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2025-03-21T14:48:30.217094
| 2020-04-04T00:28:19 |
356510
|
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|
Stack Exchange
|
Bounding the size of the conjugating elements given the Dehn function
I am learning a little bit about Dehn functions of group presentations and I came across a question that is probably pretty basic but that I was giving me trouble. I'll set some notation but essentially I want to understand why being able to compute an upper bound for the Dehn function means that the word problem is solvable.
Let $\phi: F(A) \to G$ be a surjective homomorphism of groups where $A$ is a finite set and $F(A)$ is the free group on $A$. Also assume that the kernel of this homomorphism is normally generated by a finite set $R$. In other words, we are given a finite presentation of $G$. Let $\delta : \mathbb{N} \to \mathbb{N}$ be the Dehn function of this presentation, namely, $\delta(n)$ is the smallest number such that every element $w \in F(A)$, with $\ell(w) \leq n$ and $\phi(w) = 1$, as the product of conjugates of $\delta(n)$ elements of $R$ (or there inverses). Namely, for such $w$, we have $$w = \prod_{i=1}^M p_i r_i^{\epsilon_i} p_i^{-1}$$ with $p_i \in F(A)$, where the above equality is in $F(A)$.
I was told that we can assume that there is a way to find such an expression with $M \leq \delta(n)$ and $\ell(p_i) \leq \ell(w) + M \rho$ where $\rho$ is the max length of the elements in $R$. Why is this the case?
I was thinking of this in terms of van Kampen diagrams where I see equations of the above from as "flowers". The Dehn function tells us how many "petals" we need. But I was hoping to have short "petals" (and not too many). Can you help me shorten the petals?
In addition to references given by Mark Sapir: see Lemma 2D2 here and its short proof (modulo the existence of van Kampen diagrams). It says, for a presentation $\langle S|R\rangle$ ($S,R$ not necessarily finite) with $M=\sup_{x\in R}|x|$ that is a relation $w\in F_S$ has length $n$ and area $A$ and then $w$ can be written as product of $A$ conjugates of $\le A$ elements of $R^\pm$ where the conjugating elements have length $\le n+MA$. Mark mentions the upper bound $\le n+4MA$, so I hope we made no mistake!
This is an example of a result in combinatorial group theory for which I have never seen a purely algebraic proof. That is, a proof that makes no use of van Kampen diagrams, which are guess make implicit use of the geometry of the Euclidean plane.
|
2025-03-21T14:48:30.217272
| 2020-04-04T00:59:47 |
356512
|
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|
Stack Exchange
|
How do mathematicians find coauthors?
I am totally new to academia so I am really not sure how mathematicians works together, can more experienced mathematicians here shed some light on how you find coauthors? I guess one way to do this is to attend conference. But if one doesn't have the chance to do so, are there other options?
https://academia.stackexchange.com/search?q=find+collaborator
@NateEldredge Thanks, the key word 'find coauthor' that I used was not so helpful.
How do mathematicians find coauthors?... Using MO!! :D
And, of course, keep in mind that (even today) the majority of mathematics papers have only one author. A way to find collaborators could be: publish your papers, and wait until someone notices them and wants discuss them with you.
@PraphullaKoushik: has anyone ever started a collaboration on MO?
@Taladris you have examples at https://meta.mathoverflow.net/questions/617/
Alright, I'll try to outline several ways of how mathematicians find co-authors (since I'm still quite young, I lack the experience to judge how things might change later during a career, but given that you're new to academia, I hope that what follows fits your situation):
I'll try to write my answer as a kind of classification scheme.
Preliminary remark concerning terminology:
Actually, mathematicians do not find co-authors - they find collaborators. By this I mean that it is probably not a good idea to approach people in order to "write a paper with them", but rather in order to "discuss and work with them on an interesting problem or idea." If this work turns out to be fruitful (and novel), you can then turn your results into a paper that you co-author together.
.
Your advisor:
When you're new to academia, you are most likely doing a PhD (or something related), so you will have an advisor. In some (not in all) cases your advisor is the most natural choice to discuss your ideas, seek for explanations, and so on. Depending on how large your advisor's share of your research is, you might write a paper alone or together with your advisor.
How to start a collaboration with your adviser:
Please, be advised that there are very different types of advisors who work and behave quite differently. Some might push you quite often und actively urge you to work with them, others are much more reserved and will let you do your thing unless you actively go for a collaboration with them.
Make sure that you find out and keep in mind what type of advisor yours is.
.
Other people from your institution.
Another natural choice is to work with other people from your institution. This can vary in several ways:
You might work with people at your level, with people who are a bit more advanced in their career than you are, or with people who are much more experienced than you. When you are completely new to academia, it can be helpful to work with people who have a bit more experience than you (say, more advanced PhD students, or postdocs). They can offer you a bit of advice and guidance (in addition to your advisor).
You might work with people who are very close to your field or research project, or you might discuss with people who have specialized in other things, and when you find a common question of interest, you can combine your knowledge and ideas from two different fields to attack the question. Note that this can range from people working in fields nearby to people who do things that are, at first glance, very different from yours (but I would suspect that the "younger" you are, the better it is if you work with people with similar background to yours).
How to start a collaboration with people from your institution:
First rule: Meet people, and discuss with them - at lunch, at tea, whenever. Do not sit alone in your office or at home all the time (well, Corona won't last forever...). Do not think "I'll discuss with XYZ now, maybe we can work on a project together", but rather think "I have a question in mind in which XYZ might be interested; let's see what she/he thinks about it."
Of course, details depend on which fields people are working in. Is there a PhD student or a postdoc with a field of research that is similar to yours? Go and discuss with them as often as you can! Ask them for their expertise, ask them questions you are thinking about (not necessarily research questions; also things which are probably known, but not to you, yet). Ask them what they are thinking about. Is there a postdoc in a field that is related, but not quite them same as yours? Ask them which problems they find interesting, and why. Ask them if they know how their work relates to your field, and if they have done similar things as you do.
$(*)$ Important:
After discussing a topic that you found interesting, re-think what you discussed, look things up, keep coming back to your colleagues with new information, new insights or new questions. Similarly, if somebody keeps coming back to you, and you find their questions or insights interesting, take the opportunity to discuss even more with them. After all, collaborations are not planned or directed - they grow.
One additional remark:
Sometimes collaborations which might seem a bit surprising at first glance can occur, as I can illustrate by a personal anecdote: During my PhD I gained some experience in functional analysis and operator theory, and briefly after I started my first postdoc position, a PhD student from stochastics knocked on my door: he was trying to solve an inverse problem in statistics which was given by a linear integral equation, and somebody had apparently told him that I'm the linear operator guy. Fortunately, I had time (or, say: I took the time), so we discussed the issue in several meetings and finally resolved it. So in the end we wrote a joint paper, together with his advisor, that mixed up functional analysis and mathematical statistics. What do we learn from this anecdote? Well, if a problem occurs in your research that seems to stem from another field, it can be a good idea to just knock on somebody's door. And if somebody knocks on your door, it can be a good idea to take some time for them.
.
People from other institutions, part 1: visits at your university.
Researchers usually travel a lot, so it will probably happen quite often that people from other institutions come to your university to give a talk or/and to stay for a (often short) period where they work face-to-face with people from your institute.
These researchers might have been invited by your advisor, or by somebody else, and their visits can be a good opportunity to find collaborators.
How to start a collaboration with a visiting researcher:
Well, I'm beginning to repeat myself: Talk to them, ask them questions, discuss. Most visiting researchers give a seminar talk, so attend such talks; if you find the topic of the talk interesting or useful (let alone both), ask them questions about it (after the talk, at tea, during lunch - often there's also a joint dinner after a talk, so you can go there, too). If you want to discuss even more, just knock on the research visitor's door. Most people are happy when they note that somebody is interested in their work. Besides all this: $(*)$ applies, of course.
.
People from other institutions, part 2: visit other institutions yourself (if possible).
Some PhD students have the possibility for research stays (a few days or even a few weeks) at other institutions. Since such visits are planned in advance, your host will expect you and will be prepared to spend considerable time to work with you, so this is a great chance for collaboration.
How to visit other institutions.
Of course, it depends on whether your institution (or your advisor) has sufficient funding. If funding is avaible, your advisor is most likely to know good opportunities for a visit or a research stay. Your advisor knows your research topic(s), and she/he knows other researchers in the field, so she/he might ask them whether it is possible for you to visit them.
If your advisor does not suggest a research stay on their own - just ask.
.
People from other institutions, part 3: conferences and workshops.
Of course, you can meet a lot of interesting and clever people there, and it is a good opportunity to find collaborators. However, many (not all) young academics tend to have a few misconceptions about conferences, so here is some advise:
How to find collaborators at a workshop / conference:
First, and most importantly: A conference is a social event! This means, it is not all about the talks - on the contrary, it is about the talks only as far as talks are social interactions themselves. More concretely, this means (among other things):
If you attend a talk, ask questions! Do not be afraid to ask stupid questions. (In contrast to what some people claim, there do exist stupid questions, but the point is that they do no harm at all.) I tend to ask many questions, and a considerable fraction of them turns out to be stupid afterwards. But I learn from both types of questions (the stupid and the good ones), and people tend to remember the good ones much better than the stupid ones (at least I believe so).
The coffee breaks are important! Do not waste them with the preparation of your talk (instead, have your talk prepared before the conference commences), or with reading a paper, or checking emails. Go drink coffee and talk to people (I personally don't like coffee nor tea, so I eat cookies instead). Ask the speakers additional questions there if you found their talk interesting; if you gave a talk, be there to give people the opportunity to ask you questions.
This way, you will meet many people, and you will notice that you share common interests (and expertise!) with some of them. If this happens: Keep in touch. After the conference, continue discussions that were interesting via email, or via video calls. Again, $(*)$ applies (this time by means of electronic communication rather than face-to-face).
@MarkSapir: Hmm, then we seem to have quite different approaches. I don't have access to MathSciNet right now, so I counted my co-authors in arXiv: there are currently 14, and the above answer applies to 12.5 of them. I would be very interested to learn how you find your collaborators (except for using pick up lines, as you wrote in your answer :-)) - particularly, because I can only imagine few circumstances where one just starts a project together without getting to know each other first by discussing problems, common interests, etc. (But I'd be glad to improve my imagination.)
Often collaborations happen by mutual interest ― you start talking with someone, an interesting question comes up, and you decide to pursue it. I don't think I have ever attempted to find a coauthor just for the sake of finding a coauthor (but that doesn't mean you shouldn't do this). Most of the time, it's either with someone I already know really well (this should be easy to initiate!) or with someone who has worked on very similar stuff (which is also not so hard). It's easier if you're generous sharing your own (half or full) ideas.
Just like not all of my own projects end up working out, not every collaboration ends up working out. For example because you get stuck or someone loses interest (is overwhelmed by other work, etc). That's ok; don't beat yourself up or try to force something. But do make an effort.
You might find it easier to be the initiator (coming up with the idea and finding someone suitable), or you might find it easier to join someone else's project. In the latter case you could attend a lot of talks and show interest in other people's work (especially if you are indeed interested!).
|
2025-03-21T14:48:30.218018
| 2020-04-04T03:06:08 |
356522
|
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|
Stack Exchange
|
Does Langlands use the geometric Frobenius or the classical Frobenius in his papers?
In several of Langlands' papers: Representations of Abelian Algebraic Groups, On Artin's L-functions, On the Functional Equation of Artin's L-functions, Langlands takes a finite Galois extension $K/F$ of local fields and considers an exact sequence
$$1 \rightarrow K^{\ast} \rightarrow W_{K/F} \rightarrow \operatorname{Gal}(E/F) \rightarrow 1.$$
Here $W_{K/F}$ is the group $W_F/W_K^c$, where $W_K \subset W_F$ are the Weil groups of $K$ and $F$, and $W_K^c$ is the closure of the derived group of $W_K$. To define this exact sequence, one considers the reciprocity law isomorphism $K^{\ast} \rightarrow W_K^{\operatorname{ab}} = W_K/W_K^c$ and composes it with the map $W_K/W_K^c \rightarrow W_F/W_K^c$.
For example, here is from Langlands' short paper On Artin's L-functions.
My question is, what is the reciprocity law isomorphism $K^{\ast} \rightarrow W_K^{\operatorname{ab}}$ Langlands is using here? There are two: the classical one, which sends a uniformizer to an arithmetic Frobenius, and the "modern" one, which sends a uniformizer to a geometric (inverse) Frobenius.
I can't seem to find in these papers which convention Langlands is using. I originally assumed he was following the modern approach, but now I am not so sure. When he mentions "the Frobenius substitution" in his paper on abelian algebraic groups, I can't tell whether he is talking about the arithmetic Frobenius or the geometric one.
Does anyone who has read Langlands' early papers know which one he means?
|
2025-03-21T14:48:30.218146
| 2020-04-04T03:28:39 |
356525
|
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"GH from MO",
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|
Stack Exchange
|
How does Langlands define Artin L-functions?
Let $\rho: \operatorname{Gal}(K/F) \rightarrow \operatorname{GL}_n(\mathbb C)$ be a representation for an unramified extension $K/F$ of $p$-adic fields. Let $\operatorname{Frob}_{K/F}$ be the (arithmetic) Frobenius element of $K/F$. Artin defined the local L-function $L(s,\rho)$ by
$$L(s,\rho) = \operatorname{det}(I_n - \rho(\operatorname{Frob}_{K/F})q^{-s})^{-1}$$
where $q$ is the number of elements in the residue field of $F$.
Nowadays, it's more popular to define L-functions use the geometric (inverse) Frobenius. That is, a lot of people myself included would rather say that
$$L(s,\rho) = \operatorname{det}(I_n - \rho(\operatorname{Frob}_{K/F})^{-1}q^{-s})^{-1}.$$
Which definition does Langlands use in his papers? I can't tell. Here is where he defines Artin L-functions in his paper On Artin's L-functions:
He says that for a representation $\sigma$ of dimension one, the Artin L-function $L(s,\sigma)$ is defined to be the Tate L-function $L(s,\omega)$, if $\omega$ is the quasicharacter of $C_F = F^{\ast}$ corresponding to $\sigma$. But what is the correspondence $\sigma \mapsto \omega$? Langlands does not say.
It probably comes from however Langlands is choosing to identify of $F^{\ast}$ with the abelianized Weil group $W_F^{\operatorname{ab}}$ (which I have asked about in a previous question), but I do not know for sure.
You cannot redefine Artin $L$-functions, just as you cannot redefine Dirichlet $L$-functions and the Riemann zeta function. These objects have been with us for a century or more, so you must respect them and stick to them. I am 99% sure that Langlands would agree with this, and he uses the first display as the definition of Artin $L$-functions. The second display is also legitimate, of course: it is the dual of the Artin $L$-function. My two cents.
|
2025-03-21T14:48:30.218274
| 2020-04-04T05:11:31 |
356530
|
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"Andrei Smolensky",
"Will Dana",
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|
Stack Exchange
|
Stability of infinite root systems with a long path in their Coxeter diagrams
Given a Cartan matrix associated to a Coxeter diagram, I can modify it by replacing one of the edges in the diagram with a long chain of vertices connected by simply laced edges; for example, this is where most of the infinite families of finite or affine diagrams come from.
Given a positive root of the root system associated to the original Cartan matrix, we can write it in the basis of simple roots, and thus (using the association of simple roots to vertices) as a function on the vertices. I'm interested in taking a root written in this way and similarly stretching it out to fit the extended Coxeter diagram by making the coefficients on the added vertices all the same, like so:
In particular, I'm interested in what happens to the hyperplane arrangement of shards (in the sense of Nathan Reading, introduced here) of a root constructed this way as the chain gets longer.
But before looking into that, I'd like to know in general if anything more is understood about how root systems behave under this extension.
EDIT, to expand a bit on my motivation: given a simply laced Coxeter diagram $G$ and a particular positive root $\alpha$, I can get all the Schur representations of dimension vector $\alpha$ of the preprojective algebra associated to $G$ by applying reflection functors to simple representations, in a way mimicking the construction of $\alpha$ by applying simple reflections to a simple root. In types $A$ and $\widetilde{A}$, I can cleanly describe all such representations attached to a particular root, but this gets way more complicated in other cases, even type $D$; the key distinction appears to be that all vertices in the former cases have degree 2. So I was wondering if, in the case that the Coxeter diagram has a long chain of degree 2 vertices, there are known examples of situations where some regular behavior of the root systems or anything attached to them can be extracted from that.
So I have a vague question and a specific question:
1) What previous work has been done on the behavior and stability of these families of root systems? I'm not sure how exactly to go about searching for it; the only thing my advisor and I could find was Homological stability for families of Coxeter groups, which goes in a different direction and doesn't appear to have relevant references.
2) In an attempt to narrow down the (typically) infinitely many roots at play, suppose I restrict attention to roots (viewed as functions on the Coxeter diagram) with coefficients bounded by a particular number. How will the number of such bounded roots grow as the graph lengthens? Some preliminary number-crunching suggests it grows polynomially, and my hunch would be that there are some simple restrictions on the coefficients such that counting all the functions satisfying those restrictions is asymptotically close, but I'm not sure what kind of restrictions are appropriate.
Can you elaborate a bit on the first question? Stability (of various sorts) is usually studied for the embeddings of root systems (and sometimes for diagram foldings), which is not the case. Studying the properties shared by the families of "similar" root systems that are not ordered by inclusion is often rather hard, but such questions arise rarely. So what is the motivation here?
@AndreiSmolensky, I've edited the question in an attempt to clarify this. I'm not sure if "stability" is the appropriate term to be using.
|
2025-03-21T14:48:30.218521
| 2020-04-04T05:53:11 |
356531
|
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|
Stack Exchange
|
Closed subgroup (Cartan) theorem without transversality nor Lipschitz condition within Banach algebras
Yesterday, I came across the following preliminary theorem.
Theorem Let $\mathcal{B}$ be a Banach algebra (with unit $e$) and $G$ be a closed subgroup
of $\mathcal{B}^{-1}$ (the group of multiplicative inverses). Let $L(G)$ be the tangent space of $G$ and $m:\ I\to L(G)$ be a continuous function ($I\subset \mathbb{R}$ is an open interval containing $0_\mathbb{R}$), then
i) The following system
$$
y'(t)=m(t)y(t)\ ;\ y(0)=e
$$
admits a unique solution, say $s(t)$.
ii) The trajectory of $s$ is entirely in $G$ (in other words $t\mapsto s(t)$ is a path drawn
on $G$).
My questions are the following:
Q1) Is it known? (I expect so, at least of the specialists)
Q2) If yes, is there a sound reference? (not general, but about this – very simple – precise property).
|
2025-03-21T14:48:30.218618
| 2020-04-04T05:56:52 |
356532
|
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"Burak",
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|
Stack Exchange
|
A reference to the fact that a topologically transitive action of a group on a compact metrizable space has a dense orbit
I need a proper reference to the following obvious fact:
An action of a group $G$ on a nonempty compact metrizable space $K$ is topologically transitive (= the orbit $GU$ of any nonempty open set $U$ is dense) if and only if it the orbit $Gx$ of some point $x\in K$ is dense in $K$.
For a cyclic group this characterization is proved here. I hope that some textbook in topological dynamics should contain such a basic fact.
See Theorem 9.20 of "Topological Dynamics" by Gottschalk and Hedlund. It states that, for systems $(X,G)$ whose phase space is non-empty complete separable metric, point transitivity (a point having a dense orbit) and topological transitivity (every non-empty open set having dense orbit) are equivalent.
The space should be non-empty for the equivalence to hold.
@YCor I was working with the convention that excludes empty topological spaces but let me explicitly add that.
Sure it's often implicit and quite pernicious as soon as you use operations such as considering $X^N$ ($N$-fixed points for some normal subgroup $N$ of $G$) as dynamical system too.
|
2025-03-21T14:48:30.218992
| 2020-04-04T06:24:51 |
356534
|
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|
Stack Exchange
|
A question on the Riemann zeta function
Yesterday, a certain very talented and passionate young student from Southern Africa asked me the following question about the Riemann zeta function $\zeta(s)$. He says he "thinks" he knows the answer, but he just wants to hear my views. However, I'm not a number theorist, hence I couldn't answer him. So below is the question:
Consider the Riemann zeta function $\zeta(s)$, and let $\alpha$ be the supremum of the real parts of its zeros. Let $\mu$ denote the Möbius function. Define $S(x)= \sum_{n\leq x} \frac{\mu(n)\log n}{n}$.
Note that
$$\Big(\frac{1}{\zeta(s+1)}\Big)' = -s \int_{1}^{\infty} S(x)x^{-s-1} \mathrm{d}x$$ for $\Re(s)> \alpha-1$, where the prime denotes differentiation. It is known that $S(x)=-1 + o(1)$, thus the above integral converges if and only if $\Re(s)>0$. The student's question is: what does this tell us, if anything, about the value of $\alpha$ ?
PS: Personally, i couldn't verify the above identity, neither could I verify the "known" result that $S(x) = -1 + o(1)$, hence I couldn't answer his question.
The identity
$$\sum_{n=1}^\infty\frac{\mu(n)\log n}{n}=-1$$
was conjectured by Möbius (1832) and proved by Landau (1899). It is a consequence of the prime number theorem. Not surprisingly, the rate of convergence is determined by the (known) zero-free region of $\zeta(s)$. In particular,
$$S(x)=-1+O_\epsilon(x^{\alpha-1+\epsilon})$$
holds for any $\epsilon>0$, and $\alpha$ in the exponent cannot be lowered.
Here is a sketch of the proof of the mentioned facts. By Perron's formula, we have (at least for $x\not\in\mathbb{N}$)
$$S(x)=\frac{1}{2\pi i}\int_{1-i\infty}^{1+i\infty}\left(\frac{-1}{\zeta(s+1)}\right)' \frac{x^s}{s}\,ds.$$
The integration is meant over the vertical line with abscissa $1$. By truncating the integral at some height, and applying the residue theorem appropriately, we can move the line segment of integration to the left with the benefit of $x^s$ being much smaller there. This is the same technique by which the prime number theorem was originally proven. At $s=0$, the derivative inside the integral equals $-1$, while $x^s/s$ has a simple pole with residue $1$. Therefore, as we move the curve of integration to the left of $s=0$, we pick up the main term $-1$. The error term then depends on how far to the left we can move the curve of integration without encountering further poles, i.e. where the zeros of $\zeta(s+1)$ are located. The standard zero-free region already implies my first display. If $\alpha<1$, then we have a much wider zero-free region, and the second display follows. The fact that the exponent is optimal follows by reversing this logic, namely by examining the analytic continuation of the RHS of the OP's formula to the left of $s=0$.
I hope this helps your student, or perhaps this is exactly what she/he had in mind. It is standard material, but a good way to better understand the prime number theorem and its relation to the zeros of $\zeta(s)$.
@non-numbertheorist: It does not say anything about $\alpha$. The left hand side of your display is holomorphic in the half-plane $\Re(s)>\alpha-1$, and the RHS is a particular way to represent it in the half-plane $\Re(s)>0$. The fact that this particular representation does not converge beyond the smaller half-plane, says nothing about the LHS (or the zeros of $\zeta(s)$ for that matter). It is conjectured that $\alpha$ equals $1/2$, but as of now, no value in $[1/2,1]$ has been excluded for $\alpha$. As far as we know today, $\alpha$ can be anywhere between $1/2$ and $1$.
@non-numbertheorist: How could the identity hold for $s=-0.1$ if the RHS does not converge for $s=-0.1$? The RHS has no value at $s=-0.1$, so there is no equality to talk about.
@non-numbertheorist: The identity holds at every $s$ where both sides are defined, i.e. at every $s$ with positive real part.
oh yes, now I see...thanks for the clear explanations ! The identity theorem (which facilitates the extension of the identity to a larger plane, requires both sides to be defined. But I will delete my comments in this long thread, since they seem to divert the subject of the question...
But am sure your answer would be much more helpful to the student than anything I would have told him.
@non-numbertheorist: The identity theorem also holds for meromorphic functions. Hence the RHS, which is a holomorphic function on $\Re(s)>0$, extends uniquely to a meromorphic function on $\mathbb{C}$, namely to the function given by the LHS. The poles of this merormophic extension are the zeros of $\zeta(s+1)$. In particular, the merormorphic extension of the RHS is analytic in $\Re(s)>\alpha-1$, even though the RHS only converges for $\Re(s)>0$. The last two facts are independent of each other. Continued in next remark.
@non-numbertheorist: Contination of previous remark. In general, Dirichlet series and Mellin transforms don't behave as nicely as power series. Usually, nothing special happens at the boundary of the region of convergence for a Dirichlet series or a Mellin transform. In particular, usually, there are no singularities on the boundary of the region of convergence (unlike in the case of power series).
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356537
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Stack Exchange
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What's the use of countable ordinals? (prompted by a remark of Tim Gowers)
In a typically lucid and helpful page of notes for students, A beginner’s guide to countable ordinals, Tim Gowers explains how the countable ordinals can be “constructed rigorously in a way that requires more or less no knowledge of set theory”, and this is enough for many applications. He gives some illustrations but then eventually adds
Now comes the moment to admit that my 'applications' of countable ordinals were, in a sense, a con. The application to Borel sets wasn't really solving a problem — it was just classifying the Borel sets in quite an interesting way. As for the other two results — that continuous functions on $[0,1]$ are bounded and that open games are determined — it is downright silly to use ordinals for their proofs and very easy to remove them. This is almost always true of proofs that use countable ordinals. Though there are probably several counterexamples to this assertion, I myself know of only one theorem proved with countable ordinals for which a neater ordinal-free proof has not been discovered, and even there I am convinced that it exists.
So here's the obvious question: are there nice counterexamples to Gowers's general claim? How does this claim stand w.r.t. proof theory (where I'd have thought that ordinals kinda matter!)? More generally, what theorems (outside set theory) are there for which a proof using countable ordinals is the most illuminating/most elegant/most informative?
He might be referring to Goodstein's theorem (https://en.wikipedia.org/wiki/Goodstein%27s_theorem) when he says he knows of one theorem which requires ordinals to prove. I think it's a good example.
One instance, for me, is the statement that a countably generated $\sigma$-algebra has cardinality at most $2^{\aleph_0}$. I don't know a proof of this other than by transfinite induction. (Of course, whether proofs without ordinals are "neater" is perhaps a matter of personal taste...)
Not quite an answer, perhaps, but what about theorems where ordinals are an unexpected, yet essential, part of the statement? I'm thinking of the theorem of Thurston and Jorgensen that the volumes of hyperbolic 3-manifolds have order type $\omega^\omega$.
Have I got the history wrong? Didn't Cantor invent (countable) ordinals for his own use in his work on convergence of trigonometric series? Maybe he didn't really need ordinals, maybe he could have gotten by without them if he had been more clever (and if he'd been aware that it was somehow desirable to avoid them), but that's not the same thing as saying they're useless, is it? And what's the sense of objecting specifically to countable ordinals? Since ordinals are made of smaller ordinals, how can you have uncountable ordinals if you don't have countable ones??
I find the claim about explaining the countable ordinals without set theory rather odd, since however it is that one treats ordinals, one is doing set theory. Well-founded recursion, after all, is the core set-theoretic idea.
So, we know from reverse mathematics that nearly all "bread-and-butter" theorems are, suitably encoded, provable in proof-theoretically weak subsystems of second order arithmetic. If a theorem is provable in a system whose proof theoretic ordinal is $\alpha$, then in some sense ordinals larger than $\alpha$ need not come into its proof.
Goodstein's theorem, mentioned in the comments, cannot be proven in PA, so in some way the ordinals up to $\epsilon_0$ are "needed" in its proof. But induction up to $\epsilon_0$ can be expressed in a very non-ordinal way: the consistency of PA + all true $\Pi_1$ sentences implies Goodstein's theorem, so I am not sure there is any satisfying way to formulate the claim that ordinals are "needed" in its proof, in Gowers's sense of "needed" (i.e., we have to teach someone about ordinals before they have any chance of understanding any proof of the theorem).
It sure seems like you need to use ordinals to prove the Cantor-Bendixson theorem (every closed set of reals is the union of a countable set and a perfect set), and indeed the proof-theoretic ordinal needed for reverse mathing it is relatively high, namely $\Gamma_0$. [I take this back! An ordinal-free proof is given in William's answer here.] The graph minor theorem "needs" even larger ordinals, in the sense that it cannot be proven in systems whose proof-theoretic ordinal is less then (I believe) the small Veblen ordinal.
An ordinal-free proof --- (from a lengthy late-1990s manuscript of mine that might one day see the light of day) "Three techniques have arisen to prove the Cantor-Bendixson theorem: transfinite iteration of the derived set operator (Cantor and Bendixson), the use of condensation points (Lindelöf), and the use a maximal dense in itself set (Hausdorff). Each of these methods has subsequently generated, through abstraction, its own circle of ideas and areas of applicability."
@DaveLRenfro I would like to see that paper.
FYI, I sent it to you about 11 hours ago.
@NikWeaver Is there are nice discussion out there of how "the system PA + Con(PA) is strong enough to prove Goodstein's theorem"? [Thought I knew this, but having a senior moment!]
@PeterSmith good point. The easy fact is that Con(PA + all true $\Pi_1$ sentences) implies Goodstein's theorem (since PA does prove every instance of Goodstein). Actually I doubt the stronger statement I made is right; I'll fix it.
(Or more simply, the soundness of PA implies Goodstein.)
Here is an example from operator theory that qualifies (at leat partially): the ABLV theorem.
A version of it reads as follows:
ABLV Theorem. Let $T$ be a bounded linear operator on a Banach space $E$ and assume that $T$ is power bounded in the sense that $\sup_{n \in \mathbb{N}} \|T^n\| < \infty$.
Suppose that the spectrum $\sigma(T)$ of $T$ intersects the complex unit circle $\mathbb{T}$ at most in a countable set, and that the dual operator $T'$ on $E'$ does not have any eigenvalues on $\mathbb{T}$. Then $T^nx \to 0$ as $n \to \infty$ for each $x \in E$.
The theorem is from 1988, and it is called ABLV theorem because it was proved independently by the following pairs of authors:
[1] Wolfgang Arendt & Charles Batty: "Tauberian Theorems and Stability of One-Parameter Semigroups", Transactions of the AMS, 1988.
[2] Yu. I. Lyubich & Vu Quôc Phóng: "Asymptotic stability of linear differential equations in Banach spaces", Studia Mathematica, 1988.
(Actually, both papers focussed on a $C_0$-semigroup version of it; see the remarks at the end of this post.)
I'll try to outline the main idea of the proof in [1] as Wolfgang Arendt once explained it to me: one needs to (a) use a Tauberian theorem for vector-valued Laplace transforms and (b) successively remove points from $\sigma(T) \cap \mathbb{T}$ by an induction procedure until the spectrum does not intersect the unit circle any longer.
The point about the second step is that, since the spectrum is closed, removing points from it can only work if these points are isolated in the spectrum - so just enumerating the countable set $\sigma(T) \cap \mathbb{T}$ and succesively removing each point doesn't do the job. Instead, one uses transfinite induction over countable ordinals:
In a non-empty compact countable set - such as $\sigma(T) \cap \mathbb{T}$ - there always exists at least one isolated point, so you can successively remove such isolated points $\omega$-many times (unless there are only finitely many spectral values, in which case the spectrum is empty after finitely many steps). Afterwards you're again left with a countable compact set, and repeat.
Remarks.
Of course, things are actually not quite as simple as I described them (for instance, points that are isolated in $\sigma(T) \cap \mathbb{T}$ need not be isolated in $\sigma(T)$ - so here you can already see that you cannot simply "remove" them from spectrum). Still, this somehow seems to be the intuitive idea behind it.
Arendt revisits this theorem and explicitly focusses the discussion on transfinite induction over countable sets in the following paper:
[3] W. Arendt: "Countable Spectrum, Transfinite Induction and Stability", Operator Theory: Advances and Applications, 2015.
In [3, Section 5] there is also a discussion of the uniqueness of trigonometric series that dates back to Cantor and that uses transfinite induction over countable sets, too. So this is another answer to the question of the OP.
In fact, the ABLV theorem is typically stated and proved for $C_0$-semigroups rather than for single operators (for instance, [2] deals solely with the $C_0$-semigroup case). The single operator version stated above can be found in [1, Theorem 5.1].
Why is the ABLV theorem only a partial answer? Well, there is also [2], of course - where the authors used a different technique to arrive at the same result. (By the way, in the popular $C_0$-semigroup book of Engel and Nagel, the ABLV theorem is proved by employing the technique from [2].)
Cool example! Though I somehow feel there must be an ordinal-free proof --- partly because of the superficial similarity to Cantor-Bendixson.
@NikWeaver: Indeed, I think that the proof of Lyubich and Vu is ordinal-free (see the last of my remarks).
Oh, I didn't notice that. Thanks!
I'm sure I don't quite understand this question, since my impression is that transfinite induction and ordinals are needed all the time, and often lengths are countable in practical cases.
Nevertheless, I can't resist mentioning one of my favorite results in symbolic dynamics, namely a theorem from "Structural aspects of tilings" by Ballier-Durand-Jeandel. It is one where I do not know a nice way to remove the use of countable ordinals. By basic topology, every $\mathbb{Z}^2$-subshift (closed shift-invariant subset of $A^{\mathbb{Z}^2}$ for $A$ finite set) is either finite, countably infinite or has the cardinality of the continuum. They prove an interesting result about the countable case.
Some definitions first. If $A$ is a finite set, we say $X \subset A^{\mathbb{Z}^2}$ is a subshift of finite type or SFT if there exists a clopen set $C \subset A^{\mathbb{Z}^2}$ such that
$$ X = \{x \in A^{\mathbb{Z}^2} \;|\; \forall \vec v \in \mathbb{Z}^2: \sigma^{\vec v}(x) \notin C \} $$
where $\sigma^{\vec v}(x)_{\vec u} = x_{\vec v + \vec u}$ is the shift action. (It's the same as saying it's defined by finitely many forbidden patterns.) An SFT is countably infinite if it has countably infinitely many configurations. We call elements $x \in X$ configurations. A configuration $x \in X$ is singly-periodic if the point stabilizer $\{\vec v \in \mathbb{Z}^2 \;|\; \sigma^{\vec v}(x) = x\} \leq \mathbb{Z}^2$ is nontrivial but not of finite index.
Let $X \subset A^{\mathbb{Z}^2}$ be a subshift of finite type which is countably infinite. Then $X$ contains a singly-periodic configuration.
The proof is quite interesting and I'll outline it as I remember it; there's lots of steps so this may be too quick to follow, but at least one sees from the summary that you really talk about ordinals and their successor relation on top of the Cantor-Bendixson argument. You can find the details in the Ballier-Durand-Jeandel paper.
First, you define the Cantor-Bendixson derivatives $X^{(\gamma)}$ for all ordinals $\gamma$ in the usual way. Since $X$ is countable and compact you have $X^{(\gamma)} = \emptyset$ for some $\gamma$ (or you find a perfect subset contradicting countability), and since the topology is second-countable, this happens for a countable ordinal $\gamma$.
Now let us analyze $\gamma$. It must be a successor ordinal, since otherwise $\emptyset$ is an intersection of nonempty sets contradicting compactness. So $\gamma = \beta + 1$ for some $\beta$. Since $X^{(\beta)}$ has empty Cantor-Bendixson derivative, it has to be finite. But it is classical that a finite subshift is a subshift of finite type, and it is also classical that SFTs have the "compactness" property that an SFT cannot be the intersection of a strictly descending chain of subshifts (it's a "finitely-generated group cannot be a strictly increasing union of subgroups" style argument). From this we deduce that also $\beta = \alpha + 1$ must be a successor ordinal. Clearly $X^{(\alpha)}$ is countably infinite.
Next, we analyze $X^{(\alpha)}$ (there we will find our singly-periodic configurations). It is known that a $\mathbb{Z}^2$-subshift is finite if and only if every configuration in it has finite-index stabilizer, so some configuration $X^{(\alpha)}$ has infinite-index stabilizer. Let $x$ be such a configuration, and let $V \leq \mathbb{Z}^2$ be its stabilizer.
It now suffices to show that we cannot have $|V| = 1$: Suppose for a contradiction that we had. Observe that all $\sigma^{\vec v}(x)$ are distinct, so any limit point of such shifts is in $X^{(\beta)}$. Then for every $\epsilon > 0$, every shift $\sigma^{\vec v}(x)$ with $|\vec v|$ large enough is at distance at most $\epsilon$ from the SFT $X^{(\beta)}$.
Since $X^{(\beta)}$ is finite, it has finite-index pointwise stabilizer, generated by some non-collinear $\vec u_1, \vec u_2 \in \mathbb{Z}^2$. Since all limit points of $x$ have to be in $X^{(\beta)}$, we must have $x_{\vec v} = x_{\vec v + \vec u_1} = x_{\vec v + \vec u_2}$ for all $|\vec v|$ large enough (again otherwise we have infinitely many distinct shifts $\sigma^{\vec v}(x)$ and we can extract a limit point which is not fixed by $\sigma^{\vec u_1}$ and $\sigma^{\vec u_2}$, thus is not in $X^{\beta}$).
We conclude that if $x$ does not have finite-index stabilizer, then $x$ is periodic apart from a finite "period-breaker" area, and since $X$ is an SFT, it is easy to see that it is uncountable, since we can glue these period-breakers all around $x$ and we will not see a problem with the defining clopen set $C$, as such a set will only look at the local picture. (I'm being very quick here, here you perhaps need to work a bit and draw a picture, or read the paper.)
So $x$ has infinite stabilizer which is not of finite index, i.e. $x$ is singly periodic. Square.
(There are some characterizations also for the other cardinalities. A $\mathbb{Z}^2$-subshift is finite if and only if every configuration has finite-index stabilizer. In another paper called "Structuring multi-dimensional subshifts", Ballier and Jeandel also give a characterization of uncountable SFTs, but I'll skip that.)
As mentioned in Prof. Weaver's answer, the graph minor theorem (also known as the Robertson-Seymour theorem) uses large countable ordinals. The graph minor theorem is quite relevant to non-set-theoretic "mainstream" mathematical topics, in particular it can be characterized either as "every infinite set of finite graphs contains $A,B$ such that $A$ is a graph minor of $B$", or for added relevance to forbidden minor problems, "every minor-closed class of finite graphs can be characterized by a finite set of forbidden minors". [2]
Robertson and Seymour's 2004 proof is arguably even more involved with large countable ordinals than theorems such as Cantor-Bendixson as it uses specific large countable ordinals explicitly. In particular an ordinal $\psi_0(\Omega_\omega)$ appears, this ordinal is named using one of Wilfried Buchholz's representation systems for large countable ordinals. (Buchholz's naming system originates from his 1984 paper, "A new system of proof-theoretic ordinal functions" [1], and it is a simplification of an earlier system by Feferman.)
In proving that the graph minor relation is a well-quasi-order, Harvey Friedman's gap-embedding relation on labeled finite trees is introduced, and this relation is intended to mimic Buchholz's comparison criteria for comparing two terms in his ordinal representation system. Historically, any proof of a forbidden minor criterion not using orfinals (e.g. Wagner's theorem) is a special case of Robertson-Seymour [2], however the current published proof of full Robertson-Seymour is 500 pages, involving Buchholz's system for large countable ordinals via gap-embeddibility. [3]
[1]: W. Buchholz, "A new system of proof-theoretic ordinal functions", Annals of Pure and Applied Logic 32 (1986) pp 195-207.
[2]: Kolya Malkin, "The Graph Minor Theorem", thesis (2015).
[3]: M. Rathjen, "The Realm of Ordinal Analysis", in: S. Cooper & J. Truss (Eds.), Sets and Proofs, London Mathematical Society Lecture Note Series 258 (1999) pp. 219-280, (author pdf).
Please note that links to pdfs on personal websites can be fragile, and break over time. Can you add links to the published versions as well?
I can link to a published version of "The Realm of Ordinal Analysis", thanks. Unfortunately I am not able to find a published version of "The Graph Minor Theorem".
All good. I fixed some more :-)
There is a proof in "Handbook of Categorical Algebra" vol.2 by Francis Borceux which use transfinite induction I think.
The (famous) theorem states that for every small abelian category $\mathcal{A}$ there exist a ring $R$ and a fully faithful exact functor $F: \mathcal{A} \rightarrow Mod_R$. In other words, small abelian categories can be embedded in categories of modules.
So this is the Freyd-Mitchell embedding theorem, right? I don't recall any transfinite induction in the proof, but I admit it has been a long time since I sem-learned this...
Could you please give some details of how transfinite induction is used in the proof that you are thinking of?
Well, the proof is quite long ... and difficult. I do not have any better advice for you to have a look in the book (volume 2, section 1.14). The section starts with some lemmas, where we understand that a "suitable" category $\mathcal{D}$ will help us later. In the proof of the theorem, he constructs this category $\mathcal{D}$. In order to do this, a sequence of posets indexed by the ordinals is needed. (and at some point a transfinite induction is done)
The reason I ask is that the outlines of the proof that I can find online do not mention ordinals: https://math.stackexchange.com/questions/1855549/whats-wrong-with-my-understanding-of-the-freyd-mitchell-embedding-theorem and https://mathoverflow.net/questions/47747/freyd-mitchells-embedding-theorem Could you at least comment on which parts of those outlines are where the ordinals are hidden?
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2025-03-21T14:48:30.220790
| 2020-04-04T07:56:37 |
356539
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Stack Exchange
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Singular Sturm-Liouville problems: criterion for discrete spectrum for zero potential ($q=0$) and Hermite Polynomials
There are some known criteria for the Sturm-Liouville Problem
\begin{equation} \tag{1}
\frac {\mathrm {d} }{\mathrm {d} x}\left[p(x){\frac {\mathrm {d} y}{\mathrm {d} x}}\right]+q(x)y=-\lambda w(x)y
\end{equation}
to have a spectrum discrete and bounded below (BD) in the singular case:
If singular endpoints are Limit-Circle and Non-Oscillating (LCNO), results from
(Niessen, H.-D. and Zettl, A. (1992). Singular Sturm-Liouville Problems: The Friedrichs extension and comparison of eigenvalues.)
apply.
There also is the famous criterion by Molchanov and generalizations for $w, p \neq 1$ in (Kwong and Zettl - Discreteness conditions for the spectrum of ordinary differential operators). Various other generalizations exist.
As I understand it, the Molchanov criterion and its generalizations are not applicable to the case $q = 0$ or $q$ constant. Inherently, $q$ must run to infinity at the endpoint in a certain way (details depending on $p$ and $w$). (See related questions For which type of potentials a Schrödinger operator will have discrete spectrum?, Harmonic oscillator discrete spectrum)
My questions:
Is there a theory and criteria for (1) to be BD for the case $q = 0$ if endpoints are in the Limit-Point (LP) case? I specifically appreciate pointers to literature.
Or do I overlook some tweak, allowing to apply Molchanov-style criteria to the case $q = 0$?
What I read and reasoned so far:
As far as I can assess, transformations to normal form are not suitable to work around this particular issue. Also, according to Zettl (Zettl, A. (2010). Sturm-Liouville Theory), it is questionable whether such a transformation preserves the desired properties of the spectrum (discussion in Chapter 10.13).
I looked into various books and papers and googled around, but I almost exclusively found Molchanov-style criteria, most usually setted for the half-line $[a > -\infty, \infty)$. It appears that Schrödinger operators are the main interest in this field and the Molchanov criterion is a good fit for them. I am more interested in self-adjoint operators similar to the one yielding Hermite polynomials:
\begin{equation} \tag{2}
q = 0, \quad w = p = e^{-{\frac {x^{2}}{2}}}
\end{equation}
on the real line with two singular LP endpoints. I used to perceive Hermite Polynomials as an important standard example and it puzzles me that it seems to be so hard to find criteria for their spectrum. I mean criteria that are not totally specific to equation (1) with values (2), but also hold if some other (normal-like?) probability distribution is used in (2), e.g. with adjusted variance, skewness, kurtosis etc. (not expecting the solutions to be (orthogonal) polynomials anymore, but still to have BD spectrum). (Edit: It occurred to me that the case of (2) with adjusted variance is known as the generalized Hermite polynomials.)
E.g. I skimmed through some books by L. L. Littlejohn and A.M. Krall about orthogonal polynomials in context of singular SL theory, e.g. (L. L. Littlejohn and A.M. Krall, Orthogonal polynomials and singular Sturm-Liouville Systems, I) but they do not seem to discuss the discreteness of the spectrum (sorry if I overlooked it, pointers welcome).
Secondary question:
What about the full line vs. the Half line?
It appears that in his original work, Molchanov considered the full line (A.M. Molchanov, Conditions for the discreteness of the spectrum of self-adjoint second-order differential equations.) however, I cannot read Russian and only got this impression from the math terms. Also (Inge Brinck, Self-Adjointness and spectra of Sturm-Liouville operators) assumes the full line, but (Kwong and Zettl - Discreteness conditions for the spectrum of ordinary differential operators) and (D. Hinton, Molchanov’s discrete spectra criterion for a weighted operator) assume the half line.
This gives me the impression that the distinction between full line and half line is not essential, but I did not find it discussed anywhere. A pointer to literature that specifically discusses in what sense results are transferable between the half line and full line case would be welcome. (Niessen, H.-D. and Zettl, A. (1992). Singular Sturm-Liouville Problems: The Friedrichs extension and comparison of eigenvalues.) discuss it a bit, but not in context of criteria for BD in the LP case (or I don't understand how to conclude it).
Thanks in advance, also for hints, comments and improvements!
Whole line is the same question as half line because the whole line essential spectrum is the union of the two half line essential spectra (so I guess what I'm saying is that a criterion for half line problems will settle the whole line case too).
Just by chance, do you know a reference for the whole line's essential spectrum being the union of the two half line essential spectra? This is not obvious to me as the two half line problems involve a boundary condition at the split point the full line problem does not have. But this boundary condition is not relevant for the essential spectrum I guess...
The argument is that cutting the whole line operator into two half lines is a rank $2$ (so in particular compact) perturbation of the resolvent, so the essential spectra agree by Weyl's theorem. Weidmann's Springer Lecture Notes discuss this for sure, but it must be in many other books (for example, somewhere in Teschl's books, which are available on his homepage).
For the record, I found it in Weidmann's Springer Lecture Notes, but he merely states it in a remark and refers the reader to Naimark. Finally I found it stated in (M.A. Naimark, Linear Differential Operators), §24, Theorem 1. Thanks again for all your help!
No problem, I enjoyed your question!
The case $q=0$ actually has a complete answer, and the reason this works is that we can solve the equation for $\lambda =0$ explicitly then, by $u=1$ and $v=\int_0^x \frac{dt}{p(t)}$. The spectrum is purely discrete if and only if ($w\in L^1$ and)
$$
\lim_{x\to\infty}\int_0^x wv^2\, dt \int_x^{\infty} w\, dt =0 . \quad\quad\quad\quad (1)
$$
(This holds in your example, if I managed the asymptotics of the error functions correctly.)
I obtained this by rewriting the SL equation as a canonical system $Jy'=-\lambda Hy$, and the coefficient matrix $H$ we obtain here is given by
$$
H = w\begin{pmatrix} 1 & v \\ v & v^2 \end{pmatrix} .
$$
There is a precise criterion for purely discrete spectrum for (general) canonical systems, and this gave me (1).
This will probably all sound quite cryptic if you haven't seen these things before, and there seem to be too many small details to explain them all here. You could take a look at my papers 1, 2 on my homepage for full background information.
Also, if I hadn't had that available, I would have tried to analyze the Prufer equation
$$
\varphi' = \frac{1}{p}\cos^2\varphi + (\lambda w+q)\sin^2\varphi
$$
for the phase of the solution vector $(y,py')$. Purely discrete spectrum is equivalent to $\varphi(x)$ staying bounded in $x$ for all $\lambda$. This might also be good enough to get (1).
Wow, that's a stunningly beautiful criterion. Thanks for this work! Do I understand correctly that for the full line, (1) must additionally hold with every $\infty$ replaced by $-\infty$? You defined $u=1$ but did not use it any more. Was it a typo?
Forget the question about $u$, I see, it's just another solution for $\lambda = 0$. Read it too quickly...
@stewori: Yes, that's correct (what you say about the whole line problem). The other solution $u$ does make another appearance, though I don't call it that, the general form of $H$ is $H=w \bigl( \begin{smallmatrix} u^2 & uv \ uv & v^2 \end{smallmatrix} \bigr)$.
Right :) I had just found the example in Section 5 of your paper 1 which explains the construction. Then I wrote "forget the question about $u$". The example is so hidden at the end... I had overlooked it when I skimmed through the paper at first.
|
2025-03-21T14:48:30.221468
| 2020-04-04T08:19:23 |
356542
|
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|
Stack Exchange
|
Hassan Akbar-Zadeh's mathematical legacy
Professor Hassan Akbar-Zadeh (Born: March 23, 1928- Iran), a prominent Iranian mathematician has died (March 23, 2020) in Paris after years of research and study. (I'm not sure of the exact dates.)
He has worked with the French National Center for Scientific Research and the Collège de France for many years and was director of research at C.N.R.S., Paris, France.
His famous book is "Initiation to global Finslerian geometry". He was one of the top students of André Lichnérowicz.
I am googling days and surprisingly I couldn't find a single webpage or article about him. So I want to ask:
What do you know about Hassan Akbar-Zadeh's mathematical legacy, biography, etc?
In all the papers from him I could download the author's address is a personal address, which is unusual (in 1965 in Poissy near Paris, 1987 and later in Paris). The BNF page claims he was CNRS research director in 2006 but he was then 78 so given the French rules he could only be retired (so emeritus would be plausible anyway). But I'm more inclined to believe, in the absence of other information, that he had no affiliation.
here is a photograph: https://fr.irna.ir/news/83735754/Disparition-d-un-éminent-mathématicien-iranien-à-Paris-l-ambassadeur
Thanks for information. I heard that he was nominated for French scientist medal and missed it because he did not want to change his nationality (This medal is given only to the French) and inevitably, the prize was awarded to the Marcel Berger.
I'm not aware of a "French scientist medal", nor of any French scientific distinction that would be reserved for French people. This looks like a random rumor. I'm not sure that a comparison with Berger's legacy is opportune either.
I removed the "morality". I just wanted to have a somehow complete bio.
@YCor: I am not sure what was the prize name. Perhaps it is one of these: Prix Peccot, Collège de France
Prix Maurice Audin
Prix Carrière, Académie des Sciences
Prix Leconte, Académie des Sciences
Prix Gaston Julia.
None of these is reserved to French people. This story is extremely unlikely.
Google Scholar lists: this book, 91 cit.; "Sur les espaces de Finsler à courbures sectionnelles constantes, Acad. Roy. Belg. Bull. Cl. Sci. 1988, 236 cit.; Les espaces de Finsler..., Ann. ENS 1963, 75 cit.; Espaces à tenseur de Ricci parallèle admettant des tranformations projectives with R. Couty, Rend. Mat. 1978, 14 quot., and MSN detects 13 cit. for "Generalized Einstein manifolds", J. Geom. Phys. 1995. Others are below 10 cit.
@YCor: What mathscinet says? I have no access to it.
I have inquired with one of his students, professor Behroz Bidabad; I'm hoping for some personal reminiscenses.
MSN gives 47 papers: the 2006 book quoted above with 31 cit., a 2004 CRAS note, the 1995 paper quoted above, and others between 1954 and 1988; 4 have $\ge 6$ citations: those 3 of my previous comment and a 1986 J. Math. Pures Appl. paper. 5 joint publications from 1977 to 1987 with Raymond Couty (1919-2005) who had the same PhD advisor Lichnerowicz 3 years earlier; 3 with Anna Wegrzynowska (the latter published 8 papers between 1970 and 1979), and 1 with Edmond Bonan in 1964.
@abx Actually the Prix Audin is nationality-restricted: two prices are given one to an Algerian mathematician working in Algeria and one to a French mathematician working in France. (voir https://fr.wikipedia.org/wiki/Prix_de_mathématiques_Maurice-Audin)
Abstracts of some of his publications are here: https://www.researchgate.net/scientific-contributions/2010478429_H_Akbar-Zadeh
One of his frequent collaborators was Raymond Couty, who was also the department head at the University of Limoges for a number of years. May be you can get some information from that university?
@JoséFigueroa-O'Farrill actually according to this page, the prize Audin existed from 1958 to 1963 and then since 2004, and you're quoting the rules since 2004 (it says "working in France", which is not a nationality restriction). But it seems quite distinct with the 1958-1963 prize which is obviously the one in consideration (since Berger got it in 1962).
Anyway Berger got the Audin prize in 1962, eight years after his PhD with Lichnerowicz, he had 17 published papers (including 9 CRAS) $\le 1961$ and 5 more in 1962, while Akbar-Zadeh had still not completed his PhD (1963) — he still had 6 publications $\le 1961$, all CRAS notes of 3-4 pages; his first longer paper was published in 1963 and refers to his thesis. Anyway they looked, at that time, clearly not in the same level of carrier even if Akbar-Zadeh was only 1 year younger.
@YCor what it says is "un mathématicien français exerçant en France," but I agree that this is for the new version of the prize since 2004. Anyway, I'm in no way suggesting that Berger (whom I admire and whose work I use regularly) was undeserving of the award. I was simply pointing out that there are mathematical prizes in France for which being French is a requirement.
@JoséFigueroa-O'Farrill: I admire his works and life also and my intention was not to make him look unworthy of this (and other) award.
Without going into prizes or nationalities (what's the point?): his paper "Les espaces de Finsler et certaines de leurs généralisations (1963)" was for some time the best place to understand the connections related to Finsler spaces. He made the jump between Cartan style exposition and the modern understanding of (Kozsul) connections. Moreover, his theorem that a compact Finsler space with constant negative curvature (resp. zero curvature) is Riemannian (resp. locally isometric to a normed space) is still one of the prettiest theorems in the subject.
|
2025-03-21T14:48:30.221936
| 2020-04-05T21:13:57 |
356692
|
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|
Stack Exchange
|
Proof of a discrete isoperimetric inequality
The following inequality appears in the proof of certain isoperimetric-type inequalities for analytic functions in two dimensions:
$$\sum_{m=0}^{\infty}\frac{|c_m|^2}{m+1} \leq \pi \left(\sum_{m=0}^{\infty}|a_m|^2 \right)^2,$$
where
$$c_m=a_0a_m+a_1 a_{m-1}+ \dots +a_ma_0.$$
It sounds like a basic inequality, but I wasn't able to prove it. Does this inequality have a name? Where I can find a proof or how can one prove it?
Where did you find it?
Suppose you have a power series with coefficients $a_n$
$$ f(z):= \sum_{k=1}^\infty a_k z^k .$$
Then the coefficients of $f^2$ are exactly $c_n$. Also if we denote by $\odot$ the Hadamard multiplication of powerseries (coefficient-wise or equivalently convolution of the boundary values), $c_k^2$ are the coefficients of $f^2\odot f^2$. We apply first Hardy's inequality (Theory of Hp spaces, Duren, Corollary of Theorem 3.15) and then Young's inequality for convolution
\begin{align*} \sum_{k=1}^\infty \frac{|c_k|^2}{k+1} & \leq \pi \Vert f^2 \odot f^2 \Vert_{H^1} \\
& \leq \pi \Vert f^2 \Vert_{H^1} \Vert f^2 \Vert_{H^1} \\
&= \pi \Vert f \Vert_{H^2} ^4 \\ & = \pi \Big( \sum_{k=0}^\infty |a_k|^2 \Big)^2 \end{align*}.
If your inequality is true I'm missing a constant of $\pi$.
You are right. The inequality was missing a $\pi$ which is fixed. Thanks.
|
2025-03-21T14:48:30.222065
| 2020-04-05T21:21:27 |
356693
|
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|
Stack Exchange
|
Does the Grothendieck construction produce a 2-category or a category?
Let $F : \mathcal{C} \to \mathbf{Cat}$ be a lax 2-functor. Then we can form a category $\int F $ which is the Grothendieck construction on F.
There's a number of resources detailing this construction, but none mentioning if we can get a 2-category, rather than a category this way. It seems like the natural construction would be to generate a 2-category - is this true? I'm looking for a reference.
Namely - it seems like the 2-cells in $\mathcal{C}$ in standard definitions don't really play any role at all in $\int F$, but it seems like they should correspond to 2-cells in $\int F$.
The closest constructions I found were in (https://arxiv.org/abs/2002.06055, Definition 10.7.2.) . They provide a bicategorical Grothendieck construction for a functor $F : \mathcal{C}^{op} \to \mathbf{Bicat^{ps}}$ - but of a rather strange kind.
It does not use 2-cells in $\mathcal{C}$ and it also assumes certain equality of 1-cells in $\mathcal{C}$.
Another potential candidate is https://www2.irb.hr/korisnici/ibakovic/sgc.pdf, but it seems to be on a higher level of abstraction than I'm comfortable with.
It also seems to talk about the Grothendieck construction of functors whose codomain is $\mathbf{2-Cat}$, rather than into $\mathbf{Cat}$. It seems like this is extra structure that is not needed for what I'm asking.
So, in short, if there is a lax functor $F : \mathcal{C} \to \mathbf{Cat}$, is there a way to make the Grothendieck construction into a 2-category? If so - is there reference with an explicit construction, showing in details what all the 2-cells would look like?
Bénabou studied this kind of construction: if $\mathcal C=1$ is the terminal category, $F$ consists of a monad in $\bf Cat$, on the category $A=F*$. When $\mathcal C$ is a generic bicategory, $F$ amounts to an "indexed" family of monads, i.e. to a functor $\Sigma_F : \mathcal E \to \mathcal C$ over $\mathcal C$ whose fiber over $C$ is the Kleisli category of the monad determined by $F|_{{C}} : {C} \to \bf Cat$. [tbh I'm not 100% sure of this statement. But you can try to work out this intuition and see what strictness $\hat F, \mathcal E$ have]
The usual Grothendieck construction has for $\mathcal C$ an ordinary category, so it doesn't have any 2-cells (or at least, doesn't have any non-identity 2-cells). Moreover, we actually get not only a category out of it, but an object of the slice (2-)category $\mathcal Cat/\mathcal C$. If $\mathcal C$ weren't an ordinary category, this wouldn't make as much sense since $\mathcal C$ wouldn't be an object of $\mathcal Cat$. (Just like how we can have functors $\mathcal C \to \mathcal Cat$, though, it's still possible to talk about functors from 1-categories to 2-categories).
Nonetheless, you can get a meaningful construction when adding in 2-cells. I have no idea if this construction has a name, but once you get the pattern, it's easy to extend this to any arbitrary level. For this construction, we'll assume we have a (lax) 2-functor $F : \mathcal C \to \mathcal Cat$.
0-cells of $\int F$ are pairs $(c, x)$ where $c$ is an object of $\mathcal C$ and $x$ is an object of $F(c)$.
1-cells $(c, x) \to (c', x')$ are pairs $(f, g)$ where $f : c \to c'$ and $g : x \to_f x'$. $x \to_f x'$ is the set of dependent morphisms from $x$ to $x'$ (terminology adapted from HoTT's dependent paths). Effectively, we use functorality in the types of $x$ and $x'$ to transport from one type to the other. In this case, the type of $x$ is $F(c)$ and the type of $x'$ is $F(c')$ so we can use $F(f)$ to map $x$ into the type of $x'$.
Summing up, $g$ should be a morphism $F(f)(x) \to x'$, i.e. an element of $\hom_{F(c')}(F(f)(x), x')$.
Next, our 2-cells $(f, g) \to (f', g')$ should be pairs $(\alpha, \beta)$ where $\alpha$ is a 2-cell $f \to f'$ in $\mathcal C$. $\beta$ should be a dependent morphism $g \to_\alpha g'$.
Now the types of $g$ and $g'$ are $\hom_{F(c')}(F(f)(x), x')$ and $\hom_{F(c')}(F(f')(x), x')$ respectively. This time, these type are contravariant in the variable we need to transport ($f$ and $f'$), so we'll transport $g'$ to $\hom_{F(c')}(F(f)(x), x')$ via $\hom_{F(c')}(F(\alpha)(x), x')$.
Unpacking this, $g'$ gets sent to $g' \circ F(\alpha)(x)$, so $\beta$ is a morphism $g' \circ F(\alpha)(x) \to g$. This time, though, we're talking about a morphism between morphisms in $F(c')$, which is an ordinary category. So rather than an actual morphism, we'll have an equality $g' \circ F(\alpha)(x) = g$.
Thank you for this very detailed description, I think this might be exactly what I was looking for. Clarifications:
I assume $\mathcal{C}$ and $\int F$ are both strict 2-categories here?
If this is so, is there a reason why standard references (for the lax 2-functor version, not the 1-functor one) always claim we get a category and completely ignore the 2-categorical part? (https://people.mpi-sws.org/~dreyer/courses/catlogic/jacobs.pdf, Definition 1.10.1), (https://arxiv.org/abs/2002.06055, Definition 10.1.2)
Furthermore, is it important whether $\mathsf{Cat}$ in your first and second paragraph is considered as a 1-category or a 2-category? I'm a bit confused when we mix, as you say, "functors from 1-categories to 2-categories".
@BrunoGavranovic I usually use 2-category for what might be less ambiguously called a bicategory (but I really dislike that naming scheme). For this construction, I don't think I used strictness anywhere. When defining composition and checking all the coherence diagrams, you'll likely need to insert a bunch of structural isomorphisms everywhere. I don't think that would be a problem, though.
@BrunoGavranovic A line in your second reference (at the start of chapter 10) I think answers both your other questions. "Throughout this chapter $\mathcal C$ denotes a small category, also regarded as a locally discrete 2-category." So $\mathcal C$ is an ordinary category, but for the purposes of talking about functors $\mathcal C^{op} \to \mathcal Cat$, we treat it as a 2-category whose higher morphisms are just identities.
I see - so the second reference explicitly assumes there are no non-identity morphisms, while the first one most likely does that implicitly. I am still interested in finding a nice reference for this 2-categorical construction, but I'll go ahead and accept the answer. Thank you!
However - I do have one concern. It is known that the Grothendieck construction can be seen as an oplax colimit (https://ncatlab.org/nlab/show/Grothendieck+construction#AsALaxColimit).
But here we have a 2-functor $F : \mathcal{C} \to \mathbf{Cat}$, where $\mathbf{Cat}$ is the 2-category of small categories. A colimit there would give us a category, rather than a 2-category. Does this mean that we need to specify a (?-)functor into $\mathbf{2-Cat}$?
"Is [this] construction some kind of colimit?" would be a great new question. My intuition is that it's a colimit when $F$ is interpreted as a functor $\mathcal C \to \infty$-$\mathcal Cat$ and when $\mathcal C$ is an n-category the colimit is automatically a n-category (at least for n = 1 and 2).
Let us continue this discussion in chat.
It produces a functor between categories. In fact, what is called a fibred category. The construction is detailed in Volume 2 of Borceux's Handbook on Categorical Algebra.
Also Angelo Vistolis Notes on Grothendieck topologies, fibered categories and descent theory is worth looking at.
|
2025-03-21T14:48:30.222532
| 2020-04-06T00:43:56 |
356703
|
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|
Stack Exchange
|
References and results for the eigenvalues of Ricci tensor
I am looking for references or results that gives estimates for every eigenvalue of the Ricci tensor. For example, the least eigenvalue is related to the minimum of the Ricci curvature, what can we say about the others? It was a surprise that a standard google search leaded to no satisfactory result.
If needed, one can assume that the Ricci tensor is positive definite (the case when the sectional curvature is positive).
There is a way on approaching this problem, although I don't how fruitful it its. For example, in harmonic coordinates
$$R_{ij} = -\dfrac{1}{2}\Delta g_{ij} + o(\Gamma_{ij}),$$
where $o(\Gamma_{ij})$ means that on a fixed point one can assume that
$$o(\Gamma_{ij})(p) = 0.$$
Therefore, this is somehow equivalent to solve
$$-\dfrac{1}{2}\Delta g_{ij} + o(\Gamma_{ij}) = \lambda g_{ij},$$
and the problem reduces almost to the problem of estimating the eigenvalues of the Laplace-Beltrami operator.
|
2025-03-21T14:48:30.222627
| 2020-04-06T01:44:36 |
356704
|
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|
Stack Exchange
|
Isoperimetric inequality for analytic functions on an annulus
Let $f$ be an anylytic function on the unid disk $|z|<1$. It is well known that
$$\left (\int_0^{2\pi}f(e^{i\theta})d \theta \right)^2 \geq 4\pi \iint_{|z|<1} |f(r e^{i\theta})|^2r dr d \theta.$$
I wonder if the constant $4\pi$ cound be improved on an annulus $a<z<1$. More precisely, does the following inequality
$$\left (\int_0^{2\pi}|f(e^{i\theta})|d \theta +\int_0^{2\pi}|f(ae^{i\theta}|)d \theta \right)^2 \geq C \iint_{a<|z|<1} |f(r e^{i\theta})|^2r dr d \theta$$
hold for some constant $C(a)> 4\pi$ independent of $f$? Can $C(a)$ be computed in terms of $a$? This seems to be a classical problem but I could not find a reference, and was not able to prove it after trying for a couple of days.
I think you mean absolute values in the integrals, right ? Secondly, the inequality you ask is for $f$ analytic in the disc or in the annulus ?
Yes, I just edited the question.
You can probably prove that
$$ \Big( \int_\mathbb{A_r} |f(z)|^2 \frac{dxdy}{\pi(1-r^2)} \Big)^{1/2} \leq \int_{ \mathbb{T}} |f(e^{i\theta})| \frac{d\theta}{2\pi}+\int_{ \mathbb{T_r}} |f(re^{i\theta})| \frac{d\theta}{2\pi r}, \,\,\, \forall f\in H^1(\mathbb{A_r}). $$
Where $\mathbb{A_r} $ is the annulus of internal radius $r$ with boundary $\mathbb{T} \cup \mathbb{T_r}$, and $H^1(\mathbb{A_r})$ is the Hardy space on the annulus, by following step by step the proof of Vukotic in "The isoperimetric inequality and a theorem of Hardy and Littlewood, American mathematical monthly". The only tools you need is the Parseval formula and the inner outer factorization for the Hardy space in an annulus which you can find in "Sarason, The Hp spaces of an annulus, Memoirs or the AMS". If you have problems recustructing the proof, maybe I can help.
This looks great. Do you think it would be possible to prove a similar inequality in which the right hand side is exactly what I have in the statement of the question, i.e. to remove $1/r$ in the right hand side of the inequality in your answer? Probably with a different weight in the left hand side.
Of course it is true if you replace $2\pi$ on the first term on the rhs by $2\pi r $, but in this case the inequality is not sharp.
It kind of bothers me that as $r\rightarrow 0$ we do not get the classical isoperimetric inequality on a disk as the limit of the isoperimetric inequality on an annulus. Does this mean the inequality in your answer is not sharp?
|
2025-03-21T14:48:30.222801
| 2020-04-06T04:39:47 |
356706
|
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|
Stack Exchange
|
Number of minimal elements of product order
Consider three sets: $A = \{1,2,\dotsc,n_A\}$, $B=\{1,2,\dotsc,n_B\}$, and $C=\{1,2,\dotsc,n_C\}$, where $n_A, n_B, n_C \ge 2$. Define a product order (which is a partial order) on the cartesian product $A \times B \times C$. In other words, $(a,b,c) \le (a',b',c')$ if and only if $a \le a'$, $b \le b'$, and $c \le c'$.
Let $T$ be a non-empty subset of $A \times B \times C$. We call the triple $(a,b,c) \in T$ minimal if there is no triple $(a',b',c') \in T$, $(a,b,c) \neq (a',b',c')$, such that $(a',b',c') \le (a,b,c)$.
My question is what is the maximum possible number of minimal triples in $T$?
Some particular (simplest) cases
If $(1,1,1) \in T$, then obviously the only minimal triple in $T$ is $(1,1,1)$ (all the other triples will be larger).
If $T = A \times B \times C \setminus \{(1,1,1)\}$, then there are three minimal triples: $(2,1,1)$, $(1,2,1)$, and $(1,1,2)$.
But what are potential answers in general case? And what is the answer when we have more than three sets?
You're looking for the width (i.e., size of the largest antichain) of the product of chains (btw, you mix up minimal and maximal a lot in the post). See this question on Math Stackexchange: https://math.stackexchange.com/questions/299770/width-of-a-product-of-chains.
Fixed minimal/maximal (it was late night when I was typing :) )
And thanks for the link!
Try looking at Theorem 4.1.1 in Anderson, "Combinatorics of Finite Sets."
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2025-03-21T14:48:30.222931
| 2020-04-06T06:47:41 |
356710
|
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|
Stack Exchange
|
Standard terminology for these "coarsening" and "refining" operations for compositions and ordered set partitions?
Let $[M]:=\{1,2,\dots, M\}$. (Part of the twelvefold way) as we all know, there is a bijection between surjective functions $[N] \to [B]$ and ordered set partitions of $[N]$ into $[B]$ blocks (of course $B \le N$).
There also seems to be a bijection between monotone surjective functions $[N] \to [B]$ and compositions (ordered integer partitions) of $N$ into $B$ blocks (since there is a bijection between these and contiguous set paritions of $[N]$ into $B$ blocks, e.g. $\{1\},\{2,3,4\}$ or $\{1,2\}, \{3,4\}$ for $N=4, B=2$).
Question: Are there standard terms for the operations on ordered integer partitions and/or ordered set partitions corresponding to (i) composition of functions and (ii) "function division" or "extension"?
To clarify what I mean, specifically for:
(i) given (monotone and) surjective functions $[N] \xrightarrow{P_1} [B_1]$ and $[B_1] \xrightarrow{Q} [B_2]$ (here $B_2 \le B_1 \le N$ of course), what is the term for the (ordered integer partition) ordered set partition corresponding to the (monotone and) surjective function $[N] \xrightarrow{P_2} [B_2] := [N] \xrightarrow{Q \circ P_1} [B_2]$?
(ii) given (monotone and) surjective functions $[N] \xrightarrow{P_1} [B_1]$ and $[N] \xrightarrow{P_2} [B_2]$, what is the term for the (ordered integer partition) ordered set partition corresponding to the (monotone and) surjective function $[B_1] \xrightarrow{Q} [B_2]$ such that $Q \circ P_1 = P_2$, if one exists?
For (ii), such a $Q$ exists if and only if $P_1$ is a (strong) refinement of $P_2$, in which case it is guaranteed to be unique since $P_1$ is surjective and thus an epimorphism i.e. right-cancellative. In (i) $P_1$ also refines $P_2$.
These seem like really basic operations, so I was surprised when I couldn't find them in any textbook like Stanley or Cameron. I've tried searching for papers referencing them but was not able to find anything, probably due to not using the correct terms. The only answer I got to a previous question on Math.SE was that SageMath has functions which implement both of these operations for compositions and one of them, (i), for ordered set partitions, and that (i) is a "fattening of $P_1$ using the grouping $Q$" and that (ii) are the "refinement splitting lengths" of $P_2$ with respect to $P_1$ (this being defined only when $P_1$ is finer than $P_2$ of course). However the documentation in SageMath does not give any references to sources using those terms, and searching for them on the internet I was not able to find them elsewhere, which makes me think SageMath invented them.
Are there really no more formal terms defined anywhere else besides SageMath?
|
2025-03-21T14:48:30.223113
| 2020-04-06T07:42:06 |
356711
|
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|
Stack Exchange
|
What are the tempered Gibbs measures of classical $\phi^4$-theory?
I consider classical $\phi^4$-theory on the lattice. The model is defined in finite volume with Hamiltonian
\begin{align*}
H(\phi) = - \sum_{x \sim y} J_{x,y} \phi_x \phi_y
\end{align*}
and a-priori measure
\begin{align*}
\rho(d \phi_x) = e^{- \lambda \phi_x^4 + b \phi_x^2} d\phi_x.
\end{align*}
Here the fields can be in $\mathbb{R}$, $\lambda$ is positive
and I am fine with considering only ferromagnetic interactions of the form $J_{x,y} = J > 0$.
This model fits into the framework of Lebowitz and Presutti,Statistical mechanics of unbounded spins
CMP. 50, 195—218 (1976). Which means that I can define the class of tempered (infinite volume) Gibbs measures for this particular model. That means that I get $\mu_{\pm}$ defined as limits of states on $\Lambda_R$ with boundary conditions $ \pm a \log(R)$ for an $a >0$ with all the tempered Gibbs measures lying in between $\mu_-$ and $\mu_+$.
However, as the paper is in larger generality and the $\phi^4$-model itself is very well studied.
So I wonder what additional results can be shown for this particular model. In particular I wonder
Whether all translation invariant Gibbs measures are tempered?
What happens if I choose boundary conditions $\pm a R$ instead of $ \pm a \log(R)$
In what sense is the class of tempered Gibbs measures the class that is natural to study and what classifications theorems (forming a Choquet simplex etc.) are know about tempered translation invariant Gibbs states and just translation invariant Gibbs states?
However, the litterature of the model is also a wilderness so I hope you can help me out.
Not at all an answer, but you may find useful information in Section 4.7.1 of the thesis by Ajay Chandra https://libraetd.lib.virginia.edu/public_view/2r36tx772 Another reference which can help is http://www.numdam.org/article/RCP25_1978__26__31_0.pdf a not well known article by Lebowitz.
Since I asked this question the following paper has appeared and it addresses many of the points above: https://arxiv.org/abs/2211.00319.
|
2025-03-21T14:48:30.223282
| 2020-04-06T08:24:49 |
356713
|
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"Ville Salo",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/356713"
}
|
Stack Exchange
|
Borel hierarchy and tail sets
Let $A$ be a finite set, and let $A^\infty$ be the set of all sequences $(a_n)_{n=1}^\infty$ of elements of $A$.
A set $B \subseteq A^\infty$ is a tail set if for every two sequences $\vec a, \vec b \in A$ that differ in finitely many coordinates, either both are in $A$ or both are not in $A$.
Question: How high can we find tail sets in the Borel hierarchy? That is, do tail sets can be found only in the first $\alpha$ levels of the hierarchy for some ordinal $\alpha$ (maybe even some finite ordinal $\alpha$), or can we find them as high as we look?
Let $A = {0,1}$, let $U \subset A^{\infty}$ be any Borel. Construct $t(U) \subset A^{\infty}$ by letting $x \in t(U)$ iff $\sum x = \infty$, and when $10^n1$ appears in $x$ it does infinitely many times, and the characteristic sequence of $n$ s.t. $10^n1$ appears is in $U$. Clearly $t(U)$ is tail, and is Borel because can be constructed just like $U$, replacing basic opens by $\Sigma^0_2$s stating appearance of words of form $1 0^n 1$, and recurrence. There is continuous $f : A^{\infty} \to A^{\infty}$ reducing $U$ to $t(U)$, so if $t(U)$ is $Q^0_\alpha$ then so is $f^{-1}(t(U)) = U$.
Slightly wrong, since changing finitely many can break the recurrence condition. I suppose just drop the recurrence assumption. (Also I'm not sure this is a research level question.)
@VilleSalo: I'm a little confused by your comment. But I do know that in $\mathcal P(\mathbb N)$, the set of all $A \subseteq \mathbb N$ with positive upper density is a tail set in $\Sigma^0_3 \setminus \Sigma^0_2$. (I can't tell if this fact contradicts your claim or not.)
So maybe it should be $\Sigma^0_3$ instead of $\Sigma^0_2$, I wrote $\Sigma^0_2$ and was going to think about what the correct number is but apparently didn't. We agree it is a Borel set, though, that's all we need.
What I believe is correct is: encode a set $U$ into those distances between $1$s that appear infinitely many times. The resulting set is Borel, tail, and at least as high as $U$.
Here is an example: Let $A=\{0,1\}$ and for any $x\in \{0,1\}^{\omega}$, let $n<_x m$ if $x(2^n\cdot 3^m)=1$. For any countable ordinal $\alpha$, let $x\in U_{\alpha}$ if there is some $l$ so that $<_x$ over the set $\{n\mid n>l\}$ codes a well order $\leq \alpha$.
Now for any Borel set $B$, there is real $z$ so that $B$ is $\Delta_1^1(z)$ and so there is a recursive function $\Phi$ and a $z$-recursive ordinal $\alpha$ so that $x\in B\Leftrightarrow \Phi^{x\oplus z} \mbox{codes a well order} \Leftrightarrow \Phi^{x\oplus z} \mbox{codes a well order} \leq \alpha$. Moreover $x\not\in B\Leftrightarrow \Phi^{x\oplus z} \mbox{codes a linear order with an infinite descending subsequence}$.
Now let $\Psi$ be a $z$-recursive function so that $\Phi^{x\oplus z}$ codes $n<m$ iff $\Psi^{x\oplus z}(2^n\cdot 3^m)=1$; otherwise $\Psi^{x\oplus z}(2^n\cdot 3^m)=0$.
Then it is clear that for any $x$, $x\in B$ implies $\Psi^{x\oplus z}\in U_{\alpha}$. For a contradiction, suppose that $x\not\in B$ and $\Psi^{x\oplus z} \in U_{\alpha}$ for some $x$. Then $\Phi^{x\oplus z}$ codes a linear with an infinite descending subsequence over $\{n\mid n>l\}$. Then, for arbitrarily large $l$, $\Psi^{x\oplus z}$ contains an infinite descending subsequence, a contradiction.
So $x\in B$ if and only if $\Psi^{x\oplus z}\in U_{\alpha}$.
I think that by a more complicated argument, it can be shown that this is true for any countable Borel equivalence relation.
|
2025-03-21T14:48:30.223633
| 2020-04-06T09:23:40 |
356715
|
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"David C",
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|
Stack Exchange
|
Abelian versions of straightening and unstraightening functors
Let $X$ be a quasi-category (an inner Kan complex), let $\mathfrak{C}(X)$ be its rigidification (its associated simplicial category). J. Lurie in "Higher Topos Theory" proved the following theorem <IP_ADDRESS> (I give a particular case when $\phi=Id$) :
"Straightening and unstraigntening functors determine a Quillen equivalence
$$ St_{X} : (sSets)_{/X} \rightleftarrows sSets^{\mathfrak{C}(X)} :Un_{X}$$
where $(sSets)_{/X}$ is endowed with the contravariant model structure and $sSets^{\mathfrak{C}(X)}$ with the projective model structure."
I was wondering wether one could get a similar statement for $sAb^{\mathfrak{C}(X)}$ endowed with the projective model structure (here $sAb$ is the category of simplicial abelian groups), more generally for simplicial $R$-modules. Where we replace the category $(sSets)_{/X}$ by the category of simplicial algebraic gadgets over $X$.
Morever can we stabilize this theorem, in order to identify the category $spectra^{\mathfrak{C}(X)}$ together with a category of spectra over $X$?
Let $\mathcal{C}$ be a presentable $\infty$-category, and let $X$ be a Kan complex. Let $\mathcal{C}_{/X} = \mathrm{Fun}(X, \mathcal{C})$. Then $\mathrm{Sp}(\mathcal{C}_{/X}) = \mathrm{Sp}(\mathcal{C})_{/X}$. Indeed, this follows from the facts that $\mathrm{Sp}(\mathcal{C}_{/X}) = \mathcal{C}_{/X} \otimes \mathrm{Sp}$ and $\mathcal{C}_{/X} = X \otimes \mathcal{C}$. If $\mathcal{C}$ is the $\infty$-category $\mathcal{S}$ of spaces, then this implies that $\mathrm{Sp}_{/X} = \mathrm{Sp}(\mathcal{S}_{/X})$. Googling tells me that a keyword is "parametrized spectra", and that the above argument appears in Proposition 3.4 of https://arxiv.org/pdf/1112.2203.pdf.
Thank you for your answer, can you give some references please?
@DavidC Edited.
|
2025-03-21T14:48:30.223776
| 2020-04-06T10:16:09 |
356720
|
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|
Stack Exchange
|
Explicit construction of the Jacobian of a curve
Let $k$ be an algebraically closed field (of arbitrary characteristic), and $C$ a smooth projective curve over $k$, given by defining equations in projective space. I am looking for an algorithmic construction of the Jacobian variety of $C$.
To be more precise :
I am just looking for an explicit construction method, not an efficient algorithm. Complexity issues are not relevant, as long as there is no brute-force search on an infinite set.
The algorithm should at least yield explicit affine equations of open charts covering the Jacobian as well as the transition maps between them (or of course equations of the Jacobian in projective space).
The construction should apply to any such curve, in any characteristic. There are lots of references concerning hyperelliptic curves or curves over $\mathbb{C}$, but these do not apply to the general case.
For the moment, I am only aware of the following references :
Anderson, Abeliants and their application to an elementary construction of Jacobians : this seems to answer the question but I cannot figure out the geometric intuition behind it, nor is it clear to me how to get actual equations for the projective variety constructed there.
Shepherd-Barron, Thomae's formulae for non-hyperelliptic curves and spinorial square roots of theta-constants on the moduli space of curves : the results of this paper require a restriction on the characteristic of the ground field.
|
2025-03-21T14:48:30.223903
| 2020-04-06T10:20:58 |
356721
|
{
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"Gerry Myerson",
"Seva",
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|
Stack Exchange
|
Does $g+A\subseteq A+A$ imply $g\in A$?
Suppose that $A$ is a subset of a (large) finite cyclic group such that $|A|=5$ and $|A+A|=12$. Given that $g$ is a group element with $g+A\subseteq A+A$, can one conclude that $g\in A$?
The condition "large" is weird: indeed, if there is a counterexample in a cyclic group $C_m$, then we get the same counterexample in $C_{m^n}$ for all $n$. Hence, either you want to omit it (but then you maybe know a counterexample), or you want to replace "large" with "with no small prime divisor". Or make some further assumption, e.g., that $A$ generates $C_m$.
Why $5$, Seva? why $12$? What's the real question here?
Good question, Gerry; I actually expected someone to ask it!. If the answer were positive, I would be able to exclude one of the countless cases emerging in some large project.
This is not the case. Here is a counter example.
Let $G =Z_m, \ \ m>20$ (say). Let $A = \{0, 1, 3, 4, 6\}; g=2$: Note that $g+A=\{2, 3, 5, 6, 8\} \subseteq \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12\}=A+A$ and that $g \not\in A$, $|A|=5$, and $|A+A|=12$. Note that this example works for all large finite cyclic groups, as the reduction modulo $m$ does not matter.
|
2025-03-21T14:48:30.224014
| 2020-04-06T11:20:47 |
356725
|
{
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"sort": "votes",
"url": "https://mathoverflow.net/questions/356725"
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|
Stack Exchange
|
Terminology: "sufficiently large absolute constant"
I'm currently reading the paper "Random matrices: The distribution of the smallest singular values" by '"Terence Tao and Van Vu" and have run into some terminology which I don't quite (rigorously) understand.
In Theorem 1.3, the authors state that $\mathbb{E}[|\xi|^{C_0}]<\infty$ for some sufficiently large absolute constant $C_0>0$. What does "sufficiently large absolute constant" mean? I googled it but I couldn't find a definition.
So it holds for any $\xi$ random variables in $L^2$?
"Absolute constant" usually means it does not depend on any of the fixed data.
Any random variable $\xi$ ? No. Any random variable $\xi$ as specified in Theorem 1.3. If $\xi \in L^2$, then it hods for $C_0=2$ by definition of $L^2$. Presumably the random variables in Theorem 1.3 need not be in $L^2$, so he has to use a larger constant $C_0$. Perhaps they state it that way because they do not actually compute the constant $C_0$ that works, they only prove that one exists.
"Absolute constant" means that it does not depend on anything. For example, $3, 10^{12},\pi$ and Feigenbaum number are absolute constants. They are real numbers. "Sufficiently large" means that
the authors did not care or could not compute or estimate it.
|
2025-03-21T14:48:30.224134
| 2020-04-06T12:41:33 |
356729
|
{
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"Dominic van der Zypen",
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"url": "https://mathoverflow.net/questions/356729"
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|
Stack Exchange
|
Bike lock graph
Motivation. I have a bike lock with 4 dials, and I was wondering whether I can reach any combination by always turning a fixed number $k$, say $k=2$, of the dials, by $1$ position, instead of just turning $1$ dial at the time, which makes for a boring challenge.
Formal version. For any integer $n>2$, let $[n] =\{1,\ldots, n\}$ and let $C_n = ([n], E_n)$ denote the cyclic graph on the vertex set
$[n] =\{1,\ldots, n\}$ with $$E_n = \big\{\{k, k+1\}: 1\leq k < n\big\}\cup\big\{0,n\big\}.$$
Let $d$ denote the number of dials, $n$ the number of positions that any dial can take, and let $k\leq d$ be the fixed number of dials we have to turn by $1$ position at every step.
The dial itself can be represented by $[n]^d$. For $x,y\in[n]^d$ we let the differing set $D(x,y)$ to be defined by $\{i\in[d]: x_i\neq y_i\}$ where $x_i$ denotes the $i$th component of $x\in [n]^d$.
So we can define the following bike lock graph $B(n, d, k)$ for positive integers $n,d,k>1$ with $k\leq d$:
$V(B(n,d,k)) = [n]^d$,
$E(B(n,d, k)) = \big\{\{x,y\} \in [n]^d: |D(x,y)| = k \text{, and for all } i\in D(x,y)\text{ we have } \{x_i, y_i\}\in E_n\big\}.$
Every configuration of the bike lock can be reached with the allowed moves if and only if the corresponding graph $B(n,d,k)$ is connected.
Question. Are there infinitely many integers $n>1$ such given an integer $d>2$, the graph $B(n,d,k)$ is connected for some integer $k$ with $2\leq k\leq d-1$?
If $k$ is even you preserve parity of sum, otherwise you can flip a single coordinate by flipping back and forth an odd number of times and making others cancel out, using $k+1$ coordinates?
EDIT: I've just realized that I answered a different question than was asked. The answer assumes that each of the $k$ dials needs to turn by 1 in the same direction. I'm leaving the answer as is in case someone will find the modified problem interesting.
TL;DR: the graph is connected iff $k$ and $n$ are co-prime.
I think the question can be rephrased as follows. Consider the set of all vectors $\mathbb Z_n^d$ in which exactly $k$ coordinates are 1, and the rest are 0. What is the integer span $S_n$ of this set?
Let's first consider this question in $\mathbb Z^d$ instead, and ask for the corresponding span $S$.
For distinct $i_1, ..., i_k$, let $A(i_1, ..., i_k)$ be $(x_1, ..., x_d) \in \mathbb Z^d$ such that $x_j = 1$ if $j \in \{i_1, ..., i_k\}$, and 0 otherwise. By definition, $A(i_1, ..., i_k) \in S$.
Let $B(i, j)$ be $(x_1, ..., x_d) \in \mathbb Z^d$ such that $x_i = 1$, $x_j = -1$, and all other coordinates are 0. Note that $B(1, 2) = A(1, 3, 4, ..., k+1) - A(2, 3, ..., k+1) \in S$. Since problem is invariant to co-ordinate permutations, $B(i, j) \in S$ for any $i, j$.
The coordinates $(x_1, ..., x_d)$ of any $A(i_1, ..., i_k)$ sum to $k$. Therefore any $(x_1, ..., x_d) \in S$ satisfies $\sum x_i = 0 \pmod k$. We'll now show the reverse: any $(x_1, ..., x_d) \in \mathbb Z^d$ with $\sum x_i = 0 \pmod k$ belongs to $S$. We'll show this by starting from $(x_1, ..., x_d)$ and subtracting vectors in $S$ until we reach $(0, ..., 0)$.
By assumption, $\sum x_i = mk$ for some $m$. By subtracting $m A(1, 2, ..., k)$, we obtain a new $(x_1, ..., x_d)$ such that $\sum x_i = 0$. If $(x_1, ..., x_d) \neq (0, ..., 0)$, then we take some $x_i < 0$ and some $x_j > 0$, and add $B(i, j)$ to $(x_1, ..., x_d)$. This will preserve $\sum x_i = 0$, and reduce $\sum |x_i|$. So, after repeating the last step finitely many times, you'll obtain $(0, ..., 0)$. This proves that the initial $(x_1, ..., x_d)$ was in $S$.
This solves the problem for $\mathbb Z^d$ by showing that $S = \{(x_1, ..., x_d) | \sum x_i = 0 \pmod k\}$. To find the span $S_n \subseteq \mathbb Z_n^d$, notice that $(y_1, ..., y_d) \in S_n$ iff $(y_1, ..., y_d) + n(m_1, ..., m_d) \in S$ for some $m_1, ..., m_d \in \mathbb Z$. This is equivalent to $\sum_i y_i + mn = 0 \pmod k$ for some $m$. This in turn is equivalent to $\sum_i y_i = 0 \pmod{\gcd(k, n)}$.
So the answer to your original problem is $S_n = \{(y_1, ..., y_d) | \sum y_i = 0 \pmod{\gcd(k, n)}\}$. So the graph is connected iff $k$ and $n$ are co-prime.
So in the original case $n = 10$ and $d = 4$, we have the following. For $k = 2$, because $\gcd(2, 10) = 2$, the reachable states are those where sum of the lock numbers is even. For $k = 3$, because $\gcd(3, 10) = 1$, all lock states are reachable. For $k = d = 4$, only states where all digits are the same are reachable.
Thank you - this modified problem is so interesting that I am going to accept this answer!
|
2025-03-21T14:48:30.224415
| 2020-04-06T13:04:26 |
356731
|
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|
Stack Exchange
|
Grothendieck categories and their morphisms
I am not an algebraic geometer in the first place, and I am mainly familiar with topology and category theory. Recently I am studying Grothendieck categories and I am struggling with getting acquainted with them.
What is a Grothendieck category?
This question will appear extremely naive, especially to algebraic geometers, thus please give me some time to comment and define a bit of context.
Rem 1. Indeed, following Gabriel-Popescu theorem and the general theory of lex-localizations, it follows that a Grothendieck category is no more than an $\mathbb{Ab}$-topos. By $\mathsf{V}$-topos we intend an enriched lex-reflective subcategory of an (enriched) presheaf category $\mathsf{V}^C$. Even better, following the research line lead by Borceux and later developed by Lowen, even site presentations of these abelian topoi are available.
Rem 2. Yet, in order to study Grothendieck categories with the same intuition of Grothendieck topoi, we should study geometric morphisms between them. This notion does not appear to be the correct one.
What is the correct notion of morphism between Grothendieck categories?
Cocontinuous functors. For example, if we look at the construction of Quasi-coherent sheaves over a scheme $\mathsf{Qcoh}(-)$, it looks natural to study inverse images (that is cocontinuous functors). Yet, quasicoherent sheaves over a scheme are not precisely the abelian analogous of sheaves over a manifold (for several reasons). Maybe a more natural analogous would be to study abelian group objects internal to little Zariski topos. Also in this case, the natural notion of functor seems (to me) to be cocontinuous functors.
Brandenburg's PhD thesis. The global pictures gets messier when we add Brandenburg's Phd thesis to the cocktail. To my superficial understanding, he studies Grothendieck categories equipped with a tensor product (motivated by very natural considerations that look quite convincing to me) and thus considers cocontinuous functors preserving the tensorial structure. This notion seems to me as the most promising notion of geometric morphism where lex-ity is substituted by the preservation of the tensorial structure.
Back to $\mathbb{Ab}$-topoi, flatness and intuiton. Putting all these observations together, the $2$-category of $\mathbb{Ab}$-topoi (namely Grothendieck categories and lex-cocontinuous functors) seems a strange object to look at, and yet it is naturally connected to the study of flat morphisms between schemes.
What should be my geometric intuition (!) on flat morphisms of Grothendieck categories? (And maybe more generally on flat morphisms of schemes? (I did not find satisfying this answer))
Any kind of comment to any statement I have written is extremely welcome, and I encourage everybody to correct my mistakes. As I mentioned above, I am not a specialist in this subject and I am learning.
Small comment: there are multiple sources of sheaf categories in algebraic geometry. Flatness is automatic when the structure sheaf is constant (e.g. the étale site with sheaves of $\mathbf Z/n$-modules), but in the quasi-coherent setting is related to the change of rings $f^{-1}\mathcal O_Y \to \mathcal O_X$ (see for example Tag 04JA). Which of the two are you trying to capture?
@R.vanDobbendeBruyn I like your comment and I am not sure about what I should want, I invite you to elaborate it. I know nothing about etàle site. What story do they tell?
Basically there are two kind of cohomological coefficients continuous and discrete (after Grothendieck). The former are basically quasi-coherent sheaves, the latter local systems in the classical case. As Zariski topology does not capture well the discrete coefficients one has to switch topos and consider the Étale topology and for p-torsion crystalline cohomology. Every theory has an associated Grothendieck category of abelian sheaves (with extra structure).
Is a "flat morphism" of Grothendieck categories a functor with an exact left adjoint?
Yes, that's what I mean.
|
2025-03-21T14:48:30.225055
| 2020-04-06T13:10:13 |
356732
|
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|
Stack Exchange
|
Recognizing Lipschitz functions up to change of target metric
Let $K$ be a compact subset of $\mathbb{R}^n$ (for simplicity, I am happy to take $K=\overline{B(0,1)}$ for now if it is easier).
Let $f:K \rightarrow \mathbb{R}^m$ be a continuous function.
Is there a new metric $d$ on $\mathbb{R}^m$ (compatible with the topology) such that
$$ f: K \rightarrow (\mathbb{R}^m,d) $$
is Lipschitz?
(Here $K$ is still equipped with the usual Euclidean metric, and only the metric on the target space $\mathbb{R}^m$ has changed.)
If not,
What are obstructions?
Are there sufficient conditions?
See this question and links therein https://mathoverflow.net/questions/291563/what-is-the-name-for-a-set-endowed-with-a-lipschitz-structure
Thank you erz. The reference of Luukainen-Vaisala linked there seems relevant, but I have not yet found an answer to this question there (at an admittedly brief glance). If I do, I will update.
also relevant (in fact essentially the same)
https://mathoverflow.net/questions/195611/quotient-of-metric-spaces
It can't be done. Here's a counterexample with $n = 2$ and $m = 1$.
For each $k \in \mathbb{N}$, define a function $f_k: [0,1] \to [0,1]$ by linearly interpolating between the values $f_k(\frac{i}{k}) = \frac{i}{k}$ ($0 \leq i \leq k$) and $f_k(\frac{i}{k} + \frac{1}{k^2}) = \frac{i+1}{k}$ ($0 \leq i \leq k -1$). They look like staircases.
I claim there is no metric on the target $[0,1]$ that makes all of the $f_k$ contractions. If $f_k$ is a contraction this forces the distance from $\frac{i}{k}$ to $\frac{i+1}{k}$ to be at most $\frac{1}{k^2}$, and thus the distance from $0$ to $1$ to be at most $\frac{1}{k}$. If this were true for all $k$ then the distance from $0$ to $1$ would have to be zero and the metric would not be compatible with the topology.
Now define $f(t,\frac{1}{k}) = f_k(t)$ for all $k$ and $f(t,0) = t$. Then $f$ is a continuous function defined on a compact subset of $[0,1]^2$, so by Tietze it extends to a continuous function from the unit square to $[0,1]$. If there were a change of metric on the target $[0,1]$ that made $f$ Lipschitz then by scaling there would be a change of metric that made $f$ a contraction, and this would make each $f_k$ a contraction, contradicting the claim.
Nice! Do you have any thoughts on sufficient conditions? If not, I'd still happily accept the answer, since this was the main point of my question.
I'm afraid I don't have anything intelligent to say about sufficient conditions ... I will think about it.
I guess for each distinct $s$ and $t$ in the range you want to look at the value ${\rm inf}{|x_1-y_1|+\cdots+|x_r-y_r|: f(x_1) = s$, $f(y_r) = t$, and $f(y_i) = f(x_{i+1})}$. If it's ever zero there's no hope, but if not this value describes a metric on the range which makes $f$ a contraction. It shouldn't be hard to use this to get a metric on $\mathbb{R}^m$ that makes $f$ a contraction.
Yes, this makes sense. Thanks (to both you and Pietro Majer) for the answers. I'm accepting this one just because it was first and you were kind enough to think about the sufficiency, but Pietro Majer's is also worth accepting!
You are welcome!
For instance, no equivalent distance on $\mathbb{R}^1$ can make Lipschitz the Cantor function $f:[0,1]\to\mathbb{R}^1$.
Suppose by contradiction $d$ is a distance that makes $f$ $L$-Lip.
Let $F_n$ be the closed sets in the usual construction of the cantor set $C=\bigcap_{n\ge\mathbb{N}}F_n$, that is $F_0=[0,1]$ and $F_{n+1}=\frac{1}{3}F_n \cup (\frac{1}{3}F_n+ \frac{2}{3})$.
If $0=x_1<\dots<x_{2^n}=1$ are the endpoints of the components of $F_n$, then $x_{2i}-x_{2i-1}=\Big(\frac{2}{3}\Big)^n$ and $f(x_{2i})=f(x_{2i+1})$ because $f$ is constant on the connected components of $F_n^c$.
But then for all $n$
$$d(1,0)=d(f(0),f(1))\le L\sum_{i=1}^{2^n}|x_i-x_{i-1}|=L \Big(\frac{2}{3}\Big)^n$$
so $d(0,1)=0$.
Thank you, this is also a very straightforward example.
So the idea of your example is that a Lipschitz function maps measure-zero sets into measure-zero sets, and whatever metric we choose on the target, the Hausdorff measure of an open set $(0,1)$ has to be positive. Perhaps this can be enhanced somehow to get a sufficient condition?
Exact, that was what I had vaguely in mind, though I could not find the right words :)
@erz Indeed your idea of a sufficient condition in terms of the Hausdorff measure should be the right one. I tried to expand it a bit.
Here are some suggestions trying to expand erz's and Nik Weaver's comments. I do not have time to work out the details (although stuck at home arrests like many of us), but I'd be glad if nevertheless they turn out to be useful.
Recall that a set $S\subset\mathbb{R}^m$ is $\mathcal{H}^1$- null iff for all $\epsilon>0$ it has a countable partition $\{S_j\}_{j\in\mathbb{N}}$ such that $\sum_{j\in\mathbb{N}}\rm{diam}(S_j)\le\epsilon$.
Say the that a map $f:A\subset\mathbb{R}^m\to\mathbb{R}^n$ is $\mathcal{H}^1$- Absolutely Continuous iff it is continuous and for any $\mathcal{H}^1$-null set $S\subset A$, the image $f(S)$ is $\mathcal{H}^1$-null in $\mathbb{R}^n$ .
[edit 4.24.20]: as user erz pointed out to me, this definition is too weak (the function in Nik Weaver's counterexample satisfies it). A more suitable definition is maybe: iff it is continuous and for any $\epsilon>0$ there exists $\delta>0$ such that for any $S\subset A$, $\mathcal{H}^1(S)<\delta$ implies that $\mathcal{H}^1 (f(S))<\epsilon$ .
The conjecture should be:
Any $\mathcal{H}^1$-AC map defined on a compact subset $K$ is Lipschitz
up to a choice of an equivalent distance on the target space.
We may always assume $n=m$ (via inclusion $K\subset \mathbb{R}^n\subset\mathbb{R}^m$, or $\mathbb{R}^m\subset\mathbb{R}^n$ if needed). Then
It does not seem problematic to extend an $\mathcal{H}^1$-absolutely continuous map $f$ defined on a compact $K$ to a $\mathcal{H}^1$-absolutely continuous map $f:\mathbb{R}^n\to\mathbb{R}^n$ that is also a surjective and proper map (it should sufficient to make it locally lipschitz on the complement of $K$, and equal to the identity outside a large ball.
For $x,y$ in $\mathbb{R}^n$ define
$$d(x,y):=\inf\{\mathcal{H}^1(S): f(S)\,{ \rm connected},\, \{x,y\}\subset f(S)\}$$
Then $d:\mathbb{R}^n\times \mathbb{R}^n\to[0,\infty)$ is clearly symmetric, and satisfies the triangular inequality.
To show $d(x,y)=0$ implies $x=y$, is where the hypotheses enter. The idea should be that $d(x,y)=0$ implies the existence of a $\mathcal{H}$-null subset $S$ such that $f(S)$ is connected and $\{x,y\}\subset f(S)$, which forces $x=y$ since $f(S)$ is also $\mathcal{H}$-null. To this end one should start from a minimizing bounded sequence of compact subsets $S_j$ with $\mathcal{H}^1(S_j)\to0$, and $\{x,y\}\subset f(S_j)$. Compact subsets of a given compact are a compact in the Hausdorff distance, the Hausdorff measure is lower semicontinuous, connected sets are a closed set, so I think it could be done. In fact, it seems OK that the infimum in the definition of this distance be always attained, by the same argument.
By definition of $d$, taking as $S$ the segment $[u,v]$ one has $d(f(u),f(v))\le \mathcal{H}^1([u,v])= \|u-v\|$.
It remains to show that $d$ is topologically equivalent to the Euclidean distance, which seems true, although I see it less clearly at the moment.
This goes far beyond my little comment. Very interesting!
Could you please elaborate on the extension? As for topological equivalence, here is a proof if $K$ is "nice". Assume $d(y_n,y)\to 0$ and $\varepsilon>0$. Since $f$ is uniformly continuous on $K$, there is $\delta$ such that $|x-z|<\delta\Rightarrow |f(x)-f(z)|<\varepsilon$. For large $n$ $d(y_n,y)<\delta$, and so there are connected $S_n\subset K$ and $x_n,z_n\in S_n$ such that $f(x_n)=y,~ f(z_n)=y_n$ and $|y_n-z_n|\le H^1(S_n)<\delta$, from where $|y_n-y|\to0$. Assume $|y_n-y|\to0$ but $d(y_n,y)\not\to 0$. WLOG $d(y_n,y)>a>0$. Choose $z_n\in f^{-1}(y_n)$. From compactness WLOG $z_n\to z$.
If $K$ is nice so that we can join $z_n$ and $z$ with short curves, then we get a contradiction with $d(y_n,y)\not\to 0$. So $K$ should be such that the length metric is equivalent with an Euclidean one. If you indeed can extend $f$ to a closed neighborhood of $K$ that would be enough.
I'd like to add that this condition is necessary to existence of metric on $f(K)$ such that $f$ is Lipschitz, and also the coordinate projections on $f(K)$ are Lipschitz. Perhaps it is also sufficient. What I am saying is that the condition depends on the Euclidean metric on the image, and is not purely topological, as one would like to.
@erz very good. For the issue of the extension, one can make it locally Lipschitz, or even smooth, on the complement of $K$: isn't this sufficient? For any $H^1$-null set $S$ in the domain, $f(S\cap K)$ is $H^1$-null by hypothesis, and since $S\setminus K$ is covered by countably many balls on which $f$ is Lipschitz, the set $f(S\setminus K)$ is $H^1$-null as well.
In fact, (once it is OK that $d$ is a distance), on the lines of your argument, since $(K, |\cdot|)$ is compact it is sufficient to show that the identity $(K, |\cdot|)\to (K,d)$ is continuous. That is: if $|u_j-v|\to0$ then $d(u_j,v)\to0$.
Since $f$ is 1-Lip from $ |\cdot|$ to $d$, we have for all $u,v$ $$d(u,v)\le \inf{ |x-y|: x\in f^{-1}(u), , y\in f^{-1}(y)}.$$ Since $f$ is surjective and proper, if $u_j\to v$we have a $ x_j\in f^{-1}(u_j)\subset K$, $x_j\to y\in f^{-1}(v)$, whence $d(u_j,v)\to0$
(Also, it seems to me that we may take everywhere, in the definition of $H^1$-AC map and in the definition of $d(x,y)$, all sets $S$ to be compact. Is this equivalent? It seems that everything works as well; maybe it's more clear).
sorry you lost me on your second comment. And still, i cannot see (most likely due to my ignorance), how to extend $f$ to be locally Lipschitz and surjective on the complement of $K$? (I can see why that would be sufficient). As for compactenss in the definitions - yeah, probably those are the "correct" definitions, because everywhere else in the proof the sets are compact.
(typo: above I mean: the identity map not on $K$, but on $f(K)$ , or on any closed ball)
To extend $f$ to be locally Lipschitz on the complement: we extend each coordinate of $f$, e.g. by means of a Lipschitz partition of unity in the complement of $K$; this is standard, and we can also make for the extended map $f_i(x)=x_i$ for larve $|x|$. The resulting map is proper, and it is surjective because it has topological degree $1$. (Of course, assuming $n=m$)
To be honest, i cannot reconstruct the partition of unity argument (i'd really appreciate if you point me to something analogous). What i came up with is to approximate $f_i$ with a sequence of polynomials $p_n$, while also enclose $K$ with a sequence $U_n$ of open sets that shrinks to $K$ and $\overline{U_{n+1}}\subset U_n$. Construct $g=p_1$ on $U_1$ \ $U_2$, then extend it to $U_1$ \ $U_3$ so that on $U_2$ \ $U_3$ it approximates $p_2$ and so on. In the end define $g=f_i$ on $K$. The resulting function is continuous and smooth outside of $K$.
@erz : the partition of unity argument is the usual construction of the Tietze extension thm on metric spaces (as e.g. in Dieudonné, Found. of Mod. Anal.); it gives a loc.lip. function if the partition is such. You can also google "tietze extension theorem in metric spaces"
There are other possibilities of course. The quickest is as follows: take a concave modulus of continuity $\omega$ for $f$ on $K$ (see the wiki article linked below); define the extension as in the link: $f^$ := infimum of all $\omega$-continuous functions that are above $f$ on$K$. The function $f^$ has the same mdc $\omega$ globally, and on each set $\Omega_\epsilon:= {x: d(x,K)\ge\epsilon}$ it is Lipschitz; If I'm not wrong precisely with constant $\omega'_+(\epsilon)$ (right derivative).
https://en.wikipedia.org/wiki/Modulus_of_continuity#Subadditive_moduli,_and_extendibility
... the reason is : because on $\Omega_\epsilon$ all functions in the infimum that defines $f^*$ are $\omega'+(\epsilon)$-Lip, and this because $\omega$ itself is $\omega'+(\epsilon)$-Lip on the interval $[\epsilon,+\infty)$, being concave.
|
2025-03-21T14:48:30.225950
| 2020-04-06T13:14:31 |
356733
|
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|
Stack Exchange
|
Elementary classification of division rings
Are there examples (other than the two mentioned below) of fields $K$ such that the classification of all finite dimensional division $K$-algebras is possible using only elementary theory (lets say a basic course in algebra, including field and galois theory)?
The only examples I am aware of are finite fields and $\mathbb{R}$ (and trivially algebraically closed fields).
You ask if examples exist, and then you list some, so of course you are aware they exist. Do you mean to ask whether others exist?
Probably $\mathbf{C}(!(t)!)$ is not hard either.
The classification is trivial for a $\,C_1$-field $K$ (that is, such that any homogeneous polynomial in $K[x_1,\ldots ,x_n]$ of degree $<n$ has a nontrivial zero): the only such division algebras are the finite extensions of $K$. For the proof you just need the notion of reduced norm, which can be explained in a reasonably elementary way (see Central simple algebra).
$C_1$-fields include finite fields and extensions of transcendance degree 1 of an algebraically closed field (Tsen's theorem); again, the proof in each case is relatively elementary.
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2025-03-21T14:48:30.226068
| 2020-04-06T13:47:03 |
356737
|
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|
Stack Exchange
|
Countable convergence-determining class for weak convergence of probability measures
Suppose that $E$ is a Polish space.
Portmanteau theorem asserts that a sequence $(\mu_n)$ of Borel probability measures weakly converges to a Borel probability measure $\mu$ (shortly, $\mu_n\overset{w}{\to\mu}$) if and only if $\limsup_n \mu_n(C)\le \mu(C)$ for all closed set $C\subset E$. My question is whether there exists a countable convergence-determining class of closed sets. Namely, if there exists a countable collection $\mathcal C$ of closed subsets of $E$ such that $\limsup_n \mu_n(C)\le \mu(C)$ for all closed set $C\in\mathcal C$ implies that $\mu_n\overset{w}{\to\mu}$.
Let $E$ be any separable space. The assertion is equivalent to ($*$)
$\liminf_{n\to \infty} \mu_n(U) \geq \mu(U)$ for any open $U \subset E$. Since $E$ is separable, there is a countable base $\cal{U}$, which is $\cap$- and $\cup$-stable, for the open sets in $E$. This $\cal{U}$ is "convergence-determining" for open sets. Let $U$ be an arbitrary open subset of $E$. Then there is a sequence $U_k$ in $\cal{U}$ with $U_k \uparrow U$. But then $\liminf_n \mu_n(U) \geq \liminf_n \mu_n(U_k) \geq \mu(U_k)$ for any $k \in \mathbb{N}$, since $\mu_n(U) \geq \mu_n({U_k})$. Since $\mu(U_k) \uparrow \mu(U)$ ($*$) follows.
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2025-03-21T14:48:30.226181
| 2020-04-06T14:48:46 |
356741
|
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|
Stack Exchange
|
Embedding of Cuntz algebras $O_2\subseteq O_3$?
The Cuntz algebra $O_n$ is the (universal) C*-algebra generated by n-isometries $s_1,...,s_n$ such that
$$\sum_{i=1}^n s_is_i^\ast =\mathbf{1}, \ \hbox{and}\ s_i^\ast s_j=\delta_{ij} \mathbf{1}\ (\hbox{for all}\ i,j).$$
For example if $s_1, s_2$ are the generators of $O_2$ then $s_1^2, s_1s_2, s_2$ generate a copy of $O_3$ in $O_2$ (as they satisfy the relations of $O_3$) as pointed out by Cuntz in page 184 in Comm. Math. Phys. 57(1977), 173-185.
Now i wonder if one can generate a copy of $O_2$ in $O_3$ by manipulating the generators of $O_3$?
Do you want a unital embedding?
Yes please, although $O_2$ sitting inside $O_3$ with a possible different identities.
You can always do that as you pointed out in your comment to Darth Vader's answer. You can't send the identity of $O_2$ to the identity of $O_3$ for K-theoretic reasons.
The answer is no. According to Lemma 2.1 in (K. Kawamura, Universal algebra of sectors, Int. J. Alg. Comput. 19(3)(2009), 347–371.) we have:
$Hom(O_m, O_n)$ is nonempty if and only if $m=(n-1)k+1$ for some $k \geq 1$.
Applying this to the case for $Hom(O_2, O_3)$ gives that we must have $2=2k+1$, which is impossible.
On the other hand, Prop. 4.2.3, page 75, in the textbook "Classification of Nuclear C*-algebras (Encyclopedia of Mathematical Sciences Vol 126)" says that $O_2$ embeds unitally in $pAp$, where $p$ is any (full) properly infinite projection in a given C*-algebra $A$ with $[p]_0=0$ in the K-group $K_0(A)$.
So Kawamura result implies that there is no C*-alg $A$ and proj $p$ such that $pAp=O_3$?
In any properly infinite, unital $C^\ast$-algebra $A$, you can find a full, properly infinite projection $p\in A$ with $[p]_0 = 0$ in $K_0(A)$ (this is due to Cuntz, see also Prop. 4.1.4 in Rørdams classification book). In particular, $\mathcal O_2$ embeds unitally into $pAp \subseteq A$. Such a $C^\ast$-algebra $A$ contains a unital embedding of $\mathcal O_2$ if and only if $[1_A]_0 = 0$ in $K_0(A)$.
|
2025-03-21T14:48:30.226352
| 2020-04-06T16:04:22 |
356745
|
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|
Stack Exchange
|
Taylor expansion with remainder on locally convex spaces
It is usual to introduce Fréchet and Gâteaux derivatives in Banach spaces. In this context, the familiar Taylor expansion with remainder is also at hand, as you can see on the picture below taken from this reference. Now, I'd like to know if there exists an analogous theorem for locally convex spaces and where should I go to learn it if this is the case. Do we need Fréchet and Gâteaux derivatives as well? Any comments would be helpful. Thanks in advance!
EDIT: Let $E, F$ be locally convex spaces over some field $\mathbb{K}$ and $f: E \to F$. We say that $f$ is Gâteaux differentiable at $x \in E$ if there exists a continuous linear functional $Df[x]$ such that
\begin{eqnarray}
\lim_{\epsilon \to 0}\frac{f(x+\epsilon h)-f(x)}{\epsilon} = Df[x](h) \tag{1}\label{1}
\end{eqnarray}
for every $h \in E$.
The above definition makes sense and it is the definition of Gâteaux differentiability that I know. Now, here the author states a Taylor expansion result (pag. 25, Theorem 1.4.11) but his definition of Gâteaux differentiability is a little different (Definition 1.4.7 on page 24). However, I believe my definition and his definition are equivalent and, if so, this answers my question. Am I missing something here?
Why did you put the 'distributions' tag?
I thought that if such a result exits, it'd be possible to be best known for more especific spaces like $\mathcal{S}(\mathbb{R}^{n})$.
I still don't see the link with distributions. 'schwartz-distribution' is a tag about distributions, not about Schwartz functions.
Ok. I'll replace it by a more adequate one.
Will, just an observation: in the monograph linked by prof. Michor in his answer at pages 73-77 and 116, there are some interesting insights on how the functional calculus evolved.
@DanieleTampieri thanks so much!! Seems an amazing reference!
Yes, that work is a clever and modern study on some slightly shaded aspects of functional analysis. If I could give one more advice, I’ll say to have a look at the book by Hille and Phillips, where the theory is however developed in Banach algebras, so in general the product of two functions is meaningful and belongs to the same space.
First of all the good news: There is an analogous theorem for locally convex spaces. It can be formulated in almost the same way using Gateaux derivatives (more on this notion in a moment). Obviously, the norm condition on the vanishing of the remainder makes no sense in the absence of a norm. However, you can formulate the vanishing of the remainder term of order $n$ via a homogeneity condition on the arguments of the remainder.
Before we come to the bad news, some comments on differential calculus beyond Banach spaces. Frechet derivatives make no sense beyond Banach spaces (again the norm condition breaks ones neck) and there are several (beyond Frechet spaces mutually inequivalent) ways of doing calculus in locally convex spaces. You can choose among dozens of approaches (all summarised in H.H. Keller: Calculus in locally convex spacesa), but most people stick with one of the following approaches:
1. The convenient calculus (popular since Kriegl and Michors book: The convenient setting of global analysis, from 1998 available online here 1 ). To my knowledge this does NOT feature a Taylor like theorem, though you find a section on Jet spaces (which is a more fancy way of talking about Taylor like expansions)
2. Bastiani calculus as used in most writings on infinite-dimensional Lie theory, see e.g. Section 1 of 2 for an overview and more references. In Bastiani calculus I am certain that there is a version of Taylors theorem. I give you a cartoon version since it is technical and long and I do not want to type endlessly:
Let $E,F$ be locally convex spaces, $k\in \mathbb{N}_0$ and $f\colon U \rightarrow F$ a Bastiani $C^k$-map on an open subset of $E$. Then the following holds:
(a) for each $x\in U$, there exists a unique polynomial $P_x^kf \colon E \rightarrow F$ of degree $\leq k$ such that $\delta_0^j(P_x^kf)=\delta_x^jf $ for each $j\leq k$.
(b) $\lim_{t\rightarrow 0} \frac{f(x+ty)-P_x^kf(ty)}{t^k}=0, \forall y \in E$ and the polynomial is unique with this property.
(c) There is a continuous remainder term
$R_k(x,y,t)=\frac{1}{(k-1)!}\int_0^1 (1-r)^{k-1}(\delta_{x+rty}^kf(y)-\delta_x^kf(y))\mathrm{d}r$ satisfying equation (30.14) in the picture you uploaded
On to the bad news: I am certain that it exists because I have a nice pdf in front of me stating the theorem together with all the nitty gritty details and how to prove it. The unfortunate part is that I am not at liberty to share it since this is part of the book manuscript for the forthcoming* book by Glöckner and Neeb on infinite-dimensional Lie theory. This is unfortunately also the only source I know of, where one can find it in this form. You could try to write an email to Glöckner (see 2) and see whether he can either point you towards an older source or wants to share the pdf.
*: Unfortunately, the book was forthcoming from 2005 on. I have seen many revisions but it is anybodys guess when it will finally appear in public.
Thanks so much for the enlightening comment! I don't know if you noticed my edit, but I added a hyperlink on the mentioned work of Glockner and Neeb (I found it avaiable online). So this goes in the same direction as my edit suggests.
As a final comment, I'd like to put Glockner and Neeb's Theorem in terms of my definition of Gâteaux differentiability. I think these definitions are equivalent. Do you agree?
Indeed I missed the edit. The linked page is quite like one of the draft versions for the corresponding chapter in the book I was refering to, so this should have you covered. As to the definition of the Gateaux derivative: It should be the same. If you are reading the definition of Bastiani differentiable, the chapter you linked should have the relevant Lemma which shows that the derivative has the properties you claimed for the Gateaux derivative, so in essence that should be what you want.
|
2025-03-21T14:48:30.226793
| 2020-04-06T17:25:45 |
356749
|
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Stack Exchange
|
Infinity local systems
I have seen many references in the (geometric representation theory, symplectic geometry, etc) literature to "infinity local systems".
From what I've been told, given a good cover $\{U_i\}$ of $X$, an infinity local system on a connected space $X$ assigns:
-a chain complex $C_i$ to every contractible open set $U_i$
-a chain morphism $F_{ij}: C_i \to C_j$ to every double intersection $U_i \cap U_j$
-a homotopy of chain morphisms $F_{ik} \sim F_{jk} \circ F_{ij}$ for every triple intersection $U_i \cap U_j \cap U_k$
-etc.
EDIT: actually, I think that the above definition is rather an "infinity-sheaf". An infinity local system should be a locally constant "infinity-sheaf". My guess is that the Maurer-Cartan condition mentioned below is precisely encoding the "locally-constant" condition.
Some questions:
(1) What are some good sources to learn about these objects (from the perspective of a second year graduate student with limited exposure to these ideas)?
The only paper I have found is Section 2.1 of "A Riemann–Hilbert correspondence for infinity local systems" (https://arxiv.org/pdf/0908.2843.pdf) In this paper, an infinity local system is defined as a set of maps from a simplicial set $K$ to a differential graded category $C$ which satisfy a certain Maurer-Cartan equation. However, it's not clear to me how to make sense of this definition (e.g. what is the role of Maurer-Cartan? how does this definition match up with the intuitive notion of an infinity local system described above?)
(2) The infinity local systems on a topological space $X$ are supposed to form a dg category. I recently heard a talk in which the speaker claimed (as if it was the most natural thing in the world:) )that $hom(k_X, k_X)=C^*(X;k)$, where $k$ is some field, $k_X$ is the locally constant sheaf with stalk $k$ is degree $0$ and $C^*(X;k)$ is the singular cohomology of $X$ with $k$ coefficients.
Is there a good way to understand why $hom(k_X, k_X)=C^*(X;k)$? More generally, how should one understand the dg category structure on $\operatorname{Loc}_{\infty}(X)$?
Perhaps a more concrete model for the situation you're interested in is just (dg-)modules over $C_(\Omega X)$? It's noncanonical since it involves choosing a basepoint, but it's at least concrete. Taking chains on the simplicial object $B_{\bullet}\Omega X$ gives a resolution of $k$ over $C_(\Omega X)$ which yields the computation $hom(k_X, k_X)=C^(B\Omega X)=C^(X)$, for example.
For question (2), the point is that $C^*(X;k)$ is the homotopy limit of $k_X$ seen as the constant functor from $X$ to (dg) $k$-modules, therefore the computation you want is just the universal property of the limit.
Look at proposition 5.3. in the following article https://www3.nd.edu/~wgd/Dvi/Duality.Algebra.Topology.pdf I guess you do need to assume more finiteness condition on $X$.
@GSM that proposition is referring to something else (though related). You do not need any finiteness assumptions for the statement I gave, or for what Denis said.
Also, @user142700, the "Cech" description you gave is not quite correct unless all those various intersections are also contractible (so, for example, a good cover of a manifold would be okay).
@DylanWilson: Thank you very much for your comments! Concerning your first comment, I don't understand the statement: "Taking chains on the simplicial object $BC_(\Omega X)$ gives a resolution of over $C_(\Omega X)$". Could you explain and/or give a reference which explains (1) what it means to "take chains over a simplicial object" and (2) why these chains are $C_*(\Omega X)$-modules?
@DenisNardin. Thank you very much for this comment! What do you mean by taking the homotopy limit of $k_X$? Could you also expand on your statement that "the computation you want is just the universal property of the limit."? (I'm sorry, I'm just a beginner...)
@user142700 ah sorry- I should have said to use $E_{\bullet}\Omega X$, whose $n$th term is $\Omega X^{\times n+1}$. By 'take chains' I mean apply the functor $C_$ to each term in the simplicial object. The action of $\Omega X$ on the left of the product of copies of itself gives an action of $C_(\Omega X)$ on each term. If you now Hom over $C_*(\Omega X)$ to $k$ and take alternating sums of the face maps, you get the cobar complex computing Ext.
@user142700 Local systems "really" means the functor category $\mathsf{Fun}(X, \mathsf{Ch})$ where one must interpret the space $X$ as yielding an $\infty$-category, and $\mathsf{Ch}$, likewise, as an $\infty$-category. In this setting, $k_X$ means the constant diagram $X \to \bullet \to \mathsf{Ch}$ at the chain complex $k$. The `tensor hom' adjunction tells you that $hom(k_X, -)$ is right adjoint to the functor assigning to a chain complex $C_$ the constant diagram at $C_$. On the other hand, the right adjoint to the constant diagram functor is called 'the (homotopy) limit'.
@DylanWilson You are right! Thanks
@DylanWilson Thank you so much for your explanation regarding DenisNardin's comment. I guess that my question now becomes: why is C^*(X;k) the (homotopy) limit of the constant diagram functor $X \to k$, as claimed above? (I am essentially clueless about homotopy limits of infinity categories, so in case this is hard to explain, please feel free to suggest a reference).
@user142700 : let $\hom$ denote the internal hom of $\mathsf{Ch}$ (as an $\infty$-category). Then $\hom(k,k)\simeq k$ and one may check by tensor-hom adjunction that $\hom(-,k) : \mathsf{Ch}^{op}\to \mathsf{Ch}$ preserves limits (so it sends colimits in $\mathsf{Ch}$ to limits in $\mathsf{Ch}$). Therefore the homotopy limit you're interested in is $\hom(\mathrm{colim}_X k, k)$, so you are reduced to computing the homotopy colimit of the $X$-diagram which is constant at $k$.
But now, the functor $\mathsf{Spaces}\to \mathsf{Ch}, X\mapsto \mathrm{colim}X k$ preserves colimits (it's essentially the tensorisation $X\mapsto X\otimes k$), and sends $*$ to $k$, just as $C(X;k)$ (singular homology) with coefficients in $k$. By the universal property of $\mathsf{Spaces}$, it follows that $\mathrm{colim}X k \simeq C(X;k)$; and then $\lim_X k \simeq \hom(C_(X;k),k) = C^(X;k)$
@MaximeRamzi Thanks!
One way to discuss $\infty$-local systems over a space $X$ is in terms of the fundamental $\infty$-groupoid of $X$. To motivate this, recall that for classical local systems, one has the equivalence of categories
$$LocSys(X) \simeq Rep(\pi_{1}(X)) $$
between the local systems on $X$ and representations of the fundamental group. It's easily verified that representations of the fundamental group are equivalent to functors out of the fundamental groupoid of $X$, so it's more natural (in the sense of not making any choices) to write
$$LocSys(X) \simeq Fun(\Pi_{1}(X), Vec_{k}) $$
The idea behind $\infty$-local systems is then to replace vector spaces with chain complexes, which we think of as a homotopical/derived version of vector spaces. The problem with just doing this naively though (say, by taking $Fun(\Pi_{1}(X), Ch_{k})$ where $Ch_{k}$ is the category of chain complexes over $k$), is that the fundamental groupoid only encodes information about the 1-truncated homotopy type of $X$. If we're mapping into a category that carries higher homotopical data, there's no way we're going to get a satisfactory answer unless we encode the entire homotopy type of $X$ in the domain.
So instead, we replace $\Pi_{1}(X)$ by $\Pi_{\infty}(X)$ - which is probably more familiarly written as the singular simplicial set associated to $X$, denoted $Sing(X)$. Then, we can take as our definition of infinity local systems
$$LocSys^{\infty}(X) := Fun(\Pi_{\infty}(X), Ch_{k})$$
where we view $Ch_{k}$ as an $\infty$-category. I'm not entirely sure how this definition fits in with that of the paper you cited, but it's certainly a common way to think of infinity local systems.
Some places to read:
Appendix A of Higher Algebra. This is definitely overkill for your question, as it mostly pertains to $\infty$-constructable sheaves, but it's a great place to learn some sheaf theory from the above perspective.
Kerodon, Section 2.5. This isn't directly about infinity local systems, but gives a lot of insight into the interplay between dg-categories and $\infty$-categories. (In particular, this process of 'taking chains' of a simplicial set mentioned in the comments by Dylan is explained here in detail).
As for your second question, the comments give a pretty concrete and conceptual answer as to how to perform this computation, but there's another heuristic that explains why you would expect this to be true without computing anything. The key point is that the dg-category $LocSys^{\infty}(X)$ is a differential graded enhancement of the derived category of local systems on X. This means that when we pass to the homotopy category, we recover the derived category of local systems.
Now in the (underived, 1-categorical) category of local systems, $hom(k_{X}, \mathcal{F})$ is equivalent to just taking global sections of $\mathcal{F}$. So, when we pass to the derived category, $hom(k_{X}, k_{X})$ is just taking derived global sections of the constant sheaf - i.e. singular cohomology of $X$ with coefficients in $k$. Thus, in the dg-category $LocSys^{\infty}(X)$, the mapping complex $hom(k_{X}, k_{X})$ is some chain complex whose $i^{th}$ cohomology is isomorphic to $H^{i}(X ; k)$. So it shouldn't be surprising at all that this is $C^{\star}(X; k)$.
I am a little bit confused: the derived category of local systems should be determined by the category of local systems, how can it contains information (like higher cohomology) that is not already in the category of local systems ? Am I missing something?
Since you tagged this with "symplectic geometry", I'm going to give an answer from a symplectic geometry perspective, which may not be what you're looking for, but (as a symplectic person) I find it is a helpful point of view. This will use the language of $A_\infty$-categories as well as dg-categories.
Given a manifold $X$, take its cotangent bundle $T^*X$. This is a non-compact symplectic manifold. You can consider its Fukaya category of exact embedded Lagrangian submanifolds which agree with the symplectisation of a Legendrian at infinity. The Floer cochain groups (morphisms) between $L_1,L_2$ are free k-vector spaces on intersection points between $L_1$ and $\phi(L_2)$, where $\phi$ is the time 1 map of a suitable Hamiltonian. You have to specify a suitable class of Hamiltonians, and I want to use Hamiltonians which are "quadratic at infinity", in other words they look something like kinetic energy with respect to some Riemannian metric. Since kinetic energy as a Hamiltonian generates geodesic flow, the Floer cochains are something like (k-linear combinations of) geodesics connecting the Lagrangians.
For example, if $L_1$ and $L_2$ are cotangent fibres at $x_1$ and $x_2$ then the Floer complex is something like the free k-vector space on the set of geodesics from $x_1$ to $x_2$. It is then a theorem (of Abbondandolo and Schwartz at the level of homology beefed up to the $A_\infty$ level by Abouzaid) that the Floer complex between two cotangent fibres is quasiisomorphic to chains on the space of paths between $x_1$ and $x_2$ (and to $C_{-*}(\Omega X)$ as an $A_\infty$-algebra when $x_1=x_2$ and concatenation makes sense).
Abouzaid also showed that a cotangent fibres generates this Fukaya category, so you get a fully faithful Yoneda functor from the Fukaya category to the dg-category of $A_\infty$-modules over chains on the based loop space. In other words, if you want to compute the Floer complex between two Lagrangians $L_1$ and $L_2$ and you have a cotangent fibre $F$, you can take the $CF(F,F)$-modules $CF(F,L_n)$, $n=1,2$, and take homs between them in the category of $A_\infty$ $CF(F,F)$-modules. Since $CF(F,F)\cong C_{-*}(\Omega X)$, this is quasiequivalent to the category of infinity local systems. So this category of infinity local systems is the Fukaya category of $T^*X$.
Now how do you see that $hom(k,k)=C^*(X)$? Well, there is a Lagrangian in $T^*X$ whose Floer complex is $C^*(X)$, namely the zero section. For example, a small Hamiltonian deformation of the zero section is a graph of an exact 1-form $df$, and the intersection points between this and the zero section happen at critical points of $f$; in fact Floer showed that in this case the Floer complex is the Morse complex for a suitable choice of almost complex structures.
What is the Yoneda module corresponding to the zero section? Well the zero section intersects our cotangent fibre at a single point, so $CF$ is just k (our field, considered as a trivial $A_\infty$-module over $C_{-*}(\Omega X)$). Its self homs in the category of $C_{-*}(\Omega X)$-modules should therefore compute $C^*(X)$.
The relevant papers of Abouzaid are:
https://arxiv.org/abs/0907.5606
https://arxiv.org/abs/1003.4449
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2025-03-21T14:48:30.227632
| 2020-04-06T17:44:33 |
356751
|
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|
Stack Exchange
|
Complexity of transforming a polynomial to its canonical form
Why transforming the polynomial $F(x)=\Pi_{i=1}^{d}{(x-a_i)}$ to its canonical form (@) requires $\theta(d^2)$ multiplications of coefficients?
(@) The canonical form is defined as $\Sigma_{i=0}^{d}{c_ix^i}$.
Reference: Chapter 1 in "Probability and Computing" textbook by Michael Mitzenmacher, Eli Upfal:
http://www.bioinfo.org.cn/~wangchao/maa/Probability_and_Computing.pdf
I don’t think it does. If you arrange the multiplications in a binary tree of height $\log d$, and compute the binary products (convolutions) in the tree using FFT, you will only need $d\operatorname{polylog}(d)$ coefficient multiplications.
In the textbook they specifically say “transforming $F(x)$ to its canonical form by consecutively multiplying the $i$th monomial with the product of the first $i-1$ monomials”. This inefficient method indeed requires $\Theta(d^2)$ coefficient multiplications by a trivial computation. But this is just a toy example; they do not claim that you cannot compute the product more efficiently.
To make the first comment more specific, http://cr.yp.to/papers/m3.pdf shows that for any commutative ring $R$, one can multiply two degree-$d$ polynomials from $R[x]$ using $O(d\log d)$ multiplications and $O(d\log d\log\log d)$ additions in $R$. Using a binary multiplication tree as above, it follows that given $a_1,\dots,a_d\in R$, you can compute the coefficient list of $\prod_i(x-a_i)$ using $O\bigl(d(\log d)^2\bigr)$ multiplications and $O\bigl(d(\log d)^2\log\log d\bigr)$ additions in $R$. (Presumably, the $\log\log d$ factors can be omitted for rings such as $\mathbb C$, $\mathbb R$.)
|
2025-03-21T14:48:30.227787
| 2020-04-06T17:52:26 |
356752
|
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|
Stack Exchange
|
Some questions about Hitchin's self-duality paper
I am reading this paper (The self-duality equations on a Riemann surface by N. Hitchin; DOI: 10.1112/plms/s3-55.1.59), and I don't understand a few things in page 67. In proof of Theorem 2.1 after Equation 2.4, he gives the relation for $F(B)s$ as
$$F(B)s=F(A)s-(\deg L)\omega s+ \frac12 \deg(∧^2V)\omega s.$$
In the proof, he obtains a connection $B$ on $L^*V$ using a connection $A$ on $V$, and a connection with curvature $(\deg L)\omega$. I am thinking of $L^*V$ as a pull back of $V$, or am I wrong? If I understand what is the explicit form of this connection, I may get the curvature $F(B)$, and eventually above relation. Or is there any other way to get the above result?
I tried a lot but couldn't understand how he obtains the result above for $F(B)s$. This formula is very essential in the proof, because he uses this information in Theorem 2.1 to give a condition for stability that is discussed in the third section of the paper.
I will be very glad if somebody could explain me how to derive this result.
I have another small question which maybe related to the above question:
Initially in the top of page 67 he says "Fixing a connection $A_0$ on $∧^2V$, we find that a connection $A$ on $P$ lifts to $\tilde{P}$ whose curvature is $F(A)+1/2F(A_0)1$". Here I don't see how he can get this curvature for the lift. I was thinking, maybe he uses this curvature to find the above formula. Can someone elaborates on this one as well?
I will be very grateful for your help.
$L$ is a holomorphic line subbundle of $V$, and the holomorphic inclusion $L\to V$ is the same as a holomorphic section $s$ of $L^V$ without zeros. A connection on $L$ induces a connection on its dual $L^,$ and together with the connection $A$ on $V$ you obtain the product connection whose curvature is given by the first equation in your question.
@Sebastian, Thanks for the hint! There is extra term in curvature, +1/2 deg(Λ^2V)ωs. How can I get that? Also can you please elaborate on the connection of the product?
The tensor product of two connections is just defined in a way that makes the Leibnizrule work: $\nabla\otimes\tilde\nabla (s\otimes t)=(\nabla s)\otimes t+s\otimes (\tilde\nabla t)$, and there is a simple formula for the curvature of a tensor product connection which you obtain by differentiating twice ( you can also find this in most textbooks on the topic).
The extra term in curvature is due to the assumption that the $SO(3)$-bundle $P$does not admit a lift to a $SU(2)$-bundle as $w_2(P)\neq0.$ If you prefer to think about vector bundles think of the associated real rank 3 vector bundle with metric. There is a topological invariant (over the compact surface) which tells you wether the rank 3 bundle is the skew-hermitian endomorphism bundle of a unitary rank 2 bundle with trivial determinant $w_2(P)=0$), or not ($w_2(P)\neq0$). This is analogous to the theory of $Spin^{\mathbb C}$ structures but for general principal and vector bundles.
In the case $w_2\neq0$, there exist a hermitian rank 2 vector bundle $V$ whose determinant bundle has non-zero degree. Fixing the connection on the determinant bundle, the connection $A$ on $P$ uniquely determines a connection on $V$ whose curvature satisfies the formula in the first paragraph of page 67 in Hitchin's self-duality paper.
@Sebastian: Thanks for our help! they were very helpful. I tried to use this information and derive the formula for F(B)s mentioned above. I am writing that as an answer, and I would be happy if you take a look at it.
I just have a few comments which would hopefully clarify a few points in the above discussion.
Consider the natural projection (a Lie group homomorphism) $p: U(2) \to U(2)/Z(U(2)) \simeq SU(2)/\{\pm \mathbf{1}\} \simeq SO(3)$. The kernel is $Z(U(2)) \simeq U(1)$, so we have a short exact sequence:
$$ 1 \to U(1) \to U(2) \to SO(3) \to 1. $$
Given a principal $SO(3)$ bundle, say $P$, on $M$, there is an obstruction for the existence of a principal $U(2)$ bundle $\hat{P}$ which maps onto $P$ using the above projection from $U(2)$ to $SO(3)$, which lives in $H^2(M, U(1))$, using Cech cohomology.
But I think that $H^2(M, U(1))$ itself vanishes because $M$ is complex 1-dimensional (can someone please confirm this in the comments?). As a result, for any principal $SO(3)$ bundle $P$, there always exists a principal $U(2)$ bundle $\hat{P}$ which projects onto $P$ using the projection $U(2) \to SO(3)$.
At the level of Lie algebras, we have that
$$\frak{u}(2) = \frak{so}(3) \oplus \mathbb{R}\mathbf{1},$$
where $\mathbf{1}$ denotes the $2 \times 2$ identity matrix. With formulas, this is given by
$$ x = (x - \frac{1}{2}\operatorname{tr}(x) \mathbf{1}) + \frac{1}{2} \operatorname{tr}(x) \mathbf{1}.$$
It is the decomposition of an element $x \in \mathfrak{u}(2)$ as a trace-free part plus a pure trace part.
Note that associated to $\hat{P}$ is a rank $2$ complex bundle $V$. I claim that a connection $A$ on $P$ and a connection $A_0$ on $\Lambda^2 V$ together determine a connection on $P$.
If we denote the natural bundle map $\hat{P} \to P$ by $f$, then $f^*(A)$ is itself a connection on $\hat{P}$, but it can be shown that the trace of its curvature vanishes pointwise on $M$. Indeed, its curvature is a $2$-form on $M$ with values in the trace-free part of $\hat{P}$ (refer to the above decomposition of $\mathfrak{u}(2)$).
Consider the determinant map
$$ \operatorname{det}: U(2) \to U(1). $$
This can be thought of as mapping a $g \in U(2)$ to $\Lambda^2 g$, which is a unitary map from $\Lambda^2 \mathbb{C}^2$ to itself.
At the level of principal bundles, this induces a map from $\hat{P}$ to the bundle of unitary maps from $\Lambda^2 V$ to itself. The induced hermitian inner product on $\Lambda^2 V$, coming from the hermitian inner product on $V$, determines a reduction of the group of the corresponding principal bundle from $\mathbb{C}^*$ to $U(1)$. Let us denote the resulting principal $U(1)$ bundle by $Q$, say.
A connection $A_0$ on $\Lambda^2 V$ determines a connection on $Q$, which we will also denote by $A_0$, which then determines a connection on the (trivial) line bundle $\operatorname{End}(\Lambda^2 V)$. The latter connection is associated to a connection on the bundle of unitary maps from $\Lambda^2 V$ to itself. The latter connection pulls back to a connection on $\hat{P}$ using the determinant map described previously. Let us denote this connection, by abuse of notation, by $\operatorname{det}^*(A_0)$. Note that the curvature of this connection is pointwise pure trace, with respect to the above decomposition of $\mathfrak{u}(2)$.
Given a choice of a connection $A$ on $P$ and $A_0$ on $\Lambda^2 V$, we thus see that
$$ f^*(A) + \operatorname{det}^*(A_0) $$
is a connection on $\hat{P}$, whose curvature is
$$ F(A) + \frac{1}{2} F(A_0) \mathbf{1},$$
where $F(A)$, respectively $F(A_0)$, denotes the curvature of $A$, respectively $A_0$.
Using your connection for the tensor product, we can find the curvature as
$$F_{\nabla\otimes\bar{\nabla}}(s_1\otimes s_2) =\nabla\otimes\bar{\nabla}\big(\nabla\otimes\bar{\nabla}(s_1\otimes s_2)\big)=\nabla\otimes\bar{\nabla}\big(\nabla(s_1)\otimes s_2
+s_1\otimes \bar{\nabla}(s_2)\big)=\nabla^{2}(s_1)\otimes s_2
-\nabla(s_1)\otimes \bar{\nabla} (s_2)+\nabla(s_1)\otimes \bar{\nabla} (s_2)+s_1\otimes \bar{\nabla}^{2}(s_2)\\= \nabla^{2}(s_1)\otimes s_2+s_1\otimes \bar{\nabla}^{2}(s_2)$$
the minus signs appears because $\nabla(s_1)\otimes s_2$ is (1,0) form.
Next we notice that $A_0$ is a fixed connection on $\wedge^2 V$ and on $V$ we have the connection $A+A_0$.Thus, the covariant derivative for connection $A_0+A$ is given by $d_{A+A_0}=d+A+A_0$, so the curvature on $V$ would be
$$F_{A+A_0}= d^2_{A+A_0}=dA+A\wedge A+A_0\wedge A_0\\=dA+A\wedge A+\frac{1}{2}[A_0,A_0]$$
Defining $F(A):=dA+A\wedge A$ and $F(A_0):=[A_0,A_0]$, we get the curvature on $V$ according to $F(A)+\frac{1}{2}F(A_0)$ (This is the equation given in the first paragraph of page 67) Is this derivation of $F(A)+\frac{1}{2}F(A_0)$ correct?
Furthermore, since $\wedge^2 V$ is a line bundle embedded in $V$, its curvature has the form $F(A_0)=deg(\wedge^2 V)\omega$ where $\omega$ is a positive form.
So now we have
$$F_{B}(s_1\otimes s_2)=F(V)(s_1)\otimes s_2+s_1\otimes F(L^*)(s_2)\\=(F(A)+\frac{1}{2}deg(\wedge^2 V)\omega)s_1\otimes s_2+s_1\otimes(-(deg L)\omega) s_2 \\=F(A)(s_1\otimes s_2)+\frac{1}{2}deg(\wedge^2 V)\omega (s_1\otimes s_2) - (deg L)\omega(s_1\otimes s_2)$$
Above, I used the fact that for a line bundle $L$ with some connection, the curvature is given by $(deg L)\omega$ where $\omega$ is a positive form. The minus sign is because we look at the dual space, $L^*$.
Here, I assume that in paper the section $s\in \Omega^0(M, L^*\otimes V)$ must be of the form $s:=s_1\otimes s_2$. Am I right?
If you see any errors/problems with my computations, please let me know. I will be very thankful for that!
Your computation of the curvature of the product connection is right. The remaining part does not work, for example how do you add connections on $V$ and on $\Lambda^2V.$
Okay. I didn't know that. I am new to the subject, do you recommend any book so that I can learn a similar type of arguments about connections.
You could read Nicolaesuc's 'Notes on Seiberg-Witten equations', in particular $\S 1.3$.
Thanks for the recommendation! I will check it out.
I think the remaining part you need to clear your confusion has to do with decomposing $\mathfrak{u}(2) \simeq \mathfrak{su}(2) \oplus \mathbb{R}$, by mapping $x$ to $(x - \frac{1}{2}\operatorname{tr}(x)\mathbf{1}, \frac{1}{2}\operatorname{tr}(x))$. I may write more details later if someone requests...
I edited my previous answer. I hope that it clarifies the remaining points in the above discussion!
|
2025-03-21T14:48:30.228522
| 2020-04-06T18:01:25 |
356754
|
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|
Stack Exchange
|
Principal circle bundles that are smooth foliated
Let $\xi=(\pi,E,B)$ be an orientable circle bundle, i.e., a bundle with fiber $\mathbb{S}^1$ and structural group $G=\textit{Diff}^+(\mathbb{S}^1)$.
Claim 1: The bundle $\xi$ admits a structure of $S^1$ principal bundle.
Proof: Let $g$ be a Riemannian metric on $E$, and denote by $l_b$ the length of the fiber over $b\in B$.
If $\pi(u)=b$ and $z=e^{i\theta}$, then $u\cdot z$ is the point we reach when we proceed in the direction of orientation a distance $(l_b/2\pi)$ on $E_b$ (fiber over $b$) from $u$.
We know that a principal bundle is trivial if and only if it admits a cross section. So, $\xi$ is trivial if and only if admits a cross section.
This lead us to the Euler class of $\xi$. In fact, a cohomological class $e(\xi)\in H_2(M;\mathbb{R})$ can be obtained as obstruction to the existence of a cross section for $\xi$. Moreover, there exists a one-to-one correspondence between the isomoprhism classes of $\mathbb{S}^1$ bundles over $M$ and cohomology classes in $H^2(M;\mathbb{Z})$ given by $[\xi]\rightarrow e(\xi)$, for instance see (Geometry of differential forms, Morita).
When the Euler class $e(\xi)$ is a torsion element we can obtain a codimension one smooth foliation transversal to the fibres of $\xi$, and the bundle $\xi$ with this foliation becomes a foliated bundle. Myioshi (A remark on torsion Euler classes of circle bundles) showed that the image of the holonomy homomorphism (which characterizes this foliated bundle)
$$\varphi:\pi_1(M)\rightarrow \text{Diff}^+(\mathbb{S}^1)$$
can be taken contained in $SO(2)=\mathbb{S}^1$.
Following that in the same total space $E$ and base $B$ we have two $\mathbb{S}^1$-principal bundles with the same projection. The initial bundle $\xi$, and the bundle $E(\varphi)$ induced by the representation $\varphi:\pi_1(M)\rightarrow SO(2)$ (this induced bundle is a principal bundle and admit a flat connection, see Milnor On the existence of a connection with curvature zero, Lemma 1).
Question: Are the bundles $\xi$ and $E(\varphi)$ equivalent as principal bundles?
I do not understand the question. What is $E(\phi)$? The homomorphism $\phi$ gives $\xi$ a structure of a flat $SO(2)$ bundle. Being flat is an extra structure. The underlying circle bundle has not changed. Incidentally, your claim 1 follows by saying that $Diff^+(S^1)$ deformation retracts onto $S^1$.
The bundle $E(\varphi)$ is defined as follows. Consider the left action of $\pi_1(M)$ on $\tilde{M}\times \mathbb{S}^1$ given by $\alpha\cdot(\tilde{x},g) =(\tilde{x}\cdot\alpha^{-1}, \varphi(\alpha)\cdot g)$. Then, the collapsed space $E(\varphi)$ is the total space of a principal $\mathbb{S}^1$ bundle over $M$ induced by $\varphi$. The action of $\mathbb{S}^1$ on $E(\varphi)$ is not necessarly the same as the action on $E$. These bundles are isomorphic when we consider them as $G=\text{Diff}^+(\mathbb{S}^1)$-bundles.
If bundles are isomorphic as $Diff^+(S^1)$ bundles, then they are isomorphic as $S^1$ bundles because $Diff^+(S^1)$ deformation retracts onto $S^1$. Specifically, the map of classifying spaces $BS^1\to BDiff^+(S^1)$ has contractible fiber $Diff^+(S^1)/S^1$, which induces a bijection $[M, BS^1]\to [M, BDiff^+(S^1)]$ of the sets of isomorphism classes of $S^1$ and $Diff^+(S^1)$ bundles over $M$. Of course, all this does not take into account the flat structure, but I gather you disregard it.
I have a dubt.
When a subgroup $H$ of $G$ is a deformation retract of $G$, then there exists a bijection $[M,BH]\rightarrow [M,BG]$?
In your case the map $BH\to BG$ is a homotopy equivalence, see e.g. the answer in https://math.stackexchange.com/questions/3341232/homotopy-equivalent-structure-groups-have-h-e-classifying-spaces-if-g-simeq-h?rq=1, and hence you get a bijection of the sets of homotopy classes.
I should have said "$BH\to BG$ is a weak homotopy equivalence", which is what you need to a bijection of the sets of homotopy classes.
|
2025-03-21T14:48:30.228788
| 2020-04-06T18:55:16 |
356758
|
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|
Stack Exchange
|
What is the definition of Plancherel density?
I know about the Plancherel measure, but I don't know where the term "Plancherel density" is defined.
Where did you encounter it?
@LSpice For $p$-adic groups Aubert and Plymen have some papers on explicit Plancherel measures for $GL_n$ and other groups. Their theorem gives a formula for the Plancherel density. I think it's a function of the discrete series representation of a Levi. Is it just the Plancherel measure of the singleton?
Can you give a specific paper?
https://www.sciencedirect.com/science/article/pii/S1631073X04001633. They explicitly compute the Plancherel density with respect to a given measure on the tempered dual.
the Plancherel density is derived from the Plancherel measure, see arXiv:1812.00047 for the precise definition:
|
2025-03-21T14:48:30.228878
| 2020-04-06T18:58:52 |
356759
|
{
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"zeb"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/356759"
}
|
Stack Exchange
|
Cancellation in a particular sum
In an attempt to compute cycle counts in an of a certain number theoretic graph, the following estimate was needed.
It is true that
$$\bigg|\sum_{a,b,c\in \mathbb{Z}/p\mathbb{Z}}\bigg(\sum_{d=1}^{p-1}\bigg(\frac{d}{p}\bigg)w^{-d-(a^2+b^2+c^2)/d}\bigg)^3\bigg| = o(p^{9/2}).$$
$w$ here is $\exp\big(\frac{2\pi i}{p}\big)$. Is such an estimate known? The fact that the sum in question is $O(p^{9/2})$ follows from using that the second moment is $O(p^{4})$ and the inner sum $O(p^{1/2})$ which uses Weil's work on RH for curves.
Can't you just use the triangle inequality to get the upper bound $p^3 \cdot (p^{1/2})^3$?
Also, your inner sum is a Salie sum, which can be evaluated explicitly as a gauss sum times a sum of two roots of unity. So you should be able to get even better bounds, perhaps $O(p^3)$.
Fixed the exponents; and yes that seems like it should work.
We can actually get an explicit formula for the whole sum (I will assume $p \ne 2,3$ throughout). We start with the Salié sum:
$$\sum_{d=1}^{p-1}\bigg(\frac{d}{p}\bigg)w^{-d-(a^2+b^2+c^2)/d} = \bigg(\sum_d \bigg(\frac{d}{p}\bigg)w^{-d}\bigg) \sum_{x^2 \equiv 4(a^2+b^2+c^2)} w^x.$$
The first factor is a Gauss sum, and has absolute value $\sqrt{p}$, and in fact the exact value is either $\sqrt{p}$ or $-i\sqrt{p}$ depending on whether $p$ is $1$ or $3$ modulo $4$. Write $G$ for this Gauss sum. Then the whole sum becomes
$$G^3\bigg(\sum_{x \not\equiv 0} \sum_{4(a^2+b^2+c^2) \equiv x^2} \frac{1}{2}(w^x + w^{-x})^3 + \sum_{4(a^2+b^2+c^2) \equiv 0} 1\bigg).$$
So we just need to compute the number of ways $N_x$ to write $x^2$ as a sum of three squares, for each $x$. By rescaling, we see that $N_x = N_1$ for $x \not\equiv 0 \pmod{p}$, so the sum becomes
$$G^3(N_0 + N_1\sum_{x \not\equiv 0}\frac{1}{2}(w^{3x} + 3w^x+3w^{-x} + w^{-3x})) = G^3(N_0 - 4N_1).$$
To compute $N_1$, we can use stereographic projection from the point $(1,0,0)$ to the plane $(0,x,y)$, to see that $N_2$ is $p^2$ minus the number of pairs $x,y$ with $x^2+y^2 \equiv -1$, plus the number of pairs $b,c$ with $b^2+c^2 \equiv 0$. Another stereographic projection argument (and the fact that every number is a sum of two squares modulo $p$) shows that there are exactly $p-(\frac{p}{4})$ ways to choose $x,y$ with $x^2+y^2 \equiv -1$. The number of pairs $b,c$ with $b^2+c^2 \equiv 0$ is $p+(p-1)(\frac{p}{4})$. Thus $N_1 = p^2 + p(\frac{p}{4})$.
To compute $N_0$, we can rescale to reduce to counting the number of pairs $x,y$ with $x^2 + y^2 \equiv -1$ (times a factor of $p-1$) or $b^2+c^2 \equiv 0$, and we find that $N_0 = p^2$. Thus the full sum comes out to
$$G^3\bigg(-3p^2-4p\big(\frac{p}{4}\big)\bigg),$$
and the absolute value is $3p^{7/2} \pm 4p^{5/2}$.
|
2025-03-21T14:48:30.229081
| 2020-04-06T19:38:11 |
356761
|
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|
Stack Exchange
|
When does a measure-valued map admit a continuously parametrized density function?
Let $X$ and $Y$ be Polish spaces, let $\mathcal{P}(Y)$ be the space of Borel probability measures on $Y$ endowed with the smallest $\sigma$-algebra such that all functions of the form $\nu\mapsto\nu(A)$ with $A\subseteq Y$ measurable are measurable, and let $g:X\to\mathcal{P}(Y)$ be a measurable function. If there exists a probability measure $\mu$ on $Y$ such that $g(x)$ is absolutely continuous with respect to $\mu$ for all $x\in X$, then there exists a jointly measurable function $h:X\times Y\to\mathbb{R}$ such that $h(x,\cdot)$ is a Radon-Nikodym derivative of $g(x)$ with respect to $\mu$ for all $x\in X$. This follows, for example, from [Paul André Meyeyer 1966, Probability and Potentials, Result VIII-10].
I would like to know if there is some reasonable condition on $g$ that is equivalent to, or at least implies, the existence of such a function $h:X\times Y\to\mathbb{R}$ continuous in $X$. If needed, one may assume $X$ to be compact.
Such continuously parametrized density functions are used in various areas of mathematical game theory, and I would like to know if there is a less ad hoc way to think about them.
|
2025-03-21T14:48:30.229193
| 2020-04-06T19:39:38 |
356762
|
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|
Stack Exchange
|
Eigenfunctions of the laplacian on $\mathbb{CP}^n$
I want to find explicit formulas for the eigenfunctions of the Laplacian on $\mathbb{CP}^n$ endowed with the Fubini Study metric.
For the first eigenvalue $\lambda_1 = 4(n+1)$, the eigenfunctions are given by the real and imaginary parts of $\phi^{i, j} = \frac{z_i\bar{z_j}}{\sum_{k}|z_k|^2}-\frac{\delta_{i, j}}{n+1}$, and these functions together gives the Veronese isometric embedding of $\mathbb{CP}^n$ into $S^{2(n+1)^2-1}$.
Is there an analogue of this for higher eigenfunctions? Is the $k$-th eigenspace also associated with some embedding of $\mathbb{CP}^n$? Are there explicit formulas listing the higher eigenfunctions like the $\phi$ above?
Duplicate ? https://mathoverflow.net/questions/114107/what-are-first-eigenfunctions-of-laplacian-for-cpn-with-fubini-study-metric?rq=1
Not duplicate. That question asked for the $\lambda_1$ eigenspace;
the new question gives the $\lambda_1$ eigenspace and asks for
the $\lambda_k$ eigenfunctions for $k>1$.
Note, by the way, that your claimed formula for the eigenfunctions belonging to $\lambda_1$ on $\mathbb{CP}^n$ is not quite correct. For example, your $\phi^{1,1}$ is real-valued and non-negative, so its integral over $\mathbb{CP}^n$ can't be zero, which would be so for an actual eigenfunction belonging to $\lambda_1$. You are off by an additive constant that you need to subtract in order to get the average value of the function to be $0$.
Yes, I made an edit.
The $k$-th eigenfunctions are actually easy to describe: In $\mathbb{C}^{n+1}$ with unitary complex coordinates $z_0,z_1,\ldots,z_n$, write $Z = |z_0|^2+\cdots+|z_n|^2$.
Now, for a given $k\ge0$, let $H_k$ be the (real) vector space of real-valued polynomials $p(z,\bar z)$ that are homogeneous of degree $k$ in the $z$-variables and degree $k$ in the $\bar z$-variables and that are harmonic, i.e., they satisfy
$$
\frac{\partial^2p}{\partial z_0\partial \bar z_0}+\cdots+
\frac{\partial^2p}{\partial z_n\partial \bar z_n} = 0.
$$
One easily computes that
$$
\dim_\mathbb{R}H_k = {{n+k}\choose{k}}^2
-{{n+k-1}\choose{k-1}}^2.
$$
Then, for $p\in H_k$, the function $f_p:\mathbb{CP}^n\to\mathbb{R}$ given by
$$
f_p\bigl([z]\bigr) = \frac{p(z,\bar z)}{Z^k}
$$
is well-defined and is an eigenfunction with eigenvalue $\lambda_k$. Conversely, every eigenfunction with eigenvalue $\lambda_k$ is of this form for a unique $p\in H_k$.
Added remark: I was privately asked how to prove that the description I gave of the eigenvalues and eigenfunctions of $\mathbb{CP}^n$ is correct. I'm sure it's in the literature in various places, but it's easier to just give the argument, using the known description of the eigenvalues and eigenfunctions on the standard sphere. Here's the idea:
Let $S^{2n+1}\subset \mathbb{C}^{n+1}$ be the standard unit $2n{+}1$-sphere with its induced metric and let $\sigma:S^{2n+1}\to\mathbb{CP}^n$ be the quotient mapping $\sigma(z) = [z] = \mathbb{C}z$. Then $\sigma$ is a Riemannian submersion when $\mathbb{CP}^n$ is given the Fubini-Study metric (appropriately scaled; with this choice, the area of a linear $\mathbb{CP}^1\subset\mathbb{CP}^n$ is $\pi$, not $4\pi$).
If $f$ is an eigenfunction on $\mathbb{CP}^n$ with eigenvalue $\lambda_k$, then $\sigma^*(f)$ is an eigenfunction on $S^{2n+1}$ with eigenvalue $\lambda_k$, one that is, moreover, invariant under the circle action $\mathrm{e}^{i\phi}\cdot z = \mathrm{e}^{i\phi}z$ on $S^{2n+1}$.
Now, it is known that, if $\mu_d$ is the $d$-th eigenvalue of $S^{2n+1}$, then any eigenfunction $f$ with this eigenvalue is the restriction to $S^{2n+1}$ of a harmonic polynomial on $\mathbb{R}^{2n+2}$ that is homogeneous of degree $d$. We are looking for such polynomials that are also invariant under the circle action of multiplication by $\mathrm{e}^{i\phi}$. When expressed as a polynomial in the complex coordinates $z_i$ and $\bar z_i$, a polynomial that is invariant under this circle action must have the same number of $z$'s as $\bar z$'s. This can only happen if $d$ is even. Conversely, if $d=2k$, then we get the space $H_k$ as defined above. Thus, $\lambda_k = \mu_{2k}$, and the above description is justified.
Thanks for the answer. Do you know if the higher eigenfunctions arises from some embedding of $CP^n$ like it does for the first eigenfunction?
@freidtchy: Yes. Think of the mapping $p\mapsto f_p\bigl([z]\bigr)\in\mathbb{R}$ for a given $[z]\in\mathbb{CP}^n$ as a linear function on $H_k$, so this defines a map of $\mathbb{CP}^n$ into ${H_k}^$ (the dual space of $H_k$). This is an embedding of $\mathbb{CP}^n$ into ${H_k}^$, which, when $k=1$, is the one you already know (since $H_1$ is essentially the traceless Hermitian forms on $\mathbb{C}^{n+1}$). This mapping is equivariant with respect to $\mathrm{SU}(n{+}1)$, as $H_k$ is an irreducible representation of this group (up to a covering, the isometry group of $\mathbb{CP}^n$).
|
2025-03-21T14:48:30.229503
| 2020-04-06T19:57:12 |
356765
|
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|
Stack Exchange
|
The union of two cuts is a cut?
Every poset $\langle P, \leq \rangle$ has a Dedekind-MacNeille completion, a complete lattice that embeds $\langle P, \leq \rangle$.
For $A \subseteq P$, the upset $U(A) = \{p \in P\ |\ \forall a \in A:\ a \leq p\}$.
Likewise, the downset $D(A)= \{p \in P\ |\ \forall a \in A:\ p \leq a\}$.
Then a cut is any set $A$ such that $A=D(U(A))$.
Now we take the set of cuts in $P$:
$$DM(P) = \{A\subseteq P\ |\ A=D(U(A))\}$$
The poset $ \langle DM(P), \subseteq\rangle$ consisting of the set of cuts in $P$, $ DM(P) $, ordered by set inclusion, $\subseteq$, forms a complete lattice.
I'm trying to prove this, and but can't seem to show even that for every two elements $A, B \in DM(P)$, their union is in the lattice, i.e. $A\cup B = D(U(A\cup B))$.
It's easy to show that $A\cup B \subseteq D(U(A\cup B))$: Suppose $x \in A\cup B$. Then $x \leq y$ for every $y \in U(A\cup B)$. Then $x \in D(U(A\cup B))$.
But I can't manage to prove the other direction, i.e. that $D(U(A\cup B)) \subseteq A\cup B$. The dozen or so textbooks on lattice and order theory I've looked at (and MacNeille's original paper) do not go through the proof.
When people say that cuts form a complete lattice under $\subseteq$, they don't necessarily mean that the lattice operations are the ordinary union and intersection. So verifying this claim does not require checking that $D(U(A\cup B)) = A\cup B$ when $A$ and $B$ are cuts (which is a good thing: this isn't true, if our lattice is the four element lattice $\{0,1\}^2$). Instead we have to figure out what the lattice operations on the cuts actually are, and check that they form a legitimate lattice - how do we do this?
In this case, something nice happens: the operation $C: A \mapsto D(U(A))$ is a closure operation, i.e.
$A \subseteq C(A)$,
$C(C(A)) = C(A)$, and
if $A \subseteq B$, then $C(A) \subseteq C(B)$.
In general, for any closure operation $C$, the set of "closed" sets $A$ such that $C(A) = A$ always forms a complete lattice, where the lattice meet is the usual intersection, while the lattice join is the closure of the union.
So the point is that what is denoted as a union $A \cup B$ is not to be interpreted as a set-theoretic union, but as a join within the poset of cuts.
The proof you're after is just a special case of a much more general proposition, that if you have a closure operator on a complete lattice (which in this case is the power set of the poset $P$), then the poset of fixed points of the closure operator is again a complete lattice. So in the present case, what one needs is that $D \circ U$ is a closure operator. This is not hard because there is a Galois connection between $D$ and $U$; I can explain if need be. Let me take it for granted for now.
Now suppose given a complete lattice $L$ and a closure operator $\phi: L \to L$. If $c_i$ is a collection of fixed points, then their meet in $L$ is a fixed point. We just need that $\phi(\bigwedge_i c_i) \leq \bigwedge_i c_i$, but this follows immediately by observing
$$\phi(\bigwedge_i c_i) \leq \phi(c_i) \leq c_i$$
for all $i$. This alone is enough to guarantee existence of arbitrary joins (any poset with arbitrary meets has arbitrary joins), but to be more specific about your problem, the join of two closed elements $c, d$ in the poset of closed elements is $c \vee d = \phi(c \cup d)$ where $\cup$ denotes the join in $L$. It's clear that $c, d \leq \phi(c \cup d)$, so for $e$ closed, $\phi(c \cup d) \leq e$ implies $c \leq e$ and $d \leq e$. In the converse direction, given $c \leq e$ and $d \leq e$, we have $c \cup d \leq e$ and therefore $\phi(c \cup d) \leq \phi(e) = e$.
|
2025-03-21T14:48:30.229787
| 2020-04-06T20:36:54 |
356768
|
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|
Stack Exchange
|
$L^p$ norm inequalities with respect to strongly-log-concave densities
Let $\pi(x)=\frac{e^{-f(x)}}{\int_{\mathbb{R}^d}e^{-f(u)}du}$ be a strongly-log-concave distribution, i.e., $f(x):\mathbb{R}^d\rightarrow R$ is an $m$-strongly convex function. Also, $f(x)$ has $L$-Lipschitz gradient. Define the norm of a function $g:\mathbb{R}^d\rightarrow R$:
$$\|g\|_p=\left(\int_{\mathbb{R}^d} |g(x)|^p \pi(x)dx\right)^\frac{1}{p}$$.
Question: For $p\geq q$, is there any inequality like the following:
$$C_0\|g\|_p\leq \|g\|_q\leq C_1\|g\|_p$$
where $C_0,C_1$ are constants. By Jensen's inequality, $C_1=1$ satisfies the upper bound. Is it possible to get $C_0,C_1$ which uses the fact that the density is strongly log-concave and $f(x)$ has $L$-Lipschitz gradient?
$g=1$ gives one on both sides, no?
Since you changed the question, I should change my comment: the constant functions show that $C_1=1$ is the smallest such constant.
No. Only the upper bound holds, and as Dirk pointed out the best constant is $C_1=1$.
The lower bound cannot hold for any $C_0$, since otherwise the $L^p$ and $L^q$ norms would be equivalent. This is well-known to fail.
To see this, observe that with your assumptions your reference measure $\pi(x)dx$ is locally equivalent to the Lebesgue measure $dx$ (i-e $c_0 dx\leq \pi(x)dx\leq c_1 dx$ on any ball $B_R$). The comparison between $L^p$ and $L^q$ norms on $\mathbb R^d$ is well known for the Lebesgue measure: Taking a radial function $g(x)=\chi_{B_R}(x) |x|^\alpha$ and tuning $\alpha<0$ (denpending on $p,q$) easily gives a counterexample to $C_0\|g\|_{L^p}\leq \|g\|_{L^q}$, i-e such that $\|g\|_{L^p}=+\infty$ but $\|g\|_{L^q}<\infty$. (Here I'm using a cutoff function to localize on a ball $B_R$ so that $\pi(x)dx$ and $dx$ are equivalent).
More explicitly, take $\alpha=-\frac{d}{q}+\epsilon$ (for very small $\epsilon>0$) so that $|g(x)|^qdx\sim r^{\alpha q}r^{d-1}dr=r^{\alpha q+d-1}=r^{-1+\epsilon}dr$ is bordeline integrable at the origin, while $|g(x)|^pdx\sim r^{\alpha p+d-1}dr$ is not (due to the exponent $\alpha p+d-1<-1$ since $p>q$).
I should mention that log-concavity plays absolutely no role here (as should be clear from my answer above). Unless you're working in a finite space $L^p$ and $L^q$ norms cannot be equivalent. Ever.
|
2025-03-21T14:48:30.229961
| 2020-04-06T22:15:13 |
356776
|
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|
Stack Exchange
|
Two Dehn fillings yielding the same lens space?
Let $M$ be an oriented three-manifold with $\partial M$ a torus. Suppose that two different Dehn fillings $M(r)$ and $M(r')$ are (oriented) homeomorphic to a lens space $L(p,q)$.
Does that imply that $M$ is a solid torus?
This is a case of the oriented knot complement problem in lens spaces, also called the cosmetic surgery problem. If $M(r)\cong M(r')$ preserving orientation, then these form a purely cosmetic (or truly cosmetic) pair. See Problem 1.81 (A) from Kirby's problem list (posed by Steve Bleiler).
You probably want to avoid reducible examples like that given in Mukherjee’s answer (note that this hypothesis is missing also in the statement of Problem 1.81).
Given that assumption, there is no such example which is irreducible and non hyperbolic by a result of Daniel Matignon (he classifies all cosmetic surgeries on such manifolds, and shows that all the examples are chiral cosmetic pairs, i.e. orientation-reversing).
There are examples of cosmetic surgeries on hyperbolic knots in lens spaces that reverse orientation, but I think the orientation preserving case is still open in general.
By the way, maybe the interesting case of this question is for $M$ not a homology circle. Otherwise, the two surgeries must lift to the $p$-fold cyclic cover of $M(r)\cong M(r')\cong L(p,q)$, and hence one would have two surgeries on a knot yielding $S^3$, a contradiction to the knot complement theorem.
To follow up on Ian's comment, perhaps the simplest hyperbolic manifold which admits an L(p,q) filling and an L(p,-q) filling is the figure eight sister manifold (aka m003) which admits an L(5,1) filling and an L(5,-1) filling. The homology of the figure eight sister manifold is ZxZ/5Z.
Here is an easy example. Consider M as the 3 manifolds obtained by the surgery on the two black curves ( $0$ surgery on one unknot and $p/q$ surgery on another unknot) and take a neighborhood of the pink curve out. Clearly $M$ is not a solid torus (compute the fundamental group). Now if you do any Dehn filling with integer co-efficient, the 0-framed unknot will cancel that out and you will be left with a lens space $L(p,q)$ which is $p/q$ surgey on the unknot.
That is an unknot in L(p,q), so is reducible. Probably @Adam should assume the manifold is irreducible to avoid such examples.
Yes, that question would be more interesting. But the way he asked, I think this is a quick counter example.
Agreed. The answer would be more clear if you just drew a 2 component unlink with p/q surgery on one component.
Thank you, Anubhav and Ian. Somehow I missed it, but indeed requiring M irreducible makes the question more interesting :-)
|
2025-03-21T14:48:30.230294
| 2020-04-06T22:23:25 |
356777
|
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|
Stack Exchange
|
PCP theorem to check hard proofs
Is it technically possible to check formidable proofs like Mochizuki's using PCP theorem before mathematicians spend time in understanding the mechanics of the proof? If so why have mathematicians not done that as this would have saved time and money let alone being distracted on something that might not yield new insights. Will this paradigm of checking before understanding ever stand in mathematics collectively?
Note this is not just to check IUT. Prior almost famous mistake What is the mistake in the proof of the Homotopy hypothesis by Kapranov and Voevodsky? before Vladimir Voevodsky was convinced of usefulness of proof checkers. There are several important examples and so why are the comments and answers focussed on IUT (I say 'check formidable proofs $\underline{like}$ Mochizuki's'?
Have you tried checking any proofs, using PCP?
To even attempt this you would have to first convert the natural-language proof into a formal proof candidate. That's a gigantic undertaking, and even for reasonably-well-understood arguments is currently a major task.
@NoahSchweber Perhaps then this task itself can be delegated to the prover. Is it difficult to establish the following rule? "Think you have a proof then pass it through PCP" or else the community might not investigate. Such a system may take decades to put in place but once put in place is likely useful. Also issue usually pops up in a few places which can perhaps be PCP checked (if not the entire proof).
If I am unable or unwilling to clarify one of my proofs enough so that experts in the field can understand it and see that it's right, then I am, a fortiori, unable or unwilling to formalize it or to clarify it enough so that other people can formalize it.
@AndreasBlass 'Will this paradigm of checking before understanding ever stand in mathematics collectively?'. You have provided the contemporary status quo.
@VS You're quite right that the situation might change. That would probably require big advances in computing. An analogy: 50 years ago, when I wrote my Ph.D. thesis, I couldn't imagine typing it myself --- on (at best) an IBM Selectric, switching in and out the balls that contained the various symbol sets, etc. Now, I type all my own papers. The difference is TeX and the many packages built on it. I suspect the situation that you want will have to wait for the Donald Knuth of theorem-proving.
Essentially, first we must realize the dream of the QED Manifesto: https://www.cs.ru.nl/~freek/qed/qed.html (this is a long, long way off from becoming a reality)
Does the PCP even apply? Looking at https://en.wikipedia.org/wiki/Probabilistically_checkable_proof, it doesn't seem that PCP's "proofs" have enough in common with mathematical proofs here. It seems PCP might help checking that a solution of e.g. a SAT problem is (probabilistically) correct.
On the general concept of checking formal proof automatically before understanding, we have proof assistants for that, but you definitely need to understand a proof before formalizing it. And such formalizations can easily costs millions of dollars (in researcher salaries) which grant foundations aren't willing to pay.
More positively: since Mochizuki and students insist their proof is right despite criticisms, they're welcome to try mechanizing their proof, or the contested parts.
But that seems a waste of time, after reading the summaries in https://www.reddit.com/r/math/comments/b1r4go/one_year_ago_scholze_and_stix_were_visiting/; if the contested arguments held, it could be clarified with pen-and-paper proofs.
@Blaisorblade : I don't think it would be a waste of time. Mochizuki's group is saying that everything is already clear and the people who disagree are being unreasonable. But if a computer says it doesn't understand, one can't complain that the computer is being disrespectful, so one can't dodge the job of clarification and get away with it. As for funding, the Japanese press has reported that considerable sums of money are going to be invested in IUT soon.
@AndreasBlass, this comment is wonderful. As a wannabe formal-proofs enthusiast, I'm now dreaming of a revolution in formal proofs like the one Knuth introduced. (And maybe it's out there already waiting to take its final shape, with all the great work currently going on in this space.)
As people have noted in the comments, the PCP theorem is a red herring, and it makes no sense to formalize a proof before understanding it.
Nevertheless, one could ask if it is reasonable to request that the group of people who claim to understand Mochizuki's proof (and who believe that it is complete and correct) to formalize the proof using a proof assistant.
Until recently, any such request would have been unreasonable for the simple reason that proof assistants were too cumbersome to use. They probably still are too cumbersome to use, but the technology is steadily getting better, and IMO we're close to the point where formalizing something as complicated as IUT is not out of the question. Of course, it would still take an enormous amount of effort. However, apparently it was reported in the Japanese press (Asahi Shimbun, April 4) that a new research center has been created within Kyoto University to work on IUT, and it has an annual budget of ¥40 million. So there does seem to be funding available for the project, should someone want to take it on.
Under normal circumstances, it would be far easier to resolve issues like this one by having mathematicians talk through the proof than to resort to a proof assistant. But the current circumstances are not normal. The group supporting Mochizuki seems to be taking the stance that everything has already been written down in a perfectly clear manner, and that those who object are being disrespectful. Therein lies a crucial difference between humans and computers: Humans are social creatures, and social rifts can occur that interfere with the allegedly objective nature of mathematics. When the normal process of socializing a proof breaks down, it should indeed be possible in principle for computers to "come to the rescue." The group supporting Mochizuki cannot reasonably claim that a proof assistant is being disrespectful when it complains that it doesn't understand an argument that is presented to it.
I seem to be in the minority among professional mathematicians, but I do think that it is reasonable for skeptics of the proof to request that those who claim to understand the proof, and who want to cultivate a whole new generation of younger mathematicians to pursue IUT, to formalize the proof in a proof assistant. If the proof really is correct and those people really do understand it, then the project should eventually succeed, and when it does, the skeptics should be convinced. On the other hand, if the proof has a huge gap, then eventually those tasked with formalization (I'm imagining graduate students) will be forced to confront it, and it will become increasingly hard for "believers" to make excuses for why the formalization project is stalling.
Conversely, if nobody pushes for a formalization, then I don't see any plausible way to stop millions from being invested into this new Kyoto institute. The people in charge of those funds cannot be expected to understand the actual mathematics. But they should be capable of understanding the argument that I have presented here. If the mathematical community thinks that those funds are being misallocated, then I think convincing those who hold the purse
strings that they should demand a formalization is one of the most promising avenues forward.
Well, it would be enough to formalize Corollary 3.12 or just write up its proof in more detail ;). Note that the 40 million ¥ going to the new "center" is around 370K USD, which sounds like enough to support a part time professor plus a grad student or two for a few years, not exactly a sprawling research empire. Giving the IUTT project its own funding might even partly be to free it up from possible skepticism in the regular math program.
@none That is what I am thinking.
Just for the record, I stated a similar opinion on the Foundations of Mathematics mailing list back in January 2018.
|
2025-03-21T14:48:30.231166
| 2020-04-06T22:28:07 |
356778
|
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|
Stack Exchange
|
Is there an i.i.d sequence in the unit cube $[-1,1]^d$ with $\mathbb E \left[ \Big \| \sum_{i=1}^N X_N \Big \|_\infty\right] = \sqrt {dN}$?
There are loads of concentration results for sums of scalar-valued independent sums $X_1,X_2,\ldots, X_N$ with $\mathbb E[X_n]=0$. For example Hoeffding's Inequality says if all $|X_1|\le 1$ then $\mathbb E\left[ \left|\sum_{i=1}^N X_i\right|\right] = O (\sqrt N)$.
These concentration results can be generalised for example to vector-valued random variables with $\|X_n\| \le 1$ for $\| \cdot \|$ the Euclidean norm.
Suppose instead we have $\|\cdot\|_\infty$ norm bounds. That means $X_1,X_2,\ldots, X_N \in \mathbb R^d$ are independent with $\mathbb E[X_n]=0$ and $\|X_n\|_\infty \le 1$. Since $\|X_n\|_2 \le \sqrt d \|X_n\|_\infty \le \sqrt d$ we can use concentration results for the Euclidean norm to get $\mathbb E\left[ \left\|\sum_{i=1}^N X_i\right\|_\infty\right] = O (\sqrt {dN})$.
Does anyone know an example of when this dependence on $\sqrt d$ actually occurs?
Note: This is a bit informal. What I'm really asking is if the $O(\sqrt{dN})$ bound can be improved. I imagine a negative answer would be a family of examples of i.i.d sequences $X_n^{(d)}$, one for each value of $d$, such that $\mathbb E\left[ \left\|\sum_{i=1}^N X_i^{(d)}\right\|_\infty\right] \ge F(\sqrt {dN})$ for some common $\Omega(\sqrt{dN})$ function $F$.
I suspected such an example might be $X_n = (B_1^1,\ldots, B^1_d)$ for all $B^i_j$ independent and taking values $\pm 1$ with probability $1/2$. However the expectation seems more like $O (\sqrt {\log(d)N})$. To see this apply scalar concentration to each $j$ coordinate to get, up to coefficients:
$$P\left( \Big\|\sum_{i=1}^N X_i \Big\|_\infty <t\right)= P\left(\text{all } \Big|\sum_{i=1}^N X_i(j)\Big| <t \right)=\prod_{j=1}^d P\left(\Big|\sum_{i=1}^N X_i(j)\Big| <t \right) \ge (1-e^{-t^2 /N})^d.$$
$$\mathbb E \Big\|\sum_{i=1}^N X_i \Big\|_\infty = \int_0^\infty P\left( \Big\|\sum_{i=1}^N X_i(j)\Big\|_\infty >t\right)dt \le \int_0^\infty (1-(1-e^{-t^2/N})^d )dt$$
I don't know if the integral has a closed form. What I do know is the same argument gives
$$\mathbb E \Big\|\sum_{i=1}^N X_i \Big\|_\infty ^2 \le \int_0^\infty (1-(1-e^{-t/N})^d )dt$$
which I do know how to solve. Substitute $x = e^{-t/N}$ to get $dt = - (N/x)dx$ and the integral becomes
$$ N \int_0^1 \frac{(1-(1-x)^d )}{x}dx$$
which equals $N$ times the $d$-th harmonic number, which is $O(N \log d)$. So we have $$\mathbb E \Big\|\sum_{i=1}^N X_i \Big\|_\infty ^2 \le O\left(N \log(d) \right)$$ and by the Jensen inequality $$\mathbb E \Big\|\sum_{i=1}^N X_i \Big\|_\infty \le O\left(\sqrt{\log(d) N}\right).$$
For the general case, apply concentration to each coordinate as you did for the independent case, but then use the union bound instead of taking the product. This should give you that sqrt d is impossible
Let $X_i=(X_{i,1},\dots,X_{i,d})$, $S:=(S_1,\dots,S_d)$, $S_j:=\sum_{i=1}^d X_{i,j}/\sqrt n$. Then, by Hoeffding's inequality, for $s\ge0$
$$P(|S_j|\ge s)\le2e^{-s^2/2},$$
whence
$$E\|S\|_\infty=\int_0^\infty ds\,P(\|S\|_\infty\ge s)
\le\int_0^\infty ds\,\min(1,2d\,e^{-s^2/2})
=O(1+\sqrt{\ln d});$$
here we used the inequality
$$\int_t^\infty ds\,e^{-s^2/2}\le e^{-t^2/2}/2$$
for $t\ge0$.
So, for $d\ge2$ you get
$$E\Big\|\sum_{i=1}^n X_i\Big\|_\infty=O(\sqrt{n\ln d}),$$
without the assumption of the independence of the coordinates and without any other additional assumptions.
Thanks for answering my silly probability questions again. This is a very nice trick, and the sort of thing that might come to me in a dream five or ten years down the line.
|
2025-03-21T14:48:30.231409
| 2020-04-06T22:50:36 |
356780
|
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|
Stack Exchange
|
Assuming the conformal factor is radially decreasing, prove or disprove the uniqueness of geodesic joining origin and points on the boundary of ball
Let $u$ be a radially decreasing function defined on $\mathbb{R}^n$. We consider the metric $g=e^{2u}\delta$ where $\delta$ is the standard Euclidean metric on $\mathbb{R}^n$. Let $B_r$ be the ball centered at the origin with Euclidean radius $r$. Then for any $x \in \partial B_r$, by direct computation we know that the line segment connecting the origin and $x$ must be a geodesic. My question is that, is it true that the line segment joining the origin and $x$ is the unique geodesic connecting these two points?
Intuitively this is correct, because such line segment seems to be length minimizing. However, unfortunately I'm not be able to prove this....
I'm actually not very familiar with Riemannian geometry. Any ideas, comments or references will be really appreciated!
It must be length minimizing, because it is the unique geodesic from the origin to $x$. In particular, any ray from the origin is geodesic so we know all the geodesics from the origin. Since the shortest path from the origin to $x$ is a geodesic and there is only one such geodesic, it must be length minimizing.
Note that this does not mean that the shortest path from $x$ to $-x$ is necessarily a line segment through the origin. One could imagine the surface being shaped like a light bulb with the origin being the top of the bulb. If $x$ and $-x$ are on the threading, it's much shorter to go around the threading rather than go to the top and back down.
In particular, this is a property of radially symmetric metrics. It doesn't depend on them being conformally flat or the conformal factor decreasing.
Could you show why it is the unique geodesic from the origin to $x$? I know it is a geodesic, but I don't know how to show uniqueness.
Via the exponential map, there is a correspondence between geodesics emanating from the origin and unit tangent vectors. For each unit tangent vector, there is a geodesic given by the ray based at the origin in that direction. As such, those are all of the geodesics and there is only one that reaches $x$.
Sorry I still could not get it. Why there cannot exist other unit tangent vectors along the direction of which the corresponding geodesic reaches $x$? Also how would you exclude cut locus?
I suppose I should say why the shortest path from the origin to $x$ is given by a geodesic at all, since we don't assume that the metric is complete. It's an ugly argument, but a proof of this can be found here. https://mathoverflow.net/questions/317523/minimizing-geodesics-in-incomplete-riemannian-manifolds/317547#317547
To respond to your second question, the metric is defined on $\mathbb{R}^n$ and we know that the rays based at the origin are geodesics. In Euclidean space, there is only a single ray from the origin which passes through $x$. Every other ray does not.
Thank you for your reply and I do appreciate it. Actually I don't have trouble in understanding the existence of minimizing geodesic between $0$ and $x$ even if the metric is not complete, but I do have trouble in understanding why there are no other geodesics connecting $0$ and $x$. Think about a sphere, and $0$ is the coordinate of the south pole, and $x$ be the coordinate of some point on the sphere. Even thought $\mathbb{R}^n$ is Euclidean space, with the metric $e^{2u}\delta$ the manifold is not flat.
As for the cut locus, this same reasoning shows that the origin has no cut locus. It's possible for the cut locus to be non-empty at other points, but since there is a unique geodesic from the origin to a point $x$ and the shortest paths from the origin are given by geodesics (even when the metric is not complete), this shows there cannot be any cut locus at all.
Let us continue this discussion in chat.
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2025-03-21T14:48:30.231692
| 2020-04-06T23:08:33 |
356781
|
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|
Stack Exchange
|
Elliptic estimates for self-adjoint operators
Let $A$ be a symmetric matrix in $\mathbb R^n$ such that $A$ is positive definite and hence satisfies $0< \lambda \le A \le \Lambda < \infty.$
Let $T$ be a densely defined and closed operator from some Hilbert space $H$ into $H^n$. It is a classical theorem by John von Neumann that $T^*T$ is self-adjoint with domain $D(T^*T).$
I wonder whether it is true that for some $C>0$
$$\Vert \langle AT,T \rangle x \Vert \le C (\Vert T^*T x \Vert + \Vert x \Vert) \text{ and all } x \in D(T^*T).$$
Similarly, it seems natural to ask whether we also have that
$$\Vert T^*T x \Vert \le C (\Vert \langle AT,T \rangle x \Vert + \Vert x \Vert) \text{ and all } x \in D(\langle AT,T \rangle)?$$
Is $A$ fixed, or is it a function of $x$? Do you really allow for $\Lambda = \infty$?
@MateuszKwaśnicki I don't want $\Lambda=\infty$, sorry. I am surprised it would make a difference whether $A$ depends on $x$. Does it hold only for non $x$-dependent $A$?
If $P(D)$ has constant coefficients, and all operators act on $L^2(\mathbb{R}^n)$, then the domain of $P(D)$ is indeed $W^{2,2}(\mathbb{R}^n)$: just have a look on the Fourier transform. However, if $A$ is a (non-smooth) function of $x \in \mathbb{R}^n$, then the domain of $P(D)$ can get quite unrelated to the domain of $\Delta$.
@MateuszKwaśnicki I see, is there also a proof that does not rely on the Fourier transform?-I am just wondering whether it is true in general for $T$ closed and densely-defined that $\langle A T,T\rangle $ is relatively $T^*T$ bounded under the above assumptions on $A$ (and assuming it is constant)-and such a proof could somehow give hints whether this is indeed true.
As I understand your question now, after edit, in dimension two you ask whether
$$
\|T_2^* T_1 x\| \leqslant C(\|T_1^* T_1 x + T_2^* T_2 x\| + \|x\|)
$$
whenever $T_1$, $T_2$ are densely defined closed operators; $C$ can depend on $T_1$ and $T_2$.
This need not be the case. Let $T_1$ be the identity operator acting on $\ell^2$, and let $T_2$ be given by the matrix
$$ T_2 = \pmatrix{0&1&0&0&0&0&\cdots\\0&0&0&0&0&0&\cdots\\0&0&0&2&0&0&\cdots\\0&0&0&0&0&0&\cdots\\0&0&0&0&0&3&\cdots\\0&0&0&0&0&0&\cdots\\\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots} . $$
Then
$$ T_1^* T_1 + T_2^* T_2= \pmatrix{1&0&0&0&0&0&\cdots\\0&1 + 1&0&0&0&0&\cdots\\0&0&1&0&0&0&\cdots\\0&0&0&1 + 4&0&0&\cdots\\0&0&0&0&1&0&\cdots\\0&0&0&0&0&1 + 9&\cdots\\\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots} , $$
and
$$ T_2^* T_1 = T_2^* = \pmatrix{0&0&0&0&0&0&\cdots\\1&0&0&0&0&0&\cdots\\0&0&0&0&0&0&\cdots\\0&0&2&0&0&0&\cdots\\0&0&0&0&0&0&\cdots\\0&0&0&0&3&0&\cdots\\\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots} . $$
In particular, if $e_k$ is the $k$-th vector of the canonical basis of $\ell^2$, then $$\|T_2^* T_1 e_{2n-1}\| = \|n e_{2n}\| = n,$$ but $$\|T_1^* T_1 e_{2n-1} + T_2^* T_2 e_{2n-1}\| = \|e_{2n-1} + 0\| = 1 .$$
Therefore a constant $C$ with the desired property does not exist.
thank you. What I don't quite understand at the moment is that your $A=\begin{pmatrix} 0 & 1 \ 0 & 0 \end{pmatrix}$ no? This does not look positive-definite. Or is your answer assuming that without loss of generality it suffices to show you cannot control off-diagonal entries?
@KungYao: Yes this is what I meant; sort of "abstract second-order Riesz transforms need not be bounded". The same example shows that, say, $A = \pmatrix{1&1\1&1}$ leads to an operator $\langle AT,T\rangle = 1 + T_2 + T_2^* + T_2^*T_2$ which fails to satisfy the desired inequality.
|
2025-03-21T14:48:30.231937
| 2020-04-06T23:42:06 |
356784
|
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ali Taghavi",
"Sylvain JULIEN",
"Willie Wong",
"https://mathoverflow.net/users/13625",
"https://mathoverflow.net/users/36688",
"https://mathoverflow.net/users/3948"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/356784"
}
|
Stack Exchange
|
A Fourier elliptic vector field on a Riemannian manifold
Motivation for this question:
Let $X$ be a vector field on a manifold $M$. Obviously the differential operator $D:C^{\infty}(M)\to C^{\infty}(M)$ with $D(f)=X.f$ is not an elliptic opetator when $\dim M>1$. Now consider a simple example: Put $M$ for $\mathbb{R}^2$, $X=x^2\partial_x+y^2 \partial_y$
and $D$ for the first order corresponding differential operator. Then if we dente by $\mathcal{F}$ the standard Fourier transform operator then the principal (last homogeneus) part of differential operator $\mathcal{F}^{-1} D \mathcal{F}$ is $xu_{xx} +yu_{yy}$. This operator is an elliptic operator if we restrict to the first positive quadrant $x>0, y>0$.This situation is discussed via some other quadratic system in Remark 2 and its consecutive example in page 5 of this note.
In this question we would like to globalize and generalize this situation.
Fourier transform on Riemannian manifolds and Lie groups
In this MSE post three methods of generalization of Fourier transform on a Riemannian manifold or a Lie group is discussed. So based on these definition we assume that the Fourier transform $\mathcal{F}$ is a linear isomorphism on $C^{\infty}(M)$ when $M$ is a Riemannian manifold.
Definition: A non-vanishing vector field $X$ with derivational operator $D$ on a Riemannian manifold $(M,g)$ or a Lie group $G$ is called Fourier elliptic vector field if the differential operator $\mathcal{F}^{-1}D\mathcal{F}$ gives us a global elliptic operator.
Question: What are some precise examples of both Riemannian manifold and Lie group cases which admit a Fourier elliptic vector field? Does every manifold $ M$ with a non-vanishing vector field $X$ admit a Riemannian metric such that the vector field $X$ would be a Fourier elliptic vector field?
I wrote the linked MSE post. NONE of the three notions I described give you that the "Fourier transform is a linear isomorphism on $C^\infty(M)$". The closest case is that of the second notion, and even there the Fourier transform takes function spaces on a Lie group $G$ to functions defined on its dual group $\hat{G}$. For example, if you consider the group $\mathbb{T}^k$, its dual group is $\mathbb{Z}^k$. I am not sure how you intend to even define a differential operator on the dual group.
Shouldn't the considered Fourier transform be defined on an analogue of the Schwartz class on $M$ to get an automorphism thereof?
@SylvainJULIEN When $M$ is compact, there is no distinction between $C^\infty$ and $\mathcal{S}$; on the other hand, when $M$ is compact, most notions of frequencies will require the frequency space to be discrete and not a $\mathrm{dim}(M)$ dimensional manifold. For the case of Lie groups, it is also more convenient to formulate the general theory in terms of $L^2$ rather than $\mathcal{S}$: https://mathoverflow.net/questions/37021/is-fourier-analysis-a-special-case-of-representation-theory-or-an-analogue
@SylvainJULIEN yes it was a typo. As you said it is Schartz space. I wrote in paragraph 3 of the arxived note I mentioned in the post.
@WillieWong Thank you for your MO links. With help of these two MO and MSE link(your answer) I hope that one can introduce an abstract and global version I am searching for. To be honnest I am interested to find a generalization and globalization of this situation since more than 7 years ago. In fact after I posted that note to arxive, I realized that the resulting $\mathcal{F}^{-1}D\mathcal{F}$ is not GLOBALLY elliptic. So I wonder if there exist a vector field on the plane for which $D$ preserves the Schwartz class and $\mathcal{F}^{-1}D\mathcal{F}$ is globally elliptic
For example does $D=cos x \partial_x +sin x \partial_y$ work?
What exactly is the notion of elliptic you are working with? Your operator $F^{-1}D F$ for general vector fields $D$ (on $\mathbb{R}^2$) is not going to be a differential operator. I am not even sure it can in general be a pseudodifferential operator.
@WillieWong A differential operator is a linear operator which decreases the support(according to Peetre theirem). So an elliptic one has an obvious definition. For polynomial vector field the resulting $F^{-1} D F$ is obviousely a differential operator. But as you said it is interesting to classify all $D$ whose $F^{-1}DF$ is restriction of a differential operatir from $C^{\infty}$ to Schwartz space.
@WillieWong Now I realize what you say. but when the coefficients are polynomial functions the resulting $F^{-1}DF$ is a diff. operator again. Please read example in page 5 of this note(befor Remark 3) https://arxiv.org/pdf/1302.0001.pdf
In this case: your $F^{-1} D F$ can be expressed as $P\ell$, where $\ell$ is a linear function on $\mathbb{R}^n$ and $P$ is a constant coefficient partial differential operator. Its classification is not too technical.
@What is this linear functional for $D$ say $D=x^2\partial_x+y^2\partial_y+xy \Delta$?
Delta is Laplacian
The main aim of this question is to remedy the non ellipticity of certain linear operator associated to a derivational operator.
In fact the non ellipticity which discussed in this post was the main motivation for our question: https://mathoverflow.net/questions/182415/elliptic-operators-corresponds-to-non-vanishing-vector-fields?rq=1
|
2025-03-21T14:48:30.232292
| 2020-04-07T00:57:18 |
356787
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/356787"
}
|
Stack Exchange
|
Which orthant probabilities are the largest? (For a multivariate normal distribution)
I have a $k$-dimensional multivariate normal distribution $X∼N(0,\Sigma)$ with covariance matrix $\Sigma$. $\Sigma$ has two distinct eigenvalues, say $\lambda_1 > \lambda_2$, with orthogonal eigenspaces $V_1$ and $V_2$. I am interested in the orthant probabilities; given an orthant defined by $\epsilon = (\epsilon_1,\ldots,\epsilon_k)$, with each $\epsilon_i \in \{1,-1\}$, there is an orthant probability $p_\epsilon=Pr(\forall i: \;\epsilon_i X_i \geq 0)$.
There seems to be a literature on finding closed forms for these in special cases, but I do not necessarily need a closed form, but rather I just want to know when $p_{\epsilon} \geq p_{\epsilon'}$. One can write $\epsilon$ as a sum $\epsilon = v_1 + v_2$ with $v_1 \in V_1$ and $v_2 \in V_2$, and $k = \lVert\epsilon\rVert^2 = \lVert v_1 \rVert^2+\lVert v_2\rVert^2$. My conjecture is that the greater $\lVert v_1\rVert$ is, the greater $p_\epsilon$ is (recall that $\lambda_1 > \lambda_2$). Geometrically, this means that the diagonal vector of the octant defined by $\epsilon$ is closer to the longer axes of the ellipsoids which are the equidensity contour of the distribution.
Is this true? And if so, how does one prove it? I have been stumped (though I know little about probability).
(In the particular case I am interested in, I have a complete graph $K_\ell$, and I am generating a random map from edges of $K_\ell$ to $\mathbb{R}$. The covariance matrix has, as its two eigenspaces, the cut space and the edge space of $K_\ell$.)
Surprisingly, the conjecture is false. Orthant probabilities are not always ordered by the vector norm $\lVert v_1 \rVert$.
For a counterexample, take the 4-dimensional normal distribution with the following covariance matrix:
75.990348312987877 -14.891382893378880 9.019694111954474 8.518647696984502
-14.891382893378880 21.973634189398940 31.423796140503811 -5.365552691385230
9.019694111954474 31.423796140503811 63.319136031398756 -5.748592963390807
8.518647696984502 -5.365552691385230 -5.748592963390807 2.716881466214341
The two distinct eigenvalues are $\lambda_1 = 81$ and $\lambda_2 = 1$, each one twice. (It is not a coincidence that they are integers: I created the example by stretching a standard normal distribution $9$-fold in two directions, and rotating randomly. If you re-calculate the eigenvalues from the matrix listed above, they match to more than 12 decimal places.)
Looking into the orthants (only showing first half due to symmetry):
orthant 0: ++++ |v1|=1.604343 p=0.081526 p_MC=0.081561
orthant 1: +++- |v1|=1.644290 p=0.103843 p_MC=0.103680
orthant 2: ++-+ |v1|=0.916062 p=0.003139 p_MC=0.003152
orthant 3: ++-- |v1|=0.627763 p=0.002054 p_MC=0.002070
orthant 4: +-++ |v1|=1.419708 p=0.068214 p_MC=0.068349 !!!
orthant 5: +-+- |v1|=1.268382 p=0.017151 p_MC=0.017054
orthant 6: +--+ |v1|=1.850191 p=0.198087 p_MC=0.198265
orthant 7: +--- |v1|=1.562552 p=0.025999 p_MC=0.025957 !!!
Comparing orthants 4 and 7, we observe the latter has greater $\lVert v_1 \rVert$, but clearly smaller probability. I calculated the probabilities with two methods: p is from MATLAB mvncdf which claims absolute error tolerance $10^{-4}$, and p_MC is from Monte-Carlo integration with $10^7$ points. The probabilities differ already in the second decimal, so I'm pretty confident that it is not just an artefact of numerical calculation.
More counterexamples can be generated by making random instances and checking the orthant probabilities: the conjecture fails every now and then, although it holds in most of the cases. Interestingly, in 3 dimensions I have not found any counterexample in thousands of random instances (either in the prolate case = large eigenvalue is single and small eigenvalue is double, or in the oblate case = large is double and small is single).
Figure
Here is an attempt to visualize the 4-dimensional distribution (if anyone has better suggestions, I'd like to hear). We are showing $100\,000$ random points from the distribution; points in the 4th orthant +-++ are shown red, points in the 7th orthant +--- are shown blue, and all other points cyan. Perhaps one can see that the red points are more than twice as many as the blue points.
Each subplot shows a 2D projection to two coordinate axes, in the same scale: each box ranges from $-20$ to $+20$ in both directions. Note that the two contending orthants are in the same quadrant of $X_1,X_2$, and in opposite quadrants of $X_3,X_4$.
Code
For convenience, here is Matlab code to reproduce and verify this particular instance. You can adapt it to search for other counterexamples in four or other dimensions.
Sigma = [
75.990348312987877 -14.891382893378880 9.019694111954474 8.518647696984502
-14.891382893378880 21.973634189398940 31.423796140503811 -5.365552691385230
9.019694111954474 31.423796140503811 63.319136031398756 -5.748592963390807
8.518647696984502 -5.365552691385230 -5.748592963390807 2.716881466214341
];
dim = size(Sigma,1);
[V,D] = eigs(Sigma);
Dval = diag(D);
% Find the big and small eigenvalues
Ibig = abs(Dval - max(Dval)) < 1e-3;
Ismall = abs(Dval - min(Dval)) < 1e-3;
fprintf('Big eigenvalues:\n');
Dval(Ibig)
fprintf('Small eigenvalues:\n');
Dval(Ismall)
% Generate random points from the distribution
NN = 1e7;
R = chol(Sigma);
X = R' * randn(dim,NN);
V1MAG = [];
P = [];
for i=0:2^(dim-1)-1
neg = bitget(i, dim:-1:1)';
e = ones(dim,1) - 2*neg;
signvec = repmat('+',1,dim);
signvec(e<0) = '-';
% Orthant probability from Matlab
Sigmaflip = e .* Sigma .* e';
p = mvncdf(zeros(dim,1), zeros(dim,1), Sigmaflip);
P = [P p];
% Monte Carlo orthant probability
pmc = mean(all(X .* e > 0, 1));
% Projections of e onto each eigenvector (column of V)
edots = e' * V;
v1mag = sqrt(sum(edots(Ibig).^2));
v2mag = sqrt(sum(edots(Ismall).^2));
V1MAG = [V1MAG v1mag];
fprintf('orthant %2d: %s |v1|=%.6f p=%.6f p_MC=%.6f\n', ...
i, signvec, v1mag, p, pmc);
end
% Check the order
[~,S] = sort(V1MAG);
pprev = -inf;
fprintf('\nOrthants sorted by |v1|:\n');
for i=1:2^(dim-1)
fprintf('orthant %2d : |v1|=%.6f p=%.6f', ...
S(i)-1, V1MAG(S(i)), P(S(i)));
if P(S(i)) < pprev - 0.01
fprintf(' TROUBLE!');
end
fprintf('\n');
pprev = P(S(i));
end
|
2025-03-21T14:48:30.232768
| 2020-04-07T01:15:09 |
356789
|
{
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"authors": [
"Gerry Myerson",
"Robert Israel",
"Wlod AA",
"https://mathoverflow.net/users/110389",
"https://mathoverflow.net/users/13650",
"https://mathoverflow.net/users/3684",
"https://mathoverflow.net/users/51189",
"zeraoulia rafik"
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|
Stack Exchange
|
Write $p^2$ as $x^2+2y^2+3\times 2^z$ with $x,y,z$ nonnegative integers
In April 2018, I noted that the first integer $n>1$ with $n^2\not\in\{x^2+2y^2+3\times 2^z:\ x,y,z=0,1,2,\ldots\}$ is $$5884015571=7\times17\times49445509.$$
Question. Is it true that for each prime $p$ we can write $p^2$ as $x^2+2y^2+3\times2^z$ with $x,y,z$ nonnegative integers?
I guess that this question has a positive answer. Your comments are welcome!
I think your form is suitable only for 2 to 29 , Try p=31
@zeraouliarafik $31^2 = 29^2 + 2 \cdot 6^2 + 3 \cdot 2^4 $.
OP put a related sequence online, https://oeis.org/A301472
The density of the 3-variable sequence seems fine to me but the question still feels like a smooth glass mountain. Do you see at this time any chance of climbing or even starting it?
I think all integer positive number greater than 1 can be written in your form
|
2025-03-21T14:48:30.232856
| 2020-04-07T03:17:57 |
356800
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Bernhard Boehmler",
"https://mathoverflow.net/users/12826"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/356800"
}
|
Stack Exchange
|
Can MAGMA compute almost projective $kG$-homomorphisms?
Let $G$ be a finite group and $k$ be a finite field (big enough) whith char$(k)=p$ and $p\mid |G|$.
Let $M$ be a finitely generated $kG$-module.
We denote the first syzygy of $M$ by $\Omega(M)$, i.e.
$\Omega(M):=\text{Ker}(p)$ where $P\stackrel{p}{\rightarrow} M$ is a minimal projective cover of $M$.
${}$
A $kG$-homomorphism $m: M\rightarrow \Omega(M)$ is called almost projective,
if $m$ is a generator of the simple socle of $\text{Hom}_{kG}(M,\Omega(M))/\text{Proj}_{kG}(M,\Omega(M))$ as $\text{End}_{kG}(M)$-module.
${}$
Here, $\text{Proj}_{kG}(M,\Omega(M))$ is the set of all projective $kG$-homomorphisms from $M$ to $\Omega(M)$ (i.e. those that factor through a projective $kG$-module).
${}$
I would like to ask the following question:
Can MAGMA produce / compute such an almost projective $kG$-homomorphism $m$?
Thank you very much for the help.
EDIT (9th April): The motivation is to construct almost split sequences with MAGMA and this is related to the following earlier question:
https://math.stackexchange.com/questions/3317886/can-magma-compute-auslander-reiten-sequences-in-group-algebras
I cannot answer 100%, but I can tell you what I know is there, and maybe its enough with some tweaking. AR-sequences are not something I've needed to implement in Magma yet, so I've not grappled with this one.
Magma can compute projective covers and a syzygy, first off. Then it can compute $\texttt{AHom(A,B)}$, which is simply $\mathrm{Hom}_{kG}(A,B)$. It can also compute $\texttt{PHom(A,B)}$, but this is only for basic algebras, not general group algebras. If your group is small enough that you can produce its basic algebra in Magma (there are commands for this) then you can pass between then to find $PHom$ inside $Hom$.
Finally, $Hom$ can be made into a $kG$-module itself, using $\texttt{HomMod}$. (This requires a recent version of the program.)
If you start off in a basic algebra, you can type $\texttt{SyzygyModule(M,1)}$ for $\Omega(M)$.
Thank you very much for the answer.
|
2025-03-21T14:48:30.233002
| 2020-04-07T03:39:41 |
356801
|
{
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"authors": [
"Asaf Karagila",
"https://mathoverflow.net/users/7206"
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"url": "https://mathoverflow.net/questions/356801"
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|
Stack Exchange
|
Global choice for models of complete theories in $\mathsf{ZFC}$
This is a followup to a previous question of mine. To summarize the result of that question, by a result of Kanovei and Shelah, $\mathsf{ZFC}$ is enough to show that there is a uniform procedure for taking a structure and finding a proper class sized elementary extension of it (with a definable truth predicate) which is 'class-saturated' in the sense that any type over a set of parameters is realized.
Ultimately my question is whether or not an analogous statement can be made for building such models uniformly from theories, rather than from structures. The result of my previous question is enough to reduce that to this question:
Question 1. Is there a formula $\varphi(x,y,z)$ in the language of set theory such that for any first-order language $\mathcal{L}$ and any complete $\mathcal{L}$-theory $T$, there is a unique $\mathcal{L}$-structure $\mathfrak{M}$ such that $\mathfrak{M}\models T$ and $\varphi(\mathcal{L},T,\mathfrak{M})$ holds.
What if we allow $\varphi$ to have parameters (i.e. in principle $\mathsf{ZFC}$ might not have a global choice function for models of complete theories, but it might prove that there always exists a parameter over which it has such a function)?
A few observations and thoughts.
If we add in a well-ordering of $\mathcal{L}$ then we can easily do this by adding Henkin constants and completing the resulting theory in a mechanical way (which we can do since we can well-order the set of formulas in this language).
There is such a formula if instead of requiring that $\varphi(\mathcal{L},T,\mathfrak{M})$ hold for a unique $\mathfrak{M}$ we instead require that it holds for a unique isomorphism type. Specifically, we can let $\varphi(\mathcal{L},T,\mathfrak{M})$ hold whenever $\mathfrak{M}$ is a special model of cardinality $\kappa$ with $\kappa \geq |\mathcal{L}|$ the smallest such that $T$ has a special model of cardinality $\kappa$.
Using Scott's trick we can get that down to a set of models of the same isomorphism type (i.e. we require that $\mathfrak{M}$ have minimal foundational rank among members of its isomorphism type).
If we allow $T$ to be incomplete, then this implies a global Boolean prime ideal theorem, i.e. that there is a uniform choice of prime ideal for Boolean algebras. I'll call this $\mathsf{GBPI}$.
It is easy to see that if we have $\mathsf{GBPI}$ then we can get such a formula, even for $T$ incomplete, which would imply that for $T$ incomplete this is equivalent to $\mathsf{GBPI}$. All you need to do is add Henkin constants to the theory (which can be done in a completely uniform way) and then choose a completion of the Henkinized theory using $\mathsf{GBPI}$. The term model gives the required model of the theory. That said, I doubt that $\mathsf{ZFC}$ proves $\mathsf{GBPI}$, or even fairly weak consequences of it, such as global choice for pairs, for that matter. I also suspect that $\mathsf{GBPI}$ is weaker than global choice.
Just like with global choice, despite the seemingly non-first-order nature of the statement $\mathsf{GBPI}$, it is expressible as a single first-order sentence thanks to the reflection principle: Let $\chi(x,y)$ be a formula that says '$(V_\alpha,\in) \models \psi(x,y)$ for the smallest $\alpha$ and smallest $\psi$ (in some fixed well-ordering) such that $\exists ! z ((V_\alpha,\in) \models \psi(x,z)$ and $z$ is a prime ideal for $x)$.' As long as there is a formula that uniformly selects prime ideals, this formula will as well. A similar trick will work for $\mathsf{GBPI}$ with a parameter (although $\psi$ will also need a parameter) and for a global choice function for models.
I restricted Question 1 to complete theories in the hope that it would make the question easier, but there's a chance it's equivalent.
Question 2. Is the existence of a global choice function for models of complete theories equivalent to $\mathsf{GBPI}$?
The easy way to tackle these sort of problems is starting with an inaccessible $\kappa$, then violating the axiom of choice in a myriad of ways "at $\kappa$", so $V_{\kappa+1}$ will be a model of $\sf ZFC_2$, but global choice will fail in various ways. After we understand that mechanism we can see if we can remove the inaccessible, and in all likelihood, we can.
|
2025-03-21T14:48:30.233289
| 2020-04-07T03:51:34 |
356802
|
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"url": "https://mathoverflow.net/questions/356802"
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|
Stack Exchange
|
Points where singular sum is small
We consider $x_1,..,x_N$ points in the plane $\mathbb{R}^2.$
We define the sum
$$F(x):=\frac{1}{N^2}\sum_{i=1}^N \sum_{j \neq i} \vert x_i-x_j \vert^{-2}.$$
I am looking for a statement of the following sort:
If $F(x) \le 1$ then $\vert x \vert \gtrapprox N$ where $\gtrapprox$ means that in some sense such a point configuration should have norm larger than $N.$
Here is a heuristic argument why something like this should true:
If we place points $x_1,..,x_N$ uniformly on the circle or radius $\sqrt{N}$ then arguably the points $x_1,...,x_N$ are quite far from each other and $\vert x \vert^2 =N \vert x_i \vert^2 = N^2$
In that case however
$$F(x):=\frac{1}{N^3}\sum_{i=1}^N \sum_{j \neq i} \vert y_i-y_j \vert^{-2}$$ where $y_i$ are now on the unit circle.
Since they are uniformly distributed, we have that
$$F(x):=\frac{1}{N^2} \sum_{j \neq 1} \vert y_1-y_j \vert^{-2}.$$
However, it is well-known that this last sum $\sum_{j \neq 1} \vert y_1-y_j \vert^{-2}$ is known to be of order $N^2$, compare with this one and similar questions. So in this critical case, we precisely something of order $1$.
So can we get a good description of points where $F(x) \le 1?$
EDIT: Perhaps my $\vert x \vert \gtrapprox N$ is a bit too optimistic, but then I wonder what the optimal estimate may be?
|
2025-03-21T14:48:30.233399
| 2020-04-07T04:00:36 |
356803
|
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"authors": [
"François Brunault",
"LWW",
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"https://mathoverflow.net/users/6506"
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"url": "https://mathoverflow.net/questions/356803"
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|
Stack Exchange
|
Meaning of extended principal part of weakly holomorhpic modular forms
In p.312 of 'Rhoades, Robert C., Linear relations among Poincaré series via harmonic weak Maass forms. Ramanujan J. 29 (2012), no. 1-3, 311–320', the author defines the extended principal part at infinity as follows:
A weakly holomorhpic modular form is any meromorhpic modular form whose poles are supported at the cusps. The extended principal part at infinity of a weakly holomorphic modular form $f$ is the polynomial $P_{f,\infty} \in \mathbb C[q^{-1}]$ such that $f(z)-P_{f,\infty}(q^{-1}) = O(e^{-\epsilon y})$ as $y \to \infty$.
My questions are,
what is the meaning of the extended principal part.
What's the relation of this definition of extended principal part and the 'principal part of Laurent expansion'?
Is the extended principal part of weakly holomorphic modular form determined uniquely?
In addition, the author defines the extended principal part at the cusps $x$ as follows:
If $x$ is a cusp, the extended principal part at $x$ is the finite sum of terms in the Fourier expansion around $x$ that do not have rapid decay toward $x$.
I didn't understand this sentence, but in my idea the natural definition of the extended principal part at $x$ is the following:
Let $\sigma \in \mathrm{SL}_2(\mathbb R)$ be an element satisfying $\sigma x = \infty$. Then $f|_k \sigma^{-1}$ has the Fourier expansion in $q$, where $q=e(z/h)$, $h$ is the cusp width at $x$. The extended principal part of $f$ at $x$ is a polynomial $P_{f,x}(q^{-1}) \in \mathbb C[q^{-1}]$ such that $f|_k \sigma^{-1} (z)-P_{f,x}(q^{-1}) = O(e^{-\epsilon y})$.
Is my understanding right?
The extended principal part coincides with the principal part of $f$ as a Laurent series in $q$ (except maybe for the constant term, that you may or may not want to put in the definition). For arbitrary cusp your definition is correct. The extended principal part is then unique up to $z \to z+1$ which means $q \to \exp(2\pi i/h) q$.
You have written "Let $\sigma$ be the element" but it should be "an element".
Brunault // Could you let me know why?
You can multiply $\sigma$ on the left by any matrix stabilizing $\infty$. Also, it should be $\sigma \in \mathrm{SL}_2(\mathbb{Z})$.
Brunault // I mean, the reason for the first comment you wrote.
This is just uniqueness of the Laurent expansion of a meromorphic function
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2025-03-21T14:48:30.233561
| 2020-04-07T06:40:01 |
356806
|
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Stack Exchange
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Do multiplicative Banach limits exist?
Let $(D, \succeq)$ be a directed set, and let $B$ be the space of real-valued bounded functions on $D$. A Banach limit $\ell$ on $D$ is a linear functional that satisfies
$$\sup_{d \in D} \inf_{c \succeq d} f(c) \leq \ell(f) \leq \inf_{d \in D} \sup_{c \succeq d} f(c)$$
for all $f \in B$.
Banach limits exist by the Hahn-Banach theorem. In fact, if we say instead that our functions on $D$ take values in an abstract Dedekind complete ordered vector space, then the existence of Banach limits is equivalent to the Hahn-Banach theorem. But I am primarily interested in the real-valued case.
Question. Are there Banach limits that satisfy
$$\ell(fg) = \ell(f)\ell(g)$$
for all $f,g \in B$? If so, does this require anything stronger than the Hahn-Banach theorem?
I do not think that this is the usual definition of Banach limit. (What I know under this name is linear functional on $l_\infty$ which is positive, shift-invariant and extends the usual limit, see the linked Wikipedia article. Of course, the terminology in various sources might differ.) In connection with the question it might be worth mentioning that Banach limit in this sense cannot be multiplicative.
EDIT: The OP mentioned in comments that the definition of Banach limit given in the question (i.e., with directed sets) can be found in Schechter's Handbook of Analysis and Its Foundations, see page 318 and also in Howard, Rubin: Consequences of the Axiom of Choice, see page 63, Form 372. (If we work on arbitrary directed sets, we do not have a natural way to talk about shift-invariance. So this is not a generalization of the notion of Banach limit on $\mathbb N$ mentioned in the first paragraph.)
Your conditions can be rewritten as1
$$\liminf_{c\in D} f(c) \le \ell(f) \le \limsup_{c\in D} f(c).$$
So you want a functional which is between $\liminf$ and $\limsup$ (and therefore extends the usual limit of a net) and is multiplicative.
You can simply take any ultrafilter $\mathcal U$ which contains all tail sets of the directed set $D$. (I.e., for any $d\in D$ you have $d\uparrow=\{c\in D; c\ge D\}\in\mathcal U$.) And then define $\ell$ using limit along this utlrafilter as:
$$\ell(f) = \operatorname{{\mathcal U}-\lim} f(c).$$
This functional has the properties you want. (Boundedness of $f$ guarantees that the $\mathcal U$-limit exists.2 We get multiplicativity from the fact that $\mathcal U$-limit is multiplicative. And the fact that $\mathcal U$ contains the tail filter helps with the condition about limit inferior and limit superior.)
The same construction is mentioned in the answer to: What is a generalized limit?
In case it helps to find some references for $\mathcal U$-limit (limit along an ultrafilter or, more generally, limit along a filter or a filter base), I will mention my answers to these questions: Where has this common generalization of nets and filters been written down? and Basic facts about ultrafilters and convergence of a sequence along an ultrafilter.
Since you are interested in multiplicative functionals, this might be of interest, too: Every multiplicative linear functional on $\ell^{\infty}$ is the limit along an ultrafilter.
1For more on limit superior/inferior of a net, see: Limsups of nets and About the notion of limsup and liminf
2Limit along an ultrafilter with values in a compact space always exists. The proof is given, for example, in this answer: Basic facts about ultrafilters and convergence of a sequence along an ultrafilter.
Thanks. My definition comes from Howard & Rubin's Consequences of the Axiom of Choice and Schechter's Handbook of Analysis. It's just a bit more general than yours, so I don't think there are any issues there.
I'm not sure I see why the $\mathcal U$-limit exists. Can you explain?
Also, I wonder if, in general, the existence of multiplicative Banach limits requires the ultrafilter lemma (which is stronger than Hahn-Banach).
Re: Why $\mathcal U$-limit exists: Since $\mathcal U$ is an ultrafilter and the values are all on some bounded interval, which is a compact space. A proof that in compact space limit along an ultrafilter exists is given in this answer: Basic facts about ultrafilters and convergence of a sequence along an ultrafilter. (And probably in many other resources.)
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2025-03-21T14:48:30.233825
| 2020-04-07T07:02:15 |
356808
|
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Stack Exchange
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Abelian subgroup of maximal order
Let $\mathbf{F}_q$ be a finite field of order $q$ ($q$, an odd prime, or a power of the same). I know that for the matrix algebra $M(n,\mathbf{F}_q)$ (or $gl(n,q)$), the maximal dimension for a commutative subalgebra is $\left[\frac{n^2}{4}\right] + 1$ (this is a result of Schur (for $\mathbb{C}$) and Jacobson for any field).
So, now I move to the matrix group $GL_n(\mathbf{F}_q)$, and ask if there is a formula for the order of an abelian subgroup of $GL_n(\mathbf{F}_q)$ of maximal order?
This question is related. I think the answer for $n \ge 4$ is $(q-1)q^{n^2/4}$ when $n$ is even and $(q-1)q^{(n^2-1)/4}$ when $n$ is odd. For $n \le 3$ it is $q^n-1$.
@MarkSapir It's slightly larger than a $p$-group, because you can include the scalar matrices. But yes I believe you can argue like that. I am sure this result is in the literature somewhere.
If $A$ is an abelian subgroup of the upper triangular group $U$, then $|A|\le q^{n^2/4}$ or $q^{(n^2-1)/4}$ according as $n$ is even or odd, and these upper bounds are attained -- even by elementary abelian "upper right-hand block" subgroups. If you order the above-diagonal spots $(i,j)$ lexicographically by $(j-i,j)$, and replace each element of an abelian subgroup $A$ of $U$ by just its "least" nonzero entry -- zeroes elsewhere -- I believe you obtain an abelian subalgebra $B$ of $gl(n,q)$. Combinatorial arguments show that $|A|=|B|\le q^{n^2/4}$ or $q^{(n^2-1)/4}$ as desired.
This "homogenization" trick was used by Malcev in determining the largest abelian subalgebras of all the semisimple complex Lie algebras. A. I. Malcev, “Commutative subalgebras of semi-simple Lie algebras,” Izv. Akad. Nauk SSSR, Ser. Mat., 9, No. 4, 291–300 (1945).
@DerekHolt Thank you for the help. In addition, I was able to find some stuff about this in Vdovin's papers.
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2025-03-21T14:48:30.233978
| 2020-04-07T07:18:43 |
356810
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|
Stack Exchange
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Regularity of Moore-Penrose pseudo-inverse
Let $k\in\mathbb{N}\cup\{0\}$, let $\Omega\subseteq\mathbb{R}^n$ be open, connected and let $G\in C^k(\Omega;\mathbb{R}^{n\times n})$ satisfy
$$
\operatorname*{rank}G(x)= \operatorname*{rank}G(y),\text{ for all }x,y\in\Omega.
$$
Then, is it true that the Moore-Penrose pseudo-inverse $G^+ \in C^k(\Omega;\mathbb{R}^{n\times n})$?
The case $k=0$ case seems to be true (Penrose 1955). What about $k\in\mathbb{N}$? Could you please suggest a reference? Any help in this direction is welcome.
Just for clarity, is your $G^+$ defined by $G^+(\omega):=G(\omega)^+$ for all $\omega\in \Omega$?
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2025-03-21T14:48:30.234046
| 2020-04-07T08:40:48 |
356813
|
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Stack Exchange
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Can inaccessibility be captured in relational flat set theory?
Working in a first order theory of flat sets (axioms given below) , which is a theory with a single non-trivial tier of membership, that is all sets are nonempty sets of Quine atoms, plus having some of the atoms being directional elements (which serve as ordered pairs between elements), plus having a well ordering $R$ over the whole universe of atoms; what I call as "Relational flat set theory". IF we define the cardinality of a set $A$, denoted as |A|, as the $R$-maximal element of the $\subseteq$-minimal bi-ended initial segment that is bijective to $A$, where a bi-ended initial segment is defined as a set closed under $<^R$ and that has an $R$-maximal element. And define regular cardinal as:
$$regular (\kappa) \iff (\exists X: \kappa=|X|) \land (\neg \exists S: \kappa = \lim(S) \land |S| < \kappa) $$
Lets define inaccessible cardinal as:
$$ inaccessible (\kappa) \iff \kappa > \omega \land \not \exists \lambda (\kappa = \lambda+1) \land regular(\kappa)$$
Where $``<,lim, +"$ are all defined in the usual manner but with respect to relation $R$
Now lets add the following formula to the axiom of well ordering.
Inaccessibles: $\exists \kappa : inaccessible (\kappa)$
Questions:
would those notions of regular cardinals and weak inaccessibility correspond to the same notions defined in standard set theory?
if they do so, then this would mean that this theory is NOT interpretable in ZFC, then can this theory interpret ZFC? or is it incomparable with it?
So the idea here is that if inaccessibility can be captured in a flat milieu? and if it would enable constructing a model of ZFC in that harsh restriction on set structure, perhaps by defining $L$ within it, thereby interpreting power indirectly? If that is the case then this would mean that Relational flat set theory has all the resources that ordinary set theory have? And a hierarchy of sets is not essentially needed?!
Relational Flat Set Theory:
Language: First order logic with equality $``="$, membership $``\in"$, and directional linking $L$, the latter is a partial two place function symbol, where $l=L(a,b)$ is read as $l$ links $a$ to $b$.
Axioms:
Extensionality $\forall ( z \in x \leftrightarrow z \in y) \to x=y$
Membership: $x \in y \to x=\{x\}$
Actuality: $\forall x \exists y (y \in x)$
Composition: $\exists x (\exists!y (y \in x) \phi) \to \exists s \forall x (x \in s \leftrightarrow \exists!y (y \in x) \phi)$
Let $S$ be the set of all singletons.
Naturality: $l=L(a,b) \to l,a,b \in S $
Links: $\forall a,b \in S \exists l : l=L(a,b) $
Direction: $L(a,b) = L(c,d) \to (a=c \land b=d) $
Well ordering: $\exists R: R \text{ well orders } S$
Does your axiom of Actuality mean that there is no empty set in this theory? That would already seem to break any direct interpretation of ZFC by this theory...
@StevenStadnicki, of course it would! definitely the interpretation is not direct, since after all we are having only one tier of non-trivial membership, and Quine atoms,.. so yes the interpretation is definitely indirect.
I'm just struggling to see how even an indirect interpretation would work here. Can you offer any sort of an atlas as to how you see a subset of the ZFC axioms being interpreted in this theory? My first instinct says that the $R$-minimal element of $S$ 'has to' represent the empty set, but I'm not really seeing any sensible candidates for $\in$...
@StevenStadnicki, the only idea is a vague one that if $L$ can in some way be defined in this theory, then this can indirectly interpret ZFC. The existence of an inaccessible "in the view of this theory" must add some work. But I have no detailed plan.
@StevenStadnicki, there is no problem with interpreting the empty set and the hereditarily finite sets of ZFC, this theory clearly interpret second order arithmetic and thus PA, and PA can interpret ZFC-Infinity, by constructing a model of ZFC -Infinity in which all sets are hereditarily finite (see: http://web.mat.bham.ac.uk/R.W.Kaye/publ/papers/finitesettheory/finitesettheory.pdf) But I think this theory can go way beyond that as to possibly interpret the whole of ZFC through defining L in it, but I'm not really sure of that.
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2025-03-21T14:48:30.234455
| 2020-04-07T08:41:40 |
356814
|
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|
Stack Exchange
|
Limit case of Sobolev space in $1$-D
This might look too an elementary question, but I am confined and is not able to find a textbook which answers the following question.
I have a function $f:{\mathbb R}\rightarrow{\mathbb R}$, such that $f\in L^3({\mathbb R})$ and
$$\int\int\frac{|f(y)-f(x)|^3}{|y-x|^4}dydx<\infty.$$
May I conclude that $f\in W^{1,3}({\mathbb R})$ ?
This is a limit case of Sobolev-Slobodeckij space, as $4=1\cdot3+1$. Obviously, the same integral but with exponent $s\cdot3+1$ with $s<1$ is valid, hence $f\in W^{s,3}({\mathbb R})$
If I see it correctly, then by the "modulus of continuity-characterization" of Besov spaces, $f$ will lie in $B^1_{3,3}(\mathbb{R})$. Unfortunately, this a slightly larger space than $W^{1,3}(\mathbb{R})$.
If f is smooth with nonzero derivative, the integrand is of order $1/|x-y|$, which is not integrable. I suspect the condition implies that f is constant.
@MichaelRenardy. Indeed, you are right. I had an exchange, after posting the Q, with Petru Mironescu, who confirmed the fact.
This implies that (f) is constant (see the paper by Haïm Brezis, “How to recognize constant functions. Connections with Sobolev spaces”).
Let summarize the comments there:
In order for the seminorm here to be finite, one needs at least $|f(x)-f(y)| = o(|x-y|)$ when $x-y\to 0$, and this is possible only for constants functions. Since $f∈L^3$, $f=0$ (and so $f∈W^{1,3}$ ...).
If to avoid that one uses the second order difference $f(2y-x)-2f(y)+f(x)$ instead of $f(y)-f(x)$, one could however not conclude that $f∈W^{1,3}$, and this is due to the misleading definition of fractional Sobolev(-Slobodeckij) spaces $W^{s,p}$ since when $s>0$ is not an integer $W^{s,p} = B^s_{p,p} = F^s_{p,p}$ (where the $F^s_{p,q}$ are the Triebel-Lizorkin spaces) while $W^{n,p} = F^n_{p,2}$ when $n$ is an integer. An other fractional extension of Sobolev spaces are the Bessel-Sobolev spaces $H^{s,p}$ where the seminorm is the $L^p$ norm of the fractional Laplacian. They verify $H^{s,p} = F^s_{p,2}$ (For every $s≥0$).
All these spaces are ordered in this way when $p≥ 2$ (with strict inclusion when $p>2$)
$$
B^s_{p,1} ⊂ B^s_{p,2} ⊂ F^s_{p,2} (=H^{s,p}) ⊂ F^s_{p,p} = B^s_{p,p} ⊂ B^s_{p,\infty}.
$$
This is well explained for example in the book of Hans Triebel, Theory of Function Spaces II. Springer Basel, 1992.
As a heads up, I did not claim hat the integral in the OP is exactly the $B^1_{3,3}$ seminorm. In the integer case, the usual equivalence theorem for the Besov norm (Thm. 2.5.1 in the "old Triebel") would give something like a "centered" term $f(2y-x)-2f(y)+f(x)$ in the numerator. But I figured that finiteness of the OP integral would imply finiteness of the integral with centered term, so the $B^1_{3,3}$ seminorm.
Yes you are absolutely right, there are second order differences when $n=1$ (and this is why with first order differences lead to constant functions). I edit my answer to summarize all that.
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2025-03-21T14:48:30.234680
| 2020-04-07T09:52:54 |
356818
|
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|
Stack Exchange
|
Existence of a "p-adic Mahler measure" or alternatively, the converge of a p-adic sequence
Let $f \in \mathbb Z_p[[t]]^\times$ be an invertible power series and let $\log_p$ be the p-adic logarithm with the normalization that $\log p = 0$. Consider the sequence:
$$a_n = \frac{1}{p^{n-1}}\sum_{(i,p)=1, i=1}^{p^n}\log_p f(\zeta_{p^n}^i-1).$$
This is a p-adic analogue of the Mahler measure. Does this sequence converge (perhaps under suitable conditions on $f$) and if so what is known about the value to which it converges?
I have seen a paper by Besser-Deninger but they study the sum over roots of unity with order coprime to $p$ while I am interested in the exact opposite case. Perhaps someone has also studied this variant?
A power series $f(t) \in Z_p[[t]]^\times$ can be written as $f(t) = b_0 \prod_{k \geq 1} (1-b_k t^k)$ with $b_0 \in Z_p^\times$ and $b_k \in Z_p$. Some back of the envelope calculations suggest that, if $f(t)=(1-b_k t^k)$, then it is possible that your $a_n$ converge to something like $(p-1) \log_p b_k$.
Thank you! That is a helpful observation.
|
2025-03-21T14:48:30.234783
| 2020-04-07T09:58:11 |
356819
|
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|
Stack Exchange
|
Computing powers of a special matrix fast
I've got an $n\times n$ matrix that is upper-triangular (all zeros below diagonal), diagonal entries are positive, and entries that are not on the diagonal are either $0$ or $1$. An example matrix looks like this. $$A=\begin{pmatrix}
69 & 0 & 1 & 1 & 0 \\
0 & 100 & 1 & 0 & 1 \\
0 & 0 & 15 & 0 & 1 \\
0 & 0 & 0 & 15 & 1 \\
0 & 0 & 0 & 0 & 300
\end{pmatrix}$$ I'm interested in the upper-right entry of a large power (say $A^k$) of this matrix with computations done in a finite field $\mathbb{F}_p$ for some fixed prime $p$. Is it possible to compute this faster than usual for this kind of matrices? I'm looking for runtimes something similar to $O(n^2)$ preprocessing and $O(n\lg k)$ per query.
Do you mean for a given $A$ and $p$ you want to compute the upper right entries for many different $k$? Then it may make sense to precompute the Jordan normal form (of course the eigenvalues are the diagonal entries)..
@RobertIsrael: yes for the first question. But wouldn't computing the Jordan normal form end up taking at least $O(n^3)$ precomputation time (given the eigenvalues, I still would need to compute several kernels)? That's unfortunately too much for my purpose.
Your matrix structure doesn't seem particularly special. How can you expect time linear in $n$ when the size of the output is $\Theta(n^2)$?
Actually the output size is $O(1)$. I'm interested in the upper-right entry of the power.
Right, sorry I can't read today.
You’re computing in a finite field, so when you say the diagonal entries are positive, do you just mean nonzero?
I mean they are nonzero modulo $p$. Sorry for the confusion.
|
2025-03-21T14:48:30.234930
| 2020-04-07T10:12:58 |
356820
|
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|
Stack Exchange
|
Maximal function corresponding to a family of functions
Let $(f_t)_{t\geq 0}$ be a family of measurable functions. Under which criterion we can say that $\sup_{t\geq 0}\lvert f_t\rvert$ defines a measurable function? I am working in the following framework. Let $G$ be a locally compact Hausdorff second countable topological group and $(\mu_t)_{t\geq 0}$ be a family of probability measures on $G$ with $t\mapsto \mu_t$ $w^*$-continuous. Let $X$ be a standard Borel space equipped with a probability measure invariant under a measurable action by $G$. Define $\pi(\mu_t)f(x)\mathrel{:=}\int_Gf(g^{-1}x)d\mu_t(g)$ for $f\in L^p(X)$ for a fixed $1\leq p<\infty$. Then, we define $f_t:=\pi(\mu_t)f$.
I am looking for sufficient conditions under which the pointwise supremum $\sup_{t \ge 0} \lvert f_t\rvert$ is measurable and also well-defined in the sense that it does not depend on the equivalence class of $f$. And then some more sufficient conditions so that the supremum is in $L^p$.
I'm under the impression that the question is bit vague: one can find probably find various sufficient criteria in quite different contexts (for instance, it is sufficient if $f_t$ depends monotonically on $t$). Could you add a bit more context to the question so that we can see which type of assumptions would be interesting for you?
Under the assumptions stated, $\sup_{t \geq 0} |f_t|$ doesn't even have to be a function. Let $f_t = 1/|x - t|$ for $x \neq t$ and $0$ for $x = t$. Then each element is measurable. But for any $x \geq 0$ the supremum does not exist.
@WillieWong: Well, depending on what exactly the OP is looking for, it might be alright to consider functions that take the value $\infty$. But, of course, this is one more reason why more details would be very helpful.
I have given more details.
Thank you for the clarification! So just to make sure that I understand correctly: are you really mostly interesting in measurability of the pointwise supremum? Or are you rather interested in the existence of a vector $0 \le h \in L^p$ that dominates the modulus of each $f_t$? I'm asking because if such an $h$ exists, then the equivalence classes of the $f_t$ always have a supremum in $L^p$ (which might not be a pointwise supremum, though).
If you let $X$ be the circle, and let $f(\theta) = 1/|\theta|^\epsilon$ for sufficiently small $\epsilon$ and $\theta\in (-\pi,\pi)\setminus 0$ and define $f = 0$ on the two omitted points, you can take $G$ to be the circle group $U(1)$ acting on the circle by translations. You can let $\mu_t$ be the Dirac measure centered at $t$ mod $2\pi$. This I think satisfies all your conditions and recovers essentially the example in my previous comment.
|
2025-03-21T14:48:30.235129
| 2020-04-07T10:18:53 |
356821
|
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"Stanley Yao Xiao",
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|
Stack Exchange
|
$n$-variable polynomials modulo $p$
The Hasse-Weil bound implies that for any 2-variable polynomial $P(x,y)$, there exists approximately $p$ solutions in $\mathbb{F}_p$ of $P(x,y) \equiv a \pmod p$ for sufficiently large $p$, and any integer $a$.
The Chevalley Theorem gives a sufficient condition for a homogeneous $n$-variable polynomial to have nontrivial roots in $\mathbb{F}_p$.
Are there any stronger or more general results about $n$-variable polynomials in $\mathbb{F}_p $ ?
For example, is an approximation of the numbers of solutions in $\mathbb{F}_p$ for $f(x_1 , x_2 , \cdots , x_k )=a$ known, or are there any any similar results?
Lang-Weil bounds
|
2025-03-21T14:48:30.235207
| 2020-04-07T10:48:05 |
356823
|
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"url": "https://mathoverflow.net/questions/356823"
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|
Stack Exchange
|
Equation of the Chebyshev $\psi$ function
Consider $\Psi(x)$ to be the Chebyshev function given by
$$\Psi(x)=\sum_{n\leq x} \Lambda(n)$$
where $\Lambda(n)$ is the Mangoldt function which is equal 0 unless $n $ is prime power, and let $(E)$ be the following equation:
$$\Psi(n!)=\Psi(A) + \Psi(A+2) \tag E \, ,$$
where $n , A $ are integers and $A$ is even.
Can $(E)$ have integer solution or not? Can we relate it with other conjectures or open problems in number theory?
Taking exponentials, we get $l(n!)=l(A)l(A+2)$, where $l(x)$ is the least common multiple of all natural numbers below $x$. We have $A\approx n!/2$, so there will be a prime $p$ between $A+2$ and $n!$ by Bertrand for large $n$. But then $p\mid l(n!)$ and $p\nmid l(A)l(A+2)$. If anyone feels like writing up the exact bounds, feel free to turn this into an answer.
Exact bound of what can you explain more
Exact bound on how large $n$ has to be so that $n! > 2(A+2)$, so that Bertrans can be applied.
So if we find this bound that mean when take a n and A greater than this bound the equation E does not has integers solution
This solution builds on @Wojowu's comment. The Mangoldt function $\Lambda(n)$ is defined as $\log p$ if $n=p^k$ is a prime power and as zero otherwise. The Chebyshev function
$$
\Psi(x)=\sum_{n\leq x}\Lambda(n)
$$
is thus the logarithm of the least common multiple of $1,2,\dots,\lfloor x\rfloor$ because it could be written as $\sum_{p\leq x\text{ prime}}\lfloor\log_{p}x\rfloor\log p$. Its exponential $l:={\rm{e}}^{\Psi}$ is given by
$l(x)=\log\big({\rm{lcm}}(1,\dots,\lfloor x\rfloor)\big)$.
I consider the more general equation
$$\Psi(m)=\Psi(A) + \Psi(A+2) \tag {E1},$$
where $A$ and $m$ are positive integers. Exponentiating, one needs to solve $$l(m)=l(A)\,l(A+2)\tag {E2}.$$
If $A+2\leq\frac{m}{2}$, there is no solution: By Bertrand's Postulate (a.k.a Chebyshev's Theorem) there is a prime $p$ with $\frac{m}{2}<p<m$ unless $m\leq 2$. Such a prime divides the LHS of (E2) but neither of $l(A)$ and $l(A+2)$ on the RHS since $p>A+2$. It is easy to directly check that there is no solution with $m=1$.
So suppose $A+2>\frac{m}{2}$. The exponent of $2$ in the prime factorization of $l(m)$ is $\lfloor\log_2 m\rfloor$ while the same numbers for $l(A)$ and $l(A+2)$ are given by $\lfloor\log_2 A\rfloor$ and $\lfloor\log_2 (A+2)\rfloor$ respectively. Therefore, (E2) implies
$$\lfloor\log_2 m\rfloor=\lfloor\log_2 A\rfloor+\lfloor\log_2 (A+2)\rfloor.$$
But, in view of $A+2>\frac{m}{2}$:
$$
\lfloor\log_2 (A+2)\rfloor\geq \left\lfloor\log_2 \frac{m}{2}\right\rfloor=
\lfloor\log_2 m\rfloor-1;
$$
which requires $\lfloor\log_2 A\rfloor$ to be not greater than $1$. Hence the only choices for $A$ are $1,2,3$. Directly checking them, we observe that the only solutions to (E1) is $A=1,m=3$.
Added: Changing $A+2$ to $B$ in (E1), the same idea could be used to study the equation
$$
\Psi(m)=\Psi(A) + \Psi(B)
$$
where $A\leq B$. Again, $B>\frac{m}{2}$ (unless $A=B=m=1$) which implies
$\lfloor\log_2 A\rfloor\leq 1$. So $A\in\{1,2,3\}$.
If $A=1$, then $\Psi(A)=0$ and there are infinitely many solutions, e.g. by setting $m=B$.
If $A=2$, then $\Psi(A)=\log 2$. The equation $\Psi(m)=\Psi(A)+\log 2$ admits infinitely many solutions; for instance, $m=2^k, A=2^k-1$ yields a solution since $${\rm{lcm}}(1,\dots,2^k)=2\times{\rm{lcm}}(1,\dots,2^k-1).$$
There exist solutions when $A=3$. For instance:
$$\Psi(4)=\log 12=\log 2+\log 6=\Psi(2)+\Psi(3);$$
$$\Psi(10)=\Psi(9)=\log 2520=\log 420+\log 6=\Psi(7)+\Psi(3).$$
In general, solutions $(A=3,B,m)$ could be characterized with $B>\frac{m}{2}$ and the property that there should be exactly two prime powers in $\{B+1,\dots,m\}$, one in the form of $3^j$ and the other in the form of $2^k$.
Very nice! I was thinking of a size argument, using some form of PNT, but your argument using the $2$-adic valuation is definitely nicer (and more elementary)
@Wojowu Thanks! The key idea was yours.
If p between $m$ and $2m$ how do you garante that p divide $m$
@Abdallahchaibeddrraa I am talking about the divisibility of $l(m)$ by $p$; recall that it is the l.c.m of integers $1,\dots,m$.
|
2025-03-21T14:48:30.235484
| 2020-04-07T11:36:28 |
356827
|
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|
Stack Exchange
|
Geometric theory for cohomology groups $H^p(M;\mathbb{Z})$
An excerpt from the book Loop Spaces, Characteristic Classes and Geometric Quantization by Jean-Luc Brylinski is mentioned below:
Characteristic classes are certain cohomology classes associated
either to a vector bundle or a principal bundle over a manifold. For
instance, there are the Chern classes $c_i(E)$ of a complex vector
bundle $E$ over a manifold $M$. These Chern classes can be described
concretely in two ways. In the de Rham theory, a connection on the
bundle is used to obtain an explicit differential form representing
the cohomology class. In singular cohomology, the Chern class is the
obstruction to finding a certain number of linearly independent
sections of the vector bundle; this is Pontryagin's original method.
The relation between the two approaches is rather indirect and in some
ways still mysterious. A better understanding requires a geometric
theory of cohomology groups $H^p(M, \mathbb{Z})$ for all $p$.
Question :
How does geometric theory (Geometric quantization) of cohomology groups $H^p(M;\mathbb{Z})$ helps in better understanding of the Chern-Weil theory approach and Singular cohomology approach (obstruction version) of characteristic classes?
What more clarity are we expecting between these two versions of Chern classes?
Did you try to read about Cheeger-Simons characters? In their original paper, Cheeger and Simons provide a refinement of both descriptions of Chern classes at once.
@SebastianGoette Hi.. Just to confirm, are you referring to http://numr.wdfiles.com/local--files/differential-cohomology/cheeger-simons.pdf? I have not read that, I want to read that... Can you please say in 1/2 lines in your view what refinement they are mentioning...
Yes, that's the paper. "Differential characters" are a simultaneous refinement of (say) integral cohomology and de Rham forms (cf. Deligne cohomology in arithmetic geometry). Characteristic classes of vector bundles with connections can be refined to differential characters. Easiest example: the first Chern class of a complex line bundle with connection on $S^1$ as a differential character is equivalent to $\frac\log{2\pi i}\in\mathbb{C/Z}$ of the holonomy of the connection. Note that both the integral and the de Rham version vanish for degree reasons, so the refinement is nontrivial.
@SebastianGoette Ok. I will read and try to realte to the above mentioned question :) Thank you :)
And I wondered how geometric quantisation enters the picture.
@SebastianGoette That is my usage. The book does not use that phrase at that point.. suppose I have a differential $2$-form $K$ on a manifold $M$, then, I can ask if this $2$-form is curvature of some connection on some line bundle over $M$.. This they called "quantization condition" in the book.. Is it not suitable word here?
I see. I thought you wanted to quantise $H^p(M;\mathbb Z)$.
@SebastianGoette I mean $H^2$ is a first step..the authors hoping for quantization of $H^p(M,\mathbb{Z})$... I hope I conveyed it correctly...
|
2025-03-21T14:48:30.235710
| 2020-04-07T11:42:57 |
356828
|
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|
Stack Exchange
|
Positivity of quadratic form minus linear form on the simplex
Let $a_{ij}$ be the elements of a $n$-dimensional covariance matrix. Can we prove the following?
$$ 1-\sum_{k=1}^n a_{ik} \lambda_k + \sum_{j=1}^n \sum_{k=1}^n \lambda_j a_{jk} \lambda_k > 0, \qquad i \in \{1, \dots, n\} $$
where the $\lambda_k$ are constrained by
$$ \sum_{k=1}^n \lambda_k = 1, \quad \lambda_i > 0 \text { for } i \in \{1, \dots, n\}? $$
Or, in a more general way, what are the conditions that the elements of the covariance matrix should satisfy so that the above set of inequalities hold?
NOTE: in matrix form, if $A$ is a covariance matrix and $a_i$ is a row vector having the $i$-th row of $A$, the question is:
is $1-a_i \lambda + \lambda^T A \lambda >0 $ where $\lambda=[\lambda_1 \ldots \lambda_n]^T$ and $\lambda^T e_n =1$ with $\lambda_k >0$ and $e_n$ a column vector of ones?
The question should be formulated as: find the conditions for the elements of matrix $A$ so that the inequality holds for $i=1,\dotsc, n$.
I have formulated it in matrix form. thank you for your interest Rodrigo, muito obrigado !
It would be very long to explain where this problem comes from. It arises in the context of RUM (Random Utility Maximization) model for choice modeling.
Are you acquainted with the spectrahedron?
Sorry for the delay in my answer, I have read the definiton of spectrahedron, but how can this help me?
Here is a sufficient condition. The function $f(\lambda) = 1-a_i \lambda + \lambda^T A \lambda >0$ restricted on the simplex can be written, with a change of coordinates $\lambda\rightarrow x$ given by $\lambda_i=x_i^2$, as:
$$
f(\lambda) = f(x) = \left(\sum_k x_i^2\right)^2 - \left(\sum_k a_{ik}x^2_k\right)\left(\sum_i x_i^2\right) + \sum_{i,j}A_{i,j}x_i^2x_j^2.
$$
Now, if $f(\lambda)\geq 0$ on the simplex, then this implies that $f(x)$ must be positive semidefinite on the unit sphere. Now, a homogeneous quartic PSD on the unit sphere is globally PSD. A sufficient condition for a quartic to be globally PSD is to choose a sum of squares representation, which must have a form $zPz^\top$ with $z=[x_1^2~\cdots~x_n^2]$ and $P\succeq 0$. This translates to a linear matrix inequality on elements of $A$, which is the final sufficient condition.
|
2025-03-21T14:48:30.235869
| 2020-04-07T12:00:24 |
356829
|
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|
Stack Exchange
|
Tensor algebras in the bicategory $\mathsf{2Vect}$
To my knowledge there are two main approaches to categorify the notion of a vector space. I will refer to them as BC-2-vector spaces (Baez, Crans) and KV-2-vector spaces (Kapranov, Voevodsky). Both define a symmetric monoidal bicategory (denoted by $\mathsf{2Vect}_{BC}$ and $\mathsf{2Vect}_{KV}$, resp.) and in both we can take direct sums, but a significant difference between them is that we don't have an additive inverse functor $-:V \to V$ for a KV-2-vector space, while for BC-2-vector spaces we do.
In particular, this means that for a KV-2-vs $V$, we don't have an antipode on the tensor algebra (or whatever the correct 2-categorical name is) $TV=\boxplus_{n \geq 0} V^{\boxtimes n}$. For me this is a problem, since I would like to categorify the notion of a Nichols algebra.
On the other hand, for an abelian group $G$ in the KV-picture we can easily categorify the category $\mathsf{Vect}_G$ of $G$-graded vector spaces to a bicategory $\mathsf{2Vect}_G=\oplus_{g \in G} \mathsf{2Vect}_{KV}$ and just as MacLane's third abelian cohomology classifies braided/symmetric monoidal structures on $\mathsf{Vect}_G$, the fourth abelian cohomology classifies braided/symmetric/sylleptic monoidal structures on $\mathsf{2Vect}_G$. This is now again very helpful, and actually the reason why I prefer the KV-picture.
So my question is: Is there another way to define a tensor algebra of a 2-vs $V$ in $\mathsf{2Vect}_{KV}$, so that it will be a Hopf algebra in the usual sense (of course, associativity up to bla, etc), or is there a different notion of (higher) Hopf algebras in categories like $\mathsf{2Vect}_{KV}$ which is more suitable in this context? Or both?
|
2025-03-21T14:48:30.236008
| 2020-04-07T12:15:54 |
356830
|
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|
Stack Exchange
|
An argument involving integratibility of a research paper of Rivoal
I am self studying a research paper in analytic number theory (Ball and Rivoal - Irrationalité d’une infinité de valeurs de la fonction zêta aux entiers impairs) and I am unable to think about an argument in 1 of it's lemma.
The paper is in French but I am adding images of formulas references only. Rest of text which could be useful I will write. (For more details see 7th page, logical p. 199, of research paper.)
Assume $\tilde S(z)$ to be RHS of equation 8.
Assume that it has been proved $\tilde S(z)$ is continuous on $E$ and it has also been proved that $S_n(z)$ is continuous on $E$ where $E =\{z \in \mathbb C : \lvert z\rvert \geq1\}$.
Also, $a$, $b$, $n$ are integers and $1\leq b \leq a$ and $1 \leq r < a/2$.
I am not able to think why they must be equal on the given domain. On one side is an infinite series and on other side an integral with some other terms.
Can someone please help, I have no clue regarding this and no further explanation has been given in the research paper.
I did some light editing. You mention supposing that $S_n'(z)$ is equal to the right-hand side of (8), but (8) is an equation involving $S_n(z)$, not $S_n'(z)$. Is that really what you meant?
@Lspice I wrote Tilde (~) over S(z) as S'(z) as I didn't knew how to write ~ over S(z) in math Jax.
Same as in TeX: $\tilde S(z)$ \tilde S(z).
@Gaussian has already answered it.
This question already has an accepted answer; why have you bumped it with an edit?
Since I can't make comment, let me write this as an answer. You know that the LHS and the RHS are continuous on $|z|\geq1$, so to prove that they are equal on $E$, it is enough to show that they are equal on $|z|>1$.
Actually, the condition $|z|>1$ helps you writing the integrand of $I_n(z)$ as a power series, the variable of which is $z$: then, you will have to check that you can switch the integral and the sum... and you'll have to check that this is equal to the power series defining $S_n(z)$.
can you please write a detailed answer on how it will be actually done!!
@Dxdxdade: I don't think MO is meant for people to do your work.
@DamienC , it ok I was already thinking of trying it myself once again and then if I have any questions I will ask to Gaussian.
@Gaussian , why condition |z|>1 allows to write integrand of $I_n(z) $ as power series?
|
2025-03-21T14:48:30.236451
| 2020-04-07T12:34:02 |
356831
|
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|
Stack Exchange
|
Proving the Immersion part of an Embedding
Trying to see the proof of embedding the Jacobian of a Compact Riemann Surface $X$ using Theta functions. So, using the Theta divisor we have the corresponding line bundle say $L$, we want to prove that the map $\iota_{L^3}:J(X)\rightarrow \mathbb{C}\mathbb{P}^N$ is an embedding. We have dim$H^0(J(X),\mathcal{O}(L))=1,$ the holomomorphic section induced by the Theta function. I am stuck at the immersion part.
To prove that $\iota_{L^3}$ is an immersion, we want to prove that if $J(X)=\mathbb{C}^n/\Lambda,$ and $\pi:\mathbb{C}^n\rightarrow J(X)$ be the quotient map, then given $z\in\mathbb{C}^n,$ we need to prove that $d\iota_{L^3}$ is injective at all $\pi(z).$ Now suppose $\theta_0,\dots,\theta_N$ are the functions whose corresponding sections form a basis of holomorphic sections of $L^3.$ Say we want to check the immersion condition on a chart where the first coordinate of $\mathbb{C}\mathbb{P}^N$ is nonzero. So, w.r.t the charts this means we need to check that the function $\mathbb{C}^n\rightarrow \mathbb{C}^N,z\mapsto (\frac{\theta_1}{\theta_0},\dots,\frac{\theta_N}{\theta_0})$ has full rank Jacobian at all points and similarly on other charts. The book (Griffiths and Harris) says that this condition is equivalent to say that the matrix
\begin{bmatrix}
\theta_0(z)&\dots&\theta_N(z)\\ \frac{\partial\theta_0}{\partial z_1}(z)&\dots&\frac{\partial\theta_N}{\partial z_1}(z)\\ \vdots& &\vdots\\ \frac{\partial\theta_0}{\partial z_n}(z)&\dots&\frac{\partial\theta_N}{\partial z_n}(z)
\end{bmatrix} has rank $n+1$. I can't really understand this equivalence condition. Can someone please help me?
Hint: $\theta _0^{n+1}d\left(\dfrac{\theta _1}{\theta _0}\right)\wedge\ldots \wedge d\left(\dfrac{\theta _n}{\theta _0}\right)=$ $\sum_i (-1)^i \theta _id\theta_0\wedge\ldots \wedge \widehat{d\theta _i}\wedge\ldots \wedge d\theta _n$, where the $\widehat{d\theta _i}$ means that this term is deleted. Note that this is a general fact about maps into projective space, nothing particular to do with Jacobians.
Just apply both sides to $\dfrac{\partial }{\partial z_1}\wedge \ldots \wedge \dfrac{\partial }{\partial z_n}$, this gives the equality of the Griffiths-Harris determinant with the Jacobian determinant of your map.
|
2025-03-21T14:48:30.236730
| 2020-04-07T12:41:01 |
356832
|
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|
Stack Exchange
|
Split cofibrations up to quasi-isomorphism
$R$ a ring $(1\neq 0)$, $\mathbf{Perf}(R)$ is the category of perfect complexes (of right $R$-modules).
Suppose that $A_{\bullet}\rightarrow B_{\bullet}\rightarrow B_{\bullet}/A_{\bullet}$ a short exact sequence in $\mathbf{Perf}(R)$ such that
$A_{\bullet}\rightarrow B_{\bullet}$ is a cofibration in the projective model structure of chain complexes $\mathbf{Ch}_{R}$.
The homology of the chain complexes $A_{\bullet}$, $B_{\bullet}$ and $B_{\bullet}/A_{\bullet}$ is concentrated at level $n$ (the same $n$ for all $A_{\bullet}$, $B_{\bullet}$ and $B_{\bullet}/A_{\bullet}$ complexes).
Here is my question: Is it true that $A_{\bullet}\oplus B_{\bullet}/A_{\bullet} $ is isomorphic to $B_{\bullet}$ in homotopy category $Ho(\mathbf{Ch}_{R})$.
No. You can construct counterexamples by taking projective resolutions of modules in a nonsplit short exact sequence. For example, from the short exact sequence $0\to\mathbb{Z}\stackrel{2}{\to}\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}\to0$ of abelian groups you get
$$\require{AMScd}
\begin{CD}
&&0&&0&&0\\
&&@VVV&@VVV&@VVV\\
0@>>>0@>>>\mathbb{Z}@>\text{id}>>\mathbb{Z}@>>>0\\
&&@VVV&@VV\pmatrix{1\\2}V&@VV\times2V\\
0@>>>\mathbb{Z}@>>>\mathbb{Z}\oplus\mathbb{Z}@>>>\mathbb{Z}@>>>0\\
&&@VVV&@VVV&@VVV\\
&&0&&0&&0\\
\end{CD}$$
|
2025-03-21T14:48:30.236829
| 2020-04-04T09:06:05 |
356543
|
{
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"Andreas Blass",
"Frederik Ravn Klausen",
"Nik Weaver",
"https://mathoverflow.net/users/143779",
"https://mathoverflow.net/users/23141",
"https://mathoverflow.net/users/6794"
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"url": "https://mathoverflow.net/questions/356543"
}
|
Stack Exchange
|
Is this subspace of $B(\mathcal{H})$ known?
Let $\mathcal{H}$ be a Hilbert space. Suppose that I take a fixed ONB of $\mathcal{H}$ let us call it $\{ e_i \}_{i\in \mathbb{N}}$ and then I define
\begin{align*}
\|T \|_{D} = \sup_{l_i, m_i} \sum_{i=1}^\infty \vert \langle e_{l_i}, T e_{m_i} \rangle \vert
\end{align*}
where the supremum is over all pairs of subsequences of $\mathbb{N}$.
Then look at all $ A \in B(\mathcal{H}) $ such that $\| A \|_{D} < \infty$. On this space it holds that $\| \cdot \|_{D} $ is a norm and that $ \| A \|_{D} \leq \| A \|_{1}$. By considering the matrices that have a small number in the first many element of first row in our fixed basis we can see that the norms are not equivalent.
Does this space have another characterization?
For "subsequences of $\mathbb N$" to make sense, the set $\mathbb N$ must be considered as a sequence, presumably $(0,1,2,\dots)$. So in a subsequence like $(l_i)$ the terms are in strictly increasing order. Or should we understand $\mathbb N$ as just a set and "subsequence of $\mathbb N$" as just a sequence of whose terms are in $\mathbb N$, in which case there could be repetitions? Or do you allow non-monotone sequences but prohibit repetitions?
Good point. I allow non-monotone sequences but prohibit repetitions.
It looks like a pretty crazy condition ... I'd be surprised if there was any nicer characterization.
Here is some context. The tracenorm I can estimate in the same way but where I take the supremum over all ONB. Here I am fixed in the computational basis. For normal operators the norms will be the same (namely the trace), but for non-normal operators this norm is sometimes (a lot) smaller.
|
2025-03-21T14:48:30.236977
| 2020-04-04T09:17:29 |
356544
|
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|
Stack Exchange
|
Algebra of regular functions on the quadratic cone and SU(2) representations
I was reading the paper "Short Star-Products for Filtered Quantizations" by Pavel Etingof and Douglas Stryker (MSN), where in the introduction they claim that the algebra of regular functions on the quadratic cone $X$ in $\mathbb{C}^3$ is
$$\mathbb{C}[x,y,z]/(xy-z^2)=\bigoplus_{m\geq 0}{V_{2m}}$$
where $V_m$ is the irreducible representation of $\operatorname{SU}(2)$ with highest weight $m$.
Why is that true?
The ring $\Bbb C[x,y,z]/(xy-z^2)$ is $\Bbb N$-graded because $xy-z^2$ is homogeneous. Its $m$-th component has dimension $2m+1$, because a basis is given by $\{ x^ay^b \mid a+b=m \} \cup \{ x^ay^bz \mid a+b=m-1 \}$.
It's easy to see that the representation is irreducible (I assume the action is the adjoint action of $SU(2)$ on $\mathfrak{su}(2) \otimes \Bbb C \cong \Bbb C^3$) giving the results since $V_{2m}$ is the unique irreducible representation of $SU(2)$ with dimension $2m+1$.
The quadratic cone is the quotient of $\mathbb{A}^2 = \mathrm{Spec}(\mathbb{C}[u,v])$ by the involution $\iota \colon (u,v) \mapsto (-u,-v)$. Consequently,
$$
\Gamma(\mathbb{A}^2/\iota, \mathcal{O}) =
\Gamma(\mathbb{A}^2, \mathcal{O})^{\iota} =
\left( \bigoplus V_n \right)^\iota
= \bigoplus V_{2m}
$$
since $\iota$ acts on $V_n$ by $(-1)^n$.
Apply the Borel-Weil theorem to the Lie group $G={\rm SL}_2({\mathbb C})$, with Borel $B$, the complexification of ${\rm SU}(2)$. The flag variety $G/B={\mathbb P}^1$, and we have $V_m=\Gamma({\mathbb P}^1, {\mathcal O}_{{\mathbb P}^1}(m))$. On the other hand, $X$ is the cone over ${\mathbb P}^1\subset {\mathbb P}^2$ embedded by ${\mathcal O}_{{\mathbb P}^1}(2)$. So
$\Gamma(X,{\mathcal O}_X)=\bigoplus \Gamma({\mathbb P}^1, {\mathcal O}_{{\mathbb P}^1}(2m))=\bigoplus V_{2m}$.
This is close to Balazs's answer. $X$ is the nilpotent cone for $SL(2)$. The Springer resolution $T^* \mathbb P^1 \to X$ is semismall. $T^* \mathbb P^1$ is the total space of the line bundle $O(-2)$ on $\mathbb P^1$. So global functions on $X$ are the same as global functions on the total space of the bundle $O(-2)$ on $\mathbb P^1$ whose structure sheaf is ${\rm Sym}^* O(-2)^\vee$. See this and this questions.
|
2025-03-21T14:48:30.237142
| 2020-04-04T09:35:56 |
356545
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Alexey Ustinov",
"Henri Cohen",
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"https://mathoverflow.net/users/81776",
"kodlu"
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"site": "mathoverflow.net",
"sort": "votes",
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}
|
Stack Exchange
|
Estimating the sum $\sum_{1\leq x,y\leq n} \frac{x}{ \mathrm{lcm}(x,y)}$
I am looking for approximations, or a closed form, if available, for the sum
$$S(n,a,b)=\sum_{1\leq x,y,\leq n} \frac{x^a}{\mathrm{lcm}(x,y)^{b}},$$
where $\mathrm{lcm}(x,y)$ is the least common multiple of integers $x,y$ and $a,b$ are positive quantities. I am in particular interested in $a=b=1.$ For this case numerical evidence suggests
$$
S(n,1,1)=O( n \log n)
$$
may hold. In particular, I am wondering whether by using the technique in the answer to this question here, one might obtain (as $n \rightarrow \infty$), by letting $a,b\downarrow 1,$ an estimate in terms of zeta functions. In that question the upper bound
$$
S(n,0,b)\leq\frac{\zeta(b)^3}{\zeta(2b)},\quad b>1
$$
is derived by letting $n\rightarrow \infty.$
Any pointers,comments welcome.
The original sum can be written as
$$T(\alpha,\beta,\gamma,n)=\sum_{x,y\le n}x^\alpha y^\beta(x,y)^\gamma,$$
where $(x,y)=\mathrm{gcd}(x,y)$. One can find asymptotic formula for this sum using standart approach. Let $d=(x,y)$. Then
$$T(\alpha,\beta,\gamma,n)=\sum_{d\le n}d^\gamma\sum_{{x,y\le n\atop (x,y)=d}}x^\alpha y^\beta=\sum_{d\le n}d^{\alpha+\beta+\gamma}\sum_{{x,y\le n/d\atop (x,y)=1}}x^\alpha y^\beta.$$
The condition $(x,y)=1$ can be removed using Möbius function:
$$T(\alpha,\beta,\gamma,n)=\sum_{d\le n}d^{\alpha+\beta+\gamma}\sum_{\delta\le n/d}\mu(\delta)\sum_{{x,y\le n/d\atop \delta\mid(x,y)}}x^\alpha y^\beta=\sum_{d\le n}d^{\alpha+\beta+\gamma}\sum_{\delta\le n/d}\mu(\delta)\delta^{\alpha+\beta}\sum_{x,y\le n/(d\delta)}x^\alpha y^\beta.$$
The last sum (for $\alpha,\beta>-1$) is $\sim\frac{n^{\alpha+\beta+2}}{(\alpha+1)(\beta+1)(d\delta)^{\alpha+\beta+2}},$ so (for $\gamma>1$)
$$T(\alpha,\beta,\gamma,n)\sim \frac{n^{\alpha+\beta+\gamma+1}}{\zeta(2)(\alpha+1)(\beta+1)}.$$
The special case $\gamma=1$, $\alpha=0$, $\beta=−1$ is more tricky. We can write the given sum as
$$T(n)=\sum_{x,y\le n}\frac{(x,y)}{y}=T_1(n)+T_2(n),$$
where for some $U>1$
$$T_1(n)=\sum_{{x,y\le n \atop (x,y)\le U}}\frac{(x,y)}{y},\quad T_2(n)=\sum_{{x,y\le n \atop (x,y)> U}}\frac{(x,y)}{y}.$$ The second sum will be in the error term ($y=dy_1$, $x=dx_1$):
$$T_2(n)=\sum_{d> U}d\sum_{{x,y\le n \atop (x,y)=d}}\frac{1}{y}\ll \sum_{d> U}\sum_{x_1,y_1\le n/d }\frac{1}{y_1}\ll \sum_{d> U}\frac{n}{d}\log\frac{n}{d}\ll n\log^2 \frac{n}{U}.$$
Here it is clear that for $U=n\log^{-2}n$ we get error term $O(R(n))$ with $R(n)=n\log^2\log n.$
The first sum gives main term:
$$T_1(n)=\sum_{y\le n}\frac{1}{y}\sum_{{d\le U \atop d\mid y}}d\sum_{{x\le n \atop (x,y)=d}}1=\sum_{y\le n}\frac{1}{y}\sum_{{d\le U \atop d\mid y}}d\sum_{{x_1\le n/d \atop (x_1,y/d)=1}}1.$$ The last sum is known:
$$\sum_{{x_1\le n/d \atop (x_1,y/d)=1}}1=\frac{\varphi(y/d)}{y/d}\frac nd+O(\tau(y/d)).$$
Hence
$$T_1(n)=n\sum_{y\le n}\frac{1}{y}\sum_{{d\le U \atop d\mid y}}\frac{\varphi(y/d)}{y/d}+O(R_1(n)),$$
where
$$R_1(n)=\sum_{d\le U}d\sum_{{y\le n \atop d\mid y}}\frac{\tau(y/d)}{y}=\sum_{d\le U}\sum_{y_1\le n/d }\frac{\tau(y_1)}{y_1}\ll U\log^2n\ll R(n).$$
So $$T_1(n)=n\sum_{d\le U}\frac{1}{d}\sum_{y_1\le n/d}\frac{\varphi(y_1)}{y_1^2}+O(R(n))=n\sum_{d\le U}\frac{1}{d}\left(\frac{1}{\zeta(2)}\left(\log (n/d)+\gamma-\frac{\zeta'(2)}{\zeta(2)}\right)+O\left(\frac{\log n}{n/d}\right)\right)+O(R(n))=\frac{n}{\zeta(2)}\sum_{d\le U}\frac{1}{d}\left(\log (n/d)+\gamma-\frac{\zeta'(2)}{\zeta(2)}\right)+O(R(n))=\frac{n}{\zeta(2)}\left(\left(\log n+\gamma-\frac{\zeta'(2)}{\zeta(2)}\right)\left(\log U+\gamma+O(1/U)\right)-\sum_{d\le U}\frac{\log d}{d}\right)+O(R(n)).$$
We also know that
$$\sum_{d\le U}\frac{\log d}{d}=\frac{\log^2 U}{2}+\gamma_1+O(U^{-1}\log U).$$
Collecting all together we'll have
$$T(n)=\frac{n}{\zeta(2)}\left(\frac{\log^2n}{2}+\log n\left(2\gamma-\frac{\zeta'(2)}{\zeta(2)}\right)\right)+O(n\log^2\log n).$$
Thanks. Your answer does not have a $\log n$ term, unlike Henri Cohen’s estimate, however, so the two seem inconsistent, if I let $LCM(x,y)=xy/(x,y)$.
@kodlu My answer will have another form if $\alpha=-1$ or $\beta=-1$ or $\alpha<-1$,...
Great! Since I'm specifically interested in sum of $x/LCM(x,y)=GCD(x,y)/y,$ that case would be $\gamma=1,$ $\alpha=0,\beta=-1.$ When you have a chance, can you address that case please?
Partial answer: elementary arithmetic transformations show that
$$S(n,1,1)=\sum_{1\le y\le n}\dfrac{1}{y}\sum_{d\mid y}\phi(d)\lfloor n/d\rfloor$$
which allows for much faster computation since it is essentially a single sum.
I didn't push the analysis further, but my guess is that $S(n,1,1)$ is
asymptotic to $Cn\log(n)^2$ (with a log squared), perhaps with $C=3/\pi^2=1/(2\zeta(2))$.
Complete answer: I was really lazy. From the expression above, it is immediate to show that $$S(n,1,1)=\sum_{1\le d\le n}\dfrac{\phi(d)}{d}\log(n/d)+O(n\log(n))\;,$$
and the main term is indeed asymptotic to $Cn\log(n)^2$ with $C=3/\pi^2$
if I am not mistaken.
Thanks. In Alexey Ustinov’s answer, there is no $\log n$ term, so it is a bit puzzling to me.
The two results are perfectly consistent (I write $\log(n)^2$ instead of $\log^2(n)$), but Alexey's answer is more precise.
|
2025-03-21T14:48:30.237429
| 2020-04-04T11:26:32 |
356553
|
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"D.-C. Cisinski",
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|
Stack Exchange
|
Is cup product of cycle classes on Noetherian regular excellent scheme compatible with intersection
Let $\mathcal{X}$ be a Noetherian regular integral excellent scheme. Let $Y$ and $Z$ be algebraic cycles of codimension $c$ and $d$ on $\mathcal{X}$.
Let $n$ be a positive integer invertible on $\mathcal{X}$. Assume $Y$ intersects $Z$ properly and define $Y\cap Z$ by Serre's Tor formula. Do we have
$$Cl(Y)\smallsmile Cl(Z)= Cl(Y\cap Z)$$
Where $Cl(Y)$ is the cycle class of $Y$ in $H^{2c}_{|Y|}(\mathcal{X},\mathbb{Z}/n(c))$
(By Gabber's purity theorem, it can be defined as chapter 4 2.2.10 in SGA 4 1/2).
By Remark 2.3.9 chapter 4, in Deligne's SGA 4 1/2, the claim holds for smooth varieties over algebraic closed fields. I think by localization, it should work for cycles whose supports are locally complete intersection closed subscheme.
There is no need for excellency nor regularity (if we reformulate this properly). This follows from Section 3 (applied to Example 2.11) in arXiv:1909.01332 for instance.
|
2025-03-21T14:48:30.237521
| 2020-04-04T12:25:08 |
356559
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Ilya Bogdanov",
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"https://mathoverflow.net/users/52871",
"lchen"
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|
Stack Exchange
|
Graph coloring to minimize maximum number of colors along paths
Given a graph $G$ and a pair of source-destination nodes $s$ and $t$. Each node in $G$ is to be colored. Let $C_i$ denote the available color set for node $i$. Under a coloring scheme $A$, for any $s-t$ path $p$, let $\lambda(p)$ denote the number of colors used to color the nodes in $p$ under $A$. We seek a coloring scheme that maximize $\min_{p\in P} \lambda(p)$, where $P$ denotes the set of all $s-t$ paths.
If there is no restriction on a coloring (e.g., shouldn't it be proper?), then the answer is the minimum of the number of colors and the length of the shortest $s-t$ path: just assign a color to each distance from $s$.
Thank you Ilya. Your answer holds if there are sufficient number of colors. I am more interested in the non-trivial case of limited number of colors (i.e., < path length). The coloring does not have to be proper.
in this case my coloring (of the neighborhood of $s$ with corresponding radius) already ensures that each path contains all colors.
Yes Ilya, but that is not necessarily optimal. I suppose the problem is NP-hard, but fail to figure out how to design approximation algorithms.
why it is not optimal? How can the number of cllors in the path exceed the total nimber of colors?
I forgot to put the constraint that each node can only choose a subset of colors. I have revised the problem. Pls check the revised formulation.
|
2025-03-21T14:48:30.237646
| 2020-04-04T12:25:52 |
356560
|
{
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"authors": [
"Jeremy Rickard",
"Mare",
"https://mathoverflow.net/users/22989",
"https://mathoverflow.net/users/61949"
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"url": "https://mathoverflow.net/questions/356560"
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|
Stack Exchange
|
Symmetric stable categories
Let $A$ and $B$ be Frobenius algebras that are stable equivalent.
In case $A$ is symmetric, is $B$ also symmetric? (no, see the comment of Jeremy Rickard) Does it hold in case $A$ and $B$ are quiver algebras over an algebraically closed field?
Can we characterise when the stable module category of a Frobenius algebra is symmetric (meaning coming from a symmetric algebra) using some local information? One idea might be looking when $\tau(M) \cong \Omega^2(M)$ for all indecomposables $M$, but this is not enough.
Maybe you want the ground field to be algebraically closed, but there's a counterexample over a non-algebraically closed field at the end of Ohnuki, Yosuke, Takeda, Kaoru, and Yamagata, Kunio. "Symmetric Hochschild extension algebras." Colloquium Mathematicae 80.2 (1999): 155-174.
@JeremyRickard Thanks, that counts. I added whether it also holds over algebraically closed fields.
|
2025-03-21T14:48:30.237734
| 2020-04-04T12:34:07 |
356561
|
{
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"Dominic van der Zypen",
"Wojowu",
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|
Stack Exchange
|
Minimal covering sets of continuous endomorphisms
For any topological space $(X,\tau)$, let $\text{End}(X)$ denote the set of continuous functions $f:X\to X$. We say that ${\cal C}\subseteq \text{End}(X)$ covers $\text{End}(X)$ if for every $f\in \text{End}(X)$ there is $h\in {\cal C}$ and $x\in X$ such that $f(x) = h(x)$.
I am interested in minimal covers (that is, covers that have the property that when you remove a member, it is no longer a cover). A boring example is the set of all constant functions which always has the same cardinality as $X$ - and taking away one member of the collection of all constant functions destroys the "covering" property.
Note that $\text{End}(\mathbb{R})$ does have a countable cover: For $z\in\mathbb{Z}$ let $f_z:\mathbb{R}\to\mathbb{R}$ be defined by $r \mapsto r+z$. Let $${\cal C} = \{f_z: z\in\mathbb{Z}\}\cup\{{\bf 0}\}, $$ where ${\bf 0}$ denotes the constant zero function. Clearly this cover does not have a minimal subcover.
Question. Does ${\mathbb R}$ have a countable minimal cover?
Your "countable cover" is not a cover - it doesn't cover $r\mapsto r+1/2$ for instance. A valid example is given by functions $f_z,z\in\mathbb Z$ given by $f_z(r)=zr$.
Yeah sorry I forgot to add the constant zero function. I have added this now in the question. With this I am quite certain it is a countable cover
Here is an example:
Example. Let $g$ be a continuous strictly increasing function such that $\lim_{x \to -\infty} g(x) = -1$ and $\lim_{x \to \infty} g(x) = 1$; for example
$$g(x) = \tfrac{2}{\pi}\arctan(x).$$
For $n \in \mathbf Z$, let $f_n(x) = n$ and $g_n(x) = g(x) + n + \tfrac{1}{2}$. Then $\{f_n\} \cup \{g_n\}$ form a covering: if $h$ is continuous and does not intersect any $f_n$, then there exists $n \in \mathbf Z$ such that $n < h(x) < n+1$ for all $x$, so $h$ intersects $g_n$ as $\lim_{x \to -\infty} g_n(x) < n$ and $\lim_{x \to \infty} g_n(x) > n + 1$.
On the other hand, if we remove any $g_n$, then the constant function $h(x) = n + \tfrac{1}{2}$ lies strictly between $f_n$ and $f_{n+1}$ and strictly between $g_{n-1}$ and $g_{n+1}$, so does not intersect any member. Similarly, if we remove $f_n$, then the function $h(x) = g(x) + n = g_n(x) - \tfrac{1}{2}$ lies strictly between $f_{n-1}$ and $f_{n+1}$ and strictly between $g_{n-1}$ and $g_n$, so does not intersect any member. $\square$
Remark. There are no finite covers: if $\{f_1,\ldots,f_n\}$ is a finite collection of continuous functions, then the function
$$g = \max(f_1,\ldots,f_n) + 1$$
is continuous (since $\max \colon \mathbf R^n \to \mathbf R$ is continuous) and does not intersect any of the $f_i$.
|
2025-03-21T14:48:30.237906
| 2020-04-04T13:50:34 |
356563
|
{
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"Anonymous amateur",
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|
Stack Exchange
|
Continuity of $r\mapsto\int_{\Sigma\cap B_r(x)}f^2d\mu$
Let $\Sigma$ be an embedded smooth surface in $\mathbb{R}^3$, and let $f:\Sigma\to\mathbb{R}$ be a smooth function. Suppose $f$ is square-integrable on $\Sigma$, with
\begin{align}
0<\int_{\Sigma}f^2d\mu<\infty
\end{align}
where $d\mu$ is the area element of $g$, the induced metric on $\Sigma$ from the flat metric of $\mathbb{R}^3$.
Denote the Euclidean open ball of radius $r>0$ with centre $x\in\mathbb{R}^3$ by $B_r(x)$. In this question I'm interested in
the continuity of the function $r\mapsto\displaystyle\int_{\Sigma\cap B_r(x)}f^2d\mu$
At first this seems to be continuous for any $x\in\mathbb{R}^3$. However, a simple counterexample can be found: If $\Sigma$ is a round sphere of radius $R>0$ with centre $0\in\mathbb{R}^3$, then
\begin{align}
\Sigma\cap B_r(0)=\left\{
\begin{array}{ccl}
\emptyset & \text{if} & r\leq R \\
\Sigma & \text{if} & r>R
\end{array}
\right.
\end{align}
and so we have a discontinuity at $r=R$. More generally, whenever $\Sigma$ has a region which is a spherical cap, we also have such discontinuity for some $x$.
However, in the counterexample above, it seems that the continuity only fails for a single choice of $x$. Moreover, at the discontinuity $r=R$, we still have a left-continuity. I wonder if this is true in general:
Is it true that $r\mapsto\displaystyle\int_{\Sigma\cap B_r(x)}f^2d\mu$ is continuous (or at least left-continuous) for almost every $x\in\mathbb{R}^3$?
If yes, how should we prove it? And if no, what counterexample can we construct? Also, I would be glad to know the best result that we can have in this direction.
Any comment or answer is greatly welcomed and appreciated.
What do you mean by almost every? Any monotonous function has at most countably many discontinuities..
@IlyaBogdanov By "almost every" i mean it in the usual measure-theoretical sense; that is, there exists a set $Z\subseteq\mathbb{R}^3$ of measure zero such that for all $x\in\mathbb{R}^3\setminus Z$, the function I'm interested is continuous.
@IlyaBogdanov I'm aware of the result that monotonous function has at most countably many discontinuities, which is standard in real analysis. However, here my functions have input $r\in(0,\infty)$, while being a family of functions 'parametrized' by $x\in\mathbb{R}^3$. My question is on whether the continuity holds for almost all $x$, not almost all $r$.
Ah, sorry, I misunderstood that. Thanks forthe explanation!
The function $\lambda:A\mapsto\int_{\Sigma}{\mathbf 1}_{A}f^2\mathrm d\mu$ is a measure on the Borel sets of $\mathbb R^3$, for ${\mathbf 1}_A$ the indicator function of $A$. The quantity you are interested in is $r\mapsto\lambda(B_r(x))$.
We can use the monotone convergence theorem to see that
$$ \lim_{r\uparrow r_0}\lambda(B_r(X)) = \lambda(B_{r_0}(x)), $$
so the left continuity is true for every $x$. Using Lebesgue's dominated convergence theorem, we see that in fact
$$ \lim_{r\downarrow r_0}\lambda(B_r(X)) = \lambda\big({\overline B}_{r_0}(x)\big) $$
where ${\overline B}$ denotes the closed ball, so continuity holds at $r_0$ for fixed $x$ whenever $\lambda(\partial B_{r_0}(x))=0$.
I will show that only countably many $(x,r)$ may be such that $\lambda(\partial B_{r}(x))\neq0$; in particular your function will be continuous on $\mathbb R_+$ for all but countably many $x$. I assume the surface is closed, but it is not a necessary hypothesis, as I discuss at the end.
Fix $\varepsilon,R>0$, and let $S = S_{\varepsilon,R}$ be the set of pairs $(x,r)$ such that ${\overline B}_r(x)\subset B_R(0)$ and $\Sigma\cap\partial B_{r}(x)$ has surface area larger than $\varepsilon$ (seen as a subset of $\Sigma$). For a finite collection ${(x_i,r_i)}_{0<i\leq k}$ of elements of $S$, the sum of the areas of $\Sigma\cap\partial B_{r_i}(x_i)$ (which is at least $k\varepsilon$) is the area of $\Sigma\cap\bigcup_i\partial B_{r_i}(x_i)$, because the intersection of two distinct spheres is a circle, a point or empty, hence has measure zero. In particular, the sum is less than the area of ${\overline B}_R(0)\cap\Sigma$, which is finite (because the surface is closed). This means that $S_{\varepsilon,R}$ is in fact finite.
This concludes readily: if $(x,r)$ is a point such that $\lambda(\partial B_r(x))\neq0$, then it belongs to $\bigcup_{n\geq1}S_{1/n,n}$, which is countable as a countable union of finite sets.
If $\Sigma$ is not a closed surface (I imagine this means that it is not embedded, so it goes beyond your question), one may, in the definition of $S_{\varepsilon,n}$, replace $\partial B_n(0)\cap\Sigma$ by a countable collection of compacts subsets $K_n$ of $\Sigma$ whose interiors increase to $\Sigma$. Then if $(x,r)$ is a point such that the area of $\partial B_r(x)\cap\Sigma$ is larger than $\varepsilon$, by inner regularity there exists a compact subset $K\subset\Sigma\cap\partial B_r(x)$ with area larger than $\varepsilon$ as well. This compact is included in one of the interiors of the $K_n$, so $(x,r)$ belongs to the corresponding $S_{\varepsilon,n}$.
Thanks so much for the detailed answer. One soft question: Is this particular trick of considering $S_{\epsilon, R}$ (or its variations) commonly used in analysis? I feel that it may be useful in certain context....
I don't know that it's commonly used. If the sum of non-negative terms over an arbitrary set is convergent, then only countably many of them are positive (S = indices of terms $>\varepsilon$); a compact operator has at most countably many eigenvalues, with at most one limit point (S = eigenvalues with modulus $>\varepsilon$); a discrete closed set of $\mathbb R^d$ is countable (S = elements with norm $\leq R$). I guess it is useful enough to be found here and there, but not sophisticated enough to have a name.
|
2025-03-21T14:48:30.238251
| 2020-04-04T14:08:17 |
356564
|
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|
Stack Exchange
|
Minimizing $\int\lambda({\rm d}y)\frac{\left|g(y)-\frac{p(y)}c\lambda g\right|^2}{r((i,x),y)}$ with respect to discrete parameter $i$
Let $I\subseteq\mathbb N$ be finite and nonempty, $(E,\mathcal E,\lambda)$ be a $\sigma$-finite measure space, $$\lambda f:=\int f\:{\rm d}\lambda$$ for $\lambda$-integrable $f:E\to\mathbb R$, $p:E\to(0,\infty)$ be $\mathcal E$-measurable with $c:=\lambda p\in(0,\infty)$, $r:(I\times E)\times E\to[0,\infty)$ be $(2^I\otimes\mathcal E)\otimes\mathcal E$-measurable with $\lambda r((i,x),\;\cdot\;)=1$ for all $(i,x)\in I\times E$ and $g:E\to[0,\infty)$ be $\mathcal E$-measurable.
Given $x\in E$, how can we compute for which $i\in I$ the quantity $$\sigma_i:=\int\lambda({\rm d}y)\frac{\displaystyle\left|g(y)-\frac{p(y)}c\lambda g\right|^2}{r((i,x),y)}$$ is minimal (if there is more than one minimal $i$, I want to pick the smallest)? I'm not interested in the value of $\sigma_i$.
For the moment, I'm estimating each $\sigma_i$ using Monte Carlo integration (sampling from $r((i,x),\;\cdot\;)\lambda$ is possible) and then computing the smallest index $i$ whose corresponding estimate of $\sigma_i$ is minimal. However, this is error prone (since the estimate of $\sigma_i$ might be not close enough to $\sigma_i$) and extremely slow. The latter is my actual problem, since I need to solve this problem for $\approx1e5$ different $g$ inside a loop of range $\ge1e5$.
So, is there a smarter solution to this problem?
EDIT: Feel free to assume that $g$ is of the form $g=1_Hp$, where $H\in\mathcal E$ with $\lambda(H)\in(0,\infty)$. Under this assumption, $$\sigma_i=\left(c-2\lambda(1_Hp)\right)\int_H\mu({\rm d}y)\frac{p(y)}{r((i,x),y)}+\frac{\left|\lambda(1_Hp)\right|^2}c\int\mu({\rm d}y)\frac{p(y)}{r((i,x),y)}\tag1,$$ where $\mu:=c^{-1}p\lambda$. And you can assume that $$c-2\lambda(1_Hp)\ge0\tag2$$.
|
2025-03-21T14:48:30.238389
| 2020-04-04T14:57:38 |
356567
|
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"url": "https://mathoverflow.net/questions/356567"
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|
Stack Exchange
|
Special cases of the embedding problem
Embedding problem. Let $I$ be the ideal of polynomial algebra $A=K^{[n]}$, such that $A/I$ is also a polynomial algebra with smaller number $k$ of variables. Is it true that $I$ is generated by $n-k$ variables of $A$.
Definition. Call an ideal of polynomial algebra coordinate-like if it us generated by some coordinates of a polynomial algebra.
I am interested in the following particular cases of the embedding problem.
Problem 1. Consider any polynomial algebra $A=K^{[n]}$ and its $k$ polynomials $f_1,..., f_k$. For a polynomial algebra $B=K[y_1,..., y_{n+k}]$ consider the morphism sending $ y_i$ to $n$ coordinates of $A$ for $i\leq n$ and $y_j$ to $f_{j-n}$ for $j>n$. It is obvious that thr kernel of this morphism is the ideal $I$ with $B/I\cong A$. So is it coordinate-like?
Problem 2. If $I\subset J$ are two coordinate-like ideals of polynomial algebra $A$, is it true that $J/I$ is coordinate-like ideal of $A/I$?
It is obvious that these problems follow from the embedding problem, but are they true?
First, the embedding problem has a partial affirmative answer if $k\geq n-2$ or if $n\geq 2k+2$. In these cases, $I$ is generated by $n-k$ elements, not necessarily variables.
Problem 1 has an affirmative answer. $I$ is generated by $y_j-f_{j-n}(y_1,\ldots, y_n)$ for $j>n$ and these are coordinates of $B$.
|
2025-03-21T14:48:30.238603
| 2020-04-04T16:38:46 |
356573
|
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"Bazin",
"Mateusz Kwaśnicki",
"https://mathoverflow.net/users/108637",
"https://mathoverflow.net/users/21907"
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"url": "https://mathoverflow.net/questions/356573"
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|
Stack Exchange
|
The solutions of the heat equation from $0$ datum
Consider the initial value problem
$$ \partial_t u = \Delta u$$
$$ u(0,x) = 0$$
for the heat equation in $\mathbb R^n$, where $u: [0,T] \times \mathbb R^n \to \mathbb R$ is a smooth solution up to the time $T$. Suppose there exists a large constant $N>0$ such that
$$
|u(t,x)| \le N\exp(N|x|^2t^{-1})
$$
for any $x\in \mathbb R^n$ and $t\in (0,T]$.
Can we prove that $u(t,x)$ must be identically $0$?
Notice that if $|u(t,x)| \le N\exp(N|x|^2)$, then the question is true by Tychonoff's theorem.
Have you tried the classical Tychonoff's example, $$u(t,x) = \sum_{k = 0}^\infty \frac{g^{(k)}(t) x^{2k}}{(2k)!},$$ with $g(t) = e^{-1/t^\alpha}$ and $\alpha > 1$?
As discussed, for example, in these lecture notes, in this case $$|g^{(k)}(t)| \leqslant \frac{k!}{(\theta t)^k} e^{-1/(2 t^\alpha)} \tag{$\star$}$$ for some $\theta > 0$ (edit: see below for a proof), and hence
$$ |u(t, x)| \leqslant \sum_{k = 0}^\infty \frac{k! x^{2k}}{(\theta t)^k(2k)!} e^{-1 / (2 t^\alpha)} \leqslant \sum_{k = 0}^\infty \frac{x^{2k}}{(2 \theta t)^k k!} e^{-1 / (2 t^\alpha)} = e^{x^2/(2 \theta t)-1/(2t^\alpha)} \leqslant e^{x^2 / (2 \theta t)} . $$
If I am not mistaken, this (counter-)example implies a negative question to your answer.
Edit: For completeness, here is the proof of ($\star$). This is, of course, completely standard; I reworked it only to convince myself that there is no error in my answer.
The function $g(t) = e^{-1/t^\alpha}$ is holomorphic in the right complex half-plane. By Cauchy's formula (applied to the circle $\Gamma$ centred at $t$ with radius $\theta t$), we have
$$ g^{(k)}(t) = \frac{k!}{2\pi i} \int_\Gamma \frac{g(z)}{(z - t)^{k + 1}} dz = \frac{k!}{2\pi} \int_0^{2\pi} \frac{g(t + \theta t e^{i s})}{(\theta t e^{i s})^k} ds . $$
Therefore,
$$ |g^{(k)}(t)| \leqslant \frac{k!}{2\pi (\theta t)^k} \int_0^{2\pi} |g(t + \theta t e^{i s})| ds . $$
Choose $\theta > 0$ small enough, so that the image of the disk $D(1, \theta)$ under $z \mapsto z^{-\alpha}$ is contained in the half-plane $\Re z > \tfrac{1}{2}$. Since $|g(z)| = e^{-\Re z^{-\alpha}}$, we have $$|g(t + \theta t e^{i s})| = e^{-t^{-\alpha} \Re (1 + \theta e^{i s})^{-\alpha}} \leqslant e^{-t^{-\alpha} / 2} .$$
It follows that $|g^{(k)}(t)| \leqslant \frac{k!}{(\theta t)^k} e^{-1 / (2 t^\alpha)}$, as desired.
Kwasnicki For the Tychonoff argument to work, you need $\alpha >1$. In fact, up to geometrical terms $g^{(k)}$ is bounded above by $k^{k(1+\frac{1}{\alpha})}$ and you have in the series a denominator as $k^{2k}$, so you need $1+\frac{1}{\alpha}<2$. A reference is https://doi.org/10.1017/S001309150000095X
@Bazin: Thanks for spotting the typo, and for the reference! Regarding the estimate of $g^{(k)}$, unfortunately I do not have time now to check the bound given in the lecture notes (which is left as a "good exercise").
@Kwasnicki the good exercise is to prove that your function $g$ belongs to the Gevrey class $G^{1+\frac{1}{\alpha}}$ (this is sharp). Checking the Faa di Bruno theorem is one way to get this, but complex analysis with Cauchy theorem would give you the same answer.
@Bazin: Thanks for the comment. I am not sure if I understand correctly, though: you seem to refer to uniform bounds for $g^{(k)}$, while here one needs an estimate that can deteriorate as $t \to 0^+$: a uniform bound for $t^k g^{(k)}(t)$. I believe both estimates follow from Faà di Bruno's formula, but, as said above, I did not attempt to check it. If you did, or if you can improve the answer in any other way, please feel free to edit it.
@Kwasnicki you do not need anything else than $\vert g^{(k)}\vert\le C^{k+1} k^{k(1+\frac{1}{\alpha })}$ with $\alpha >1$ for the Tychonoff example to work: you have applying the heat operator a telescoping series and the convergence follows from the estimate $\vert g^{(k)}\vert\frac{x^{2k}}{(2k)!}\lesssim R^{2k}k^{k(-1+\frac{1}{\alpha})}$.
@Bazin: OK, I understand now: your comments aim at explaining what is needed to check that Tychonoff's example is a solution to the heat equation (as long as $\alpha > 1$), while mine were focused on how one gets the (slightly simpler) bound now tagged ($\star$). I should have been more careful and write "for Tychonoff's example to answer the original question" rather than "here" in my previous comment!
|
2025-03-21T14:48:30.238866
| 2020-04-04T16:57:55 |
356576
|
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|
Stack Exchange
|
A sectionwise fiber sequence is homotopy fiber sequence?
Let $\mathscr{C}$ be a site and $\mathsf{sPre}(\mathscr{C})$ the category of simplicial presheaves on $\mathscr{C}$ equipped with Jardine's local model structure. Let $E\to B$ is a sectionwise Kan fibration and $F\to E\to B$ a sectionwise fiber sequence of simplicial presheaves on $\mathscr{C}$ (i.e. for every $U\in\mathscr{C}$, $F(U)\to E(U)\to B(U)$ is a homotopy fiber sequence of simplicial sets). Then is $F\to E\to B$ always a homotopy fiber sequence in $\mathsf{sPre}(\mathscr{C})$ (with respect to Jardine's local model structure)? This is asking whether the map from $F$ to the homotopy fiber of $E\to B$ is a local weak equivalence.
I am trying to understand some of these same notions: wouldn't a counterexample be provided by any sectionwise fibration that is not a global fibration? So say something that is sectionwise fibrant but not globally fibrant?
|
2025-03-21T14:48:30.238942
| 2020-04-04T16:59:11 |
356577
|
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"Conjecture",
"https://mathoverflow.net/users/101411"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/356577"
}
|
Stack Exchange
|
Connections on vector bundles over elliptic curves - concrete computations
This is linked to my question on math.Stackexchange for which I had no answer.
I have found nowhere a historical account on connections (or rather called logarithmic connections?) that would help me fully grasp this object.(except maybe Katz's paper) that links the notion of connection on a vector bundle to:
solvability of homogeneous systems of first-order differential equations(a question first raised by Grothendieck).
Let $X$ be an elliptic curve and let $(U_i)_i$ be an open covering of $X$.
A connection on a vector bundle $\nabla|_{U_i} = d_{X} + A_i$ where $A_i$ is a $2\times 2$ matrix with entries $1$-form in $\Omega^1_X$.
How can we make sense of $\nabla|_{U_i}$, in other words, how can we add a scalar ($\Omega^1$ is of dimension $1$ as $X$ is smooth) with a matrix?
I would be happy if you have any references on this subject.
Mnay thanks for your help.
Perhaps the following very elementary discussion will help you to understand the relationship between systems of first order ODE's, connections, and vector bundles. If what I write below isn't helpful, I apologize for wasting your time.
Let $u: \mathbb{C}\rightarrow \mathbb{C}^{n}$ be a vector valued holomorphic function. Let $A: \mathbb{C}\rightarrow M_{n}(\mathbb{C})$ be a holomorphic function valued in $n \times n$ matrices.
Consider the differential equation
$\frac{\partial}{\partial z} u(z)= A(z)u(z),$
where the juxtaposition on the right hand side is matrix multiplication.
This is a system of first order holomorphic differential equations. This equation can be solved using fundamental theorems in ODE, but it's not so important for the discussion.
Now, consider the trivial holomorphic vector bundle $\mathbb{C}\times \mathbb{C}^{n}$ over $\mathbb{C}.$ Then, the vector valued holomorphic function $u: \mathbb{C}\rightarrow \mathbb{C}^{n}$ becomes a section of this vector bundle.
Furthermore, the above mentioned system of first order differential equations can be viewed as a (holomorphic) connection $D_{A}$ on this bundle.
To wit, given a (local) holomorphic section $u,$ we declare that
$D_{A}u=\frac{\partial}{\partial z} u(z) \ dz-A(z)u(z)\ dz.$
In particular, $D_{A}u=0$ if and only if it solves the aforementioned ODE.
So, we see that $D_{A}$ sends a section of the trivial bundle to a holomorphic $1$-form with values in sections of the trivial bundle. Checking the Leibniz rule is straightforward, and hence $D_{A}$ is a connection.
In order words, a (holomorphic) connection on the trivial bundle $\mathbb{C}\times \mathbb{C}^{n}$ over $\mathbb{C}$ is just a first order system of homogeneous holomorphic differential equations. The general $1$-dimensional complex manifold is not biholomorphic to $\mathbb{C},$ and therefore the trivial bundle becomes a rank $n$-vector bundle, and the first order system of differential equations becomes a "local system." It's a shame, in my opinion, that this reason for the terminology is not more well known, though it is certainly known to experts.
Deriving the transformation law you state for a connection is now a reasonably simple matter of making sense of how coordinate changes of a nontrivial vector bundle act on the relevant ODE that is specifying a connection.
I admit, I haven't even begun to answer your specific question, but I hope this explanation isn't totally useless.
Thank you @Andy Your explanation makes sense to me and is very helpful
|
2025-03-21T14:48:30.239171
| 2020-04-04T17:45:22 |
356583
|
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"LSpice",
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|
Stack Exchange
|
Help with a definition of a two-person game in a referenced paper
In the paper "Finding Mixed Nash Equilibria of Generative Adversarial Networks" the authors write in equation (1) on page 2:
Consider the classical formulation of a two-player game with
finitely many strategies:
\begin{equation*}
\tag1\label1
\min_{\boldsymbol{p} \in \Delta_m} \max_{\boldsymbol{q} \in \Delta_n} \langle \boldsymbol{q},\boldsymbol{a} \rangle -
\langle \boldsymbol{q},A\boldsymbol{p} \rangle ,
\end{equation*}
where $A$ is a payoff matrix, $\boldsymbol a$ is a vector, and $ \Delta_d :=
\{\boldsymbol{z} \in \mathbb{R}_{\geq 0}^d \mid \sum\nolimits_{i=1}^d z_i = 1\}$ is
the probability simplex, representing the mixed strategies (i.e.,
probability distributions) over $d$ pure strategies. A pair
$(\boldsymbol{p}_{\text{NE}},\boldsymbol{q}_{\text{NE}})$ achieving the min-max
value in (\ref{1}) is called a mixed NE.
I was wondering:
What does this formulation mean?
The formulation seems to result in a parametrized by a vector $\boldsymbol a$ pair of strategies $(\boldsymbol p, \boldsymbol q)$. What is the role of vector $\boldsymbol a$ in the above equation?
Thank you
After further contemplating: I guess they want to align their application (GAN) with the game theory framework. To that end, they write on page 3:
[W]e consider the set of all probability distributions
over $\Theta$ and $\mathcal{W}$, and we search for the optimal distribution that solves the following program:
\begin{equation*}
\tag4\label4
\min_{\nu \in \mathcal{M}(\Theta)} \max_{\mu \in \mathcal{M}(\mathcal{W})}
\mathbb{E}_{\boldsymbol{w} \sim \mu} \mathbb{E}_{X \sim \mathbb{P}_{real}} [f_\boldsymbol{w}(X)] -
\mathbb{E}_{\boldsymbol{w} \sim \mu} \mathbb{E}_{\boldsymbol{\theta} \sim \nu} \mathbb{E}_{X \sim \mathbb{P}_{\boldsymbol{\theta}}} [f_\boldsymbol{w}(X)] .
\end{equation*}
They then show that the above can be cast as
\begin{equation*}
\tag5\label5
\min_{\nu \in \mathcal{M}(\Theta)} \max_{\mu \in \mathcal{M}(\mathcal{W})} \langle \mu,g \rangle -
\langle \mu,G\nu \rangle ,
\end{equation*}
with $g$ defined as $g : \mathcal{W} \rightarrow \mathbb{R}$ by $g(w) := \mathbb{E}_{X \sim \mathbb{P}_{real}} [f_\boldsymbol{w}(X)]$, the operator $G : \mathcal{M}(\Theta) \rightarrow \mathcal{F}(\mathcal{W})$ as $(G\nu)(w) := \mathbb{E}_{\boldsymbol{\theta} \sim \nu} \mathbb{E}_{X \sim \mathbb{P}_{\boldsymbol{\theta}}} [f_\boldsymbol{w}(X)]$ and denoting $\langle \mu,h \rangle := \mathbb{E}_{\mu}h$ for any probability
measure $\mu$ and function $h$ (where $\langle \mu,h \rangle$ is NOT an inner product, but a dual pairing in Banach spaces),
which looks like (\ref{1}) (for finitely many strategies). Notice that (\ref{4}) has a free parameter $\mathbb{P}_{real}$ (hidden in $g$ in (\ref{5})), which $\boldsymbol{a}$ in (\ref{1}) seems to have been introduced to account for.
Also,
\begin{equation*}
\min_{\boldsymbol{p} \in \Delta_m} \max_{\boldsymbol{q} \in \Delta_n} \langle \boldsymbol{q},\boldsymbol{a} \rangle -
\langle \boldsymbol{q},A\boldsymbol{p} \rangle =
\min_{\boldsymbol{p} \in \Delta_m} \max_{\boldsymbol{q} \in \Delta_n} \langle \boldsymbol{q}, (\boldsymbol{a} \otimes \boldsymbol{1} - A)\boldsymbol{p} \rangle
\end{equation*}
This is because $\boldsymbol{p}$ is a probability simplex and therefore each row $m$ of vector $\boldsymbol{a}$ increases row $m$ of payoff matrix $A$. Therefore, the above game is equivalent to a standard zero-sum game with payoff matrix $\tilde{A}=(\boldsymbol{a} \otimes \boldsymbol{1} - A)$.
Please be consistent with typography: if you use $\boldsymbol a$ $\boldsymbol a$ once, don't switch to a ***a***for the same variable later. I have edited accordingly.
The classical formulation of a two player zero sum game with finitely many strategies is $\langle q, Ap \rangle$. There is no need to introduce the vector $a$, as it can be incorporated in the matrix $A$. In the paper you mention, the authors assume that it is too costly to evaluate $A$; maybe the formulation they provide is useful in their analysis.
Is $A$ assumed non-singular? If not, why can $a$ be incorporated into it?
This non-standard formulation matches better the intended application - (Wasserstein) GAN. It can be mapped to the standard formulation $\langle \boldsymbol{q}, \tilde{A}\boldsymbol{p} \rangle$ through a new payoff matrix $\tilde{A} = \boldsymbol{a} \otimes \boldsymbol{1} - A$, where $\otimes$ is the outer product operation ($\boldsymbol{a} \otimes \boldsymbol{b} = \boldsymbol{a} \boldsymbol{b}^T)$
|
2025-03-21T14:48:30.239429
| 2020-04-04T18:40:28 |
356586
|
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|
Stack Exchange
|
$m$-regularity of sheaves
This is Lemma 1.4 on Green and Lazarsfeld's Some results on the syzygies of finite sets and algebraic curves. Let $X$ be a closed subscheme of $\mathbb{P}^r$. Suppose the ideal sheaf $\mathcal{I}$ of $X$ is 3-regular, meaning that $H^i(\mathbb{P}^r,\mathcal{I}(3-i))=0$ for $i>0$. Then for $p\leq codim(X,\mathbb{P}^r)$, $(N_p)$ holds for $X\subset\mathbb{P}^r$ iff $Tor^S_p(R,k)_{p+2}=0$, where $S=k[x_0,\cdots,x_r]$, $R=\bigoplus\limits_{l\geq 0}H^0(\mathbb{P}^r,\mathcal{O}/\mathcal{I}(l))$. The proof is short. First observe that $m$-regularity implies that $\Gamma_*(\mathcal{I})$ is generated by elements of degree $\leq m$. Apply this to the sheafification $0\to\mathcal{E}_{r+1}\to\cdots\to\mathcal{E}_1\to\mathcal{I}\to 0$ of a minimal resolution of $I=\Gamma_*(\mathcal{I})$. One finds that $\mathcal{E}_i$ is $(i+2)$-regular, which means that $Tor_i^S(R,k)_j=0$ for $j\geq i+3$ and therefore $(N_p)$ only requires $Tor^S_p(R,k)_{p+2}=0$.
I don't know how to deduce the $(i+2)$-regularity of $\mathcal{E}_i$. For example, let $r=2$. Consider the homogeneous ideal $I=(xy^2,z^3)$. Let $E_1=S(-3)\oplus S(-3)$, with the map $E_1\to I$ given by $(xy^2,z^3)$. Then the kernel is $S(-6)$ with the inclusion to $S(-3)\oplus S(-3)$ given by $(-z^3,xy^2)^T$. This is a minimal resolution but $\mathcal{O}(-6)$ is not 4-regular. Can anyone tell me what's wrong with my observation?
This just mean that your ideal $\mathscr{I}$ is not 3-regular -- in fact $H^1(\mathscr{I}(2))\cong H^2(\mathscr{O}(-4))\neq 0$.
|
2025-03-21T14:48:30.239549
| 2020-04-04T19:17:58 |
356587
|
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|
Stack Exchange
|
Age of the most recent common ancestor for the neutral Wright-Fisher model
The neutral Wright-Fisher model with $n$ individuals is a genealogical model often used in population genetics that can be described as follows: at all generations, there are exactly $n$ individuals, and each child chooses its parent uniformly at random. One can then follow genealogies backward in time.
In this simple model, one can take interest in the age of the most recent common ancestor $T_n$, defined as the first time when, looking for the ancestors of the ancestors of the ancestors... of all individuals of the first generation, only one ancestor remains.
My question is the following : is there a good estimate existing on the mean of $T_n$? I'm after not only an equivalent, but also a development up to $O(1)$ as $n \to \infty$.
For context, I'm aware of the Kingman coalescent, obtained when considering a finite subset of $k$ individuals in the current generation, and that writing $T_{n,k}$ for the age of the most recent common ancestor one has
$$ \lim_{n \to \infty }T_{n,k}/n = \sum_{j=2}^k \frac{e_j}{j(j-1)} \quad \text{ in law.}$$
It hints that $\mathbb E(T_n)$ should grow at least as fast as $2n$.
I also tried to apply the result of Kimura and Osha (1969) on the fixation of a mutation in a Wright-Fisher process. I did not succeeded in applying it to this question as that result essentially considers the fixation of a mutation starting from $\epsilon n$ individuals instead of 1, and then letting $n \to \infty$ then $\epsilon \to 0$.
|
2025-03-21T14:48:30.239684
| 2020-04-04T19:31:29 |
356591
|
{
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"Kevin Casto",
"Moishe Kohan",
"Pii_jhi",
"https://mathoverflow.net/users/141502",
"https://mathoverflow.net/users/39654",
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|
Stack Exchange
|
Packing a Riemannian manifold with disjoints balls
Let $M$ be a smooth Riemannian manifold with Riemannian measure $\mu$. I don't suppose that $M$ is complete. Can we find a finite or countable disjoint collection of open (or closed) and relatively compact geodesic balls$(B_n)_{n\in\mathbb{N}}$ such that:
$$
\mu\left(M\setminus\bigsqcup_{n\in\mathbb{N}}B_n\right) = 0\ \ \ \ ?
$$
We might assume that $M$ has a bounded curvature. I'm interested in the case when $M$ is the intersection of a (complete) submanifold of $\mathbb{R}^d$ with a ball. The compactness restriction ensures that each ball is a "real" ball, and hasn't been cropped by the "edge" of $M$.
It seems to me we should be able to actually cover $M$ by a countable collection of closed (compact) balls.
I forgot the disjoint hypothesis in my question..
In the case of $E^n$'s, such packings do exist. For instance, Apollonian packing does the job for the Euclidean plane. Not sure what happens if one requires zero Hausdorff dimension of the residual set.
The residual set will contain the boundary of the ball (if we consider open balls), so it will be of Hausdorff dimension $d-1$ at least and I think it has to be strictly greater than $d-1$. In the case of Apollonian packing, which feels close to an"optimal" packing, has Hausdorff dimension $\simeq 1.3$..
@Pii_jhi: Of course, I meant closed balls, as in your question. In all packing constructions I know, the residual set has positive H.D.
I believe that this is true. If you look at Lemma 1.10. Of Introduction to Smooth Manifolds By John Lee,
Lemma 1.10. Every topological manifold has a countable basis of precompact coordinate balls.
If you follow the proof using an open cover of $M$ by Riemannian normal coordinate charts, you should end up with a countable basis of precompact geodesic balls of $M$ so that the union of all such balls is equal to $M$. Bounded curvature and completeness are not necessary for the proof.
I'm not sure what you mean by the last two sentences, but the fact that $M$ is topologically embedded into $\mathbb{R}^d$ should imply that the intersection of $M$ with a ball in $\mathbb{R}^d$ is an open set in $M$.
Thank you for your answer. In fact I forgot the disjointness hypothesis in my question, which makes it less trivial. I believe it is true on R^n and belong to the family of Vitali covering theorem.
For any smooth Riemannian manifold $(M,g)$ there is a countable disjoint union of balls with complement of measure $0$.
Let $\mu$ be Riemannian measure and for each $p\in M$ let $B_p$ be a small precompact ball centered at $p$ such that $\mu(\partial B_p)=0$ (we can do this because $\{r>0;\mu(\partial B(p,r))>0\}$ is countable).
Claim 1: There is some $k>0$ such that for small $\delta>0$ and for any $q\in B_p$, $\frac{\mu(B(q,\delta))}{\mu(B(q,2\delta))}>k$.
Proof: Let $\varphi:U\to\mathbb{R}^n$ be a chart, where $U$ is a nhood of $\overline{B_p}$. Let $g'$ be the pullback from the metric of $\mathbb{R}^n$ in $U$. Then in some neighborhood $V$ of $\overline{B_p}$, the Riemannian distances $d,d'$ from $g,g'$ respectively are bilipschitz (so the volumes $\mu,\mu'$ are also related by a constant). So if $\delta_0$ is small enough that $d$-balls centered in $B_p$ of radius $\delta_0$ are contained in $V$, then for some big constant $\alpha$ we have that for any $q\in B_p$ and $\delta<\delta_0$,
$$B'\left(q,\frac{\delta}{\alpha}\right)\subseteq B(q,\delta)\subseteq B(q,2\delta)\subseteq B'(q,\alpha\delta).$$
So, as $\frac{\mu(B'(q,\alpha\delta))}{\mu(B'(q,\frac{\delta}{\alpha}))}$ is uniformly bounded, we have proved the claim. $\square$
Claim 2: For any open subset $A$ of $B_p$ there is a finite set of disjoint balls $B_1,\dots,B_m$ contained in $A$ such that $\mu(\cup_{i=1}^m B_i)>\frac{1}{2k}\mu(A)$.
Proof: Let $\delta$ be so small that it satisfies the previous claim and such that, if $B:=\{x\in A;d(x,M\setminus A)>\delta\}$, then $\mu(B)>\frac{1}{2}\mu(A)$. Now consider a maximal $2\delta$-separated set $\{x_1,\dots,x_m\}$ in $B$, and let $B_i:=B(x_i,\delta)$. These balls are disjoint, and the balls $B(x_i,2\delta)$ cover $B$, so $\sum_i\mu(B_i)>\frac{1}{k}\sum_i\mu(B(x_i,2\delta))\geq\frac{1}{k}\mu(B)\geq\frac{1}{2k}\mu(A)$. $\square$
We can also ensure that the boundaries of the balls $B_i$ of claim $2$ have measure $0$: if not, note that for each $q\in M$, the set $\{r>0;\mu(\partial B(q,r))>0\}$ is countable, so we can reduce the radii of the balls just a little bit so that the sum of their volumes is still $>\frac{1}{2k}\mu(A)$.
Claim 3: We can cover any open set $X\subseteq B_p$ up to measure $0$ by a disjoint collection of balls contained in $X$.
Proof: Take $A=X$ in the previous claim, and find balls $B_{0,1},\dots,B_{0,n_0}$ with boundary of measure $0$ such that $\mu(\cup_{i=1}^m B_i)>\frac{1}{2k}\mu(X)$. Now let $X_1=X\setminus\cup_i\overline{B_{0,i}}$, so that $\mu(X_1)\leq(1-\frac{1}{2k})\mu(X)$. Applying the same to $X_1$ we can remove from it finitely many balls $B_{1,1},\dots,B_{1,n_1}$ to obtain some open $X_2$ with $\mu(X_2)\leq(1-\frac{1}{2k})\mu(X_1)$. Repeating this step to obtain spaces $X_n$ for each $n$, we get that the balls $\{B_{i,j}\}_{i\in\mathbb{N};j=1,\dots,n_i}$ are pairwise disjoint, and $\mu(X\setminus\bigcup_{i,j}B_{i,j})=\lim_{m\to\infty}\mu(X_m)\leq \lim_{m\to\infty}(1-\frac{1}{2k})^m\mu(X)=0$. $\square$
Claim 4: We can cover $M$ up to measure $0$ using a countable collection of disjoint compact balls.
Proof: Consider the collection of balls $\mathcal{B}:=\{B_p;p\in M\}$. As $M$ is second countable, we can find a countable subcover of $\mathcal{B}$, $(B_n)_{n\in\mathbb{N}}$. Moreover, for each $n$, we can cover $B_n\setminus\bigcup_{i=1}^{n-1}\overline{B_i}$ up to measure $0$ with a countable collection of disjoint compact balls. The union of these countable collections of balls covers $B_n$ up to measure $0$ for all $n$, thus it covers all $X$ up to measure $0$. $\square$
|
2025-03-21T14:48:30.240050
| 2020-04-04T20:03:50 |
356593
|
{
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"authors": [
"Iosif Pinelis",
"dohmatob",
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|
Stack Exchange
|
Tail bounds for the absolute difference of a coupled pair of sub-Gaussian random variables
Let $P$ and $Q$ are sub-Gaussian distributions on $\mathbb R$, and $(X,X')$ be a coupling of $P$ and $Q$, i.e $(X,X') \sim \pi$ for some distribution on $\mathbb R^2$ with marginals $P$ and $Q$.
Question. Does $|X-X'|$ have any concentration properties ?
Can one reasonably bound $\mathbb P(|X-X'| > \epsilon)$ for $\epsilon > 0$ ?
Thanks in advance for any help.
For $X_1:=X$, $X_2:=X'$, some positive real $c_1,c_2,a_1,a_2$, and all positive real $t$ we have
$$P(|X_j|>t)\le c_j e^{-a_j t^2} \tag{1}$$
for $j=1,2$. So, for $t:=\epsilon>0$,
$$P(|X_1-X_2|>t)\le P(|X_1|>t/2)+P(|X_2|>t/2) \\
\le c_1 e^{-a_1 t^2/4}+c_2 e^{-a_2 t^2/4}.$$
The OP commented that the notion of a sub-Gaussian distribution was meant not in the usual sense. Namely, according to that comment, condition (1) should be replaced by
$$P(|Y_j|>t)\le c_j e^{-a_j t^2}, \tag{2}$$
where
$$Y_j:=X_j-m_j$$
for some real $m_j$. Then, for $t:=\epsilon>|m_1-m_2|$,
$$P(|X_1-X_2|>t)\le P(|Y_1-Y_2|>t-|m_1-m_2|) \\
\le P(|Y_1|>(t-|m_1-m_2|)/2)+P(|Y_2|>(t-|m_1-m_2|)/2) \\
\le c_1 e^{-a_1(t-|m_1-m_2|)^2/4}+c_2 e^{-a_2(t-|m_1-m_2|)^2/4}.$$
Thanks. I meant sub-Gaussian in the general sense, with different means $\mu_1 \ne \mu_2$, e.g as in the pair $P=\mathcal N(\mu_1,\sigma_1^2)$ and $P'=\mathcal N(\mu_2,\sigma_2^2)$.
Was just about to write down the same addendum. Thanks! Any hope for the case $\epsilon \le |m_1-m_2|$ ?
@dohmatob : Of course, no hope for that case: If the $X_j$'s are concentrated near the $m_j$'s, then $X_1-X_2$ is naturally concentrated near $m_1-m_2$ -- rather than near $0$. E.g., let the $X_j$'s be iid normal with means $m_1=-m_2=m$, where $m$ large compared with the standard deviation -- then you have almost no concentration for $X_1-X_2$ near $0$.
Yes, indeed. Thanks again!
|
2025-03-21T14:48:30.240195
| 2020-04-04T20:05:55 |
356594
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/356594"
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|
Stack Exchange
|
On a formula for the Auslander-Reiten translate
For an Artin algebra $A$ and an indecomposable non-projective module $M$ we should have that $\tau(M) \cong \nu \Omega^2(M)$ iff $Ext_A^i(M,A)=0$ for $i=1,2$. ($\nu$ being the Nakayama functor)
Applying this to the indecomposable module $A$ as an $A^e$-module ($A^e$ being the enveloping algebra of A) and noting that $Ext_{A^e}^i(A,A^e)=Ext_A^i(D(A),A)$, we get that in case $Ext_A^i(D(A),A)=0$ for $i=1,2$, we have that $\tau(A) \cong \nu \Omega^2(A)$ as $A^e$-modules.
Now tensoring this with an arbitrary $A$-module $M$, we get $\tau(A) \otimes_A M \cong \nu \Omega^2(A) \otimes_A M$.
Now we should have that up to projective summands $P$ that $\tau(A) \otimes_A M = \tau(M) \oplus P$ as $A$-modules.
Now I wonder whether we can also write $\nu \Omega^2(A) \otimes_A M$ in a nice form (avoiding the tensor product) to get a non-trivial global formula for $\tau(M)$?
I know that $\Omega^2(A) \otimes_A M \cong \Omega^2(M) \oplus P'$ for some projective module $P'$, but the Nakayama functor coming from the enveloping algebra seems to be more complicated.
It can not be true that $\tau(M) \cong \nu \Omega^2(M)$ in general except for $A$ being selfinjective, since for example $A$ could have finite global dimension and then the formula is wrong for $M$ having projective dimension equal to one in general.
|
2025-03-21T14:48:30.240422
| 2020-04-04T20:49:18 |
356595
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Andy Sanders",
"Angelo",
"Ben McKay",
"Lev Soukhanov",
"Nadia SUSY",
"abx",
"https://mathoverflow.net/users/103164",
"https://mathoverflow.net/users/125941",
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"user347489"
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|
Stack Exchange
|
Vector bundle over compact complex manifold which is not holomorphic?
A vector bundle over a complex manifold is said to be holomorphic if its trivialization maps are biholomorphic maps. What is a "natural" example example of a vector bundle over compact complex manifold which is not holomorphic? I guess by "natural" I mean that one would be interested in such examples for reasons besides them being a counter example.
TO CLARIFY: I am interested in
i) COMPLEX VECTOR BUNDLES
ii) NOT ADMITTING ANY HOLOMORPHIC STRUCTURE
Related: https://mathoverflow.net/questions/7304/complex-vector-bundles-that-are-not-holomorphic
@Thomas: So what are interesting examples of bundles that do not admit almost complex structures?
The introduction to the paper paper https://arxiv.org/pdf/1506.08111.pdf contains a good survey of the problem.
Nadia, if you are interested in real vector bundles that don't admit the structure of a complex vector bundle, then you should say so. Because, while not trivial, this can be investigated by methods of algebraic topology. Meanwhile, since you use the word holomorphic in your title, you should specify that you are interested in complex vector bundles, since otherwise it's not clear what you are interested in. I think you might have stumbled into a very deep question on accident.
Complex line bundles on a manifold $X$ are classified by the first Chern class $c_1\in H^2(X,\mathbb{Z})$. If $X$ is a compact Kähler manifold, such a line bundle is holomorphic if and only if $c_1$ is of type $(1,1)$.
Real line bundles don't admit a complex structure, for example the trivial real line bundle.
I interpret the question as "Are there any complex bundles over the complex manifold which do not admit the holomorphic structure". The answer to this is "yes", but I'm not sure if it is known in full generality for what manifolds the answer is positive and for what it is negative.
Line bundles case. Suppose X is compact Kahler, L - complex line bundle over X. Then it has a holomorphic structure iff it's 1st Chern class is (1,1). See the comment by @abx
Proof: the "only if" part of the statement is the fact that holomorphic bundle admits so-called Chern connection, which has curvature of type (1,1). "If" part comes from the following consideration:
Complex line bundles are classified by $H^1(X, C^{\infty}(X, \mathbb{C}^*))$. This sheaf is subject to the following exponential exact sequence:
$0 \rightarrow \mathbb{Z} \rightarrow C^{\infty}(X, \mathbb{C}) \rightarrow C^{\infty}(X, \mathbb{C}^*) \rightarrow 0$
Similarly, holomorphic bundles are classified by $H^1(X, \mathcal{O}^*)$. There is also an analogous exact sequence (which is also embedded in the previous one by the inclusion of the holomorphic functions into the smooth ones):
$0 \rightarrow \mathbb{Z} \rightarrow \mathcal{O} \rightarrow \mathcal{O}^* \rightarrow 0$
Now, middle term in the former exact sequence is acyclic, so $H^1(X, C^{\infty}(X, \mathbb{C}^*)) \simeq H^2(X, \mathbb{Z})$ (this delta homomorphism is one of the equivalent formulations of the 1st Chern class). The latter exact sequence, however, gives (after standard identification $H^k(X, \mathcal{O}) \simeq H^{0,k}(X)$ the following description for holomorphic bundles:
$0 \rightarrow H^{0,1}(X)/H^1(X, \mathbb{Z}) \rightarrow H^1(X, \mathcal{O}^*) \rightarrow H^2(X, \mathbb{Z}) \cap (H^{1,1}(X) \oplus H^{2,0}(X)) \rightarrow 0$
The third term of this sequence is actually $H^2(X, \mathbb{Z}) \cap H^{1,1}(X)$ (because $H^2(X, \mathbb{Z})$ is real, and $H^{2,0}(X)$ is conjugate to $H^{0,2}(X)$.
This gives the following obstruction for line bundle - it's first Chern class should be (1,1), and provided this obstruction holds there is a complex torus $H^{0,1}(X)/H^1(X, \mathbb{Z})$ of different holomorphic structures.
General case. I think in the general case the similar classification is not known. The criterion that all Chern classes should be of (p,p) type clearly holds, but I think it is not enough. I'm not up to date with the current state of this field but you can see from here http://www.numdam.org/item/?id=SB_1978-1979__21__80_0 that even for projective spaces when this criterion is empty the question is far from trivial (and if I've understood correctly has a negative answer). Also see @Angelo 's comment which has a reference to the recent survey of this problem https://arxiv.org/pdf/1506.08111.pdf
Maybe it is also worth noting that over a manifold of complex dimension 1 any complex bundle admits holomorphic structure: indeed, one can choose an almost holomorphic structure (take any connection and take it's (0,1) part), and then it is automatically integrable by dimensional reasons).
Angelo's comment above is relevant here: he gives a reference to some results on obstructions.
Thanks, I didn't see it, will add to answer. This survey also has a reference to the text I linked.
The bundles of (0, 1)-forms are not holomorphic. (the transition functions are anti-holomorphic) In fact the bundles of $(p, q)$-forms are all not holomorphic if $q>0$. (and if both $p$ and $q$ are not 0, then the transition functions are neither holomorphic nor anti-holomorphic)
But these bundles admit a holomorphic structure, namely the opposite complex complex. I suppose I am looking for a smooth bundle that doesn't admit any holomorphic structure.
|
2025-03-21T14:48:30.240755
| 2020-04-04T21:13:59 |
356597
|
{
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"Asaf Karagila",
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|
Stack Exchange
|
Inner models with all sets generic
Question: Under large cardinal axioms, what is the intersection of all inner models $M$ of ZFC such that every set in $V$ is set-generic over $M$?
Every set belongs to a generic extension of HOD, and we hope that HOD is canonical in the true $V$, but ordinary large cardinal axioms do not imply that HOD is a canonical well-behaved model. Every countable model $M$ of ZFC is HOD of some model of ZFC (obtained using class forcing over $M$).
However, let $M_∞$ (or $M_\text{Ord}$) be the minimal iterable inner model with a proper class of Woodin cardinals. There is an ordinal definable iterate $M'_∞$ of $M_∞$ such that every set in $V$ is set-generic over $M'_∞$. Specifically, pick an OD set of ordinals $X_0$; iterate the first Woodin cardinal of $M_∞$ to make $X_0$ generic; then pick $X_1$ and iterate the second Woodin cardinal to make $X_1$ generic, and so on. Also, using genericity over local HODs, we can choose $M'_∞$ such that $M'_∞∩H(λ)$ is definable in $H(λ)$ (for $λ>c$) and with every $X⊂λ$ being $M'_∞$-generic for a poset in $M'_∞∩H((2^λ)^+)$ (as usual, $H(λ)=\{x:|\mathrm{tc}(x)|<λ\}$).
But is this bound optimal? For every $M$ in the question and a set of ordinals $s∈M$, is $M_∞(s)$ elementarily embeddable into an $M$-definable submodel of $M$? Does the intersection of all such $M$ equal $M_∞$ with the least measurable cardinal iterated away? And what kind of large cardinals must such $M$ have?
Using $ω$ steps of core model induction, every $M$ as in the question satisfies projective determinacy (PD) in all generic extensions of $M$ (assuming PD in all generic extensions of $V$), but I do not know how far further core model induction can go here.
Note that every set $S$ is generic over some (dependent on $S$) iterate of $M_1$ (the minimal iterable inner model with a Woodin cardinal). Thus, for example, if there is a superstrong cardinal (and every set has a sharp), then there is a generic extension of $M_1$ with a superstrong cardinal. This is analogous to existence of complicated transitive models in $L$; and more Woodin cardinals give genericity over models with more closure.
Formalization note: The answer is presumably the same regardless of whether $M$ is $Σ_2$ definable using parameters in $V$, or we use NBG (plus large cardinal axioms) and treat $M$ as a class. Also, allowing choice to fail in $M$ likely does not change the answer. A likely sufficient large cardinal assumption is that $M_∞$ (above) exists and is fully iterable.
Local versions: A variation is to consider inner models $M$ with (for a specific $λ$) every element of $H(λ)$ generic over $M$ using a forcing in $H(λ)$. Examples include:
- for countable cofinality strong limit $λ$, some iterates of $M_ω$
- for singular strong limit $λ$ of uncountable cofinality, some iterates of the minimal iterable inner model with a measurable number of Woodin cardinals (i.e. $κ$ Woodin cardinals with $κ$ measurable in the model)
- for inaccessible $λ$, some iterates of $M_∞$.
Class forcing:
While some class forcing is set-like, in general class forcing lacks the same kind of closure. For example, even preserving ZFC, we can code the universe into a real even if every set has a sharp. The lack of closure makes it easier to make $V$ generic, and if I understand correctly, it suffices to use an appropriate iterate of the minimal inner model satisfying "Ord is Woodin" (not sure if we need its sharp, or if there are definability isssues), with the class forcing satisfying Ord chain condition (and thus well-behaved). An analogous relation should also hold for a number of extensions to the language of set theory, with both "Ord is Woodin" (and the inner model) and closure properties of classes (and class forcing, or Ord-cc class forcing) strengthened in the same way.
While many iterates should work, a particularly elegant choice and encoding of an iterate is (conjecturally) the 'stability' predicate $S=\{n,α,β: n<ω ∧ H(α) ≺_{Σ_n} H(β)\}$. $(L[S],∈,S)$ is called the stable core (see The Stable Core and Structural Properties of the Stable Core). Caveat (conjectural): The iterate encoded by $S$ (there are different encodings, but if it works, among iterates in which $S$ is definable, the unique iterate that is definable from every iterate in which $S$ is definable) is outside of $(L[S],∈,S)$, though a further iterate is $\text{HOD}^{L[S]}=K^{L[S]}$. Without large cardinal assumptions, the theory of the stable core is not canonical, but we still get the genericity. While the specific choice of $S$ is somewhat arbitrary, I think the use of the cumulative hierarchy is important for the genericity, and presumably a different definition would simply lead to a different iterate that works, or be insufficient or suboptimal.
Is "every set in V is set-generic over M" just mean that M is a ground of V?
@AsafKaragila No, unlike in a ground, the poset and the generic extension of $M$ may depend on the set in $V$.
@AsafKaragila And consistently V is not even a class-generic extension of HOD (see this paper by Hamkins and Reitz), while every set in V is set-generic over HOD by Vopěnka.
|
2025-03-21T14:48:30.241102
| 2020-04-04T21:42:44 |
356598
|
{
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"Mateusz Kwaśnicki",
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|
Stack Exchange
|
Find distribution that minimises a function of its moments
Imagine a probability density function $f(x)$, defined for positive $x$, and let's note its $n$th non-centred moment $x_{n}$. The mean $x_{1}$ is fixed (and positive).
How can I find $f(x)$ that minimises some given function of its moments? In my case, $$\frac{ x_{3}+x_{1}^{3}-2x_{1}x_{2} }{ (x_{2}-x_{1}^{2})^{2} }$$
I tried to take the Gateaux derivative of that expression in the direction of a test function $h(x)$, and setting the result to be zero for any $h(x)$. In the end, I find a relation involving a few moments of $f(x)$ and the variable $x$, which makes no sense. Would you have any idea of the correct approach here?
Many thanks!
Let $X$ be a positive random variable (r.v.) with probability density function $f$. The exact lower bound on
$$r(X):=\frac{x_3+x_1^3-2x_1x_2}{(x_2-x_1^2)^2}$$
is $0$, and it is not attained at any $f$.
Indeed, by the Cauchi--Schwarz inequality, $x_2\le x_3^{1/2}x_1^{1/2}$, and $x_2=x_3^{1/2}x_1^{1/2}$ only if the r.v. $X$ is a constant. Since $X$ has a pdf $f$, it is not discrete and hence not a constant. So, $x_2<x_3^{1/2}x_1^{1/2}$ and hence
$$x_3+x_1^3-2x_1x_2>x_3+x_1^3-2x_1x_3^{1/2}x_1^{1/2}=(x_3^{1/2}-x_1^{3/2})^2\ge0.$$
So, $$r(X)>0.$$
Note also that, for any real $t$ and any natural $k$, if we replace $X$ by $tX$, then $x_k$ gets replaced by $t^k x_k$. So,
$$r(tX)=\frac{t^3}{t^4}\,r(X)=\frac{r(X)}t\to0$$
as $t\to\infty$.
Therefore and because $r(X)>0$, we see that indeed the exact lower bound on
$r(X)$
is $0$, and it is not attained at any $f$.
I believe $x_1$ is fixed in the question, while here $x_1$ goes to infinity.
This is to complete Mateusz Kwaśnicki's answer by proving that
$$EY^2(1+Y)\ge(EY^2)^2\tag{1}$$
if $Y\ge-1$ and $EY=0$.
Since $Y\ge-1$, for any real $v$ we have
\begin{align}
Y^3=(Y+1)(Y-v)^2&+(2v-1)Y^2+(2v-v^2)Y-v^2 \\
&\ge (2v-1)Y^2+(2v-v^2)Y-v^2.
\end{align}
So, choosing now $v=EY^2$, we have
$$EY^3
\ge (2v-1)EY^2+(2v-v^2)EY-v^2
=(2v-1)v+(2v-v^2)0-v^2=v^2-v,
$$
so that $EY^3
\ge v^2-v$, which is equivalent to (1).
(This is not an answer, rather an extended comment.)
If $X = a \frac{n}{n-1}$ with probability $\frac{n-1}{n}$ and $X = 0$ otherwise, then $x_1 = \mathbb{E}X = a$, $x_2 = \mathbb{E}X^2 = a^2 \frac{n}{n-1}$ and $x_3 = \mathbb{E}X^3 = a^3 (\frac{n}{n-1})^2$, so that $$\frac{x_3 + x_1^3 - 2 x_1 x_2}{(x_2 - x_1^2)^2} = \frac{1}{a} .$$
Of course, one can smooth out $X$ a little bit to get an absolutely continuous distribution with the above ratio arbitrarily close to $\frac{1}{a}$.
My guess would be that $\dfrac{1}{a}$ is the lower bound for $\dfrac{x_3 + x_1^3 - 2 x_1 x_2}{(x_2 - x_1^2)^2}$ if $x_1$ is required to be equal to $a$.
Let $X$ have density function $f(x)$, and let $Y = X/a - 1$, so that $Y \geqslant -1$ and $\mathbb{E} Y = 0$ (recall that we assume that $x_1 = a$). Observe that
$$x_3 + x_1^3 - 2 x_1 x_2 = \mathbb{E}(X^3 + a^3 - 2 a X^2) = a^3 \mathbb{E}(Y^2 + Y^3)$$
and
$$ x_2 - x_1^2 = \mathbb{E}(X^2 - a^2) = a^2 \mathbb{E} Y^2 . $$
Thus,
$$ \frac{x_3 + x_1^3 - 2 x_1 x_2}{(x_2 - x_1^2)^2} - \frac{1}{a} = \frac{1}{a} \, \frac{\mathbb{E}Y^2 + \mathbb{E}Y^3}{(\mathbb{E}Y^2)^2} - \frac{1}{a} = \frac{1}{a} \, \frac{\mathbb{E}(Y^2 (1 + Y)) - (\mathbb{E}Y^2)^2}{(\mathbb{E}Y^2)^2} . $$
My guess is therefore equivalent to
$$ \mathbb{E}(Y^2 (1 + Y)) \geqslant (\mathbb{E}Y^2)^2 $$
whenever $\mathbb{E} Y = 0$ and $Y \geqslant -1$.
I do not an immediate proof of the above inequality, nor do I see a counter-example. I though I would share it anyway, perhaps someone else can help. Edit: the proof is completed in Iosif Pinesis's answer.
|
2025-03-21T14:48:30.241325
| 2020-04-04T22:48:00 |
356602
|
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|
Stack Exchange
|
A Riemannian manifold with a non-degenerate metric and an inner product $u_{\beta} u^{\beta}=1$
The question is: given a Riemannian manifold with a non-degenerate metric g and an inner product $u_{\beta}u^{\beta}=1$, is $\nabla_{\mu} (u_{\alpha}u_{\beta})=0$ without demanding the trivial solution $\nabla_{\mu}u_{\alpha}=0$? The equivalent question can be raised on a Lorentzian manifold with $u_{\beta}u^{\beta}=-1$. I recently asked this seemingly simple question on Math Exchange but am getting little feedback and would appreciate the expertise on this site.
Given the norm as stated, it is clear that $\nabla_{\mu}(u_{\beta}u^{\beta})=0$ which means $ g^{\alpha\beta}\nabla_{\mu}(u_{\alpha}u_{\beta})=0 $. Since g is non-degenerate, $ \mid \nabla_{\mu}(u_{\alpha}u_{\beta})\mid=0 $ and $\nabla_{\mu}(u_{\alpha}u_{\beta})=0 $ is a solution (the singular solution of the corresponding matrix which would be represented by a sum of linear dependent terms) that does not require the trivial solution $\nabla_{\mu}u_{\alpha}=0 $. The trivial solution is sufficient but not necessary. In fact, $ u_{\beta}\nabla_{\mu}u_{\alpha}+u_{\alpha}\nabla_{\mu}u_{\beta}=0 $ means $ u^{\beta}\nabla_{\mu}u_{\beta}=0 $ so in general, the vector is orthogonal to the covariant derivative of its covector, and not constant.
The only comment I have had against taking $\nabla_{\mu}(u_{\alpha}u_{\beta})=0 $ to be consistent with $u_{\beta}u^{\beta}=1$ and with the vector u being orthogonal to the covariant derivative of its covector, is to require the vector u to be constant. That makes no sense to me. Please give me your valued opinions on this issue.
I don't understand the phrase "an inner product $u_\beta u^\beta=1$. This doesn't characterise the inner product (it characterises $u$).
Yes It means g(uu)=1.
I have trouble understanding the question. You seem to want to prove two things: 1) $\nabla_{\mu}(u_{\alpha}u_{\beta})=0$ and 2) the assumptions are not sufficient to prove $\nabla_{\mu}u_{\alpha}=0$. For 1), it seems that your second paragraph gives the correct argument. For 2), I think you need to provide a counterexample. I would suggest taking a metric connection $\nabla'$ with nonzero torsion and then choosing $u$ such that $\nabla'u$=0.
Thank you for your valid comment. I am restricting the connections to those being torsionless. Please note that $\nabla_{\mu}(u_{\alpha}u_{\beta})=0 $ is the equivalent of the sum of two orthogonal terms, neither of which can have $\nabla_{\mu}u_{\alpha}=0 $ for orthogonality to hold; for then, u is just a constant vector. In fact $\nabla_{\mu}u_{\alpha} $ can be expanded into the vorticity and shear tensors (GR and the Einstein eqn Choquet-Bruhat (pp409-411). So requiring $\nabla_{\mu}u_{\alpha}=0 $ is definitely not a necessary condition for $\nabla_{\mu}(u_{\alpha}u_{\beta})=0 $ to hold.
I meant that the connection with torsion could be used to construct a vector field with constant length but which 'rotates', making it non-parallel for the Levi-Civita connection. There may be simpler ways to construct counterexamples.
Got it. Thanks.
|
2025-03-21T14:48:30.241527
| 2020-04-04T22:57:07 |
356603
|
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|
Stack Exchange
|
Monomorphisms in homotopy categories
In hTop, insertions into coproducts $i : X \to X \amalg Y$ are monomorphisms, since, for any $f,g : Z \to X$ such that $if \simeq ig$, the homotopy factors through $i$ and thus $f \simeq g$.
This meshes well with the fact the monomorphisms in the $\infty$-category of spaces are precisely those whose class in in hTop are of the above form. Or a similar statement for the homotopy monomorphisms of Top (or other model). This prompts the question
Question 1: Are all of the monomorphisms in hTop in the form? (i.e. the insertion into a coproduct or equivalently the class of a homotopy monomorphism from Top or equivalently the class of a monomorphism in the $\infty$-category of spaces)
Assuming the answer is affirmative, my next question is when we can make similar assertions in the more general case:
Question 2: For a model category M, when can we say the monomorphisms in Ho(M) are precisely the (classes of) homotopy monomorphisms in M?
Question 3: For an $\infty$-category M, when can we say the monomorphisms in Ho(M) are precisely the (classes of) monomorphisms in M?
"...of this form" -- of which form? ##### Could you write your questions in a stand-alone style?
If $X$ is the terminal object in hTop (a one-point space), then any map $X\to Y$ is a monomorphism
@TomGoodwillie Ah, of course! I'm now glad I was skeptical of my calculation enough to ask... now to figure out where it went wrong....
I can't self-delete the question.
Can you edit the question?
|
2025-03-21T14:48:30.241656
| 2020-04-04T23:34:06 |
356608
|
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|
Stack Exchange
|
Help with $\mathbf{Q}_{\ell}$ sheaves
Let $X\to S$ be a morphism of smooth connected varieties over an algebraically closed field $k$; let $j:\eta\to S$ be the inclusion of the generic point into $S$ (not a geometric generic point) and let $\mathscr F$ be a constructible étale sheaf of $\Lambda$-modules on $X_\eta$. If $\Lambda=\mathbf{Z}/\ell^n\mathbf{Z}$, for example, then (letting $j$ also denote its base extension $X_\eta\to X$), $j_*\mathscr F$ is a constructible sheaf. But if $\Lambda=\mathbf Z_\ell$ or $\mathbf Q_\ell$, is $j_*\mathscr F$ still constructible? When I think about applying $j_*$ to a projective system (of constructible sheaves of modules for finite coefficients), the result is a projective system of constructible sheaves. When I think about it when $X=S$ is a smooth curve and think about $\mathscr F$ as a $\mathbf Q_\ell$-representation $V$ of $\operatorname{Gal}(\overline\eta/\eta)$, however, then the condition is equivalent to the condition that the inertia of all but finitely many points of $S$ act trivially on $V$—this seems dodgy. Thanks for your help.
A necessary condition is that for each stratum $X_{\eta}^i$ of some stratification of $X_{\eta}$ on which $\mathcal F$ is lisse, the associated representation of $\pi_1(X_{\eta}^i)$ is unramified away from a closed subset $Z^i$ of the closure $\overline{X_\eta^i}$ of $X_{\eta}^i$ in $X$, such that the projection of $Z_i$ to $S$ is not dense.
The fact that this condition is necessary is clear - if the pushforward is constructible, it's lisse on a stratification, and then intersecting each stratum with the generic fiber, you get a stratification with this property.
This can indeed fail, so the very general statement you wrote is not true. Your idea is precisely the right one - it's easy to construct a Galois representation that is not finitely ramified, for instance by constructing a suitable extension of $\mathbb Z_\ell$ by $\mathbb Z_\ell(1)$ using Kummer theory.
What will go wrong in the limiting proof you want to write down is that the pushforward of the associated $\mathbb Z/\ell^n$ pushforward, modulo $\ell$, will not be the pushforward of the associated $\mathbb Z_\ell$-sheaf, and will not even stabilize - it will be a sheaf with zero stalk at more and more ramification points as $n$ goes to $\infty$.
I think this condition is also suffiicent, but I didn't check it outside the case where $X =S$, where it can be proven by first pushing forward to the open set where the Galois representation is unramified, and then pushing tot he whole space.
Kudos, Will; thanks!
|
2025-03-21T14:48:30.242110
| 2020-04-04T23:43:20 |
356610
|
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|
Stack Exchange
|
Irrationality measure of arctan(1/3)
I recently came across the concept of the irrationality measure. It really fascinated me and when I was looking for known values $\mu(x)$ for mathematical constants $x$, I also came across this paper: On the irrationality measure of arctan 1/3
Unfortunately it could't help me out quite well.
What is the irrationality measure of $\arctan(1/3)$? Are there upper and lower bounds? Which famous constant have known irrationality measures or upper/lower bounds? (Except for the ones, that can easily be found on mathworld e.g. $\pi, \ln(2),\ln(3), \pi^2, \zeta(3)$)?
The irrationality measure of $\arctan(1/3)$ is not known. It lies between $2$ and $6.096755\dots$. The lower bound is trivial (it holds for every irrational number), while the upper bound is the main result of the paper you quote. It is reasonable to guess that the irrationality measure of $\arctan(1/3)$ is exactly $2$, because almost all irrational numbers (in Lebesgue measure) have this property.
Regarding your last, rather open ended question, I recommend that you use standard tools such as MathSciNet and Zentralblatt to search the mathematical literature.
As mentioned by GH from MO, the irrationality measure
for almost all real numbers is 2. However,
computing it for a particular number
is a notoriously difficult problem. For an irrational algebraic number this measure is indeed 2, but this is a pretty hard theorem by Roth for which he got a Fields medal.
Other then that, to my knowledge the only "famous constant" with the known irrationality measure is the number $e$. It is, again, equal to 2 and this is a very old result essentially known to Euler. (It follows from the expansion of $e$ into a continued fraction.)
For other "famous", or even not that famous, constants only upper bounds are known.
[EDIT] There is actually a pretty good answer here on MO which I missed somehow:
Numbers with known irrationality measures?
This looks like a comment to me.
You are right, but it's too long for a comment.
I think this is a pretty good answer.I upvoted it.
Champernowne's constant 0.123456789101112... is famous though slightly artificial. It has irrationality measure exactly 10.
|
2025-03-21T14:48:30.242301
| 2020-04-05T02:18:32 |
356617
|
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|
Stack Exchange
|
Upper bound on the condition number of the product of a random sparse matrix and a semi-orthogonal matrix
Let $G \in \mathbb{R}^{n \times m}$ (m > n, m = O(n)) whose all entries are i.i.d. distributed as $\mathcal{N}(0, 1) * \text{Ber}(p)$. Let $V \in \mathbb{R}^{m \times n}$ be a fixed semi-orthogonal matrix i.e. the columns of $V$ are orthonormal vectors. Define $A=GV$, then what is the smallest value of $p$ so that we can give a polynomial upper bound with probablity $1-o(1)$ on the condition number of $A$ i.e. $\kappa(A) \leq \text{poly}(n)$?
Interesting subcase/related problems:
Let $V$ be defined as $V_{i, j} = 1$ if $i = j$ and $V_{i, j} = 0$ otherwise. Let $G = [g_1, g_2, \ldots, g_m]$, in this case $A = GV = [g_1, g_2, \ldots, g_n]$. Hence in this case $A$ has the same distribution as $G$ except $m = n$. This was studied by Basak and Rudelson who proved that $\kappa(A) \leq \text{poly}(n)$ for $p = \Omega(\log n)/n$.
For $p = 1$, $G$ is just a random Gaussian matrix and $A = GV$ can also be seen to be random gaussian matrix are gaussian vectors are isotropic. This is just a subcase of 1.
For $m = n$, $V$ is just an orthonormal matrix hence $\kappa(V) = 1$ and from 1. we have $\kappa(G) \leq \text{poly}(n)$ if $p = \Omega(\log n)/n$. Hence we get $\kappa(A) \leq \kappa(G)*\kappa(V) \leq \text{poly}(n)$.
$G$ is a random matrix, but you seem to be looking for a deterministic upper bound? Please clarify.
I meant a w.h.p. bound, edited.
The Basak rudelson result is difficult because they handle a general class of distributions (they only assume finite fourth moments ...). In this case since you know so much about $G$, it might be just easier to expand the product $Ax$ and use a simple net argument, especially since you only want a 1-o(1) probability bound
|
2025-03-21T14:48:30.242461
| 2020-04-05T04:04:35 |
356618
|
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|
Stack Exchange
|
What is the precise relationship between pyknoticity and cohesiveness?
Pyknotic and condensed sets have been introduced recently as a convenient framework for working with topological rings/algebras/groups/modules/etc. Recently there has been much (justified) excitement about these ideas and the theories coming from them, such as Scholze's analytic geometry. (Small note: the difference between pyknotic and condensed is essentially set-theoretic, as explained by Peter Scholze here.)
On the other side, cohesion is a notion first introduced by Lawvere many years ago that aims to axiomatise what it means to be a category of "spaces". It has been developed further by Schreiber in the context of synthetic higher differential geometry (and also by Shulman in cohesive HoTT and by Rezk in global homotopy theory, to give a few other names in this direction).
Recently, David Corfield started a very interesting discussion on the relation between these two notions at the $n$-Category Café. The aim of this question is basically to ask what's in the title:
What is the precise relation between pyknoticity and cohesiveness?
Along with a few subquestions:
(On algebraic cohesion) It seems to me that the current notion of cohesion only works for smooth, differential-geometric spaces: we don't really have a good notion of algebraic cohesion (i.e. cohesion for schemes/stacks/etc.) or $p$-adic variants (rigid/Berkovich/adic/etc. spaces). Is this indeed the case?
(On the relevance of cohesion to AG and homotopy theory) Despite its very young age, it's already clear that condensed/pyknotic technology is very useful and is probably going to be fruitfully applied to problems in homotopy theory and algebraic geometry. Can the same be said of cohesion?
(On "condensed cohesion") Cohesion is a relative notion: not only do we have cohesive topoi but also cohesive morphisms of topoi, which recover the former in the special case of cohesive morphisms to the punctual topos. Scholze has suggested in the comments of the linked $n$-CatCafé discussion that we should not only consider cohesion with respect to $\mathrm{Sets}$, but also to condensed sets. What benefits does this approach presents? Is this (or some variant of this idea) a convenient notion of "cohesion" for algebraic geometry?
The idea of using relative cohesion in this case is mentioned by Schreiber back here: https://nforum.ncatlab.org/discussion/5473/etale-site/?Focus=43431#Comment_43431.
@DavidCorfield Great! Are you aware if Schreiber kept working on this after that discussion?
Also, do you know if he has since found an appropriate notion of cohesion for schemes or spaces appearing in rigid-analytic geometry?
@Sofia no, Urs has been working on matters related to M-theory these days (with great results!). And I don't think anyone has, else we would see some attention on it in eg the nCafé or the nForum (or it's published in some really obscure place).
Perhaps working out some consequences in the relative cohesion case would lead somewhere, such as https://ncatlab.org/nlab/show/differential+cohomology+diagram.
For a case of relative cohesion over stacks on arithmetic schemes, see https://ncatlab.org/nlab/show/differential+algebraic+K-theory.
@DavidRoberts Thanks for clarifying that! (By the way I agree with you: I've been following Fiorenza–Sati–Schreiber's program on M-theory and I'm finding it very interesting!)
Concerning the first subquestion: simplicial sets and the Zariski topos are usually mentioned as examples; not just categories of smooth, differential-geometric spaces. So I think that the answer to that subquestion is 'no; it is not the case'.
The work on analytic geometry is all joint with Dustin Clausen!
Your main question seems a little vague to me, but let me try to get at it by answering the subquestions. See also the discussion at the nCatCafe. Also, as David Corfield comments, much of this had been observed long before: https://nforum.ncatlab.org/discussion/5473/etale-site/?Focus=43431#Comment_43431
Yes, I think cohesion does not work in algebraic or $p$-adic contexts. The issue is that schemes or rigid-analytic spaces are just not locally contractible.
Cohesion does not seem to have been applied in algebraic or $p$-adic contexts. However, I realized recently (before this nCatCafe discussion), in my project with Laurent Fargues on the geometrization of the local Langlands correspondence, that the existence of the left adjoint to pullback ("relative homology") is a really useful structure in the pro-etale setting. I'm still somewhat confused about many things, but to some extent it can be used as a replacement to the functor $f_!$ of compactly supported cohomology, and has the advantage that its definition is completely canonical and it exists and has good properties even without any assumptions on $f$ (like being of finite dimension), at least after passing to "solid $\ell$-adic sheaves". So it may be that the existence of this left adjoint, which I believe is a main part of cohesion, may play some important role.
As I already hinted in 2, this relative notion of cohesiveness may be a convenient notion. In brief, there are no sites relevant in algebraic geometry that are cohesive over sets, but there are such sites that are (essentially) cohesive over condensed sets; for example, the big pro-etale site on all schemes over a separably closed field $k$. So in this way the approach relative to condensed sets has benefits.
All of these questions sidestep the question of why condensed sets are not cohesive over sets, when cohesion is meant to model "toposes of spaces" and condensed sets are meant to be "the topos of spaces". I think the issue here is simply that for Lawvere a "space" was always built from locally contractible pieces, while work in algebraic geometry has taught us that schemes are just not built in this way. But things are OK if instead of "locally contractible"(="locally contractible onto a point") one says "locally contractible onto a profinite set", and this leads to the idea of cohesion relative to the topos of condensed sets.
Let me use this opportunity to point out that this dichotomy between locally contractible things as in the familar geometry over $\mathbb R$ and profinite things as codified in condensed sets is one of the key things that Dustin and I had to overcome in our work on analytic geometry. To prove our results on liquid $\mathbb R$-vector spaces we have to resolve real vector spaces by locally profinite sets!
I just learned about your and Dustin Clausen's work on analytic geometry; it would be an understatement to say that it is amazing! Would it be ok to ask a question about it? From what I understand, this version of analytic geometry includes many other theories as special cases, such as smooth and complex manifolds as well as Berkovich spaces or schemes. Q: How is your and Clausen's theory related to other recent approaches to combining algebraic and p-adic geometries, such as e.g. Paugam or Poineau's work on global analytic geometry?
Thanks for your excitement :-)!
Let me also point to work of Ben-Bassat and Kremnizer. These works are generally based on Banach or (dual) Frechet algebras, and I would expect that all of these theories can be embedded into our category of analytic spaces. However, these theories are generally less expressive than adic spaces (and as is well-known, I really like those) as they cannot talk about the second component of a Huber pair $(A,A^+)$, while our notion of analytic spaces is able to handle adic spaces in a very clean way.
Actually, when I gave my course on condensed mathematics I was quite amazed how abstractly developing the desired $6$-functor formalism for coherent duality was almost automatically leading one to rediscover (discrete) adic spaces, in particular the seemingly obscure conditions on $A^+\subset A$ -- an open and integrally closed subring of powerbounded elements.
'condensed sets are meant to be "the topos of spaces" '. What conclusion was reached as to pyknotic/condensed sets and being a Grothendieck/elementary topos? And then similarly for pyknotic/condensed anima/infinity-groupoids and Grothendieck/elementary infinity-toposes?
We have a case of relative cohesion used in an algebraic geometric setting discussed at the nLab. The entry for differential algebraic K-theory interprets
Ulrich Bunke, Georg Tamme, Regulators and cycle maps in higher-dimensional differential algebraic K-theory (arXiv:1209.6451)
via cohesion over the base $Sh_\infty\left(Sch_{\mathbb{Z}}\right)$, ∞-stacks over a site of arithmetic schemes.
See also Urs Schreiber's entry, differential cohesion and idelic structure, and arithmetic elements of his research proposal, Higher theta functions and higher CS-WZW holography.
This sounds very exciting; thank you for the references!
Maybe it's worth pointing out that in this example the base topos is arithmetic, but that the relevant "locally contractible" objects are still smooth manifolds (if I understand this correctly), so it's of a somewhat orthogonal nature.
Yes, I was interpreting sub-question 2 generously, "the relevance of cohesion to AG".
Recent work by Qi Zhu, Fractured Structure on Condensed Anima, is looking to approach condensed mathematics via Jacob Lurie's concept of a fractured structure, seen as "local cohesive structure", cf. Remark 2.20.
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2025-03-21T14:48:30.243180
| 2020-04-05T06:20:04 |
356621
|
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|
Stack Exchange
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Map which is null-homotopic on compacts
This is the missing ingredient towards answering my previous question.
Let $M$ and $N$ be path connected locally compact, locally contractible metric spaces (you may assume that they are manifolds). It seems the "correct" condition on $N$ is absolute neighborhood retract. Let us also assume that $M$ is $\sigma$-compact, i.e. a union of a sequence of compact sets (and then we can even assume that every compact set in $M$ is contained in an element of that sequence).
Let $\varphi:M\to N$ be such that for every compact $K\subset M$ the map $\varphi|_{K}$ is null-homotopic. Does it follow that $\varphi$ is in fact null-homotopic?
The intuition says that if there is a hole in $N$ such that $\varphi$ is wrapped around it, it should be wrapped already on some compact set.
Let me also add a specific case when $\varphi$ is identity map.
If $N$ is such that the inclusion of every compact $K$ is null-homotopic (meaning $K$ is contractible within $N$), does it follow that $N$ is contractible?
There exist phantom maps $\mathbb{C}P^\infty\to S^3$. These are non-null homotopic maps which become null-homotopic when restricted to each finite skeleton. Doesn't this answer your question in the negative?
ANRs are locally contractible.
@MarkGrant Unfortunately, I am not competent to see if this answers my question. Is $\mathbb{C}P^\infty$ locally compact and $\sigma$-compact? Is every compact subset of it included in a finite skeleton?
A trivial remark: homomorphisms of homotopy groups which are induced by $\ \phi\ $ are trivial.
@erz Ah, OK, it is more subtle than I thought. It appears that $\mathbb{C}P^\infty$ with the weak (CW) topology may not be metrizable, which would imply that it's also not locally compact. I haven't checked, but I assume that the constructions of these phantom maps use the weak topology.
There is however a metric on $\mathbb{C}P^\infty$ for which each inclusion $\mathbb{C}P^n\subseteq\mathbb{C}P^\infty$ is an isometric embedding (Fubini-Study). One could ask if the construction of the phantom maps still works with this coarser topology.
You can replace $\mathbb CP^\infty$ by the telescope of the inclusions $\mathbb CP^n \hookrightarrow \mathbb CP^{n+1}$ which is locally compact and $\sigma$-compact.
@MarkGrant Fubini-Study metric is probably also highly non-compact being а quotient of the Hilbert sphere with respect to $S^1$.
@GustavoGranja could you please elaborate? Especially for somebody who is used to general topology as opposed to the constructions common in algebraic topology.
The mapping telescope of a sequence of maps is defined in section 3F in Hatcher and a relevant special case is used in the proof of Lemma 2.34 in Hatcher. It's homology groups or homotopy groups are computed by applying the homology or homotopy group functor to the sequence of maps and taking the colimit. Thus when $M$ has the homotopy type of a cell complex, the canonical map from the mapping telescope to $M$ is a homotopy equivalence.
The extent to which the answer to your question is no is analysed by Milnor's exact sequence. You can write $M$ as the colimit of a sequence $M_n \subset M_{n+1}$ of cofibrations with $M_n$ compact (at least if $M$ is a manifold but much more generally). Then there is a "short exact sequence" of pointed sets
$$
\{1\} \to \textstyle{\lim^1_n} [\Sigma M_n, N]_* \to [M,N]_* \to \lim_n [M_n,N]_* \to \ast
$$
(in the usual sense that the map of pointed sets on the right is surjective and its fibers are orbits of the action of the group $\lim^1$ which acts on the set in the middle). Brayton Gray used this sequence to construct the example that Mark Grant mentions in the comments above in this paper (since $S^3$ is simply connected there is no difference between pointed and unpointed homotopy classes).
Another reference for the Milnor exact sequence is Bousfield and Kan, Homotopy Limits, Completions and Localizations, Corollary IX.3.3.
Edit Regarding the second question: under the assumptions, $N$ has trivial homotopy groups, i.e. it is weakly contractible. Therefore, if it has the homotopy type of a cell complex (for instance if it is a manifold) then it is contractible.
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2025-03-21T14:48:30.243494
| 2020-04-05T06:43:57 |
356622
|
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|
Stack Exchange
|
What is the relationship between Hecke algebras and the enveloping algebra of Lie groups?
Here is the story as I see it.
Let $G$ be an abelian locally compact group. Then the (spherical) Hecke algebra for $K=1$ is by definition the endomorphism algebra of $l^2(G)$ as a $G$-module, where the $G$ action on $l^2(G)$ is defined by taking $f(x)$ to $f(x-g)$. In turn, this Hecke algebra can be identified with $l^1(G)$ (by $\rho\in l^1(G)$ mapping to the $G$-invariant endomorphism $f(x)\mapsto\int_G \rho(s)f(x+s)$), where composition in $End_G(l^2(G))$ becomes convolution in $l^1(G)$.
I'm trying to connect the above story with the following story. If $G$ is real Lie group, then the enveloping algebra of its Lie algebra can be identified with left-invariant differential operators.
I think I'm missing out on the big story here... Somehow simultaneously diagonalizing the center of $End_G(l^2(G))$ or the center of the enveloping algebra of the Lie algebra of $G$ (are they one and the same?!) is related to the irreducible representations of $G$? I'm pretty confused about how all this ties together.
So here are a few concrete questions:
How is $End_G(l^2(G))$ related to the universal enveloping algebra of the Lie algebra of $G$? Are they isomorphic? (The center of the universal enveloping algebra can be identified with left $G$-invariant differential operators; differential operators are a kind of function from $l^2(G)$ to itself, but it also operates on functions that are not in $l^2(G)$, and it is certainly not all the functions from $l^2(G)$ to itself... So the relationship here isn't quite clear to me...)
It looks to me like if instead of $l^2(G)$ we looked at the space of all functions $G\rightarrow\mathbb{C}$, then the center of $End_G(\mathbb{C}^G)$ would exactly be the functions that are homomorphisms from $G$ to $\mathbb{C}^{\times}$. But I also feel like this is the wrong way to think about this... So my concrete question here is: In the classical Fourier Analysis situation example specifically, how do the eigenfunctions of the Laplacian fit into the two stories above? I feel like the Laplacian, being a differential operator, belongs to the story about the center of the universal enveloping algebra; but I also see it sometimes in the context of an endomorphism of $l^2(G)$...
Given my level of understanding, what facts do you think would help elucidate to me how all the pieces fit together? Do you think I have any misconceptions?
I think "automorphism group" should be "endomorphism algebra" in your description of the spherical Hecke algebra for $K=1$. Automorphisms are the invertible endomorphisms.
"abelian td-group": do you mean "totally disconnected locally compact abelian group"? If so, why "abelian td-group or abelian lc-group" since one encompasses the other?
"the Hecke algebra is by definition $Aut_G(\ell^2(G))$: this sounds weird, this is not an algebra. Don't you mean $End_G(\ell^2(G))$ instead?
Edited appropriately, thanks!
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2025-03-21T14:48:30.243711
| 2020-04-05T06:50:29 |
356623
|
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"Denis Nardin",
"Maxime Ramzi",
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|
Stack Exchange
|
Contractible chain complex from non-contractible space
Recall that a chain complex $(C_*,d)$ of abelian groups is contractible if it is homotopic to the zero map. Or equivalently: there exists a degree 1 map $F: C_* \to C_*$ such that $\operatorname{Id}= dF+ Fd$.
Question: does there exist a topological space $X$ which is not contractible (in the sense of topology), but whose complex of singular chain $C_*(X)$ is contractible?
More generally, one can ask whether the functor $X \mapsto C_*(X)$ from the category of topological spaces to the homotopy category of chain complexes of abelian groups forgets any information. I assume the answer is "yes", but I can't seem to come up with a counterexample.
I've decided to interpret the question as asking whether the reduced singular complex is trivial, or equivalently if the canonical map $C_*(X)→\mathbb{Z}$ is a homotopy equivalence, since that's the property that's analogue to the contractibility of $X$ (i.e. $X\to \ast$ is a homotopy equivalence).
These are known as acyclic spaces (note that since $\tilde C_*(X)$ is a bounded below complex of projectives, it being contractible is equivalent to its homology being trivial).
There's an extensive literature about them, starting with this Emanuel Farjoun's paper.
In general, yes $C_*(X)$ does forget some information about $X$: first of all it factors through the stable homotopy type $\Sigma^\infty_+X$ of $X$, and it forgets further information from there (the slogan is that it remembers only the "$\mathbb{Z}$-linear" information contained in $\Sigma^\infty_+X$). For example it is unable to distinguish between $\mathbb{CP}^2$ and $S^2\vee S^4$ (since the attaching map $\eta:S^3\to S^2$ is sent to a map homotopic to 0 by $C_*(-)$).
You can get much closer to reconstructing the full homotopy type of $X$, by remembering the coalgebra structure on $C_*(X)$ induced by the diagonal, although in general that is still not enough.
For more information relating to the last paragraph, one can look up Mandell's theorem that gives some conditions for $C^*(X)$ to lose no information
@MaximeRamzi I think this paper by Allen Yuan is the state-of-the art on the subject (reconstructing unstable data from stable data plus algebra structures), although of course there's a vast literature on this too :).
|
2025-03-21T14:48:30.243905
| 2020-04-05T06:52:21 |
356624
|
{
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"A. Lerario",
"Jérémy Blanc",
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|
Stack Exchange
|
Approximating zero sets of real polynomials with "less complicated" polynomials
Let $K \subset \mathbb{R}^n$ be a compact subset, and let $P(x_1,\dots,x_n)$ be a real multivariable polynomial of degree $d$, whose vanishing set we denote by $Z_P$. Is it plausible to approximate $Z_P$ within $K$ (in the Hausdorff sense, for example) with the zero set of another polynomial $Q(x_1,\dots, x_n)$ which satisfies the following properties:
$Q$ is "simpler" than $P$ in the sense that a lot more of the coefficients of $Q$ are zero? For example, $x^4 - 1$ is simpler than $x^4 - x^3 + 1$.
The degree of $Q$ is preferably $\leq d$.
If yes, is there is a constructive method for finding $Q$? Any pointer/reference would be highly appreciated!
I do not know exactly what you want here by "approximate", but my guess is that if $P$ is the sum of monomials of degree $d$ with all coefficients near $1$, then any polynomial $Q$ of degree $\le d$ which is such that $Z_P$ is "near" $Z_Q$ would have to have all coefficients nonzero, no?
You can take a look at Remark 30 in this paper: https://arxiv.org/pdf/2010.14553.pdf
(Or more generally at the whole paper.)
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