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2025-03-21T14:48:31.800697
| 2020-08-17T21:05:52 |
369418
|
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|
Stack Exchange
|
Is there a Geometric/Smooth version of Homotopy Hypothesis using the path $\infty$-Groupoid of a Smooth Space?
A version of Homotopy Hypothesis says that the Fundamental $n$-grupoids model Homotopy $n$-types... and if we continue upto $\infty$, then the Fundamental $\infty$- groupoids or Kan Complexes model Topological spaces upto Weak Homotopy.
Now in Higher Gauge Theory there is a notion of Path Groupoid of a smooth space https://ncatlab.org/nlab/show/path+groupoid which is a refinement of fundamental groupoid of a topological space and which remembers more information(Thin homotopy class) than the fundamental Groupoid (homotopy class). Analogous to the notion of Higher Fundamental Groupoids of a topological space one can define higher Path Groupoids of a smooth space https://ncatlab.org/nlab/show/path+n-groupoid. Also one can go upto $\infty$ and can analogously define Path $\infty$-Groupoid of a smooth space https://ncatlab.org/nlab/show/path+infinity-groupoid.
My Questions are the following:
What do Path $\infty$-Groupoid of a smooth space models?
Is there an analogous smooth or geometric version of Homotopy Hypothesis which deals with Path $\infty$-Groupoid of a smooth space?
Now according to Remark 4.3 in https://ncatlab.org/nlab/show/diffeological+space#OnTopologicalHomotopyTypeAndDiffeologicalShape one can present Path $\infty$-Groupoid of a smooth space as a diffeological singular simplicial set. Also they mentioned that the diffeological singular simplicial set $Sing_{diff}(X)$ of a smooth space $X$ is equivalent to the "Shape" of $X$. (Though I could not understand clearly what they actually meant by "shape" of $X$).
Now according to https://ncatlab.org/nlab/show/smooth+infinity-stack#in_terms_of_groupoids_internal_to_smooth_spaces an $\infty$-groupoid internal to diffeological spaces models Smooth $\infty$-stack. Now I guess Path $\infty$-Groupoid of a smooth space is also an $\infty$-groupoid internal to diffeological spaces. So I am expecting Path $\infty$-Groupoid to model Smooth $\infty$-stack.
So in that case I am expecting the answer to my questions is:
Path $\infty$-Groupoid of a smooth space $X$ models Smooth $\infty$-stack over the site $O(X)$ and a geometric/smooth version of Homotopy Hypothesis will say that the Path $\infty$-Groupoid of $X$ is same as Smooth $\infty$-stack over $O(X)$ which is same as a diffeological singular simplicial set $Sing_{diff}(X)$.
But still my answer is incomplete in the sense that I still do not know completely what extra information does the Path $\infty$-Groupoid of $X$ captures about the smooth space $X$ other than the homotopical information of the underlying topological space of $X$.
Also I am feeling little positive about my expectations because of the following reason:
Roughly the usual Homotopy Hypothesis makes sense because $Top_{Quillen}$ is Quillen Equivalent to $sSets_{Quillen}$ via the total singular complex functor $Sing$ and the geometric realization functor $|-|$. Now according to https://ncatlab.org/nlab/show/model+structure+on+diffeological+spaces I am guessing that we may have different model structures on the category of diffeological spaces(Diff). Then for my claim of "Smooth or Geometric Homotopy Hypotheis" to be more meaningful I should expect a Quillen equivalence of Model categories between Diff(with some Model structure) and sSets(with some Model Structure). Now according to https://ncatlab.org/nlab/show/model+structure+on+diffeological+spaces#details there exists a Quillen equivalence between Diff(with some appropriate model structure) and $Top_{Quillen}$.(Though it is mentioned in the same page that there may exist some gap in the proof of this Quillen Equivalence). Now If I assume for the time being that there is no gap in the proof then in that case we get that Diff(with appropriate Model Structure) is Quillen equivalent to $sSets_{Quillen}$(because $sSets_{Quillen}$ is Quillen equivalent to $Top_{Quillen}$). But I am not sure whether this Quillen equivalence is given via $Sing_{diff}$ and $|-|$ or not as mentioned in https://ncatlab.org/nlab/show/diffeological+space#OnTopologicalHomotopyTypeAndDiffeologicalShape.
If this is not the case then I can't say anything. But if it is the case then this is what I expected. Even if it is true still this Quillen Equivalence is not a sufficient condition for my claim of Smooth or Geometric Homotopy Hypotheis" to hold true but definitely it can give a positive feeling about my claim.
Am I misunderstanding anything? Or are my expectations not making much sense?
I am not an expert, so apology in advance if I sound stupid.
I'm not convinced that the smooth path $\infty$-groupoid is different from the smooth fundamental $\infty$-groupoid, since the whole point of eg the path 1-groupoid is that associativity is forced on path concatenation by a quotient of the space of all paths. Note also that the page https://ncatlab.org/nlab/show/path+infinity-groupoid has only been updated once in the past 11 years, to fix a bug with page rendering, and to remove material that looks, to me at least, a bit dubious. That's not to say one can't ask about a diffeological/Lie $\infty$-groupoid $\Pi_\infty$ and the Homotopy Hypoth
People do study homotopy types of diffeological spaces, for instance Dan Christensen (see eg https://theoreticalatlas.wordpress.com/2011/06/10/dan-christensen-on-diffeological-spaces-and-homotopy-theory/ and Christensen–Wu The homotopy theory of diffeological spaces https://arxiv.org/abs/1311.6394 New York J. Math. 20 (2014), 1269-1303)
@DavidRoberts Thank you Sir for the comment. But I guess the usual path groupoid is different from smooth fundamental groupoid in the sense that the parallel transport over a Principal $G$ bundle(with non trivial curvature) is not invariant under homotopy but is invariant under Thin homotopy. So from this observation isn't it counter intuitive to believe that smooth fundamental $\infty$ groupoid is same as Path $\infty$ groupoid? (I am not an expert, so apology in advance if I sound stupid).
@DavidRoberts Also I guess there is already a notion of Principal $\infty$ bundles https://ncatlab.org/nlab/show/principal+infinity-bundle#smooth_principal_bundles with (an appropriate notion of Smoothness ) with a possibly higher notion of Parallel transport in it . Now taking account of this fact and my previous comment on parallel transport of Classical Principal $G$ bundles isn't it counter intuitive to think that Smooth Fundamental $\infty$ groupoid is same as Path $\infty$ groupoid?
@DavidRoberts Thank you Sir for the reference on Homotopy types of Diffeological Spaces.
The page you link to about the path $\infty$-groupoid has no definition, it just mentions an analogy, and then gives no hints as to how it might even work. The definition at the other page linked there is the same as the ordinary fundamental $\infty$-groupoid, just with the induced diffeological structure. The underlying Kan complex is the same, unlike the path 1-groupoid case.
@DavidRoberts Yes I agree that the prsentation of Path $\infty$ groupoid as a diffeological singular simplicial set is same as the usual singular simplicial set (If we talk about only the underlying Kan Complex). So are you saying that the induced diffeological structure is playing no role at all?
@DavidRoberts Then in that case , from the intuition we developed from Path 1-Groupoid case it seems to me that when we are moving upto to infinity level then we are loosing a lot of information if we present them as a Diffeological singular simplicial set. So it seems to me that the presentation of Path $\infty$ groupoid as a diffeological singular simplicial set is not an appropriate one then . So can we expect to find a much suitable presentation in future? Or there are some technical reasons due to which we cannot expect to improve the presentation of Path $\infty$ groupoid ?
I don't know how to answer your question, but I strongly suggest the 1-groupoid case is too degenerate to get a proper intuition. I will point out that even the underlying algebraic structure of the path 1-groupoid and the fundamental groupoid are completely different, even ignoring the diffeological structures that both naturally carry.
@DavidRoberts Thanks!! I got your point. I guess you are saying that though Path 1-groupoid of a smooth space is very much different from smooth fundamental 1- groupoid but if we "move up to infinity" and present them as Kan complexes then they are becoming same object. Am I right?
@DavidRoberts We know that both smooth fundamental 1-groupoid $\pi_{\leq 1}$ and path 1-groupoid $\pi'{\leq 1}$ constructions induce functors from $Man \rightarrow Groupoids$. Now, $ |N \circ \pi{\leq 1}(X)|$ will tell us everything about the 1st homotopy groups of $X$.(Where $N$ is the nerve). Now if we repeat the same with $\pi'{\leq 1}(X)$ then should I expect $ |N \circ \pi{\leq 1}(X)|= |N \circ \pi'i_{\leq 1}(X)|$ or something different? I got this intuition from https://mathoverflow.net/questions/360118/what-is-the-geometric-realization-of-the-the-nerve-of-a-fundamental-groupoid-of
@DavidRoberts Here $"="$ is in appropriate sense.
These should be new and separate questions.
@DavidRoberts I am apologising for repeating the same mistake again.
@DavidRoberts As you suggested I asked a separate question here https://mathoverflow.net/questions/369541/what-is-the-difference-between-path-infty-groupoid-and-smooth-fundamental-i
|
2025-03-21T14:48:31.801240
| 2020-08-18T00:07:02 |
369425
|
{
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"Aaron Meyerowitz",
"Alex M.",
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"HJRW",
"Joseph O'Rourke",
"M.G.",
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|
Stack Exchange
|
Theorems with many proofs
Q. What are the characteristics of theorems that seem to invite (or possess) several or even many distinct proofs?
What I have in mind are examples such as these:
Proofs that there are infinitely many primes.
Proofs of irrationality of $\sqrt{2}$.
Pythagorean Theorem.
Twenty Proofs of Euler's Formula: V-E+F=2 (David Eppstein's collection).
The last is particularly striking to me, as it took some time for an accurate proof to emerge.1
I'm sure there are many other more modern examples;
suggestions welcomed.
Is there some characteristic of these theorems that lend
themselves to often rather distinct proofs?
Or is it where within the network of mathematical connections
these theorems reside?
Or is it that these theorems are so useful that researchers keep inventing new proof approaches?
1Imre Lakatos, Proofs and Refutations.
Wikipedia link.
Another prominent example is quadratic reciprocity. Does the fundamental theorem of algebra count?
The claim that two things are not equal (or variants of it) often has many fundamentally different proofs. This is because two things that are not equal are usually equal for many reasons - i.e. two integers are not equal because they are unequal mod $2$, unequal mod $3$, etc. Usually for an equality, there will be similarities between different proofs, and you can argue about whether they are really the same proof.
I don't think Lakatos' thesis was that the proof of V-E+F=2 holds for "polyhedra" was difficult to get right. It was the determining of what should be considered a polyhedron (for these purposes, not toroidal ones, for example) . GrunBaum said: ""The Original Sin in the theory of polyhedra goes back to Euclid, and through Kepler, Poinsot, Cauchy and many others ... at each stage ... the writers failed to define what are the polyhedra".
There are a plethora of proofs that there are $n^{n-2}$ trees on $n$ labelled points. I'd say an common feature of most such theorems is to be easy understand, hard enough to be challenging but not so hard as to be unapproachable. Then other criteria come into play,
I don't have an answer, but two more examples: Robin Chapman collected $14$ proofs of $\zeta(2)=\pi^2/6$ at https://empslocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf And $14$ is a good number – Stan Wagon won a prize for "Fourteen proofs of a result about tiling a rectangle." https://www.maa.org/programs/maa-awards/writing-awards/fourteen-proofs-of-a-result-about-tiling-a-rectangle
The example theorems all have one thing in common, which makes it far more likely that they will have many proofs: they are OLD. (Euler's formula is by far the newest of them, but even that is over 200 years old.) I would be surprised to learn of a >200-year-old theorem with fewer than, say, five proofs.
@AaronMeyerowitz: Nice Grünbaum quote!
@HJRW the Wagon paper about tiling a rectangle, I believe the result is considerably less than 200 years old.
@GerryMyerson: sure, many of the examples in the comments are newer. I meant the ones listed in the question.
183 proofs! Meštrović, Romeo. "Euclid's theorem on the infinitude of primes: a historical survey of its proofs (300 BC--2017) and another new proof." arXiv:1202.3670 (2012). arXiv abstract.
Around a year after posting of this question a different, similar one was posted, which gained a lot more traction: https://mathoverflow.net/q/401493/30186
I think two important characteristics are
The importance of the theorem itself
The extent of connections within the subject
These two are not, of course, mutually exclusive!
This book is a recent discussion of mathematical style which has 99 proofs. A really great read.
@BOOK{Ording2019,
title = {99 Variations on a Proof},
author = {Ording, P.},
publisher = {Princeton University Press},
address = {Princeton and Oxford},
year = {2019},
isbn = {9780691158839},
}
and this book collects together proofs of $\sqrt{2}$ is irrational.
@BOOK{Duchene2010,
author = {Duch\^{e}ne, L. and LeBlanc, A.},
title = {Rationnel mon Q: 65 exercices de styles},
publisher = {Hermann},
year = {2010},
}
"Rationnel mon Q" ^^
Indeed, the title of the second book is a pun that does not excel in taste; its French pronunciation can be translated as "rational my ass" (the letter "Q" and the French word "cul" have the same French pronunciation, with "cul" meaning "ass").
Does this book credit Tom Apostol with a proof he published in 2000 and that I presented in a classroom some years before that after I learned it from a book published in about 1960?
|
2025-03-21T14:48:31.801634
| 2020-08-18T00:10:34 |
369426
|
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|
Stack Exchange
|
Analytic p-adic functions that take an algebraic value
Suppose it exists $r\in\mathbb R$ such that the non constant p-adic function $f(z)=\sum_{n\ge0}a_nz^n$ ($a_n\in\mathbb C_p$) is defined on $\mathcal D=\{z\in\mathbb C_p\mid v_p(z)>r\}$. Does it exist $\alpha\in\overline{\mathbb Q}\cap f(\mathcal D)$? If the answer is yes, does it exist $\alpha\in{\overline{\mathbb Q}}\cap f(\mathcal D\cap\mathbb Q_p)$?
For the first, yes. Without loss of generality by shifting we may assume $a_1 \neq 0$.
For $\alpha\in \mathbb Q$, let $x_0=0$ and $x_{n+1} = x_n + \frac{\alpha-f(x)}{a_1}$. To check that $x_n$ converges as $n$ goes to infinity to a root of $\alpha-f(x)$, it suffices to check that $v_p ( \alpha - f(x_{n+1})) \geq v_p ( \alpha - f(x_n)) + 1$.
To do this, it suffices to have $v_p(x_n) \geq s$ and $v_p( x_{n+1}- x_n) \geq s$ for some $s \in \mathbb R$ such that $v_p (a_n)+n s> v_p (a_1) + s + 1$ for all $n > 1$, as then the contributions of $a_2$ and higher to $\alpha - f(x_{n+1})$ will be dominated by the contribution of $a_1$.
This is easy to ensure by first choosing such an $s < \frac{ v_p (a_n) - v_p(a_1) - 1}{ n-1}$ for all $n>1$ (checking that this series is bounded below) and then choosing $alpha$ such that $v_p ( \alpha- a_0) > a_1 + s$, so that $v_p(x_1-x_0)>s$ and thus inductively $v_p(x_{n+1} -x_n) >s$ for all $n$.
For the second, no. Just take $f(z) = z + b$ where $b\notin \mathbb Q_p + \overline{\mathbb Q}$. Since $\mathbb C_p/\mathbb Q_p$ is uncountable, such $b$ exists.
|
2025-03-21T14:48:31.801758
| 2020-08-18T00:59:45 |
369430
|
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|
Stack Exchange
|
Fermat's little theorem, Poulet numbers, Carmichael numbers, and primes
To begin with, i would like to apologize if my question is not up to the level of this forum.
I have tried asking a variant of the following question on math.stackexchange.com and my question generated some comments (even one upvote) but no answers, so i decided to give it a shot over here.
My original question was:
Fermat's primality test for base 2 permits Poulet numbers to pass the test, as follows: ($2^x$−2)/$x$. Fermat's primality test in different bases will act as a sieve for eliminating most pseudo primes from passing the test, unless the numbers are Carmichael numbers.
I ran an experiment for the following formula ($5^x$−$3^x$−$2^x$)/$x$ and it seems to eliminate all but Carmichael numbers, without having to check different bases.I was capable of running the experiment until 10000 only (due to my lack of computing calculation power).Does anyone know about this formula and whether it still holds forever?
One of the comments mentioned that "25326001 is a (strong) pseudoprime for the bases 2,3,5, thus it will pass your test. But it is not a Carmichael number."
I have then asked if that will be the smallest number that is not a carmicahel number to pass the test?
And i received this following comment: "If you check larger numbers, more pseudoprimes that aren't Carmichael numbers should turn up alongside the Carmichael numbers. But that takes more computational power"
So my question is whether anyone knows if 25326001 is the first non carmichael number to pass the test or no?
Again, my apologies if i am interrupting the level of this forum, but my goal is simply to get an answer.
Thanks,
Link to the question on math.stackexchange, please? And please leave a link there to the question here.
$25326001$ is the smallest number that's a strong pseudoprime to all three bases $2$, $3$, and $5$ (which doesn't quite answer your question).
I will do the same there, here is the link: https://math.stackexchange.com/questions/3789932/fermats-little-theorem-poulet-numbers-carmichael-numbers-and-primes
The answer is No. E.g., see OEIS A153580. for smaller examples.
|
2025-03-21T14:48:31.801950
| 2020-08-18T01:03:58 |
369431
|
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|
Stack Exchange
|
Deformation of Display on Zink's paper
I am going to compute some intersection numbers on certain RZ spaces and therefore need to fully understand the deformation of $p$-divisible groups. This can be understood as deformation of displays, where displays are generalized notions of Dieudonné modules.
The question is about the paper
Thomas Zink, A Dieudonné Theory for p-Divisible Groups, in: Class Field Theory – Its Centenary and Prospect, K. Miyake, ed. (Tokyo: Mathematical Society of Japan, 2001), 139-160, doi:10.2969/aspm/03010139, author pdf.
On page 8, Theorem 4, it says it follows immediately from Theorem 3, which I didn't see why and confusing for a long time. Let me put my question clear by writing the following.
The definition of a display is the following
Definition A Dieudonne display over $R$ is a quadruple $(P, Q, F, V^{−1})$, where $P$ is a finitely generated free $\widehat{W}(R)$-module, $Q\subset P$ is a submodule and $F$ and $V^{−1}$ are $F$-linear maps
$$
F \colon P \longrightarrow P,\qquad V^{ −1}: Q \longrightarrow P.
$$
The following properties are required to be satisfied:
$\widehat{I}_RP \subset Q \subset P$ and $P/Q$ is a free $R$-module.
$V^{−1}\colon Q\longrightarrow P$ is an $F$-linear epimorphism.
For any $x \in P$ and $w \in \widehat{W} (R)$, we have
$$
V^{−1}(\;^Vwx) = wF x.
$$
The above was the definition of the display, one can look at the reference for more details, here $\widehat W(R)$ is a subring of the Witt vectors $W(R)$ and $\widehat{I_R}$ is the kernel of the first ghost map $\widehat W(R)\longrightarrow R$.
I wish to understand the deformation of the display, the set up is a surjective ring homomorphism
$$
S\longrightarrow R
$$
with ideal $\mathfrak a$. We assume $p$ is nilpotent in $\mathfrak a$ and $\mathfrak a$ has equipped with a $p$-divided power structure. Zink has proved the following Theorem
Theorem 3 Let us consider a pd-thickening $S\longrightarrow R$ as above. Let
$$
\mathcal P_i =(P_i, Q_i, F, V^{−1})
$$
for $i = 1, 2$ be Dieudonne displays over $S$. Let
$$
\overline{\mathcal P_i }=(\overline{P_i}, \overline{Q_i}, F, V^{−1})=P_{i,R}
$$
be the reductions over $R$. Assume we are given a
morphism of Dieudonne displays $u \colon \overline{\mathcal P_1} \to \overline{\mathcal P_2}$. Then there exists a unique
morphism of quadruples
$$
u \colon (P_1, \widehat Q_1, F, V^{−1})\longrightarrow (P_2, \widehat Q_2, F, V^{−1}),
$$
which lifts the morphism $u$.
Here $\widehat Q_i$ is the preimage of $\overline{Q_i}$ in the map $P_i\longrightarrow \overline{P_i}$
Then Zink just said that Theorem 4 in the paper and says it follows immediately from Theorem 3.
Let me say an equivalent form of Theorem 4
Equivalent of Theorem 4 Let
$$
\overline{\mathcal P }=(\overline{P}, \overline{Q}, F, V^{−1})
$$
be a display over $R$. And let $P$ be the base change of $\overline {P}$ from $\widehat{W}(R)$ to $\widehat{W}(S)$. For any
$$
L\subset P/\widehat I_S P
$$
such that it reduces to $\overline Q\subset \overline P/\widehat I_R P$. Let $Q_L$ be the preimage of $L$ via the map $P\longrightarrow P/\widehat I_S P$. Then there exists a unique display up to (*)-isomorphisms
$$
\mathcal P = (P,Q_L,F,V^{-1})
$$
such that it reduces to the display $\overline{\mathcal P}$ in $R$. Here a (*)-isomorphism means it reduces to identity map in $R$.
Zink says it follows immediately from Theorem 3, which I feel not obvious at all.
I am interested in Theorem 4, it is saying to lift a display is the same as lifting a Hodge filtration. But it is extremely confusing because Zink did not mention why any lifting of a Hodge filtration there exists a display corresponds to it? Zink also did not explain why the display corresponding to a lifting of Hodge filtration is unique. Both of the questions is not obvious from Theorem 3. I guess uniqueness part might come from Theorem 3.
|
2025-03-21T14:48:31.802203
| 2020-08-18T01:20:58 |
369432
|
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|
Stack Exchange
|
Source request: Optimal bounds on signings of points from a convex body
I recently came across an old survey of problems in discrete geometry: https://pdfs.semanticscholar.org/c350/f4d4a9466fa6708d99ec1187c63d89bed20f.pdf
Problem 2.1 from the list caught my eye. It states the following (more or less):
Let $y_1, y_2, ... y_m \in \mathbb{R}^k$, $y_i \neq 0$. Let $K$ be the convex body consisting of all points $z = \sum_{i=1}^m c_i y_i$ for $c_i \in [-1, 1]$. Does there exists a universal constant $C$ such that given $x_1, x_2, ... x_k \in K$, there are signs $\delta_1, \delta_2, ... \delta_k \in \{-1,1\}$ such that $\sum_{i=1}^k \delta_i x_i \in C \sqrt{k} K$?
Has any progress been made on this problem? I have not been able to find anything in the literature. Further, Schectman claims that a bound of $C \sqrt{k \log \log k}$ is possible: I have not been able to find this in the literature. Does anyone know a source for this claim?
|
2025-03-21T14:48:31.802301
| 2020-08-18T01:53:27 |
369433
|
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|
Stack Exchange
|
Reference request: Diophantine equations
I am looking for a textbook, or preferably lectures, on the subject of Diophantine equations. I am familiar with the basic principles of modular arithmetic, conics and the Hasse Principle, and the basics of elliptic curves, Mordell's Theorem etc (though I'm not up to the point where I can understand the proof).
What I need is something that takes me beyond the basics. Something which will teach me the advanced theory, and also teach me about diophantine surfaces (not just curves).
I don’t think a single textbook or lecture series will provide you with a robust picture of the “advanced theory” of half of the topics you mentioned in the post. For example, Hasse principle is still a subject of intense research.
Fair enough. I'm just looking for anything which can push forward my knowledge in this area.
This may be a good choice for someone who (like yourself) is already superficially acquainted with some of the definitions and methods of Diophantine geometry:
Marc Hindry, Joseph H. Silverman -- Diophantine Geometry: An Introduction, Graduate Texts in Mathematics 201, Springer (2000), https://doi.org/10.1007/978-1-4612-1210-2.
The following two are great expository articles (especially the first), which provided me with plenty of inspiration back in the day:
Mazur, Barry. Arithmetic on curves. Bull. Amer. Math. Soc. (N.S.) 14 (1986), no. 2, 207--259. https://projecteuclid.org/euclid.bams/1183553167
Mazur, Barry. On the passage from local to global in number theory (link)
Henri Darmon has a couple of nice articles on the topic of rational points on curves:
Rational points on curves (link)
Rational points on modular elliptic curves (link)
Anthony Varilly-Alvarado has a number of very good introductions to the topic of rational points on different types of surfaces:
Lectures on the Arithmetic of del Pezzo surfaces (link)
Arithmetic of K3 surfaces (link)
Alexei Skorobogatov taught a course in 2013 on the topic of rational points on surfaces and higher-dimensional varieties. The notes strike a great balance between accessibility and generality:
Arithmetic geometry: rational points (link)
Then there are these notes by Yonatan Harpaz on rational points on elliptic surfaces:
Rational points on elliptic fibrations -- Course notes (link)
Finally (for now), Brendan Hassett has a nice article on the topic of potential density of rational points on varieties, which is very interesting as well:
Potential density of rational points on
algebraic varieties (link)
E.g.
Number Theory: Volume I: Tools and Diophantine Equations, Graduate Texts in Mathematics 239, https://doi.org/10.1007/978-0-387-49923-9; and
Number Theory: Volume II: Analytic and Modern Tools, Graduate Texts in Mathematics 240, https://doi.org/10.1007/978-0-387-49894-2
by Henri Cohen.
It is difficult to get far in the modern theory without some algebraic geometry.
This is the approach taken in the book:
Bjorn Poonen, Rational points on varieties, Graduate Studies in Mathematics 186 (2017), publisher page, Author pdf.
If you are interested in applications of Baker's method, Schmidt's subspace theorem etc., then you might like the following recent books by Evertse and Győry:
Discriminant equations in Diophantine number theory, New Mathematical Monographs, 32, Cambridge University Press, Cambridge, 2017.
Unit equations in Diophantine number theory, Cambridge Studies in Advanced Mathematics, 146, Cambridge University Press, Cambridge, 2015.
To the books mentioned above I would add one more:
Rational and Nearly Rational Varieties (Cambridge Studies in Advanced Mathematics) by J. Kollár, K. E. Smith, and A. Corti.
The authors present a more or less elementary approach to the rationality questions using a mix of classical and modern methods.
|
2025-03-21T14:48:31.802675
| 2020-08-18T02:10:17 |
369435
|
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|
Stack Exchange
|
a question about vector valued Banach spaces
I wonder the difference between $L^1(\mu\times\nu)$ and $L^1(\mu;L^1(\nu))$, as if partial derivatives can be exchanged with integration in the second spaces in many articles. In Folland's real analysis, Fubini-Tonelli theorem can't be used without the assumptions $L^1(\mu\times\nu)$ or $L^+(X\times Y)$.
Precisely, does the following statement hold:
If
$$\partial_s f(s,x)=b(s,x)\in L^1([0,T];L^1(B_R)),$$
where $B_R\subset \mathbb{R^d}$, then
$$\frac{d}{ds}\int_{B_R}f(s,x)\,dx=\int_{B_R}b(s,x)\,dx.$$
It is exactly true if $b\in L^1([0,T]\times B_R)$ by Fubini-Tonelli theorem.
Your question is a little hard to understand. Are you asking if these two spaces are the same (or, at least, isomorphic in a natural way)? I think this is true for arbitrary measure spaces, without any sigma-finiteness assumptions, by a Fubini-type theorem
The reference I have is Section 1.2.b of the book of Hytonen -- van Neerven -- Veraar -- Weiss, Analysis in Banach spaces, vol 1. An alternative might be somewhere in the background sections of Defant+Floret's Tensor norms and operator ideals -- basically, many texts that define the Bochner--Lebesgue spaces ought to include this isomorphism
Sorry, I didn't make the question clear
OK, I don't know what Folland's book says (and your updated question is still unclear about what he uses as the assumptions for Fubini--Tonelli) but yes, $L^1$-valued $L^1$ is isomorphic to $L^1$ on the product space. The proof works by approximating an $L^1$-valued Bochner-integrable function by $L^1$-valued simple functions
I'm fuzzy about the definition of $L^1$-valued $L^1$, e.g., $u\in L^1_tL^1_x$, does it imply $u$ is measurable w.r.t. both $t$ and $x$.
It is stronger than that. When dealing with $X$-valued $L^p$-spaces, the correct notion of measurable is not the naive Borel-to-Borel, but instead you define the strongly measurable $X$-valued functions to be those which are pointwise limits of simple $X$-valued functions. Then a function $f:\Omega \to X$ is said to be Bochner-integrable if it is strongly measurable and $\omega\mapsto \Vert f(\omega)\Vert$ is integrable in the usual sense. I recommend you look online for definitions of "Bochner integral" and "Bochner-Lebesgue space"
That is what I need, thank you very much!
There are two parts to your questions and the second hasn’t been touched on so far. Before bringing some suggestions which I hope will be useful, let me add to the information alrady given in the comments in the first part.
The fact that a function on a product $S\times T$ can be regarded as a function on $S$ with values in a space of functions on $T$ or as a member of a suitable tensor product of function spaces on $S$ and $T$ has a long history and attempts to extend this circle of ideas to Banach function spaces goes back at least to Grothendieck.
For simplicity, it is convenient to start with Banach spaces of continuous functions on compacta. In this situation, it is fairly transparent that the space $C(K\times L)$ is naturally identifiable with the space of continuous functions on $K$ with values in $C(L)$ and with a suitable Banach space tensor product of $C(K)$ and $C(L)$.
The case of $L^1$-spaces is rather more delicate, not least because we are dealing with equivalence classes of functions. However, a more or less complete analogy was established about 60 to 70 years ago.
There is, however, one crucial difference between the above cases—the tensor product norms used (the so-called inductive and projective norms respectively).
Now to the second part. You mention conditions on $b$ but none on $f$ for your statement to be true. In fact the only condition you require on the latter is that it be $L^1$ (no smoothness is required!). The reason for this is that such a function defines a distribution and so is differentiable in the distributional sense. The euler-type result you are looking for is then always true—but taking all operations (integration and differentiation) in the sense of distributions.
The usual application of this kind of result is to establish explicit formulae and there it is irrelevant whether the operations are carried out in the classical or in the distributional sense. If, however, you require the former, then one has to add more restrictive conditions and live with much less elegant formulations.
The crucial results that you require can easily be found online, in an elementary text (in english) by J. Sebastião e Silva at the site jss100.campus.ciencias.ulisboa.pt. The text you require is III.1 “Theory of Distributions” which can be found under “publicações,
Textos Didácticos”. The results you require are Th. 8.4.1 (Fubini) and 8.2.4 (Euler).
|
2025-03-21T14:48:31.803014
| 2020-08-18T02:22:20 |
369436
|
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|
Stack Exchange
|
Understanding the Exercise 9.9.5 of Weibel homological algebra
The exercise 9.9.5 of Weibel's homological algebra states that
$\textbf{Exercises 9.9.5}$ If $Z$ is a nilpotent ideal of $R$ and $k$ is a field of $char(k) = 0$, show that $H_{dR}^{\ast}(R) \cong H_{dR}^{\ast}(R/I)$.
where $R$ is commutative positively graded $k$-algebra. Here, I don't understand what is $Z$? He used the $Z(R)$ to denote the cyclic module associated to ring $R$, but which is nilpotent does not make sense. So what I am understanding is that this $Z$ should be $I$ in above exercise. But in that case exercise becomes:
$\textbf{Exercises}$ If $I$ is a nilpotent ideal of $R$ and $k$ is a field of $char(k) = 0$, show that $H_{dR}^{\ast}(R) \cong H_{dR}^{\ast}(R/I)$.
which looks very suspicious. Even though this is true when we replace $H^{\ast}_{dR}$ by periodic cycle homology $HC^{per}_{\ast}$ but for de Rham cohomology it seems very odd. Can someone clear this confusion whether this is typo error (Z instead of I) or the original exercise really mean something else.
There is a correction to the exercise listed here: https://sites.math.rutgers.edu/~weibel/Hbook.errors.edition2.pdf
Are you looking at the hardback version? My paperback copy of Weibel's Intro to HA says: if $I$ is a nilpotent ideal of $R$ and $k$ has characteristic zero, show that $H^_{dR}(R)=H^_{dr}(R/I)$
@ZhenLin Oops, I didn't notice that! Guess this shows that I didn't study this book as diligently as I should have during my PhD studies ...
Nor did I, but I remembered seeing comprehensive errata for this book. Lesson: check errata first!
@zhen this correction says that this exercise is wrong and replace by some different exercise.
@Yemen Can you share the link of your version where it is written as in form of our new exercise.
|
2025-03-21T14:48:31.803177
| 2020-08-18T02:50:21 |
369438
|
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Stack Exchange
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$H_n(Z \otimes C) = (Z \otimes C)_n$
(This is based on Heller's simple proof, according to Rotman)
$A, B$ are a complex and its subcomplex of boundaries. Assume all the modules in $A, B$ are flat.
We have an exact sequence of complexes:
$0\to Z \to A \to B[-1] \to 0$ and after tensoring $C$ where $C$ is any complex.
$0\to Z\otimes C \to A\otimes C \to B[-1] \otimes C \to 0$.
The corresponding long exact sequence is:
$H_{n+1}(B[-1]\otimes C) \to H_n(Z\otimes C) \to H_n(A\otimes C)\to H_{n}(B[-1]\otimes C) \to ...$
Then $H_{n+1}(B[-1]\otimes C) = (B \otimes C)_n$ and $H_n(Z \otimes C) = (Z \otimes C)_n$ since $Z, B[-1]$ have zero differentials. Thus, we may rewrite the long exact sequence as:
$(B \otimes C)_n \to (Z \otimes C)_n \to H_n(A\otimes C)\to (B \otimes C)_{n-1} \to ...$
I got a bit confused. I think as a total complex, the differential of $Z \otimes C$ should be of the form $a_{p}\otimes b_{q} \mapsto \Delta'a_{p}\otimes b_{q} + (-1)^p a_{p}\otimes \Delta''_{q}b_{q}$. So $\Delta'=0$ but we still have $\Delta''$ here, so it should not be a zero-map...
Any help would be appreicated!
Could you tell us which source you are using for this notation/terminology? I am not sure what a "cycle subcomplex" of $A$ is supposed to be: do you mean that you are taking $Z_n$ to be the kernel of $A_n \to A_{n-1}$, and then viewing $Z_{\bullet}$ as a complex with zero differential?
Yes! It is from page 679 of Rotman's introduction to homological algebra. Thx!
Something doesn't look right here: if I start with A already having zero differential, so that A=Z, then the claim you mention seems to assert that the homology of the total complex $A\otimes C$ doesn't depend on the differential of $C$. Since I don't have a copy of Rotman's book at hand, could you add some of the surrounding context? Is this something to do with a K"unneth formula?
yes! it is the proof of his theorem! I will add all from the beginning up to this point! There's not much! Thx again!
Just to let you know it is 4am here so I am unlikely to respond for some time ...
OK. I really appreciate it!
Since $Z$ has $0$ differential, the differential of $Z\otimes C$ is $(-1)^a_Z\otimes \Delta_C$, and since everything is flat, you should get $H(Z\otimes C)\cong Z\otimes H(C)$. I cannot figure out what else might be going on.
Crossposted on MSE
|
2025-03-21T14:48:31.803356
| 2020-08-18T04:13:04 |
369441
|
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|
Stack Exchange
|
Is the asymptotic density of positive integers $n$ satisfying $\gcd(n, \sigma(n^2))=\gcd(n^2, \sigma(n^2))$ equal to zero?
(This post is an offshoot of this MSE question.)
Let $\sigma(x)$ denote the sum of divisors of $x$. (https://oeis.org/A000203)
QUESTION
Is the asymptotic density of positive integers $n$ satisfying $\gcd(n, \sigma(n^2))=\gcd(n^2, \sigma(n^2))$ equal to zero?
I tried searching for examples and counterexamples to the equation
$$\gcd(n, \sigma(n^2))=\gcd(n^2, \sigma(n^2))$$
via Sage Cell Server, it gave me this output for the following Pari-GP script:
for(x=1, 100, if(gcd(x,sigma(x^2))==gcd(x^2,sigma(x^2)),print(x)))
All positive integers from $1$ to $100$ (except for the integer $99$) satisfy $\gcd(n, \sigma(n^2))=\gcd(n^2, \sigma(n^2))$.
Generalizing the first (counter)example of $99$ is trivial.
If ${3^2}\cdot{11} \parallel n$, then $11 \parallel \gcd(n,\sigma(n^2))$ and $11^2 \parallel \gcd(n^2,\sigma(n^2))$. So the asymptotic density in question is less than
$$1-\frac{2}{3^3}\cdot\frac{10}{11^2} = \frac{3247}{3267} \approx 0.993878.$$
Also, if $3 \parallel n$, then with probability $1$ there exist two distinct primes $y$ and $z$ congruent to $1$ modulo $3$ such that $y \parallel n$ and $z \parallel n$. In this case, we get $3 \parallel \gcd(n,\sigma(n^2))$ and $3^2 \parallel \gcd(n^2,\sigma(n^2))$. So the asymptotic density in question is less than
$$1-\frac{2}{3^2} = \frac{7}{9} \approx 0.\overline{777}.$$
The real open problem is whether the asymptotic density is $0$.
For what it's worth, $\sigma(n^2)$ is tabulated at http://oeis.org/A065764
I think the density does go to zero, but quite slowly. If $p \equiv 1 \bmod 6$ is prime then there are two solutions $0<r<s<p-1$ of $$x^2+x+1=0 \bmod p$$
If $p\parallel n$ then, with probability $1,$ there are two distinct primes $x $ and $ y,$ each congruent to $r \bmod p,$ with $x \parallel n$ and $y \parallel n.$ ( Either or both could be congruent to $s$ as well.)
Then $p \parallel \gcd(n,\sigma(n^2))$ while $p^2 \parallel \gcd(n^2,\sigma(n^2)).$ So the asymptotic density for this not to happen is $1-\frac{p-1}{p^2}<1-\frac{1}{p+2}$
If we can argue that the chance that none of these events happen is asymptotically $\prod(1-\frac{p-1}{p^2})$ over the primes congruent to $1 \bmod 6,$ then that asymptotic density is $0.$
Thank you for your answer, @AaronMeyerowitz. Assuming
$$1 - \frac{p-1}{p^2} < 1 - \frac{1}{p-1}$$
is correct, then we obtain
$$\frac{1}{p-1} < \frac{p-1}{p^2}$$
which implies that
$$p^2 < (p-1)^2 = p^2 - 2p + 1$$
resulting in the contradiction
$$p < \frac{1}{2}.$$
Hence, I am led to conclude that there must be a typo in your upper bound for
$$1 - \frac{p-1}{p^2}.$$
OK, I fixed it. Though I hadn't used it. The idea was, roughly $1-\frac{1}{p}$
|
2025-03-21T14:48:31.803540
| 2020-08-18T04:16:47 |
369442
|
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Stack Exchange
|
What transformation is required to find a unique solution of this problem instead of multiple solutions?
$$
\max\limits_{\mathbf{f},\ \|\mathbf f\|=1 } \log_2\left(\prod^K_{i=1} \ \frac{ \mathbf{f}^H {\mathbf E} (\mathbf{W}_i, \Theta, \tau_i) \mathbf{f}} { \mathbf{f}^H \mathbf{G}_i ( \mathbf{W}_i, \Theta, \tau_i) \mathbf{f} } \right)
$$
For the product of Rayleigh quotients in the above problem,
solution $\psi_i\in (M\times1)$, corresponding to the largest eigenvalue $\zeta_i$, for the $N \times N$ matrix pairs $E$ and $G_i$ is obtained. So, for $K$ product terms, $K$ solutions instead of one is obtained.
How to find an unique solution for this problem?
May I say that if what was in this question is the LaTeX in your own articles, please, please learn good coding style and improve how you set mathematics,
yea. Sure. Thanks .
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2025-03-21T14:48:31.803630
| 2020-08-18T04:35:34 |
369444
|
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Stack Exchange
|
Balls in Hilbert space
I recently noticed an interesting fact which leads to a perhaps difficult question. If $n$ is a natural number, let $k_n$ be the smallest number $k$ such that an open ball of radius $k$ in a real Hilbert space of sufficiently large dimension or infinite dimension contains $n$ pairwise disjoint open balls of radius 1. (The dimension of the
Hilbert space is irrelevant as long as it is at least $n-1$ since it can be replaced by the affine subspace spanned by the centers of the balls.) We obviously have $k_1=1$ and $k_2=2$, and it is easy to see that $k_3=1+\frac{2}{\sqrt{3}}\approx 2.1547$. The interesting fact is that $k_n\leq 1+\sqrt{2}\approx 2.414$ for all $n$, since in an infinite-dimensional Hilbert space an open ball of this radius contains infinitely many pairwise disjoint open balls of radius 1 [consider balls centered at points of an orthonormal basis]. The obvious questions are: (1) What is $k_n$? This may be known, but looks difficult since it is related to sphere packing. (2) Is $k_n$ even strictly increasing in $n$? (3) Is $k_n<1+\sqrt{2}$ for all $n$, or are they equal for sufficiently large $n$? (4) Is it even true that $\sup_n k_n=1+\sqrt{2}$? It is not even completely obvious that $k_n$ exists for all $n$, i.e. that there is a smallest $k$ for each $n$, but there should be some compactness argument which shows this. I find it interesting that the numbers $1+\frac{2}{\sqrt{3}}$ and $1+\sqrt{2}$ are so close but the behavior of balls is so dramatically different.
I suppose the question is also interesting in smaller-dimensional Hilbert spaces: let $k_{n,d}$ be the smallest $k$ such that an open ball of radius $k$ in a Hilbert space of dimension $d$ contains $n$ pairwise disjoint open balls of radius 1. Then $k_{n,d}$ stabilizes at $k_n$ for $d\geq n-1$. What is $k_{n,d}$? (This my be much harder since it is virtually the sphere-packing question if $n>>d$.)
Perhaps $k_n=1+\sqrt{2(1-1/n)}$?
For convenience of notation, let me write the expectation $\mathop{\mathbb{E}}_i t_i$ to denote the average $(\sum_{i=1}^n t_i)/n$.
If I understand your construction correctly, you have disjoint balls of radius $1$ centered at $x_i = \sqrt{2} e_i$ contained in a ball of radius $1+\sqrt{2}$ centered at $y = 0$. This construction, which places $n$ balls tightly packed at the vertices of a regular simplex, is optimal in terms of the positions $x_i$. For the exact optimal bound for your problem, you should pick $y=\mathop{\mathbb{E}}_i x_i$ to get the radius $$\boxed{k_n = 1+\sqrt{2 (1-1/n)}}.$$
The claim that placing the $x_i$ at the vertices of a regular $(n-1)$-simplex and $y$ at the centroid of this simplex is optimal has been proven many times before in many different contexts. For example, it is implied by a bound known by various substrings of "the Welch-Rankin simplex bound" in frame theory. Here's a simple direct proof:
By the triangle inequality, a ball of radius $1+r$ centered at $y$ contains a ball of radius $1$ centered at $x_i$ iff $\lVert x-y\rVert \le r$. Two balls of radius $1$ centered at $x_i$ and $x_j$ are disjoint iff $\lVert x_i - x_j \rVert \ge 2$. Therefore, your problem asks to minimize $1 + \max_i \lVert y-x_i\rVert$ subject to $\min_{i\ne j} \lVert x_i - x_j\rVert \ge 2$.
Working with squared distances is easier. The maximum squared distance $\max_i \lVert y-x_i\rVert^2$ is surely at least the average $\mathop{\mathbb{E}}_i \lVert y-x_i\rVert^2$. This average is minimized when $y$ is itself the average $\mathop{\mathbb{E}}_i x_i$, in which case it equals $\mathop{\mathbb{E}}_i \mathop{\mathbb{E}}_j \lVert x_i-x_j\rVert^2/2$. Each term where $i=j$ contributes $0$ to this expectation, while each term where $i\ne j$ contributes at least $2$, so overall this expectation is at least $2(n-1)/n$. Thus the maximum squared distance $\max_i\lVert y-x_i\rVert^2$ is at least $2(n-1)/n$ and thus $1+r \ge 1+\sqrt{2(n-1)/n}.$ We can check that the optimal configuration mentioned before achieves this bound either by direct calculation or by noting that it achieves equality in every step of our argument.
|
2025-03-21T14:48:31.803906
| 2020-08-18T05:20:33 |
369446
|
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Stack Exchange
|
A question about asymptotic affinity and strict convexity with unbounded means
Let $F:[0,\infty) \to [0,\infty)$ be a $C^1$ strictly convex function.
Let $\lambda_n \in [0,1],a_n\le c<b_n \in [0,\infty)$ satisfy
$$ \lambda_n a_n +(1-\lambda_n)b_n=c_n \tag{1}$$ and assume that
$c_n \to ֿ\infty$. (which implies $b_n \to ֿ\infty$). $c>0$ is just some constant, to make $a_n$ bounded.
Set $D_n=\lambda_nF(a_n)+(1-\lambda_n)F(b_n)-F\big(c_n\big) $, and assume that $\lim_{n \to \infty}D_n=0$
Question: Does $\lambda_n \to 0$?
My intuition is that even if $F$ becomes "less convex" (closer to being affine) when $x \to \infty$, then we cannot put to much weight on the $a_n$-since otherwise we get hit by the "convexity gap" between $a_n$ and $b_n$ by a non-negligible amount, which should make $D_n$ large.
Edit:
This is an attempt to understand Ron P's answer:
We have
$D(a_n,c_n,b_n)=\lambda_n F(a_n)+(1-\lambda_n)F( b_n)-F(c_n)$, where $ \lambda_n a_n +(1-\lambda_n) b_n=c_n$.
Similarly,
$D(a,c_n,b_n)=\tilde \lambda_n F(a)+(1-\tilde \lambda_n)F( b_n)-F(c_n)$, where $ \tilde\lambda_n a +(1-\tilde \lambda_n) b_n=\tilde c_n$.
Suppose that $a_n \to a$. (This implies $\lambda_n-\tilde \lambda_n \to 0$). We have
$$D(a_n,c_n,b_n)-D(a,c_n,b_n)=\lambda_n F(a_n)-\tilde \lambda_n F(a)+(\tilde \lambda_n-\lambda_n)F(b_n). \tag{2}$$ The first term tends to zero, since $F(a_n) \to F(a)$ and $\lambda_n-\tilde \lambda_n \to 0$.
Why does the second term tend to zero? we don't have control over $F(b_n)$, right?
It should be easy to make a strictly convex function with $F(2n)/2+F(0)/2 -F(n)=1/n$, as envelope of a family of lines. (that is a counterexample with $a_n=0, c_n=n, b_n=2n, \lambda_n=1/2$)
One can also just do it for powers of 2, b_n=2^n
Dear Pietro, it seems that your suggestion cannot work. You may see Iosif Pinelis's answer below which proves that the answer is in fact positive.
Yes! I realized it with the tentative counterexample in the answer then deleted. The fact it does not work may be turned into a positive proof
Is assumed that $F$ is increasing?
@RonP No, I don't assume that, why? Do you have a simpler proof for the case where $F$ is increasing?
@AsafShachar. Got it. See another proof below. Thanks!
The answer is yes.
Indeed, by rescaling, without loss of generality (wlog) $c=1$. To simplify the notations, let $f:=F$, $a:=a_n$, $b:=b_n$, $c:=c_n$, $t:=\lambda_n$, $D:=D_n$. Passing to a subsequence, wlog $a\to a_*\in[0,1]$ and $t\to t_*\in(0,1]$. Also, wlog $a+2\le c$, since $a\le1$ and $c\to\infty$. Also, wlog $b>c$, since wlog $t>0$ and $c>a$.
By the convexity of $f$ and inequalities $a+1\le a+2\le c$,
\begin{equation*}
f(a+1)\ge f(c)+\frac{a+1-c}{b-c}\,(f(b)-f(c)).\tag{1}
\end{equation*}
Using now the convexity of $f$ again together with the inequality $a+2\le c$ and (1), we have
\begin{align*}
0\le d&:=\frac{f(a)+f(a+2)}2-f(a+1) \\
&\le \frac{f(a)}2+\frac12\,\frac{(c-a-2)f(a)+2f(c)}{c-a}-f(a+1) \\
&=\tilde d:=\frac{(c-a-1)f(a)+f(c)}{c-a}-f(a+1) \\
&\le\frac{(c-a-1)f(a)+f(c)+D}{c-a} \\
&\ \ -\Big(f(c)+\frac{a+1-c}{b-c}\,(f(b)-f(c))\Big) \\
&=\tilde D:=\frac{b-a-1}{b-a}\frac Dt\sim\frac D{t_*}\to0,
\end{align*}
so that
\begin{equation*}
d\to0. \tag{2}
\end{equation*}
On the other hand,
\begin{equation*}
d\to\frac{f(a_*)+f(a_*+2)}2-f(a_*+1)>0
\end{equation*}
by the strict convexity of $f$. This contradicts (2). $\Box$
Remark: As seen from the above, condition $c\to\infty$ can be relaxed to $\liminf(c-a)>0$.
The above solution might look somewhat mysterious. In fact, the idea is a rather simple geometric one. For any real $A,B,C$ such as $A\le B\le C$, let the "gain" $g(A,B;C)$ denote the distance between the point on the graph of the convex function $f$ with abscissa $C$ and the point with the same abscissa on the chord connecting the points on the graph of $f$ with abscissas $A$ and $B$.
So (see the picture below), $D=g(a,b;c)$ and $\tilde d=g(a,c;a+1)$, where $\tilde d$ is as defined in the above multi-line display. In that display, it was shown that $\tilde d\le\tilde D$, which is clear from the picture. Also, if $t$ is bounded away from $0$ -- that is, if $c/b$ is bounded away from $1$, then, as it is clear from the picture by looking at the similar triangles, we have $\tilde D\asymp D\to0$; cf. the last line of the above multi-line display. This and the inequality $\tilde d\le\tilde D$ imply $\tilde d\to0$.
By the convexity of $f$, for any fixed real $A,C$ such as $A\le C$, the gain $g(A,B;C)$ is nondecreasing in $B\in[C,\infty)$ (here you may want to draw another picture). Therefore and because $a+2\le c$, we have $d=g(a,a+2;a+1)\le g(a,c;a+1)=\tilde d$, so that $d\le\tilde d$, which was shown in the first three lines of the above multi-line display.
This is the geometric explanation of (1) and the above multi-line display.
wow! This is an amazing answer, really. I liked a lot your geometric explanation and picture-very clear. (and indeed it was a bit mysterious before you added them). Just to be sure- in your third (last) inequality you have added the non-negative term $\frac{D}{c-a}$? I am also not sure how did you deduce that this was the "right thing" to add in order to get $\tilde D$. Did you also use the visual intuition from the picture here, or did you just take the difference and then wrote it "backwards"?
First let's reformulate the question. For $0\leq a\leq c\leq b$, let $\lambda=\lambda(a,c,b)\in[0,1]$ be the number such that $c=\lambda a + (1-\lambda)b$, and for $f\colon \mathbb R_+\to\mathbb R$ define
$$
D_f(a,c,b)= \lambda f(a)+(1-\lambda)f(b)-f(c).
$$
Lemma 1. Let $f\colon \mathbb R_+\to\mathbb R$ be strictly convex and continuously differentiable. Let $0\leq a_n\leq c_n\leq b_n$ be sequences such that $a_n$ is bounded, $c_n-a_n$ is bounded away from 0, and $\limsup \lambda(a_n,c_n,b_n)>0$. Then, $\limsup D_f(a_n,c_n,b_n)>0$.
We first apply a sequence of reduction steps that allow us to assume wlog that $a_n=0$, $c_n\geq 1$, , for all $n$, and $\liminf\lambda(a_n,c_n,b_n)>0$. If you trust that that is possible, you may skip directly to Lemma 5 below.
By taking a sub-sequence $n'$ on which $\liminf \lambda(a_{n'},c_{n'},b_{n'})>0$, Lemma 1 follows from Lemma 2.
Lemma 2. Let $f\colon \mathbb R_+\to\mathbb R$ be strictly convex and continuously differentiable. Let $0\leq a_n\leq c_n\leq b_n$ be sequences such that $a_n$ is bounded, $c_n-a_n$ is bounded away from 0, and $\liminf \lambda(a_n,c_n,b_n)>0$. Then, $\limsup D_f(a_n,c_n,b_n)>0$.
By further taking a sub-sequence $n'$ on which both $a_{n'}$ converges, Lemma 2 follows from Lemma 3.
Lemma 3. Let $f\colon \mathbb R_+\to\mathbb R$ be strictly convex and continuously differentiable. Let $0\leq a_n\leq c_n\leq b_n$ be sequences such that $a_n\to a$, $c_n-a_n$ is bounded away from 0, and $\liminf \lambda(a_n,c_n,b_n)>0$. Then, $\limsup D_f(a_n,c_n,b_n)>0$.
For any fixed $\epsilon>0$, the functions $\lambda(a,c,b)$ is continuous in $a$ uniformly in $c$ and $b$ over the domain $\epsilon\leq a +\epsilon\leq c\leq b$; therefore, under the assumptions of Lemma 3, $0<\liminf\lambda(a_n,c_n,b_n)=\liminf\lambda(\lim a_n,c_n,b_n)$. Furthermore, for $\lim a_n <a<\liminf c_n$ small enough, we have $\liminf\lambda(a,c_n,b_n)>0$. Since $D_f(a,c,b)$ is decreasing in $a$, $\limsup D_f(a_n,c_n,b_n)\geq\limsup D_f(a,c_n,b_n)$.
Therefore, Lemma 3 follows from Lemma 4.
Lemma 4. Let $f\colon \mathbb R_+\to\mathbb R$ be strictly convex and continuously differentiable. Let $0\leq a\leq c_n\leq b_n$ be sequences such that $c_n-a$ is bounded away from 0, and $\liminf \lambda(a,c_n,b_n)>0$. Then, $\limsup D_f(a,c_n,b_n)>0$.
Let $T\colon \mathbb R\to\mathbb R$ be the affine transformation that maps $a$ to $0$ and $\inf c_n$ to $1$. Replacing $f$ by $F=f\circ T^{-1}$, and $a,c_n,b_n$ by $T(a),T(c_n),T(b_n)$ respectively, Lemma 4 follows from Lemma 5.
Lemma 5. Let $F\colon \mathbb R_+\to\mathbb R$ be strictly convex and continuously differentiable. Let $1\leq c_n\leq b_n$ be sequences such that $\liminf \lambda(0,c_n,b_n)>0$. Then, $\limsup D_F(0,c_n,b_n)>0$.
Proof of Lemma 5.
We assume wlog that $F(0)=0$ and denote $\lambda_n=\lambda(0,c_n,b_n)$ and $D_n=D_F(0,c_n,b_n)$.
Define a function $G\colon [1,\infty)\times (1,\infty)\to \mathbb R$ by
$$
G(x,y)=\tfrac 1 y F(xy)-F(x).
$$
Claim 6. $G$ is positive and increasing in both $x$ and $y$.
Proof of Claim 6.
Since $F$ is strictly convex, $F(0)=0$, and $x = 1/y(xy)+(1-1/y)0$, $G(xy)>0$. Since $F'$ is increasing, we have $\frac {d}{dx}G(xy)=F'(xy)-F'(x)>0$, so $G$ increases in $x$. Since $F'$ is increasing and $G(x,y)=1/y\int_0^yF'(xt)x\,dt - F(x)$, $G$ increases in $y$, completing the proof of Claim 6.
Suppose there is $\lambda_0>0$ such that $\lambda_n\geq \lambda_0$ for all $n$. Then,
$$
D_n = G(c_n,1/(1-\lambda_n))\geq G(1,1/(1-\lambda_0)>0, \quad\text{for all $n$.}
$$
QED
@AsafShachar you're right. I should elaborate about this point. I will do it soon. The idea is to replace $a_n$ with its limit in the expression of $D_n$. By continuity, such a modification does not change whether it converges to zero or not.
@AsafShachar I've elaborated on "re-scaling." Note that the answer does not assume neither that $F$ is non-negative, nor that $c_n\to\infty$, rather just that it is bounded away from $a_n$.
Thanks for the elaboration. I am still somewhat troubled tough. It seems to me that the most crucial non-trivial step is in passing from Lemma 3 to Lemma 4 (from $a_n$ to $\lim a_n$). I understand why $\liminf\lambda(a_n,c_n,b_n)=\liminf\lambda(\lim a_n,c_n,b_n)$ (because the difference between the 'lambdas' tend to zero.) However, I don't see why $\limsup D_f(a_n,c_n,b_n)= \limsup D_f(\lim a_n,c_n,b_n)$. I think it reduces to $\lim_{n \to \infty }( \lambda(a_n,c_n,b_n)-\lambda(\lim a_n,c_n,b_n))F(b_n)=0, $ but I don't see why this limit should be zero.
I have edited the question to make my misunderstanding clear. Thanks.
@AsafShachar, you're right. I've corrected the reduction from Lemma 3 to Lemma 4. Instead of using the continuity of $D_f$ in $a$, I now use monotonicity.
|
2025-03-21T14:48:31.804573
| 2020-08-18T07:00:03 |
369453
|
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"Calvin McPhail-Snyder",
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|
Stack Exchange
|
Embedded ribbons and regular isotopy
I'm reading Kauffman's 1990 paper "An Invariant of Regular Isotopy" about knots that are isotopic through only Reidemeister Type II and III moves, which is known as a regular isotopy. His paper claims there is a relationship between regular isotopy and embedded bands ($S^1 \times [0,1]$) in $S^3$. He refers to Burde's Knots textbook, but I can't find any mention of regular isotopy, because it seems Kauffman coined that phrase in the paper, after Burde's text was written.
I think a regular isotopy of knots corresponds to an embedded band. However, I'm worried there might be a pathology I'm overlooking. Does anyone have a precise statement of the relationship?
From any knot diagram, one can obtain a framed knot by taking the "blackboard framing." The point of regular isotopy of knot diagrams is that it preserves this blackboard framing. Since framed knots and embedded bands are the same thing, regular isotopy will also preserve the embedded band corresponding to the blackboard framing of the knot diagram.
I assume this is discussed in more detail in Burde, maybe in terms of framed knots. It's also possible Burde doesn't discuss framed knots at all, since I think people became much more interested them after the discovery of the Jones polynomial/the Chern-Simons TQFT. And I agree: I think Kauffman coined the term "regular isotopy," so it is probably not used in Burde.
Thank you for the helpful comment! By the way, I searched through an electronic copy of Burde, and "regular isotopy" does not appear anywhere in the text.
This is more of a comment than an answer, but I hope it's helpful. There is a much older and better-studied notion of regular homotopy. Let $X$ and $Y$ be smooth manifolds and let $f,g\colon X \rightarrow Y$ be immersions. Then $f$ and $g$ are regularly homotopic if they are homotopic through immersions.
Let's focus on regular homotopy classes of immersions $S^1 \rightarrow \mathbb{R}^2$. Such an immersion is what you get from a knot diagram by forgetting the over/under crossings. It is not hard to see that if $f,g\colon S^1 \rightarrow \mathbb{R}^2$ are regularly homotopic immersions with transverse self-intersections, then $f$ can be transformed into $g$ by a sequence of the obvious analogues of Reidemeister II/III moves. However, you can't perform an analogue of a Reidemeister I move since at the moment you pull your loop tight the derivative has to vanish, so it is not a regular homotopy.
My guess is that this is what Kauffman was thinking about. By the way, regular homotopy classes of immersions $S^1 \rightarrow \mathbb{R}^2$ can be completely classified. Taking the derivative of such an immersion and rescaling to make the derivative have unit length, you get an associated map $S^1 \rightarrow S^1$. The degree of this map is called the degree of the immersion, and the Whitney-Graustein theorem says that this degree is a complete invariant. This theorem is an early precursor to the Hirsch-Smale immersion theorem, which for the special case of immersions $S^2 \rightarrow \mathbb{R}^3$ includes Smale's famous "sphere eversions" that turns the sphere inside-out.
I was always confused by Kauffman calling it "regular isotopy" because I think of isotopy as a relation on knots, not knot diagrams. Your answer clarifies why he might have chosen that term.
A diagram is drawn in the plane. Restrict to knots (not links). Orient the curve, & associate to each crossing a (+/-) via a Right hand rule: palm along over-crossing with pinky pointing towards orientation curl to +under-crossing. Thumb up= +sign. Sum over all crossings. This is the writhe. Writhe determines the self-linking of the knot with a push-off. Draw \infty+,\infty-, and 0. The \infty+ has the arc with +slope as the over-arc. Draw a push off curve in the plane, and compute the linking number <--tricky calc, best done by using RI moves to form Hopf link. The knot & a push-off bound an annulus. If the self-linking # of the knot is 0, then the annulus extends to a Seifert surface. The push-off defines a preferred longitude. But in general, the black-board framed curve has self-linking = writhe. With an \alpha -\gamma curve you can draw this in 4 ways. 2 have 0 writhe, 1 has +2, the other -2. The ones with 0 writhe are regularly homotopic to unknots. The other 2 require type I moves. Somewhere in Kauffman you'll see a Whitney trick.The alpha-gamma curve has 1 kink outward and 1 kink inward. There are alpha-alpha curves and gamma-gamma curves: two out or two in resp. In either case, the writhe can be arranged like a phone cord, or can cancel.
The canceling cases are tricky. There the diags are on S^2.E.g. the bigon bounded in the gamma gamma case is on the outside. That's why you need to perform the framed isotopy in S^3 rather than R^3. [![0 and -/+ infinity curves
|
2025-03-21T14:48:31.804915
| 2020-08-18T07:45:36 |
369459
|
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|
Stack Exchange
|
Embedding any graph in a regular graph with the same chromatic number
If $G=(V,E)$ is a simple, undirected graph, is there a regular graph $G_R$ such that $G$ is isomorphic to an induced subgraph of $G_R$ and $\chi(G) = \chi(G_R)$?
Yes, you can find such a $G_R$ of any degree greater than or equal to the maximum degree of $G$. This is the main theorem of the paper "On regular bipartite-preserving supergraphs" by G. Chartrand and C. E. Wall .
|
2025-03-21T14:48:31.804979
| 2020-08-18T07:58:31 |
369461
|
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"authors": [
"Daniele Tampieri",
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"url": "https://mathoverflow.net/questions/369461"
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|
Stack Exchange
|
Variational problem where the end point cannot be held constant
I want to solve a variational problem for paths $y:[0,1]\to \mathbb{R}^n$ where $F$ is a functional on the paths and additional parameter, and the end points are fixed, $y(0)=0$ and $y(1)=b$. However, the problem is that the paths are functions of the parameter space $E$, and the functional $F$ depends on the parameter. The big problem is that it is extremely difficult to keep the end point $b$ fixed.
To be more precise we have a function $\mathrm{path}:[0,1]\times E\to \mathbb{R}^n$, where $\mathrm{path}(0,e)=0$ but $\mathrm{path}(1,e)$ is not fixed. The functional is $\mathrm{Fun}:E\to [0,\infty)$ (which requires a calculation of the path). The variational problem to solve is to minimise $\mathrm{Fun}:E_b\to [0,\infty)$, where
$$
E_b=\big\{ e\in E : \mathrm{path}(1,e)=b \big\}\ .
$$
The problem is because $\mathrm{path}$ is only specified as a solution to differential equations with parameters that it is difficult to describe $E_b$.
The original problem is a length minimisation in noncommutative geometry, where the parameter space is the noncommutative vector fields, and the paths are given by solving a differential equation given the vector fields (this the problem in keeping the end point fixed). The functional depends on both path and the vector field.
Any suggestions would be gratefully received. I would be happy to supply more details if required.
Only a comment, but have you thought about trying to adapt the tools of variational inequalities to your problem? These problems involve boundary conditions that are a priori not known, and thus, at least from a conceptual point of view, similar to the one you are dealing with.
|
2025-03-21T14:48:31.805381
| 2020-08-18T08:36:43 |
369465
|
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"authors": [
"Emil Jeřábek",
"Louis D",
"Mike Sanz",
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"https://mathoverflow.net/users/163905",
"https://mathoverflow.net/users/17798"
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|
Stack Exchange
|
Does every primitive digraph have a directed cycle?
A digraph is a directed graph.
A directed cycle or simple directed circuit is a directed circuit in which the only repeated vertices are the first and last vertices.
A digraph is primitive if its adjacency matrix is primitive.
A square non-negative matrix $A$ is said to be primitive if there exists a positive integer $k$ such that $A^k >0$ (all entries of $A^k$ are positive).
I need only the existence of a path with the structure $i_0 i_1...i_k i_0$ (sequence of distinc edges) with $k\geq 1$.
Simultaneously cross-posted to https://math.stackexchange.com/q/3794785 . Please, do not do that.
Thanks for the observation it will not happen again
Does your definition of digraph allow for loops?
If no loops are allowed, then yes. Since $(A^k)_{ii}$ is the number of walks of length $k$ from $v_i$ to $v_i$.
Thanks LouisD. Loops are allowed.
Yes.
For all $i,j$, $(A^k)_{ij}>0$, so there is at least one walk of length $k$ from $v_1$ to $v_2$ and there is at least one walk of length $k$ from $v_2$ to $v_1$. This closed directed walk which goes from $v_1$ to $v_2$ and then back to $v_1$ must contain a non-trivial directed cycle (i.e. a cycle of length at least 2).
Thank you very much.
|
2025-03-21T14:48:31.805501
| 2020-08-18T08:42:17 |
369467
|
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"dohmatob",
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"url": "https://mathoverflow.net/questions/369467"
}
|
Stack Exchange
|
Reference for "the algebra of multiplication by all measurable bounded functions acts in Hilbert space in a unique manner"
I read a paper of Alain Connes on "Duality between shapes and spectra" and in page 4, he says
Due
to a theorem of von Neumann the algebra of
multiplication by all measurable bounded functions acts in Hilbert space in a unique manner,
independent of the geometry one starts with.
Question.
What is a precise statement and reference for this "mysterious" theorem of von Neumann ?
What are nontechnical explanations and justifications of this phenomenon ?
One possible interpretation of Connes's statement is that
up to an isomorphism, there is a unique faithful indecomposable representation of any commutative von Neumann algebra on a Hilbert space.
Indeed, the category of von Neumann algebras is contravariantly
equivalent to the category
of compact strictly localizable enhanced measurable spaces.
After extracting an enhanced measurable space $(X,M,N)$ from a commutative von Neumann algebra $A$
in this manner,
elements of $A$ can be identified with equivalence classes of bounded
measurable functions on $X$ modulo equality almost everywhere.
We can now easily describe isomorphism classes of representations
of $A$ on a Hilbert space.
Such an isomorphism class is specified by partitioning $X$
into almost disjoint (up to a negligible set) nonnegligible measurable subsets $\{X_i\}_{i∈I}$,
and assigning a distinct cardinal number $a_i$ to each element of the partition.
The corresponding Hilbert space is $$\bigoplus_{i∈I} {\rm L}^2(X_i,M_{X_i},N_{X_i})⊗{\bf C}^{a_i}$$
and $A$ acts on each summand by restricting the corresponding
bounded measurable function on $X$ to $X_i$ and then acting via multiplication
on the corresponding ${\rm L}^2$-space.
Here ${\bf C}^{a_i}$ denotes any complex Hilbert space of dimension $a_i$.
Such a representation is faithful if $a_i≥1$ for all $i$.
It is indecomposable if $a_i≤1$ for all $i$.
Thus, a faithful indecomposable representation must have $a_i=1$ for all $i$,
and there is a unique such a representation, namely ${\rm L}^2(X,M,N)$,
also known as the Haagerup standard from of $A$.
The point of all this is that although the (abstract) commutative von Neumann
algebra appears to know nothing about an enhanced measurable space
or its ${\rm L}^2$-space, all this data can be reconstructed
in a unique manner, i.e., it is unique up to a unique isomorphism.
If, furthermore, we know the Dirac operator, we can proceed to refine
$X$ to a smooth manifold, as described by Connes.
Thanks for the detailed answer.
|
2025-03-21T14:48:31.805682
| 2020-08-18T09:18:11 |
369468
|
{
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"Max Alekseyev",
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|
Stack Exchange
|
Testing the primality of Mersenne and Fermat numbers using third order recurrence relation
Can you prove or disprove the claims given below?
Inspired by generalization of Lucas-Lehmer test I have formulated the following claims:
Claim 1 Let $M_p=2^p-1$ where $p$ is an odd prime number , let $S_k=3S_{k-1}-5S_{k-2}+3S_{k-3}$ with $S_0=0 , S_1=1 , S_2=2$ , then $$M_p \text{ is prime iff } \operatorname{GCD}\left(S_{(M_p+1)/2},M_p\right)=1 \text{ and } S_{M_p+1} \equiv 0 \pmod{M_p}$$
You can run this test here.
Claim 2 Let $F_n=2^{2^n}+1$ where $n>1$ , let $S_k=3S_{k-1}-5S_{k-2}+3S_{k-3}$ with $S_0=0 , S_1=1 , S_2=2$ , then $$F_n \text{ is prime iff } \operatorname{GCD}\left(S_{(F_n-1)/2},F_n\right)=1 \text{ and } S_{F_n-1} \equiv 0 \pmod{F_n}$$
You can run this test here.
Note that I verified these claims only for small values of $M_p$ and $F_n$ because computation of $\operatorname{GCD}$ takes a lot of time.
Notice that $\gcd(a,b) = \gcd(a\bmod b,b)$. That is, you can restrict computation of $S_k$ only to modulo $M_p$ (or $F_n$) and still compute the gcd.
@MaxAlekseyev Thanks!
In fact, the sequence $S_k$ satisfy a second-order recurrence: $$S_k = 2S_{k-1} - 3S_{k-2}.$$
|
2025-03-21T14:48:31.805809
| 2020-08-18T10:16:24 |
369472
|
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"Jason Starr",
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|
Stack Exchange
|
Line bundle on product scheme
Let $k$ be a field, $X$ be a complete variety over $k$, $V$ be an open subvariety of $X$, $Y$ be a scheme over $k$. Suppose $L$ is a line bundle on $V\times Y$. If $L|_{V\times\lbrace y\rbrace}$ extends to a line bundle on $X\times\lbrace y\rbrace$ for every closed point $y$ of $Y$, does the line bundle $L$ extend to $X\times Y$?
What if a stronger condition is assumed ,namely for any functor $\phi\colon\operatorname{Pic}(V\times Y) \to \operatorname{Pic}(V)$ (here $\operatorname{Pic}$ denotes the Picard functors), the line bundle $\phi(L)$ on $V$ extends to $X$. Does $L$ extends to $X\times Y$?
Edit: $X$ is assumed to be smooth i.e. a smooth complete variety.
Welcome new contributor. That is not true. Let $X$ be a nodal plane quartic curve, let $Y$ be the normalization of $X$, let $V$ be the open complement of the node, i.e., the maximal open subscheme over which $Y\to X$ is an isomorphism, and let $L$ be the invertible sheaf of the effective Cartier divisor in $V\times Y$ that is the image of the graph of the open inclusion of $V$ in $Y$.
Hi @JasonStarr , thank you for your answer. I should've said $X$ is smooth. Would my statements be true?
Welcome new contributor. This is not true, even if $X$ is smooth. One example permutes the role of $X$ and $Y$ in my previous example.
Let $X$ be a smooth, geometrically connected, projective curve of genus $g>0$. Let $f:X\to Y$ be the normalization of a nodal curve with a single node $p$ that is a $k$-rational point. For instance, $Y$ could be a nodal plane quartic, and $X$ could be the normalization (a genus $3$ curve). Assume that the preimage of $\{p\}$ in $X$ is split, i.e., $\{r',r''\}$ for $k$-rational points $r',r''$ of $X$.
Let $V$ be the open complement of $\{r',r''\}$ in $X$. Denote the graph morphism of the restriction of $f$ to $V$ as follows, $$\Gamma:V\to V\times Y.$$ The image of this graph morphism is a prime Cartier divisor in $V\times Y$. Denote by $L$ the invertible sheaf on $V\times Y$ associated to this Cartier divisor.
The pullback of this Cartier divisor to $V\times X$ does extend to a Cartier divisor on $X\times X$. Every such extension is of the form $$D_{c',c''} = \underline{\Delta} + \text{pr}_1^*\left(c' \underline{r'} + c''\underline{r''}\right).$$
For each of these extended Cartier divisors, the restrictions over $X\times \{r'\}$ and over $X\times \{r''\}$ are not rationally equivalent. Indeed, if they were, then $\underline{r'}$ and $\underline{r''}$ would be rationally equivalent, so that the genus $g$ equals $0$. (This was my reason for working with smooth curves of positive genus.) Since $X\times X$ is smooth, the homomorphism from the group of rational equivalence classes of Cartier divisors to the Picard group is an isomorphism. Thus, every invertible sheaf on $X\times X$ that extends the pullback of $L$ has non-isomorphic restrictions over $X\times\{r'\}$ and over $X\times\{r''\}$. Therefore each extended invertible sheaf on $X\times X$ is not isomorphic to the pullback of an invertible sheaf on $X\times Y$.
Edit. In the example above, for every Zariski cover $Y'\to Y$, the same result holds. However, there is an étale cover $Y'\to Y$ such that the invertible sheaf extends to $X\times Y'$. For an example where there is no such extension even after an étale cover, instead of letting $X\to Y$ be the normalization of a nodal curve, let is be the normalization of a cuspidal curve. Then the same construction gives an invertible sheaf $L$ on $V\times Y$ such that for every étale cover $Y'\to Y$, there is no extension of the invertible sheaf to $X\times Y'$.
Thank you very much for the example and detailed elaboration! I realize I only want to consider the extension result locally on $Y$. So I have a further question: Can the line bundle $L$ (in your example and in general) extends at least locally on $Y$ i.e. there exists a (Zarisiki, etale or fppf) cover $\lbrace Y_i \to Y \rbrace$ such that $L|{V \times Y_i}$ extends to $L|{X \times Y_i}$ for each $i$ ?
That still does not work. Let me modify the example above to explain this.
Thx @JasonStarr. I don't see your modification. But I guess you were trying to say that there is no neighborhood $U_p$ around $p$ such that $L|{V \times U_p} $extends to $L|{X \times U_p}$ for similar reason? I'm not entirely sure though...
Sorry about that; yesterday was hectic. I just added the modification now.
Thx for patiently answering my questions! I need to think about how the argument in the etale situation works.. Anyway does it also imply there is no extension after an fppf cover?
I believe that there is an extension after an fppf cover . . .
Uhhh.. Why? Do you mean in general the line bundle extends fppf locally on Y? Do you have a proof for that?
Yes, I believe that there is an fppf cover of $Y$ such that the pullback invertible sheaf extends. Essentially, I believe the two issues identified in my answer are the only obstructions to extension. These obstructions are eliminated by etale covers and ramified fppf covers.
So are you expecting that if $X$ is any smooth complete curve ( i.e. 1-dimensional case) then any line bundle $L$ on $V \times Y$ does extend to $X \times Y$ fppf locally on $Y$? I've got no clue how to prove this though..
Yes, that is precisely what I expect. I will try to write a proof soon.
That'll be great. Thx a lot!
I am probably not going to have time to write a proof soon. There are three key steps. First, by Zariski's Main Theorem, for the normalization $\widetilde{Y}$ of $Y$, every extension of the Cartier divisor on $X\times \text{Spec}\ k(Y)$ extends uniquely to $X\times \widetilde{Y}$. Next, by Artin approximation (or something simpler), after base change by an etale extension, the irreducible containing a specified closed point of $Y$ are geometrically irreducible. Finally, a sufficiently ramified fppf cover factors through the normalization.
"geometrically irreducible" --> "unibranch"
I haven't really figured out your sketched proof tbh.. But are you putting any assumption on $Y$?
So this is what I understand so far (only the outline) : given a line bundle $L$ on $V \times Y$, by step 1, the pullback $\tilde{L}$ of $L$ to $V \times \tilde{Y}$ extends to $X \times \tilde{Y}$. By step 3, there exists an fppf cover $Y' \to Y$ s.t. it factors through $\tilde{Y} \to Y$, so the pullback $L'$ of $L$ to $V \times Y'$ extends. But what's the significance of Step 2? Also which version of Zariski' main theorem are you referring to and is $Y$ arbitrary scheme (i.e. without further assumption) ? Excuse my ignorance. Anyway, thx again for your continuous answers.
|
2025-03-21T14:48:31.806254
| 2020-08-18T11:23:22 |
369477
|
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|
Stack Exchange
|
On the link between homology and homotopy
In the last semester I learned homological algebra and higher category theory/homotopy theory.
But I am kind of confused when I try to really understand the link between the two subjects (this is really not my comfort zone ...)
Therefore I try to write (a kind of self-exercise) a text on homological algebra and homotopy theory but really introduce from $0$ the two subjects.
I would like to introduce the following concepts in homological algebra:
chain complex
1$\frac{1}{2}$. Grothendieck group
homotopy of a complex
derived category
t-structures
And also I would like to introduce the following concepts in homotopy theory:
Model categories
Homotopy category of a model category
Derivation in the setting of model categories
Quasi-categories
4.5. simplicial object in a category and homotopy in this context
Dold-Kan equivalence
Now the "hard" part start:
How to organize these concepts in a good way? For 1-3 (either in homology/homotopy) I think that I know how to do that but for 3-5 especially in homotopy I don't have any idea ...
This gives rise to my questions:
How to motivate infinity categories, or more generally homotopy theory/higher category theory but from a homological point of view. I read somewhere a maybe good idea:
For an abelian category $\mathcal{A}$, the derived category $\mathcal{D(A)}$ is not defined by a universal property.
I read somewhere that in some sense higher category theory resolves the problem. Okay but why? And, do we need quasi categories, or would model categories be sufficient for doing that?
If someone have some idea to organize this text I open to any suggestion.
I will be grateful if someone could give me some clues for doing this self exercise.
As for why someone with a homological background might want to know about higher categories, there is this very annoying fact that the cone of a map is not a functorial construction. DG and infinity categories fix this
@NoahRiggenbach the fact that the derived category is not given by a universal property is equivalent in some sense to the fact that the cone is not functorial so i will be grateful if you could explain me how infinite category fix this ? and what do you mean by DG? thank you in advance
DG means differential graded. This means it's enriched over chain complexes. By the Dold-Kan correspondence this will give you an infinity category. As for why infinity categories help here, the idea is they make homotopy cofibers, and homotopy colomits in general, functorial. This is because they let you treat commuting up to homotopy as commuting, and let you treat rather complicated classes of weak equivalence like they were homotopy equivalences.
@NoahRiggenbach If you could give me a reference it's will help me a lot
probably Lunts and Orlovs paper has something about this. I'll check and get back to you
Just a remark: for functorial mapping cones, 2-categories suffice.
@FernandoMuro Stable derivator ? Or something else
@anonyme not that much, just the homotopy 2-category. The homotopy category is constructed by taking sets of connected components on mapping spaces. For the homotopy 2-category you take fundamental groupoids instead.
@FernandoMuro Seem very interesting, could you please send reference ? thanks in advance.
@Anonyme have a loot at Baues, Hans-Joachim. 2006. “Triangulated Track Categories.” Georgian Mathematical Journal 13 (4): 607–634.
I'd encourage the OP to read the writings of others on this topic, before trying to write something from scratch. I attended lectures at OSU where Aaron Mazel-Gee motivated $\infty$-categories very much as the OP suggests in Question 1. It appears some of the ideas from those lectures have now appeared here.
As for Question 2, Weibel's book Introduction to Homological Algebra does a great job with the first collection of topics, then Hovey's book (or Dwyer-Spalinski) gives the first three items in the second collection, and Lurie's books give you everything you could want about quasicategories and their connection to model categories and homological algebra (seriously, the introductions he writes for each chapter are phenomenal). As for the Dold-Kan correspondence, while I'm sure it appears somewhere in Lurie's writings, the clearest exposition I've read anywhere is by Akhil Mathew.
I agree with Arthur that, if you were more categorically minded, you could reverse the order (e.g., starting with Lurie, if you knew about simplicial sets already). For myself, I'd rather start with something concrete and then build the abstraction on top of that, little by little, as this ordering suggests. Weibel's book really is written in such a way to make it easy to step from there into triangulated categories, model categories, and quasi-categories. But it starts at a place very accessible to algebraists.
The two answers was very good the attribution of bounty/accepted answer was arbitrary
I'll answer your second question first. To some degree, the ordering you choose will largely depend on whether you want to lead with examples or with full abstraction. As an example, you can introduce projective resolutions and the derived category using only facts about $\text{Ch}(\mathcal{A})$ and Ore's calculus of fractions (see Weibel's book for a treatment like this) or you can introduce model categories, prove their properties, prove that $\text{Ch}(\mathcal{A})$ admits a projective model structure using a small object argument (see this nLab page for an outline of the argument), and arrive thus at a description of the derived category as a homotopy category.
Personally I think the second account would be unnecessarily convoluted and it would make more sense to introduce some homological algebra first, not least because that way you can introduce the projective model structure as an example of a model structure, projective resolution as an example of a cofibrant resolution, derived category as an example of a homotopy category et cetera; these concepts can be difficult to gain an intuition for without several examples! But both orderings are available to you.
On the question of model categories and quasicategories: model categories can be viewed as "presentations" for quasicategories (see this nLab page for this perspective, and Appendices A.2 and A.3 of Lurie's Higher Topos Theory for a development of the theory of model categories with this explicit goal). Quasicategories have several advantages over model categories: for example, there is a quasicategory of functors from any quasicategory to another, whereas the analogous statement doesn't hold for model categories. Model structures are heavily involved in many of the foundational proofs regarding quasicategories, however, so there's no two ways of ordering these topics.
On your first question: personally I don't believe homological algebra is sufficient motivation for introducing either model categories or infinity-categories. As raised in the comments, the triangulated category $\mathcal{D}(\mathcal{A})$ doesn't allow functorial cones and this is annoying in some applications, but people mostly got on fine with applying homological algebra for decades before people started talking about dg- and quasicategories. A stronger order for your text, in my opinion, would be to introduce basic concepts from homological algebra, then use these as examples when you start talking about model categories and finally quasicategories.
On the question of a universal property for $\mathcal{D}(\mathcal{A})$ using infinity categories, you might find section 1.3.3 of Lurie's Higher Algebra helpful. Note, however, that $\mathcal{D}(\mathcal{A})$ certainly does have a universal property in ordinary 1-categorical language: it is the localization of $\text{Ch}(\mathcal{A})$ at the quasi-isomorphisms.
The two answers was very good the attribution of bounty/accepted answer was arbitrary
|
2025-03-21T14:48:31.806890
| 2020-08-18T12:05:39 |
369479
|
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|
Stack Exchange
|
Eilenberg–Zilber-type theorem for Map([n],A), where the degeneracy maps for [n] are forgotten
The following statement should be immediately implied by Eilenberg–Zilber theorem if the sequences $(i_0,\ldots,i_k)$ below are only monotone. But I need the strict monotone version which I believe to be true, as I have checked it for degree 0 and 1.
Let $A=(A_k)_{k\geqslant 0}$ be a cosimplicial object in an abelian category. Denote by $\partial_k^r:A_{k-1}\rightarrow A_k$ and $\sigma_k^r:A_{k+1}\rightarrow A_k$ its coface and codengeneracy maps respectively. We have the associated unnormalized cochain complex
$$C^{*}(A):~ 0\xrightarrow{} A_0\xrightarrow{\delta_1} A_1\xrightarrow{\delta_2} A_2\xrightarrow{\delta_3} \cdots $$
where $\displaystyle \delta_k=\sum_{r=0}^k (-1)^r\partial_k^r$.
Now let $n$ be a positive integer. Denote by $\Delta_k^n$ the set of sequences $(i_0,\ldots,i_k)$ of integers satisfying $0\leqslant i_0<\cdots <i_k\leqslant n$. Consider the following cochain complex
$$C^{*}_{\Delta^n}(A):~ 0\xrightarrow{} A_0^{\oplus\Delta^n_0}\xrightarrow{\delta'_1} A_1^{\oplus\Delta^n_1}\xrightarrow{\delta'_2} \cdots \xrightarrow{\delta'_n} A_n^{\oplus\Delta^n_n}\xrightarrow{} 0 $$
where for each element $\mathbf{a}=(a_{i_0,\ldots,i_k})\in A_k^{\oplus\Delta^n_k} $,
$$\left(\delta'_{k+1}(\mathbf{a})\right)_{j_0,\ldots,j_{k+1}}:= \sum_{r=0}^{k+1}(-1)^r \partial_{k+1}^r(a_{j_0,\ldots,\hat{j_r},\ldots,j_{k+1}}).$$
Finally, let $Tr^{\leqslant n}C^*(A)$ be the truncation of $C^*(A)$ by forgetting the terms $C^k(A)$ for all $k>n$.
We have the obvious diagonal map $D:Tr^{\leqslant n}C^*(A)\rightarrow C^{*}_{\Delta^n}(A)$ which is a chain map.
Is $D$ a quasi-isomorphism?
|
2025-03-21T14:48:31.807013
| 2020-08-18T12:14:37 |
369480
|
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|
Stack Exchange
|
Write $(4x+1)^n$ as the linear combination of certain polynomials
Let
$$
P_m(x):=\begin{cases}4x+1\quad&\ \text{if}\ m=1,\\
0\quad&\ \text{if}\ m=2,\\
8x^m+(x+1)^{m-3}(2x+1)^3\quad&\ \text{if}\ m\geq3.\end{cases}
$$
How to prove that for any positive odd integer $n$, there exist integers $a_1^{(n)},a_2^{(n)},\ldots,a_n^{(n)}$ such that
$$
(4x+1)^n=\sum_{k=1}^n a_k^{(n)}P_k(x).
$$
Can you say a few things about what led you to this question?
I need to reduce the sum $\sum_{k=0}^n(4k+1)^{2l+1}A_k$, where $A_k$ are usually some hypergeometric terms. I have found the closed forms of $\sum_{k=0}^nP_n(k)A_k$. Now I want to find another form of the sum $\sum_{k=0}^n(4k+1)^{2l+1}A_k$ by establishing the identity as stated in the question.
I enjoyed very much this question! My solution contains two ideas, each of which addresses one of two distinct subproblems:
show that the coefficients $a_m^{(n)}$ are integers;
show that the coefficients $a_m^{(n)}$ exist.
The subproblem (1) is not completely obvious because the polynomials $P_m$ are not monic.
However we are lucky here! The simple idea here is to renormalize the variable as $x=y/4$ and clear the denominators appropriately, noticing that there is a fortunate 2-adic coincidence that makes this approach work.
The subproblem (2) is not obvious and quite annoying because $P_0=P_2=0$. In other words, there is no nonzero $P_m$ of degree $0$ and $2$, so a priori the linear elimination may leave us with a remainder of degree at most 2. The idea to go here is more hidden (and nicer!). It starts by essentially rewriting the problem as
$$
(4x+1)^n = 8x^3 A(x) + (2x+1)^3 A(x+1)
$$
and finding some hidden symmetry. It may even be the case that this observation may be useful to find the coefficients $a_m^{(n)}$ explicitly.
So, let's start the proof. After the change of variable $x=y/4$ the polynomial $(4x+1)^n$ becomes $(y+1)^n$, which has integer coefficients in the variable $y$.
Instead the polynomials $P_m(x)$ for $m\geq 3$ become:
$$
P_m(y/4) = \frac 8 {4^m} y ^m + \left(\frac{y+4} 4 \right)^{m-3} \left(\frac{y+2} 2 \right)^3.
$$
Now it is tempting to multiply over by $4^m/8$ to clear all the denominators, but instead we will multiply by $4^{m-2}$, that is, one less factor of 2 than the greedy one.
So, for $m\geq 3$ define $Q_m(y):=4^{m-2} P_m(y/4)$ and we compute that
$$
Q_m(y) = \frac 1 2 y^m + (y+4)^{m-3}\frac{(y+2)^3} 2
$$
Now, you should note, expanding the binomials and the products, that all the terms $y^k$ with $k<m$ get multiplied by some positive power of 2 or 4, before being divided by the second 2 at denominator. Therefore they are all integers.
Moreover the leading coefficient of $Q_m$ (the one multiplying $y^m$) is equal to $1/2 + 1/2=1$ (here is the fortunate 2-adic coincidence). Summing up, we have:
Lemma 1 For all $m\geq 3$ we have that $Q_m(y)$ is a monic polynomial in $y$ with integer coefficients and degree $m$.
We conclude the following:
Corollary 1 For all $n$ there exist (unique!) integers $b_m^{(n)}$ such that
$$
(y+1)^n = \sum_{m=3}^n b_m^{(n)} Q_m(y) + R(y)
$$
for some polynomial $R(y)$ with integer coefficients of degree at most 2.
Note that up to now we did not use any information on the given polynomial $(y+1)^n$. Note also that if only we could prove that $R(y)$ is a multiple of $P_1(y/4)=y+1$ , which we ignored so far, that would finish the proof.
To prove this, I came up with the following argument. First note that
$$
Q_m(y) = \frac {y^3} 2 y^{m-3} + \frac {(y+2)^3} 2 (y+4)^{m-3},
$$
so the linear combination of $Q_m$ can be written as
$$
\sum_{m=3}^n b_m^{(n)} Q_m(y) = \frac {y^3} 2 S(y) + \frac {(y+2)^3} 2 S(y+4),
$$
where $S$ is the polynomial of degree at most $n-3$ given by
$$
S(y) :=\sum_{m=3}^n b_m^{(n)} y^{m-3}.
$$
It is convenient (to "increase the symmetry") to write $S(y)=T(y-1)$ for some other polynomial $T$ of the same degree. Then we will again change variable $z=y+1$, so the polynomial equation in Corollary 1 reads as follows:
$$
z^n = \frac {(z-1)^3} 2 T(z-2) + \frac {(z+1)^3} 2 T(z+2) + R(z-1).
$$
Since $T$ is a polynomial of degree at most $n-3$, we have that $z^n-R(z-1)$ is equal to a linear combination of terms of the form
$$
F_m(z) = \frac {(z-1)^3} 2 (z-2)^{m-3} + \frac {(z+1)^3} 2 (z+2)^{m-3}
$$
for $3\leq m\leq n$.
We now exploit the symmetry! We note that $F_m(z)$ is a monic polynomial (with integer coefficients, but we don't need it here) of degree $m$ which satisfy the odd/even equation:
$$
F_m(-z) = (-1)^m F_m(z).
$$
Therefore, $F_m(z)$ either has only monomials of odd degree, or only monomials of even degree.
Write
$$
z^n-R(z-1) = \sum_{m=3}^n c_m^{(n)} F_m(z).
$$
We are going to prove that $c_m^{(n)}=0$ for all even $m$. Suppose the contrary, and let $M$ be the largest even number $3\leq M\leq n$ such that $c_M^{(n)}\neq 0$. By what we said before (each $F_m$ is monic of degree $m$ and odd/even) we see that only $c_m^{(n)}F_m(z)$ with $m=M$ contributes a nonzero multiple of the monomial $z^M$. Therefore $c_M^{(n)}z^M$ appears as the unique monomial of degree $M$ in the power expansion of the polynomial $z^n-R(z-1)$. However, $n$ is odd by assumption (in the question) and $R(z-1)$ has degree at most 2 by construction.
Therefore $z^n-R(z-1)$ does not contain any term of degree $M$; that's a contradiction. This shows that only $F_m$ with odd $m$ appear in the expansion in display above. Summing up:
Lemma 2 Let $R$ be as in Lemma 1. Then $z^n-R(z-1)$ is an odd polynomial function.
Since $z^n$ itself is odd, we get that $R(z-1)$ is odd. Since $R$ is a polynomial of degree at most 2, it can be odd only if it is a scalar multiple of the linear monomial $z$. We conclude
Corollary 2 $R(y)=b_1^{(n)} (y+1)$ for some scalar $b_1^{(n)}$. (a fortiori, $b_1^{(n)}\in\mathbb Z$)
From Corollary 1 and Corollary 2 we get the wanted solution, with $a_1^{(n)}=b_1^{(n)}$ and $a_m^{(n)}= 4^{m-2}b_m^{(n)}$ for all $m\geq 3$.
Thanks for your nice solution！
Thanks for sharing the problem! As for explicit formulae, maybe I was too optimistic. The reformuation symmetrizes the problem but I don't have ideas that are better than writing down the recursion. Maybe some expert in binomial sums may help in recognize known recursions? Another idea is to write the coefficients b_m^n for, say, n=7 and put these numbers into OEIS. If you're lucky you may get something
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2025-03-21T14:48:31.807430
| 2020-08-18T12:15:52 |
369481
|
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|
Stack Exchange
|
Structures for random graphs with structure
Background[You may skip this and go immediately to the Definitions.]
Crucial features of a (random) graph or network are:
the degree distribution $p(d)$ (exponential, Poisson, or power law)
the mean degree $\bar{d}$
the mean clustering coefficient $\bar{C}$
the mean distance $L$ and diameter $D$
Randomly generated graphs often are required to exhibit the small-world property, i.e. $L\propto \log N$ and $\bar{C}$ is “not small”. There are several random graph models that address at least one of these conditions:
The Watts-Strogatz model (with underlying regular ring lattice)
The Barabasi-Albert model (with preferred attachment)
The Configuration model (with given degree sequences, resp. distributions)
The Newman model (incorporating community structure)
While the Watts-Strogatz and the Barabasi-Albert model are modifications of the Erdős–Rényi model, and the Newman model is a specific generalization of the configuration model, I wonder if there is already a "meta-model" that tries to incorporate the best of all of these models. (Reference request.)
Generalizing both Watts-Strogatz's and Newman's model, I'd like to investigate random graphs that "interpolate between a randomized structure close to ER graphs and [some arbitrary regular graph]" (quote from Wikipedia).
For this, I'd like to have at hand a multitude of regular graphs which can
be systematically symbolized and enumerated,
be easily generated from their symbol (i.e. their adjacency matrices), and
possibly have closed form expressions for the small-world characteristics $L$ and $\bar{C}$
Which regular graphs I have in mind can most easily be explained by an example.
Definitions
Let a vertex configuration be a graph that represents a vertex $\nu$ with a number of immediate neighbours $\nu_0,\nu_2,\dots,\nu_{d-1}$ and a shortest path (of arbitrary length) between each pair of consecutive neighbours $\nu_i, \nu_{i+1}$. A vertex configuration can be codified by the symbol $(n_1.n_2.\dots.n_k)^m$ which tells, that $\nu$ has degree $d = m \cdot k$ and is surrounded by an $m$-periodic sequence of $n_i$-faces resp. shortest cycles. (This is nothing but the standard definition of vertex configurations in geometry in the language of graph theory.)
Example:
$(4)^4$
A vertex is said to have a given vertex configuration $\Gamma$ when its neighbourhood together with one shortest path between neighbours is isomorphic to $\Gamma$. A graph is said to have a given vertex configuration $\Gamma$ when all of its vertices have vertex configuration $\Gamma$. A vertex configuration is said to be realizable when there is a graph that has it.
Now consider finite graphs in which all vertices have the same vertex configuration.
Questions
Are all vertex configurations $\Gamma$ realizable by graphs of more or less arbitrary size? How to prove or disprove this?This has to do with the question if all vertex configurations (in the sense of geometry) which don't define a periodic tiling of the sphere (i.e. a regular polyhedron) define a periodic tiling of the Euclidean or hyperbolic plane.
If there are non-realizable vertex configurations: How do I check if a given vertex configuration is realizable?
Does a graph with a given vertex configuration $\Gamma$ have to be vertex-transitive?
Since the (equal) number of vertices of two vertex-transitive graphs with the same vertex configuration doesn't guarantee that they are isomorphic: By which general means can their "shape" be defined, so that two equally defined graphs must be isomorphic? (For an example: see below.)
Is there a systematic way to generate an adjacency matrix for a given realizable vertex configuration and "shape"?
With "shape" I mean what Dolbilin and Schulte call "neighborhood complexes (coronas)" in their paper The Local Theorem for Monotypic Tilings.
Examples
Consider the vertex configuration $(4)^4$ and a "shape" defined by numbers $(4, 6)$
When linking vertices on opposite sides of the shape all vertices have the same vertex configuration $(4)^4$, moreover the resulting graph is vertex-transitive:
We find diameter $D = 5$, clustering coefficient $\bar{C} = 0$, and mean distance $L =\frac{1}{23}(4\times 1 + 7 \times 2 + 7 \times 3 + 4 \times 4 + 1 \times 5) \approx 2.61$ for which to find a closed or recursive explicit expression (depending on $(n,m)$) seems to be feasible.
For the "shape"
with the same vertex configuration and number of vertices we find $D = 5$ and mean distance $L =\frac{1}{23}(4\times 1 + 6 \times 2 + 6 \times 3 + 5 \times 4 + 2 \times 5) \approx 2.78$
For the "shape"
with roughly the same number of vertices we find $D = 4$ and mean distance $L =\frac{1}{24}(4\times 1 + 8 \times 2 + 8 \times 3 + 4 \times 4 ) \approx 2.5$.
If you want a cluster coefficient $\bar{C} = 1/2$ you can start with a vertex configuration $(3.n)^m$, e.g. $(3.4)^2$:
Unfortunately, this configuration does not qualify because it doesn't tile a plane but the sphere (giving rise to the cuboctahedron). So you have to choose $(3.4)^3$ at least. To draw a nice "shape" of some size that can be made into a finite graph with vertex configuration $(3.4)^m$, $m > 2$, requires hyperbolic geometry. To find an adjacency matrix is even harder, as I guess (see question 5). Also the diameter $D$ and mean distance $L$ (as closed expressions).
Alternatively, one can add an edge to half of the $n\cdot m$ $4$-cycles (randomly chosen) of the $(4)^4$ graph - thus reducing diameter $D$ and mean distance $L$.
The following vertex configuration has notation $(<IP_ADDRESS>)^1$ and should provide counterexamples to question 1 (existence of graphs of arbitrary size) and question 3 (vertex-transitivity).
There are only finitely many graphs that realize this configuration, and all of them are finite with at most 24 vertices. Exactly two of them are planar, the edge-graph of rhombicuboctahedron (left), and the edge-graph of the closely related pseudo-rhombicuboctahedron (right).
Only the first one is vertex-transitive.
All other graphs can be obtained from these by identifying vertices.
For example, identifying antipodal vertices in the left graph gives a "projective polyhedron":
I highlighted the vertex configuration in the right image because it is not obvious in this drawing.
I think these are all the graphs with this configuration. I might be wrong, but there are certainly no such graphs with more than 24 vertices.
More generally, you might be interested in the Local Theorem from
"The Local Theorem for Monotypic Tilings" by Dolbilin and Schulte
which is concerned with the question when certain local restrictions imply global symmetry. Usually, it gives uniqueness and vertex-transitivity, but it only applies if the topology is "simply connected" (so, for tilings of the sphere, Euclidean/hyperbolic plane, but not for the torus, as you have seen in your question that the graph is not unique for $(4)^4$).
In the start of Section 3 (below Theorem 3.1) they state that the configuration $(<IP_ADDRESS>)^1$ can be realized as an infinite graph, but not as a vertex-transitive one. I have tried to track down this claim, but they only refer to the book "Tilings and Patterns" which contains literally thousands of tilings, and I was not able to find the desired one.
Finally, the following configuration $(3.4.5)^1$ should not be realizable at all:
To see this, note that the graph must contain a "triangular face" (since the configuration does).
Each of the three edges of that triangle is shared either with a quadrangle, or a pentagon. W.l.o.g. assume that two edges are shared with a quadrange. But these two edges share a vertex, and so this vertex cannot be of type $(3.4.5)^1$.
In general it seems quite tricky to distinguish the realizable from the non-realizable configurations.
As a rule of thumb, it seems that odd-faces pose a problem, similarly as they did in the previous example.
So, e.g. a configuration $(\mathbf 5.8.10)^1$ cannot exist either for the same reason, as there is a pentagonal face that bounds two different kinds of faces, and there is no face-type repeated at a vertex.
Since you mention (in the comments) that you are mostly interested in $(3.n)^m$ (assuming $n\ge 3$, $m\ge 2$):
This configuration always exists, is unique and vertex-transitive (assuming a "simply connected topology", which we can translate as "the graph is planar").
It is finite only for $(3.3)^2$ (octahedron), $(3.4)^2$ (cuboctahedron) and $(3.5)^2$ (icosidodecahedron). You can consider it "planar" for $\smash{(3.3)^3}$ (triangular tiling) and $\smash{(3.6)^2}$ (trihexagonal tiling), and hyperbolic in all other cases.
The uniqueness and symmetry is essentially a consequence of the Local Theorem (and the related Extension Theorem) mentioned before. But in easy terms: if you try to build a graph with such a vertex configuration, and you start from any vertex, and then you try to complete the vertex configuration around any of the other vertices, you can do this only in a unique way (really, try it on paper). Since you make no choice in any (of the possibly infinitely many) steps, the result is unique.
Thanks, your answer is relevant in two respects: First, it shows that a graph that realizes a configuration doesn't have to be vertex-transitive. (That's important for me to know, and it answers question 3) Second it shows that there are (of course) configurations that tile only the sphere (defining a polyhedron) but not a plane. This case was excluded in the remark to question 1.
Do you have a tool at hand that draws graphs for given configurations?
@Hans-PeterStricker Sadly no, I always do it by hand.
As I do .......;-)
Thanks for the hint to the Dolbilin-Schulte paper!
And thanks a lot for the counter example! This was my impression, too: that it is tricky to check a configuration for realizability. By the way: I always tended to consider only "tame" configurations of the form $(3.n)^m$ - these should work. Or do some of them don't pass your rule-of-thumb test?
@Hans-PeterStricker The graph with configuration $(3.n)^m$ exists, is unique and vertex-transitive. I edited my answer.
Note that $(3.3)^3 = (3)^6$. Note further that $(6)^3$ gives rise to the hexagonal tiling (which you know for sure).
@Hans-PeterStricker Okay, uniqueness is a lie. Imagine that you can build finite graphs of type $(3.3)^3$ by wrapping around the tiling to a torus as you did with $(4)^4$. The Local Theorm assumes that the "topology is simply connected" (e.g. sphere, Euclidean/hyperbolic plane). Similar wrapping might be possible in the spherical and hyperbolic case. I am not sure what this does to vertex-transitivity, but I am optimistic that it is preserved. I might edit the answer later.
|
2025-03-21T14:48:31.808148
| 2020-08-18T13:16:49 |
369485
|
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|
Stack Exchange
|
A random variable whose characteristic function decreases the fastest
A random variable $X$ is "good" for $(a_0, b_0) \in (0,1)^2$ if its characteristic function $\varphi_X(t)$ satisfies the following constraints:
$\forall t : \varphi_X(t) \geq 0$.
$\varphi_X$ is monotonically decreasing on $\mathbb{R}_{+}$.
$\varphi(a_0) = b_0$.
Given $(a_0, b_0)$ and $a_0 < x < 1$, I would like to find a random variable $X^*$ which is "good" for $(a_0, b_0)$ and whose characteristic function $\varphi_X$ has the lowest value at $x$, i.e. for any random variable $X'$ which is good for $(a_0, b_0)$ it holds that $\varphi_X(x) \leq \varphi_{X'}(x)$.
My conjecture is that a Gaussian satisfies this optimality criterion. My idea of proving it was thinking about the connections between $\varphi_X(t)$ and the entropy of $X$ ($h(X):= \mathbb{E}[-\log(X)]$): it is known that a Gaussian maximizes the entropy, so if it can be shown that $\varphi_X(t)$ decrease in a rate depends on the entropy, perhaps the claim would follow. However I couldn't obtain a more rigorous claim.
What do you mean by characteristic function? Usually it is $\varphi_X(t) = \mathbb{E}e^{itX}$? But then $\varphi_X(0) = 1 \geq |\varphi_X(t)|$ for any $t \in \mathbb{R}$.
I mean exactly that, why doesn't it fit with what I wrote?
What does it mean for a complex-valued function to be "monotone"? What does $\phi_X(x)\leq \phi_{X'}(x)$ mean?
@JamesMartin: I think (1) is implicitly asking for the characteristic function to take only nonnegative real values, in which case everything else makes sense.
ah ok right. And monotone on $\mathbb{R}_+$.
Indeed condition (1) implicitly implies that $\varphi_X : \mathbb{R} \rightarrow \mathbb{R}_{+}$
@Ernie: Ok, to clarify, $\varphi$ is increasing on $\mathbb{R}-$ and decreasing on $\mathbb{R}+$, correct? Calling this behaviour "monotone" is irritating for me.
Yes. I agree this is not the correct term to use here - I will edit the question.
You know Polya's criterion ?
@mike - I did not, reading about it now - thanks! At first glance, I am not sure how to use it though
Your minimization problem is clearly a problem of infinite-dimensional linear programming. Usually, such problems do not have a closed-form solution. Your problem, too, seems unlikely to have such a solution in full generality; however, see Remark 1 at the end of this answer.
One thing that can be definitely said is that your conjecture "that a Gaussian satisfies this optimality criterion" is false. Indeed, for any $(a_0,b_0)\in(0,1)^2$, let $c:=c_{a_0,b_0}:=-a_0^{-2}\ln b_0$, so that $c$ is the positive real root of the equation $\phi(a_0)=b_0$, where $\phi(t):=\phi_c(t):=e^{-ct^2}$. Then $\phi$ is a characteristic function and your conditions 1--3 are satisfied with $\phi$ in place of $\varphi=\varphi_X$.
Let now $\psi(t):=\psi_k(t):=\max(0,1-|t|/k)$ for all real $t$, where $k:=k_{a_0,b_0}:=a_0/(1-b_0)$ is the only root of the equation $1-a_0/k=b_0$. Then $\phi$ is a characteristic function and your conditions 1--3 are satisfied as well with $\psi$ in place of $\varphi=\varphi_X$.
However, $k>a_0$ and $\psi(k)=0<\phi(k)$. $\Box$
Remark 1:
This also shows that for each $t\ge k=k_{a_0,b_0}$ the function $\psi_k$ is a minimizer of $\varphi(t)$ over all characteristic functions $\varphi$ satisfying your conditions 1--3.
Moreover, note that $t=a_0$ is a positive real root of the equation $\phi_c(t)=\psi_k(t)$. The latter equation may have one or two positive real roots. Let $a_{\max}$ be the largest root of the equation $\phi_c(t)=\psi_k(t)$. It follows that $a_{\max}<k$ and for each $t>a_{\max}$ we have $\psi_k(t)<\phi_c(t)$. In particular, if $a_{\max}=a_0$ (that is, if $a_0$ is the largest root of the equation $\phi_c(t)=\psi_k(t)$), then $a_0<k$ and for each $t\ge a_0$ we have $\psi_k(t)<\phi_c(t)$.
The condition $a_{\max}=a_0$ can be rewritten as $\psi'_k(a_0)\le\phi'_c(a_0)$ (with $c=c_{a_0,b_0}$ and $k=k_{a_0,b_0}$), and then further as $0<b\le b_*$, where $b_*=0.28466\dots$ is the only root $b\in(0,1)$ of the equation $1-b=2b\ln\frac1b$; the necessary and sufficient condition $0<b\le b_*$ for $a_{\max}=a_0$ does not involve the scaling parameter $a_0$ (and of course it should not).
For $a_0=1$, here are the graphs $\{(t,\phi_c(t))\colon t\in[0,k+1)\}$ (blue) and $\{(t,\psi_k(t))\colon t\in[0,k+1)\}$ (gold) (i) for $b=b_*$ and hence $a_{\max}=a_0[=1]$ (left); (ii) for $b=b_*-2/10$ and hence $a_{\max}=a_0[=1]$ (middle); and (ii) for $b=b_*+2/10$ and hence $a_{\max}>a_0[=1]$ (right):
Thank you so much. Do you think the solution to my problem can be approximated somehow? Can I even prove there is a solution?
@Ernie : I have added that for $t\ge a_0/(1-b_0)$ there is the (rather trivial) minimizer. For $t\in(a_0,a_0/(1-b_0))$, the main problem is that there seems to be no good description of the set of all nonnegative characteristic functions -- see e.g. shorturl.at/twDJK and, in particular, shorturl.at/rDFPZ
@ Iosif Pinelis thanks. Is there a trivial way to see why the Gaussian fails even for $t \in (a_0, a_0/1-b_0)$? Is there a way of saying something meaningful about the minimizer?
May you please explain the part after the remark? thanks!
@Ernie : As stated now in the remark, $\psi_k$ is better than $\phi_c$ for all $t\ge a_0$ in the case $b\le b_=0.28\dots$. At this point, I cannot say anything definite for $a\in(a_0,a_{\max})$ and $b>b_$.
I see. Do you have an idea about how to generalize it for $t < a_0$?
@Ernie : I don't have anything definite to say about $t<a_0$ at the moment. You may want to post this question separately.
|
2025-03-21T14:48:31.808521
| 2020-08-18T13:24:10 |
369486
|
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|
Stack Exchange
|
bracketing number vs covering number
Just want to double check if the lemma on page 9 of this slides is correct:
http://www.math.leidenuniv.nl/~avdvaart/talks/09hilversum.pdf
Lemma: $N(\epsilon,\cal F,||\cdot||)\leq N_{[]}(2\epsilon,\cal F,||\cdot||). $
Proof: If $f$ is in the $2\epsilon$-bracket $[l,u]$, then it is in the ball of radius $\epsilon$ around $(l+u)/2$.
I think what the proof means is that, if a set of $2\epsilon$-brackets covers $\cal F$, then this set is also a set of balls of radius $\epsilon$ that can cover $\cal F$. Since there may be other sets of balls of radius $\epsilon$ that can cover $\cal F$, the covering number is no larger that the bracketing number.
I haven't found the same conclusion in any textbook I can find so far (not sure if it is because this conclusion is much too trivial), so I am not quite confident to say if it is right or wrong. I'd really appreciate it if anyone can enlighten me!!
Page 17 of the following notes shows that this lemma is correct: http://www.stat.columbia.edu/~bodhi/Talks/Emp-Proc-Lecture-Notes.pdf
Your elaboration is essentially right, except the brackets themselves are not $\|\cdot\|$-balls.
If $[l,u]$ is a $2\epsilon$-bracket, then it is contained in the $\|\cdot\|$-ball of radius $\epsilon$ centered at $(l+u)/2$, since $l \le f \le u$ implies
$$\|f - (l+u)/2\| \le \frac{1}{2} \|f-l\| + \frac{1}{2} \|f - u\| \le \|u-l\| = \epsilon.$$
Thus a cover of $2\epsilon$-brackets can be replaced by a cover of larger $\epsilon$-$\|\cdot\|$-balls of the same cardinality.
$$
\begin{aligned}
\|f - (l+u)/2\| & = \| (f - l)/2 - (u - f)/2 \| \\
& \le \max \left\{ \frac{1}{2} \|f-l\| ,\frac{1}{2} \|u - f\| \right\}\\
& \le \frac{1}{2} \|u-l\| \\
&\le\epsilon.
\end{aligned}$$
since $(f - l)/2 \ge 0$ and $(u - f)/2 \ge 0$.
Since my reputation is too low I could not add a comment. However, I recently met with this problem when I was reading Van der Vaart& Wellner (1996). According to this book, a hidden assumption for @angryavian's answer to hold is that the norm must satisfy the Riesz property ($L_r(Q)$-norms do possess this property), i.e.,
$$
|f| \le |g| \Rightarrow \|f\| \le \|g\|.
$$
so that
$$
\begin{aligned}
|f - l| \le |u - l| &\Rightarrow \|f - l\| \le \|u - l\| \\
|f - u| \le |u - l| &\Rightarrow \|f - u\| \le \|u - l\|
\end{aligned}
$$
I hope this could serve as a supplement.
|
2025-03-21T14:48:31.808714
| 2020-08-18T13:30:44 |
369487
|
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|
Stack Exchange
|
Is $C^{*}$-algebra the most modern way to study QFT?
I am not an expert on either QFT or $C^{*}$-algebras, but I'm trying to learn the basics of QFT. In all books/papers and other materials that I know, QFT is studied mainly using a lot of functional analysis and distribution theory, but I know that some algebraic constructions are also being used, and in this context $C^{*}$-algebras seem to be the most modern tool. So, what should an inexperienced student like me know about these approaches to QFT and statistical mechanics? What's the role of $C^{*}$-algebras and other algebraic methods in those theories? What are the problems they fit better? If I'd like to study QFT, do I have to learn $C^{*}$-algebra? Are there problems in which algebraic methods don't fit well? Are there problems in which either approach is fruitful? What does one lose by not knowing these algebraic constructions?
ADD: I work with rigorous statistical mechanics but I'm trying to learn some QFT because...well, these are two related areas at some level. However, I don't know yet what or how much I need to learn about QFT. I have a background in functional analysis and distribution theory, but not in $C^{*}$-algebra. As an unexperienced student, it will be very useful to get a general picture, i.e. what are the problems one is trying to solve in QFT and where do each of these approaches come into play. I think each of these tools are applicable for different kinds of problems or even different subareas of the theory, but I don't know for sure.
Short answer: no. Long answer/clarifying question : what do you want to study QFT for?
@AaronBergman I think I am unexperienced enough to say that I don't know yet. My research area is statistical mechanics but QFT ideas end up being important at some level. What level? Still don't know for sure. I think this is one of the points that motivated my question in the first place. I'm having enough trouble trying to learn QFT on my own, and I know some people deal with it by using $C^{*}$-algebra and other tools I've never studied either.... But, on the other hand, I have a background on functional analysis and distribution theory. I wonder if this is enough to some extent.
And, also, it would be very clarifying to know why people use it, to what kind of problems, the differences between these approaches and so on.
In addition, I know people are using $C^{*}$-algebra to study statistical mechanics as well. But I see this as a reflection of the fact that statistical mechanics has some strong connections to QFT. Don't know if this reasoning is accurate, tho.
Are you approaching this as a mathematician or as a physicist? Are you looking to prove rigorous theorems about stat mech? Do you want to do stat mech like the physicists do? For the latter, I’d say a vanishingly small fraction use C* algebras.
I'm approaching this as a mathematician. But I know it's hard to dodge the physics behind it. But I'm really interested in rigorous approaches to both areas.
I'm gonna add more information to the post.
I am (was?) a physicist, so I’d always suggest learning QFT the way the physicists do particularly because it hasn’t been made rigorous. In that sense, you will rarely if ever see a C*-algebra. I don’t think there’s any harm in learning about C*-algebras — they’re pretty cool — but they’re certainly not required and I would not call them the most modern way of looking at things. Cont’d...
On the other hand, operator algebras do seem to hang around QFT a lot, and just because something hasn’t proved particularly fruitful in the past (algebraic QFT say), it doesn’t mean it won’t be in the future.
My PhD work used C*-algebras quite heavily, so I guess I can claim some expertise there, but I'm not an expert in QFT. That will be the main perspective of my answer.
A good starting point for this discussion is the Stone-von Neumann theorem, a foundational result in both operator algebras and quantum mechanics. The setup is basically the Heisenberg uncertainty principle, which asserts that the operations of measuring the position $x$ and the momentum $p$ of a quantum system don't commute:
$$[x,p] = 2\pi i h$$
An important mathematical question about quantum mechanics in its early history was: what kind of objects are $x$ and $p$? Physicists want them to be self-adjoint operators on some Hilbert space, but you can prove rigorously that no pair of bounded operators have this property. This result belongs to the representation theory of Lie algebras - essentially, the Lie algebra with two generators and the relation above has no representation by bounded self-adjoint operators on Hilbert space.
Stone and von Neumann's idea was to focus on the Lie group rather than the Lie algebra; the relation above is the derivative at 0 of the following relation between time evolution operators $U(t)$ and $V(s)$:
$$U(t) V(s) = e^{-ist} V(s) U(t)$$
The Lie group generated by such $U$ and $V$ is called the Heisenberg group, and the Stone-von-Neumann theorem asserts that that this group has a unique unitary representation on Hilbert space, up to unitary equivalence (and some adjectives that I won't go into here). This provides a nice foundation for basic quantum mechanics which unifies the Heisenberg and Schrodinger pictures of the theory into one set of axioms.
To handle more complicated quantum systems, we need to generalize to more operators satisfying possibly more complicated relations. Here's how this generalization works:
Start with a locally compact group $G$; for the original Stone-von-Neumann theorem, $G = \mathbb{R}$.
The Fourier transform determines and isomorphism $C^*(G) \to C_0(\hat{G})$, where $C^*(G)$ is the group C*-algebra and $\hat{G}$ is the Pontryagin dual.
Such an isomorphism is equivalent to a unitary representation of the crossed-product algebra $C_0(G) \rtimes G$.
All irreps of this C*-algebra are unitarily equivalent.
So now we have quantum mechanics for systems with many particles. But what about QFT? The basic reason why QFT is hard, as I understand it, is that the Stone-von-Neumann theorem is no longer true.
For ordinary quantum mechanics, the classical phase spaces are finite dimensional manifolds - for instance, the classical phase space of a single particle flying around in $\mathbb{R}^3$ is $\mathbb{R}^6$. The classical analog of the phase space in quantum field theory, however, is the space of paths in $\mathbb{R}^3$, which is some sort of infinite dimensional manifold. This means infinitely many operators with infinitely many commutation relations, and the corresponding infinite dimensional Lie groups, to the extent that they even exist, have a much more complicated representation theory.
So now I can try to answer your question. Operator algebras were more or less invented in order to provide a nice model for quantum mechanics. The nice property that this model has - namely, that there is only one realization of it up to unitary equivalence - is no longer true in QFT. So one (implicit) goal of a lot of work in QFT is to cope with this situation and search for better foundations. I have no idea if C*-algebras are the best or most modern way to think about QFT - probably not - but a good place to start for a student is to learn the Stone-von-Neumann theorem in some reasonable generality since we can blame a lot of the difficulty of QFT on its absence.
Again, a provisional answer from a non-expert: likely someone who is a real Jedi Master in Mathematical Physics/Operator Algebras will chime in.
In classical QM, one starts from a Hilbert space of states $H$, and builds from there by looking at special types of operators acting on $H$ (unitary for simmetries, and hermitians for observables). So, in a sense, operator algebras are right there from the start, though in classical QM it looks and feels as if the basic entities are (quantum) states, and the secondary ones are processes (operators).
But I think it is fair to say that the movement has been toward inverting the order, in a sense beginning with the algebra of abstract operators and then modeling the set of states using the infamous Gelfand duality. What I just sketched is a supermarket chat on Algebraic Quantum Field Theory (you can find a condensate here).
You may ask why: I am not sure, but to me it seems that the movement toward processes as opposed to states makes sense
mathematically (for instance it connects with Non-Commutative Geometry of Connes, where one works directly on non-commutative algebras as if they were the algebras of functions over a ghost non commutative space). The algebras are good enough to capture the topology and geometry of the ghost space, and also lend itself to more abstract machinery
physically. There is a growing awareness that QM/QFT is about processes/interactions, rather than a world in which systems exist by themselves. See for instance Rovelli's Relational Interpretation, to just quote one option.
ADDENDUM: so, are C* algebras the newest tool for QFT? The answer is: which QFT do you have in mind? For instance, in Quantum Gravity the answer is definitely no. There folks play with all sorts of goodies, running from higher category theory, to the already mentioned non commutative geometry, to ... pretty much anything under the sun, and even a tad more.
Man, you really like Star Wars. I thought I liked Star Wars, but you really like Star Wars. :)
I do my fellow in the Force. You know why? Because after many many centuries studying pretty much anything under the sun, I discovered the basic simple truth: fables are the only thing which does not lie....PS I like both Jedis and Siths, see my last post :)
|
2025-03-21T14:48:31.809743
| 2020-08-18T16:20:23 |
369496
|
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|
Stack Exchange
|
Uniformization theorem with boundary in the non-compact case
Let $\Sigma$ be a simply connected (and therefore orientable) smooth $2$-manifold with non-empty and connected boundary. Suppose that the interior $\operatorname{int}(\Sigma)$ is endowed with a Riemannian metric $g$ (which does not necessarily extend to all of $\Sigma$). Furthermore, let $u:\Sigma\to\mathbb{R}$ be a smooth function which is harmonic and strictly positive on $\operatorname{int}(\Sigma)$, and which vanishes identically on $\partial\Sigma$. Now let $H$ be the open upper half-plane $\{z\in\mathbb{C}\mid\operatorname{Im}z>0\}$, and let $\overline{H}$ be the closed upper half-plane. Is there then a conformal equivalence $f:\operatorname{int}(\Sigma)\to H$, which extends to a homeomorphism $\Sigma\to\overline{H}$?
We can certainly find a conformal equivalence $f:\operatorname{int}(\Sigma)\to H$: apply the uniformization theorem (for Riemann surfaces without boundary) to $\operatorname{int}(\Sigma)$ to conclude that $\operatorname{int}(\Sigma)$ is conformally equivalent to either $\mathbb{C}$ or $H$. Letting $v:\operatorname{int}(\Sigma)\to\mathbb{R}$ be a harmonic conjugate for $u$, we see that $\exp(-u-iv)$ is a non-constant, bounded and holomorphic function on $\operatorname{int}(\Sigma)$. A conformal equivalence with $\mathbb{C}$ would therefore contradict Liouville's theorem, and we thus have a conformal equivalence $f:\operatorname{int}(\Sigma)\to H$.
However, as far as I can see, the above argument does not in any way say anything about the boundary behavior of $f$. Is there a version of uniformization which takes into account the boundary?
You have to explain what you exactly mean by "$2$-manifold with boundary". A 2-manifold (surface) has no boundary. There is the notion of "bordered surface". Its boundary is a 1-manifold. But this $1$ manifold may consist of circles and open intervals. For example, an open disk with half of the circle added is a bordered surface whose boundary is homeomorphic to an interval.
|
2025-03-21T14:48:31.809898
| 2020-08-18T17:23:07 |
369500
|
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|
Stack Exchange
|
Computing the number of elementary abelian p-subgroups of rank 2 in $GL_{n}(\mathbb{F}_{p})$
Let $p$ be a prime number, and let $\mathbb{F}_{p}$ be a finite
field of order $p$. Let $GL_{n}(\mathbb{F}_{p})$ denote the general linear
group and $U_{n}$ denote the unitriangular group of $n\times n$ upper
triangular matrices with ones on the diagonal, over the finite field $%
\mathbb{F}_{p}$. In fact $U_{n}$ is a Sylow $p$-subgroup of $GL_{n}(\mathbb{F}_{p})$ of order $p^{\frac{n(n-1)}{2}}$. Given $n_{p^{2}}$ be the number of elementary abelian p-subgroups of rank $2$ in $U_{n}$. How can we deduce the number of elementary abelian p-subgroups of rank $2$ in the whole linear group $GL_{n}(\mathbb{F}_{p})$?.
Conversely, given $N_{p^{2}}$ be the number of elementary abelian p-subgroups of rank $2$ in $GL_{n}(\mathbb{F}_{p})$. Is there a criterion deduces the number of elementary abelian p-subgroups of rank $2$ in $U_{n}$?.
In other words, what is the relationship between $N_{p^{2}}$ and $n_{p^{2}}$?.
Any help would be appreciated so much. Thank you all.
Wouldn't $n_{p^2}$ be a better notation than $n_p^2$, which looks like the square of $n_p$?
Yes, thank you. I made a change on the notation.
I think that counting the number of conjugacy classes of such subgroups may be difficult for large $n$. For the number of subgroups, disregarding conjugacy, it is a matter of counting the number of elements of order $p$ in $C_{G}(x)$ for each element $x$ of order $p$., and doing an elementary computation after that. However, those centralizers do not have a very transparent structure as the Jordan form of the element of order $p$ gets more complicated.
Yes, in general, it is complicated to find $N_{p^{2}}$ or $n_{p^{2}}$, so for this moment, I am not interested in this question. Here, I want to know if we assume that one of them is given, then how to find the other.
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2025-03-21T14:48:31.810043
| 2020-08-18T18:02:36 |
369502
|
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|
Stack Exchange
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4-dimensional Lie subalgebras of $o(4,\mathbb{C})$ and of $o(3,\mathbb{C})\ltimes\mathbb{C}^3$
I would like to know all complex 4-dimensional Lie subalgebras of $o(4,\mathbb{C})$ and of $o(3,\mathbb{C})\ltimes\mathbb{C}^3$.
This looks like an exercise in a first course in representation theory. Have you tried using the standard tools and run into difficulty?
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2025-03-21T14:48:31.810112
| 2020-08-18T18:38:53 |
369505
|
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"David E Speyer",
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|
Stack Exchange
|
Representing iteration of a function in PA
Let $\mathscr{L}$ be a (recursive) FOL language, with numeral symbols $\underline{0},\underline{1},\ldots$. Let $T$ be a recursive, consistent theory, containing PA (or even just Robinson arithmetic). It is well-known that all computable functions are strongly represented. This means that for any computable function $f\colon \mathbb{N}^k\to \mathbb{N}$ there exists a formula $\varphi(x_1,\ldots, x_k,y)$ in $\mathscr{L}$ (with the free variables as given) such that the following two conditions hold:
(1) $\vdash_T \forall x_1 \cdots \forall x_k \exists y\, \varphi(x_1,\ldots, x_k,y)$.
(2) For any $(m_1,\ldots, m_k,n)\in \mathbb{N}^{k+1}$, we have $f(m_1,\ldots, m_k)=n$ if and only if $\vdash_T \varphi(\underline{m_1},\ldots, \underline{m_k},\underline{n})$.
There are many ways to build new computable functions from old ones. The standard way to build all computable functions is to start with a few initial functions, and take the closure under addition, composition, basic recursion, and $\mu$-minimization; e.g., see this wiki article about primitive recursive functions.
Thus, the proof that all computable functions are strongly represented seems clear enough. Prove it for the initial functions and for those four types of operations. The hardest step seems to be the basic recursion operation. Looking through the literature, it appears that the standard proof is to use a clever trick of Godel to encode sequences of numbers. I wonder if there is any easier proof.
In particular, suppose that $f\colon \mathbb{N}\to \mathbb{N}$ is a computable function, and $\varphi(x,y)$ strongly represents it (relative to PA, say). Define a new function $f^{\square}\colon \mathbb{N}\to \mathbb{N}$ by the rule that $f^{\square}(m)=\underbrace{f\circ f\circ \cdots \circ f}_{m \text{ times}}(0)$. In other words, we just iterate $f$ a total of $m$ times, with initial value $0$. (This is about the simplest version of recursion I can think of.)
Questions: What is the simplest way to create a formula that strongly represents $f^{\square}$? Is there a formula that is uniform, in the sense that we can just plug in $\varphi(x,y)$?
In what sense is the standard Godel coding trick not uniform?
Use sequence coding, such as https://en.wikipedia.org/wiki/G%C3%B6del%27s_%CE%B2_function .
I.e. write $\bmod(x,y)=z$ as shorthand for $y | x-z$ and $0 \leq z < y$. Then encode $f^m(0)=n$ as $\exists (L,a,b)$ such that $\bmod(L,a)=0$, $\bmod(L,a+mb)=n$ and, for $0 \leq k < m$, we have $\bmod(L,a+(k+1)b) = f(\bmod(L,a+kb))$.
@DavidESpeyer That certainly answers my question over PA, where one can easily prove that there is a unique remainder. Is it difficult to modify the idea for Robinson arithmetic? (I looked at the link Emil gave, which cites Mendelson's textbook. As far as I can tell, Mendelson works in PA in that book.)
Representability in $Q$ follows easily from $\Sigma_1$-definability in $\mathbb N$.
@EmilJeřábek Yes, but what about strong representability?
That makes no difference. If $\phi(\vec x,y)$ represents $f$, then $(\phi(\vec x,y)\land\exists!z,\phi(\vec x,z))\lor(y=0\land\neg\exists!z,\phi(\vec x,z))$ strongly represents $f$.
You two are using language I don't understand, but it doesn't seem to me that is there is a good way to define $f^m(0)$ as a total function in $Q$. Let $R = \mathbb{Z}[t]$ and let $R_+$ be the elements in $R$ which have positive leading term. I believe that $R_+$ is a model of $Q$. The function $f(x) = 2x$ is then a perfectly good function in $R_+ \to R_+$, but there is no function $g : Q_+ \to Q_+$ with $g(y+1) = 2 g(y)$. You can still do the Godel encoding trick above, but it doesn't define a complete function anymore. But perhaps I misunderstand the question.
@EmilJeřábek I think I'm starting to understand. If we take, as part of the definition of representability, that $\vdash_T \forall y, (\varphi(\underline{m_1},\ldots, \underline{m_k},y)\rightarrow y=\underline{n})$, then the formula you gave works just fine. What I'm not seeing is how this part of representability "follows easily from $\Sigma_1$-definability in $\mathbb{N}$". From my understanding, $\Sigma_1$-definability is the condition that, for a relation (not necessarily a function) $R\subseteq \mathbb{N}^{k+1}$, then for any $(m_1,\ldots, m_{k+1})\in \mathbb{N}^{k+1}$, we have ...
$\vec{m}\in R$ if and only if $\vdash_T \psi(\vec{m})$ (for some formula $\psi$). It isn't clear to me how that function-like behavior on numerals would follow easily from this definability condition, under the additional condition that $\psi$ is $\Sigma_1$.
The predecessor of $t$ is $t-1$. Since $t-1$ has positive leading term, it is in $R_+$. Indeed, Wikipedia says "there is a computable model of Q consisting of integer-coefficient polynomials with positive leading coefficient, plus the zero polynomial, with their usual arithmetic. "
And yes, I failed to include the $0$ polynomial in my description of $R_+$, sorry.
@DavidESpeyer For some reason I was confusing "leading term" with "constant term" (had my ordering backwards), sorry. I'm assuming that by $Q_{+}$ you meant $R_{+}$? I'm a little confused about $g$ though. Doesn't the function $g(x)=0$ satisfy the recursion $g(y+1)=2g(y)$?
@PaceNielsen This should be explained in any textbook where these concepts are introduced. If $\psi(\vec x,y)=\exists z,\theta(\vec x,y,z)$ defines the graph $f$, where $\theta$ is $\Delta_0$, then $\exists w,(y\le w\land\exists z\le w,\theta(\vec x,y,z)\land\forall y'\le w,\forall z'\le w,(\theta(\vec x,y',z')\to y'=y)))$ represents $f$ in $Q$.
@DavidESpeyer If you want the result to be a provably total function in $Q$, you don’t use sequence coding directly as is; you further need to embellish the formula with witness comparison as in my previous comment to make it univalued on standard arguments, and then use the trick in https://mathoverflow.net/questions/369505/representing-iteration-of-a-function-in-pa#comment932996_369505 to make it total. Note that this will give a formula that strongly represents $f^\Box$ in $Q$, but in general it will not provably in $Q$ satisfy the recurrence $f^\Box(m+1)=f(f^\Box(m))$.
... The latter is not part of the deal, and is in general impossible, as it would lead to an interpretation of PRA in $Q$ (though it does trivially work in your example, as pointed out by Pace Nielsen).
@EmilJeřábek Thank you! Everything is clear now. I think that perhaps part of the problem was that I started by reading the early papers proving that all partial computable functions are strongly represented, and your specific trick for making a representation strong doesn't work that easily in that case (as it makes the function defined where it shouldn't be). If there is an easy way to account for that possibility too, I'd be interested.
I don’t think I understand what you mean. The difference between representability and strong representability as you defined it in the question is exactly that the latter is required to be provably total in the theory. That’s not possible if at the same time you want it to be undefined on specific inputs.
@EmilJeřábek https://www.sciencedirect.com/science/article/pii/0020025569900164
I see. Well, in that case, I don’t think you can iterate it in a uniform way. For example, consider a function $f$ that is undefined on $0$, but maps all other numbers to $1$. Then $(f\circ f)(0)$ should be undefined, but if you just take a strong representation $\phi$ of $f$, then the canonical definition of $f\circ f$ using $\phi$ as a black box may well make it provably the constant $1$ function (as $\phi$ may happen to provably map all nonzero inputs to $1$, and $0$ to nonzero).
@EmilJeřábek It might be possible to rule out that type of behavior using Corollary 3 of https://mathscinet.ams.org/mathscinet/search/publdoc.html?loc=refcit&r=1&refcit=258630&sort=Newest&vfpref=html to make a (non-strong) representation $\phi(x,y)$ such that $\phi(0,0)$ is independent of the theory, and then try to "strongify" that formula. However, I think I'm getting out of my depth now. Thank you again for clearing up my confusions in respect to total functions!
|
2025-03-21T14:48:31.810613
| 2020-08-18T19:31:55 |
369510
|
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|
Stack Exchange
|
Does there exist an arbitrarily large bipartite graph that is critical of dimension 4?
Here's a question in geometric graph theory that I recently kicked the tires on but was unable to resolve.
For a finite, simple graph $G$, say that $G$ is of dimension $n$, and write $\dim(G) = n$, if $n$ is the smallest integer such that $G$ can be drawn as a unit-distance graph in $\mathbb{R}^n$. Define $G$ to be dimension-critical if $\dim(G') < \dim(G)$ for any proper subgraph $G'$ of $G$. It is known that any bipartite graph $G$ has $\dim(G) \leq 4$, with examples of this upper bound being given by complete bipartite graphs of the form $G = K_{r,s}$ where $r,s \geq 3$.
My question is this:
Does there exist an arbitrarily large bipartite graph $G$ that is dimension-critical and has $\dim(G) = 4$?
Even if a full resolution is not given for the above question, I'd be happy to see some examples of a $K_{3,3}$-free bipartite graph of dimension 4, should such a thing exist.
What are the largest critical 4-dimensional bipartite graphs you know of?
Other than $K_{3,3}$, I can't guarantee the existence of any critical 4-dimensional bipartite graphs. One seemingly likely candidate that I considered is $K_{5,5}$ minus a 1-factor. Ultimately, however, I could find neither a unit-distance representation of this graph in $\mathbb{R}^3$ nor a proof that no such representation is possible.
Sorry if this is a silly question, but can't any bipartite graph be embedded into $\Bbb R^1$ with unit distance. Just put all vertices in the one partition class at $1$, and the others at $0$. Or do we need that the vertices are mapped to distinct points?
I should have been more clear on that. In all cases, the vertices of the graph must be distinct points of $\mathbb{R}^n$.
I think it may be useful to have a characterization of bipartite graphs of dimension 2. Is that available?
@Luca: I doubt an all-purpose characterization exists describing bipartite graphs $G$ with $\dim(G) = 2$. It's immediate that any such example cannot have $K_{2,3}$ as a subgraph, but that alone isn't sufficient. Consider $G$ formed by starting with two disjoint paths on $n$ vertices for odd $n \geq 5$, with those vertices labeled $a_1, ..., a_n$ and $b_1, ..., b_n$. Place edges $a_ib_i$ for $i = 1, ..., n$ along with the additional edges $a_1b_n$ and $a_nb_1$. It turns out that $\dim(G)=3$, and incidentally, this graph is what I used to answer the posted question in the case of $n=3$.
|
2025-03-21T14:48:31.810910
| 2020-08-18T21:41:10 |
369515
|
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|
Stack Exchange
|
Factorisation of a polynomial from the Boolean algebra
Let $B_n$ denote the Boolean algebra of a set with $n \geq 2$ elements and $C_n$ the matrix with entries $c_{i,j}=1$ if $i \leq j$ and $c_{i,j}=0$ else, where $i,j\in B_n$.
Let $M_n:=C_n+C_n^T$ and $f_n$ the characteristic polynomial of $M_n$ (this is the Cartan matrix of a Frobenius algebra associated to the incidence algebra of $B_n$, thus is appears quite naturally).
Question: Is it true that $f_n$ factors over $\mathbb{Q}$ in exactly $n+1$ different irreducible factors? What are the numbers appearing in the multiplicity of the irreducible factors?
This is true for $n \leq 14$.
For example for $n=6$ the 7 irreducible factors are:
(x - 2)^5 * (x + 1)^10 * (x - 1)^15 * (x^2 + 7x + 2) * (x^2 - 11x + 2)^5 * (x^2 - 5x + 2)^9 * (x^4 - 20x^3 - 161x^2 - 40x + 4)
For $n=8$ the 9 irreducible factors are:
(x - 2)^14 * (x + 1)^42 * (x - 1)^56 * (x^2 + 25x + 2) * (x^2 + 7x + 2)^7 * (x^2 - 11x + 2)^20 * (x^2 - 5x + 2)^28 * (x^4 - 44x^3 - 1313x^2 - 88x + 4) * (x^4 - 20x^3 - 161x^2 - 40x + 4)^7
The next two questions indicate that something interesting might be going on for the powers of the irreducible factors:
Question 2: The maximal power $p_n$ such that $(x-2)^{p_n}$ is a factor of $f_n$ starts for even $n \geq 2$ with $1,2,5,14,42,132$. Is it the Catalan sequence?
Question 3: The maximal power $q_n$ such that $(x-3)^{q_n}$ is a factor of $f_n$ starts for odd $n \geq 1$ with $1,2,5,14,42,132$ . Is it the Catalan sequence?
Sadly I can only test those things for $ n \leq 14$ at the moment.
One might find many more sequences appearing, for example for even $n$ the maximal power of (x^2 - 5*x + 2) appearing starts with 1, 3, 9, 28, 90 and might be given by https://oeis.org/A000245 .
The matrix $M_n$ commutes with the action of the symmetric group $S_n$ on $B_n$. In this action, the irrep indexed by the partition $(n-i,i)$, $0\leq i\leq \lfloor n/2\rfloor$, has multiplicity $n+1-2i$ (and no other irreps appear). One should try to compute the characteristic polynomial of the action of $M_n$ on each isotypic component. The multiplicity of each irreducible factor of $f_n$ must be a sum of dimensions of the irreps indexed by $(n-i,i)$. For instance, when $n=6$ the multiplicity 9 of $x^2-5x+2$ can only correspond to $\dim( 4,2)$, but in general there will be ambiguity.
Continuing my comment, the matrix $M_n$ actually commutes with the action of the larger group $G=S_n\times(\mathbb{Z}/2\mathbb{Z})$, where the generator of $\mathbb{Z}/2\mathbb{Z}$ is complementation. Exactly $n+1$ distinct irreps of $G$ appear in this action. This suggests that the characteristic polynomial of the action of $G$ on each isotypic component will have a "dominant" factor. All other factors are dominant factors of other isotypic components.
There is also a galois action of $Gal(\bar{\mathbb{Q}} /\mathbb{Q} ) $ on the eigenvalues of each isotypic component. Grouping eigenvalies in orbits produces exactly an irreducible factor over $\mathbb{Q}$. Since @RichardStanley pointed out there are exactly $(n+1) $ components, the thesis is equivalent to such galois action to be transitive on each (eigenvalues set of) isotypic component.
Maybe https://oeis.org/A000344 for $(x^2 -11 x+ 2)$ ? Part of https://oeis.org/A009766
For n=10, one gets $(x - 2)^{42} \cdot (x + 1)^{170} \cdot (x - 1)^{212} \cdot (x^{2} + 79 x + 2) \cdot (x^{2} + 25 x + 2)^{9} \cdot (x^{2} + 7 x + 2)^{35} \cdot (x^{2} - 11 x + 2)^{75} \cdot (x^{2} - 5 x + 2)^{90} \cdot (x^{4} - 100 x^{3} - 9601 x^{2} - 200 x + 4) \cdot (x^{4} - 44 x^{3} - 1313 x^{2} - 88 x + 4)^{9} \cdot (x^{4} - 20 x^{3} - 161 x^{2} - 40 x + 4)^{35}$.
For n=12, one gets $(x - 2)^{132} \cdot (x + 1)^{682} \cdot (x - 1)^{814} \cdot (x^{2} + 79 x + 2)^{11} \cdot (x^{2} + 25 x + 2)^{54} \cdot (x^{2} + 7 x + 2)^{154} \cdot (x^{2} - 11 x + 2)^{275} \cdot (x^{2} - 5 x + 2)^{297} \cdot (x^{4} + 220 x^{3} - 3041 x^{2} + 440 x + 4) \cdot (x^{4} - 100 x^{3} - 9601 x^{2} - 200 x + 4)^{11} \cdot (x^{4} - 44 x^{3} - 1313 x^{2} - 88 x + 4)^{54} \cdot (x^{4} - 20 x^{3} - 161 x^{2} - 40 x + 4)^{154} \cdot (x^{6} - 247 x^{5} - 63659 x^{4} + 797003 x^{3} - 127318 x^{2} - 988 x + 8)$.
|
2025-03-21T14:48:31.811426
| 2020-08-18T21:42:10 |
369516
|
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|
Stack Exchange
|
Comparing sets of twin primes with other sets. Why is there a max and min value?
I have taken 2 sets: The first is a consecutive list of the first prime of twin pairs. The second is a consecutive list of numbers as follow 1, 1+2, 1+2+3, 1+2+3+4, 1+2+3+4+5 ....
I have then compared between the lists by dividing the numbers of the second list with the numbers of the first list, and a steady growth rate of distribution occurs (as seen in the pictures below).
If you analyze the data (as seen in the pictures below), you will notice that:
If the fluctuation of column E is too high (usually above 1.1), then the "next" twin pair will have to be smaller than the "current:" pair, thus producing an error.
You can also notice that the fluctuation of column E is never too low (probably not less than 0.99 after the first few hundreds).
The same phenomenon happens if I replace Column C with the squares 1,4,9,16,… or with an arbitrary quadratic polynomial.
When replacing column C with a constant equal to 1, the max value never passes 1 (obviously). However, after the first few hundreds the min value again is probably not less than 0.99
Can anyone provide me with a theoretic explanation for why this might be?.
List of first 100,000 with column C: 1, 1+2, 1+2+3, 1+2+3+4 ....
List of first 100,000 with column C: with the squares 1,4,9,16,25 ...
List of first 100,000 with column C: constant = 1
Thanks.
These numbers look way too small to predict any long-term behaviour.
I have a list of the first 100,000, here is the link: http://numbersprime.com/crawlx.php
You do know there is a 'formula' (closed form) for the numbers in column C, right?
Yes, thank you. I would like to clarify: my question is not whether column C grows infinitely or not. My question is what could possibly explain the minimum of 0.99 after the first 100?
You might experiment to see whether you get the same phenomenon if you replace Column C with the squares $1,4,9,16,\dots$ or with an arbitrary quadratic polynomial
Here are the results for C with the squares 1,4,9,16: http://numbersprime.com/crawlxz.php
Here are the results if C would of been a constant equal to 1: http://numbersprime.com/crawlxzy.php
What is the motivation of this tangle of computations?
Let $B_2=3,B_3=5,\cdots $ be your sequence of "first member of a twin prime pair." For some reason starting at index $2.$ We don't know that this is an infinite sequence but strongly suspect that it is with $B_n \approx k n (\ln n)^2$ for some constant $k.$ There are conjectures on $k$ but that hardly matters here. So for a plausible explanation we can say that $\frac{B_n}{B_{n-1}}$ is definitely greater than $1$ but approaching it at a steady average pace. Perhaps with $1<\frac{B_n}{B_{n-1}}<\frac{n+8}{n-1}.$ Or, to be especially reckless, $\frac{B_n}{B_{n-1}} \approx \frac{n}{n-1}.$
The numbers $E_n$ you are analyzing are exactly $\frac{B_n}{B_{n-1}}\frac{n-1}{n+1}$ so there is your explanation for why they are sometimes above $1$ and sometimes below, with convergence to $1.$
Digression: After the first few pairs, every member of the sequence is $11,17$ or $29 \bmod 30.$ Perhaps this introduces a little clumpiness. I don't know. You might check if the over vs under $1$ behavior correlates to congruence class $\bmod 30$ being $11$ vs $17$ or $29.$ If so, does this behavior seem to continue or die out?
The sequence $C_1=1,C_2=3,\cdots $ of triangular numbers has $C_n=\frac{n(n+1)}2$ so $\frac{C_{n-1}}{C_{n}}=\frac{n-1}{n+1}$ exactly.
You define $D_n=\frac{C_n}{B_n}$ and then, for $n \ge 3,$ $$E_n=\frac{D_n}{D_{n-1}}=\frac{B_n}{B_{n-1}}\frac{C_{n-1}}{C_n}=\frac{B_n}{B_{n-1}}\frac{n-1}{n+1}\approx\frac{n}{n-1}\frac{n-1}{n+1} \rightarrow 1$$
If instead of twin primes you used primes, with $p_n \approx n\ln n,$ results should be about the same, possibly less choppy. If instead of triangular numbers you used squares you would have $\frac{(n-1)^2}{n^2}\approx \frac{n-2}{n}$ which is very close to $\frac{n-1}{n+1}$
The further steps of adding successive terms of a previous column or taking ratios gives sequences which converge to one or grow like $n.$
I understand your answer and accept your answer; yet I am sorry for my following naive question: I still don’t understand why the notion is that “Twin primes become increasingly scarce among larger numbers”, If the numbers in column E convergence to 1?
Squares are more predictable but MUCH rarer than twin primes. According to your data $B_{2000}=182,009$ $C_{2000}=2,003,001>10 B_{2000}.$ So for sure the squares get sparse. But their ratio goes to $1$ like $1+\frac2n.$ And the twin primes, while increasingly sparse (but not as sparse), have ratio going to $1$ like $1+\frac1n.$ Up to $x$ there are about $\sqrt{x}$ squares, $\frac{x}{\ln x}$ primes and $\frac{x}{\ln(x)^2}$ twin primes.
|
2025-03-21T14:48:31.811752
| 2020-08-18T21:58:03 |
369517
|
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|
Stack Exchange
|
Non-alternating knot diagrams
(I asked this on
MSE
a few days ago without a clear resolution.)
Start with a closed, self-intersecting curve, where every crossing is
transverse. Now form something like the opposite of an alternating knot
diagram as follows. Starting anywhere, traverse the curve, and
at each previously unvisited crossing, go over/above. If the crossing has been previously visited, leave the assigned crossing designation.
Two examples are shown below. (a) is clearly the unknot.
(b) is also the unknot, perhaps not as obviously.
Red circle indicates starting point, arrow the traversal direction.
I expected these diagrams to obviously represent the unknot, but
I am not seeing a clear proof. So:
Q. Prove (or disprove) that such a knot diagram always represents
the unknot.
This construction can be found in some textbooks on knot theory in the context of the unknotting number of a knot. For example it is described on pp. 58-59 of The Knot Book by Colin Adams and on p. 133 of Knot Theory by Charles Livingston. I have a dim memory of seeing it somewhere else as well. In Adams' book he gives an argument which is essentially the one in the answer by Wojowu below, except that Adams makes the unstated hypothesis that the starting point lies on the boundary of the convex hull of the knot projection, which simplifies his argument significantly.
I believe this fact was also used in the original proof that the skein-theoretic definition of the Homfly polynomial is well-defined.
Let us parametrize the plane curve by $\gamma:[0,1]\to\mathbb R^2$ and assume $\gamma(0)=\gamma(1)=(0,0)$. Then your curve is the knot diagram of the knot which is parametrized by $K:[0,2]\to\mathbb R^3$ given by
$$K(t)=\begin{cases}(\gamma(t),1-t)&\text{if }t\leq 1,\\(0,0,t-1)&\text{if }t>1.\end{cases}$$
(essentially, imagine suspending your knot on a stick, such that the rope goes down at a uniform speed.) Then we can "unwind" this knot. Namely, since $\gamma$ only goes through $(0,0)$ at the endpoints, we can write $\gamma(t)$ in polar coordinates by $(r(t),\phi(t))$ with $r,\phi$ continuous on $(0,1)$. We can then unknot $K$ by the following sequence of knots $K_s$, which starts with an unknot and ends with $K$, written in cylindrical coordinates:
$$K_s(t)=\begin{cases}(r(t),s\phi(t),1-t)&\text{if }t\leq 1,\\(0,0,t-1)&\text{if }t>1.\end{cases}$$
Likely this is my understanding. I am a bit concerned that your $\gamma(t)$ rises consistently, $t \le 1$, but in fact in (b) of the figure the curve encounters a crossing previously assigned, and so then it must go under; and then later above.
I think the parametrization here is in the opposite direction than in the figure.
I think this is essentially correct. We can construct $K$ as the union of a finite number of arcs ${K_i}{i=1}^n$ such that the $z$-coordinate of $K_i$ is monotonically increasing for each $i$, and then $K{i+1}$ is strictly below $K_i$ for every $i$, as well as arcs connecting the "top" point of $K_i$ with the "bottom" point of $K_{i+1}$ for every $1 \leq i \leq n$, with $K_{n+1} = K_1$. Each of the $K_i$ can be isotoped rel endpoints to a straight line, so the result should be unknotted.
@JanKyncl Oh yes, good point. I meant the curve to go down, but wrote it as going up.
@RohilPrasad I have slightly mixed up the formula, I meant the entirety of the curve to go down rather than up in the $z$-coordinate, and then the beginning and end going up to join the endpoints. I believe with this construction I do not need to split $K$ into arcs?
As far as I can tell, the parameterization of $K$ needs to change $z$-direction at least as many times as you have undercrossings. So you shouldn't be able to have it go up once and then go down if there is more than one undercrossing.
@RohilPrasad If we start drawing our knot in $\mathbb R^3$ at $z=1$, and then continually go down, then for any intersection point on the diagram, we will go over every intersection point of the original diagram twice, and the first visit will always be at a higher $z$ coordinate. If you then project the resulting knot onto the $z=0$ plane, the resulting diagram will be the non-alternating diagram as defined by the OP.
Oh, I finally get it! Thanks, your approach is cleaner than what I wrote in the comments above.
|
2025-03-21T14:48:31.812069
| 2020-08-18T22:08:04 |
369518
|
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|
Stack Exchange
|
Faltings' height theorem for isogenies over finite fields
For an Abelian scheme over a ring of integers in a number field, Faltings has a theorem that describes how the Faltings' height changes through an isogeny. There are multiple references for this statement and proof. See Proposition 4.1 in the notes Faltings height by Daniele Agostini (pdf), for instance.
I am interested in the analogous statement over curves over a finite field, in particular when the isogeny has degree a power of $p$, the characteristic of the finite field.
Is any such statement true in this case and if so, what's a reference?
Quick note: the mathematician's name is "Gerd Faltings", so it should be "Faltings's" or "Faltings'", not "Falting's". The notes you link consistently make that mistake in the references.
And so did I! Thanks for the correction.
I changed the link reference, because a blank 'this' sending the user to a pdf on a personal webpage is not stable.
Thanks, that's a good point! @davidroberts
|
2025-03-21T14:48:31.812190
| 2020-08-18T22:21:49 |
369520
|
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"url": "https://mathoverflow.net/questions/369520"
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|
Stack Exchange
|
Definability of ordinals in various signatures
Recently, I've been studying what the definable subsets of the countable ordinals "look like" from the perspective of bare-bones first order logic (not set theory) equipped with various ways to "access" the structure of the ordinals.
For example, we may have a signature consisting only of a 2-arity relational symbol $S$ which we interpret in a structure $\mathcal{A}$ with underlying set $\omega_1$ as the set of $(\alpha,\beta)$ such that $\beta$ is the successor of $\alpha$. We can then ask questions about which subsets of $\mathcal{A}$ are definable by first-order logic sentences with this signature, where a subset $S\subset\mathcal{A}$ is considered definable if there is a first order logic sentence $\phi(x)$ for which the set of satisfying assignments of $x$ is $S$. In our example, we can define the set of all countable successor ordinals via the formula $\exists y:S(y,x)$.
We can also ask questions like "what is the smallest ordinal $\alpha$ such that $\alpha$ is undefinable in the sense that $\{\alpha\}$ is undefinable" and such. In the example above, it's clear to see that in fact no ordinal is definable, so the smallest undefinable ordinal is zero. I am particularly interested in how the smallest undefinable ordinal grows as we have stronger and stronger signatures. For example, I have been able to convince myself that with the signature $\{<\}$ with the obvious interpretation in $\omega_1$ as the "less than relation", the smallest undefinable ordinal is $\omega^\omega$ (though I haven't written my argument out formally yet).
My question is: has anyone studied questions like these? Is it known what the smallest definable ordinal is for various other signatures, like $\{ADD(x,y,z)\}$ which is true on all $x,y,z$ so that $x+y=z$, or even other signatures with multiplication, exponentiation, veblen functions, or more? Are there any known generalizations of these ideas? Any help or related literature would be appreciated.
Thanks! Do you think if I extended his results I could publish a paper on it, or does that not seem paper worthy?
I was given the impression that Ehrenfeucht’s work was all that was done on this area, so extending it would already be saying something new and interesting. Are there more recent papers that you didn’t mention that I’d have to build on?
If I may ask, what keywords did you use? Logic is not my area of research, so I when I tried I couldn't find anything other than "normal" ordinal definability.
For the signature ${<}$, see https://mathoverflow.net/questions/35971/ , particularly the paper of Doner, Mostowski, and Tarski cited there.
I do not have enough reputation to add a comment. The following paper may be useful for you. It contains results extending the work of Tarski, Mostowski, and Doner, as well as some very nice historical overview and references.
Buchi, Siefkes - The Complete Extensions of the Monadic Second Order Theory of Countable Ordinals.
Weak monadic second order logic appears already in Ehrenfeucht's original work. Even if you are exclusively interested in first order results, (weak) monadic second order logic can play a role.
For example the first order theory of ordinal addition coincides with the first order theory of ordinal addition inside $\omega^{\omega^{\omega}}$ (by Ehrenfeuct), while $(\omega^{\omega^{\omega}},+)$ is a reduct of a generalised power of $(\omega,+)$ with 'exponent' being the weak monadic second order version of $(\omega^{\omega},<)$ (the Feferman-Vaught theorem is the correct tool to understand this). For more details there is Thomas - Ehrenfeucht, Vaught, and the decidability of the weak monadic theory of successor, the details here are all correct but I think the conclusions have some issues.
There is also more recent work on the automata side such as Cachat - Tree Automata Make Ordinal Theory Easy. I know nothing about the content of this but if you want a comprehensive overview of the area, this is maybe a starting point.
|
2025-03-21T14:48:31.812484
| 2020-08-18T23:55:27 |
369522
|
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|
Stack Exchange
|
Codimension two foliations with transverse surfaces
Suppose I have some closed $4$-manifold $X$ and a codimension-two foliation $\mathcal{F}$, as well as a closed surface $\Sigma$ of nonnegative self-intersection that is everywhere transverse to $\mathcal{F}$.
Then what kind of restrictions are there on the foliation $\mathcal{F}$? This question gives some answers in the case where $X$ is a complex surface and $\mathcal{F}$ is holomorphic, but I'm more interested in what happens in the real case.
For the special case that $\Sigma$ is the fiber of a surface bundle structure for $X$, the foliations you ask about are completely classified. That probably barely scatches the surface of the problem though.
What is the classification in this case?
Foliations transverse to the fibering of a fiber bundle with compact fiber, in any dimension and codimension, correspond to conjugacy classes of representations of $\pi_1(B)$, into $Homeo(F)$. Where $B$ is the base space and $F$ is the fiber. Let $\rho$ be such a representation, and let $\tilde{B}$ be the universal cover of $B$ and let $G = {(\alpha, \rho(\alpha)}$ where $\alpha$ ranges over $\pi_1(B)$. To construct the foliation, mod out $\tilde{B} \times F$ by $G$. The leaves descend from the product foliation of $\tilde{B} \times {point}$, and leafwise cover $B$ under bundle projection.
In this real case, there are few restrictions. Indeed, choose $\Sigma\subset X$ such that $X$ admits a smooth 2-plane field $\xi$ (not necessarily integrable) transverse to $\Sigma$. Then, it is easy to perturb slightly $\xi$ to make it integrable on a small neighborhood of $\Sigma$. Then, by a theorem of Thurston (Commentarii 1974), $\xi$, being of real dimension $2$, can be homotoped rel. $\Sigma$ to become integrable everywhere. You can even begin with extending $\xi$ to a partial foliation of your choice over any regular subset of $X$. So, the possibilities are enormous.
|
2025-03-21T14:48:31.812659
| 2020-08-19T01:12:53 |
369524
|
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|
Stack Exchange
|
Stability of displacement interpolation in optimal transport
Let $(X,d)$ be a complete separable metric space, and let $(\mathcal{P}_2 (X), W_2)$ be the space of probability measures on $X$ with finite second moments, equipped with the 2-Wasserstein distance. It is known that discrete measures are dense inside $(\mathcal{P}_2 (X), W_2)$ - namely, given any $\mu \in \mathcal{P}_2 (X)$, and $\delta>0$, one can find a discrete measure $\mu_\delta$ with $W_2 (\mu, \mu_\delta)<\delta$.
Now, let $\mu_0, \mu_1 \in \mathcal{P}_2 (X)$, and let $\mu_t$ be a $W_2$ geodesic connecting $\mu_0$ and $\mu_1$ (a.k.a. $\mu_t$ is a [not necessarily unique] displacement interpolation between $\mu_0$ and $\mu_1$). Is the displacement interpolation stable under discrete approximation? That is, can one pick discrete $\mu_{0,n}, \mu_{1,n}$ such that $\mu_{t,n}$ is close to $\mu_t$ for all $t\in[0,1]$?
The displacement interpolation $\mu_t$ should not be fixed a priori, due to nonuniqueness of Wasserstein Geodesics. Thus, the correct question should be: fix the approximating sequences $(\mu_{0,n}),(\mu_{1,n})$ and $W_2$ geodesics $\mu_{t,n}$, and ask if there exists one $\mu_t$ close to $\mu_{t,n}$ for $t \in [0,1]$.
This is a comment rather than an answer, but I could not post it as a comment. Anyway, something useful in this direction can be found in Lemma 4.4 https://arxiv.org/pdf/1609.00782.pdf which, combined with Proposition 4.8 of https://arxiv.org/pdf/1311.4907.pdf gives you $W_2$ close $\mu_{t,n}$'s.
Certainly there are non-uniqueness issues. I actually meant something like: given a Wasserstein geodesic $\mu_t$, can we produce sequences $(\mu_{0,n})$ and $(\mu_{1,n})$ such that $(\mu_{t,n})$ converges to $\mu_t$ in some suitable sense.
These links do look quite helpful, thank you!
I think is very hard to fix $\mu_t$ and produce, afterwards, approximating marginals $(\mu_{0,n}),(\mu_{1,n})$. If you stick to the other way, as in my answer, you can try to argue essentially by tightness to get $\mu_t$ from $\mu_{t,n}$. Finally, you just need to show that the limit $\mu_t$ is a $W_2$ geodesic. The first link I gave you follows this path. With this approach, is very hard to control whether you are converging to your fixed a priori Wasserstein geodesic, or to another.
There is not only the problem of uniqueness, but also of existence. Under the present assumptions, $\mu_{0,n}$ and $\mu_{1,n}$ may not be connected by a $W_2$-geodesic.
Absolutely right: either you fix $\mu_{0,n},\mu_{1,n}$ and $t\mapsto \mu_{t,n}$ as hypothesis of your problem, and and try to exhibit $\mu_t$, or you impose some additional properties on the base space $X$ (e.g. Geodesic Space) not to deal with existence issues. It depends on which statement you seek.
|
2025-03-21T14:48:31.812847
| 2020-08-19T01:19:13 |
369525
|
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|
Stack Exchange
|
To see that the fundamental class of a local complete intersection is independent of choice of regular sequence
In SGA 4½ ‘Cycle,’ Grothendieck defines (among other things) the fundamental class of a local complete intersection $Y\subset X$ ($X$ simply a noetherian scheme) of codimension $c$ locally as the cup-product of the fundamental classes of each divisor $D_i$ in a regular sequence. To argue that this cup-product does not depend on the choice of the regular sequence, he claims that the product only depends on the ‘flag’ $D_1\supset D_1\cap D_2\supset\ldots\supset Y$. To show this, he asserts the existence of a cup-product
$$H^*_{D_1}(X)\otimes H^*_{D_1\cap D_2}(D_1)\otimes\ldots\otimes H_Y^*(D_1\cap\ldots\cap D_{c-1})\longrightarrow H^*_Y(X).\tag{*}$$
To try to see the existence of such a product, first I begin to consider a product (for sheaves $\mathcal F,\mathcal G$ on $X$; we can take $\mathcal F=\mathcal G=\mathbf{Z}/n(1)=\mu_n$)
$$\Gamma_{D_1}(X,\mathcal F)\otimes\Gamma_{D_1\cap D_2}(D_1,\mathcal G)\longrightarrow\Gamma_{D_1\cap D_2}(X,\mathcal F\otimes\mathcal G).$$
This is fine: take a global section $s$ of $\mathcal F$ with support in $D_1$ and any section $t$ representing a global section of $\mathcal G$ restricted to $D_1$ so that the restriction of $t$ to $D_1$ has support in $D_1\cap D_2$; then the tensor product $s\otimes t$ does not depend on the choice of $t$ and indeed has support in $D_1\cap D_2$. My problem is when I try to derive this arrow to give a map
$$R\Gamma_{D_1}(X,\mathcal F)\otimes^L R\Gamma_{D_1\cap D_2}(D_1,\mathcal G)\longrightarrow R\Gamma_{D_1\cap D_2}(X,\mathcal F\otimes^L\mathcal G).$$
The matter of finding complexes adapted to both $\Gamma$ and $\otimes$ is not the issue: it’s the $R\Gamma_{D_1\cap D_2}(D_1,\mathcal G)$ term. To compute this I need to find an injective resolution $\mathcal I^\bullet$ of the restriction of $\mathcal G$ to $D_1$. But then $\mathcal I^\bullet$ is just a complex of sheaves on $D_1$ not given as the restriction of a complex on $X$, and so I can’t follow the recipe outlined above. How to see the product (*)? Thank you.
|
2025-03-21T14:48:31.813013
| 2020-08-19T04:04:39 |
369528
|
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|
Stack Exchange
|
Rigid non-algebraic manifolds
The famous Kodaira problem asks: whether a compact Kähler manifold can always be deformed to a projective manifold? In order to provide a counterexample, one way is trying to construct a rigid compact Kähler manifold which is non-algebraic, then it can never be deformed to a algebraic one, thus provide a counterexample to Kodaira problem, do anyone have such kind of examples?
That is a good question, but I doubt there is a known counterexample. I believe that Voisin's counterexamples to the Kodaira problem use blowing up of products of complex tori along special subvarieties (such as graphs of endomorphisms) that deform to families of complex tori (none of which is an Abelian variety by careful analysis of the cup product of the Hodge structures of the blowings up). You might double-check her work: perhaps for very special collections of endomorphisms, the "complex torus with endomorphisms" is rigid.
@Jason Starr this is where my question originated from, I vaguely remember I have seen Voisin or someone else has mentioned such an example, but I forget where it is, and I'm looking for it.
Huybrechts' paper about Voisin's work says:Inspired by Voisin’s examples, Oguiso studies simply-connected compact Kähler manifolds of dimension ≥ 4 which are not projective, but rigid, see https://arxiv.org/pdf/math/0312515.pdf
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2025-03-21T14:48:31.813243
| 2020-08-19T05:29:06 |
369529
|
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|
Stack Exchange
|
Can we pin down the "recursively closed" levels of $L$ by a first-order theory?
I'm broadly interested in problems of the form
Characterize those ordinals $\alpha$ which are not computable from any smaller ordinal
for some meaning of "computable." For example, the admissible ordinals arise as such a class: we have $L_\alpha\models\mathsf{KP\omega}$ iff $Col(\omega,\alpha)$ forces "For every $\beta<\alpha$ there is a copy of $\beta$ not computing any copy of $\alpha$." This is basically due to Sacks (+ a simple absoluteness argument).
The above result, however, uses a somewhat odd notion of "computable." Personally I think it's quite natural, but there's definitely a simpler one coming from higher recursion theory:
Say that an ordinal $\gamma>\omega$ is recursively closed iff for every admissible $\alpha<\gamma$ the supremum of the $\alpha$-recursive well-orderings of $\alpha$ is $<\gamma$.
In general recursively closed $\gamma$s need not be admissible (although of course the converse does hold except for $\omega$); see here. An admissible ordinal $\alpha$ is Gandy if the smallest recursively closed $\gamma>\alpha$ is admissible.
My question is whether we can characterize the recursively closed ordinals by a first-order theory, analogously to how we characterized the "generically Muchnik-incomputable-from-below" ordinals at the beginning of this question:
(Main question) Is there a first-order theory $T$ such that for all $\gamma>\omega$ we have $L_\gamma\models T$ iff $\gamma$ is recursively closed?
It would be enough if - uniformly in a non-Gandy admissible ordinal $\alpha$ - there were an $L_\alpha$-definable well-ordering of $\alpha$ of ordertype the supremum of the $\alpha$-recursive well-orderings of $\alpha$. As such, the following arises as a sub-question of the main question:
(Secondary question) Suppose $\alpha$ is a non-Gandy admissible ordinal. Must there be an $L_\alpha$-definable well-ordering of $\alpha$ which is longer than any $\alpha$-recursive well-ordering?
An affirmative answer to the secondary question would not yield an affirmative answer to the main question immediately, due to the lack of uniformity, but it would probably help. I personally suspect that even the secondary question has a negative answer, but I cannot prove it.
Note that the secondary question has a very strong positive answer for the specific case $\alpha=\omega_1$, since every ill-founded relation on $L_{\omega_1}$ has a descending sequence which is an element of $L_{\omega_1}$, and so in general we get a uniformly positive answer for "sufficiently $\omega_1$-like" ordinals - but this still leaves a large gap between "$\omega_1$-like" and Gandy.
I don't fully understand the question so I might be missing something. But writing $\beta_\alpha$ to denote the $\alpha$-th ordinal which is admissible (or a limit of those), it seems that by your definition $\beta_{\omega+1}$ would be "recursively closed". Same for $\beta_{\omega \cdot 2+1}$ etc. Is this intended (in the sense that it doesn't matter for the question)? Or perhaps you wanted to replace "iff for every admissible $\alpha<\gamma$" with "iff for every admissible (or limit of admissibles) $\alpha<\gamma$"?
Also, given the definition, it seems you want all ordinals (and obviously many more too) of the form $\beta_l$ (with $l$ being a limit ordinal) to be categorized as "recursively closed".
Also, regarding your secondary question, I have a (likely) trivial question about terminology. Is "$L_\alpha$-definable well-ordering of $\alpha$" synonymous to "$\alpha$ (or $L_\alpha$?)-arithmetic well-orderings of $\alpha$"? Anyway, one observation is that there exist countable, recursively inaccessible and non-gandy ordinals $\alpha$ such that, even though $\alpha$ is non-gandy, there exists an ordinal $\alpha<p<\alpha^+$ such that the supremum of $p$-computable well-orderings of $\alpha$ would be exactly equal to $\alpha^+$. But I don't know how $L_\alpha$-definable [continued]
would relate to any of this, since I don't have any understanding of it. That's why I don't know how this observation relates to the secondary question. But essentially $p$ would only be "slightly" larger than $\alpha$ though (e.g. $p$ would be very small compared $\alpha$-recursive well-orderings of $\alpha$). Note: Instead of posting another comment (with only one sentence), I deleted the last comment and copy-pasted with an added sentence towards the end.
|
2025-03-21T14:48:31.813531
| 2020-08-19T06:34:47 |
369533
|
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|
Stack Exchange
|
Property of triangle centers
$M$ is the intersection of 3 cevians in the triangle $ABC$.
$$AB_1 = x,\quad CA_1 = y,\quad BC_1= z.$$
It can be easily proven that for both Nagel and Gergonne points the following equation is true:
$$S = xyz / r,$$
where $S$ is the area of the triangle $ABC$ and $r$ is the radius of the inscribed circle.
I wonder what other triangle centers might possibly have the same property and what is the geometric place for them?
Also, please note that for the case where point $M$ is the centroid the formula looks as follows: $S = 2xyz/R$, where $R$ is the radius of the circumcircle. Substitution $x = b/2$, $y = a/2$, $z = c/2$ brings it back to the classic $S = abc/4R$. Perhaps, some other triangle centers might exist, so that this equation $S = 2xyz/R$ holds true for them as well. I wonder in what particular relation these hypothetical points might be to the centroid of $ABC$?
Fixed $xyz$ defines a cubic curve. There are some known triangle-related cubics, possibly the cubics $xyz=Sr$ and $xyz=SR/2$ were also studied.
So it must be a cubic that is passing through Nagel and Gergonne points and some other known triangle centers are probably lying on it as well.
I checked that the Triangle Center X(883) satisfies the condition =/, so that its isotomic conjugate X(885) must also satisfy the same condition and the curve in question is inevitably Tucker-Gergonne-Nagel cubic: https://bernard-gibert.pagesperso-orange.fr/Exemples/k013.html
"It is the locus of point M such that the cevian triangles of X(7) and M have the same area." This interpretation is a bit different from mine though. I wonder whether it is trivial or not that both geometric interpretations of Tucker-Gergonne-Nagel cubic are the same.
Please use TeX like $S = x y z/r$, not Markdown fakery like S = xyz/r, which reads badly (e.g., compare $a$ $a$ vs. a *a*). I have edited accordingly.
GeoGebra found X(7) X(8) X(506) X(507) and some more if you let outlying intersections of cevians.
PS: a bug was found in GeoGebra.
I hope it is fixed soon. [Edit: now fixed]
Geogebra seems to show that X(7) is the same point as X(506) and X(507)... it must be a bug! It shows X(883), X(885) correctly though.
You can use the Contact Us form to have your accounts merged, so you can freely edit your post.
This is just a coda to the above comments but too long for a comment. If $M$ has barycentric coordinates $(\lambda,\mu,\nu)$ (not necessarily positive and normalised so that $\lambda+\mu+\nu=1$), then both conditions reduce to a cubic equation of the form
$$ \frac{\lambda\mu\nu}{(\mu+\nu)(\nu+\lambda)(\lambda+\mu)} $$
is a constant which depends on the (shape of the) triangle and can easily be computed explicitly.
In order to verify if a given centre (with centre function $f$ from the Encyclopedia of Triangle Centers, normalised to be homogeneous with $f(a,b,c)+f(b,a,c)+f(c,a,b)=1$), it should be easy to write a small programme, say in Mathematica, to check this on the spot.
|
2025-03-21T14:48:31.813759
| 2020-08-19T06:40:00 |
369534
|
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|
Stack Exchange
|
What does $f^!$ do for line bundles on a curve?
Sorry for the rather basic question, I asked this on mathstackexchange but didn't get an answer.
I'm trying to understand how does the inverse exceptional image $f^!$ of a coherent sheaf look like.
Let's say for the sake of this question that $f:X\rightarrow Y$ is a finite flat morphism between schemes. In this case, my understanding is that $f^!$ is a functor from coherent sheaves on $Y$ to coherent sheaves on $X$, and that if $\mathcal{F}$ is a coherent sheaf on $Y$ then $f^!\mathcal{F}$ can be described as $\underline{Hom}_{\mathcal{O}_Y}(f_*(\mathcal{O}_X),\mathcal{F})$.
If $X$ and $Y$ are affine then it's not hard to work out what this does for a coherent sheaf on Y.
As another basic case, I'm trying to understand what happens if $X$ and $Y$ are smooth projective curves, and $\mathcal{L}$ is a line bundle on $Y$, in terms of divisors. Say $D=\sum{P_i}$ is a divisor on $Y$, and $\mathcal{L}(D)$ is the corresponding line bundle. My understanding is that if $f$ is etale then $f^! = f^*$, so that in that case $f^!(\mathcal{L}(D))=\mathcal{L}(D')$ where $D'=\sum_i\sum_{f(Q)=P_i}{Q}$.
Can we understand $f^!(\mathcal{L}(D))$ in general (by which I mean, when $f$ is assumed finite flat but not necessarily etale)? (can we say something more specific than $f^!\mathcal{L}(D)=\underline{Hom}_{\mathcal{O}_Y}(f_*(\mathcal{O}_X),\mathcal{L}(D))$, in terms of the divisor $D$)? And how should I think of the functor $f^!$ for general coherent sheaves, in this setting?
Thanks!
Your question is imprecise, I will assume that $X$ and $Y$ are smooth curves. Then $f^{!}\mathscr{F}=K_X\otimes f^*(\mathscr{F}\otimes K_{Y}^{-1})$ for any coherent sheaf $\mathscr{F}$ on $Y$. See any book on Grothendieck duality.
Whoops, meant to assume $X$ and $Y$ are smooth curves in the second part of the question. I added the edit now. Thanks!
Any recommended books on Grothendieck duality?
|
2025-03-21T14:48:31.813904
| 2020-08-19T07:22:29 |
369536
|
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|
Stack Exchange
|
Are two definitions of "smooth planar domain" related?
Let's consider non-empty open subsets of the real plane.
Definition 1. Suppose $\gamma: S^1 \to \mathbb R ^2$ is a $C^{k,\alpha}$ Jordan curve ($0 \leq \alpha<1$) with $\gamma ' $ non-vanishing. The bounded component of $\mathbb R ^2 - \gamma(S^1 )$ may be called a $C^{k,\alpha}$ planar domain.
Definition 2.
However, there is a "standard" definition of "smoooth domains", namely, a non-empty open set $U \subseteq \mathbb R^2$ is called $C^{k,a}$ if for each $z \in \partial U$ there exists an $r>0$ and a $C^{k,\alpha}$ function $f:\mathbb R \to \mathbb R$ such that $B(z,r) \cap U = B(z,r) \cap \{(x,y):y>f(x)\} $(upto rotations and reflections of the coordinate axes).
It is clear that the latter may not imply the former, since the former will require the domain to be simply connected. I am interested in the following questions.
Does the former implies the latter? (I am particularly interested in this direction.)
Does the latter implies the former, provided that $U$ is simply connected?
I tried googling with keywords something like "planar domain smoothness" "smooth Jordan domains" etc. but I could not find definitive answers to this question, although this question seems to be very natural.
(I suspect that some form of Caratheodory's extension theorm for conformal mappings may be useful... )
Thanks to everyone in the community.
1 is always true due to the implicit function theorem. 2 is true is you add the assumption that $U$ is bounded (and simply connected), otherwise a half-plane is a counter-example.
@MoisheKohan Thanks for your comment. Maybe I am being stupid, but how do I use the implicit function theorem? Could you please elaborate your comment as an answer?
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2025-03-21T14:48:31.814047
| 2020-08-19T08:54:25 |
369540
|
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|
Stack Exchange
|
Deformations of vector bundles and tubular neighborhood
I had a number of questions that are somewhat related to each other. I decided to post them altogether instead of separately. I'd appreciate any kinds of answers, ideas or sources regarding any of them. Throughout these questions $X$ is a smooth variety and $Z$ is a closed smooth subvariety (you can assume it is of codimension 1). The complement is denoted by $U$.
Q1) This is my main question which is a special case of Q4 but I bring it first because of its importance to me. Here assume $X=Y\times \mathbb{A}^1$ where $Y$ is a smooth projective variety over a field. $Z=Y\times \{0\}$. Given two vector bundles on $X$ where they are isomorphic on $U$ and $Z$, is it possible to deform one of them to other, where throughout the deformation the restriction to $U$ and $Z$ remain the same vector bundles? (In other words find a vector bundle on $X\times \mathbb{A}^1$ where its restrictions at times $0$ and $1$ gives us the original two vector bundles and the restriction to $U\times \mathbb{A}^1$ and $Z\times \mathbb{A}^1$ are just extended vector bundles from $U$ and $Z$ respectively.). Note that here the obstruction in $K_0$ or chern classes vanishes automatically.
Q2) Let's denote the formal completion of $X$ along $Z$ by $\hat{X}_Z$. If $E$ is a vector bundle on $Z$ is there any nice obstructions for the extension of $E$ to $\hat{X}_Z$? If so and this obstruction vanishes is there any nice description of all these lifts? (For example: Do they form certain vector space or not?)
If we replace the formal completion with order 1 thickening there is such a description where the obstruction lives in $H^2(Z,I\otimes \mathcal{E}nd(E))$ where $I$ is the ideal corresponding to $Z$. You can find more details here page 34 Theorem 5.3.
Q3) Given two extensions of $E$ to $\hat{X}_Z$, denoted by $E_1$ and $E_2$, is it possible to deform $E_1$ to $E_2$ while the restriction to $Z$ should be $E$ throughout the deformation? This means finding a vector bundle on $\hat{X}_Z\times \mathbb{A}^1$ such that it is the extension of $E$ on $Z\times \mathbb{A}^1$ and gives us vector bundles $E_1$ and $E_2$ at times $0$ and $1$. I'd imagine something like this should follow by some sort of affineness of all extensions of $E$ given the obstruction is zero(Basically Q1). Somewhat similar to deforming any two extensions to each other, in the $Ext$ group.
Q4) Given two vector bundles $V_1$ and $V_2$ on $X$ such that $V_1|_{Z}\cong V_2|_{Z} \cong E$ and $V_1|_{U}\cong V_2|_U \cong F$. Is it possible to deform $V_1$ to $V_2$ while throughout the deformation the restriction to $Z$ should be isomorphic to $E$ and the restriction to $U$ should be isomorphic to $F$? (With the similar definition of deformation given in Q2). An obvious obstruction is $K_0$ so let's assume $V_1$ and $V_2$ have the same image in $K_0(X)$
There is a descent that is given in this paper, basically claims any coherent sheaf can be uniquely determined by its restriction to $U$, the tubular neighborhood $\hat{X}_Z$ and some sort of patching data in the punctured tubular neighborhood. The punctured neighborhood is defined using Berkovich spaces which I'm not too familiar with them. But I think it should boil down to Q2 and deforming in the punctured nbhd (if true).
|
2025-03-21T14:48:31.814259
| 2020-08-19T09:03:27 |
369541
|
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|
Stack Exchange
|
What is the difference between Path $\infty$-groupoid and Smooth Fundamental $\infty$-groupoid of a smooth space?
A couple of days back I asked a question Is there a Geometric/Smooth version of Homotopy Hypothesis using the path $\infty$-Groupoid of a Smooth Space? in MO about the existence of a possible Smooth/Geometric version of Homotopy Hypothesis using the notion of Path $\infty$-groupoid of a smooth space.
After a discussion in the comments section with @David Roberts I got a feeling (but not completely convinced) that though Path 1-groupoid and smooth fundamental 1-groupoid of a smooth space are quite different objects but "if we move upto infinity level" and present them as Kan Complexes then they are becoming the same object.
3 months back I asked the following MO question What is the geometric realization of the the nerve of a fundamental groupoid of a space?.
From the discussions in
Is there a Geometric/Smooth version of Homotopy Hypothesis using the path $\infty$-Groupoid of a Smooth Space?
What is the geometric realization of the the nerve of a fundamental groupoid of a space?
now I have the following Questions/Doubts:
We know that the construction of Smooth Fundamental 1-Groupoid and Path 1-Groupoid of a smooth space induce natural functors $Man \rightarrow Groupoids$. Now from the discussion in What is the geometric realization of the the nerve of a fundamental groupoid of a space? I expect that $|N \circ \pi_{\leq 1}(X)|$ contains all informations of the 1st Homotopy groups of the smooth space $X$ where $N$ is the Nerve functor, $\pi_{\leq 1}$ is the Smooth Fundamental 1-Groupoid functor and $|-|$ is the Geometric realization functor. Now we can repeat the same procedure with Path 1-Groupoid functor $\pi'_{\leq 1}: Man \rightarrow Groupoids$.
My questions are the following:
Is $|N \circ \pi_{\leq 1}(X)|= |N \circ \pi'_{\leq 1}(X)|$? (where "$=$" is in an appropriate sense)
Is there a way to present a Path $\infty$-groupoid of a smooth space such that it is different from Smooth Fundamental $\infty$-groupoid of the space? (So that it matches our intuition for $n=1$ case)
(By "$n$" I mean "Groupoids in the level 1").
I can only answer your first question, and the answer is no. Take for example $X=\mathbb{R}^2$, so that the fundamental groupoid is trivial, but the path groupoid contains distinct arrows represented by circles of every positive radius passing through a fixed basepoint (and many many more besides). This is ignoring all questions of topology or smooth structure on the set of arrows, which I think is your intent. And so the geometric realisations of the nerves of these cannot even be weakly homotopy equivalent, as one is contractible and one has fundamental group that is not even finitely generated.
Thank you Sir for answering my first question.
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2025-03-21T14:48:31.814438
| 2020-08-19T09:05:36 |
369542
|
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|
Stack Exchange
|
Tor and inverse images
Let $X$ and $Y$ be two schemes over $\mathbb C$, $E$ a coherent sheaf on $X$, $F$ a coherent sheaf on $Y$, and $p:X\times Y\to X$, $q:X\times Y\to Y$ the projections. Is it always true that $\text{Tor}^i(p^*(E),q^*(F))=0$ for $i>0$ ?
Yes, that is true. There is much more about "Tor independence" in SGA 6 (and even more in Thomason-Trobaugh). In this case, you can work locally on affine schemes. Since $F(Y)$ is a free $k$-module (every $k$-module is free), also $q^*F(X\times Y)$ is a free $\mathcal{O}_X(X)$-module. Thus, when you form a free $\mathcal{O}_X(X)$-resolution of $E(X)$ and tensor this over $\mathcal{O}_X(X)$ with $q^*F(X\times Y)$, this complex is still acyclic. This complex is precisely the sections over $X\times Y$ of the tensor product of $q^*F$ and the pullback by $p$ of the free resolution of $E$.
@Jason Starr. Thank you very much.
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2025-03-21T14:48:31.814534
| 2020-08-19T09:30:07 |
369543
|
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|
Stack Exchange
|
Maximum number of tuples from $n$ numbers such that no pair is repeated
What is the maximum number of $k$-tuples($3\le k\le n$) of $n$ numbers such that no pair is repeated in any of the tuples?
The maximum of number of $k$-tuples occur when $k=\lfloor\frac{n}{2}\rfloor$ as $n\choose \lfloor\frac{n}{2}\rfloor$ is maximum among all binomial coefficients. But, when we give the additional constraint that no pair is repeated twice, I have a hard time in getting the answer. For example, for $n=6$, I think the answer is $4$: the tuples desired are $(123),(145),(246),(356)$. Is it possible that the maximum in this case occurs for $k\neq\lfloor\frac{n}{2}\rfloor$? Any hints? Thanks beforehand.
See https://en.wikipedia.org/wiki/Block_design
@MaxAlekseyev thanks. what is $r$ of the block design here? I dont put any condition on number of blocs containing any number. And, by the way, isnt my answer right?
@vidyarthi: Block design represents a uniform construction. Like in your example for $n=6$, you have $r=2$.
@MaxAlekseyev but, $r=2$ does not satisfy $\lambda(n-1)=r(k-1)$ right?
@vidyarthi: It does not because not every pair appears in your example. E.g., there is no pair $1,6$.
@MaxAlekseyev so then, the maximum given by my answer is right?
@vidyarthi: Yes. Each number appear in at most $\lfloor (n-1)/(k-1)\rfloor$ blocks. Then the number of blocks is at most
$$\left\lfloor \frac{n}{k}\left\lfloor \frac{n-1}{k-1}\right\rfloor\right\rfloor.$$
You did achieve this bound in your example for $n=6$.
@MaxAlekseyev no, that was ok. I asked about my answer posted below. If not, could you give me an example for, say $n=8$
This question is a special case of https://mathoverflow.net/questions/175969/an-upper-bound-on-families-of-subsets-with-a-small-pairwise-intersection/176462#176462 .
@BrendanMcKay ...which further links to https://mathoverflow.net/q/161159/41291 which in turn links to https://mathoverflow.net/q/160787/41291
|
2025-03-21T14:48:31.814686
| 2020-08-19T09:56:44 |
369548
|
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"Arno Fehm",
"Nicolás",
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|
Stack Exchange
|
Prove that a translate of a polynomial has roots without using FTA
Let $F\in\mathbf C[Z]$ be a polynomial of degree $d$. Without using the Fundamental Theorem of Algebra, is it true that there is some $u\in\mathbf C$ such that $F + u$ has $d$ roots (counted with multiplicity)?
Thanks!
Nicolás
What you mean by "without using the fundamental theorem of algebra"? What properties of $\mathbb{C}$ do you want to use? Certainly the claim will not be true over arbitrary fields.
The statement is a weak form of the FTA. I am wondering if there is a simpler proof of this statement, without proving the full FTA.
This seems quite unlikely to me. The proofs that I know of FTA show that a polynomial $F(Z)$ has one root $\alpha$. One then applies induction on degree to $F(Z)/(Z-\alpha)$. Now perhaps one can show simply that $F(Z)+u$ has a root $\alpha$, but then one needs to show that $(F(Z)+u)/(Z-\alpha)$ itself (rather than $(F(Z)+u)/(Z-\alpha)+v$) has a root.
Thanks for the answers!
|
2025-03-21T14:48:31.814790
| 2020-08-19T11:23:51 |
369552
|
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|
Stack Exchange
|
Fastest algorithm for finding an expression of a number as a sum of 4 or less primes
We know from the proof of the ternary Goldbach problem, that every positive integer $\ge 2$ can be expressed as a sum of $4$ or lesser number of primes.
Can anyone share some fast algorithms that are known for writing a number as a sum of $4$ or less primes ?
Anyone knowing any algorithm which works faster than the naive algorithm for computing sums of primes, is welcome to share the details, or reference.
How many primes? You can write it as a bunch of $2$s, and a $3$ if the number is odd.
I meant $ \le 4$ many primes. I have fixed the question. Thanks !
|
2025-03-21T14:48:31.814887
| 2020-08-19T11:29:51 |
369553
|
{
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"authors": [
"Fernando Muro",
"Gabriel C. Drummond-Cole",
"Javi",
"LSpice",
"Tim Campion",
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|
Stack Exchange
|
Conceptual explanation for the sign in front of some binary operations
In several situations, I've seen that given a binary operation on a graded module $m:A\otimes A\to A$, a new operation $M(x,y)=(-1)^{|x|}m(x,y)$ is defined so that it satisfies some properties.
One example of this happens in Homotopy G-algebras and moduli spaces, where for a binary operation $m\in\mathcal{O}(2)$ such that $m\circ m=0$ for some operad $\mathcal{O}$, an associative product is defined by $xy=(-1)^{|x|+1}m\{x,y\}$, where the brace notation stands for the brace algebra structure on $\mathcal{O}$. In this case, the explanation I've been able to deduce is that this is necessary for the brace relation (equation (2) in the paper) to imply associativity of the product $xy$. In this case the sign $(-1)^{|x|}$ works for this purpose too.
Another more direct instance of this situations occurs in Cartan homotopy formulas and the Gauss-manian connection in cyclic homology, where given an $A_\infty$-algebra with $m_i=0$ for $i>2$, one gets a dg-algebra by defining again $xy=(-1)^{|x|}m_2(x,y)$. In this case this is because the author uses a convention for $A_\infty$-algebras in which the equations only have plus signs, so some extra sign is needed to produce the associativity relation and the Leibniz rule. So the reasons are very similar to the previous case even though the construction is simpler because there is no brace algebra here.
And another extra example for which I don't have any reference is in the case of Lie algebras. When one defines a generator of the operad of graded Lie algebras, often one takes $l(x,y)=(-1)^{|x|}[x,y]$ instead of directly defining $l$ as the bracket. If I remember correctly this was needed to obtain the Jacobi identity in purely operadic terms.
So it looks like it's very common to add that sign in order to make some relations hold. What I would like to know if there is a more conceptual explanation of why this holds systematically. Maybe it's just that it works when writing down the equations, but I'm looking for a more general intuition.
My motivation is generalizing this idea to maps of higher arity. More precisely, given an $A_\infty$-multiplication $m\in\mathcal{O}$ such that $m\circ m=0$, I want to define an $A_\infty$-structure $M$ on $\mathcal{O}$ that satisfies the sign convention
$$\sum_{n=r+s+t}(-1)^{rs+t}M_{r+1+t}(1^{\otimes r}\otimes M_s\otimes 1^{\otimes t})=0.$$
(There is also another possible convention where $rs+t$ is replaced by $r+st$)
So this is very similar to Getzler's paper where he defines $M_j(x_1,\dots, x_j)=m\{x_1,\dots x_j\}$, and this structure maps satisfy the relation $M\circ M=0$ but with all plus signs. So I need to modify these maps by some signs in a similar way as the associative case. Of course I can try to sit down and write the equations and find some necessary conditions for the signs and maybe find a pattern. But if there is a conceptual explanation for the associative case and the lie algebras, then maybe there is an easier way to find out what the signs I need are.
I once heard Raghuram quote Harder as saying "when you are dealing with signs, you are doing very difficult mathematics." After a life of signs, that made me feel good about my work. It doesn't seem directly related to your question, but I reflectively link Lawson - In which I try to get the signs right for once in any such question.
@LSpice I think I've already seen your work before, maybe having a look will help me find some ideas, even if it's not for this particular question, probably for future questions, so thank you.
In order to avoid some other signs, sometimes people define $A$-infinity algebras not using the $A$-infinity operad, but its operadic suspension. That causes those signs you ask about.
(Re, just to be clear, the work I linked is by Tyler Lawson, not me. I also meant that I reflexively link it, not that I reflectively link it ….)
@FernandoMuro I know, and I've done that, I just didn't write it like that because I didn't want to introduce more notation. I actually want my $M_j$'s to satisfy $M\tilde{\circ}M=0$ where $\tilde{\circ}$ is the composition in the operadic suspension of the endomorphism operad, but that's the same in this case as satisfying the $A_\infty$-equation with signs. And indeed I've defined them as maps taking elements of $\mathfrak{s}\mathcal{O}$ (a shift $\Sigma\mathfrak{s}\mathcal{O}$ actually). I'm doing this because of degree reasons, I can tell you in more detail if you want.
@FernandoMuro so put it in a different way, the motivation of my question would be: assuming that I have $A_\infty$-maps $M_j\in \End_{\Sigma\mathfrak\mathcal{O}}$ (so they satisfy $M\circ M=0$ here), how can I deform them to obtain maps $M_j'\in\mathfrak{s} \End_{\Sigma\mathfrak\mathcal{O}}$ satisfying $M'\tilde{\circ}M'=0$?
It seems to me you're asking for some explanation of the Koszul sign rule. Maybe this question helps? Or perhaps I misunderstand.
@TimCampion I am not asking about that and that question is indeed mine. In this case it doesn't seem to be any Koszul rule being applied since no symbols are being permuted. It is just adding a factor sign to make some things work.
@Javi Ah, I see, sorry. I think I was confused by the Lie algebra example. I was sure the definition of a graded Lie algebra was just an application of the Koszul sign rule...
@TimCampion Yes, I think it is. And the Koszul rule does have a role in the relations thad need to hold. So maybe there is a rule in disguise, but it's not obvious if that's the case. The thing about the graded Lie algebra is that the operation $l$ satisfies the relation $l\circ_1 l+(l\cdot(132))\circ_1 l+(l\cdot (123))\circ_1 l=0$, where the $\cdot (abc)$ is the right action of the cycle $(abc)$. And if you define $l(x,y)=[x,y]$ I think you need to make reference to elements in that relation (which is the Jacobi identity but expressed in operadic language).
I think that your example is indeed the Koszul sign rule and it has to do with the conventions one chooses for dealing with a suspension. Signs like those ones arise when you compose two operations of the form s^{-1} m_i s^{\otimes i} and s^{-1} m_j s^{\otimes j} and compare them to s^{1} (m_i \circ_k m_j) s^{\otimes i+j-1}. Further differences arise because e.g. you might realize "s" as tensoring with something on the right while I realize it as tensoring on the left.
Yes, that's more or less how I derived the sign convention for $A_\infty$-algebras. Maybe the explanation of the factor sign is also there, kind of like the differential of a chain complex when it's shifted,
As Gabriel C. Drummond-Co commented, it has to do with suspensions that are implicit. I'll do it with the example of Gerstenhaber and Voronov and the others should follow similarly. Let us denote $M_2(x,y)=x\cdot y$ the product that we want to define based on the brace $m\{x,y\}$. If we define it as a map $(s\mathcal{O})^{\otimes 2}\to s\mathcal{O}$ (suspension as graded vector spaces), then the natural thing to do is using the brace $m\{-,-\}:\mathcal{O}^{\otimes 2}\to \mathcal{O}$, but to do so one has to compose with suspensions and desuspensions. Namely, $M_2(x,y)=s(m\{(s^{-1}x,s^{-1}y)\})$. And it's applying $(s^{-1})^{\otimes 2}(x,y)$ what makes the sign $(-1)^{|x|}$ appear. If we use $(s^{\otimes 2})^{-1}$ instead then we get the original sign $(-1)^{|x|+1}$.
I find the question quite interesting (in the sense that similar questions related to sign factors appearing in various different algebraic structures with no apparent reason, have been going through my studies for quite some time in the past..)
Although i am not really familiar with most of your examples, since you are also mentioning associative and Lie algebras, i will refer to a similar "phenomenon" from graded algebras: This has to do with the $\mathbb{Z}_2$-graded tensor product, between two associative superalgebras ($\mathbb{Z}_2$-graded algebras) $A$ and $B$. If $b$, $c$ are homogeneous elements of $B$ and $A$ respectively, then the so-called super tensor product algebra or $\mathbb{Z}_2$-graded tensor product algebra, of superalgebras, is the superalgebra $A\underline{\otimes} B$, whose multiplication is given by
$$
(a \otimes b)(c \otimes d) = (-1)^{|b| \cdot |c|}ac \otimes bd
$$
with $|b|, |c|\in\mathbb{Z}_2$. Here the sign factor, reflects the braiding of the monoidal category of representations of the group hopf algebra $\mathbb{CZ}_2$: Recall that, superalgebras can be alternatively viewed as algebras in the the braided monoidal Category ${}_{\mathbb{CZ}_{2}}\mathcal{M}$ (i.e. the Category of $\mathbb{CZ}_{2}$-modules) and that the above multiplication can be abstractly written as:
$$
m_{A\underline{\otimes} B}=(m_{A} \otimes m_{B})(Id \otimes \psi_{B,A} \otimes Id): A \otimes B \otimes A \otimes B \longrightarrow A \otimes B
$$
Here, the braiding is given by the family of natural isomorphisms $\psi_{V,W}: V\otimes W \cong W\otimes V$ explicitly written:
$$
\psi_{V,W}(v\otimes w)=(-1)^{|v| \cdot |w|} w \otimes v
$$
where $V$, $W$ are any two $\mathbb{CZ}_2$ modules.
Furthermore, this braiding is induced by the non-trivial quasitriangular structure of the group Hopf algebra $\mathbb{CZ}_{2}$, given by the $R$-matrix:
\begin{equation}
R_{\mathbb{Z}_{2}} =\sum R_{\mathbb{Z}_{2}}^{(1)} \otimes R_{\mathbb{Z}_{2}}^{(2)}= \frac{1}{2}(1 \otimes 1 + 1 \otimes g + g \otimes 1 - g \otimes g)
\end{equation}
through the relation: $\psi_{V,W}(v \otimes w) = \sum (R_{\mathbb{Z}_{2}}^{(2)} \cdot w)
\otimes (R_{\mathbb{Z}_{2}}^{(1)} \cdot v)=(-1)^{|v| \cdot |w|} w \otimes v$.
For yet another point of view, the above mentioned $R$-matrix can be considered to be "generated" by the corresponding bicharacter (or: commutation factor) of the $\mathbb{Z}_2$ group.
There are bijections between $R$-matrices, braidings and bicharacters (which here are actually commutation factors) in the braided, graded setting for either assoc or Lie braided ("colored" is another name), graded algebras.
All these can be generalized for graded algebras, gradings and braidings, or $R$-matrices, or bicharacters of the corresponding groups, for any finite, abelian group. Also for $\mathbb{G}$-graded, $\theta$-colored Lie superalgebras, to produce more complicated bicharacters $\theta:\mathbb{G}\times\mathbb{G}\to k$ (which in the example above where $\mathbb{G}=\mathbb{Z}_2$ is exactly the sign factor of the $\mathbb{Z}_2$ abelian group).
To conlude: the sign factors here, are an "implicit" appearance of the corresponding group bicharacters. And they can also be viewed as braidings of the corresponding category of representations or as $R$-matrices for the corresponding quaitriangular group hopf algebras (of the fin, abelian, grading group).
If you are interested in these examples and you consider them relevant to your question, you can also take a look at the description in this answer: https://mathoverflow.net/a/261466/85967 and my linked paper therein.
Thank you for your answer. In the case of that tensor product I think it's natural to obtain that sign factor, as well as in many other situation in which the braiding acts. But in the case of my question I don't see how the braiding could pop up, since no elements are being permuted.
|
2025-03-21T14:48:31.815703
| 2020-08-19T12:02:05 |
369558
|
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|
Stack Exchange
|
Problem of Thickening an Arc in a Topological $ 2 $-Manifold
Let $ M $ be a topological $ 2 $-manifold (possibly with boundary), $ C $ an arc in the interior of $ M $ (i.e., an injective continuous function from $ [- 1,1] $ into $ \operatorname{Int}(M) $), and $ K $ a compact subset of $ M $ that is disjoint from $ \operatorname{Range}(C) $.
Problem. Prove that there exists a continuous embedding $ f: [- 2,2] \times [- 1,1] \to M $ with the following properties:
For all $ x \in [- 1,1] $, we have $ f(x,0) = C(x) $.
$ \operatorname{Range}(f) \cap K = \varnothing $.
I think that the Jordan-Schoenflies Theorem is needed to solve this problem, but I do not know how to do it. Thank you!
See D.B.A. Epstein, Curves on 2-manifolds and isotopies, Acta Math., 1966.
@MoisheKohan: Thanks! However, Epstein does assume that the topological $ 2 $-manifold comes with a triangulation, which I can’t.
Every surface admits a triangulation. This is a theorem due to Rado. A standard reference is "Riemann Surfaces" by Ahlfors and Sario.
@MoisheKohan: Ah! I forgot to mention that I’m supposed to solve the problem without assuming that topological $ 2 $-manifolds (with or without boundary) can be triangulated. The reason for my post is that a similar result was used by P. H. Doyle and D. A. Moran to establish the main objective of their paper A Short Proof that Compact $ 2 $-Manifolds Can Be Triangulated, but they neglected to prove that result, merely saying that it was a standard result in geometric topology.
I see. Then (unless they/you are doing something seriously wrong) you should not need the result about topological arcs in full generality, only for arcs contained in coordinate neighborhoods, which are already known to be homeomorphic to open planar sets in which case you can quote the planar simple arc theorem.
|
2025-03-21T14:48:31.815869
| 2020-08-19T13:32:19 |
369565
|
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|
Stack Exchange
|
Rota's cross-cut theorem - special case?
A question on accurate attribution/description:
Lemma.- Let $Q$ be a collection of subsets of a finite set $V$, and let $S_0\subset Q$. Then $$\left|\mathop{\sum_{S_0\subset S\subset Q}}_{\bigcup S = V} (-1)^{|S|}\right|\leq 2^{|V|},$$
where we write $|X|$ for the number of elements of a finite set $X$.
The proof isn't hard (basically three lines - use inclusion-exclusion, then change the order of summation, then use inclusion-exclusion again). Taking a look at the proof of Rota's cross-cut theorem, it seems to me that the idea of the proof is the same in both cases.
Question: is it accurate to call this Lemma a special case of Rota's cross-cut theorem? (Question 2: is it really just a rephrasing of Rota's cross-cut theorem, i.e., equivalent to it?)
I should add that the connection to Rota's cross-cut theorem was pointed out to me in an answer to https://mathoverflow.net/questions/364743/alternating-sum-over-collections-of-sets
This might depend on what you mean by the cross-cut theorem. The version for the Mobius function is a consequence of the topological cross cut theorem which says the order complex of a poset is homotopy equivalent to a certain simplicial complex built from the cross cut. The Mobius function result then follows from computing Euler characteristics.
I meant the cross-cut theorem for the Möbius function.
I normally think of Rota's crosscut theorem not as an inequality but as an equality (giving a formula for the Möbius function), so for that reason alone I would hesitate to call your lemma a "special case" of Rota's theorem. As for whether it's equivalent, Rota's theorem applies to an arbitrary geometric lattice, and unless I'm missing something, it doesn't seem that an arbitrary geometric lattice can be easily reduced to the setting of your lemma as stated. You might try emailing Bruce Sagan directly because he has thought a lot about generalizations of Rota's theorem.
So, should I call it an immediate consequence of Rota's crosscut theorem (to which I might as well give a self-contained proof)? Or is it not really an immediate consequence?
|
2025-03-21T14:48:31.816039
| 2020-08-19T14:16:00 |
369568
|
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|
Stack Exchange
|
A functional integral inequality
Suppose $f:I=(0,1)\to \mathbb R$ is a continuous function that satisfies
$$ \int_I f(t) e^{at}\,dt \geq 0\quad \text{for all $a \in \mathbb R$}.$$
Does it follow that $f\geq 0$ on $I$?
No. Just take $f=-1$ near $t=1/2$, $f$ large and positive everywhere else. Then either $t\simeq 0$ or $t\simeq 1$ compensates the contribution from $t\simeq 1/2$ (which one depends on whether $a>0$ or $a<0$).
In fact $f(t)=\sin(3\pi t)$ is already sufficient to see it. Since $f(t)$ is $\pi/3$-anti-periodic, and $e^{at}$ is monotone, the negative contribution on the middle third interval $[1/3,2/3]$ is compensated by either the first or the the third interval, according whether $a\ge0$ or $a\le0$.
(typo: "$1/3$-anti-periodic", sorry)
The answer is no. E.g., take
$$f(t):=1-\frac{3}{2} \max \left(0,1-2 \left| t-\frac{1}{2}\right| \right).$$
Then $f(1/2)=-1/2<0$, but
$$ \int_0^1 f(t) e^{at}\,dt=
\frac{2 e^{a/2}}{a^2}\,\left(a \sinh\frac{a}{2}-3 \left(\cosh
\frac{a}{2}-1\right)\right)\ge0$$
for all real $a$. (The inequality here follows from the inequality $u\sinh u\ge2(\cosh u-1)$ for $u\ge0$, which is easily checked by calculus.)
I don't think so. Roughly, the condition given should force $f$ to be nonnegative at least at around the endpoints $0$ and $1$, but it could be negative on the midpoint of the interval.
Consider a continuous function $g(t): [1/2, 1) \to \mathbb{R}$ such that $g$ satisfies the following properties.
$g(1/2) < 0$ and $g(1/2 + \epsilon) = 0$ for some small $\epsilon > 0$.
$g$ is negative on $(1/2, 1/2 + \epsilon)$ and nonnegative on $[1/2 + \epsilon, 1)$.
$\int_{1/2 + \epsilon}^1 g(t) dt + 2 \int_{1/2}^{1/2 + \epsilon} g(t) dt > 0$.
Now define $f(t)$ to be equal to $g(t)$ on $[1/2, 1)$ and $g(1-t)$ on $(0,1/2]$.
If $a \geq 0$, then
\begin{align*}
\int_0^1 f(t)e^{at} dt &\geq \int_{1/2+\epsilon}^1 g(t)e^{at} dt + \int_{1/2}^{1/2+\epsilon} g(t)e^{at} dt + \int_{1/2 - \epsilon}^{1/2} g(1-t)e^{at}dt \\
&\geq e^{a(1/2+\epsilon)}(\int_{1/2+\epsilon}^1 g(t) dt + \int_{1/2}^{1/2+\epsilon} g(t) dt) + e^{a/2} \int_{1/2}^{1/2 + \epsilon} g(t) dt \\
&\geq e^{a/2}(\int_{1/2+\epsilon}^1 g(t)dt + 2\int_{1/2}^{1/2+\epsilon}g(t) dt) \\
&> 0.
\end{align*}
The first inequality just throws away the manifestly non-negative part
$$\int_0^{1/2 - \epsilon} f(t)e^{at} dt.$$
The second inequality replaces the $e^{at}$ term in each integral by its infimum or supremum on that interval, depending on whether $f$ is positive or negative on the interval.
If $a < 0$, then we reduce to the above estimate by using the symmetry $f(t) = f(1-t)$. Note that $e^{at} = e^{-a(1-t)} \cdot e^a$, so then we get
$$\int_0^1 f(t)e^{at} dt = e^a\int_0^1 f(1 - t) e^{-a(1-t)} dt = e^a \int_0^1 f(t) e^{-at} dt,$$
which is non-negative as well.
Note that by making $f(1/2)$ much less than zero and $\epsilon$ much smaller depending on the value of $f(1/2)$, this gives examples where $f$ is negative on a set of arbitrarily small measure in $I$ but takes on arbitrarily large negative values.
The answer is no. For instance, let $A$ and $B$ be any random variables with continuous density functions $g$ and $h$ supported on $[0,1]$ such that $g\ne h$ and $A$ dominates $B$ in terms of the convex stochastic ordering. E.g., it is enough that $g$ and $h$ be symmetric about $1/2$ and such that $g/h$ is increasing on $[1/2,1]$; in particular, $g(t)=4|t-1/2|$ and $h(t)=1$ for all $t\in[0,1]$ will do. Letting now $f:=g-h$ and noting that $\int_0^1 f=0$ and $f\ne0$, we see that $f(t)<0$ for some $t\in(0,1)$. On the other hand, since the exponential functions $t\mapsto e^{at}$ are convex, by the convex stochastic domination we have
$$\int_0^1 f(t)e^{at}\,dt=\int_0^1 g(t)e^{at}\,dt-\int_0^1 h(t)e^{at}\,dt=Ee^{aA}-Ee^{aB}\ge0$$
for all real $a$.
|
2025-03-21T14:48:31.816295
| 2020-08-19T14:20:37 |
369569
|
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|
Stack Exchange
|
Two curious series for $1/\pi$
On Jan. 18, 2012 I conjectured that for any prime $p>3$ we have
$$R_p^2\equiv\frac1{10}\left(512\left(\frac{10}p\right)-27\left(\frac{-15}p\right)-475\right)\pmod p,$$
where $(\frac{\cdot}p)$ denotes the Legendre symbol and
$$R_p:=\frac1p\sum_{n=0}^{p-1}\frac{6n+1}{(-1728)^n}\binom{2n}n\sum_{k=0}^n\binom nk\binom{n+2k}{2k}\binom{2k}k(-324)^{n-k}.$$
Motivated by this I believed in 2011 that
$$\sum_{n=0}^{\infty}\frac{6n+1}{(-1728)^n}\binom{2n}n\sum_{k=0}^n\binom nk\binom{n+2k}{2k}\binom{2k}k(-324)^{n-k}=\frac{c}{\pi}$$
for some algebraic number $c$, but I could not figure out the exact value of $c$ at that time.
On August 19, 2020 I learned from Dr. Deyi Chen that Maple has a function to justify the form of an algebraic number if we know enough digits of this number. Inspired by this, here I pose the following conjecture.
Conjecture 1. We have
$$\begin{aligned}&\sum_{n=0}^{\infty}\frac{6n+1}{(-1728)^n}\binom{2n}n\sum_{k=0}^n\binom nk\binom{n+2k}{2k}\binom{2k}k(-324)^{n-k}\\&\qquad=\frac{24}{25\pi}\sqrt{375+120\sqrt{10}}.
\end{aligned}\tag{1}$$
On June 16, 2011 I conjectured that for any odd prime $p\not=5$ we have
$$\sum_{n=0}^{p-1}\frac{4n+1}{(-160)^n}\binom{2n}n\sum_{k=0}^n\binom nk\binom{n+2k}{2k}\binom{2k}k(-20)^{n-k}\equiv 0\pmod p.$$
Motivated by this I formulated the following conjecture on August 20, 2020.
Conjecture 2. We have
$$\begin{aligned}&\sum_{n=0}^{\infty}\frac{4n+1}{(-160)^n}\binom{2n}n\sum_{k=0}^n\binom nk\binom{n+2k}{2k}\binom{2k}k(-20)^{n-k}\\&\qquad=\frac{\sqrt{30}}{5\pi}\cdot\frac{5+\root 3\of{145+30\sqrt6}}{\root 6\of{145+30\sqrt6}}.
\end{aligned}\tag{2}$$
I have checked $(1)$ and $(2)$ for the first $100$ decimal digits.
Question. How to prove the identities $(1)$ and $(2)$?
Note that my two other similar conjectural identities
$$\sum_{n=0}^\infty\frac{357n+103}{2160^n}\binom{2n}n\sum_{k=0}^n\binom nk\binom{n+2k}{2k}\binom{2k}k(-324)^{n-k}=\frac{90}{\pi}\tag{3}$$
and
$$\sum_{n=0}^\infty\frac{n}{3645^n}\binom{2n}n\sum_{k=0}^n\binom nk\binom{n+2k}{2k}\binom{2k}k486^{n-k}=\frac{10}{3\pi}\tag{4}$$
found in 2011-2012 and published in this paper also remain open.
What is the relation between the congruence and the identity for pi? Is there an explanation for why we would expect such things?
You may look at my talk available from http://maths.nju.edu.cn/~zwsun/CNT-talk3.pdf and my recent paper available from http://dx.doi.org/doi:10.3934/era.2020070.
Thank you, this is very interesting!
|
2025-03-21T14:48:31.816439
| 2020-08-19T15:20:10 |
369571
|
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"Enzo Creti",
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|
Stack Exchange
|
Are there infinitely many primes which are the difference of two 3-smooth numbers plus one?
Consider numbers N of the form:
$(2^a\cdot 3^b)-(2^{(a-3)}\cdot 3^{(b-3)})+1$, with a, b integers >3.
Is it known if there are infinitely many N which are prime?
I doubt the answer is known. The set of numbers of this form is quite sparse, way beyond the capabilities of current methods. Also let me note that the way the title is phrased it seems to be asking a different question, which may be easier
So these are numbers of the form $215k+1$ where $k$ is 3-smooth. Have you done any calculations?
No I did not do any calculation
Heuristically speaking, there should be infinitely many such primes, by the same sort of heuristic as one would use to expect there to be infinitely many Mersenne primes. That is, by the Prime Number Theorem, the "chance" that a given number is prime is about 1/log n, and 1/log (s) where s ranges over all of the numbers in your series diverges. So we should expect that there are infinitely many such primes. Your set is in fact denser than the Mersenne primes, since for a given x, there are about (log x)^2 numbers in your set which are at most x, while there are about only (log x) powers of 2 less than x.
However, proving that such a thin set has infinitely many primes is in general beyond us at this time for any set which isn't more or less constructed to have infinitely many primes explicitly. The best state of the art now is results is much weaker. For example, Iwaniec proved there are infinitely many numbers of the form $n^2+1$ which have at most two prime factors. See here. Similarly, it has been proven that there are infinitely many primes of the form $x^4 +y^2$. See here. Both of the corresponding sets are much denser than the set of numbers in your question and are close to what the state of the art can do.
|
2025-03-21T14:48:31.816602
| 2020-08-19T16:14:41 |
369576
|
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}
|
Stack Exchange
|
Relation between signal derivative and frequency spectrum
I want to sample a signal whose derivative I know to be bounded by physical constraints. The sampling is disturbed by gaussian noise, hence I need to filter the sample with a lowpass filter.
Since I know precisely the bound on the derivative magnitude, I was wondering if there is a way to translate this bound in a frequency bound, in order to determine the cutoff-frequency of the filter.
My intuitive idea is that the low frequencies make up the "smooth" part of the sample (i.e. the part with a derivative that is compatible with the bound, the signal), while the higher frequencies are responsible for the sudden changes in the sample (i.e. the part with a slope that exceeds the bound, the noise); so, I think there should be a relation between the derivative and the frequency components of the sample. I'm looking for something that formalizes this concept. Thanks!
A lot depends on how you want to formalize your question. Here is one possible approach. Let's say that the signal can be any function on $\mathbb Z$ with the derivative bounded by $1$ and the noise has the standard deviation $\sigma$ and its values at different samples are independent. You apply a linear filter and you want to minimize the expectation of the $L^2$ norm of the error over a long period of time in the worst case scenario. What should this filter be?
Taking the Fourier transform of everything, as usual, we see that the question reduces to finding a function $\varphi$ on the circle with unit measure that minimizes $\sigma^2\int|\varphi|^2+\sup_{a}\frac 1N\int{|1-\varphi|^2\left|\sum_{k=1}^N a_k z^k\right|^2}$ where $a_k$ is an arbitrary sequence of real numbers with $|a_{k+1}-a_k|\le 1$ and $a_0=a_{N+1}=0$ ($N$ is the duration of the signal). Note that any such sum is just $\frac{1}{1-z}\sum_{k=0}^N b_kz^k$ where $|b_k|\le 1$ and $\sum_k b_k=0$.
This brings us to the problem of finding the supremum of $\frac 1N\int\psi^2\left|\sum_{k=1}^N b_k z^k\right|^2$ for a given $\psi=\frac{1-\varphi}{|1-z|}$.
This supremum is, of course, not greater than $\sup\psi^2$, but it is also not much less than that since if we allow complex coefficients instead of real ones, we can approximate the delta-measure at any point we want. Thus, if we do not care too much about factors like $2$, we can restate our problem as follows:
Minimize $\sigma^2\int(1-M|1-z|)_+^2+M^2$. If we pass to the continuous case of the line (which makes a decent approximation if you sample frequently enough, so in this normalization $\sigma\gg 1$) and assume that our Fourier transform is given by $\widehat f(\omega)=\int f(t)e^{-2\pi i \omega t}$ (so that the $L^2$ norm is preserved, which corresponds to $z=e^{2\pi i \omega}$), we see that we are to minimize $\sigma^2\int(1-2\pi M|\omega|)_+^2+M^2=\frac{\sigma^2}{3\pi M}+M^2$, which results in the minimum at $M=\sqrt[3]{\frac{\sigma^2}{6\pi}}$. Thus, from this point of view, the optimal filter should pass $e^{2\pi i\omega t}$ for $|\omega|\le \omega_0=\sqrt[3]{\frac{3}{4\pi^2\sigma^2}}$ with linear decline in amplification from frequency $0$ (amplification $1$) to frequences $\pm\omega_0$ (amplification $0$).
Now about scaling. Suppose that you sample at time intervals $\tau$, your time derivative is bounded by $D$ and the noise standard deviation at each sample is $\Sigma$. Then $\sigma=\frac{\Sigma}{D\tau}$ and the final answer should become $\Omega_0=\omega_0/\tau=\sqrt[3]{\frac{3 D^2}{4\pi^2 \Sigma^2\tau}}$.
Note once more that it is the worst case scenario optimization under the only restriction concerning the derivative with the objective to minimize the average square error. If you have more restrictions on your signal (say, some amplitude bound in addition to the derivative bound), or wish to optimize for a "typical signal" (which has then to be defined) and do not care much about outliers, or prefer a different objective, the answer may change. Also I believe that my logic is correct but I'm notoriously bad in algebra after midnight, so check the numbers involved before applying the final answer.
Thank you for your answer! I have two questions:
I can't figure out how you came to the first expression of the minimum problem (the one with |\varphi|^2..);
I don't understand the relation between the noise standard deviation of the sample \Sigma and the general noise standard deviation\sigma: \sigma = \frac{\Sigma}{D \tau};
Thanks!
@LucioPhys 1) If you have pure Gaussian noise at each sampling with variance $\sigma^2$ and use the filter corresponding to the function $\varphi$ on the circle, you'll get the Gaussian noise with the variance $\sigma^2\int|\varphi|^2$ at each sample (it will now no longer be independent at different samples but it will still be independent of the signal, so we can just add the squares to get the mean square error).
@LucioPhys 2) I normalized to sampling time interval $1$ and the derivative $1$ in the main part. If you sample more frequently, the allowed change of the signal between samples diminishes proportionally to the sampling time interval (if the derivative is $1$ unit/second and you sample $10$ times per second, the allowed change between samples becomes $1/10$ unit, so you should scale everything accordingly. The same with the size of the derivative. My $\sigma$ is just the ratio of the observed noise level at a sample to the allowed change in signal between two consecutive samples.
@LucioPhys I should also say that I assume that you are doing purely digital processing, i.e., you have a device that produces some number every $\tau$ seconds and you can only process this sequence of numbers (by doing FFT to get to the frequency side, multiplying there and doing inverse FFT to get the filtered signal). If you have a real physical device producing electric signal with noise and want to put an actual low pass filter between it and the measuring device, then the story gets a bit different: the idea is the same, but the meaning of the parameters and the formula will change.
Perfect, now it's clear! Thank you!!
|
2025-03-21T14:48:31.817263
| 2020-08-19T17:33:58 |
369580
|
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"Mikhail Bondarko",
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|
Stack Exchange
|
Is there an exponential map in $\mathbb A^1$ homotopy theory?
Let $k$ be a field, and let $Z \subset X$ be a smooth subscheme of a smooth scheme $X$.
When $k = \mathbb C$, there is a distinguished isotopy class of (topological) open embeddings $N_Z \to X$. In particular, these induce a homotopy class of maps $N_Z - Z \to X - Z$.
Is there a similar construction of a homotopy class of maps $N_Z -Z \to X - Z$ in the $\mathbb A^1$ homotopy category?
My hope was that such a map could be constructed using deformation to the normal cone. However, I found the paper "Motivic Tubular Neighborhoods" by Levine which may suggest that an elementary construction does not exist. In section 4.2 an exponential map is constructed stably. Is it possible to construct an unstable version using similar methods? I would also be interested in a natural map $|N_Z - Z|^{et} \to |X-Z|^{et}$ between the etale realizations.
Is Theorem 2.23 of Morel-Voevodsky's "$A^1$-homotopy theory of schemes" fine for you? I suspect that Levine actually needed something more complicated for his own purposes.
@MikhailBondarko No, I should clarify that I am aware of the purity theorem and I would really like the exponential map itself, and not the equivalence $Th(N_Z) \simeq X/(X-Z)$.
There is a canonical isomorphism between these two things in the $A^1$-homotopy category. If this is not what you want, then may you explain what do you actually need?
@MikhailBondarko As I said in my question, I want to construct a map $N_Z - Z \to X-Z$ unstably. My question is not about $X/(X-Z)$ or the Thom space.
|
2025-03-21T14:48:31.817524
| 2020-08-19T18:17:07 |
369585
|
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"Will Sawin",
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}
|
Stack Exchange
|
Which schemes are divisors of an abelian variety?
Let $X$ be a smooth, projective ireducible scheme over an algebraically closed field $k$. I'm trying to understand when there exists an abelian variety $A$ such that $X$ is isomorphic to a prime divisor on $A$.
There are some simple cases, of course. If $X$ is zero-dimensional, i.e. a point, then it is isomorphic to the identity of any elliptic curve $E$ over $k$, hence it is a divisor of $E$. If $X$ is of genus $1$, then if we choose a $k$-point, then $X$ is an elliptic curve. Then $X$ is isomorphic to the diagonal $\Delta\subset X\times X$, which is a divisor. Since $X$ is an elliptic curve, $X\times X$ is also an abelian variety. If $X$ is a curve of genus $2$, then the Jacobian of $X$ is 2-dimensional, and thus $X$ is of codimension one and thus the embedding $X\rightarrow \text{Jac}(X)$ lets us identify $X$ with a divisor of $\text{Jac}(X)$.
However, these simple cases do not give me an idea for the general case. The Jacobian only works for the genus $2$ case etc. The Albanse Variety also doesn't help, as the codimension might be to big. Are there any counter-examples of a smooth, projective ireducible scheme over an algebraically closed field which is not a divisor of an abelian variety?
The moduli space of smooth genus $g$ curves which are divisors on an abelian surface is an open subset of a $\mathbb P^{g-2}$-bundle on the moduli space of abelian surfaces with a polarization of degree $g-1$, hence has dimension $\leq 3 + (g-2) = g+1$, while the moduli space of all curves has dimension $3g-3$.
Any rational variety, e.g. $\Bbb{P}^n$ for $n≥1$...
@WillSawin Wow, thats a cool way of thinking about it!
Any curve of genus greater than two, whose Jacobian $J$ is simple, will do. If it were a divisor on an abelian surface $S$, then there would be a surjection $J\to S$ with positive dimensional kernel, contradicting the simplicity of $J$. Most curves of genus larger than two have this property; a randomly chosen example is $y^3 = x^4 - x$.
Thanks. Apologies for the dumb question.
An obvious class of counterexamples are uniruled varieties. In fact, abelian varieties contain no rational curves.
More generally, and for the same reason, if $X$ is any algebraic variety that contain a (possibly singular) rational curve, then $X$ is not a subvariety of an abelian variety, in particular it is not a divisor there.
Here's another answer using the Albanese that's of a slightly different flavor. Let $X$ be $n$-dimensional and suppose that $h^0(X,\Omega^1_X)<n$. Then any map $X\rightarrow A$ where $A$ is an abelian variety factors through the Albanese, which is of dimension less than $n$, so $X$ can't be a divisor on any abelian variety. So as an example you could take any simply connected variety. Of course, $\mathbb{P}^1$ does the trick.
I just want to point out that "adjunction+translation" tells us quite a bit:
Let $A$ be an abelian variety, say of dimension $n>1$ and let $D \subset A$ be a (let's say smooth) divisor. Since $\omega_A = \mathcal{O}_A$, the adjunction formula
$$
\omega_D = \omega_A(D)|_D = \mathcal{O}(D)|_D,
$$
the normal bundle of $D$. By differentiating the translation action of $A$, we can obtain non-0 global sections $0 \neq \sigma \in H^0(D,\omega_D)$, in which case the powers $\sigma^d$ show $H^0(D, \omega_D^d) \neq 0$ for all $d>0$. This shows that $D$ has non-negative Kodaira dimension: $\kappa(D) \geq 0$.
Remark: it's known that $D$ uniruled $\implies$ $H^0(D, \omega_D^d)=0$ for all $d > 0$ (and the converse is a conjecture), so the above is more-or-less an elaboration of Polizzi's observation that $D$ can't be uniruled.
|
2025-03-21T14:48:31.817780
| 2020-08-19T18:32:32 |
369588
|
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|
Stack Exchange
|
Proving that a model exhibits either a first or second order phase transition
Motivating example:
Take the (wired) random cluster model $\phi^1_{p,q}$ with parameter $q$ (see http://arxiv.org/abs/1707.00520 for an introduction).
It is now known on $\mathbb{Z}^2$ that it has a critical point (for example in the sense that the free energy is not analytic or equivalently in the sense that $ \phi^1_{p,q}( 0 \leftrightarrow \infty) $ changes from $0$ to a nonzero value) exactly at $p_c(q) = \frac{ \sqrt{q}}{1+ \sqrt{q}}$.
Further, it has been proven that the phase transition is second order (i.e. the free energy is $C^1$ but not $C^2$ or equivalently $\phi^1_{p_c(q),q}(0 \leftrightarrow \infty) = 0$ for $ 1 \leq q \leq 4$ and first order for $q > 4$ (meaning that the free energy is not differentiable at $p_c$ or equivalently $\phi^1_{p_c(q),q}(0 \leftrightarrow \infty) > 0$. Further, it is known that $\phi^1_{p_c(q),q}(0 \leftrightarrow \infty) \to 0$ for $q \to 4$ which means that the quantity $\phi^1_{p_c(q),q}(0 \leftrightarrow \infty)$ is continuous in $q$.
All this is now well established, but suppose I was interested in claims like:
i) Continuity of $ q \mapsto \phi^1_{p_c(q),q}(0 \leftrightarrow \infty)$.
ii) Proving that the phase transition is either first or second order.
iii) Proving that the area of the first order phase transition (which now know is $ (4,\infty) $) is open.
Do you know any older references or models where statements like that is proven?
The following paper contains results related to this question: https://arxiv.org/abs/2206.07033
|
2025-03-21T14:48:31.817891
| 2020-08-19T19:14:27 |
369596
|
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|
Stack Exchange
|
Conditions on triangle inequality for integral kernel
Consider $\int_RK(x,y)f(y)dy$, where $K(x,y) \in M_+(R^2)$.
Let $L(t,s)$ be an iterated rearrangement of $K$. Let also $$
A(t,v)=\int_0^{1/v}L(1/t,s)ds,
$$
which is decreasing with $v$ and increasing with $t$.
What conditions on $L$ would guarantee that $A(t,v)\leq A(t,u)+A(u,v)$ with $v<u<t$?
For example, if $L=1/t+s$ the condition would be specified. I am looking for more general answer.
|
2025-03-21T14:48:31.817957
| 2020-08-19T19:23:47 |
369598
|
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"Gerry Myerson",
"Max Alekseyev",
"Sylvain JULIEN",
"Thomas Browning",
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"https://mathoverflow.net/users/158000",
"https://mathoverflow.net/users/7076",
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"idmercer"
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|
Stack Exchange
|
Primality of Fermat numbers associated with "tetration"
A lot is known about the Fermat numbers $2^{2^k}+1$. For example, the first few
$$
2^1+1=3,\;2^2+1=5,\;2^4+1=17,\;2^8+1=257,\;2^{16}+1=65537
$$
are known to be prime, and Euler showed that the next ($2^{32}+1$) is not prime, being divisible by 641.
But what about the subset of special Fermat numbers formed by "tetration" or "power towers"?
$$
2+1 = 3\\
2^2+1 = 5\\
2^{2^2}+1 = 2^4+1=17\\
2^{2^{2^2}}+1 = 2^{16}+1=65537\\
2^{2^{2^{2^2}}}+1 = 2^{65536}+1
$$
I thought I had a vague memory of once reading that someone had conjectured that these are all prime (I guess people at one point conjectured that more generally all Fermat numbers were prime) and I also thought I had a vague memory of once reading that $2^{65536}+1$ had been proved composite.
However, I've been finding it difficult to get an answer to my question via a Google search. So, is it known whether $2^{65536}+1$ is prime or composite?
I would not say that "a lot is known" about Fermat numbers. As far as I know, the existence of other Fermat primes has neither been proven nor ruled out.
It's composite. It's divisible by 825753601.
Edit: It's also divisible by<PHONE_NUMBER>75021318420037633.
Factors of $2^{2^{16}}+1$ must be of the form $k\cdot2^{18}+1$. Here $k=3150$ which was found with a short computer search.
...And the number 825753601 is very searchable by Google.
It looks like this may have been first noticed by Selfridge in 1953?
In any case, thanks for your reply!
See also http://factordb.com/index.php?query=2%5E65536%2B1
That 27-digit factor was found by Brent, Crandall, and Dilcher in 1997, https://maths-people.anu.edu.au/~brent/pd/rpb175tr.pdf
|
2025-03-21T14:48:31.818097
| 2020-08-19T20:42:01 |
369601
|
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"Abdelmalek Abdesselam",
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"url": "https://mathoverflow.net/questions/369601"
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|
Stack Exchange
|
Generalized Young symmetrizers, why not?
For $n$ a positive integer, let $[n]=\{1,2,\ldots,n\}$. Consider two set partitions $\mathcal{A}=\{A_1,\ldots,A_p\}$ and $\mathcal{B}=\{B_1,\ldots,B_q\}$ of the set $[n]$.
We will denote by $G(\mathcal{A})$ the Young subgroup of permutations $\sigma$ of $[n]$ which preserve the blocks of $\mathcal{A}$ (i.e., satisfy $\sigma(A_i)\subset A_i$ for all $i$), and similarly for $\mathcal{B}$.
We will say that $\mathcal{A}$ and $\mathcal{B}$ are transverse iff for all $i,j$, the set $A_i\cap B_j$ has at most one element. Now define the group algebra element
$$
Y_{\mathcal{A},\mathcal{B}}=\left(\sum_{\sigma\in G_{\mathcal{A}}}\sigma\right)\times
\left(\sum_{\rho\in G_{\mathcal{B}}}{\rm sgn}(\rho)\ \rho\right)\ .
$$
These elements generalize the Young symmetrizers, the latter corresponding to the greedy case where, given $\mathcal{A}$, the blocks of $\mathcal{B}$ are as big as allowed by transversality?
Q1: Has there been a systematic study of these more general group algebra elements?
Q2: If not, why not?
Of course, with hindsight, one could say the interesting pairs of transverse partitions are the greedy ones (i.e., corresponding to rows and columns of a Young tableau), because they successfully explain the representation theory of the symmetric group. I am looking more for a reason to a priori predict this success, like: "among all pairs of transvers set partitions, the greedy ones exactly are the ones such that $Y_{\mathcal{A},\mathcal{B}}$ satisfies magic property $X$".
I was trying to read Young's QSA 1 article, and his definition comes out of the blue in Section 15.
Q3: Does anyone have any clue as to how Young got there?
Q1: Ricky Ini Liu's thesis https://dspace.mit.edu/handle/1721.1/60196 comes to mind.
Possibly related: https://mathoverflow.net/questions/254782/a-schur-positivity-conjecture-related-to-row-and-column-permutations
Thanks to both. I did not know about the terminology of general "diagram" and hence did not know what to google. Indeed, this is exactly the situation I am describing with general transverse pairs of set partitions. So Q1's answer is yes, although perhaps the relevant literature might be a bit small. My question about property $X$ still stands as well as Q3.
I suppose that in the extreme case when $\mathcal{B}$ is made of singletons, the left ideal is a Young module which has been studied before.
An interesting case is when $D$ is the Rothe diagram of a permutation: then you get the corresponding Stanley symmetric function out: see https://hal.inria.fr/hal-01229705/document.
I should have added that Ricky Liu's thesis is interesting for giving some results on the general case, but much more has been done in some more restrictive situations, such as the "%-avoiding diagrams" of Shimozono and Reiner. I would love to see an actual text (as opposed to a bunch of papers) exposing this material.
I deleted a comment about Q3 because I realized now that you're asking specifically- how did Young know to use partition diagrams as opposed to others. Again, extremely naive answer, but: he knew the irreps had to be indexed by partitions, right?
@SamHopkins: "he knew the irreps had to be indexed by partitions,right?" probably not for $S_n$ since he was not really following the work of Frobenius, but for GL_n certainly since they were figured out by Clesbch in 1872 https://gdz.sub.uni-goettingen.de/id/PPN250442582_0017?tify=%7B%22pages%22:[167],%22view%22:%22toc%22%7D
|
2025-03-21T14:48:31.818369
| 2020-08-19T20:58:36 |
369603
|
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"Henry",
"Iosif Pinelis",
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|
Stack Exchange
|
Asymptotics of $\chi_m$-distribution where the degree of freedom $m \to \infty?$
I'm interested to see a result where for large degree of freedom $m,$ the chi distribution $\chi_m$ is increasingly well approximated by a family of normal distributions with parameters depending on $m.$ The motivation comes from the fact that for large $m, \chi^2_m \sim \mathcal{N}(m,2m)$ approximately and the difference between the corresponding CDF's approach zero as the d.f. $m \to \infty$, as asked and proved in this question. The problem is: we can't apply some sort of continuous mapping theorem here to pass onto square root. I said "some sort of" because we'd be looking here for a case where $X_m - Y_m \to_d 0 \implies g(X_m) - g(Y_m)\to_{d}0$ with $g$ being a continuous, real-valued function (in our case $X_m:=\chi^2_m, Y_m:=\mathcal{N}(m,2m), g:= \sqrt{}$), and this is not the continuous mapping theorem.
Regardless, the wiki page of chi-squared distribution seems to refer to this theorem, proved by Ronald Fisher:
Approximately, $\sqrt{2\chi^2_m} \sim \mathcal{N}(\sqrt{2m-1},1) \implies \chi_m \sim \mathcal{N}(\sqrt{m-\frac{1}{2}},\frac{1}{\sqrt{2}}).$
Despite the reference to a book on that wiki page, I'm unable to access it and am also unable to otherwise get a reference to this theorem. Do you happen to know the proof or could you refer me to it please? Thank you!
Related: https://stats.stackexchange.com/questions/241504/central-limit-theorem-for-square-roots-of-sums-of-i-i-d-random-variables which (since $\chi^2_1$ has a mean of $1$ and variance of $2$) implies $\dfrac{\chi_m -\sqrt{m-1/2}}{1/2}$ converges in distribution to a standard normal distribution $\mathcal N(0,1)$ as $m$ increases, the same result as you attribute to Fisher
@Henry Thanks for pointing this out - this is indeed of help here - appreciate it!
@Henry BTW: a little typo I think: your denominator would be $1/\sqrt{2}$, not $1/2.$
Learning math: you are correct but I cannot edit comments. I should have written $\dfrac{\chi_m -\sqrt{m-1/2}}{\sqrt{1/2}}$ converges in distribution to a standard normal distribution $\mathcal N(0,1) $ as $m$ increases
Such a result can be obtained by the so-called delta method, which yields, in particular, the following: if $X_m\sim\chi^2_m$, then the distribution of $\sqrt X_m$ is approximately $N(\sqrt m,1/\sqrt2)$ (for large $m$).
Details: By the central limit theorem, $\overline X_m:=X_m/m$ is approximately normal with mean $\mu:=1$ and standard deviation $\sqrt{2/m}$. Hence, by the delta method, for $g(x)\equiv\sqrt x$, $\sqrt{\overline X_m}$ is approximately normal with mean $g(\mu)=\sqrt1=1$ and standard deviation $|g'(\mu)|\sqrt{2/m}=1/\sqrt{2m}$. Thus, $\sqrt X_m=\sqrt m\,\sqrt{\overline X_m}$ is approximately normal with mean $\sqrt m$ and standard deviation $1/\sqrt2$, as claimed.
On the uniform and nonuniform bounds on the rate of convergence to normality in the (possibly multivariate) delta method, see this and references there.
The convergence of the distribution of $\sqrt X_m$ to normality is in a sense monotonic; cf. formula (2.6).
@Ioself: thank you, I'll check out the delta method! Quick question before I do though: will this method imply that the distribution of the two CDF's converge to zero?
@Learningmath : The uniform (Kolmogorov) distance between the two cdf's does converge to $0$, actually with rate $O(1/\sqrt m)$; stronger than this nonuniform bounds also hold. See the first linked paper of the two added ones.
thanks again! I do wonder if we can also say that the $\sqrt{X_m} = \chi_m$ converges in density to the density of $N_m \sim N(\sqrt{m-1/2}, 1/\sqrt{2}), i.e. lim_{m \to \infty} ||f_{\chi_m} - f_{N_m}||_{L^{\infty}(\mathbb{R})} \to 0, m \to \infty?$ Here $f_Z$ denotes the PDF of the random variable $Z.$
@Learningmath : Yes, the pdf convergence also holds, as can be checked directly.
|
2025-03-21T14:48:31.818637
| 2020-08-19T21:44:11 |
369605
|
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|
Stack Exchange
|
Existence of probability measure on the circle with given Fourier coefficients
We say that a Hermitian symmetric (i.e., $f_{-n} = f_n^*$ for any $n \in \mathbb{Z})$ sequence $(f_n)_{n\in \mathbb{Z}}$ is positive-definite if, for any $N \geq 0$ and any $z_0 , \ldots, z_N \in \mathbb{C}$,
\begin{equation}
\sum_{n,m =0}^N f_{n-m} z_n z_m^* \geq 0. \tag{1}
\end{equation}
According to the Herglotz-Bochner theorem, a Hermitian symmetric sequence $(f_n)_{n\in \mathbb{Z}}$ with $f_0 = 1$ is positive-definite if and only if there exists a probability measure $\mu$ in the circle $\mathbb{T} = \mathbb{R} / \mathbb{Z}$ such that
\begin{equation}
f_n = \hat{\mu}_n := \int_{\mathbb{T}} \mathrm{e}^{2\pi \mathrm{i} n x } \mathrm{d}\mu (x).
\end{equation}
Assume now that I am given a vector $(f_{-N_0} , \ldots, f_0 , \ldots , f_{N_0}) \in \mathbb{C}^{2N_0+1}$ such that $f_0 = 1$ and $f_{-n} = f_n^*$ for any $|n|\leq N_0$ and such that (1) holds for any $N \leq N_0$. Is it always possible to complete the vector $(f_n)_{|n|\leq N_0}$ into a positive-definite sequence $(f_n)_{n\in \mathbb{Z}}$, or, equivalently, is there always a probability measure $\mu$ in $\mathbb{T}$ such that $\hat{\mu}_n = f_n$ for $|n|\leq N_0$?
Yes, this works. Condition (1) says that $\int |p(e^{ix})|^2\, d\mu(x)\ge 0$ for every polynomial $p(z)=\sum_{n=0}^N p_n z^n$. By the Fejer-Riesz theorem, these squares $|p|^2$ range exactly over the trigonometric polynomials $f=\sum_{|n|\le N} f_n z^n$ with $f\ge 0$ on $|z|=1$.
So we have a positive linear functional on this vector space $\{ f = \sum_{|n|\le N} f_n z^n \}$. This can be extended to a positive linear functional on $C(T)$; see here for background. This extension gives us the desired measure.
Very nice. The extension of positive linear functional indeed do the job. Thanks a lot!
@Goulifet: By the way, my notation in the first line is a bit sloppy, I really mean what you get if you assume that $\mu$ has the right Fourier coefficients, and then you multiply out $\int |p|^2, d\mu$ (in other words, this depends on the $f_n$ only and not on a hypothetical $\mu$). But I leave it in this form, it seems clear from the context.
I agree, I got your point so I think the context is indeed clear enough.
@Goulifet: One more afterthought: In the analogous problem for moments (instead of Fourier coefficients), there is a rather explicit description of all such measures $\mu$ ("Nevanlinna parametrization"). There probably must be something similar here.
|
2025-03-21T14:48:31.818814
| 2020-08-19T21:56:20 |
369606
|
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"Martin Seysen",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/369606"
}
|
Stack Exchange
|
Monster group as automorphism group of a distributive lattice
It is known that every finite group is the automorphism group of a finite distributive lattice.
Question: What is the minimal order of a distributive lattice $L$ such that the automorphism group of $L$ is isomorphic to the monster group?
Are there particular nice distributive lattices (with any order) realising the monster group in that way?
By Birkoff's FTFDL, realizing a group as an automorphism group of a finite distributive distributive lattice is the same as realizing it as an automorphism group of a finite poset, right?
Anyways here is a reference to the result about automorphism groups of posets: https://people.kth.se/~jbarmak/automorphismgroups.pdf. I think in principle it gives you an algorithm to construct the poset in question.
You may take the union of the set of the 2A involutions in the Monster (in the sense of the ATLAS) and the set of triples of 2A involutions such that their product is the neutral element. Here the order is given by inclusion. Does this qualify as a 'nice' poset on which the Monster acts as the automorphism group?
|
2025-03-21T14:48:31.818910
| 2020-08-19T22:53:19 |
369607
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/369607"
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|
Stack Exchange
|
The conformal map from interior of ellipse to interior of the unit disk (property check)
Based on example 5 (page 546) and example 7 (page 550) of the book "Applied and computational complex analysis. Volume 3, Wiley, 1986" written by Peter Henrici, if $a,b>0$ satisfies $a^2-b^2 = 1$ and we define $\rho>1$ through $\rho+\rho^{-1} = 2a$ (or through $\rho - \rho^{-1} = 2b$), the conformal map sending the interior of the ellipse $R:=\{(x,y)\in \mathbb{R}^2 \mid \frac{x^2}{a^2}+\frac{y^2}{b^2} \leq 1\}$, which in polar coordinates reads $$R = \left\{(\theta,r) \mid r \leq \frac{ab}{\sqrt{a^2\sin^2\theta + b^2\cos^2\theta}}:=r(\theta)\right\},$$ to the interior of the unit circle is given by
\begin{equation}\label{E2D}
h(z) = \frac{4}{h'(0)}\sum_{m=0}^\infty \frac{(-1)^mT_{2m+1}(z)}{\rho^{4m+2}-\rho^{-4m-2}},\quad z\in R,
\end{equation}
where $(T_n)_{n\geq 0}$ are the Chebyshev polynomials of the first kind and $$h'(0) = \left(4\sum_{m=0}^\infty \frac{2m+1}{\rho^{4m+2}-\rho^{-4m-2}}\right)^{\frac 12}.$$ But I get stuck when trying to check this is indeed the correct conformal map. Specifically, I do not know how to show $$|h(z)| < 1 \quad \text{for $z = re^{i\theta} \in R$}.$$ Also, another key question that is of my interest concerns the boundary behavior of $h$: what is the image under $h$ of the boundary "elliptical" arc $\{(\theta,r(\theta)) \in E \mid |\theta| \leq \delta \}$ (where $E$ is boundary of the elliptical region enclosed by $R$ and $\delta$ is small)? Thanks for any help or suggestions!
|
2025-03-21T14:48:31.819036
| 2020-08-19T23:13:56 |
369608
|
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"AmorFati",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/369608"
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|
Stack Exchange
|
Prescribing the discriminant locus of fiber spaces
Let $X$ be a projective manifold with $\dim_{\mathbb{C}} X \geq 3$. Assume $X$ is the total space of a fiber space, i.e., there is a proper surjective holomorphic map $f : X \to Y$ with connected fibers. The discriminant locus of $f$ is the set of all points $y \in Y$ such that $f^{-1}(y)$ is singular. We will assume $Y$ is a normal irreducible projective variety with $0 < \dim_{\mathbb{C}} Y < \dim_{\mathbb{C}} X$.
The following general question interests me:
Given a divisor $D \subset Y$, can we construct a fiber space $f : X \to Y$ whose discriminant locus is $D$?
This question is too big, so I will ask a more specific question:
Can we construct an example of a fiber space $f : X \to Y$ where the codimension-one part of the discriminant locus does not have normal crossings?
Notice that if we look at surfaces $\dim_{\mathbb{C}} X =2$, then the discriminant locus is a finite set of discrete points on a curve.
Edit: I would like to add that I would appreciate as many examples/references as one has. I think it is worth cultivating a bank of examples.
Just take any fiber bundle over $Y$ of fiber dimension $\geq 1$, then take a section over $D$, then form the blowing up of the total space along the image of $D$ under the section, and finally resolve any singularities of the new total space.
Let me just give an example. Let $V$ be a vector space, set $Y = \mathbb{P}(S^2V^\vee)$, and let
$$
X \subset \mathbb{P}(V) \times \mathbb{P}(S^2V^\vee)
$$
be the universal quadric, i.e., the natural divisor of bidegree $(2,1)$. The projection $X \to \mathbb{P}(V)$ is a projectivization of a vector bundle, hence $X$ is smooth. On the other hand, the projection
$$
X \to Y = \mathbb{P}(S^2V^\vee)
$$
is a quadric bundle, its discriminant divisor $D \subset \mathbb{P}(S^2V^\vee)$ is the symmetric determinantal hypersurface, and it is not a normal crossing divisor (it is normal, but singular) when $\dim(V) \ge 3$.
Thank you for the example. I'm not familiar with the construction, do you have any references? Or perhaps, even a heuristic for constructing such examples?
Essentially, there is no reason for the discriminant divisor to be normal crossing. So, if you take a general morphism $f$ it will give you an example similar to the above.
|
2025-03-21T14:48:31.819205
| 2020-08-19T23:54:49 |
369611
|
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|
Stack Exchange
|
Surjectivity of Étale cohomology
I've come across https://mathoverflow.net/q/172523, which says that if $D\subset Y$ is a contractible divisor, then we have a surjection
$$H^n(Y,\mathbb{C})\rightarrow H^n(D,\mathbb{C})$$
for $n\geq \text{dim }Y$. I don't quite know the proof to this statement and would like to know. In a remark to this post, it is mentioned that the decomposition theorem is used in the proof, and I don't know how.
Furthermore, I'm curious if this is true for higher codimensions, i.e. if $C$ is an algebraic cylce of arbitrary codimension, do we have a similar statement? One way one could get such a statement is if we have a sequence
$$(*)\quad C\subset D_1\subset \ldots \subset D_c\subset Y$$
of subspaces of $Y$ such that $D_{i-1}$ is a contractible divisor of $D_i$, which would give us a surjection
$$H^n(Y,\mathbb{C})\rightarrow H^n(C,\mathbb{C})$$
even though I have no idea how to get such a chain $(*)$ or how to describe those cycles for which such a chain exists.
The proof applies to a closed subscheme $Z$ such that there is a map out of $Y$ that is an isomorphism away from $Y$ and contracts $Z$ to a point.
|
2025-03-21T14:48:31.819300
| 2020-08-19T23:55:49 |
369612
|
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"authors": [
"Abdelmalek Abdesselam",
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|
Stack Exchange
|
On a certain expansion in term of Schur functions
This question is related to this other one
A Schur positivity conjecture related to row and column permutations
by Richard Stanley (thanks to Sam Hopkins for letting me know about it).
Consider a Young subgroup $S_{\lambda}$ of the symmetric group $S_n$, corresponding to some integer partition $\lambda$ of $n$. Let $\tau$ be some permutation and define the symmetric function
$$
F(\tau)=\sum_{\sigma\in S_{\lambda}}p_{c(\tau\sigma)}
$$
where $p_{\mu}$ is the usual power sum symmetric function and $c(\rho)$ denotes the integer partition given by the cycle-type of the permutation $\rho$.
Q: What is known about the Schur function expansion of $F(\tau)$, given the double coset class of $\tau$ for the Young subgroup?
One fact is that $F(\tau)$ is Schur positive if and only if $\tau\in S_\lambda$. More generally, if $K$ is any coset (left or right) of any subgroup $G$ of $S_n$, then $\sum_{\sigma\in K}p_{c(\sigma)}$ is Schur positive if and only if $K=G$. The only known proof for the "if" part requires representation theory; see Enumerative Combinatorics, vol. 2, page 396. For the "only if" part, it is easy to see that if a nonzero linear combination $\sum_{\lambda\vdash n} a_\lambda p_\lambda$ of power sums is Schur positive, then $a_{1^n}>0$.
Thank you for letting me know about this iff characterization. Do you know any articles where the $F(\tau)$ have been discussed?
@AbdelmalekAbdesselam: I don't know of any such articles.
|
2025-03-21T14:48:31.819538
| 2020-08-20T00:36:16 |
369613
|
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"Thomas Lesgourgues",
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|
Stack Exchange
|
Oriented path in a graph
Let $G=(\mathcal{V}_G,\mathcal{A}_G)$ be an oriented acyclic graph. Assume that $G$ has a unique source $s\in \mathcal{V}_G$ and a unique sink $t\in \mathcal{V}_G$. Now, fix $u,v\in \mathcal{V}_G$ such that $(v,u)\in \mathcal{A}_G$. Is it true that there exists an oriented path in $G$ of type: $v_0=s, v_1, \cdots, v_i=v, v_{i+1}=u, \cdots, v_n=t$? In other words, is there an oriented path containing the arrow $(v,u)$, starting in the source and ending in the sink?
What if $\mathcal V_G={s,t,u,v,w}$ and $\mathcal A_G={st,vu,uw,wv}$? Why isn't that a counterexample? Are you assuming some kind of connectedness for your graph? Does your definition of "path" allow repeated vertices?
you could event add the edge $(sv)$ making your graph weakly connected without containing the desired path. On a side note, I don't see what does the condition $(st)\in\mathcal{A}_G$ do. A path from $s$ to $t$ through another vertex $u$ will never use this edge.
I forgot to mention, but the graphs I'm dealing with are acyclic. I think that the acyclic condition together with the uniqueness of the source/sink implies that $G$ is connected. Therefore, on the paths here, the vertices are not repeated, because this would generate a cycle.
I don't know if condition $(s, t)\in\mathcal{A}_G$ is necessary (I don't think so). But, in the problem I'm working on, it is satisfied.
With the acyclic condition, the answer is yes. Starting at $v$ and repeatedly following any edge exiting the current vertex, you will eventually end up at $t$, by acyclicity and uniqueness of the sink. Thus there exists a path from $v$ to $t$ and, similarly, there exists a path from $s$ to $u$, which shows what you want.
|
2025-03-21T14:48:31.819687
| 2020-08-20T01:56:33 |
369618
|
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|
Stack Exchange
|
Probability a near universal hash function $ax \bmod p \bmod m$ produces an output from inputs equal modulo $m$
For one of the near universal hash functions $f(x) = ax \bmod p \bmod m$ where $p$ is prime and $m < p, m>1$ and $x$ ranges over $1 \dots p-1$ , what is the probability that given $x_r \in \{ x | x \bmod p \bmod m = x \bmod m = r\}$, $f(x_r) = s$? That is, find $Pr_{x_r}(f(x_r)=s)$. The probability is the fraction of $x$s such that $x \bmod p \bmod m = x \bmod m= r$ that have $f(x)=s$.
I asked this question a few months ago on the Mathematics Stack Exchange and never got a reply. It's simple enough, but it is either research level or prior research on the question is very hard to search for. The section below is just how I've thought about it and shouldn't be taken too seriously as part of the question. Actually, I suspect that there's a simpler way of approaching it.
This question differs from the usual question about the probability of collisions in near universal hash functions because $a$ is constant and $x$ is variable and only those $x$ with the same double modulus are considered. The answer appears to be a question of counting off-by-one errors. The answer is very diffent, too. For instance, if $a=1$, $Pr_{x_r}(f(x_r)=s) = \delta_{r,s}$ and if $a = m+3$, $Pr_{x_r}(f(x_r)=s) \approx \frac{1}{m+3} or \frac{2}{m+3}$.
I've thought of three sources of fence post errors in counting the number of solutions.
Firstly, without taking $\mod p$ or $\mod m$, $f(x_r)=s$ occurs only in certain ranges of length $p$ which repeat every $mp$. (Specifically, $s = (a \bmod m)(x_r \bmod m) + \lfloor \frac{ax_r}{p} \rfloor p \bmod m$.) At the end (and the beginning) of the range of $f(x)$ before taking the moduli there can be an additional region of length of $p$ filled with extra solutions. This produces deviations from uniformity on the order of $\frac{p}{ma}$ solutions (and differences of probability on the order of $\frac{1}{a}$).
Secondly, at a 'higher order' there can be two regions at the beginning and end of the range of $f(x_r)$ (again before taking the moduli) where a number $(f(x_r)=s)$-regions of length $p$ have an additional solution each. (That is, there is an additional fence post in the length $p$.) With the addition of each $mp$, the first solution rolls back by $m(p \bmod a)$. This produces on the order of $\frac{a}{p \bmod a}$ $mp$ lengths which may contain an additional solution. (The number of $mp$s is about $\frac{a}{p \bmod a}$ and some fraction of them rounded has an extra solution.) For there to be an extra solution the first $ax_r$ at or after the correct multiple of $p$ must be less than $p - \lfloor \frac{p}{am} \rfloor am = p \bmod am$. So the actual number of extra solution will be at most $\lfloor \frac{p \bmod am}{mp \bmod a} \rfloor + 1$ on each end of the range.
Thirdly, however, since the period is usually not an integer, it seems that there can be higher order fence post errors. If you look at a sequence of large multiples of $\frac{a}{p \bmod a}$ $mp$s that are smaller than the total range of $f(x_r)$ (that is multiples of $mp$ on the order $\frac{(p \bmod a)(ap)}{a(mp)} = \frac{p \bmod a}{m}$ ) (for instance from the continued fraction expansion of $\frac{a}{p \bmod a}$ or from powers of 10) there should be fence post errors at the edges of $[0,a(p-1)]$ for each approximation in the sequence. The length of the edge regions where off-by-one errors should be longer for each multiple, but the fraction of off-by-one errors should be proportionally smaller. Thus fence post errors should occur at some constant fraction of the ratio of the lengths of adjacent members of the sequence, so, provided the ratio between the accuracies doesn't vary too much, the total deviation from uniformity should be around $\log p$ solutions.
This means that the average over $a$ deviation from uniformity should be of order $\frac{m\log p}{p}$. So that for a randomly chosen $a$, most of the deviation from a uniform distribution will be in this third higher order source of fence post errors. Since $ax \bmod p \bmod m$ is only a near universal hash function this shouldn't be a problem, but I'm worried that I might be over thinking it and there might be a simpler way of solving the problem. The question is motivated by computing the hashed modular exponent in hashed variants of Shor's discrete logarithm algorithm from the unitary matrices represent multiplications by power-of-two exponents of the bases as outlined here: https://arxiv.org/abs/1905.10074 and https://quantumcomputing.stackexchange.com/questions/12354/shors-discrete-logarithm-algorithm-with-a-qft-with-a-small-prime-base/
What is the domain for $f(x)$ and how do you define the probability?
The domain can be taken to be $p-1$ $x$'s $1 \dots p-1$ and the probability is the $#(x_r$ s.t. $f(x_r)=s)/#(x_r)$ or the fraction of $x$ s.t. $x \mod m =r$ that $f$ takes to $s$.
No. That's true for $a=1$, but in general $ax \mod p \mod m \neq ax \mod m$.
Then in $x_r \in { x | x \mod p \mod m =r}$ "mod p" is redundant. Since $x_r$ belong to the domain of $f(x)$, one can simply say that $x_r \in { x | x\mod m =r}$.
True, but keeping the $\mod p$ makes it easier to pick up the meaning at a glance. The range of $x$, however, does need clarification since it should not include $0$ or $0 \mod p$.
@botsina: Use \bmod, not \nod. Compare the spacing in $x\bmod p$ with $x\mod p$.
Denote $\mathbb Z_p^*:=\{1,2,\dots,p-1\}$ and $\mathbb Z_m:=\{0,1,2,\dots,m-1\}$.
I assume $p\nmid a$. Then $f(x) = g(h(x))$, where $h:\mathbb Z_p^*\to \mathbb Z_p^*$ is a bijection defined by $h(x):=ax\bmod p$, and $g:\mathbb Z_p^*\to \mathbb Z_m$ is defined by $g(x):=x\bmod m$.
Let $b:=(p-1)\bmod m$ and $q:=\left\lfloor\frac{p-1}m\right\rfloor=\frac{p-1-b}m$. It follows that $p=qm+b+1$. Let $B:=\{1,2,\dots,b\}\subset\mathbb Z_m$ and $I_B:\mathbb Z_m\to\{0,1\}$ be the indicator function for the set $B$.
Now, for given $r,s\in \mathbb Z_m$, we have the sample space
$$X_r := \{ x\in Z_p^*\mid x\bmod m=r\} = \{ cm+r\mid \delta_{r0}\leq c\leq q-1+I_B(r)+\delta_{r0}\},$$
where $\delta$ is Kronecker delta.
In particular, we have $|X_r| = q + I_B(r)$. This is the denominator of the probability $\mathrm{Pr}(f(x_r)=s)$. Getting the numerator is trickier.
Sampling of $x_r\in X_r$ corresponds to sampling an integer $c\in [ \delta_{r0}, q-1+I_B(r)+\delta_{r0} ]$, and setting $x_r=cm+r$.
We have
$$(1)\qquad 1\leq cm+r\leq p-1.$$
Then $h(x_r) = acm + ar - kp$ for some $k$ (depending on $c$) satisfying
$$(2)\qquad 0\leq acm + ar - kp\leq p-1.$$
Finally, $g(h(x_r))=s$ is equivalent to
$$(3)\qquad ar - kp = s + mt$$
for some integer $t$ (again, depending on $c$).
The (in)equalities (1), (2), (3) define a polyhedron in the 3D space of $(c,k,t)$, and the numerator of $\mathrm{Pr}(f(x_r)=s)$ equals the number of integer points in this polyhedron.
I don't think there is a simple expression for this number in terms of the given parameters $p,m,a,r,s$.
A search shows that Alexander Barvinok found an algorithm in the 1990s to exactly count the number of integer points in convex polytopes of fixed dimensions (if you vary the dimension the problem is NP-hard). The algorithm involves finding a short rational generating function for the number of integer points and (because the solution is at a pole) calculating its limit with a finite Taylor series. Barvinok's algorithm has been implemented a couple of times (LattE, barvinok), even a parametric version.
Unfortunately, the parametric version only parametrizes the right-hand side of the polytope equation $Ax=b$, so there's no program to find a closed form solution. Unless there's a huge blow up of complexity when the left-hand side is parametrized, it should be possible to write the exact count as a small number of polynomials dealing with special cases of the parameters. A practical choice would be to precompute $m^2$ special cases for fixed $p$ and $m$ for each pair $r,s$ and substitute for different $a$s.
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2025-03-21T14:48:31.820198
| 2020-08-20T03:07:44 |
369624
|
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"authors": [
"Alexey Ustinov",
"GH from MO",
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"მამუკა ჯიბლაძე"
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|
Stack Exchange
|
Finding the asymptotic of the function $\Lambda(x):=\sum_{1 \leq m,n \leq x \,\land \,\gcd(m,n)=1} \frac{1}{mn}$
Inspired by this question Is there a known asymptotic for $A(x):=\sum_{1\leq i,j \leq X} \frac{1}{\text{lcm}(i,j)}$? I tried to find the asymptotic of the following function.
$$
\Lambda(x)=\sum_{\substack{ 1 \leq m,n \leq x \\ \text{gcd}(m,n)=1}} \frac{1}{mn}.
$$
My approach:
$$
\left(\sum_{1\leq k \leq x} \frac{1}{k}\right)^2=\sum_{1\leq l \leq x} \frac{\Lambda\big(\frac{x}{l}\big)}{l^2}\label{1}\tag{1}
$$
Now,
$$
f(x)=\left(\sum_{1\leq k \leq x} \frac{1}{k}\right)^2≈(\ln(x)+\gamma)^2
$$
From, \eqref{1} can establish the approximate identity
$$
2f(x)-\Lambda(x)≈ 2\int_{1}^{x} \frac{\Lambda(\frac{x}{t})}{t^2} dt
\label{2}\tag{2}$$
or,
$$
2f(x)-\Lambda(x)≈ \frac{2}{x}\int_{1}^{x} {\Lambda(\varphi)} d\varphi
$$
Using the Newton-Leibniz rule we get
$$x\Lambda'(x)+3\Lambda(x)≈4(\ln(x)+\gamma)+2(\ln(x)+\gamma)^2$$
Solving this differential equation we get,
$$
\Lambda(x)≈\frac{2}{3}\ln^2(x)+\left(\frac{8}{9}+\frac{4}{3}\gamma\right)\ln(x)+\left(\frac{2}{3}\gamma^2+\frac{8}{9}\gamma-\frac{8}{27}\right)+\frac{c_1}{x^3}
$$ ($c_1$ is the integral constant, for large $x$ this term can be neglected).
My question: Is the asymptotic formula correct? If not, then how to find the asymptotic of the function $\Lambda(x)$?
Is the method correct?
Edit: Though the answer comes wrong with the relation \eqref{2} , but if we use the identity involving the equation $A(x)$ instead of ${\zeta_x}^2(1)=\tau(x)$, then we get the correct answer (the leading term). The approximation \eqref{2} works well here. See my answer below.
Experimentally (with Mathematica, defining \[CapitalLambda][x_]:=Sum[1/Times@@mn,{mn,Select[Tuples[Range[x],2],CoprimeQ@@#&]}])$$x^3\left(\Lambda(x)-\left(\frac{2}{3}\ln^2(x)+\left(\frac{8}{9}+\frac{4}{3}\gamma\right)\ln(x)+\left(\frac{2}{3}\gamma^2+\frac{8}{9}\gamma-\frac{8}{27}\right)\right)\right)$$goes like$$0.561097,0.72963,-1.77106,-22.5681,-41.879,-140.577,-202.573,-384.69,-575.653,-962.996,-1180.86,-1858.96,-2198.3,-3077.97,-3986.3,-5207.03,-5898.68,-7809.84,-8719.26,-10982.,-13018.1,-15792.3,-17282.4,-21169.4,-23506.3,...$$ Does not look like this will tend to a finite limit $c_1$
@მამუკაჯიბლაძე: The asymptotic formula in the original post is incorrect, even the leading term is incorrect. For the correct asymptotics, see my post.
@GHfromMO I see, thanks. Would it be difficult to determine the constant in your $O$ term?
In fact, again experimentally, it looks like it might be $o(\log x)$, something like $O(\log^cx)$ with $c\approx0.98...$
@მამუკაჯიბლაძე: I updated my post to include the second main term. It is $C\log x$, where $C=1.3947995\dots$.
@GHfromMO Spectacular! Thanks
We have, for $x\geq 2$,
\begin{align*}
\sum_{\substack{ 1 \leq m,n \leq x \\ \mathrm{gcd}(m,n)=1}} \frac{1}{mn}
&=\sum_{1 \leq m,n \leq x}\frac{1}{mn}\sum_{k\mid\mathrm{gcd}(m,n)}\mu(k)\\
&=\sum_{1\leq k\leq x}\mu(k)\sum_{\substack{ 1 \leq m,n \leq x \\ k\mid\mathrm{gcd}(m,n)}} \frac{1}{mn}\\
&=\sum_{1\leq k\leq x}\frac{\mu(k)}{k^2}\left(\sum_{1\leq m\leq x/k}\frac{1}{m}\right)^2\\
&=\sum_{1\leq k\leq x}\frac{\mu(k)}{k^2}\left(\log\frac{x}{k}+\gamma+O\left(\frac{k}{x}\right)\right)^2\\
&=\sum_{1\leq k\leq x}\frac{\mu(k)}{k^2}\left(\log^2\frac{x}{k}+2\gamma\log\frac{x}{k}+O(1)\right)\\
&=\sum_{1\leq k\leq x}\frac{\mu(k)}{k^2}\left(\log^2 x-2\log x\log k+2\gamma\log x+O(\log^2 k)\right)\\
&=S_1(x)(\log^2 x+2\gamma\log x)-S_2(x)(2\log x)+O(1),
\end{align*}
where
\begin{align*}
S_1(x)&:=\sum_{1\leq k\leq x}\frac{\mu(k)}{k^2}=\frac{6}{\pi^2}+O\left(\frac{1}{x}\right),\\
S_2(x)&:=\sum_{1\leq k\leq x}\frac{\mu(k)\log k}{k^2}=\frac{36\zeta'(2)}{\pi^4}+O\left(\frac{\log x}{x}\right).
\end{align*}
We conclude that, for $x\geq 2$,
$$\sum_{\substack{ 1 \leq m,n \leq x \\ \mathrm{gcd}(m,n)=1}} \frac{1}{mn}=
\frac{6}{\pi^2}\log^2 x+C\log x+O(1),$$
where
$$C:=\frac{12\gamma}{\pi^2}-\frac{72\zeta'(2)}{\pi^4}=1.3947995\dots$$
One can also replace the first $O(1)$ by $O(k\log x/x).$ It will give the constant term in asymptotic formula and the error term $O(\log^2 x/x).$
@AlexeyUstinov: Indeed, since $\sum_{k=1}^\infty\mu(k)\log^2 k/k^2$ converges.
|
2025-03-21T14:48:31.820448
| 2020-08-20T04:43:30 |
369628
|
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|
Stack Exchange
|
Asymptotic expansion of hypergeometric function ${}_3F_2$ for large parameters
I encountered the following hypergeometric function in my research: $${}_3F_2(2,1+n,1+n;1,2+n;z)$$ where $0<z<1$. I'm interested in its behavior for large $n$. Semilog plot suggests exponential increase in $n$, however, I'm having trouble deriving the expression for the asymptotic expansion. There are lots of references to 1969 volume by Yudell Luke. I scanned it to no avail for a result that fits the above formula. I am a mere computer scientist, and am unfamiliar with literature on hypergeometric functions (which seems quite extensive). I would appreciate any help.
Below is a closed-form evaluation in terms of simple functions. Let $y=z/(z-1).$ Then
$${}_3F_2(2,n,n;1,n+1;z)=n(-z)^{-n}\Big((n-1)\big(\log(1-y)+\sum_{k=1}^{n-1}\frac{y^k}{k} \big) + y^n \Big)$$
I proved this by using Pochhammer symbol properties, which gets me to a linear combination of ${}_2F_1.$ Then I used a linear transformation to get from the argument $z$ to $y.$ That series can be manipulated to give the logarithm and a finite sum. Variable $y$ is always negative, but if it is small, the sum will converge rapidly. If you put this on a computer, watch out for large negative $y,$ which occurs for $z$ close to 1. In the finite sum, I'd probably add terms pairwise. If the $z\sim 1$ case is your most important case, then it's probably worth thinking about this some more.
Added: The sum for $z\sim 1$ can be done by an expansion found in 'Asymptotic Expansions Pertaining to the Logarithmic Series and Related Trigonometric Sums,' G. Fikioris & P. Andrianesis, J. Class. Analysis vol 7 #2, (2015) 113-127.
$$ \sum_{k=1}^{n-1}\frac{y^k}{k} \sim y^n\sum_{k=0}^\infty \frac{A_k(y)}{(y-1)^{k+1}} \frac{1}{n^{k+1}} \, ,n \to \infty $$
where the $A_k(y)$ are the Eulerian polynomials and begin with $A_0(y)=1$ and $A_1(y) = y.$
|
2025-03-21T14:48:31.820584
| 2020-08-20T05:33:18 |
369630
|
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"Asvin",
"Venkataramana",
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"url": "https://mathoverflow.net/questions/369630"
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|
Stack Exchange
|
What is the p-adic Plancherel measure?
What I know as the Plancherel measure for a group is a measure on the spectrum of $G$, aka the set of irreducible representations - at least for finite groups, this makes perfect sense.
Now, this paper by Conrey, Duke and Farmer defines a measure $\mu_p(\theta) = \frac{2}{\pi}(1+\frac1p)\frac{\sin^2(\theta)}{(1-\frac1p)^2 + \frac4p\sin^2(\theta)}d\theta$ and mentions that this is known as a p-adic Plancherel measure.
What does this mean? What is the group for which this is the Plancherel measure and how is it's spectrum identified with $S^1$? I suspect that the group is $PGL_2(\mathbb Q_p)$ but I don't know what it's spectrum looks like.
After looking at the paper (they thank Peter Sarnak for the remark that it is the "p-adic Plancherel measure"), it seems that the Plancherel measure is not of the group $PGL(2,{\mathbb Q}_p)$ but that of the quotient space $PGL(2,{\mathbb Q}_p)/PGL(2,{\mathbb Z}_p)$. The irreducibles (unramified principal series which are unitary) occurring here are indeed parametrised by $S^1$, namely characters on the split torus $T({\mathbb Q}_p)$ modulo $T({\mathbb Z}_p)$.
Thank you for the comment! Do you have a reference where I can learn more about this?
any book which has representations of GL(2,{\mathbb Q}_p)$ would consider unramified representations (and Satake parameters). Maybe Bump's book?
|
2025-03-21T14:48:31.820696
| 2020-08-20T06:12:26 |
369632
|
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"authors": [
"Angelo",
"Denis Nardin",
"Hans",
"https://mathoverflow.net/users/36563",
"https://mathoverflow.net/users/43054",
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"url": "https://mathoverflow.net/questions/369632"
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|
Stack Exchange
|
Complete target and complete fibers imply complete source?
Let $f:X\to Y$ be a surjective morphism of smooth irreducible varieties over $\mathbb{C}$. Assume further that $Y$ is complete and that every fiber $f^{-1}(y)$ for $y\in Y$ is complete and irreducible. Does it necessarily follow that $X$ is complete as well? If no, what additional assumptions can we put so that this follows?
Edit: According to the remark of vrz, I added the assumption that each fiber is irreducible (which implies connectedness). If I don't require equidimensionality, is there also a counterexample?
Yes, thanks. I actually want the fibers to be irreducible. Do you have a counterexample when the fibers are not equidimensional?
Suppose that $y$ is a point of $Y$, and $X$ is the disjoint union of $Y$ and ${y}$.
@Angelo Yes, I assume $X$ to be irreducible. Also, I am now assuming that the fibers are irreducible which is in your example not the case (fiber over $y$ is two points). Finally, I think that in your example $X$ is in fact complete as the union of complete varieties.
@Hans I suspect Angelo might have meant $X$ to be the disjoint union of $Y\smallsetminus{y}$ and ${y}$. Asking for $X$ to be irreducible eliminates this counterexample though
I apologize for the typo, and for not reading the post carefully enough.
Let $k$ be a field. Let $Y$ be a separated, finite type $k$-scheme that is geometrically connected and normal. Let $f:X\to Y$ be a separated, finite type morphism from a geometrically connected and reduced $k$-scheme to $Y$ such that the fiber over every geometric point of $Y$ is connected and proper.
Proposition. The morphism is proper.
Proof. By Nagata compactification, there exists a dense open immersion $i:X\hookrightarrow \overline{X}'$ into a proper $k$-scheme. The product morphism, $$(i,f):X\to \overline{X}'\times_{\text{Spec}\ k}Y,$$ is also an open immersion between separated, finite type $k$-schemes. Thus, the closure of the image is also a separated, finite type $k$-scheme. Denote this closure by $\overline{X}$. Also denote the restriction to $\overline{X}$ of the second projection by
$$\overline{f}:\overline{X}\to Y.$$
By construction $\overline{f}$ is a proper, surjective morphism. Thus, there exists a Stein factorization, $$\overline{X}\xrightarrow{h} Z\xrightarrow{g} Y,$$ where $h$ has geometrically connected fibers and where $g$ is a finite surjective morphism. Since $X$ is a dense open subscheme of $\overline{X}$ that is connected, also $\overline{X}$ is connected. Therefore $Z$ is also connected. Similarly, since $X$ is reduced, also $Z$ is reduced. Since the geometric generic fiber of $f$ is connected and dense in the geometric generic fiber of $\overline{f}$, also the geometric generic fiber of $\overline{f}$ is connected. Thus, $g$ is birational. Since $Y$ is normal, by Zariski's Main Theorem, the morphism $g$ is an isomorphism. In other words, the fiber of $\overline{f}$ over every geometric point is connected and proper.
The fiber of $f$ over that same geometric point is an open subscheme of the fiber of $\overline{f}$. By hypothesis, it is also proper, and thus it is a closed subscheme of the fiber of $\overline{f}$. Since the fiber of $\overline{f}$ is connected, the fiber of $\overline{f}$ equals the fiber of $f$ for every geometric point of $Y$. For every geometric point of $\overline{X}$, for the image geometric point of $Y$, the geometric point of $\overline{X}$ is a point of the fiber of $\overline{f}$ over that geometric point of $Y$. Thus, it is also a point of $X$. In other words, the open subscheme $X$ of $\overline{X}$ equals all of $\overline{X}$. Therefore, also $f$ equals $\overline{f}$, so that the morphism $f$ is proper. QED
|
2025-03-21T14:48:31.820932
| 2020-08-20T07:04:05 |
369636
|
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|
Stack Exchange
|
Uniqueness in Mare combinatorics and bounds on Gromov-Witten invariants
Let $R$ be the root system of a Weyl group $W$. Let $\tilde{R}^+$ be the set of $B$-cosmall roots, i.e. the set of positive roots $\alpha$ such that $\ell(s_\alpha)=2\operatorname{ht}\alpha-1$. Based on the combinatorics developped in Mare's paper, Section 3, I have the following conjecture.
Conjecture. Let $\alpha_1,\ldots,\alpha_r,\gamma_1,\ldots,\gamma_r\in\tilde{R}^+$ be such that $s_{\alpha_1}\cdots s_{\alpha_r}=s_{\gamma_1}\cdots s_{\gamma_r}$, $$\ell(s_{\alpha_1}\cdots s_{\alpha_r})=\ell(s_{\gamma_1}\cdots s_{\gamma_r})=\ell(s_{\alpha_1})+\cdots+\ell(s_{\alpha_r})=\ell(s_{\gamma_1})+\cdots+\ell(s_{\gamma_r})\,,$$ and such that $\alpha_1^\vee+\cdots+\alpha_r^\vee=\gamma_1^\vee+\cdots+\gamma_r^\vee$. Then, we have $\{\alpha_1,\ldots,\alpha_r\}=\{\gamma_1,\ldots,\gamma_r\}$.
I proved that this conjecture is true whenever $r=2$, $\alpha_1\not\perp\alpha_2$, $\gamma_1\not\perp\gamma_2$. While the general conjecture above might be too optimistic, I would like to see a counterexample preferably for $r=2$, but could not find one, and a counterexample for $r>2$ is also welcome. Any help is appreciated!
Motivation. Concerning the motivation for this conjecture, I would like to add that an affirmative solution to the above conjecture would have implications on numerical bounds on three point genus zero Gromov-Witten invariants. For example, in type $\mathsf{A}_n$, one might conjecture that $$N_{u,v}^{w,d}\leq\left\lceil\frac{n}{2}\right\rceil!$$ for all $u,v,w\in\mathbb{S}_{n+1}$ and all degrees $d$, where $N_{u,v}^{w,d}$ is the three point genus zero Gromov-Witten invariant of degree $d$ passing through three Schubert cycles parametrized by $u,v,w^*$, in other words, the structure constant of (small) quantum cohomology.
Edit. You are also welcome to give a counterexample to the bounds in the motivation.
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2025-03-21T14:48:31.821056
| 2020-08-20T08:26:00 |
369639
|
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"https://mathoverflow.net/users/100231",
"vidyarthi"
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|
Stack Exchange
|
List coloring of a graph corresponding to a Steiner triple system
Consider the Steiner triple system $S(2,3,n)$ for a suitable integer $n$. We define a graph $G$ with all the vertices as precisely the blocks of the above steiner triple system and any two points adjacent iff the intersection of the blocks corresponding to them in the STS are non-empty.
Are there results on the choosability of such graphs? Or, at least, is anything known about the chromatic number of such graphs? I think these numbers are closely related to the chromatic number and choosability of the generalized Kneser graph $K(n,3,2)$. The maximum degree of these graphs seems to be $\frac{3(n-2)}{2}$ and the clique size is $\frac{n-2}{2}$. Any hints? Thanks beforehand.
@bof thanks and sorry! edited
The said graph can be seen as the line graph of a rank three linear hypergraph. Some results on list coloring of this is known here. The list chromatic number of this is bounded above by $n$.
|
2025-03-21T14:48:31.821152
| 2020-08-20T08:42:10 |
369640
|
{
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"bean",
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"url": "https://mathoverflow.net/questions/369640"
}
|
Stack Exchange
|
Number of unramified quadratic extensions of a number field
Is there a general formula for the number of unramified quadratic extensions of a number field $K$?
When $K$ is quadratic, this is known (by genus theory) to be $2^{\omega(\Delta_K)-1}$, where $\omega(n)$ denotes the number of distinct prime factors of $n$ and $\Delta_K$ is the discriminant of $K$. I'm interested in results for when $K$ is of higher degree.
It seems like this problem might be much harder and is maybe adjacent to understanding the two-torsion of the class group $\text{Cl}_K$ (which seems hard when $K$ is not quadratic), but I'm pretty new to the area and could be totally off-base. Is there any hope of a more direct approach?
"is maybe adjacent": it is in fact completely equivalent.
The unramified abelian extensions of $K$ are in bijection with the subgroups of $\text{Cl}_K$.
@abx In particular, it is "maybe adjacent".
@GHfromMO I agree, by HIlbert Class Field I guess.
@abx Perhaps I'm missing something obvious, but it's not clear to me what the two-torsion of the class field group tells us about the index-two subgroups. If I'm not mistaken, the latter is what I want but the former is what seems very hard?
@bean However, #A[2] equals #Hom(A,Z/2Z) by duality.
Oh I see - thanks that makes sense! Would one of you want to write this up as an answer or shall I?
The answer seems to be no.
The number of unramified quadratic extensions of $K$ is equal to the number of index-two subgroups of the ideal class group $\text{Cl}_K$ by class field theory.
The index-two subgroups of $\text{Cl}_K$ correspond to the non-zero elements of $\text{Hom}(\text{Cl}_K, \mathbb{Z}/2\mathbb{Z})$.
$\#\text{Hom}(\text{Cl}_K, \mathbb{Z}/2\mathbb{Z}) = \#\text{Cl}_K[2]$ by Pontryagin duality, as pointed out to me by @RP_ and @abx in the comments.
The problem of computing (or even bounding) the size of $\#\text{Cl}_K[2]$ when $K$ is not a quadratic extension appears to be under active study and seems to be a challenging problem in general.
|
2025-03-21T14:48:31.821420
| 2020-08-20T09:15:04 |
369641
|
{
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"Andrey",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/369641"
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|
Stack Exchange
|
Commutativity of $\operatorname{Hom}$ and $\varprojlim$
Is it true the formula$\newcommand{\Hom}{\operatorname{Hom}}$
$\Hom(\varprojlim G_\alpha ;I) \simeq \varinjlim \Hom(G_\alpha ;I),$
if the group $I$ is injective? We can assume that $G_\alpha$ is finite generated as well.
Could you clarify what you mean by an injective group? Do you mean an injective abelian group? The only injective objects in the category of (not necessarily abelian) groups are trivial groups, so if that’s what you mean then the answer to your question is trivially “yes”!
Thank you. Yes, I mean an injective abelian group, but I do not understand why is an injective group trivial. For example, groups of rational numbers $\mathbb{Q}$ is an injective or sometimes is called infinity divisible group. Can you tell me any classical book where I can find it in details.
There are various proofs in, or referred to in, this thread.
Re, in particular, as @JeremyRickard says, the key object is in what category you're working. If you embed $\mathbb Q$ in $G = \langle\mathbb Q, x, y \mathrel: [x, y] = 1\rangle$, then the identity map on $\mathbb Q$ does not extend to $G \to \mathbb Q$—we left the category of Abelian groups.
No.
$\varprojlim_{n\in\mathbb{N}}\mathbb{Z}/p^n\mathbb{Z}=\mathbb{Z}_p$, the $p$-adic integers.
Take $I=\mathbb{Q}_p$, the $p$-adic rationals.
There is a nonzero map $\mathbb{Z}_p\to\mathbb{Q}_p$, the inclusion, but $\operatorname{Hom}(\mathbb{Z}/p^n\mathbb{Z},\mathbb{Q}_p)=0$ for every $n\in\mathbb{N}$, since $\mathbb{Q}_p$ is torsion-free.
Or alternatively, take $G_n=\mathbb{Z}$, with $G_{n+1}\to G_n$ multiplication by two. Then $\varprojlim G_n=0$ but $\varinjlim\operatorname{Hom}(G_n,\mathbb{Q})\cong\mathbb{Q}$.
Thank you for this examples!
|
2025-03-21T14:48:31.821568
| 2020-08-20T09:54:15 |
369644
|
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"Dror Speiser",
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|
Stack Exchange
|
A global code for the character table of PSL(2,q)
We can easily get the character table of $\mathrm{PSL}(2,q)$ for some fixed small prime power $q$, we can just do (for example):
gap> Display(CharacterTable(PSL(2,q)));
I don't know how the software is doing, I guess it uses some Atlas database for $q$ small enough or computes directly. The point is that there exists a global understanding of the character table of $\mathrm{PSL}(2,q)$, see for example Section 5.2 in Fulton & Harris, or page 12 in this note of J. Adams (up to typos). I need a code computing this character table following this global understanding (in order to interpolate it for $q$ non prime-power, see why here). I tried to write it myself but I got troubles with some ambiguities. I asked J. Adams but he does not have such a code. Now this global understanding is a well-known result, so the code I am looking for should already exist somewhere (in the source of some software, or as a private material).
Question: If you have such a code (that you would be willing to share), could you put it as an answer of this post?
Otherwise, if you know somewhere (or someone) susceptible to have such a code, could you please mention it (or her/him)?
Remark: My laptop needed 5min36s to compute the following:
gap> Display(CharacterTable(PSL(2,163)));
whereas it should be instantaneous be above global understanding. So the computation of the character table of $\mathrm{PSL}(2,q)$ is suboptimal on gap, and the expected global code would fix that.
Checkout the GAP3 package CHEVIE.
The quickest way to display the character table for a fixed prime-power $q$ is:
Display(CharacterTable( "PSL", 2, q)
The generic character table is in fact available on GAP4 (as pointed out in private by Frank Lübeck) with the following commands:
For $q$ even:
gap> Print(CharacterTableFromLibrary("SL2even"));
For $q \equiv 1 \mod 4$:
gap> Print(CharacterTableFromLibrary("PSL2even"));
For $q \equiv 3 \mod 4$:
gap> Print(CharacterTableFromLibrary("PSL2odd"));
|
2025-03-21T14:48:31.821713
| 2020-08-20T10:07:04 |
369646
|
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|
Stack Exchange
|
Fréchet derivative of evaluation-like functional (multivariate)
I'm fairly new to functional calculus but and posting here since the question seems more appropriate than for MSE. When coming across this post I could not help but wonder the following.
Let $H$ be the reproducing-kernel Hilbert space obtained by completing the set of all $C^1(\mathbb{R}^n,\mathbb{R}^m)$ with finite norm finite:
$$
\|f(x)\|_H:= \|f(0)\|_{\mathbb{R}^m} + \int_{x \in \mathbb{R}^n} \|(\nabla f)(x)\|_{\mathbb{R}^m} e^{-\|x\|} dx.
$$
If $w:C^1(\mathbb{R}^n,\mathbb{R}^m)\times \mathbb{R}^n\rightarrow [0,\infty)$ is the functional
$$
(f,x) \mapsto \|f(x)\|_{\mathbb{R}^m},
$$
what is its Fréchet derivative? Thinking analogously to the linked post and appealing to the chain-rule for Fréchet derivatives, I would guess it is
$$
Dw(f,x) (g,y)= \frac1{\|g(x)+J_f(y)\|}\left(g(x) + (J_f)(y)\right).
$$
However, I don't know how to show more than this (if even it is a correct ansatz).
The general procedure for the identification of a Fréchet derivative is the following
Calculate the functional derivative of the given functional, then
verify its linearity and
verify its continuity respect to the topology that is considered on the domain of the given functional i.e., for a Banach or Hilbert space topology, verify that the norm of the derivative does not depend on the structure of the variation but only on its size (norm).
The functional $w$ is defined on a vector space structure defined on $C^1(\Bbb R^n, \Bbb R^m)\times \Bbb R^n$, since we should be able to give a meaning to the word "linear", and the topology considered on this vector space is the product topology between the Banach space topology on $\Bbb R^n$ and the topology by $H$ on $C^1(\Bbb R^n,\Bbb R^m)$: following the above list we have
$$
\begin{split}
Dw\big[(f,x)\big]\big((g,y)\big) & \triangleq \frac{\mathrm{d}}{\mathrm{d}\varepsilon}w\big[(f,x)+\varepsilon(g,y)\big]\bigg{|}_{\varepsilon = 0}\\
&=\frac{\mathrm{d}}{\mathrm{d}\varepsilon}w\big[(f+\varepsilon g, x+\varepsilon y)\big]\bigg|_{\varepsilon = 0}\\
&=\frac{\mathrm{d}}{\mathrm{d}\varepsilon}\big\|f(x+\varepsilon y)+\varepsilon g(x+\varepsilon y)\big\|\bigg|_{\varepsilon = 0}\\
&=\left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon}\bigg[\sum_{i=1}^m \Big(f_i(x+\varepsilon y)+\varepsilon g_i(x+\varepsilon y)\Big)^2\bigg]^\frac{1}{2}\right|_{\varepsilon = 0}\\
&=\frac{1}{2}{\big\|f(x+\varepsilon y)+\varepsilon g(x+\varepsilon y)\big\|}^{-1} \left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon}\bigg[\sum_{i=1}^m \Big(f_i(x+\varepsilon y)+\varepsilon g_i(x+\varepsilon y)\Big)^2\bigg]\right|_{\varepsilon = 0}\\
&=\frac{1}{2}{\big\|f(x+\varepsilon y)+\varepsilon g(x+\varepsilon y)\big\|}^{-1}\\
&\qquad\cdot \left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon}\bigg[\sum_{i=1}^m f_i^2(x+\varepsilon y)+ 2\varepsilon f_i(x+\varepsilon y)g_i(x+\varepsilon y) +\varepsilon^2 g_i^2(x+\varepsilon y)\bigg]\right|_{\varepsilon = 0}\\
&={\big\|f(x+\varepsilon y)+\varepsilon g(x+\varepsilon y)\big\|}^{-1}\\
&\qquad\cdot \bigg[\sum_{i=1}^m \langle\nabla f_i(x+\varepsilon y),y\rangle+ f_i(x+\varepsilon y)g_i(x+\varepsilon y) \\
&\qquad\qquad +\varepsilon\langle\nabla f_i(x+\varepsilon y),y\rangle g(x+\varepsilon y) +\varepsilon f_i(x+\varepsilon y) \langle \nabla g_i(x+\varepsilon y),y\rangle \\
&\qquad\qquad\qquad + \varepsilon g_i^2(x+\varepsilon y)
+ \left.\varepsilon^2 g_i(x+\varepsilon y)\langle \nabla g_i(x+\varepsilon y),y\rangle\bigg]\right|_{\varepsilon = 0}\\
&={\big\|f(x)\big\|}^{-1}\bigg[\sum_{i=1}^m \langle\nabla f_i(x),y\rangle
+f_i(x)g_i(x)\bigg]=\frac{\langle 1, \mathbf{J}_f(x)y\rangle+\langle f(x),g(x)\rangle }{\big\|f(x)\big\|}
\end{split}
$$
Thus, apart from errors, we have done step 1 and checked the linearity as required by step 2. Regarding step 3, we see that that if
$$
\|f(x)\|>0 \iff f(x)\neq 0
$$
for the given $x\in\Bbb R^n$, then the functional derivative norm depend only on the value $\|g(x)\|_H+\|y\|_{\Bbb R^n}$ and not on the structure of the element $(g,y)$. Thus the functional derivative of $w$ is a Fréchet derivative.
Great answer Daniele. May I ask, have you ever come across this space/ is it a well-studied object?
Thank you, @AnnieLeKatsu. Regarding the linear product space $C^1(\Bbb R^n, \Bbb R^m)\times\Bbb R^n$ I am not aware of other uses and I haven’t seen it used in the literature. It’s a very particular space since it is the product between a infinite dimensional Hilbert space and a finite dimensional one, and I constructed (or better guessed)the structure of its topology from the one you gave to $C^1$ and the standard $\Bbb R^n$ one.
Oh I meant $C^1$ with topology induced by the inner product I provided.
@AnnieLeKatsu the answer is still no: however, the inner product you provided reminds me of (as it is very similar to) the one defined on the Segal-Bargmann space, which is a Hilbert space of entire functions and has recently been investigated extensively by Brian C. Hall.
@AnnieLeKatsu, after thinking a bit, I noticed that the structure of Hilbert space you provided on $C^1(\Bbb R^n,\Bbb R^m)$ is quite common: indeed it is a weighted Hilbert space with weight $|x|$. This kind of structure was used also by Hörmander for the solution of the $\bar\partial$-equation, therefore is quite "popular": by googling that name, you'll probably find a lot of information on spaces closer to the one you use than Segal-Bargmann ones I associated by analogy.
Ah this is nice, you mean googling Hörmander spaces? I've only ofund some russian texts...
@AnnieLeKatsu, better "Weighted Hilbert space". In general Hörmander spaces are (weighted) Hilbert spaces only if their associated Lebesgue space is $L^2$: otherwise they are only Banach. By googling "Weighted Hilbert space" you should be able to choose from a wide range of constructions and find what you need.
How much would it change if $n=m$ and we evalue $f\circ f$ instead of $f$ at x? We only switch $J_f$ for $J_{f\circ f}?$
@AnnieLeKatsu I suspect it is not so: as a matter of fact $$ (f+\varepsilon g)\circ (f +\varepsilon g)\neq f\circ f+\varepsilon g\circ g.$$ However, despite the presumably higher complexity of calculations, the same method for the calculation of functional derivative is applicable. Just the computations are more tedious.
|
2025-03-21T14:48:31.822055
| 2020-08-20T10:09:17 |
369647
|
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"url": "https://mathoverflow.net/questions/369647"
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|
Stack Exchange
|
Khovanov-Rozansky/HOMFLY-PT homology in type B
The picture, as I understand it, is that Hecke algebras (of type A, i.e. associated to the symmetric group $S_n$) lead to the HOMFLY-PT polynomials, and the categorified version of this says that the category of Soergel bimodules (again of type A) leads to Khovanov-Rozansky homology. The main paper outlining this is "Triply-graded link homology and Hochschild
homology of Soergel bimodules" (Khovanov).
However Hecke algebras, and Soergel bimodules can be defined for any Coxeter group. I'm particularly interested in type B, and the paper "Markov traces and knot invariants related to
Iwahori-Hecke algebras of type B" (Geck, Lambropoulou) seems to show the connection between Hecke algebras and HOMFLY-PT polys extends to this case. Is there a categorification of this? i.e. a type B version of Khovanov-Rozansky homology? references/papers would be great if so.
Maybe take a look at the following paper of Rose and Tubbenhauer
https://arxiv.org/abs/1908.06878.
|
2025-03-21T14:48:31.822139
| 2020-08-20T10:12:18 |
369648
|
{
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"authors": [
"Federico Barbacovi",
"Jan Grabowski",
"https://mathoverflow.net/users/13215",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/369648"
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|
Stack Exchange
|
Is there a notion of projective dg category?
Since the paper Smooth and proper noncommutative schemes and gluing of DG categories by Orlov, dg categories are considered the non-commutative counterpart of algebraic geometry. More specifically, we call a dg category a non-commutative scheme if it is an admissible dg subcategory of the dg category $\mathfrak{Perf}(X)$ for a smooth projective scheme $X$. Now, many properties of a scheme $X$ defined over a field $k$ can be translated into properties of the category $\mathfrak{Perf}(X)$ (a dg enhancement of $Perf(X)$), e.g.
(1) the scheme $X$ is proper over $k$ if and only if for any $E,F \in \mathfrak{Perf}(X)$ we have (here I am identifying $E$ and $F$ with their image in the homotopy category of $\mathfrak{Perf}(X)$, which by definition is $Perf(X)$)
$$ \sum_{n \in \mathbb{Z}} \text{dim} \, \text{Hom}_{Perf(X)}(E,F[n]) < +\infty$$
(2) the scheme $X$ is smooth over $k$ if and only if the diagonal bimodule associated to $\mathfrak{Perf}(X)$ is perfect in the derived category of $\mathfrak{Perf}(X)-\mathfrak{Perf}(X)$ bimodules.
From the above, one can then generalise these notions to that and smooth and proper dg categories. My question is whether there exists a similar analogy for the notion of projective scheme, and therefore the notion of a "projective dg category".
Thanks in advance.
If you want an approach to noncommutative projective geometry, this survey https://arxiv.org/abs/math/9910082 of Stafford-Van den Bergh describes work following the philosophy due to Artin-Tate-van den Bergh of using a category of graded modules modulo torsion to stand as a noncommutative projective space. This ultimately relies on a noncommutative version of Serre's theorem, which in the commutative setting says that this category is equivalent to that of (quasi-) coherent sheaves. I know this doesn't quite answer your question as posed but hope it's helpful.
@Riza That is what I was trying to make sense of actually, but I have not had luck so far.
@Jan thanks, I'll take a look at the reference!
If $X$ is a smooth projective threefold with a flopping curve $C$ then typically the variety $Y$ obtained from $X$ by a flop in $C$ is not projective, but smooth, proper, and derived equivalent to $X$. This shows that projectivity is not invariant under derived equivalence, hence does not correspond to a property of the derived category.
|
2025-03-21T14:48:31.822317
| 2020-08-20T11:01:41 |
369652
|
{
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"Hans-Peter Stricker",
"Richard Stanley",
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"https://mathoverflow.net/users/2807",
"Łukasz Grabowski"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/369652"
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|
Stack Exchange
|
Distance spectra of uniform tilings
Let a uniform tiling be defined by a vertex configuration $(n_1.n_2.\cdots.n_k)^m$, which is either spherical, Euclidean or hyperbolic. Assume that the tiling is vertex-transitive, especially that each vertex has the same number of vertices at all distances $d=1,2,3,\dots$ (not the geometrical distance but the number of edges between two vertices).
Let the distance spectrum $\delta$ of such a tiling be the vector with $\delta_d$ = number of vertices at distance $d$ (which is by assumption the same for all vertices).
For the spherical uniform tilings, i.e. the Platonic polyhedra, the distance spectrum is finite:
tetrahedron $(3)^3 \rightarrow$ $\delta = \langle 3 \rangle$
cube $(4)^3 \rightarrow \delta = \langle 3,3,1 \rangle$
octahedron $(3)^4 \rightarrow \delta = \langle 4,1 \rangle$
dodecahedron $(5)^3 \rightarrow \delta = \langle 3,6,6,3,1 \rangle$
icosahedron $(3)^5 \rightarrow \delta = \langle 5,5,1 \rangle$
I am more interested in uniform tilings of the plane (Euclidean or hyperbolic), which of course have infinite distance spectra which have to be expressed by a function $\delta(d)$.
For the three regular tilings of the Euclidean plane this function can be given easily:
triangular $(3)^6 \rightarrow \delta(d) = 6\cdot d$
square $(4)^4 \rightarrow \delta(d) = 4\cdot d$
hexagonal $(6)^3 \rightarrow \delta(d) = 3\cdot d$
which seems to depend on $k$ and $m$ in $(n_1.n_2.\cdots.n_k)^m$, but not on the specific value of $n_1$. The general pattern is $\delta(d) = k\cdot m \cdot d$.
But what about arbitrary vertex configurations?
For which (especially hyperbolic) vertex configurations is $\delta(k)$ known in explicit form – or is
there even a chance to derive it systematically from $(n_1.n_2.\cdots.n_k)^m$?
I especially wonder if generating functions may be of help.
For my favorite configuration $(3.4)^3$ a manual count gives $\delta(2) = 21$, but it's hard to determine even $\delta(3)$, even with this picture at hand:
At least some kind of pattern looms: $\delta(2) = 21 = \color{red}{\mathbf{3}} + \color{green}{\mathbf{6}} \cdot \color{blue}{\mathbf{3}}$ which is
$\color{red}{\mathbf{3}} = m \cdot \min(n_2 - 3, 2)$, i.e. the number of quadrilaterals $m = 3$ times $1$ (would be $2$ if $n_2 - 3 > 1$).$2$ and $3$ are fixed numbers and don't depend on the vertex configuration.
$\color{green}{\mathbf{6}} = m \cdot k$ which is the degree of the vertices
$\color{blue}{\mathbf{3}} = m\cdot k - 3$ $3$ is a fixed number and doesn't depend on the vertex configuration.
To sum it up (adding the term $\min(n_1 - 3, 2) = 0$):
$$\delta(2) = m \big(m k^2 - 3 k + \min(n_1 - 3, 2) + \min(n_2 - 3, 2) \big)$$
In general – i.e. for arbitrary configurations $(n_1.n_2.\cdots.n_k)^m$ with $k>1$ – it should hold that
$$\delta(2) = m \big(m k^2 - 3 k + \sum_{i=1}^k \min(n_i - 3, 2)\big)$$
Among the non-hyperbolic tilings above, the formula holds for all but
$(3)^5$ where it yields $10$ instead of $5$
$(3)^6$ where it yields $18$ instead of $12$
$(6)^3$ where it yields $15$ instead of $6$
How does the formula for $\delta(2)$ have to be adjusted to cover also these cases?
A related paper (but mostly about more than two dimensions) is https://arxiv.org/pdf/cond-mat/9706122.pdf.
@RichardStanley: Thanks for the hint. The paper is mainly concerned with "graphs based upon the root systems" - whatever that means.
My understanding is that in the case of, say, cayley graphs of hyperbolic groups (so tilings by octagons etc.), there's no hope to obtain a closed formula. It seems to be a research area, see e.g. https://link.springer.com/article/10.1007/s006050200043
|
2025-03-21T14:48:31.822891
| 2020-08-20T11:06:41 |
369653
|
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"Erel Segal-Halevi",
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"url": "https://mathoverflow.net/questions/369653"
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|
Stack Exchange
|
Birkhoff's theorem for hypergraphs
Birkhoff's theorem says that, in a bipartite graph $G$ in which both sides have size $n$, any fractional matching of size $n$ can be presented as a convex combination of integral matchings of size $n$ (at most $n^2-2 n +2$ such matchings are needed).
Consider a tripartite hypergraph $H$ in which all three sides have size $n$. In contrast to the bipartite case, a fractional matching of size $n$ does not imply the existence of an integral matching of size $n$, so a direct generalization of Birkhoff's algorithm is not true. Is there a weaker theorem that holds in the case of tripartite hypergraphs? For example, is it true that a fractional matching of size $n$ can be decomposed into integral matchings of size smaller than $n$?
Sounds like Problem 2 of https://imc-math.org.uk/imc2011/imc2011-day2-solutions.pdf Also, there's a discussion of Hall's Theorem for hypergraphs at https://gilkalai.wordpress.com/2012/11/25/happy-birthday-ron-aharoni/
@GerryMyerson Interesting, but this is about finding a single perfect matching - not about decomposition into a convex combination of perfect matchings.
I don't know if this is relevant: Aharoni and Haxell, "Hall's Theorem for Hypergraphs," http://www.mathcs.emory.edu/~whalen/Papers/Hall/matchingsinhyps.pdf
|
2025-03-21T14:48:31.823023
| 2020-08-20T11:35:46 |
369655
|
{
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"Alex Ravsky",
"Joseph O'Rourke",
"Penelope Benenati",
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|
Stack Exchange
|
Trade-off between hypervolume and diameter of $d$-dimensional shapes having a hypercubic smallest bounding box
Given any $d$-dimensional shape $X$, let $V(X)$ be its $d$-dimensional volume, and let $\ell(X)$ be the length of the longest line segment connecting two points of $X$.
Let $\mathcal{S}_C$ be the set of all $d$-dimensional shapes such that their minimum bounding box is a $d$-dimensional cube $C$. I am interested in quantifying the trade-off between $\frac{V(X)}{V(C)}$ and $\frac{\ell(X)}{\ell(C)}$ over $X\in\mathcal{S}_C$ (informally, how much $\frac{V(X)}{V(C)}$ can be large while $\frac{\ell(X)}{\ell(C)}$ is small).
Question: Can we prove that for $d\gg 1$ and for all $X\in\mathcal{S}_C$ there exists a constant $c$ such that the following inequality always holds?
$$\left(\frac{V(X)}{V(C)}\right)^{\tfrac1d}\le c\cdot\frac{\ell(X)}{\ell(C)}$$
A candidate for the extreme is $X=$ the sphere.
Thank you @JosephO'Rourke. I corrected the typo. Yes, the sphere, but I am realizing now that I am especially interested in the last question I just added: I would like to be sure that it is not possible to have a solid $X\in\mathcal{S}_C$ with a $\ell(\cdot)$ ratio small (like the one for the sphere) and a $V(\cdot)$ ratio large (that the sphere does not have).
For $X$ a $d$-ball, $\frac{\ell(X)}{\ell(C)} = d^{-1/2}$ and $\frac{V(X)}{V(C)} = \frac{\pi^{d/2}}{2^d\Gamma(d/2 + 1)}$. For large $d$, this is about $\frac{1}{\sqrt{\pi d}} (\pi e/2d)^{d/2}$
Thank you @S.Carnahan, is it true for all solids $X\in\mathcal{S}_C$ with such a low value for $\frac{\ell(X)}{\ell(C)}$ (e.g., like the one of the hypersphere), we always have a very small value for $\frac{V(X)}{V(C)}$? My first two questions are basically: what is the trade-off between these two ratios over all solids $X\in\mathcal{S}_C$?
I cannot find a good reference with a proof, but some web searching suggests that if $\frac{\ell(X)}{\ell(C)} = d^{-1/2}$, then $X$ is contained in a body of constant width, and that volumes of such bodies are bounded above by that of the ball.
Thank you @S.Carnahan, this is very interesting and corroborates my opinion about the conjecture in the third question. Do you have any idea about how to prove it?
I removed the first two questions, because the third one is sufficiently significant (and I guess it is an easier problem). Hence, there is only one question now.
@S.Carnahan basically I am interested in the maximum value of $\frac{V(X)}{V(C)}$ for the whole the range $\frac{\ell(X)}{\ell(C)}\in[d^{-1/2},1]$. What if, for instance, $\frac{\ell(X)}{\ell(C)}=\log(d)$? It is still true that $\left(\frac{V(X)}{V(C)}\right)^{1/d}\le \frac{\ell(X)}{\ell(C)}$?
Are you sure you want $\log d$ there? It's only less than 1 for $d=2$.
@S. Carnahan, sorry, there was a typo in my comment. By writing $\log(d)$ I meant $\log(d)^{-1}$, which is in fact in the range I wrote I am interested in: $[d^{-1/2}, 1]$. Thank you for your comment.
From what I said it follows that you always have $(V(X)/V(C))^{1/d}\le (1+o(1))K(\ell(X)/\ell(C))$ where the best $K$ is $\sqrt{\pi e/2}$ if $\ell\le\sqrt{\frac d{2\log d}}$, when the intersection is essentially the entire ball of diameter $\ell$, and then it gradually drops to $\sqrt 3$ at $\ell=\sqrt{d/3}$, when the intersection becomes essentially the entire unit cube. After that nothing interesting happens but the exact value of $K$ in the range $[\sqrt{\frac d{2\log d}},\sqrt{\frac d 3}]$ is still unclear to me though, I suspect, it is not too important for the question as asked.
Thank you a LOT @fedja! Is it possible, in your opinion, to state that $\sqrt{\pi e / 2}$ is un upper bound for $K$ which holds for all values of $\ell$? As you perhaps already understood, I am interested in finding a general upper bound of $\frac{\left(V(X)/V(C)\right)^{1/d}}{\left(\ell(X)/\ell(C)\right)}$. Thanks!
Yes, Penelope, you cannot go worse than if you just consider an unrestricted ball of diameter $\ell$ against the unit cube, so $\sqrt{\pi e/2}$ is a sure upper bound. The point was rather that it usually doesn't get much better than that even with all the extra conditions you imposed.
OK, great! Hence, if $\frac{\left(V(X)/V(C)\right)^{1/d}}{\ell(X)/\ell(C)}\le \sqrt{\pi e /2}$ always holds, could you please add this part in your answer draft so that I can validate it?
This is a bit too long for the comment box, so I'm posting it as an answer.
The worst case scenario is when $X$ is the intersection of a ball of radius $r\ge 1$ with the cube $C=[-1,1]^d$. Indeed, if we take the difference body $\frac{X-X}{2}$ of any body $X$ contained in the cube and of diameter $\ell=2r$, we'll get a body contained in the cube and also in the ball of radius $r$ and the volume will not decrease by Brunn-Minkowski. Also, since any such body contains the unit ball, the standard cube is, indeed, the minimal box for it. Since $\frac{\sqrt n}r X\supset C$, we see that for that body the reverse inequality always holds.
It would be nice to find a decent approximation for the volume of that intersection to see what happens in the regime when $r/\sqrt d$ stays fixed and $d\to\infty$, say.
I guess when $1\le r\le\sqrt{2}$ then the intersection is the ball with cut $2d$ hats of height $r-1$, so its volume can be easily calculated. When $\sqrt{2}\le r\le \sqrt{d}$ then the intersection seems to be the cube with spherically cut $2^d$ corners. Maybe there is a way to calculate a volume of such corner too, for instance, by induction with respect to $d$.
@AlexRavsky your intuition is correct and is explained here: https://math.stackexchange.com/questions/1996000/intersection-of-hypercube-and-hypersphere However, it is not clear how to formally exploit this recursion for the final goal of the problem (for instance by using an induction based on this recursion).
|
2025-03-21T14:48:31.823408
| 2020-08-20T12:45:13 |
369659
|
{
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|
Stack Exchange
|
Existence of finite limits of quasi-coherent modules on a scheme
Defining a quasi-coherent module $\mathcal{M}$ on a scheme $X$ to be a compatible family of modules $(\mathcal{M}(x))_{x \in X(A), A \in \textbf{Rings}}$ (as in here), is there a straightforward way to show the existence of (finite) limits (and that it forms an abelian category)?
One possible way, of course, should be to show that this definition gives rise to a category equivalent to the category of quasi-coherent sheaves of modules on the small Zariski-site associated to $X$, but that feels like a rather dirty solution.
The problem, I guess, is that taking pullbacks of sheaves of modules (generally) doesn't commute with taking limits so that the limit isn't defined "fibrewise"; colimits work fine for exactly that reason.
Another argument that a friend of mine explained to me seems to be that, denoting the in the above way defined category of modules as $\textbf{Mod}(X)$, one has
$$\textbf{Mod}(X) = \varprojlim_{A \in \textbf{Aff}/X} \textbf{Mod}(A)$$
where the ($2$-)limit is taken in the $(2,1)$-category of categories, functors and natural isomorphisms.
Now the argument would be that $\textbf{Mod}(A)$ is a locally presentable category, (certain?) limits of locally presentable categories are locally presentable, and locally presentable categories admit arbitrary limits.
I was still wondering whether there wouldn't be a more elementary way to for example directly construct kernels and finite products of modules when defined this way.
I'd appreciate any thoughts!
//Edit: Ok another way seems to be to first show that one can glue quasi-coherent modules along Zariski-coverings and then do everything locally. I guess that's fine for me, but I'd still be interested in seeing other elementary arguments if anyone has one!
The 2-category of accessible categories and accessible functors between them is closed under PIE limits in the 2-category of categories, yes. This isn’t entirely trivial to prove. If you only care about finite limits then I would focus on the fact that a scheme is a colimit of a nice kind of diagram where all the arrows are flat morphisms of affine schemes.
@RizaHawkeye Is that correct though? That sounds like you're suggesting to take the limit fiber by fiber which should be wrong as pullbacks of modules don't commute with arbitrary limits...
@lush By avoiding the use of Zariski covers, you are more or less trying to prove this statement for presheaves $X$ rather than schemes $X$. As you noted, the statement still holds true in this setting by generalities on presentable categories. However, the category in question is badly behaved for general presheaves - to construct a good category you need to work in a dg setting - and to the best of my knowledge is never used in this generality. So I recommend not avoiding Zariski covers.
@dhy Lurie uses it in this generality quite a bit in SAG in the chapters around Tannaka duality and some stuff much later on. You need to be able to talk about QCoh(X) for a space-valued functor on connective rings in order to even make sense of the generalized Artin representability theorem.
@HarryGindi I'm not seeing where in SAG he uses the abelian category QCoh for a general prestack X (as opposed to in specific cases e.g. X is an actual geometric stack.) Can you give me the number?
@dhy So in SAG, this gets used in chapters 16,17,18 to formulate the condition of existence of a relative cotangent complex for the purpose of Artin representability.
@HarryGindi Can you point to a specific lemma/theorem? I am seeing many uses of the derived category QCoh, but I cannot find any part where he is using the abelian category for a general prestack.
@dhy Oh, I think we were talking at cross-purposes. I don't mean the abelian category, I mean the whole derived category. Sorry! I was responding to the sentence fragment "to construct a good category you need to work in a dg setting - and to the best of my knowledge is never used in this generality.", but I see that I mis-parsed it now.
So I was lush's friend who he had originally asked this question, and I had some concerns, specifically because I gave the same answer as Riza, then realized that it gave incorrect answers if you follow the direct nLab construction. The point is that the limit of a diagram in the limit has to be computed first pointwise in the lax limit as above, then you have to apply a coreflector into the actual limit.
For example, if I have a cartesian square of locally presentable categories
$$\begin{matrix}
P&\xrightarrow{f^\prime_!}&C_1\\
g^\prime_!\downarrow &\ulcorner&\downarrow g_!\\
C_2&\xrightarrow{f_!}&C_0
\end{matrix}$$
and a diagram $d:D\to P$, I can compute $P$ as a colocalization of the lax limit of this diagram (the category of not-necessarily-cartesian sections of the associated cartesian fibration over the span category $\operatorname{Span}$). Let's denote this lax limit by $L$. Then we have an adjunction $P\leftrightarrows L$, where the left adjoint $P\to L$ is fully faithful. This tells us that the limit in $P$ is computed as the image under the coreflector $L\to P$ of the limit in $L$, which is actually indeed the pointwise limit together with the connecting maps
$$g_! \lim (f^\prime_! \circ d)\to \lim (g_! \circ f^\prime_! \circ d)=\lim (f_! \circ g^\prime_! \circ d) \leftarrow f_!\lim(g'_!\circ d).$$
So to form the true limit, I have to apply the coreflector to this formal diagram (viewed as an object of the lax limit).
This gives you a formula to compute the limit now of such a diagram, but actual existence of limits is following from the fact that this fibre product is presentable (plus the thing about arbitrary products still being presentable).
To finish working out the example, the coreflector then gives you the fibre product in $P$
$$ \lim(f^{\prime \ast}\lim (f^\prime_! \circ d)\to f^{\prime\ast}g^\ast\lim (g_! \circ f^\prime_! \circ d)=g^{\prime\ast}f^\ast\lim (f_! \circ g^\prime_! \circ d) \leftarrow g^{\prime\ast}\lim(g'_!\circ d)).$$
but in order for this formula to make sense, you first needed to know that limits in $P$ existed, and that's because $\operatorname{Pr}^L$ admits limits that agree with limits in $\mathbf{Cat}$.
Note: I've used the categorical convention for left and right adjoints (lower shriek and upper star, rather than upper star and lower star) in $\operatorname{Pr}^L$ rather than the algebro-geometric convention, because it is clearer in this case.
Edit: It looks like lush's question here was slightly different from the one we discussed in private. My mistake. Riza's answer is correct for flat covers (this is a theorem, but it is completely obvious for open immersions, as desired).
The existence of limits in $\operatorname{Pr}^L$ is probably somewhere in Adamek-Rosicky, but the proof in HTT <IP_ADDRESS> should be the same for 1-categories or ∞-categories.
|
2025-03-21T14:48:31.823965
| 2020-08-20T12:48:22 |
369660
|
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"url": "https://mathoverflow.net/questions/369660"
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|
Stack Exchange
|
Uniform closure of a neighbourhood complex in the tritetragonal tiling
Consider a neighbourhood complex of eight vertices (red) with vertex configuration $(3.4)^3$ which gives rise to the tritetragonal tiling of the hyperbolic plane:
Not knowing if this complex can be uniformly closed – by adding edges such that all vertices in the resulting graph have vertex configuration $(3.4)^3$ – one might start trying to do so. Start "complementing" the first three vertices (from the top right one down). First create the missing faces virtually:
You end up with a sequence of connected virtual vertices which now have to be identified with a free sequence on the border of the complex. ("Free" means that each vertex in the sequence on the border must have degree $d \leq 6-k$ where $k$ is the number of new edges attached to the virtual vertex (black) – except for the dummy vertices (pale green).)
You can continue this process – complementing one vertex with $d < 6$ after the other and identifying the sequence of new virtual vertices with a free sequence – but most probably you will get stuck and not find a free sequence before you are done.
What does this mean?
Will this procedure never be successful for this specific neighbourhood complex?
If it can be successful: How do I have to choose a correct free sequence in each step? (There may be several.)
If it can not be successful: Are there other neighbourhood complexes for which it is?
Edit: This is a slightly larger neighbourhood complex containing $36$ vertices with configuration $(3.4)^3$ (exactly the ones in the original complex above).
Just some thoughts.
Suppose you can close a complex $(4.3)^3$ somehow to a finite complex with, say, $v$ vertices, $e$ edges, and $f=f_3+f_4$ faces ($f_3$ triangles and $f_4$ quadrangles), then by double counting
$$3 f_3 = 3 v \implies f_3= v,$$
$$4 f_4 = 3v \implies f_4=3v/4.$$
Also, every vertex has degree six, so $e=3v$.
If we assume that this complex has Euler characteristic $\chi$, then by Euler's formula
$$\chi = v-e+f=v(1-3+1+3/4) = -v/4\quad\implies\quad v=-4\chi.$$
So the characteric must be negative and $v$ must be a multiple of $4$.
Similarly, we find $f=-7\chi$ and $e=-12\chi$.
For example, if you want to close this complex to a double torus, then $\chi=-2$, $v=8$ and $f=14$. These numbers feel too small. So we need a larger genus (more holes, smaller characteristic).
But the vertex count grows only linearly with the number of holes, and right now I have a hard time imagening how this can work out nicely.
Here I read that the solid double torus has $\chi = -1$ and the double torus = genus 2 torus = boundary of solid
double torus $\chi = -2$. This means that our "torificated" graphs always come as (or on) boundaries of solid double tori?
Note that the complex I want to close (e.g. to a torus or some other surface) has $36$ vertices (necessary condition on the number of vertices fulfilled!) , so the characteristic has to be $-9$, i.e. a $9$-whole torus, right? $f = 63$ and $e=108$? I have to count.
@Hans-PeterStricker It does not have to be the (surface of the) double torus, it can potentially be any $n$-holed torus. All I was able to show was that (if embedded) it must lie on a surface of negative Euler characteristic, and this is (in the oriented case) a torus with at laest two holes. You can have unoriented variants too but I do not know their Euler characteristic off the top of my head.
I counted $e$: Since $64$ edges are already present and $24 \times 3 + 4 \times 4 = 88$ have to be added, a total number of $152$ results, which far exceeds the number of $108$. Is this a strong proof, that my neighourhood complex cannot be uniformly closed?
@Hans-PeterStricker Where is this bound 108 from? Based on my answer, a large number of edges means that we need a large negative genus. But is not a proof of non-existence per se.
$108 = -12\chi$, $\chi = -9$ (I was only talking about my specific complex. For this one, non-existence per se can be proved in principle.)
@Hans-PeterStricker Why $\chi=-9$?
Because my complex has $v = 36$ vertices.
Let us continue this discussion in chat.
|
2025-03-21T14:48:31.824249
| 2020-08-20T13:59:10 |
369666
|
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|
Stack Exchange
|
Density of gaussian primes inside consecutive disks centered along the real axis of complex plane
Let's define the family of consecutive subsets of $\mathbb{N}$:
$$S_n =\{x \in \mathbb{N}\,:\,|x-n^2|\le n\}$$
With the previous definition we have that
$$U_n=\bigcup_{k=1}^n S_k=\{x \in \mathbb{N}\,:\,0\le x \le n^2+n\}$$
and
$$\pi(U_n)\sim\frac{n^2}{2\log n}$$
while
$$\pi(S_n)\sim\pi(U_n)-\pi(U_{n-1})\sim\frac{n}{\log n}$$
Therefore, the density of primes in $S_n$ is given by:
$$\rho_n=\frac{\pi(S_n)}{2n}\sim \frac{1}{2\log n}$$
Now let's extend all the previous arguments to the complex plane:
$$D_n =\{z \in \mathbb{C}\,:\,|z-n^2|\le n\}$$
$$V_n=\bigcup_{k=1}^n D_k$$
If we indicate with $\pi_G(X)$ the number of gaussian primes inside the subset $X$ of $\mathbb{C}$, numerical investigation suggests that
$$\pi_G(V_n)\sim\frac{n^3}{3\log n}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(1)$$
while
$$\pi_G(D_n)\sim\pi_G(V_n)-\pi_G(V_{n-1})\sim\frac{n^2}{\log n}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(2)$$
So the density of gaussian primes in $D_n$ is given by:
$$\rho_n^G\sim\frac{\pi_G(D_n)}{\pi n^2}\sim \frac{1}{\pi \log n}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(3)$$
I would appreciate any suggestion about theoretical validation of
asymptotic behaviors (1), (2), (3).
What you observe can be explained heuristically, based on the Riemann hypothesis for the Dedekind zeta function of $\mathbb{Q}(i)$, and the expectation that $D_n$ is a not too special subregion of the annulus
$$A_n:=\{z\in\mathbb{C}:n^2-n\leq|z|\leq n^2+n\}.$$
Indeed, assuming the Riemann hypothesis for $\zeta_{\mathbb{Q}(i)}$, we get that the density of Gaussian primes in $A_n\cap\mathbb{Z}[i]$ is
$$\sim \frac{4((n^2+n)^2-(n^2-n)^2)/\log n^4}{\text{area of $A_n$}}=\frac{1}{\pi\log n}.$$
The factor $4$ is the size of the unit group $(\mathbb{Z}[i])^\times$. Perhaps this result already follows from a proven zero density theorem for $\zeta_{\mathbb{Q}(i)}$, since the analogous result for rational primes is an old result of Ingham's. At any rate, the absolute values of $z\in D_n$ vary between $n^2-n$ and $n^2+n$, and they are not too concentrated around $n^2$, so it is reasonable to expect that the density of Gaussian primes in $D_n\cap\mathbb{Z}[i]$ is asymptotically the same as in $A_n\cap\mathbb{Z}[i]$; this is what your $(3)$ records. The statements $(1)$ and $(2)$ follow readily from $(3)$. Proving $(3)$ seems nontrivial even under the Riemann hypothesis for $\zeta_{\mathbb{Q}(i)}$; but again, the known zero density theorems might be sufficient for this purpose.
|
2025-03-21T14:48:31.824422
| 2020-08-20T14:37:47 |
369670
|
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|
Stack Exchange
|
Is this (somewhat specific) moment problem treated somewhere?
Suppose I have a measure $\mu$ over $\mathbb R_+$ given by its moments $\mu_0,...,\mu_n$, defined as :
$$\mu_k = \int x^{k} \partial\mu(x),\; k \in 1,...,n$$
Using Faà di Bruno's formula, I can obtain the corresponding cumulants $\kappa_0,...,\kappa_n$.
Say that there exists another measure $\nu$ that happens to have the set of moments $\kappa_0,...,\kappa_n$.
Is there some work somewhere about the relationship between $\mu$ and $\nu$ ?
Yes, this is a very classical topic. The Nevanlinna parametrization describes all such measures (try a search for this perhaps).
@ChristianRemling Is'nt Nevanlinna parametrization a way to solve the Hamburger moment problem ? I am not asking for a solution to my moment problem, but for a relationship between the two measures. Please tell me if i did not get what you meant.
$\newcommand\ka\kappa$$\newcommand\R{\mathbb R}$Without loss of generality, $\mu_0=1$, so that $\mu$ is a probability measure. Let $\ka$ be a measure on $\R$ with moments $\ka_0,\dots,\ka_n$ and $|\ka|:=\ka(\R)$.
Of course, only in exceptional cases (such as a case with $\mu_0=\mu_2=\mu_4=1$ and $\mu_1=0$) is a measure determined by finitely many of its moments. Therefore, usually the measure $\mu$ will not be determined by $\ka$ (nor $\ka$ will be determined by $\mu$).
However, for each measure $\ka$ with moments $\ka_0,\dots,\ka_n$ we can construct a measure $\nu$ with the same moments as $\mu$ for all moment orders from $0$ through $n$.
Indeed, the accompanying (infinitely divisible) compound Poisson probability distribution
$$\nu:=\nu_\ka:=e^{-|\ka|}e^{*\ka}=e^{-|\ka|}\sum_{j=0}^\infty\frac{\ka^{*j}}{j!}$$
will have the same moments as $\mu$ for all moment orders from $0$ through $n$; this is easily seen by considering the Fourier transform of $\nu_\ka$.
Coming back a little late but this is still up in my mind. First thanks for the leads you gave me, very usefull stuff. However, what i wanted (without clearly knowing it at the time) was the other way around: Is there some simplification in the reconstruction of a representing measure for $\kappa$ due to the fact that $\kappa_0$,..., $\kappa_n$ are cumulants of another distribution ?
@lrnv : I am afraid I don't understand the question in your comment. In particular, what do you mean by "a representing measure for $\kappa$"? In my answer, $\kappa$ itself is measure.
Yes, i meant 'a representing measure for $\kappa_0,...,\kappa_n$'. My 'Big goal' is to find a measure that have moments $\kappa_0,...,\kappa_n$, and the question is: Will the fact that these moments are also cumulants of some other measure (that i completely know) help me in any way ?
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2025-03-21T14:48:31.825048
| 2020-08-20T15:04:55 |
369672
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Stack Exchange
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Prominent non-mathematical work of mathematicians
First of all, sorry if this post is not appropriate for this forum.
I have a habit that every time I read a beautiful article I look at the author's homepage and often find amazing things.
Recently I read a paper of Andrew Hicks and when I opened his homepage I found many links about his invention: Flawless wing mirrors (car mirror).
(Image Source)
I would not be surprised if this invention was made by a non-mathematician. His mirror is an amazing invention to me because every day I see it, but didn't know its inventor is a mathematician! Anyway, I want to ask
Question 1: Are there mathematicians who have done outstanding/prominent non-mathematical work like inventions, patents, solving social/economical/etc. problems, papers in these areas, etc.?
Of course, one can say that almost all technology nowadays is based on the work of mathematicians, but I'm asking for specific contributions/innovations.
I want to ask a similar question (Maybe it will be useful for those who are looking for a job!):
Question 2: Which mathematicians are working in non-mathematical areas/companies?
Note: Please add to your answers the name and the work of the mathematician.
Unabomber has to be #1 on this list.
Not clear if it fits, but Emanuel Lasker was a Chess World Champion and a mathematician. I understand that he proved some important results in Commutative Algebra.
You may consider Sir Isaac Newton. Apart from his mathematical discoveries he did many non mathematical inventions. He built telescope, worked with Alchemy, was involved in politics.
The question is too vague. Before 18 century, all mathematicians had non-mathematical jobs since there were no mathematical jobs.So probably you mean modern times. It is also not clear who exactly is counted as a mathematician.
@Piyush if we put Unabomber on the list, then we should also make room for André Bloch, mathematician and triple murderer. https://en.wikipedia.org/wiki/André_Bloch_(mathematician)
@GerryMyerson, not necessarily. Bloch was a murderer first, and did his mathematics while institutionalized (not to mention that murder isn't usually considered "non-mathematical work"). On the other hand, Kaczynski was a mathematician first, and his later non-mathematical work was a sociological thesis on the inevitability of human enslavement by industrial society.
@Ray Are you saying murder IS considered mathematical work?!
@HarryWilson There’s mathematical work, there’s non-mathematical work, and then there are other things than work.
I’m not sure whether Jean van Heijenoort would qualify. He was a Trotskyist activist and https://en.wikipedia.org/wiki/Jean_van_Heijenoort?wprov=sfti1 says he also was personal secretary and bodyguard of Trotsky.
@HarryWilson, The bombings and deaths were mostly an irrelevant side effect (from TK's point of view). That was simply a means of getting his work published. Remember, he was abused by his parents and as an under-aged student by his university professors, and was tortured and brainwashed by the CIA's "MK Ultra" mind-control program. His mathematical publications were legitimate, though not major breakthroughs, his Industrial Society and its Future is a not unreasonable sociological analysis, its purpose being to present the world with its only possible chance of avoiding inevitable slavery.
@RayButterworth: This comment thread is rapidly getting both off-topic and speculative. (For instance, the accusation that your comment contains about Kaczynski's parents is not supported by the Wikipedia article about Ted Kaczynski, and neither by a quick google search that I did for his name.) I do not want to be rude, but I would really appreciate it if we could keep this comment thread focussed on the question rather than on claims that are speculative or off-topic (or both).
Elena Ventsel (Russian: Елена Сергеевна Вентцель) was a mathematician working in probability theory and operations research. Besides her scientific output, she also authored several monographs and textbooks. And she was also an accomplished fiction writer publishing under the pen name of Irina Grekova (Russian: Ирина Грекова). By the way, this pen name is mathematical: if you write this name as the first name initial and the last name, it's I.Grekova (Russian: И.Грекова) -- from "igrek" (Russian: "игрек"), which is the Russian name of the letter "y".
I wonder if Brendan McKay's work on debunking the Bible Code counts. See: John Safran vs God - Bible Code.
I think Eric Lander deserves a mention
No pun intended, but isn't Persi Diaconis quite high up in Magic Circles?
@Geoff I thought of Persi, I agree he'd be a good answer. (But I've already posted four.)
Why no John von Neumann? One of the most prominent and versatile mathematician of 20th centure. https://en.wikipedia.org/wiki/John_von_Neumann
Surely there must be mathematicians who are also famous for their work as translators? I can well-imagine mathematicians translating old writings just for fun, and I feel like I've heard stories, but can't think of any off the top of my head.
@Beta: it has been posted before here!!
Perhaps, you could append your Question by the alphabetic list of mathematicians that were already mentioned in this thread.
Samuel Eilenberg, one of the key mathematicians of the XX Century (co-created category theory, systematized homological algebra, opened new roads in topology, etc), was a good example:
he had (at least) TWO LIVES, with only one thing in common, his short name, Sammy.
In the first one he was the mathematician, whereas in the other he was a formidable expert and collector of Chinese and far-eastern ancient pottery (and other artifacts as well). He was world-famous in his second life just like he was in his first one (when he died he donated his immense collection to NYC, you can still admire it here).
What is funny (and a bit odd ) is this: Sammy did not like to mix his two lives at all. At his funeral, the two groups (mathematicians and art collectors) collided for the first and last time. Nobody could believe that Prof. Eilenberg, the Math Genius, and Prof. Eilenberg, one of the greatest authorities in ancient eastern arts, were one and the same man.
POST SCRIPT
I have done some research on Sammy's life as art collector: apparently, he was struck by the beauty of indian art during a trip to India. From that point on, he decided that he had to assemble a collection of eastern art and craftmanship, which he did in the next 30 + years. Now I finally see what the two Eilenberg had in common: a passion for aesthetics, for the formal beauty of structures. Alex Heller wrote the following words to honor his Teacher Sammy:
As I perceived it, then, Sammy considered that the highest value in mathematics was to be found, not in specious depth nor in the overcoming of overwhelming difficulty, but rather in providing the definitive clarity that would illuminate its underlying order. This was to be accomplished by elucidating the true structure of the objects of mathematics. Let me hasten to say that this was in no sense an ontological quest: the true structure was intrinsic to mathematics and was to be discerned only by doing more mathematics. Sammy had no patience for metaphysical argument. He was not a Platonist; equally, he was not a non-Platonist. It might be more to the point to make a different distinction: Sammy’s mathematical aesthetic was classical rather than romantic.
One might say that, because of his contributions to the theory of finite automata, he was also a computer scientist which, if we don't consider this part of mathematics, is a third domain of specialty.
@Gro-Tsen Then I might also add Edward Spanier to the list. Also more known as a mathematician; but published fundamental, and much cited, work in theoretical computer science; mostly with S. Ginsburg. Eilenberg's work in automata theory is clearly written and still inspirational after almost 50 years, but most TCS-Researchers know that he is a famous mathematician too. However, Spanier, I guess most TCS people do not know that he is a mathematician at all. At least it is not that well-known.
@StefanH A small thing, but it's Edwin, not Edward, Spanier.
@DannyRuberman Oh yes, I am sorry. Thank you!
Charles Lutwidge Dodgson (1832-1898), better known as Lewis Carroll.
please expand, Noam :-). surely you can name a few things done/invented/made by Dodgson which you find outstanding...
Well, probably his universally-known fictional books qualify for "outstanding/prominent non-mathematical work", or am I missing something?
I think the issue is whether this is a prominent non-mathematical work of a professional mathematician, or a prominent mathematical work of a non-mathematician, or if that matters for this question. Certainly in this case his novels are more famous than his math.
@BradyGilg His day job was as a mathematician. That most people don't know it doesn't make him any less of one :).
Fun fact: The mad hatters in Alice in Wonderland were meant to make fun of other mathematicians he knew of.
If we had to quibble, it would be by saying that a mathematics lecturer at Oxford at the time wasn't expected to be a mathematician in the modern sense (i.e., a research mathematician). He did do a bit of research (I once used a lemma on his on determinants), but apparently didn't really try to be part of the research community (and there was such a thing in the UK at the time, though nothing compared to France or Germany). At any rate, that's really quibbling - Noam's answer is of course right, and we should be happy to call Dodgson a mathematician.
I've also used Dodgson condensation https://en.wikipedia.org/wiki/Dodgson_condensation in one of my own papers. He certainly counts as a research mathematician in practice even if he technically wasn't one on paper.
The White Knight in Looking Glass is often thought to be his caricature of himself. But the part of the White Knight's speech beginning with 'The name of the song is called "Haddocks' Eyes"' I've seen used as a good illustration of pointers and indirection in computer science.
I have noticed, after reading all previous answers, that no woman mathematician was listed.
My first impression is that most women mathematician born before 1960 (or maybe until today), are outstanding in non mathematics areas as they must have been activist for just having the chance of studying. Many of them were not even recognized or could work as mathematicians, because universities didn't hire women professors. A sad example is that of Emmy Noether who taught for seven year without income from the university. But besides this, there are many women contributing for medicine and biology, social equity, and inclusions of minorities in mathematics. But they are ghosts, as we do not see or know them. Here is a list with some of them: https://en.wikipedia.org/wiki/List_of_women_in_mathematics
Here are two examples:
Sophie- Germain, who, besides being a important mathematician of her time, contributed to elasticity theory and to philosophy - her philosophical work were admired and cited by Auguste Comte. She has struggled to study in her time but were recognized by the great mathematicians Lagrange and Gauss, even after they discovered that she was not Monsieur LeBlanc. ( Her history is worth reading: https://en.wikipedia.org/wiki/Sophie_Germain )
More recently, we have Eugenia Cheng, who is a category theorist and an accomplished pianist. She is also a writer engaged in mathematical popularization and has a column in The Wall Street Journal.
I went to a mathematics lecture by Eugenia once and she played the keyboard in the middle of it ;)
Kovalevskaya had manifold interests and wrote a novel, but apparently it was not very good (to go by my memories of what is stated in Ann Hibner Koblitz's excellent biography, which I unfortunately don't have at hand). Perhaps we need a separate question on "first-rate mathematicians who can't focus and keep doing non-outstanding work in other fields", taking care of course not to forget elements of the set who are women.
Wouldn't Ada Lovelace qualify? Does writing software fall under non mathematical work?
"Kovalevskaya had manifold interests" Pun intended, @HAH?
Lilian Pierce has also introduced musical interludes in her mathematical talks. She won the 2018 Sadosky Prize for research that "spans and connects a broad spectrum of problems ranging from character sums in number theory to singular integral operators in Euclidean spaces" including in particular "a polynomial Carleson theorem for manifolds". She is an associate professor of mathematics at Duke University, and a von Neumann Fellow at the Institute for Advanced Study. By age 11 she began performing professionally as a violinist. https://en.wikipedia.org/wiki/Lillian_Pierce
We will probably need a separate, (very) long list of "mathematicians who are talented instrumentalists".
@HAHelfgott I see, you like being in lists:)
Noam D. Elkies, who gave an answer to this question, is himself also an accomplished composer.
Interesting. Is he currently active in mathematics?
OMG. He is also winner of the 1996 World Chess Solving Championship!! duo to wiki.
I think it would be topologically satisfying if Noam could write something about the achievement in physics of some mathematician gmvh ;-).
Although he states his rating is nothing to brag about, he's also a chess master. Take it from a guy that has been playing casually since age 5, that isn't an easy feat.
Fields medallist Cédric Villani is a French politician.
https://en.wikipedia.org/wiki/List_of_mathematician-politicians has a long list of people who "achieved notability both as academically-trained mathematicians (with a graduate degree, or published in mathematical journals) and also as elected politicians (at a state or national level)." Daniel Biss was a member of the Illinois Senate. Kathleen Ollerenshaw was Lord Mayor of Manchester. Paul Painlevé was Prime Minister of France.
The mathematician Borel was a cabinet member under the mathematician Painleve, who was Prime Minister of France.
I don't believe that a mathematician can work as politician successfully. because if you take a glimpse of what most of politicians have done during the history by deceiving, lying, non-logical offering against other countries, hypocrisy, etc (e.g. USA and Br or even Fr) while a mathematician cannot left his/her logic.
@C.F.G there is no shortage of deceiving, lying, hypocritical and/or illogical mathematicians.
@C.F.G There is nothing stopping a mathematician from asserting something he or she believes to be untrue in the hope of ultimately gaining something from this deception ("suppose for a contradiction that ... ")
https://en.wikipedia.org/wiki/Ahmed_Chalabi#Western_education
Paul Painlevé was also the first person to be an airplane passenger in France.
@GerryMyerson also worth noting that there do exist some quite successful non-deceiving, non-lying, non-hypocritical, non-illogical politicians (although to be fair, there is in fact a shortage of such).
@C.F.G: Mathematicians are human beings, with biases and beliefs like every other human being. For example about the current economical system: is it efficient? How can we fix/improve it? What fraction of mathematicians read peer-reviewed academic literature on economics?
@C.F.G It's very easy to do that without leaving logic: Axiom 1: I want money; Axiom 2: I want power; Axiom 3: Me > anybody else. ¯_(ツ)_/¯
Of course they are human but watch this for a moment.
The politic is not similar to math. A mathematician can work alone successfully, but being successful politician depend on the other powerful countries. Of course it is not too hard being successful politician if you are US alliance. I don't want to to turn discussion into politic.
But he hasn't done anything as a politician. Seating in the parliamant to sleep and gulp down your 5000 euros (after tax) every month can't count as "proeminent non-mathematical work".
@Libli I confess that I only have an existence proof of his political career, and know nothing about its success or otherwise.
Omar Khayyam was a Persian mathematician, astronomer, philosopher, and poet. As a mathematician, he is most notable for his work on the classification and solution of cubic equations, where he provided geometric solutions by the intersection of conics. Khayyam also contributed to the understanding of the parallel axiom. As a poet, he gave us the Rubaiyat of Omar Khayyam – "A Jug of Wine, a Loaf of Bread—and Thou."
+1. Rubaiyat in MIT website?
Yes, that's the one.
He was also one of the contributors to Solar Hijri calendar which is more accurate than the Gregorian calendar. (The GC has 1 day non-accurate after 3320 years while SHC has 1 day non-accurate after 3770 years!!)
I think it is not clear that the mathematician Khayyam is the same as poet Khayyam. at least in Iran, I hear arguments about it between historians.
Jim Simons - probably best known to mathematicians for his work on the secondary characteristic classes developed various mathematical models for market trading that have been highly successful. He is an outstanding philanthropist, using much of the income from his financial engineering work to support lots of research activities through e. g. the Simons foundation: the Simons investigators awardees list includes such names as Aaronson, Aganagic, Bhargava, Daubechies, Eskin, Kapustin, Katzarkov, Kitaev, Mirzakhani, Okounkov, Ooguri, Poonen, Rouquier, Seidel, Tao, ...
By "best known" I assume you mean best known published mathematics? The Medallion Fund, Renaissance Tech, and the Simons foundation are more well known overall by a factor of thousands.
@BradyGilg Thanks, tried to clarify. Feel free to edit it, it is community wiki
From the known histories, the early strategies used by Medallion Fund (which became the basis for Renaissance Technologies) were derived basically by James Ax. Later strategies were created by Elwyn Berlekamp. It seems Simons' contributions were more in the business and management aspects. Not saying this to lessen him. Realistically, given the lack of success Berlekamp had later in finance, management (and hiring) practices may be more significant than actual strategies. I think this is also more inline with the intent of the question.
@Chan-HoSuh Robert Mercer is an electrical engineering PhD who also made significant contributions to Renaissance Tech, but much greater contributions to digital image processing. I did not know about James Ax. Thank you!
Anna Kiesenhofer is a mathematician at Lausanne. She works in PDE, and has nine publications listed at researchgate.
She just won a gold medal in cycling at the Tokyo Olympics.
see also https://meta.mathoverflow.net/a/5102/11260
This deserves a separate thread, on mathematicians who are into sports. Incredible achievement!
Emanuel Lasker was a great mathematician and is regarded as one of the best chess players ever. If it counts.
Lasker is mentioned in one of the comments on the original post.
I think it's appropriate for that comment to be turned into an answer, especially since it's CW.
And indeed World Chess Champion from 1894 to 1921, the longest period of time anyone has been World Chess Champion.
Interesting! I knew him first as a chess player - I didn't know that he was also a mathematician.
Emanuel Lasker was a great chess player, one of the best ever, and a not insignificant mathematician: he proved a basic result on ideal theory - it's familiar (in a more general form due to Noether) to anybody who has taken a course in algebraic number theory.
Another way to put it is that he was someone who had an excellent paper a couple of years after his thesis, unfortunately never got a permanent job, and decided to focus on other things (including one at which he was the greatest of his age).
Max Euwe was also a wolrd champion in Chess, had a PhD in mathematics and was a professor of Computing Science at Rotterdam U.
Here is a link to another answer mentioning Emanuel Lasker.
Tom Lehrer published a couple of papers in mathematical statistics, and taught Mathematics at University of California, Santa Cruz for many years, but is undoubtedly better known for his three albums of humorous songs.
Let us not forget his work on analytic and algebraic topology of locally Euclidian metrization of infinitely differentiable Riemannian manifolds.
@Aurelio : And this, he knew from nothing!
Incidentally, Lehrer has recently put all his lyrics in the public domain, and made them available for download until 2024. https://tomlehrersongs.com
Bertrand Russell was a mathematician well-known for his philosophical and political work.
Bertrand Russell was a philosopher who did also some math. His work in philosophy is still taught, his work in mathematics is not.
@Michael The Principia are still a well-researched subject. The interest in it is by no means just historical.
@MichaelGreinecker : Russell's type theory is still sometimes taught since type theory is the foundation of certain programming languages and proof assistants.
@TimothyChow: those programming languages and proof assistants are based on descendants of Church’s type theory. Russell’s type theory was very different, and I’ve never heard of it being taught in this connection. That said, I agree that Principia continues to be very relevant.
@AndrésE.Caicedo Can you point me to some modern mathematical research on PM?
@MichaelGreinecker The question is about "prominent non-mathematical works of mathematicians", not "non-mathematical works of prominent mathematicians". If someone dismissed your own work as "not being taught anymore" in 2070, would you think this were appropriate? He wrote a three-volume book on the foundations of mathematics, he was a mathematician (among other things).
@NajibIdrissi My point is that by all reasonable standards Russell was a philosopher who also did some math (with Whitehead) that grew out of his philosophical work. TO me this answer reads a bit like "Edward Witten is a mathematician well-known for his work in physics."
@MichaelGreinecker If a Fields medalist does not deserve the title "mathematician" by your standards, I am not sure we are going to get anywhere with this discussion. I don't know why you are so insistent that if someone is to be described as a mathematician then that's all they should be.
@MichaelGreinecker Surely Russell's paradox is still relevant (even if it may be the Russell-Zermelo paradox...).
@NajibIdrissi I have no problem with calling Witten a mathematician in many contexts, his Fields medal is obviously why I have chosen him as an example. I have a problem with calling someone a "a mathematician who also did X" if X was what the person did primarily. I would find it less objectionable to describe Russell as a mathematician and a philosopher instead of describing him as a mathematician that also did philosophy. It is belittling other fields and that is what I object to.
@Peter-ReinstateMonica You are right. Proving Frege's system inconsistent was of major mathematical importance.
@MichaelGreinecker No one else but you wrote that anyone is "a mathematician who also did X"... The OP wrote that Russell was "mathematician well-known for his philosophical and political work" which makes it very clear that Russell is... well-known for his philosophical and political works, implicitly not his math. Since your objection is about phrasing more than substance, at least quote things correctly...
@NajibIdrissi My objection is about substance. If we consider all instances in which someone whose primary work lied outside of mathematics but who did also math as an example, the list would certainly explode. I am sorry if understood what I wrote as a direct quote.
I don't understand the distinction, can a person not be both a philosopher and a mathematician? (Leibniz is the best example).
You know you’ve made it when people spend inordinate amounts of time arguing whether you were a mathematician who did philosophy or a philosopher who did mathematics, all the while completely ignoring your Nobel Prize in Literature.
@DanRomik: Not when it is idolization rather than actual acknowledgement of real contribution.
Russell himself was apparently insecure about his status as a mathematician. According to Littlewood (in the posthumous edition of his Miscellany) "He had a secret craving to have proved some straight mathematical theorem."
Littlewood goes on: "As a matter of fact there is one: $2^{2^a} > \aleph_0$ if $a$ is infinite. Perfectly good mathematics."
Felix Hausdorff wrote philosophical works, essays, poems and plays under the pseudonym Paul Mongré. Let me quote from the information of the Hausdorff Center for Mathematics in Bonn:
https://www.hcm.uni-bonn.de/about-hcm/felix-hausdorff/about-felix-hausdorff/
Hausdorff pursued, especially during the early years in Leipzig, a
kind of double identity: as Felix Hausdorff, the productive
mathematician, and as Paul Mongré. Under this pseudonym, Hausdorff
enjoyed remarkable recognition within the German intelligentsia at the
end of the 19th century as a writer, philosopher and socially critical
essayist....Between 1897 and 1904, Hausdorff reached the peak of
his literary-philosophical accomplishment: during this period, 18 of a
total of 22 works were published under his pseudonym. These included
the volume of aphorisms Sant’ Ilario: Thoughts from Zarathrustra’s
Country, his critique Das Chaos in kosmischer Auslese, a book of poems
entitled Ekstases, the farce Der Arzt seiner Ehre, as well as numerous
essays....The play was Hausdorff’s greatest literary success, as it was
performed over 300 times in 31 cities.
From Wikipedia: Hermann Günther Grassmann (German: Graßmann, pronounced [ˈhɛʁman ˈɡʏntɐ ˈɡʁasman]; 15 April 1809 – 26 September 1877) was a German polymath, known in his day as a linguist and now also as a mathematician.
May be it should be said that he left Mathematics disappointed by the reception of his work and because his posibilities of teaching math were very low due to really bad reports, for example by Kummer. His book was almost ignored for more than two decades; now it is considered the father of linear algebra.
@Xarles, have you read Kummer's report, and if so, do you know if it can be found online? What you wrote is a common narrative; I also heard another one - that Kummer's report was completely accurate, and whatever in Grassmann's writing we now deem important was buried in the many pages of his unclear, unrigorous and hardly readable text. I have no opinion of my own though.
When I looked into Grassmann's book Die Lineale Ausdehnungslehre, ein neuer Zweig der Mathematik from 1844, my impression was, that he aimed to explain his ideas and results to philosophers like Kant. The second edition Die Ausdehnungslehre. Vollständig und in strenger Form begründet from 1862 seems already like a standard text on linear algebra.
I think that the name of Archimedes immediately springs to mind. Among its inventions are the block-and-tackle pulley systems based on the lever, the screw, the parabolic reflectors used to burn Roman ships attacking Syracuse, the mechanical planetarium.
The story of the parabolic reflectors used to burn Roman ships attacking Syracuse is not true. It is a legend.
The story is in Polybius, Plutarch and Lucian of Samosata. The Wikipedia page on Archimedes says that there are still an ongoing debate about its truth; however, some models of reflectors have been built demonstrating that the construction of such weapons was actually possible with the technology that Archimedes had at his disposal.
https://en.wikipedia.org/wiki/Archimedes
The wikipedia page mentions some experiments indeed but the weapon requires a lot of luck. A catapult or flaming arrows are far much easier to use for the same effect.
Paul Painlevé was briefly prime minister of France on two separate occasions, as well as holding many other government posts including Minister of Defense. In mathematics, he is best remembered for his contributions to nonlinear differential equations.
I mentioned Painlevé in a comment on the answer that proposed Villani.
Before being mainly involved in politics (mostly $\ge 1910$), Painlevé had (1897-1906) a significant involvement in defending Alfred Dreyfus, see Des mathématiciens dans l'affaire Dreyfus (French)
Gauß contributed to the development of the telegraph.
Gauß was also an astronomer, metrology engineer and land surveyor.
Really?????????
@C.F.G Well, I don't think it was the fundamental theorem of algebra or quadratic reciprocity that got his name attached to the unit of magnetic flux density.
Major inventions are not usually made by a single person (there were lightbulbs before Edison), and the telegraph is no exception. But Gauss made an essential contribution.
@C.F.G : Is it a coincidence that your initials are the same as Gauss's? If not, your astonishment is a bit amusing!
Persi Diaconis is an accomplished magician, in addition to his mathematical career.
Frank Ryan earned a math PhD from Rice for the dissertation "Characterization of the set of asymptotic values of a function holomorphic in the unit disc." He published two fundamental papers on the set of asymptotic values of a function holomorphic in the unit disc in Duke Mathematical Journal. [the papers, not the unit disc, were in the journal] He was an assistant professor at the Case Institute of Technology, 1967-71. He has an Erdős number of 3. He was also a lecturer in mathematics at Yale University, 1977-86.
But he is best-known as a quarterback in the National Football League, 1958-70. He led the Cleveland Browns to an NFL title in 1964.
John Urschel played in the NFL for a couple of seasons, and has some publications in Mathematics.
Frank Ryan died earlier this week (thought people might like to know): https://en.wikipedia.org/wiki/Frank_Ryan_(American_football)
Martin Hairer has written a widely used professional audio editing software "Amadeus Pro" (https://www.hairersoft.com/).
It is amusing to note that if you find the message board for Amadeus you'll see that Martin still personally replies to every user's troubleshooting request (e.g. https://www.hairersoft.com/forum/viewtopic.php?f=1&t=7697)
Martin Hairer is mentioned also here: Mathematicians with both “very abstract” and “very applied” achievements.
I'm surprise no-one has mentioned Isaac Newton. He spent almost half his career outside academia fighting against forgery at the Royal Mint. He was also an MP and wrote a huge amount on biblical chronology, alchemy, and theology.
Harald Bohr was a member of the Danish national football team for the 1908 Summer Olympics, where he won a silver medal. The semi-final against France was 17-1.
Interesting result!!!
Interestingly, Harald Bohr played professional football also together with his older brother, Niels, who also had a prominent science career.
Ron Graham's mathematical work is probably familiar to most users of this website.
In his youth, he won a title as California state trampoline champion. In 1972 he was elected president of the International Jugglers' Association.
I think later on he also won regional titles in table tennis after getting destroyed by Erdos in a table tennis game and then deciding that he had to improve.
@Hol, D. J. Albers (1996) A Nice Genius, Math Horizons, 4:2, 18-23, DOI: 10.1080/10724117.1996.11974993, "[in 1963] I saw this rather senior guy of 50 [Erdos], already quite famous, playing ping-pong during one of the breaks. He asked me if I wanted to play and I agreed. He absolutely killed me! I had played casual ping-pong but I couldn't believe that this old guy had beaten me. ... I went back to New Jersey ... I bought a table, joined a club, started playing at Bell Labs, and in the State league. I eventually became the Bell Labs champion at ping-pong, and won one of the New Jersey titles."
Also, I think mathematics is the only subject where someone aged 50 is an ''old guy''.
@Hollis I think maybe Ron meant it was (relatively) old for ping-pong.
Nikolai Durov, who launched Telegram with his brother Pavel Durov, has some works on Arakelov geometry.
I think Greg Egan (sci-fi author) qualifies. He has a few papers on arxiv, and made the news for his work on superpermutations, which motivates the title 'Mathematician'. He is also an occasional MO-user.
I think Egan would be a great answer to a different question, of people whose profession is not "mathematician" doing good maths
@YemonChoi why don't you ask that as a separate question? I'd answer Aubrey de Grey.
@NikWeaver : That question arguably already exists.
@TimothyChow and Egan and de Grey are both answers there! Never mind, then.
I am on the fence regarding whether this question should stay open, but while it does, I thought the following example might be of interest:
Peter Rosenthal, perhaps best known to mathematicians for his work on subspace lattices of operator algebras (see e.g. his book with Heydar Radjavi), developed a 2nd career/mission as a lawyer.
Wikipedia link: https://en.wikipedia.org/wiki/Peter_Rosenthal
Link from comments: 2014 profile in the Toronto Star
While he definitely seems to be vocal about being left-wing, the causes he represent are more often regarding basic human rights than left-wing. E.g., he has represented many cases of police brutality; see this beautiful piece on Toronto Star: https://www.thestar.com/news/crime/2014/01/05/peter_rosenthals_passions_for_law_and_math_make_for_a_beautiful_if_different_life.html
@auniket thank you for the clarification. I must admit that originally I omitted the adjective and then added it at the last-minute in a confused effort at sign-posting, since I couldn't remember the details that I'd heard from friends on the Canadian scene
Per Enflo, a Swedish mathematician known for solving the so called ``basis problem'', one of the problems from Scottish book (about the existence of Schauder basis). Besides being a mathematician he is also known as a talented pianist.
I will add a link to another answer about Per Enflo.
Joseph Fourier was also an Egyptologist.
He was a bit more than just that. Excerpts from hiw biography on wikipedia include: "He took a prominent part in his own district in promoting the French Revolution, serving on the local Revolutionary Committee. ... Fourier accompanied Napoleon Bonaparte on his Egyptian expedition ... Cut off from France by the British fleet, he organized the workshops on which the French army had to rely for their munitions of war ... In 1801, Napoleon appointed Fourier Prefect (Governor) of the Department of Isère in Grenoble, where he oversaw road construction and other projects."
@HJRW Equally, one should also mention Gaspard Monge, who was on the same expedition to Egypt, similarly held political appointments during the Revolution and all the while also is known for his results in geometry and as being the father of optimal transport.
Mathematician Eric Temple Bell
President of the MAA, 1931-32
Author of the book Men of Mathematics
Bell numbers, Bell polynomials
etc.
Also was a successful science fiction author under the pseudonym John Taine.
See the fascinating book
Reid, Constance, The Search for E.T. Bell, MAA spectrum, The Mathematical Association of America, 1993
Otto Iulievich Schmidt is now best remembered for his polar expeditions and a geophysical institute in Russia bears his name. The English wikipedia does not mention it, but he was also one of the founders of the modern theory of groups and had a strong influence on A.G. Kurosh.
Oh -- that's the Schmidt from Krull-Remak-Schmidt!
In the late 1950s, I had a textbook on group theory written by O.J.Schmidt (in Russian). It was very nice.
Richard Garfield is a mathematician and former math professor who is nowadays famous as the inventor of the wildly successful card game Magic: The Gathering, and many other card and board games. Here is what Wikipedia says about his math background:
After college, he joined Bell Laboratories, but soon after decided to continue his education and attended the University of Pennsylvania, studying combinatorial mathematics for his PhD. Garfield studied under Herbert Wilf and earned a Ph.D. in combinatorial mathematics from Penn in 1993. His thesis was On the Residue Classes of Combinatorial Families of Numbers. Shortly thereafter, he became a professor of mathematics at Whitman College in Walla Walla, Washington.
A colleague of mine who is somewhat knowledgeable about Magic: The Gathering once claimed that the game became as successful as it was because it was invented by a mathematician who systematically set out to create the most addictive game possible. I cannot say how much truth there is to that claim.
It is odd that there are 4 magician in the answers!!
Years ago, I visited a Penn math professor who complained that the graduate students were spending all their time in the department lounge playing card games.
Years ago when I was at Cambridge the graduate students spent their time in the department lounge playing Conway games.
In addition to his work in number theory, Carl Størmer made important contributions to the study of the aurora borealis. Here he is conducting an experiment rather far from his blackboard:
Størmer was awarded the 1922 Janssen Medal of the Paris Academy of Sciences for his work on the aurora. https://en.wikipedia.org/wiki/Janssen_Medal_(French_Academy_of_Sciences)
Alexander Esenin-Volpin was a mathematician well-known for his work as a Soviet political dissident.
EDIT: I'm sorry for writing the above so glibly. Among other things, Esenin-Volpin was repeatedly imprisoned or else confined to mental institutions for political reasons -- according to wikipedia he spent 6 years cumulatively in either of those situations. I'd love if somebody more competent than I would write something more informative and fitting.
The same can be said about Shafarevich too
Can't resist mentioning a non-answer to the question: someone prominent for non-mathematical work who was a maths+physics student at university: https://www.nobelprize.org/prizes/literature/1970/solzhenitsyn/biographical/
There was a gulf of difference between Esenin-Volpin and Schafarevich; it'd be simpler to remember Schafarewich as an outstanding mathematician and let's mercifully forget about the rest of his story.
@WlodAA I did not say that they are similar in any other respect, except that "was a mathematician well-known for his work as a Soviet political dissident" applies to both; maybe it'd be also simpler to remember only part of what we happen to remember, so let's mercifully forget about the rest of it?
Gian-Carlo Rota in addition to being an influential combinatorialist was a philosopher, and his philosophical writing was not in the tradition often thought of as being closest to math ('analytic philosophy') but was rather inspired by phenomenology. Apparently this heterodoxy caused some consternation from e.g. his colleagues in the philosophy department at MIT.
Emily Riehl is a professor of mathematics at Johns Hopkins. She does transformative work in abstract homotopy theory, and has won many grants. She is also a professional Australian rules football player.
René Descartes
Mathematically you're likely to know him for Cartesian geometry.
Philosophers will know him for "Cogito, ergo sum"/"Je pense, donc je suis"
Quoting wikipedia: "Descartes is also widely regarded as one of the founders of modern philosophy." I suspect, as with Russell, people might argue Descartes was a philosopher who did a little mathematics, but ignoring the eponym of Cartesian coordinates is too hard to do.
Rene Descartes walks into a bar and orders a drink. When he finishes his drink, the bartender asks him if he would like another. Descartes replies, “No, I think not,” and disappears in a puff of logic.
I think Descartes did more than "a little" Mathematics
I agree, I was just attempting to preempt similar comments to the Bertrand Russell (https://mathoverflow.net/a/369713/164087) answer.
Descartes did much more than systematising the $(x,y)$-coordinates. It is worthwile to read his essay "La géométrie", where among other things he studies some plane curves and without much overstating prefigures algebraic geometry.
Ruggero Freddi is an Italian mathematics lecturer (holding a PhD) and a former gay pornographic film actor known professionally as Carlo Masi. Here you can read his thesis.
this is not work of a mathematician.
@C.F.G Sorry, but now you will have to justify and explain to me why exactly do you consider that Freddi or its previous work do not qualify to be answers to what you asked. I pretty much think that it fits in both of your questions (specially the second). He is a mathematician and did "famous" work out of mathematics (unless you think that gay porn is mathematics). There is a dichotomy that I do not understand (in fact, I do but I want to expose you). So either you consider that acting in gay porn is a job or you consider that Freddi is not a mathematician. You will have to be more clear...
He was not a mathematician during in his previous work.
@C.F.G He was actually forming himself to be one and in your question you never mentioned that these works have to happen at the same time. In fact, this restriction would not allow several previous answers. E.g., this would eliminate some you own answer of Glen Bredon. Villani would not qualify either because he stopped doing math (for the moment).
S
So either your question is ill posed and thus it should be closed because it already attracted answers that you did not mean or your problem is other.
@C.F.G I am going to tell you what I think. If you consider that being gay porn actor is not a job you are plainly wrong. Otherwise, the possibility is that you consider that Freddi is not a mathematician, which is quite problematic given the fact that he holds a PhD degree based in the linked thesis. Having a PhD generally mean that you did research work in Maths and it is more than enough to be considered as a mathematician. So, unless you found a flaw in the thesis of Freddi (and even in this hypothetical case), he qualifies to your question. However, I think that your problem is different.
All posts you mentioned first were mathematician then they become or did other thing.
@C.F.G I really think that your question is educative and deserve to stay so I will not vote to close. Your question is a place to see that one can become mathematician from other interests and can be a mathematician and something else. However, I think that having a gay porn actor in this wiki is problematic for you. I really hope that you explain yourself better. Your downvote and your comment have not being correctly justified yet and I wish that the problem does not relate to the fact that he is gay or porn actor (sadly I think that it does). It is pretty sad seeing this mentality here.
This reminds me that Catalan mathematician and logician Angel García-Cerdaña (Cerdanya) is also a known TV actor. Am I to understand that for some reason he does not qualify for this question either?
@C.F.G Again, in your question you never mentioned that this order was necessary to keep. Your question is literally "Are there mathematicians who have done outstanding/prominent non-mathematical work like inventions, patents, solving social/economical/etc. problems, papers in these areas, etc.?" Where is the restriction that are now mentioning exactly? I don’t see it. Freddi, mathematician and former gay porn actor and gay porn actor is a non-mathematical job as I see it.
You can ask a new post like "prominent non-mathematicians later become mathematician" if you are interested in. But this is distinct from mine. so pls do not comment anymore.
@EmilJeřábek Sadly, I think that "being gay" and "appearing naked in TV" is the real problem here.
It's hard to interpret the downvotes on this answer as anything other than narrow-minded prejudice. Would it have been downvoted if Freddi had been a Shakespearean stage actor?
There are other people listed who did much of their non-mathematical work before they became mathematicians. This includes Persi Diaconis, Danica McKellar, and Frank Ryan
@TomLeinster : I did not downvote or upvote. Narrow-minded prejudice is certainly one possible explanation for the downvote. However, I just checked several other answers, and there are downvotes in several cases which cannot be explained by prejudice (in the sense you mean it). The pattern seems to be that if the achievement is either not obviously "prominent and outstanding" or if it deviates too far from "inventions, patents, solving social/economical/etc. problems, papers" then it might attract a downvote. For example, as I write this, I note that Grothendieck attracted a downvote.
Benjamin F. Logan was primarily an electrical engineer who spent his career at Bell Labs, but he has 37 publications listed in Mathematical Reviews. His best known mathematical work is his 1977 paper with Larry Shepp, A variational problem for random Young tableaux, in which they proved that if $L_n$ is the expected length of the longest increasing sequence in a randomly chosen permutation of $\{1,\dots,n\}$ then $\lim_{n\to\infty} L_n/\sqrt{n}\ge 2$.
Long ago when I used to go to bluegrass festivals, I sometimes saw a bluegrass fiddler named Tex Logan who played with Peter Rowan then, but had earlier played with such greats as Mike Seeger, Bill Monroe, and The Lilly Brothers & Don Stover. Tex Logan also wrote songs recorded by Johnny Cash, Emmylou Harris, and Bob Dylan.
It wasn't until many years later that I learned that B. F. Logan and Tex Logan were the same person.
Alexander Grothendieck was a political activist and spiritualist.
this is what I was looking for
Raymond Smullyan was a
mathematician, magician, concert pianist, logician, Taoist, and philosopher.
Another politician would be Éamon de Valera, who graduated in mathematics and taught at various schools (and applied for a professorship, but without success), but then became a rather influential Irish politician.
I think, one can add Francesco Faa di Bruno to this list, see https://en.wikipedia.org/wiki/Francesco_Fa%C3%A0_di_Bruno. He is to my knowledge the only beatified mathematician.
Boris Berezovsky, a Russian oligarch and government official, was a professional mathematician.
Jerry McNerney (Wiki page) is a US congressperson from California, with a PhD in differential geometry.
Nowadays he's more known as a congressman than as a mathematician, but every now and then he will give quick floor speeches about math or mathematicians.
See, e.g., his tribute to Mirzakhani.
Dutch mathematician Alexander Rinnooy Kan is also a politician and businessman. He used to be a member of the board of directors of ING Group, served as the Chairman of the Social and Economic Council that advises the government, and was a member of the senate.
According to a national newspaper, he was the most influential person in the Netherlands in 2007, 2008, and 2009.
Daniel Biss received his PhD in mathematics at MIT in 2002, then was an Assistant Professor at the University of Chicago until 2008. He won the 1999 Morgan Prize for outstanding research as an undergraduate. However, in 2007, a serious flaw was discovered that destroyed the main results of papers he had published in the Annals of Mathematics and in Advances. He is now a State Senator in Illinois. In this position, he has worked on legislation to "allow for automatic voter registration," to "elect a number of statewide offices by ranked-choice ballot," and on healthcare, among other things.
The numerical analyst, Manil Suri, is also an accomplished novelist. He has written a trilogy of novels. The first of which, The Death of Vishnu, was long listed for the Booker
Prize.
I don't understand why this is down voted. He's a quite accomplished mathematician (https://scholar.google.com/citations?user=lFWFsSkAAAAJ&hl=en&oi=ao) and an award winning novelist. Seems to fit well.
I've seen a couple of mentions of Emanuel Lasker, a former world chess champion. Here's his non-mathematical academic works, which include a play "History of Mankind" cowritten with his brother:
Kampf (Struggle), 1906.
Das Begreifen der Welt (Comprehending the World), 1913.
Die Philosophie des Unvollendbar (sic; The Philosophy of the Unattainable), 1918.
Vom Menschen die Geschichte ("History of Mankind"), 1925 – a play, co-written with his brother Berthold.
The Community of the Future, 1940.
In his "Kampf" he foresaw the application of game theory in 20th century social sciences. He also wrote on other games besides chess as well. Here's a list of those books.
Encyclopedia of Games, 1929.
Das verständige Kartenspiel (Sensible Card Play), 1929 – English translation published in the same year.
Brettspiele der Völker (Board Games of the Nations), 1931 – includes sections about Go and Lasca.
Das Bridgespiel ("The Game of Bridge"), 1931.
Also, don't forget Max Euwe. Max Euwe was also world chess champion and president of FIDE (the international chess body). Dr. Euwe wrote on the Thue-Morse sequence and it implying that, according to the rules of chess at the time, a game could be played as an infinite game without resolution under certain circumstances. He taught mathematics at one point and was a professor of computer science at the Universities of Rotterdam and Tillberg.
I read somewhere that Max Euwe won some kind of amateur boxing title. Is that true?
http://members.tripod.com/HSK_Chess/euwe.html
I'd never heard of this before, but it is true. The link is confirmation.
He also didn't play chess professionally, but still won the world championship.
It was the amateur boxing championship of Europe that Euwe won. I don't have a year though.
Tony Scholl is also a bassist with the Cambridge Philharmonic.
Sergio Fajardo, the former mayor of the Colombian city of Medellín, wrote a dozen papers in model theory before switching to politics.
Crap, I accidentally duplicated this answer in 2021
Sergio Fajardo is a mathematician who was also Mayor of Medellin after the death of Pablo Escobar, was Governor of Antioquia (the state that Medellin is in) from 2012–2016, ran for President of Colombia in 2018, and has announced plans to run again for President. In mathematics, he earned his PhD from UW-Madison and was a professor at the Universidad de los Andes and the Universidad Nacional of Colombia.
In politics, he completely revitalized the city of Medellin, building new parks and libraries, and rebranding the city. The Park of Lights is a great example. That park used to be (figuratively) the darkest place in the city. Escobar had complete control of it and any (non-corrupt) police officer who came nearby would be killed. The park contains two of the oldest buildings in the city, and they were full of squatters. Now, the park is full of life, with Colombian bamboo (Guada kindiana), tons of people, and these giant lightsaber-looking lights that run all night. It's never dark there. It's the safest place in the city. The buildings that used to house squatters now house the department of education. Fajardo also built beautiful libraries to serve the poorest neighborhoods of the city, to give children a chance at a life based around something other than drugs. He was named "best Mayor of Medellin" in 2007. Regarding his time as governor, according to Wikipedia
During his administration, Antioquia experienced the best national performance in open government, transparency and investment of oil royalties (according to the National Planning Department and the Anti corruption Office of Colombia). He was named the best governor of the country in 2015 by the organization Colombia Líder.
Claude Elwood Shannon was also an inventor. I recall he also invented a rocket-powered pair of boots, but I cannot seem to find the source anymore.
Vladimir Voevodsky. He won a Fields Medal for inventing motivic homotopy theory and proving the Milnor Conjecture and in his later years worked on homotopy type theory.
He was a world-class photographer. This is mentioned in his "in memoriam" by Friedlander and you can also see some of his photos hosted by IAS. On that page, it mentions that his photography was included in a 2011 exhibit. I've heard that some of his photos were published by top magazines, like National Geographic, but I don't know all the details. If anyone else does, please add!
Nasīr al-Dīn al-Tūsī was a persian polymath of thirteenth century. His contribution to trigonometry includes the plane law of sines. He had a prominent place in the court of Hulagu Khan (a grandson of Genghis Khan and the founder of the Ilkhanate Empire), and benefited from Khan's patronage to found the Maragheh Observatory. The legend even has it that he was influential in persuading Hulagu to siege Baghdad in 1258 which ended the 500 years old Abbasid Caliphate.
Gunnar Carlsson is a professor at Stanford famous for his work in K-theory, and on the Segal Conjecture in homotopy theory. He is also the President and Founder of the company Ayasdi, described as
A machine intelligence software company that offers a software platform and applications to organizations looking to analyze and build predictive models using big data or highly dimensional data sets. Organizations and governments have deployed Ayasdi's software across a variety of use cases including the development of clinical pathways for hospitals, anti-money laundering, fraud detection, trading strategies, customer segmentation, oil and gas well development, drug development, disease research, information security, anomaly detection, and national security applications.
Patrick Billingsley, author of two well-known books in probability theory, was also a stage and screen actor.
Whoa. [Preston & Logan, 1989]
Although John Urschel is already mentioned in a comment to the answer about Frank Ryan, I think he deserves his own answer. Urschel was a football player first at Penn State and then with the Baltimore Ravens. He majored in math at Penn State and then, allegedly without the knowledge of the Baltimore Ravens, enrolled in the MIT math PhD program. Urschel will receive his PhD in spring 2021 and already has an impressive list of publications
He wasn't a mathematician before or during playing football?
Based on the years of his publications, it appears that he did write research math papers while he was playing football at Penn State and then for the Baltimore Ravens. Given the physical and mental energy, as well as the time needed each week to prepare for a football game, this is quite remarkable.
Merely meant as an interesting and amusing fact. People are not born as mathematicians. At the age of 14, long time before his mathematical career and winning the Fields medal, Wendelin Werner played a role in 'The Passerby' at the side of Romy Schneider. She died a few weeks after the movie premiere.
"People are not born as mathematicians." -- Some do.
We disagree. That's fine.
Art Benjamin is a professor of mathematics at Harvey Mudd College, with more than 100 publications. He is also an accomplished magician and was the 1997 American Backgammon Tour Player of the Year.
Baruch Spinoza was a mathematician, philosopher, and physicist "involved in important optical investigations of the day." His masterpiece, The Ethics, is "written with a forbidding mathematical structure modeled on Euclid's geometry." He was an early Enlightenment thinker of the Dutch Golden Age.
Alan Turing is a mathematician famous for his contributions to the foundations of computer science, for being a codebreaker in WW2, and for being persecuted by the UK government for being homosexual. Readers have probably heard of the Turing test in artificial intelligence. I would argue that his contributions to computer science don't count as "non-mathematical work" but that his contributions to biology do. For example, his paper The Chemical Basis of Morphogenesis is highly cited and is the reason the name "Turing patterns" is used to describe zebrafish embryos.
Hans Freudenthal was a famous topologist. Indeed, the Freudenthal suspension theorem is the foundational result you need to get stable homotopy theory off the ground. He also invented a language, Lincos, "to make possible communication with extraterrestrial intelligence." And he invented a famous puzzle. And an asteroid is named after him.
The famous puzzle looks pretty mathematical (though not topological) itself.
Per Enflo is famous for solving Banach's basis problem, Grothendieck's approximation problem, and the invariant subspace problem for general Banach spaces, and has other fundamental research in linear and non linear geometric functional analysis. He also has done research in population dynamics. A child prodigy as a pianist, he continues to give concerts in Europe and the United States, but you can hear him at home by creating a Per Enflo station on Spotify.
There is such an answer (although less detailed) already: https://mathoverflow.net/a/369738/4312
James Garfield submitted an original proof of Pythagoras's theorem to the New-England Journal of Education and which was published in the April 1, 1876 issue.
He taught the liberal arts and practised law, he was also a brigadier-general in the Civil War.
He was also the 20th President of the USA, inaugurated in 1881. He developed the proof whilst a member of Congress.
I'm not sure that publishing an original proof of Pythagoras qualifies someone as a mathematician.
@Gerry Myerson: What does these days? Some people think that only if you dedicate your life to research level maths makes you a mathematician. I take a broader view. But then I'm a democrat.
Doctor Ahmed Chalabi earned a PhD from the University of Chicago in 1969, founded Petra Bank in 1977, was sentenced in absentia in Jordan for bank fraud in 1992, and (allegedly) very successfully lobbied for the 2003 Invasion of Iraq.
I think this one falls more under the "prominent figure who has a PhD in mathematics" category (as do a number of the other posts, to be fair)- but, wow! this guy has a wild life story!
@SamHopkins Indeed. However, I assume that the University of Chicago would not have granted him a doctoral degree had he not made a non-trivial contribution. Perhaps I assume too much.
Danica McKellar may not qualify as a mathematician, but (all quotes below are from Wikipedia)
McKellar studied at the University of California, Los Angeles where she earned a Bachelor of Science degree summa cum laude in Mathematics in 1998. As an undergraduate, she coauthored a scientific paper with Professor Lincoln Chayes and fellow student Brandy Winn titled "Percolation and Gibbs states multiplicity for ferromagnetic Ashkin–Teller models on ${\bf Z}^2$." Their results are termed the "Chayes–McKellar–Winn theorem". Later, when Chayes was asked to comment about the mathematical abilities of his student coauthors, he was quoted in The New York Times, "I thought that the two were really, really first-rate." For her past collaborative work on research papers, McKellar is currently assigned the Erdős number four, and her Erdős–Bacon number is six.
Also, she
wrote six non-fiction books, all dealing with mathematics: Math Doesn't Suck, Kiss My Math, Hot X: Algebra Exposed, Girls Get Curves: Geometry Takes Shape, which encourage middle-school and high-school girls to have confidence and succeed in mathematics, Goodnight, Numbers, and Do Not Open This Math Book.
Her acting career, in brief:
She played Winnie Cooper in the television series The Wonder Years from 1988–1993, and since 2010 has voiced Miss Martian in the animated superhero series Young Justice.
In 2015, McKellar was cast in the Netflix original series Project Mc2. She appears in several television films for Hallmark Channel. She is the current voice of Judy Jetson from The Jetsons since 2017 following Janet Waldo's death in 2016.
Glen E. Bredon is the author of the programs DOS.MASTER for Apple II computers, Merlin (a macro assembler) and ProSel for Apple machines. He was the professor of mathematics at Berkley and IAS and author of worthwhile math books like
Bredon, Glen E., Topology and geometry., Graduate Texts in Mathematics. 139. Berlin: Springer. xiv, 557 p. (1997). ZBL0934.55001.
and
Bredon, Glen E., Introduction to compact transformation groups, Pure and Applied Mathematics, 46. New York-London: Academic Press. XIII,459 p. $ 21.00 (1972). ZBL0246.57017.
Not so prominent I think but I'd like to mention that the MathTime Professional 2 (MTPro2) fonts were designed by Michael Spivak of Publish or Perish Inc., see here.
It has been noted already that Noam Elkies is an accomplished composer.
What I find at least as extraordinary about him is that he can hum-whistle some of Bach's two-part inventions. (Anecdotal "evidence", will remove this answer if it is false.)
Marcel-Paul Schuetzenberger (https://en.wikipedia.org/wiki/Marcel-Paul_Sch%C3%BCtzenberger) is also an interesting example (he started as a physician).
Andrew Blumberg, in addition to being an accomplished homotopy theorist, a full professor at UT Austin, and having published more than 60 papers, also works on issues related to geolocation data and privacy. For example, he worked on an amicus brief for a Supreme Court case.
I learned this answer from another thread where the OP said it didn't quite fit.
Frank Garside did important work in the 1960s related to braid groups. He also invented what is now known as the Garside element, leading to Garside groups. Later, he was the mayor of Oxford.
In a related vein, George Reid was an algebraist who was later mayor of Cambridge in 1990-91.
My memory from my (long ago) days in Cambridge is that George Reid was more of an analyst (Banach algebras?) than an algebraist.
I confess I just copied the other answer. I don't know much about Reid's work at all. His thesis was about topological groups: https://www.genealogy.math.ndsu.nodak.edu/id.php?id=26731 Also, here's a link to his later work in government: https://aru.ac.uk/about-us/governance/strategy-and-leadership/board-of-governors/george-reid
Which Cambridge?
Here is some mathematical work by someone better known for other work.
Isn't this the exact reverse of what the question is asking for?
@YemonChoi Indeed, this answer should be reformulated as “Karl Marx wrote political texts that have gained some measure of celebrity”.
Does any of this go beyond philosophical ramblings about basic definitions? Legit question -- I haven't seen the entire work.
Karl Marx and the Foundations of Differential Calculus by Hubert Kennedy gives a nice discussion of Marx's investigations into calculus. At the time, the received view of the derivative (following Leibniz) was a ratio of infinitesimals, which today we would find lacking in rigor. Marx perceived this difficulty and took some nontrivial steps toward a modern rigorous account.
Abu Rayhan al-Biruni (973 – after 1050) was an Iranian polymath, physicist, astronomer, natural sciences, historian, chronologist and linguist. One of the best his achievements is a method proposed and used by him to estimate the radius and circumference of the Earth almost 1000 years ago.
He was also inventor of minutes and seconds of time.
Charles Kalme, a mathematician (Contributions to the Theory of Discontinuous Groups of Mobius Transformations, Ph.D. thesis, NYU, 1967) and pioneering chess programmer. His achievements as a chessplayer include winning the United States Junior Championship and playing twice in the United States Chess Championship. He was also a master of contract bridge.
The well-known mathematician Henry Peter Francis Swinnerton-Dyer was also an outstanding chess and bridge player in his student days. A variation in Ponziani's Opening that he played in 1949 is cited in 10th edition of Modern Chess Openings.
There is Bertrand Russell, a logician, who had another career as a philosopher as well as an anti-war protestor. He helped publicise what was happening in Vietnam - in fact, according to Paul McCartney of the Beatles, at one point he lived on the same square or street with Russell and he had knocked on his door to introduce himself and Russell proceeded to tell him all about the war. This is, according to Paul McCartney, how the Beatles became involved in the anti-war protest movement, in particularly against the war of aggression by the USA on Vietnam.
Laurent Schwartz, a french mathematician known for his work was on distributions, was also an anti-war activist and who focused on labour activism as well on the colonial war in Algeria.
Grothendieck, who needs no introduction, was also an ati-war activist, focusing on what was then occuring Vietnam. His fater was a revolutionary socialist.
Smale was also another anti-war activist; also Solzhenitsyn and many others.
Sofia Kovalevskaya was one of the first handful of female mathematicians. She was also a writer, having written a well-received autobiography, A Russian Childhood in where she reveals she met Dostoyevsky as a young woman and did not think much of him.
Likewise, Lewis Carroll, also a logician and who wrote the best-selling Alice's Adventures in Wonderland.
Also Abdus Salam, who was a mathematical physicist and had a second career setting up the Triest Centre of Mathematical Physics to help te Third World get on to its feet.
And a special mention for Walter Sisulu, who studied a science degree, but after joining the ANC, was imprisoned several times, and was finally imprisoned on Robbens Island with Nelson Mandela and sentenced to twenty-five years - the same sentence as Nelson. He rose to become the ANC's Secretary-General and Deputy President of the organisation.
https://mathoverflow.net/a/369713/90655
Russell is previously mentioned here. Carroll is mentioned here.
@Noah Schweber: But with very little detail - they deserve better than that. Besides Kovalevskaya and Sisulu are not mentioned.
Solzhenitsyn was not a mathematician. Carroll hardly one, to be honest. Sisulu only studied science, but never worked as a mathematician. There are many students in mathematics (the majority, in fact) who pursue careers not related to mathematics after graduation, so according to your view they would all qualify as answers to this question.
@AlexM: Carroll nevertheless was a logician.
@AlexM: Wot about Abdus Salam - are you going to complain about him too?
This is a community wiki big-list question for a reason. (1) If you feel some other answer has insufficient detail, add that detail to that answer, do not put it in a new answer. (2) Ideally, each answer should only give one example. If you have more people, put each in a separate answer.
Stan Wagon, whose work was featured in several Mathoverflow posts (e.g. an unexpected image, Gaussian prime spirals), has also an entry in Ripley's Believe It or Not due to his Square Wheel Bike.
I think, John F. Nash fits in this category, because of 1994 Nobel Memorial Prize in Economic Sciences. (I am not aware the details of his work, I just know this because of "A beautiful mind" movie!)
The work Nash received the Nobel memorial prize for was from his Princeton PhD-thesis in mathematics. This was not non-mathematical work.
@MichaelGreinecker: thanks for comment. So I should delete this answer but I think it is better to not to do that for those don't know this fact like me!!
Although it is difficult to say exactly what Thales accomplished, if you take what he's given credit for it is immense. By your criteria he would qualify as an answer.
Although he lacked the rigor of proof, he was the first mathematician we attribute a mathematical result to - namely Thales' Theorem used to get the distances of ships at sea.
He was also the first philosopher. With his contention that everything is water we have the first theory that says everything is one.
Also with his contention that everything is water we have a testable claim. This makes him the first scientist.
He also was an astronomer, predicting the eclipse of May 28th, 585 BC.
He even dabbled in business, by legend. There is a famous story attributed to Aristotle, and others, about him. If true, Thales predicted the weather and foresaw a good olive harvest for the coming year. He then bought all the olive presses at a discount, and rented them out during the harvest. This would be the first recorded use of futures to turn a profit.
Legend has it that Thales was an engineer.
I think Joseph of Genesis preceded Thales in the use of futures. Seven good years, followed by seven lean years, and all that. But there's no record of Joseph making any mathematical contributions, Thales has him beat there.
Wikipedia is incorrect then. I've also seen other sources that cite Thales as well.
The old testament reached its current form in the Persian period (538-332 BC). Thales came before that. I suppose if you believe the Bible is to be taken as a real account, then that reference would be older. However, I don't think you'll find a scholar that would argue that Thales wasn't real.
Wikipedia? Who said anything about Wikipedia?
From wikipedia, with their source on the topic: "This first version of the story would constitute the first historically known creation and use of futures, whereas the second version would be the first historically known creation and use of options."
George Crawford, Bidyut Sen – Derivatives for Decision Makers: Strategic Management Issues, John Wiley & Sons, 1996 ISBN<PHONE_NUMBER>943
Albert Einstein's contribution to differential geometry was significant. One could mention also Einstein's tensor notation.
On the other hand, Einstein was an inventor, he had several inventions (patents) to his credit.
He wasn't a mathematician I think.
If you forget that Albert Einstein was a physicist, you will see that he was a strong research mathematician (more than so many mentioned in this thread).
One of his papers on math?
Mathematics can be recorded also outside of papers in mathematical journals. Differential geometers would direct you to Einstein's results on the affine connection.
The probability theory, classical mechanics, the special relativity theory, ... ... ... are all mathematical theories these days.
What is the definition of a mathematician? Who is educated in math or did some math research. Isn't?
Forget "educated". It's the mathematical research impact that counts. Thus, it'd be silly to skip Einstein in this thread.
Professionally, Pierre de Fermat was a lawyer; he had also contributed to physics.
Henry Poincare, Solomon Lefschetz, and Raoul Boot were into engineering when they were young.
Hugo Steinhaus was serious about applications, e.g. about the different feet shapes in order to help shoe designers. The famous Steinhaus slogan was
A mathematician will do it better.
In particular, Steinhaus patented longimeter.
Karol Borsuk invented a successful game during WWII, and it helped him to survive in those hardship years.
Stan (Stanisław) Ulam was the main inventor of the H bomb. He has also invented cellular automata.
Israel Gelfand at his mature stage turned his interests toward biology and medicine.
Rene Thom in his later years became seriously interested in too many things to mention here.
Several American mathematicians, including John Milnor, worked on DOD contracts.
Topologist James Munkres made a significant contribution to the assignment problem -- it's even called Munkres assignment algorithm or Kuhn–Munkres algorithm.
A parallel processor invented by a mathematician in the US, and the technology that followed, had an impact on the fall of communism.
Greg Kuperberg was an early pioneer in the field of computer games when he was in his teens; this helped him to pay his tuition at MIT.
Did Fermat do anything prominent as a lawyer? Did Poincare, Lefschetz, Bott do anything prominent when they were "into" engineering? Did Thom do anything prominent in the unmentionable areas that interested him? Did Milnor et al do nonmathematical things on those DoD contracts? Isn't the assignment problem mathematics? Any details on the mathematician who invented a parallel processor, and how it impacted communism?
@GerryMyerso, it's mixed. OP said outstanding/prominent. But there were over 30 answers before mine, and the average level for these 30+ is way below outstanding/prominent, and overall I believe that also clearly below my answer. ### My weakest case was wonderful Bott who indeed was simply an engineering student. ... Munkres did an algorithm (Erdos would call it computer science, not mathematics).
I see nobody mentioned about John Forbes Nash, who obtained Nobel prize in Economic science for his contribution to economy. Further German Mathematician Carl Fredric Gauss, Von Neumann etc both were not only prominent mathematicians but also prominent physicist as well.
Gauss was already mentioned, https://mathoverflow.net/posts/369701/revisions (I think von Neumann was already mentioned, too).
@GerryMyerson, But Prof John Nash was not mentioned.
Actually, Nash was mentioned in this answer of C.F.G: https://mathoverflow.net/a/369703
@JeremyRickard, I am confused who answered first , he or me.I answered 10 hours ago. During my answer i searched with keyword nash but got no link. That is why i answered as you see my first sentence. Anyway let it remain.I will help
@M.A.SARKAR: I posted my answer 2 month ago.
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2025-03-21T14:48:31.831447
| 2020-08-20T15:07:40 |
369673
|
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|
Stack Exchange
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Periods of the harmonic conjugate and a Dirichlet problem on a multiply connected domain
Any harmonic function $u$ on a simply connected domain in $\mathbb{R}^2$ is the real part of a holomorphic function. If the domain is multiply connected, then this is no longer true: the harmonic conjugate of $u$ may have periods.
I wonder if the following is true: let $C_1, \ldots, C_n$ be the components of the boundary of a domain, and let functions $f_i \colon C_i \to \mathbb{R}$ be given. Do there exist constants $a_1, \ldots, a_n$ such that the solution to the Dirichlet problem $\Delta u = 0$, $u \mid_{C_i} = f_i + a_i$ is the real part of a holomorphic function?
This is equivalent to the non-degeneracy of a matrix (periods of the conjugates of the solutions to several Dirichlet problems, the $i$-th problem being $f_j = \delta_{ij}$), and maybe follows from some classical facts about Riemann surfaces, but I am a stranger to the field...
The answer is yes and you can find it in the book [1], chapter 1, §4, theorem 4.3, pp. 20-22. Precisely Wen, by constructing a suitable harmonic function and its harmonic conjugate, proves that on a $(N+1) $-connected domain in $\Bbb C$ whose connected components of the boundary $\Gamma$ are $C_0,\ldots,C_N$, there exists a unique analytic function $F(z)$ such that
$$
\begin{cases}
\Re F(t) =\Phi(t) + h(t) & t\in\Gamma=\bigcup_{i=0}^NC_i\\
\Im F(a) = b & a\in C_0
\end{cases}
$$
where
$\Phi(t)\in C^{1,\alpha}(\Gamma)$ is a real valued function on the boundary of $D$ (the requirement can be relaxed in $\Phi(t)\in C^{0,\alpha}(\Gamma)$ by a suitable approximation as explained in [1], p. 20 footnote 2)
The function $h(t)$ is a simple function defined as
$$
h(t)=
\begin{cases}
0, & t\in C_0 \\
h_j & t\in C_j, \; j= 1,\ldots, N.
\end{cases}
$$
where the $\{h_j\}_{j=1,\ldots,N}$ are indeterminate real constants
$b$ is a real constant.
Reference
[1] Guo-Chun Wen, Conformal mappings and boundary value problems, translated from the Chinese by Kuniko Weltin (English), Translations of Mathematical Monographs 106, Providence, RI: American Mathematical Society (AMS). viii, 303 p. (1992), ISBN 0-8218-4562-4, MR1187758, Zbl 0778.30011
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2025-03-21T14:48:31.831641
| 2020-08-20T15:56:33 |
369681
|
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Stack Exchange
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Non-counital coalgebras
For any unital algebra $A$, we have an associated dual coalgebra $A^{\circ}$. (Recall that it is defined to be the largest subalgebra of the $\mathbf{C}$-linear dual of $A$ such that the coproduct $\Delta(f)(a,b) = f(ab)$ is well-defined.) What is the corresponding construction for a non-unital algebra. The coproduct part still works, but we have no counit, since this should arise as the dual of the unit. So are non-counital coalgebras studied in the literature? If so what are some references?
Non-counital coalgebras are used in Aguiar's Infinitesimal Hopf algebras (they form part of the structure of the latter).
The finite dual of a non-unital algebra has been introduced in Semiperfect and coreflexive coalgebras, S. Dăscălescu, M. C. Iovanov, Forum Math. 27 (2015), No. 5, 2587--2608. See also: arXiv:1512.09344 [math.RT].
For the (more general) notion of noncounital corings (and some relevant examples) you can also see ch.4, sect. 26 of Corings and comodules, T. Brzezinski, R. Wisbauer
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2025-03-21T14:48:31.831743
| 2020-08-20T16:07:59 |
369683
|
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Stack Exchange
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Ring with different graded and ungraded global dimensions
Let $A$ be a $\mathbb N$-graded ring. One can consider the two categories $M_A^g$ and $M_A^u$ of graded and ungraded modules over $A$. Both have, say, enough projectives, hence one can compute various $\operatorname{Ext}^i$ groups, define projective dimensions and so on. In particular, one can define the graded global dimension $\operatorname{gr.gl.dim} A := \sup_{M \in M_A^g} \operatorname{pd} M$, where the projective dimension is computed in $M_A^g$, and the (ungraded) global dimension $\operatorname{gl.dim} A := \sup_{M \in M_A^u} \operatorname{pd} M$, where the projective dimension is computed in $M_A^u$.
For a graded $A$-module $M$ it is known that the projective dimensions in $M_A^g$ and in $M_A^u$ are actually the same. This immediately implies that $\operatorname{gr.gl.dim} A \leq \operatorname{gl.dim} A$. With a little more work, one can show that $\operatorname{gl.dim} A \leq 1 + \operatorname{gr.gl.dim} A$ (Theorem $II.8.2$ in Nastasescu, van Oystaeyen - Graded ring theory).
One can wonder whether is an example where the two notions actually differ (hence $\operatorname{gl.dim} A = 1 + \operatorname{gr.gl.dim} A$). Theorem $I.3.4$ in Nastasescu-van Oystaeyen ensures that $\operatorname{gr.gl.dim} A = \operatorname{gl.dim} A_0$ if $A$ is strongly graded over $\mathbb{Z}$, that is $1 \in A_i A_{-i}$ for all $i$. This provides us examples when the grading is over $\mathbb{Z}$, but of course it never applies with a grading over $\mathbb{N}$.
Hence my question:
is there an example of a ring $A$ graded over $\mathbb{N}$ such that $\operatorname{gl.dim} A = 1 + \operatorname{gr.gl.dim} A$?
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2025-03-21T14:48:31.831869
| 2020-08-20T16:12:29 |
369684
|
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|
Stack Exchange
|
Completion w.r.t. ideal generated by a part of regular system of parameter
Let $R$ be a $d$-dimensional Noetherian regular local $k$-algbera ($k$ any field of char($k$) = 0, $d \geq$ 2). Let $x, y$ be a part of regular system of parameter for $R$. Let $I = (x, y)$ be a ideal generated by $x$ and $y$ and $\hat{R}$ denote the completion of $R$ with respect to $I$.
Is it true that $\hat{R} = \frac{R}{I}[[x, y]]$?
I know the very special case of it is true when $d = 2$ and $I$ is the maximal ideal. The above statement seems correct but I am not entirely sure. Any help would be great.
Is the "$n=1$" case known, i.e. whether the $(x)$-adic completion of $R$ is isomorphic to $R/(x)[[s]]$?
I am not entirely sure. But isn't it follows from $n =1$. Since if you go mod one parameter, we still get regular local ring with less number of regular parameter and then again apply your result which will imply required result. So basically this first case is key step.
Suppose additionally that $R$ is essentially of finite type over $k$.
Then $\hat{R}$ contains a subring that maps isomorphically on to $R/I$.
Proof: Note that $R/I$ is regular and essentially of finite type over $k$. Hence it is $0$-smooth over $k$ (Matsumura, Commutative Ring Theory, Theorem 30.3). We can therefore lift the identity map of $R/I$ to a $k$-algebra map $f_2 : R/I \to R/I^2$. Repeating the argument, we get a $k$-algebra map $f_j : R/I \to R/I^j$ lifting $f_{j-1}$, for every $j \geq 3$. Thus we get a map $R/I \to \hat{R}$ such that the composite $R/I \to \hat{R} \to R/I = \hat{R}/I\hat{R}$ is the identity map of $R/I$.
Older answer, left here so that the comments below this make sense.
$\hat{R}$ is a flat $R$-module (Matsumura, Commutative Ring Theory, Theorem 8.8) but $\frac{R}{I}[[x,y]]$ is not, since every non-zero element of $I$ is a zero-divisor on it.
I think the question is whether they are isomorphic as rings, not as $R$-modules.
@Manoj. Yes I am ask whether they are isomorphic as rings.
Sorry, I misunderstood your question. Are you asking whether $\hat{R}$ contains a subring which maps isomorphically to $R/I$? (As one would have, if $I$ were the maximal ideal.)
I am asking whether these two rings are isomorphic rings or not and the weaker question is whether $\hat{R}$ contains a subring isomorphic to R/I.
|
2025-03-21T14:48:31.832045
| 2020-08-20T16:29:16 |
369686
|
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|
Stack Exchange
|
Serre vanishing on one-point blow-ups
This is basically the last step of problem 5.3.7 in Huybrechts' Complex Geometry.
Let $X$ be a complex manifold, $x \in X$, $E$ a holomorphic vector bundle on $X$ and $L$ a positive line bundle. Denote by $
\hat{X} = \operatorname{Bl}_{\{x\}} X$ the blow-up of $X$ at $x$ and let $\sigma: \hat{X} \to X$ be the projection map and $F = \sigma^{-1}(x)$ the exceptional divisor. I'm trying to show that
$$
H^1\left(\hat{X}, \sigma^* E \otimes \mathcal{O}_{\hat{X}} \left(-F\right) \otimes \sigma^*L^k\right) = 0
$$
for $k$ large enough. Any ideas?
This is of course almost Serre's vanishing theorem: $\sigma^*L$ is positive for every tangent vector except those of $TF$ and one can show that $\mathcal{O}_{\hat{X}} \left(-F\right) \otimes \sigma^*L^k$ is positive for $k$ large enough, but the problem is that we can't take powers of $\mathcal{O}_{\hat{X}} \left(-F\right)$...
This question was migrated from math.SE
I think the following works: I think $R^1 \sigma_\mathcal O_{\widehat X}(-F)=0$, so by the Leray--Serre spectral sequence and projection formula we have that the thing you're trying to show is 0 is equal to $H^1(E\otimes \sigma_\mathcal O_{\widehat X}(-F)\otimes L^k)$; now using Serre vanishing on $X$ this is 0 for $k$ large.
To see $R^1 \sigma_\mathcal O_{\hat X}(-F)=0$: take $0\to \mathcal O_{\hat X}(-F)\to \mathcal O_{\widehat X}\to \mathcal O_F\to 0$, push it forward, and use that $\sigma_ \mathcal O_{\widehat X}=\mathcal O_X\to \sigma_* \mathcal O_F=\mathcal O_x$ is surjective and $R^1\sigma_* \mathcal O_{\hat X}=0$.
|
2025-03-21T14:48:31.832179
| 2020-08-20T16:38:09 |
369689
|
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|
Stack Exchange
|
Confusion around the reflection equation algebra
I have encountered several occurrences of the so called reflection equation algebra (REA) but depending on where I find them, I feel like I get slightly different objects. In all cases there is a quasi-triangular Hopf algebra lurking in the background. In what follows $V$ will always be a vector space of dimension $n$. Here is a list of the different occurrences I encountered:
Let $H$ be a quasi-triangular Hopf algebra with $R \in H \otimes H$ its universal $R$-matrix (here we possibly have completions but it does not really matters). The reflection algebra is then as vector space the restricted dual $H^\circ$. That is the subalgebra of the full dual spanned by the so called matrix coefficients. The algebra structure comes from the algebra structure of the full dual but twisted by the universal $R$-matrix. I think this is sometimes called the braided dual of $H$.
See for example definition 4.12 of https://arxiv.org/pdf/math/0204295.pdf
Let $R: V \otimes V \rightarrow V \otimes V$ be an endomorphism of $V \otimes V$ satisfying the Yang-Baxter equation. Then the reflection equation algebra if the algebra generated by elements $(a_{ij})_{1 \leq i,j \leq n}$ with relation
$$RA_1RA_1 = A_1RA_1R$$
where $A$ is the matrix $n \times n$ having the generating elements as coefficients and $A_1 = A \otimes Id$.
I think here the generating elements are somewhat thought to be elements of $V^{\ast} \otimes V$. This was found at the very beginning of the introduction of https://arxiv.org/pdf/1806.10219.pdf
This one is a special example. Here the Hopf algebra lurking in the background is $U_q(\frak{sl_2})$ and the $R$-matrix is given by
$$
\begin{pmatrix}
q & 0 & 0 & 0\\
0 & 1 & q-q^{-1} & 0 \\
0 & 0 & 1 & 0\\
0 & 0 & 0& q
\end{pmatrix}.$$ In this case it is the algebra generated by the elements $(a_{ij})_{1 \leq i,j \leq 2}$ with relation :
$$R_{21}A_1RA_2 = A_2R_{21}A_1R$$ and also
$$ a_{11}a_{22}- q^2a_{12}a_{21} = 1.$$
This algebra is often denoted by $\mathcal{O}_q(SL_2)$ or sometimes $\mathcal{O}(Rep_q(SL_2))$. This appeared as Example 1.23 in https://arxiv.org/pdf/1908.05233.pdf and also as Definition 2.1. in https://arxiv.org/pdf/1811.09293.pdf (be aware of the foot note to get back what I wrote).
I can see how some of those are related, for example the third one is almost a specific case of the second one but there is one more relations.
In the first one matrix elements can be thought as being in $W^{\ast} \otimes W$ for any representation $W$ of $H$. In the case where any finite dimensional representation of $H$ can be seen as subrepresentation of a tensor product of the standard representation $V$, then it is actually generated only by the matrix coefficients coming from $V$.It then looks a lot like what we have in 2). However, there is still the a relation missing if one specializes to the case $H = U_q(\frak{sl2})$ to get the same as in 3). And what if there is a representation of $H$ that is not a subrepresentation of a tensor product of the standard one?
QUESTION:
Are all of those actually the same thing or am I missing something? I am a bit confused on what people actually call the reflection equation algebra. Is there some kind of nice definition for any quasi-triangular Hopf algebra $H$ that englobes all the above "examples"?
This is an interesting and nicely formulated question. And it can become even better, if you add some citation for each one of the occurrences you mention. (This would also be very helpful for potential answerers).
Thank you for your suggestion. I have added references for each occurences.
The only reasonable definition of the REA associated with a quasi-triangular Hopf algebra is 1). This is, of course, a somewhat abstract definition but provides a solution of the RE which is universal in a precise sense.
is reminiscent of the so-called Faddeev-Reshetikhin-Takhtajan (usually abbreviated as FRT) construction. Its main advantage is that it does not require an Hopf algebra to start with (rather, in the original FRT construction, the goal was to produce a Hopf algebra starting from an arbitrary solution of the QYBE). EVen if $R$ do come from a quasi-triangular Hopf algebra, it won't give the same answer as 1) except in the case of $U_q(\mathfrak{gl}_n)$ (even then this is not quite true, you get some deformation of $\mathcal O(\mathfrak{gl}_n)$ rather than $\mathcal O(GL_n)$). In general there will be a map from 2) to 1).
On the other hand as you say, you can run this construction in the case $V$ is some representation that generates all the other. Indeed this approach is useful to find a presentation of the REA, since it is indeed generated by matrix coefficients: roughly speaking this will give you a set of generators but not in general all the relations. This is what happens here: if you run the FRT-like reconstruction for the R-matrix of $\mathfrak{sl}_n$ you get some algebra, but then you need to add this extra relation you mention which, as you probably know, is nothing but a $q$-analog of $\det(A)=1$. Again this already shows up in the original situation, see Definition 4 in http://math.soimeme.org/~arunram/Resources/Reshetikhin/QuantizationOfLieGroupsAndLieAlgebras.html.
Edit It's useful to think about universal properties: 1) is universal for algebras $A$ with a solution of the RE in $A\otimes H$, while 2) is universal for algebras $A$ with a solution of the RE in $A\otimes End(V)$. Of course, composing with the algebra map $H\rightarrow End(V)$ given by the action of $H$ on $V$ every solution of the first equation gives you a solution to the second one, so applying this to the case $A$ is the REA itself you get a map from the algebra constructed in 2) to the one constructed in 1).
Thank you for your answer. One thing I am not sure sure to get from your answer is what happens if you plug in $U_q(\frak{sl_2})$ into case 1). Do you get exactly the same algebra as in 3) or there is still a relation missing?
Yes, sorry if that was unclear.
I am sorry but is this "yes it is the same algebra" or "yes there is still a relation missing"?
Yes this is the same algebra. My point was that 2 gives you in nice circumstances a set of generators and a bunch of correct relations, but not all of those. 3) is an example of that, where you need to add that particular relation to get 1).
Let me first note that the reflection matrix, which you denote by $A$, is often called K-matrix, cf its graphical representation with | a 'wall' and < the 'worldline' of particle bouncing off the wall. The graphical form of the equation can already be found in Cherednik, Factorizing particles on a half-line and root systems (1984) https://link.springer.com/article/10.1007/BF01038545. The notation $K$ might be due to Sklyanin, Boundary conditions for integrable quantum systems (1988), https://iopscience.iop.org/article/10.1088/0305-4470/21/10/015.
The reflection (equation) algebra is the reflection-equation analogue of the Yang--Baxter algebra: to any choice of finite-dimensional vector space and R-matrix obeying the Yang--Baxter equation (and suitable other properties, such as braiding unitarity and an 'initial condition') one can associate a unital associative algebra generated by the operator-valued (noncommutative) entries of the K-matrix obeying the reflection equation.
If one would replace the reflection (`$RKRK$') equation by the $RLL$-equation one instead arrives at the Yang--Baxter algebra, which is the operator algebra closely related to the FRT (or R-matrix) presentation of quantum affine algebras.
Re 3: The FRT presentation says nothing about the quantum determinant, so to get $SL_n$ you need to impose $qdet = 1$ separately, which is your last equation in 3. The version of the reflection equation that you give there can sometimes be simplified: Suppose that the R-matrix is symmetric in the sense that $P R P = R$ with $P$ the permutation. Then $R_{21} = R_{12}$ in the usual tensor-leg notation. In such cases all R-matrices in the reflection equation can be written using just $R$. (Graphically the need for $R_{21}$ is clear, though.)
Re 2: These authors work with the braid-like version of the R-matrix, often denoted by $\check{R}$. Namely, suppose that $R$ obeys the YBE
$$ R_{12}(u,v) \ R_{13}(u,w) \ R_{23}(v,w) = R_{23}(v,w) \ R_{13}(u,w) \ R_{12}(u,v) \ , $$
where I have assumed that the R-matrix might depend on a spectral parameter associated to each copy of the auxiliary space in general. (This is for the affine case, but helps highlighting the structure of the equation.)
Then both of $P \ R$ and $R \ P$ obey the braid-like version of the YBE
$$ \check{R}_{12}(u,v) \ \check{R}_{23}(u,w) \ \check{R}_{12}(v,w) = \check{R}_{23}(v,w) \ \check{R}_{12}(u,w) \ \check{R}_{23}(u,v) \ . $$
You always have to check which version is used. In the paper you cite in 2 it's the latter, which is why both $A$s have the same subscript.
Re 1: I believe that the proper algebraic interpretation of Sklyanin's construction of representations of the K-matrix as the double-row monodromy matrix, constructed from a K-matrix with scalar entries and an L-operator, is as a coideal subalgebra, see Kolb and Stokman, Reflection equation algebras, coideal subalgebras, and their centres, https://arxiv.org/abs/0812.4459.
You might also be interested in the recent paper by Appel and Vlaar, Universal k-matrices for quantum Kac-Moody algebras, https://arxiv.org/abs/2007.09218
PS. In your reference in 2 the authors write $T$ for what I denoted by $L$. In physics literature this is called the monodromy matrix.
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