprove) that $\mathbb{R}$ is complete with $d(x,y)=|x^{\frac{1}{3}}-y^{\frac{1}{3}}|$.
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http://mathhelpforum.com/advanced-math-topics/174027-what-cyclonical-curve.html
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# Thread:
1. ## What is a Cyclonical curve?
Hi everyone,
Recently in a problem I encountered the word cyclonical curve. Do you have any idea what this is? One of the ideas that I thought of is a cyclone can be approximated to a spiral. But then there are many different kind of spirals.......
2. Originally Posted by Sudharaka
Hi everyone,
Recently in a problem I encountered the word cyclonical curve. Do you have any idea what this is? One of the ideas that I thought of is a cyclone can be approximated to a spiral. But then there are many different kind of spirals.......
Cyclonical - pertaining to a cyclone, so in this case a curve that spirals (inward if direction has a sense)
Is there more context available?
CB
3. Originally Posted by CaptainBlack
Cyclonical - pertaining to a cyclone, so in this case a curve that spirals (inward if direction has a sense)
Is there more context available?
CB
Yes. The question is,
"A uniform string of weight W rests on a rough cyclonical curve which has its axis vertical and vertex upwards, the string extending from the vertex to a cusp. Show that the least horizontal force applied at the vertex that will cause the string to slip upwards is $\displaystyle\frac{W(3e^{\frac{\mu\pi}{2}}-\mu^2-1)}{\mu^2+4}$, where $\mu$ is the coeffecient of friction."
So to solve the problem we need to know the intrinsic equation (equation which relates arc length to the angle of the tangent) of a cyclonical curve. I think the problem does not give enough details to solve it. For there are many different spirals such as, logarithmic spiral, Archimedean spiral etc. What is your idea?
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http://mathoverflow.net/revisions/92747/list
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Return to Answer
2 corrected two typos
The answer is no: for example, $k[[x]]\otimes_k k((x))$ is not noetherian.
Indeed, if it were, so would be $k((x))\otimes_k k((x))$.
But this would contradict the following interesting general theorem of Vámos:
Given an extension of fields $K/F$ the tensor product $K\otimes_F K$ is noetherian if and only if the $K$ is finitely generated as a field over $K$. F\$.
Full confession
I have only read an abstract of Vámos's article because I have no access to it. Anyway, here is the reference:
P. Vámos, On the minimal prime ideal of a tensor product of two fields, Math. Proc. Cambridge Philos. Soc. 84 (1978), no. 1, p.25-35.
1
The answer is no: for example, $k[[x]]\otimes_k k((x))$ is not noetherian.
Indeed, if it were, so would be $k((x))\otimes_k k((x))$.
But this would contradict the following interesting general theorem of Vámos:
Given an extension of fields $K/F$ the tensor product $K\otimes_F K$ is noetherian if and only if the $K$ is finitely generated as a field over $K$.
Full confession
I have only read an abstract of Vámos's article because I have no access to it. Anyway, here is the reference:
P. Vámos, On the minimal prime ideal of a tensor product of two fields, Math. Proc. Cambridge Philos. Soc. 84 (1978), no. 1, p.25-35.
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http://mathoverflow.net/questions/61185/are-there-any-results-about-equation-over-rational-field-or-the-extension-qx
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## are there any results about equation over rational field or the extension Q[x]?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Given a polynomial $p(x)\in \mathbb{Q}[x]$, it is known that its roots can be obtained in terms of the coefficients of the polynomial by formulas involving the usual algebraic operations (addition, subtraction, multiplication, division), application of radicals (square roots, cube roots, etc), and application of automorphic functions.
What is known when we go up one level, to a polynomial $p(y)\in \mathbb{Q}(x)[y]$? Meaning, what operations on the coefficients (lying in $\mathbb{Q}(x)$) do we need in order to express the roots (lying in $\overline{\mathbb{Q}(x)}$)?
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This is within the scope of a classical theorem on Riemann surfaces. See en.wikipedia.org/wiki/Uniformization_theorem. – Charles Matthews Apr 10 2011 at 8:19
1
+1 to Dror for editing. – Gerry Myerson Apr 10 2011 at 12:30
Dror:Thank you for your editing.I think I had described my question badly – XL Apr 10 2011 at 14:31
:) I edited because I didn't want to see this question closed. In fact, I still want to see the question answered! I don't know much about this theory, let alone how the Uniformization Theorem solves it. Since I'm a graduate student, I think the normal MO rules apply and say that someone should answer. So, please answer! – Dror Speiser Apr 10 2011 at 19:54
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http://mathoverflow.net/revisions/19724/list
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## Return to Answer
1 [made Community Wiki]
Vague definition of canonical:
Let $X$ and $Y$ be collections (often sets) for which assumptions has been made (has been given structures and/or are related somehow). A function $f\colon X \to Y$ is canonical if it is given by a rule using only the already given structure.
This explains the relation to the greek word rule (kanon). The precise meaning of the above are open for enterpretation: how much structure can the rule itself contain (maybe this can be made precise)! This "definition" somewhat contradicts many of the other answers, which for some reason is under the impression that canonical implies unique (or almost unique), which in my point of view is very wrong since different rules may define different maps. E.g. if we let $X$ be the objects in the category of abelian groups and $Y$ the morphisms then the definitions makes all the group homomorphisms $A \to A$ given by multiplication with an element in $\mathbb{Z}$ canonical, which to me is not a problem.
Usually when there is an especially simple rule it is often assumed without mentioning that this is the rule defining the function. E.g. most will understand the following:"there is a canonical endemorphism of any object in a category". This emphasizes the multiplication with 1 above as somehow speciel or "more canonical" than the rest. This is simply because the rule works in much greater generality and is shorter.
Usually if a rule is very simple the function will have nice properties. E.g. simply rules in category theory often define functors, natural transformations, e.t.c. This leeds many people to confuse the notion of canonical with "something behaving nicely".
I am somewhat puzzled by the use of the word uniform in one of the answers. The nature of the word uniform is "of the same form" and relates more to symmetries and things looking the same every where. This often leeds to canonical maps, since a choice at one point can sometimes be extended to a choice at every point. Please someone comment on this since maybe this is just a use of the word I have not seen before!
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http://terrytao.wordpress.com/tag/russells-paradox/
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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
# Tag Archive
You are currently browsing the tag archive for the ‘Russell’s paradox’ tag.
## The “no self-defeating object” argument, and the vagueness paradox
2 November, 2010 in expository, math.GM, math.LO | Tags: cantor's theorem, Russell's paradox, set theory, sorites paradox, Zorn's lemma | by Terence Tao | 29 comments
This is the third in a series of posts on the “no self-defeating object” argument in mathematics – a powerful and useful argument based on formalising the observation that any object or structure that is so powerful that it can “defeat” even itself, cannot actually exist. This argument is used to establish many basic impossibility results in mathematics, such as Gödel’s theorem that it is impossible for any sufficiently sophisticated formal axiom system to prove its own consistency, Turing’s theorem that it is impossible for any sufficiently sophisticated programming language to solve its own halting problem, or Cantor’s theorem that it is impossible for any set to enumerate its own power set (and as a corollary, the natural numbers cannot enumerate the real numbers).
As remarked in the previous posts, many people who encounter these theorems can feel uneasy about their conclusions, and their method of proof; this seems to be particularly the case with regard to Cantor’s result that the reals are uncountable. In the previous post in this series, I focused on one particular aspect of the standard proofs which one might be uncomfortable with, namely their counterfactual nature, and observed that many of these proofs can be largely (though not completely) converted to non-counterfactual form. However, this does not fully dispel the sense that the conclusions of these theorems – that the reals are not countable, that the class of all sets is not itself a set, that truth cannot be captured by a predicate, that consistency is not provable, etc. – are highly unintuitive, and even objectionable to “common sense” in some cases.
How can intuition lead one to doubt the conclusions of these mathematical results? I believe that one reason is because these results are sensitive to the amount of vagueness in one’s mental model of mathematics. In the formal mathematical world, where every statement is either absolutely true or absolutely false with no middle ground, and all concepts require a precise definition (or at least a precise axiomatisation) before they can be used, then one can rigorously state and prove Cantor’s theorem, Gödel’s theorem, and all the other results mentioned in the previous posts without difficulty. However, in the vague and fuzzy world of mathematical intuition, in which one’s impression of the truth or falsity of a statement may be influenced by recent mental reference points, definitions are malleable and blurry with no sharp dividing lines between what is and what is not covered by such definitions, and key mathematical objects may be incompletely specified and thus “moving targets” subject to interpretation, then one can argue with some degree of justification that the conclusions of the above results are incorrect; in the vague world, it seems quite plausible that one can always enumerate all the real numbers “that one needs to”, one can always justify the consistency of one’s reasoning system, one can reason using truth as if it were a predicate, and so forth. The impossibility results only kick in once one tries to clear away the fog of vagueness and nail down all the definitions and mathematical statements precisely. (To put it another way, the no-self-defeating object argument relies very much on the disconnected, definite, and absolute nature of the boolean truth space $\{\hbox{true},\hbox{ false}\}$ in the rigorous mathematical world.)
Read the rest of this entry »
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http://mathinsight.org/conservative_vector_field_determine
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# Math Insight
• Top
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• Notation
When printing from Firefox, select the SVG renderer in the MathJax contextual menu (right click any math equation) for better printing results
### How to determine if a vector field is conservative
#### Suggested background
A conservative vector field (also called a path-independent vector field) is a vector field $\dlvf$ whose line integral $\dlint$ over any curve $\dlc$ depends only on the endpoints of $\dlc$. The integral is independent of the path that $\dlc$ takes going from its starting point to its ending point. The below applet illustrates the two-dimensional conservative vector field $\dlvf(x,y)=(x,y)$.
The line integral over multiple paths of a conservative vector field. The integral of conservative vector field $\dlvf(x,y)=(x,y)$ from $\vc{a}=(3,-3)$ (cyan square) to $\vc{b}=(2,4)$ (magenta square) doesn't depend on the path. Path $\dlc$ (shown in blue) is a straight line path from $\vc{a}$ to $\vc{b}$. Paths $\adlc$ (in green) and $\sadlc$ (in red) are curvy paths, but they still start at $\vc{a}$ and end at $\vc{b}$. Each path has a colored point on it that you can drag along the path. The corresponding colored lines on the slider indicate the line integral along each curve, starting at the point $\vc{a}$ and ending at the movable point. Moving each point up to $\vc{b}$ gives the total integral along the path, so the corresponding colored line on the slider reaches 1 (the magenta line on the slider). This demonstrates that the integral is 1 independent of the path.
What are some ways to determine if a vector field is conservative? Directly checking to see if a line integral doesn't depend on the path is obviously impossible, as you would have to check an infinite number of paths between any pair of points. But, if you found two paths that gave different values of the integral, you could conclude the vector field was path-dependent.
Here are some options that could be useful under different circumstances.
1. As mentioned in the context of the gradient theorem, a vector field $\dlvf$ is conservative if and only if it has a potential function $f$ with $\dlvf = \nabla f$. Therefore, if you are given a potential function $f$ or if you can find one, and that potential function is defined everywhere, then there is nothing more to do. You know that $\dlvf$ is a conservative vector field, and you don't need to worry about the other tests we mention here. Similarly, if you can demonstrate that it is impossible to find a function $f$ that satisfies $\dlvf = \nabla f$, then you can likewise conclude that $\dlvf$ is non-conservative, or path-dependent.
For this reason, you could skip this discussion about testing for path-dependence and go directly to the procedure for finding the potential function. If this procedure works or if it breaks down, you've found your answer as to whether or not $\dlvf$ is conservative. However, if you are like many of us and are prone to make a mistake or two in a multi-step procedure, you'd probably benefit from other tests that could quickly determine path-independence. That way, you could avoid looking for a potential function when it doesn't exist and benefit from tests that confirm your calculations.
2. Another possible test involves the link between path-independence and circulation. One can show that a conservative vector field $\dlvf$ will have no circulation around any closed curve $\dlc$, meaning that its integral $\dlint$ around $\dlc$ must be zero. If you could somehow show that $\dlint=0$ for every closed curve (difficult since there are an infinite number of these), then you could conclude that $\dlvf$ is conservative. Or, if you can find one closed curve where the integral is non-zero, then you've shown that it is path-dependent.
Although checking for circulation may not be a practical test for path-independence, the fact that path-independence implies no circulation around any closed curve is a central to what it means for a vector field to be conservative.
3. If $\dlvf$ is a three-dimensional vector field, $\dlvf : \R^3 \to \R^3$ (confused?), then we can derive another condition. This condition is based on the fact that a vector field $\dlvf$ is conservative if and only if $\dlvf = \nabla f$ for some potential function. We can calculate that the curl of a gradient is zero, $\curl \nabla f = \vc{0}$, for any twice continuously differentiable $f : \R^3 \to \R$. Therefore, if $\dlvf$ is conservative, then its curl must be zero, as $\curl \dlvf = \curl \nabla f = \vc{0}$.
For a continuously differentiable two-dimensional vector field, $\dlvf : \R^2 \to \R^2$, we can similarly conclude that if the vector field is conservative, then the scalar curl must be zero, $$\pdiff{\dlvfc_2}{x}-\pdiff{\dlvfc_1}{y} = \frac{\partial f^2}{\partial x \partial y} -\frac{\partial f^2}{\partial y \partial x} =0.$$
We have to be careful here. The valid statement is that if $\dlvf$ is conservative, then its curl must be zero. Without additional conditions on the vector field, the converse may not be true, so we cannot conclude that $\dlvf$ is conservative just from its curl being zero. There are path-dependent vector fields with zero curl. On the other hand, we can conclude that if the curl of $\dlvf$ is non-zero, then $\dlvf$ must be path-dependent.
4. Can we obtain another test that allows us to determine for sure that a vector field is conservative? We can by linking the previous two tests (tests 2 and 3). Test 2 states that the lack of “macroscopic circulation” is sufficient to determine path-independence, but the problem is that lack of circulation around any closed curve is difficult to check directly. Test 3 says that a conservative vector field has no “microscopic circulation” as captured by the curl. It's easy to test for lack of curl, but the problem is that lack of curl is not sufficient to determine path-independence.
What we need way to link the definite test of zero “macroscopic circulation” with the easy-to-check test of zero “microscopic circulation.” This link is exactly what both Green's theorem and Stokes' theorem provide. Don't worry if you haven't learned both these theorems yet. The basic idea is simple enough: the “macroscopic circulation” around a closed curve is equal to the total “microscopic circulation” in the planar region inside the curve (for two dimensions, Green's theorem) or in a surface whose boundary is the curve (for three dimensions, Stokes' theorem).
Let's examine the case of a two-dimensional vector field whose scalar curl $\pdiff{\dlvfc_2}{x}-\pdiff{\dlvfc_1}{y}$ is zero. If we have a closed curve $\dlc$ where $\dlvf$ is defined everywhere inside it, then we can apply Green's theorem to conclude that the “macroscopic circulation” $\dlint$ around $\dlc$ is equal to the total “microscopic circulation” inside $\dlc$. We can indeed conclude that the “macroscopic circulation” is zero from the fact that the “microscopic circulation” $\pdiff{\dlvfc_2}{x}-\pdiff{\dlvfc_1}{y}$ is zero everywhere inside $\dlc$.
According to test 2, to conclude that $\dlvf$ is conservative, we need $\dlint$ to be zero around every closed curve $\dlc$. If the vector field is defined inside every closed curve $\dlc$ and the “microscopic circulation” is zero everywhere inside each curve, then Green's theorem gives us exactly that condition. We can conclude that $\dlint=0$ around every closed curve and the vector field is conservative.
The only way we could run into trouble is if there are some closed curves $\dlc$ where $\dlvf$ is not defined for some points inside the curve. In other words, if the region where $\dlvf$ is defined has some holes in it, then we cannot apply Green's theorem for every closed curve $\dlc$. In this case, we cannot be certain that zero “microscopic circulation” implies zero “macroscopic circulation” and hence path-independence. Such a hole in the domain of definition of $\dlvf$ was exactly what caused in the problem in our counterexample of a path-dependent field with zero curl.
On the other hand, we know we are safe if the region where $\dlvf$ is defined is simply connected, i.e., the region has no holes through it. In this case, we know $\dlvf$ is defined inside every closed curve $\dlc$ and nothing tricky can happen. We can summarize our test for path-dependence of two-dimensional vector fields as follows.
If a vector field $\dlvf: \R^2 \to \R^2$ is continuously differentiable in a simply connected domain $\dlr \in \R^2$ and its curl is zero, i.e., $$\pdiff{\dlvfc_2}{x}-\pdiff{\dlvfc_1}{y}=0,$$ everywhere in $\dlr$, then $\dlvf$ is conservative within the domain $\dlr$.
It turns out the result for three-dimensions is essentially the same. If a vector field $\dlvf: \R^3 \to \R^3$ is continuously differentiable in a simply connected domain $\dlv \in \R^3$ and its curl is zero, i.e., $\curl \dlvf = \vc{0}$, everywhere in $\dlv$, then $\dlvf$ is conservative within the domain $\dlv$.
One subtle difference between two and three dimensions is what it means for a region to be simply connected. Any hole in a two-dimensional domain is enough to make it non-simply connected. But, in three-dimensions, a simply-connected domain can have a hole in the center, as long as the hole doesn't go all the way through the domain, as illustrated in this figure.
The reason a hole in the center of a domain is not a problem in three dimensions is that we have more room to move around in 3D. If we have a curl-free vector field $\dlvf$ (i.e., with no “microscopic circulation”), we can use Stokes' theorem to infer the absence of “macroscopic circulation” around any closed curve $\dlc$. To use Stokes' theorem, we just need to find a surface whose boundary is $\dlc$. If the domain of $\dlvf$ is simply connected, even if it has a hole that doesn't go all the way through the domain, we can always find such a surface. The surface can just go around any hole that's in the middle of the domain. With such a surface along which $\curl \dlvf=\vc{0}$, we can use Stokes' theorem to show that the circulation $\dlint$ around $\dlc$ is zero. Since we can do this for any closed curve, we can conclude that $\dlvf$ is conservative.
The flexiblity we have in three dimensions to find multiple surfaces whose boundary is a given closed curve is illustrated in this applet that we use to introduce Stokes' theorem.
Macroscopic and microscopic circulation in three dimensions. The relationship between the macroscopic circulation of a vector field $\dlvf$ around a curve and the microscopic circulation of $\dlvf$ (illustrated by small green circles) along a surface in three dimensions must hold for any surface whose boundary is the curve. No matter which surface you choose (change by dragging the blue point on the slider), the total microscopic circulation of $\dlvf$ along the surface must equal the circulation of $\dlvf$ around the curve, as long as the vector field $\dlvf$ is defined everywhere on the surface.
Of course, if the region $\dlv$ is not simply connected, but has a hole going all the way through it, then $\curl \dlvf = \vc{0}$ is not a sufficient condition for path-independence. In this case, if $\dlc$ is a curve that goes around the hole, then we cannot find a surface that stays inside that domain whose boundary is $\dlc$. Without such a surface, we cannot use Stokes' theorem to conclude that the circulation around $\dlc$ is zero.
#### Cite this as
Nykamp DQ, “How to determine if a vector field is conservative.” From Math Insight. http://mathinsight.org/conservative_vector_field_determine
Keywords: conservative, gradient, gradient theorem, path independent, vector field
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http://mathoverflow.net/questions/24181?sort=oldest
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## Representations of Pin vs. Representations of Clifford
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This may be total nonsense. But I need to know the answer quickly and I am too tired to think about it thoroughly. Let $k$ be a positive integer. Roe's "Elliptic Operators" claims that there is a 1-to-1 correspondence between:
• representations of the Clifford algebra $\mathrm{Cl}\mathbb R^k$ of the vector space $\mathbb R^k$ with the standard inner product;
• representations of the Pin group of this vector space (i. e., of the subgroup of the multiplicative group of $\mathrm{Cl}\mathbb R^k$ generated by vectors from $\mathbb R^k$) on which the element $-1$ of the Pin group acts as $-\mathrm{id}$;
• representations of the subgroup $\left\lbrace \pm e_1^{i_1}e_2^{i_2}...e_k^{i_k} \mid 0\leq i_1,i_2,...,i_k\leq 1 \right\rbrace$ of the Pin group (where $\left(e_1,e_2,...,e_n\right)$ is the standard orthogonal basis of $\mathbb R^k$) on which the group element $-1$ acts as $-\mathrm{id}$.
I do see how representations restrict from the above to the below, and also how there is a 1-to-1 correspondence between the first and the third kind of representations. But is it really that obvious that there are no "strange" representations of the second kind? I mean, why is a representation of the Pin group uniquely given by how it behaves on $-1$, $e_1$, $e_2$, ..., $e_k$ ?
Any help welcome, I'd already be glad to know whether it's really that obvious or not.
EDIT: This seems to have caused some confusion. Here is the core of the question:
Assume that we have a representation $\rho$ of the Pin group $\mathrm{Pin}\mathbb R^k$ such that $\rho\left(-1\right)=-\mathrm{id}$. This, in particular, means an action of each unit vector. By linearity, we can extend this to an action of every vector. Is this always a representation (i. e., does the sum of two vectors always act as the sum of their respective actions)?
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For 3 -> 1, extend the induced map R^k --> End(V) using the universal property of Cl(R^k). – Michael May 10 2010 at 23:55
If I understand correctly, the question basically boils down to whether every rep of Pin is the restriction of a rep of Cl? – Eric O. Korman May 11 2010 at 1:21
Every representation of Pin which has -1 acting as -id. (The others are clearly taboo.) – darij grinberg May 11 2010 at 6:44
I remember being really confused by this when I read that stuff awhile back, and I wound up turning to Spin Geometry by Lawson and Michelsohn in order to understand the construction of the spinor bundle. I guess I'm relieved I didn't dwell on it too much! Still, the only part of this discussion that is needed for the rest of the book is the existence and uniqueness of the spin representation in even dimensions, and uniqueness even survives the mistake. I think I'll discuss this with Prof. Roe when I see him again next week. – Paul Siegel May 11 2010 at 23:52
Actually, Roe uses this to characterize the irreducible representations of the Clifford algebra. In my eyes this is quite an overkill, since it is easy to see that the representation map from the Clifford algebra to the endomorphism ring of the half-spin representation (or, in the odd case, to the sum of the endomorphism rings of the two half-spin representations) is bijective. – darij grinberg May 12 2010 at 7:13
## 5 Answers
I'm not sure whether a representation of an algebra $A$ means a representation of the unit group of $A$, or an $A$-module. With the second interpretation, the statement is false.
Let's take $k=2$ and use the negative definite inner product. (This example will occur inside any larger example.) So the Clifford algebra is generated by $e_1$ and $e_2$, subject to $e_1^2=e_2^2=1$ and $e_1 e_2 = - e_2 e_1$. Let $S$ be the $2$-dimensional representation `$$\rho_S(e_1) = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \quad \rho_S(e_2) = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.$$` Let $V=S^{\otimes 3}$. I claim that $V$ is a $\mathrm{Pin}$ representation where $-1$ acts by $- \mathrm{Id}$, but $V$ is not a module for the Clifford algebra.
For all $\theta$, the vector $v(\theta) := \cos \theta e_1 + \sin \theta e_2$ is in the Pin group. Clearly, `$$\rho_S( v(\theta)) = \begin{pmatrix} \cos \theta & \sin \theta \\ \sin \theta & - \cos \theta \end{pmatrix}.$$` Then $\rho_V(v(\theta))$ is an $8 \times 8$ matrix I don't care to write down, whose entries are degree $3$ polynomials in $\sin \theta$ and $\cos \theta$. The point is, `$$ \rho_V( v(\theta) ) \neq \cos \theta \rho_V(e_1) + \sin \theta \rho_V(e_2).$$` So $V$ is not an $A$-module. It is easy to build similar examples for the other signatures.
I'm not sure what happens if we read "representation of $A$" as "representation of the unit group of $A$."
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Representations of the unit group $\Gamma$ are representations of Pin with some rubbish attached. In general, you have a central extension $0 \to K^\times \to \Gamma \to O(V) \to 1$, and Pin is the subgroup of $\Gamma$ with spinor norm 1. In the case above, a representation of Pin becomes a representation of $\Gamma$ once you choose a commuting action of the multiplicative group of positive reals. – S. Carnahan♦ May 11 2010 at 13:23
Thanks, David! I assumed that "representation of an algebra" is generally understood as "representation of the algebra" and not "representation of the unit group", because otherwise it would be a bit pointless to talk about representations of path algebras of acyclic quivers, and similar nilpotent stuff... I'm still wondering: does this work if the Clifford algebra is over a real vector space with a real, positive-definite inner product? – darij grinberg May 11 2010 at 14:24
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Correction: My previous comment holds when $\Gamma$ is the Clifford group, which is not in general the group of units. Invertible elements need to satisfy a twisted conjugation condition to lie in the Clifford group, and this condition is typically nontrivial. Since the Clifford algebra is basically a matrix algebra, its group of units looks a lot like a general linear group (which has a lot more representations than a matrix algebra). – S. Carnahan♦ May 11 2010 at 17:12
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I think this argument works in the case that the inner product is definite (which is the case you are considering anyways). In this case, the Pin group is generated by the set of all normalized non-zero vectors. Therefore, by the universal property of Cl, a rep of Pin gives a unique rep of Cl and restricting this rep gives you back the rep of Pin you started with since they agree on unit vectors. Thus every rep of Pin is the restriction of a rep of Cl. If you know where all the $e_i$'s get mapped, you then get a unique rep of Cl which gives a unique rep of Pin.
If the inner product is not definite you will still get a rep of Cl but I can't see how you know that it will agree with the original rep of Pin.
EDIT: I have assumed that the map we get from the set of all unit vectors (by restricting the map from Pin), is the restriction of a linear map from $R^n$. This doesn't seem to be obviously the case, but is equivalent to the statement that every rep of Pin is the restriction of a rep of Cl.
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Sorry, but could you be more precise about "by the universal property of Cl"? – darij grinberg May 11 2010 at 6:43
en.wikipedia.org/wiki/… – Orbicular May 11 2010 at 6:57
So...? Why do we have a linear map from V to the endomorphism space to begin with? How do we know that the normalization of $v+w$ acts as a scalar multiple of the sum of the actions of $v$ and $w$? – darij grinberg May 11 2010 at 7:04
This has nothing to do with the signature of the inner product. – José Figueroa-O'Farrill May 11 2010 at 9:18
For the equivalence between (3) and (2), let $(\rho,V)$ denote a representation of type (3). Therefore, we can define the action of the generator $a = \sum \alpha_i e_i \in \mathbb{R}^n$ of $Pin$ as follows: $\tilde{\rho}(a) = \sum \alpha_i \rho(e_i)$. To do this, we used that $\rho(-1) = -id$, because then there is no need to insert $\rho(-1)$ if $\alpha_i < 0$. Checking that these satisfy the relations of the generators of $Pin$ is then not hard (note that the standard inner product on $\mathbb{R}^n$ makes $\rho(e_i)$ and $\rho(e_j)$ on $V$ commute and use the relations in the group in (3)). It is clear that restriction $\tilde{\rho}$ to the group in (3) gives back $\rho$ again.
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My difficulty is proving that (2) -> (3) -> (2) gives the same representation that we started with - not (3) -> (2) -> (3). – darij grinberg May 11 2010 at 10:53
David Speyer gave a very nice counterexample, so I'd like to follow up with some holistic reasons for why we should not expect such a bijection:
1.Clifford algebras are basically matrix algebras with some extra noise (see the wikipedia page), and matrix algebras have very poor representation theory. In particular, all (finite dimensional) representations of a Clifford algebra are direct sums drawn from a finite set of irreducibles, while you can take any representation of Pin where -1 acts as -Id, and take a tensor product with a representation of Pin where -1 acts as Id (i.e., any representation of the orthogonal group) to get something new. David used the particular example of the tensor square of a spinor rep. These group representations are therefore parametrized by something about as big as the monoid semiring $\mathbb{N}[\Lambda^+]$ on the dominant integral weights.
2.If you want a representation of Pin to come from a Clifford algebra, you need the restriction to a maximal torus to have a very specific set of characters. Otherwise, linearity gets violated. You can in fact easily classify such characters, since Clifford algebras have very few irreducible representations. The characters span a finite dimensional vector space, unlike the characters of suitable representations of Pin.
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Your reason 1 was exactly the reason why I had doubts about the statement. Otherwise I would have accepted Roe's "The proof is obvious"... – darij grinberg May 11 2010 at 15:43
It would have been nice if you had included some indication to that effect in the statement of the question. It looks like many people were thrown off by the expectation that the statement was true. – S. Carnahan♦ May 11 2010 at 16:27
You are right: Reading my first post again, I don't see an indication that I disbelieved the assertion. I didn't expect I could ever fail to be negative enough... – darij grinberg May 11 2010 at 17:11
Thanks to everyone who posted here. It is not "obvious" to me what I was thinking of here, and I'm embarrassed that this argument has stood unchanged in the book since the first edition in 1988 or so. I appreciate your pointing the issue out. There aren't any plans currently for a new edition of the book - I just dragged the old TeX files out and found that I can no longer compile them - but if one does come to be I will ensure that appropriate corrections are incorporated.
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I thought that one of the biggest advantages of TeX over proprietary typesetting systems was that it would be forever compilable (i.e. backward compatible). Is it not true??? – Victor Protsak Jul 9 2010 at 22:44
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http://mathoverflow.net/questions/57252/convex-polynomial-homogenization-and-convexity
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## Convex polynomial homogenization and convexity
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I have a polynomial that I know to be convex. If I homogenize the polynomial, is the resulting homogeneous polynomial also convex? I know that the perspective of a convex function is convex, but cannot find a citation for the homogenization of a convex function.
The function whose convexity I would like to show in the variables $(a,p_1,\dots,p_N)$ for $a > 0$ and under some bound on the magnitudes of the $p_n$'s (I suspect the bound will be $\vert p_n \vert < a$) is
$a^2 \prod_{n=1}^N \left(1 + \vert p_n \vert^2/a^2\right)$.
The function $\prod_{n=1}^N \left(1 + \vert p_n \vert^2\right)$ is log-convex and therefore convex in the $p_n$'s, but I am having difficulty showing that the original function is convex. However, the original function is a homogenized version of this convex function.
Thank you!
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Keep trying. It should work. – Deane Yang Mar 3 2011 at 14:52
But are you trying to prove convexity directly or by computing the Hessian of the homogeneous function in terms of the Hessian of the inhomogeneous function. It seems to me that the latter is a straightforward computation. – Deane Yang Mar 3 2011 at 16:25
Hi Deane, thanks for your response. I would like to prove convexity directly by invoking a result (that I presume to exist somewhere) about convexity of homogenized convex polynomials. This would be the most elegant way to show it I think. But I have also tried to show that x'Hx is >= 0 for the homogeneous function, which of course can be decomposed into a sum involving the Hessian of the inhomogeneous function plus the column/row for the variable a, but I was not successful in doing that, because it was not clear that the inner products involving x and that column+row were > 0. – Will Mar 3 2011 at 16:36
Will, I agree that there should be a direct proof but I don't know it. With the Hessian, I don't recall all the details offhand, but I believe that you can show that the last diagonal term dominates the two terms arising from the last row and column. – Deane Yang Mar 3 2011 at 18:11
## 2 Answers
It is definitely not true that the homogenization of a convex polynomial is convex. In fact, any convex polynomial that is not nonnegative will no longer be convex after homogenization. (This is because homogenization preserves lack of nonnegativity and convex homogeneous polynomials are always nonnegative.)
But even if the polynomial is nonnegative, the statement is still not true. Take e.g. the univariate polynomial $10x_1^4-5x_1+2$, which is convex and nonnegative, but its homogenization $10x_1^4-5x_1x_2^3+2x_2^4$ is not convex.
(Of course, for your specific polynomial the statement can be true, but I'm answering the general question that you raised.)
What is true generally is that the homogenization of a polynomial of degree d is convex if and only if the d-th root of the polynomial is convex. See Proposition 4.4 on p. 13 of the following nice paper of Reznick for the precise statement and a proof: http://www.math.uiuc.edu/~reznick/blenders2.pdf
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Amir - thanks for your nice answer. I did end up figuring this one out:
The function $a_0^2 \prod_{j=1}^{N_t} \left( 1 + \vert p_j \vert^2/a_0^2 \right)$ is convex under certain bounds on $a_0$ and the $p_j$. Consider the function \begin{equation} \label{eq:sqrtabmag} \sqrt{\prod_{j=1}^{N_t} \left( 1 + \vert p_j \vert^2 \right)}, \end{equation} whose logarithm is: \begin{equation} \label{eq:logsqrtabmag} \frac{1}{2}\sum_{j=1}^{N_t} \log\left(1 + \vert p_j \vert^2 \right). \end{equation} The logarithm's Hessian matrix is diagonal with entries \begin{equation} \begin{array}{ll} \frac{1 - \vert p_j \vert^2}{( 1 + \vert p_j \vert^2 )^2}, & j=1,\dots,N_t, \end{array} \end{equation} which are greater than or equal to zero if $\vert p_j \vert \leq 1$, so the logarithm is convex if $\vert p_j \vert \leq 1$. Because log-convex functions are themselves convex (Ref. \cite{Boyd:ConvexOpt}, \textsection 3.5.1), $\sqrt{\prod_{j=1}^{N_t} \left( 1 + \vert p_j \vert^2 \right)}$ is also convex if $\vert p_j \vert \leq 1$. Convexity of the function $a_0 \sqrt{\prod_{j=1}^{N_t} \left( 1 + \vert p_j \vert^2 / a_0^2 \right)}$ for $a_0 > 0$ and $\vert p_j \vert \leq a_0$ follows from the fact that the perspective of a convex function is convex (Ref. \cite{Boyd:ConvexOpt}, \textsection 3.2.6). Finally, because a nonnegative convex function raised to a power $b \geq 1$ is convex (Ref. \cite{Boyd:ConvexOpt}, \textsection 3.2.4), the original function $a_0^2 \prod_{j=1}^{N_t} \left( 1 + \vert p_j \vert^2 / a_0^2 \right)$ is convex.
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http://en.wikipedia.org/wiki/Inertial_frames
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# Inertial frame of reference
(Redirected from Inertial frames)
It has been suggested that be merged into this article. (Discuss) Proposed since March 2013.
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In physics, an inertial frame of reference (also inertial reference frame or inertial frame or Galilean reference frame) is a frame of reference that describes time and space homogeneously, isotropically, and in a time-independent manner.[1]
All inertial frames are in a state of constant, rectilinear motion with respect to one another; an accelerometer moving with any of them would detect zero acceleration. Measurements in one inertial frame can be converted to measurements in another by a simple transformation (the Galilean transformation in Newtonian physics and the Lorentz transformation in special relativity). In general relativity, in any region small enough for the curvature of spacetime to be negligible, one can find a set of inertial frames that approximately describe that region.[2][3]
Physical laws take the same form in all inertial frames.[4] By contrast, in a non-inertial reference frame the laws of physics vary depending on the acceleration of that frame with respect to an inertial frame, and the usual physical forces must be supplemented by fictitious forces.[5][6] For example, a ball dropped towards the ground does not go exactly straight down because the Earth is rotating. Someone rotating with the Earth must account for the Coriolis effect—in this case thought of as a force—to predict the horizontal motion. Another example of such a fictitious force associated with rotating reference frames is the centrifugal effect, or centrifugal force.
## Introduction
The motion of a body can only be described relative to something else - other bodies, observers, or a set of space-time coordinates. These are called frames of reference. If the coordinates are chosen badly, the laws of motion may be more complex than necessary. For example, suppose a free body (one having no external forces on it) is at rest at some instant. In many coordinate systems, it would begin to move at the next instant, even though there are no forces on it. However, a frame of reference can always be chosen in which it remains stationary. Similarly, if space is not described uniformly or time independently, a coordinate system could describe the simple flight of a free body in space as a complicated zig-zag in its coordinate system. Indeed, an intuitive summary of inertial frames can be given as: In an inertial reference frame, the laws of mechanics take their simplest form.[1]
In an inertial frame, Newton's first law (the law of inertia) is satisfied: Any free motion has a constant magnitude and direction.[1] Newton's second law for a particle takes the form:
$\mathbf{F} = m \mathbf{a} \ ,$
with F the net force (a vector), m the mass of a particle and a the acceleration of the particle (also a vector) which would be measured by an observer at rest in the frame. The force F is the vector sum of all "real" forces on the particle, such as electromagnetic, gravitational, nuclear and so forth. In contrast, Newton's second law in a rotating frame of reference, rotating at angular rate Ω about an axis, takes the form:
$\mathbf{F}' = m \mathbf{a} \ ,$
which looks the same as in an inertial frame, but now the force F′ is the resultant of not only F, but also additional terms (the paragraph following this equation presents the main points without detailed mathematics):
$\mathbf{F}' = \mathbf{F} - 2m \mathbf{\Omega} \times \mathbf{v}_{B} - m \mathbf{\Omega} \times (\mathbf{\Omega} \times \mathbf{x}_B ) - m \frac{d \mathbf{\Omega}}{dt} \times \mathbf{x}_B \ ,$
where the angular rotation of the frame is expressed by the vector Ω pointing in the direction of the axis of rotation, and with magnitude equal to the angular rate of rotation Ω, symbol × denotes the vector cross product, vector xB locates the body and vector vB is the velocity of the body according to a rotating observer (different from the velocity seen by the inertial observer).
The extra terms in the force F′ are the "fictitious" forces for this frame. (The first extra term is the Coriolis force, the second the centrifugal force, and the third the Euler force.) These terms all have these properties: they vanish when Ω = 0; that is, they are zero for an inertial frame (which, of course, does not rotate); they take on a different magnitude and direction in every rotating frame, depending upon its particular value of Ω; they are ubiquitous in the rotating frame (affect every particle, regardless of circumstance); and they have no apparent source in identifiable physical sources, in particular, matter. Also, fictitious forces do not drop off with distance (unlike, for example, nuclear forces or electrical forces). For example, the centrifugal force that appears to emanate from the axis of rotation in a rotating frame increases with distance from the axis.
All observers agree on the real forces, F; only non-inertial observers need fictitious forces. The laws of physics in the inertial frame are simpler because unnecessary forces are not present.
In Newton's time the fixed stars were invoked as a reference frame, supposedly at rest relative to absolute space. In reference frames that were either at rest with respect to the fixed stars or in uniform translation relative to these stars, Newton's laws of motion were supposed to hold. In contrast, in frames accelerating with respect to the fixed stars, an important case being frames rotating relative to the fixed stars, the laws of motion did not hold in their simplest form, but had to be supplemented by the addition of fictitious forces, for example, the Coriolis force and the centrifugal force. Two interesting experiments were devised by Newton to demonstrate how these forces could be discovered, thereby revealing to an observer that they were not in an inertial frame: the example of the tension in the cord linking two spheres rotating about their center of gravity, and the example of the curvature of the surface of water in a rotating bucket. In both cases, application of Newton's second law would not work for the rotating observer without invoking centrifugal and Coriolis forces to account for their observations (tension in the case of the spheres; parabolic water surface in the case of the rotating bucket).
As we now know, the fixed stars are not fixed. Those that reside in the Milky Way turn with the galaxy, exhibiting proper motions. Those that are outside our galaxy (such as nebulae once mistaken to be stars) participate in their own motion as well, partly due to expansion of the universe, and partly due to peculiar velocities.[7] (The Andromeda galaxy is on collision course with the Milky Way at a speed of 117 km/s.[8]) The concept of inertial frames of reference is no longer tied to either the fixed stars or to absolute space. Rather, the identification of an inertial frame is based upon the simplicity of the laws of physics in the frame. In particular, the absence of fictitious forces is their identifying property.[9]
In practice, although not a requirement, using a frame of reference based upon the fixed stars as though it were an inertial frame of reference introduces very little discrepancy. For example, the centrifugal acceleration of the Earth because of its rotation about the Sun is about thirty million times greater than that of the Sun about the galactic center.[10]
To illustrate further, consider the question: "Does our Universe rotate?" To answer, we might attempt to explain the shape of the Milky Way galaxy using the laws of physics.[11] (Other observations might be more definitive (that is, provide larger discrepancies or less measurement uncertainty), like the anisotropy of the microwave background radiation or Big Bang nucleosynthesis.[12][13]) Just how flat the disc of the Milky Way is depends on its rate of rotation in an inertial frame of reference. If we attribute its apparent rate of rotation entirely to rotation in an inertial frame, a different "flatness" is predicted than if we suppose part of this rotation actually is due to rotation of the Universe and should not be included in the rotation of the galaxy itself. Based upon the laws of physics, a model is set up in which one parameter is the rate of rotation of the Universe. If the laws of physics agree more accurately with observations in a model with rotation than without it, we are inclined to select the best-fit value for rotation, subject to all other pertinent experimental observations. If no value of the rotation parameter is successful and theory is not within observational error, a modification of physical law is considered. (For example, dark matter is invoked to explain the galactic rotation curve.) So far, observations show any rotation of the Universe is very slow (no faster than once every 60·1012 years (10−13 rad/yr)[14]), and debate persists over whether there is any rotation. However, if rotation were found, interpretation of observations in a frame tied to the Universe would have to be corrected for the fictitious forces inherent in such rotation. Evidently, such an approach adopts the view that "an inertial frame of reference is one where our laws of physics apply" (or need the least modification).
## Background
A brief comparison of inertial frames in special relativity and in Newtonian mechanics, and the role of absolute space is next.
### A set of frames where the laws of physics are simple
According to the first postulate of special relativity, all physical laws take their simplest form in an inertial frame, and there exist multiple inertial frames interrelated by uniform translation: [15]
Special principle of relativity: If a system of coordinates K is chosen so that, in relation to it, physical laws hold good in their simplest form, the same laws hold good in relation to any other system of coordinates K' moving in uniform translation relatively to K.
—Albert Einstein: The foundation of the general theory of relativity, Section A, §1
The principle of simplicity can be used within Newtonian physics as well as in special relativity; see Nagel[16] and also Blagojević.[17]
The laws of Newtonian mechanics do not always hold in their simplest form...If, for instance, an observer is placed on a disc rotating relative to the earth, he/she will sense a 'force' pushing him/her toward the periphery of the disc, which is not caused by any interaction with other bodies. Here, the acceleration is not the consequence of the usual force, but of the so-called inertial force. Newton's laws hold in their simplest form only in a family of reference frames, called inertial frames. This fact represents the essence of the Galilean principle of relativity:
The laws of mechanics have the same form in all inertial frames.
—Milutin Blagojević: Gravitation and Gauge Symmetries, p. 4
In practical terms, the equivalence of inertial reference frames means that scientists within a box moving uniformly cannot determine their absolute velocity by any experiment (otherwise the differences would set up an absolute standard reference frame).[18][19] According to this definition, supplemented with the constancy of the speed of light, inertial frames of reference transform among themselves according to the Poincaré group of symmetry transformations, of which the Lorentz transformations are a subgroup.[20] In Newtonian mechanics, which can be viewed as a limiting case of special relativity in which the speed of light is infinite, inertial frames of reference are related by the Galilean group of symmetries.
### Absolute space
Main article: Absolute space and time
Newton posited an absolute space considered well approximated by a frame of reference stationary relative to the fixed stars. An inertial frame was then one in uniform translation relative to absolute space. However, some scientists (called "relativists" by Mach[21]), even at the time of Newton, felt that absolute space was a defect of the formulation, and should be replaced.
Indeed, the expression inertial frame of reference (German: Inertialsystem) was coined by Ludwig Lange in 1885, to replace Newton's definitions of "absolute space and time" by a more operational definition.[22][23] As referenced by Iro, Lange proposed:[24]
A reference frame in which a mass point thrown from the same point in three different (non co-planar) directions follows rectilinear paths each time it is thrown, is called an inertial frame.
A discussion of Lange's proposal can be found in Mach.[21]
The inadequacy of the notion of "absolute space" in Newtonian mechanics is spelled out by Blagojević:[25]
• The existence of absolute space contradicts the internal logic of classical mechanics since, according to Galilean principle of relativity, none of the inertial frames can be singled out.
• Absolute space does not explain inertial forces since they are related to acceleration with respect to any one of the inertial frames.
• Absolute space acts on physical objects by inducing their resistance to acceleration but it cannot be acted upon.
— Milutin Blagojević: Gravitation and Gauge Symmetries, p. 5
The utility of operational definitions was carried much further in the special theory of relativity.[26] Some historical background including Lange's definition is provided by DiSalle, who says in summary:[27]
The original question, "relative to what frame of reference do the laws of motion hold?" is revealed to be wrongly posed. For the laws of motion essentially determine a class of reference frames, and (in principle) a procedure for constructing them.
## Newton's inertial frame of reference
Figure 1: Two frames of reference moving with relative velocity $\stackrel{\vec v}{}$. Frame S' has an arbitrary but fixed rotation with respect to frame S. They are both inertial frames provided a body not subject to forces appears to move in a straight line. If that motion is seen in one frame, it will also appear that way in the other.
Within the realm of Newtonian mechanics, an inertial frame of reference, or inertial reference frame, is one in which Newton's first law of motion is valid.[28] However, the principle of special relativity generalizes the notion of inertial frame to include all physical laws, not simply Newton's first law.
Newton viewed the first law as valid in any reference frame that is in uniform motion relative to the fixed stars;[29] that is, neither rotating nor accelerating relative to the stars.[30] Today the notion of "absolute space" is abandoned, and an inertial frame in the field of classical mechanics is defined as:[31][32]
An inertial frame of reference is one in which the motion of a particle not subject to forces is in a straight line at constant speed.
Hence, with respect to an inertial frame, an object or body accelerates only when a physical force is applied, and (following Newton's first law of motion), in the absence of a net force, a body at rest will remain at rest and a body in motion will continue to move uniformly—that is, in a straight line and at constant speed. Newtonian inertial frames transform among each other according to the Galilean group of symmetries.
If this rule is interpreted as saying that straight-line motion is an indication of zero net force, the rule does not identify inertial reference frames, because straight-line motion can be observed in a variety of frames. If the rule is interpreted as defining an inertial frame, then we have to be able to determine when zero net force is applied. The problem was summarized by Einstein:[33]
The weakness of the principle of inertia lies in this, that it involves an argument in a circle: a mass moves without acceleration if it is sufficiently far from other bodies; we know that it is sufficiently far from other bodies only by the fact that it moves without acceleration.
—Albert Einstein: The Meaning of Relativity, p. 58
There are several approaches to this issue. One approach is to argue that all real forces drop off with distance from their sources in a known manner, so we have only to be sure that we are far enough away from all sources to ensure that no force is present.[34] A possible issue with this approach is the historically long-lived view that the distant universe might affect matters (Mach's principle). Another approach is to identify all real sources for real forces and account for them. A possible issue with this approach is that we might miss something, or account inappropriately for their influence (Mach's principle again?). A third approach is to look at the way the forces transform when we shift reference frames. Fictitious forces, those that arise due to the acceleration of a frame, disappear in inertial frames, and have complicated rules of transformation in general cases. On the basis of universality of physical law and the request for frames where the laws are most simply expressed, inertial frames are distinguished by the absence of such fictitious forces.
Newton enunciated a principle of relativity himself in one of his corollaries to the laws of motion:[35][36]
The motions of bodies included in a given space are the same among themselves, whether that space is at rest or moves uniformly forward in a straight line.
—Isaac Newton: Principia, Corollary V, p. 88 in Andrew Motte translation
This principle differs from the special principle in two ways: first, it is restricted to mechanics, and second, it makes no mention of simplicity. It shares with the special principle the invariance of the form of the description among mutually translating reference frames.[37] The role of fictitious forces in classifying reference frames is pursued further below.
## Separating non-inertial from inertial reference frames
Main article: Fictitious force
See also: Non-inertial frame, Rotating spheres, and Bucket argument
Figure 2: Two spheres tied with a string and rotating at an angular rate ω. Because of the rotation, the string tying the spheres together is under tension.
Figure 3: Exploded view of rotating spheres in an inertial frame of reference showing the centripetal forces on the spheres provided by the tension in the tying string.
Inertial and non-inertial reference frames can be distinguished by the absence or presence of fictitious forces, as explained shortly.[5][6]
The effect of this being in the noninertial frame is to require the observer to introduce a fictitious force into his calculations….
—Sidney Borowitz and Lawrence A Bornstein in A Contemporary View of Elementary Physics, p. 138
The presence of fictitious forces indicates the physical laws are not the simplest laws available so, in terms of the special principle of relativity, a frame where fictitious forces are present is not an inertial frame:[38]
The equations of motion in a non-inertial system differ from the equations in an inertial system by additional terms called inertial forces. This allows us to detect experimentally the non-inertial nature of a system.
—V. I. Arnol'd: Mathematical Methods of Classical Mechanics Second Edition, p. 129
Bodies in non-inertial reference frames are subject to so-called fictitious forces (pseudo-forces); that is, forces that result from the acceleration of the reference frame itself and not from any physical force acting on the body. Examples of fictitious forces are the centrifugal force and the Coriolis force in rotating reference frames.
How then, are "fictitious" forces to be separated from "real" forces? It is hard to apply the Newtonian definition of an inertial frame without this separation. For example, consider a stationary object in an inertial frame. Being at rest, no net force is applied. But in a frame rotating about a fixed axis, the object appears to move in a circle, and is subject to centripetal force (which is made up of the Coriolis force and the centrifugal force). How can we decide that the rotating frame is a non-inertial frame? There are two approaches to this resolution: one approach is to look for the origin of the fictitious forces (the Coriolis force and the centrifugal force). We will find there are no sources for these forces, no associated force carriers, no originating bodies.[39] A second approach is to look at a variety of frames of reference. For any inertial frame, the Coriolis force and the centrifugal force disappear, so application of the principle of special relativity would identify these frames where the forces disappear as sharing the same and the simplest physical laws, and hence rule that the rotating frame is not an inertial frame.
Newton examined this problem himself using rotating spheres, as shown in Figure 2 and Figure 3. He pointed out that if the spheres are not rotating, the tension in the tying string is measured as zero in every frame of reference.[40] If the spheres only appear to rotate (that is, we are watching stationary spheres from a rotating frame), the zero tension in the string is accounted for by observing that the centripetal force is supplied by the centrifugal and Coriolis forces in combination, so no tension is needed. If the spheres really are rotating, the tension observed is exactly the centripetal force required by the circular motion. Thus, measurement of the tension in the string identifies the inertial frame: it is the one where the tension in the string provides exactly the centripetal force demanded by the motion as it is observed in that frame, and not a different value. That is, the inertial frame is the one where the fictitious forces vanish.
So much for fictitious forces due to rotation. However, for linear acceleration, Newton expressed the idea of undetectability of straight-line accelerations held in common:[36]
If bodies, any how moved among themselves, are urged in the direction of parallel lines by equal accelerative forces, they will continue to move among themselves, after the same manner as if they had been urged by no such forces.
—Isaac Newton: Principia Corollary VI, p. 89, in Andrew Motte translation
This principle generalizes the notion of an inertial frame. For example, an observer confined in a free-falling lift will assert that he himself is a valid inertial frame, even if he is accelerating under gravity, so long as he has no knowledge about anything outside the lift. So, strictly speaking, inertial frame is a relative concept. With this in mind, we can define inertial frames collectively as a set of frames which are stationary or moving at constant velocity with respect to each other, so that a single inertial frame is defined as an element of this set.
For these ideas to apply, everything observed in the frame has to be subject to a base-line, common acceleration shared by the frame itself. That situation would apply, for example, to the elevator example, where all objects are subject to the same gravitational acceleration, and the elevator itself accelerates at the same rate.
## Newtonian mechanics
Main article: Newton's laws of motion
Classical mechanics, which includes relativity, assumes the equivalence of all inertial reference frames. Newtonian mechanics makes the additional assumptions of absolute space and absolute time. Given these two assumptions, the coordinates of the same event (a point in space and time) described in two inertial reference frames are related by a Galilean transformation.
$\mathbf{r}^{\prime} = \mathbf{r} - \mathbf{r}_{0} - \mathbf{v} t$
$t^{\prime} = t - t_{0}$
where r0 and t0 represent shifts in the origin of space and time, and v is the relative velocity of the two inertial reference frames. Under Galilean transformations, the time t2 − t1 between two events is the same for all inertial reference frames and the distance between two simultaneous events (or, equivalently, the length of any object, |r2 − r1|) is also the same.
## Special relativity
Main articles: Special relativity and Introduction to special relativity
Einstein's theory of special relativity, like Newtonian mechanics, assumes the equivalence of all inertial reference frames, but makes an additional assumption, foreign to Newtonian mechanics, namely, that in free space light always is propagated with the speed of light c0, a defined value independent of its direction of propagation and its frequency, and also independent of the state of motion of the emitting body. This second assumption has been verified experimentally and leads to counter-intuitive deductions including:
• time dilation (moving clocks tick more slowly)
• length contraction (moving objects are shortened in the direction of motion)
• relativity of simultaneity (simultaneous events in one reference frame are not simultaneous in almost all frames moving relative to the first).
These deductions are logical consequences of the stated assumptions, and are general properties of space-time, typically without regard to a consideration of properties pertaining to the structure of individual objects like atoms or stars, nor to the mechanisms of clocks.
These effects are expressed mathematically by the Lorentz transformation
$x^{\prime} = \gamma \left(x - v t \right)$
$y^{\prime} = y$
$z^{\prime} = z$
$t^{\prime} = \gamma \left(t - \frac{v x}{c_0^{2}}\right)$
where shifts in origin have been ignored, the relative velocity is assumed to be in the $x$-direction and the Lorentz factor γ is defined by:
$\gamma \ \stackrel{\mathrm{def}}{=}\ \frac{1}{\sqrt{1 - (v/c_0)^2}} \ \ge 1.$
The Lorentz transformation is equivalent to the Galilean transformation in the limit c0 → ∞ (a hypothetical case) or v → 0 (low speeds).
Under Lorentz transformations, the time and distance between events may differ among inertial reference frames; however, the Lorentz scalar distance s between two events is the same in all inertial reference frames
$s^{2} = \left( x_{2} - x_{1} \right)^{2} + \left( y_{2} - y_{1} \right)^{2} + \left( z_{2} - z_{1} \right)^{2} - c_0^{2} \left(t_{2} - t_{1}\right)^{2}$
From this perspective, the speed of light is only accidentally a property of light, and is rather a property of spacetime, a conversion factor between conventional time units (such as seconds) and length units (such as meters).
Incidentally, because of the limitations on speeds faster than the speed of light, notice that a rotating frame of reference (which is a non-inertial frame, of course) cannot be used out to arbitrary distances because at large radius its components would move faster than the speed of light.[41]
## General relativity
Main articles: General relativity and Introduction to general relativity
See also: Equivalence principle and Eötvös experiment
General relativity is based upon the principle of equivalence:[42][43]
There is no experiment observers can perform to distinguish whether an acceleration arises because of a gravitational force or because their reference frame is accelerating.
—Douglas C. Giancoli, Physics for Scientists and Engineers with Modern Physics, p. 155.
This idea was introduced in Einstein's 1907 article "Principle of Relativity and Gravitation" and later developed in 1911.[44] Support for this principle is found in the Eötvös experiment, which determines whether the ratio of inertial to gravitational mass is the same for all bodies, regardless of size or composition. To date no difference has been found to a few parts in 1011.[45] For some discussion of the subtleties of the Eötvös experiment, such as the local mass distribution around the experimental site (including a quip about the mass of Eötvös himself), see Franklin.[46]
Einstein’s general theory modifies the distinction between nominally "inertial" and "noninertial" effects by replacing special relativity's "flat" Euclidean geometry with a curved metric. In general relativity, the principle of inertia is replaced with the principle of geodesic motion, whereby objects move in a way dictated by the curvature of spacetime. As a consequence of this curvature, it is not a given in general relativity that inertial objects moving at a particular rate with respect to each other will continue to do so. This phenomenon of geodesic deviation means that inertial frames of reference do not exist globally as they do in Newtonian mechanics and special relativity.
However, the general theory reduces to the special theory over sufficiently small regions of spacetime, where curvature effects become less important and the earlier inertial frame arguments can come back into play. Consequently, modern special relativity is now sometimes described as only a "local theory". (However, this refers to the theory’s application rather than to its derivation.)
## References
1. ^ a b c Landau, L. D.; Lifshitz, E. M. (1960). Mechanics. Pergamon Press. pp. 4–6.
2. Albert Einstein (2001) [Reprint of edition of 1920 translated by RQ Lawson]. Relativity: The Special and General Theory (3rd ed.). Courier Dover Publications. p. 71. ISBN 0-486-41714-X.
3. Domenico Giulini (2005). Special Relativity. Cambridge University Press. p. 19. ISBN 0-19-856746-4.
4. ^ a b Milton A. Rothman (1989). Discovering the Natural Laws: The Experimental Basis of Physics. Courier Dover Publications. p. 23. ISBN 0-486-26178-6.
5. ^ a b Sidney Borowitz & Lawrence A. Bornstein (1968). A Contemporary View of Elementary Physics. McGraw-Hill. p. 138. ASIN B000GQB02A.
6. Amedeo Balbi (2008). The Music of the Big Bang. Springer. p. 59. ISBN 3-540-78726-7.
7. Abraham Loeb, Mark J. Reid, Andreas Brunthaler, Heino Falcke (2005). "Constraints on the proper motion of the Andromeda galaxy based on the survival of its satellite M33". The Astrophysical Journal 633 (2): 894–898. arXiv:astro-ph/0506609. Bibcode:2005ApJ...633..894L. doi:10.1086/491644.
8. John J. Stachel (2002). Einstein from "B" to "Z". Springer. pp. 235–236. ISBN 0-8176-4143-2.
9. Peter Graneau & Neal Graneau (2006). In the Grip of the Distant Universe. World Scientific. p. 147. ISBN 981-256-754-2.
10. Henning Genz (2001). Nothingness. Da Capo Press. p. 275. ISBN 0-7382-0610-5.
11. J Garcio-Bellido (2005). "The Paradigm of Inflation". In J. M. T. Thompson. Advances in Astronomy. Imperial College Press. p. 32, §9. ISBN 1-86094-577-5.
12. Wlodzimierz Godlowski and Marek Szydlowski (2003). "Dark energy and global rotation of the Universe". General Relativity and Gravitation 35 (12): 2171. arXiv:astro-ph/0303248. Bibcode:2003GReGr..35.2171G. doi:10.1023/A:1027301723533.
13. P Birch Is the Universe rotating? Nature 298, 451 - 454 (29 July 1982)
14. Einstein, A., Lorentz, H. A., Minkowski, H., & Weyl, H. (1952). The Principle of Relativity: a collection of original memoirs on the special and general theory of relativity. Courier Dover Publications. p. 111. ISBN 0-486-60081-5.
15. Ernest Nagel (1979). The Structure of Science. Hackett Publishing. p. 212. ISBN 0-915144-71-9.
16. Milutin Blagojević (2002). Gravitation and Gauge Symmetries. CRC Press. p. 4. ISBN 0-7503-0767-6.
17.
18. Richard Phillips Feynman (1998). Six not-so-easy pieces: Einstein's relativity, symmetry, and space-time. Basic Books. p. 73. ISBN 0-201-32842-9.
19. Armin Wachter & Henning Hoeber (2006). Compendium of Theoretical Physics. Birkhäuser. p. 98. ISBN 0-387-25799-3.
20. ^ a b
21. Lange, Ludwig (1885). "Über die wissenschaftliche Fassung des Galileischen Beharrungsgesetzes". Philosophische Studien 2.
22. Julian B. Barbour (2001). The Discovery of Dynamics (Reprint of 1989 Absolute or Relative Motion? ed.). Oxford University Press. pp. 645–646. ISBN 0-19-513202-5.
23. L. Lange (1885) as quoted by Max von Laue in his book (1921) Die Relativitätstheorie, p. 34, and translated by Harald Iro (2002). A Modern Approach to Classical Mechanics. World Scientific. p. 169. ISBN 981-238-213-5.
24. Milutin Blagojević (2002). Gravitation and Gauge Symmetries. CRC Press. p. 5. ISBN 0-7503-0767-6.
25. NMJ Woodhouse (2003). Special relativity. London: Springer. p. 58. ISBN 1-85233-426-6.
26. Robert DiSalle (Summer 2002). "Space and Time: Inertial Frames". In Edward N. Zalta. The Stanford Encyclopedia of Philosophy.
27. C Møller (1976). The Theory of Relativity (Second ed.). Oxford UK: Oxford University Press. p. 1. ISBN 0-19-560539-X.
28. The question of "moving uniformly relative to what?" was answered by Newton as "relative to absolute space". As a practical matter, "absolute space" was considered to be the fixed stars. For a discussion of the role of fixed stars, see Henning Genz (2001). Nothingness: The Science of Empty Space. Da Capo Press. p. 150. ISBN 0-7382-0610-5.
29. Robert Resnick, David Halliday, Kenneth S. Krane (2001). Physics (5th ed.). Wiley. Volume 1, Chapter 3. ISBN 0-471-32057-9.
30. RG Takwale (1980). Introduction to classical mechanics. New Delhi: Tata McGraw-Hill. p. 70. ISBN 0-07-096617-6.
31. NMJ Woodhouse (2003). Special relativity. London/Berlin: Springer. p. 6. ISBN 1-85233-426-6.
32.
33. William Geraint Vaughan Rosser (1991). Introductory Special Relativity. CRC Press. p. 3. ISBN 0-85066-838-7.
34. Richard Phillips Feynman (1998). Six not-so-easy pieces: Einstein's relativity, symmetry, and space-time. Basic Books. p. 50. ISBN 0-201-32842-9.
35. ^ a b See the Principia on line at Andrew Motte Translation
36. However, in the Newtonian system the Galilean transformation connects these frames and in the special theory of relativity the Lorentz transformation connects them. The two transformations agree for speeds of translation much less than the speed of light.
37. V. I. Arnol'd (1989). Mathematical Methods of Classical Mechanics. Springer. p. 129. ISBN 978-0-387-96890-2.
38. For example, there is no body providing a gravitational or electrical attraction.
39. That is, the universality of the laws of physics requires the same tension to be seen by everybody. For example, it cannot happen that the string breaks under extreme tension in one frame of reference and remains intact in another frame of reference, just because we choose to look at the string from a different frame.
40. LD Landau & LM Lifshitz (1975). The Classical Theory of Fields (4rth Revised English ed.). Pergamon Press. pp. 273–274. ISBN 978-0-7506-2768-9.
41. David Morin (2008). Introduction to Classical Mechanics. Cambridge University Press. p. 649. ISBN 0-521-87622-2.
42. Douglas C. Giancoli (2007). Physics for Scientists and Engineers with Modern Physics. Pearson Prentice Hall. p. 155. ISBN 0-13-149508-9.
43. A. Einstein, "On the influence of gravitation on the propagation of light", Annalen der Physik, vol. 35, (1911) : 898-908
44. National Research Council (US) (1986). Physics Through the Nineteen Nineties: Overview. National Academies Press. p. 15. ISBN 0-309-03579-1.
45. Allan Franklin (2007). No Easy Answers: Science and the Pursuit of Knowledge. University of Pittsburgh Press. p. 66. ISBN 0-8229-5968-2.
## Further reading
• Edwin F. Taylor and John Archibald Wheeler, Spacetime Physics, 2nd ed. (Freeman, NY, 1992)
• Albert Einstein, Relativity, the special and the general theories, 15th ed. (1954)
• Henri Poincaré, (1900) "La théorie de Lorentz et le Principe de Réaction", Archives Neerlandaises, V, 253–78.
• Albert Einstein, On the Electrodynamics of Moving Bodies, included in The Principle of Relativity, page 38. Dover 1923
Rotation of the Universe
• Julian B. Barbour, Herbert Pfister (1998). Mach's Principle: From Newton's Bucket to Quantum Gravity. Birkhäuser. p. 445. ISBN 0-8176-3823-7.
• PJ Nahin (1999). Time Machines. Springer. p. 369; Footnote 12. ISBN 0-387-98571-9.
• B Ciobanu, I Radinchi Modeling the electric and magnetic fields in a rotating universe Rom. Journ. Phys., Vol. 53, Nos. 1–2, P. 405–415, Bucharest, 2008
• Yuri N. Obukhov, Thoralf Chrobok, Mike Scherfner Shear-free rotating inflation Phys. Rev. D 66, 043518 (2002) [5 pages]
• Yuri N. Obukhov On physical foundations and observational effects of cosmic rotation (2000)
• Li-Xin Li Effect of the Global Rotation of the Universe on the Formation of Galaxies General Relativity and Gravitation, 30 (1998) doi:10.1023/A:1018867011142
• P Birch Is the Universe rotating? Nature 298, 451 - 454 (29 July 1982)
• Kurt Gödel An example of a new type of cosmological solutions of Einstein’s field equations of gravitation Rev. Mod. Phys., Vol. 21, p. 447, 1949.
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http://stats.stackexchange.com/questions/21237/calculating-statistical-power
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# Calculating statistical power
As I understand it, I need to know at least three aspects (out of four) of my proposed study in order to conduct power analysis, namely:
• type of test - I intend to use Pearson's r and ANCOVA/Regression - GLM
• significance level (alpha) - I intend to use 0.05
• expected effect size - I intend to use a medium effect size (0.5)
• sample size
Could anyone recommend a good online power calculator that I can use to do a priori power calculation. (Can SPSS do a priori power calculation?)
I have come across GPower but I am looking for a simplier tool!
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Unfortunately SPSS package does not include a module for power analysis. IBM SPSS company sells a separate program for power analysis. – ttnphns Jan 17 '12 at 14:51
3
I'd give GPower a chance. With 20 or 30 minutes exploring it, you'll probably find it's very manageable--at least for procedures like correlation, not necessarily for a complicated regression model. – rolando2 Jan 17 '12 at 22:52
Thanks! Is there a user-friendly guide available on GPower? – Adhesh Josh Jan 18 '12 at 3:45
This looks like it is for a grant application. These are vexing to produce and to evaluate. For well used experimental designs (genome-wide association studies for example) there may be well documented specialised calculators. Otherwise, I think G. Jay Kerns answer is the right way to go with the following addition: while you are at it you should simulate a range of the most important parameters and present a graph. – Leo Schalkwyk Jan 24 '12 at 9:55
## 1 Answer
This isn't an answer you are going to want to hear, I am afraid, but I am going to say it anyway: try to resist the temptation of online calculators (and save your money before purchasing proprietary calculators).
Here are some of the reasons why: 1) online calculators all use different notation and are often poorly documented. It is a waste of your time. 2) SPSS does offer a power calculator but I've never even tried it because it was too expensive for my department to afford! 3) Phrases like "medium effect size" are at best misleading and at worst just plain wrong for all but the simplest research designs. There are too many parameters and too much interplay to be able to distill effect size down to a single number in [0,1]. Even if you could put it into a single number, there's no guarantee that Cohen's 0.5 corresponds to "medium" in the context of the problem.
Believe me - it is better in the long run to bite the bullet and teach yourself how to use simulation to your benefit (and the benefit of the person(s) you're consulting). Sit down with them and complete the following steps:
1) Decide on a model that is appropriate in the context of the problem (sounds like you've already worked on this part).
2) Consult with them to decide what the null parameters should be, the behavior of the control group, whatever this means in context of the problem.
3) Consult with them to determine what the parameters should be in order for the difference to be practically meaningful. If there are sample size limitations then this should be identified here, as well.
4) Simulate data according to the two models in 2) and 3), and run your test. You can do this with software galore - pick your favorite and go for it. See if you rejected or not.
5) Repeat 4) thousands of times, say, $n$. Keep track of how many times you rejected, and the sample proportion $\hat{p}$ of rejections is an estimate of power. This estimate has standard error approximately $\sqrt{\hat{p}(1 - \hat{p})/n}$.
If you do your power analysis this way, you are going to find several things: A) there were a lot more parameters running around than you ever anticipated. It will make you wonder how in the world it's possible to collapse all of them into a single number like "medium" - and you will see that it isn't possible, at least not in any straightforward way. B) your power is going to be a lot smaller than a lot of the other calculators advertise. C) you can increase power by increasing sample size, but watch out! You may find as I have that in order to detect a difference that's "practically meaningful" you need a sample size that's prohibitively large.
If you have trouble with any of the above steps you could collect your thoughts, well-formulate a question for CrossValidated, and the people here will help you.
EDIT: In the case you find that you absolutely must use an online calculator, the best one I've found is Russ Lenth's Power and Sample Size page. It's been around for a long time, it has relatively complete documentation, it doesn't depend on canned effect sizes, and has links to other papers which are relevant and important.
ANOTHER EDIT: Coincidentally, when this question came up I was right in the middle of writing a blog post to flesh out some of these ideas (otherwise, I might not have answered so quickly). Anyway, I finished it last weekend and you can find it here. It is not written with SPSS in mind, but I'd bet if a person were clever they might be able to translate portions of it to SPSS syntax.
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5
+1 Good answer. It's worth pointing out the drawbacks of simulation. (The alternative is that power curves can be computed mathematically.) Simulation becomes unwieldy when many parameters (such as effect size and sample size) have to be manipulated or when you are seeking some threshold value, such as a minimum sample size. Even an approximate exact expression for the power can be valuable for indicating in general how the power behaves and for identifying initial solutions that can be polished with a little bit of simulation. – whuber♦ Jan 17 '12 at 16:19
2
@whuber Thanks, and you are absolutely right. Your comment reminds me that there's often addt'l uncertainty in the null/alt parameters (scant info, crummy pilot studies, etc.) which adds another layer of complexity to the simulation approach. This is another benefit of the mathematical approach. – G. Jay Kerns Jan 17 '12 at 17:01
1
Instead of fixing the values of the unknown parameters it is useful to simulate them by assigning a prior distribution on these parameters and then to get a "prior power" (this is not a Bayesian approach, in spite of the concept of prior distribution, because we simulate the result of the frequentist test) – Stéphane Laurent Jan 17 '12 at 19:52
4
There are two problems with simulation: Learning it (this one is soluble) and getting step 3 done. In my experience, none of my clients would be willing to do 3). Many have trouble specifying ANY effect size whatsoever. To ask them to specify the parameters in (say) a multiple regression equation would be .... well, they wouldn't know how to answer, even if they know the meaning, they won't be willing to specify. – Peter Flom Jan 17 '12 at 20:36
1
Stephane yes, you are right, and that was what I meant by the extra layer I was trying to communicate. @Peter Sigh! yes, I've encountered this, too. I try to talk about means, standard errors, etc and then work out as much of the math as I can afterward. Part of it is a communication barrier which is sometimes a challenge. The unwillingness part is even tougher, though. It used to be that I would give up and try to fill in the blanks myself, but it rarely worked out well. That is, the answer's essentially a shot in the dark with a blindfold on and standing backward. – G. Jay Kerns Jan 17 '12 at 21:28
show 3 more comments
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http://en.wikipedia.org/wiki/Binary_symmetric_channel
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# Binary symmetric channel
A binary symmetric channel (or BSC) is a common communications channel model used in coding theory and information theory. In this model, a transmitter wishes to send a bit (a zero or a one), and the receiver receives a bit. It is assumed that the bit is usually transmitted correctly, but that it will be "flipped" with a small probability (the "crossover probability"). This channel is used frequently in information theory because it is one of the simplest channels to analyze.
## Description
The BSC is a binary channel; that is, it can transmit only one of two symbols (usually called 0 and 1). (A non-binary channel would be capable of transmitting more than 2 symbols, possibly even an infinite number of choices.) The transmission is not perfect, and occasionally the receiver gets the wrong bit.
This channel is often used by theorists because it is one of the simplest noisy channels to analyze. Many problems in communication theory can be reduced to a BSC. Conversely, being able to transmit effectively over the BSC can give rise to solutions for more complicated channels.
## Definition
A binary symmetric channel with crossover probability p denoted by $BSC_{p}$, is a channel with binary input and binary output and probability of error p; that is, if X is the transmitted random variable and Y the received variable, then the channel is characterized by the conditional probabilities
Pr( Y = 0 | X = 0 ) = 1 − p
Pr( Y = 0 | X = 1) = p
Pr( Y = 1 | X = 0 ) = p
Pr( Y = 1 | X = 1 ) = 1 − p
It is assumed that 0 ≤ p ≤ 1/2. If p > 1/2, then the receiver can swap the output (interpret 1 when it sees 0, and vice versa) and obtain an equivalent channel with crossover probability 1 − p ≤ 1/2.
### Capacity of BSCp
The capacity of the channel is 1 − H(p), where H(p) is the binary entropy function.
The converse can be shown by a sphere packing argument. Given a codeword, there are roughly 2n H(p) typical output sequences. There are 2n total possible outputs, and the input chooses from a codebook of size 2nR. Therefore, the receiver would choose to partition the space into "spheres" with 2n / 2nR = 2n(1 − R) potential outputs each. If R > 1 − H(p), then the spheres will be packed too tightly asymptotically and the receiver will not be able to identify the correct codeword with vanishing probability.
## Shannon's channel capacity theorem for BSCp
Shannon's noisy coding theorem is general for all kinds of channels. We consider a special case of this theorem for a binary symmetric channel with an error probability p.
### Noisy coding theorem for BSCp
We denote (and would continue to denote) noise $e$ from $BSC_{p}$ as "$e \in BSC_{p}$". The noise $e$ is essentially a random variable with each of its indexes being a $1$ with probability $p$ and a $0$ with probability $1-p$.
Theorem 1 For all $p$ < $\frac{1}{2}$ and all $\epsilon$ such that $0 < \epsilon < \frac{1}{2} - p$, there exists a pair of encoding and decoding functions $E$: $\{0,1\}^k \rightarrow \{0,1\}^n$ and $D$: $\{0,1\}^n$ $\rightarrow$ $\{0,1\}^{k}$ respectively, and $k$ $\leq$ $\lfloor$ $(1 - H(p + \epsilon))n$ $\rfloor$ such that every message $m$ $\in$ $\{0,1\}^{k}$ has the following property: $\Pr_{e \in BSC_p}[D(E(m) + e) \neq m] \leq 2^{-{\delta}n}$ for a sufficient large integer $n$.
What this theorem actually implies is, a message when picked from $\{0,1\}^k$, encoded with a random encoding function $E$, and send across a noisy $BSC_{p}$, there is a very high probability of recovering the original message by decoding, if $k$ or in effect the rate of the channel is bounded by the quantity stated in the theorem. The decoding error probability is exponentially small.
We shall now prove Theorem 1 .
Proof We shall first describe the encoding function $E$ and the decoding function $D$ used in the theorem. Its good to state here that we would be using the probabilistic method to prove this theorem. Shannon's theorem was one of the earliest applications of this method.
Encoding function $E$: The encoding function $E$: $\{0,1\}^k \rightarrow \{0,1\}^n$ is selected at random. This means, given any message $m \in \{0,1\}^k$, we choose $E(m) \in \{0,1\}^n$ uniformly and independently at random.
Decoding function $D$: The decoding function $D$: $\{0,1\}^n \rightarrow \{0,1\}^k$ is given any received codeword $y$ $\in$ $\{0,1\}^n$, we find the message $m$ $\in$ $\{0,1\}^{k}$ such that $\Delta(y, E(m))$ is minimum (with ties broken arbitrarily). This kind of a decoding function is called a maximum likelihood decoding (MLD) function.
Now first we show, for a fixed $m \in \{0,1\}^{k}$ and $E$ chosen randomly, the probability of failure over $BSC_p$ noise is exponentially small. At this point, the proof works for a fixed message $m$. Next we extend this result to work for all $m$. We achieve this by eliminating half of the codewords from the code with the argument that the proof for the decoding error probability holds for at least half of the codewords. The latter method is called expurgation. This gives the total process the name random coding with expurgation.
A high level proof: Given a fixed message $m \in \{0,1\}^{k}$, we need to estimate the expected value of the probability of the received codeword along with the noise does not give back $m$ on decoding. That is to say, we need to estimate: $\mathbb{E}_{E}[\Pr_{e \in BSC_p}[D(E(m) + e) \neq m]]$.
Let $y$ be the received codeword. In order for the decoded codeword $D(y)$ not to be equal to the message $m$, one of the following events must occur:
• $y$ does not lie within the Hamming ball of radius $(p+\epsilon)$ for any $\epsilon$ greater than $0$, centered at $E(m)$. This condition is mainly used to make the calculations easier.
• There is another message $m^{\prime} \in \{0,1\}^k$ such that $\Delta(y, E(m^{\prime})) \leq \Delta(y, E(m))$. In other words the errors due to noise take the transmitted codeword closer to another encoded message.
We can apply Chernoff bound to ensure the non occurrence of the first event. By applying Chernoff bound we have, $Pr_{e \in BSC_p}$ $[\Delta(y, E(m)) > (p+\epsilon)] \leq 2^{-{\epsilon^2}n}$. Thus, for any $\epsilon > 0$, we can pick $n$ to be large enough to make the above probability exponentially small.
As for the second event, we note that the probability that $E(m^{\prime}) \in B(y,(p+\epsilon)n)$ is $Vol(B(y,(p+\epsilon)n)/2^n$ where $B(x, r)$ is the Hamming ball of radius $r$ centered at vector $x$ and $Vol(B(x, r))$ is its volume. Using approximation to estimate the number of codewords in the Hamming ball, we have $Vol(B(y,(p+\epsilon)n)/2^n \approx 2^{H(p)n}$. Hence the above probability amounts to $2^{H(p)n}/2^n = 2^{H(p)n-n}$. Now using union bound, we can upper bound the existence of such an $m^{\prime} \in \{0,1\}^k$ by $\le 2^{k +H(p)n-n}$ which is $2^{-\Omega(n)}$, as desired by the choice of $k$.
A detailed proof: From the above analysis, we calculate the probability of the event that the decoded codeword plus the channel noise is not the same as the original message sent. We shall introduce some symbols here. Let $p(y|E(m))$ denote the probability of receiving codeword $y$ given that codeword $E(m)$ was sent. Denote $B(E(m),(p+\epsilon)n)$ by $\text{Ball}$. $\Pr_{e \in BSC_p}[D(E(m) + e) \neq m] = \sum_{y \in \{0,1\}^{n}} p(y|E(m))\cdot 1_{D(y)\neq m} \leq \sum_{y \notin \text{Ball}} p(y|E(m)) \cdot 1_{D(y)\neq m} + \sum_{y \in \text{Ball}} p(y|E(m))\cdot 1_{D(y)\neq m} \leq 2^{-{\epsilon^2}n} + \sum_{y \in \text{Ball}} p(y|E(m)) \cdot 1_{D(y)\neq m}.$
We get the last inequality by our analysis using the Chernoff bound above. Now taking expectation on both sides we have, $\mathbb{E}_E$[$Pr_{e \in BSC_p}$[$D(E(m) + e)$ $\neq$ $m]$] $\leq$ $2^{-{\epsilon^2}n}$ $+$ $\sum_{y \in \text{Ball}}$ $p(y|E(m))$.$\mathbb{E}[1_{D(y)\neq m}]$.
Now we have $\sum_{y \in \text{Ball}} p(y|E(m)) \leq 1$. This just says, that the quantity $\mathbb{E}[1_{D(y)\neq m}] \leq 2^{k +H(p + \epsilon)n-n}$, again from the analysis in the higher level proof above. Hence, taking everything together we have $\mathbb{E}_{E}[\Pr_{e \in BSC_p}[D(E(m) + e) \neq m]] \leq 2^{-{\epsilon^2}n} + 2^{k +H(p + \epsilon)n-n} \leq 2^{-\delta n}$, by appropriately choosing the value of $\delta$. Since the above bound holds for each message, we have $\mathbb{E}_m[\mathbb{E}_E[\Pr_{e \in BSC_p}[D(E(m) + e)] \neq m]] \leq 2^{-\delta n}$. Now we can change the order of summation in the expectation with respect to the message and the choice of the encoding function $E$, without loss of generality. Hence we have $\mathbb{E}_E[\mathbb{E}_m [\Pr_{e \in BSC_p}[D(E(m) + e)] \neq m]] \leq 2^{-\delta n}$. Hence in conclusion, by probabilistic method, we have some encoding function $E^{*}$ and a corresponding decoding function $D^{*}$ such that $\mathbb{E}_m[\Pr_{e \in BSC_p}[D^{*}(E^{*}(m) + e)\neq m]] \leq 2^{-\delta n}$.
At this point, the proof works for a fixed message $m$. But we need to make sure that the above bound holds for all the messages $m$ simultaneously. For that, let us sort the $2^k$ messages by their decoding error probabilities. Now by applying Markov's inequality, we can show the decoding error probability for the first $2^{k-1}$ messages to be at most $2.2^{-\delta n}$. Thus in order to confirm that the above bound to hold for every message $m$, we could just trim off the last $2^{k-1}$ messages from the sorted order. This essentially gives us another encoding function $E^{\prime}$ with a corresponding decoding function $D^{\prime}$ with a decoding error probability of at most $2^{-\delta n + 1}$ with the same rate. Taking $\delta^{\prime}$ to be equal to $\delta - \frac{1}{n}$ we bound the decoding error probability to $2^{-\delta^{\prime}n}$. This expurgation process completes the proof of Theorem 1.
## Converse of Shannon's capacity theorem
The converse of the capacity theorem essentially states that $1 - H(p)$ is the best rate one can achieve over a binary symmetric channel. Formally the theorem states:
Theorem 2 If $k$ $\geq$ $\lceil$ $(1 - H(p + \epsilon)n)$ $\rceil$ then the following is true for every encoding and decoding function $E$: $\{0,1\}^k$ $\rightarrow$ $\{0,1\}^n$ and $D$: $\{0,1\}^{n}$ $\rightarrow$ $\{0,1\}^{k}$ respectively: $Pr_{e \in BSC_p}$[$D(E(m) + e)$ $\neq$ $m]$ $\geq$ $\frac{1}{2}$.
For a detailed proof of this theorem, the reader is asked to refer to the bibliography. The intuition behind the proof is however showing the number of errors to grow rapidly as the rate grows beyond the channel capacity. The idea is the sender generates messages of dimension $k$, while the channel $BSC_p$ introduces transmission errors. When the capacity of the channel is $H(p)$, the number of errors is typically $2^{H(p + \epsilon)n}$ for a code of block length $n$. The maximum number of messages is $2^{k}$. The output of the channel on the other hand has $2^{n}$ possible values. If there is any confusion between any two messages, it is likely that $2^{k}2^{H(p + \epsilon)n} \ge 2^{n}$. Hence we would have $k \geq \lceil (1 - H(p + \epsilon)n) \rceil$, a case we would like to avoid to keep the decoding error probability exponentially small.
## Codes for BSCp
Very recently, a lot of work has been done and is also being done to design explicit error-correcting codes to achieve the capacities of several standard communication channels. The motivation behind designing such codes is to relate the rate of the code with the fraction of errors which it can correct.
The approach behind the design of codes which meet the channel capacities of $BSC$, $BEC$ have been to correct a lesser number of errors with a high probability, and to achieve the highest possible rate. Shannon’s theorem gives us the best rate which could be achieved over a $BSC_{p}$, but it does not give us an idea of any explicit codes which achieve that rate. In fact such codes are typically constructed to correct only a small fraction of errors with a high probability, but achieve a very good rate. The first such code was due to George D. Forney in 1966. The code is a concatenated code by concatenating two different kinds of codes. We shall discuss the construction Forney's code for the Binary Symmetric Channel and analyze its rate and decoding error probability briefly here. Various explicit codes for achieving the capacity of the binary erasure channel have also come up recently.
## Forney's code for BSCp
Forney constructed a concatenated code $C^{*}$ $=$ $C_\text{out} \circ C_\text{in}$ to achieve the capacity of Theorem 1 for $BSC_p$. In his code,
• The outer code $C_\text{out}$ is a code of block length $N$ and rate $1-\frac{\epsilon}{2}$ over the field $F_{2^k}$, and $k = O(log N)$. Additionally, we have a decoding algorithm $D_\text{out}$ for $C_\text{out}$ which can correct up to $\gamma$ fraction of worst case errors and runs in $t_\text{out}(N)$ time.
• The inner code $C_\text{in}$ is a code of block length $n$, dimension $k$, and a rate of $1 - H(p) - \frac{\epsilon}{2}$. Additionally, we have a decoding algorithm $D_\text{in}$ for $C_\text{in}$ with a decoding error probability of at most $\frac{\gamma}{2}$ over $BSC_p$ and runs in $t_\text{in}(N)$ time.
For the outer code $C_\text{out}$, a Reed-Solomon code would have been the first code to have come in mind. However, we would see that the construction of such a code cannot be done in polynomial time. This is why a binary linear code is used for $C_\text{out}$.
For the inner code $C_\text{in}$ we find a linear code by exhaustively searching from the linear code of block length $n$ and dimension $k$, whose rate meets the capacity of $BSC_p$, by Theorem 1.
The rate $R(C^{*}) = R(C_\text{in}) \times R(C_\text{out}) = (1-\frac{\epsilon}{2}) ( 1 - H(p) - \frac{\epsilon}{2} ) \geq 1$ $-$ $H(p)$ $-$ $\epsilon$ which almost meets the $BSC_p$ capacity. We further note that the encoding and decoding of $C^{*}$ can be done in polynomial time with respect to $N$. As a matter of fact, encoding $C^{*}$ takes time $O(N^{2})+O(Nk^{2}) = O(N^{2})$. Further, the decoding algorithm described takes time $Nt_\text{in}(k) + t_\text{out}(N) = N^{O(1)}$ as long as $t_\text{out}(N) = N^{O(1)}$; and $t_\text{in}(k) = 2^{O(k)}$.
### Decoding error probability for C*
A natural decoding algorithm for $C^{*}$ is to:
• Assume $y_{i}^{\prime} = D_\text{in}(y_i), \quad i \in (0, N)$
• Execute $D_\text{out}$ on $y^{\prime} = (y_1^{\prime} \ldots y_N^{\prime})$
Note that each block of code for $C_\text{in}$ is considered a symbol for $C_\text{out}$. Now since the probability of error at any index $i$ for $D_\text{in}$ is at most $\frac{\gamma}{2}$ and the errors in $BSC_p$ are independent, the expected number of errors for $D_\text{in}$ is at most $\frac{\gamma N}{2}$ by linearity of expectation. Now applying Chernoff bound, we have bound error probability of more than $\gamma N$ errors occurring to be $e^\frac{-\gamma N}{6}$. Since the outer code $C_\text{out}$ can correct at most $\gamma N$ errors, this is the decoding error probability of $C^{*}$. This when expressed in asymptotic terms, gives us an error probability of $2^{-\Omega(\gamma N)}$. Thus the achieved decoding error probability of $C^{*}$ is exponentially small as Theorem 1.
We have given a general technique to construct $C^{*}$. For more detailed descriptions on $C_\text{in}$ and $C{out}$ please read the following references. Recently a few other codes have also been constructed for achieving the capacities. LDPC codes have been considered for this purpose for their faster decoding time.[1]
## Notes
1. Richardson and Urbanke
## References
• David J. C. MacKay. Information Theory, Inference, and Learning Algorithms Cambridge: Cambridge University Press, 2003. ISBN 0-521-64298-1
• Thomas M. Cover, Joy A. Thomas. Elements of information theory, 1st Edition. New York: Wiley-Interscience, 1991. ISBN 0-471-06259-6.
• Atri Rudra's course on Error Correcting Codes: Combinatorics, Algorithms, and Applications (Fall 2007), Lectures 9, 10, 29, and 30.
• Madhu Sudan's course on Algorithmic Introduction to Coding Theory (Fall 2001), Lecture 1 and 2.
• G. David Forney. Concatenated Codes. MIT Press, Cambridge, MA, 1966.
• Venkat Guruswamy's course on Error-Correcting Codes: Constructions and Algorithms, Autumn 2006.
• A mathematical theory of communication C. E Shannon, ACM SIGMOBILE Mobile Computing and Communications Review.
• Modern Coding Theory by Tom Richardson and Rudiger Urbanke., Cambridge University Press
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http://mathoverflow.net/questions/47504?sort=votes
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Intersection Cohomology of Coordinate Hyperplanes
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I'm trying to learn how to compute stalks of IC sheaves, and I was wondering about the following example:
Fix $n$. Let $X \subset \mathbb{C}^n$ be the variety cut out by the equation $x_1 \cdots x_n =0$, i.e. the coordinate hyperplanes. What are the stalks of $\mathrm{IC}(X)$ at the various points of $X$, in particular at the origin?
This seems like a natural toy example, but if the general answer is difficult, I'd be happy to know how to compute this for small $n$.
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1 Answer
Let $Y$ denote the disjoint union of the coordinate hyperplanes in $\mathbb{C}^n,$ and let $f:Y \to X$ denote the corresponding resolution of singularities.
1) Show that $f_{\ast}\mathbb{C}_Y[n-1] \simeq IC_X$ (consider, for example, the support conditions and the fact that both sheaves are isomorphic to $\mathbb{C}_U[n-1]$ when restricted to the nonsingular open $U \subset X$).
Edit (some details added): Letting $U$ denote the complement of the set where any two coordinate planes intersect, $f$ is an isomorphism when restricted to $U.$ We therefore have that the restriction (i.e., pullback) of $f_{\ast}\mathbb{C}_Y[n-1]$ to $U$ coincides with $\mathbb{C}_U[n-1]$ (by proper base change if you like).
In order to conclude that $f_{\ast}\mathbb{C}_Y[n-1] \simeq IC_X,$ we now just need to check the support and cosupport conditions which uniquely define the intersection cohomology sheaf (together with the fact that its restriction to $U$ is the (shifted) constant sheaf). These conditions are similar to, but more restrictive than, the support and cosupport conditions for perverse sheaves.
I recommend looking at page 21 of the wonderful article by de Cataldo and Migliorini, which can be found at http://arxiv.org/abs/0712.0349 for a statement of these support and cosupport conditions (and figure 1 on page 25 for a visual illustration of the definition).
Since the fibers of $f$ consist of a finite number of points, the cohomology of the fibers is non-zero only in degree zero. This shows that the first condition (the support condition) is satisfied.
For the second condition (the cosupport condition), you can either derive it from the support condition using Verdier duality and the properness of $f,$ or you can simply note that an open ball in $\mathbb{C}^{n-1}$ has non-zero compactly supported cohomology only in degree $2n-2.$
2) Now it's straightforward to compute any of the stalks since the fiber of $x \in X$ consists of anywhere between one point and n points, depending on how many hyperplanes $x$ lives inside of.
Alternatively, it is also possible to do this by using only basic definitions. To compute the stalk at $x,$ intersect a sufficiently small open ball around $x$ in $\mathbb{C}^n$ with $X$ and then calculate the intersection cohomology by considering intersection cochains (just like you would for singular cohomology, but now with a less restrictive notion of cochain).
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Can you elaborate on the argument in part 1? I'm sorry to belabor it, but I think I need it spelled out to me. – Dinakar Muthiah Nov 27 2010 at 18:35
A small detail, which makes me nervous: In "D-modules, perverse sheaves and Representation theory", they don't even define IC-complexes on non irreducible varieties. Given a not necessarily irreducible variety with equidimensional components. Is there still a unique complex which coincides on a dense open subset with a given shifted local system, and satisfies the (co)support condition for IC complexes? – Jan Weidner Nov 28 2010 at 8:36
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Jan, that's certainly a good point to bring up. I think they assume irreducible throughout because it makes things less messy and I think it's necessary for the decomposition theorem. However, if you look at Goresky and MacPherson's "Intersection Homology II," you'll see that the IC sheaf can be defined for a fairly large class of topological spaces known as pseudomanifolds (maybe you knew this already) and there is still a uniqueness result like the one you ask for (see 4.1 and 6.1 of "Intersection Homology II"). In the OP's question, $X$ is pure dimensional which means we should be ok. – Mike Skirvin Nov 28 2010 at 16:36
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One further comment: On an irreducible variety, IC sheaves are indecomposable in the derived category, as stated in Corollary 2 of section 4.1 of "Intersection Homology II." This would certainly seem to be a good reason for usually dealing with irreducible varieties. – Mike Skirvin Nov 28 2010 at 16:44
Okay, thanks very much Mike! – Jan Weidner Nov 28 2010 at 20:48
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http://mathoverflow.net/revisions/43984/list
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Return to Answer
3 Corrected a switch between N and n in the Khovanskii bound
To expand on David's answer, the bound given by Khovanskii's theorem is of the form $2^{\binom{N}{2}} (N+1)^n$ n+1)^N\$ per quadrant (more or less). Incremental improvements on this bounds have been obtained http://arxiv.org/abs/1010.2962 being the latest, but nothing revolutionary and we're nowhere near realistic bounds.
As far as I know, there have been no new attempts on a multivariable Descartes (something that would take signs into consideration) since Itenberg and Roy's paper, and it remains a major open problem in the area.
Added Later: As far as an algorithm is concerned, I don't think you can count without solving, in which case the standard one to use would be the Cylindrical Algebraic Decomposition pioneered by Collins. It is more or less based on Sturm and induction, but it is a lot more involved than the 1-dimensional case. For reference, I recommend the book by Basu Pollack and Roy Algorithms in Real Algebraic Geometry (free download).
(More on counting without solving: Marie-Françoise Roy mentions the problem of determining if a semi-algebraic set is non-empty without producing one point per connected components as one of the major open problems in algorithmic real algebraic geometry).
2 Added the paragraphs about algorithm.
To expand on David's answer, the bound given by Khovanskii's theorem is of the form $2^{\binom{N}{2}} (N+1)^n$ per quadrant (more or less). Incremental improvements on this bounds have been obtained http://arxiv.org/abs/1010.2962 being the latest, but nothing revolutionary and we're nowhere near realistic bounds.
As far as I know, there have been no new attempts on a multivariable Descartes (something that would take signs into consideration) since Itenberg and Roy's paper, and it remains a major open problem in the area.
Added Later: As far as an algorithm is concerned, I don't think you can count without solving, in which case the standard one to use would be the Cylindrical Algebraic Decomposition pioneered by Collins. It is more or less based on Sturm and induction, but it is a lot more involved than the 1-dimensional case. For reference, I recommend the book by Basu Pollack and Roy Algorithms in Real Algebraic Geometry (free download).
(More on counting without solving: Marie-Françoise Roy mentions the problem of determining if a semi-algebraic set is non-empty without producing one point per connected components as one of the major open problems in algorithmic real algebraic geometry).
1
To expand on David's answer, the bound given by Khovanskii's theorem is of the form $2^{\binom{N}{2}} (N+1)^n$ per quadrant (more or less). Incremental improvements on this bounds have been obtained http://arxiv.org/abs/1010.2962 being the latest, but nothing revolutionary and we're nowhere near realistic bounds.
As far as I know, there have been no new attempts on a multivariable Descartes (something that would take signs into consideration) since Itenberg and Roy's paper, and it remains a major open problem in the area.
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http://physics.stackexchange.com/questions/35644/do-we-live-in-a-world-with-4-or-more-dimension
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# Do we live in a world with 4 or more dimension?
A NOVA show have told the audience that we are live in 3 dimensional world, the world we lived in is compose by 3 element: the energy, matter, space. By the time Einstein have invented the 4-dimensional model including the time element, why no one claim that we actually live in 4-dimensional world
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1
Energy and matter are not dimensions and neither is space, although space has dimensions - at least three of them. In physics "dimension" simply refers to a direction you can move in: left-right, forward-backward, up-down. Hence three dimensions. – Michael Brown Apr 2 at 1:32
## 2 Answers
Victor, This is a good question. Probably one of the greatest tragedies surrounding modern science in popular literature is the portrayal of dimensions and our understanding of them.
Many popular shows are inconsistent in when they talk about dimensions, in some cases they refer to the 3 spatial and 1 time dimension that we commonly observe as separate things, and in some cases they talk about the 4-dimensional spacetime that is used in modern physics on a routine basis without any real discussion about what they mean.
Spacetime is a combination of the 3 space dimensions and 1 time dimension, e.g. 3+1 dimensions which equals 4 dimensions.
In the early days of relativistic mechanics, in order to represent the 3 spatial dimensions and 1 time dimension as one entity, they would give the time dimension an imaginary value. So the 4 dimensional spacetime would be represented as:
$$x + y + z + it$$
This covenient, because when one takes the inner product (sometimes called the dot product) of the representation of space time one gets:
$$x^2 + y^2 + z^2 - t^2$$
This sort of representation is convenient because one can begin to understand the relationship between space and time in terms of the Pythagorean theorem.
This is the simplest example of how one can begin to relate the nature of spacetime to geometry. The broader theory of how geometry of spacetime changes as you move in spacetime is Einstein's Theory of General Relativity.
General Relativity assumes that there are 3 dimensions of space and 1 dimension of time that make up the 4-dimensional spacetime that programs sometimes reference.
There are other theories that propose there are more than 3 space dimensions of space and 1 dimension of time. However, these are very speculative still.
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Yes, We live in a world with four or more dimensions.
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## protected by Qmechanic♦Apr 2 at 3:18
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http://physics.stackexchange.com/questions/tagged/homework+velocity
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3answers
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http://math.stackexchange.com/questions/57153/conditional-density-of-two-jointly-gaussian-random-vectors
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# Conditional density of two jointly gaussian random vectors
In my estimation theory textbook, the following is stated as a reminder without any further explanation:
Consider the gaussian random vector $\bf{z} = \begin{pmatrix}\bf{x} \\ \bf{y}\end{pmatrix}$ with mean $\hat{z} = \begin{pmatrix}\hat{x} \\ \hat{y}\end{pmatrix}$ and covariance matrix $\bf{C}_z = \begin{pmatrix} \bf{C}_{xx} & \bf{C}_{xy} \\ \bf{C}_{yx} & \bf{C}_{yy} \end{pmatrix}$
If a measurement $y^*$ is given, the conditional density of $x$ conditioned on that measurement $f\left(x | y^*\right)$ is gaussian with mean
$\hat{x} + \bf{C}_{xy}\bf{C}_{yy}^{-1} \left( y^* - \hat{y} \right)$
and covariance matrix
$\bf{C}_{xx} - \bf{C}_{xy} \bf{C}_{yy}^{-1} \bf{C}_{yx}$
(Pseudo-inverses replace inverses when necessary)
Having never worked with conditional densities before, I don't see how to derive these formulas, or what the intuition behind them is.
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## 1 Answer
A detailed proof with carefully presented computations is here, see Part b of Theorem 4.
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http://math.stackexchange.com/questions/113250/the-area-of-the-superellipse?answertab=active
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# The area of the superellipse
I'm watching this video, where D. Knuth explains the connection of $\pi$ and factorials, and other matters (it is very interesting). Almost at the end of the talk he says the area of the superellipse
$$x^{\frac{1}{\alpha}}+y^{\frac{1}{\alpha}}=1$$
is given by
$$A(\alpha) = \frac{2 \alpha \cdot\Gamma{(\alpha)}^2}{\Gamma{(2 \alpha)}}$$ whic would be
$$A(\alpha) = 2 \alpha B(\alpha,\alpha) = 2 \alpha\int_0^1(1-u)^{\alpha-1}u^{\alpha-1}du$$
I was trying to check this so I put
$$A\left( \alpha \right) = \int\limits_0^1 {{{\left( {1 - {x^{1/\alpha }}} \right)}^\alpha }dx}$$
Now let $x = {u^\alpha }$
$$A\left( \alpha \right) = \alpha \int\limits_0^1 {{{\left( {1 - u} \right)}^\alpha }{u^{\alpha - 1}}du}$$
What's going on?
The $2$ in Knuth's formula probably comes from the fact he considers the full figure and not only a fourth, as I am, but I don't know what I'm doing wrong here. If you want to check, it is at $1:21:00$ aproximately.
PS: Just as a curiosity, does Knuth have a stutter or is it he is just thinking about too many things in too little time?
So it was just OK:
$$A\left( \alpha \right) = \alpha \int\limits_0^1 {{{\left( {1 - u} \right)}^\alpha }{u^{\alpha - 1}}du} = \frac{{\alpha \Gamma \left( {\alpha + 1} \right)\Gamma \left( \alpha \right)}}{{\Gamma \left( {2\alpha + 1} \right)}} = \frac{{\Gamma {{\left( {\alpha + 1} \right)}^2}}}{{\Gamma \left( {2\alpha + 1} \right)}}$$
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I did the case $\alpha = 2,$ a sort of nonconvex star shape. In the first quadrant it is just $y = 1 + x - 2 \sqrt x,$ with area $1/6,$ so the whole figure has area $2/3.$ Agrees with Dirichlet. – Will Jagy Feb 25 '12 at 20:37
@WillJagy It does. Thanks for you help. Did you receive my response to your mail? – Peter Tamaroff Feb 25 '12 at 21:05
## 1 Answer
I wouldn't call it a stutter. It strikes me as closer to someone very famous trying to seem folksy; not quite the same as saying "um," more "I'm as uncertain as you."
Dirichlet wrote a generalization of this in Über eine neue Methode zur Bestimmung vielfacher Integrale. Collected Works, volume 1, page 389.
For integrating the constant 1 on $$x \geq 0, \; y \geq 0, \; x^{1/\alpha} + y^{1/\alpha} \leq 1,$$ the result is $$\frac{\alpha^2 \Gamma(\alpha)^2}{\Gamma(1 + 2 \alpha)} = \frac{ \Gamma(1 + \alpha)^2}{\Gamma(1 + 2 \alpha)},$$ multiply by 4 to get the whole thing, $$\frac{ 4 \; \Gamma(1 + \alpha)^2}{\Gamma(1 + 2 \alpha)}.$$ Sample points, $\alpha = 1/2$ is the circle, $\Gamma(3/2) = (1/2) \sqrt \pi$ and $\Gamma(2) = 1,$ so we get $\pi.$ With $\alpha = 1,$ we have a tilted square, $\Gamma(3) = 2,$ area is indeed $2.$ As $\alpha \rightarrow \infty,$ area goes to $0,$ it is not necessary to quote Stirling's to believe that $(\alpha!)^2 / (2 \alpha)! \rightarrow 0.$ Finally, as $\alpha \rightarrow 0,$ we approach the entire square, and the area approaches $4.$
I understand part of the Dirichlet technique is in Whittaker and Watson.
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I see. So I was on the right path. It just looked like a stutter, and although I watched the whole hour and 40 minutes, it made me a little nervous while watching... he just changed is words too many times. However, I was delighted by the exposition. – Peter Tamaroff Feb 25 '12 at 20:20
@joriki, very nice, I have never known how to do umlauts. I do know better, but the article title is spelled incorrectly in my notes, including an illegible final e on integrale. I made my best estimate on which words are nouns. – Will Jagy Feb 25 '12 at 21:58
– joriki Feb 25 '12 at 22:25
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http://mathoverflow.net/questions/2615?sort=newest
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Least number of charts to describe a given manifold
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Hello, I'm wondering if there is a standard reference discussing the least number of charts in an atlas of a given manifold required to describe it.
E.g. a circle requires at least two charts, and so on (I couldn't manage to get anything relevant neither on wikipedia nor on google, so I guess I'm lacking the correct terminology).
Edit: in the case of an open covering of a topological space by n+1 contractible sets (in that space) then n is called the Lusternik-Schnirelman Category of the space, see Andy Putman's answer. The following book seems to be the standard reference http://books.google.fr/books?id=vMREfNN-L4gC&pg=PP1
Great, now I'm still interested by the initial question: does anybody know of another theory without this contractibility assumption (hoping that it allows more freedom)? e.g. would it lead to different numbers say for genus-g surfaces?
Final edit: yes different numbers for genus-g surfaces (see answers below), but not sure there is a theory without contractibility. Right, really lots of interesting literature on the LS category nevertheless, hence the accepted answer. For example there are estimates for non-simply connected compact simple Lie groups like PU(n) and SO(n) in Topology and its Applications, Volume 150, Issues 1-3, 14 May 2005, Pages 111-123.
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6 Answers
It's not quite the same thing, but a related object is the Lyusternik–Schnirelmann category of a topological space. See
http://en.wikipedia.org/wiki/Lyusternik-Schnirelmann_category
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It's pretty close, thanks! – Thomas Sauvaget Oct 26 2009 at 14:25
Is there some reason this has the same name as the "category" that has objects and morphisms, or is this purely a case of the same English word getting used for two mathematical objects? – Michael Lugo Oct 26 2009 at 14:34
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The concepts are unrelated. The notion of the L-S category predated the "category-theoretic" notion of a category. I suspect that its inspiration comes from the Baire category theorem. – Andy Putman Oct 26 2009 at 14:47
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.
After "dimension" this is the most basic numerical invariant of a manifold and the least explored. I found this reference some years ago: I. Bernstein, "On Imbedding Numbers of Differentiable Manifolds", Topology, Vol. 7, pp. 95-109.
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Wow, thanks for pointing out this reference! Mike Hopkins has also written on this concept: springerlink.com.ezproxy.webfeat.lib.ed.ac.uk/… In particular I think he answers the OP's question for (almost) all projective spaces. – Mark Grant Jun 17 2011 at 12:52
I do not know if the following exactly answers your question.
I have found on the second page of Michor "Topics in Differential Geometry": "Note finally that any manifold $M$ admits a finite atlas consisting of $\dim{M}+1$ not connected charts. This is a consequence of topological dimension theory [cf. Nagata, Modern Dimension Theory]; a proof for manifolds may be found in [cf. Greub, Halperin, Vanstone, Connections, curvature and cohomology.I]."
I hope to have been useful.
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Yes, sounds like interesting references, and the result is intuitively satisfying, thanks! I'll try to find a library that has them... – Thomas Sauvaget Feb 5 2011 at 21:19
I believe that Cech cohomology could yield the sort of answer you're looking for (at least in the contractible case). The general idea is that it computes cohomology based on nothing but the so-called "incidence data" of a good cover (that is, which n-fold intersections of open sets in the cover are nonempty) -- in fact, a n-chain is nothing more than a formal sum of (nonempty) (n+1)-fold intersections, with R coefficients. Of course, this cohomology theory is usually isomorphic to singular cohomology, de Rham cohomology, et al. (in particular, they agree in the case of manifolds). So if your manifold has a high h1=rank(H1), for example, then there must be lots of different 2-fold intersections to generate H1, and also if your manifold has nonzero Hk then there must exist a (k+1)-fold intersection of sets in the open cover, which means that you must have at least that many sets in any good cover. (Note that if M is a k-dimensional orientable manifold, then Hk(M)=R by Poincare duality.)
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Cech cohomology is too weak to give a full answer, though, because it only gives you a bound on how to cover a space such that all nonempty intersections are contractible, rather than just the individual open sets themselves. For example, a sphere S^n can be covered by 2 contractible sets, even though it takes n+1 to give a good cover. Also, you could have an acyclic space where you see no cohomological obstructions to the space itself being contractible, but \pi_1 is nonzero so you certainly need at least two contractible sets to cover. – Eric Wofsey Oct 26 2009 at 19:48
Right. (I also just realized that my last statement is clearly false; e.g., the unit disk has trivial cohomology.) But maybe there's a weaker Cech-ish cohomology that can address this anyways. What happens if we work with a complex where only the individual open sets must be contractible? It seems possible that there's a statement along the lines of "the cohomology of this complex injects into the cohomology of a good cover", and maybe we could get a lower bound on the difference in rank (based on higher-dimensional cohomology of M and/or the cohomology of the nonempty intersections...?). – Aaron Mazel-Gee Oct 27 2009 at 5:03
To answer your last question, the least number of charts needed to cover any orientable 2-manifold is 2. Consider the usual embedding of an orientable surface Σ in R3 which is symmetric across the plane z = 0 (as shown here), and let ε > 0 be sufficiently small. The open subsets Σ ∩ {z > -ε}, Σ ∩ {z < ε} form a covering of Σ by charts: by Morse theory Σ ∩ {z > -ε} is diffeomorphic to Σ ∩ {z > ε}, which is diffeomorphic to an open subset of R2 by projecting onto the xy-plane.
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Note that the two open sets in Reid's answer may not be made contractible in general. Indeed, for orientable 2-manifolds, 3 contractible sets are required, except for S^2, where 2 are required. – jc Oct 26 2009 at 17:53
Yeah, I figured out after asking, so really the LS category measures something different, thanks! – Thomas Sauvaget Oct 26 2009 at 18:06
Orthogonal question: Does the (minimum) number of charts needed to describe a manifold tell you anything about the manifold?
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Is this appropriate as an answer? I would have commented on the question if I had the rep. – Sonia Balagopalan Oct 26 2009 at 16:15
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For the L-S category, having an upper bound on the category can bound the complexity of your manifold (e.g. in terms of cohomology) from above – jc Oct 26 2009 at 17:31
The L-S category gives a lower bound on the number of critical points of a functional on your manifold. That's the big abstract theorem that Lusternik and Schnirelmann proved. – Jeff Strom Jul 29 2010 at 13:38
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http://mathoverflow.net/questions/113858?sort=votes
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## Is there an algebraic geometry analogue of the closed graph theorem?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
In functional analysis, the closed graph theorem asserts that if a linear map $T: X \to Y$ between two Banach spaces $X, Y$ has a closed graph $S := \{ (x,Tx): x \in X \}$, then the map is continuous. Thus, it gives a criterion for regularity of a map in terms of regularity of the graph of that map.
I am curious as to whether there is any analogous statement in algebraic geometry. The naive formulation would be: if $T: X \to Y$ was a function (in the set-theoretic sense) between algebraic varieties $X, Y$ over an algebraically closed field $k$ whose graph $S := \{ (x,Tx): x \in X \}$ was also an algebraic variety, then $T$ would be a regular map. (Here I will be vague as to whether I want varieties to be affine, projective, quasiprojective, or abstract.) But this is false, even in characteristic zero: for instance, the coordinate function $(t^2,t^3) \mapsto t$ from the cuspidal curve $\{ (t^2,t^3): t \in k \}$ to $k$ has a graph which is an algebraic variety, but is not a regular map (it is not given by a rational function in a neighbourhood of the origin). In characteristic $p$, the inverse of the Frobenius map $x \mapsto x^p$ provides another counterexample. Somehow the difficulty is that regular functions in $S$ need not come from pullback from regular functions in $X$, even though the vertical line test suggests that such maps should be "degree 1" in some sense.
Still, I feel like there should be some positive statement to be made here, though I was not able to find one after searching through a few algebraic geometry texts. For instance, if one demands that $X, Y, S$ are all smooth and that the field has characteristic zero, does the claim now hold? Ideally, I would like to only have conditions on the varieties $X,Y,S$ and not on the various maps between these varieties; for instance, I would prefer not to have to assume that the projection map from $S$ to $X$ is finite (though perhaps this is automatic?).
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3
It has nothing to do with $Y$, right? The projection $S\to X$ is a regular map that is one to one and onto, and the question is: when can you say that the inverse is also regular? Isn't it basically when $X$ is normal? – Tom Goodwillie Nov 19 at 19:30
There is a closed graph theorem in real algebraic geometry, or more generally in an $o$-minimal setting. See Exercise 7, Sec. 6.1, page 97 in L. Van den Dries' book Tame topology and $o$-minimal structures, London Math. Soc. Lecture Notes Series 248 Cambridge Univ. Press, 1998. – Liviu Nicolaescu Nov 19 at 20:11
## 1 Answer
You might be rediscovering Zariski's Main Theorem, which implies your statement in case $X$ is normal (or just weakly normal) and the projection from the graph $\Gamma$ to $X$ is proper and separable. What you really need is the map $\Gamma\to X$ to be an isomorphism, so the question is equivalent to "when is a bijective map an isomorphism"?
You already explained why we need (weak) normality (example with the cuspidal curve) and separability (the Frobenius map). To see why properness is also necessary, look at the map $\mathbb{A}^1\to\mathbb{A}^1$ sending $x$ to $1/x$ for $x\neq 0$ and sending $0\to 0$, whose graph is a union of $(0,0)$ and a hyperbola $xy=1$.
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Ah, yes, that makes sense. So, in particular, if we insist on $X$ and the graph being smooth projective varieties over characteristic zero, the analogue of the closed graph theorem should hold, as the hypotheses of smoothness, projectiveness, and characteristic zero should guarantee normality, properness, and separability respectively? (I guess in retrospect it seems reasonable to have "smooth projective variety over characteristic zero" be the analogue to "complete normed vector space over the reals".) – Terry Tao Nov 19 at 19:56
1
Or perhaps this formulation is even closer to the functional analysis closed graph theorem: if $X,Y$ are smooth projective varieties over characteristic zero, then $f: X \to Y$ is regular iff its graph is Zariski closed. – Terry Tao Nov 19 at 20:02
Yes, that's a good point. I like your analogy :). – Piotr Achinger Nov 19 at 20:02
3
this reminds me that once years ago when i substituted one day for a colleague teaching real analysis, i recall i thought of zariski's main theorem as motivation for the closed graph theorem. – roy smith Nov 19 at 20:25
1
There was a question mathoverflow.net/questions/78696/… with much discussion of different ways of thinking about Zariski's Main Theorem. – Tom Goodwillie Nov 20 at 0:18
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http://physics.stackexchange.com/questions/tagged/calabi-yau
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# Tagged Questions
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I've been very impressed to learn about kaluza-klein theory and compactification strategies. I would like to read more about this but in the meantime i'm curious about 2 different points. I have the ...
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http://math.stackexchange.com/questions/2543/square-root-of-symmetric-matrix-and-transposition
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square root of symmetric matrix and transposition
I have a symmetric matrix A. How do I compute a matrix B such that $B^tB=A$ where $B^t$ is the transpose of $B$. I cannot figure out if this is at all related to the square root of $A$.
I've gone through wikimedia links of square root of a matrix.
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3
– alex Aug 15 '10 at 22:22
2 Answers
As J. M. says, you need your matrix $A$ to be positive definite. Since $A$, being symmetric, is always diagonalizable, this is the same as saying that it has non-negative eigenvalues. If this is the case, you can adapt alex's comment almost literally for the real case: as we've said, $A$ is diagonalizable, but, also, there exists an orthonormal base of eigenvectors of $A$. That is, there is an invertible matrix $S$ and a diagonal matrix $D$ such that
$$D = SAS^t , \quad \text{with} \quad SS^t = I \ .$$
Since
$$D = \mathrm{diag} (\lambda_1, \lambda_2, \dots , \lambda_n) \ ,$$
is a diagonal matrix and has only non-negative eigenvalues $\lambda_i$, you can take its square root
$$\sqrt{D} = \mathrm{diag} (\sqrt{\lambda_1}, \sqrt{\lambda_2}, \dots , \sqrt{\lambda_n} ) \ ,$$
and then, on one hand, you have:
$$\left( S^t \sqrt{D} S \right)^2 = \left( S^t \sqrt{D} S\right) \left(S^t \sqrt{D} S \right) = S^t \left( \sqrt{D}\right)^2 S = S^t D S = A \ .$$
On the other hand, $S^t \sqrt{D} S$ is a symmetric matrix too:
$$\left( S^t \sqrt{D} S \right)^t = S^t (\sqrt{D})^t S^{tt} = S^t \sqrt{D^t} S = S^t \sqrt{D} S \ ,$$
so you have your $B = S^t \sqrt{D} S$ such that $B^t B = A$.
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What you apparently want here is the Cholesky decomposition, which factors a matrix A into $BB^T$ where $B$ is a triangular matrix. However, this only works if your matrix is positive definite.
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2
The times between this and the comment are miraculously close :) – BBischof Aug 15 '10 at 22:24
if the matrix is not positive definite, then is there another property I could use? – user957 Aug 15 '10 at 22:25
if the matrix is not a positive definite, there may not be a real solution. for example, the $1 \times 1$ matrix $A=-1$ does not have a real square root. – alex Aug 15 '10 at 22:27
1
On the other hand, if you are OK with complex answers, then if $A$ is diagonalized as $A=U D U^{T}$ with diagonal $D$ and unitary $U$, then take $B=U D^{1/2} U^{T}$. $B$ will have complex entries, since some of the entries of $D$ will be negative. However, $B B^T = U D U^T = A$. – alex Aug 15 '10 at 22:30
1
alex - I don't think Q D^{1/2} Q going to be triangular matrix. take for example A = ((2,-2),(-2,5)) – user957 Aug 16 '10 at 1:48
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http://math.stackexchange.com/questions/94691/why-is-the-pullback-completely-determined-by-d-f-ast-f-ast-d-in-de-rham-co
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# Why is the pullback completely determined by $d f^\ast = f^\ast d$ in de Rham cohomology?
Fix a smooth map $f : \mathbb{R}^m \rightarrow \mathbb{R}^n$. Clearly this induces a pullback $f^\ast : C^\infty(\mathbb{R}^n) \rightarrow C^\infty(\mathbb{R}^m)$. Since $C^\infty(\mathbb{R}^n) = \Omega^0(\mathbb{R}^n)$ (the space of zero-forms) by definition, we consider this as a map $f^\ast : \Omega^0(\mathbb{R}^n) \rightarrow \Omega^0(\mathbb{R}^m)$. We want to extend $f^\ast$ to the rest of the de Rham complex in such a way that $d f^\ast = f^\ast d$.
Bott and Tu claim (Section I.2, right before Prop 2.1), without elaboration, that this is enough to determine $f^\ast$ . I can see why this forces e.g.
$\displaystyle\sum_{i=1}^n f^\ast \left[ \frac{\partial g}{\partial y_i} d y_i \right] = \sum_{i=1}^n f^* \left[ \frac{\partial g}{\partial y_i}\right] d(y_i \circ f)$,
but I don't see why this forces each term of the LHS to agree with each term of the RHS -- it's not like you can just pick some $g$ where $\partial g/\partial y_i$ is some given function and the other partials are zero.
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I claim that the function $y_j$ has the property that $\partial y_j / \partial y_i$ is the constant $1$ if $i = j$ and is the constant $0$ otherwise. – Zhen Lin Dec 28 '11 at 15:09
Right, but you can't get a function with the property that $\partial g / \partial y_i$ is an arbitrary smooth function and the other partials are zero, because this forces $\partial(\partial g/\partial y_i)/\partial y_j = \partial^2 g / \partial y_i \partial y_j = \partial(\partial g / \partial y_j)/\partial y_i = 0$ for all $i \neq j$. (I was trying to do this without the assumption that $f^\ast$ preserved multiplication, which is apparently a mistake.) – Daniel McLaury Dec 28 '11 at 15:47
## 2 Answers
$\newcommand\RR{\mathbb{R}}$I don't have the book here, but it seems you are asking why there is a unique extension of $f^*:\Omega^0(\RR^n)\to\Omega^0(\RR^m)$ to an appropriate $\overline f^*:\Omega^\bullet(\RR^n)\to\Omega^\bullet(\RR^m)$ such that $f^*d=df^*$. Here appropriate should probably mean that the map $\overline f^*$ be a morphism of graded algebras.
Now notice that the since $f^*$ is fixed on $\Omega^0(\RR^n)$ and the commutation relation with $d$ tells us that it is also fixed on the subspace $d(\Omega^0(\RR^n))\subseteq\Omega^1(\RR^n)$. The uniqueness follows from the fact that the subspace $\Omega^0(\RR^n)\oplus d(\Omega^0(\RR^n))$ of $\Omega^\bullet(\RR^n)$ generates the latter as an algebra.
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Dear Mariano: You wrote "morphism of graded algebras", but I think you meant "morphism of differential graded algebras". – Pierre-Yves Gaillard Dec 28 '11 at 15:36
Actually, I separated the 'differential' part because it was singled out in the question itself, leaving for 'appropriate' simply the graded-algebra part :) Of course, the resulting map will be a morphism of differential graded algebras. – Mariano Suárez-Alvarez♦ Dec 28 '11 at 15:37
Yes, you're right! I hadn't read your answer carefully enough... Sorry! +1 – Pierre-Yves Gaillard Dec 28 '11 at 15:39
Ah, so $f^\ast$ should not only be $\mathbb{R}$-linear, but should preserve multiplication as well -- that's what I was missing. For some reason it didn't occur to me that the original pullback function preserved multiplication, so I never considered that I should want such a property for the extension. – Daniel McLaury Dec 28 '11 at 15:41
(There was supposed to be a "Thanks!" at the end of the previous comment.) – Daniel McLaury Dec 28 '11 at 15:50
Consider $h = g \circ Q_{i}$ where $Q_{i}$ sends $(a_1,...,a_n)$ to $(0,\dotsc, a_i, \dotsc, 0)$. Apply your observation to $h$.
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This seems to establish the desired result for functions with a rank-one Jacobian. I tried this before, but I couldn't see a useful way to extend that to all functions. – Daniel McLaury Dec 28 '11 at 17:35
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http://math.stackexchange.com/questions/228954/an-inequality-like-the-triangle-inequality
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# an inequality like the triangle inequality
Its easy question, but I cannot find the name of the inequality. Please provide me with it.
I am doing the following, let $a$ be $n$-dimentional vector. Let $b_i, i=1, \ldots, n$ be positive numbers. Then, $$\left(\left|\sum_{i=1}^n a_i\right|^p\right)^{1/p}=\left(\left|\sum_{i=1}^n(a_i-b_i)+\sum_{i=1}^nb_i\right|^p\right)^{1/p}\leq \left(\left|\sum_{i=1}^n(a_i-b_i)\right|^p\right)^{1/p}+\left(\left|\sum_{i=1}^nb_i\right|^p\right)^{1/p}$$ Which inequality did I use here?
Thank you.
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2
The way you wrote it, this is just the triangle inequality in $\mathbb{R}$. Maybe you misplaced some brackets? – IHaveAStupidQuestion Nov 4 '12 at 16:37
Indeed, notice that the exponents just cancel out, i.e. $(x^p)^{1/p}=x$. – nonpop Nov 4 '12 at 17:15
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http://mathhelpforum.com/algebra/18321-polynominal-print.html
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# Polynominal
Printable View
• September 1st 2007, 07:54 AM
Ununuquantium
Polynominal
Determine all polynominals $P(x)$ (real coeffitients of course) such that for arbitrary number $x$ there is:
$P(x^{2})\times P(x^{3}) = (P(x))^{5}$
• September 2nd 2007, 08:42 AM
ThePerfectHacker
Quote:
Originally Posted by Ununuquantium
Determine all polynominals $P(x)$ (real coeffitients of course) such that for arbitrary number $x$ there is:
$P(x^{2})\times P(x^{3}) = (P(x))^{5}$
Here is an partial solution. We shall determine all real polynomial $P(x)$ having real zeros that satisfy this.
First $P(x)=C$ implies $C^2 = C \implies C=1\mbox{ or }C=0$.
Now say $P(x)$ is non-constant so $P(x) = (x-a_1)(x-a_2)...(x-a_n)$. Where $a_i$ are not necessrily distinct. (Note $P(x)$ must be a monic polynomial. This is easily to see. Just expand LHS and RHS to the leading term).
Thus,
$(x^2-a_1)...(x^2-a_n)(x^3-a_1)...(x^3-a_n) = (x-a_1)^5...(x-a_n)^5$.
Now here is the result that we need: The equation $x^3 - b = 0$ has complex zeros if and only if $b\not = 0$.
Look at the LHS at the factors $x^3 - a_i$. These has complex zeros if $a_i\not = 0$. But if this is the case then the RHS cannot have the same zeros because it has only real solutions. A contradiction! Thus, $a_i = 0$ for all $1\leq i \leq n$. And hence the only possibility is $P(x) = (x-0)(x-0)...(x-0) = x^n$.
And this works because if $P(x)=x^n$ we have $(x^2)^n (x^3)^n = (x^n)^5$.
Hence, $P(x) = 0\mbox{ or }1\mbox{ or }x^n$.
• October 19th 2007, 09:20 AM
Ununuquantium
I have some questions to this solution:
1. I cannot see why $P(x)$ is monic polynominal - I mean I see that it must have first coefficient equal to 1, but I cannot see why it must have integer coefficients!
2. I do not understand that $x^{3}-b$ have complex zeros when $b$ is not equal to $0$.
3. I do not understand why you mean that RHS has got only real zeros! - It has when $P(x)$ is monic??(why??) For example $x^{2}+x+1$ is monic polynominal - but is does not have real solution. That is why you cannot say that $a_{i}$ are ALL real zeros of $P(x)$
4. Why do you say that it is only a partial solution - maybe I did not said the $x$ is not real? So I am sorry, $x$ is REAL.
--------
Mayby someone has another solution??
• October 19th 2007, 09:49 AM
ThePerfectHacker
Quote:
Originally Posted by Ununuquantium
1. I cannot see why $P(x)$ is monic polynominal - I mean I see that it must have first coefficient equal to 1, but I cannot see why it must have integer coefficients!
The word "monic" means the first coefficient is 1 it does not say anything about the other coefficients.
Note: I made a mistake, I should have been more careful. If the first coefficient is $A$ then this equality leads to $(Ax^2+...)(Ax^3+....)=(A^5x^5+....) \implies (A^2x^5+...)=(A^5x^5+...)$. Thus, $A^2 = A^5 \implies A^2(A^3-1)=0$. We omit the solution $A=0$ because we are working with a polynomial of non-zero coefficient. Thus, $A^3 - 1 = 0$ and hence $A=1,\zeta , \zeta^2$ where $\zeta$ is the 3rd-root of unity.
Hence the first coeffcient can be $1,\zeta , \zeta^2$. But once you factor that coefficient you have a monic polynomial (the coefficients are not necessarily integers).
Quote:
2. I do not understand that $x^{3}-b$ have complex zeros when $b$ is not equal to $0$.
Do you something about complex numbers? The solution to the equation $x^3 = b$ is $\sqrt[3]{b} , \sqrt[3]{b}\zeta,\sqrt[3]{b}\zeta$ where $\zeta$ is the 3rd-root of unity. Now if $b \not = 0$ these solutions are non-zero and hence complex (provided that $b\in \mathbb{R}$).
Quote:
4. Why do you say that it is only a partial solution?????
We are considering all the polynomials that have only real zeros. The polynomials that have complex zeros are not being considered here.
Quote:
3. I do not understand why you mean that RHS has got only real zeros! - It has when $P(x)$ is monic - but I cannot see it, as I have written above.
By what is answered to #4 the polynomial $P(x)$ here has only real zeros. So $P^5(x)$ (the RHS) has only real zeros.
So with my minor mistake fixed the solutions should be this time (for polynomials having zero reals) $0,1,\zeta,\zeta^2,x^n,\zeta x^n,\zeta^2 x^n$.
EDIT: Ignore what I just said on the bottom. I forgot that you said "polynomials with real coefficients". So my original solution is correct.
• October 19th 2007, 10:04 AM
Ununuquantium
#1 - ok.
#2 :), my stupid mistake, ignore it, i did not see sth, sorry
#3What if we have $x^{2}+x+1$ - this polynominal has only complex solutions, and: it's coefficients are real, and x is real, and it is monic.
??
#4 ok
I just do not understand why you have written:
$P(x) = (x-a_1)(x-a_2)...(x-a_n)$
maybe some
$a_{i}$ are complex??
• October 19th 2007, 10:49 AM
ThePerfectHacker
Quote:
Originally Posted by Ununuquantium
#3What if we have $x^{2}+x+1$ - this polynominal has only complex solutions, and: it's coefficients are real, and x is real, and it is monic.
I just do not understand why you have written:
$P(x) = (x-a_1)(x-a_2)...(x-a_n)$
maybe some
$a_{i}$ are complex??
[/QUOTE]
What I did does not apply to this polynomial. Because I was only doing ones with real zeros not complex.
• October 19th 2007, 11:36 AM
Ununuquantium
Ok, sorry, I forgot what was the question exactly.
Thank you very much
• October 19th 2007, 12:26 PM
CaptainBlack
Quote:
Originally Posted by Ununuquantium
Determine all polynominals $P(x)$ (real coeffitients of course) such that for arbitrary number $x$ there is:
$P(x^{2})\times P(x^{3}) = (P(x))^{5}$
Let $P(x)=a_nx^n + a_mx^m + .. ,\ n>m \ge 0$ be the highest order
terms, then:
$<br /> (a_nx^n + a_mx^m + ..)^5=(a_nx^{2n} + a_mx^{2m} + ..)(a_nx^{3n} + a_mx^{3m} + .. ) <br />$
expanding these as far as the two highest order terms on each side:
$<br /> a_n^5x^{5n} + 5 a_n^4 a_m x^{4n+m}+..= a_n^2x^{5n}+a_na_mx^{3n+2m}+..<br />$
So by equating exponents of the second heighest order terms we conclude
that $n=m$ a contradiction. So $P(x)=a x^n$ if $n \ge 1$, and then we must
have $a^5=a^2$, so if $a$ is real $a=1$.
This all assumed that $n>1$, so we need also to check
if $P(x)\equiv 1$ and $P(x) \equiv 0$ are solutions, which they are.
RonL
• October 19th 2007, 12:29 PM
CaptainBlack
Quote:
Originally Posted by ThePerfectHacker
Hence, $P(x) = 0\mbox{ or }1\mbox{ or }x^n$.
This is overdone as $P(x)=1$ is a case of $P(x)=x^n$.
RonL
All times are GMT -8. The time now is 07:15 AM.
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http://physics.stackexchange.com/questions/44826/proca-equation-question?answertab=votes
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# Proca equation question
i'm having trouble understanding the Proca equation
$$(g_{\mu\nu}(\Box+\mu^2)-\partial_{\mu}\partial_{\nu})\varphi^{\nu}=0$$
where $g_{\mu\nu}$ is the metric and $\mu$ is a parameter. Where is the wrong assumption?
$$(g_{\mu\nu}(\partial_{\mu}\partial^{\mu}+\mu^2)-\partial_{\mu}\partial_{\nu})\varphi^{\nu}=0$$ $$((\partial_{\mu}g_{\mu\nu}\partial^{\mu}+g_{\mu\nu}\mu^2)-\partial_{\mu}\partial_{\nu})\varphi^{\nu}=0$$ $$((\partial_{\mu}\partial_{\nu}+g_{\mu\nu}\mu^2)-\partial_{\mu}\partial_{\nu})\varphi^{\nu}=0$$
which implies $$\mu^2=0$$ not the case. The problem is using $\Box =\partial_{\mu}\partial^{\mu}$ instead of other index, say $\alpha$?. In that case how can I show that
$$\partial_{\nu}\varphi^{\nu}=0$$
Any hit will be appreciated.
-Edit- Reference Bjorken et.al Relativistic Quantum Fields page 23
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## 2 Answers
Hint: Contract with derivative $\partial^{\mu}$ on both sides of your first equation (v1) to deduce that
$$\mu^2 \partial_{\nu}\varphi^{\nu}~=~0.$$
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2
$$\partial^{\mu}(g_{\mu\nu}(\Box+\mu^2)-\partial_{\mu}\partial_{\nu})\varphi^{\nu}=0$$ $$\partial^{\mu}g_{\mu\nu}\Box\varphi^{\nu}+\partial^{\mu}g_{\mu\nu}\mu^2 \varphi^{\nu}-\partial^{\mu}\partial_{\mu}\partial_{\nu}\varphi^{\nu}=0$$ $$\partial_{\nu}\Box\varphi^{\nu}+\partial_{\nu}\mu^2 \varphi^{\nu}-\Box\partial_{\nu}\varphi^{\nu}=0$$ Given that derivatives commute: $$\Box\partial_{\nu}\varphi^{\nu}+\mu^2\partial_{\nu} \varphi^{\nu}-\Box\partial_{\nu}\varphi^{\nu}=0 \Rightarrow \mu^2\partial_{\nu} \varphi^{\nu}=0$$ – Nivalth Nov 22 '12 at 11:24
I still don't have the reputation to comment everywhere so I leave this as an answer although it's a comment.
Be careful when defining the sum indexes. There are two things that you did wrong in your calculation.
1. You said correctly that $\square=\partial_\mu \partial^\mu$ but those are indexes to express the sum so wrinting$$g_{\mu\nu}\square=g_{\mu\nu}\partial_\mu \partial^\mu,$$ is wrong, you should write something like$$g_{\mu\nu}\square=g_{\mu\nu}\partial_\rho \partial^\rho.$$
2. The other thing you should be careful with is in the second step you put the metric between the partial derivatives but then you're are taking the partial derivative of the metric and that is not correct.
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About 1: yes, thanks, this is what I tried to mean with my $\alpha$ index About 2: There is no problem because the metric is constant, but thanks for the hint! – Nivalth Nov 22 '12 at 12:06
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http://mathhelpforum.com/number-theory/56077-gcd-polynomials.html
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# Thread:
1. ## GCD of polynomials
Hello,
could you help me understand the procedure of finding the GCD of two polynomials step by step, using the Euclidean Algorithm?
For example
$a = x^3 - 3x +2 , b = x - 1$
Thanks.
2. Originally Posted by drthea
Hello,
could you help me understand the procedure of finding the GCD of two polynomials step by step, using the Euclidean Algorithm?
For example
$a = x^3 - 3x +2 , b = x - 1$
Thanks.
There is no need to use Euclidean here and besides it does not work since it produces the answer in the first line.
Therefore, it does not give any infromation in this problem.
Just note $(x^3 - 3x + 2) = (x-1)(x^2 + x - 2)$ by long division.
Thus, $\gcd = x-1$
3. Hi and thanks for your answer.
Could you give me an example of where the Euclid's algorithm can be used,
concerning polynomials? Should they be polynomials of higher degree?
4. Originally Posted by drthea
Hi and thanks for your answer.
Could you give me an example of where the Euclid's algorithm can be used,
concerning polynomials?
When using the Euclidean algorithm you repeating apply the division algorithm in several steps until you get rid of the remainder. The problem here is that first application already produces no remainder. That is why this is not a good example.
Should they be polynomials of higher degree?
Look at this.
It has examples for both numbers and polynomials.
Both are extremely cases similar.
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http://mathhelpforum.com/advanced-algebra/130887-prove-there-matrix-b-print.html
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# Prove there is a matrix B
Printable View
• February 26th 2010, 09:24 AM
Runty
Prove there is a matrix B
Let $A$ be an $m\times n$ matrix and suppose that $X$ is an $n\times p$ matrix such that every column of $X$ is in the column space of $A$. Prove there is a matrix $B$ such that $AX=B$
[Hint: we know that the equation $Ax=v$ has a solution if and only if $v\in col(A)$]
I don't know if this is exactly how the question is meant to be said. Our Prof. made a severe typo in the original, so I'm asking him to simply rewrite the question in its intended form rather than just provide us with a quick correction (which could be easily misinterpreted).
• February 28th 2010, 07:52 AM
Runty
I've consulted a few other sources for a possible answer to this question, but I am still in need of a formulaic answer. One such answer read as this:
"For every column b of B, you get a column solution to Ax=b.
Assemble all of those columns x into X (in the same order that you got them from b in B). For example, the third column of X will be a solution to Ax=b, where b is the third column of B."
How would I show this in terms of mathematical formulas? Or is that thinking the wrong way?
All times are GMT -8. The time now is 09:50 AM.
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http://math.stackexchange.com/questions/198404/is-this-scenario-possible-and-if-so-find-the-equation-of-the-hill/198454
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# Is this scenario possible and if so, find the equation of the hill
Assume a ball is placed on a hill and the time is measured for the ball to reach the ground (y=0). If the time taken for the ball to reach the ground is independent on the position the ball is placed on the hill, determine the equation or set of possible equations for the hill (y as a function of x).
If this is not possible can you provide a proof of why this is not possible?
Also assume gravity is 9.8m/s.
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6
– Brian M. Scott Sep 18 '12 at 7:08
Yes that does seem to answer what I was asking. Thanks! – Chris Sep 18 '12 at 7:10
@Brian: That's what might be called the winter-games version of the problem; but the summer-games version has the same form of solution, just a different travel time; see my answer. – joriki Sep 18 '12 at 10:14
## 1 Answer
Brian's solution remains valid if we make a slightly more realistic assumption by ignoring dissipative friction but taking friction to be strong enough to cause the ball to roll rather than slide. In this case the ball's energy is given by
$$\begin{align} E &=\frac12mv^2+\frac12I\omega^2+mgz \\ &=\frac12(1+\alpha)mv^2+mgz\;, \end{align}$$
where the factor $\alpha=I/(mR^2)$, with $I$ the ball's moment of inertia and $R$ its radius, depends on the mass distribution and is $\frac25$ for a solid ball of constant density.
Thus the rotational energy effectively increases the ball's inertial mass, or equivalently reduces the gravitational acceleration. The solution is still a cycloid; only the time required to reach the ground is now
$$T=\pi\sqrt\frac{r(1+\alpha)}g$$
(where $r$ is the radius of the cycloid), i.e. the rolling ball takes slightly longer than the sliding ball, by a factor of $\sqrt{7/5}\approx1.18$ in the case of a solid ball of constant density.
Note that this treatment ignores the fact that for a ball with non-zero radius $R$ the ball's centre of mass moves on a different curve than its surface; this approximation is only valid for $r\gg R$.
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I’m glad that someone did this; the last time I did a physics problem of this kind was in 1965, and I’m just a wee bit rusty! – Brian M. Scott Sep 18 '12 at 18:01
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http://math.stackexchange.com/questions/18890/how-to-visualize-points-on-a-high-dimensional-3-manifold
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# How to Visualize points on a high dimensional (>3) Manifold?
Are there any ways to visualize(plot/draw) points on a high dimensional (ex: dimension = 5) manifold?
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4
Heh, there's a joke about that. An engineer goes with a mathematician friend to a geometry seminar. Afterwards, the mathematician asks him what he thinks. "Well, when he gave the examples in one and two dimensions, I followed it. But how do you guys picture high dimensional manifolds, say 5?" "Simple, start by picturing an $N$-dimensional manifold, then set $N= 5$." – Willie Wong♦ Jan 25 '11 at 12:08
1
Exactly the same ways with which you can visualize 2- and 3-manifolds: considering projections, level sets for various functions, etc. – Mariano Suárez-Alvarez♦ Jan 25 '11 at 12:22
## 2 Answers
A usual way to work is to take slices, CAT-scan style.
Have you seen a CAT-scan? It represents a 3-dimensional object as sequence of two-dimensional slices of the object, in a way that you can choose which directions your slices cut (after computer processing).
There are two usual ways of representing a 3-dimensional object on a plane. You can take projections (think a picture of the "outside" of the object) or slices. Similarly, to represent higher ($N$) dimensional objects on a plane (draw it on a piece of paper), you must throw away $N-2$ degrees of freedom. Each degree of freedom you can choose to throw away by projection or by slicing. So a typical way of visualizing a 4-dimensional object is to take a sequence of 3-dimensional slices, and project each slice onto a computer screen, and display this sequence one-by-one over time. This way you get the illusion of a 4 dimensional object "passing through" our three dimensional world.
Higher dimensional objects can be done similarly, you just need to take more slices.
One fun thing you can do with this is to try to play 4-dimensional tic-tac-toe. To make the game not a trivial win for the first player, you need a $5\times 5\times 5 \times 5$ grid, and first to make 5 in a row wins. So on paper you need to draw a $5\times 5$ matrix of $5\times 5$ grids, each grid represent the intersection of two orthogonal hyper-slices of the full 4-dimensional game grid.
Another fun way to develop some intuition on higher dimension spaces is to spend some time playing Magic Cube 4D, a higher dimensional analogue of Rubik's cube. (It could also get you dizzy pretty quickly).
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4D object passing through.. ie time! – bobobobo May 20 '12 at 22:47
Well, here's how I think about it.
Every time you add a dimension, you need an infinite number of the previous dimension.
1D is the x-axis. You need a line to visualize.
1D -> 2D means you need an infinite number of additional x-axes (call them y-axes). You need a plane to visualize (flatland picture).
2D -> 3D means you need an infinite number of additional xy planes. You need a space to visualize (3d picture).
3D -> 4D means you need an infinite number of additional xyz spaces. You need a movie to visualize.
4D -> 5D means you need infinitely many vhs cassettes, each with a movie on it to visualize.
So buy some VHS! They are cheap now.
(really, like Willie Wong said, to take a "slice" of the 5d space, by fixing one of the variables. Then, you can create a "movie" of the other 4 variables (use 3 for space, and 1 for time). to visualize the whole space though, you would need an infinite number of these "movies" (and would each be infinite in duration), for each value you want to the 5th variable to).
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http://mathoverflow.net/questions/59812/are-gits-good-categorical-quotients-just-locally-ringed-space-coequalizers
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## Are GIT’s good categorical quotients just locally ringed space coequalizers?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Introduction: The definition of "good categorical quotient" in geometric invariant theory (given below) seems fairly ad hoc to me, except that it looks very similar to the coequalizer of the action in the category of locally ringed spaces. I'm trying to understand this similarity.
Some background: Suppose we have a group action $G\times X \to X$ in the category $\bf Sch$ of schemes. The category $\bf LRS$ of locally ringed spaces is cocomplete (see propositiion 1.12 in A. Vezzani's thesis), so in particular, $G\times X \rightrightarrows X$ has a $\bf LRS$-coequalizer; call it $Q$. One can construct it as the coequalizer in the category of topological spaces equipped with the sheaf of $G$-invariant functions to make it a locally ringed space, as Vezzani's proposition explains more carefully.
If $G\times X \rightrightarrows X$ has a $\bf Sch$-coequalizer $Y$, then GIT calls $Y$ a "categorical quotient"; see, for example, R. Birkner's Introduction to GIT. Note that here we get a universal map $u: Q\to Y$ in $\bf LRS$.
$Y$ is called a "good categorical quotient" if it satisfies 3 properties, also found in Birkner's notes, which I prefer to re-express here in terms of the universal map $u$:
i) $u^\sharp : {\cal O}_Y \to u _* {\cal O}_Q$ is an isomorphism.
ii) $u: Q\to Y$ is a closed map (at the level of topological spaces).
iii) $u$ takes disjoint closed sets to disjoint closed sets.
(These together imply that $X\to Y$ is a categorical quotient, i.e. $\bf Sch$-coequalizer; see Mumford's GIT I.2 remark 6.)
All of these properties are clearly satisfied if $u$ is an isomorphism. That is, if the $\bf LRS$-coequalizer is a scheme, that scheme is a good categorical quotient.
Towards a converse, (i) implies that $u$ must be dominant, and hence by (ii) it is surjective. And (iii) implies it is injective on closed points, so it's looking a lot like an isomorphism. Can anything more be said?
1) When is a good categorical quotient an $\bf LRS$-coequalizer of the group action? I.e., are there nice and general conditions where $u$ must be an isomorphism?
2) If the answer is not always, what's a counterexample? (Answered by Anton)
Thanks!
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1
The cocompletness of LRS is not due to Vezzani, it's already contained in Groupes algebriques (Demazure, Gabriel), 1970. I don't know the answer to your problem, but usually the notions concerning colimits of schemes are motivated by explicit applications, and not by abstract nonsense. I doubt that there is any connection. – Martin Brandenburg Mar 28 2011 at 10:19
2
I agree with Martin (and also don't know the answer). The goodness is a kind of wish list of properties that make the quotient useful in applications. Even if it (by chance) should be the same as the LRS quotient that would be more or less useless as there does not seem to be any way of really using that a scheme is the LRS quotient of a schematic equivalence relation. – Torsten Ekedahl Mar 28 2011 at 10:56
@ Martin, I prefer to give links to online references when possible. Thanks for the book recommendation. – Andrew Critch Mar 28 2011 at 16:56
## 1 Answer
The answer to the title question is NO. If $X\rightrightarrows Y\to Z$ is a coequalizer in $\bf LRS$, then the underlying topological space of $Z$ is the coequalizer of the underlying topological spaces (and the structure sheaf is the equalizer of pushforwards of the structure sheaves). On the other hand, good categorical quotients identify any two points whose orbit closures intersect. So one of my favorite good quotients, $\mathbb A^1/\mathbb G_m$, provides a counterexample. The good categorical quotient in schemes has a single point as its underlying topological space, but the quotient in $\bf LRS$ has two points.
I think the bad behavior on underlying topological spaces is probably the only problem. That is, a good categorical quotient is a coequalizer in $\bf LRS$ precisely when it is a geometric quotient (i.e. when we impose the additional condition that each fiber is a single orbit).
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Thanks, Anton! But the relationship between (good) geometric quotients and categorical quotients is not clear to me... The LRS coequalizer Q, being a topological coequalizer, has as its topological fibres the "G orbits" in the sense that if there exists a (g,x,p) \in |GxX| (here p \in Spec(k(g)\otimes k(x)) mapping via the action to y \in X, then x and y map to the same point of Q, and conversely. But I don't see what this means for the geometric fibres ... can this be cleared up? – Andrew Critch Mar 28 2011 at 18:30
...When X and G are finite type over an algebraically closed field, at least over closed points the topological fibres and geometric fibres are the same... but even for varieties, over non-closed points these may differ. For example when flipping a parabola about its axis: Say G is Z/2 acting on X=k[x,y]/(y=x^2) by flipping sign of the x coordinate, and X->Y is projection to the Y axis... the topological fibre over the generic point is 1 point, but a geomteric fibre over the generic point has 4 points... hmm... – Andrew Critch Mar 28 2011 at 18:54
Your question is only about the second paragraph, right? Say $q\in Q$ is a point of the LRS coequalizer, and $F=X\times_Q Spec(k(q))$ is the fiber (note: I think this is the topological fiber ; does the proof of Hartshorne, Ex. II.3.10 work in LRS?). Saying that $F$ is a single $G$-orbit is the same as saying that the map $G\times F\to F\times F$ (given by $(g,f)\mapsto (g\cdot f,f)$) is surjective. Surjectivity of this map can be checked on geometric points, so the topological fibers of $X\to Q$ are single $G$-orbits if and only if the geometric fibers are single $G$-orbits. – Anton Geraschenko♦ Mar 28 2011 at 19:06
Whoops, I missed your second comment while composing the above. I think a geometric fiber over the generic point in your example only has 2 points, and that $G=\mathbb Z/2$ swaps them. – Anton Geraschenko♦ Mar 28 2011 at 19:13
Bleh ... it's not true that a fiber is a single $G$-orbit topologically iff $G\times F\to F\times F$ is surjective. In the example of the parabola mapping to the line, the trivial group $G$ acts on the fibers. Over the generic point, the fiber $F$ is one point topologically, but $G\times F\to F\times F$ is the inclusion of one point into two points. Nevertheless, the direction I actually used is true: if $G\times F\to F\times F$ is surjective (which it is for a good geometric quotient), then $F$ is topologically a single $G$-orbit. – Anton Geraschenko♦ Mar 28 2011 at 19:20
show 2 more comments
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http://mathhelpforum.com/statistics/203850-proof-semi-graphoid-symmetry-axiom.html
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# Thread:
1. ## Proof of semi-graphoid symmetry axiom
I have some trouble understanding why this proof is a proof:
Lemma:
$I_p(X, Z, Y) \iff I_p(Y, Z, X)$
Proof:
$I_p(X, Z, Y) \iff P(C_x | C_y \land C_z) = P(C_x | C_z)$
Ok, this is just conditional independence, x independent of y given z.
$\iff \frac{P(C_x \land C_y \land C_z)}{P(C_y \land C_z)} = \frac{P(C_x \land C_z)}{P(C_z)}$
This is rewritten according to the rule of conditional independence P(A|B) = P(A & B) / P(B).
Bottom up the same is done for the right hand side of the lemma, but I don't get how the following and previous step are equal.. Why does this prove anything?
$\iff \frac{P(C_x \land C_y \land C_z)}{P(C_x \land C_z)} = \frac{P(C_y \land C_z)}{P(C_z)}$
$\iff P(C_y | C_x \land C_z) = P(C_y | C_z)$
$\iff I_p(Y, Z, X)$
2. ## Re: Proof of semi-graphoid symmetry axiom
Hey Lepzed.
Is the random variable X independent to Z?
3. ## Re: Proof of semi-graphoid symmetry axiom
The independence relation $I_p(X, Z, Y)$ expresses that in the context of information about Z, information about Y is irrelevant with respect to Y or put differently: X is condtionally independent of Y given Z.
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http://mathoverflow.net/revisions/76500/list
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## Return to Answer
2 edited body
The dimension of $S_{21}(V)$ is $(n+1)n(n-1)/3$; it is only $8$ if $n=3$. The other half of Schur Weyl duality says that $S_{21}(V)$ is a $GL_n$ (not $S_n$) S_3$) irrep; namely, the one which is indexed by the partition$(2,1)$. Similarly,$\mathrm{Sym}^3(V)$and$\bigwedge^3 V$have dimensions$(n+2)(n+1)n/6$and$n(n-1)(n-2)/6$and are$GL_n$, not$S_3\$, irreps.
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The dimension of $S_{21}(V)$ is $(n+1)n(n-1)/3$; it is only $8$ if $n=3$. The other half of Schur Weyl duality says that $S_{21}(V)$ is a $GL_n$ (not $S_n$) irrep; namely, the one which is indexed by the partition $(2,1)$. Similarly, $\mathrm{Sym}^3(V)$ and $\bigwedge^3 V$ have dimensions $(n+2)(n+1)n/6$ and $n(n-1)(n-2)/6$ and are $GL_n$, not $S_3$, irreps.
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http://mathhelpforum.com/calculus/133800-vector-problem.html
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# Thread:
1. ## vector problem
Describe the shape of the intersection of the plane z=-3 and the plane y=z in three space. I drew out a diagram of z=-3 but im not sure what to do with y=z i know that from what i've learned, every point on y is equal to a point on z and that the x coordinates can be any value.
2. Originally Posted by surffan
Describe the shape of the intersection of the plane z=-3 and the plane y=z in three space. I drew out a diagram of z=-3 but im not sure what to do with y=z i know that from what i've learned, every point on y is equal to a point on z and that the x coordinates can be any value.
Dear surffan,
Since $z=-3~ and~ y=z\Rightarrow{z=y=-3}$
This is a straight line for any x value. Please see the attachement so you will understand this correctly.
Attached Thumbnails
3. If you look at just the y-z plane, the equation y=z describes a line. In 3-space, the plane y=z is that same line extended in the positive-x and negative-x direction. You have y=z and x=anything.
The intersection of 2 planes is normally a line. The exceptions are when the two planes are parallel (intersection is empty) or coincident (intersection is a plane).
Ignore the above - Sudharaka's post and picture show it much better. The purple plane going diagonally is the plane y=z, and the pink horizontal plane is z=-3.
4. thank you so much, i couldn't visualize it at all, this is a huge help
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http://math.stackexchange.com/questions/178741/additive-quotient-group-mathbbq-mathbbz-is-isomorphic-to-the-multiplicat/178771
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# Additive quotient group $\mathbb{Q}/\mathbb{Z}$ is isomorphic to the multiplicative group of roots of unity
I would like to prove that the additive quotient group $\mathbb{Q}/\mathbb{Z}$ is isomorphic to the multiplicative group of roots of unity.
Now every $X \in \mathbb{Q}/\mathbb{Z}$ is of the form $\frac{p}{q} + \mathbb{Z}$ for $0 \leq \frac{p}{q} < 1$ for a unique $\frac{p}{q} \in \mathbb{Q}.$ This suggest taking the map $f:\mathbb{Q}/\mathbb{Z} \mapsto C^{\times}$ defined with the rule $$f(\frac{p}{q} + \mathbb{Z}) = e^{\frac{2\pi i p}{q}}$$ where $\frac{p}{q}$ is the mentioned representative.
Somehow I have problems showing that this is a bijective function in a formal way. I suspect I do not know the properties of the complex roots of unity well enough.
Can someone point me out (perhaps with a hint) how to show that $f$ is injective and surjective?
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– Martin Sleziak Aug 4 '12 at 10:45
I will say every $\mathbb{Q/Z}$ is isomorphic to $X = \left\{x : x \in [0,1] \text{ and }x \in \mathbb{Q} \right\}$ – Jayesh Badwaik Aug 4 '12 at 10:46
From the geometrical form ($e^{i\varphi}$ corresponds to angle $\varphi$) you have: $e^{(2\pi p)/q}=1$ implies $2\pi p/q=2\pi k$ for some $k\in\mathbb Z$; i.e. $\frac pq\in\mathbb Z$. Was this what you had problem with? – Martin Sleziak Aug 4 '12 at 10:49
You probably mean i.e $\frac{p}{q} \in Q$ right? I don't quite follow your argument here. Could you please elaborate a bit more? Thanks. – Jernej Aug 4 '12 at 11:59
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BTW you can read here how to reply in comments. (It was only accident that I came back to this question and I saw your comment directed to me. If you want to get attention of other users who previously left comment, you can use `@username`.) – Martin Sleziak Aug 4 '12 at 13:23
show 1 more comment
## 3 Answers
To prove it is a bijection, one can use rather "primitive" methods. suppose that:
$f\left(\frac{p}{q} + \Bbb Z\right) = f\left(\frac{p'}{q'} + \Bbb Z\right)$,
then: $e^{2\pi ip/q} = e^{2\pi ip'/q'}$, so $e^{2\pi i(p/q - p'/q')} = 1$.
This, in turn, means that $\frac{p}{q} - \frac{p'}{q'} \in \Bbb Z$, so the cosets are equal. Hence $f$ is injective.
On the other hand, if $e^{2\pi i p/q}$ is any $q$-th root of unity, it clearly has the pre-image $\frac{p}{q} + \Bbb Z$ in $\Bbb Q/\Bbb Z$ (so $f$ is surjective).
One caveat, however. You haven't actually demonstrated $f$ is a function (i.e., that it is well-defined, although if you stare hard at the preceding, I'm sure it will come to you).
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This is what I thought. But somehow I wasn't sure that 1. Every q-th root of unity is of the form $e^{\frac{2 \pi i p}{q}}$ and consequently that for every n-th root of unity z, $e^k=1$ if and only if $n|k$. As for the well defined remark isn't that implied by the uniqueness of the representative? – Jernej Aug 4 '12 at 18:40
Be canonical!
You have a morphism of groups $ex:\mathbb R \to S^1: r\mapsto e^{2i\pi r}$, where $S^1$ is the multiplicative group of complex numbers with $\mid z\mid=1$. This morphism is surjective and has kernel $\mathbb Z$.
[The wish to have kernel $\mathbb Z$ instead of $2\pi \mathbb Z$ dictated the choice of $ex(r)=e^{2i\pi r}$ instead of $e^{ir}$].
Restricting the morphism to $\mathbb Q$ induces a morphism $res(ex):\mathbb Q\to S^1$ with kernel $\mathbb Q\cap \mathbb Z=\mathbb Z$ and image $\mu_\infty\stackrel {def}{=} e^{2i\pi \mathbb Q} \subset S^1$.
The crucial observation is that this image is $\mu_\infty=\bigcup_n \mu_n$, where $\mu_n$ is the set of $n$-roots of unity $e^{\frac {2i\pi k}{n}}\quad (k=1,2,...,n)$.
Hence $\mu_\infty$ is the set of all roots of unity i.e. the set of complex numbers $z \in \mathbb C$ with $z^n=1$ for some $n\in \mathbb N^*.$
Applying Noether's isomorphism you finally get the required group isomorphism (be attentive to the successive presence and absence of a bar over the $q$ in the formula) $$Ex: \mathbb Q/\mathbb Z \xrightarrow {\cong} \mu_\infty:\overline {q}\mapsto e^{2i\pi q}$$
A cultural note
This elementary isomorphism is actually useful in quite advanced mathematics.You will find it, for example, in Grothendieck's Classes de Chern et représentations linéaires des groupes discrets.
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Define $\,f: \Bbb Q\to S^1:=\{z\in \Bbb C\;:\;|z|=1\}\,\,,\,f(q):=e^{iq}\,$ , show this is a homomorphism of groups, find its kernel and use the fist isomorphism theorem.
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http://mathoverflow.net/questions/66954/collapsing-cyclic-quotient-singularities
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## Collapsing Cyclic quotient Singularities
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $X$ be a surface with cyclic quotient singularities, and let $f:X\rightarrow Y$ be a birational morphism with exceptional locus $E$. Assume $E\cong\mathbb{P}^{1}$, $x_{1},x_{2}\in E$ two singular points of $X$, of type $\frac{1}{r}(a_{1},a_{2})$ and $\frac{1}{m}(b_{1},b_{2})$ respectively. Let $y = f(E)\in Y$ be the point on which $E$ is contracted.
What can one say on the type of singularity of $y\in Y$ ?
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http://physics.stackexchange.com/questions/21793/entropy-change-in-a-cycle-with-two-isochoric-and-two-adiabatic-processes?answertab=active
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# Entropy change in a cycle with two isochoric and two adiabatic processes [closed]
Prove that the change of Entropy in a cycle with two isochoric and two adiabatic processes is 0. How can I prove that? Thanks!
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Welcome to Physics! This is a site for conceptual questions about physics, not general homework help. If you can edit your question to ask about the specific physics concept that is giving you trouble, I'll be happy to reopen it. For instance, describe how you tried to prove it and what went wrong in the process. See our FAQ and homework policy for more information. – David Zaslavsky♦ Mar 3 '12 at 19:31
## closed as too localized by David Zaslavsky♦Mar 3 '12 at 19:29
This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, see the FAQ.
## 2 Answers
We don't give full solutions here. But here is a hint: Write $dS=\frac{q}{T}$ and apply first law.
Or just say: Entropy is a state function, so it does not change in a cyclic process.
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Entropy is a state function, so if your system ends up in the same state as it started and there has been no net heat flow the entropy change must be zero.
Are you allowed to assume it's an ideal gas? If so draw a PV diagram of the cycle. The two isochoric stages are vertical lines, and the two adiabatic stages are the usual curves of PV = nRT. Using this relation you can show the temperature change in the two isochoric stages is the same, and therefore that the $\Delta Q$ is the same (assuming $C_v$ is constant). This means the net entropy change in the cycle is zero. To do the calculation start at one point ($P_1$, $V_1$) and work your way round the cycle using PV = nRT to calculate the pressure at the remain three points.
If you're not allowed to assume it's an ideal gas I'd have to think about it because you can no longer use PV = nRT.
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http://nrich.maths.org/5952/solution?nomenu=1
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## 'Escape from Planet Earth' printed from http://nrich.maths.org/
### Show menu
Steven from City of Sunderland College sent us in a wonderful, complete solution to this problem, which we recommend reading in full to any budding problem solver. The main points are as follows:
The gravitational potential energy of a cannon ball of mass $m$ at a distance $r$ from the centre of a planet of mass $M$ is
V = -\frac{GMm}{r}
The kinetic energy of a cannon ball launched atspeed $v$ is
KE = \frac{1}{2}mv^2
Suppose that a cannon ball just escapes the pull of a planet and makes it to infinity. At this point, both its potential and kinetic energies will be zero. Thus, the initial kinetic and potential energies must sum to zero. So, if launched from a planet of radius $R$ we must have
\frac{GMm}{R} = \frac{1}{2}mv^2
This gives the escape velocity $v$ as
v =\sqrt{\frac{2GM}{R}}
Putting in the numbers for Earth gives
v=\sqrt{\frac{2\times 6.674\times 10^{-11}\times 5.9763\times 10^{24}}{6.378\times 10^6}}=11.2km s^{-1}
For the moon, Jupiter and the sun the escape velocity changes by the relative change in the factor $\sqrt{\frac{M}{R}}$. For the moon, Jupiter and the sun these are 0.2122, 5.32 and 55.26, giving rise to escape velocities of
Moon: 2.37 km s$^{-1}$
Jupiter: 59.5 km s$^{-1}$
Sun: 619 km s$^{-1}$
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http://mathoverflow.net/questions/81963/regular-functions-on-slodowy-slices
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## Regular Functions on Slodowy Slices
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I recently learned that there is a natural identification, given a simple Lie algebra,
Polynomial functions on the Slodowy slice of a regular nilpotent orbit $\simeq$ polynomial functions on the Cartan subalgebra invariant under the Weyl group.
I guess there's a generalization of this statement for subregular or other nilpotent orbits. Which book/article should I have a look at? For example, I guess there's a equivalence of the form
Polynomial functions on the Slodowy slice of a subregular nilpotent orbit $\simeq$ polynomial functions on the Levi subalgebra of the form $sl_2\times \mathbb{C}^{r-1}$ satisfying conditions *
What are the conditions * I need to fill in the statement above?
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## 1 Answer
The Slodowy slice to a nilpotent $e$ is by definition the set of elements $x$ such that $x-e$ commutes with some fixed $f$ such that $e$ and $f$ generate an $\mathfrak{sl}_2$. Thus, it is an affine space modeled on the commutant of $f$, a Lie subalgebra of $\mathfrak{g}$.
In the regular case, this subalgebra isn't a torus (it's nilpotent), though it is a flat limit of tori (just as a regular nilpotent is a limit of semi-simples). In $\mathfrak{sl}_n$, if we take the principal nilpotent, its commutant is the space of matrices with 0's on or below the diagonal, and with entries constant on every shifted diagonal.
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Thank you, I learned the definition of Slodowy slice recently. I'm now more interested in understanding the "Miura map", see Premet's eprints.ma.man.ac.uk/495/02/GGG.pdf , p. 28, in the case of sub regular orbit. This should give a homomorphism from $\mathbb{C}[S]$ to $\mathbb{C}[sl_2\times \mathbb{C}^{r-1}]$... – Yuji Tachikawa Nov 27 2011 at 6:51
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http://math.stackexchange.com/questions/117351/the-lebesgue-differentiation-theorem-holds-for-any-measure
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# The Lebesgue differentiation Theorem holds for any measure?
What are the assumptions to a measure to the Lebesgue Differentiation Theorem be True?
What is(if there is) a practical way to find the decomposition of a measure $\lambda$ in relation to a measure $\mu$?
OBS: I am only interested on non negative measure, but answers dealing with signed measures are welcomed as well.
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Radon-Nikodym theorem will be a nice reference. – sos440 Mar 7 '12 at 3:28
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http://mathoverflow.net/questions/53988/what-is-the-motivation-for-a-vertex-algebra/54029
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## What is the motivation for a vertex algebra?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
The mathematical definition of a vertex algebra can be found here:
http://en.wikipedia.org/wiki/Vertex_operator_algebra
Historically, this object arose as an axiomatization of "vertex operators" in "conformal field theory" from physics; I don't know what these phrases mean.
To date, I haven't been able to gather together any kind of intuition for a vertex algebra, or even a precise justification as to why anyone should care about them a priori (i.e. not "they come from physics" nor "you can prove moonshine with them").
As far as I am aware, theoretical physics is about finding mathematical models to explain observed physical phenomena. My questions therefore are:
What is the basic physical phenomenon/problem/question that vertex operators model?
What is the subsequent story about vertex operators and conformal field theory, and how can we see that this leads naturally to the axioms of a vertex algebra?
Are there accessible physical examples ("consider two particles colliding in an infinite vacuum...", etc.) that illustrate the key ideas?
Also, are there alternative, purely mathematical interpretations of vertex algebras which make them easier to think about intuitively?
Perhaps people who played a role in their discovery could say a bit about the thinking process that led them to define these objects?
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The main problem for mathematicians who want to learn about QFT is that there is no universal definition of what a QFT is. I think Wightman axioms are a big help to start. The first chapter of Kac's "Vertex algebras for beginners" shows how the notion of vertex algebra arises naturally from Wightman axioms. – YBL Feb 1 2011 at 16:11
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But is there a "fundamental example" that captures what a QFT is supposed is to do (or be)? I would be happy just to hear a story about a particular physical system, in a way that illustrates the important features of QFT. – AH Feb 1 2011 at 16:19
Have you seen ncatlab.org/nlab/show/vertex+operator+algebra ? – Qiaochu Yuan Feb 1 2011 at 16:26
You ask what physical problem vertex operators model but you actually give the answer yourself! :-) They can be used to answer questions about "two particles colliding in an infinite vacuum". A pair of strings coming from infinity, interacting "once", and going off to infinity, say, sweep out a surface that is topologically a sphere with 4 tubes sticking out out of it. String theory is (sort of) conformally invariant and this surface is conformally a Riemann sphere with 4 punctures in it. Vertex operators arise when studying quantum fields on Riemann spheres in the vicinity of these punctures. – Dan Piponi Feb 1 2011 at 19:13
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Addendum to Qiaochu's comment: Also see ncatlab.org/nlab/show/conformal+field+theory and its reference Martin Schottenloher: "A mathematical introduction to conformal field theory", for a pre-vertex-algebra introduction to conformal field theory. – Tim van Beek Feb 2 2011 at 13:07
## 8 Answers
Vertex algebras precisely model the structure of "holomorphic one-dimensional algebra" -- in other words, the algebraic structure that you get if you try to formalize the idea of operators (elements of your algebra) living at points of a Riemann surface, and get multiplied when you collide.
Our geometric understanding of how to formalize this idea has I think improved dramatically over the years with crucial steps being given by the point of view of "factorization algebras" by Beilinson and Drinfeld, which is explained (among other places :-) ) in the last chapter of my book with Edward Frenkel, "Vertex algebras and algebraic curves" (second edition only). This formalism gives a great way to understand the algebraic structure of local operators in general quantum field theories -- as is seen in the recent work of Kevin Costello -- or in topological field theory, where it appears eg in the work of Jacob Lurie (in particular the notion of "topological chiral homology").
In fact I now think the best way to understand a vertex algebra is to first really understand its topological analog, the structure of local operators in 2d topological field theory. If you check out any article about topological field theory it will explain that in a 2d TFT, we assign a vector space to the circle, it obtains a multiplication given by the pair of pants, and this multiplication is commutative and associative (and in fact a commutative Frobenius algebra, but I'll ignore that aspect here). It's very helpful to picture the pair of pants not traditionally but as a big disc with two small discs cut out -- that way you can see the commutativity easily, and also that if you think of those discs as small (after all everything is topologically invariant) you realize you're really describing operators labeled by points (local operators in physics, which we insert around a point) and the multiplication is given by their collision (ie zoom out the picture, the two small discs blend and look like one disc, so you've started with two operators and gotten a third).
Now you say, come on, commutative algebras are SO much simpler than vertex algebras, how is this a useful toy model? well think about where else you've seen the same picture -- more precisely let's change the discs to squares. Then you realize this is precisely the picture given in any topology book as to why $\pi_2$ of a space is commutative (move the squares around each other). So you get a great model for a 2d TFT by thinking about some pointed topological space X.. to every disc I'll assign maps from that disc to X which send the boundary to the basepoint (ie the double based loops of X), and multiplication is composition of loops -- i.e. $\Omega^2 X$ has a multiplication which is homotopy commutative (on homotopy groups you get the abelian group structure of $\pi_2$). In homotopy theory this algebraic structure on two-fold loops is called an $E_2$ structure.
My claim is thinking about $E_2$ algebras is a wonderful toy model for vertex algebras that captures all the key structures. If we think of just the mildest generalization of our TFT story, and assign a GRADED vector space to the circle, and keep track of homotopies (ie think of passing from $\Omega^2 X$ to its chains) we find not just a commutative multiplication of degree zero, but a Lie bracket of degree one, coming from $H^1$ of the space of pairs of discs inside a bigger disc (ie from taking a "residue" as one operator circles another). This is in fact what's called a Gerstenhaber algebra (aka $E_2$ graded vector space). Now all of a sudden you see why people say you can think of vertex algebras as analogs of either commutative or Lie algebra (they have a "Jacobi identity") -- -the same structure is there already in topological field theory, where we require everything in sight to depend only on the topology of our surfaces, not the more subtle conformal geometry.
Anyway this is getting long - to summarize, a vertex algebra is the holomorphic refinement of an $E_2$ algebra, aka a "vector space with the algebraic structure inherent in a double loop space", where we allow holomorphic (rather than locally constant or up-to-homotopy) dependence on coordinates.
AND we get perhaps the most important example of a vertex algebra--- take $X$ in the above story to be $BG$, the classifying space of a group $G$. Then $\Omega^2 X=\Omega G$ is the "affine Grassmannian" for $G$, which we now realize "is" a vertex algebra.. by linearizing this space (taking delta functions supported at the identity) we recover the Kac-Moody vertex algebra (as is explained again in my book with Frenkel).
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For a nice exposition of the topological 2d picture there is Kock's book: mat.uab.es/~kock/TQFT.html – Qiaochu Yuan Feb 1 2011 at 18:16
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The original motivation for vertex algebras is explained briefly in the original paper http://www.jstor.org/stable/27441 as follows. For any even lattice one can construct a space $V$ acted on by vertex operators corresponding to lattice vectors. More generally one can write down a vertex operator for every element of $V$. These vertex operators satisfy some complicated relations, which are then used as the definition of a vertex algebra. In other words, the original example of a vertex algebra was the vertex algebra of an even lattice, and the definition of a vertex algebra was an axiomatization of this example.
This was motivated by my attempt to understand Igor Frenkel's work on the Lie algebra with Dynkin diagram the Leech lattice. Frenkel had constructed a representation of this Lie algebra using vertex operators acting on the space $V$, and I was trying to use his work to understand its root multiplicities. I did not use any insights from conformal/quantum/topological field theory or operator product expansions when defining vertex algebras (as implied by some of the other answers), for the simple reason that I had barely heard of these concepts and had almost no idea what they were.
This is not all that helpful for understanding what a vertex algebra really is. A better view is to regard them as something like a commutative ring with an action of the (formal) 1-dimensional additive group. In particular any such ring is canonically a vertex algebra. The difference between rings with group actions and vertex algebras is that the "multiplication" of a vertex algebra from $V\times V$ to $V$ is not defined everywhere: it can be thought of as a "rational map" rather than a "regular map" in some sense. More precisely if we write $u^g$ for the action of a group element $g$ on $u$, then the product $u^gv^h$ is not defined for all group elements $g$ and $h$, but behaves formally like a meromorphic function of $g$ and $h$, which in particular may have poles when $g$ and $h$ are trivial. Making sense of this informal idea produces the definition of a vertex algebra. (For more details see the unpublished paper http://math.berkeley.edu/~reb/papers/bonn/bonn.pdf.) This means that vertex algebras behave like commutative rings: for example, one should define modules over them, tensor products of modules, and so on.
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The answers here have focused on the mathematical aspects of VOAs and the motivation coming from QFT, the specialization to Conformal Field Theory, and then the further specialization to two-dimensional holomorphic CFT. Two-dimensional CFT's did arise from string theory where the fields on the 2d world-sheet of the string define a CFT. However there is another important part of the story which has not been mentioned and is more directly tied to physical phenomenon and that is the theory of critical phenomenon. Many systems, such as water-ice, magnetic systems and so on undergo phase transitions as a thermodynamic parameter such as temperature is varied. Typically these are first order transitions, meaning that there is a latent heat associated to the transition. Sometimes one can vary an additional parameter and find a line of first order transitions which terminates at a second order transition. The second order transition is characterized by fluctuations on all scales: the theory becomes scale and conformally invariant at that point. It also turns out that the behavior of thermodynamic quantities as one approaches the critical point are characterized by numbers called critical exponents which are universal for systems with the same symmetry structure. These exponents are related to what are called the conformal dimensions of operators in CFT and they are directly measurable in the lab for a variety of systems. One important tool which was used in the study of critical phenomenon is the operator product expansion or operator algebra of K. Wilson and L. Kadanoff. There is a huge literature on this. Here is a reference to an early paper on the operator algebra for the Ising model: http://prb.aps.org/abstract/PRB/v3/i11/p3918_1 . VOA's are a rigorous mathematical formalization of this kind of algebraic structure. For someone who wants to learn about CFT starting from a particular physical system (or at least a mathematical idealization of a physical system) the Ising Model is a good place to start.
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Speaking as a string theorist, I would say Victor Kac's book "Vertex Algebras for Beginners" followed fairly closely what physicists originally thought. So you might want to have a look at it.
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"*As far as I am aware, theoretical physics is about finding mathematical models to explain observed physical phenomena. My questions therefore are:
What is the basic physical phenomenon/problem/question that vertex operators model?*"
Please have a look at the nLab at the pages that have already been mentioned:
ncatlab.org/nlab/show/vertex+operator+algebra and
ncatlab.org/nlab/show/conformal+field+theory
There is also a lot of material about QFT there.
Here is a quick and dirty and oversimplified introduction to some aspects of QFT:
In the Wightman framework we can think of quantum fields as operator valued distributions, that map test functions living on a given spacetime to (maybe unbounded) operators. Essentially selfadjoint operators represent observables that can be measured, in principle, by some experiment or device. Physicsists are mostly interested in systems that have interactions. We can think of elementary particles as localized excitations of quantumf fields that are solutions to specified wave equations (in the distributional sense). In order to have any interactions, these wave equations should have non-linear terms (actually you can define the notion of "free field" (a field with no interactions whatsoever) this way: A free field in the sense of physicists is a quantum field that is a solution of a linear equation).
Non-linear terms entail products of quantum fields, which are undefined, in general, because products of distributions are undefined, in general. The history of QFT is about the struggle of physicists to dodge this problem in one way or another. One way to dodge this problem is to introduce, as an axiom, so called "(associative) operator product expansions". This is a way to formulate, as an axiom, (handwaving:) that the "severity of the singularity of products of distributions" is under control.
operator product expansion
An operator product expansion (OPE) for a family of fields means that there is for all fields and all $z_1 \neq z_2$ a relation of the form $$\psi_i(z_1) \psi_j(z_2) \sim \sum_{k \in B_0} C_{ijk} (z_1 - z_2)^{h_k - h_i - h_j} \psi_k(z_2)$$
Here $\sim$ means modulo regular functions. (See ncatlab.org/nlab/show/conformal+field+theory for an explanation of the notation.)
In a handwaving way, this axiom says that we assume we know something about the kind of singularity of the product of two fields, as their support comes closer and closer, and it is not so bad.
A rigorous interpretation of an OPE would interpret the given relation as a relation of e.g. matrix elements or vacuum expectation values.
An OPE is called associative if the expansion of a product of more than two fields does not depend on the order of the expansion of the products of two factors. Warning: Since the OPE has no interpretation as defining products of operators, or more generally the product in a ring, the notion of associativity does not refer to the associativity of a product in a ring, as the term may suggest.
The axioms of vertex operator algebras are an axiomatization of OPEs. In this sense there is reason to expect that vertex operator algebras will play a key role, on one way or another, in a rigorous construction of interacting QFTs in four dimensions.
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Name-dropping here, but I can claim that Richard Borcherds discussed vertex algebras with me, before anyone knew what they were (not that I was able to be of any help). At that stage they were purely algebraic objects, with some kind of "Jacobi identity". At some point he said he had a definition as Lie algebra in an internal sense in a category - this idea was binned as not in fact useful. Much later he said something about a relationship with the Wightman axioms, but I gather that isn't really watertight when it comes down to it. So I doubt whether the emergence of an axiomatic theory can be "rationally reconstructed" without doing some damage to the history.
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I'll join in the name-dropping. Over a period of years, I saw Borcherds give several talks on vertex algebras, and my impression was that he was putting a lot of energy into massaging the definition so that it seemed mathematically inevitable. I believe he began one talk with some words such as "the aim of this talk is to make the theory of vertex algebras trivial". I also remember him emphasizing that the theory of vertex algebras is not an algebraic theory (in the sense of universal algebra). So although I have no answer for Rex, I have reason to hope there are good answers out there. – Tom Leinster Feb 1 2011 at 17:16
This may hold you off until a REAL answer comes along. I found something resembling intuition on this subject when I first studied Quantum Field Theory. I'd recommend looking there before trying to tackle the many-headed beast that is conformal field theory. Many (but not all) of the VOA axioms can be seen in the properties of a QFT. As far as a "mathematician-friendly" place to learn about QFT goes I remember liking the book by Ryder.
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I'll take a more physical POV. Although, we don't have any mathematically concrete definition of QFT in general, but we can define 2d Conformal Field theory (this is attributed to infinite symmetries and exact solvability), other that CFTs we can also define Topological QFTs (Atiyah).Greame Segal proposed a geometric definition of CFT. In Conformal field theory we deal with vertex operators analog to operators in QFT, we can write a Taylor like expansion of two vertex operators, k.a Operator product expansions (OPEs) which gives the QFT analog of two fields interacting. All these notions are captured by the axiomatization based on Vertex Algebras. To see a better picture,look at the classic paper of Belavin, Polyakov, and Zamolodchikov, where an algebraic approach of CFT was proposed.
Recently Kapustin and Orlov proposed a more general definition of Vertex algebras and they showed the relation between their algebraic definition and Segal's geometric one.
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@J Verma: I have been wondering if there were rigorous algebraic constructions which capture both the holomorphic and anti-holomorphic sectors of a conformal field theory. Is the paper you mentioned by Kapustin and Orlov (I guess CMP Vol 233, No 1 (2003), 79-136, right?) the answer to this question? – Abdelmalek Abdesselam Oct 29 at 17:37
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http://mathoverflow.net/revisions/19911/list
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## Return to Answer
3 added 62 characters in body
Here is a general trick that you can use to make yourself look like you have an amazing memory.
Start with a finite abelian group $(G,+)$ in which you are comfortable doing arithmetic. Be sure to know the sum $$g^* = \sum_{g \in G} g.$$
Take a set $S$ of $|G|$ physical objects with an easily computable set isomorphism $$\varphi : S \longrightarrow G.$$ Allow your audience to remove one random element from $a \in S$ and then shuffle $S$ without telling you which one it what $a$ is. [Shuffling means we need $G$ to be abelian.]
Now inform your audience that you are going to look briefly at each remaining element of $S$ and remember exactly which elements you saw, and determine by process of elimination which element of $S$ was removed.
Now glace through all the remaining elements of $S$ one by one and keep a "running total" to compute $$\varphi(a) = g^* - \sum_{s \in S-{a}} \varphi(s).$$
Finally apply $\varphi^{-1}$ and obtain $a.$
Note that $\varphi$ is not "canonical" in the sense there are definitely choices to be made. On the other hand in should be "natural" in the sense that you should be very comfortable saying $s = \varphi(s).$
The prototypical example is to take $G$ to be $\Bbb Z / 13 \Bbb Z \times V_4,$ $S$ to be a standard deck of 52 cards, and $\varphi(s)$ to be $( \text{rank}(s) , \text{suit}(s) )$.
2 deleted 1 characters in body
Here is a general trick that you can used use to make yourself look like you have an amazing memory.
Start with a finite abelian group $(G,+)$ in which you are comfortable doing arithmetic. Be sure to know the sum $$g^* = \sum_{g \in G} g.$$
Take a set $S$ of $|G|$ physical objects with an easily computable set isomorphism $$\varphi : S \longrightarrow G.$$ Allow your audience to remove one random element from $a \in S$ without telling you which one it is.
Now inform your audience that you are going to look briefly at each remaining element of $S$ and remember exactly which elements you saw, and determine by process of elimination which element of $S$ was removed.
Now glace through all the remaining elements of $S$ one by one and keep a "running total" to compute $$\varphi(a) = g^* - \sum_{s \in S-{a}} \varphi(s).$$
Finally apply $\varphi^{-1}$ and obtain $a.$
Note that $\varphi$ is not "canonical" in the sense there are definitely choices to be made. On the other hand in should be "natural" in the sense that you should be very comfortable saying $s = \varphi(s).$
The prototypical example is to take $G$ to be $\Bbb Z / 13 \Bbb Z \times V_4,$ $S$ to be a standard deck of 52 cards, and $\varphi(s)$ to be $( \text{rank}(s) , \text{suit}(s) )$.
1 [made Community Wiki]
Here is a general trick that you can used to make yourself look like you have an amazing memory.
Start with a finite abelian group $(G,+)$ in which you are comfortable doing arithmetic. Be sure to know the sum $$g^* = \sum_{g \in G} g.$$
Take a set $S$ of $|G|$ physical objects with an easily computable set isomorphism $$\varphi : S \longrightarrow G.$$ Allow your audience to remove one random element from $a \in S$ without telling you which one it is.
Now inform your audience that you are going to look briefly at each remaining element of $S$ and remember exactly which elements you saw, and determine by process of elimination which element of $S$ was removed.
Now glace through all the remaining elements of $S$ one by one and keep a "running total" to compute $$\varphi(a) = g^* - \sum_{s \in S-{a}} \varphi(s).$$
Finally apply $\varphi^{-1}$ and obtain $a.$
Note that $\varphi$ is not "canonical" in the sense there are definitely choices to be made. On the other hand in should be "natural" in the sense that you should be very comfortable saying $s = \varphi(s).$
The prototypical example is to take $G$ to be $\Bbb Z / 13 \Bbb Z \times V_4,$ $S$ to be a standard deck of 52 cards, and $\varphi(s)$ to be $( \text{rank}(s) , \text{suit}(s) )$.
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http://math.stackexchange.com/questions/174020/prove-or-disprove-lim-limits-n-to-infty-p-n1-p-n-sqrtp-n-0/175112
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Prove or disprove $\lim\limits_{n \to \infty} (p_{n+1} - p_{n})/\sqrt{p_n} = 0$
Can anyone prove or disprove the following statement?
$$\lim_{n \to \infty} \frac{p_{n+1} - p_{n}}{\sqrt{p_n}} = 0.$$
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I'm guessing $p_n$ is meant to the be $n$th prime? Ideally, one should not be forced to guess, though, and the government doesn't like it when I read minds without a warrant. – Arturo Magidin Jul 23 '12 at 0:38
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– sdcvvc Jul 23 '12 at 1:01
3 Answers
At present, no one can prove or disprove this statement. It is very famous and still open. The best unconditional result is Baker-Harman-Pintz:
$$\lim_{n \to \infty} \frac{p_{n+1} - p_{n}}{p_n^{0.525}} < \infty.$$
Quite likely it can be shown that this limit is exactly $0$, but I haven't read enough of the paper to be certain.
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The limit is $0$ if you change the exponent to $0.526$ – Will Jagy Jul 23 '12 at 1:09
I don't think anyone knows, although I am looking up stuff just in case. Meanwhile, what people suspect is the Cramer-Granville conjecture, $$\lim \sup \frac{p_{n+1}-p_n}{\left( \log p_n \right)^2} = 2 e^{- \gamma} = 1.1229\ldots,$$ where the logarithm is to base $e = 2.718281828459\ldots$ and $\gamma = 0.5772156649\ldots$ is the Euler-Mascheroni constant. This conjecture, and the Baker result mentioned in the other answer, are in GRANVILLE PDF and WookiePedia. Hmmm, not quite, Granville mentions the earlier $0.535$ result of Baker and Harman. With Pintz they later got it to $0.525.$
This is consistent with a (mostly) stronger conjecture that I made up for no good reason except that it also applies to small numbers, $$p_{n+1} \, - \, p_n < \; 3 \; \log^2 \, p_n.$$ For example, $$p_1 = 2,\; \log 2 = 0.693147\ldots, \log^2 \, 2 = (0.693147\ldots)^2 = 0.480453\ldots, \; 3 \,\log^2 \, 2 = 1.441359\ldots,$$ and $$2 + 1.441359\ldots > 3 = p_2.$$
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Was "WookiePedia" intentional? – Potato Jul 23 '12 at 2:17
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@Potato, such a good feeling to have my work noticed. I work so hard for my fans. Also for everyone else. – Will Jagy Jul 23 '12 at 2:20
This is not known even under the Riemann hypothesis, which gives only $$\frac{p_{n+1}-p_n}{\sqrt{p_n}\log p_n}<\infty.$$
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http://crypto.stackexchange.com/questions/754/is-the-following-statement-about-prg-true-or-false/789
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# Is the following statement about PRG true or false?
Is the following statement true?
If $G: \{0,1\}^k \rightarrow \{0,1\}^n$ is a PRG, then so is $G':\{0,1\}^{k+l} \rightarrow \{0,1\}^{n+l}$ defined as $$G'(x.x')=G(x).x'$$ where $x \in \{0,1\}^k$ and $x' \in \{0,1\}^l$. We are using $.$ to denote concatenation.
My thought is that since $x'$ is not processed through the pseudo number generator, then this statement is false. Is my analysis correct?
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What about using bar $|$ for concatenation? Is $.$ commonly used? – Ethan Heilman Sep 23 '11 at 13:12
I suppose the formal proof depends on your definition of PRG, but your intuition looks right. – Paŭlo Ebermann♦ Sep 23 '11 at 13:43
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This sounds like a homework question. Is it a homework question? – D.W. Sep 26 '11 at 4:27
## 2 Answers
If you use a concrete-security definition of security for a PRG, then this statement is true. The proof is a good exercise. If you know enough to pose the problem and to understand the definition of security for a PRG, you should be able to find the reduction proof without difficulty. Start by tracing out what the definition is saying.
A general comment on your analysis: Your analysis makes it sound like you are trying to guess at the answer. Guesses are good, but then you should follow them up with proof. In cryptography, our intuition can often be wrong. Therefore, the accepted standard is a proof. In particular, if you guess that the statement is true, your next step should be to try to prove it; if you guess that the statement is false, your next step should be to find an attack that disproves it. Give it a try! This is a fun problem, and not too hard.
P.S. This looks very much like a standard homework question. I don't know whether it is or not, so I'll just say this. If you are using this web site to answer homework questions, you are not only cheating your fellow students, you are also cheating yourself of the chance to learn the material. To learn cryptography, you must grapple with the problems and try to solve them on your own. You'll only learn through undergoing the process yourself; you won't learn nearly as much by reading other people's solutions. Think of it like learning to ride a bicycle: you can't learn to ride a bicycle by watching someone else ride it. Instead, you have to get on the bicycle yourself, accept that you'll fall or stumble a few times, and work through it.
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A few definitions could assist: are $x$ and $x'$ random elements of $\{0,1\}^k$ and $\{0,1\}^l$?
Is $\mathsf{G}$ a cryptographically-secure PRG or a PRG with some statistical properties?
Let $\mathcal{D}$ be a PPT-distinguisher and $r$ be a uniformly random bitstring of length $n + \ell$. A common definition ($e.g.,$ from Katz-Lindell book) of a PRG includes the following condition :
$| \mathrm{Pr}[\mathcal{D}(r)=1] - \mathrm{Pr}[\mathcal{D}(\mathsf{G}'(x||x'))=1] | \leq \mathsf{negl}(k+\ell)$
Essentially it says, can you tell the difference between the output of the PRG and a uniformly random string?
$\mathsf{G}'$ is not secure under this definition.
For example, consider if $\mathcal{D}$ mounts an exhaustive search. This initially appears to not work since it takes exponential time. But $\mathcal{D}$ only needs to search for values of $x$ since it knows $x'$ from the output $\mathcal{G}(x||x')$. Therefore the time is exponential in only $k$, not in $k+\ell$. That is not good enough for $\mathsf{negl}(k+\ell)$.
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This answer is technically correct if you use the asymptotic definition of PRG, but only if $\ell$ is superpolynomially larger than $k$. Moreover, in practice the asymptotic definition of PRG is not a good one; and if you use a concrete-security definition of PRG, then you find the construction is secure. – D.W. Sep 26 '11 at 4:29
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http://mathoverflow.net/questions/14376?sort=votes
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## Why is addition of observables in quantum mechanics commutative?
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I am no expert in the field. I hope the question is suitable for MO.
Background/Motivation
I once followed a quantum mechanics course aimed at mathematicians. Instead of the usual motivations coming from experiment at the turn of the 19th century, the following argument (more or less) was given to show that the QM formalism is in some sense unavoidable.
When one does physics, he is interested in measuring some quantity on a given state of the universe. The quantity (say the speed of a particle) is defined experimentally by the tool used to do the measure. We define such an instrument, with a given measure unit, an observable. So for every state and every observable we get a real number.
We can now define a sum and a product of observables. These are obtained by performing the two measures and then adding or multiplying their values. Similarly we can define scalar multiplication. These operations are then associative, but there is no reason why they should be commutative, since performing the first measure can (and indeed does) change the state of the universe. For some reason I cannot understand, anyway, addition is assumed commutative. I also see no reason why multiplication should distribute over addition. We can now also consider observables with complex values, by linearity.
At this point observables form an $\mathbb{R}$-algebra. We intoduce a norm it as follows. The norm of an observable is the sup of the absolute values of the quantities which can be measured. Every instrument will have a limited scale, so this is a real number. By definition this is a norm. Moreover it satisfies $\|A B \| \leq \|A\| \| B \|$. We can now formally take the completion of our algebra and obtain a Banach algebra.
Finally we define an involution * on our algebra by complex conjugation of observables. This yields a Banach * -algebra, and the third assumption which is mysterious to me is that the $C^*$ identity holds.
Finally we can use the Gelfand-Naimark theorem to represent the given algebra as an algebra of operators on a Hilbert space. If this turns out to be separable, it is isomorphic to $L^2(\mathbb{R}^3)$ and we recover the classical Schrodinger formalism.
The problems
In this approach I see three deductions which seems arbitrary: addition is commutative, multiplication is distributive and the $C^*$ identity holds. Is there any kind of hand-waving which can jusify these? In particular
Why is addition of observables commutative, while multiplication is not?
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Hurriedly looking over your question, don't you really want to ask why addition of observables is commutative? I mean, we know why addition of operators is commutative, but that doesn't seem to be what you're asking. – Yemon Choi Feb 6 2010 at 11:48
Sorry, I edit it. – Andrea Ferretti Feb 6 2010 at 11:50
It seems I cannot edit the title: it gets changed here, but not in the main page. Of course addition of observables was what I had in mind. – Andrea Ferretti Feb 6 2010 at 11:52
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I don't think that is how addition and multiplication are defined. In fact it is definitely not that way, since we know the answer for quantum mechanics. Consider the finite dimensional case. Observables are modeled by certain operators and the numbers you get from measurement are eigenvalues. But operators' addition is not defined as addition of their eigenvalues unless you can simultaneously diagonalize them. – MBN Feb 6 2010 at 16:43
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@Georges I did not mean for that to come across as condescending, or imply that I am in any way so brilliant that I find this book simple. I simply meant, that reading the book in the way I did, it did not require the level of effort that serious reading normally does, and most of what I read of it, took place walking across campus, lying in bed, or in a coffee shop. I hope I didn't give the wrong impression. – B. Bischof Feb 6 2010 at 22:00
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## 6 Answers
Your description of the structure of the algebra of observables isn't quite how I'm used to it being. Indeed, I believe that in the best algebraic descriptions of quantum mechanics, addition is a formal operation, rather than a physical operation as you've described. The best reference I know for this point of view is L.D. Faddeev and O.A. Yakubovskii, 2009, Lectures on Quantum Mechanics for Mathematics Students. I don't have my copy handy right now, so I will describe my memory of how they set up the algebra of observables.
The first thing to point out is that in the real world, there is no such thing as pure states. This has nothing to do with quantum mechanics, and everything to do with an experimenter's inability to perfectly measure the initial set-up. For your notion of "state" to make sense physically, it must be something like "repeatable initial set-up for an experiment". Once this is your notion of state, you are perfectly able to run your experiment 1000 times, make your measurements (each individual run may give a different answer, but you can look at the distribution), and process them as you want.
So really an observable assigns a probability distribution on $\mathbb R$ to each state. We now demand the following axiom: the (good) functions $\mathbb R \to \mathbb R$ act on the set of observables by composition. So if $X: \{\text{states}\} \to \{\text{probability distributions}\}$ is an observable, so is $X^2$: the probability that the observable $X^2$ assigns to an interval $[a,b]$ is the same as the probability that $X$ assigns to the interval $[\sqrt a,\sqrt b]$. In particular, suppose you compose your observable $X$ with a step function $\Theta(x - \xi)$, where $\xi \in \mathbb R$. Then the observable $\Theta(X-\xi)$ measures the whether the value of $X$ is more than $\xi$. Then you can check that the full distribution $X$ is recoverable from the knowledge of all the $\Theta(X-\xi)$. In particular, it's recoverable from the expected values of $\Theta(X-\xi)$ on each state. So to set up the algebra of observables, it's enough to know only the expectation values for observable at each state.
Now you should realize the following. The previous paragraph makes sense even for classical mechanics, and in fact is the correct formalism (as there are no pure states). But in quantum mechanics, it's worse than that. A definite state is one that gives a delta distribution for each observable. Classically, we believe that a sufficiently good experimenter can approximate definite states to whatever desired accuracy. But there is very good evidence that this fails in the quantum world: no matter what tools you use, there are absolute bounds preventing states from approximating definite states. So the language of distributions and expectations is absolutely necessary to formalize quantum mechanics, whereas in classical mechanics you could say that there are idealized definite states, observables are functions on definite states, and states are probability distributions on the space of definite states.
Finally, the question is how to assign algebraical operations to the collection of observables. And here I admit that I don't have a great answer. One possibility is simply to convolve probability distributions: this gives a commutative addition, for example. Then you could define a commutative associative multiplication by taking logs and adding and exponentiation, but my memory is that this does not distribute over addition in general. F&Y define a commutative nonassociative multiplication by $(X,Y) = \frac12\bigl((X+Y)^2 - X^2 - Y^2\bigr)$. Oh, right. The problem is the following: do you add, multiply, etc. the distribitions, or the expectation values? For addition, adding expectation values is the same as the usual convolution of distributions and then taking expectation. But for multiplication it is not. I don't remember what F&Y do, but I think it might at the level of expectation values.
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Wait, so it is in theory possible to have pure states? I thought that QM was nondeterministic. – Harry Gindi Feb 6 2010 at 23:29
@HG: I left out a definition. Given two states $\mu, \nu$, you can mix them with proportion $p, 1-p$, where $0 \leq p \leq 1$ is some probability, by flipping a (classical) coin that lands Tails with probability $p$ and Heads with probability $1-p$, and depending on how the coin falls, set up your experiment in state $\mu$ or $\nu$. You can call the resulting state $p\mu + (1-p)\nu$. Then a pure state is a state $\alpha$ so that if $\alpha = p\mu + (1-p)\nu$, then either $\alpha = \mu$ or $\alpha = \nu$. Classically, pure states are the same as definite states, but this fails in QM. – Theo Johnson-Freyd Feb 6 2010 at 23:53
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The failure of this is the "nondeterminacy" of QM. Whether you can actually set up a pure state (really you cannot), you can (presumably) approximate pure states with arbitrary accuracy. But in quantum mechanics, there are absolute bounds preventing you from approximating definite states: for any state, there is an observable that does not give a single $\delta$-distribution as output when it eats that state, i.e. it assigns a nontrivial probability distribution. – Theo Johnson-Freyd Feb 6 2010 at 23:56
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Of course, in every theory, the mechanics is deterministic. In QM, and also in the classical world, "dynamics" or "mechanics" describes how the "value" (a probability distribution or equivalently an expectation) of an observable on a given state evolves over time. It's reasonable to insist that the $\mathbb R\to \mathbb R$ action on observables is time-independent; then in particular any theory set up the way I've described has linear, deterministic evolution at the level of probability distributions. – Theo Johnson-Freyd Feb 6 2010 at 23:59
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@HG: Time evolution in QM is always deterministic. Any state, mixed or pure, evolves deterministically via the Schrodinger equation. The formalism I outlined above fits better with the Heisenberg picture (states don't evolve), in which case the observables evolve deterministically (and linearly). The only non-deterministic part is the measurements: the pairing between states and observables. In classical mechanics, the value of an observable on a pure state is determined. In quantum mechanics, even on a pure state the value of an observable is not determined. – Theo Johnson-Freyd Feb 7 2010 at 17:52
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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.
We may think of a state $\omega$ as a functional on the algebra of observables $\mathcal O$ which is interpreted as giving the expected value of each observable. With this in mind, it is natural to require $\omega$ to be linear (as well as two other usual properties, positivity and normalization).
Thus given two observables $A, B \in \mathcal O$, if we are going to have a sum $A + B$ it should be true that $\omega(A + B) = \omega(A) + \omega(B)$ for any state $\omega$. Since this gives the values of $A + B$ on every state, it suffices to define it. Since $B + A$ has the same values on every state, $B + A$ is the same observable.
On the other hand, there is no natural way to say what $\omega(AB)$ should be, since states (like expectation values) need not be multiplicative.
So: commutativity of observables reduces to commutativity of $\mathbb C$ since expectations are linear, but nothing analogous applies to multiplication.
This is based on what I've read in F. Strocchi, An Introduction to the Mathematical Structure of Quantum Mechanics.
Note that you can eventually interpret states as arising out of probability distributions, leading to Theo's comments.
Personally I am still a bit hazy on why we postulate a multiplication on observables at all, when (unlike the classical case) there is not a clear physical interpretation of what such an operation should mean. However, given the full structure of a $C^*$-algebra, one can show that the uncertainty principle (or the existence of complementary observables) requires noncommutative multiplication, et voila, you have quantum mechanics.
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I followed myself the course from Strocchi, and it is exactly his argument I'm trying to fill. Defining a state as a linear functional seems as arbitrary as defining it as a multiplicative linear functional. The question is: how con we derive these properties from the description of the process of measure alone? I have tried to explain the full reasoning in my question. – Andrea Ferretti Jul 28 2010 at 10:20
Basically, a state is the current situation of the universe, an observable is some physical quantity we have some instrument to measure. They have an obvious pairing: in a given state, apply your instrument and read the result. If you now define sum and multiplication of observables by doing the two measures one after the other, the result may depend on the order. And indeed it does for the multiplication. Why not for the addition? – Andrea Ferretti Jul 28 2010 at 10:23
If you say $\omega(A + B)$ is the expected value of preparing the universe in state $\omega$, then measuring $A$, then measuring $B$ in the state the universe was left in after the first measurement, then summing the results, then indeed $\omega(B + A)$ may be different. Instead, $A + B$ represents a single measurement we can do that results (in expectation) in the sum of $A$ and $B$. That is a more sensible requirement than the corresponding one for multiplication, since even if $A$ and $B$ are somehow linked (as they are in QM), expectation is still linear. – Jonathan L Long Jul 29 2010 at 8:41
As for a purely physical definition of addition of observables, I don't have one, and neither does Strocchi; he even notes that the introduction of the sum leads to an extension of the physically defined observables. The situation for multiplication seems even worse, so I am not sure how much more physical motivation can be given at the purely algebraic level. – Jonathan L Long Jul 29 2010 at 8:47
While Theo's comments are excellent, yours (Jonathan) get to the heart of the difference between addition and multiplication: states are expectation values, expectations values add, they don't multiply. (Obviously there is more to be said.) – Toby Bartels Jan 3 2012 at 17:54
The introduction you outlined is basically reminiscent of the old quantum mechanics, anyway in the approach you depicted it is the culmination and not the premise of the construction, and the comment was surely intended to be explanatory of the far origin of these choice. I now try to resume the history.
There are basically two approaches to mathematical quantum mechanics. The first one very complex and stratified in its development, but simple in the premise was discussed by John Von Neumann in a lot of papers after the "Foundation of quantum mechanics", the second one is basically conceveid to be an extension for the first, and is this second approach you are referring to: the GNS approach.
Anyway both of them are surely derived after an abstraction process very far beginning on the methods of classical mechanics, joint to the newest evidence from atomic and particle physics of the first quantum mechanics.
Just as in classical mechanics we define functions of observables dynamical quantity so the founders of quantum mechanics conceived it is possible in quantum mechanics, anyway we need to clarify in which sense this is possible and explanation isn't fully depleted from the naive extension of classical theory of the measure, based on real numbers, but it need of a clear axiomatic and this was furnished from John Von Neumann (and in some way from Heisenberg, Dirac, and Schroedinger before him formulated this axiomatic)
Anyway, just as in classical mechanics there is a notion of repeatibility and regularity, so there is in quantum mechanics. The true difference is in the outcome of the measures, deterministic in classical, probabilistic in quantum mechanics. So that measure processes are conceived deterministic in a statistical sense, and, for example, the component energies of the isotropic harmonic oscillator sums exactly in mean value, but the variance is zero if the considered states are eigenstates. Old quantum mechanics can be founded on few axioms about the measures and led Von Neumann, in a natural way to linear operators acting, like a non commutative algebra, on Hilbert spaces.
In order to grant correspondence principle we, following the founders of quantum mechanics, need to hypothesize the existence of intrinsically deterministically evolving observable, and just the measure process make the difference, because these dynamical "quantities" with respect to the measures doesn't appear as real numbers, this point was the first time realized some time after the Copenaghen interpretation was developed.
So they are assumed, after Heisenberg (speaking of non commutative numbers) and Jordan (speaking of matrices), and Schroedinger (speaking of operators acting on functional space of probability) all these three point of view were showed to be in a certain strict framework to be equivalent, from Dirac assuming they are algebraic elements obeying to canonical commutation relation generalizing the Poisson algebra.
In brief the Dirac point can be summarized in assuming an Hilbert space structure for the states, and in developing step by step a theory of observables compatible with the Copenaghen interpretation spirit and with the correspondence principle.
Anyway Von Neumann felt the need to obtain an axiomatic foundation based on more general operators algebras, and an axiomatic of measure, unifying from scratch the theoretical framework, in fact obtaining a more general theory with respect to the Heisenberg and Dirac theoretical "prejudices". The Von Neumann point was in fact based on the general representation theory in the geometrical framework of Banach operator algebras of operators in Hilbert space, and in particular on the CCR irreducible representation theory, but from this point the research of Von Neumann continued in search of an intrinsic point of view based on the geometry of observable.
After time and time was in fact recognized that part of quantum theory of measure is nothing else then a generalized probabilistic theory in a Banach algebra and the general setting of Gelfand Najmark Segal construction rebuild intrinsically the Hilbert spaces. Anyway the field extension of this setting is very problematic and a hierarchy of Hilbert spaces appears. Anyway in this way a circle is closed and a new loop is opened: in the GNS approach to quantum mechanics we postulate that operators are living in an abstract algebra, obeying familiar rules for an algebra with an involution (the * operation). Via Gelfand theorem the commutative case led to the algebra of complex valued continuous functions in an Hausdorf space, the spectrum of the algebra (which will led the ordinary numerical set of coordinates of classical mechanics), and more in general to a spectral theory, culminating in the GNS construction, which associate to a given linear form an Hilbert space and a representation for the algebra.
Anyway the true achievement of this approach is the net of algebras, that is very more general with respect to the Hilbert space interpretation of quantum mechanics,this achievement is useful in relativistic field theory and leads to very far reaching results firstly partially discovered by Von Neumann in some papers, and after then developed from Araky, Haag, Kastler. In this full setting is now possible to address in more precise terms the question of the cluster decomposition principle implicit in the deterministic evolutionary scheme, and the question of repeatability principle of classical and quantum mechanics, and to understand quantitatively something about the limitation, arising from the change of the state of the universe, to this principle, which can be espressed, for example, in term of a change of representation, becaused from the change of the linear form representing the thermokinetic state of "universe", without any change in the postulates of quantum field theory and the derived quantum mechanics. This is perhaps the perspective of the search about KMS theorem.
I'm not very satisfied from this resume, anyway I think you can correct and integrate it, and I hope to read and write something else more precise and delimited.
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This looks very interesting and covers a lot of ground. Would you object if I made some small edits for language? – Yemon Choi Feb 7 2010 at 0:14
It looks very interesting indeed, but sadly it does not touch in any way the original question. – Andrea Ferretti Feb 8 2010 at 19:56
@ YC sure you can edit it, I'm sure there are a lot of mistaken. Thanks in advance. – Tetis - Gianmarco Bramanti Feb 9 2010 at 20:59
@AF I think the points "touching" the very interesting question you pointed out are two: the choice of a algebra structure for operator is IMHO based on both: conservative approach with regard to the construction of quantum mechanics ( in order to safe a predictive theory, so that the effect of measure in the state of universe was at first neglected at all) an empiric base with respect to the measure of quantum quantities corresponding to additive classical quantities. E.g. the energies and spin components. Anyway in old QM these were partly "theorems" nowadays, at all, definitions. – Tetis - Gianmarco Bramanti Feb 9 2010 at 21:17
Since you raised your question I'm uncomfortable about some question, and I re-read Von Neumann, in order to becalm myself, anyway in Von Neumann the problem isn't solved in a deductive way neither is justified at all, only is asserted the additivity, as customary, between commutating operators (so there the correspondence principle is granted) and then the additivity is extnded to non commutating operators.
Anyway reflecting on the practical use of non commutating linear combination of operators I realize that the arguments of linear combination are generally elements of some Lie Algebra, and their powers, and I remember a point in the first book of Landau about classical mechanics that I like to quote:
Conservation laws.
"Not all integrals of motion have an equally relevant role in mechanics. Among these there are some whose invariance over time has an origin very deep, related to fundamental properties of space and time and that is their homogeneity and isotropy. These quantity, these conservative, have an important general property, they are additive, that is, their value for a system composed of several elements, whose interaction can be neglected, is equal to the sum of the values for each of the elements."
It seems just like Landau is mixign two unrelated points: the isotropy and the commutativity. In fact this isn't the additivity we are thinking to. Anyway there is an important point: and this is the role of simmetry and in mathematic the role of the Stone-Neumann Theorem, and the role of parallel transport, and in mathematic the role of gauging the space and time, so we can perahps re-connect the two point arised by Landau to the additivity of observables.
Generalizing, perhaps we need to relate an observable to an infinitesimal of a continous symmetry group: an elementar generator of lie algebra in order to justify additivity. Just handwaving.
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This question has bothered me for a long time! Although I don't have an answer, I'd like to mention an approach that looks promising at first, but turns out not to work.
First, recall that in quantum mechanics, you can think of a "state" as a way of preparing a physical system. Theo Johnson-Freyd pointed out in a comment that if you have two states $\rho$ and $\sigma$, you can construct a state that intuitively deserves to be called $\tfrac{1}{2}(\rho + \sigma)$:
Flip a fair coin. If the coin comes up heads, prepare the system in state $\rho$. If the coin comes up tails, prepare the system in state $\sigma$.
This state deserves the name $\tfrac{1}{2}(\rho + \sigma)$ because if $\rho[X]$ is the expectation value of the observable $X$ for a system prepared in state $\rho$, and $\sigma[X]$ is the expectation value of $X$ for a system prepared in state $\sigma$, the expectation value of $X$ for a system prepared in state $\tfrac{1}{2}(\rho + \sigma)$ should be $\tfrac{1}{2}(\rho[X] + \sigma[X])$, by the laws of classical probability.
Now, what happens if we use the same trick to define the sum of two observables? Given two observables $X$ and $Y$, let's define $X + Y$ to be the observable:
Flip a fair coin. If the coin comes up heads, measure $X$ and double the result. If the coin comes up tails, measure $Y$ and double the result.
The laws of classical probability tell us that if $\rho[X]$ and $\rho[Y]$ are the expectation values of $X$ and $Y$ for a system prepared in state $\rho$, the expectation value of $X + Y$ for a system prepared in state $\rho$ should be $\rho[X] + \rho[Y]$, just as you would hope.
Here's where things go pear-shaped. Given an observable $Z$, it makes sense to define $Z^2$ to be the observable:
Measure $Z$ and square the result.
So what's the expectation value of $(X + Y)^2$ for a system prepared in state $\rho$? The laws of classical probability tell us that it's $\rho[X^2] + \rho[Y^2]$. In the formalism of quantum mechanics, however, $X$ and $Y$ are operators and $\rho$ is a linear functional on the operator space, so
$\rho[(X + Y)^2] = \rho[X^2] + \rho[Y^2] + \rho[XY + YX]$.
If $\rho[XY + YX]$ is nonzero, this formula disagrees with the expectation value for $(X + Y)^2$ that follows from our definitions of $X + Y$ and $Z^2$, according to the laws of classical probability!
In practice, it's not hard to find observables $X$ and $Y$ for which $\rho[XY + YX]$ can be nonzero. For example, let $X$ and $Y$ be the x-spin and z-spin of a spin-1 particle, represented by the operators
$X = \frac{1}{\sqrt{2}}\left[\begin{array}{ccc}0&1&0\\1&0&1\\0&1&0\end{array}\right],\qquad Y = \left[\begin{array}{ccc}1&0&0\\0&0&0\\0&0&-1\end{array}\right].$
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'Given an observable Z, it makes sense to define Z<sup>2</sup> to be the observable Measure Z and square the result' - I think this is where that argument falls over, and not just on a quantum but even a classical level. I expect to be able to define the observable X+Y readily simply because linearity is a 'plausible' thing to have, but I see no reason to believe that I'll be able to ascribe any sort of invariant meaning to Z<sup>2</sup> in general. – Steven Stadnicki Jul 28 2010 at 17:12
@Steven Stadnicki: To me, given an observable $Z$ and a function $f \colon \mathbb{R} \to \mathbb{R}$, it seems totally natural to define $f(Z)$ as the observable you measure by measuring $Z$ and then applying $f$ to the result. What do you mean by an "invariant meaning," and why doesn't this definition give an "invariant meaning" to $f(Z)$ for any $Z$ and $f$? – Vectornaut Jul 28 2011 at 23:05
On the contrary: in quantum mechanics as it exists, measuring $Z$ and squaring the result does measure $Z^2$, but flipping a coin (then making an appropriate measurement and doubling the result) does not measure $X + Y$. – Toby Bartels Jan 3 2012 at 17:58
If I correctly understand, you are misinterpreting the meaning of the product and sum of observables.
When you say "We can now define a sum and a product of observables. These are obtained by performing the two measures and then adding or multiplying their values."
This cannot possibly describe the usual sum A+B and product AB of operators. For the product, it is not even hermitian unless A and B commute. Agreed, A+B is hermitian, but the spectrum of A+B does not contain the result of the sum of a measurement of A followed by a measurement of B (in either way), again unless A and B commute. For a counter-example take $A=\pmatrix{1& 0\cr 0&-1}$ and $B=\pmatrix{0&1\cr 1&0}$.
I hope I correctly understood your question.
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Well, the point of my question is in big part metamathematical. That is, the explanation I received was meant as a motivation to study quantum mechanics via C*-algebras, and a posteriori, via the Gelfand-Naimark-Segal theorem, via operator theory. I admit that the explanation does not convince me completely, but I think it has some points. The reason why I asked here was trying to fill the gaps. Instead you assume that observable are represented by operators, which is meant to be the conclusion – Andrea Ferretti Jan 26 2012 at 22:14
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http://physics.stackexchange.com/questions/34063/what-is-the-maximum-surface-charge-density-of-aluminum?answertab=active
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# What is the maximum surface charge density of aluminum?
I understand that the maximum free charge carrier density for aluminum has been measured using the Hall effect (in the case of electric current). However, I'm not clear how to determine the maximum surface charge density to which aluminum (or any conductor) can be charged, assuming the neighboring medium does not breakdown.
Say for instance we had a parallel plate capacitor with an idealized dielectric that could withstand infinite potential across it. What is the max surface charge density that the plates could be charged to? I assume that at some point all of the surface atoms are ionized.
Is this simply the volumetric free carrier density multiplied by the atomic diameter?
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Do you meant the maximum surface charge density one can charge aluminium to? That is you imply aluminium specimen is charged, not neutral, right? – Yrogirg Aug 13 '12 at 8:52
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Right, under idealized conditions, how much surface charge is aluminum capable of being charged to? – thrusty Aug 13 '12 at 16:01
Anybody, how does Hall effect helps to measure maximum free charge carrier density? I though one measures just free charge carrier density. By the way, I'm starting a bounty, to answer the question "What is the maximum potential one can charge a conductor to?" – Yrogirg Aug 15 '12 at 7:40
## 3 Answers
I think you can estimate the maximal surface charge density as follows. The energy needed to remove an electron from a solid to a point immediately outside the solid is called work function $W$. For aluminum $W$ is about $4.06-4.26$ eV. The thickness of the charged layer on the surface of a conductor is about several Fermi lengths $$\lambda_{F}=\left( 3\pi^{2}n_{e}\right) ^{-1/3},$$ where $n_{e}=N/V$ is the total electron number density for the conductor. I think that the charge starts to drain from the surface of a conductor when $E\lambda_{F}$ is of the order of the work function, where $E=4\pi\sigma$ is the electric field near the surface: $$eE\lambda_{F}=4\pi\sigma\left( 3\pi^{2}n_{e}\right) ^{-1/3}\sim W,$$ hence $$\sigma\sim\frac{W}{4e}\left( \frac{3n_{e}}{\pi}\right) ^{1/3}.$$
About the question Yrogirg. I think that the question is not quite correct. The charge are distributed on the surface of a conductor in such a way that the electric potential is a constant in the body of the conductor. The «stability» of charge on the surface is greatly dependent on the geometry of the object in question. Sharp points require lower voltage levels to produce effect of charge «draining» from the surface, because electric fields are more concentrated in areas of high curvature, see, e.g. St. Elmo’s fire.
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Oh, how could I forget about the work function. And yes you are right, it is the potential difference between the surface and the point just right outside it that matters (electric field outside the surface), not the potential difference between the surface and infinity as I though first. If the electron is just out it will be repelled away. – Yrogirg Aug 15 '12 at 14:23
what's $n_e$? "total electron number density for the conductor" --- is that the free charge density? I want to find value for $\sigma$. Might there be some kind of spin repulsion at such densities in addition to the electric one? – Yrogirg Aug 15 '12 at 18:04
No it is not free charge density. It is what I wrote $N/V$, where $N$ is the total number of electrons in the metal. According to Landau theorem the spectrum of low energy excitations of strongly interacting Fermi liquid coincides with the free fermion gas (which is strongly degenerate for low temperatures) that is how you can estimate the Fermi momenta and thus Fermi length. In fact, for this estimation you can use the experimental Fermi energy for aluminum. – Grisha Kirilin Aug 15 '12 at 18:17
However, the question "what is $n_e$?" is not important. The only important point is the thickness of charged layer. For plasma, the thickness of the charged layer on the surface is several Debye radii, for metal — several Fermi lengths. – Grisha Kirilin Aug 15 '12 at 18:25
@Yrogirg See e.g. A. Abrikosov "Fundamentals of the theory of metals" § 2.2. Quasiparticles in an isotropic Fermi liquid – Grisha Kirilin Aug 15 '12 at 18:37
show 9 more comments
There is no sensible answer to this question.
You can put any amount of charge on a blob of aluminum sitting in a vacuum, or surrounded by an ideal insulator. Why not?
If you put an awful lot of electrons on a blob of aluminum sitting in a vacuum, the electrons will eventually start shooting off by thermionic emission, and most of the excess charge will be gone after, let's say, 1 day. If you put even more electrons, most of them will be gone after 1 millisecond. But there is no "maximum" really, just a gradual speed-up of the discharging. Even 1 excess electron will not be stable for eternity.
If you subtract electrons instead of adding them, certainly nothing will happen. Well, I guess positively-charged atomic nuclei could fly off if the charge was significant enough. Again, this process does not let you say that a certain amount of charge is "the maximum possible", it's just a process that happens more and more frequently as the charge increases.
If you add or subtract an awful lot of electrons from a blob of aluminum surrounded by insulator, the insulator will eventually break down. If you have an ideal insulator that cannot break down, then nothing will happen no matter how many electrons are there.
You seem to have the idea that all electrons must come from surface atoms, so if you take away every electron from every surface atom, then it will be impossible to take away any more charge. But if you think about it, that's sort of a weird idea, when there's still all those electrons inside the metal! In fact, the idea is not correct. The "surface" where an insulator can store charge is not infinitesimal, nor necessarily exactly one atom thick. It's actually a depth equal to the so-called Debye length. If you subtract loads of electrons -- every electron from every "surface" atom -- the Debye length will just increase allowing you to scrape electrons out of atoms residing farther and farther from the surface.
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Wouldn't it be difficult to scrape off anything but valence electrons from the surface atoms? This would leave inner-shell electrons on the surface atoms to continue screening the external field, limiting the penetration depth. – thrusty Aug 17 '12 at 6:26
Yes ... you need an extra 1500 volts to pull off a 1s electron from aluminum compared to a valence electron at the same spot. (I got that number from x-ray data.) So, as voltage increases, you will pull valence electrons out of a thicker and thicker depth from the surface, until the only remaining valence electrons are so far inside that there is a 1500V drop between the remaining valence electrons and the surface atoms. Then as the voltage increases even more, you will simultaneously pull off valence electrons from the inside and core electrons from the outside. – Steve B Aug 17 '12 at 13:29
"If you put even more electrons, most of them will be gone after 1 millisecond." So you admit that with more electrons it would be easier for an electron to emit? At some point (charge) it would be so easy that no additional energy, say from heat, would be required, that's what is meant by the maximum charge. Though your remark about thermoemission is important. – Yrogirg Aug 18 '12 at 5:30
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@Yrogirg, you're saying "Maximum charge is the point where the potential energy of an electron monotonically decreases as it goes from inside the metal to the vacuum". OK, with this definition, I think the maximum charge is some large and finite value. But I don't think this criterion corresponds to what thrusty was asking about. It is, after all, pretty irrelevant in the real world. A bit below this threshold, the charge spits out almost just as fast as it does a bit above the threshold, due to thermal fluctuations and quantum tunneling. – Steve B Aug 18 '12 at 12:32
Given the work function from the answer above, no electron can get as close as
$\frac{1}{4\pi\epsilon_0} e^2/r = 4.06eV$, at the range it would rather leave the aluminium than get closer to another electron.
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it gives the distance of $3.5 \times 10^{-10}$ meters, the same estimation as one derived from the Grisha's model. – Yrogirg Aug 21 '12 at 8:20
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http://mathoverflow.net/questions/80949/monodromy-representations-of-families-of-hyperelliptic-jacobians/118601
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## monodromy representations of families of hyperelliptic jacobians
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $k$ be a number field , $f \in k[x]$ of degree $2g$ with distinct roots, and $X$ the complex plane with the roots of $f$ removed. Then define the abelian scheme $J \rightarrow X$ where for each $t \in X$, $J_{t}$ is the jacobian of the hyperelliptic curve $y^{2} = f(x)(x - t)$. Then there is a monodromy representation of the topological fundamental group $\rho: \pi_{1}(X) \rightarrow \mathrm{Sp}_{2g}(\mathbb{Z})$ which preserves the Riemann form on the homology elements of the fiber over the basepoint.
I know that the image of this representation is the subset of $\mathrm{Sp}(\mathbb{Z})$ whose image modulo 2 is the identity but don't know of any "elementary" proof of it. In particular, I've seen it claimed that for genus $g = 1$, the image of the representation is simply $\Gamma(2) \cap \mathrm{SL}_{2}(\mathbb{Z})$. Can anyone tell me how to prove it for this elliptic curve case, in an intuitive way? I'm looking for something similar to http://rigtriv.wordpress.com/2010/02/25/monodromy-representations/, which I can't quite follow.
I would very much appreciate any kind of argument, but especially a relatively intuitive, visual one.
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## 3 Answers
I think I do have an informal argument in the $g = 1$ case that can be obtained by drawing pictures of one copy of the Riemann sphere with $\infty$ and three other points chosen, and drawing homology loops and looking at what would happen to them if one of the finite points is rotated around the others. But this is very rough, and I'd still appreciate someone else's insight.
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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.
I believe the following link answers your question (this is apparently an unpublished result of J.K Yu)
http://mathoverflow.net/questions/105048/the-image-of-the-point-pushing-group-in-the-hyperelliptic-representation-of-the-braid group
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This isn't actually true for $g=1$. $-I$ is an element of $\Gamma(2) \cap SL_2(\mathbb Z)$ which satisfies the equation $a \cdot a = 1$. No element in the fundamental group of the complex plane minus two points satisfies this, so for it to be hit the map must have some kernel. But it's easy to check that the map has no kernel. Indeed, the complex plane minus two points has universal cover the upper half plane, which is the moduli space for elliptic curves with a pair of generators for their homology, and all the deck transformations lie in the group which acts by changing the generators and thus acts nontrivially on homology.
But it is true up to $\pm 1$. Here is a slightly offbeat answer, which is intuitive to me but might not be to anyone else. The glib summary is "because $\mathbb P^1$ minus $3$ points is the moduli space of elliptic curves with full level $2$ structure."
Suppose you have an element in $SL_2(\mathbb Z) \cap \Gamma(2)$. I am not sure how this differs from just $\Gamma(2)$. This is a mapping class of the torus. It is easy to see that you can realize this by a family of complex tori, i.e., elliptic curves, over the cirle. Take a lattice in $\mathbb C$ and move the generators around so they form a new pair of generators for the lattice that realizes your chosen element in $\Gamma(2)$.
Since the mapping class is in $\Gamma(2)$, the three $2$-torsion points will be preserved by this mapping class, so we can uniquely identify and label them.
Embed each elliptic curve in the family in a Weierstrauss form of type $y^2=f(x)(x-t)$ such that the first $2$-torsion point is the first root of $f$, the second two-torsion point is the second root of $f$ and the third $2$-torsion point is $t$. This embedding is possible and is unique up to the automorphism $y \to \pm y$. This gives a function, $t$, from the circle to the complex plane minus the roots of $f$. If you follow the family $y^2=f(x) (x-t)$ and the original family around this loop simultaneously, you see they remain the same up to $\pm y$, so they give the same element of $SL_2(\mathbb Z)$ up to $\pm 1$.
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http://mathoverflow.net/revisions/102934/list
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## Return to Question
Post Made Community Wiki by aglearner
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# Pedagogical notes on line bundles on complex projective manifolds
I would like to find some notes (or book), that explains on a very basic level what is a line bundle on a complex projective manifold. Maybe even, what is a line bundle on $\mathbb CP^n$. It seems to me that in a "usual" algebro-geometric approach line bundles come quite late, only after one defines what is a sheaf, ect. I wonder if this can be explained in "quicker" way?
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http://math.stackexchange.com/questions/39309/complete-lattices-and-sublattices-which-requirement-is-more-stringent?answertab=oldest
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# Complete lattices and sublattices – which requirement is more stringent?
I'm studying from Michael Carter's "Foundations", and on page 29 he makes the comment, "Note that the requirement of being a sublattice is more stringent than being a complete lattice in its own right".
This seems counterintuitive. Some lattice $L$ is complete if every nonempty subset $S\subseteq L$ has a least upper bound and greatest lower bound in $L$. This is a more stringent requirement than merely needing bounds for every combination of two points in $L$. By the same token, requiring that all subsets $T \subseteq S$ have meets and joins in $S$ (thereby making $S$ complete) ought to be more stringent than merely requiring $S$ to be a lattice, right? In what sense are the requirements for being a sublattice more stringent on $S$?
One more related question – suppose $X= \{ (1,1),(2,1),(3,1),(1,2),(1,3),(3,3) \}$. The book says $X$ is a complete lattice, but not a sublattice of $\{1,2,3\}\times \{1,2,3\}$. The point $(3,3)$ is an upper bound for all pairs of points in $X$, so shouldn't that be enough to qualify $X$ as a sublattice of the set above? Why is $X$ not considered a sublattice?
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Take $\mathcal{L} = [0,2]$ with the usual total order on the reals, which is a complete lattice. Let $L=[0,1)\cup\{2\}$; this is a lattice under the usual order of the reals, and as a lattice it is a sublattice of $\mathcal{L}$; it is also a complete lattice, but it is not a sublattice of $\mathcal{L}$ because the least upper bound of $[0,1)$ in $\mathcal{L}$ does not agree with the least upper bound in $L$. – Arturo Magidin May 16 '11 at 0:36
## 1 Answer
The meet and join operations in a sublattice must agree with the meet and join operations in the lattice. Consider your example. The least upper bound, in $X$, of $(1,2)$ and $(2,1)$ is $(3,3)$. In the space $\{ 1,2,3 \} \times \{ 1, 2, 3 \}$ the least upper bound of $(1,2)$ and $(2,1)$ is $(2,2)$.
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Oh wow, that's a pretty big difference. Thanks for the help. – jefflovejapan May 16 '11 at 0:36
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This is a very common point of confusion: there is a difference between being a sublattice and being a subposet that happens to be a lattice. The same problem comes up in the context of subtrees. – Carl Mummert May 16 '11 at 0:36
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http://mathoverflow.net/revisions/60458/list
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## Return to Answer
2 not knot
Note: This is a generalization in a different direction from what the question asked for (these references generalize in terms of finding volumes, but Koundinya Vajjha wanted a generalization in terms of finding the centroid). The higher dimensional version by Gray and Miquel linked to below might yield this, but I haven't read their paper yet.
A quick google search yielded the following reference: "Generalizations of the Theorems of Pappus" by A. W. Goodman and Gary Goodman.
They show that the second theorem you stated has a generalization when the circle that the centroid of F travels on is replaced with any sufficiently smooth simple closed space curve whose curvature never vanishes, however, the first theorem does not generalize easily.
I had to share Figure 1 from the paper - you can apparently compute the volume of the solid you get when you transport the "face" $\mathcal{D}$ bounded by $\mathcal{C}_P$ around the embedded trefoil knot curve $\mathcal{C}_S$. [$\mathcal{C}_S$ is not actually knotted, but the authors state that "... the curve $\mathcal{C}_S$ may even have knots in it, as we have tried to indicate in Figure 1".]
Now, for even more generalizations, you can look at the papers on google scholar that cite this:
There's a paper of Alfred Gray and Vicente Miquel that discusses volumes of similar constructions in space forms. Their motivation was to try to better understand Weyl's famous tube formula. M. Carmen Domingo-Juan, Ximo Gual and Vicente Miquel have since written several more papers on the subject.
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Note: This is a generalization in a different direction from what the question asked for (these references generalize in terms of finding volumes, but Koundinya Vajjha wanted a generalization in terms of finding the centroid). The higher dimensional version by Gray and Miquel linked to below might yield this, but I haven't read their paper yet.
A quick google search yielded the following reference: "Generalizations of the Theorems of Pappus" by A. W. Goodman and Gary Goodman.
They show that the second theorem you stated has a generalization when the circle that the centroid of F travels on is replaced with any sufficiently smooth simple closed space curve whose curvature never vanishes, however, the first theorem does not generalize easily.
I had to share Figure 1 from the paper - you can apparently compute the volume of the solid you get when you transport the "face" $\mathcal{D}$ bounded by $\mathcal{C}_P$ around the embedded trefoil knot $\mathcal{C}_S$.
Now, for even more generalizations, you can look at the papers on google scholar that cite this:
There's a paper of Alfred Gray and Vicente Miquel that discusses volumes of similar constructions in space forms. Their motivation was to try to better understand Weyl's famous tube formula. M. Carmen Domingo-Juan, Ximo Gual and Vicente Miquel have since written several more papers on the subject.
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http://en.wikipedia.org/wiki/Akaike_information_criterion
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# Akaike information criterion
The Akaike information criterion is a measure of the relative quality of a statistical model, for a given set of data. As such, AIC provides a means for model selection. AIC is founded on information entropy: it offers a relative estimate of the information lost when a given model is used to represent the process that actually generates the data.
AIC deals with the trade-off between the complexity of the model and the goodness of fit of the model.
AIC does not provide a test of a model in the sense of testing a null hypothesis; i.e. AIC can tell nothing about how well a model fits the data in an absolute sense. If all the candidate models fit poorly, AIC will not give any warning of that.
## Definition
In the general case, the AIC is
$\mathit{AIC} = 2k - 2\ln(L)\,$
where k is the number of parameters in the statistical model, and L is the maximized value of the likelihood function for the estimated model.
Given a set of candidate models for the data, the preferred model is the one with the minimum AIC value. Hence AIC not only rewards goodness of fit, but also includes a penalty that is an increasing function of the number of estimated parameters. This penalty discourages overfitting (increasing the number of free parameters in the model improves the goodness of the fit, regardless of the number of free parameters in the data-generating process).
AIC is founded in information theory. Suppose that the data is generated by some unknown process f. We consider two candidate models to represent f: g1 and g2. If we knew f, then we could find the information lost from using g1 to represent f by calculating the Kullback–Leibler divergence, DKL(f ‖ g1); similarly, the information lost from using g2 to represent f would be found by calculating DKL(f ‖ g2). We would then choose the candidate model that minimized the information loss.
We cannot choose with certainty, because we do not know f. Akaike (1974) showed, however, that we can estimate, via AIC, how much more (or less) information is lost by g1 than by g2. It is remarkable that such a simple formula for AIC results. The estimate, though, is only valid asymptotically; if the number of data points is small, then some correction is often necessary (see AICc, below).
## How to apply AIC in practice
To apply AIC in practice, we start with a set of candidate models, and then find the models' corresponding AIC values. There will almost always be information lost due to using one of the candidate models to represent the "true" model (i.e. the process that generates the data). We wish to select, from among R candidate models, the model that minimizes the information loss. We cannot choose with certainty, but we can minimize the estimated information loss.
Denote the AIC values of the candidate models by AIC1, AIC2, AIC3, …, AICR. Let AICmin be the minimum of those values. Then exp((AICmin−AICi)/2) can be interpreted as the relative probability that the ith model minimizes the (estimated) information loss.[1]
As an example, suppose that there were three models in the candidate set, with AIC values 100, 102, and 110. Then the second model is exp((100−102)/2) = 0.368 times as probable as the first model to minimize the information loss, and the third model is exp((100−110)/2) = 0.007 times as probable as the first model to minimize the information loss. In this case, we would omit the third model from further consideration. We could take a weighted average of the first two models, with weights 1 and 0.368, respectively, and then do statistical inference based on the weighted multimodel;[2] alternatively, we could gather more data to distinguish between the first two models.
The quantity exp((AICmin−AICi)/2) is the relative likelihood of model i.
If all the models in the candidate set have the same number of parameters, then using AIC might at first appear to be very similar to using the likelihood-ratio test. There are, however, important distinctions. In particular, the likelihood-ratio test is valid only for nested models whereas AIC (and AICc) has no such restriction.[3]
## AICc
AICc is AIC with a correction for finite sample sizes:
$AICc = AIC + \frac{2k(k + 1)}{n - k - 1}$
where n denotes the sample size. Thus, AICc is AIC with a greater penalty for extra parameters.
Burnham & Anderson (2002) strongly recommend using AICc, rather than AIC, if n is small or k is large. Since AICc converges to AIC as n gets large, AICc generally should be employed regardless.[4] Using AIC, instead of AICc, when n is not many times larger than k2, increases the probability of selecting models that have too many parameters, i.e. of overfitting. The probability of AIC overfitting can be substantial, in some cases.[5]
Brockwell & Davis (1991, p. 273) advise using AICc as the primary criterion in selecting the orders of an ARMA model for time series. McQuarrie & Tsai (1998) ground their high opinion of AICc on extensive simulation work with regression and time series.
AICc was first proposed by Hurvich & Tsai (1989). Different derivations of it are given by Brockwell & Davis (1991), Burnham & Anderson, and Cavanaugh (1997). All the derivations assume a univariate linear model with normally distributed errors (conditional upon regressors); if that assumption does not hold, then the formula for AICc will usually change. Further discussion of this, with examples of other assumptions, is given by Burnham & Anderson (2002, ch. 7). In particular, bootstrap estimation is usually feasible.
Note that when all the models in the candidate set have the same k, then AICc and AIC will give identical (relative) valuations. In that situation, then, AIC can always be used.
## Relevance to chi-squared fitting
Often, one wishes to select amongst competing models where the likelihood functions assume that the underlying errors are normally distributed (with mean zero) and independent. This assumption leads to $\chi^2$ model fitting.
For $\chi^2$ fitting, the likelihood is given by
$L=\prod_{i=1}^n \left(\frac{1}{2 \pi \sigma_i^2}\right)^{1/2} \exp \left( -\sum_{i=1}^{n}\frac{(y_i-f(\mathbf{x}))^2}{2\sigma_i^2}\right)$
$\therefore \ln(L) = \ln\left(\prod_{i=1}^n\left(\frac{1}{2\pi\sigma_i^2}\right)^{1/2}\right) - \frac{1}{2}\sum_{i=1}^n \frac{(y_i-f(\mathbf{x}))^2}{\sigma_i^2}$
$\therefore \ln(L) = C - \chi^2/2 \,$,
where C is a constant independent of the model used, and dependent only on the use of particular data points. i.e. it does not change if the data do not change.
The AIC is therefore given by $AIC = 2k - 2\ln(L) = 2k - 2(C-\chi^2/2) = 2k -2C + \chi^2 \,$. As only differences in AIC are meaningful, the constant C can be ignored, allowing us to take $AIC = \chi^2 + 2k$ for model comparisons.
Another convenient form arises if the σi are assumed to be identical and the residual sum of squares (RSS) is available. Then we get AIC = n ln(RSS/n) + 2k + C, where again C can be ignored in model comparisons.[6]
## History
The Akaike information criterion was developed by Hirotugu Akaike, under the name of "an information criterion". It was first published by Akaike in 1974.[7]
The original derivation of AIC relied upon some strong assumptions. Takeuchi (1976) showed that the assumptions could be made much weaker. This work, however, was in Japanese, and was not widely known outside Japan for many years.
AICc was originally proposed for linear regression (only) by Sugiura (1978). That instigated the work of Hurvich & Tsai (1989), and several further papers by the same authors, which extended the situations in which AICc could be applied. The work of Hurvich & Tsai contributed to the decision to publish a second edition of the volume by Brockwell & Davis (1991), which is the standard reference for linear time series; the new edition states, "our prime criterion for model selection [among ARMA(p,q) models] will be the AICc".[8]
The volume by Burnham & Anderson (2002) was the first attempt to set out the information-theoretic approach in a general context. It includes an English exposition of the results of Takeuchi. The volume led to far greater use of the information-theoretic approach, and now has over 17000 citations on Google Scholar.
Akaike originally called his approach an “entropy maximization principle”. Burnham & Anderson (2002, ch. 2) discuss and expand on this, and trace the approach back to the work of Ludwig Boltzmann on thermodynamics. Briefly, minimizing AIC in a statistical model is essentially equivalent to maximizing entropy in a thermodynamic system. In other words, the information-theoretic approach in statistics is essentially applying the Second Law of Thermodynamics.
## Bayesian information criterion
The AIC penalizes the number of parameters less strongly than does the Bayesian information criterion (BIC), which was independently developed by Akaike and by Schwarz in 1978, using Bayesian formalism.[9] Akaike's version of BIC was originally denoted ABIC (for "a Bayesian Information Criterion") or referred to as Akaike's Bayesian Information Criterion.[10]
A comparison of AIC/AICc and BIC is given by Burnham & Anderson (2002, § 6.4). The authors argue that AIC/AICc has theoretical advantages over BIC. Firstly, because AIC/AICc is derived from principles of information. Secondly, because the (Bayesian) derivation of BIC has a prior of 1/R (where R is the number of candidate models), which is "not sensible", since the prior should be a decreasing function of k. The authors also show that AIC and AICc can be derived in the same Bayesian framework as BIC, just by using a different prior. Additionally, they present a few simulation studies that suggest AICc tends to have practical/performance advantages over BIC. See also Burnham & Anderson (2004).
Further comparison of AIC and BIC, in the context of regression, is given by Yang (2005). In particular, AIC is asymptotically optimal in selecting the model with the least mean squared error, under the assumption that the exact "true" model is not in the candidate set (as is virtually always the case in practice); BIC is not asymptotically optimal under the assumption. Yang further shows that the rate at which AIC converges to the optimum is, in a certain sense, the best possible.
## References
• Akaike, Hirotugu (1974), "A new look at the statistical model identification", IEEE Transactions on Automatic Control 19 (6): 716–723, doi:10.1109/TAC.1974.1100705, MR 0423716 .
• Akaike, Hirotugu (1980), "Likelihood and the Bayes procedure", in Bernardo, J. M.; et al., Bayesian Statistics, Valencia: University Press, pp. 143–166 .
• Anderson, D. R. (2008), Model Based Inference in the Life Sciences, Springer .
• Brockwell, Peter J.; Davis, Richard A. (1987), Time Series: Theory and Methods, Springer, ISBN 0387964061 .
• Brockwell, Peter J.; Davis, Richard A. (1991), Time Series: Theory and Methods (2nd ed.), Springer, ISBN 0387974296 . Republished in 2009: ISBN 1441903194
• Burnham, K. P.; Anderson, D. R. (2002), Model Selection and Multimodel Inference: A Practical Information-Theoretic Approach (2nd ed.), Springer-Verlag, ISBN 0-387-95364-7 .
• Burnham, K. P.; Anderson, D. R. (2004), "Multimodel inference: understanding AIC and BIC in Model Selection", Sociological Methods and Research 33: 261–304 .
• Cavanaugh, J. E. (1997), "Unifying the derivations of the Akaike and corrected Akaike information criteria", Statistics and Probability Letters 31: 201–208 .
• Claeskens, G.; Hjort, N. L. (2008), Model Selection and Model Averaging, Cambridge .
• Fang, Yixin (2011). "Asymptotic equivalence between cross-validations and Akaike Information Criteria in mixed-effects models", Journal of Data Science, 9:15-21.
• Hurvich, C. M.; Tsai, C.-L. (1989), "Regression and time series model selection in small samples", 76: 297–307 .
• Lukacs, P.M., et al. (2007). "Concerns regarding a call for pluralism of information theory and hypothesis testing", Journal of Applied Ecology, 44:456–460. doi:10.1111/j.1365-2664.2006.01267.x.
• McQuarrie, A. D. R.; Tsai, C.-L. (1998), Regression and Time Series Model Selection, World Scientific, ISBN 981-02-3242-X .
• Sugiura, N. (1978), "Further analysis of the data by Akaike’s information criterion and the finite corrections", A7: 13–26 .
• Takeuchi, K. (1976), "???", Suri-Kagaku (Mathematical Sciences) (in Japanese) 153: 12–18 .
• Yang, Y. (2005), "Can the strengths of AIC and BIC be shared?", Biometrika 92: 937–950 .
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http://mathhelpforum.com/calculus/60749-partial-derivatives.html
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# Thread:
1. ## Partial derivatives..
I'm looking for the indicated second-order partial derivative for the function f(x,y).
f(x,y)=(1+2xy^2)^8
What I have is:
fx=(2y^2)^8 fy=(4xy)^8
fxx=0 fyy=(4x)^8
fxy=(4y)^8 fyx=4^8
I'm not good at the language of math so I'm not sure if I need to go on and if so, if I have even done what I have right. Any help is more than appreciated.
2. Originally Posted by halborse
I'm looking for the indicated second-order partial derivative for the function f(x,y).
f(x,y)=(1+4xy^2)^8
What I have is:
fx=(2y^2)^8 fy=(4xy)^8
fxx=0 fyy=(4x)^8
fxy=(4y)^8 fyx=4^8
I'm not good at the language of math so I'm not sure if I need to go on and if so, if I have even done what I have right. Any help is more than appreciated.
When you take partial derivative w.r.t x, consider y as a constant.
So $f_x=8(1+4xy^2)^74y^2$
$f_y=(8(1+4xy^2)^78xy$
Can you continue from here?
3. ## Thanks
I made an error when I wrote it down. I have since corrected this, but thank you for your help thus far. I just needed to check my numbers against someone else and I think you helped enough for me to do so. thank you once again.
4. ## my math is weak
I still can't see how to make this work. and when I use my Ti-89 I just end up with y=0. Wish it would show me the steps. Maybe you can. Please?
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http://terrytao.wordpress.com/2010/07/10/cayley-graphs-and-the-geometry-of-groups/
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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
# Cayley graphs and the geometry of groups
10 July, 2010 in expository, math.CO, math.GR | Tags: Cayley graphs, fibre bundles, normal subgroups, semi-direct products
In most undergraduate courses, groups are first introduced as a primarily algebraic concept – a set equipped with a number of algebraic operations (group multiplication, multiplicative inverse, and multiplicative identity) and obeying a number of rules of algebra (most notably the associative law). It is only somewhat later that one learns that groups are not solely an algebraic object, but can also be equipped with the structure of a manifold (giving rise to Lie groups) or a topological space (giving rise to topological groups). (See also this post for a number of other ways to think about groups.)
Another important way to enrich the structure of a group ${G}$ is to give it some geometry. A fundamental way to provide such a geometric structure is to specify a list of generators ${S}$ of the group ${G}$. Let us call such a pair ${(G,S)}$ a generated group; in many important cases the set of generators ${S}$ is finite, leading to a finitely generated group. A generated group ${(G,S)}$ gives rise to the word metric ${d: G \times G \rightarrow {\bf N}}$ on ${G}$, defined to be the maximal metric for which ${d(x,sx) \leq 1}$ for all ${x \in G}$ and ${s \in S}$ (or more explicitly, ${d(x,y)}$ is the least ${m}$ for which ${y = s_1^{\epsilon_1} \ldots s_m^{\epsilon_m} x}$ for some ${s_1,\ldots,s_m \in S}$ and ${\epsilon_1,\ldots,\epsilon_m \in \{-1,+1\}}$). This metric then generates the balls ${B_S(R) := \{ x \in G: d(x,\hbox{id}) \leq R \}}$. In the finitely generated case, the ${B_S(R)}$ are finite sets, and the rate at which the cardinality of these sets grow in ${R}$ is an important topic in the field of geometric group theory. The idea of studying a finitely generated group via the geometry of its metric goes back at least to the work of Dehn.
One way to visualise the geometry of a generated group is to look at the (labeled) Cayley colour graph of the generated group ${(G,S)}$. This is a directed coloured graph, with edges coloured by the elements of ${S}$, and vertices labeled by elements of ${G}$, with a directed edge of colour ${s}$ from ${x}$ to ${sx}$ for each ${x \in G}$ and ${s \in S}$. The word metric then corresponds to the graph metric of the Cayley graph.
For instance, the Cayley graph of the cyclic group ${{\bf Z}/6{\bf Z}}$ with a single generator ${S = \{1\}}$ (which we draw in green) is given as
while the Cayley graph of the same group but with the generators ${S = \{2,3\}}$ (which we draw in blue and red respectively) is given as
We can thus see that the same group can have somewhat different geometry if one changes the set of generators. For instance, in a large cyclic group ${{\bf Z}/N{\bf Z}}$, with a single generator ${S = \{1\}}$ the Cayley graph “looks one-dimensional”, and balls ${B_S(R)}$ grow linearly in ${R}$ until they saturate the entire group, whereas with two generators ${S = \{s_1,s_2\}}$ chosen at random, the Cayley graph “looks two-dimensional”, and the balls ${B_S(R)}$ typically grow quadratically until they saturate the entire group.
Cayley graphs have three distinguishing properties:
• (Regularity) For each colour ${s \in S}$, every vertex ${x}$ has a single ${s}$-edge leading out of ${x}$, and a single ${s}$-edge leading into ${x}$.
• (Connectedness) The graph is connected.
• (Homogeneity) For every pair of vertices ${x, y}$, there is a unique coloured graph isomorphism that maps ${x}$ to ${y}$.
It is easy to verify that a directed coloured graph is a Cayley graph (up to relabeling) if and only if it obeys the above three properties. Indeed, given a graph ${(V,E)}$ with the above properties, one sets ${G}$ to equal the (coloured) automorphism group of the graph ${(V,E)}$; arbitrarily designating one of the vertices of ${V}$ to be the identity element ${\hbox{id}}$, we can then identify all the other vertices in ${V}$ with a group element. One then identifies each colour ${s \in S}$ with the vertex that one reaches from ${\hbox{id}}$ by an ${s}$-coloured edge. Conversely, every Cayley graph of a generated group ${(G,S)}$ is clearly regular, is connected because ${S}$ generates ${G}$, and has isomorphisms given by right multiplication ${x \mapsto xg}$ for all ${g\in G}$. (The regularity and connectedness properties already ensure the uniqueness component of the homogeneity property.)
From the above equivalence, we see that we do not really need the vertex labels on the Cayley graph in order to describe a generated group, and so we will now drop these labels and work solely with unlabeled Cayley graphs, in which the vertex set is not already identified with the group. As we saw above, one just needs to designate a marked vertex of the graph as the “identity” or “origin” in order to turn an unlabeled Cayley graph into a labeled Cayley graph; but from homogeneity we see that all vertices of an unlabeled Cayley graph “look the same” and there is no canonical preference for choosing one vertex as the identity over another. I prefer here to keep the graphs unlabeled to emphasise the homogeneous nature of the graph.
It is instructive to revisit the basic concepts of group theory using the language of (unlabeled) Cayley graphs, and to see how geometric many of these concepts are. In order to facilitate the drawing of pictures, I work here only with small finite groups (or Cayley graphs), but the discussion certainly is applicable to large or infinite groups (or Cayley graphs) also.
For instance, in this setting, the concept of abelianness is analogous to that of a flat (zero-curvature) geometry: given any two colours ${s_1, s_2}$, a directed path with colours ${s_1, s_2, s_1^{-1}, s_2^{-1}}$ (adopting the obvious convention that the reversal of an ${s}$-coloured directed edge is considered an ${s^{-1}}$-coloured directed edge) returns to where it started. (Note that a generated group ${(G,S)}$ is abelian if and only if the generators in ${S}$ pairwise commute with each other.) Thus, for instance, the two depictions of ${{\bf Z}/6{\bf Z}}$ above are abelian, whereas the group ${S_3}$, which is also the dihedral group of the triangle and thus admits the Cayley graph
is not abelian.
A subgroup ${(G',S')}$ of a generated group ${(G,S)}$ can be easily described in Cayley graph language if the generators ${S'}$ of ${G'}$ happen to be a subset of the generators ${S}$ of ${G}$. In that case, if one begins with the Cayley graph of ${(G,S)}$ and erases all colours except for those colours in ${S'}$, then the graph foliates into connected components, each of which is isomorphic to the Cayley graph of ${(G',S')}$. For instance, in the above Cayley graph depiction of ${S_3}$, erasing the blue colour leads to three copies of the red Cayley graph (which has ${{\bf Z}/2{\bf Z}}$ as its structure group), while erasing the red colour leads to two copies of the blue Cayley graph (which as ${A_3 \equiv {\bf Z}/3{\bf Z}}$ as its structure group). If ${S'}$ is not contained in ${S}$, then one has to first “change basis” and add or remove some coloured edges to the original Cayley graph before one can obtain this formulation (thus for instance ${S_3}$ contains two more subgroups of order two that are not immediately apparent with this choice of generators). Nevertheless the geometric intuition that subgroups are analogous to foliations is still quite a good one.
We saw that a subgroup ${(G',S')}$ of a generated group ${(G,S)}$ with ${S' \subset S}$ foliates the larger Cayley graph into ${S'}$-connected components, each of which is a copy of the smaller Cayley graph. The remaining colours in ${S}$ then join those ${S'}$-components to each other. In some cases, each colour ${s \in S\backslash S'}$ will connect a ${S'}$-component to exactly one other ${S'}$-component; this is the case for instance when one splits ${S_3}$ into two blue components. In other cases, a colour ${s}$ can connect a ${S'}$-component to multiple ${S'}$-components; this is the case for instance when one splits ${S_3}$ into three red components. The former case occurs precisely when the subgroup ${G'}$ is normal. (Note that a subgroup ${G'}$ of a generated group ${(G,S)}$ is normal if and only if left-multiplication by a generator of ${S}$ maps right-cosets of ${G'}$ to right-cosets of ${G'}$.) We can then quotient out the ${(G',S')}$ Cayley graph from ${(G,S)}$, leading to a quotient Cayley graph ${(G/G', S \backslash S')}$ whose vertices are the ${S'}$-connected components of ${(G,S)}$, and the edges are projected from ${(G,S)}$ in the obvious manner. We can then view the original graph ${(G,S)}$ as a bundle of ${(G',S')}$-graphs over a base ${(G/G',S\backslash S')}$-graph (or equivalently, an extension of the base graph ${(G/G',S\backslash S')}$ by the fibre graph ${(G',S')}$); for instance ${S_3}$ can be viewed as a bundle of the blue graph ${A_3}$ over the red graph ${{\bf Z}/2{\bf Z}}$, but not conversely. We thus see that the geometric analogue of the concept of a normal subgroup is that of a bundle. The generators in ${S \backslash S'}$ can be viewed as describing a connection on that bundle.
Note, though, that the structure group of this connection is not simply ${G'}$, unless ${G'}$ is a central subgroup; instead, it is the larger group ${G' \rtimes \hbox{Aut}(G')}$, the semi-direct product of ${G'}$ with its automorphism group. This is because a non-central subgroup ${G'}$ can be “twisted around” by operations such as conjugation ${g' \mapsto sg's^{-1}}$ by a generator ${s \in S}$. So central subgroups are analogous to the geometric notion of a principal bundle. For instance, here is the Heisenberg group
$\displaystyle \begin{pmatrix} 1 & {\bf F}_2 & {\bf F}_2 \\ 0 & 1 & {\bf F}_2 \\ 0 & 0 & 1 \end{pmatrix}$
over the field ${{\bf F}_2}$ of two elements, which one can view as a central extension of ${{\bf F}_2^2}$ (the blue and green edges, after quotienting) by ${{\bf F}_2}$ (the red edges):
Note how close this group is to being abelian; more generally, one can think of nilpotent groups as being a slight perturbation of abelian groups.
In the case of ${S_3}$ (viewed as a bundle of the blue graph ${A_3}$ over the red graph ${{\bf Z}/2{\bf Z}}$), the base graph ${{\bf Z}/2{\bf Z}}$ is in fact embedded (three times) into the large graph ${S_3}$. More generally, the base graph ${(G/G', S \backslash S')}$ can be lifted back into the extension ${(G,S)}$ if and only if the short exact sequence ${0 \rightarrow G' \rightarrow G \rightarrow G/G' \rightarrow 0}$ splits, in which case ${G}$ becomes a semidirect product ${G \equiv G' \rtimes H}$ of ${G'}$ and a lifted copy ${H}$ of ${G/G'}$. Not all bundles can be split in this fashion. For instance, consider the group ${{\bf Z}/9{\bf Z}}$, with the blue generator ${1}$ and the red generator ${3}$:
This is a ${{\bf Z}/3{\bf Z}}$-bundle over ${{\bf Z}/3{\bf Z}}$ that does not split; the blue Cayley graph of ${{\bf Z}/3{\bf Z}}$ is not visible in the ${{\bf Z}/9{\bf Z}}$ graph directly, but only after one quotients out the red fibre subgraph. The notion of a splitting in group theory is analogous to the geometric notion of a global gauge. The existence of such a splitting or gauge, and the relationship between two such splittings or gauges, are controlled by the group cohomology of the sequence ${0 \rightarrow G' \rightarrow G \rightarrow G/G' \rightarrow 0}$.
Even when one has a splitting, the bundle need not be completely trivial, because the bundle is not principal, and the connection can still twist the fibres around. For instance, ${S_3}$ when viewed as a bundle over ${{\bf Z}/2{\bf Z}}$ with fibres ${A_3}$ splits, but observe that if one uses the red generator of this splitting to move from one copy of the blue ${A_3}$ graph to the other, that the orientation of the graph changes. The bundle is trivialisable if and only if ${G'}$ is a direct summand of ${G}$, i.e. ${G}$ splits as a direct product ${G = G' \times H}$ of a lifted copy ${H}$ of ${G/G'}$. Thus we see that the geometric analogue of a direct summand is that of a trivialisable bundle (and that trivial bundles are then the analogue of direct products). Note that there can be more than one way to trivialise a bundle. For instance, with the Klein four-group ${{\bf Z}/2{\bf Z} \times {\bf Z}/2{\bf Z}}$,
the red fibre ${{\bf Z}/2{\bf Z}}$ is a direct summand, but one can use either the blue lift of ${{\bf Z}/2{\bf Z}}$ or the green lift of ${{\bf Z}/2{\bf Z}}$ as the complementary factor.
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## 30 comments
First off, I LOVE these “by the way” lecture posts of yours,Professor Tao. They’re a great resource for the entire mathematical community. It’s very comforting to know there’s a top notch researcher who seems just as passionate about teaching as you are. Keep up the great work.
The serious study of the geometric properties of groups and thier associated graph embeddings in Euclidean spaces-which was initiated by Klein and developed by so many great mathematicians,most notably Wilhelm Magnus and his students-is a fantasticaly rich topic that isn’t incorporated into traditional courses in algebra and geometry anywhere near as fully as it should be. The classic introduction is of course, GRAPHS AND THIER GROUPS by Grossman and Magnus. A much more advanced text that covers the graph theoretic aspects-including the deeper topological properties of the Cayley graph and it’s fundamental group structure touched on in your post-can be found in GRAPHS OF GROUPS ON SURFACES by Arthur T.White,currently in it’s 3rd edition.
Again,thanks for the great introduction to such an important topic.
It’s a well-known fact that a connected topological group is generated by any open neighborhood of its identity element. (So it is tempting to think of a Lie group as _generated_ by its Lie algebra, after identifying the Lie algebra with an infinitesimal neighborhood of the identity. This is reasonable in models of mathematics where infinite numbers really exist. Moreover, the Baker-Campbell-Hausdorff series identifies the Lie algebra with the formal group, and over $\mathbb R$ the BCH series converges for finite-dimensional Lie algebras in a non-trivial neighborhood of the origin, and so you really can identity a neighborhood of the Lie algebra with a neighborhood of the Lie group.) So it is tempting to take some sort of “formal neighborhood” of the identity element as my generating set $S$. Can I then think of the topology on \$G\$ as coming from some sort of “formal Cayley graph”? Maybe I will ask this over on MathOverflow.
Yes, one can view topological groups and Lie groups as being continuous analogues of generated groups, and indeed a significant portion of geometric group theory is devoted to viewing the former as asymptotic limits of the latter (e.g. Gromov’s original proof of Gromov’s theorem on groups of polynomial growth definitely falls into this category). The literature on Hilbert’s fifth problem (which, roughly speaking, asks whether all locally compact topological groups come from Lie groups) don’t explicitly use Cayley graphs, but it seems to be lurking beneath the surface (particularly with those solutions to that problem that use nonstandard analysis).
Fascinating stuff. I didn’t realize that groups were so closely connected with certain graphs. Thanks for posting this.
I like the commonsense explanation of normal groups that comes out of this presentation. By contrast, the usual definition ( g n g^{-1} \in G) is unintuitive despite its simplicity.
So a Cayley graph is not associated with a group, but with a presented group. It’d be nice to have something that only depended on the group itself, and not the presentation. (And yes, I do know where this goes…)
To be utterly pedantic, a presented group is a slightly richer object than a generated group (which, in turn, is richer than a raw group); a presented group is a group with a distinguished set of generating elements and a distinguished set of generating relations, whereas a generated group only has the former and not the latter. (So there are forgetful functors from the category of presented groups to the category of generated groups to the category of groups.) Cayley graphs are more naturally associated with generated groups than with presented groups.
In the case of infinite finitely generated groups, changing the choice of generators only affects the word metric by a bilipschitz distortion. As such, the asymptotic (coarse) geometry of the finitely generated group does not depend on the choice of generators (e.g. the concept of a finitely generated group of polynomial growth does not depend on the choice of generators). Much of geometric group theory focuses on this asymptotic geometry and so indeed one can focus primarily on groups themselves rather than on generated groups. However, if one wants to work more quantitatively and non-asymptotically (and specifically, if one wants to talk about results that are non-trivial even for finite groups) then the choice of generators becomes more important. For instance, as discussed above, a cyclic group can look one-dimensional, two-dimensional, or many-dimensional at various scales, depending on how many generators one selects and how commensurate they are to each other.
11 July, 2010 at 10:00 am
Anonymous
typo in the second paragraph: \$y = s_1^{\epsilon_1} \ldots s_m^{\epsilon_m} y\$ should be \$y = s_1^{\epsilon_1} \ldots s_m^{\epsilon_m} x\$. Thanks for the post.
[Corrected, thanks - T.]
12 July, 2010 at 1:35 pm
Anonymous
Could you elaborate on the third property (Homogeneity) of the Cayley Graph?
Is this saying that there aren’t two paths between x and y of the same color and direction along each step?
Thanks.
No, this is already a consequence of the regularity hypothesis.
One way to view homogeneity is that it asserts that any graph theoretic statement that is true if one uses x as a starting vertex, is also true if one uses y as a starting vertex, so that all vertices “look the same” from a graph theoretic perspective. For instance, in the Z/9Z graph, no matter where one starts from, the operation of following three blue arrows is always equal to following one red arrow. If three blues equaled a red for one starting vertex x but not for another y, then the graph would not be homogeneous.
The formal definition of homogeneity requires the notion of a graph isomorphism,
http://en.wikipedia.org/wiki/Graph_isomorphism
«Latex path not specified» in place of several formulas. About eight hours ago everything was OK.
Thanks for interesting post!
15 July, 2010 at 9:30 am
Anonymous
I am pleased by this post, and I would like to understand better why the group properties discussed here are geometric?
Is it because they are (geo)metric? But it’s a word metric, why isn’t it algebraic?
The notion of ‘ball’ is not geometric either. Is ‘(polynomial) growth’ a geometric concept? you just count how many words there are of length 1,2,3,.. in your generated group.
That a group is abelian is visible from a multiplication table (anyway visualisation of Cayley graphs is restricted to ‘small’ groups).
“Nevertheless the geometric intuition that subgroups are analogous to foliations is still quite a good one.”
It would be interesting to know how it could help. Why is it really geometric? The notion of ‘metric’ is not engaged here. Is it geometric because it is visualised and explained by means of symbolism embedded in Euclidean plain? Yes, this method is used in geometry, but not only in geometry, in analysis, for example, too.
The fact that “one can view topological groups and Lie groups as being continuous analogues of generated groups” gives another view on group but does not add geometric properties, it seems to me. please, tell me if I am wrong.
Groups are both algebraic objects and geometric objects; it is not a dichotomy.
While metric geometry is certainly an important component of modern geometry, metrics are certainly not the only geometric objects of study. Riemannian geometry, for instance, is definitely very much concerned with volumes of balls, among other things. Differential geometry and its relatives (e.g. contact geometry) is similarly quite concerned with foliations, bundles, and connections. Kleinian geometry focuses on the notion of symmetry, which is very much a group-theoretic concept also. (The classical Euclidean geometry of points, lines, and planes can be viewed as a special case of all of the above modern types of geometry.)
Traditionally, geometry has focused mostly on continuous structures, but nowadays discrete geometries are also an important area of study, and Cayley graphs form a basic example of such.
16 July, 2010 at 6:23 am
Anonymous
Is it not a dichotomy in the sense that groups have algebraic and geometric properties or in the sense, the same properties of groups can be seen as algebraic and as geometric properties? or are there properties of groups which are certainly geometric but not algebraic? Because what is ‘geometric’ at the first glance, seems to be perfectly algebraic: e.g. metric, hyperbolicity can be formulated algebraically, Cayley graphs -aren’t they algebraic objects?
Almost every fundamental concept or object in mathematics can be interpreted in a number of different ways (algebraic, geometric, dynamical, analytical, etc.), and it is important to internalise as many of these as one can in order to truly understand these concepts. I recommend the discussion in Thurston’s article “On proof and progress in mathematics” for further reading. (Section 2 is the most relevant here, but the entire article is worthwhile.)
15 July, 2010 at 11:16 am
Anonymous
Naive question: How do you yourself distinguish algebraic objects from geometric objects? I didn’t find the relevant wikipedia article especially helpful in this regard.
15 July, 2010 at 7:09 pm
xiaodong xu
Dear Terry, you wrote
•(Connectedness) The graph is connected.
This may not be true for some Cayley graphs. For instance, Z/8Z and S={2}.
Yes?
{2} does not generate Z/8Z.
Dear Terrence,
I thought it would be interesting to mention the relationship between Cayley graphs and tilings. I have some pictures on thi on my web page:
http://www.xs4all.nl/~westy31/Geometry/Geometry.html#Cayley
Basically, you can take any Cayley graph with 2 generators, and turn it into a tiling of a Riemann surface. Each tile corresponds to a vertex of the Cayley graph. Since any finite simple group can be generated by 2 generators, all finite simple groups are a tiling of a Riemann surface. Also, you can interpret the tiling itself as a Cayly graph, by labeling rotations around the vertices as generators.
One question that comes to mind if the genus of the surface has some group theoretic meaning.
Gerard
Interesting article. In the second paragraph after the picture of S_3 you mention it can be split into “two red components” and “three blue components”. I believe you mean “two blue components” and “three red components”.
[Corrected, thanks - T.]
19 July, 2010 at 1:21 am
student
You wrote:
“In some cases, each colour {s \in S\backslash S’} will connect a {S’}-component to exactly one other {S’}-component; this is the case for instance when one splits {S_3} into three red components. In other cases, a colour {s} can connect a {S’}-component to multiple {S’}-components; this is the case for instance when one splits {S_3} into two blue components. ”
which should probably (???) read:
In some cases, each colour {s \in S\backslash S’} will connect a {S’}-component to exactly one other {S’}-component; this is the case for instance when one splits {S_3} into two blue components. In other cases, a colour {s} can connect a {S’}-component to multiple {S’}-components; this is the case for instance when one splits {S_3} into three red components.
[Oops! I implemented a previous fix incorrectly. Thanks, it should be correct now - T.]
[...] 3….y un post muy informativo sobre grafos de Cayley como “representación geometrica” de grupos: http://terrytao.wordpress.com/2010/07/10/cayley-graphs-and-the-geometry-of-groups/ [...]
Hi,
I am newly interested in cayley graph. I read note on cayley graphs. may be it is very clear but I could not understand how can I draw ,i.e. how can ı draw directed edge? you gave example for 6Z and S={1}. as you say, I use from x to sx. but only one element exists in S and then edge is from 1 to 1, 2 to 2… and so on. Can I misunderstand?
In the case of an additive group such as ${\bf Z}/6{\bf Z} = ({\bf Z}/6{\bf Z}, +)$ rather than a multiplicative group $G = (G,\cdot)$, one would connect x to s+x, rather than x to sx.
Dear Prof. Tao,
firstly, thanks for answering my previous question.
I wonder that all inverse element of S is also S. Is this necessary or not?
in your note about cayley graphs you did not wrote but some books add this condition.
and you said that word metric is the maximal metric .What reason is maximal?
best regards,
There are two types of Cayley graphs: directed Cayley graphs (the ones discussed here) and undirected Cayley graphs. The latter require a symmetric set of generators, and the former does not.
The word metric is maximal with respect to the pointwise order on metrics (in which one writes $d \leq d'$ for two metrics d,d’ if one has $d(x,y) \leq d'(x,y)$ for all x,y) subject to the constraint that $d(x,sx) \leq 1$ for all x in G and s in S. This is a succinct way to define the word metric, but perhaps not the most enlightening, which is why I also added a more explicit description of that metric.
Dear prof. Tao,
Would $d(x,y)\in \{0,1\}$ only? Only that would make sense to me because this is kind of defining a metric of orbit of $x$. What we care here is to establish the connection between $x$ and its orbits. If the action of $s\in S$ on $x$ results in $x$, $d(x,sx)=0$. Otherwise, it is $1$.
If that’s the case, $d(x,y)=d(x,s_m^{\epsilon_m}\dots s_1^{\epsilon_1}x)\le d(s_2^{\epsilon_2}\dots s_m^{\epsilon_m}x,s_1^{\epsilon_1}\dots s_m^{\epsilon_m}x)+\dots+d(x,s_1^{\epsilon_1}x)\le m$.
That’s how I can make sense of this. Please correct me if I’m wrong.
Dear prof. Tao,
you said that a directed coloured graph is a Cayley graph (up to relabeling) if and only if it obeys the above three properties. I understand one way of the proof. conversely for showing that cayley graph satisfy property of Homogeneity ,you use right multiplication xg. Can we choose left multiplication instead of right?
best regards,
[...] describes the geometry of the Cayley graph on formed by connecting to for each and . (See this previous post for more discussion of using Cayley graphs to study groups [...]
[...] is a sequel to my previous blog post “Cayley graphs and the geometry of groups“. In that post, the concept of a Cayley graph of a group was used to place some geometry on [...]
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http://mathhelpforum.com/calculus/66450-mathematical-analysis-problem-please-help.html
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# Thread:
1. ## mathematical analysis problem. please help.
suppose {x(n)} is a sequence such that
x(1)=1 and
x(n+1)=1+x(n)/(1+x(n+1)) ; n>=1
Prove that this sequence converges and find the limit.
I know how to find the limit of this sequence but dont know how to show that {x(n)} is a convergent sequence. Please help me on this.
2. Originally Posted by Kat-M
suppose {x(n)} is a sequence such that
x(1)=1 and
x(n+1)=1+x(n)/(1+x(n+1)) ; n>=1
Prove that this sequence converges and find the limit.
I know how to find the limit of this sequence but dont know how to show that {x(n)} is a convergent sequence. Please help me on this.
First, multiplying the series by $1 + x_{n+1}$ gives
$x_{n+1} (1+x_{n+1})=1 + x_{n+1}+x_n$ or $x^2_{n+1} =1 + x_n$
To prove that the sequence converges we will prove two things
(i) the sequence is increasing and
(ii) it is bounded above
First off, I think it's clear that $x_n > 0$
(i) writing out the first few terms suggests its increasing. To prove this assume that
$x_{k} < x_{k+1}$
thus,
$x_{k} + 1 < x_{k+1} + 1\;\;\; \Rightarrow\;\;\;x^2_{k+1} < x^2_{k+2}$
so
$x_{k+1} < x_{k+2}$
and by induction, $x_{n} < x_{n+1}\;\; \forall n$
(ii) It is bounded by 2
The first few terms suggest this. Assume that
$x_{k} < 2$
then
$x_{k} + 1 < 2 + 1 = 3 < 4$
thus,
$x^2_{k+1} < 4\;\;\; \Rightarrow \;\;x_{k+1} < 2$
so again by induction, it is ture for all n. Since the sequence is increasing and bounded above, it must converge.
3. Here's one very similar that I think can be solved exactly
$2 x^2_{n+1} = x_n + 1,\;\;\;x_0 = 0.$
Any ideas?
4. ## thanks
i made a typo in the original question and i fixed it which is posted below.
5. ## oops i had a typo in the question.
here is the correct problem. In the original question i had 1+x(n)/(1+x(n+1)) but it should be 1+x(n)/(1+x(n)).
suppose {x(n)} is a sequence such that
x(1)=1 and
x(n+1)=1+x(n)/(1+x(n)) ; n>=1
Prove that this sequence converges and find the limit.
I know how to find the limit of this sequence but dont know how to show that {x(n)} is a convergent sequence. Please help me on this.
6. Originally Posted by Kat-M
here is the correct problem. In the original question i had 1+x(n)/(1+x(n+1)) but it should be 1+x(n)/(1+x(n)).
I wondered about that! What you had originally seemed a strange way to write the formula.
suppose {x(n)} is a sequence such that
x(1)=1 and
x(n+1)=1+x(n)/(1+x(n)) ; n>=1
Prove that this sequence converges and find the limit.
I know how to find the limit of this sequence but dont know how to show that {x(n)} is a convergent sequence. Please help me on this.
x(n+1)= 1+ x(n)/(1+ x(n)) which we can rewrite as [tex]\frac{1+ 2x(n)}{1+ x(n)}= 2- 1/(1+ x(n)).
Assuming that the limit exists and is "x", we must have lim x(n+1)= 2- 1/(1+ lim x(n) or x= 2- 1/(x+1). Multiplying on both sides by x+1, $x^2+ x= 2x+ 2- 1$ or $x^2- x- 1= 0$. The quadratic formula gives two possible values for x, $\frac{1- \sqrt{5}}{2}$ and $\frac{1+\sqrt{5}}{2}$. Since x(n) is positive for all x, the limit must be $\frac{1+ \sqrt{5}}{2}$. I presume that is what you got.
That is assuming the limit exists. We still need to prove that. Since x(1)= 1 is less than $\frac{1+ \sqrt{5}}{2}$, the sequence has to increase to it and it seems reasonable to show that the sequence is increasing and has an upper bound- that is sufficient to show that it converges.
So, first we want to prove that the sequence is increasing: that x(n+1)> x(n) for all n. x(n+1)= 2- x(n)/(1+ x(n))> x(n), multiplying on both sides by the positive number 1+ x(n), is equivalent to $2+ 2x(n)- x(n)> x(n)^2+ x(n)$ or $x(n)^2- x(n)- 1< 0$. We have already seen that $x^2- x- 1= 0$ only for $x= \frac{1- \sqrt{5}}{2}$ and $x= \frac{1+ \sqrt{5}}{2}$. If x= 0, which is between those two values, then $(0)^2- 0 - 1= -1< 0$ so $x^2- x- 1< 0$ for all x between those values. x(n) is positive for all n so if we can prove that $x(n)< \frac{1+ \sqrt{5}}{2}$ for all n, we will have proved both that the sequence is increasing and that it has an upper bound and so converges.
Let's try induction. Clearly x(1)= 1< $\frac{1+\sqrt{5}}{2}$, which is about 1.47. Suppose x(k)< $\frac{1+\sqrt{5}}{2}$
Then x(k)+ 1< $\frac{3+ \sqrt{5}}{2}$, $\frac{1}{x(k)+1}> \frac{2}{3+ \sqrt{5}}$, and $x(k+1)= 2- 1/(x(k)+1)< 2- \frac{2}{3+ \sqrt{5}}= \frac{4+\sqrt{5}}{3+ \sqrt{5}}$
Rationalizing the denominator, we have $x(n+1)< \frac{4+ 2\sqrt{5}}{3+ \sqrt{5}}\frac{3-\sqrt{5}}{3-\sqrt{5}}= \frac{2+ 2\sqrt{5}}{4}= \frac{1+ \sqrt{5}}{2}$, exactly what was needed.
That is, since x(n) lies between $\frac{1- \sqrt{5}}{2}$ and $\frac{1+\sqrt{5}}{2}$, it is, as we saw above, an increasing sequence. Further, since it has $\frac{1+ \sqrt{5}}{2}$ as upper bound, and is increasing, it is a convergent sequence.
7. Originally Posted by Kat-M
here is the correct problem. In the original question i had 1+x(n)/(1+x(n+1)) but it should be 1+x(n)/(1+x(n)).
suppose {x(n)} is a sequence such that
x(1)=1 and
x(n+1)=1+x(n)/(1+x(n)) ; n>=1
Prove that this sequence converges and find the limit.
I know how to find the limit of this sequence but dont know how to show that {x(n)} is a convergent sequence. Please help me on this.
Another way is to solve your problem directly. The substitution
$x_n = \frac{a y_n}{y_n + 1}$
for suitable $a$ (guess what number see HallsofIvy reply) the difference equation will become a linear difference equation which can be solved exactly.
8. ## Thank you very much
HollsofIvy and danny arrigo. you two helped me a lot. Thank you so much.
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http://mathoverflow.net/questions/60641?sort=oldest
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## In what sense is the étale topology equivalent to the Euclidean topology?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I have heard it said more than once—on Wikipedia, for example—that the étale topology on the category of, say, smooth varieties over $\mathbb{C}$, is equivalent to the Euclidean topology. I have not seen a good explanation for this statement, however.
If we consider the relatively simple example of $\mathbb{P}^1_\mathbb{C}$, it seems to me that an étale map is just a branched cover by a Riemann surface, together with a Zariski open subset of $\mathbb{P}^1_\mathbb{C}$ that is disjoint from the ramification locus. (If there is a misconception there, small or large, please let me know) The connection to the Euclidean topology on $\mathbb{P}^1_\mathbb{C}$, however, is not obvious to me.
What is the correct formulation of the statement that the two topologies are equivalent, or what is a good way to compare them?
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2
It has to do with the fact that the étale cohomology groups are closely related (they might be isomorphic, but I don't remember) to the singular cohomology groups in the case of smooth complex varieties. – Harry Gindi Apr 5 2011 at 4:01
2
Vague answer: I think there are two parts. First, by some GAGA principle, algebraic etale covers = Euclidean etale covers, say when they are both finite. The Euclidean etale topology is equivalent just to the Euclidean topology, since every covering map is a local homeomorphism. To see that the GAGA correspondence really does capture the fineness of the Euclidean topology, you need to check that you can make enough refinements; the "refinement" of a covering is measured in some sense by its Cech cohomology, and as Harry says, etale and singular cohomology are isomorphic, morally. – Ryan Reich Apr 5 2011 at 4:46
## 4 Answers
Saying that the étale topology is equivalent to the euclidean topology is vastly overstating the case. For example, if you compute the cohomology of a complex algebraic variety with coefficients in $\mathbb Q$ in the étale topology, typically you get 0. On the other hand, it is a deep result that the étale cohomology of such a variety with coefficients in a finite abelian group coincides with its cohomology in the euclidean topology.
Similarly, you can't capture the whole fundamental group with the étale topology, but only its finite quotients (and the fact that you can indeed describe the finite quotients of the fundamental group via étale covers is, again, a deep result).
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This might be a stupid question, but is there any Grothendieck topology that you can put on an arbitrary scheme (say, fppf, fpqc, nisnevich, etc) that recovers the euclidean cohomology groups when restricted to complex algebraic varieties? If not, is there a proof that no such topology could exist? – Harry Gindi Apr 5 2011 at 8:32
1
It is not stupid at all. An indication of the fact that this should not be possible is the fact that $\ell$-adic cohomology of elliptic curves over an algebraically closed field of positive characteristic cannot be defined over $\mathbb Z$. At least, I believe this is known and due Serre, but I can't recall the precise argument (I would guess it has to do with the existence of elliptic curves with exotic complex multiplication, but I am not sure). – Angelo Apr 5 2011 at 10:04
3
Harry, there's no purely algebraic definition of the $\mathbb{Q}$ cohomology of an algebraic variety $X$ over $\mathbb{C}$. The choice of a model of $X$ over a subfield $K$ of $\mathbb{C}$ will define an action of $Aut(C/K)$ on $X$ and hence on the cohomology. But we can choose $K$ so that such that $\mathbb{C}/K$ is an infinite Galois extension. Then $Gal(C/K)$ is uncountable, and should typically act through an uncountable quotient on the $\mathbb{Q}$ cohomology (because it does on the etale cohomology). But the $\mathbb{Q}$ cohomology is countable. – mephisto Apr 5 2011 at 11:34
Thanks, guys! – Harry Gindi Apr 5 2011 at 15:18
1
I think that the argument that Angelo is referring to is the second explanation of why there is no universal cohomology theory over $\mathbf Q$ given in Milne's notes on motives: jmilne.org/math/xnotes/mot.html (The first explanation in Milne's notes is the one given above by mephisto.) – Dan Petersen Apr 5 2011 at 15:26
### 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 good start is to read Chapter 21 of Milne's textbook on etale cohomology. I suspect the results there are not stated with maximal power, but it at least gives you a feeling for what's true.
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Your final question, i.e. in what way does one compare the classical and the étale topology of a scheme, is answered for instance in section 2 of Mumford's classical "Picard Groups of Moduli Problems". There are also some nicely written parts in Vistoli's notes on descent on how to compare Grothendieck topologies in general.
As you say there is no direct way to compare the two. What you do is introduce an auxilliary topology which refines both of them. If you take a sep. finite type scheme X over $\mathbf C$, then you can put a topology on complex analytic spaces over X by taking for open subsets those maps $U \to X(\mathbf C)$ that form a covering space over an open subset of $X(\mathbf C)$. Coverings are jointly surjective. We call this site $X_{cx}^\ast$. Then every every open set in both the étale and classical topologies are also open sets of this site, so there are maps $\alpha : X_{cx}^\ast \to X_{cx}$ and $\beta : X_{cx}^\ast \to X_{ét}$. Moreover, $\alpha$ is an equivalence of topologies. (I see now that most of what I said here was already stated in the comment by Ryan Reich.)
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Dear Andrew, here is another take on your question, in the direction of (ramified) covers.
1. Riemann There is an analytification functor $X\mapsto X^{an}$ from the category of $\mathbb C$-schemes locally of finite type to that of (non-reduced) complex analytic spaces. Its introduction is due principally to Riemann, Chow, Serre and Grothendieck. It has all the desirable properties and of course the set of points of $X^{an}$ is $X(\mathbb C)$. Riemann's existence theorem states that this functor induces an equivalence of categories between the finite étale covers of $X$ and the finite étale analytic covers of $X^{an}$.
This is the deep result (alluded to in Angelo's fine answer) which, in particular, yields the identification of the topological fundamental group of $X^{an}$ with a completion of the scheme-theoretic fundamental group.
2. Grauert-Remmert In a sense the classification of algebraic covers has been reduced to that of analytic ones. We can then apply the following result, due to Grauert and Remmert :
Let $X$ be a normal analytic space and $U\subset X$ an open subset with analytic complement. Then any (ramified) finite normal cover of $U$ uniquely extends to a normal finite cover of $X$.
(Grothendieck, in SGA 1, gave a slick proof of this theorem by invoking Hironaka's resolution of singularities)
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http://mathoverflow.net/questions/45036?sort=votes
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## Spectral sequences: opening the black box slowly with an example
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
My friend and I are attempting to learn about spectral sequences at the moment, and we've noticed a common theme in books about spectral sequences: no one seems to like talking about differentials.
While there are a few notable examples of this (for example, the transgression), it seems that by and large one is supposed to use the spectral sequence like one uses a long exact sequence of a pair- hope that you don't have to think too much about what that boundary map does.
So, after looking at some of the classical applications of the Serre spectral sequence in cohomology, we decided to open up the black box, and work through the construction of the spectral sequence associated to a filtration. And now that we've done that, and seen the definition of the differential given there... we want some examples.
To be more specific, we were looking for an example of a filtration of a complex that is both nontrivial (i.e. its spectral sequence doesn't collapse at the $E^2$ page or anything silly like that) but still computable (i.e. we can actually, with enough patience, write down what all the differentials are on all the pages).
Notice that this is different than the question here: http://mathoverflow.net/questions/23297/simple-examples-for-the-use-of-spectral-sequences, though quite similar. We are looking for things that don't collapse, but specifically for the purpose of explicit computation (none of the answers there admit explicit computation of differentials except in trivial cases, I think).
For the moment I'm going to leave this not community wikified, since I think the request for an answer is specific and non-subjective enough that a person who gives a good answer deserves higher reputation for it. If anyone with the power to thinks otherwise, then feel free to hit it with the hammer.
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+1 for the precisely formulated, interesting question and the creative use of a venerable metaphor in the title. – Georges Elencwajg Nov 6 2010 at 7:05
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One case which is very concrete but also very simple is the Bockstein spectral sequence: Start with a torsion free complex $C^\bullet$ and filter it by $p^kC^\bullet$ where $p$ (of course!) is a prime. This is non-trivial alread for a simple example such as $\mathbb Z \to \mathbb Z$ where the differentail is multiplication by $p^n$. – Torsten Ekedahl Nov 6 2010 at 7:09
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Nice question. However, I would not say that the degeneration at the $E^2$-level (or before) is "silly". After all, for compact Kaehler manifolds the degeneration of the Frohlicher spectral sequence at the $E^1$-level implies Hodge decomposition! – Francesco Polizzi Nov 6 2010 at 11:04
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Hopkins said last year in Oberwolfach that there are two kinds of people dealing with spectral sequences: 1) the spirituals, who pray and pray and pray, when they see a spectral sequence, that all differentials vanish - they use the grading convention of the Serre SS, where it is easy to put in the E^2-term 2) the gladiators: every time they turn around they kill another element with a differential. They use the grading convention for the Adams spectral sequence. I think, Hopkins identified himself more with the second type. – Lennart Meier Nov 6 2010 at 16:13
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To everyone: thank you for the wonderful suggestions! I had no idea there would be so many! After I work through as many as I can, perhaps I'll write up some notes with the exercises in them and post them here for other gladiators in training. :) – Dylan Wilson Nov 6 2010 at 17:27
show 6 more comments
## 10 Answers
Two simple examples with lots of interesting differentials are given by the Serre spectral sequences for integer homology (rather than cohomology) for the fibrations $$K({\mathbb Z}_2,1) \to K({\mathbb Z}_4,1)\to K({\mathbb Z}_2,1)$$ and $$K({\mathbb Z}_2,1) \to K({\mathbb Z},2) \to K({\mathbb Z},2)$$ where in the second case the map $K({\mathbb Z},2) \to K({\mathbb Z},2)$ induces multiplication by $2$ on $\pi_2$. In both cases one knows the homology of all three spaces and this allows one to work out what all the differentials must be. The differentials give a real shoot-out, with nontrivial differentials on more than one page, and in the second case there are nontrivial differentials on infinitely many pages. The best thing is to work everything out oneself, but if you want to check your answers these two examples are worked out as Examples 1.6 and 1.11 in Chapter 1 of my spectral sequence notes, available on my webpage.
These examples may not really be the sort of thing you're looking for since they involve computing differentials purely formally, not by actually digging into the construction of the spectral sequence. But of course a lot of spectral sequence calculations have to be formal if one is to have any chance of succeeding.
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Speaking of your notes on spectral sequences: I was looking at your treatment of the transgression and noticed that you proved its equivalence with a differential by running the Serre sseq of a pair. Are there examples of this method in action to compute other differentials? – Dylan Wilson Nov 6 2010 at 17:17
Oh- and thank you for the answer! This exercise looks very instructive. – Dylan Wilson Nov 6 2010 at 17:18
### 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 is more than you bargained for, but it's too good an opportunity to pass up plugging a couple cool and readable papers with impact beyond containing computable spectral sequences. In the 1980s Ravenel and Wilson famously used Hopf rings to compute some extraordinary homologies of a variety of families of infinite loopspaces. In the specific case where the loopspaces are Eilenberg-Mac Lane spaces, they used a bar spectral sequence (which arises as a filtration spectral sequence), together with Hopf ring information, to compute $E_* K(G, *)$ for various $E$ and $G$. One of the cool features of their work is that everything involved is explicit and identifiable.† You might try:
Doug Ravenel and Steve Wilson, The Morava $K$-theories of Eilenberg-Mac Lane spaces, published in 1980 in the American Journal of Mathematics, vol. 102, no. 4, pages 691-748
for a mildly complicated but very rewarding example. Or, the algebras $H_*(K(\mathbb{Z}/p, *); \mathbb{Z}/p)$ (along with a million other things!) are computed in Wilson's exceptionally nice book
Steve Wilson, Brown-Peterson Homology: An Introduction and Sampler, published in 1980, no. 48 in the CBMS series of conference notes,
which uses all the same machinery as the Morava $K$-theory paper but employs singular homology and tells you about some algebras which you already understand. The familiarity of these two things will probably ease digestion of the ideas.
† -- Well, this is kind of a lie, since they argue the existence or nonexistence of some of their differentials by knowing what the $E^\infty$ page must look like together with some kind of sparseness of the $E^2$ page. However, the way they build the spectral sequence does actually give you a formula for the differentials, and it's certainly possible, if difficult, for you to make the relevant calculations once you read their argument so you know whereabouts to look. At small primes (and small heights, in the Morava K-theory paper), this is probably even accessible.
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Very well thought-out response. Couldn't have said it better myself. – Quadrescence Nov 6 2010 at 7:33
This sounds difficult, but definitely would be rewarding! Perhaps an example I can build up to after gaining some confidence with a few of the other suggestions. In any case, thank you for the links to some interesting looking papers! – Dylan Wilson Nov 6 2010 at 17:23
There are some examples at http://www.shef.ac.uk/nps/courses/bestiary/ss.pdf that you might or might not find useful.
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Ah, just got around to glancing at these notes, and they look like a great resource. It's basically like a spectral sequence cheat sheet with definitions and examples- I look forward to reading through the notes carefully! Side-question: what package do you use to draw spectral sequences in LaTex? – Dylan Wilson Nov 6 2010 at 17:40
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In the document that I linked to, the spectral sequences are just done in picture mode, which is very pedestrian. Tilman Bauer has a specific spectral sequence package, which you can find here: few.vu.nl/~tilman/tex.html. I haven't tried it yet, however. – Neil Strickland Nov 6 2010 at 22:07
Computing differentials in spectral sequences can be fun and very useful, and I absolutely agree with OP that the literature does not cover this adequately -- surely not in a book, but even articles using this are few and far between (Ravenel-Wilson being an excellent one, as Eric mentioned). I think there are a number of reasons for this:
• Like diagram chases (but on a higher level), print form without the time coordinate doesn't lend itself to a good presentation of what's happening from page to page in a spectral sequence
• Spectral sequence charts are hard to draw (but, if you don't know it, take a look at the "sseq" TeX package)
• for many people, computing higher differentials in spectral sequences is the epitome of technicality, and they don't want to read it.
So, it seems to be considered more of a craft than something that can be laid out in a textbook. Additionally, correct me if I'm wrong, this is only done in homotopy theory. I don't really understand why that is, there are plenty of spectral sequences in algebraic geometry, for example, but they tend to collapse. I notice you're based at Seattle, so my strong recommendation would be to go ask an expert to teach you -- you've got Ethan Devinatz and John Palmieri right there, right?
I'll add a few random comments about things to learn. One of my favorite (simple) examples is to compute the homotopy groups of connective K-theory by 1) producing a free resolution of $F_2$ over $A(1)$, the subalgebra of the mod-2 Steenrod algebra generated by $Sq^1$ and $Sq^2$. Then notice that this is $H^*(bo;F_2)$ and run the Adams spectral sequence on this. You can also compute the homotopy groups of real K-theory as the fixed points of the conjugation action on complex K-theory using the homotopy fixed point spectral sequence. If you liked that, you can take this one chromatic step up (compute the homotopy groups of topological modular forms) and see lots of techniques in action -- you'll find that in some readily available notes by Charles Rezk or (excuse me for advertising again) my paper "Computation of the homotopy of the spectrum tmf".
One thing that happens very frequently when there's more than just a few differentials is that there is a strong interplay between multiplicative extensions, Massey products/Toda brackets and differentials. Often, one needs to compute all three pieces of data at once to inductively derive longer differentials/longer extensions. This is what happens for tmf. By the way, there is something completely absent in any treatment of spectral sequences I know: multiplicative extensions in $E^r$ terms (with $r<\infty$).
I realize this is turning into some fairly unstructured rant, so I'll stop and reiterate: go ask the masters to teach you in person.
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At the moment Steve Mitchell is heading up our reading course- and doing a wonderful job! But you're right, it's good to get other points of view. Thanks for the advice! I'm afraid I'll have to do some reading up on the adams spectral sequence before attempting your exercises- at the moment I only have the spectral sequence of a double complex, the spectral sequence of a filtration, and the Serre's spectral sequence in my toolkit. – Dylan Wilson Nov 6 2010 at 17:10
One problem with the Adams spectral sequence is that there are no baby examples with non-trivial differentials. In fact, in the case of $ko$ there is only "one" possible non zero differential for degree reasons. But maybe $A(2)$ is easier than I thought. – Sean Tilson Feb 3 2012 at 18:40
Here's another answer that addresses the question more directly. Take $A^\ast=\mathbb{Z}[x]\otimes E[a]$ with $d(x^k)=kx^{k-1}a$ and $d(x^ka)=0$, and filter it by $F_kA^\ast=2^kA^\ast$. This gives a Bockstein spectral sequence with $E_1=(\mathbb{Z}/2)[x,a,t]/a^2$ (where $t$ represents $2$). It converges to the cohomology of the $2$-adic completion of $A^\ast$ rather than $A^\ast$, so it gives a natural example where the target is not the group you first thought of. Moreover, you can work out everything very explicitly from the definitions, the main point being that $x^{2^r(2k+1)}$ survives to the $E_r$ page and then $d_r(x^{2^r(2k+1)})=t^rx^{2^r(2k+1)-1}a$.
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Thank you for the specific example! I'll tackle this one during next week at the chalkboard with my friend. (And this is one where, as Tilman said, it's helpful to be based in Seattle... I think Prof. Palmieri wrote a paper all about using the Bockstein spectral sequence!) – Dylan Wilson Nov 6 2010 at 17:13
One fruitful example of differentials is the Lyndon-Hochschild-Serre spectral sequence associated to a short exact sequence of groups $1 \to N \to G \to G/N \to 1$, which is of the form `$$H^p(G/N, H^q(N, M)) \Rightarrow H^{p+q}(G, M)$$` and similarly for homology. Since you know specific interpretations for low-level groups in terms of crossed homomorphisms and extensions, you might try working out differentials (to put together the inflation-restriction sequence) more explicitly. You might also try full computations with mod-p cohomology for $G$ a small p-group, and $N$ something like its center or its commutator subgroup.
But as most of the previous examples have concentrated on examples arising from topological considerations, let me advocate the idea that you should build examples yourself.
When learning about chain complexes it's helpful to write down small, illustrative examples from algebra when learning to compute homology, and this is no different. One of the simplest ways to do this is to work with the spectral sequence associated to a double complex. This has two different associated spectral sequences, one for filtering the total complex by rows and one for filtering by columns, and this gives you two ways to compute the result that you can play off each other to determine differentials (and extensions).
For example, if you write down any double complex where the rows are exact, the "vertical differentials first" spectral sequence must ultimately result in the death of all classes.
Other double complex examples are "limit/colimit" spectral sequences where you only have nonzero entries for two values of $p+q$, and where the nonzero horizontal differentials are all isomorphisms. These ones are simple enough that you can get some intuition for chasing higher differentials.
Finally, another suggested exercise is to look carefully into the exact couple formalism. If you have an exact triangle $D \to D \to E \to D$ where you understand all the groups and maps, you can trace explicitly through the derived couples and see how they're building up the limiting object. (Unfortunately a lot of exact couples seem somewhat artificial in nature at first because they are formed by summing up a large number of separate long exact sequences. They're much more common than one might think.)
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Yes, I realized when we were trying to come up with complexes that had an interesting filtration, we could only think of things coming from topology. Double complexes seem to be more natural in the algebraic realm. What I haven't thought about doing is the exact couple approach- that sounds very interesting! Do you happen to know any nice examples off the top of your head? – Dylan Wilson Nov 6 2010 at 17:21
@Tyler: Do you know of any references that explicitly show calculations of differentials in the LHSS? – Josh Roberts Nov 6 2010 at 21:00
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@Josh: I was thinking specifically of Huebschmann's "Automorphisms of group extensions and differentials in the Lyndon-Hochschild-Serre spectral sequence": ams.org/mathscinet-getitem?mr=641328 – Tyler Lawson Nov 6 2010 at 21:57
@Dylan: The Bockstein sequence is one that was already mentioned. Also, associated to any filtered chain complex you get a collection of long exact sequences for adjacent pairs of filtrations; these give an "unrolled" exact couple (a collection of long exact sequences D_{p-1} -> D_p -> E_p -> D_{p-1} as p varies) that can be assembled into an actual exact couple if you desire by summing up. Chasing elements through the underlying unrolled exact couple to determine differentials was something that I always found gave an enlightening perspective on the filtered complex level. – Tyler Lawson Nov 6 2010 at 22:08
@Josh: Also: S.F. Siegel, On the cohomology of split extensions of finite groups, Trans. Amer. Math. Soc. 349 (1997), 1587-1609, dx.doi.org/10.1090/S0002-9947-97-01747-9 and references therein. – Pasha Zusmanovich Jan 3 2011 at 18:52
Let me elaborate on part of Tyler's answer, and "second" the importance of constructing examples yourself in a setting where that is feasible, namely that of bicomplexes. Even though as you say they seem more algebraic, they do come up in topology (as I note below) and you can "see everything" in first examples. Here's what I usually suggest to my students (often in or right after a first-year course).
Exercise: Show that the bicomplex with entries of ${\bf k}$ (the ground ring) at (0,1), (1,1), (1,0) and (2,0) and all horizontal and veritcal differentials given by identity maps is acyclic, but the spectral sequence obtained by taking homology vertically first has a non-trivial $E^2$ page and a non-trivial $d_2$ differential which kills that $E^2$ page.
Exercise: Construct bicomplexes with arbitrarily long differentials.
Playing with these for a while you see what's going on, but they might seem too algebraic/ artificial. Here's something which you can understand at almost the same level of detail but which computes something interesting.
Definition: Let $A \bigoplus A^i$ be a differential graded associative algebra (that is, a cochain complex with an associative multiplication for which the differential is governed by the Leibniz rule), defined in non-negative degrees with say $H^0(A) = {\bf k}$. Then Bar(A), the bar construction on $A$, is the bicomplex which in bidegree (-p, q) is the degree $q$ part of $\overline{A}^{\otimes p}$ - which we denote $a_1 | a_2 | \cdots | a_p$ (you can pretend at first that $\overline{A}$ is just $A$ - really, $\overline{A}$ agrees with $A$ in degrees two and greater, is 0 in degree zero, and in degree one we replace $A^1$ by $A^1/ im(d)$). The vertical differential is the standard one on $A^{\otimes p}$ (using the Leibniz rule to extend to the tensor product), and the horizontal differential is defined as a sum obtained by "removing bars and multiplying." Let ${\bf k}$ be $F_2$ if you don't want to worry about signs.
Exercise: Show this is a bicomplex. If you're brave, do so with signs.
Exercise: The spectral sequence for this bicomplex obtained by taking homology vertically first has $E^1$ given by $Bar(H_*(A))$, where the homology of $A$ is a differential graded algebra with zero differential. (This is an immediate application of one of the main theorems from a first algebraic topology course.)
Theorem (Eilenberg-Moore, after Adams-Hilton): If $X$ is simply connected and $A$ is (quasi-isomorphic to) the cochains of $X$, then the homology of $Bar(A)$ is isomorphic to the cohomology of the loopspace of $X$.
Exercises: first compute for $A$ with all products zero. Then compute when $A$ is a free associative algebra, and then free commutative algebra. Finally, try to make examples with higher differentials.
(I might elaborate more later... and we can talk at the Cascade seminar this week-end).
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In Jim Stasheff's original papers on $A_\infty$-algebras he generalises the bar construction and its spectral sequence to $A_\infty$-algebras, and then he identifies all of the differentials explicitly in terms of things that look like (duals of) Massey products - he calls them Yessam products.
I realise that this isn't exactly an explicit compution, but it is at least a general explicit description of the higher differentials that is not formal, and to get it requires a bit of the guts of the spectral sequence.
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is that in his H-spaces from a htopy point of view? – Sean Tilson Nov 7 2010 at 2:24
There's a paper of Fadell and Hurewicz (in the Annals, mid 1950's) identifying certain differentials with cap products. I can't recall the precise result.
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You mean probably the paper jstor.org/stable/1970250 It identifies the first non-trivial differential in the Serre spectral sequence if the base space is highly connected. For a specific computation of a differential in the Serre spectral sequence in a similar vein see also the following paper by McCleary: springerlink.com/content/443j202m17562714/… – Lennart Meier Nov 6 2010 at 13:34
Verdier and Deligne introduced more terms in a spectral sequence of a filtered complex (indexed by 4+1 indices instead of 2+1), thus in particular factoring the differentials into epis and monos. This enables one to splice certain short exact sequences to obtain Massey triangles or to obtain H(E_r) = E_{r+1}. Not sure whether this helps for calculations, but it does help with understanding what's going on with differentials. Cf. Deligne, Décompositions dans la catégorie dérivée, appendix, MR1265526; cf. also Verdier, Des catégories dérivées des catégories abéliennes, MR1453167.
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http://physics.stackexchange.com/questions/17024/energy-non-conservation-for-time-dependent-potentials
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# Energy non-conservation for time-dependent potentials
Written in a book I read that the "total energy is not preserved when the potential depends explicitly on time", i.e. $U=U(x,t)$. Is there any proof or explanation for this?
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– Qmechanic♦ Nov 16 '11 at 18:17
## 3 Answers
It is easy to understand on a "ball & wall" problem. If you throw a ball in the wall, the ball total energy is conserved during reflection: $E=\frac{mv^2}{2}+U(x)$. The potential energy $U(x)$ is explicitly time-independent here.
If you keep the ball at rest but hit it with a wall, the ball energy changes. Now the wall is moving and its position explicitly depends on time. It transfers some energy to the ball. This is the case of $U=U(x-Vt)$ .
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Yes, generically it will then not be conserved. On the other hand, if there is no explicit time dependence, then time translations $t\to t+a$ will be a symmetry of the action. The corresponding Noether charge is the total energy $E$, and it will be conserved due to Noether's first Theorem.
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The Noether argument is true, but it is also simple to see by thinking about it--- if you start with a free particle at position -50, turn on a spring potential kx^2/2 for a short time, it will start moving towards the center. Then you turn off the spring, and the particle is moving. In general, any force you apply to the particle will have a description in terms of a time dependent potential of the form
$$U(t) = - F(t)x$$
So whatever force you apply to a particle, you have a time dependent potential. The notion of potential is then only really mathematically useful when there are further conditions, most importantly that it is time independent.
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http://unapologetic.wordpress.com/2010/06/17/the-fatou-lebesgue-theorem/?like=1&source=post_flair&_wpnonce=39170bc4ce
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# The Unapologetic Mathematician
## The Fatou-Lebesgue Theorem
Now we turn to the Fatou-Lebesgue theorem. Let $\{f_n\}$ be a sequence of integrable functions (this time we do not assume they are non-negative) and $g$ be some other function which dominates this sequence in absolute value. That is, we have $\lvert f_n(x)\rvert\leq g(x)$ a.e. for all $n$. We define the functions
$\displaystyle\begin{aligned}f_*(x)&=\liminf\limits_{n\to\infty} f_n(x)\\f^*(x)&=\limsup\limits_{n\to\infty} f_n(x)\end{aligned}$
These two functions are integrable, and we have the sequence of inequalities
$\displaystyle\int f_*\,d\mu\leq\liminf\limits_{n\to\infty}\int f_n\,d\mu\leq\limsup\limits_{n\to\infty}\int f_n\,d\mu\leq\int f^*\,d\mu$
Again, this is often stated for a sequence of measurable functions, but the dominated convergence theorem allows us to immediately move to the integrable case. In fact, if the sequence $\{f_n\}$ converges pointwise a.e., then $f_*=f^*$ a.e. and the inequality collapses and gives us exactly the dominated convergence theorem back again.
Since $g$ dominates the sequence $\{f_n\}$, the sequence $\{g+f_n\}$ will be non-negative. Fatou’s lemma then tells us that
$\displaystyle\begin{aligned}\int g\,d\mu+\int f_*\,d\mu&=\int g+f_*\,d\mu\\&=\int\liminf\limits_{n\to\infty}(g+f_n)\,d\mu\\&\leq\liminf\limits_{n\to\infty}\int g+f_n\,d\mu\\&=\int g\,d\mu+\liminf\limits_{n\to\infty}\int f_n\,d\mu\end{aligned}$
Cancelling the integral of $g$ we find the first asserted inequality. The second one is true by the definition of limits inferior and superior. The third one is essentially the same as the first, only using the non-negative sequence $\{g-f_n\}$.
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