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2025-03-21T14:48:29.580133
| 2020-01-04T17:24:17 |
349711
|
{
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"Victor Petrov",
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|
Stack Exchange
|
Category of representations of a tensor product algebra
Given two semisimple unital algebras $A$ and $B$, defined over $\mathbb{R}$ or $\mathbb{C}$, denote their categories of representations by $_A\mathcal{M}$ and $_B\mathcal{M}$ respectively. Can one describe the category of representations of $A \otimes_{\mathbb{C}} B$ as some type of "tensor product" of the categories $_A\mathcal{M}$ and $_B\mathcal{M}$?
There is a notion of Deligne tensor product of tensor categories, probably that's what you need.
See also some discussion here: https://mathoverflow.net/questions/335810/the-tensor-product-of-two-monoidal-categories?rq=1
Yes, it will be exactly Deligne's tensor product of abelian categories. See https://ncatlab.org/nlab/show/Deligne+tensor+product+of+abelian+categories
|
2025-03-21T14:48:29.580209
| 2020-01-04T17:30:38 |
349712
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Philip Ehrlich",
"Tobias Fritz",
"https://mathoverflow.net/users/18939",
"https://mathoverflow.net/users/27013"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/349712"
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|
Stack Exchange
|
Hahn's approach to Hilbert's 17th problem?
The Wikipedia article on Hahn Series mentions mentioned that these were studied by Hahn "in his approach to Hilbert's seventeenth problem".
Is this correct? If so, what was this approach, and where can I read about it?
I have read most of Hahn's paper über die nichtarchimedischen Größensysteme, where Hahn series were introduced, but I have not seen Hilbert's 17th problem on positive polynomials being mentioned there. I have also skimmed his list of publications and not found anything else that looks relevant.
Like you, I'm confident that Hahn does not discuss Hilbert's problem in the paper you cited, and, like you, I am not aware of any of his publications that apply that work to Hilbert's 17th problem. However, in the paper you mention, Hahn does discuss Hilbert's (arithmetic) completeness condition, which is discussed by Hilbert in his famous list of problems.
Very interesting, @PhilipEhrlich! So what you're suggesting is that the Wikipedia reference to Hilbert's 17th problem should rather be pointing to Hilbert's 2nd problem, right? I would like to check but can't find a pdf of Hahn's paper right now. But I see that your paper talks about this!
Yes, that is precisely what I am suggesting.
Thank you! I've corrected the statement on the Wiki page. And in case that you want to make this into an answer to the question, I'll be happy to accept it.
I will do as you suggest later. For the time being, let me simply clarify that Hahn did not see his work as directly applicable to Hilbert's second problem, but rather to Hilbert's completeness condition that Hilbert mentions in his discussion.
As you suspect, Hahn does not discuss Hilbert's 17th problem in the paper you cite, and, like you, I am not aware of any of Hahn's publications that applies his work on non-Archimedean ordered systems to Hilbert's 17th problem.
However, in the paper you mention Hahn does discuss Hilbert's (arithmetic) completeness condition, which is discussed by Hilbert in the second problem of his famous list of problems. There, however, Hahn is not concerned with the consistency of the real number system, which is the subject of the second problem, but rather with generalizing Hilbert's completeness condition so as to be applicable to non-Archimedean as well as Archimedean ordered Abelian groups and ordered fields.
|
2025-03-21T14:48:29.580362
| 2020-01-04T17:34:25 |
349714
|
{
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"Jonas",
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"site": "mathoverflow.net",
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"url": "https://mathoverflow.net/questions/349714"
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|
Stack Exchange
|
About residually finite solvable groups
Let $G$ be a fixed group. for each property $P$ of groups, $G$ is said to be residually-$P$ if for each $1\neq g\in G$ there exists $N\unlhd G$ such that $g\notin N$ and $G/N$ has the property $P$. If $P$ is the property of to be a finite solvable group, a group residually-$P$ is called residually-(finite solvable) group. And if $P$ is the property of to be a finite nilpotent group, a residually-$P$ group is said to be a residually-(finite nilpotente) group.
Let $G$ be a finite solvable group. Then $G$ has a series $1=G_{0}\leq \ldots\leq G_{n}=G$ of normal subgroups whose of each factors are finite nilpotent groups. The least $n$ such that $G$ has a chain with this property is called the Fitting height of the group $G$, denoted bt $h(G)$.
In a text i found the proof of the following statement: "Let $G$ be a residually-(finite solvable) group and suppose that for every normal subgroup $N$ of finite index, $G/N$ is a solvable group and $h(G/N)\leqslant h$. Then $G$ has a chain $G=G_{1}\geq\ldots\geq G_{h+1}=1$ whose factors are residually-(finite nilpotent)". Later, the author uses this result in a version where the subgroups $G_{i}$ are all characteristic subgroup.
How can i obtain this version?
Link of the text: https://arxiv.org/abs/1706.07963
I got it for finitely generated groups. Fortunately i just need of this case. Thank you very much!
One question remains: Is the statement true without the suposition that $G$ is finitely generated?
|
2025-03-21T14:48:29.580599
| 2020-01-04T17:41:58 |
349715
|
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"M. Dus",
"Nate Eldredge",
"Nik Weaver",
"Wlod AA",
"aglearner",
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"url": "https://mathoverflow.net/questions/349715"
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|
Stack Exchange
|
Choosing a convergent sub-sequence from a sequence of bi-Lipschitz homeomorphisms
Let $X,Y$ be two compact metric spaces. Suppose there is a sequence of bi-Lipschitz homeomorphisms $f_n: X\to Y$, and $c\in(0,1]$, satisfying
$$c\cdot d(x_1,x_2)\le d(f(x_1),f(x_2))\le \frac{1}{c}\cdot d(x_1,x_2),$$
for any $x_1,x_2\in X$.
Is is true that one can choose a sub-sequence $f_{n_i}$ that uniformly converges to a bi-Lipschitz homeomorphism $f$ (with the same constant $c$)?
Such a statement would imply that in case the Lipschitz distance between two compact metric spaces is bounded, it is actually realised by a bi-Lipschitz homeomorphism (which is better than taking an inf over all bi-Lipschitz homeomorphisms).
This should follow from an appropriate version of the Arzela-Ascoli theorem for maps between metric spaces. Even if you haven't seen such a version, it would be a good exercise to adapt the proof for real-valued functions.
Yes: all you need is a subsequence that converges uniformly (or even just pointwise). The fact that the limit function must be bi-Lipschitz is easy.
Dear Nate and Nik, I guess you are right. I wonder though why the proof of Theorem 7.2.4 in Burago, Burago, Ivanov http://citeseerx.ist.psu.edu/viewdoc/download?doi=<IP_ADDRESS>8.3775&rep=rep1&type=pdf doesn't use this fact. It is a tiny bit more involved than that. This theorem states, in particular, that two compact metric spaces on Lipschitz distance $0$ from each other are isometric.
@aglearner That's a striking application ! Did not see that at first.
Consider a subsequence $\,g_n\,$ of $\,f_n\,$ which is uniformly convergent. Then consider a subsequence $\,h_n\,$ of $\,g_n\,$ such that sequence $\,h^{-1}_n\,$ is uniformly convergent. Then the limit function $\,h\,$ of $\,h_n\,$ is bilipschitz with the bi-constant equal the same $\,c.$ Great
REMARK This proves what @NikWeaver has already said in his comment under the OP's Question. (Yes, I agree with the comment by @NatEldredge).
Note that the result also holds not assuming that the $f_n$ are homeomorphisms and in this case, you cannot use $h_n^{-1}$. The $f_n$ certainly are one-to-one but necessarily onto in general. However, you don't need to play this trick to obtain a bi-Lipschitz limit. Since $g_n$ converges pointwise (note that this is enough) to a limit $g$, fixing $x_1$ and $x_2$, you get $1/c d(x_1,x_2)\leq d(g_n(x_1),g_n(x_2))\leq c d(x_1,x_2)$ for every $n$, so that the same holds with $g$.
@aglearner Actually the answer proves that and I think that's the point of the answer (although not explicitely stated). Denote by $g$ the limit of $g_n$ and by $h$ the limite of $h_n^{-1}$. Then, $h_n$ also converges to $g$. Also, $h_n \circ h_n^{-1}$ converges pointwise to $Id$ and converges pointwise to $g\circ h$, so that $g\circ h=Id$. Similarly, $h\circ g=Id$. Since both $g$ and $h$ are Lipschitz, they are in particular continuous and so they are inverse homeomorphism.
You are right M. Dus, thanks!
@Dirk, did I say "bilinear"? :) === seriously, thank you Dirk for the correction.
|
2025-03-21T14:48:29.580822
| 2020-01-04T18:52:09 |
349719
|
{
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"Carlo Beenakker",
"Claude Chaunier",
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"neverevernever"
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"url": "https://mathoverflow.net/questions/349719"
}
|
Stack Exchange
|
Random matrix with given singular values
Let $\sigma_1\geq\sigma_2\geq...\geq\sigma_n\geq0$ be any deterministic sequence of positive real numbers such that $\sum_{i=1}^n\sigma_i^2=1$. Let
$$D=diag\{\sigma_1,...,\sigma_n\}\in\mathbb{R}^{n\times n}$$
be a diagonal matrix of size $n\times n$.
Let $U$ and $V$ be two independent random matrices uniformly distributed on the orthogonal group $O(n)$. Then we form a random matrix as follows:
$$A=UDV^T$$
Therefore, $A$ is a random matrix with given singular values and the distribution is orthogonally invariant. Now we define a quantity that measures how far away is $A$ from a diagonal matrix:
$$f_{\sigma}(A)=\sum_{1\leq i\neq j\leq n}A_{ij}^2=\sum_{1\leq i\neq j\leq n}\left(\sum_{k=1}^n\sigma_kU_{ik}V_{jk}\right)^2=1-\sum_{i=1}^n\left(\sum_{j=1}^n\sigma_jU_{ij}V_{ij}\right)^2$$
which is just the sum of squares of all off-diagonal elements of $A$, the smaller $f_{\sigma}(A)$ is, the "more diagonal" $A$ is. Note the dependency of $f$ on the given singular values $\sigma$.
I'm interested in the following quantity:
$$g_{\sigma}(t)=\frac{\mathbb{P}\left(f_{\sigma}(A)\leq 2t\right)}{\mathbb{P}\left(f_{\sigma}(A)\leq t\right)}$$
where $0<t<1/2$.
I have the following 2 conjectures:
For $n\geq 5$, for any $0<t<1/2$, $g_{\sigma}(t)$ is maximized when $\sigma_1=...=\sigma_n=1/\sqrt{n}$.
There exists a constant $C>1$ independent of $t, \sigma$, such that $g_{\sigma}(t)\leq C^{n^2}$
I believe they should be correct but I have no clue of how to prove them. Any suggestions and discussions are appreciated. What kind of tools could possibly be useful?
I have a feeling that existing literature in random matrix mainly focus on going from the matrix to eigenvalues or singular values, ignoring the eigenvectors or singular vectors. I do not see results about going from given spectrum to the matrix.
Is uniform distribution on $O(n)$ defined by the euclidean measure on spheres ?
@ClaudeChaunier It is the Haar measure on the orthogonal group.
Conjecture 1 is false. Here is the counterexample for $n=2$.
this is the conjecture 1 as originally given by the OP; I see that it has now been changed.
I take $n=2$, set $\sigma_1=\cos\alpha$, $\sigma_2=\sin\alpha$, with $0\leq\alpha\leq\pi/4$, and parameterize the orthogonal matrices as
$$U=\begin{pmatrix}
\cos\phi&\sin\phi\\
-\sin\phi&\cos\phi
\end{pmatrix},\;\;V=\begin{pmatrix}
\cos\phi'&\sin\phi'\\
-\sin\phi'&\cos\phi'
\end{pmatrix}.$$
The Haar measure on $\text{SO}(2)$ is a uniform distribution of the angles $\phi,\phi'\in(0,2\pi)$, with $\phi$ independent of $\phi'$. I calculate $A=U\,\text{diag}\,(\sigma_1,\sigma_2)V^T$ and evaluate
$$f_\alpha=A_{12}^2+A_{21}^2=\tfrac{1}{2} (1-\sin 2 \alpha \sin 2\phi \sin 2\phi'-\cos 2\phi \cos 2\phi').$$
Let me now compare the two extreme cases $\alpha=\pi/4$ and $\alpha=0$,
$$f_{\pi/4}=\sin^2(\phi-\phi'),\;\;f_0=\tfrac{1}{2}(1-\cos 2\phi\cos 2\phi').$$ The corresponding probability distributions are
$$p_{\pi/4}(f)=\frac{1}{\pi}f^{-1/2}(1-f)^{-1/2},$$
$$p_0(f)=\frac{4}{\pi^2} \int_0^{\arccos|1-2f|}\frac{d\phi}{\sqrt{\cos^2 \phi-(1-2f)^2}}.$$
(The expression for $p_0(f)$ is an elliptic integral.) I checked both distributions numerically (by generating random $\phi,\phi'$) and they do seem to be correct, see the histograms:
Because $p_{\pi/4}(f)$ has a peak at $f=0$, while $p_0(f)$ has a peak at $f=1/2$, the ratio $g_\alpha(t)$ of cumulative distributions at $2t$ and $t$ is larger for $\alpha=0$ than it is for $\alpha=\pi/4$, for all $0<t<1/2$. Here is a plot that compares the two, blue is for $\alpha=\pi/4$ and gold is for $\alpha=0$.
So this is a counter example to conjecture 1, because $\alpha=\pi/4$ corresponds to $\sigma_1=\sigma_2=1/\sqrt n$ for $n=2$.
I got a somewhat different expression for $f$... However, from a geometric perspective, if $\alpha=0$, the dimension of the problem is essentially reduced, the enlargement of the sublevel set of $f$ on the orthogonal group should be smaller.
This is very counter intuitive and it is surprising that $\alpha=0$ consistently beats $\alpha=\pi/4$. Could the conjecture be the other way around, which means the most unbalanced case, $\sigma_1=1$ will dominate all other $\sigma$?
here is my intuition: If I partition an orthogonal matrix $M$ into four blocks, $M_{11}$, $M_{12}$, $M_{21}$, and $M_{22}$, and consider the singular values of an off-diagonal block, then the distribution of these singular values peaks at 0 and at 1 when $M$ is uniformly distributed in ${\rm O}(n)$. Now if you choose the balanced case, then $M=UV^T/\sqrt n$ has a uniform distribution in $\text{O}(n)$. This means that $M$ is either largely off-diagonal (peak at 0) or largely diagonal (peak at 1), with equal probability. The peak at 0 skewes your $g$-ratio to small values.
I also did some simulation. However, for $n>2$, the peak at $0$ no longer exists. Also, as $n$ gets larger, the distribution of $f$ for the two extreme cases gets closer and closer.
Also, my simulation result shows that as $n$ even larger, say, > 5, the distribution of balanced case is more concentrated while the unbalanced case is slightly more spread out.
What you want for 2- seems to me similar to a large deviation result (https://en.wikipedia.org/wiki/Large_deviations_theory) of speed $n^2$ for the diagonal term
$$\frac{1}{n^2}\log\mathbb{P}(\sum_i A_{ii}^2\geq s)\approx I(s) $$
with some function $I(s)<0$. Indeed if so we have
$$\log (g(t)) \approx n^2 (I(1-2t)-I(1-t)) $$ and then you can choose $\log C=\inf (I(1-2t)-I(1-t))$
This is not true for the case $\sigma=(1,0,\cdots,0)$ but we have a better upper bound. Here for $A_{11}^2\geq s$ it is enough to ask that $U_{11}\geq s^{1/4}$ and $V_{11}\geq s^{1/4}$.
The first column of $U$ is a random vector chosen uniformly on the sphere $\mathbb{S}^{n-1}$ and a good way to construct it is using a set of iid Gaussian variables $Y_i$ and write $$ U_{i1}:=\frac{Y_i}{\sqrt{\sum_{j}Y_j^2}} $$
We use here that the direction of a Gaussian vector is isotropic on $\mathbb{S}^{n-1}$. Then
$$ \mathbb{P}(U_{11}\geq s^{1/4})\geq \mathbb{P}(Y_1\geq 2s^{1/4},\forall j\geq2: |Y_j|\leq s^{1/4}/\sqrt{n})\geq C\left(\frac{c}{\sqrt{n}}\right)^{n-1}$$ for some constant $c$ and $C$. And then $$\log(\mathbb{P}(A_{11}^2\geq s)\geq \log\left( \mathbb{P}(U_{11}\geq s^{1/4})\mathbb{P}(V_{11}\geq s^{1/4})\right)\geq (n-1)(\log n+c')$$ which is a slower speed that $n^2$.
As a conclusion we have $$g_{(1,0,\cdots,0)}(t) \leq (C')^{n\log n} $$
for some $C'>1$.
For the case $\sigma = (\frac{1}{\sqrt{n}},\cdots,\frac{1}{\sqrt{n}})$, I think one should be able to obtain the large deviation principle with speed at least $n^2$. Here we only consider $U\in \mathcal{O}(n)$ chosen uniformly as multiplication by $V^{T}$ doesn't modify the Haar measure. The construction of the fist column of $U$ is similar as above and we should at least obtain
$$ \mathbb{P}(U_{11}^2\geq s)\leq \exp(n\tilde{I}(s))$$
for some $\tilde{I}<0$ as large deviation principle for $\sum_j Y_j^2$ are well known (Cramer Theorem). We now condition on $U_{11}$ and write the matrix
$$\begin{pmatrix} U_{11} & Y \\ \tilde{Y} & U'\end{pmatrix} $$
Because $Y$ is rotation invariant, so is $U'$. Moreover because for any vector $v=(0,v_2,\cdots,v_n)^T$ we have $\|Uv\|^2=|\langle Y,(v_2,\cdots,v_n)\rangle|^2+\|U'(v_2,\cdots,v_n)\|^2\geq \|U'(v_2,\cdots,v_n)\|^2 $, we are reduce to the original problem on $U'$ for some $\sigma'=\frac{1}{\sqrt{n}}(\sigma_1',\cdots,\sigma_{n-1}')$ with $\sigma_i'\leq 1$ for all $i$. Therefore $$\mathbb{P}(\sum_{j\leq n-1}(U_{jj}')^2\geq (n-1)s)\leq \mathbb{P}(\sum_{j\leq n-1}(U_{jj}'')^2\geq (n-1)s)$$ where $U''$ is chosen uniformly on $\mathcal{O}(n-1)$ and we are reduce to system of size $n-1$. By direct iteration we get
$$ \mathbb{P}(U_{11}^2\geq s)\mathbb{P}(U_{22}^2\geq s)\cdots \mathbb{P}(U_{nn}^2\geq s)\leq \exp(n\tilde{I}(s))\exp((n-1)\tilde{I}(s))\cdots \exp(\tilde{I}(s))\\ =\exp(n(n-1)\tilde{I}(s)/2) $$
which reveal the large deviation with $n^2$ speed.
One can use $\mathbb{P}(\sum_{i\leq n}U_{ii}^2\geq s n)\leq \mathbb{P}(\exists K\subset [N]:|K|\geq sn/2:\forall i\in K : U_{ii}^2\geq s/2)$ to finish the proof with an union bound
|
2025-03-21T14:48:29.581546
| 2020-01-04T19:00:39 |
349721
|
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|
Stack Exchange
|
Prove that this integral operator is onto
Let us consider the operator $T$ defined by$$\eqalign{
& T:{L^2}((a,b) \times (c,d)) \to {L^2}((c,d)) \cr
& Tf(s,x) \mapsto \int\limits_{q(x)}^{p(x)} {f(\alpha (s,x),s)ds} \cr} $$
where $c \leqslant q(x) \leqslant p(x) \leqslant d$ for all $x\in (c,d)$ and $\alpha$ has a range in $(a,b)$ fo all $(s,x)\in ((a,b) \times (c,d))$.
What are the assumptions on the functions $p$,$q$ and $\alpha$ to ensure that $T$ is onto.
Thank you.
|
2025-03-21T14:48:29.581621
| 2020-01-04T19:15:49 |
349722
|
{
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"authors": [
"Ben Webster",
"Joseph O'Rourke",
"Robert Israel",
"Sylvain JULIEN",
"https://mathoverflow.net/users/13625",
"https://mathoverflow.net/users/13650",
"https://mathoverflow.net/users/142929",
"https://mathoverflow.net/users/6094",
"https://mathoverflow.net/users/66",
"user142929"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/349722"
}
|
Stack Exchange
|
Remarkable articles about the distribution of prime numbers that were written by contemporary physicists
I would like to ask about if you know papers containing remarkable achievements that were written by contemporary physicists concerning the distribution of prime numbers (or closely related, maybe the distribution of the zeros of the Riemann zeta function) in the spirit of the modern analytic number theory.
I am asking about it since I don't know how to search it, and I don't know how to know if a paper is remarkable for the community of professional mathematicians.
Question. Can you as professional number theoretician or professor tell me what are important and remarkable articles/papers due to contemporary physicists concerning the distribution of prime numbers in the spirit of the modern analytic number theory? Then add the references answering this question as a reference request. Many thanks.
Thus I assume that there are professors who, apart from studying their discipline in some field of physics or as a natural consequence of his/her investigations, provided relevant results in the study of the distribution of prime numbers.
I would nominate Sir Michael Berry, FRS. According to Wikipedia, "He is known for the Berry phase, a phenomenon observed e.g. in quantum mechanics and optics, as well as Berry connection and curvature. He specialises in semiclassical physics (asymptotic physics, quantum chaos), applied to wave phenomena in quantum mechanics and other areas such as optics."
His papers using tools of quantum chaos in analytic number theory include "The Riemann Zeros and Eigenvalue Asymptotics" (with J. Keating), SIAM REVIEW Vol. 41, No. 2, pp. 236–266, among others.
Many thanks now I am going to read the Wikipedia, and your answer. Many thanks again for your attention and patience.
The connection between the distribution of the Riemann zero's and the distribution of the eigenvalues of random unitary matrices was noted by the physicist Freeman Dyson, in a conversation with the mathematician Hugh Montgomery. Here is a this description of the conversation:
Hugh Montgomery: "[Indian number theorist Daman Chowla] said, 'Have
you met Dyson?' and I said, 'No,' and he said, 'I'll take you and
introduce you to Dyson,' and I said 'No, no, that's OK, I don't need
to meet Dyson.' This went back and forth and it ended up with Chowla
dragging me across the room. I didn't really want to bother [Dyson]. I
didn't think of having anything useful to say to him, but when Chowla
introduced me Dyson was very cordial and asked me what I'd been
working on, and so I told him that I'd been looking at the zeros of
the zeta function."
It was when Montgomery mentioned the formula he had found for this
distribution that Dyson's ears pricked up. At the mention of $1 - [(\sin
\pi u)/(\pi u)]^2$, Dyson said something like, 'Well, that's the density
of the pair correlation of eigenvalues of random matrices in the
Gaussian Unitary Ensemble.'
"I'd never heard any of these terms before," Montgomery went on. "I
don't know exactly what his words were because I have heard all of
these terms many times since, but he said 'pair correlation' and
something resembling 'random matrices'..."
"As far as I know," Montgomery said, "he didn't think about it after
this five-minute conversation. I haven't spoken with him since, so
I've had one conversation with him in my life. But it was quite a
fruitful conversation.
Many thanks for your answer and links.
It should maybe be mentioned that besides being a physicist, Dyson is also very accomplished as a pure mathematician. At Cambridge, while he studied physics under Dirac, he also studied pure mathematics under Hardy and Besicovich. He published quite a few papers in mathematics throughout his career.
Marek Wolf, a Polish physicist, is the author of several articles about the distribution of prime numbers. He studied jumping champions and provided a heuristic formula refining Cramer's conjecture, and is also the author with Kourbatov, whose given name I forgot, of a paper about maximal prime gaps. Googling his name with such keywords should point to the relevant papers.
Many thanks, I don't know what are the most remarkable papers on primes of professor Marek Wolf.
Here comes his homepage: http://pracownicy.uksw.edu.pl/mwolf/
Alexei Kourbatov. "Predicting maximal gaps in sets of primes." arXiv.
Alexei, da! Thank you for reminding me.
|
2025-03-21T14:48:29.581940
| 2020-01-04T20:47:02 |
349727
|
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|
Stack Exchange
|
Can a non-compact manifold become compact by cutting it?
I'm trying to understand a step in a proof, where one starts with a non-compact manifold $V$ containing a trapped (2-sided, closed) surface $\Sigma$ that's non-separating. In order to complete the proof, one needs to somehow double that manifold such that in the end one has a (still non-compact) manifold where the surface now separates it.
They do this by cutting the manifold $V$ along $\Sigma$, thus obtaining a (still connected) manifold $\tilde{V}$ with two separate boundaries diffeomorphic to $\Sigma$, and then gluing $\mathbb{Z}$ such copies of $\tilde{V}$ end-to-end along the boundaries. Then each ex-boundary ($\Sigma$) separates the new manifold.
My question is: why do I need $\mathbb{Z}$ such copies instead of just two? Can it happen that $\tilde{V}$ turns out to be compact, although $V$ wasn't?
Thanks guys.
If you used two copies the surface wouldn’t separate the resulting manifold.
I mean 2 copies glued together along just one of the boundaries instead of both?
Then you get a manifold with boundary (I assume you do not want it). Anyway, could you tell what do you need to prove with this trick (it might help).
Its the second part of the proof for a Variation of the Penrose singularity theorem, switching the condition that the Cauchy surface $V$ contains a trapped surface to $V$ contains a MOTS $\Sigma$ and additionally the generic condition holds on each future and past inextendible null geodesic normal to $\Sigma$. It's Theorem 3.2 from the paper "Topological censorship from the initial data point of view" from Eichmair, Galloway and Pollack.
What does MOTS mean? Anyway, if I understood it correctly, the answer to your second, more specific, question is “no” — if $V$ is non-compact then neither is $\tilde V$, however you construct it.
@HJRW A MOTS $\Sigma$ is a "marginally outer trapped surface", i.e. a trapped surface $\Sigma$ in a spacelike Cauchy hypersurface $V$ of a spacetime on which the null mean curvature w.r.t. the outward normal of $\Sigma$ in $V$ vanishes. (So it builds on a whole lot of other definitions of general relativity.)
|
2025-03-21T14:48:29.582119
| 2020-01-04T21:27:36 |
349730
|
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|
Stack Exchange
|
The Fano plane, stericated 6-simplex, and pentallated 6-simplex
According to this link:
https://en.wikipedia.org/wiki/Stericated_6-simplexes
the stericated 5-simplex "scal" has 105 vertices defined as permutations of (0,0,1,1,1,1,2).
In the course of my team's research into the structure of a particular biomolecular energetic space, we have discovered that there are exactly 21 different ways in which (0,0,1,1,1,1,2) can be assigned to the seven points of the Fano plane so that the sum of the three integers assigned to each line is even. (Of these 21, there are seven sets such that no assignment in any set can be obtained from any assignment in the other six sets.)
Hence, the vertices of "scal" can be used to orgnaizze verticces of the pentallated 6-simplex "staf", since it is clear from this projection of "staf":
https://en.wikipedia.org/wiki/Pentellated_6-simplexes#Pentellated_6-simplex
that "heptagonal Fano planes" can be inscribed in "staf" in many various ways.
Has this result (or any portion thereof) been previously reported? If so, can you provide a link or citation?
Also, since the 42 vertices of "staf" occur as a subset of the roots of the algebraic group E8, does anyone know of any discuession of staf and the Fano plane within the context of the sub-algebras of E8?
Thanks as always for considering these questions.
See also this link for a related question regarding coordinates https://math.stackexchange.com/questions/3503457/coordinates-for-stericated-6-simplex-relative-to-pentellated-6-simplex-in-4-21
As already was outlined in one of the answers to https://math.stackexchange.com/questions/2070413/for-which-dimensions-does-it-exist-a-regular-n-polytope-such-that-the-distance-o/3492945?noredirect=1#3492945, it is possible to understand the expanded simplex of any dimension as an axial stack of 3 vertex layers of the mere simplex atop the expanded simplex atop the dual mere simplex (each of one dimension less), i.e.
$$x3o3o3o...o3x=\text{hull}(x3o3o...o3o\ ||\ x3o3o...o3x\ ||\ o3o3o...o3x)$$
Thus esp. each of the following is nothing but the midsection of the following:
$x3x$ = 2D hexagon, with $6$ vertices
$x3o3x$ = 3D cuboctahedron ("co"), with $2*3+6=12$ vertices
$x3o3o3x$ = 4D small prismated decachoron ("spid", aka: runcinated pentachoron), with $2*4+12=20$ vertices
$x3o3o3o3x$ = 5D small cellated dodecateron ("scad", aka: stericated hexateron), with $2*5+20=30$ vertices
$x3o3o3o3o3x$ = 6D small terated tetradecapeton ("staf", aka: pentellated heptapeton), with $2*6+30=42$ vertices
$x3o3o3o3o3o3x$ = 7D small petated hexadecaexon ("suph", aka: hexicated octaexon), with $2*7+42=56$ vertices
$x3o3o3o3o3o3o3x$ = 8D small exiated bi-enneazetton ("soxeb", aka: heptellated enneazetton), with $2*8+56=72$ vertices
etc.
(As was outlined in that very cited answer as well, all these expanded simplices have the common interesting feature of having circumradius = edge size.)
"scal" OTOH is a further Wythoffian member of the 6D simplex family. It is the small cellated heptapeton, aka: stericated heptapeton with Dynkin symbol $x3o3o3o3x3o$ and 105 vertices. It features for facets: $7\ $ $x3o3o3o3x\ .$ (scads) + $35\ $ $x3o3o\ .\ x3o$ (tratets) + $35\ $ $x3\ .\ o3x3o$ (trocts) + $21\ $ $x\ .\ o3o3x3o$ (rappips) + $7\ $ $.\ o3o3o3x3o$ (rixes).
Thus indeed scal has scads for facets. And each scad (on its own) can be seen as an axial projection of staf.
It should be noted furthermore that scal has 105 vertices. Scad OTOH has 30 vertices, and there are 7 scad facets in scal, so there will be exactly 2 scads incident at each vertex of scal.
--- rk
|
2025-03-21T14:48:29.582348
| 2020-01-04T21:46:21 |
349731
|
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|
Stack Exchange
|
When is Thom isomorphism a ring map?
Let $R$ be an $E_{\infty}$-ring spectrum and $B$ be an $E_\infty$-space. Suppose we have an $E_\infty$-map $$ f: B \to BGL_1(S^0)$$ such that the composite $$f_R: B \to BGL_1(S^0) \to BGL_1(R) $$
is null, then a choice of null-homotopy produces Thom isomorphism which is a weak-equivalence
$$u: Mf \wedge R \simeq B_+ \wedge R,$$
where $Mf$ is the Thom spectrum associated to $f$.
Note that, both sides of the Thom isomorphism are ring spectra.
Q: I wonder when $u$ is a ring-map?
A possible guess is that if $f_R$ is homotoped to null via infinite-loop space maps, then maybe $u$ is a ring map. I am not quite sure if this is true or how to see this.
One comment is that you have to be careful, because to be an $E_{\infty}$-map is not a property but rather additional structure.
You are exactly right in that what is needed is a null-homotopy of $B \rightarrow BGL_{1}(R)$ as a map of $E_{\infty}$-spaces. This is Proposition 3.16 in Omar and Toby's "A simple universal property of Thom ring spectra" when you set $A = R$.
Cool, this is exactly what I was looking for!
|
2025-03-21T14:48:29.582449
| 2020-01-04T22:41:38 |
349733
|
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|
Stack Exchange
|
Understanding a subset of linear transformations with 2 DOF produced by 3 shear transformations
I'm interested in a subset of linear transformations of the plane that has only two degrees of freedom. They must have determinant equal to 1, thus a subset of the special linear group. The other constraint is that, apart from a pure scaling, it does not contain the pure squeeze either. This subset contains pure rotations and shearing.
I'm trying to understand what subsets like this can be defined, and how they can be complemented to form the complete set of linear transformations. While I don't think such a set can form a closed group like rotations, I would like to understand what other interesting properties it may have. Could it be related to a matrix decomposition? I'm especially interested in finding out ways to parameterize these transformations.
Some alternatives I have explored were a shear followed by a rotation, or vice-versa. I am not sure these definitions create the same subsets, or which is the most complete. So that's the first question.
The most interesting definition I have found is a shear on the x dimension followed by a shear on y and then on x again repeating the first operation:
$$
A(\alpha,\beta) = \left[\matrix{ 1 & \alpha \cr 0 & 1}\right]
\left[\matrix{ 1 & 0 \cr \beta & 1}\right]
\left[\matrix{ 1 & \alpha \cr 0 & 1}\right]
$$
This is an orientation and area-preserving transformation, that becomes a pure rotation if $\beta=\sin(\theta)$ and $\alpha=-\tan(\theta/2)$. It can also produce pure shearing at either the x or y directions, although I'm not even sure about other axes. It has the nice property that the inverse transformation is obtained by just changing the sign of the parameters, although like I said it does not seem to form a group.
The question is then: what exactly is this subset of transformations represented by the formula above? Can it be made more complete while still having only two degrees of freedom? And how it relates to the complete set of linear transformations?
A little bit more context: I'm basically looking for the largest set of transformations with 2DOF that can be parameterized in such a way that changing one parameter necessarily affects the angles of lines anywhere on the plane (except for lines exactly at the shearing axis). Therefore rotations and shearing, excluding scaling and pure squeezing. What is this set of transformations?
This seems to be the set of unitary Toeplitz matrices.
|
2025-03-21T14:48:29.582752
| 2020-01-04T22:59:59 |
349734
|
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|
Stack Exchange
|
Implications of Geometrization conjecture for fundamental group
Hempel proved that Haken manifolds have residually finite fundamental groups. With the Geometrization conjecture, this now holds for any compact and orientable 3-manifold.
How exactly does the Geometrization conjecture imply that the only non-Haken compact orientable irreducible 3-manifolds are compact hyperbolic manifolds with no cusps?
Thanks a lot.
I guess even compact and orientable are not required for residual finiteness, only finite-generation of the fundamental group.
The statement you make is not quite correct (you can also have "small" Seifert-fibered spaces): see Theorem 1.7.6 of this book: https://www.ems-ph.org/books/book.php?proj_nr=195
@RyleeLyman Sure, just invoking Scott's compact core theorem and that subgroups of residually finite groups are r.f, whatever the correct statement is will extend to that level of generality.
"compact with no cusp" sounds funny: a hyperbolic manifold with a cusp can't be compact.
There are a few ways to look at this, but let me give a synopsis of Peter Scott's discussion in Section 6 of his 8 geometries paper, because it seems to directly address the question:
Scott, Peter, The geometries of 3-manifolds, Bull. Lond. Math. Soc. 15, 401-487 (1983). ZBL0561.57001.
(Note: there is also an errata for this reference: http://www.math.lsa.umich.edu/~pscott/errata8geoms.pdf. The corrections mentioned there do not affect this summary. )
Let's set up some standard 3-manifold terminology (anything left out is clearly defined in the reference).
A compact orientable 3-manifold is irreducible if is either $S^1 \times S^2$ or every embedded $S^2$ bounds a 3-ball. From now on, assume $M$ is a compact, orientable, irreducible 3-manifold (unless otherwise specified).
A surface $S$ (smoothly) embedded in $M$ is incompressible if every embedded curve $\gamma$ in $S$ which bounds a disk in $M$ also bounds a disk in $S$. A compact, orientable, irreducible 3-manifold is toroidal if it contains an incompressible torus.
The most natural thing to say is that the affirmative solution to the Geometrization Conjecture implies that a compact, irreducible 3-manifold is either toroidal or is homeomorphic to a quotient of the form $X/\Gamma$ where $X$ is one of the following geometric spaces: $S^3,S^1 \times \mathbb{R},E^3,Nil,Sol, H^2 \times \mathbb{R}, \widetilde{PSL(2,\mathbb{R})}$ or $H^3$ and $\Gamma \subset Isom^+(X)$ acting properly and discontinuously.
If a space admits a Sol geometry it is Haken. In fact, it is both toroidal and has positive first Betti number, (see [Scott, Theorem 4.17]).
The remaining geometries are either $H^3$ or Seifert fibered. Of course, some Seifert fibered manifolds like $T^3 \cong S^1 \times S^1 \times S^1$ are both toroidal and Seifert fibered and there are other minor pathologies: for example, $RP^3 \# RP^3$ is Seifert fibered and reducible and in the wake of Geometrization manifolds with $S^3$ geometry are exactly those with finite fundamental group.
Happily, Scott gives a clean statement of what you want in the conjecture on page 484 (of course the affirmative solution to the Geometrization Conjecture implies this conjecture is now known to be true):
Conjecture (now theorem): Let $M$ be a closed, irreducible, non-Haken 3-manifold with infinite fundamental group. Then $M$ is either a Seifert fibered space or admits a hyperbolic structure.
To connect this statement to the comments mentioned above, if we assume $M$ is Seifert fibered, non-Haken, irreducible, and has infinite fundamental group, then it is small Seifert fibered, which implies $M$ that the base orbifold of $M$ is $S^2(a_1,a_2,a_3)$ with $1/a_1 + 1/a_2 + 1/a_3 \leq 1$.
Is it also true that your quoted conjecture implies Geometrization pre-Perelman?
No. The conjecture (due to Scott) is only part of the Geometrization conjecture. It does not say anything about simply connected 3-manifolds for example.
The conjecture is enough to imply residual finiteness of all (finitely generated) 3-manifold groups, though.
|
2025-03-21T14:48:29.583023
| 2020-01-04T23:17:50 |
349735
|
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|
Stack Exchange
|
Counting Gaussian integers with an angle condition
Consider elements $x,z \in \mathbb{Z}[i]$, the Gaussian integers. Let $S(T_1, T_2)$ be the subset of $\mathbb{Z}[i] \times \mathbb{Z}[i]$ consisting of those elements $(x,z)$ such that $T_1 < |x| \leq 2T_1, T_2 < |z| \leq 2T_2$. I want to estimate the cardinality of the set
$$\displaystyle S_k(T_1, T_2) =\{(x,z) \in S(T_1, T_2) : \left \lvert \Re(xz^k) \right \rvert \leq M \}, k \geq 2.$$
Note that if $M$ is close in size to $T_1 T_2^k$, then the restriction that $\lvert \Re(xz^k) \rvert \leq M$ is hardly a restriction at all and the obvious bound of $|S(T_1, T_2)|$ will give the correct order of magnitude. However when $M$ is very small compared to $T_1 T_2^k$ then $|S_k(T_1, T_2)|$ should be substantially smaller than $|S(T_1, T_2)|$.
If we write $x = r_1 e^{i \theta_1}$ and $z = r_2 e^{i \theta_2}$, then the condition $|\Re(xz^k)| \leq M$ can be written as $|r_1 r_2^k \cos(\theta_1 + k \theta_2)| \leq M$. If $M$ is small compared to $T_1 T_2^k$, then this is essentially saying that the angle $\theta_1 + k\theta_2 \pmod{\pi}$ is close to $\pi/2$.
Is there a good way to estimate the size of $S_k(T_1, T_2)$?
|
2025-03-21T14:48:29.583126
| 2020-01-05T01:36:24 |
349737
|
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|
Stack Exchange
|
Classification of "extensions" of algebraic varieties
Let $k$ be an algebraically closed field, and $Z,U$ be smooth varieties over $k$. A variety is by definition integral (irreducible + connected).
Can we classify all smooth projective varieties $X$ over $k$ with a closed subvariety isomorphic to $Z$ and the complement isomorphic to $U$ (assume there is one)? Of course they are all birational and are the same in the Grothendieck group. In general, it seems too difficult. I hope there is a simple condition for uniqueness, and classification in low dimension case (for curve it's trivial).
Examples:
$Z$ is a point, this can be thought as whether the one-point compactification is unique.
$Z=\mathbb P^{n-1}$, $U=\mathbb A^n$, $X=\mathbb P^n$. Are there any other examples?
$Z=\mathbb P^{n-1}$, $U=\mathbb P^1 \times \mathbb A^{n-1}$, then consider any rank $2$ projective bundle over $\mathbb P^{n-1}$. Are there any other examples?
$X_0$ is an abelian variety, $Z$ an abelian subvariety, $U=X_0-Z$.
The general type case.
In another words you are asking about classifications of "projective compactifications'' of $U$. It is not even known for your second example $Z = \mathbb{P}^{n-1}$, $U = \mathbb{A}^n$. For this example the complete answer for $n = 2$ and $k = \mathbb{C}$ can be found from this paper of mine: https://arxiv.org/abs/1307.5577 (combine Proposition 3.3 and Corollary 3.5). In particular there are examples other than $X = \mathbb{P}^2$, e.g. the weighted projective surfaces of the form $\mathbb{P}^2(1,p,q)$.
|
2025-03-21T14:48:29.583256
| 2020-01-05T02:33:56 |
349741
|
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|
Stack Exchange
|
Can the intermediate Chern classes be expressed as Euler classes?
General question: We know that the top Chern class $c_n(\xi)$ of an $n$-dimensional complex vector bundle $\xi$ is its Euler class, while the first Chern class, $c_1(\xi)$, is the Euler class of its determinant bundle. Can the intermediate classes also be expressed as Euler classes of bundles related to $\xi$?
Let me also ask a much more specific question: We know that $H^\ast(BU(3);{\mathbb Z}) = {\mathbb Z}[c_1, c_2, c_3]$. Is there a complex 2-plane bundle over the classifying space $BU(3)$ (or $BU(n)$ for $n\geq 3$) whose Euler class is $c_2$?
Motivation: Long story short, I'm trying to construct an analogue of $c_2$ in a context where I know that the usual constructions I'm familiar with won't work. But I do have Euler classes in this context, so if I could express $c_2$ as an Euler class as above, I might be able to use that in my context. On the other hand, if anyone has a proof that there is no such bundle over $BU(3)$, then that will save me time chasing down that particular rabbit hole.
In Milnor and Stashev's "Characteristic Classes" (pg. 158 for me), they define all the Chern classes via the Euler class. Maybe this is not surprising since the kth Chern class of the tautological bundle over $BU(n)$ pulls back to the Euler class of $BU(k)$ for $k<n$.
That's one of the constructions I'm familiar with, but note that only the top class is defined directly as an Euler class of a bundle on the given space. The construction carries over to my context, but gives the "wrong" classes for reasons too complicated to explain in this comment.
If I am not mistaken, reduced mod 2 we have $Sq^2 c_2 = c_1 c_2 + c_3$ so the answer to the specific question is no.
@GustavoGranja: I believe you're right about the formula, but perhaps I'm blanking on something obvious: How does this show that such a bundle doesn't exist? If there were such a bundle $\xi$, it need not be true that $c_3(\xi)$ maps to the element $c_3$ in the cohomology of $BU(3)$.
Maybe I misunderstood the question. Such a bundle would be classified by a map to $BU(2)$ which would pullback $c_2 \in H^2(BU(2))$ to $c_2 \in H^2(BU(3))$. That is precluded by the formula for the action of $Sq^2$ on $H^*BSU(3)$.
@GustavoGranja: Agreed about $c_2$, but we know little about what the map would do to $c_1$. So I'm still not seeing how the formula precludes such a map.
The image of the map on cohomology would be contained in the algebra generated by $c_1,c_2$ but this is not closed under the action of the Steenrod squares.
@GustavoGranja: Again, we know little about what the map would do to $c_1$ in $H^\ast(BU(2))$: There's no reason to expect it maps to $c_1$ in $H^\ast(BU(3))$. So I don't see that the image need be contained in that subalgebra.
It will map $c_1$ to some multiple of $c_1$.
@GustavoGranja: Yes, of course, now I see. If you want to submit your comment as an answer I'll accept it. Many thanks.
Are you ok with virtual bundles in your generalized setup?
@JohnGreenwood: Maybe. What do you have in mind?
@SteveCostenoble I thought there might a virtual bundle operation (a la Adams operations, etc) generalizing the "top exterior power" and "identity" operations that gives you the first and last chern class, but i tried a few things and nothing worked...it seems there's something special about the first and last symmetric polynomials!
@JohnGreenwood: Thanks for looking at it, anyway. With the search for such a bundle quashed, I think I've found a different approach to construct the elements I need. So it's all good in the end.
@SteveCostenoble Glad to hear it!
@johngreenwood Indeed, I have thought this for quite a while too. For instance the first and last symmetric polynomial define the trace and norm (in finite ring maps) and are additive and multiplicative and very important. But the other symm polynomials don't give any algebraic compatibilities as far as I can tell. Curious!
|
2025-03-21T14:48:29.583535
| 2020-01-05T02:39:53 |
349742
|
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"authors": [
"Christian Remling",
"Iosif Pinelis",
"Nate Eldredge",
"Piero D'Ancona",
"geometricK",
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"sort": "votes",
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|
Stack Exchange
|
"Square root" of multiplication operator on Sobolev space
Let $f:\mathbb{R}^n\rightarrow\mathbb{R}$ be a non-negative, smooth, uniformly bounded function with uniformly bounded first derivative. Then $f$ defines a bounded operator on $L^2(\mathbb{R}^n)$ as well as on $H^1(\mathbb{R}^n)$, the first Sobolev space. In $L^2(\mathbb{R}^n)$ the following equality holds: for $e_1,e_2\in C_c^\infty(\mathbb{R}^n)$,
$$\langle fe_1,e_2\rangle_{L^2}=\langle\sqrt{f}e_1,\sqrt{f}e_2\rangle_{L^2}.$$
Question: Is there a way to do this in $H^1$, ie does there exist a function $g_f$ such that
$$\langle fe_1,e_2\rangle_{H^1}=\langle g_f(e_1),g_f(e_2)\rangle_{H^1}?$$
Follow-up question: Is the adjoint $f^*:H^1\rightarrow H^1$ also given by function multiplication?
Such an identity would imply that multiplication by $f$ is a symmetric operator, which is not
As to the followup question about the adjoint, I previously asked it on Math.SE, but never got an answer.
@NateEldredge: I think your MSE question can be answered by taking Fourier transforms. The multiplication becomes convolution, or we can say we have an integral operator. Of course, the adjoint of a (bounded) integral operator on $L^2(\mathbb R)$ with kernel $K(s,t)$ is the integral operator with kernel $\overline{K(t,s)}$, and we can handle the actual situation (on a weighted $L^2$ space) in the same way by absorbing the weight into the kernel temporarily.
In particular, the answer to the "follow-up question" is no.
$\newcommand{\R}{\mathbb R}$
The square root of the multiplication operator does not exist unless the function $f$ is constant (and, obviously, the square root exists if $f$ is constant).
Indeed, suppose that there exists a function $g_f$ such that
\begin{equation*}
\langle fx,y\rangle_{H^1}=\langle g_f(x),g_f(y)\rangle_{H^1} \tag{1}
\end{equation*}
for all $x$ and $y$ in $C_c^\infty(\R^n)$. The simple but crucial observation is that $\langle g_f(x),g_f(y)\rangle_{H^1}=\overline{\langle g_f(y),g_f(x)\rangle_{H^1} }$, so that (1) implies
$$\langle fx,y\rangle_{H^1}=\overline{\langle fy,x\rangle_{H^1}},
$$
which can be rewritten as
\begin{equation*}
\int_{\R^n}\sum_{j=1}^n (D_jf)\,(x\,D_jz-z\,D_jx)=0, \tag{2}
\end{equation*}
where $z:=\bar y$ and $D_j$ denotes the partial derivative with respect to the $j$th argument.
Take any $x\in C_c^\infty(\R^n)$ and any $a=(a_1,\dots,a_n)\in\R^n$, and then let $z:=xe_a$, where the function $e_a$ is defined by the formula
$e_a(t_1,\dots,t_n):=e^{i(a_1t_1+\dots+a_nt_n)}$ for $t=(t_1,\dots,t_n)\in\R^n$. Then $x\,D_jz-z\,D_jx=ia_jx^2e_a$ for all $j=1,\dots,n$. So, (2) yields $\sum_{j=1}^n a_jD_jf=0$ for all $(a_1,\dots,a_n)\in\R^n$, which means that $f$ is constant.
Many thanks for your answer. I've added a second part to the original question that now seems relevant, about adjoints.
@ougoah : To keep things in good order, any additional questions should be asked in separate posts, especially when the original question has been answered.
@ougoah : Your additional question has now been answered at https://mathoverflow.net/questions/349853/adjoint-of-the-multiplication-operator-on-a-sobolev-space/349854#349854
|
2025-03-21T14:48:29.583738
| 2020-01-05T04:34:00 |
349745
|
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"url": "https://mathoverflow.net/questions/349745"
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|
Stack Exchange
|
Protocategory generates category uniquely (up to isomorphic)?
I am reading Johnstone's "Sketches of an Elephant", and I get stuck in the construction of an idea of a category B is structured over a category C.
Johnstone introduces protocategory to give this idea, and he mentions the key axiom that
if $g: A \rightarrow B$ and $f: B \rightarrow C$, there is exactly ont $h$ satisfying $h: A \rightarrow C$ and $ f \circ g = h $.
But I don't get how protocategory generates category and the uniqueness of the generated category too.
I mean, if a category B is generated by a protocategory C, is a category B unique (up to isomorphism or something like that)?
The category generated by a protocategory is unique by construction, just like any other explicit mathematical construction of an object: you write down what the objects, the hom-sets, and the operations are. What are you worried about?
Maybe the n-lab description will help? There it's quite clear that each protocategory really does generate a unique category: as Mike says, it lays out specifically what the objects, morphisms, and composition operation are (see the last bulletpoint in section 2).
Okay, I misunderstand the idea of protomorphism and source-target predicate.
I considered "doing generate category" as "taking a kind of closure of source-target predicate. But now I have just noticed that protomorphisms are the same as morphisms except they are not directly related to (source-target) specific objects.
|
2025-03-21T14:48:29.583864
| 2020-01-05T07:33:22 |
349750
|
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"sort": "votes",
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|
Stack Exchange
|
survey paper on the construction of hyperbolic manifolds
Is there a good survey paper which discusses the common ways of building hyperbolic $n$-manifolds?
Section 3.1 here.
See Dave Witte-Morris' book for arithmetic groups, and section 6.5 for the non-arithmetic Gromov-Piatetskii-Shapiro examples. http://deductivepress.ca/ These and variants (together with Selberg's Lemma) are essentially the only known general method for proving the existence of finite volume hyperbolic $n$-manifolds (variants are given here https://arxiv.org/abs/1802.04619). In 3 dimensions, in some sense the geometrization theorem gives a construction of all finite volume hyperbolic 3-manifolds. The Deligne-Mostow construction gives some examples as moduli spaces of polygons.
For recent constructions of finite volume hyperbolic manifolds with various properties, see:
https://arxiv.org/abs/1008.2646
https://arxiv.org/abs/1507.02747
https://arxiv.org/abs/1703.10561
https://arxiv.org/abs/1812.06536
https://arxiv.org/abs/1904.12720
Not a paper, but a book : Foundations of Hyperbolic Manifolds by John Ratcliffe. Chapters 10 and 11 might contain what you're looking for. Also available here.
Concerning 4-manifolds, you have a survey by Bruno Martelli here.
Also, I'm sure you can find what you want typing construction of hyperbolic manifolds on your favorite search engine.
|
2025-03-21T14:48:29.583976
| 2020-01-05T10:01:33 |
349753
|
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"authors": [
"Jochen Glueck",
"Pietro Majer",
"erz",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/349753"
}
|
Stack Exchange
|
Dual fixed point
Let $E$ be a Banach space, let $T:E\to E$ have norm $1$ and let $\nu\in E^*\setminus\{0\}$ be such that $T^*\nu=\nu$. Under which conditions there is $e\in E$ such that $Te=e$ and $\langle e,\nu\rangle\ne 0$?
This is true for reflexive spaces. Indeed, WLOG $\|\nu\|=1$. Consider $D=\nu^{-1}(1)\cap \overline{B}_{E}$, which is weakly compact and non-empty. Then $T_{D}$ is weak-to-weak continuous self-map of $D$. Therefore, there is a fixed point due to Tychonoff fixed point theorem. Thus, I am looking for conditions significantly weaker than reflexivity.
REMARK: In fact, the argument works if there is a linear topology on $E$ such that $T$ and $\nu$ are continuous with respect to that topology and $\overline{B}_{E}$ is compact in that topology.
The existence of such an $e$ for each $\nu$ is equivalent to $T$ being mean ergodic.
@JochenGlueck do you mean for every $\nu$ that is a fixed point of $T^*$? If so, could you please elaborate?
Yes, I mean for every $\nu$ that is a fixed point of $T^*$. I added an answer with details and a reference.
Part 1 of the answer. In terms of $T$, the property you are looking for is characterized by the mean ergodic theorem:
Theorem. Let $E$ be a Banach space and let $T$ be a bounded linear operator on $E$ that is power-bounded in the sense that $\sup_{n \in \mathbb{N}_0}\|T^n\| < \infty$. Then the following assertions are equivalent:
(i) $T$ is mean ergodic, i.e. the sequence of Cesàro means $(\frac{1}{n}\sum_{k=0}^{n-1} T^k)_{n \in \mathbb{N}}$ converges strongly.
(ii) The fixed space of the dual operator $T^*$ is separated by the fixed space of the operator $T$, i.e. for every $0 \not= \nu \in \ker(1 - T^*)$ there exists $0 \not= e \in \ker(1-T)$ such that $\langle \nu, e \rangle \not= 0$.
(iii) The space $E$ is the topologically direct sum of the fixed space $\ker(1-T)$ and the closure of the range of $1-T$.
This is a classical theorem in operator theory. You can find it, for instance, in Theorem 8.20 of [T. Eisner, B. Farkas, M. Haase, R. Nagel: Operator Theoretic Aspects of Ergodic Theory (2015)] (actually, the theorem there contains several more equivalent assertions and is not only stated for power-bounded operators but for the slightly larger class of Cesàro bounded operators).
The result for reflexive spaces that is stated in the question is a special case of the above theorem since every power-bounded operator on a reflexive Banach space is mean ergodic [op. cit., Theorem 8.22].
Part 2 of the answer. The question on which spaces every power-bounded (or contractive) operator is mean-ergodic is a very subtle one.
In the paper [V. P. Fonf, M. Lin, P. Wojtaszczyk: Ergodic Characterizations of Reflexivity of Banach Spaces (Journal of Functional Analysis, 2001)] it is shown that a Banach space with a Schauder basis is necessarily reflexive if every power-bounded operator on it is mean ergodic.
In the paper [V. P. Fonf, M. Lin, P. Wojtaszczyk: A non-reflexive Banach space with all contractions mean ergodic (Israel Journal of Mathematics, 2010)] it is shown that there exist non-reflexive Banach spaces an which every operator $T$ of norm $\|T\| \le 1$ is mean ergodic.
We may add that the limit in (i) is the linear projector onto $\ker(I-T)$ associated with the decomposition in (iii)
Thank you! A small correction: in (ii) the latter is separated by the former.
This is however, not exactly what I was looking for; I was more interested in conditions on $E$ and $\nu$...
@erz: Thank you for the correction; I corrected (ii) accordingly. Concerning your second comment, I added a second part to the answer: for most Banach spaces, one cannot weaken the reflexivity assumption if one wants all power-bounded operators to be mean ergodic.
The thing is, I don't need this property to hold for every $\nu$, I need it to hold for a given $\nu$, and so it may depend on $\nu$. This should be a geometric condition that takes into account the way the kernel of $\nu$ dissects the unit ball of $E$. Thank you nonetheless!
|
2025-03-21T14:48:29.584261
| 2020-01-05T10:23:36 |
349754
|
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|
Stack Exchange
|
Postprojective components of quiver algebras
Let $A=kQ/I$ be a quiver algebra with acyclic quiver $Q$.
An indecomposable module $M$ is called postprojective in case $M \cong \tau^{-1}(P)$ for an indecomposble projective module $P$. A component of the Auslander-Reiten quiver of $A$ is called postprojective in case every module in the component is postprojective. Assume $A$ has at least one postprojective component.
Question 1: Is there a (quick) way to obtain the number of postprojective components of such an algebra?
Question 2: Is there an easy way to see whether any indecomposable projective module is postprojective?
Im especially interested wheter such a thing is possible using the GAP-package QPA.
One place to start is to look at the paper "An algorithm for finding all preprojective components of the Auslander-Reiten quiver" by Peter Draexler and Klara Kögerler.
|
2025-03-21T14:48:29.584344
| 2020-01-05T10:35:36 |
349755
|
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"Dattier",
"Wlod AA",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/349755"
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|
Stack Exchange
|
Regularize continuous functions with bounded variation
Is it true that :
$\forall f,g \in C([0,1],\mathbb R), \exists h \in C([0,1],[0,1])$ $f,g,h$ strictly increasing and $h([0,1])=[0,1]$ with $(f \circ h, g\circ h) \in C^{\infty}([0,1],\mathbb R)^2$?
It was kind of you to add the homeomorphism assumption about $h$.
The answer is NO:
Let $\,f:[0;1]\to[0;1]\,$ be the identity function. Let $\,g:[0;1]\to[0;1]\,$ be such that the set of $\,D\subseteq[0;1]\,$ of points $\,x\in[0;1]\,$ for which derivative of $\,g\,$ doesn't exist is dense.
Then, if $\,f\circ h\,$ is smooth then $\,h\,$ is smooth.
But then $\,g\circ h\,$ cannot be smooth. Otherwise, the derivative of $\,h\,$ at every $\,y\in h^{-1}(D)\,$ would be $0,\,$ and $\,h^{-1}(D)\,$ would be dense ($\,h\,$ is really assumed to be a homeomorphism), hence $\,h\,$ would be constant. A contradiction. Great!
For the sake of mine and everybody's peace of mind let me add the ultimate detail:
Assume that $\,g\circ h\,$ is smooth, and that the derivative of smooth $h$ at point $\,h^{-1}(x)$ is not $0\,$ for a certain point $x$. Then, $g$ is differentiable at $x$.
Indeed, the derivative of $\,g\,$ at $\,x\,$ is the derivative at $x$ of $\,g\,=\,(g\circ h)\circ h^{-1}.\,$
Happy New Year!
why the derivative of $h$ at every $y\in h^{-1}(D)$ would be $0$ ?
For instance, if $h$ were constant then $,g\circ h,$ would be smooth.
it does not seem to me that you are answering my question.
@Dattier, I've misread your question (have imagined a question which could have been "natural" to ask). You already know at the stage of the proof that $h$ is smooth. Thus at the critical points that derivate (would exist and) being different from zero would not compose with $g$ to a smooth function.
I may add a small but truly detailed note at the end of the Answer if you still would like me to do so.
|
2025-03-21T14:48:29.584495
| 2020-01-05T10:54:08 |
349757
|
{
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"authors": [
"Mark Wildon",
"Timothy Budd",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/349757"
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|
Stack Exchange
|
On the sum $\sum_{k=b}^{a-1} \binom{2k+1+n}{2n+1}\binom{2a-1}{a+k}$
Let $n$ be a non-negative integer.
Does there always exist a polynomial $P_n(a,b)$ such that for all integers $a > b \geq n/2$ we have
$$
\sum_{k=b}^{a-1} \binom{2k+1+n}{2n+1}\binom{2a-1}{a+k} = \binom{2a-1}{a+b} P_n(a,b)\quad ?
$$
For small values of $n$ this is easily verified using Gosper's algorithm, for example
$$P_0(a,b) = a+b,\qquad P_1(a,b) = \tfrac{2}{3}(a+b)(b^2 +a -1), $$
but I am struggling to prove the general case.
Any suggestions or literature references on this problem?
Here are some details on Gosper's algorithm. Let us denote the summand by $t(k) = \binom{2k+1+n}{2n+1}\binom{2a-1}{a+k}$ and let
$$p(k) = (k-\tfrac12n+1)_n(k-\tfrac12n+\tfrac12)_{n+1}, \quad q(k) = a-k-1, \quad r(k)=a+k$$
where $(x)_n = x(x+1)\cdots(x+n-1)$.
Then we have
$$ \frac{t(k+1)}{t(k)} = \frac{q(k)}{r(k+1)}\frac{p(k+1)}{p(k)}.$$
According to Gosper, $T(k+1)-T(k)=t(k)$ has a hypergeometric term solution $T(k)$ iff there exists a polynomial $s(k)$ of degree $2n$ (with coefficients that depend on $a$) that solves
\begin{equation}
\tag{1} p(k) = q(k) s(k+1) - r(k)s(k).
\end{equation}
If such a solution exists, then $$T(k) = \frac{r(k)}{p(k)} t(k)s(k)\quad \text{and}\quad P_n(a,b) = \frac{T(a)-T(b)}{\binom{2a-1}{a+b}} = -\frac{2^{2n+1}}{(2n+1)!}(a+b)\,s(b)$$
fulfilling my request. The linear mapping $s(k) \mapsto q(k) s(k+1) - r(k)s(k)$ from polynomials of degree $2n$ to those of rank $2n+1$ is easily seen to be injective, but it seems difficult to show that $p(k)$ lies in its image for general $n$.
Update: I have included a proof in an answer below, but I'd still be interested in literature references.
Do you know if the condition $b \ge n/2$ is necessary? And/or, could you give some motivation for it?
For $k<n/2$ the summand vanishes, so I suppose the condition is necessary (but have not thought deeply about it).
Thanks. Sorry, I now see this is quite obvious.
Actually, checking whether $p(k)$ lies in the image of $s(k) \mapsto q(k) s(k+1) - r(k)s(k)$ turns out to be not too difficult after all. We need to determine a single linear condition that spans the cokernel of the linear mapping. A convenient way to do this is by turning $q(k) s(k+1) - r(k)s(k)$ into a differential operator and seeking a (formal) power series $V(x)$ solving
\begin{align}
0 &= \left[s(1+\partial_x)q(\partial_x) - s(\partial_x)r(\partial_x) \right] V(x)\big|_{x=0} \\
&=\left[e^{-x}s(\partial_x)e^{x}q(\partial_x) - s(\partial_x)r(\partial_x) \right] V(x)\big|_{x=0} \\
&=s(\partial_x)\left[e^{x}q(\partial_x) - r(\partial_x) \right] V(x)\big|_{x=0}.
\end{align}
Hence, a solution to
\begin{equation}
0 = \left[e^{x}q(\partial_x) - r(\partial_x) \right] V(x) = (e^x(a-1)-a)V(x)-(e^x+1)V'(x)
\end{equation}
will do for any $n\geq 0$ and any polynomial $s(k)$. One easily finds
\begin{equation}
V(x) = e^{-ax} (1+e^{x})^{2a-1} = e^{-x/2} (2\cosh(x/2))^{2a-1}.
\end{equation}
It remains to show that $p(\partial_x) V(x)|_{x=0}=0$. This follows from
\begin{equation}
p(\partial_x) V(x)\big|_{x=0} = p(\partial_x-\tfrac{1}{2}) e^{x/2}V(x)\big|_{x=0} = p(\partial_x-\tfrac{1}{2}) (2\cosh(x/2))^{2a-1}\big|_{x=0},
\end{equation}
which vanishes because $p(\partial_x-1/2)$ is odd in $\partial_x$, while $(2\cosh(x/2))^{2a-1}$ is an even power series in $x$.
This shows that the recursion equation (1) has a unique polynomial solution $s(k)$. Moreover the coefficients of $s(k)$ are seen to be polynomials in $a$, since the linear mapping $s(k) \mapsto q(k) s(k+1) - r(k)s(k)$ is essentially upper-triangular (in the monomial basis) with a diagonal that is independent of $a$. QED.
In fact this demonstrates an equivalence:
Let $p(k)$ be a polynomial. Then $\sum_{k=b}^{a-1} p(k)\binom{2a-1}{a+k}$
is given by a hypergeometric term in $b$ if and only if $p(k-1/2)$ is odd in $k$. In that case there exists a polynomial $P(a,b)$ such that
\begin{equation}\sum_{k=b}^{a-1} p(k)\binom{2a-1}{a+k} = \binom{2a-1}{a+b} P(a,b).\qquad (a>b)
\end{equation}
|
2025-03-21T14:48:29.584835
| 2020-01-05T10:54:16 |
349758
|
{
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|
Stack Exchange
|
Analog of Reed's conjecture for hypergraph
Reed's $\omega$,$\Delta$, $\chi$ conjecture proposes that every graph has $\chi \leq \lceil \tfrac 12(\Delta+1+\omega)\rceil$. Here $\chi$ is the chromatic number, $\Delta$ is the maximum degree, and $\omega$ is the clique number.
Is there any similar theorem or conjecture that has to do with hypergraph colorings? That is where the vertices can be colored in $\chi$ colors such that no edge is monochromatic.
More helpful for me is the case of coloring where every k-edge has k different colors (Rainbow coloring).
https://mathoverflow.net/questions/47663/hypergraph-chromatic-number-vs-degree-clique-size?rq=1 answers this for the rainbow case.
|
2025-03-21T14:48:29.584913
| 2020-01-05T11:39:23 |
349760
|
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"url": "https://mathoverflow.net/questions/349760"
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|
Stack Exchange
|
A kind of isomorphicity of vector bundles
Let $X$ be a connected topological space. Let $E$ be a $k$ dimensional sub vector bundle of the trivial vector bundle $X\times \mathbb{R}^n$. Then $E$ defines an idempotent with trace $k$ in $M_n(C(X))$. Conversely every trace $k$ idempotent of this algebra determines a $k$ dimensional sub bundle of $n$ domensional trivial bundle over $X$.
Two idempotents associated to two isomorphism bundles are Murray von Neumann equivalent.
Are there two non isomorphic $k$ dimensional sub bundle of $X\times \mathbb{R}^n$ for which their corresponding idempotents $e,f$ admit an automorphism $\alpha$ of $M_n(C(X))$ with $\alpha(e)=f$?
Note: The above question actualy defines an equivalent relation on the space of all $k$ dimensional subbundles of the $n$ dimensional trivial bundle.
The antipodal map of $S^{2}$ sends $L$ to $L^{-1}$, where $L$ is the line bundle constructed via clutching along an equatorial $S^1$ with the "identity" map $S^1\rightarrow U(1)$.
So let $\alpha$ be the automorphism of $M_{n}(C(S^2))$ induced by the antipodal map, let $e$ be the projection corresponding to $L$, and let $f=\alpha(e)$.
|
2025-03-21T14:48:29.585011
| 2020-01-05T12:41:22 |
349762
|
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|
Stack Exchange
|
Free Lie algebra and nilpotent groups in Rothschild and Stein's paper
In
Rothschild, Linda Preiss; Stein, Elias M., Hypoelliptic differential operators and nilpotent groups, Acta Math. 137(1976), 247-320 (1977). ZBL0346.35030. PDF at archive.ymsc.tsinghua.edu.cn
they mentioned the free nilpotent Lie algebra $\mathfrak{R}_{F,s}$ in Part I, section 3, example 4. And they introduced $n_s$ and $m_s$ in Part II, section 7, where
$n_s$ is the dimension of the free nilpotent Lie algebra $\mathfrak{R}_{F,s}$ of step $s$ on $n$ generators,
$m_s$ is the dimension of the linear space spanned by all commutators of the vector fields $\{W_k\}_{k=1}^n$ ($n$ is the number of vector fields) of lengths $\leq s$ restricted to $\xi$.
I can understand $m_s$ easily but It's difficult to understand $n_s$ to me. I think I need some example for $n_s$. Consider the vector fields:
$$X=(X_1,X_2)=(\partial_1,x_1\partial_2)$$
Its $m_1=1$ restricted to point $(0,x_2)$ and $m_1=2$ at others, while $m_2=2$ at any point in $\mathbb{R}^2$. What is $n_1$ and $n_2$ for this example? why?
And by the theorem 4 and introduction in this paper, if we lift the vector fields as
$$\widetilde{X}=(\widetilde{X_1},\widetilde{X_2})=(\partial_1,x_1\partial_2+\partial_3)$$
then, $m_2=3$ and $m_2=n_2$. Why does $m_2=n_2$?
Thanks for your help!
You don't introduce $W_k$. Also I don't see the meaning of "its $m_1=1$"... Also calling $n_s$ a function of $n$ and $s$ is utterly confusing.
Sorry. ${W_k}_{k=1}^{m}$ are vector fields, so are the $X$ below. The definition of $n_s$ and $m_s$ is the original words in his paper. In the first example, when $s=1$, then at the point $(0,x_2)$ we can see $X=(\partial_1,0)$. Thus $m_1=dim(span{\partial_1,0})=1$. The other cases are similar. But I don't understand the definition of $n_s$ in his paper, even don't know how to calculate $n_s$ for this example.
More explanations. When $s=2$, we have the commutator $[X_1,X_2]=\partial_2$, thus by the definition, we have $m_2=dim(span{\partial_1,x_1\partial_2,\partial_2})=2$ at any point in $\mathbb{R}^2$.
@YCor: The notation is indeed a bit unfortunate. I think in context, R&S are treating $n$ as fixed throughout, and the $n$ in $n_s$ is not a reference to the same $n$.
First, the notation is a little confusing. I think you are supposed to understand that $n$ is always the number of vector fields in the fixed set $\{X_1, \dots, X_n\}$. But the symbol $n$ in the notation $n_s$ is just an arbitrary letter and isn't a reference to the number of vector fields. So in your example, $n$ is $2$, and $n_1$ just means "the dimension of the free nilpotent algebra of step 1 on 2 generators". $n_2$ is the dimension of the free nilpotent algebra of step 2 on 2 generators. If we were working with a set of 47 vector fields, then $n_2$ would refer to the dimension of the free nilpotent algebra of step 2 on 47 generators.
In small examples, this is easy to compute. $n_1$ will always equal $n$, because the free nilpotent Lie algebra of step $1$ on $n$ generators, call them $Y_1, \dots, Y_n$, is simply the abelian Lie algebra spanned by $Y_1, \dots, Y_n$ with all brackets vanishing, and its dimension is $n$. In your example with $n=2$, we have $n_2 = 3$; the free nilpotent Lie algebra of step $2$ on $2$ generators is the Heisenberg Lie algebra spanned by $Y_1, Y_2, Z$, where $[Y_1, Y_2]=Z$ and $[Y_1, Z]=[Y_2, Z]=0$.
When $s \ge 3$ it gets harder. You can't just count all possible brackets of $Y_1, \dots, Y_n$ of order up to $s$, because Jacobi's identity implies some linear dependence among them. But in general, the value of $n_s$ can be found from Witt's formula; see Corollary 4.14 of
Reutenauer, Christophe, Free Lie algebras, London Mathematical Society Monographs. New Series. 7. Oxford: Clarendon Press. xvii, 269 p. (1993). ZBL0798.17001.
Indeed, we have $$n_s = \sum_{k=1}^s \frac{1}{k} \sum_{d \mid k} \mu(d) n^{k/d}$$
where $\mu$ is the Möbius function.
The Witt's formula seems that ns only depends on the number of the generators. By the description in R&S's paper, I try to explain like the following. For given "generator": $Y_1,Y_2,Y_3$, let them generate the "free nilpotent Lie algebra" $\mathfrak{R}_{F,2}$ (let, say $s=2$,) $={Y_1,Y_2,Y_3,[Y_1,Y_2],[Y_1,Y_3],[Y_2,Y_3]}$. Thus $n_2=6$, which is consistent with the result made by Witt's formula. Once we set $Y_i$ to be a specific vector field, like $Y_1=\partial_1$ etc. then we can calculate $m_2$ at some point and obviously $m_2\leq n_2$.
@xixixi: The value of $n_s$ given by Witt depends on both the number of generators $n$ and the step $s$ (the sum on $k$ is up to $s$). But yes, your explanation is basically right. $n_s$ is the "worst case" dimension, where none of the brackets coincide with each other, and so $m_s$ can be no larger.
|
2025-03-21T14:48:29.585314
| 2020-01-05T12:42:20 |
349763
|
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|
Stack Exchange
|
When is the pushout of projective varieties along embeddings a projective variety?
From Karl Schwede's paper "Gluing schemes and a scheme without closed points'', I know that there exists a pushout of schemes for closed embeddings.
Now, if I start in the projective world I would like to be able to stay there under pushout. However, it seems that the answer to this questions is a counterexample in which one glues two copies of $\mathbb{P}^3$ along two different embedded curves. So naturally, I have a question if there is a condition under which the pushout of embeddings of projective varieties is again projective.
The main example that I would like to understand is as follows. Let $i:D\hookrightarrow X$ be a particular embedding of a normal crossing divisor into a smooth projective variety $X$ over $\mathbb{C}$. Is $X\cup_{i,D,i} X$ projective again?
Section 5 here, especially Example 36. It shows that you need to be careful with how you glue the two copies of $D$.
It seems that in the example 36, they reparametrize the embeddings of the divisor. Is there any indication why a problem might occure if I take equal embeddings?
See also D. Ferrand, section 6.3 in https://smf.emath.fr/publications/conducteur-descente-et-pincement
@ArkadijBojko: In this case, there is no problem and the amalgam will be projective. You can use Ferrand's paper as a reference. I suggest you revise your question clarifying the definition of amalgams you are interested in.
|
2025-03-21T14:48:29.585438
| 2020-01-05T15:03:54 |
349767
|
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"Frederik vom Ende",
"Robert Furber",
"https://mathoverflow.net/users/116991",
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|
Stack Exchange
|
Is the set of weak*-continuous operators closed in the weak*-operator topology?
I recently came across this unanswered MO question an answer to which I would also be interested in. However the formulation of said question is somewhat imprecise and lacking detail in my opinion so I figured I would post this question in my own words (if this is against this site's etiquette please let me know!).
Given normed spaces $X,Y$ (as usual over a complete field $\mathbb F$) one can consider the map ${}^*:\mathcal B(X,Y)\to\mathcal B(Y^*,X^*)$ which maps any bounded operator $T$ to its adjoint operator $T^*$ (defined via $T^*(y)=y\circ T$ for all $y\in Y^*$). The map ${}^*$ is known to be a linear isometry but in general it is not surjective. In fact one can show that
$$
{}^*(\mathcal B(X,Y))=\{S\in\mathcal B(Y^*,X^*)\,|\,S\text{ is weak${}^*$-continuous}\}
$$
so $\mathcal B(X,Y)\simeq \{S\in\mathcal B(Y^*,X^*)\,|\,S\text{ is weak${}^*$-continuous}\}$ by means of ${}^*$. Here weak${}^*$-continuity refers to continuity of above $S$ as a map
$
S:(Y^*,\sigma(Y^*,Y))\to (X^*,\sigma(X^*,X))
$ (i.e. continuity when equipping domain and codomain with the respective weak${}^*$-topology).
To ask about this set being closed we quickly have to recall some available topologies on $\mathcal B(Y^*,X^*)$: aside from the usual operator norm, strong operator and weak operator topology on this space one can equip it with the weak${}^*$-operator topology $\tau_w^*$ which is locally convex topology induced by the seminorms $\{T\mapsto |(Ty)(x)|\}_{x\in X,y\in Y^*}$. Equivalently $\tau_w^*$ is the coarsest topology on $\mathcal B(Y^*,X^*)$ such that all maps $\{T\mapsto (Ty)(x)\}_{x\in X, y\in Y^*}$ are continuous and a neighborhood basis of $\tau_w^*$ at $T\in\mathcal B(Y^*,X^*)$ is given by
$$
\{N^*(T,A,B,\varepsilon)\,|\,A\subset X\text{ and }B\subset Y^*\text{ both finite, }\varepsilon>0\}\quad\text{ where}\\
N^*(T,A,B,\varepsilon):= \{S\in\mathcal B(Y^*,X^*)\,|\,|(Ty)(x)-(Sy)(x)|<\varepsilon\text{ for all }x\in A,y\in B\}\,.
$$
The idea behind this construction is to obtain a topology $\tau_w^*$ which is weaker than the weak operator topology (on $\mathcal B(Y^*,X^*)$) which is indeed the case; as expected these topologies coincide if $X$ is reflexive.
Now for some applications a desirable result would be the following: if a net $(T_i)_{i\in I}$ in $\mathcal B(Y^*,X^*)$ of weak${}^*$-continuous operators converges to $T\in\mathcal B(Y^*,X^*)$ with respsect to $\tau_w^*$ then $T$ is weak${}^*$-continuous itself.
In other words: is ${}^*(\mathcal B(X,Y))=\{T^*\,|\,T\in\mathcal B(X,Y)\}$ closed in $(\mathcal B(Y^*,X^*),\tau_w^*)$?
This was also asked on math.SE in 2016 but the only answer given there is flawed because there is no reason for weak${}^*$-convergent nets to be bounded. In fact from my own attempts that seems to be the only thing which prevents a direct proof (e.g., showing that $\mathcal B(Y^*,X^*)\setminus{}^*(\mathcal B(X,Y))$ is open in $\tau_w^*$ using the neighborhood basis).
If this were true this would---as an immediate consequence---imply that the weak${}^*$-operator topology "mirrors" the weak operator topology (on $\mathcal B(X,Y)$) in the following sense:
Consider a subset $A\subset {}^*(\mathcal B(X,Y))$ with pre-dual set $A_0\subset\mathcal B(X,Y)$, i.e. $(A_0)^*=A$. Then $A$ is closed in the weak${}^*$-operator topology if and only if $A_0$ is closed in the weak operator topology.
Thanks in advance for any answer or comment!
I've given a negative answer to the original MathOverflow question: https://mathoverflow.net/a/352265/61785
The answer is "no", in general.
An easy counterexample can be found as follows: Let $X = \mathbb{F}$ and let $Y$ be a non-reflexive Banach space. Then $\mathcal{B}(Y^*,X^*)$ is simply the bi-dual $Y^{**}$, and ${}^*(\mathcal{B}(X,Y))$ is precisely the image $j(Y)$ of $Y$ in $Y^{**}$ under the evaluation map $j: Y \to Y^{**}$.
The topology $\tau^*_w$ on $\mathcal{B}(Y^*,X^*) = Y^{**}$ is simply the weak${}^*$-topology on $Y^{**}$, so $j(Y)$ is $\tau^*_w$-dense in $Y^{**}$, but not equal to $Y^{**}$ (since $Y$ is non reflexive). Hence, $j(Y)$ is not $\tau^*_w$-closed in $Y^{**}$.
This is a simple but quite elegant counterexample, thank you!
|
2025-03-21T14:48:29.585699
| 2020-01-05T17:17:05 |
349773
|
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|
Stack Exchange
|
Ordered $m$-tuples with fixed number of changes
Given $1\leq k\leq m$, $2\leq d\leq c i\ln i$ and $2\leq i\leq c'\ln(mi\ln i)$ at some $c,c'>0$ how many sequences (lower and upper bounds) are of form $$z_1,\dots,z_m$$ on the condition that
$$0\leq z_1\leq\dots\leq z_m\leq 2^d$$
$$|\{i\in\{1,\dots,m\}: z_i\neq z_{i+1}\}|=k$$
are there?
Is there a standard combinatorics problem associated with such problems?
References would be very helpful.
Apparently, you mean $|\{i\in\{1,\dots,m-1\}: z_i\neq z_{i+1}\}|=k$ as $z_{m+1}$ is undefined.
This condition implies that among $z_1, \dots, z_m$ there are $k+1$ district values. Since the number of subsets $\{1,\dots,m-1\}$ of size $k$ equals $\binom{m-1}{k}$, we conclude that the number of sequences of interest equals
$$\binom{m-1}{k}\binom{2^d+1}{k+1}.$$
|
2025-03-21T14:48:29.585785
| 2020-01-05T18:03:12 |
349776
|
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"Greg Hurst",
"Kung Yao",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/349776"
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|
Stack Exchange
|
$\sum_{k =1, k \neq j}^{N-1} \csc^2\left(\pi \frac{k}{N} \right)\csc^2\left(\pi \frac{j-k}{N} \right)=?$
It is well-known that one can evaluate the sum
$$\sum_{k =1}^{N-1} \csc^2\left(\pi \frac{k}{N} \right)=\frac{N^2-1}{3}.$$
The answer to this problem can be found here
click here.
I am now interested in the more difficult problem of evaluating for some $j \in \{1,...,N-1\}$ the sum (we do not sum over $j$)
$$\sum_{k =1, k \neq j}^{N-1} \csc^2\left(\pi \frac{k}{N} \right)\csc^2\left(\pi \frac{j-k}{N} \right)=?$$
Does this one still allow for an explicit answer?
Looking at the first 20 values of $N$, it looks like the sum equals $\frac{4}{45} (n-2) (n-1) (n^2 + 3n + 2)$.
You think it is independent of j?
Oh sorry, I thought this was a sum over both $k$ and $j$.
No it is only over k. Sorry for the ambiguous notation
Start from the well known formula
\begin{equation}
2^{N-1} \prod _{k=1}^N \left[\cos (x-y)-\cos \left(x+y+\frac{2\pi k}{N}\right)\right]=\cos N(x-y)-\cos N(x+y),
\end{equation}
take logarithmic derivative
\begin{equation}
\sum _{k=1}^N \frac{\sin(x-y)}{\cos (x-y)-\cos \left(x+y+\frac{2\pi k}{N}\right)}=\frac{N\sin N(x-y)}{\cos N (x-y)-\cos N(x+y)},
\end{equation}
rewrite it as
\begin{equation}
N+\sum _{k=1}^N \frac{\cos(x-y)+\cos \left(x+y+\frac{2\pi k}{N}\right)}{\cos (x-y)-\cos \left(x+y+\frac{2\pi k}{N}\right)}=\frac{2N\sin N(x-y)}{\cos N (x-y)-\cos N(x+y)}\cot(x-y),
\end{equation}
and simplify to get
\begin{equation}
\sum_{k=1}^N\cot\left(x-\frac{\pi k}{N}\right)\cot\left(y-\frac{\pi k}{N}\right)=\frac{2N\cot(x-y)\sin N(x-y)}{\cos N(x-y)-\cos N(x+y)}-N.
\end{equation}
Then take derivatives wrt $x$ and $y$
$$
\sum_{k=1}^N\csc^2\left(x-\frac{\pi k}{N}\right)\csc^2\left(y-\frac{\pi k}{N}\right)=\frac{N}{\sin ^2(x-y)} \left(\frac{N}{\sin ^2Nx}+\frac{N}{\sin ^2Ny}-\frac{2\sin N(x-y)}{\sin Nx\sin Ny}\,\cot (x-y) \right).
$$
Yes, it may be simplified. The text below is not probably the shortest way, but it explains how to calculate many similar sums.
We start with $$\sin^2 x=-\frac14(e^{ix}-e^{-ix})^2=-\frac14e^{-2ix}(e^{2ix}-1)^2.$$
So, denoting $\omega_k=e^{2\pi i k/N}$ for $k=0,\ldots,N-1$ we get $$S:=\sum_{1\leqslant k\leqslant N-1,k\ne j}\csc^2\left(\pi \frac{k}{N} \right)\csc^2\left(\pi \frac{j-k}{N} \right)=
16\sum_{k\ne 0,j}\frac{\omega_k \omega_{k-j}}{(\omega_k-1)^2(\omega_{k-j}-1)^2}=\\
16\omega_j \sum_{k\ne 0,j}\frac{\omega_k^2}{(\omega_k-1)^2(\omega_{k}-\omega_j)^2}.$$
Consider the rational function $f(x):=\frac{x^2}{(x-1)^2(x-a)^2}$ (where $a=\omega_j$). We have
$$f(x)=
\frac{a^2}{(a - 1)^2 (x - a)^2} + \frac{2 a}{(a - 1)^3 (x - 1)} - \frac{2 a}{(a - 1)^3 (x - a)} + \frac1{(a - 1)^2 (x - 1)^2}.$$
Each of four summands may be easily summed up over $x\in \{\omega_0,\ldots,\omega_{N-1}\}\setminus \{1,a\}$. Indeed, we have
$$
\sum_{k} \frac 1{t-\omega_k}=\frac{(t^n-1)'}{(t^n-1)}=\frac{nt^{n-1}}{t^n-1},\,\,\,
\sum_{k} \frac 1{(t-\omega_k)^2}= -\left(\frac{nt^{n-1}}{t^n-1}\right)'=n\frac{t^{n-2}(t^n+n-1)}{(t^n-1)^2}.
$$
If we want to substitute $t=\omega_j$, we get something like
$$
\sum_{k:k\ne j} \frac1{\omega_j-\omega_k}=\left(\frac{nt^{n-1}}{t^n-1}-\frac1{t-\omega_j}\right)_{t=\omega_j}=\left(\frac{(n-1)t^{n}-n\omega_j t^{n-1}+1}{(t^n-1)(t-\omega_j)}\right)_{t=\omega_j}=(n-1)\omega_j^{-1}
$$
by L'Hôpital, analogously for the squares.
|
2025-03-21T14:48:29.585960
| 2020-01-05T19:25:25 |
349777
|
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"authors": [
"Jochen Wengenroth",
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|
Stack Exchange
|
Convergence of sesqui-holomorphic kernels on the diagonal
Let $X\subset \mathbb{C}^d$ be a domain.
A function (kernel) $K:X\times X\to \mathbb{C}$ is called sesqui-holomorphic if it is holomorphic in the first variable, and anti-holomorphic in the second variable
A kernel $K:X\times X\to \mathbb{C}$ is called positive (semi-)definite if for every $x_1,...,x_n\in X$ the matrix $[K(x_i,x_j)]_{i,j=1}^n$ is positive (semi-)definite. Note that this condition implies that $K$ is Hermitean, i.e. $K(y,x)=\overline{K(x,y)}$.
For a kernel $K:X\times X\to \mathbb{C}$ define it's diagonal $\widehat{K}:X\to\mathbb{C}$ by $\widehat{K}(x)=K(x,x)$. It is known that if for sesqui-holomorphic kernels $K$ and $L$ we have $\widehat{K}=\widehat{L}$, then in fact $K=L$ (this follows from Theorem 7, section II.4 in the book Bochner, Martin - Several Complex Variables). I am interested in a strengthening of that fact:
Let $\{K_n\}_{n\in\mathbb{N}}$ and $K$ be sesqui-holomorphic Hermitean (positive definite if it helps) kernels on $X$ such that $\widehat{K_n}\to\widehat{K}$ uniformly on compacts sets in $X$. Does it follow that $K_n\to K$ uniformly on compacts sets in $X\times X$?
A reference for the known fact might be helpful.
@JochenWengenroth I've added a reference to a result from which the known fact follows immediately (with an addition of Hartogs' theorem)
|
2025-03-21T14:48:29.586072
| 2020-01-05T19:46:39 |
349779
|
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|
Stack Exchange
|
Gentzen's result on PA
The Wikipedia states that Gentzen proved that "in modern terms, the proof-theoretic ordinal of PA is $\varepsilon_0$." Further down, the article defines what the "proof theoretic ordinal" of a theory means. However, I'm not sure what this means regarding PA, since PA can only make finitary statements.
Let me elaborate. Define some encoding of the ordinals ${}<\varepsilon_0$ as natural numbers. This encoding allows us to express statements involving ordinals ${}<\varepsilon_0$ in PA. Then, allegedly PA cannot prove transfinite induction using this encoding.
But what does this mean exactly? One option would be using sets: "Suppose $S$ is a set of ordinals such that, for every $\beta<\varepsilon_0$, whenever $\alpha\in S$ for every $\alpha<\beta$, we also have $\beta\in S$. Then $\beta\in S$ for all $\beta<\varepsilon_0$."
But this is an infinitary statement, as far as I understand, so it cannot be stated in PA.
Another option is to have a schema of infinitely many statements, one for each possible formula $\varphi$ (just like the "induction schema" contains infinitely many axioms, one for each possible formula):
"Suppose that for every $\beta<\varepsilon_0$, whenever $\varphi(\alpha)$ holds for every $\alpha<\beta$, we also have $\varphi(\beta)$. Then $\varphi(\beta)$ holds for every $\beta<\varepsilon_0$."
So what is it exactly that Gentzen proved? Presumably PA can prove the above statement for some formulas $\varphi$. So did Gentzen find some specific $\varphi$ for which PA cannot prove the above statement? Or what?
I saw question https://mathoverflow.net/questions/5065/ , and it does not answer my question. Over there, the OP wanted to know how you encode ordinals <eps_0 as natural numbers. I understand how you do that
To answer the first part of the question: yes, transfinite induction is stated as a schema, just like usual induction. When talking about theories weaker than PA we may want to restrict which formulas we include (e.g. only quantifier-free or bounded ones), but for PA and stronger it doesn't matter.
Noah Schweber has answered your question, but note that the existence of some $\varphi$ follows from Gentzen's consistency proof. Namely, if there were no such $\varphi$, then by mimicking Gentzen's consistency proof, we would be able to prove the consistency of PA within PA itself.
I found an online draft of this book (Proof theory, by Herman Ruge Jervell) to be a very accessible introduction to Gentzen's proof and the concepts leading up to it. The draft is no longer at its old location but
Yes, Gentzen found a single "induction instance" which is PA-unprovable: more-or-less $\varphi(\alpha)$ = "Every sentence with a proof of cut-rank $\alpha$ has a cut-free proof."
Now, this $\varphi$ is a $\Pi^0_2$ formula. If memory serves, this is suboptimal: with some coding work this can be improved from a $\Pi^0_2$ formula to a $\Sigma^0_1$ formula. The basic idea is to assign in a primitive recursive way to each ordinal $\alpha<\epsilon_0$ a sentence $p_\alpha$ and a candidate proof $s_\alpha$ of cut rank $<\alpha$ such that each pair occurs cofinally often, and then look at the formula $\psi(\alpha)$ = "Either $s_\alpha$ is not a proof of $p_\alpha$ or there is a cut free proof of $p_\alpha$."
And I think even that's suboptimal - that we can get to the level of $\Delta_0$ - but I'm not sure.
What's a reference for the claim that things can be improved to a $\Sigma_1$ or $\Delta_0$ formula?
|
2025-03-21T14:48:29.586292
| 2020-01-05T19:49:36 |
349780
|
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|
Stack Exchange
|
When the image under a real morphism is constructible?
Given a morphism $f:\Bbb{A}^n_{\Bbb{C}}\rightarrow\Bbb{A}^m_{\Bbb{C}}$ of complex affine spaces, Chevalley's theorem states that the image of a closed subvariety is constructible, i.e. could be described with finitely many equations $f(y_1,\dots,y_m)=0$ and inequations $g(y_1,\dots,y_m)\neq 0$ combined with "and" & "or". On the other hand, over real numbers one must allow inequalities as well, e.g. the image of $$f:\Bbb{A}^1_{\Bbb{R}}\rightarrow\Bbb{A}^1_{\Bbb{R}}:x\mapsto x^2$$
is given by $x\geq 0$. Indeed, it is known that the image of a closed subvariety under a real morphism $f:\Bbb{A}^n_{\Bbb{R}}\rightarrow\Bbb{A}^m_{\Bbb{R}}$ is semialgebraic.
My question: Is there any criteria guaranteeing that the image of a closed subvariety under $f:\Bbb{A}^n_{\Bbb{R}}\rightarrow\Bbb{A}^m_{\Bbb{R}}$ is constructible so that one can avoid inequalities when it comes to describing the image?
Remark: I am interested only in the closed points; that's why I am discussing varieties.
|
2025-03-21T14:48:29.586632
| 2020-01-05T20:37:39 |
349783
|
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|
Stack Exchange
|
Invertible bimodules and projectivity
Let $A$ be a noncommutative algebra over a field, say $\mathbb{C}$ or $\mathbb{R}$, and let $L$ be a bimodule over $A$. If $L$ is invertible, that is, if the dual right $A$-module $L^*$ satisfies
$$
L^* \otimes_A L \simeq A
$$
then will $L$ necessarily be projective a left module over $A$?
Yes. If $\varphi :L^*\otimes _A L\rightarrow A$ is an isomorphism, there exists an element $t =\sum\limits_{i=1}^{n} x_i^*\otimes x_i$ of $L^*\otimes _A L$ such that $\varphi (t )=1$. For any $x\in L$, we have then $x\varphi (t)=x$, that is, $\sum_i \langle x,x_i^*\rangle x_i=x$. Now consider the homomorphisms $u:A^n\rightarrow L$ and $v:L\rightarrow A^n$ defined by $u(e_i)=x_i$ and $v(x)=(\langle x,x_1^*\rangle,\ldots ,\langle x,x_n^*\rangle)$; we have $u\circ v=\operatorname{Id}_L $, thus $L$ is a direct summand of $A^n$, hence projective.
|
2025-03-21T14:48:29.586843
| 2020-01-05T20:39:34 |
349785
|
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|
Stack Exchange
|
On Exercise 2.5.10 in Ram. M. Murty's book, "Problems in Analytic Number Theory."
I have just been told about this result, available as Exercise 2.5.10 in Ram. M. Murty's book, "Problems in Analytic Number Theory (2nd edition)". It says:
Let $\alpha>0$. Suppose $a_n \ll n^{\alpha}$ and $A(x) \ll x ^{\delta} $ for some $\delta<1$, where $A(x) = \sum_{n\leq x} a_n$. Define $b_n = \sum_{d|n} a_d$. Then one has
$$\sum_{n\leq x} b_n = cx + O\Big(x^{(1-\delta)(1+\alpha)/(2-\delta)}\Big)$$ for some constant $c$.
Does anyone know who first came up with this result, or maybe it's just too straightforward to be attributed to anyone ?
The reason why i'm particularly interested in knowing the originator of this result is that, Murty's proof (which is on pages 262-263 of the aforementioned book) doesn't look quite right to me (but of course, i may be mistaken).
What do you think is wrong with the proof presented?
I quickly looked at the proof, and I could find no fault with it. I would submit that the result is straightforward enough so that particular attribution is necessary.
@Craig France, thanks for your comment. My concern is on the Dirichlet hyperbola formula that Murty's used. Is it the correct formula ?
@2734364041, so may you please explain how the choice $y=x^{(1-\delta)/(2-\delta)}$ minimises the error term $x^{\delta}y^{1-\delta} + xy^{\delta-1}$ ? Also, plugging this choice of $y$ into the error term doesn't seem to yield $x^{(1-\delta-\delta^2)/(2-\delta)}$ ?
I see the problem, the second sum on the right should run up to $\frac{x}{y}$, not $y$.
@Craig France, yes...the formula running up to $y$ would be correct if $y=√x$, but that would be inconsistent with Murty's later choice of $y$.
Well, it just means that both of the following error terms are of the same order of magnitude, in this case $O(xy^{1-\delta})$. The other error that arises from extending the first sum is $O(y^{1+\alpha})$. So, in the end, you're left with the same error term, $O(xy^{\delta-1}+y^{1+\alpha})$. Thus, the choice the author made for $y$ leads to the same conclusion, although I'll admit it seems that you could've chosen it differently. Does that make sense?
the result in the book is incorrect as the computation of the powers is incorrect
It seems there is a typo in the application of the hyperbola method. Since $b_n=\sum_{d\mid n}a_d$, we have
\begin{align}
\sum_{n\le x}b_n &=\sum_{n\le x}\sum_{de=n}a_d=\sum_{de\le x}a_d\\
&=\sum_{\substack{de\le x\\ d\le y}}a_d+\sum_{\substack{de\le x\\ e\le x/y}}a_d-\sum_{de\le x\\ d\le y, e\le x/y}a_d.
\end{align}
If $A(x)=\sum_{n\le x}a_n$, then this can be written as
\begin{align}
\sum_{d\le y}a_d\left[\frac{x}{d}\right]+\sum_{e\le x/y}A\left(\frac{x}{e}\right)-A(y)\left[\frac{x}{y}\right].
\end{align}
The author's assumption that $A(x)\ll x^{\delta}$ then gives the last two terms as $O(xy^{\delta-1})$, and the proof can be continued in the usual manner outlined in the text. Although, it appears that a sign was missed later on as well, which has some bearing on the final stated error (see Conrad's answer below).
But isn't this a special case of Theorem 2.4.1 in the text?
Yes, you'll get the same thing if you set $f(n)=b_n$, $g(n)=a_n$, and $h(n)=1$ in Theorem 2.4.1.
Thanks Craig ! So, the result is true but Murty's proof has a minor flaw, right ?
Yeah, it happens. Was this result used in a paper?
@as noted below, the result in the book is incorrect as the powers are not computed correctly
The computation of the powers is wrong and the result stated in the book is incorrect and it should be $cx+ O(x^{(1+\delta-\delta^2)/(2-\delta)})+O(x^{\frac{(1-\delta)(1+\alpha)}{2-\delta}})$
If $y=x^{(1-\delta)/(2-\delta)}$, $y^{\delta-1}=x^{-(1-\delta)^2/(2-\delta)}$, so $xy^{\delta-1}=x^{(1+\delta-\delta^2)/(2-\delta)}$ and it is not true that $\frac{1+\delta-\delta^2}{2-\delta} \le \frac{(1-\delta)(1+\alpha)}{2-\delta}$ in general, only for $\alpha \ge \delta + \frac{\delta}{1-\delta}$
Note that as stated, the result doesn't make sense because one can always increase $\delta$ in the hypothesis, while keeping $\alpha$ fixed, so in particular if $A(x) << x^{\delta}$ for a given $\delta < 1$, then $A(x) <<_{\epsilon} x^{1-\epsilon}$ for arbitrary $\epsilon >0$, hence we would get that $B(x) =cx + O_\epsilon(x^{\epsilon})$ under very general conditions and I am sure lots of counterxamples to that can be found
You're right, it looks like a sign was missed later on. Also, the $O(y^{\alpha+1})$-term looks like it could have been replaced by $O(y^{\delta})$ since $\delta<1<1+\alpha$ ... that would make the error term $O(xy^{\delta-1})$.
|
2025-03-21T14:48:29.587154
| 2020-01-05T20:46:20 |
349787
|
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|
Stack Exchange
|
Construct Torsion element in $H^2(X,\mathbb{Z})$ with Ambrose–Singer theorem
Let $X$ be a Kahler manifold. Using the exponential sequence one obtains a homomorphism $c_1:H^1(X,\mathcal{O}_X^*)\rightarrow H^2(X,\mathbb{Z})$. This is associating to a holomorphic line bundle $L$ its Chern class $c_1(L)$.
Take any connection $D$ on a holomorphic line bundle $L$, if we do $D_L\circ D_L = \Omega_L$ we obtain a curvature and then take its class in $H^2(X,\mathbb{C})$ we get a curvature form independent of the choice of the connection. After multiplying by $-i/(2\pi)$ we run into $c_1(L)$ when identifying $H^2(X,\mathbb{Z})$ as a subspace of $H^2(X,\mathbb{C})$, if we know that $\phi: H^2(X,\mathbb{Z}) \to H^2(X,\mathbb{C})$ is injective. This is not always true but if $H^2(X,\mathbb{Z})$ has no torsion using universal coefficient thm we see $H^2(X,\Bbb C) \cong H^2(X,\Bbb Z)\otimes_{\Bbb Z}\Bbb C$ and thus $\phi$ is in that case injective.
Looking for examples where $\phi$ is not injective, so where $H^2(X,\mathbb{Z})$ has torsion I encountered the class of flat bundles, which provide by definition connections with zero curvature.
What I still not understand is how to see explicitely that these flat bundles provide a curvature what induces in $H^2(X,\mathbb{Z})$ a torsion class. Technically, one can simply say that because the class of cuvatures of flat bundles are zero in $H^2(X,\mathbb{C})$, it must be torsion.
The point is that I wan't to have a better intuitive inside look what really is going on in the torsion. That is I aim to consider a representative of the curvature of a flat line bundle "alive" in $H^2(X,\mathbb{Z})$ instead simply kill it by tensoring with $C$ and say "thus it was torsion".
The promissing tool for this seems to lie in the Ambrose–Singer theorem which allows to "measure" the infinitesimal Holonomy by curvature. Especially that allows to see that for prinzipal $G$-bundles with flat curvature the infinitesimal holonomy has to be trivial (not globally in general) . That is homotopical paths induce same holonomy action and we obtain well defined representation
$$\pi _{1}M\rightarrow G$$
More precisely a consequence of Ambrose–Singer is that we obtain a bijection between flat $G$-Bundles ($G$ certain group) over connected manifolds and representations
$$\rho : \pi_1(M) \to G $$
More concretly we associate to a given representation a flat bundle using monodromy action of $\pi_1(M)$ on the universal cover $\tilde{M}$ of $M$ and set
$$E_{\rho }:=\widetilde {M}\times G/\sim $$
with equivalence relation $(\gamma x,g)\sim (x,\rho (\gamma )g)$ for ${\displaystyle \gamma \in \pi _{1}M,x\in {\widetilde {M}},g\in G}$.
If $\pi(M)$ is for exaple finite that it looks promising since this gives a non trivial action of a finite group on the line bundle.
Problem/Question: can this action be "transfered" in certain way to the curvature induced by concerned principal line line bundle or it's class in $H^2(X,\mathbb{Z})$. My goal is if I find out that this action is still non trivial and $\pi_1(M)$ is finite group, then the class of the curvature is torsion and we have constructed a desired example of a torsion element explicitly. What I still not understand is how to expand the action of fundamnetal group via $\rho$ to the curvature form of $E_{\rho}$ and why it stays non trivial?
I do not understand most of your question (for a flat bundle the action of the fundamental group on the curvature form is zero, because the curvature form is zero...) but you can see how torsion classes arise explicitly. One obtains the isomorphism $H^1(X;U(1)) \cong \text{Hom}(\pi_1 X, U(1))$ in the usual fashion. This splits as a sum of $U(1)^{b_1(X)}$ and $\text{Hom}(H_1(X;\Bbb Z)_{\text{tors}}, U(1))$ (the free part and the torsion part). The free part is what dies in the boundary map $H^1(X;U(1)) \to H^2(X;\Bbb Z)$ from the exponential exact sequence.
So the desired classes in $H^2$ are just those coming from the boundary map in the exact sequence associated to $\Bbb Z \to \Bbb R \to U(1)$, and those that survive come from homomorphisms $\pi_1(X) \to U(1)$ with finite image.
@MikeMiller: two questions: how do you obtain the isomorphism $H^1(X;U(1)) \cong \text{Hom}(\pi_1 X, U(1))$?
And secondly: this concerns the case if a $G$ bundle $P$ is not flat, ie. the the curvature $\Omega_P$ not vanish, is it possible to extend the action of $G$ to the curvature in canonical way?
That isomorphism holds for any abelian group in the place of $U(1)$ and follows from the universal coefficient theorem. I thought you were talking about the action of $\pi_1$ earlier. There is certainly an action of $G$ on the bundle $\Lambda^2 T^* M \otimes T\text{Aut}(G)$, which is one way to describe where curvature lives. This gives you an action of $G$ on the curvature form at each point, but not globally. I still don't know why you would want this anyway.
the background arosed in this discussion https://math.stackexchange.com/questions/2073846/neron-severi-group-h1-1x-mathbbz. Primary I wanted to understand the torsion part in $H^2(X,\mathbb{Z})$ better. Probably I misunderstood a hint from that thread in order to approach them by flat line bundles.
|
2025-03-21T14:48:29.587483
| 2020-01-05T23:46:23 |
349792
|
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|
Stack Exchange
|
Diagonal of a diagram of codescent objects
Given the following diagram in a $2$-category, in which squares of the same "type" commute, where each column and each row is a strong codescent diagram (Edit: it should be reflexive as well), is then the diagonal a codescent diagram as well?
Actually, for each row and each column we also have six $2$-isomorphisms (for example, $\xi^h_i : x^h_i d^h_i \to x^h_i c^h_i$ for the $i$th row), which I didn't write down here.
This question is motivated by Mike Shulman's comment here. In fact, this statement would probably be an important step to construct bicategorical pushouts of symmetric pseudomonoids.
I tried to find a proof, but basically get lost because of this huge amount of data.
Notice that this statement (if it is true) is a generalization of the corresponding $1$-dimensional statement about reflexive coequalizers (Sketches of an elephant, Lemma A.1.2.11). I would be very happy for a reference to the literature where this is proven. Instead of writing down a proof on my own, which will be probably too long anyway, I would like to cite and then use this result in a paper.
This follows from the fact (due to Steve Lack) that (reflexive) codescent objects are sifted flexible colimits (see Proposition 4 in John Bourke’s paper http://arxiv.org/abs/1206.1203 for a proof). I can explain how when I get home.
I should also point out that for this result to hold, each row and column of your diagram must be a reflexive codescent diagram (see the paragraph before Proposition 4.3 of Lack's paper to which you link).
Thanks a lot! This looks very promising. I am looking forward to your answer. Actually Prop 4.3. in Lack's paper is then a special case of this statement (for the 2-category of small categories), and perhaps the proof can be generalized? Unfortunately Lack only sketches the proof.
As I wrote in a comment above, for this result to hold, each row and column of your diagram must be a reflexive codescent diagram. I do not know of any place in the literature where this result is explicitly stated, but, as I will explain below, it follows without difficulty from a result of Steve Lack.
(For simplicity, let me deal only with strict reflexive codescent objects. Since these are flexible colimits, one can deduce the fully weak bicategorical version of this result from the strict version by standard arguments.)
Definition. Let $\Delta_{\leq 2}$ denote the full subcategory of the simplex category $\Delta$ containing the objects $[0]$, $[1]$, and $[2]$, and let $W : \Delta_{\leq 2} \longrightarrow \mathbf{Cat}$ denote the composite of the full inclusion $\Delta_{\leq 2} \longrightarrow \mathbf{Cat}$ with the groupoid reflection functor $\mathbf{Cat} \longrightarrow \mathbf{Cat}$. For any $2$-category $\mathcal{K}$, the reflexive codescent object of a functor $X : \Delta_{\leq 2}^\mathrm{op} \longrightarrow \mathcal{K}$ is the colimit $W \ast X$ of $X$ weighted by $W$.
We will deduce the "diagonal lemma" of your question for reflexive codescent objects from the fact that reflexive codescent objects are sifted colimits (in the $\mathbf{Cat}$-enriched sense), i.e. that the functor $$W \ast (-) : [\Delta_{\leq 2}^\mathrm{op},\mathbf{Cat}] \longrightarrow \mathbf{Cat}$$ preserves finite products. This fact is due to Steve Lack -- see Proposition 4.3 of
Lack, Stephen. Codescent objects and coherence. J. Pure Appl. Algebra 175 (2002), no. 1-3, 223--241. doi
and Proposition 4 of
Bourke, John. A colimit decomposition for homotopy algebras in Cat. Appl. Categ. Structures 22 (2014), no. 1, 13--28. doi
Thanks to the "Fubini theorem" for iterated weighted colimits, we may state the diagonal lemma for reflexive codescent objects in the following form.
Lemma (diagonal lemma for reflexive codescent objects). Let $\mathcal{K}$ be a $2$-category and let $X \colon \Delta_{\leq 2}^\mathrm{op} \times \Delta_{\leq 2}^\mathrm{op} \longrightarrow \mathcal{K}$ be a functor. Then we have an isomorphism of weighted colimits in $\mathcal{K}$
$$W \ast (X \circ \delta) \cong (W \times W) \ast X,$$
either side existing if the other does. (Here $\delta$ denotes the diagonal functor $\Delta_{\leq 2}^\mathrm{op} \longrightarrow \Delta_{\leq 2}^\mathrm{op} \times \Delta_{\leq 2}^\mathrm{op}$).
Remark. It is also worth displaying the isomorphism of this lemma in coend form:
$$\int^{[k]} W^k \times X_{k,k} \cong \int^{[n],[m]} W^n \times W^m \times X_{n,m}.$$
Proof of lemma. The preservation of binary products of representables by the functor $W \ast (-) : [\Delta_{\leq 2}^\mathrm{op},\mathbf{Cat}] \longrightarrow \mathbf{Cat}$ implies, via the weighted colimit formula for left Kan extensions, that the functor $W \times W : \Delta_{\leq 2} \times \Delta_{\leq 2} \longrightarrow \mathbf{Cat}$ is the left Kan extension of $W \colon \Delta_{\leq 2} \longrightarrow \mathbf{Cat}$ along the diagonal functor $\delta \colon \Delta_{\leq 2} \longrightarrow \Delta_{\leq 2} \times \Delta_{\leq 2}$. Hence the lemma follows from Theorem 4.38 of Kelly's Basic concepts of enriched category theory. $\Box$
It is worth mentioning that, when working bicategorically (i.e. "up to equivalence"), the codescent object of a (pseudo)functor $X : \Delta_{\leq 2}^\mathrm{op} \longrightarrow \mathcal{K}$ is simply its bicolimit. Hence the bicategorical version of the diagonal lemma for reflexive codescent objects -- which follows from the strict version by standard arguments -- is simply the statement that the diagonal functor $\Delta_{\leq 2}^\mathrm{op} \longrightarrow \Delta_{\leq 2}^\mathrm{op} \times \Delta_{\leq 2}^\mathrm{op}$ is 2-final.
Thank you for this answer. Can you perhaps say more about why strict reflexive codescent objects are sufficient? I don't know the "standard arguments" you are mentioning. And by the way, I am actually interested in bicategorical codescent objects, so your last paragraph is very helpful.
Essentially it’s a combination of the following facts. To prove the “weak” diagonal lemma in a general bicategory, it suffices to prove it for “weak” coreflexive descent objects (the dual bilimit notion) in Cat, since representable functors jointly detect bilimits. Next, any pseudofunctor from $\Delta_{\leq 2}$ to Cat is equivalent to a strict one. Finally, the strict descent object of a functor from $\Delta_{\leq 2}$ to Cat is also its bilimit, since the weight $W$ defined above is a flexible (i.e. projective cofibrant) replacement of the terminal weight.
Your first sentence seems to suggest that the reflexivity hypothesis is not just sufficient but required — that without it, the result could fail. Did you intend to suggest this, and if so could you elaborate?
I did intend to suggest that. The diagonal lemma is very nearly equivalent to (reflexive) codescent objects being sifted colimits, so any example showing that non-reflexive codescent objects are not sifted (I don’t know any off the top of my head, but the literature asserts their existence) should also yield a counterexample to the diagonal lemma for non-reflexive codescent objects.
In Lack's paper, in the description of reflexive codescent data, a $2$-cell $ue \to ve$ is missing, right?
In your definition of the weight $W$, you use the groupoid reflection $\mathbf{Cat} \to \mathbf{Cat}$. But this is not mentioned in Bourke's paper (also not in his thesis on codescent objects). Can you explain this difference? This must also be related to the category $\mathscr{D}_r$ in Lack's paper.
Bourke describes the same weight in the first sentence of Section 3.2 of his paper; he denotes it by $W_i$.
Certainly there should be an invertible $2$-cell $ue \to ve$, and I don't see at the moment that it can be built up from the other $2$-cells he lists.
|
2025-03-21T14:48:29.587922
| 2020-01-06T03:26:47 |
349795
|
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|
Stack Exchange
|
Suggestions for infinite horizontal optimization
I have been looking at this question for a while without any progress.
Question. Maximize
$$ I[\eta] = \int_0^\infty e^{-s} \Big[\sin\big(\eta(s)\big) + \sin\big(\sqrt{2}\eta(s)\big)\Big]\;ds$$
subjected to $\eta:[0,\infty) \rightarrow \mathbb{R}$ is absolutely continuous with $\eta(0) = 0$ and $\dot{\eta}(s) \in [-1,1]$.
What are the tools (if any) available to study this kind of question (both analytically and numerically)?
As written, the maximizer is, obviously, $\eta\equiv 0$. Check the statement of the question; I strongly suspect that you meant something else.
You are right, I meant $\sin$ instead of $\cos$.
The first maximum of $f(s)=\sin s+\sin\sqrt 2 s$ is attained at some $s_0\in [1.2,1.3]$ and exceeds $1.9$. The derivative of $f$ is at most $1+\sqrt 2\le\frac 52$. One obvious strategy is to go at the highest speed $1$ to $s_0$ (i.e., to put $\eta(s)=s$ on $[0,s_0]$) and then stay there forever. Any attempt to go to another maximum will force you to go through $0$ first. Up to that point, say, $S\ge s_0$, you can only lose compared to the obvious strategy, but beyond that point, even if you increase $f$ at the maximal available rate $\frac 52$ and reach the maximal possible value $2$, you'll get only $\int_0^{4/5}\frac 52se^{-s}ds+\int_{4/5}^{\infty}2e^{-s}ds=2\frac{1-e^{-4/5}}{4/5}<1.4$ times $e^{-S}$ compared to at least $1.9e^{-S}$ the obvious strategy yields. So the obvious strategy is the best one here. Of course, this is rather ad hoc and heavily based on the properties of the exponential weight, but it solves the problem as posed, so I'll stop here.
|
2025-03-21T14:48:29.588071
| 2020-01-06T04:23:45 |
349799
|
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|
Stack Exchange
|
Entropy rate problem of ergodic Markov process with non-ergodic joint
I have a problem with the entropy rate when two ergodic Markov processes who are independent of each other having a non-ergodic joint. More specifically let us consider two finite-state Markov processes $\mathscr{P}_1$ and $\mathscr{P}_2$ with transition matrices $\Pi_1$ and $\Pi_2$, respectively. Let us assume that $\Pi_i$, $i=1, 2$, is a irreducible row stochastic matrix of period $p_i$, and $\gcd(p_1, p_2) = p > 1$. Because $\Pi_i$, $i=1, 2$, is irreducible, we know that $\mathscr{P}_i$ is ergodic.
Assume that $\mathscr{P}_1$ and $\mathscr{P}_2$ are independent, we have their joint process has transition matrix $\Pi =\Pi_1\otimes\Pi_2$, where $\otimes$ means the Kronecker product. However, since $p>1$, $\Pi$ is not irreducible. In fact, we can reorder the columns and rows of $\Pi$ simultaneously so that it becomes $\textrm{diag}(A_1, \ldots, A_p)$ where $A_i$, $i=1,\dots, p$, is a irreducible row stochastic matrix.
My question is: Let us denote entropy rate of the Markov process given by transition matrix $A$ by $\mathcal{H}(A)$, do we have $$\mathcal{H}(A_i) = \mathcal{H}(\Pi_1) + \mathcal{H}(\Pi_2)$$ for all $i = 1,\dots, p$?
The statement holds for a few examples I tried (some of them are not even Markov process, but were generated by a little bit more complicated models). But I wasn't successful in comping up either a proof or disproof.
Please help! Thank you in advance.
Yes. That’s a general result of ergodic theory. See for example Walters book.
Great! Thank you, Anthony! Just to confirm, by Walters' book, you mean "An Introduction to Ergodic Theory (Graduate Texts in Mathematics)" by Peter Walters, right? If so, would you please point to me the chapter/section that I can find relevant information?
Yes. Exactly...
The easiest way to see this is by using the Shannon-McMillan-Breiman equidistribution theorem and noticing that the space of sample paths of the product chain is the product of the sample path spaces of the original chains, so that for a.e. sample path of the product chain the logarithms of the measures of the corresponding cylinder sets are asymptotically $-n(H(\Pi_1)+H(\Pi_2))$.
Hi R W, thank you for your answer! I think the part I don't quite get is why the statement holds for each individual component. My numerical experiment told me that each individual component is actually NOT the independent product of the two processes. Would you please give me a little more detailed help? Thank you so much!
You apply the Shannon theorem to each of the chains $\mathcal P_1,\mathcal P_2$, and then pass to the product of the path spaces. What do you mean by "each individual component"?
|
2025-03-21T14:48:29.588272
| 2020-01-06T08:40:10 |
349809
|
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|
Stack Exchange
|
Is there a classification of reflection groups over division rings?
I asked a version of this question in Math StackExchange about a week ago but I've received no feedback so far, so following the advice I received on meta I decided to post it here.
Details
The irreducible finite complex reflection groups were classified by Shephard and Todd. The list consists of a three-parameter family of imprimitive groups and 34 exceptional cases. If $\mathbb{K}$ is any field of characteristic zero, it is known (see e.g. Section 15-2 here) that if a group has a representation as a $\mathbb{K}$-reflection group then it also has a representation as a complex reflection group, so we get no new examples for fields. I'm interested to know if an analogue of this classification exists when we allow $\mathbb{K}$ to be a division ring of characteristic zero.
I have only found partial results. The irreducible finite quaternionic reflection groups were classified in this paper by Cohen. Finite $\mathbb{K}$-reflection groups of rank $1$, or equivalently finite subgroups of division rings of characteristic $0$, were classified in this paper by Amitsur. The classification for rank $2$ can perhaps be extracted from the classification of finite subgroups of $GL(2,\mathbb{K})$ by Banieqbal using a case-by-case check.
In view of the complex and quaternionic lists, I would expect the full classification (if there is one) to follow a form similar to this:
An infinite family $G_n(M,P,\alpha)$ of imprimitive reflection groups of rank $n$, where $M$ is a finite subgroup of a division ring $\mathbb{K}$ (i.e. an Amitsur group) and $[M,M]\le P \trianglelefteq M$, possibly with some extra data $\alpha$ in low rank. Algebraically it should correspond to something like the group of generalized permutation matrices with entries in $M$ whose determinant is in $P$, like in the cases of fields and quaternions (note that order doesn't matter when computing the determinant, since $[M,M]\le P$).
A family or families of examples in rank $2$ (or perhaps in rank $\le m$ where $m^2$ is the dimension of $\mathbb{K}$ as a division algebra over its center). In the quaternionic case they are the primitive reflection groups whose complexification is imprimitive, and are all constructed from certain $2$-dimensional primitive complex reflection groups; it's not clear to me how this construction could generalize to arbitrary $\mathbb{K}$.
A number of exceptional cases of small rank. These are the ones I'm most interested in. In the quaternionic case these are precisely the primitive reflection groups whose complexification is also primitive; in the general case they might correspond to primitive reflection groups which remain primitive after tensoring the representation with a splitting field.
Question
My main question is thus:
Is there a classification of groups representable as a $\mathbb{K}$-reflection group over some division ring $\mathbb{K}$ of characteristic zero?
I would also appreciate any references dealing with this problem or with particular cases. If the classification turns out to be intractable, or currently out of reach, I would ask if at least an example can be found of a new exceptional reflection group of rank $\ge 3$ (see details above).
Update
After much searching, I finally found a brief reference to this problem in the literature. The mention occurs at the end of Section 3 in this 1981 paper by Kantor, which I quote here for convenience:
[...] For example, consider the problem of determining all finite primitive reflection groups $G$ in $GL(n,D)$, for $D$ an arbitrary noncommutative division ring of characteristic $0$. If $n=1$, this is just the famous problem solved by Amitsur (1955) (and independently and almost simultaneously by J. A. Green). If $n=2$ and $G$ is solvable, the problem seems to involve even more difficult number theory than Amitsur used. But if $n\ge 3$, and if simple group classification theorems are thrown at the problem, no new nonsolvable examples arise. [...]
If the last sentence is true, it means that the new examples of "exceptional groups" I asked for must necessarily be solvable. However, the author does not provide any in-text citation for that statement, and I haven't been able to find which result is being alluded to.
|
2025-03-21T14:48:29.588558
| 2020-01-06T10:50:29 |
349812
|
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|
Stack Exchange
|
Recurrence relation for the number of partitions of an integer with distinct summands
Let $Q(n)$ give the number of ways of writing the integer $n$ as a sum of positive integers without regard to order with the constraint that all integers in a given partition are distinct. Equation $(11)$ on this page mentions (without proof) a recurrence relation for $Q(n)$,
$$Q(n) = s(n) + 2\sum_{k=1}^\sqrt{n}(-1)^{k+1}Q(n-k^2)$$
where
$$s(n) = \begin{cases}
(-1)^j,& \text{if } n= j(3j \pm 1)/2\\
0, & \text{otherwise}
\end{cases}$$
I also came across this identity (again no proof) in Abramowitz and Stegun's book on mathematical formulas (pg. 826). What is the proof of this fact?
You asked the same question MSE in https://math.stackexchange.com/questions/3498499/recurrence-relation-for-the-number-of-partitions-of-an-integer-n-with-distinct Perhaps you want to decide for one site, since then there will be no duplicate answers.
@EFinat-S I apologize. I had originally intended to post on MSE but there was no activity there. I had heard of MO being for research level math but never posted here . Since this question seemed to fit the guidelines, this became my first post on this site. I was unfortunately not aware of the protocol for migrating questions between MSE and MO.
Gauss showed that
$$ \prod_{n\geq 1}\frac{1-q^n}{1+q^n} = 1+2\sum_{n\geq 1}(-1)^n q^{n^2}. $$
We also have $\sum_{n\geq 0} Q(n)q^n = (1+q)(1+q^2)\cdots$ and $\sum_{n\geq 0} s(n)q^n = (1-q)(1-q^2)\cdots$ (Euler's pentagonal number formula). The recurrence follows from equating coefficients of $q^n$ on both sides of
$$ \prod_{n\geq 1}(1-q^n) = \left(1+2\sum_{n\geq 1}(-1)^n q^{n^2}\right)\prod_{n\geq 1}(1+q^n). $$
Could you include a reference for the first identity you wrote?
See Gauss's proof at https://books.google.com/books?id=uDMAAAAAQAAJ, page. 447, equation (14).
A proof of Gauss' identity also appears in http://www.personal.psu.edu/gea1/pdf/48.pdf.
Richard Stanley already aswered your question about the proof of the identity. But, if you are looking for references, this two articles prove similar (equivalent?) identities:
Ewell, John A., Recurrences for two restricted partition functions,
Fibonacci Quart. 18 (1980), no. 1, 1–2.
Ono, Ken; Robbins, Neville; Wilson, Brad, Some recurrences for arithmetical functions,
J. Indian Math. Soc. (N.S.) 62 (1996), no. 1-4, 29–50.
|
2025-03-21T14:48:29.588853
| 2020-01-06T11:33:25 |
349813
|
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|
Stack Exchange
|
Alternative definitions of Weibel's homotopy K-theory
Consider a sort of $\mathbb{A}^1$-homotopy-stable algebraic $K$-theory for rings constructed as follows.
For $K_0$ we take a symmetrization subject to natural direct sum operation of $\mathbb{A}^1$-homotopy classes of finite-dimensional idempotents in $\mathbb{M}_\infty(B)$. Higher and lower groups are defined using looping/suspension. I hypothesize that this construction coincides with Weibel's homotopy algebraic K-theory $KH$, which is originally defined via spectra. This my be a trivial fact not even worth describing in the literature, or quite possibly I'm terribly wrong and missing some simple counterexample.
(Sorry if the question is too simple or too unclear - I haven't been doing real math for years.)
What kind of $\mathbb{A}^1$-homotopy are we talking about? Do you mean the $\mathbb{A}^1$-localization of the sheaf or are you just taking naive $\mathbb{A}^1$-classes? The second approach doesn't seem very likely to work out. And what do you mean by "suspension"?
It is the naive approach - my rings are in general not commutative, so there is no way to build sheaves. Suspension in this terms is a particular $\Sigma$ fitting into the short exact sequence $\Sigma A \to A[t, t^{-1}] \xrightarrow{ev_1} A$.
Thing is, I get this kind of theory as a byproduct of another construction, and it seems to be related to Cortinas-Thom bivariant algebraic theory $kk$, for which, given an $R$-aglebra $B$, one has $kk(E,B)\cong KH(B)$. I'm trying to prove the coincidence of my group with $KH_0$ via this sideway, but wonder if I'm inventing a bicycle by downgrading a jet fighter :)
|
2025-03-21T14:48:29.588994
| 2020-01-06T13:06:01 |
349818
|
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|
Stack Exchange
|
local UFD with dimension less than or equal 3 is catenary
Let $R$ be a commutative ring with identity. Then $R$ is $\textit{catenary}$ if for each pair of prime ideal $p \subsetneq q$, all maximal chains of prime ideals $p = p_0 \subsetneq p_1 \subsetneq \dots \subsetneq p_n = q$ have the same length.
In some (informal) texts the author conclude that (without further explanation) the ring is catenary since it is a local UFD and its dimension is less than or equal 3.
Is this considered trivial?
thank you.
This result appears as Proposition II.3 in
Hamet Seydi, Anneaux henséliens et conditions de chaînes. III, Bull. Soc. Math. France 98 (1970), 329–336. Numdam: BSMF_1970__98__329_0. DOI: 10.24033/bsmf.1706. MR: 276222.
Namely, Seydi proves that "every (Noetherian) UFD of dimension three is catenary." The Noetherian assumption is necessary due to a counterexample of Fujita.
There is no proof provided, but I think that Seydi is pointing out that this result is a consequence of the proofs of previous results, namely, Théorème II.2 and Corollaire II.2.4. We give a version of Seydi's proof here, by showing that Noetherian UFD's of dimension at most three are catenary.
Proof. Let $A$ be a Noetherian UFD of dimension at most three. Since the property of being catenary can be checked after localizing at every maximal ideal, it suffices to consider the case when $A$ is local. We recall that Ratliff's criterion [Matsumura, Theorem 31.4] says that a Noetherian local domain $B$ is catenary if and only if
$$\operatorname{ht} \mathfrak{p} + \dim(B/\mathfrak{p}) = \dim B\tag{$*$}\label{eq:ratliff}$$
for every prime ideal $\mathfrak{p} \subseteq B$. We also recall that Noetherian domains of dimension $\le 2$ are catenary [Matsumura, Corollary 2 to Theorem 31.7], and so it suffices to consider the case when $\dim A = 3$.
Consider a prime ideal $\mathfrak{p} \subseteq A$. If $\operatorname{ht} \mathfrak{p} = 0$, then $\mathfrak{p} = 0$, in which case \eqref{eq:ratliff} trivially holds for $B$ replaced by $A$. Otherwise, suppose that $\operatorname{ht} \mathfrak{p} \ge 1$. Then, there exists a prime ideal $\mathfrak{q} \subseteq \mathfrak{p}$ such that $\operatorname{ht} \mathfrak{q} = 1$ and such that $\operatorname{ht}(\mathfrak{p} \cdot A/\mathfrak{q}) + 1 = \operatorname{ht}\mathfrak{p}$. Since $A$ is a local UFD, the ideal $\mathfrak{q}$ is principal, and we have $\dim(A/\mathfrak{q}) = 2$. We then have
$$\operatorname{ht}(\mathfrak{p} \cdot A/\mathfrak{q}) + \dim(A/\mathfrak{p}) = \dim(A/\mathfrak{q}) = 2$$
by Ratliff's criterion, since $A/\mathfrak{q}$ is a Noetherian domain of dimension $2$. But $\operatorname{ht}(\mathfrak{p}\cdot A/\mathfrak{q}) + 1 = \operatorname{ht}\mathfrak{p}$, and hence by adding $1$ to both sides of the equation above, we obtain
$$\operatorname{ht}\mathfrak{p} + \dim(A/\mathfrak{p}) = \dim(A/\mathfrak{q}) + 1 = 3 = \dim A.\tag*{$\blacksquare$}$$
thank you very much Takumi
|
2025-03-21T14:48:29.589201
| 2020-01-06T13:33:39 |
349819
|
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|
Stack Exchange
|
Integer points of one Mordell equation
How can I determine all integer points of the following equation
$$y^2=x^3+10546$$
I tried Magma with
IntegralPoints(EllipticCurve([0,10546]));
but got the answer that it "could not determine the Mordell-Weil group." What are my options here?
I have edited the tags a bit - it is recommended to use at least one top level tag on [mathoverflow.se]. I wasn't entirely sure whether or not to include the tag ([tag:magma]) - feel free to remove it if you think it does not fit.
There are some nice papers containing data related to the question: Gebel, Pethő and Zimmer: On Mordell’s equation, Compos. Math. 110, No. 3, 335-367 (1998) and Bennett and Ghadermarzi: Mordell’s equation: a classical approach,
LMS J. Comput. Math. 18, 633-646 (2015) (here the technique is applied in case of $|k|<10^7$ to the equation $y^2=x^3+k.$)
This curve has rank 0 over $\mathbb{Q}$. The 2-descent fails to determine this, because the $2$-torsion subgroup of the Tate-Shafarevich group is non-trivial. Instead, one can compute the $L$-value. One can prove that $L(E,1) = 16 \Omega_{+}$. By Kolyvagin, this implies that the rank is $0$.
Now one just needs to compute the torsion order and since there are no non-trivial torsion points. One gets $E(\mathbb{Q})=\{O\}$ and hence there are no integral points either.
Nice! Can you please point to the origin of $L(E,1)$.
sage, magma, pari,... they can all compute $L$-function with sufficient precision to convince you that $L(E,1)\neq 0$. For the precise proven value of 16 I used my own code.
|
2025-03-21T14:48:29.589348
| 2020-01-06T14:11:16 |
349822
|
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|
Stack Exchange
|
How to prove the rational polynomial is nonincreasing?
Suppose $p(x)$ is a univariate real-rooted polynomial. It is easy to see that the following rational polynomial
$$\Psi_p(x) = \frac{\partial^2p}{p}(x)=\sum_{1\leq j<k\leq r}\frac{2}{(z_i-\lambda_j)(z_i-\lambda_k)}$$
is nonincreasing when $x>maxroot(p)$, where $\lambda_i,i=1,\cdots,r$ are the roots of $p$.
Now I want a more general case.
Suppose $p(x_1,\cdots,x_n)$ is a multivariate real polynomial and all univariate restrictions are real-rooted. We say $x$ is above the roots of $p$ if $$p(x+t)>0
\text{ for all non-negative vector t}.$$Define
$$\Psi_p^i(x) = \frac{\partial^2_{x_i}p}{p}(x).$$
It is obvious that $\Psi_p^i$ is nonincreasing in $i$ direction when $x$ is above the roots of $p$ for we can consider the univarate restriction of $p$ in that direction and regard it as a univariate polynomial.
What about other direction? Is the rational polynomial above is nonincreasing in any direction when $x$ is above the roots of $p$?
The problem is discribed in a general case. In fact, we can consider the case of only two variables. The problem is about the root of polynomial. Maybe some knowledege of complex analysis is needed.
Another thing I want to mention is that the problem is an variant of the Lemma 17 in this post https://terrytao.wordpress.com/2013/11/04/real-stable-polynomials-and-the-kadison-singer-problem/#more-7109.
|
2025-03-21T14:48:29.589464
| 2020-01-06T14:33:51 |
349824
|
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|
Stack Exchange
|
Fredholm elements of a Lie algebra
An element $a$ of a Lie algebra $L$ is called a Fredholm element if the adjoint operator $\mathrm{ad}_a:L \to L$ is a Fredholm linear map. That is: its kernel is a finite-dimensional space and its range $\mathrm{ad}_a(L)$ is a finite-codimensional subspace.
Is there an infinite-dimensional Lie algebra with at least one Fredholm element? Is there a Lie algebra $L$ whose only non-Fredholm element is $0$?
Do you have an application—for Fredholm elements generally, or for these questions in particular—in mind (or does it already exist), or is this just a curiosity?
@LSpice I don't know if OP has a particular motivation, but I find the question very interesting and natural. Since we cannot expect $\mathrm{ad}_a$ to be invertible (for $L\neq 0$) the closest possible to invertibility would be being Fredholm. To each Lie algebra we can consider the group generated by Fredholm adjoint maps in the Fredholm automorphism group.
PS: by "Fredholm automorphism group" (of $L$, viewed as vector space only), I mean, the group $\mathrm{Fred}_L$ obtained as follows: the quotient of the monoid consisting of all Fredholm linear endomorphisms of $L$, by identifying $f,g$ if $f-g$ has finite rank. If the Lie algebra $L$ has infinite dimension and has the property that all nonzero elements are Frehdolm, then the map $L\smallsetminus{0}\to \mathrm{Fred}_L$ induced by $x\mapsto\mathrm{ad}_x$, is injective.
@LSpice it was just a curiosity. In fact this question was not a part of a research question or a project.
The infinite-dimensional Lie algebra with basis $(e_n)_{n\in\mathbf{N}}$ and brackets $[e_i,e_j]=(i-j)e_{i+j}$, over a field of characteristic zero, satisfies the required condition: every nonzero element is Fredholm.
First let me mention it's immediate that $\mathrm{ad}_{e_i}$ is Fredholm for every $i$. Denote by $V_k$ the subspace generated by $(e_i)_{i\le k}$.
Let $f$ be a nonzero element, say $f=\sum_{i=0}^ka_ie_i$ with $a_k\neq 0$ and $T=\mathrm{ad}_f$ (let's call $k$ the degree of $f$). Then the kernel of $T$ is contained in $V_k$ (indeed, if $g\notin V_k$, say of degree $\ell$, then $[f,g]$ has degree exactly $k+\ell$). Next computing $[f,e_i]$ for each $i\ge k+1$, we see by induction that $V_i/V_k$ is contained in the image of $p_k\circ T$, where $p_k$ is the projection $L\to L/V_k$. Hence $p_k\circ T$ is surjective. Hence $\mathrm{Im}(T)+V_k=L$, thus $\mathrm{Im}(T)$ is finite-codimensional.
Thanks for this answer and very helpful information in your comments. I was not aware of this Lie structure and Fredholm automorphism group.
This Lie algebra is quite well-known (as 1-dimensional Witt Lie algebra). Here $e_n$ corresponds to the derivation $x^{n+1}\frac{\partial}{\partial x}$. One can also consider it with $n\in\mathbf{Z}$; I think the arguments can be adapted, and the latter has a central extension (with 1-dimensional center) called Virasoro Lie algebra.
|
2025-03-21T14:48:29.589676
| 2020-01-06T14:35:30 |
349825
|
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|
Stack Exchange
|
A characterization of root systems via their intersections with halfspaces
In a recent preprint I obtained a nice characterization of root systems as a side product.
I can imagine that this was known before, and that a source for this statement can shorten the proof of my main result.
Question: Can you point me to a source, or a simple proof of the facts I list below?
In the following, let $R\subset \Bbb R^d$ be finite and centrally symmetric, i.e. $R=-R$.
Some definitions:
$R$ is a root system if for all $r\in R$ the set $R$ is invariant w.r.t. reflection on the hyperplane $r^\bot$.
A half (or semi-star in the preprint) is the intersection of $R$ with a halfspace, so that the intersection contains exactly half the elements of $R$. (see figure)
The norm of a half is the norm of the sum of the vectors it contains.
$\qquad\qquad\qquad\qquad\qquad\qquad$
The two characterizations now read as follows:
Result.
(Corollary 5.1) All halves of $R$ are congruent (i.e. they are related by orthogonal transformations) if and only if $R$ is a root system.
(Theorem 5.2) If all vectors in $R$ are of the same length, and all halves of $R$ are of the same norm, then $R$ is a root system.
In point 1. the direction that proofs that $R$ is a root system is the non-trivial part.
That's a nice result! I had a quick look, but, without tracing the chain of implications all the way back, it's not apparent: do you use a classification result, or is this from first principles? (Also, you're missing a period at the end of Section 5.)
@LSpice Thank you! I would say it is from first principles, just with a detour over zonotopes, so probably using some well-known facts about polytopes (but no classification). In particular, the classification of root systems plays no role for this characterization.
|
2025-03-21T14:48:29.589819
| 2020-01-06T14:39:26 |
349826
|
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|
Stack Exchange
|
Definition of Morphisms of algebraic stacks smooth of relative dimension n
There is a notion of smooth morphism of algebraic stacks e.g. Tag 075U and a notion of
relative dimension of a locally of finite type morphism $T\to \mathcal{X}$ from an algebraic space into an algebraic stack e.g. Tag 0DRG.
In the case of locally Noetherian algebraic stacks, there is also a notion of relative dimension of a locally of finite type morphism $\mathcal{T}\to \mathcal{X}$ between algebraic stacks e.g. Tag 0DRL.
What is the (commonly used) definition of a locally of finite type morphism between (locally Noetherian) algebraic stacks being "smooth of relative dimension $n$"?
The most natural approach would be to define it as being a smooth morphism such that the relative dimension in every point is $n$. However I couldn't find this definition anywhere in the literature.
Another possible Definition that I can think about is to follow the definition of smoothness given in stacks project:
A locally of finite type morphism of algebraic stacks $\mathcal{X}\to \mathcal{Y}$ is called "smooth of relative dimension $n$" if and only if there is a commutative diagram
$$\begin{array}{c} U & \overset{h}{\rightarrow} & V \\ \overset{a}{}\downarrow && \overset{b}{}\downarrow \\ \mathcal{X} & \overset{f}{\rightarrow} & \mathcal{Y} \end{array}$$, where $U,V$ are algebraic spaces with smooth vertical arrows such that $U\to \mathcal{X}\times_{\mathcal{Y}} V$ is smooth and $a$ surjective, such that $h$ is smooth of relative dimension $n$.
Compare Tag 06FM. In this case one would need to define what it means for a morphism of algebraic spaces to be smooth of relative dimension $n$ using a similar commutative diagram with etale vertical arrows, as is done in stacks project for smoothness of morphisms of algebraic spaces Tag 03ZC.
Another possible definition that I can think of is using a presentation $[U/R]\to \mathcal{X}$.
Is there a commonly used definition for the notion of smooth of relative dimension $n$ morphisms of algebraic stacks?
Thank You for Your help.
Excuse me if this is a bad question. I am only a graduate master student, not an expert at all.
You cannot follow the definition given in the stacks project directly, because the property "relative dimension $n$" is not smooth local on source-and-target. In particular, this property is not stable under precomposing with smooth maps (condition 1 in https://stacks.math.columbia.edu/tag/06F8), because the relative dimension may change.
|
2025-03-21T14:48:29.589973
| 2020-01-06T14:44:34 |
349828
|
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|
Stack Exchange
|
Looking for the multiplicity-free paper by N.Inglis
I'm looking for the paper "Multiplicity-free permutation characters, distance-transitive graphs and classical groups, PhD Thesis, University of Cambridge, 1986" by Nicholas Francis John Inglis. The paper is cited several times (Google Scholar says 10), e.g. John van Bon's "Finite primitive distance-transitive graphs".
The Google Scholar link directs to the EThOS service, but the EThOS service gives the message "Full text unavailable from EThOS".
Is it possible to find the paper?
I guess there is a copy at the University of Cambridge, but no idea if it is possible to get a copy without going to the library (the website there says "not borrowable"). In cases like this it is sometimes useful to look at other papers of the author, for example some results of the PhD thesis are in "Multiplicity-free permutation representations of finite linear groups" (joint w/ Liebeck and Saxl) link. You might find whatever you are looking for there as well. Also, you could email and ask some people who are still active and have cited the thesis.
Update: the conclusion of the paper can be found in p.353 of Investigations in Algebraic Theory of Combinatorial Objects by I. A. Faradzev et al. and that's enough for my research, so there's no need to post an answer.
|
2025-03-21T14:48:29.590103
| 2020-01-06T15:28:45 |
349832
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/349832"
}
|
Stack Exchange
|
Subset of $[\omega]^\omega$ that can be "colored" with $3$, but not $2$ colors
Let $[\omega]^\omega$ denote the set of infinite subsets of $\omega$.
Let $S\subseteq [\omega]^\omega$. We say that a map $c:\omega \to \{0,\ldots,n-1\}$ is a coloring for $S$ with $n$ colors if for all $s\in S$ the restriction $c|_s$ of $c$ to $s$ is non-constant.
What is an example of a set $S\subseteq [\omega]^\omega$ such that $S$ has a coloring with $3$ colors, but not with $2$ colors?
Partition $\omega$ into three infinite subsets $A_0,A_1,A_2$. Let $S$ consists of subsets which intersects precisely two of the $A_i$ at infinitely many elements. It can obviously be $3$-colored. Suppose there was a $2$-coloring, with color classes $c_0,c_1$. Then either $c_0$ or $c_1$ contains infinitely many elements of some two of $A_0,A_1,A_2$, say $c_0\cap A_0,c_0\cap A_1$ are infinite. Then $c_0\cap(A_0\cup A_1)\in S$ is monochromatic.
This can be generalized to a hypergraph $S\subseteq[\omega]^\omega$ with chromatic number $n$: partition $\omega$ into $A_0,\dots,A_{n-1}$ and let $S$ consists of subsets which intersect two of those at infinitely many elements.
This is cool! Is this construction in the literature? Gerhard "If So, Where Is It?" Paseman, 2020.01.06.
Amazing - thanks @Wojowu!
@GerhardPaseman I'm afraid I do not know any literature which could contain it. My motivation for coming up with it was basically to "emulate" a complete graph on $3$ (respecively $n$) vertices.
|
2025-03-21T14:48:29.590227
| 2020-01-06T15:37:18 |
349834
|
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|
Stack Exchange
|
Minimize spectral radius with orthogonal matrix
Let $A$ be a real square and invertible matrix. I would like to find
$$
s(A) = \min_U \rho(U A),
$$
Where $U$ is orthogoal, i.e. $U U^T = I$ and $\rho(A)$ is the spectral radius, i.e. the largest eigenvalue of $A$ in absolute values.
I am interested in a numerical solution to determine $s(A)$.
Isn't the answer just "singular value decomposition"? In other words, $U$ is the left matrix in SVD.
For an $n\times n$ matrix, the answer is
$$|\det A|^{\frac1n}.$$
Explanation: on the one hand, $\rho(UA)\ge|\det (UA)|^{\frac1n}=|\det A|^{\frac1n}$. On the other hand, singular value decomposition gives $A=PDQ$ where $P,Q$ are orthogonal and $D>0$ is diagonal. Then
$\rho(UA)=\rho(QUPD)$ and
$$\min_U\rho(UA)=\min_V\rho(VD)$$
where $V$ runs over the orthogonal group. Take for $V$ the permutation matrix associated with the cycle $(12\ldots n)$. We have
$$VD=\begin{pmatrix} 0 & \cdots & \cdots & 0 & s_1 \\ s_2 & \ddots & & & 0 \\ 0 & \ddots & \ddots & & \vdots \\ \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & \cdots & 0 & s_n & 0 \end{pmatrix},$$
whose spectrum is made of the $n$-roots of $s_1\cdots s_n=\det D=|\det A|$.
Remark that the answer is valid for an arbitrary matrix, singular or not. It is valid for complex matrices too, provided one takes the minimum over unitary matrices $U$.
Thanks. That is a very helpful answer. I suppose there is a small typo: it should read $\rho(QUPD)$, right?
@SebastianSchlecht Of course ; I'll fix it. Besides, feel free to accept the answer :).
|
2025-03-21T14:48:29.590361
| 2020-01-06T16:04:04 |
349838
|
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"url": "https://mathoverflow.net/questions/349838"
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|
Stack Exchange
|
Given $\theta$, find $f$ such that $\int_{\mathbb{T}} \text{e}^{i\theta} \cos(h \cdot f) = 0,$ for all $h \in \mathbb{N}$
Let $\theta$ be a $C^{\infty}$ (resp. analytic) real-valued function on $\mathbb{T}=[0,2\pi]/\{0,2\pi\}$.
When can one find $f \neq 0$, $C^{\infty}$ (resp. analytic) real-valued function on $\mathbb{T}$ such that
$$ \tag{1}
\int_{[0,2\pi]} \text{e}^{i\theta(s)} \cos(h \cdot f(s)) \text{d}s = 0, \text{ for all } h\in \mathbb{N}\cup \{0\}?
$$
Setting $n=0$, one obvious necessary condition is $\int_{[0,2\pi]} \text{e}^{i\theta(s)} \text{d}s = 0$.
A sufficient condition I found, which is far from being explicit, is the existence of a reparametrization $v:\mathbb{T}\rightarrow \mathbb{T}$ that switches off all the harmonics multiple of a certain $k$. More precisely, given the Fourier expansion
$$
v'(t) \text{e}^{i\theta(v(t))}= \sum_{j\in\mathbb{Z}} c_j e^{i(jt)},
$$
if there exists $k\in \mathbb{N}$ such that $c_{k \cdot h}=0$ for all $h\in \mathbb{Z}$, then
$$
\int_{[0,2\pi]} \text{e}^{i\theta(s)} \cos(h \cdot k \cdot v^{-1}(s)) \text{d}s=\int_{[0,2\pi]} v'(t) \text{e}^{i\theta(v(t))} \cos(h \cdot k \cdot t) \text{d}t=0.
$$
How can one more explicitely characterize/construct (or fail to construct) such $f$'s in the $C^\infty$ (resp. analytic) case? Does this problem connect to something you know from the literature?
MORE EQUIVALENT CONDITIONS
The existence of a function as above is equivalent to the existence of a function $\phi$ such that
$$ \tag{2}
\int_{[0,2\pi]} \text{e}^{i\theta(s)} \phi(s)^n \text{d}s = 0, \text{ for all } n\in \mathbb{N}\cup \{0\}.
$$
One direction follows from the Weierstrass approximation theorem: if (2) holds, then it must hold for all the composition of $g\circ \phi$, with $g$ an $L^2$ function, and in particular for $g(x)=\cos(h\cdot x)$. The other direction is obtained by observing that $\cos(f)^n$ is a linear combination of $\cos(h\cdot f), h \in \mathbb{N}$, and therefore if (1) holds then (2) holds for $\phi=\cos(f)$.
Knowing that, I attempted, without success for the moment, to construct such a $\phi$ by finding suitable Fourier coefficients. If $\phi(t)=\sum_{j\in\mathbb{Z}} \widehat{\phi}(j) e^{i(jt)} $, then, by convolution,
$$
\phi(t)^n=\sum_{j}\;\sum_{j_{n-1}}\sum_{j_{n-2}} ... \sum_{j_{1}} \widehat{\phi}(j-\sum_{r=1}^{n-1} j_r) \prod_{r=1}^{n-1} \widehat{\phi}(j_r) \; \text{e}^{i (j t)},
$$
and hence (2) is translated into a sequence of homogeneus equations in $l^2$. The regularity of $\phi$ should then be deduced by bounding the decay of its smartly chosen Fourier coefficients.
Do you see how this other approach could work?
The two conditions have very interesting geometrical implications. In fact, another equivalent characterization is the existence of $\phi$ such that
$$
\int_{[0,2\pi]} e^{i(\theta(s)+\lambda\phi(s))} ds=0, \text{ for all } \lambda \in \mathbb{R}.
$$
The equivalence can be obtained by observing that the term on the left is analytic in $\lambda$ and by computing its $n$-th derivative at $\lambda=0$. This would provide a one parameter family of closed curve associated to an affine line in the space of turning angle functions.
Note. The first part of this question had been already posted without receiving an answer. Since edits were major I just decided to repost it from scratch. Thank you
|
2025-03-21T14:48:29.590571
| 2020-01-06T16:53:44 |
349840
|
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|
Stack Exchange
|
Sign in May’s General algebraic approach to Steenrod operations
In the first section of J. P. May’s General algebraic approach to Steenrod operations, May defines for $\pi\subseteq\Sigma_r$ an integer $q\in\mathbb{Z}$ and a commutative ring $\Lambda$, the $\Lambda\pi$-module $\Lambda(q)=\Lambda$ with sign action $\sigma\lambda = (-1)^{qs(\sigma)}\cdot \lambda$ where $(-1)^{s(\sigma)}$ is the sign of $\sigma$.
For a $\Lambda$-chain complex $K$ we consider $K^{\otimes r}(q)= K^{\otimes r}\otimes \Lambda(q)$ with the diagonal action. I assume that this translates to the explicit sign rule for the transposition $\sigma_{i,i+1}$ in $K^{\otimes r}(q)$
$$\sigma_{i,i+1}\cdot (a_1\otimes\dotsb\otimes a_r) = (-1)^{q+|a_i|\cdot |a_{i+1}|}\cdot (a_1\otimes\dotsb \otimes a_{i+1}\otimes a_i\otimes\dotsb\otimes a_r).$$
Am I correct with this? He later considers cycles $a,b\in K_q$ and $c\in K_{q+1}$ with $dc=a-b$. Now let $I$ be the cellular chain complex of the intervall, i.e. $I_1=\Lambda\langle e\rangle$ and $I_0=\Lambda \langle e_0,e_1\rangle$, and $de=e_1-e_0$. We consider the chain map of degree $q$
$$f:I\to K, e\mapsto (-1)^qc, e_1\mapsto a, e_2\mapsto b.$$
This satisfies $dfe=(-1)^qfde$ and hence is a chain map. May now claims that $f^{\otimes r}:I^{\otimes r}\to K^{\otimes r}(q)$ is $R\pi$-equivariant. I don’t see why: We have
$$f^{\otimes r}(\sigma_{i,i+1}\cdot (a_1\otimes\dotsb\otimes a_r)) = (-1)^{|a_i|\cdot |a_{i+1}|} f(a_1)\otimes\dotsb\otimes f(a_{i+1})\otimes f(a_i)\otimes\dotsb\otimes f(a_r),$$
whereas on the other side, we have
$$\sigma_{i,i+1}\cdot f^{\otimes r}(a_1\otimes\dotsb\otimes a_r)=(-1)^{q+(|a_i|+q)\cdot(|a_{i+1}|+q)}\cdot f(a_1)\otimes\dotsb\otimes f(a_{i+1})\otimes f(a_i)\otimes\dotsb\otimes f(a_r).$$
There are equal iff $q\cdot (|a_i|+|a_{i+1}|)$ is even, but this is not necessarily the case. Concretely, if $q=1$ and $\sigma=\sigma_{1,2}$, then
$$f^{\otimes 2}(\sigma\cdot e\otimes e_1) = f^{\otimes 2}(e_1\otimes e)=a\otimes (-c)\ne a\otimes c=\sigma\cdot (-c)\otimes a=\sigma\cdot f^{\otimes 2}(e\otimes e_1).$$
What am I missing?
A totally reflexive comment on any question about signs in algebraic topology: have you read Tyler Lawson's "In which I try to get the signs right for once"?
Ah, thatʼs cool! Thank you! It also solves my confusion: I forgot to use the sign rule $(f\otimes g)(a\otimes b) = (-1)^{|g|\cdot |a|} \cdot f(a)\otimes g(b)$.
When in doubt, invoke the Koszul rule of signs: if one symbol passes through another, multiply by $-1$ raised to the product of their degrees.
@FKranhold You mean I got it right? You had me fooled. I should apologize for leaving that detail to the reader, but let me give two excuses. First, one does not actually need that detail to prove Lemma 1.1(iv). It just answers an obvious question the reader might have about the proof. Second, that paper was from the good old days when things went fast. There was a conference, March 30 to April 4, 1970, for Steenrod's 60th birthday. Its Proceedings were submitted June 15, 1970. No time to polish the writing. I remember Steenrod telling me that mine was the only paper in the Proceedings that he understood. He died the next year.
Hopefully there is no causal relationship between the last two sentences.
@LSpice Thanks, Loren. I did need a laugh.
Yes, I think that everything should be fine. I was just confused because I missed the sign rule for $(f\otimes g)(a\otimes b)$.
|
2025-03-21T14:48:29.590908
| 2020-01-06T18:00:17 |
349844
|
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|
Stack Exchange
|
Spectral norm of difference of quadratic matrices restricted to a subspace
Say that we have two matrices $X$ and $Y$ of dimensions $(T \times N)$ with $N < T$ and $rank(X)=rank(Y)=N$. Furthermore, define a $(T \times k)$ dimensional matrix $D$ with $k<N$ and $rank(D)=k$. Then, construct the idempotent matrix $M = I - D(D^\prime D)^{-1}D^\prime$, with $I$ being a $T$-dimensional identity matrix.
Let $|| \cdot ||_2$ denote the spectral norm. Is it true that
$||X^\prime M X - Y^\prime M Y||_2 \leq ||X^\prime X - Y^\prime Y||_2$ ?
Simulations definitely indicate that the above inequality holds, no matter how I randomly generate $X,Y,D$. If the above claim does not hold, is it then at least true that
$||X^\prime M X - Y^\prime M Y||_2 \leq ||X^\prime X - Y^\prime Y||_F$,
where $||\cdot||_F$ represents the Frobenius norm?
My own reasoning only brings me to the following. It is straightforward to show that
$||X^\prime M X||_2 = ||M XX^\prime M||_2\leq ||M||_2^2||X^\prime X||_2 \leq ||X^\prime X||_2$,
since $M$ is an idempotent matrix with the first $T-k$ eigenvalues equal to 1 and the remaining 2 equal to 0. The same obviously holds for $||Y^\prime M Y||$, but that does not seem sufficient for the claim to hold?
For the Frobenius norm proof, I tried to write it as the trace of the inner product and use the inequality that $tr(AB) \leq ||A||_2\sum_i \sigma_i(B)$, but that got me nowhere.
I hope any of you can help. Please let me know if you need more information.
Thanks!
In particular, you claim that $X'X=Y'Y$ implies $X'MX=Y'MY$. Either I misunderstand your setup, or that sounds extremely fishy.
@fedja Ah, I'm afraid you are right.. For some reason I did not check this border case, but it clearly demonstrates that the claim is infeasible. Thanks!
|
2025-03-21T14:48:29.591044
| 2020-01-06T18:02:02 |
349845
|
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|
Stack Exchange
|
Reference request - parallel rectangles discrepancy theory
I've been reading about discrepancy theory and trying to understand some of the open problems in the field. Wikipedia has a list of some of the open problems, but the descriptions are terrible. In particular, I am curious about:
"Axis-parallel rectangles in dimensions three and higher (folklore)"
Curious if anyone knows what problem the author was referring to, i.e. if this is some famous topic that everyone in discrepancy knows of and if there is a good reference to look at? (Or is the above quote just nonsense?)
Perhaps the issue is to establish a tight lowerbound for the discrepancy of $n$ points in dimension $d$ with respect to $d$-dimensional boxes.
In
Matousek, Jiri, ed. Geometric discrepancy: An illustrated guide. Vol. 18. Springer Science & Business Media, 2009,
after discussing the planar case, where the optimal $\Omega( \log n)$ bound is established, Matousek says for $d$-dimensional boxes (p.176),
In particular, perhaps the $\tfrac{1}{2}$ in the exponent $\tfrac{d-1}{2}$ is not needed
(as it is not needed in $d=2$).
But this book is now a decade old. I am not certain of the current status.
|
2025-03-21T14:48:29.591149
| 2020-01-06T18:37:20 |
349849
|
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|
Stack Exchange
|
When do blow-ups at $N$ points give isomorphic varieties?
Let $X$ be a smooth projective variety over an algebraically closed field $k$. For any length $n$ ideal sheaf $P$ of $X$ (e.g $N$ different points $P=(P_1)(P_2)..(P_N)$), we can consider $Bl_{P}(X)$, the blow up of $X$ at $P$. Assume $Bl_{P}(X) \cong Bl_{Q}(X)$ for a length $n$ ideal sheaf $P$ and a length $m$ ideal sheaf $Q$ , what can we say about $P$ and $Q$ ? If neccessary, one can assume $P,Q$ are radical.
Note blow up at $I$ and $I^2$ are the same. So there is a trivial case i.e when there exists an automorphism $\phi:X \rightarrow X$ s.t $\phi^*(P^k)=Q^l$.
I am interested in some examples e.g projective spaces, abelian varieties, surfaces.
The question has more content and a different emphasis over non-algebraically closed fields. Let $P$ and $Q$ be reduced zero-dimensional subschemes of $X$ such that the blow ups are isomorphic. Are $P$ and $Q$ isomorphic schemes?
Note that in spite of blow ups of $I$ and $I^2$ being the same, ideals with the same radical as $I$ do not in general lead to isomorphic blow up. Indeed, if $I = (x,y)$ in the plane, then $I^2 \subset (x,y^2) \subset I$ but blowing up the intermediate ideal leads to a singular point (a node).
The answer in general is a mess, because isomorphisms between the blow-ups do not necessarily descend to automorphisms of $X$. If you take $X = \mathbb P^2$ and fix a general configuration $P$ of $n \geq 9$ points, then the set of $Q$ for which $Bl_P(X) \cong Bl_Q(X)$ is a countable union of subvarieties of $(\mathbb P^2)^n$, which is probably Zariski dense (though I'm not sure whether anybody has actually tried to prove this).
In some sense the point is that such blow-ups have infinitely many $(-1)$-curves on them, and so you can blow down to $\mathbb P^2$ in infinitely many different ways by choosing which curves to contract. The resulting configurations in $\mathbb P^2$ that you get as the images of the contracted curves do not simply differ by elements of $PGL(3)$, as you might hope.
If $X$ is not uniruled, things are probably easier. For example, if $X$ is abelian, any isomorphism $f : Bl_P(X) \to Bl_Q(X)$ descends in the obvious way to an isomorphism $g : X \to X$, and so the answer is that the blow-ups are isomorphic if and only if $P$ and $Q$ differ by some element of $Aut(X)$.
Just a comment on your last paragraph for those interested: If $X$ is smooth projective and has no rational curves, then every isomorphism $Bl_P(X) \to Bl_Q(X)$ descends to an isomorphism $X\to X$.
|
2025-03-21T14:48:29.591328
| 2020-01-06T18:45:39 |
349850
|
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"Derek Holt",
"Zhiyu",
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|
Stack Exchange
|
Group cohomology of modular representations for finite groups of Lie type
$GL_n(\mathbb F_q)$ naturally acts on the vector space $V=\mathbb F_q^n$. As $GL_n(\mathbb F_q)$ is a finite group, the cohomology group $H^i(GL_n(\mathbb F_q),V)$ are all finite abelian groups. Can we compute those cohomology groups explicitly?
This is a baby example, and for odd $p$ one can use the trick in Cohomology of SL(2,R) with coefficients given by linear action. In general, Let $G=\mathbb G(\mathbb F_q)$ where $\mathbb G$ is a connected reductive group over $\mathbb F_q$ (or more generally a finite group of Lie type), $V$ be an irreducible algebraic representation of $\mathbb G$ defined over $\mathbb F_{q^n}$ (or more generally any irreducible equal characteristic modular representation), can we compute $H^i(G,V)$ explicitly or at least give some bounds?
In the baby example you ask about in the first paragraph, the cohomology groups are zero whenever $q>2$, because there are nontrivial central elements of the group acting fixed-point-freely on the module.
@DerekHolt Yes, this is exactly the trick in the link. How about the case $p=2$?
You mean how about $q=2$. They are known for $i=1$ and $2$, but I am afraid that I don't know of any results for higher $i$ (although some smaller examples could be computed).
@DerekHolt Thanks, can you give a reference for $i=1,2$?
They are done in papers by G.W. Bell, On the cohomology of the special lienar groups I and II in Journal of Algebra, Volume 54, Issue 1, September 1978. That might not be the earliest reference for these specific results.
As Derek Holt comments, cohomology has complications even for fimite general linear groups. Probably you are using the term "reductive" too casually and should replace it by "simple" or perhaps "semisimple" to get a finite group of Lie type: for example, an algebraic torus (direct product of copies of the multiplicative group of the field involved) is reductive but does not lead to a group of Lie type. Here "simple" refers to a connected algebraic group with no proper normal connected subgroups except the trivial group.
For example, SL$_2$ is simple in this sense and the associated finite group of Lie type has tricky cohomology to compute: see the old paper by Jon Carlson, which has not been improved on much, here .
See also the many papers of Cline-Parshall-Scott, which lead to some bounds on dimension of cohomology and some other approaches to computation. A summary and references are given in my LMS Lecture Notes 326 (2006) here .
Thank you for references, now I see this is a difficult problem in general.
|
2025-03-21T14:48:29.591513
| 2020-01-06T19:06:06 |
349853
|
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|
Stack Exchange
|
Adjoint of the multiplication operator on a Sobolev space
Let $f\colon\mathbb{R}^n\rightarrow\mathbb{C}$ be a bounded function with a bounded first derivative. Then the multiplication operator $H^1(\mathbb{R}^n)\ni x\mapsto A_f x:=fx\in H^1(\mathbb{R}^n)$ is bounded, where $H^1(\mathbb{R}^n)=W^{1,2}(\mathbb{R}^n)$, a Sobolev space. In a comment, user ougoah asked whether the adjoint operator $A_f^*$ is a multiplication operator, too. Here this question will be answered.
$\newcommand{\R}{\mathbb R}$
The answer is: $A_f^*$ is a multiplication operator iff the function $f$ is constant.
Indeed, recall that for all $x$ and $y$ in $H^1:=H^1(\mathbb R^n)$
\begin{equation*}
\langle x,y\rangle:=\langle x,y\rangle_{H^1}
=\int\Big(x\bar y+\sum_{j=1}^n(D_jx)\,(D_j\bar y)\Big),
\end{equation*}
where $\int:=\int_{\mathbb R^n}$ and $D_j$ denotes the partial derivative with respect to the $j$th argument.
The condition that $A_f^*$ is a multiplication operator $A_{\bar g}$ for some bounded function $g\colon\mathbb{R}^n\rightarrow\mathbb{C}$ with a bounded first derivative means that
\begin{equation*}
\langle fx,\bar z\rangle=:L(x,z)=R(x,z):=\langle x,\bar g\bar z\rangle \tag{1}
\end{equation*}
for all $x$ and $z$ in $H^1$.
So, the "if" part of our iff claim is obvious: if $f$ is constant, then $A_f^*=A_{\bar f}$.
To prove the "only if" part, take any $c=(c_1,\dots,c_n)$ and $a=(a_1,\dots,a_n)$ in $\R^n$ and let
\begin{equation*}
x(t):=e^{ia\cdot t-|t|^2/2}\quad\text{and}\quad z(t):=e^{i(c-a)\cdot t-|t|^2/2}
\end{equation*}
for $t=(t_1,\dots,t_n)\in\R^n$, where $a\cdot t:=\sum_1^n a_j t_j$ and $|t|:=\sqrt{t\cdot t}$. Then
\begin{align*}
R(x,z)&=\int\Big(gzx+\sum_{j=1}^n D_j(gz)\,D_jx\Big) \\
&=\int gz\Big(x-\sum_{j=1}^n D_j^2x\Big) \\
& =\int dt\,g(t)e^{ic\cdot t-|t|^2/2}\sum_{j=1}^n(a_j^2+2ia_jt_j+\tfrac{n+1}n-t_j^2)
\end{align*}
and, similarly,
\begin{equation*}
L(x,z)=\int dt\,f(t)e^{ic\cdot t-|t|^2/2}\sum_{j=1}^n((c_j-a_j)^2+2i(c_j-a_j)t_j+\tfrac{n+1}n-t_j^2).
\end{equation*}
So, in view of (1), $L(x,z)$ and $R(x,z)$ are equal quadratic polynomials in $a_1$, for each $c\in\R^n$. Equating the coefficients of $a_1^2$ in these two polynomials, we see that
$$\int dt\,f(t)e^{ic\cdot t-|t|^2/2}=\int dt\,g(t)e^{ic\cdot t-|t|^2/2}
$$
for all $c\in\R^n$, which implies $f=g$, which in turn means that (1) can be rewritten as (2) in the previous answer, whence, according to what follows formula (2) in that answer, $f$ is indeed constant.
Thank you for this detailed answer. I have asked another question about properties of $A_f^*$ over on StackExchange: https://math.stackexchange.com/questions/3502022/properties-of-the-adjoint-of-multiplication-on-sobolev-space.
|
2025-03-21T14:48:29.591681
| 2020-01-06T19:23:06 |
349855
|
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"Leo Alonso",
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|
Stack Exchange
|
Representable diagonal map $\Delta: \mathcal{X} \to \mathcal{X} \times \mathcal{X}$ for DM-Stacks/algebraic spaces
Following the standard definitions of a algebraic space or Deligne–Mumford stack one imposed condition is that the diagonal morphism $\Delta: \mathcal{X} \to \mathcal{X} \times \mathcal{X}$
has to be representable. The latter means that if $X,Y$ are schemes, $h_X,h_Y$ their Yoneda representations and we have natural maps $h_X \to \mathcal{X}, h_Y \to \mathcal{X}$ then the fiber product $h_X \times_{\mathcal{X}} h_Y$ has to be representable in "usual" sense by a scheme. That's a clear formulated technical condition that one can start to verify.
Question: What is the intuition/philosophy one should have in mind associating with this condition on representability of the diagonal $\Delta: \mathcal{X} \to \mathcal{X} \times \mathcal{X}$?
What makes this condition so "interesting" in order to study stacks? What is the wittness of this condition?
Recall then we talking about schemes a scheme $S$
is called separated if the diagonal map $S→S×S$ is a closed immersion. It allows to "transfer" in certain way a "category theoretical Hausdorff axiom" the algebraic geometry, since the classical one fails for Zariski topology.
For schemes the property "be separated" is is often used in following sence: Let $f: X\to Y, g:Y \to Z$ morphisms, composition $g \circ f$ has certain property P well behaving under pullbacks and $g$ separated. Then $f$ has also P.
Is the the motivation for representability of $\Delta: \mathcal{X} \to \mathcal{X} \times \mathcal{X}$ of similar manner?
That is my question is which intuitive property is "transfered" by imposing the representability for the diagonal map to the world of stack?
The basic issue is the following
Theorem. If the diagonal is representable (by schemes, algebraic spaces, etc.), then any morphism $S\rightarrow \mathcal{X}$ with $S$ a scheme, algebraic space, etc. is representable (by schemes, algebraic spaces, etc.).
For a proof, see
MathOverflow "Diagonal is representable then any morphism is representable".
Note, for instance that if the diagonal is affine, then on a scheme the intersection of two open affines is open (which is a kind of weak separability that is often useful in cohomological situations.)
A representability condition on the diagonal gives some information on the wildness of the (possible) non-separability of the stack/space considered. Being representable by a closed embedding is as good as it gets (i.e. the Hausdorff condition) but as long as you are willing to consider patchings with bad properties, you need some degree of control of its behavior. So some condition on the diagonal tames this. You might weaken from closed embeddings to affine schemes or to just schemes (this would make the patching scheme-like) or even to algebraic spaces (then the patching would look schematic form the etale toposic point of view). Beyond that, the rough idea is that very complicated behaviors might arise and one should avoid it. Thus the condition.
yes probably I see the issue. That allows to transfer terminology from schematic world (like smooth, etale, flat and so on) to morphisms $S\rightarrow \mathcal{X}$ with $S$ scheme. For example lot of interesting stacks like DM-stack come by defintion with a cover $p:U \to \mathcal{X}_{DM}$ and one want to impose an extra schematic condition $p$ (eg in case of DM-stack $p$ has to be etale). And the most naive way might be to demand that the scheme that represents this covering map is etale and so on. I think that was the point. Thank you a lot!
Yes, I could not have said it better. Glad my comment helped you.
|
2025-03-21T14:48:29.591924
| 2020-01-06T19:23:59 |
349856
|
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|
Stack Exchange
|
The Fock space in Costello's paper "Higher genus Gromov-Witten invariants as genus zero invariants of symmetric products"
Let $X$ be a smooth projective variety. In this Annals paper, Costello expressed the descendent genus $g$ Gromov-Witten (GW) invariants of $X$ in terms of genus zero GW invariants of the symmetric product stack $S^{g+1}(X)$. In the introduction of the paper, he defined the Fock space as follows
$$ \text{Sym}^* \left( H^*(X, \mathbb{C} ) \otimes_{\mathbb{C}} t \mathbb{C}[t] \right) = \oplus_{d \geq0} H^*_{\text{orb}}\left( S^d(X),\mathbb{C}\right).$$
Why is the LHS same as the RHS?
What does the parameter $t$ in the LHS correspond to in the RHS? Also I'm not sure why it is $t \mathbb{C}[t]$ rather than $\mathbb{C}[t]$. An arbitrary element of order $k$ in the LHS has the form
$$\sum_{j=1}^{n_1} a_{1j} t^j \cdots \sum_{j=1}^{n_k} a_{kj} t^j,$$
where $a_{ij} \in H^*(X, \mathbb{C} )$. Does this correspond to an element in $H^*_{\text{orb}}\left( S^k(X),\mathbb{C}\right)$ in the RHS? If everything is obvious, then it means I have a gap somewhere. I would appreciate it if someone could point that out for me.
|
2025-03-21T14:48:29.592272
| 2020-01-06T19:24:57 |
349857
|
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"Noam D. Elkies",
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|
Stack Exchange
|
References for the computation of the Mordell-Weil group of an elliptic curve
I am reading about the Mordell-Weil group of an elliptic curve over a number field using primarily Silverman's AEC. While the book is excellent in discussing materials prior to Chapter 8, I think that the exposition in Chapter 10 could have been better. But I don't know if there is any approachable text that discusses this topic in as much detail as Silverman does. Hence, the need to make this post.
In short, I am looking for algorithms to compute the Mordell-Weil group for elliptic curves over number fields or over just $\mathbb{Q}$.
From what I understand by skimming over Silverman and John Cremona's Algorithms for Modular Elliptic Curves book, there is $2$-descent algorithm (what Silverman and other texts like Husemoller, Knapp explain) that works when the set of rational points $E(\mathbb{Q})$ has a $2-$torsion point. And the other is 'general descent'. As far as I know, Silverman does not discuss this. But I would really like to read about it.
Could someone suggest some references for it? I know about Cremona's text (but I guess it discusses it in some special case? I'm not sure so correct me if I'm wrong) and Birch and Swinnerton-Dyer's Notes on Elliptic Curves Volume 1. Could someone tell me what are their prerequisites or are they somewhat self-contained? Also, are there any other references anyone would like to suggest?
Thank you very much.
Cremona describes 2-descent for both the special case where there's a rational 2-torsion point and the general case of a curve y^2=P(x) with P irreducible.
oh, thanks! Could you mention some (or all) of the prerequisites one needs to understand the general descent given in Cremona's text? And how does the exposition given there differs from the one in Birch, Swinnerton-Dyer's Notes? My guess is that Cremona explains BSD's thing in a special case you mentioned above?
|
2025-03-21T14:48:29.592422
| 2020-01-06T19:40:22 |
349859
|
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|
Stack Exchange
|
Engel residually finite groups
I know that every finite Engel group is a nilpotent group. Then, if $G$ is a residually finite Engel group, every finite quotient group of $G$ is a nilpotent group. Is necesseraly true that $G$ is a nilpotent group?
The examples by Golod, of groups on $d\geq 3$ generators, whose $(d-1)$-generator subgroups are all nilpotent (see example 18.3.2 in "Fundamentals of the theory of groups" by Kargapolov and Merzljakov for an exposition of the construction and properties) give $p$-groups that are residually finite, Engel, yet not nilpotent.
See this survey by G. Traustason on Engel groups, for other conditions one might add to guarantee nilptency.
Thank you for your response!
|
2025-03-21T14:48:29.592493
| 2020-01-06T20:12:39 |
349863
|
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|
Stack Exchange
|
A different version of list coloring
Consider a non-regular bipartite graph $G$ . We consider list edge coloring the edges of the graph by giving lists of cardinality $max(deg(v_i),deg(v_j))+2$ for each edge $e=v_iv_j$ where $deg(v_i)$ denotes the degree of the vertex $v_i$.
Then, is the graph $G$ properly colorable. If so, would the maximum cardinality of the lists be $\Delta+2$ where $\Delta$ be the maximum degree of $G$? Is this version of list coloring of $G$ similar to online list coloring? I also see it is exactly somewhat similar to the concept of $f-$ chhosabality of the graphs, where $f$ is the choice function for edge. Thanks beforehand.
This problem can indeed be solved using online list coloring. The stronger result that you can use lists of size $\max(\deg(v_i),\deg(v_j))$ is proven as Theorem 3.3 in this paper of Uwe Schauz.
|
2025-03-21T14:48:29.592582
| 2020-01-06T20:19:31 |
349864
|
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|
Stack Exchange
|
Strongly graded algebras with no zero divisors
Let $A = \bigoplus_{i \in \mathbb{Z}} A_i$ be a strongly graded unital algebra over $\mathbb{C}$, with no zero divisors. Is it always true that
$$
m: A_i \otimes_{A_0} A_j \to A_{i+j}
$$
is an isomorphism?
For context, for an abelian group $G$ a $G$-graded algebra $A$ is called strongly graded if $A_tA_u=A_{t+u}$ for all $t,u\in G$. (I'm not sure what's the implicit meaning of "algebra": I just guess "associative unital algebra".)
and the group $G$ does not necessarily have to be abelian.
yes, algebra means associative unital algebra
Yes this is always an isomorphism of $A_0$-bimodules.
It is a general result for strongly graded rings. It holds for an arbitrary grading group $G$ (not necessarily $\mathbb{Z}$) and does not depend on the presence of zero divisors. For a proof, see Corollary 3.1.2, p.82, from Methods of Graded Rings.
|
2025-03-21T14:48:29.592670
| 2020-01-06T21:16:41 |
349869
|
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"Mateusz Kwaśnicki",
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|
Stack Exchange
|
Courant nodal domain theorem for fractional Laplacian
Let $\lambda_k$ and $\varphi_k$ be the $k$-th eigenvalue and a corresponding eigenfunction of the fractional Laplacian in a bounded domain $\Omega \subset \mathbb{R}^N$, $N \geq 2$.
That is, $\lambda_k$ and $\varphi_k$ satisfy
$$
\left\{
\begin{aligned}
(-\Delta)^{\alpha/2} \varphi_k &= \lambda_k \varphi_k &&\text{in}~ \Omega,\\
\varphi_k&=0 &&\text{in}~ \mathbb{R}^N \setminus \Omega.
\end{aligned}
\right.
$$
Here the fractional Laplacian $(-\Delta)^{\alpha/2}$ is given by
$$
(-\Delta)^{\alpha/2} u = - C\lim_{\varepsilon \to 0+} \int_{\mathbb{R}^N \setminus B_\varepsilon(0)} \frac{u(y)-u(x)}{|y-x|^{N+\alpha}} \, dy,
$$
where $C>0$ is a some explicit constant and $\alpha \in (0,2)$.
If $\alpha=2$, then the fractional Laplacian becomes the classical Laplace operator and it is very well known that the number of nodal domains of $\varphi_k$ is bounded from above by $k$. This is the content of the Courant nodal domain theorem.
However, in the purely fractional case $\alpha \in (0,2)$ much less seems to be known in this respect. Some upper bounds on the number of nodal domains in 1D-case and $\alpha=1$ can be found in [Bañuelos, Kulczycki, 2004].
I'm interested in the following very particular case of a weak version of the Courant nodal domain theorem:
Does any second eigenfunction $\varphi_2$ have a finite number of nodal domains?
Maybe the answer is well-known for experts or, at least, can be relatively easy to obtain (e.g., by using some variational properties of $\lambda_2$)? Perhaps, the answer is known in the radially-symmetric case?
I'll appreciate any hint.
As far as I can tell, although it is widely believed that the Courant–Hibert estimate holds for the fractional Laplacian, even finiteness of the number of nodal domains of the second eigenfunctions is open for a general domain. The only known result is that the "harmonic extension to the upper half-space" (in the sense of Caffarelli–Silvestre for $\alpha \ne 1$) has at most two nodal domains.
For a ball (and – if I am not mistaken – more generally, for any domain invariant under rotations), the second eigenfunction is either radial or antisymmetric. For a ball, it is conjectured that it is always antisymmetric, but the proof is only known for low dimensions; see my paper with Bartłomiej Dyda and Alexey Kuznetsov, DOI:10.1112/jlms.12024, and a follow-up by Rui A.C. Ferreira DOI:10.1007/s00030-019-0554-x.
If the second eigenfunction is antisymmetric, a rather standard Perron–Frobenius-type argument shows that it has two nodal domains. If it happens to be radial, it could have up to three nodal domains, again by the harmonic extension argument and an application of the classical Courant–Hilbert theorem. In any case, the number of nodal domains of the second eigenfunction is at most three.
Thanks a lot for the fast and comprehensive answer!
Mateusz, could you, please, indicate a reference for the statement that "harmonic extension to the upper half-space has at most two nodal domains"?
This is proved by Bañuelos and Kulczycki (perhaps only when $\alpha = 1$, I do not remember well). A closely related result (for operators with potentials rather than in a domain) is used by Frank, Lenzmann and Silvestre in DOI:10.1002/cpa.21591, and there is a short discussion after the statement of Theorem III in my paper with Jacek Mucha, DOI:10.1007/s00028-018-0444-4.
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2025-03-21T14:48:29.593018
| 2020-01-07T00:57:07 |
349876
|
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|
Stack Exchange
|
Why is a convex variety called convex?
Let $X$ be a smooth projective variety. By Definition 24.4.2 in the 2003 book Mirror Symmetry, $X$ is called convex if $h^1\left( \Sigma, f^*T_X \right) = 0$ for every genus zero stable map $f:\Sigma \to X$.
Question: Why is a convex variety called convex?
On page 505 of the book, it was said that $H^1\left( \Sigma, f^*T_X \right) $ measures obstructions to the deformations of the map $f$ (when the structure of the source curve $\Sigma$ is fixed). But I don't know what the obstructions mean, or what zero obstruction means. So I'm also asking the following
Related question: What do zero and nonzero obstructions to the deformations of the map $f$ mean?
My background and what answer I'm looking for: I have close-to-zero background in deformation theory. So I don't expect to fully understand the precise mathematical meaning over one night. I would appreciate it very much if someone could explain what zero and nonzero obstructions mean in an intuitive way. If this can be further supported by concrete examples for zero and nonzero obstructions, then it will be perfect.
If there are no obstructions to deformation, i.e. $h^1=0$, then Kodaira's deformation theorem says that the map admits a family of deformations, forming a manifold, near that given map, with tangent space given by $H^0$. You think of genus zero curves as like lines, or line segments, and deforming a line while staying in the variety should be like trying to move around on a set of points in Euclidean space while still being able to move your line segment with you, something you can do when points of the set are connected by line segments, i.e. in a convex set.
For example, if $f$ is the identity map of the projective line, then the space of deformations of $f$ is the space of maps of degree 1, i.e. the space of projective linear transformations of the projective line. The tangent space at the identity element of the group of projective linear transformations is the Lie algebra of vector fields on the projective line, i.e. $H^0$ is the space of linear vector fields, i.e. the space of $2 \times 2$ traceless matrices, so 3 dimensional.
Another example: if $f$ is constant, then the deformations are constant maps, and if the target $X$ is smooth at the image point $x=f(\mathbb{P}^1)$, then $H^0=T_x X$.
Kodaira's book on deformation theory is where I learned about this stuff. Maybe there are better references, especially in the algebraic category.
Thanks for the explanation. It is very intuitive and helpful. Also thanks for the reference. Can you say something about $H^1(\Sigma, f^*T_X)$ in the following two cases? Case 1: The map $f$ is a constant, i.e., it maps the source curve to a point. Case 2: The map $f$ is an identity from the line $\mathbb{CP}^1$ to itself.
So both cases have zero obstructions, right?
In case one, $f^T_X$ is trivial, so $H^1 = g(\Sigma) = 0$. In case 2, $f^ T_X = \mathcal O_{\mathbb P^1}(2)$ so $H^1 = 0$ by standard calculations. So yes, both of these are unobstructed.
@TabesBridges In case one, did you get the result from the Riemann-Roch theorem, i.e., $h^0(\Sigma,L) -h^1(\Sigma,L) = \deg L -g-1$ where the line bundle $L = f^*T_X \cong \Sigma \times \mathbb{C}$? In case 2, excuse my ignorance but what did you mean by standard calculations?
@YuhangChen First, the last term on the RHS of Riemann-Roch is $+1$, not $-1$. If you plug $h^0(\Sigma, L) = 1$ and $\deg L = 0$ (which uniquely determine that $L$ is trivial), you get $1 - h^1(\Sigma,L) = 0 - g + 1$, which does indeed simplify to $h^1 = g$, but in general $h^1(C,\mathcal O_C)$ is one way of computing the genus of a smooth, projective curve. In case 2 I'm referring to the computation of the cohomology of line bundles on projective space. This particular fact can also be recovered from Serre duality, which identifies $h^1(C,L) = h^0(C, \omega_C\otimes L^\vee)$ which....
@YuhangChen... in this case is $h^0(\mathbb P^1, \mathcal O_{\mathbb P^1}(-2)\otimes \mathcal O_{\mathbb P^1}(-2)) = h^0(\mathbb P^1, \mathcal O_{\mathbb P^1}(-4)) = 0$, where $\omega_C$ is the canonical bundle (here the same as the holomorphic cotangent bundle), and we use that negative degree line bundles have no sections.
@YuhangChen Sorry, I suppose a constant map to a variety $X$ will actually have $f^* T_X$ trivial of rank $\dim X$ (too late to edit previous comment). So instead the RR calculation reads $\dim X - h^1(\Sigma,f^* T_X) = (\dim X)(1-g)$, which reduces again to $h^1(\Sigma, f^* T_X) = 0$.
@TabesBridges Thanks for the detailed explanations. The application of Serre duality to compute $h^1$ is pretty neat!
@YuhangChen this is a trick you should get familiar with. The only way I can remember the full calculation of cohomology for line bundles on $\mathbb P^n$ is via the slogan "line bundles of non-negative degree have global sections. the only other cohomology is $H^n$ forced by those global sections and Serre duality."
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2025-03-21T14:48:29.593339
| 2020-01-07T01:30:14 |
349877
|
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|
Stack Exchange
|
A binary operation on vector bundles that adds Chern classes?
Let $E$ and $F$ be two complex vector bundles over a space $X$. There's a fairly well-known binary operation called the direct sum, written $E\oplus F$, which has the property that its first Chern class is the sum of the Chern classes of the constituents: $c_{1}(E\oplus F)=c_{1}E+c_{1}F$.
My question is: For $k\geq 1$, are there binary operations $B^{k}(E,F)$ with the property that $c_{k}(B^{k}(E,F))=c_{k}E+c_{k}F$?
I am happy to allow the binary operation to take values in virtual bundles if need be.
Motivation: These operations would of course be pretty useful for contstructing (virtual) vector bundles with desired Chern classes.
Let's work with virtual bundles. Your question is equivalent to the following:
If we fix a $k \geq 1$, does the map $BU \times BU \rightarrow K(\mathbb{Z},2k)$ representing $c_k \otimes 1 + 1 \otimes c_k$ factor through some map $BU \times BU \rightarrow BU$ composed with the map $BU \rightarrow K(\mathbb{Z},2k)$ representing $c_k$. Recall that $BU$ has a CW structure with cells in every even dimension, and that $c_k$ is represented by a single cell. From this description of $c_k$, the second map $BU \rightarrow K(\mathbb{Z},2k)$ can be constructed by extending the map $BU_{2k}/BU_{2k-1} \rightarrow K(\mathbb{Z},2k)$ ($BU_{2k}$ meaning the 2k-skeleton) which on the summand corresponding to $c_k$ represents a generator of $\pi_{2k}(K(\mathbb{Z},2k))$ and elsewhere is constant, to $BU/BU_{2k-1}$ and then precomposing with the quotient $BU \rightarrow BU/BU_{2k-1}$.
Similarly, $c_k \otimes 1 + 1 \otimes c_k$ can be represented in the same manner where the nontrivial maps in the bouquet of spheres correspond to the cells $c_k \otimes 1$ and $1 \otimes c_k$.
For a CW complex X with cells in only even dimensions and a single 0-cell, there is a one to one correspondence between $[X,BU]_*$ and the direct product generated by by the non-basepoint cells of $X$. This is given by letting the coordinate corresponding to the d-cell, which we name e, take $[f]:X \rightarrow BU$ to the class $S^d \xrightarrow{e} X_d/(X_{d-1}-e) \xrightarrow{f} BU$ (see Atiyah's K-theory Proposition 2.5.2).
In this case we consider $BU \times BU/ (BU \times BU)_{2k-1}$, and we choose an element of the product which in the coordinates corresponding to $c_k \otimes 1$ and $1 \otimes c_k$ is represented by the generator of $\pi_{2k}(BU)$. Let's call the corresponding map $\theta$ from $BU \times BU \rightarrow BU \times BU/ (BU \times BU)_{2k-1} \rightarrow BU$. Up to homotopy we then have a factorization of $c_k \otimes 1 + 1 \otimes c_k$ as $c_k \circ \theta$ because the maps $\pi_{2k}(BU \times BU/ (BU \times BU)_{2k-1}) \rightarrow \pi_{2k}(K(\mathbb{Z},2k))$ are the same. This relies on the (I think true) statement that generator of $\pi_{2k}(BU)$ under the composite $\pi_{2k}(BU) \rightarrow\pi_{2k}(BU /BU_{2k-1}) \rightarrow H_{2k}(BU /BU_{2k-1})$ lands on the cell which represents $c_k$.
$\bf{Edit:}$ Unfortunately, I've read that this is untrue. Instead it lands on some multiple of the dual of the Chern class. Denote this multiple $a_k$.
So to apply the operation to the virtual bundles $E,F$ you take the composite $\theta \circ (E,F)$. As well as having the property $c_k(\theta \circ (E,F))=a_k (c_k (E)+c_k(F))$, we also have that $c_l=0$ for $0<l<k$.
Thanks for the answer!. Is $a_{k}$ equal to the euler class of the bundle on $S^{2k}$ represented by the generator of $\pi_{2k}BU$?
That seems to be correct.
For anyone who's interested, $a_{k}=1$ if $k=1,2,4$ and $a_{k}=2$ otherwise, as computed in "Realizing cohomology classes as Euler classes" by A. Naoleka. Always a pleasure to see the Hopf-invariant -1 pattern showing up
It seems to me that $a_k$ is instead the $k$-th Chern class of the generator of $\pi_{2k}(BU)$, which is $(k-1)!$. This is essentially why the number $6 = (4-1)!$ shows up in my answer.
@BertramArnold By euler class I had actually meant $c_{k}$. Could you explain why that's $(k-1)!$ for the generator of $\pi_{2k}BU$?
@ConnorMalin: Is the conclusion of your answer that such a binary operation $B^k(E, F)$ exists if and only if $a_k = 1$?
@MichaelAlbanese I admit I am skeptical of my answer now. For one $S^d \xrightarrow{e} X_d/(X_{d-1}-e) \xrightarrow{f} BU$ does not make sense and should rather be $S^d \xrightarrow{e} X_d/(X_{d-1}-e) \xrightarrow{f} BU/BU_{d-1}$. So it seems like there is some flaw in my understanding of Atiyah's proposition. I'll add that as written in Atiyah, the proposition cannot be correct since it says that instead of a direct product, it is a direct sum over all the cells.
But after talking with Mike Miller, I think the construction I gave (upon correction) is the following: take your virtual bundles $E,F$over $X$ and use the Bott isomorphism to get virtual bundles over $\Sigma^2 X$. Now over a suspension it is true that $c_k(E \oplus F)=c_k(E)+c_k(F)$, so add these bundles in $\Sigma^2 X$ and then transport them back down to $X$ via the Bott isomorphism. Then the result is that the chern class of this new bundle over $X$ is $a_k(c_k(E)+c_k(F))$ for all $k>0$. And this $a_k=(k-1)!$
@JohnGreenwood In case you want to give one of the other posts the title of "answer".
@ConnorMalin Ah that's a very very nice way to say it. Could you add that to your answer maybe? Also could you explain the $a_{k}=(k-1)!$ ? Somehow that's still baffling me...
@MichaelAlbanese I would also like to know a reference for it. But the two people I talked to (aside from you) agreed that the $c_k$ of the generator of $\pi_{2k}(BU)$ was indeed $(k-1)!$. Perhaps it follows from some manipulations with the Chern character.
Ah, I meant to tag John in that last comment. I will think more about the $a_k$ and see if I can justify it.
@ConnorMalin I thought so :). I think i figured out why the $c_{k}$ of the generator is $(k-1)!$ - it's given by the $k-1$-fold (external) tensor product with the Bott class $[L]-1$ in $\tilde{K}(S^2)$. The you plug and chug with the rules for chern classes of direct sums and tensor products of line bundles.
The factor shows up since the $k$-th component of the Chern character is $(-1)^{k+1}\frac{c_k}{(k-1)!} + $polynomial in $c_1,\dots,c_{k-1}$. The factor is $\frac{1}{k!}$ times the coefficient of $x_1\dots x_n$ in the expression of $x_1^k+\dots+x_k^k$ in terms of symmetric polynomials, which is $(-1)^{k+1} k$ (compare https://en.wikipedia.org/wiki/Newton%27s_identities#Formulation_in_terms_of_symmetric_polynomials ). Thus $\frac{c_k((L-1)^{\boxtimes k})}{(k-1)!}$ generates the image of the Chern character in $H^{2n}(S^{2n};\mathbb Q)$ which by multiplicativity is $H^{2n}(S^{2n};\mathbb Z)$.
There is such an operation for $k = 2$ using virtual bundles.
Note that $c_2(E\oplus F) = c_2(E) + c_1(E)c_1(F) + c_2(F)$ so
\begin{align*}
c_2(E) + c_2(F) &= c_2(E\oplus F) - c_1(E)c_1(F)\\
&= c_2(E\oplus F) - c_1(\det E)c_1(\det F)\\
&= c_2(E\oplus F) - c_2(\det E \oplus \det F).
\end{align*}
Recall that the second Chern class of a virtual vector bundle $A - B$ is given by
$$c_2(A - B) = c_2(A) - c_2(B) - c_1(A)c_1(B) + c_1(B)^2$$
which reduces to $c_2(A) - c_2(B)$ if $c_1(A) = c_1(B)$. As
$$c_1(E\oplus F) = c_1(E) + c_1(F) = c_1(\det E) + c_1(\det F) = c_1(\det E\oplus\det F),$$
the above computation shows
$$c_2(E) + c_2(F) = c_2(E\oplus F - \det E\oplus\det F).$$ That is, we can take $B^2(E, F) = E\oplus F - \det E\oplus \det F$.
Thanks, this is wonderfully concrete and easy to apply in practice!
Such an operation with values in bundles does not exist for $k = 4$ and the base space $\mathbb{HP}^2$. For virtual vector bundles, it depends how you extend the definition of the Chern classes; for the natural extension such that the total Chern class is multiplicative under sums, it also doesn't exist (in fact, this extension is used in the proof).
Pullback along the evident inclusion $\mathbb{HP}^n\hookrightarrow \mathbb{HP}^\infty\cong BSU(2)$ gives a ring homomorphism $R(SU(2))\to K^0(\mathbb{HP}^n)$ preserving the augmentation ideals $I$. Since $\mathbb{HP}^n$ is covered by $n+1$ contractible sets (the standard coordinate charts), we have $I^3_{K^0(\mathbb{HP}^n)} = 0$, and therefore an induced morphism $\phi_n:R(SU(2))/I_{R(SU(2))}^{n+1}\to K^0(\mathbb{HP}^n)$. Now the representation ring $R(SU(2))$ is additively generated by the representations $[V_n]$, of dimension $n+1$, subject to the relations $[V_m]\otimes [V_n] = [V_{|m-n|}] + [V_{|m-n|+2}] + \dots + [V_{m+n}]$. Setting $v = [V_1]-2$, it follows easily that $I_{R(SU(2))}^n/I_{R(SU(2))}^{n+1}$ is generated by $v^n$; multiplicativity of the Atiyah-Hirzebruch spectral sequence then shows that $\phi_n$ is an isomorphism. (This calculation is a special case of the Atiyah-Segal completion theorem.)
Now $I_{R(SU(2))}/I_{R(SU(2))}^3$ is generated by the classes of $V_1-2$ and $V_2-3$, and their images $E_1-2$ and $E_2-3$ define virtual vector bundles over $\mathbb{HP}^2$ whose classes generate $\widetilde K^0(\mathbb{HP}^2)$. Let's calculate their total Chern classes: $E_1$ is just the canonical $2$-dimensional bundle over $\mathbb{HP}^2$, so its total Chern class is $c(E_1) = 1 + x$ where $x\in H^4(\mathbb{HP}^2;\mathbb Z)$ is a generator. Since $E_2$ is $3$-dimensional, only its second Chern class can be nonzero, and one easily calculates $c(E_2) = 1 + c_2(E_2) = 1 + 4x$: we have $E_2\oplus \mathbb C = E_1^{\otimes 2}$, so pulling back along the map $\iota:\mathbb{CP}^2\to \mathbb{HP}^2$, noting that $\iota^*E_{1} = L\oplus L^{-1}$ splits as a sum of the canonical line bundle and its dual and setting $u = c_1(L)$ we have
\begin{align*}
\iota^*c(E) &= (1+u)(1-u) = 1-u^2\\
\implies \iota^*x &= -u^2\implies \iota^*|_{H^4}\text{ is injective}\\[12pt]
\iota^*E_2 &= (\iota^*E_1)^{\otimes 2} - 1 \\
&= (L\oplus L^{-1})^{\otimes 2} - 1 \\
&= L^{\otimes 2} + L^{\otimes -2} + 1\\
\iota^*c(E_2) &= (1 + 2u)(1 - 2u)\\
&= 1 - 4u^2
\end{align*}
The total Chern class can be extended to a natural transformation $K^0(X) \to \bigoplus H^{2*}(X;\mathbb Z)$ by $[E] - [F]\mapsto c(E)c(F)^{-1}$. We immediately obtain
\begin{align*}
c(k\cdot E_1 + l\cdot E_2) &= (1+x)^k(1+4x)^l \\
&= 1 + (k + 4l)x + \left(\frac{k(k-1)}{2} + 4kl + 8l(l-1)\right)x^2\\
&= 1 + (k+4l)x + \left(\frac{(k+4l)(k+4l-1)}{2} - 6l\right)x^2
\end{align*}
In particular, no cohomology class of the form $(3n+2)x^2$ with $n\in\mathbb Z$ can be the fourth Chern class of a (virtual) bundle over $X$, so there is no (virtual) bundle $F$ with $c_4(F) = c_4(E_1^{\oplus 2}) + c_4(E_1^{\oplus 2})$.
Thanks, this is nice! I took the liberty to change some minor typos.
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2025-03-21T14:48:29.593973
| 2020-01-07T03:23:37 |
349880
|
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|
Stack Exchange
|
Classifying space for fibrations with Eilenberg-MacLane space fibers and nontrivial fundamental group actions
Let $A$ be an abelian group and let $n \geq 2$. For any connected CW complex $X$, it is standard that a fibration $f\colon E \rightarrow X$ whose fibers are homotopy equivalent to a $K(A,n)$ is fiberwise homotopy equivalent to the pull-back of the loop-space fibration over a $K(A,n+1)$ if and only if $\pi_1(X)$ acts trivially on the fibers of $f$ (up to homotopy equivalence).
Question: Consider the more general set of fibrations $f\colon E \rightarrow X$ whose fibers are homotopy equivalent to a $K(A,n)$, but with a possibly nontrivial monodromy action, up to fiber homotopy equivalence. Is this functor representable?
From the first paragraph, if it is representatable by a space $Z$ then the universal cover of $Z$ is a $K(A,n+1)$.
Of course, there are many variants here (e.g. of the definition of a fibration, or the equivalence relation on the set), and I'd be interested in any of these.
Do the hypotheses of Brown representability not apply?
@JohnGreenwood: It's not clear to me. The issue is that you have to be able to glue fibrations that are isomorphic on a subcomplex, and since the gluing map is just a fiberwise homotopy equivalence (and thus not a homeomorphism) this seems subtle.
(if we could assume that these fibrations were actually locally trivial fiber bundles and isomorphisms were fiber bundle isomorphisms, then this would not be a problem; however, I don't know if this is possible).
My mistake! Gluing fibrations seems tricky...what if you glue with the homotopy equivalence anyway, and then force the result to be a fibration?
Mark Grant's excellent answer already resolves the question. However, let me sketch how this arises as a special case of the more general problem of classifying fibrations with a given fiber.
For any space $X$, the homotopy automorphisms $\operatorname{hAut}(X)$ are defined as the self-homotopy equivalences of $X$ (topologized with the compact-open topology; note that they form a union of path components of the space of all self-maps of $X$). They form a group-like monoid, so there is a connected pointed space $B\operatorname{hAut}(X)$ such that there is an equivalence of $A_\infty$-spaces $\operatorname{hAut}(X)\simeq \Omega B\operatorname{hAut}(X)$. In fact, $B\operatorname{hAut}(X)$ can be built as the geometric realization of the simplicial space $B(*,\operatorname{hAut}(X),*) = * \leftarrow \operatorname{hAut}(X) \Leftarrow \operatorname{hAut}(X)\times\operatorname{hAut}(X) \Lleftarrow \dots$. Since $\operatorname{hAut}(X)$ acts on $X$ from the left, we can also build the simplicial space $B(*,\operatorname{hAut}(X),X) = X\leftarrow \operatorname{hAut}(X)\times X\Leftarrow\dots$, and the map $X\to *$ induces a fibration $X\to E_X\to B\operatorname{hAut}(X)$. This is the universal fibration with fiber $X$, that is, for every fibration $X\to F\to Y$ there is a unique homotopy class of maps $f: Y\to B\operatorname{hAut}(X)$ such that $F\simeq f^*E_X$.
If $X = K(A,n)$ is an Eilenberg-MacLane space, the grouplike monoid $\operatorname{hAut}(X)$ can be described quite explicitly: In this case, $X$ can also be given the structure of a grouplike monoid with identity $e$, so that the map $\operatorname{hAut}(X)\to X, f\mapsto f(e)$ has a homotopy right inverse given by sending $x$ to (left, say) translation by $x$. Thus there is a homotopy equivalence $\operatorname{hAut}(X)\simeq \operatorname{hAut}_*(X)\times X$, where $\operatorname{hAut}_*(X)$ is the submonoid of self homotopy equivalences that preserve the basepoint (note, however, that this is not compatible with the monoid structure). Now it is straightforward to show that $\operatorname{hAut}_*(K(A,n))\simeq K(\operatorname{Aut}(A),0)$ is homotopy equivalent to a discrete space, and there is a retraction $\operatorname{hAut}(X) \to \pi_0(\operatorname{hAut}(X))\cong \operatorname{Aut}(A)$. All in all, we obtain an equivalence of grouplike monoids
$$
\operatorname{hAut}(K(A,n))\simeq \operatorname{Aut}(A)\ltimes K(A,n)
$$
It follows that $B\operatorname{hAut}(K(A,n))$ has exactly two nonvanishing homotopy groups, namely $\pi_1(\operatorname{hAut}(K(A,n)))\cong \operatorname{Aut}(A)$ and $\pi_{n+1}(\operatorname{hAut}(K(A,n)))\cong A$, with the evident action of $\pi_1$ on $\pi_{n+1}$. In particular, a map $f:Y\to B\operatorname{hAut}(K(A,n))$ is determined by $\rho\in H^1(Y;\operatorname{Aut}(A))$, which determines a local system $A_\rho$ on $Y$, and a cohomology class $[f]\in H^{n+1}(Y;A_\rho)$.
As a side note, for bundles whose fibers are $K(G,1)$ with $G$ is a nonabelian groupthere there is a similar analysis. But then $\pi_0(hAut)$ turns out to be the outer automorphism group $Out(G)$ and $\pi_1(hAut)$ is the center $Z(G)$.
There's also work of Wirth, summarised in this paper joint with Stasheff: https://arxiv.org/abs/math/0609220 on homotopy locally trivial fibrations.
So is it right to think that, morally, the difference between fiber bundles and fibrations is the structure "group" being an actual group versus a group-up-to-homotopy?
This is wonderful! Do you know a good place to read about this? (ps: sorry for only replying now -- I am not on the internet very often).
(pps: I'll accept your answer once I get a reference.)
For the (nontrivial) fact that $B\operatorname{hAut}(X)$ classifies fibrations with fiber $X$, the paper linked by David Roberts seems like a good starting point. A more modern perspective is to consider the base $B$ as an $\infty$-groupoid; by (un)straightening, a fibration over $B$ corresponds to a functor from $B$ to spaces, and the fiber over a point is the space it is sent to. It follows formally that fibrations with fixed fiber are represented by a connected component of the object classifier, the maximal sub-$\infty$-groupoid of spaces.
References for these concepts can be found here: https://ncatlab.org/nlab/show/%28infinity%2C1%29-Grothendieck+construction#GrpdFibsOverGrpds https://ncatlab.org/nlab/show/%28sub%29object+classifier+in+an+%28infinity%2C1%29-topos As for the actual calculation of $\operatorname{hAut}(K(A,n))$, see this questions and its answer: https://mathoverflow.net/questions/222914/automorphisms-of-eilenberg-mac-lane-spaces-and-semidirect-products-and-the-odd
Ugh, I have to learn about $\infty$-groupoids to understand something this classical? I'm a little skeptical that this is really necessary (or that it really gives the same answer when you're dealing with ordinary spaces rather than simplicial objects), but David Robert's comment seems like a good place to start learning the old-fashioned way to think about this, so I'll accept this answer.
Denote $\pi=\pi_1(X)$ and fix the monodromy action $\rho:\pi\to \operatorname{Aut}(A)$. There is a generalized Eilenberg-Mac Lane space $L_\rho(A,n+1)$, whose only non-trivial homotopy groups are $\pi_1(L_\rho(A,n+1))=\pi$ and $\pi_{n+1}(L_\rho(A,n+1))=A$ and such that the action of $\pi_1$ on $\pi_{n+1}$ is given by $\rho$. The construction appears in Section 7 of
Gitler, Samuel, Cohomology operations with local coefficients, Am. J. Math. 85, 156-188 (1963). ZBL0131.38006.
Essentially, $L_\rho(A,n+1)$ is the Borel construction $E\pi\times_\rho K(A,n+1)$ done with simplicial sets, so that everything is sufficiently natural (and the action of $\pi$ on $A$ gives an action of $\pi$ on $K(A,n)$ for all $n$). The path loop fibration is equivariant, giving a fibration sequence
$$
K(A,n)\to E\pi\times_\rho PK(A,n+1) \to E\pi\times_\rho K(A,n+1)=L_\rho(A,n+1).
$$
I would bet that fibrations over $X$ with fibre $K(A,n)$ and monodromy $\rho$ are classified by maps (Edit: from $X$) into $L_\rho(A,n+1)$ (Edit: inducing the identity on $\pi_1$) using this construction.
Arbitrary maps? Maybe rather only those inducing isomorphism on $\pi_1$, or something like that?
You're right, of course. For example, pulling back by the trivial map would give a fibration with trivial monodromy. In fact, the monodromy of the pullback should (I think) be the pullback of the monodromy, so we'd want the map to induce the identity on $\pi_1$; see my edit.
Bertram's answer is more general, and probably does a better job of answering the quesiton asked.
One place this was done but in greater generality was by Blakers, Annals of Matyh.2nd series, 49 (2) 428-461. He defines what he calls "group systems". These are now called "crossed complexes". An account is in the book Nonabelian Algebraic Topology (EMS, 2011). A particular case $C$ is where the crossed complex has $C_0$ a singleton, $C_1$ is a group, say $G$, for some $n>1$, $C_n$ is a $G$-module, and otherwise $C_n$ is trivial.
Any crossed complex $C$ has a classifying space $BC$ whose simplicial definition in the case $C_0$ is a point is due to Blakers. Any filtered space, $X_*$, and in particular any, CW-complex $X$ with skeletal filtration $X_*$, has a fundamental crossed complex $\Pi X_*$, (also due to Blakers). In particular the cubical nerve $NC$ of $C$ can be defined in dimension $n$ by $$(NC)_n = Crs(\Pi I^n_*, C), $$ where $I^n$ is the standard $n$-cube. (The simplicial version of this was published with P.J. Higgins as Math. Proc. Camb. Phil. Soc. 110 (1991) 95-120.)
(I confess I rather suspect the homotopy classification theorems of our book have to be restricted to $X$ finite dimensional or $BC$ having only a finite number of non trivial homotopy groups,)
|
2025-03-21T14:48:29.594547
| 2020-01-07T05:36:11 |
349882
|
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|
Stack Exchange
|
Factor group isomorphic with Klein four-group
Let $G$ be a finite solvable group and $N$ be a normal subgroup of it which is an elementary abelian 2-group. Suppose that $G/N\cong \mathbb{Z}_2\times \mathbb{Z}_2$ and $|C_G(x)|=16$ for any $x\in G-N$. What can we say about $G$? Is it abelian? Is it elementary?
We know that since $G$ is a 2-group, it is nilpotent.
Could you give some motivation for this question? It looks rather specific. Certainly there is no reason for $G$ to be abelian, e.g. you could take a dihedral or quaternion group of order 8 and then stick on an elementary abelian direct factor in order to make the centralizers have the specified order.
I would provide some examples, on which you may build some expectations:
$G=C_2\times C_2\times D_4$, where $D_4$ is the dihedral group of $8$ elements. $N=C_2\times C_2\times C_2$ is the center of $G$. Since $D_4$ mod its center is $C_2\times C_2$, the factor group condition holds. For $x \in G-N$, $|C_G(x)|$ contains the $N$ and $x$, so its size is strictly larger than $8$. Also, $|C_G(x)|\neq 32$, otherwise $x$ is in the center. So $|C_G(x)|=16$ for any element not in the center.
$G$ can be replaced by $C_2\times C_2\times Q_8$ ($Q_8$ is the quarernion group). The reasons are the same.
There are three more groups which can play the role of $G$, and $N$ their (respective) centralizers. The structure of $N$ and $G/N$ can be read off from the webpage; and for the size of the centralizer, you can use the formula $|C(x)|=\sum_{χ}|χ(x)|^2$, where $x$ is a conjugacy class and $χ$ runs over the group characters.
|
2025-03-21T14:48:29.594691
| 2020-01-07T06:04:39 |
349883
|
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|
Stack Exchange
|
Numerically solve a specific saddle-point problem
Let $(\Omega,\mathcal E,\mu)$ be a probability space, $k\in\mathbb N$, $$W:=\left\{w:E\to[0,\infty)^k:\sum_{i=1}^kw_i=1\;\mu\text{-almost surely}\right\},$$ $G$ be a finite nonempty set and $a^{(g)}:E\to[0,\infty)^k$ be $\mathcal E$-measurable for $g\in G$.
How can we solve the saddle-point problem $$\inf_{w\in W}\sup_{g\in G}\sum_{i\in I}\int a^{(g)}_iw_i\:{\rm d}\mu\tag1$$ (at least numerically)?
Note that $W$ is a closed, convex subset of $\mathcal L^r(\mu;\mathbb R^k)$, with empty interior, contained in the unit closed ball, for all $r\ge1$.
Note that for fixed $g\in G$, a minimizer $w\in W$ of $\sum_{i\in I}\int a^{(g)}_iw_i\:{\rm d}\mu$ is given by $$w^{(g)}_i(x):=\delta_{ij^{(g)}(x)}\;\;\;\text{for }x\in E\text{ and }i\in\{1,\ldots,k\},$$ where $\delta$ denotes the Kronecker delta and $$j^{(g)}(x):=\min\underset{i\in I}{\operatorname{arg\:min}}\:a^{(g)}_i(x)\;\;\;\text{for }x\in E$$ ($\operatorname{arg\:min}$ is treated as being set-valued and we break ties by selecting the smallest index).
Remark: I'm actually not interested in the $G$-component of a saddle-point, i.e. I'm only interested in a minimizer $w\in W$ of $\sup_{g\in G}\sum_{i\in I}\int a^{(g)}_iw_i\:{\rm d}\mu$.
The formulation of your question is too general. Numeric optimizers deal with concrete functions only.
@user64494 I've previously asked for the specific problem: https://mathoverflow.net/q/348766/91890. But since I didn't get an answer (even after a bounty), I've tried to simplify the description. Actually, only the concrete shape of the $a^{(g)}$ is missing, $$a_i^{(g)}(x):=\int_{{:p:>:0:}}\lambda({\rm d}y)\frac{|g(y)|^2}{q_i(x,y)};;;\text{for }x\in E\text{ and }i\in{1,\ldots,k},$$ and it is assumed that $\mu$ has a density $p$ with respect to a reference measure $\lambda$.
|
2025-03-21T14:48:29.594941
| 2020-01-07T08:59:07 |
349888
|
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|
Stack Exchange
|
maximum weighted matching with weights being sets
Given a set $S$ and a bipartite graph $G$, each edge $v\in E(G)$ covers a subset $S_v$ of $S$. My problem is to find a matching maximizing the number of covered elements, i.e., denote $V$ the set of edges in the matching, we seeks to maximize $|\bigcup_{v\in V} S_v|$.
Your problem is NP-complete as we can translate instances of set-covering problems to your problem:
Let $S$ be a family of sets, its members be subsets of $\{1,2,...,n\}$. The set-covering problem is:
Does there exists $m$ members of $S$ whose union is $\{1,2,...,n\}$?
To translate the instance to your problem, construct a bipartite graph with vertices $\{1,2,...,m\} \cup S$ and edges the elements of $\{1,2,...,m\}\times S$. Assign the set $s$ to the edge $(k,s)$, where $k$ is a number and $s$ is a member of $S$. Thus any matching corresponds to $m$ members of $S$, and maximum number of covered elements can be achieved if and only if there exists $m$ members of $S$ with union $\{1,2,...,n\}$.
As the set-covering problem is NP-complete, your problem is also NP-complete.
Thank you. It seems logical to me that the problem is NP-complete. Any clue on designing approximation algorithm?
Using linear programming may be possible, by combining the LP formulations of bipartite maximum weight matching and set covering.
does edmonds-blossom or karp algorithm help here?
Thank you. edmonds-blossom or karp algorithm cannot be directly applied here as the construction of augmenting path is by nature different.
Just wonder whether a related formulation is tractable or not: now I am interested in deciding whether there exists a matching of $G$ that covers all the elements of $S$.
|
2025-03-21T14:48:29.595096
| 2020-01-07T09:21:51 |
349892
|
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|
Stack Exchange
|
Integration of a particular rational expression
I am trying to solve the following integration, where $a,b,c,d,e$ and $f$ are constants:
$$I=\int\frac{x^4+ax^3+bx^2+cx+d}{x^3(x^3+ex+f)}dx$$
I tried to solve the integral using the following two methods, but both seemed to be very much complicated:
Method 1:
Using partial fraction decomposition by calculating the three roots of the denominator. Among the three roots, one root is real and the other two roots are complex (both are complex conjugates of each other). However, the roots are too much complicated and are as follows:
Method 2:
I tried to expand the denominator using binomial series as follows:
$$I=\int\frac{\frac{1}{x^2}+\frac{a}{x^3}+\frac{b}{x^4}+\frac{c}{x^5}+\frac{d}{x^6}}{1+\frac{e}{x^2}+\frac{f}{x^3}}dx$$
Then writing $\epsilon=\frac{e}{x^2}+\frac{f}{x^3}$, the above integral becomes
$$I=\int\left(\frac{1}{x^2}+\frac{a}{x^3}+\frac{b}{x^4}+\frac{c}{x^5}+\frac{d}{x^6}\right)\left(1+\epsilon\right)^{-1}dx$$
For a quite good approximation, it is required to expand the binomial series up to $\epsilon^{11}$, i.e. $40$ terms in the expression for the integral and this is too much cumbersome.
Even Mathematica expresses the result as a conditional expression and it is required to provide the range of the constants $a,b,c,d,e$ and $f$ to obtain the exact expression.
QUESTION:
Is it possible to solve the integration analytically using any other suitable method? If no such methods are possible, do there exist any approximation technique that might work?
You can start by reducing the degree of the numerator using polynomial division.
@Nemo Should I divide the numerator by the cubic polynomial in the denominator? Otherwise, the degree of the denominator is more than the numerator.
Yes, that's what I meant.
There is simplified form of the roots of the cubic equation that might help https://en.wikibooks.org/wiki/Trigonometry/The_solution_of_cubic_equations
there is really no hope for a compact closed-form expression; if some of your parameters are small, you might exploit that, but otherwise why not just evaluate the integral numerically over the desired interval?
Too long for a comment. You should discuss on the denominator, which has 0 as a triple root if $f\not=0$. Assuming this, you have three non-zero roots for $X^3+eX+f$ and you have explicit formulas to calculate them. Reducing the rational fraction in simple elements is then easy.
|
2025-03-21T14:48:29.595291
| 2020-01-07T09:26:45 |
349893
|
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|
Stack Exchange
|
Absolutely continuous coupling of probability measures
I have a Borel probability measure $\pi$ on $\mathbb{R}^{n+1}$ such that $\pi_1=\mu_1, \ldots, \pi_{n+1}=\mu_{n+1}$ for some fixed Borel probability measures $\mu_1, \ldots, \mu_{n+1}$ (where each $\mu_i$ is absolutely continuous with respect to Lebesgue measure). I want to construct a sequence of probability measures $\pi^{(k)}$ such that $\pi^{(k)}$ converges weakly to $\pi$ and has the same marginals as $\pi,$ that is, $\pi^{(k)}_j=\mu_j$ for $j=2,\ldots, n+1.$
Now of course, I can take $\pi^{(k)}=\pi.$ But, I am trying to prove some inequality where I need $\pi^{(k)}$ to be absolutely continuous with respect to Lebesgue measure (on $\mathbb{R}^n$).
Now I have a twofold goal: I can make $\pi^{(k)}$ absolutely continuous but then in the process I end up messing with the marginals. Can someone point it out to me if it is possible at all to cook up such a sequence $\pi^{(k)}$ which is absolutely continuous with respect to Leb, and has the fixed marginals and which weakly converges to $\pi.$
All the answers below are pretty good, but unfortunately I can accept only one answer. I am taking some time to go through all the references I have got in the answer before I accept one answer as the final.
Let me formulate and prove it in greater generality (which actually makes your question easier). Let $X$ be a metric space, and $\mu$ be a probability measure on $X\times X$ (for simplicity I consider the product of two copies of $X$ only; the general case is precisely the same). You want to obtain a sequence of measures $\nu_n$ on $X\times X$ which
(1) have the same marginals $\mu_1,\mu_2$ as $\mu$;
(2) are absolutely continuous with respect to the product measure $\mu_1\times\mu_2$;
(3) weakly converge to $\mu$.
For each $n$ take a countable partition of $X$ into measurable sets $X^i$ with diameter $\le 1/n$ (presuming the space $X$ is such that partitions like this exist for any $n$) and denote by $\mu_\epsilon^i$ the normalized restriction of the measure $\mu_\epsilon$ to $X^i$. Then put
$$
\nu_n = \sum_{i,j} \mu(X^i\times X^j) \mu_1^i\times \mu_2^j \;.
$$
Nice and simple. I wish I had thought about this.
Thank you, Iosif
This is the same construction as used in the second paper I linked to in my answer.
There are various papers where this question occurs. I guess a paper which directly covers the case you are interested in is https://arxiv.org/pdf/1901.07407.pdf . Note that here, the marginals don't have to be one dimensional.
A more general case in polish spaces is studied in https://arxiv.org/pdf/1811.00304.pdf Proposition 2. Since there is no Lebesgue measure on polish spaces, here the question is about absolute continuity with respect to the product measure of the marginals. Since in your case the product measure is absolutely continuous with respect to the Lebesgue measure, this of course also answers your problem.
There are other papers which incorporate this problem, mostly in the context of regularized optimal transport, but I do not have the references at hand.
Note also that in your case, with one dimensional marginals, things are simple as the concept of quantile functions and copulas can be used (similar in spirit to Iosif's answer) .
$\newcommand{\R}{\mathbb R}$
Here is an idea of the desired construction.
You have a Borel measure $\pi$ on $\R^n$ with marginals $\mu_1,\dots,\mu_n$. Let $X:=(X_1,\dots,X_n)$ be a random point in $\R^n$ with distribution $\pi$. For each $j=1,\dots,n$, let $U_j:=F_j(X_j)$, where $F_j$ is the cumulative distribution function of $X_j$, so that $U_j$ is uniformly distributed on the interval $(0,1)$, which we will identify with the additive group $G:=\R/\mathbb Z$. So, $U:=(U_1,\dots,U_n)$ is a random element of the group $G^n$ with uniform marginals. Let $V:=(V_1,\dots,V_n)$ be an independent copy of $U$. For each natural $k$, let $W_k=(W_{k,1},\dots,W_{k,n}):=U\oplus(V/k)$, where $\oplus$ denotes the addition in the group $G^n$, so that $W_k$ is a random element of the group $G^n$ with an absolutely continuous distribution and uniform marginals.
Finally, let $\pi_k$ be the distribution of the random point $X_k:=(X_{k,1},\dots,X_{k,n})$ in $\R^n$, where $X_{k,j}:=F_j^{-1}(W_{k,j})$ and
$$F_j^{-1}(u):=\inf\{x\in\R\colon F_j(x)\ge u\}=\min\{x\in\R\colon F_j(x)\ge u\}
$$
for $u\in(0,1)$. Then for each $k$ the probability measure $\pi_k$ will be absolutely continuous with marginals $\mu_1,\dots,\mu_n$, and $\pi_k$ will weakly converge to $\pi$ as $k\to\infty$.
If you work on compact domains instead of $\RR$ (or if you assume that all marginals have compact support), you can use couplings from regularized optimal transport: For $n=1$ (i.e. a coupling of just two marginals) there are two popular ways.
Entropic regularization:
Use $\pi_k$ defined by
$$\newcommand{\RR}{\mathbb{R}}
\min_\pi \int_{\RR^2} cd\pi + \gamma_k\int_{\RR^2}\pi(\log(\pi)-1)d\lambda
$$
where $\lambda$ denotes the Lebesgue measure, $c$ is some cost function (e.g. continuous and non-negative), and $\gamma_k$ is a sequence of regularization parameters converging to zero and the minimum is taken over all measures $\pi$ which have a density with respect to the Lebesgue-measure (and the density has finite entropy), which have their support in the product of the supports of the marginals, and which have the marginals $\mu_1$ and $\mu_2$. If $\gamma_k$ tends to zero it should hold that $\pi_k$ converges weakly (in the sense of measures) to $\pi$.
Quadratic regularization:
Use $\pi_k$ defined by
$$\newcommand{\RR}{\mathbb{R}}
\min_\pi \int_{\RR^2} c d\pi + \gamma_k\int_{\RR^2}\pi^2 d\lambda
$$
where $\lambda$ denotes the Lebesgue measure, $c$ is again some cost function (e.g. square integrable and non-negative) and $\gamma_k$ is a sequence of regularization parameters converging to zero and the minimum is taken over all measures $\pi$ which have a square integrable density with respect to the Lebesgue-measure, having their support in the product of the supports of the marginals and which have the marginals $\mu_1$ and $\mu_2$. I guess that weak convergence is also true here.
For entropic regularization I recommend Computational Optimal Transport (free version here) and for quadratic regularization I only have my own paper Quadratically Regularized Optimal Transport (also on the arXiv).
You have both $\int c, d\pi$ and $\int\pi^2 d\lambda$. Is your $\pi$ a measure or a density?
Also, in your paper you say that a unique minimizer exists "since the objective is continuous, coercive, and strongly convex". But how can you ensure that the minimizer is a probability density (noting that the constant $1$ is not in $L^2(\mathbb R^2)$)?
For the first comment: I use $\pi$ for both. For the second: Oops, we needed compact domains (at least finite with finite Lebesgue measure), so not sure if anything works here. I adapt my answer!
@Dirk It's great that you mentioned regularization. The problem I have at hand is related somewhat to entropic regularization. I was wondering if this approach will allow us to control $H(\pi_k|\nu)$? What I mean is that if $\pi$ is not absolutely continuous wrt to $\nu$ then $H(\pi_k|\nu)$ will also go to infinity (because of lower semi continuity of $H(\cdot|\nu)$) but can we say choose $\pi_k$ cleverly so that $\frac{1}{k}H(\pi_k|\nu)\to 0$?
I would like to know the answer to this, too!
|
2025-03-21T14:48:29.595743
| 2020-01-07T09:38:32 |
349895
|
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|
Stack Exchange
|
Usage of étale cohomology in algebraic geometry
I'm a student interested in arithmetic geometry, and this implies I use étale cohomology a lot. Regarding its definition, étale cohomology is a purely algebro-geometric object. However, almost every material I found on étale cohomology focus on its number-theoretic applications, such as the Weil conjectures and Galois representations. So, this is my question:
Are there some applications of étale cohomology on pure algebro-geometric problems?
Here, "pure algebro-geometric problems" means some problems of algebraic geometry without number-theoretic flavors, such as birational geometry, the minimal model program, classifying algebraic varieties (curves, surfaces, etc..), especially over algebraically closed fields.
Since étale cohomology coincides with singular cohomology over $\mathbb{C}$, there must exist such problems over the complex numbers. Hence, I am looking for applications of étale cohomology which are also useful over algebraically closed fields which are not $\mathbb{C}$.
Well, one can use étale cohomology to solve geometric problems using arithmetic methods. The idea is that a complex algebraic variety is in fact defined by polynomial equations with coefficients which are in a ring of finite type $R$ over $\mathbb Z$. Moereover, what is true for our variety is in fact true on a dense open subscheme of $R$. Playing with étale local systems, we see that many properties over $\mathbb C$ can be detected by pulling back over a closed point of $Spec (R)$. In other words, many theorems in algebraic geometry over finite fields produce theorems in complex geometry.
Explicit instances of the principle I skteched above are Deligne's proof of the relative hard Lefschetz theorem (thm. 6.2.5 in Astérisque 100). The main tools are the version over finite fields (which uses the theory of weights) and Lemma 6.2.6 in loc. cit. which explains how to relate complex geometry and geometry over finite fields using suitable derived categories of étale l-adic sheaves.
It's not algebraic geometry, but an important application of $\ell$-adic cohomology which is not "number theory" is the representation theory of classical groups over finite fields, a.k.a., Deligne-Lusztig theory (https://en.wikipedia.org/wiki/Deligne%E2%80%93Lusztig_theory)
There are many, for example, Artin's proof in nonzero characteristic of Castelnuovo's criterion for the rationality of a surface, and a proof that the Neron-Severi group is finitely generated. Both of these are in Chapter V of Milne's book (3.25, 3.30).
Another interesting (at least to me) application of étale cohomology is the following. The étale cohomological dimension of the complement of a variety $V$ gives a lower bound on the arithmetical rank of $V$, i.e., the minimum number of equations needed to define $V$ set-theoretically. For example, Bruns and Schwänzl [BS90] used this to prove that over an algebraically closed field the determinantal variety defined by $t$-minors of a generic $m\times n$ matrix has arithmetical rank $mn-t^2+1$.
[BS90] Bruns, Winfried, Schwänzl, Roland - The number of equations defining a determinantal variety. Bull. London Math. Soc. 22 (1990), no. 5, 439–445.
The Betti numbers of many (complex) moduli spaces have been computed by counting points over finite fields, using the Weil conjectures, as proved by Deligne, and comparison theorems for étale and singular cohomology. The first example of which I am aware is Lothar Göttsche's calculation of the Betti numbers of Hilbert schemes of points on a smooth projective surface.
The work of Harder and Narasimhan on the moduli of vector bundles on curves predates Gottsche's work by about 20 years.
Right, I had forgotten about that.
Borel has a nice \'etale cohomological proof of Matsushima's criterion: A homogeneous space of a connected reductive group is affine if and only if the isotropic group is reductive.
(The paper is https://doi.org/10.1007/BF01194008)
|
2025-03-21T14:48:29.596036
| 2020-01-07T10:30:46 |
349897
|
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|
Stack Exchange
|
Absolutely irreducible representations of affine group schemes of finite type over a field
Let $G$ be an affine group scheme of finite type over an algebraically closed field $k$. Suppose that $V$ is a finite dimensional representation of $G$.
For every $k$-algebra $A$ we have the base change representation $V_A = V\otimes_kA$ of the group $G_A = \mathrm{Spec}\, A\times_{\mathrm{Spec}\,k}G$.
Definition 1. $V$ is absolutely irreducible if for every $k$-algebra $A$ there are no nontrivial subspaces $W\nsubseteq V$ such that $W_A = A\otimes_kW$ is $G_A$-stable in $V_A$.
Definition 2. $V$ is absolutely irreducible if for every field extension $L$ of $k$ the representation $V_L$ of the group $G_L$ is irreducible.
Two definitions above capture the same notion. Indeed, if $W_A$ is $G_A$-stable for some $k$-algebra $A$, then pick a nonzero morphism $A\rightarrow L$ into a field $L$ over $k$. Then $W_L$ is $G_L$-stable. So if there are no stable subspaces in the second sense, there are also no stable subspaces in the first. The other implication is trivial.
Fact.
Suppose that $G$ is reduced. If $V$ is irreducible, then $V$ is absolutely irreducible.
Sketch of proof.
Since $G$ is reduced and $k$ is algebraically closed, we deduce that $V$ is an irreducible representation of the abstract group $G(k)$. Suppose that the dimension of $V$ is $n$ and consider a morphism $\rho:G(k)\rightarrow \mathbb{M}_n(k)$ inducing action of $G(k)$ on $V$. Extend $\rho$ to a morphism of $k$-algebras $\tilde{\rho}:k[G(k)]\rightarrow \mathbb{M}_n(k)$. By Jacobson's density theorem we derive that $\tilde{\rho}$ is surjective. This property is stable under base change. This implies that $V_L$ is irreducible over $G_L(L)$ and hence $V_L$ is irreducible over $G_L$.
Now my question:
Suppose that $G$ is nonreduced. If $V$ is irreducible, then is $V$ absolutely irreducible?
Your definition does not make much sense to me: if $I$ is a proper ideal of $A$, then $V\otimes_kI$ is non-trivial $G_A$-stable submodule of $V\otimes_kA$. The standard definition assumes that $V\otimes_kA$ is irreducible for all field extensions $A$ of $k$; but the fact that every irreducible representation is absolutely irreducible is standard, since $k$ is algebraically closed
@Angelo The definition I wrote was incorrect and was not the one I had in mind. Thank you for your comment. Can you please provide reference for irreducible $\Rightarrow$ absolutely irreducible for every group scheme of finite type over $k$?
@JimHumphreys It seems that this answers my question. Can you provide reference or maybe write an answer?
Suppose $G$ is an affine group scheme over an algebraically closed field $k$, and $V$ is a finite dimensional representation of $G$. Let $K$ be an extension of $k$, and assume that $V_K$ is reducible as a representation of $G_K$. Then I claim that $V$ is reducible as a representation of $G$.
Let $r$ be an integer with $0 < r < \dim V$, such that $V_K$ has a $G_K$-invariant $r$-dimensional sub-vector space. Let $\mathbb{G}(r,V)$ the Grassmannian of $r$-dimensional vector spaces of $V$, with its induced action of $G$. Since formation of scheme-theoretic fixed point loci commutes extension of base field, the fix point locus $\mathbb{G}(r,V)^G$ has a $K$-valued point. This means that is non-empty; since it is of finite type, and $k$ is algebraically closed, this means that $\mathbb{G}(r,V)^G$ has a $k$-valued point, so $V$ is reducible.
Thank you. One more question. Can the proof be generalized to arbitrary group functors on the category of $k$-algebras?
Slup: the problem with actions of arbitrary group functors is that the fixed points functor might be not representable.
The classical viewpoint is captured well in the 1962 book Representation Theory of Finite Groups and Associative Algebras by Curtis and Reiner (Wiley), Corollary 29.15. This of course doesn't directly answer the question about non-reduced representations of affine group schemes. But I suspect this generalizes. See for example II, $\S2$ of the 1970 book by Demazure and Gabriel, Groupes Algebriques (Masson, Paris). This book is written in the language of affine group schemes.
(By the way, my comment on the question here was being edited when I was interrupted by a phone call and didn't finish the editing.)
@LSpice: Yes, I've now deleted the comment.
|
2025-03-21T14:48:29.596321
| 2020-01-07T10:39:05 |
349899
|
{
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"authors": [
"Dominic van der Zypen",
"GH from MO",
"Martin Sleziak",
"bof",
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"url": "https://mathoverflow.net/questions/349899"
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|
Stack Exchange
|
A balancing property of infinite subsets of $\mathbb{N}$
Let $\omega$ denote the set of non-negative integers and let $[\omega]^\omega$ be the collection of infinite subsets of $\omega$.
If $S\in [\omega]^\omega$ and $A\subseteq \omega$ we say that $A$ is well-balanced with respect to $S$ if $$\lim_{n\to\infty}\frac{|A\cap S\cap \{1,\ldots,n\}|}{|S\cap\{1,\ldots,n\}|+1} = 1/2.$$ Intuitively speaking, this means that $A$ contains "half of the members of $S$" (which also implies that $A$ is infinite.)
Question. Given ${\frak S}\subseteq [\omega]^\omega$ with $|{\frak S}| = \aleph_0$, is there $A\in[\omega]^\omega$ such that $A$ is well-balanced with respect to every member of ${\frak S}$?
If we take ${\frak S}$ to be the collection of infinite recursive sets of $\omega$, then any such $A$ would correspond to a bitstream that is computationally random.
$\liminf a_n=\limsup a_n=1/2$ can be abbreviated as $\lim a_n=1/2$.
Oh right :-) Will simplify the post accordingly, thanks @GHfromMO
A random subset of $\omega$ (meaning that $n\in A$ iff the $n^{\text{th}}$ toss in an infinite sequence of fair coin tosses comes of heads) will do the job with probability $1$.
Isn't this sometimes called relative density (of the set $A$ with respect to the set $S$)?
By the strong law of large numbers, if $S$ is an infinite subset of $\omega$, a random subset of $\omega$ will be well-balanced with respect to $S$ with probability one.
By the countable additivity of Lebesgue measure, if $\mathfrak S$ is a countable collection of infinite subsets of $\omega$, a random subset of $\omega$ will be well-balanced with respect to every member of $\mathfrak S$ with probability one.
If Lebesgue measure is $\kappa$-additive (the union of fewer than $\kappa$ measure zero sets has measure zero) then the same holds for a collection $\mathfrak S$ of infinite subsets of $\omega$ with $|\mathfrak S|\lt\kappa$.
|
2025-03-21T14:48:29.596468
| 2020-01-07T13:08:39 |
349906
|
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"ABIM",
"Yuval Peres",
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"https://mathoverflow.net/users/7691"
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}
|
Stack Exchange
|
Convergence rate estimates of Monte-Carlo first-passage time estimates
Setup
Let $X_t$ be a $d$-dimensional diffusion process solving the Ito-stochastic differential equation
$$
X_t = x+ \int_0^t f(X_t,u_t)dt + \int_0^t \sigma dW_t,
$$
where $x \in \mathbb{R}^d$, $u_t$ is predictable, $f(\cdot,\cdot)$ is locally Lipschitz, $\sigma \sigma^T$ is positive-definite, and $W_t$ is a $d$-dimensional Browinian motion.
If $D$ is a connected open neighbourhood of $x$ with smooth boundary, and suppose that we would like to estimate the distribution of the first exit time
$$
\tau \triangleq \left\{
t >0: \, X_t \not\in D
\right\} =
\left\{
t >0: \, X_t \in \partial D
\right\}
,
$$
via Monte-Carlo. Let us denote $\tau^n$ the empirical distribution approximating $\tau$ from producing $n$ sample paths of $X_t$.
Question:
Are there known estimates on the convergence speed of $\tau^n$ to $\tau$ (in any reasonable sense)? For example an estimate on the first moment the form:
$$
\|\tau^n - \mathbb{E}[\tau]\| \leq C r(n),
$$
where $r$ is some nice lsc rate function and $C>0$ is some universal constant.
Assuming $D$ is bounded, the best estimate that holds almost surely is given by the law of the iterated logarithm https://en.m.wikipedia.org/wiki/Law_of_the_iterated_logarithm
Two problems:
$n$ is not wrt the discritization but the number of samples
In this case the increments are independent but are not identically distributed... (even if it were discussing the discritization of iid rvs).
You wrote “$n$ sample paths.” They are not independent? How are they produced?
|
2025-03-21T14:48:29.596616
| 2020-01-07T15:39:36 |
349913
|
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"Robert Furber",
"Taras Banakh",
"YCor",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/349913"
}
|
Stack Exchange
|
Compact-open limit of continuous functions is continuous?
Let $X$ be a topological space and $Y$ a metric space.
A classical result states that compact-open topology on the space $C(X,Y)$ of continuous functions is the same as the topology of uniform convergence on compact sub-sets.
In general one may define compact-open topology on the whole space $Y^X$ of all functions from $X$ to $Y$.
Is it true that if $f_n\to f$ with respect to the compact-open topology, and all $f_n$ are continuous, then the limit $f$ is continuous?
I never seen any theorem stated that way, and I suspect that it may be false.
If so, is there some condition to put on $X$ so to make the above statement true?
If $X$ is a Cantor space (or any locally compact space with a non-isolated point $x_0$ with a basis $(K_i)$ of clopen neighborhoods) and $Y={0,1}$, the sequence (or net) $(1_{K_i})$ of continuous functions tends to the non-continuous function $1_{{x_0}}$ in the compact-open topology.
Sequences of characeristic functions seldom converge uniformly
@YCor Your example seems to me correct. Could you please post it as a (more detailed) answer?
@HennoBrandsma yes $X$ is compact, so what? you might have in mind that for locally compact $X,Y$ the compact-open topology is that of uniform convergence on compact subsets, but this is within continuous functions.
@HennoBrandsma I'm afraid I gave a counterexample to this statement.
@HennoBrandsma could you please post a proof or a precise reference to the fact that when X is compact, then continuous functions form a closed subset? I did not find that neither in Kelley's nor in Engelking's books. Also, the example I gave and that given by Ycor seem to me both correct. Could you please point out "where" such examples are wrong?
All you can say about the limit function is that its restrictions to compacta are continuous. In general, this does not imply global continuity.
In fact spaces which DO have the required property have been much studied. They are called $k$– spaces (sometimes Kelley spaces). The defining property is that a subset is open if its intersection with each compact subset is open in the corresponding induced topology. You can find a great deal of information on such spaces in standard textbooks on general topology (I would recommend Engelking‘s monograph with precisely this title). Examples are metric or locally compact spaces.
Could you post a proof (or a reference) that limit functions are continuous on Compacta? (Even if I guess that I likely find a proof in the book you mentioned)
The example of Ycor seems to contradict continuity of the limit function on Compacta
It is a very elementary result of topology that a uniform limit of a sequence of continuous functions is continuous (usualy proved in a beginner‘s analysis course for functions on a subset of the reals).
@user131781 but compact-open topology agree with uniform convergence on Compacta in $C(X,Y)$. Is it true in general that compact-open convergence implies uniform convergence on Compacta? (Even if the limit is not continuous). The Ycor example sounds correct to me.
@HennoBrandsma I checked Kelley book. Theorem 11 at page 230 explicity state the result that compact-open coincides with uniform convergence on compacta, only for continuous functions.
Another name for these spaces is "compactly generated spaces".
@HennoBrandsma I also checked Engelking book, and I did not found any statement that compact-open convergence imly uniform convergence on compacta, without assuming that the limit function is continuous
As already mentioned, it's false: compact-open convergence to a non-continuous function doesn't imply uniform convergence on compact subsets. Always better to have the proof in mind (or reprove it) to figure out that it uses continuity of the limit.
Spaces with the required property include not only $k$-spaces, but $k_R$-spaces, i.e., spaces $X$ such that a function $f:X\to \mathbb R$ is continuous iff for any compact set $K\subset X$ the restriction $f|K$ is continuous. The class of $k_R$-spaces was studied by Michael and is strictly wider than the class of $k$-spaces.
It seems to me that Ycor example is correct, but I propose another counterexample:
$X=[0,1]$, $Y=[-1,1]$, $f=\sin(1/x)$ for $x\neq 0$ and $f(0)=0$.
$f_n=\sin(1/x)$ for $x>1/2\pi n$ and $f_n(x)=0$ for $x\in[0,1/2\pi n]$
The functions $f_n$ are continuous, $f$ is not, but $f_n\to f$ with respect to the compact open topology.
Proof.
Given a compact $K$ and an open $U$ let $V(K,U)=\{g:g(K)\subseteq U\}$. We prove that for any $V(K,U)$ if it containis $f$, then it contains also $f_n$ for $n$ big enough.
There are two cases:
1) $0\notin K$. In this case $f_n$ and $f$ coincide on $K$ for $n$ big enough.
2) $0\in K$. In this case, if $f\in V(K,U)$ then $0\in U$. Since $f_n(x)=0$ for $x\in[0,1/2\pi n]$ and $f_n=f$ elsewhere, if follows that $f_n\in V(K,U)$ (for any $n$).
So, for any open set $A$ in the compact-open topology containing $f$, then $A$ must eventually contain $f_n$. I.e. $f_n\to f$.
Indeed for functions $[0,1]\to\mathbf{R}$, it seems that a compact-open limit of continuous functions maps intervals to intervals (hence it can't be a non-continuous piecewise continuous function, for instance). It's nice you found an example in this setting.
|
2025-03-21T14:48:29.597084
| 2020-01-07T15:47:55 |
349917
|
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|
Stack Exchange
|
Generalization of semi-hereditarity
Let $R$ be a ring. A left $R$-module $K$ is called an $N$-th kernel if there are projective left $R$-modules $P_1, \ldots P_N$ and a short exact sequence
$$ 0\rightarrow K \rightarrow P_N \rightarrow \ldots \rightarrow P_1\,.$$
The last arrow need not be surjective. A $0$-th kernel is an arbitrary module, a $1$-st kernel a submodule of a projective module, and so on. The ring $R$ has global dimension $\le N$ if and only if all $N$-th kernels are projective.
Having global dimension $1$ is called hereditary and $R$ is called semi-hereditary if all finitely generated submodules of projective modules, i. e. all finitely generated $1$-st kernels, are projective.
Question: Is there a name for the property that all finitely generated $N$-th kernels are projective?
|
2025-03-21T14:48:29.597164
| 2020-01-07T15:58:15 |
349919
|
{
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"authors": [
"Boris Sklyar",
"Gerhard Paseman",
"https://mathoverflow.net/users/3402",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/349919"
}
|
Stack Exchange
|
Matrix sieve algorithm for finding prime numbers
I have derived the following theorem:
An odd positive integer $N=6n−1$ is a prime iff neither of two diophantine equations
$6x^2+(6x−1)y=n$
$6x^2+(6x+1)y=n$
has solution.
An odd positive integer $N=6n+1$ is a prime iff neither of two diophantine equations
$6x^2−2x+(6x−1)y=n$
$6x^2+2x+(6x+1)y=n$
has solution.
$x=1,2,3,\ldots;y=0,1,2,\ldots;n=1,2,3,\ldots$
Theorem allows to substitute the task: "Find all primes in the range $(N_1;N_2)$" by the task: "Find positive integers which do not appear in the range $(n_1;n_2)$ in two pairs of $2$-dimensional arrays
$P_1(i,j)=6i^2+(6i-1)(j-1)$
$P_2(i,j)=6i^2+(6i+1)(j-1)$
$i,j = 1,2,3,\ldots$
for primes in the sequence $N=6n-1$.
$P_3(i,j)=6i^2-2i+(6i-1)(j-1)$
$P_4(i,j)=6i^2+2i+(6i+1)(j-1)$
$i,j = 1,2,3,\ldots$
for primes in the sequence $N=6n+1$.
Since all primes (except $2$ and $3$) are in one of two forms $6n−1$ or $6n+1$,
so we can find primes simply by picking up positive integers which do not appear in these arrays and we need not to perform operations of dividing.
See http://www.planet-source-code.com/vb/scripts/BrowseCategoryOrSearchResults.asp?lngWId=3&blnAuthorSearch=TRUE&lngAuthorId=21687209&strAuthorName=Boris%20Sklyar&txtMaxNumberOfEntriesPerPage=25
Can proposed algorithm be regarded as an alternative to sieve of Eratosthenes?
Assuming the assertions are formally proved correct, it would seem then to be an alternative. Even so, it is unclear to me what benefits there are. Can you use it to test for a prime quickly? Can you say something new about the distribution of primes with it? Until you are able to convince me of its potential, I will focus on my own prime finding algorithms. Gerhard "Not Sure Of The Applicability" Paseman, 2020.01.07.
Theoretical background and C++ code - see https://www.academia.edu/13890086/Matrix_sieve_the_easiest_way_to_find_prime_numbers program calculates primes in interval of 10^6 for numbers up to 10^19 for run time of 10 sec
You can simplify by using just one matrix
$$P(i,j)=6ij+i-j$$
with $i\geq 2$ and $j\geq 1$
We can find primes (except $2$ , $3$ and $5$) simply by picking numbers $k>1$ which do not appear in this array (except $21$) with:
$p= \frac{(6k-1)}{5}$ if $k\equiv 1 \bmod 5$
and
$p=6k-1$ otherwise
However a faster method is obtained using this code in python:
n=10000000
primes5m6 = [True] * (n//6+1)
primes1m6 = [True] * (n//6+1)
for i in range(1,int((n**0.5+1)/6)+1):
if primes5m6[i]:
primes5m6[6*i*i::6*i-1]=[False]*((n//6-6*i*i)//(6*i-1)+1)
primes1m6[6*i*i-2*i::6*i-1]=[False]*((n//6-6*i*i+2*i)//(6*i-1)+1)
if primes1m6[i]:
primes5m6[6*i*i::6*i+1]=[False]*((n//6-6*i*i)//(6*i+1)+1)
primes1m6[6*i*i+2*i::6*i+1]=[False]*((n//6-6*i*i-2*i)//(6*i+1)+1)
where for $i>0$
$ p = 6i-1 $ is prime if $ primes5m6 [i] = True $
and
$ p = 6i + 1 $ is prime if $ primes1m6 [i] = True $
|
2025-03-21T14:48:29.597348
| 2020-01-07T16:41:42 |
349922
|
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"url": "https://mathoverflow.net/questions/349922"
}
|
Stack Exchange
|
Basis for a lattice in a subspace of $\Bbb R^n$
Let $S$ be a linear subspace of $\Bbb R^n$ having dimension $k<n$ and assume $S$ is described by $n-k$ linear equations with integer coefficients. Look at now the intersection $\Lambda=S\cap \Bbb Z^n$ - such an intersection is a lattice in $S$. Theoretically speaking, I already know the existence of a basis $\{v_1,\dots,v_k\} \in S$ such that $\Lambda=\big\{\lambda_1v_1+\cdots+\lambda_kv_k\,|\, \lambda_i\in \Bbb Z \text{ for all } i\big\}$. Of course, any $v_i$ has integer coefficients.
I am wondering if the basis $\big\{v_1,\dots,v_k\big\}$ can be completed to a basis $\big\{v_1,\dots,v_k,v_{k+1},\dots,v_{n}\big\}$ of $\Bbb R^n$ by adding other $n-k$ vectors (with integer coefficients) such that there is a matrix $g\in \text{SL}(n,\Bbb Z)$ such that $g(e_i)=v_i$ for any $i=1,\dots,n$, where $\{e_1,\dots,e_n\}$ denotes the standard basis.
In other words, I am wondering if $\{v_1,\dots,v_k\}$ can be completed to a primitive set of vectors generating the standard lattice $\Bbb Z^n$. By setting $V$ the $k\times n$ matrix having the vectors $v_i$ as columns, if I am not mistaken, this condition is equivalent to say that the gcd of the $k^{th}$ order minors is one. In this case, the matrix $V$ can be completed to a square matrix with determinant one - hence a matrix in $\text{SL}(n,\Bbb Z)$.
When $n=2$, this is always true. Indeed, let $qx-py=0$ a one-dimensional space in $\Bbb R^2$ - $p,q$ are taken coprime. Clearly $(p,q)$ satisfies the equation above. Taking any solution $(x_o,y_o)$ of the associated diophantine equation $qx-py=1$ (the solution exists because $p,q$ are coprime) we have two vectors $(p,q), \, (x_o,y_o)$ and they form a basis for the standard lattice. In other words, the basis $(p,q)$ is completed to a basis of $\Bbb Z^2$.
When $n=3$, the problem seems more subtle and I don't have an answer. I began with this example. I considered the plane $S$ given by the equation $x+y-2z$ in $\Bbb R^3$. The vectors $(2,0,1),\, (0,2,1)$ form a basis for $S$ and they belong $\Bbb Z^3$, clearly. However, they don't form a basis for the lattice $\Lambda=S\cap \Bbb Z^3$. Indeed, the vector $(1,1,1)$ doesn't belong to $\Bbb Z$-span of them. By chance, I noticed that $\big\{(2,0,1),\,(1,1,1)\big\}$ is a basis for the lattice $\Lambda$. Not only, by adding the vector $(0,-1,0)$, for instance, I can complete the latter basis to a basis of $\Bbb R^3$ and there is $g\in\text{SL}(3,\Bbb Z)$ such that $g(e_1)=(2,0,1),\, g(e_2)=(1,1,1),\, g(e_3)=(0,-1,0)$. I have made some other examples and they work similarly.
For a generic $n\ge3$, I don't know if there is an algorithmic method to find a basis for $\Lambda=S\cap \Bbb Z^n$ and if such a basis can be always complete to a basis for $\Bbb Z^n$, like in the $n=2$ case.
If I recall correctly, a basis of sublattice can be completed to a basis of the full lattice if and only if the sublattice is of the form $\mathbb{Z}^n\cap W$, where $W$ is a linear subspace.
There is a (classic?) book on lattices that treats all this type of questions in detail, but I can't remember which one.
@EFinat-S, Thanks for your answer. I think you are right but I am not able to find a proof. I have checked two books: Integral Matrices by Newmann and Lectures on the Geometry of numbers by Siegel. Is the book you have mentioned one of them? I didn't find a similar result but I may miss it.
It was the book of Cassels on Geometry of numbers.
I see, I check that book! Thanks a lot!
See Cassels "An Introduction to the Geometry of Numbers", Corollary 3 in page 14. Also, you may want to look at Theorem 1.28 here: https://books.google.com/books?id=i5AkDxkrjPcC&pg=PA13&lpg=PA13&source=bl&ots=Pz00ZD6qFZ&sig=ACfU3U0u37_G6qfpiAIyVmLeJUAkSe220g&hl=es-419&sa=X&redir_esc=y#v=onepage&q&f=false (I hope the link works for you)
@EFinat-S yes, thank a lot for point me out the reference!
The answer to your question, with a full proof, appears in "An Introduction to the Geometry of Numbers" by J.W.S. Cassels, Corollary 3 in page 14. It is also stated as Theorem 1.28 here but without a proof.
|
2025-03-21T14:48:29.597612
| 2020-01-07T17:49:04 |
349926
|
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|
Stack Exchange
|
Finite groups with unbounded commutator width and no nontrivial central chief factor
The commutator width of a group is the smallest $n$ such that every product of commutators is a product of $n$ commutators.
My initial question was:
Do there exist finite perfect groups with trivial center with unbounded commutator width?
It is known without the center-free assumption: the constructions I'm aware of crucially use that the center is large, namely producing for every $n$ a finite group $G$ with center $Z$ such that $|G/Z|^{2n}<|G|$ (so that the commutator width of $G$ is $>n$, using that the commutator map $G^2\to G$ factors through a map $(G/Z)^2\to G$).
Let me also mention the well-known fact that finite nonabelian simple groups have bounded commutator width ($=1$ by the somewhat recent solution to Ore's conjecture).
Edit Jan 8 '20:
Richard Lyons and Derek Holt showed in the comments the above question has a positive answer, observing that center-free groups are obtained by a simple trick: if $G$ is any finite group and $p$ is a prime not dividing $|G|$ then there exists a faithful $\mathbf{F}_pG$-module $V$ with trivial invariant submodule/quotient (just take $\mathbf{F}_pG$ modulo constants). Then if $G$ is perfect, then $G\ltimes V$ is a perfect group with at least as large commutator width, and in addition is center-free. (I hadn't thought of the trick of viewing $G$ as quotient of a center-free group!)
As suggested by Richard Lyons, let me therefore ask the following question:
Do there exist finite groups with no central chief factor (that is, all of whose quotients have a trivial center) and unbounded commutator width?
The commutator width of $G$ is at least the commutator width of any quotient $G/N$. So there are examples where $G/N$ is perfect with arbitrarily large commutator width and $N$ is a faithful $F_p[G/N]$-module without trivial constituent, $p$ a prime not dividing the order of $G$. To avoid this you could require that there be no central chief factors.
@RichardLyons well, you seem to almost answer my question. It seems you're using the fact that every perfect finite group is quotient of a center-free one, or at least the in the known cases. In the cases in mind, the center $Z$ of $G$ is an elementary abelian $p$-group, and $G/Z$ is center-free, in case it helps.
But for any finite group $G$ and prime $p$ not dividing $|G|$, we can find a module $M$ over ${\mathbb F}_p$ on which $G$ acts faithfully and without fixed points. Take, for example, the quotient of the regular module (i.e. ${\mathbb F}_pG$) with the $1$-dimensional fixed point space factored out. Then $Z(M \rtimes G) = 1$ and $M \rtimes G$ is perfect if $G$ is.
@YCor Sorry I was not clear. The answer to your question is 'yes,' but you could change the question by replacing "trivial center" by "no central chief factors." Then the answer is not clear (to me anyway).
Or if the revised question is too hard, does there exist a finite group with no central chief factors in which not every element of the commutator subgroup is a commutator.
@DerekHolt I'll not change the question once more, but it's a good starting point!
|
2025-03-21T14:48:29.598198
| 2020-01-07T18:27:25 |
349929
|
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|
Stack Exchange
|
Polynomials and complete set of algebraic conjugate numbers
Let $P,Q$ be two polynomials of $\mathbb Q[X]$, $\alpha,\beta$ be elements of $\overline{\mathbb Q}$. Denote by $\alpha_1=\alpha,\cdots,\alpha_{\mu}$ (resp. $\beta_1=\beta,\cdots,\beta_\nu$) the conjugates of $\alpha$ (resp. of $\beta$). The question:
Is $\prod_{i=1}^\mu\prod_{j=1}^\nu(P(\alpha_i)+Q(\beta_j))$ a rational number?
Thanks in advance for any answer.
Let $A$ and $B$ be square matrices over $\mathbb{Q}$ such that $\alpha_1, \alpha_2, \ldots, \alpha_\mu$ are the eigenvalues of $A$ and such that $\beta_1, \beta_2, \ldots, \beta_\nu$ are the eigenvalues of $B$. (Such matrices exist, since any set of conjugate algebraic numbers is the set of roots of an irreducible polynomial.) Now, the $P\left(\alpha_i\right) + Q\left(\beta_j\right)$ are the eigenvalues of the matrix $P\left(A\right) \otimes I + I \otimes Q\left(B\right)$. Thus, their product is the determinant of this matrix, and therefore rational.
You can check that this expression is invariant under the action of the absolute Galois group, so is rational. You could also apply the fundamental theorem of symmetric polynomials.
Wojowu: thanks. I see what you mean
danj. Nice proof !!
|
2025-03-21T14:48:29.598332
| 2020-01-07T19:15:36 |
349933
|
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|
Stack Exchange
|
Representing spaces of $\infty$-stacks
In The stable moduli space of Riemannsurfaces: Mumford’s conjecture, Madsen and Weiss introduce the representing space $|\mathcal{F}|$ of a sheaf of sets $\mathcal{F}$ on the site $\mathscr{X}$ of smooth manifolds as the geometric realization of the simplicial set $[n]\mapsto \mathcal{F}(\Delta^n)$ and prove that concordance classes of elements in
$\mathcal{F}(X)$ bijectively correspond to homotopy classes of maps from $X$ to $|\mathcal{F}|$. As also the classifying space of a topological group $G$ admits a description as geometric realization (but now of a simplicial topological space rather than a simplicial set) and concordance classes of principal $G$-bundles on are the same thing as isomorphism classes, this suggests that more generally one should have a representing space for not only sheaves, but also for stacks and higher stacks on $\mathscr{X}$, similarly defined by a topological realization construction, and representing concordance classes. Unfortunately, I have not been able to precisely locate such a statement in the literature, so I'm asking for a reference here (in case such a result has some hope to be correct).
The case of sheaves valued in $\infty$-groupoids is the content of a recent preprint of (Berwick-Evans)-[Boavida de Brito]-Pavlov: https://arxiv.org/pdf/1912.10544.pdf
Thanks, Dylan! If you can promote your comment to an answer (just the same text) I'll accept it and consequently close the question: that's precisely what I was looking for!
@domenicofiorenza There's no reason to close the question.
Right, "close" was not the correct term here, as it means a precise thing on MO. What I meant was: I'll make the question appear as "answered" so that MO users won't think I'm not satisfied with Dylan comment and struggle to find a more satisfying answer.
As requested, the comment as an answer:
The case of sheaves valued in ∞-groupoids is the content of a recent preprint of (Berwick-Evans)-[Boavida de Brito]-Pavlov: arxiv.org/pdf/1912.10544.pdf.
|
2025-03-21T14:48:29.598497
| 2020-01-07T20:11:59 |
349935
|
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|
Stack Exchange
|
Having a paper published via both Conference Proceedings and via a refereed journal
Forgive me if this isn't the right place to pose this question. I do need guidance on this.
In 2018 I had submitted a paper to a refereed journal. It had gotten accepted for publication by said refereed journal modulo some minor revisions and I am in the process of revising it. I also had presented this result at the 2019 Symposium On The Theory Of Computing [STOC 2019] in June. So far so good at this point, plenty of papers have appeared in both the conference proceedings--to disseminate quickly, and then in polished revised form in a refereed journal a bit later.
Here is where my dilemma is. This past fall 2019 I also received an invitation to submit this same result to a special edition of STOC 2019, which will appear in the SIAM Journal Of Computing. I had said yes thinking that this was just like the Conference Proceedings, without realizing at that time, that there may be a moral dilemma. I am not sure if I would be double-publishing my paper, which I understand is a no-no.
I am not sure where to go from here. I explained the situation to both the editors of the refereed journal and to the committee that invited me to submit this result to the special edition of STOC 2019, but have not received guidance back.
So this is my question: What do I need to do. I want this paper to go into the refereed journal, as they were the ones who reviewed it first. I admit to not being that familiar with the rules of publishing in academic journals. Thanks in advance for any guidance!
This seems simple: Withdraw your acceptance to the special edition of the SIAM Journal Of Computing. (I am assuming your Yes came after the submission to the refereed journal.)
Hi @JosephO'Rourke yes you are correct.
You might ask at a computer science site as well since the conference publication culture is a bit different in computer science and in math.
It is not a good practice to publish the same paper twice, (or more). Except the publication on the arXiv (I have a strong opinion that ALL mathematical papers should be published on the arXiv). I can tell you what I do in similar circumstances. If I talk on a conference on a result which is sent to a journal, or is already published, and the conference publishes proceedings, I publish my TALK in the proceedings, but this is a different text from the paper itself.
|
2025-03-21T14:48:29.598706
| 2020-01-07T21:04:58 |
349939
|
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|
Stack Exchange
|
Limits (growth rates) of power series coefficients
Take two positive integers $m$ and $n$ and consider the rational function
$$G_{m,n}(x,t)=\frac{d}{dx}\left(\frac1{(1-x^m)(1-tx^n)}\right)$$
and the corresponding Taylor expansion as
$$G_{m,n}(x,t)=u_0(t)+u_1(t)x+u_2(t)x^2+\cdots$$
where $u_j(t)$ itself is expanded as a polynomial in $t$ (note: coefficients in each $u_j(t)$ are all equal). Now, read-off the coefficients of $G_{m,n}(x,t)$ in the exact order they appear and be listed (including multiplicities) as $\beta_{\ell}(m,n)$.
QUESTION. Is this limit true?
$$\lim_{\ell\rightarrow\infty}\frac{\beta_{\ell}^2(m,n)}{\ell}=2mn.$$
EXAMPLE 1. If $m=n=1$ then
$$G_{1,1}(x,t)=1+t+(2+2t+2t^2)x+(3+3t+3t^2+3t^3)x^2+\cdots$$
and hence $\beta_{\ell}(1,1)$ starts with (keep in mind: $\beta_1=1, \beta_2=1, \beta_3=2$, etc)
$$1,1,2,2,2,3,3,3,3,\dots.$$
Hence
$$\beta_{\ell}(1,1)=\left\lfloor\frac{\sqrt{8\ell+1}-1}2\right\rfloor \qquad
\Longrightarrow \qquad \lim_{\ell\rightarrow\infty}\frac{\beta^2_{\ell}(1,1)}{\ell}=2.$$
EXAMPLE 2. If $m=1$ and $n=2$ then
$$G_{1,2}(x,t)=1+(2+2t)x+(3+3t)x^2+(4+4t+4t^2)x^3+\cdots$$
and hence $\beta_{\ell}(1,2)$ starts with (keep in mind: $\beta_1=1, \beta_2=2, \beta_3=2$, etc),
$$1,2,2,3,3,4,4,4,\dots.$$
$$\beta_{\ell}(1,2)=\left\lfloor\sqrt{4\ell+1}-1\right\rfloor \qquad
\Longrightarrow \qquad \lim_{\ell\rightarrow\infty}\frac{\beta^2_{\ell}(1,2)}{\ell}=4.$$
I take it $\beta_\ell^2(m,n)$ means $\beta_\ell(m,n)^2$?
Don't you need $n,m$ relatively prime? Because if $(n,m)=d>1$ then the power series expansion of $\frac{1}{1-x^m)(1-tx^n)}$ is a series in $x^d$, so that, deriving in $x$, $G_{m,n}(t,x)$ has $u_j\neq0$ only for $j=-1\mod d$, which implies the $\beta_\ell(m,n)$ are frequently zero...
Maybe in the definition of the $\beta_\ell(m,n)$ only the non-zero coefficients are listed?
@Wojowu: yes that is right.
@PietroMajer: yes, you are right, the non-zero terms are listed. Thanks.
Yes, the conjectured limit is true. Let $d=\gcd (m,n)$ and $m=m_1d, n=n_1d$. Suppose $a_k$ denotes the number of solutions to $k=m_1r+n_1s$ with $r,s\geq 0$, so that
$$a_0+a_1x+a_2x^2+\cdots =\frac{1}{(1-x^{m_1})(1-x^{n_1})}.$$
We have $\beta_{\ell}(m,n)=Nd$ if and only if $\sum_{i=0}^{N-1}a_i< \ell\le \sum_{i=0}^{N}a_i$.
Notice that $a_i\in \{0,1\}$ for $0\le i\le m_1n_1-1$, and the rest satisfy $a_{mn+i}=1+a_i$.So we have
$$\sum_{i=0}^{N-1}a_i\geq \sum_{i=0}^{m_1n_1\lfloor \frac{N-1}{m_1n_1}\rfloor}a_i\geq m_1n_1\binom{\lfloor \frac{N-1}{m_1n_1}\rfloor}{2}=\frac{N^2}{2m_1n_1}+O(N)$$
and similarly
$$\sum_{i=0}^{N}a_i\le \sum_{i=0}^{m_1n_1\lceil \frac{N}{m_1n_1}\rceil}a_i\le m_1n_1\binom{1+\lceil \frac{N}{m_1n_1}\rceil}{2}=\frac{N^2}{2m_1n_1}+O(N).$$
So we have $$\lim_{\ell\to \infty} \frac{\ell}{\beta_{\ell}^2(m,n)}=\lim_{N\to \infty}\frac{\frac{N^2}{2m_1n_1}}{N^2d^2}=\frac{1}{2mn}$$
as desired.
Thank you for an elementary argument.
|
2025-03-21T14:48:29.598891
| 2020-01-07T22:49:56 |
349943
|
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|
Stack Exchange
|
Question about eigenvalues of connectivity matrices for graphs
I'm a computer science student working on a research project that deals with computational study of atomic clusters. I'm using a graph based representation of the clusters using a binary connectivity matrix. My question is, are the eigenvalues invariant to a transformation in particle indices?
For example, if I have a 3 particle cluster with particle 1 attached to 2 and 2 attached to 3. The connectivity matrix will be different if I swapped the particle index even though it is the same structure. My question is will the eigenvalues be the same? I checked it using python and indeed, the values were same, but I'm trying to look for a formal established proof (or an article to cite).
I really appreciate the help and I apologize if it is a stupid question.
Hi! First, welcome to MathOverflow: this website is intended for mathematicians to ask research questions to each other. MathStackExchange (or MathSE) is for other questions about mathematics. Second, the eigenvalues should not depend on how you choose to name your vertices. Lastly, your question is not stupid, this is just not the right place to ask it. (I dare say that many of us once asked how to compute the volume of a sphere and where the formula comes from; it's far from being a stupid question, but it's not a research question of current mathematics.)
If $P$ is a permutation matrix, i.e., $P$ is binary and doubly stochastic, then $P^\top = P^{-1}$. Thus, any matrix $A$ is co-spectral to $P^\top A P$. Furthermore, if the matrix $A$ is the incidence matrix of a graph, then a straightforward exercise shows that the latter is a relabeling of the indices of the graph.
If you permute the indices i.e., rename each of the particles, the eigenvalues will stay the same. The eigenvectors will be what they were before but with the indices permuted.
i.e., if [2,8,6] was an eigenvector before, with 2 corresponding to the first particle, 8 the 2nd particle, and 6 the 3rd, and you switch the 2nd and third indices so that what was before the 2nd particle is now the 3rd and what was the 3rd particle is now the 2nd, then [2,6,8] will be the new eigenvector.
|
2025-03-21T14:48:29.599099
| 2020-01-07T23:04:01 |
349945
|
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|
Stack Exchange
|
Should the "L" in the term latin/Latin square be capitalized?
In Denes and Keedwell's book the word "latin" is not capitalized, and there seems to be some precedent in the literature for this usage. However, the vast majority of work on the subject capitalizes the term "Latin square."
Indeed, most English dictionaries and computer spell checkers treat the word Latin as a proper noun. To me, this doesn't seem a particularly compelling argument for its usage in mathematics as, for example, neither Merriam Webster nor Dictionary.com contain an entry for the word "quasigroup."
Perhaps this is a pedantic question, but I have encountered it in personal correspondence several times in the past couple of months. Nonetheless, it may also be worth asking if there is any value to having a consistent convention within the mathematical community.
Whatever you choose, do it consistently.
English adjectives that derive from proper nouns are usually capitalised. However, over time, such an adjective can lose its capitalisation provided that it sufficiently departs from its origins in the speaker's mind.
The word 'latin' derives from the central western Italian region of Latium, but its mathematical meaning has little to do with Latium, so it seems sensible to de-capitalise it, much like the term 'roman' in 'roman numerals'.
Wow, I didn't even know that "roman numerals" and "hindu-arabic numerals" are written without capitals. Is it the "chinese remainder theorem" too?
isn't it the Chinese Remainder Theorem? :)
The "Quadratic Reciprocity Theorem" tells you about quadratic reciprocities. By analogy, I believe that the "Chinese Remainder Theorem" tells you about chinese remainders. Definition: A chinese remainder is the unique remainder mod $mn$, where $m$ and $n$ are coprime, which has remainders $a$ and $b$ respectively upon division by $m$ and $n$.
A close mathematician friend of mine used to try to stick to the rule of capitalizing any word that derives from a person's name: Noetherian (not "noetherian") ring, Abelian (not "abelian") group, etc. My response: Narcissism?
What a Dunce . . .
Hah! Dunce comes from John Duns Scotus, who was seen as a hairsplitting scholastic (for those, like me, who had to look it up).
|
2025-03-21T14:48:29.599408
| 2020-01-07T23:19:59 |
349947
|
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|
Stack Exchange
|
A plausible hyperbolic link
This link is hyperbolic according to SnapPy's computation. There is an obvious non-boundary parallel annulus spanned by two components at the very top in the diagram. If this annulus is essential, then the link is not hyperbolic. Is this annulus inessential in the link exterior? If yes, how can one see it?
This link is not hyperbolic. Let $K_1$ and $K_2$ be the link components that cobound the annulus $A$. Let $T$ be the boundary of a regular neighbourhood of $A$ in $S^3$. Let $K_3$ and $K_4$ be the other link components. Let $V$ be the solid torus bounded by $T$ containing $K_3$ and $K_4$.
If $A$ were inessential, then it would compress so that each of $K_1$ and $K_2$ bound disks. But then such a disk would contain a subdisk $D$ which is a compressing disk for $V$. But then $K_3$ and $K_4$ could be isotoped into a ball contained in $V$, and this is impossible. Hence $A$ is essential in the link exterior.
I don't see how SnapPy is telling you that this link is hyperbolic. It does give a nonzero volume, but that is a simplicial volume. It is the hyperbolic volume of the complement of $K_3$ and $K_4$ inside $V$ since $T$ is an essential torus which gives the JSJ decomposition of the link complement. In fact the SnapPy command M.splitting_surfaces() shows that this torus $T$ exists.
As a complement (sorry) to Josh Howie's answer, here is my Snappy session.
In[1]: M = Manifold()
Starting the link editor.
Select Tools->Send to SnapPy to load the link complement.
New triangulation received from PLink!
In[2]: M.volume()
Out[2]: 7.327724753
In[3]: M.solution_type()
Out[3]: 'contains degenerate tetrahedra'
Snappy is "guessing" that the manifold is not hyperbolic, but does have a hyperbolic piece in its JSJ decomposition. (There is a lot of accuracy in the volume computation!) As Josh Howie points out, you can actually use Snappy (running under Sage) to prove that the manifold is not hyperbolic, and that that piece is a Borromean rings complement. (The .identify() method is your friend here.)
Added later: So I got snappy running under sage. Here is that session.
sage: %gui tk
sage: M = snappy.Manifold()
Starting the link editor.
Select Tools->Send to SnapPy to load the link complement.
New triangulation received from PLink!
sage: M.volume()
7.32772475341777
sage: M.solution_type()
'contains degenerate tetrahedra'
sage: M.splitting_surfaces()
[Orientable two-sided with euler = 0]
sage: M.split(0)
[unnamed link.a(0,0)(0,0)(0,0), unnamed link.b(0,0)(0,0)(0,0)]
sage: A, B = M.split(0)
sage: A.identify()
[t12067(0,0)(0,0)(0,0),
6^3_2(0,0)(0,0)(0,0),
L6a4(0,0)(0,0)(0,0),
ooct02_00005(0,0)(0,0)(0,0)]
sage: B.identify()
[6^3_3(0,0)(0,0)(0,0)]
sage: B.volume()
-1.45994327738208e-14
sage: A.volume()
7.32772475341775
and we are done. Snappy identified the two pieces (after splitting) as link complements. Thus the splitting torus is incompressible. So the manifold A (which is the Borromean rings complement) is a geometric piece of the JSJ decomposition. The other piece, a three-times punctured sphere crossed with the circle, is Seifert fibered and so is the other piece of the JSJ decomposition.
Thanks for sharing the SnapPy session. This is very helpful.
@shashankmarkande - you are welcome. If I may suggest: you should accept Josh's answer, as it answers the question you asked. You do this by clicking the "check mark" just to the left of the answer (and just below the voting buttons). best,
|
2025-03-21T14:48:29.599642
| 2020-01-07T23:35:02 |
349949
|
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|
Stack Exchange
|
Hom spaces in (∞, 1)-categories
In ordinary category theory it is a well-known and important fact that the $\hom$ bifunctor into $\text{Set}$ preserves limits.
I am unable to find a reference for the corresponding fact in infinity category theory nor am I able to write down a proof myself.
I am looking for a statement of the following form:
Let $\mathcal{C}, \mathcal{D}$ be $(\infty, 1)$-categories (considered as a weak Kan complex preferably) and let $F: \mathcal{D} \to \mathcal{C}$ be a diagram. Then, for any object $X \in \mathcal{C}$, we have a natural equivalence between $\lim \hom(X, F(D))$ and $\hom (X, \lim F(D))$ where each $\hom$ is considered as a weak Kan complex.
I would presume that HTT would have a statement to this effect, but I am unable to find one.
I thought this was the definition of "limit".
@TheoJohnson-Freyd Limits can be defined as terminal cones, in which case something has to be proved about the relationship between $\infty$-categories of cones and the mapping space functor. They can be defined as certain Kan liftings, in which case the connection to hom is still less obvious. I am not sure whether anyone writing foundational material literally defines a limit cone as one sent to a limit in spaces under hom. This would depend on giving an explicit definition of limits in spaces, which is not so easy as in sets.
This is Cisinski, Corollary 6.3.5. The proof is essentially to show that cocontinuous functors out of presheaf $\infty$-categories admit right adjoints given by the usual formula, so that $Hom_C$ has a left adjoint if $C$ is cocomplete, and then to use the Yoneda embedding for a general $C$.
This is explained in the opening paragraph of HTT.5.5.2; it’s a combination of the fact that both the Yoneda embedding and evaluation in functor categories preserve limits.
|
2025-03-21T14:48:29.599796
| 2020-01-07T23:52:42 |
349950
|
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|
Stack Exchange
|
Average value of $\prod_{p|d}{p-1\over p-2}$ for $d=nq$, $n\in{\mathbb N}$, with $p$ odd prime
$\newcommand{\mean}{\mathop{\mathrm{mean}}}$
Define
$$
S(d) = \prod_{p|d\atop p>2}{p-1\over p-2}.
$$
Bombieri and Davenport (1966) proved that
$$
\mean\limits_{d\in{\mathbb N}} S(d) =
\mean\limits_{d\in{\mathbb N}} \prod_{p|d\atop p>2}{p-1\over p-2} ~=~
\Pi_2^{-1} = 1 / 0.66016\ldots = 1.51478\ldots\,, \tag{1}
$$
where $\Pi_2=0.66016\ldots$ is the twin prime constant.
It is not difficult to check that the average value of $S(d)$ remains unchanged ($=\Pi_2^{-1}$) if $d$ runs through an arithmetic progression $d=qn$ with $q=2^k$; in particular the average of $S(d)$ is $\Pi_2^{-1}$ when $d$ runs through all positive even integers.
Some experimentation with PARI/GP leads me to the following
Generalization of $(1)$ for an arithmetic progression $d=nq$, $n\in{\mathbb N}$:
$$
\mean\limits_{d=nq\atop n\in{\mathbb N}} S(d) =
\mean\limits_{d=nq\atop n\in{\mathbb N}}
\prod_{p|d\atop p>2}{p-1\over p-2} ~=~
\Pi_2^{-1} \prod_{p|q\atop p>2}{p\over p-1}. \tag{2}
$$
For example, if $q=5$, then average of $S(d)$ over the progression $d=5n$ is $$\mean\limits_{d=5n\atop n\in{\mathbb N}} S(d) = {5\over4}\Pi_2^{-1}.$$
Can anyone please point me to existing proofs of $(2)$?
An idea/sketch of a proof would also be greatly appreciated!
Note: Let $q'$ be the largest odd divisor of $q$, then the "extra factor" in $(2)$ is
$$\prod_{p|q\atop p>2}{p\over p-1} = \prod_{p|q'}{p\over p-1} =
{q'\over\varphi(q')}.
$$
In particular, if $q$ itself is odd, then the "extra factor" is $q/\varphi(q)$, as was noted in comments.
Note that the extra factor equals $q/\varphi(q)$.
You may also find https://mathoverflow.net/questions/61842/about-goldbachs-conjecture insightful.
$(1)$ follows from the same Tauberian theorem as in the proof of the PNT. Then for $(2)$ use that $S(nq)= S(q/\gcd(n,q))S(n)$
Many thanks! Yes, for odd $q$ the extra factor is $q/\varphi(q)$. For even $q$, it's slightly less pretty.
Thanks a lot @reuns for your insight! Could you possibly expand your comment into an answer, especially the $(1)\Rightarrow(2)$ part, so I can upvote... :) It looks less and less likely that someone comes up with a reference for proof of $(2)$...
|
2025-03-21T14:48:29.599957
| 2020-01-08T06:37:29 |
349963
|
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|
Stack Exchange
|
Riemann's bilinear relations
I am reading the paper [1], which states
Haupt showed that a vector with complex entries $(w_1, \cdots, w_g, z_1, \cdots, z_g)$ is the period row of some holomorphic differential with respect to a canonical homology basis on some closed Riemann surface of genus $g$ if and only if
(a) $i\sum_j (w_j\bar{z_j} - \bar{w_j}z_j) > 0$
and
(b) no transformation corresponding to a change of canonical basis will bring the vector to the form $(w,0,\cdots,0,z,0,\cdots,0)$
I am looking at the necessity of these conditions for the time being. Condition (a) is standard and is covered in standard textbooks. However, I do not see any mention of part (b) in these textbooks, nor am I able to see why (b) should hold.
The only related thing I can think of are conditions on $g \times 2g$ period matrices which state that when the left $g \times g$ block of a period matrix is identity then the right $g \times g$ block is symmetric and has positive definite imaginary part. This seems to allow period vectors of the form specified in (b).
Why does (b) hold and how does one interpret this condition?
The author gives a reference [2] to the paper where this is proved.
It's in German and I don't know how to read German.
|
2025-03-21T14:48:29.600086
| 2020-01-08T08:47:12 |
349968
|
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|
Stack Exchange
|
Is unit ball in 2-Wassestein metric weakly compact?
This might be a trivial question, but I am trying to prove equi-coerciveness of some family of functions on the space of Probability measures on some space. I could reduce the problem to showing that $$\{\nu:\mathcal{W}_2^2(\mu, \nu)\le t\}$$ is compact (or is at least contained in some compact subset of $\mathcal{P}(\mathbb{R})$.)
The space of probability measures on $\mathbb{R}$ is equipped with weak topology. Now, I know from Banach Alaglou theorem unit ball in weak topology is compact. But here I am taking the ball in $\mathcal{W}_2$ metric. Can anyone tell me if this is correct or I am just wasting time?
Do you mean "contained in some compact subset of the space of probability measures", or do you mean "contained in some compact subset of the space of probability measures for which $\cal W_2$ is defined"?
Yes, it is true. It follows from Prokhorov's theorem that in order to prove (pre-)compactness, it suffices to prove tightness. However, if we define $K$ to be the compact set such that $\mu(\mathbb{R}\setminus K)<\varepsilon$, and $K_T:=\{x\in \mathbb{R}:\mathrm{dist}(K,x)\leq T\}$, then $\nu(\mathbb{R}\setminus K_T)>2\varepsilon$ implies that $\mathcal{W}_2^2(\mu,\nu)>\varepsilon T^2$ (since you have to move a mass $\geq\varepsilon$ over a distance $\geq T$), and so it is impossible for $\nu$ in your set provided that $\varepsilon T^2>t$. This proves tightness.
Since we care about probability measures, a sufficient condition for compactness is that the supports of the respective probability measures are compact. Here is the reasoning:
Let $X$ be a compact subset of $\mathbb{R}^d$. Denote by $C_0^\ast(\mathbb{R}^d)$ the space of all continuous functions on $\mathbb{R}^d$ that vanish at infinity. Then the Banach-Alaoglu theorem implies that the unit ball of $C^\ast_0(\mathbb{R}^d)$ is compact in the weak$^\ast$-topology. This implies that the set of probability measures is compact in the weak$^\ast$ topology.
In general, the weak$^\ast$ and the weak topology do not coincide for measures on $\mathbb{R}^d$, but they do for measures on the compact $X$, as then $C_0^\ast(X)=C_b^\ast(X)$, where $C_b(X)$ is the space of all bounded and continuous functions on $X$. This means that the set $\mathcal{M}(X)$ of all probability measures on $X$ equipped with the total variation norm is weak$^*$- and therefore weakly compact.
On compact sets the $p$-Wasserstein distance (for $p\in[1,\infty)$) metrizes weak convergence (Theorem 5.10 in Santambrogio 2015), from which it follows that the set you defined is compact since it is closed.
In my opinion, a great reference for these sorts of questions is:
Santambrogio, F (2015): "Optimal Transport for Applied Mathematicians", Birkhäuser
It is not true since $R^d$ is not compact. See page 34 of Ambrosio and Gigli's User's Guide (https://www.math.umd.edu/~yanir/OT/AmbrosioGigliDec2011.pdf).
To quote:
...if X is unbounded, then P2(X) is not locally compact. Actually,
for any measure $\mu\in P_2(X)$ and any $r > 0$, the closed ball of radius $r$ around µ is not compact. To see this, fix $\bar{x}$ ∈ X and find a sequence $x_n$ ⊂ X such that $d(x_n, \bar{x})$ → ∞. Now define the measures $\mu_n := (1 − \epsilon_n)\mu + \epsilon_n \delta_{x_n}$, where $\epsilon_n$ is chosen such that $\epsilon_n d^2(\bar{x},x_n)=r^2.$ To bound from above $W_2^2(\mu,\mu_n)$, leave fixed $(1-\epsilon_n)\mu$, move $\epsilon_n \mu$ to $\overline{x}$ and then move $\epsilon \delta_{\overline{x}}$ into $\epsilon_n \delta_{x_n},$ this gives $$W_2^2(\mu,\mu_n)\leq \epsilon_n \left(\int d^2(x,\overline{x})d\mu(x) + d^2(x_n,\bar{x})\right),$$
so that $\limsup W_2(\mu, \mu_n) ≤ r$. Conclude observing that $$\liminf_{n\rightarrow\infty}\int d^2(x,\bar{x})d\mu_n =\liminf_{n\rightarrow\infty}(1-\epsilon_n)\int d^2(x,\bar{x})d\mu+\epsilon_n d^2(x_n,\bar{x})=\int d^2(x,\bar{x})d\mu+r^2,$$
thus the second moments do not converge. Since clearly $\mu_n$ weakly converges to $\mu$, we proved that there is no local compactness.
The OP wanted compactness in the weak convergence topology (which is true), not in the Wasserstein-2 topology.
|
2025-03-21T14:48:29.600369
| 2020-01-08T08:54:15 |
349969
|
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|
Stack Exchange
|
Multiplicity of an irrep of SO(n-1) in SO(n)
I am trying to prove the following fact.
Let $V$ be a unitary irreducible representation of $SO(n)$. How to prove that, if we reduce $V$ as unitary irreducible representation with respect to SO(n-1) then each irreducible representation of $SO(n-1)$ occurs at most once in $V$.
Kindly share your thoughts.
Thank you.
This certainly is true: https://en.m.wikipedia.org/wiki/Restricted_representation. I believe a proof can be found somewhere in Knapp’s book on semisimple Lie algebras, though I don’t have a copy on me.
As far as I remember, the point is that $SO(n-1)$ is a (connected component) of fixed point group of an involution on $SO(n)$ (so the quotient is spherical).
@VictorPetrov, that is so; see @FriedrichKnop's answer.
It need to specify what is SO(n) (over what field, and defined by what quadratic forms). Anyway, there are two Annals papers on this question. https://arxiv.org/abs/0903.1413 https://annals.math.princeton.edu/wp-content/uploads/annals-v172-n2-p14-p.pdf
The statement is true and well known. See e.g. Thms. 8.1.3 and 8.1.4 in Goodman-Wallach: Symmetry, representations and invariants, Springer GTM 255.
In fact, much more is known. Let, more generally, $H\subseteq G$ be a subgroup (everything is compact). Then it follows easily from Peter-Weyl that $G$-to-$H$ branching is multiplicity free if and only if $\mathbb C[G]$ is multiplicity free as a $G\times H$-module. This means that $G\times H/{\rm diag}\, H$ is a spherical $G\times H$-variety. There are (up to trivial modifications of the centers) only two series of indecomposable pairs $(H,G)$ with multiplicity free branching namely $U(n-1)\subset U(n)$ and $SO(n-1)\subset SO(n)$. This was known much earlier to Kostant but also follows easily from independent classifications classifications of Brion and Mikityuk.
The spaces $G\times H/H$ for $G$ unitary or orthogonal can also be defined for indefinite scalar products or over local fields. Then they are called Gross-Prasad spaces. Gross and Prasad stated a conjecture on the multiplicity freeness of these spaces which has attracted a lot of attention in recent years.
I voted to close as not a research-level question, but this answer is so good it makes me re-consider.
|
2025-03-21T14:48:29.600549
| 2020-01-08T09:18:34 |
349971
|
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|
Stack Exchange
|
Survey papers on spectral radius
Let $M$ be a $n\times n$ matrix.
Are there any survey papers which give lower and upper bounds on its spectral radius?
What are the ways to find some lower bounds and upper bounds on $\rho(M)$ except using the fact that $\rho(A)\ge \frac{x^TMx}{x^Tx}$?
If someone could provide any way to find bounds or share any recent survey paper, I would be really grateful
Typo? Is $A = M$?
@GeoffRobinson; yes
The "fact" that $\rho(M)\ge \frac{x^TMx}{x^Tx}$ is not correct for general non-normal matrices; consider for instance $M = \begin{pmatrix} 0& 1 \ 0 & 0 \end{pmatrix}$.
I think your question is based on a false premise. There is not going to be a survey paper on something so general, especially since you seem to have overlooked nilpotent matrices (as @JochenGlueck has pointed out). If your underlying question is about getting lower bounds on tthe spectral radius, then you should edit your question to remove the stuff about survey papers
@YemonChoi: Small remark: The problem is not so much about nilpotent matrices only, but rather about the fact that the supremum of $\frac{\overline{x}^T M x}{\overline{x}^Tx}$ over all non-zero complex $x$ is the numerical radius rather than the spectral radius - and for non-normal matrices the numerical range can be considerably larger than the convex hull of the spectrum, so the numerical radius can be much larger than the spectral radius. (Of course, this comment does not contradict your remark that the question is very broad.)
If the matrix is not positive definite, this lower bound might be useful [source]:
An overview of bounds of positive definite matrices is given in Some new estimations for the upper and lower bounds for the spectral radius of nonnegative matrices (2015). A survey focusing on the adjacency matrix of graphs is the book Spectral Radius of Graphs (2014).
|
2025-03-21T14:48:29.600712
| 2020-01-08T10:54:43 |
349979
|
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|
Stack Exchange
|
How do I check if two linear binary codes are equivalent?
Suppose I have a list of generator matrices $G_i$, $i=1,\ldots N$, of the same size (each defines an $n$-bit linear binary code encoding $k$ logical bits).
I consider two codes to be equivalent if they contain the same codewords, or if they differ by permutation of the physical bits. This means that, if $G'$ is a generator matrix obtained from $G$ by permutation of columns, or elementary row operations (adding and subtracting rows), then $G'$ defines a code equivalent to the code defined by $G$. I believe this is the standard notion of equivalence.
I want to prune my list of generator matrices to contain only inequivalent codes. One way to partially prune is to put each matrix in reduced row echelon form, and then just delete duplicates, but that doesn't take into account column swaps. Is there a simple way to also take into account column swaps?
If the best solution is exponential time, that is also OK, but I'd like to know what it is.
I would have thought that the (or at least a) standard notion of equivalence is that the code subspaces are the same. In this case you could just combine pairs of matrices and compute ranks, and compare to the ranks of individual matrices.
Can you clarify what you mean?
There is a nice overview of this problem in the beginning of the following paper by Sendrier and Simos which is a good place to start.
Essentially the problem is harder than Graph Isomporphism (see p.5 of linked paper). The following is a quote.
Due to its relation to Graph Isomorphism, some researchers have tried to
solve the Permutation Code Equivalence problem by interpreting graph
isomorphism algorithms to codes. This approach, was followed in [5] using the
fact that ELC orbits of a bipartite graph correspond to equivalence classes of
binary linear codes. Mapping codes to graphs and using the software Nauty
by B. D. McKay has been used in [19], for binary, ternary and quaternary
codes where the permutation, linear and semi-linear equivalence was considered, respectively
References cited above:
Danielsen, L.E., Parker, M.G.: Edge local complementation and equivalence of binary linear codes. Designs Codes Cryptography: 49, 161–170 (2008)
Ostergard, P.R.J.: Classifying subspaces of hamming spaces. Designs Codes Cryptography: 27, 297–305 (2002)
|
2025-03-21T14:48:29.600884
| 2020-01-08T11:41:46 |
349981
|
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|
Stack Exchange
|
WKB expansion for NLS
We consider the equation (NLS)
\begin{eqnarray}\label{gnls}
i \epsilon\partial_t u^{\epsilon} + \frac{\epsilon^2}{2}\Delta_{\eta}u^{\epsilon} = \epsilon |u^{\epsilon}|^{2}u^{\epsilon}, \quad x \in \mathbb R^d
\end{eqnarray}
$d\geq 1$ The initial data is supposed to be given by a
superposition of highly oscillatory plane waves, i.e.
\begin{eqnarray}\label{icscm}
u_0^{\epsilon}(x) = \sum_{j\in J_0} \alpha_j(x) e^{ik_j\cdot x/ \epsilon},
\end{eqnarray}
where for some index set $J_0 \subset \mathbb N$ we are given initial wave vectors
$\Phi_0= \{ k_j \in \mathbb R^d: j\in J_0\}$ with corresponding smooth, rapidly decaying amplitudes $\alpha_j \in \mathcal{S}(\mathbb R^d).$
We seek an approximation of the exact solution $u^{\epsilon}$ in the following form
\begin{eqnarray}\label{as}
u^{\epsilon}(t,x)\sim_{\epsilon \to 0} u^{\epsilon}_{app} (t,x) = \sum_{j\in J} a_{j} (t,x) e^{i \phi_j(t,x)/\epsilon} \ \quad (*)
\end{eqnarray}
I'm trying to understand the following paragraph, p. 8-9:
Formally, we plugging the approximation (*) into (NLS), and comparing equal powers of $\epsilon$, we find that the leading order term is of order $ \mathcal{O}(\epsilon^0).$ It can be made identically zero, if for all $j\in J$
\begin{eqnarray}
\partial_t\phi_j + \frac{1}{2} |\Delta \phi_j|^2=0.
\end{eqnarray}
Questions are: (A) What is the leading order term of equation that is formed after plugging the approximation (*) into NLS ? (B) Why it can be made zero if if for all $j\in J$
\begin{eqnarray}
\partial_t\phi_j + \frac{1}{2} |\nabla \phi_j|^2=0.
\end{eqnarray}
A. Note that the time derivative $\partial u^\epsilon/\partial t$ and the spatial derivative $\partial u^\epsilon/\partial x$ are both of order $1/\epsilon$, since $u^\epsilon\propto e^{i\phi(t,x)/\epsilon}$. Since in the NLS the first-order time derivative has a prefactor $\epsilon$ and the second-order spatial derivative has a refactor $1/\epsilon^2$, the powers of $\epsilon$ cancel and to leading order the NLS is of order $\epsilon^0$.
B. This term of order $\epsilon^0$ is
$$\sum_j a_j(t,x)e^{i\phi_j(t,x)/\epsilon}\left[\frac{\partial}{\partial t}\phi_j+\frac{1}{2}\left(\frac{\partial}{\partial x}\phi_j\right)^2\right],$$
so to make it vanish the term between square brackets should vanish for all $j$.
I think that $\Delta$ should be $\nabla$.
indeed it should, thanks.
|
2025-03-21T14:48:29.601046
| 2020-01-08T11:48:05 |
349982
|
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|
Stack Exchange
|
When does a graph have a circular orientation? Or equivalently can anyone help me characterize this particular class of $3$-colorable perfect graphs?
Call an oriented digraph $D=(V,A)$ circular when for all $\small x,y,z\in V$ if $(x,y)\in A$ and $(y,z)\in A$ then $(z,x)\in A$ or equivalently if $D$ is any oriented digraph whose arc set is a circular relation.
With that said, when does an undirected graph have a circular orientation?
I can prove these graphs are perfect i.e. graphs with circular orientations have identical clique and chromatic numbers for all their induced subgraphs, also I can show they are $3$-colorable and there exists a family of forbidden induced subgraphs which characterizes them. Its also easy to see these graphs look similar in their definition to comparability graphs i.e. graphs which have an orientation $D=(V,A)$ such that for all $x,y,z\in V$ if $(x,y)\in A$ and $(y,z)\in A$ then $(x,z)\in A$ (this last arc in the definition of circular orientations is flipped) also like graphs with circular orientations, these (graphs with transitive orientations) are similarly perfect graphs. Now Gallai proved the countable set $S$ below, of forbidden induced subgraph isomorphism types characterized every comparability graph:
$$\small S=\{(G_k)^{\complement}:1\leq k\leq 8\}\cup\{B_1^{\complement},B_2^{\complement}\}\cup\bigcup_{n=2}^{\infty}\{C_{2n+1},J_n,J'_{n+1},J''_n,(K_n)^{\complement},(C_{n+4})^{\complement},(L_{n-1})^{\complement},(L'_{n-1})^{\complement}\}$$
Where the indexed types $G,B,K,L,C,J,J',J''$ are each defined diagrammatically as follows:
Thus surely characterizing those graphs with circular orientations by a family of forbidden induced subgraphs can not be harder then this? I mean they have bounded clique/chromatic number so I'd expect them not to be that complicated, not like the above graph types. Perhaps even looking for a forbidden induced subgraph characterization is overkill, though there must be some simple way to characterize these graphs, if fifty years ago Tibor Gallai managed to do this for the class of far more complicated, comparability graphs. Right? If so I'd appreciate any help.
There have been numerous edits in a short span of time. Please be aware that every edit bumps the post to the top of the stack, and pushes off the front page other questions that are vying for attention.
@ToddTrimble Sorry I noticed some errors and then fiddled a lot with the formatting. Isn't there a stop gap which prevents bumps unless successive edits take a long enough time span? It would prevent this sort of problem for people obsessing over details.
There's a 5-minute window for combining multiple edits into one.
The class is exactly the class of bipartite graphs ∪ complete 3-partite graphs.
Such a graph must be paw-free. To check this, note that the orientation of the triangle in the paw must be cyclic, and check the two cases where the remaining edge points to or away from the triangle.
According to ISGCI, a graph both paw-free and perfect is a bipartite graph or a complete multipartite graph.
As such graphs have chromatic number≤3, they are bipartite graphs or complete 3-partite graphs. And it's easy to check that these two types of graphs are circularly orientable.
EDIT: There's a proof that a 3-partite circularly orientable graph is complete.
Proof. It's easy to check the case where the graph have 3 vertices. Suppose $H$ is the smallest counterexample. As $H$ is 3-partite, call the three color classes $A$, $B$ and $C$. Without loss of generality, assume $A$ has more that 2 vertices and $a$ is a vertex of $A$, and every vertex of $A\backslash a$ is connected to every vertex of $B$, and that holds for $B$ to $C$ and $C$ to $A\backslash a$.
Since the graph is connected, $a$ is connected either to some vertex in $B$ or some vertex in $C$. If $a$ is connected to some vertex in $C$, then by circularity, every vertex in $A\backslash a$ is connected to $a$, which is a contradiction. So $a$ is connected to some vertex in $B$. By circulatity, every vertex in $C$ is connected to $a$, and by the same reasoning, $a$ is connected to every vertex in $B$. Thus $H$ is complete, and is not a counterexample.
So there are no counterexamples, which proves the claim.
Why is a circular orientation of a complete 3-partite graph Hamiltonian? You can try that with the 3 parts having size 1,2,1.
Sorry I made an error, you are correct. The only part I don't understand, is how a connected graph $G$ with $\chi(G)=3$ that is not a complete tripartite graph, will not have a circular orientation. How would you prove this, say by contradiction somehow? I do not mean to sound ungrateful, but could you write me a short sketch/proof of this? I just don;'t understand why this is the case.
The new edit may answer your question.
What do you mean by $H$ is the smallest counter example? $H$ is a graph with the least number of vertices that is a counter example? Or edges? I'm confused with how you are describing connectivity between the color classes. I think what you're saying is that if $H$ is a graph with the least number of vertices serving as a counter example, then for any vertex $a\in V(H)$ the induced subgraph $G=H[V(H)\setminus{a}]$ must be a complete tripartite graph, and then from there you are somehow deriving that $H$ is a complete tripartite graph from $G$ which is a contradiction. Is this what you mean?
Yes. "Smallest" means the graph with the least number of vertices, and the phase "$x$ is connected to $y$" means there is an edge $x → y$.
Let us continue this discussion in chat.
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2025-03-21T14:48:29.601525
| 2020-01-08T11:51:14 |
349983
|
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"Markus Lange-Hegermann",
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|
Stack Exchange
|
Relation between Gaussian processes and RKHSs with tensor product kernels
For sets $\cal X$ and $\cal Y$, let $a:{\cal X}\times{\cal X}\rightarrow \mathbb{R}$ and $a:{\cal Y}\times{\cal Y}\rightarrow \mathbb{R}$ be positive definite symmetric kernels. Define the tensor product $a\otimes b$ as usual via
$$ (a\otimes b)((x,x'),(y,y')) = a(x,x')a(y,y') $$
for all $x,x'\in \cal X$ and $y,y'\in \cal Y$.
A positive definite kernel defines both a Gaussian process (GP) and a reproducing kernel Hilbert space (RKHS), also known as the Cameron-Martin space of the GP (and defined as the set of shift vectors for the GP that preserve equivalence).
Define ${\cal R}_k$ as the RKHS with reproducing kernel $k$.
Defining
$$ {\cal R}_a\otimes{\cal R}_b = \langle\{f\otimes g|f\in{\cal R}_a,g\in{\cal R}_b,\}\rangle ,$$
the vector product tensor product, it is well-known that
$$ {\cal R}_{a\otimes b} \subset {\cal R}_a\otimes{\cal R}_b $$
where the subset is strict in the infinite dimensional case, and
\begin{equation}\tag{1}\label{closure} {\cal R}_{a\otimes b} = \overline{{\cal R}_a\otimes{\cal R}_b} \end{equation}
where the closure is with respect to the tensor product norm.
My question is what can we say about the supports of the corresponding Gaussian processes? Denoting an appropriately defined support of a Gaussian process $GP(0,a)$ by ${\cal S}_a$, is it true that
$$ {\cal S}_{a\otimes b} \subset {\cal S}_a\otimes{\cal S}_b $$
and if so is it possible to define some kind of closure operation so that an analogue of \eqref{closure} holds?
Background: GPs with tensor product kernels are used a lot in GP regression, so an answer could help to understand why it may make sense to use GPs with tensor product kernels.
Is the tensor product with ${\cal R}_{a\otimes b} \subset {\cal R}_a\otimes{\cal R}b$ the tensor product in the category of vector spaces?
Because, if you take the tensor product in the category of Hilbert spaces (which is the composition of the vector space tensor product, followed by the closure monad), then ${\cal R}{a\otimes b} = {\cal R}_a\otimes{\cal R}_b$.
If I remember correctly, the support of a GP was just the closure of its RKHS.
This closure depends on the topology.
Then, one would "just" need to check which tensor product is the correct one in the category where the supports live (Banach spaces, Frechet spaces, ...) and check whether the closure RKHS$\to$support commutes with the tensor product.
This might again depend on the topology.
Thanks for the comments, I clarified the tensor product definition in the post (so this should be in the category of vector spaces).
Thanks, that seems to be the right direction (support of a GP is the closure of its RKHS), but I do not intuit this at all yet. According to Lukic and Beder (https://www.ams.org/journals/tran/2001-353-10/S0002-9947-01-02852-5/S0002-9947-01-02852-5.pdf), in the infinite dimensional case the probability that a sample path of a GP lies in the corresponding RKHS is zero. So some definitions of support of the GP (say, roughly speaking, the smallest set such that sample paths are in it with probability one) should exclude the RKHS I guess...
The RKHS is dense in the support but still has measure zero. This is similar to the rationals, which are dense in the support of the standard Gaussian but still have measure zero.
For simplicity, I assume that the Gaussian process $g$ has mean zero such that it is completly specified by its covariance function $k$.
Note that the support of a (Borel) measure depends on the chosen topology.
So I cannot give a general answer, which is independet of the chosen topology.
For non-Borel measures, the question is probably much more complicated.
Let $\mathcal{F}$ be a topological $\mathbb{R}$-vectorspace of functions $X\to\mathbb{R}$, where $X\subseteq\mathbb{R}^d$.
We assume furthermore that $\mathcal{F}$ is a Banach space (much of the following also holds for Frechet spaces, I think) and that all realizations of $g$ are contained in $\mathcal{F}$.
We have to distinguish Tensor products and closures.
Denote by
$\operatorname{clo}_\mathcal{R}$ the closure in the category of Hilbert spaces.
$\operatorname{clo}_\mathcal{F}$ the closure in the category of Banach spaces.
Denote by
$\otimes_\mathbb{R}$ the tensor product in the category of $\mathbb{R}$-vectorspaces.
$\otimes_\mathcal{R}$ the tensor product in the category of Hilbert spaces, which is the tensor product $\otimes_\mathbb{R}$ in the category of vector spaces followed by the closure in the Hilbert space, i.e. $\otimes_\mathcal{R}=\operatorname{clo}_\mathcal{R}\circ\otimes_\mathbb{R}$.
$\otimes_\mathcal{F}$ the tensor product in the category of Banach spaces, which is the tensor product $\otimes_\mathbb{R}$ in the category of vector spaces followed by the closure in the Banach space, i.e. $\otimes_\mathcal{F}=\operatorname{clo}_\mathcal{F}\circ\otimes_\mathbb{R}$
Now, by Lemma 5.1 from ``Reproducing kernel Hilbert spaces of Gaussian priors'' (https://arxiv.org/abs/0805.3252v1) the closure (w.r.t. its own norm) $\operatorname{clo}_\mathcal{R}\mathcal{R}(k)$ of the RKHS $\mathcal{R}(k)$ of $k$ is the support $\operatorname{sup}(g)$ of the Gaussian process $g$ in $\mathcal{F}$.
As the question already mentions, given two covariance functions $k_1$ and $k_2$ with corresponding Gaussian processes $g_1$ and $g_2$, we have $\mathcal{R}(k_1\otimes k_2)=\mathcal{R}(k_1)\otimes_\mathcal{R}\mathcal{R}(k_2)$, where the first tensor product $k:=k_1\otimes k_2$ is just the pointwise product of the covariance functions $k_1$ and $k_2$.
This implies for the Gaussian process $g$ with covariance $k$ that
\begin{align*}
\operatorname{supp}(g)
&=
\operatorname{clo}_\mathcal{F}(\mathcal{R}(k_1\otimes k_2))
\\
&=
\operatorname{clo}_\mathcal{F}(\mathcal{R}(k_1)\otimes_R\mathcal{R}(k_2))
\\
&=
\operatorname{clo}_\mathcal{F}(\operatorname{clo}_\mathcal{R}(\mathcal{R}(k_1)\otimes_\mathbb{R}\mathcal{R}(k_2)))
\\
&=
\operatorname{clo}_\mathcal{F}(\mathcal{R}(k_1)\otimes_\mathbb{R}\mathcal{R}(k_2))
\\
&=
\mathcal{R}(k_1)\otimes_\mathcal{F}\mathcal{R}(k_2)
\\
&=
\operatorname{clo}_\mathcal{F}(\mathcal{R}(k_1))\otimes_\mathcal{F}\operatorname{clo}_\mathcal{F}(\mathcal{R}(k_2))
\\
&=
\operatorname{supp}(g_1)\otimes_\mathcal{F}\operatorname{supp}(g_2)
\end{align*}
This answers your question in the positive, at least for the above assumptions (mainly: $\mathcal{F}$ is Banach).
Probably, your question still has a positive answer without out such assumptions.
Another good reference, which shows how complicated these questions are, is https://arxiv.org/abs/1604.05251.
Many thanks! Just two queries if I may. For the fourth equality, so if $\cal F$ is a Banach space and $\cal R$ an RKHS, then $\text{clo}{\cal F}\circ \text{clo}{\cal R}=\text{clo}_{\cal F}$? Similarly the sixth equality, this is valid because the closure of the tensor product of two spaces is the closure of the tensor product of their closures?
@Wicher: yes, you are right.
|
2025-03-21T14:48:29.601905
| 2020-01-08T13:09:59 |
349987
|
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|
Stack Exchange
|
Strict transform by normalization
I am trying to understand the following definition in C. Sabbah's paper
(Quelques remarques sur la géométrie des espaces conormaux), page 186, Numdam link.
Let $\phi\colon X\to \mathbb{C}^2$ be a map such that
X is irreducible
$\phi^{-1}(t)$ is of dimension $\dim X-2$ for $t\neq 0$.
$\phi$ factors through a closed imbedding $X\to M$ and a smooth morphism $M\to \mathbb{C}^2$.
Now, let $(C,0)\subset (\mathbb{C}^2,0)$ be a germ of irreducible curve, and let $p:\hat{C}\to C$ be a normalization. Sabbah used the following phrase:
"The strict transform of $X$ by $p$" (I denote it by $X_C$ here.)
My question is, what is the definition of this phrase? In the appendix, he was trying to show that, let $D_C$ be the fiber of $X_C$ over $0$, then $[D_C]$ is independent of the choice of $C$ in $H_*(|\phi^{-1}(0)|)$.
I think I got what it means by strict transform in the paper. In the Stacks Project 31.33, the strict transform is defined as follows
Let $f\colon X\to B$ be a morphism, let $p\colon \hat{B}\to B$ be the blowup of $B$ along a subscheme $Z$, then the strict transform $\hat{X}$ of $X$ by $p$ is just the blowup of $X$ along $f^{-1}(Z)$.
In this situation, this $X_C$ is just the strict transform of $\phi^{-1}(C)$ by $p$.
|
2025-03-21T14:48:29.602028
| 2020-01-08T13:55:08 |
349990
|
{
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"Dominic van der Zypen",
"Gerald Edgar",
"Martin Sleziak",
"Mateusz Kwaśnicki",
"https://mathoverflow.net/users/108637",
"https://mathoverflow.net/users/454",
"https://mathoverflow.net/users/8250",
"https://mathoverflow.net/users/8628"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/349990"
}
|
Stack Exchange
|
Continuous function $f:\mathbb{R}\to\mathbb{R}$ with fixed size finite fibers
During a business meeting, I was trying to find a continuous function $f:\mathbb{R}\to\mathbb{R}$ such that $|f^{-1}(\{y\})| = 2$ for all $y\in \mathbb{R}$, and after some experimentation I found $$f:\mathbb{R}\to\mathbb{R}, \, x\mapsto \log(x^2).$$
For which integers $n>2$ is it possible to find a continuous function $f:\mathbb{R}\to\mathbb{R}$ such that $|f^{-1}(\{y\})| = n$ for all $y\in \mathbb{R}$?
The function $f(x)=\log(x^2)$ which you suggested is undefined for $x=0$. (And I don't think it can be continuously extended.)
Can be done with fibers of size $1$ or $3$ or $5, \dots$
BTW a reformulation of this problem is that you want Banach indicatrix to be constant and finite-valued.
If $n$ is odd, $x+\alpha_n \sin x$ will work for an appropriate $\alpha_n$. And this is not possible for $n$ even, by the intermediate value theorem. We necessarily have $f(x)\to \pm\infty$ as $x \to \pm\infty$ or $f(x)\to \mp\infty$ as $x \to \pm\infty$. Assume the former, and write $f^{-1}(a)={x_1,\ldots,x_n}$. Then at least one of $x_1,\ldots,x_n$ is a (strict) local extremum. This means that $f$ has uncountably many strict local extrema, which is not possible.
Apologies for this incredibly stupid non-example!
There is no continuous function $f:\,\mathbb{R}\to \mathbb{R}$ such that $|f^{-1}(x)|=2k$ for any $x$ (here $k$ is some fixed positive integer.)
Indeed, let $p_1<p_2<\ldots<p_{2k}$ be preimages of $0$. Fix small $\delta$ so that the $4k$ intervals $(p_i-\delta,p_i)$ and $(p_i,p_i+\delta)$ are mutually disjoint. Note that if $\varepsilon>0$ is small enough, then each of them contains a preimage either of $+\varepsilon$ or $-\varepsilon$. But $+\varepsilon$ and $-\varepsilon$ have $4k$ preimages in total, therefore exactly $2k$ intervals contain preimages of $+\varepsilon$ (call them ''positive'') and $2k$ intervals contain preimages of $-\varepsilon$ (call them ''negative''). Note that the intervals $(p_{i+1}-\delta,p_{i+1})$ and $(p_i,p_i+\delta)$ are simultaneously positive or simultaneously negative, since $f$ has constant sign on $(p_i,p_{i+1})$. Thus by parity of the number of positive intervals we conclude that the intervals $(p_1-\delta,p_1)$ and $(p_{2k},p_{2k}+\delta)$ also have the same sign. Since $f$ preserves the sign on the rays $(-\infty, p_1)$ and $(p_{2k},+\infty)$, we see that $f$ must have the same sign on these rays, and therefore $f$ is bounded either from above or from below. This contradicts to surjectivity.
Ah, I just realize that this is proved by Mateusz Kwaśnicki in the comments, but slightly differently (above reasoning does not use cardinality).
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