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2025-03-21T14:48:31.769943
2020-08-14T03:17:16
369136
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Deformation of p-divisible groups along nilpotent thickening Let $S_0 \rightarrow S$ be a nilpotent thickening of schemes (no divided power provided) where $p$ is nilpotent, let $G$ be a $p$-divisble group over $S_0$, how to describe all liftings of $G$ to $S$ as $p$-divisible groups (using linear algebra data)? I am curious about the existence in good cases which is trivial for etale ones. In general the original Grothendieck-Messing theory is not available (as no divided power structure is introduced), unless we're in some good cases like $\mathbb Z/p^n \rightarrow \mathbb Z/p^m$ ($p>2$).
2025-03-21T14:48:31.770010
2020-08-14T05:41:06
369141
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Arno Fehm", "Mehta", "Mikhail Borovoi", "https://mathoverflow.net/users/122801", "https://mathoverflow.net/users/4149", "https://mathoverflow.net/users/50351" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632079", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369141" }
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Splitting of simply connected algebraic group Let $k$ be a number field and let $G$ be a connected semisimple, simply connected algebraic group defined over $k$. Let $k'$ be a finite Galois extension over which $G$ splits. By the Chebotarev Density Theorem, there are an infinite number of places $v$ of $k$ such that $v$ splits completely in $k'$. The paper I'm reading states without proof that over all such $v$, $G \times_k k_v$ is split. Why is this true? Is the simply connected or even semisimple hypothesis necessary? If $v$ splits completely in $k'$, then $k'$ is contained in $k_v$, hence if $G\times_kk'$ is split, so is $G\times_kk_v$. (I assume $k_v$ denotes completion.) Am I missing something? @Arno Fehm: Why is $k'$ contained in $k_v$? I must be missing something basic in algebraic number theory. $k'$ embeds into $k'\otimes_k k_v$. Since $v$ splits completely in $k'$, we have $k'\otimes_k k_v=k_v\times\dots\times k_v$. Thus $k'$ embeds into $k_v$.
2025-03-21T14:48:31.770221
2020-08-14T06:19:30
369142
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Maxime Ramzi", "Tanimura", "https://mathoverflow.net/users/102343", "https://mathoverflow.net/users/163737" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632080", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369142" }
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When does $\operatorname{Ext}_C^1(M,N_i)=0$ imply $\operatorname{Ext}_C^1\left(M,\lim\limits_\longleftarrow N_i\right)=0$? Let $C$ be an abelian category. Suppose that $(N_i)_{i\in I}$ is an inverse system of objects in $C$. Under which conditions does the hypothesis that $$\operatorname{Ext}_C^1(M,N_i)=0\quad\forall i\in I\tag{1}$$ imply $$\operatorname{Ext}_C^1\left(M,\lim\limits_\longleftarrow N_i\right)=0?\tag{2}$$ The functor $\operatorname{Ext}^1_C$ is the Yoneda $\operatorname{Ext}^1$-functor. So we don't have to worry whether $C$ has enough injectives/projectives. However, I would be satisfied if there are some results when $C$ is the category of (unitary) left modules over a ring $R$ with unity. In the case that $C$ is the category of (unitary) left modules over a ring $R$ with unity, every object in $C$ is an inverse limit of injective modules due to this paper. Therefore, an object $M\in C$ satisfies $$\operatorname{Ext}_C^1\left(M,\lim\limits_\longleftarrow N_i\right)=\lim\limits_\longleftarrow \operatorname{Ext}_C^1(M,N_i)\tag{3}$$ for any inverse system $(N_i)_{i\in I}$ in $C$ iff $\operatorname{Ext}^1_C(M,N)=0$ for all $N\in C$ iff $M$ is projective. However, I'm not imposing that $(3)$ should be true. I only suppose that $(1)$ is true, and I'd like to know when $(2)$ is also true. Counterexamples in which $(1)$ is true but $(2)$ is false will also be very helpful. Thank you in advance. An answer to the dual problem below will also be greatly appreciated. If there are cases where $(1')$ is true but $(2')$ is not, I would also like to see examples. Let $C$ be an abelian category. Suppose that $(N_i)_{i\in I}$ is an directed system of objects in $C$. Under which conditions does the hypothesis that $$\operatorname{Ext}_C^1(N_i,M)=0\quad\forall i\in I\tag{$1'$}$$ imply $$\operatorname{Ext}_C^1\left(\lim\limits_\longrightarrow N_i,M\right)=0?\tag{$2'$}$$ In the case where $I=\mathbb N$ and over $R-\mathbf{Mod}$, under niceness assumptions on your inverse system, you have the Milnor exact sequence : $0\to \varprojlim^1_n \mathrm{Ext}^2_R(M,N_i) \to \mathrm{Ext}^1_R(M,\varprojlim_i N_i) \to \varprojlim_i \mathrm{Ext}^1_R(M,N_i)\to 0$; e.g. if your system satisfies the Mittag-Leffler condition; so the conditions might involve conditions on your system $(N_i)$, as well as on $\mathrm{Ext}^2_R$ @MaximeRamzi Thank you. Do you have a reference for this? I have seen something similar in Weibel's book, but it has the assumption $R=\Bbb Z$, and not for a general $R$. Sorry, actually it should be $\hom$, not $\mathrm{Ext}^2_R$, I messed up between homological and cohomological grading. For a reference, I'm not exactly sure. The point is that under these conditions, you $\varprojlim_i N_i$ is an actual homotopy limit, and so (with everything derived) $Hom(M,\varprojlim_i N_i)\simeq \varprojlim_i Hom(M,N_i)$. Then you can replace this inverse system $(Hom(M,N_i))$ with a good inverse system (which doesn't change its homology) and apply the Milnor sequence as seen here (https://ncatlab.org/nlab/show/lim%5E1+and+Milnor+sequences#for_chain_homology) You just have to be careful (unlike I was) about homological vs cohomological grading : If you want $H_q$ (with their notation) to be $\mathrm{Ext}^1$, you will have $H_{q+1}$ being $\hom$ (not $\mathrm{Ext}^2$) The standard result in this direction is the dual Eklof lemma (for your first problem) or the Eklof lemma (for your dual problem). Any version of the Eklof lemma presumes that your direct/inverse system is indexed by a well-ordered set. For an inverse system, it should be a smooth chain of epimorphisms with the kernels which also satisfy the $\operatorname{Ext}^1_C(M,-)$ vanishing. For a direct system, it should be a smooth chain of monomorphisms with the cokernels which also satisfy the $\operatorname{Ext}^1_C(-,M)$ vanishing. Then the answer to your question is positive. References: P.C. Eklof, J. Trlifaj, "How to make Ext vanish", Bull. London Math. Soc. 33, #1, p. 41-51, 2001, https://doi.org/10.1112/blms/33.1.41 . Lemma 1 is the Eklof lemma (for a direct system), Proposition 18 is the dual Eklof lemma (for an inverse system). This is a paper about modules over associative rings. L. Positselski, J. Rosický, "Covers, envelopes, and cotorsion theories in locally presentable abelian categories and contramodule categories", Journ. of Algebra 483, p. 83-128, 2017, https://doi.org/10.1016/j.jalgebra.2017.03.029 , https://arxiv.org/abs/1512.08119 . Lemma 4.5 is the Eklof lemma for abelian categories. This is for direct systems, but you can pass to the inverse systems by inverting the arrows.
2025-03-21T14:48:31.770738
2020-08-14T06:39:45
369143
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Brendan McKay", "Denis Serre", "Emil Jeřábek", "GA316", "Mark Wildon", "https://mathoverflow.net/users/100231", "https://mathoverflow.net/users/12705", "https://mathoverflow.net/users/33047", "https://mathoverflow.net/users/7709", "https://mathoverflow.net/users/8799", "https://mathoverflow.net/users/9025", "vidyarthi" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632081", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369143" }
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Eigenvalues of the complement of a graph Let $A$ and $\widetilde A$ be the adjacency matrices of a graph $G$ and of its complement, respectively. Is there any relation between the eigenvalues of $A + \widetilde A$ and the eigenvalues of $A$ and $\widetilde A$? Also, do $A$ and $\widetilde A$ have the same set of eigenvectors? Thank you. $A+\tilde A$ is the adjacency matrix of the complete graph regardless of what $G$ is. On your second question, "yes" if $G$ is regular, otherwise "not usually". @BrendanMcKay If $G$ is regular, how to show they have the same set of eigenvectors? Thank you. I edited my answer. In particular, I give a complete answer for regular graphs. @DenisSerre Thank you. Edit (bis). There are two answers, depending on whether loops about vertices are allowed or not. In addition, the case of regular graphs is completely described. If loops are allowed The relation between matrices is $$A+{\widetilde A}=J$$ where $J={\bf1}{\bf1}^T$ is the all-ones matrix. The first consequence is that the sum of the eigenvalues of $A$ and ${\widetilde A}$ equals $|V|$ where $V$ is the set of vertices. A second consequence concerns multiple eigenvalues. If $\lambda$ is an eigenvalue of $A$, of multiplicity $m\ge2$, then $-\lambda$ is an eigenvalue of ${\widetilde A}$, of multiplicity larger than or equal to $m-1$. Just consider the intersection of the eigenspace with the hyperplane ${\bf1}^\bot$. In particular, this is a case where $A$ and ${\widetilde A}$ share common eigenvectors. If loops are not allowed Here $$A+{\widetilde A}=K:=J-I_V$$ The sum of the eigenvalues of $A$ is the opposite of that of ${\widetilde A}$. If $\lambda$ is an eigenvalue of $A$, of multiplicity $m\ge2$, then $-1-\lambda$ is an eigenvalue of ${\widetilde A}$, of multiplicity larger than or equal to $m-1$. Again, this is a case where $A$ and ${\widetilde A}$ share common eigenvectors. Regular graphs If a graph is regular and connected (thanks to Emil for having pinned the point), then $\bf1$ is an eigenvector, with eigenvalue $d$, the degree of each vertex. It is a simple eigenvalue because $A$ is irreducible (connectedness). The other eigenspaces are contained in $\bf1^\bot$, because $A$ is symmetric. Thus eigenvectors of $A$ remain eigenvectors for $\widetilde A$, with the same multiplicity. The correspondance between eigenvalues is $\lambda\rightarrow-1-\lambda$. Remark also that $d$ is the Perron eigenvalue of $A$, $n-1-d$ being that of $\widetilde A$. We thus deduce $$\lambda\in D(0;d)\cap D(-1;n-1-d)$$ for all the other eigenvalues of $A$. but $A+\widetilde{A}$ equals the adjacency matrix of the complete graph and not $J$, right? @vidyarthi. This is a matter of convention. The question should mention whether loops about a vertex are allowed, or not. @vidyarthi. See my edits. thanks! deleted @DenisSerre Thanks for the nice answer. What happens in the case when $\lambda$ is an eigenvalue of $A$ with multiplicity $1$? Can you share some references for the part regarding common eigenvectors? Thank you again. The orthogonality trick with $\mathbf{1}\mathbf{1}^t - A = \widetilde{A}$ and so $-Av = \widetilde{A}v$ if $\mathbf{1}^t v$ is really nice! (And perhaps even a counterexample to many answers, including mine, on a recent popular question about the virtues of writing $\langle u,v \rangle$ rather than $u^t v$ for inner products.) I don't see how you get that for regular graphs, the eigenvectots of $A$ remain eigenvectors of $\tilde A$. Yes, $\vec1$ remains an eigenvector, and eigenvectors perpendicular to $\vec1$ as well, and in particular, there is an orthogonal basis of common eigenvectors. However, assume that the graph is disconnected, hence the eigenspace of $d$ has dimension $\ge2$. Then some vectors from this space (namely $\vec1$) will be eigenvectors of $\tilde A$ of value $n-d-1$, and some (those perpendicular to $\vec1$) will be eigenvectors of value $-d-1$. But there are also vectors neither parallel ... ... nor perpendicular to $\vec1$, and as far as I can see, these will not remain eigenvectors of $\tilde A$. In fact, eigenvectors of $A$ remain eigenvectors of $\tilde A$ if and only if $G$ is regular and connected (hence $A$ and $\tilde A$ havce the same eigenvectors iff $G$ is a regular connected graph with connected complement). To see this, if $Av=\lambda v$, then $\tilde Av=(\vec 1^Tv)\vec1-\lambda v$ by your argument, which shows that all eigenvectors remain eigenvectors iff all eigenvectors of $A$ are parallel or orthogonal to $\vec1$. Since eigenvectors of a symmetric matrix span the whole space, they can’t all be orthogonal to $\vec1$, hence $\vec1$ is an eigenvector, ... ... i.e., $G$ is regular of some degree $d$. Then, as I already pointed out, the condition also implies that the eigenvalue $d$ must have multiplicity $1$, thus $G$ is connected. Conversely, if $G$ is regular and connected, then the eigenspace of eigenvalue $d$ consists of multiples of $\vec1$, and other eigenspaces are orthogonal to it.
2025-03-21T14:48:31.771075
2020-08-14T06:48:44
369144
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "GH from MO", "Pedja", "https://mathoverflow.net/users/11919", "https://mathoverflow.net/users/88804" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632082", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369144" }
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Primality test for $N=4p+1$ Can you prove or disprove the following claim: Let $N=4p+1$ where $p$ is an odd prime number , let $T_n(x)$ be the nth Chebyshev polynomial of the first kind and let $F_n(x)$ denote an irreducible factor of degree $\varphi(n)$ of $T_n(x)$ . If there exists an integer $a$ such that $F_{p}(a) \equiv 0 \pmod{N} $ then $N$ is a prime. You can run this test here. An incomplete list of primes $p$ such that $4p+1$ is prime can be found here . I have verified this claim for $p \in [3,30000)$ with $a \in [1,100]$ . The claim is true, and it holds more generally for every odd integer $p\geq 3$; the assumption that $p$ is prime is not needed. By the known factorization of Chebyshev polynomials, $$F_p(x/2)=\prod_{\substack{1\leq m\leq 2p-1\\(m,4p)=1}}(x-\zeta^m-\zeta^{-m}),$$ where $\zeta\in\mathbb{C}$ is a primitive $4p$-th root of unity. The splitting field of $F_p(x/2)$ is the subfield of the $4p$-th cyclotomic field fixed by complex conjugation; it is of degree $\varphi(p)$. Assume that $q\nmid 2p$ is a prime number such that the reduction of $F_p(x/2)$ mod $q$ has a root in $\mathbb{F}_q$. The roots of $F_p(x/2)$ in $\overline{\mathbb{F}_q}$ are of the form $\xi^m+\xi^{-m}$, where $\xi\in\overline{\mathbb{F}_q}$ is a primitive $4p$-th root of unity. By assumption, the Frobenius automorphism $t\mapsto t^q$ fixes one of these roots, which is only possible when $q\equiv\pm 1\pmod{4p}$. It follows that, for any $a\in\mathbb{Z}$, the prime factors of $F_p(a)$ coprime to $2p$ are congruent to $\pm 1$ modulo $4p$. In particular, if $4p+1$ divides $F_p(a)$, then the only prime factor of $4p+1$ can be itself, i.e., $4p+1$ is prime. Is it possible in your opinion to prove generalization of this claim (for $N=4n+1$ , where $n$ is an odd natural number) . @PeđaTerzić: Yes, the generalized claim is also true. I updated the first and third sentences to reflect this.
2025-03-21T14:48:31.771242
2020-08-14T07:09:44
369145
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Deane Yang", "Liviu Nicolaescu", "Romain Gicquaud", "Yuxiao Xie", "https://mathoverflow.net/users/20302", "https://mathoverflow.net/users/24271", "https://mathoverflow.net/users/613", "https://mathoverflow.net/users/90076" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632083", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369145" }
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Reference request: $L^p$ regularity for elliptic differential operators I am looking for a book that proves the following: Let $M$ be a closed manifold, $L$ an elliptic differential operator on $M$ of order $k$ and with smooth coefficients. Suppose $u$ is a distribution on $M$ such that $Lu\in L^p$ ($1<p<\infty$), i.e., there exists $v\in L^p$ such that $\langle u,L^*\phi\rangle=\int_Mv\,\phi$ for all smooth $\phi$, where $L^*$ is the formal adjoint of $L$, $\langle\cdot,\cdot\rangle$ is the dual pairing of distributions and test functions. Then $u\in W^{k,p}$ and $\|u\|_{W^{k,p}}\leq C(\|Lu\|_{L^p}+\|u\|_{L^p})$. The case $k=2$ is discussed in many books on elliptic PDEs. Also, it seems that the case $p=2$ appears in many books, for example the first volume of M. E. Taylor's trilogy Partial Differential Equations. However, I have not found a proof for the general case. Here are my questions: Does this result necessarily rely on the theory of pseudodifferential operators? What are some accessible books for learning its proof, with or without pseudodifferential operators? (For example, I believe this result is contained in Hörmander's tetralogy on linear PDEs, but it could be very difficult to study from those books...) In general it follows from local elliptic regularity in Euclidean space by freezing the coefficients. Have you tried to look at "Fredholm operators and Einstein metrics on conformally compact manifolds" by John Lee. Lemma 4.8 seems to be what you need. See section 10.3.2 of https://www3.nd.edu/~lnicolae/Lectures.pdf The regularity part is proved only for first order operators. The general strategy is outlined and references are provided. @LiviuNicolaescu In fact, this question arose when I was reading your notes (which by the way are very well written!). However, it seems that one has to read hundreds of pages on pseudodifferential operators to get this specific result. I wonder if there are easier proofs available... Perhaps this is in the work of Agmon-Douglis-Nirenberg. @Colescu That was my problem. I was not able to produce a reasonably succinct proof of the regularity other than in those two cases. As Deane Yang pointed out, the first complete proof of $L^p$-regularity is in the papers of Agmon-Douglis-Nirenberg that deal with operators acting on vector valued functions. For scalar operators you can find a shorter proof in the book Bers_John-Shechter Partial Differential Equations. Both sources construct a parametrix of sorts. Everything reduces to the Calderon-Zygmund inequality. Thanks for all the helpful comments! I'm very surprised that there does not seem to be any textbook account of the complete proof of this elegant result...
2025-03-21T14:48:31.771445
2020-08-14T09:46:02
369153
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Jason Starr", "https://mathoverflow.net/users/13265" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632084", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369153" }
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Perfect complexes on stacks are strict? As mentioned in the comments of this question, on a quasiprojective scheme over a field, every perfect complex is globally a complex of vector bundles. I have some question about the extension of this to stacks: Is this true $G$ equivariantly? i.e. is it true for $X/G$ where $X$ is quasiprojective? As a slight extension of this, is this true for $\text{Bun}_G(X)$ where $X$ is a curve? (and 3. it would be nice to have a modern reference to the original question about quasiprojective schemes, which I've not actually been able to find yet). Yes, 1 is true, at least for bounded perfect complexes. The key point is that for every perfect complex of amplitude $[a,b]$, for a $G$-equivariant complex of vector bundles of amplitude $[a+1,b]$ and a $G$-equivariant morphism from that complex to the original complex that is a quasi-isomorphism on the "good truncations" $\tau_{>a}$, the cone is a perfect complex concentrated in one degree, which then has a $G$-equivariant homology sheaf that is a $G$-equivariant vector bundle. So now you can "adjoin" this to the $G$-equivariant complex of vector bundles. Surely there is a reference . . .
2025-03-21T14:48:31.771570
2020-08-14T10:04:46
369154
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Anton Klyachko", "David E Speyer", "Donu Arapura", "Fedor Petrov", "Geoff Robinson", "Jeremy Rickard", "LSpice", "darij grinberg", "https://mathoverflow.net/users/14450", "https://mathoverflow.net/users/22989", "https://mathoverflow.net/users/2383", "https://mathoverflow.net/users/24165", "https://mathoverflow.net/users/2530", "https://mathoverflow.net/users/297", "https://mathoverflow.net/users/4144", "https://mathoverflow.net/users/4312" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632085", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369154" }
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Is simultaneous similarity of matrices independent from the base field? Suppose that $F$ is a subfield of a field $E$ and, for $n\times n$ matrices $A_1,\dots,A_m, B_1,\dots,B_m$ over $F$, there exists a matrix $T\in{\rm GL}_n(E)$ such that $T^{-1}A_iT=B_i$ for all $i$. Does this imply that such a matrix $T$ can be chosen from ${\rm GL}_n(F)$? It is easy to see that the answer is yes if $m=1$; and yes if the field $F$ is infinite. I have edited that. In these cases the answer is yes. I think that if the $A_{i}$ span ${\rm M}_{n}(F)$, then I the answer is yes, essentially as in the Noether-Doering Theorem, there is a solution to the linear equations in the smaller field, because there is one in the bigger field. This solution is then unique up to non-zero scalar multiples, so since the solution in the bigger field gives an invertible matrix, so does the one in the smaller field. @Geoff why is it unique? @FedorPetrov :It's Schur's Lemma I think. The $A_{i}$ span $M_{n}(F)$ over $F$ and certainly span $M_{n}(G)$ over $G$. Hence the $B_{i}$ span $M_{n}(G)$ as well. Thus, over $G$ , if there is a non-zero matrix $S$ with $A_{i}S = SB_{i}$ for each $i$. Then $ImS$ is an invariant subspace of the vectors space $G^{n}$ of column vectors., so is the whole space $G^{n}$. Thus $S$ is invertible, and the we have $TS^{-1}A_{i}ST^{-1} = A_{i}$ for each $i$, so $ST^{-1}$ is scalar. Doesn't the Noether-Deuring theorem work anyway? The hypotheses give us two finite-dimensional modules over the free algebra $F\langle x_1,\dots,x_m\rangle$ that become isomorphic when you extend scalars to $G$, so by Noether-Deuring they are already isomorphic over $F$. @Jeremy Rickard: I think it probably does, but I always remember the irreducible case Irreducible in my proof above meant under left multiplication by all of $M_{n}(G)$. @GeoffRobinson ah, $Im(S)$ is invariant for all $A_i$, thus for all matrices, thus it is the whole space $F^n$, and $S$ is invertible. This seems to avoid the uniqueness argument. I like this question, but I have never before seen someone so notationally bold as to use $G$ for a field. (My personal field alphabet has $E$ or $K$ as the next letter after $F$.) Now what do we call algebraic groups over $G$? // Also, is it any help to look at a matrix $\widetilde T$ over $F$ conjugating $A_1 \oplus \dotsb \oplus A_m$ to $B_1 \oplus \dotsb \oplus B_m$? Yes, this follows from the Noether-Deuring theorem, as @JeremyRickard has said. You don't even need to consider infinite-dimensional algebras for that; it suffices to use the subalgebra of the matrix ring generated by the matrices $A_1, A_2, \ldots, A_m$. As everyone is saying, this follows from Noether-Deuring. See https://mathoverflow.net/questions/28469/hilbert-90-for-algebras for a quick proof. I also asked this question a while back https://math.stackexchange.com/questions/305696 . @LSpice, I have changed the letter:) I’m voting to close this question because @DavidESpeyer points out it's a duplicate of https://math.stackexchange.com/questions/305696 . @LSpice I don't agree. While it's true this was asked and answered before, the sites are different. Some of us don't usually look at the other one. This question is answered in comments: "As everyone is saying, this follows from Noether-Deuring. See mathoverflow.net/questions/28469/hilbert-90-for-algebras for a quick proof. I also asked this question a while back math.stackexchange.com/questions/305696."         – David E Speyer Thanks to all!
2025-03-21T14:48:31.771847
2020-08-14T11:06:24
369156
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Qmechanic", "https://mathoverflow.net/users/13917" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632086", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369156" }
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What is known about the density of states for the Anderson Model? The Anderson Model is given by the random Hamiltonian (as an operator on $l^2(\mathbb{Z}^d)$) $$ H_\omega = - \triangle + V(\omega) $$ where $V(\omega) \mid x \rangle = \omega(x) \mid x \rangle$ with $\{ \omega(x) \}_{x \in \mathbb{Z}^d}$ independent and uniformly distributed in $[-L, L]$. It is known for example by Kunz and Soulliard that the spectrum $ \sigma(H_\omega) = [-L, 4d +L ]$. This is also easily verified by simulating finite dimensional approximations. By simulation it is also easy to look at the density of states. Below I have plotted the density of states for $L=0,1,5$ for a $500 \times 500$ approximation. One can see that the high probabilities towards the end of the spectrum wash out. So far this is all numerics, but what is known analytically about this particular density of states? $L=0,1,5$ for 500x500 approximatio" /> Crossposted to https://physics.stackexchange.com/q/574809/2451
2025-03-21T14:48:31.771944
2020-08-14T12:12:23
369159
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Anonymous", "Arkady", "Henno Brandsma", "https://mathoverflow.net/users/163753", "https://mathoverflow.net/users/2060", "https://mathoverflow.net/users/89233" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632087", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369159" }
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adding one point from the Stone-Cech compactification Let $X$ be any non-compact Tychonoff space and $\beta X$ be its Stone-Čech compactification. The following fact is known: any point $p$ from the reminder $\beta X \setminus X$ is not a $G_{\delta}$-set in $\beta X$. Consider a topological space $X^{\ast}$ which is the space $X \cup \{p\}$ equipped with the topology induced from $\beta X$. It is easy to see that if $X$ is a $\sigma$-compact space, i.e. $X$ is a countable union of compact subspaces, then $p$ is $G_{\delta}$ in $X^{\ast}$. I ask the following question: is the converse true? In other words a general question is: does there exist a non- $\sigma$-compact Tychonoff space $X$ such that $\{p\}$ is a $G_{\delta}$-set in $X \cup \{p\}$ for any point $p$ from the reminder $\beta X \setminus X$? One characterization of realcompactness is that $X$ is realcompact if and only if for each $p \in \beta X \setminus X$, there is a zero-set $Z$ of $\beta X$ such that $p \in Z \subseteq \beta X \setminus X$. Therefore, taking $X$ to be any realcompact space which is not $\sigma$-compact gives an example. You are right, thank you. So $\Bbb P$, the irrationals, will do. It's Lindelöf so realcompact. And from its homeomorphic representation $\omega^\omega$ it's easily seen to be non-$\sigma$-compact.
2025-03-21T14:48:31.772063
2020-08-14T13:05:47
369164
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "AG learner", "Feng Hao", "Hacon", "Jason Starr", "https://mathoverflow.net/users/13265", "https://mathoverflow.net/users/19369", "https://mathoverflow.net/users/63996", "https://mathoverflow.net/users/74322" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632088", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369164" }
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Singularities of a morphism from a smooth projective variety to an abelian variety Let $f: X\to A$ be a (flat) morphism from a smooth complex projective variety $X$ to an abelian variety $A$. Consider the following natural diagram: $$T^*X\overset{df}{\longleftarrow}X\times H^0(A, \Omega_A^1)\overset{f\times id}{\longrightarrow}A\times H^0(A, \Omega_A^1)$$ where $T^*A$ and $T^*X$ are the cotangent bundles, and $df$ is the usual differential. Notice that $$(f\times id)(df^{-1}(o_X))=\{(a, \omega)| \ \text{there is a point}\ x\in f^{-1}(a)\ \text{such that}\ f^*\omega(x)=0\},$$ where $o_X$ is the zero section of $T^*X.$ My question is: Is there an example that there is an irreducible component $Z$ of $(f\times id)(df^{-1}(o_X))$ such that $\dim Z<\dim A$? I am curious about this question, because I noticed in Schnell's notes (Lemma in page 3), we always have $\dim (f\times id)(df^{-1}(o_X))\leq\dim A$. I think such an example will come from a projective example of your previous question. It should be easy to make local examples global and then generically finite over an AV, see for example the proof of Prop 3.13 of https://arxiv.org/pdf/1212.5105.pdf @AGlearner Sorry I messed up example and counterexample. I think an projective example of my previous question, i.e., a proper map $f: U\to V$ with $\dim f(Z)<r$ cannot give a example of above question, since the image of the tangent map at singular points of a singular fiber may move around downstairs. This a priori can make each irreducible of $(f\times id)(df^{-1}(o_X))$ is of the right dimension $\dim A$. However, if one can prove the previous question is a right statement for $f$ being proper and flat. Then each irreducible component has dimension $\dim A$ if $f: X\to A$ is flat. @Hacon Thank you very much for your comment! Right now I don't think I have a local example. For the example of my previous question $f: \mathbb{A}^5\to \mathbb{A}^3; (x_1, x_2, x_3, x_4, x_5)\mapsto (x_1, x_2, x_1x_3+x_2x_4+x_5^2)$. The ``$(f\times id)(df^{-1}(o_X))$'' has dimension 3. One local example is a modification of the earlier example, namely $f(x,y,z)=(x,x^2z+y^2).$ @JasonStarr Thank you for the interesting example! Indeed, the image of the tangent map in this example is fixed. However, if I understand Prof. Hacon's comment and the paper correctly, I think to do the procedure in Prop 3.13, I need to find a local example (local from the base), i.e., it should be a proper morphism to a variety. For the example $f(x,y,z)=(x, x^2z+y^2)$, one should first projectivize the fibers of it and make sure the whole family smooth. After this, it is not clear to me that I will not add more covectors, so that $\dim (f\times id)(df^{-1}(o_X))$ does not increase.
2025-03-21T14:48:31.772253
2020-08-14T13:35:00
369165
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Christian Remling", "Goulifet", "LSpice", "Michael Renardy", "Robert Israel", "https://mathoverflow.net/users/12120", "https://mathoverflow.net/users/13650", "https://mathoverflow.net/users/2383", "https://mathoverflow.net/users/39261", "https://mathoverflow.net/users/48839" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632089", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369165" }
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Riesz Representation Theorem for $L^2(\mathbb{R}) \oplus L^2(\mathbb{T})$? The spaces $L^2(\mathbb{R})$ (square-integrable functions) and $L^2(\mathbb{T})$ (1-periodic square-integrable functions, considered over the real line $\mathbb{R}$) are two subspaces of the space of tempered distributions $\mathcal{S}'(\mathbb{R})$ and one can easily show that the sum $L^2(\mathbb{R}) \oplus L^2(\mathbb{T})$ is direct. The duals of $L^2(\mathbb{R})$ and $L^2(\mathbb{T})$ are isometrically isomorphic to $L^2(\mathbb{R})$ and $L^2(\mathbb{T})$, respectively (Riesz representation theorem). Therefore, the continuous dual of the direct sum is simply $L^2(\mathbb{R}) \oplus L^2(\mathbb{T})$ in the sense that (1) an element $g_1 + g_2 \in L^2(\mathbb{R}) \oplus L^2(\mathbb{T})$ defines a continuous linear functional over $L^2(\mathbb{R}) \oplus L^2(\mathbb{T})$ via $$(f_1 + f_2) \mapsto \langle f_1 , g_1 \rangle_{L^2(\mathbb{R})} + \langle f_2 , g_2 \rangle_{L^2(\mathbb{T})}$$ (which uses that both decompositions $f = f_1 + f_2$ and $g = g_1+g_2$ are unique), and that (2) any element of $(L^2(\mathbb{R}) \oplus L^2(\mathbb{T}))')$ is of this form. I would like to identify the subset $\mathcal{X}\subset \mathcal{S}'(\mathbb{R})$ of functions $g$ such that $$ L^2(\mathbb{R}) \oplus L^2(\mathbb{T}) \ni f_1 + f_2 \mapsto \int_{\mathbb{R}} g(x) (f_1 + f_2)(x)\mathrm{d}x$$ specifies a continuous linear functional over $L^2(\mathbb{R}) \oplus L^2(\mathbb{T})$. Clearly, by restricting it to $L^2(\mathbb{R})$ (i.e. setting $f_2=0$), we need to have $g \in L^2(\mathbb{R})$. Moreover, $\mathcal{X}$ contains any square-integrable compactly supported functions, but also functions that are not compactly supported but that have sufficiently nice asymptotic properties such that the integral $\int_{\mathbb{R}} g (x) f_2(x)\mathrm{d}x$ is well-defined for any square-integrable periodic $f_2$ and defines a continuous functional over $L^2(\mathbb{T})$. Question: Is there a way to identify the space $\mathcal{X}$ I am depicting? Can we reach any linear functionals over $L^2(\mathbb{R}) \oplus L^2(\mathbb{T})$ by doing so? I am also interested by generalization to other direct sums between spaces of periodic and non-periodic functions (e.g., $L^p$-spaces, or spaces of continuous-functions for the supremum norm). To regard $\mathcal L^2(\mathbb T)$ as a subspace of the tempered distributions, you pull back to periodic functions on $\mathbb R$ and then integrate them against Schwartz functions? $\mathcal L^2(\mathbb R) \oplus \mathcal L^2(\mathbb T)$ can be considered as $\mathcal L^2$ of the disjoint union of a line and a circle, with Lebesgue measure on each. What do you mean by $f_1+f_2$ in the last displayed equation? There doesn't seem to be an obvious way to define this sum that gives an element of $L^2(\mathbb R)$ (which is what you need here). @LSpice Yes, that's right, I will had a precision about that. @ChristianRemling Good point, $f_1 + f_2$ is not in $\mathcal{L}^2(\mathbb{R})$ but still, so the notation $\langle g, f_1+f_2\rangle$ is a bit ambiguous, but I am considering the integration of the product $g \times (f_1+f_2)$. I am updating accordingly. @MarkWildon It is not a typo but my way of writing is not very good, as Christian Remling also pointed out. I try to update for a better version. @Goulifet: Thanks for clarifying. You can certainly not get all functionals $F$ on $L^2(\mathbb R)\oplus L^2(\mathbb T)$ in this way because $g$ is already determined by $F$ restricted to the first summand. @ChristianRemling I completely agree and this is part of the problem. The first displayed equation specifies a linear functional, but the $g$ in the second one is definitely not in the direct sum. I hope to find the class of functions such that the second displayed equation "works" (gives a continuous linear functional over $L^2(\mathbb{R}) \oplus L^2(\mathbb{T})$). A good candidate would be the set of all functions $g\in L^2(R)$ for which $\sum_{k\in Z} g(\cdot+2\pi k)$ is in $L^2(0,2\pi)$. @MichaelRenardy This seems reasonable: it would be all the square-integrable functions such that the periodization is well-defined and specifies a square-integrable periodic function. The big question is then to clarify which functions can be periodized in this way. Just so you are aware, any edit, even for a change of notation like $\mathcal L^2 \rightarrow L^2$, still bumps a question. @LSpice Sure, I will remember that.
2025-03-21T14:48:31.772648
2020-08-14T14:09:13
369167
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Ashutosh", "Chris Lambie-Hanson", "Johannes Schürz", "https://mathoverflow.net/users/134910", "https://mathoverflow.net/users/26002", "https://mathoverflow.net/users/2689" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632090", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369167" }
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The stationary reaping number $\mathfrak{r}_{cl}$ Let $\kappa$ be at least inaccessible (but measurable is what I am primarily interested at the moment). Let $x,y \in [\kappa]^\kappa$ both be stationary. We say that $y$ stationary-splits $x$ iff $x \cap y$ and $x\setminus y$ are both stationary. Define $\mathfrak{r}_{cl}:=\min \{\vert \mathcal{R}\vert \colon \,\, \forall x \in \mathcal{R} \,\, x \, \text{is stationary} \, \land \neg \exists y \in [\kappa]^\kappa \,\, \mathcal{R} \, \, \text{is stationary-split by} \,\, y\}$. Of course, $\mathfrak{r}_{cl}$ has to be infinite and $\mathfrak{r}_{cl}= 2^\kappa$ in the $\kappa$-Sacks model. But can ZFC prove a nontrivial lower bound ($\omega_1,\, \kappa \,\, \text{or even} \,\, \kappa^+$) for $\mathfrak{r}_{cl}$ ?? Note that this question is motivated by investigating cardinal characteristics on the 'higher' Cantor/ Baire space modulo the non-stationary (not the bounded) ideal. A theorem of Balcar and Vojtas (Theorem 3.14 in Chapter I of Handbook of Boolean algebras Vol. I) says that if $\mathbb{B}$ is a nowhere $\lambda^+$-saturated boolean algebra, then every $F \in [\mathbb{B}^+]^{\leq \lambda}$ has a disjoint refinement. It follows that $\mathfrak{r}_{cl}(\kappa) \geq \kappa$ for every regular limit $\kappa$. If you also have diamond on every stationary subset of $\kappa$, then this can be improved to $\kappa^+$. But I am not sure about the general case. @Ashutosh You should write this as an answer, since it gives a stronger result than my answer gives. @Ashutosh Can you please elaborate how diamond would help to get $\kappa^+$ as a lower bound. I think that $\omega_1$ is certainly a lower bound, and that this is the case for any regular, uncountable $\kappa$. To see this, suppose that $\langle x_n \mid n < \omega \rangle$ is a sequence of stationary subsets of $\kappa$. We'll find a $y$ that stationary-splits every $x_n$. We'll do this by constructing a sequence $\langle x_n^* \mid n < \omega \rangle$ such that $x_n^*$ is a stationary subset of $x_n$ and $x_n^* \cap x_m^* = \emptyset$ for all $n < m < \omega$. Then we can let $y_n$ be a stationary subset of $x_n^*$ such that $x_n^* \setminus y_n$ is also stationary, and let $y = \bigcup_{n < \omega} y_n$. The construction of $\langle x_n^* \mid n < \omega \rangle$ can be done by recursion on $n$, where we maintain the following additional recursion hypothesis: for all $n \leq m < \omega$, the set $x_{m,n} := x_m \setminus \bigcup_{k < n} x_k^*$ is stationary. Fix $n < \omega$ and suppose that $\langle x_k^* \mid k < n \rangle$ has been constructed. By hypothesis, $x_{n,n}$ is stationary; partition it into $\omega_1$-many disjoint stationary subsets, $\langle x^\alpha_{n,n} \mid \alpha < \omega_1 \rangle$. For each $m > n$, there is at most one $\alpha < \omega_1$ such that $x_{m,n} \setminus x^\alpha_{n,n}$ is nonstationary, so we can choose an $\alpha^* < \omega_1$ such that $x_{m,n} \setminus x^{\alpha^*}_{n,n}$ is stationary for all $m > n$, and then set $x_n^* = x^{\alpha^*}_{n,n}$, and continue to the next step. It's not immediately clear to me how to achieve a better lower bound. This particular construction breaks down at limit steps, but it seems conceivable that a more clever version of the argument might yield a lower bound of $\kappa$ or even $\kappa^+$. Your proof can be generalized as follows: Let $(x_\alpha){\alpha < \beta}$ be a family of stationary reals of size $\beta < \kappa$. The point is that when we choose the $\langle x{\alpha, \alpha}^{\gamma} \colon \gamma < \beta_\alpha \rangle$ ($\beta_\alpha > \beta$) we make sure that it refines $\langle x_{\alpha', \alpha'}^{\gamma} \colon \gamma < \beta_{\alpha'} \rangle$ for all $\alpha' < \alpha$ and partions every $x_{\alpha''}$ (if not disjoint modulo non-stationary) into at least $\beta^+$ many pieces for every $\alpha''> \alpha$. Here we use that $\kappa$ is inaccessible (for the limit steps) and that a stationary set cannot be split into $< \kappa$ non-stationary sets. This way we can assure that the $x_{\alpha', \alpha}$ (for $\alpha' \geq \alpha$) remain stationary even in limit steps. The point is that the $x_\alpha^{*}$ are only a very small piece (one of at least $\beta^+$ many) of an $x_{\alpha'}$ (if not event totally disjoint) for any $\alpha' > \alpha$.
2025-03-21T14:48:31.772945
2020-08-14T14:25:12
369168
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Louis D", "https://mathoverflow.net/users/17798" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632091", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369168" }
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Two more counterexamples to a conjecture from 1975 about hamiltonicity of digraphs Question from 2013 gives one counterexample to Nash-Williams's conjecture 1975 about hamiltonicity of dense digraphs. In the linked answer, @LouisD "reverse engineered" the counterexample and pointed out it is special case in another paper. We found 2 more digraph counterexamples with directed edges: g1=[(0, 3), (0, 4), (1, 3), (1, 5), (2, 4), (2, 5), (3, 0), (3, 1), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4, 5), (5, 0), (5, 2), (5, 3), (5, 4)] g2=[(0, 2), (0, 3), (0, 4), (0, 5), (1, 3), (1, 4), (2, 0), (2, 4), (3, 0), (3, 1), (3, 4), (3, 5), (4, 0), (4, 3), (5, 0), (5, 1), (5, 2), (5, 3), (5, 4)] g3=[(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 5), (2, 0), (2, 1), (2, 4), (2, 5), (3, 1), (3, 5), (4, 1), (4, 2), (5, 0), (5, 1), (5, 2), (5, 3)] Q1 Are these counterexamples true? Q2 Can we get infinitely many counterexamples from the small ones? All three counterexamples are planar. These two examples are closely related in the sense that g3 can be obtained from g2 by deleting the edge (5,4). One can verify that both of these satisfy the Nash-Williams condition and that g2 does not have a Hamiltonian cycle. Thus g3 does not have a Hamiltonian cycle. Since they are both planar, it would probably be helpful to draw them that way so the pictures can be read more clearly. I suggest redrawing the first picture with the vertices in three rows: 5 centered in the first row, 2,0,3,1 in the second row in that order, 4 centered on the bottom. Similarly with the second picture.
2025-03-21T14:48:31.773059
2020-08-14T15:06:49
369170
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Christian Remling", "https://mathoverflow.net/users/48839" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632092", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369170" }
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When does a one-dimensional Schrödinger operator have a threshold resonance? Consider the operator $$ L = -\partial_x^2 + V(x),$$ for some bounded, decaying potential, i.e. $V(x)\to 0$ as $x\to \pm \infty$. I'm interested in the $L^2(\mathbb R)$ spectrum of $L$. We know that $L$ has continuous spectrum $[0,\infty)$, and in general, $L$ may or may not have negative discrete eigenvalues, depending on $V$. My question: Are there any known conditions on $V$ that guarantee the existence or non-existence of a threshold resonance, i.e. a pole of the resolvent $(L-\lambda)^{-1}$, at $\lambda = 0$? Actually, $L$ need not have continuous spectrum on $(0,\infty)$, the spectrum can be pure point. What is true is that the essential spectrum equals $[0,\infty)$.
2025-03-21T14:48:31.773151
2020-08-14T15:26:33
369172
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Anthony Quas", "Kernel", "Mateusz Kwaśnicki", "https://mathoverflow.net/users/108637", "https://mathoverflow.net/users/11054", "https://mathoverflow.net/users/52960" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632093", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369172" }
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Hitting measure/overshoot for random walk in $\mathbb{Z}$ with heavy-tail Let $\alpha \in (0,2)$, (or for simplicity just $\alpha \in (1,2)$) and let $X_1,X_2,\dots$ be an i.i.d collection of random variables with common distribution $$ p(x,y)= \frac{c_\alpha}{|x-y|^{1+\alpha}}1_{[x\neq 0]} $$ where $c_\alpha = \frac{1}{2\zeta(1+\alpha)}$ is the normalising constant. Now, consider the random walk $\{S_n\}$ given by $S_n = \sum_{i=1}^n X_i$ and $S_0=0$. For a given $n\ge 1$, consider $\Lambda_n = [-n,n]$, I would like to know whether good bounds for the hitting measure defined as $$ H_{\Lambda_n}(0,x)= P_0[S_{\tau_n}=x], $$ where $\tau_n = \inf\{n\ge 0, |S_n|\ge n\}$. I have seen such problems under the name of overshoot and undershoot bounds. But I have not found an article dealing with the setting without a drift and with heavy tail random walks. It is easy to get very bad bounds, but then they are far from matching. From a few simulations I made, it seems like it decays as a power of $(|x|-n)$, but I couldn't figure out what was the exponent for an arbitrary $\alpha$. Are there good bounds known for $H_{\Lambda_n}(0,x)= P_0[S_{\tau_n}=x]$? EDIT: As mentioned in the comments bellow, one can use that for $x>n$ $$ (1) \qquad \frac{c_\alpha}{(x+2n)^{1+\alpha}} \le H_{\Lambda_n}(0,x)\le \frac{c_\alpha}{(x-n)^{1+\alpha}} $$ and therefore for $x>cn$ for some $c>1$, we have that $H_{\Lambda_n}(0,x)\asymp n^{-1-\alpha}$. So the question is really about the remaining $x \in [n+1,cn]$. As the bounds given in $(1)$ would lead to upper and lower bounds that that do not match. If you're interested in the regime $|x|\ge 2n$, then it's obvious that the decay is like $1/(|x|)^{1+\alpha}$: for any point in $\Lambda_n$, the probability of jumping to $x$ is within a constant factor of $p(0,x)$. Agreed, the problem is really between $n+1$ and $cn$ for some $c>1$. Maybe I should edit the question to reflect that. But I really mean good bounds taking small $x$ into account. So I think editing questions to change what they’re really asking is frowned upon. In this case, I don’t think the question is being changed. I’d see it as a clarification rather than a new question, but I think it is preferred to mark the edit as such (e.g. by adding a paragraph at the end labeled EDIT) in order that future readers can make sense of the comments. So I have been doing some back-of-the-envelope calculations (I don't think I have the technical expertise to prove this). I believe that the hitting measure looks like $1/N(|x|-N)^\alpha$ for $|x|<2N$ and $1/|x|^{1+\alpha}$ for larger $x$. My rough argument is to consider jumps of size up to $N/1000$ as "small" and anything else as large. Then the variance of a single small step is something like $N^{2-\alpha}$. Thus the standard deviation coming from the small steps is of size $N$ after $N^\alpha$ steps. Heuristically after this order number of steps, the position is uniformly distributed on $[-N,N]$. It is then the bigger jumps that are responsible for leaving the interval. If a jump is of size $N/L$ for some $L$, then my heuristic would say you have a $1/L$ chance of leaving the interval. Summing the contributions leads to my back-of-the-envelope answer. This is an interesting approach, I will definitely try. Thank you for your suggestion! Hmmm - it seems my hitting measure doesn't integrate to anything close to 1... OK - I see what I did wrong - if you make a jump of size N/L and don't leave the interval, I stopped keeping track of what happens. What I should instead assume you re-randomize over the entire interval in that case. I think this means my first comment (and hence your EDIT) is wrong: you are looking at where you go conditional on leaving, where I was looking at unconditional probabilities in each step. But I now have a new guess! Because you re-initialize if you don't escape, this just corresponds to normalizing the probability distribution I guessed before So my revised version is the hitting density should look like $1/(|x|-n)^\alpha$ on $[n,2n]$ and $n/|x|^{\alpha+1}$ beyond. I noticed another problem. Both of us seemed to have used that $P_0(X_{\tau_n}=x)= \sum_{y\in[-n,n]}P_0(X_{\tau_n-1}=y)P_0(X_{\tau_n}=x|X_{\tau_n-1}=y) $ $=\sum_{y\in[-n,n]}P_0(X_{\tau_n-1}=y)p(x,y)$. But the last identity is false. For instance one can check for the simple random walk that this fails. But maybe it is still asymptotically true. By this, I mean that the expressions could be bounded above and below by each other up to multiplicative constants. Your random walk is in the domain of attraction of a stable law/process, and thus for large $n$, the harmonic measure of the interval $\Lambda_n$ will approximate that of the stable process, which is well-known. Thus, $H_{\Lambda_n}(x,y)$ should be roughly equal to $c_n (n^2 - x^2)^{\alpha/2} (y^2 - x^2)^{-\alpha/2} |x - y|^{-1-\alpha}$. This is heuristic only, and working out the details can be quite messy if you need a bound uniform near the boundary. I think I have seen something similar in the literature, but unfortunately I do not have a reference. @MateuszKwaśnicki that was precisely the type of argument I was looking for. But I have not found a reference even for the hitting measure of the stable process. Do you know where I could find this? @Kernel: For the stable process, see: R. M. Blumenthal, R. K. Getoor, and D. B. Ray, On the distribution of first hits for the symmetric stable processes, Trans. Am. Math. Soc., 99 (1961), 540–554. Related references: M. Kac, Some remarks on stable processes, Publ. Inst. Stat. Univ. Paris, 6 (1957), 303–306; and: M. Riesz, Intégrales de Riemann–Liouville et potentiels, Acta Sci. Math. Szeged, 9 (1938), 1–42.
2025-03-21T14:48:31.773533
2020-08-14T15:43:45
369173
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Families of arithmetic fields arising from arithmetic varieties Suppose we have a polynomial $f(x,y)$ in two variables over a number field $k$, e.g., $y^2 - x^3 + x +1$. As we put different $k$-values of $y$ in the polynomial and adjoin the roots of the resulting polynomial in $x$ to $k$, we get different field extensions of $k$. We can think of this as a one-parameter algebraic family of number fields generated by $f$ over $k$, and ask how the structures and invariants associated with this family vary. For example, we can take normal closures and ask how their Galois groups vary, or how their zeta functions behave, or even the associated "reciprocity laws". We'd expect that the variation is related to the geometry of the curve $C: f(x,y) = 0$. We can, of course, get another family from specializing $x$ instead of $y$, and there would be other ways of fibering $C$ over $\mathbb{A}^1_k$ to get other families of number fields as well. The collection, $\mathcal{F}(C, k)$, of all such (isomorphism classes of) families of number fields arising from a single curve $C$ may itself have an interesting structure and a relationship to $C$ over $k$ worth looking at. (Is $\mathcal{F}$ a representable functor?) We can carry out this process for any $n$-dimensional variety $V$ over an arbitrary field $k$, i.e., specialize to $k$ in all but one dimensions, and thereby, get an $(n-1)$ -parameter family of field extensions of $k$, and then ask how the properties of the family, and of the family of such families are determined by the geometry of $V$, but I am interested in the arithmetic setting, primarily when $k$ is a number field, but also when it is a finite field or $p$-adic field, and curves are enough to begin with. The case of elliptic curves would be specially interesting where the group structure and modularity can play significant roles. General study of these of families of arithmetic fields arising from varieties over arithmetic fields must be well-known. Where can one begin to learn about this theory and its implications? This is essentially covered by "Hilbert's Irreducibility Theorem". Serres Topics in Galois Theory is a good reference for this. Some invariants of the field will emerge from Hilbert Irreducibility as mentioned above. On the other hand, even if you take a simple equation like $y^2-x$, you cannot expect, for different values of $x$, too much relation between the class numbers of the corresponding fields (which are all quadratic fields!). BTW: Your use of the term "arithmetic variety" is non-standard and should be avoided.
2025-03-21T14:48:31.773714
2020-08-14T16:04:52
369175
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Uniform distribution in Ball with radius $\sqrt{n}$ is sub-gaussian I have to show that a random vector $X$ who ist uniformly distributed on the Ball with Radius $\sqrt{n}$ is sub-gaussian with $$\lVert X \rVert_{\psi_2}\leq C$$ I already know that the same result does hold for a random vector on the sphere with radius $\sqrt{n}$ (1). I tried to show that $r Y$ is uniformly distributed on the Ball with Radius 1 for r uniformly distributed in $[0,1]$ and $Y$ uniformly distributed on the sphere with radius 1. Then I could use this to prove the claim in a similar way (1) is proved. However I am not sure if this is the right way? Let $X$ be uniformly distributed on the ball $B_{\sqrt n}$ of radius $\sqrt n$ in $\mathbb R^n$. Then $$X=RY,$$ where $R:=|X|/\sqrt n$ and $Y:=\sqrt n\,X/|X|$ is uniformly distributed on the sphere $S_{\sqrt n}$ of radius $\sqrt n$ in $\mathbb R^n$. Note that $0\le R\le1$ and hence $E\exp\{c(X\cdot t)^2\}=E\exp\{cR^2(Y\cdot t)^2\}\le E\exp\{c(Y\cdot t)^2\}$ for all real $c>0$ and $t\in\mathbb R^n$. So, $\|X\|_{\psi_2}\le\|Y\|_{\psi_2}$. Also, you know that $\|Y\|_{\psi_2}\le C$. Thus, $$\|X\|_{\psi_2}\le C,$$ as desired. Could you please explain how you get $\lVert X \rVert_{\psi_2} \leq \lVert Y \rVert_{\psi_2}$ from $E(X\cdot t)^2 \leq E(Y \cdot t)^2$? @Hugo10T : Oops! I forgot to insert $\exp$. This is now fixed.
2025-03-21T14:48:31.773821
2020-08-14T16:44:02
369177
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Are nuclear spaces used in creating variant theories of distributions? Laurent Schwartz proved his Kernel Theorem in 1952  to justify extending his theory of distributions to several variables. Then he and Jean Dieudonne gave Alexander Grothendieck the assignment to explain what was really going on in the Kernel Theorem.  Grothendieck did that to their very great satisfaction by creating the concept of nuclear spaces and generalizing the Kernel Theorem to them.   I suppose that one motivation for better understanding,and then generalizing, the Kernel Theorem was to help in finding other variant theories of distributions, based on other  spaces of test functions than Schwartz's,  that might have other good features than Schwartz distributions.  Certainly, very many people have used other spaces of test functions for various purposes.  My question is, does the theory of nuclear spaces get used in developing these variant theories of generalized functions?  I do not find it prominently used in the sources I find. This is just a suggestion which might interest you. If $T$ is an unbounded, self adjoint operator on Hilbert space, then there is a canonical (and functorial) way to embed $H$ into a locally convex space on which $T$ operates continuously and which is nuclear if and only if $T$ has a discrete spectrum with eigenvalues such that $|\lambda|^\alpha$ is summable for some negative $\alpha$. By choosing many classical differential operators (Sturm-Liouville, Laplace-Beltrami, Schrödinger ...), one obtains a whole palette of nuclear distribution-type spaces on various manifolds. Since this question has not been answered, I have decided to expand my comment. A well documented construction allows one to associate a generalised space of distributions to each unbounded self-adjoint operator $T$ on a Hilbert space $H$. It is motivated by the elementary and axiomatic approach of the portuguese mathematician J. Sebastião e Silva who showed that one can correct the basic “fault” of the differential operator on spaces of continuous functions by embedding them (in a unique manner) into larger superspaces on which it is everywhere defined. These are the Schwartzian distributions. In the general situation described above, a simple construction shows we can embed $H$ in a (in a suitable sense) unique manner into a vector space $H^{-\infty}$ on which $T$ is everywhere defined. If we choose for $T$ any of the classical self-adjoint differential operators, we obtain a unified approach to many known, but also many new, spaces of distribution, including those introduced by Schwartz. We can also vary the construction in simple ways to obtain other variants, including some which involve differential operators of infinite order. The space $H^{-\infty}$ has a natural lc structure of a well-studied type (as the inductive limit of a sequence of Banach spaces with weakly compact linking mappings—Komatsu). If $T$ has a discrete spectrum and its sequence $(\lambda_n)$ of eigenvalues converges to infinity (i.e., it is genuinely unbounded), then it is a Silva space (defined as above but with compact links). Finally, if $(|\lambda_n|^\alpha)$ is summable for some negative $\alpha$, then it is nuclear, with all resulting benefits. This latter condition is fulfilled if the eigenvalues are asymptotically like a positive power of $n$ as is the case for most of the classical differential operators which are relevant here. References as requested: Komatsu spaces: Projective and injective limits of weakly compact sequences of locally convex spaces, J. Math. Soc. Japan 19 (1967), 366-383. Silva spaces (not under that name of course): J. Sebastião e Silva, Su certe classi di spazi localmente convessi importanti per le applicazioni, Rend. Mat. e Appl. 14 (1955), 388-410. (Now in the secondary literature as Silva spaces, notably in the first volume of Köthe’s monumental “Topological Vector Spaces”). His works on distribution theory can be found at the site “jss100.campus.ciencias.ulisboa.pt” For an elementary overview (in english), go to “publicações”, then “Textos Didáticos”. The abstract construction is in “Normal operators and spaces of distributions”, Collectanea Math. (1975), 257-284. This contains the criterion for nuclearity. You can then construct ready made distribution spaces by consulting the long tradition of computing the spectral properties of the classical self-adjoint differential operators, a tradition which goes back to Fourier, Thomson and Tait. For the Sturm-Liouville operators, you could use the good old Courant and Hilbert. A more modern treatment with emphasis on connections with functional analysis is in Triebel’s “Höhere Analysis”. For the spectral properties required for distributions on manifolds, see Berger et al, “Le Spectre d’une Variété Riemannienne”. Corresponding results for Schrödinger operators have been investigated by Barry Simon amongst others.
2025-03-21T14:48:31.774289
2020-08-14T16:52:32
369179
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Small ideas that became big I am looking for ideas that began as small and maybe naïve or weak in some obscure and not very known paper, school or book but at some point in history turned into big powerful tools in research opening new paths or suggesting new ways of thinking maybe somewhere else. I would like to find examples (with early references of first appearances if possible or available) of really big and powerful ideas nowadays that began in some obscure or small paper maybe in a really innocent way. What I am pursuing with this question is to fix here some examples showing how Mathematics behave like an enormous resonance chamber of ideas where one really small idea in a maybe very far topic can end being a powerful engine after some iterations maybe in a field completely different. I think that this happens much more in mathematics than in other disciplines due to the highly coherent connectedness of our field in comparison with others and it is great that Mathematics in this way give a chance to almost every reasonable idea after maybe some initial time required to mature it in the minds, hands and papers of the correct mathematicians (who do not necessarily have to be the same that first found that idea). Summarizing, I am looking for ideas, concepts, objects, results (theorems), definitions, proofs or ways of thinking in general that appeared earlier in history (it does not have to be very early but just before the correct way of using the idea came to us) as something very obscure and not looking very useful and that then, after some undetermined amount of time, became a really powerful and deep tool opening new borders and frontiers in some (maybe other) part of the vast landscape of mathematics. Edit: I really do not understand the aim in closing this question as it is actually at research level. I am clearly asking for tools that developed into modern research topics. I recognize that some answers are not research level answers, but then you should downvote the answer, not the question. I am really surprised by this decision as one of the persons that vote to close suggested it for publication in a place where it is clear that some of the most valuable answers that this question has received would have never occur precisely because the site that this person suggested is not research oriented. I do not imagine people on HSM answering about species or pointfree topology sincerely as these topics are really current research and not history (and I am interested mainly in current research topics). I do not agree with the fact that a limitation in reading understanding of some people can be enough to close a legitimate question, a question that it is worth for us as mathematicians to do and to show to other people that think that mathematics is useful and powerful the day after being published ignoring thus the true way mathematics is done, with its turnabouts and surprises; a discipline where a simple idea has the power to change the field as $0$ did, as the positional systems did, as sheaves did, or as species did. I am really sad for this decision. It is a pity that so many mathematicians regret the actual way in which their field develops, reject to explain and expose this behavior and hide themselves from this kind of questions about the internal development of ideas in mathematics. I challenge all those who voted to close this question as off-topic to look in HSM for any mention about "locale theory" there. Are you sure that your title matches with the body of the question? I see only one direction The meaning is the same. I ask for ideas that began as small and maybe naïve or weak but at some point turned into big powerful tools maybe somewhere else. Where is the discussion of the part "big ideas that became small"? It is "big ideas that began small". Sorry, I misread. I laughed out loud at big ideas that became small. I have had lots of those. @MattF. I appreciate your comment but I think that this question has (maybe in a low level if you feel so) some philosophical and sociological background in the sense that, first, it can be considered as a question about gnoseology and epistemology of ideas within mathematics and how this knowledge is able to resonate and grow inside the mathematical box until it forms a wide highway among the mathematical paths even when the first appearance of the idea looked innocent... ... and second, it is also slightly sociological in the sense that it asks in a way of how the way of sharing mathematics between mathematicians make ideas grow sequentially in each iteration (or generation). This is a great question for HSM, where I think it better belongs than here. You sound like someone who has a meeting with a national funding agency... @LSpice Strictly speaking you are almost right but in practice putting that question in HSM would not have had the same effect. The question is certainly historic in a sense but not just that: it is asking for mathematics that right now could be at a research level. I doubt that people from HSM could have come with, for example, the answer of Michael Greinecker (one of my favorites so far) and I do not say this to diminish HSM but just because I understand that they like to discuss aspects that are farther in history getting thus far from the research front level discussed just in this site. @JPMcCarthy Mmm...? Why do you say that? @MattF. I think that this aspect is implicit in the question. I do not know if it good to edit the question again as then it will jump to the front page again annoying some people maybe. @Hvjurthuk because this is exactly the kind of thing that national funding agencies should be hearing. @JPMcCarthy Well, then we have to explain how the process of doing mathematics works more often to them ;) I wonder to what extent most or all of mathematics has this property? I wonder this seriously because it has implications for mathematical practice. If 97% of math arises from unravelling some little thing that is interesting (I'll bet this is so) then this kind of inquiry should dominate research effort, and grad students should be told this. Please comment if I am dead wrong about this... @JonBannon What you say is not necessarily true. There are open problems and programs of research that are well known, established and developed thoroughly. Solving problems within some of these programs do not have at all the property that I am looking for in most cases. For example, Wiles' or Perelman's proofs are certainly extraordinary achievements, but the point is that they were always (or long before these proofs were made) considered extraordinary in the sense that we as a community already knew that solving these problems was going to be something formidable. ...On the other hand, introducing, for example, complex numbers to solve cubic equations and realizing that these numbers allow us to pursue a better description of many different mathematical topics would fill the bill. I am just looking more modern phenomena having this characteristic. @Hvjurthuk: I'm not sure I agree. Fermat's Last Theorem originally involved a pretty innocuous looking statement. The high powered stuff that followed (Shimura-Taniyama, for example) appear as examples below. Many big industries in math started small. I'd love to see an example of one that obviously did not. (Maybe Weil approaching RH? But RH itself came from looking at Euler products etc..) I would love to be convinced of what you say, but the examples you give do not convince me... I guess I see. You mean little things against the backdrop of modern mathematical architecture with this property. (That filter may help me, since certainly all of math started small...) I miss the picturesque old title, with the multiple readings required to understand it, and its apparent evocation of what @VilleSalo points out is a familiar phenomenon. @LSpice: I do too! Quadratic Reciprocity was big, but a curiosity. It ultimately morphed into Langlands Reciprocity, which is gigantic and central to modern number theory. This is likely mostly an apocryphal origin story, but often the birth of probability theory is traced to the Fermat/Pascal analysis of the "problem of points" in gambling (https://en.wikipedia.org/wiki/Problem_of_points). In a letter to Frobenius, Dedekind made the following curious observation: if we see the multiplication table of a finite group $G$ as a matrix (considering each element of the group as an abstract variable) and take the determinant, then the resulting polynomial factors into a product of $c$ distinct irreducible polynomials, each with multiplicity equal to its degree, where $c$ is the number of conjugacy classes of $G$. This is now known as Frobenius determinant theorem, and it is what led Frobenius to develop the whole representation theory of finite groups (https://en.wikipedia.org/wiki/Frobenius_determinant_theorem). Amazing answer! Really good one. This works in a very concrete layer of mathematics and moreover it opened a huge branch of mathematics! In Introduction to Representation Theory by Etingof et al., Gerovitch explains how Frobenius used group determinants with great facility to prove what we would nowadays call fundamental theorems of representation theory. Frobenius's methods have been largely forgotten since they have been replaced by other methods, but I wonder if there remain undiscovered treasures lying in wait for someone who decides to take a fresh look at group determinants. The problem of the seven bridges of Königsberg is surely one of the best-known examples of this. Euler apparently didn't even consider this problem to be mathematical when he solved it, but in doing so he introduced the basic concepts of graph theory (a field which did not really begin to take off until a century and a half later). ''Euler apparently didn't even consider this problem to be mathematical when he solved it.'' Is there some evidence for this? He noted that in order to move across each bridge only once, the number of bridges adjacent to each vertex must be an even number apart from the start point and the end point, resolving the Konigsberg problem in the negative. This was very simple for him, but surely he must have seen it as a mathematical argument? In a letter to the mayor of Danzig, Euler wrote that "this type of solution bears little relationship to mathematics, and I do not understand why you expect a mathematician to produce it". Ah interesting, that makes sense though. I always wondered why he would even sit down to give that a formal argument. This example qualifies then, as it introduced graph theory and can also be seen as the first gestures towards topology apart from Leibniz (cutting out all the geometric information from the problem etc.) This still seems strange to me though. The result was presented at the St. Petersburg Academy in 1735 and then published in a journal article ''The solution of a problem relating to the geometry of position'' in 1741 so he must have known its significance and explicitly referred to the geometric nature of the problem in the title of the article. Perhaps he considered it more of a scientific problem. I've found the original article in Latin https://scholarlycommons.pacific.edu/cgi/viewcontent.cgi?article=1052&context=euler-works. Although I can't really read Latin, it seems to mention Leibnitz (presumably in the topological context), and looks from the detailed drawings and 21 sections of the paper that he is taking it seriously as a geometric problem (at least at the time of writing). It's possible that he changed his mind about it. I found the article "The Truth About Konigsberg" by Brian Hopkins and Robin Wilson which includes the quote I gave, but upon reading more closely it seems he wrote that letter before he'd solved the problem, and perhaps felt differently after seeing how it related to Leibniz's proto-topological ideas. It still feels like a good example given that Euler did not to my knowledge do any further work along the same lines and neither did anyone else for some time. As I remember it, his initial reaction to number theory was also lukewarm and we all know he must have changed his mind on that, although yes I accept that your answer is a good answer to the question regardless. Also, I don't mean to argue, but I'm not convinced either that he did no further work on the same lines (ie. graph theory). In his paper Recherches sur un nouvelle espèce de quarrés magiques' (Researches on a new kind of magic squares)", he studied the problem of moving around a chessboard with a knight and visiting each square only once (known as the problem of the knight's tour). This is basically a Hamiltonian path problem in graph theory. This is not to mention his study of plane mazes, which are basically graphs. @Hollis I believe Euler considered his solution to be "pure logic", not science. Again, I can't really read Latin, but if you read the original article by Euler which I posted it really does seem that he regarded this as a geometric problem, referring to it as such in the title and including several detailed geometric drawings. @HollisWilliams While we now see the bridges of Königsberg, knight's tours, and plane mazes all as graph theory, there's no indication Euler ever connected these three topics. The generalizations in the 1741 paper, as you saw, were to more complicated systems with more rivers and more bridges. @HollisWilliams Euler's introduction says (more or less), that Leibniz had mentioned a new branch of geometry, called Geometria Situ, which did not use concepts like numbers, but nobody knew really what it was about. Euler now had found a problem that certainly belonged to Geometria Situ and wanted to present it and its solution in his article. From this I conclude that Euler thought that there was a new, undiscovered branch of mathematics, and he wanted to find out what it was. That the new branch became as big as it is now, he might not have expected, but clearly he expected something. @rimu Thanks for this, I thought that something along those lines had to be true and the idea that he thought the problem to be beneath him was almost certainly something of an urban myth. Cantor's monumental investigation of the infinity started very innocently as a method to understand the uniqueness of the representation of a function by trigonometric series. I wonder if there is a book about Cantor that I can read. He was seriously mistreated at the time and even fell ill to depression if I heard correctly. @dezdichado: The standard biography in English is Dauben's biography. A well known biography in German is this book by Purkert and Ilgauds. There are also numerous published papers in many different languages on all phases of Cantor's life and work written throughout the period since his death (I've intentionally left the quantifer order somewhat ambiguous), but either of these two books should be more than enough for someone starting out. @DaveLRenfro I see, thanks! Integration by parts would seem like a good example. Whoever first used it to integrate a function such as $x\exp(x)$ could certainly not have anticipated the fundamental role it would once play in the theory of PDEs. "When Peter Lax went to receive the national medal of science, he was asked by the other recipients about his merits. His answer was (apocryph) I integrated by parts." https://mathoverflow.net/posts/60908/revisions Surely 'once' in "… it would once play …" should be 'eventually' or so (since I assume it's not done playing that role)? And the linear algebra viewpoint helps make it clear why integration by parts is so fundamental: integration by parts states that the adjoint of $\frac{d}{dx}$ is $-\frac{d}{dx}$ (or the adjoint of $\nabla$ is $-\text{div}$) in a setting where boundary terms vanish. Oh, great Peter Lax, always so charismatic! Totally accidental downvote on the phone app in my pocket last night, my apologies. (If you edit so I can change it I’ll upvote) @GerryMyerson always chuckled at this quote. But after I took a seminar in PDE and then tried to work on Vlassov-Poisson and Boltzmann equations for research, this quote started giving me PTSD. Pick's theorem states that the area $A$ of a simple polygon $P$ in $\mathbb{R}^2$, whose vertices are in the lattice $\mathbb{Z}^2$, can be computed by means of the formula $$A=I+\frac{B}{2}-1,$$ where $I$ is the number of lattice points in the interior of $P$, and $B$ is the number of points in the boundary of the polygon $P$. George Pick published this theorem in 1899, in his article "Geometrisches zur Zahlenlehre" [Sitzungsberichte des deutschen naturwissenschaftlich-medicinischen Vereines für Böhmen "Lotos" in Prag. (Neue Folge). 19: 311–319]. Apparently, this result remained unknown until the middle of the 20th century when Hugo Steinhaus included it in his book "Mathematical Snapshots". This beautiful result is a precursor of theories about "counting lattice points in polyhedra" (e.g., Ehrhart's theory, and generalized Euler-MacLaurin summation formulas) which intersect, as far as I know, with problems in linear programming, values of number theoretic zeta functions, toric varieties, and even physics (I've heard). Nice, didn't know that Pick's theorem was mor ethan a simple curiosity! Pointfree topology originated in some sense in terms of "local lattices" (lokale Strukturen) in a 1957 paper of Charles Ehresmann, but the topic was of little interest until it took off with a 1972 paper of John Isbell who argued that the pointfree approach to topology is in some ways superior to the usual approach. A great overview of the history is given in the entry Elements of the History of Locale Theory by Peter Johnstone in the 3rd volume of the Handbook of the History of General Topology. This is actually a really good answer! I imagine that when the branch is more abstract it is even easier to find these examples of ideas that appear at first with low relevance and then suddenly at some point they explode when the correct person find them again in the correct moment. @Hvjurthuk The same paper by Ehresmann also introduced species, which arguably only became significant in the 80s with the work of André Joyal on combinatorial species. I don't know if this post will answer your question but it's about the higher homotopy groups of topological spaces. The story is told here: https://ncatlab.org/nlab/show/homotopy+group. In 1932, E. Čech proposed a definition of higher homotopy groups using maps of spheres, but the paper was rejected for the Zurich ICM since it was found that these groups $\pi_n(X,a)$ were abelian for $n \geq 2$, and so do not generalize the fundamental group in the way that was originally desired. Nonetheless, they have proved to be extremely important in homotopy theory, although more difficult to compute in general than homology groups. The higher homotopy groups of spheres are central in many problems and conjectures in mathematics and they are known to be very difficult to calculate. This one is also a great example although being rejected for the ICM is not an obscure beginning at all: it is more like a failed shiny beginning. In his book An Introduction to Combinatorial Analysis, Riordan observed that the number of ways to choose $k$ objects from $n$ objects, allowing repetition and disregarding order, can be written $(-1)^k{-n\choose k}$, while ${n\choose k}$ is the number of ways without repetition. This was the first inkling of the vast subject of combinatorial reciprocity. See for instance the book Combinatorial Reciprocity Theorems by Matthias Beck and Raman Sanyal. Julia sets were relatively obscure and little known until the advent of personal computing, when the ability to graph them in detail made it clear how amazing they are. They now command a global audience of fans, even with little or no knowledge of mathematics. Counting: one, two, three, four, ... This originated and was put to practical use in prehistoric times. Do you think that this discovery was "something very obscure and not looking very useful"? @FrancescoPolizzi : Perhaps not, but no one at the time could have anticipated the life of that idea. @MichaelHardy Okay, you went too far in (pre)history (even in biology itself). I am mostly interested in more modern things. However this answer is in the line of the famous quotation of Grothendieck about simple ideas. Legit though then. Obligatory: https://www.smbc-comics.com/comic/a-new-method @AmirSagiv, whether or not it's what @‍Hvjurthuk had in mind, this certainly doesn't seem like an attempt at a new question. I'm sure that numbers were "not looking very useful" early on. Somebody came up with a collection of words that can be used to determine whether one multitude is more numerous, equally numerous, or less numerous than another multitude. Well, gee, who spends so much time comparing multitudes that they need special words in order to do it? Useful, perhaps, but obviously not nearly as useful as knowing how to make a knife. I'd say giving a name to numbers, as abstract objects, and listing them, was already advanced mathematics. Maybe even before that, it appeared the idea of bijection e.g. fingers $\leftrightarrow$ goats, opening the the way to a whole mathematical world. Introduction of $0$ in the place value system of counting. It was not given proper consideration in other places. So perhaps was pre-medieval mystery. What about: from Euler characteristic (special problem, thus "small") to homology theory of cw-complexes from Galois (special problem, thus "small") to group theory and modern algebra? The problem of solving polynomial equations by radicals was considered a central one, so I would not say that Galois theory began "small" I actually was going to comment the same as @FrancescoPolizzi. In a sense these problems were actually central to mathematics then and I would not in general consider work done by Gauss or Euler small in any sense as they are two of the brightest minds in mathematics. Anyway, thanks a lot for your answer! @Hvjurthuk I do not think that Euler's theorem on polytopes was considered a central subject. @FrancescoPolizzi I feel like Galois theory is an interesting case in that the more we learned about mathematics, the less important solving polynomials by radicals has seemed, but the more important Galois's insights have become (or at least they've stayed proportionally equally important by becoming relevant to more and more areas). Like it was lucky that people were initially so focused on solving a polynomial equation by radicals. @WillSawin: Obviously I agree with you. However, the OP is asking for ideas that started "in a real innocent way" and "not looking very useful". In this respect, I am not sure that Galois theory qualifies, because it went straight to the heart of a central problem of that time. In fact, Galois theory (at least, in my opinion) is more an example of a mathematical discovery that killed a research field (solving polynomial equations over a field by radicals) or, better, radically transformed it in something much more deep and general. I want to mention Selberg's integral, an $n$-dimensional generalization of Euler's beta integral. Selberg published it 1944 in Norwegian in the journal Norsk Matematisk Tidsskrift. Not surprisingly, it did not get a lot of publicity there. Later it was key to results in random matrix theory and other areas. There is an excellent article by Forrester and Warnaar summarizing the history and applications of the integral. I think the Durfee square is a good fit. The idea is incredibly simple - distinguish integer partitions based on the largest square that fits inside of them. The use of a square has since been extended to rectangles, staircases and likely some other shapes. The idea is indisputably a small one, but has proven itself an indispensable tool for working with integer partitions, hence also with hypergeometric functions and related topics. A quick search on Google Scholar shows many papers with Durfee square in the title, some of them quite recent. Note the top hit, which points out the Durfee square is identical to the h-index! One strike against the Durfee square for your question is that Durfee's advisor Sylvester recognized almost immediately how powerful a tool the Durfee square is in partition analysis. As he wrote in a letter to Cayley in 1883 (see Wikipedia), "Durfee's square is a great invention of the importance of which its author has no conception." Ramsey Theory has to be mentioned in this context I think. This is a somewhat obscure but interesting branch of combinatorics that is named after the mathematician/philosopher Frank Ramsey who proved its first result through Ramsey's theorem. Interestingly, Ramsey only proved this theorem in passing as a minor lemma. He was actually trying to prove a decision problem for a particular model of first order logic, namely the Bernays–Schönfinkel class. This lemma ended up spawning and entire sub-branch of mathematics and is mostly known for Ramsey Numbers, a class of numbers that are known to exist but are ridiculously hard to compute. Two instances come to mind in digital signal processing (applied mathematics). The Fast Fourier Transform (FFT) computes the Discrete Fourier Transform in $O(N \log N)$ instead of $O(N^2)$. Supposedly, Gauss had a version of the FFT long before (electronic) computers made their impact. The second is the original wavelet transform, by A. Haar in 1909. Research in wavelet transforms has exploded since. Neural networks are a great example right now in machine learning. They were around for decades before the computing power to actually train them properly became available. A similar candidate would be the Relu function $f(x) = max(0, x)$. Using this as an activation function instead of the sigmoid made it easier to train neural networks with more than 2 layers, which turn out to perform better in practice, for unknown reasons. This (together with computing power) has made all sorts of advances possible. There can't be many examples of such a simple mathematical object having such a massive impact on society. I don't think "lack of computing power" made them into a "small idea" If memory serves, in James Gleick's book Chaos, he describes the origins of this field as attempts to find numerical bugs and rounding errors in PDE solvers -- before it was realized that something far more profound was happening. Maybe this is a reference to https://en.wikipedia.org/wiki/Fermi%E2%80%93Pasta%E2%80%93Ulam%E2%80%93Tsingou_problem ? Richard Stanely's 1973 paper "Linear homogeneous Diophantine equations and magic labelings of graphs" was the first time commutative algebra was used to study convex polytopes. But the paper is not really about polytopes per se. Rather, its main focus is on resolving the Anand-Dumir-Gupta conjecture about "magic squares," specifically, about the number $H_n(r)$ of $n\times n$ nonnegative integer matrices having all row and column sums equal to $r$. Part of the Anand-Dumir-Gupta conjecture was that for fixed $n$, the function $H_n(r)$ is a polynomial in $r$, which Stanley showed as a consequence of some basic theorems in commutative algebra going back to Hilbert. As Stanley says in his personal account "How the Upper Bound Conjecture was Proved", In this paper appears a geometric interpretation of MacMahon’s algorithm which, among other things, relates the polynomials $H_n(r)$ (and some more general polynomials) to certain triangulations of polytopes, in particular, the number $f_i$ of $i$-dimensional faces of such triangulations for all $i$. At that time I had no interest in the $f_i$’s themselves. The situation changed not too long after that, however, when Stanley was able to use the commutative algebra connection to prove remarkable results in polytopal combinatorics, like the Upper Bound Conjecture for simplicial spheres. Nowadays there is a whole subfield of combinatorial commutative algebra: commutative algebra is a basic tool in the study of polytopes (e.g., their face numbers); and conversely polytopal combinatorics provides commutative algebraists with interesting questions and examples. Another possibility could be the problem of the brachistochrone, a famous but one might think relatively innocent problem, which then led to the development of the calculus of variations. Does Fermat's last theorem count? I mean, it was a small idea at first. For which he thought he had a proof, but it didn't fit in a small margin of paper. At the time, who would have thought this theorem will have such a deep impact on mathematics? I wouldn't describe Fermat's last theorem as an idea; more of a problem. @Ben it was a conjecture actually. Then became a theorem. I think it's fair to classify conjectures as ideas. @Bumblebee what do you expect me to say? :) I think the implicit function theorem fits very well. The idea of solving an implicit equation is simple, an for examples like the circle one might call it a small idea. However, the implicit function theorem is still very useful and can be applied in various situation, for example to prove existence in complicated situations.
2025-03-21T14:48:31.776939
2020-08-14T16:52:51
369180
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Frêchet differentiability of the composition on a suitable Banach space Let $E$ be a real Banach space included in the space of functions $C^1$ of $\mathbb R\to \mathbb R $ and and $T:E\to E$ defined by $T(f)=f\circ f$. I am looking for an example of space E such that T is differentiable on E. By developing, $T(f+h)$ we find $$T(f(x)+h(x))=f(f(x))+h(f(x))+f'(f(x)).h(x)+h'(f(x)).h(x)+o(h(x))$$ but the term $h'(f(x)).h(x)$ is not linear on h and it seems that $DT_{f}\left(h\right)=\left(f'\circ f\right)\times h+\left(h\circ f\right)$ but we need a space where $(h'\circ f).h=o(h)$ I thought of the subspace of $C^{\infty}(\mathbb R, \mathbb R)$ formed of bounded functions of which all derivatives are bounded, provided with the norm $f\mapsto \|f\|_{\infty} + \|f'\|_{\infty}$ but this space is not complete space edit As suggested by Giorgio Metafune , I start with Gateaux differentiability . Let $E=C_b$ (bounded and continuous functions with the sup-norm). Let $g\in E$ and f be a function derivable on $\mathbb R$. By developping $$ (T(f+tg)-T(f))(x)=f(f(x)+tg(x))+tg(f(x)+tg(x))-f(f(x)) $$ then $$\lim_{t\to0}\frac{(T(f+tg)-T(f))(x)}{t}=\lim_{t\to0}\frac{f(f(x)+tg(x))-f(f(x)) }{t}+g(f(x))\\= f'(f(x)).g(x)+g(f(x)) $$ Then $$\qquad D_G(T)f(g)=g (f'\circ f)+g\circ f$$ and $g\to D_G(T)f(g)$ is linear. It's must be continuous to deduce Frechet differentiability Update question Giorgio Metafune suggests considering $E=C_0(\mathbb R)$ Let $E=C_0(\mathbb R)$ and $T: E\to E$ defined by $T(f)=f\circ f$. Show ( if true) that T is Fréchet differentiable at all $f\in C^1(\mathbb R)$. Can some one help me or give answer Is there anything preventing you from using finite dimensional examples, like the subspace of linear functions? the linear case no interesting to me Why not use $C^1_b$, the space of all bounded diferentiable functions, with bounded derivatives, with the norm you wrote in the last line? @Giorgio Metafune $ C^1_b$ is not complete with this norm. I need Banach space! If by $C^1_b$ we understand the space of bounded and differentiable functions with bounded first derivatives, it Is complete under your norm. sorry, I wanted to say that $C^{\infty}(\mathbb R, \mathbb R)$ is not complete. there is a problem with your space $C^1_b$ because $\left(f'\circ f\right)\times h$ is not necessarily in E (f must be twice differentiable) there is also another problem, f is differentiable $\iff $ $h'\to 0$ if $h\to 0$ but the fact that $h \to 0$ according to this norm does not imply that $h'\to 0$, because nothing is known about $h''$. I see. Maybe it is better to look first at Gateaux differentiability and it is possible that your map has a derivative only at certain points $f$ and along certain "directions" $h$. If we consider $C_b$ (bounded and continuous functions with the sup-norm), maybe it is differentiable when $f \in C^1_b$ and along directions $h \in C^1_b$, too. The choice of $C_b$ should simplify the handling of the non linear part. @Giorgio Metafune you can see my update for Gateaux-dfferentiability and after ? Seen. It seems that the map is Gateaux differentiable exactly at points $f \in C_1$. Maybe it is Frechet differentiable at the same points. It could be easier to work in $C_0$ instead of $C_b$ (functions vanishing at $\infty$), since uniform continuity holds for free. I changed the question of the thread following your last remark For reasons already pointed out, you would need to control the norm of the derivative by the norm of the function (to deal with the $h'(f(x))$ term). This essentially rules out most of the usual function spaces. You can come up with examples in spaces of analytic functions. For instance, if you consider bounded analytic functions on the disk, and f maps to a smaller disk, you can indeed bound $h'(f(x))$ by the norm of h.
2025-03-21T14:48:31.777357
2020-08-14T16:54:56
369181
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Geodesics and potential function I try to assemble concepts of differential geometry for my own comprehension of the subject. I understand a manifold is a higher dimensional surface. It has a metric which perform inner product in the tangent space. A curve on the respective manifold has a covariant derivative, which remains on the tangent bundle. A geodesic is a manner to comprehend straight lines on a manifold. They might be closed like on a sphere. They are the main tool of physicists to comprehend the universe through the lagrangian and hamiltonian framework. On the lagrangian framework, there is a kinetic $K = g_{ij} \dot{x}_i \dot{x}_j$ and potential $U = V(x)$ energies which computes the lagrangian $L = K - U$. The ausence of potential energy coincide with the geodesic equation. The Einstein notation is in force here. In mathematical terms, I do not comprehend the role of the potential or dissipative term on the geodesic equation and further explanations on similar manner as I will explain shortly. As far I comprehend, the geodesic statement is: given two points A and B, the geodesic which binds both points on a simply connected non-compact smooth manifold is the solution to the boundary value problem of former equation below. The same statement but for the latter equation entertains other interpretation, no longer a geodesic at strict sense. I comprehend from physical perspective that given the manifold endowed by a metric, the second and third are relative to the actuation of forces on the motion particle. But for this, one defines the particle, which is merely an abstract conception, a trick to better comprehend intuitively. \begin{equation} \ddot{x}^j + \Gamma^j_{i k} \dot{x}^i \dot{x}^k = 0 \end{equation} \begin{equation} \ddot{x}^j + \Gamma^j_{i k} \dot{x}^i \dot{x}^k + g^{ji} \frac{\partial V}{\partial x^i} = 0 \end{equation} \begin{equation} \ddot{x}^j + \Gamma^j_{i k} \dot{x}^i \dot{x}^k + g^{ji} \frac{\partial V}{\partial x^i} + g^{ji} \frac{\partial R}{\partial \dot{x}^i} = 0 \end{equation} What is your question here? From the looks of it, $R$ is just another potential term just like $V$, that could be absorbed into it. The questions are: What is the geodesic equation when the boudary values are not points A and B but initial position and velocity? What is the influence of the potential term on the geodesic equation when employed as a Point A B condition? The same for with Rayleigh i.e . dissipative function? The role of those terms is the same as their role in Newtonian classical mechanics, where the Riemannian manifold is flat Euclidean space. So you should ask a physicist or look in a classical mechanics textbook to see how such terms appear. Your questions have no mathematical answer, only an answer in physical intuition. Why doesn't it have a mathematical answer? The physical statement relies on the mathematical toolset to perform its calculation. @BrunoPeixoto I'm not sure what the question actually is. If I re-interpret it as asking for a geometrical set-up to discuss lagrangian/hamiltonian systems with potentials, then I have two comments: (1) you may be able to rewrite the equation as a geodesic equatoin for an affine connection which is not necessarily metric (cf. Newton-Cartan theory) and (2) you may be able to view it as a geodesic equation in an auxiliary space relative to the Levi-Civita connection (cf. the Eisenhart lift). I'm also not sure I really understand the issue, but my reaction is that the geodesic equation and its generalizations on the one side, and the boundary conditions on the other side are separate ingredients. Ideally, you would solve the geodesic equation in general, leaving you with integration constants, and then you fix the integration constants to the boundary data, in whatever way they're given. So the geodesic equation, and additional potential, etc. terms don't depend on the form of the boundary data, these are separate pieces of information. Maybe I'm missing something. Perhaps you could revise what you wrote and be more specific and precise about what you want to know? I would observe, however, if there is a non constant potential, then there is a force acting on the particle’s motion and therefore the path with least energy is no longer a geodesic. The second equation can be viewed geometrically as saying something about the curvature of the path. Does the last paragraph clarify my intentions regarding my doubt? I enjoy José Figueroa's answer. Michael provides some intuition, but it lacks rigor. Deane, R regards the Rayleigh function and not the curvature notation. I was referring to the second equation, which has no $R$.
2025-03-21T14:48:31.777739
2020-08-14T17:15:51
369183
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Euler Systems and Coleman’s Conjecture I’m trying to work on Coleman’s conjecture for abelian extensions of imaginary quadratic fields. I’ve read most papers by Seo regarding circular distributions. However, I’m a still confused about what it means that “all Euler systems come from cyclotomic units in an easy way”; what does this mean and what is that easy way? Also, what is the extend to which similar thing can be done if the field Q of rationals is replaced by an imaginary quadratic field? I have no idea what this question means, and I've been working on Euler systems for half my career. Can you make it a bit more precise, maybe with some links and/or references? The last paragraph of page 7 of Seo's "Circular Distributions and Euler Systems" states that "there are almost no other Euler systems except the cyclotomic units". How do I prove this? re there any other examples apart from the cyclotomic units? With the commonly understood definition of "Euler system" (as in Karl Rubin's classic book "Euler Systems"), this assertion of Seo is simply untrue. Presumably Seo is using some other, more specific definition -- maybe he means only Euler systems for the trivial representation and $K = \mathbf{Q}$ -- but even so, the use of "almost" suggests this is supposed to be a rough guideline rather than a rigorous theorem. Soogil Seo appears to be currently mathematically active, so why don't you ask him directly what the phrase meant? Seo actually describes cyclotomic units of an abelian number filed F and then uses these cyclotomic units to construct Euler systems of F, which then suggests that there are almost no other Euler systems except these. Is this assertion untrue? The assertion is not precise enough to have a well-defined truth value. Maybe what Seo is aiming at is the assertion that the module of Euler systems for $K = \mathbf{Q}$ and $T = \mathbf{Z}_p(1)$ is generated by the image of the cyclotomic units under the Kummer map; this sounds plausible, and I think it may follow from the Mazur--Wiles theorem (former Iwasawa main conjecture), but I'd have to check. But there are lots of other Galois representations for which it is interesting to study Euler systems, and these other Euler systems have nothing much to do with cyclotomic units. @david I deleted my previous comments as they had lots of typing mistakes. I’m trying to get the hang of how to type in mathoverflow. I'm afraid you're probably wasting your time trying to get an answer to this question on MathOverflow. The quote you give from Seo's work is quite imprecise, as well as relying on non-standard terminology and definitions; hence it's not really possible for anyone but Seo to deduce what he meant by it. If you absolutely must know, your best bet is to contact Soogil Seo directly. @david I know and I tried. I emailed him but hasn’t got a reply yet; I understand that he must be very busy that he hasn’t replied yet. I’m trying my best to read as many papers as possible to understand the following statement: Coleman’s conjecture concerning circular distribution imply that Euler systems over abelian number fields arise in “an elementary” way from the theory of cyclotomic units. What is that “elementary way”? I would appreciate if you could help. You've asked your question 3x already, you don't need to repeat it. My guess is that this sentence of Seo is intended to be an informal description of the two (related) results formally stated and proved as Theorem 4.2 and Theorem A of the same paper, both of which assert that (for the particular, nonstandard notion of "Euler system" used in op.cit.) the Euler system of cyclotomic units generates the module of all Euler systems -- i.e. any Euler system can be gotten from the cyclotomic-unit Euler system by applying an element of the ring $R$. @DavidLoeffler The above explanation is quite helpful, however, you have said "any Euler system can be gotten from the cyclotomic unit Euler system by applying an element of the ring $R$". What does that element of $R$ look like and how doe s it act on the cyclotomic unit? As this question came up again today, let me add that Coleman's conjecture has been proven in this article by Bullach, Burns, Daoud, Seo. Probably this answers this vague question.
2025-03-21T14:48:31.778105
2020-08-14T17:43:29
369185
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Alexandre Eremenko", "David Roberts", "R. van Dobben de Bruyn", "Stanley Yao Xiao", "Will Sawin", "Yemon Choi", "anon", "https://mathoverflow.net/users/10898", "https://mathoverflow.net/users/163772", "https://mathoverflow.net/users/18060", "https://mathoverflow.net/users/25510", "https://mathoverflow.net/users/4177", "https://mathoverflow.net/users/763", "https://mathoverflow.net/users/82179" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632101", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369185" }
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Australian Mathematical Society journal rankings I apologize for a question that is not about mathematics, but I believe it is of interest to research mathematicians, and I believe there may be people on MathOverflow who can answer it objectively. If it is deemed unacceptable, I can survive. For many years, I (and many others I know) have used a ranking of mathematics journals produced by the Australian Mathematical Society in 2009, formerly found here https://austms.org.au/Rankings/AustMS_final_ranked.html. Without wishing to get into a debate on the usefulness or methodology of ranking journals, I found the list useful for judging what journals to submit papers to, and for justifying to deans or hiring committees the quality of journals that I or others (e.g. job, tenure applicants) have published in. I understand that it was getting out of date and I certainly did not agree with every grade there. But since it was made by mathematicians, rather than some trite formula, and since it used a simple A*/A/B/C grading scheme, it was solid and easy to reference and cite. For example, I planned to use it to help justify to deans an upcoming tenure decision. It was also the top rated answer to this MO question about journal rankings. Unfortunately, the link above is now dead. Question 1: Does anyone know if this ranking is permanently gone from the internet? If the society has "disavowed" it as incorrect or out of date? If they are revising it? Or if the link has simply changed? I was able to find a "cached" version, so I still have access to the information. But I do not know how long this will be up, and it detracts from any semblance of authority if it is no longer hosted on a reputable website. If this list is gone forever, Question 2: Do people have suggestions for a replacement with similar features? (Made by a reputable institution, with input from mathematicians rather than a trite formula, and easy to explain to non-mathematicians.) I understand that the second question has some overlap with the prior MO question linked above. The article https://theconversation.com/journal-rankings-ditched-the-experts-respond-1598 linked in the previous question seems to answer that this has been disavowed as out of date, and won't be replaced in any kind of similar form. Thanks. I understand the decision and don't wish to criticize them. The simple hosting of the old list was useful to me and others, and it would be nice to know if it is being scrubbed from the web. But given that, I suppose my question 2 is more salient at this point. I appreciate it. The Wayback machine does have an archived version. Besides general legal ambiguity, I think this is supposed to be a somewhat permanent secondary source. Thank you R. van Dobben de Bruyn, that is also helpful. Journal ranking discussion is out of scope of this site. The MO question you refer to is closed, and I voted to close this as well. Apologies, Alexandre Eremenko. I have accepted an answer since it seems the issue is as resolved as it will be. Some additional context : this ranking was done for all (reasonable) journals, at the behest of the Australian Research Council. Mathematicians complained that journal rankings are not as simple as people think and so (I think to shut us up) the AustMS got given the job of doing something reasonable using research community-sourced information. These rankings were for the purpose of 'grading' Australian universities' research output, and the whole exercise was abandoned after a few years (about a decade ago). There are several sources online that rank math journals by impact factor (and, this comes from a professor's webpage) or journal citation reports. However, it is important to realize that impact factor is a highly unstable metric, as discussed here. Thomson Reuters has a list ranking by JCR, which appears to now be behind a paywall called InCites (though you might be able to access it through your library). If I remember correctly there was also an option to rank by impact factor. The Institute for Scientific Information (ISI) is another source that maintains a database. And there's Scimago mentioned by Yemon. And you can look up impact factors over the years at bioxbio.com. Another ranking by impact factor was SciJournal.org, but their website also appears to be down. A general way to solve the issue of disappearing website is the WayBack machine. For SciJournal, here is a recent snapshot from 2020. For the Australian list, here is a July, 2020 snapshot. In the same spirit as the Australian list, I am aware of a Chilean list that breaks journals into categories such as "Muy Buena," "Buena," and "aceptable" (I'm not sure if this is actually the third category). I've heard that this breakdown is not super well-regarded. Lastly, an alternative to impact factor, that I've read about but have not yet looked into, is scite. Thank you, I appreciate your taking time to compile this list. It is versatile and I did not know about many of these. It seems that none of these satisfy my desire that the list not be simply based on citation statistics, but there is apparently no other such thing (except perhaps the Chilean one you mention unfavorably.) Impact factor is a terrible metric, and in experimental sciences has been shown to be positively correlated with retractions, and there is no correlation between IF and statistical power (which indicates studies with good evidence) Source: http://bjoern.brembs.net/2016/01/even-without-retractions-top-journals-publish-the-least-reliable-science/. One should also look at the International Mathematical Union's position on citation metrics: https://www.mathunion.org/fileadmin/IMU/Report/CitationStatistics.pdf Here's a better citation for the rather average performance of high-IF science journals: Brembs B (2018) Prestigious Science Journals Struggle to Reach Even Average Reliability. Front. Hum. Neurosci. 12:37. https://doi.org/10.3389/fnhum.2018.00037 Needless to say, three decimal place accuracy on a noisy quantity that for most mathematics journals is less than 2 is a ridiculous way to break ties. One would be better placed just rounding to the IF of maths journals to nearest integer and grouping journals that way, if one absolutely must use such a metric. One neat alternative to impact factor for slower moving fields like mathematics is the normalised lifetime impact factor described in this answer. (It uses Acta as a baseline to normalise for differences in overall research output throughout recent history in order to produce outcomes independent of age of a journal.) This is just one example with its own flaws, but it shows that alternatives exist, and we could be thinking about even better systems. (Like the OP I believe that rankings ― although ultimately inevitably flawed ― do serve a useful purpose. But the IMU's stance that @DavidRoberts shared is also a very healthy one, and just like any other scientist using data we should at least justify our choice and interpretation of metrics.) @R.vanDobbendeBruyn yes, if only to prove to university administrators that one's article published in Annals of Mathematics is really that much better than one's colleague's article published in Mathematische Annalen. @DavidRoberts I can't resist mentioning one of my over-used stories which is that when I was hired at USask 10 years ago (https://www.youtube.com/watch?v=avtK9Bz0sPg ) I was stunned to learn that someone there had just had a paper published in Annals and that no one was making a big deal. I then learned over the next 3 years that all deans etc outside the dept neither knew nor cared about some particular maths journal that some particular maths prof chose to publish in @Yemon I was in a group convo with someone who thought it was no big deal they themselves had published in the Annals. Like it was somehow ordinary?? (BTW, my comment should really have said "Math. Ann." and "Ann. Math.") @Yemon in the other direction, my faculty (Engineering + Comp Sci + Maths) recently singled out a paper I published as being high-impact and praised it up, when it was not much more than a bunch of uncomplicated long exact sequence calculations in a physics context. The only reason I could guess is that the IF of the journal (a physics journal) is higher than the IF of basically every other maths journal, so I think they'd normalised IF by field of research, then found I was top of the pack? This should perhaps be a comment rather than an answer, but I thought I should post it so that others can offer corrections or further detail. One of my colleagues has sometimes justified his opinions/suggestions to me with references to the Scimago journal rankings in mathematics: https://www.scimagojr.com/journalrank.php?area=2600&type=j Unfortunately I don't know if these rankings carry the appropriate "weight" in the eyes of deans and similar creatures. (I also don't know about accuracy, but your post suggests that this is not really the main issue which you want to address; the ranking puts JAMS at number 3 and Annals at number 4, for what it's worth.) There's a lot of finer-grained detail one can extract from this database, by selecting a particular country or a particular area of mathematics. Some information which may be of interest to readers other than the OP For background on this ranking or system of metrics, see https://www.scimagojr.com/aboutus.php The SJR number is described in technical detail https://www.scimagojr.com/files/SJR2.pdf and it appears to be a metric based on a version of the Google PageRank algorithm: some "alt text" on the webpage claims tha the SJR number ... expresses the average number of weighted citations received in the selected year by the documents published in the journal in the last three years. Thanks, this is another option. I am of course concerned with "accuracy" too, hence my preference for a ranking or grading that is based on the opinions of real live mathematicians, rather than just citation statistics, which we all know are manipulable, vary by sub-discipline, and tell an incomplete story (though I guess SciMago makes an attempt to correct for some of these issues). I appreciate your adding this response. It may be that there is no good response to my Question 2, and that this is the best available. I will wait to see what others say, and if nothing better, I will accept. Understood -- to some extent, my answer was just meant to get the ball rolling. My experience has been that sometimes the best prompt for a mathematician to give a good answer is for them to see someone else give what they think is a poor answer :) For what it's worth, I have had papers rejected by journals only for the same paper get accepted by a journal higher up on SJR. The system officially used in Finland is Jufo. It has four grades in the ranking, from 0 (lowest) to 3 (highest). The funding of Finnish universities partly depends on the amount of publications in those journals, with coefficients per publication being 0.1 - 1 - 3 - 4 (i.e., a publication in a level 3 journal is worth 4 publications in a level 1 journal, and 40 publications in a level 0 one.) The ranks are re-allocated every 5 years by a committee based on suggestions from the academics; also there are some constraints (if one journal goes up, some other has to go down). This naturally creates some distortions, e. g., a new strong journal will lag in ranking, which means that nobody has incentive to publish there, which means nobody will be strongly pushing to move it up the ranking etc. For example, "Forum of mathematics, Pi" is only level 1. But other than that, the ranking is reasonable. I think all the other questions have been answered. The reason the link you posted is now dead is that the Australian Mathematical Society updated its (very old) website recently. Yes, you can find this link on web.archive.org. For what it's worth, your link is the 2008 list. The 2010 list, the final edition of the ERA rankings, can be found here http://www-users.math.umn.edu/~arnold//math-journal-ratings/ - this is "only" 10 years out of date rather than 12.
2025-03-21T14:48:31.779000
2020-08-14T18:10:58
369186
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Zeroes of the Euler series Consider a prime $p$. Let $f$ be the Euler series defined by $f(z)=\sum_{n\ge0}n!z^n\}$. It is defined and analytic over $\mathcal D=\{z\in\mathbb C_p\mid v_p(z)>-\frac1{p-1}\}$. I try to check if this function admits zeroes on $\mathcal D$. I tried to compute the Newton polygon of $f$ in order to apply Proposition 1 of Chapter 5 Section 2 of Bruhat lectures (here: http://www.math.tifr.res.in/~publ/ln/tifr27.pdf). But no luck. I did not manage. My question: Does $f$ admit zeroes in $\mathcal D$? This power series converges on the closed unit disc ${z : |z|_p \leq 1}$, so why are you looking only in the disc ${z : |z|_p < (1/p)^{1/(p-1)}}$? And a power series in $1 + z\mathbf Z_p[[z]]$ can't have any zeros with $|z|_p < 1$ since all terms have absolute value less than 1 besides the constant term 1. To count zeros in the closed unit disc, use Strassmann's theorem (all zeros in the closed unit disc will be algebraic over $\mathbf Q_p$).
2025-03-21T14:48:31.779100
2020-08-14T18:48:37
369190
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Francesco Polizzi", "R. van Dobben de Bruyn", "https://mathoverflow.net/users/7460", "https://mathoverflow.net/users/82179" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632103", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369190" }
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Automorphism induced by an automorphism of the base Let us consider a closed Riemann surface $\Sigma_b$ of genus $B$, and let $\Delta \subset \Sigma_b \times \Sigma_b$ be the diagonal. If $G$ is a finite group, then any group epimorphism $$\varphi \colon \pi_1( \Sigma_b \times \Sigma_b - \Delta) \to G$$ induces, by Grauert-Remmert Extension Theorem, the existence of a compact complex manifold $X$ (actually, a complex projective surface), endowed with a Galois cover $$\pi \colon X \to \Sigma_b \times \Sigma_b$$ branched at most over $\Delta$. Let us now denote by $a$ the involutory automorphism of $\Sigma_b \times \Sigma_b$ given by $a(x, \, y)=(y, \, x)$; it leaves $\Delta$ (pointwise) invariant, so we may ask the following Question. Under which conditions on $\varphi$ the automorphism $a \colon \Sigma_b \times \Sigma_b \to \Sigma_b \times \Sigma_b $ lifts to an automorphism $\bar{a} \colon X \to X$? Are you claiming that $X$ is smooth? I think it is constructed as the normalisation of $C \times C$ in the étale Galois cover $U \to (C\times C)\setminus\Delta$, so wouldn't it a priori only be normal? (And conversely, any normal surface with a finite Galois map to $C \times C$ is generically a finite étale Galois cover, right?) Or is there something special about this situation that makes it smooth? If $X \to Y$ is a (branched) analytic covering of complex analytic spaces, with $Y$ smooth, the singularities of $X$ lie above the singularities of the branch locus. In our situation, since $\Delta$ is smooth, the same is true for $X$. Of course, the existence of a complex structure on the closure of $U$ is strongly non-trivial (this is ensured by Grauert-Remmert, indeed). Ah, you're right; in the type of example I had in mind the branch locus is singular (I was thinking of a cone $V(z^2-xy)$ projecting onto the $(x,y)$-plane, by way of toy example). Right. In this example of double cover, the projective branch locus is given by two lines in $\mathbb{P}^2$, and the conical singularity lies over their cross point. In this post I will work in the greatest generality I can think of. In particular, fundamental group means étale fundamental group, but for varieties over $\mathbf C$ the same argument carries through using the topological fundamental group instead. Lemma. Let $X$ and $Y$ be separated normal integral schemes, let $f \colon Y \to X$ be a finite and finitely presented separable Galois cover with group $G$, and let $a \colon X \to X$ be an automorphism. Let $U \subseteq X$ be the dense open locus where $f$ is étale, let $V = f^{-1}(U)$, let $\bar y \to V$ be a geometric point with image $\bar x \to U$, and let $\phi \colon \pi_1(U,\bar x) \twoheadrightarrow G$ be the surjection corresponding to the $G$-cover $V \to U$. Then the following are equivalent: There exists an automorphism $b \colon Y \to Y$ lifting $a$; There exists a dominant rational map $b \colon Y \to Y$ lifting $a$; The isomorphism $a$ takes $U$ to itself, and the pullback $V' \to U$ of $V \to U$ along $a$ is isomorphic to $V \to U$ (as étale $G$-covers of $U$); For any choice of path $[\gamma] \in \pi_1(U,\bar x, a^*\bar x)$, the subgroups $\ker \phi$ and $\ker(\phi a_* \gamma_*)$ of $\pi_1(U,\bar x)$ are conjugate (see proof for precise statement). Moreover, the set of such lifts is a $G$-bitorsor via pre- and post-composition of deck transformations. Proof. For (1) $\Leftrightarrow$ (2), note that a dominant rational lift $b$ is automatically an automorphism. Indeed, given a commutative diagram $$\begin{array}{ccc}Y & \stackrel{b}\dashrightarrow & Y \\ \downarrow & & \downarrow \\ X & \underset a\to & X,\!\end{array}\tag{1}\label{1}$$ multiplicativity of function field degrees shows that $b$ is birational. Since $Y$ is the integral closure of $X$ in $K(Y)$, we conclude that $b$ is an isomorphism since normalisation is a functor. Thus for (2) $\Leftrightarrow$ (3), we know that $b$ gives an isomorphism $V \to V$ lifting $a|_U \colon U \to U$. This is exactly the same thing as an isomorphism $V \to V'$ over $U$, where $V'$ is the pullback $$\begin{array}{ccc}V' & \to & V \\ \downarrow & & \downarrow \\ U & \stackrel a\to & U.\!\end{array}$$ Finally, for (3) $\Leftrightarrow$ (4), we note that the cover $V' \to U$ corresponds to the surjection $$\pi_1(U,a^*\bar x) \stackrel{a_*}\to \pi_1(U,\bar x) \stackrel \phi\twoheadrightarrow G.$$ Any choice of path $[\gamma] \in \pi_1(U,\bar x, a^*\bar x)$ gives an identification \begin{align*} \gamma_* \colon \pi_1(U,\bar x) &\stackrel\sim\longrightarrow \pi_1(U,a^* \bar x)\\ [\alpha] &\longmapsto [\gamma^{-1}] \cdot [\alpha] \cdot [\gamma], \end{align*} well-defined up to conjugation. Under this identification, the surjection $\pi_1(U,a^*\bar x) \twoheadrightarrow G$ above corresponds to the surjection $\pi_1(U,\bar x) \twoheadrightarrow G$ given by $\phi a_* \gamma_*$. The induced cover is isomorphic to the cover $V \to U$ given by $\phi$ if and only if the kernels are conjugate (see e.g. [Munkres, Thm. 79.4] in the topological setting), proving (3) $\Leftrightarrow$ (4). The final statement follows for example because $\operatorname{Isom}_X(Y,Y')$ is naturally a $G$-bitorsor, as $G$ agrees with both $\operatorname{Aut}_X(Y)$ and $\operatorname{Aut}_X(Y')$. (See also this post for a general discussion of Galois covers of normal schemes.) $\square$ References. [Munkres] J. R. Munkres, Topology (second edition). Pearson, 2018. Thank you for the nice answer. I have a couple of remarks: (1) In the ramified case one does not have (as far I can see) an epimorphism $\pi_1(X) \to G$, but an epimorphism $\pi_1(X-B) \to G$, where $B$ is the branch locus. Are you sure that your reduction to the étale case is correct in this respect, or am I missing something? (2) Do you have a reference for the result on "Galois theory of the fundamental group" that you quote in the proof of equivalence $(1)-(3)$? You're absolutely right, and in looking up good references I realised that my criterion wasn't quite right because I ignored basepoints.
2025-03-21T14:48:31.779629
2020-08-14T19:40:01
369191
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On the functional equation $f(xf(y))=\frac{f(f(x))}y$ on arbitrary groups In this answer, it was shown that there is no function $f\colon\mathbb Q_{+}^{*}\to\mathbb Q_{+}^{*}$ such that \begin{equation} f(xf(y))=\frac{f(f(x))}y \label{1}\tag{1} \end{equation} for all $x$ and $y$ (in $\mathbb Q_{+}^{*}$), where $\mathbb Q_{+}^{*}$ is the set of all (strictly) positive rational numbers. The only properties of $\mathbb Q_{+}^{*}$ used in the proof were that $\mathbb Q_{+}^{*}$ is a abelian group with respect to the multiplication and $x^{-1}\ne x$ for some $x\in\mathbb Q_{+}^{*}$. The question now is this: Can the stated result be extended to non-abelian groups? A somewhat similarly looking functional equation was discussed here. For not necessarily abelian groups, we can interpret the division in (1) as the right or left division. Let $G$ be any group. The answer to the question is given by Theorem: The following three conditions are equivalent to one another: (right): there is a function $f\colon G\to G$ such that \begin{equation*} f(xf(y))=f(f(x))y^{-1} \label{r}\tag{$r$} \end{equation*} for all $x$ and $y$ (in $G$); (left): there is a function $f\colon G\to G$ such that \begin{equation*} f(xf(y))=y^{-1}f(f(x)) \label{l}\tag{$l$} \end{equation*} for all $x$ and $y$; (involutive) $x^{-1}=x$ for all $x$ (and hence $G$ is abelian). Proof of implication ($x^{-1}=x$ for all $x$)$\implies$ $G$ is abelian: For all $x$ and $y$ we have $xy=(xy)^{-1}=y^{-1}x^{-1}=yx$, so that $xy=yx$, as claimed. $\Box$ Proof of implications (involutive)$\implies$(right) and (involutive)$\implies$(left). Suppose the (involutive) property holds, so that $G$ is abelian. Let $f(x)=x$ for all $x$. Then $f(xf(y))=xy=f(f(x))y^{-1}$ for all $x,y$, so that condition \eqref{r} holds. Since $G$ is abelian, condition \eqref{l} holds as well. $\Box$ Proof of implication (right)$\implies$ (involutive): Substituting $x=1$ in \eqref{r}, we get $f(f(y))=f(b)y^{-1}$ (for all $y$), where \begin{equation*} b:=f(1). \end{equation*} So, $f(f(x))=f(b)x^{-1}$, and now \eqref{r} yields $f(xf(y))=f(b)x^{-1}y^{-1}$. Substituting here $y=1$, we get $f(xb)=f(b)x^{-1}$ or, equivalently, \begin{equation*} f(z)=cz^{-1} \label{2}\tag{2} \end{equation*} for all $z$, where $c:=f(b)b$. Now \eqref{r} can be rewritten as \begin{equation*} yc^{-1}x^{-1}=xc^{-1}y^{-1}. \label{r'}\tag{$r'$} \end{equation*} Substituting here $y=1$ and $x=c$, we get $c^{-2}=1$, that is, $c^{-1}=c$. Taking now any $z\in G$ and letting $y=zxc$, we rewrite \eqref{r'} as $z=z^{-1}$, which means $G$ has the (involutive) property. $\Box$ Proof of implication (left)$\implies$ (involutive): Substituting $x=1$ in \eqref{l}, we get $f(f(y))=y^{-1}f(b)$, where $b=f(1)$, as before. So, $f(f(x))=x^{-1}f(b)$, and now \eqref{l} yields $f(xf(y))=y^{-1}x^{-1}f(b)$. Substituting here $y=1$, we get $f(xb)=x^{-1}f(b)$ or, equivalently, \begin{equation*} f(z)=bz^{-1}d \end{equation*} for all $z$, where $d:=f(b)$. Now \eqref{l} can be rewritten as \begin{equation*} bd^{-1}yb^{-1}x^{-1}=y^{-1}bd^{-1}xb^{-1}. \label{l'}\tag{$l'$} \end{equation*} Substituting here $y=1$ and $x=b$, we get $b^{-2}=1$, that is, $b^{-1}=b$. Substituting in \eqref{l'} $y=b$ and $x=d$, we get $bd^{-2}=b^{-1}=b$ and hence $d^{-2}=1$, that is, $d^{-1}=d$. Now \eqref{l'} becomes \begin{equation*} bdybx^{-1}=y^{-1}bdxb. \label{l2}\tag{$l''$} \end{equation*} Substituting here $y=b$ and $x=1$, we get $bd=db$. Taking now any $z\in G$ and letting $y=bdz$, so that $z=bdy$, we rewrite \eqref{l2} as \begin{equation*} zbx^{-1}=z^{-1}xb. \end{equation*} Substituting here $x=b$, we get $z=z^{-1}$, which means $G$ has the (involutive) property. $\Box$ The theorem is now completely proved. Remark: As seen from \eqref{2}, if any one of the three equivalent conditions -- (right), (left), or (involutive) -- holds, then the solutions of equation \eqref{r} and/or, equivalently, \eqref{l} are precisely those of the form $f(x)=cx$ for some $c$ and all $x$. Corollary: Neither the (right) nor the (left) property can hold for any non-abelian group $G$.
2025-03-21T14:48:31.779841
2020-08-14T23:19:22
369202
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Are countable graphs with infinite minimal degree $1$-factorizable? We say that a simple, undirected graph $G=(V,E)$ is $1$-factorizable if there is a partition of $E$ such that every member of the partition is a perfect matching of $G$. It is easy to see that any $1$-factorizable graph is regular (every vertex has the same degree). If $G =(\omega, E)$ is a graph on the vertex set $\omega$ such that every vertex has degree $\aleph_0$, does this imply that $G$ is $1$-factorizable? (It is a fact that such a graph has a perfect matching.) Does this answer your question? Cardinality of a set of mutually disjoint perfect matchings of $K_\omega$. See my answer (not the accepted answer) to that question, in particular the sentence "More generally, if $G$ is a countable graph in which each vertex has infinite degree, then $E(G)$ is the union of $\aleph_0$ disjoint perfect matchings." Yes, any $\aleph_0$-regular graph $G=(V,E)$ is $1$-factorizable. (By "graph" I mean "simple graph" as in the question. Actually a loopless multigraph is OK provided no two vertices are joined by an infinite number of edges.) Let $C$ be a set of colors, $|C|=\aleph_0$. We will color the edges of $G$ with colors from $C$ so that each vertex is incident with just one edge of each color. Since each connected component of $G$ is countable, we may assume that $|V|=|E|=\aleph_0$. The coloring will be done in $\omega$ steps, at most one edge being colored at each step, so that at each step only a finite number of edges will have been colored. Fix enumerations $E=\{e_i:i\in\omega\}$ and $V\times C=\{(v_i,c_i):i\in\omega\}$. At step $2i$ consider the edge $e_i$. If it has already been colored, do nothing; otherwise, give it a color which has not been given to any edge adjacent to $e_i$. At step $2i+1$ consider the pair $(v_i,c_i)$. If the vertex $v_i$ is already incident with an edge of color $c_i$, do nothing; otherwise find an edge incident with $v_i$ which has not yet been colored and is not incident with any edge of color $c_i$, and give it the color $c_i$. It is easy to see that this construction can be carried out, and the resulting coloring has the desired properties. A similar argument shows that every $\aleph_\alpha$-regular graph is $1$-factorizable.
2025-03-21T14:48:31.780015
2020-08-15T00:49:30
369204
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Nikolay", "RobPratt", "https://mathoverflow.net/users/141766", "https://mathoverflow.net/users/163177" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632106", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369204" }
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How to solve MILP problem on several linear subspaces I have a set of close mixed-integer programming problems. More exactly, all the problems share the same set of (binary and continuous) variables, the same set of linear inequality constraints, and the same linear objective function. However, linear equality constraints are specific to each problem. Is there an algorithm to solve the set of problems that is more efficient than separate solving of each of the problems? Both algorithms and references to solvers that do this are interesting. Thanks in advance. Is everything linear (MILP)? One equality constraint per problem? Yes, linear (fixed title and description), No, typically multiple equality constraints. You could dualize the (complicating) equality constraints and use the branch-and-price algorithm. The resulting problems all have the same subproblem, and so you can use a common pool of columns. More explicitly, solve the first problem, use the resulting columns as an initial pool for the second problem, and repeat for each subsequent problem.
2025-03-21T14:48:31.780124
2020-08-15T03:29:05
369211
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "LSpice", "YCor", "https://mathoverflow.net/users/14094", "https://mathoverflow.net/users/163784", "https://mathoverflow.net/users/2383", "user163784" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632107", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369211" }
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Best texts on Lie groups for number theorists What are the most comprehensive textbooks on the structure of Lie groups and their infinite-dimensional representations if one is interested in their applications to number theory (so covering discrete subgroups and automorphic representations)? It seems to me that (most?) number theorists are not interested by general discrete subgroups but some very special ones (arithmetic subgroups). I think the most comprehensive reference would be the following conference proceedings (Proceedings in Symposia in Pure Mathematics) : Automorphic Forms, Representations, and L-functions, Parts 1&2, vol. 33 Motives, Parts 1&2, vol.51 Representation Theory and Automorphic Forms, vol. 61 However you might be also interested in the following books. An introduction to the Archimedean representation theory is given in "Representation Theory of Semisimple Groups" by Knapp. A slightly more advanced book is "Continuous Cohomology, Discrete Subgroups, and Representations of Reductive Groups" by Borel and Wallach. There is also a book "An Introduction to Automorphic Representations with a view toward Trace Formulae" by Getz and Hahn. Another recent introductory book is "Eisenstein Series and Automorphic Representations with Applications in String Theory" by Fleig, Gustafsson, Kleinschmidt, Persson. Thank you! I was aware of the older texts and monographs, but not Getz and Hahn which looks wonderful in the freely available version, nor Fleig et al., again freely available on arXiv and full of concrete examples. PSPM 9, Algebraic groups and discontinuous subgroups, is another great in that series.
2025-03-21T14:48:31.780263
2020-08-15T04:15:13
369213
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Felipe Voloch", "Mark Schultz-Wu", "Sylvain JULIEN", "https://mathoverflow.net/users/101207", "https://mathoverflow.net/users/13625", "https://mathoverflow.net/users/2290" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632108", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369213" }
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Why is a prime being even a bigger problem than a prime being divisible by 3? Many results in number theory start with "Let $p$ be an odd prime...", but it's rare to see something like "Let $p$ be a prime that is not divisible by $3$..." or similar. Why is that so, fundamentally? Is it just psychology? Evenness and oddness are more familiar concepts than residues with arbitrary moduli, so statements involving them tend to be more concise, and thus are easily perceived as being more elegant and interesting. Is it a consequence of the fact that mathematics is heavily biased towards $2$-ary operations, which align more naturally with a modulus of $2$ to yield results about the residue class of operands? Or is there something else going on? It's hard to imagine how an explanation of the type "$\rm GF(2)$ is the only field that..." could be sufficient, because of course dozens of unique, potentially relevant properties can be found for every small prime. The fact that quadratic reciprocity only works for odd primes also doesn't seem like it would quite be enough to explain the colossal weight given to odd primes in number theory, compared to primes of any other residue class. https://mathoverflow.net/questions/915/is-there-a-high-concept-explanation-for-why-characteristic-2-is-special "Every prime number is odd and 2 is definitely the oddest". J. W. S. Cassels. @Felipe Voloch or "2 is even the oddest prime" :-)
2025-03-21T14:48:31.780421
2020-08-15T04:16:51
369214
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Alejandro Tolcachier", "Derek Holt", "Francesco Polizzi", "HenrikRüping", "Luc Guyot", "YCor", "https://mathoverflow.net/users/14094", "https://mathoverflow.net/users/150901", "https://mathoverflow.net/users/35840", "https://mathoverflow.net/users/3969", "https://mathoverflow.net/users/7460", "https://mathoverflow.net/users/84349" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632109", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369214" }
Stack Exchange
Deciding if $\mathbb{Z}\ltimes_A \mathbb{Z}^5$ and $\mathbb{Z}\ltimes_B \mathbb{Z}^5$ are isomorphic or not I asked this in this MSE question but I didn't get answers. I think maybe here someone can help me. I have the two following groups $G_A=\mathbb{Z}\ltimes_A \mathbb{Z}^5$, where $A=\begin{pmatrix} 1&0&0&1&0\\0&-1&0&0&0\\0&0&-1&0&0\\0&0&0&0&-1\\0&0&0&1&0\end{pmatrix}$ and $G_B=\mathbb{Z}\ltimes_B \mathbb{Z}^5$, where $B=\begin{pmatrix} 1&0&0&1&0\\0&-1&0&1&0\\0&0&-1&0&0\\0&0&0&0&-1\\0&0&0&1&0\end{pmatrix}$. The product is given (for example in $G_A$) by $(k,m)\cdot(\ell,n)=(k+\ell, m+A^k n)$. Problem: Decide if $G_A$ is isomorphic to $G_B$ or not. My thoughts: I think strongly that they are not isomorphic but I couldn't prove it. The matrices $A$ and $B$ are both of order 4, they're not conjugate in $\mathsf{GL}(n,\mathbb{Z})$ (neither $B$ and $A^{-1}$) but they are conjugate in $\mathsf{GL}(n,\mathbb{Q})$. In some other cases, I've seen that they're not isomorphic by computing the abelianization, but in this case both have the same abelianization, namely $\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2$. Even worse, both have 1 as an eigenvalue. In my previous MO question there is a counterexample for the implication "$G_A\cong G_B\Rightarrow A\sim B^{\pm 1}$" so I cannot use that. Thanks! Let me just remark that the fact that these matrices are conjugate over $\mathbb{Q}$ gives rise to an embedding of $G_A$ into $G_B$ as a finite index subgroup and vice versa. In general this is still not enough to conclude that $G_A$ and $G_B$ are isomorphic. I did a quick Magma calculation, and the exponent 2 class 2 central factors (these are calculated by the well-known $p$-quotient algorithm) are different in the two groups, which means that they are not isomorphic. They were order $2^9$ in the first group and $2^8$ in the second. Oh, great, thanks. Could you please give me some reference about "exponent 2 class 2 central factors"? I've never seen such thing before I think. I'm also puzzled by the terminology. My guess would be: define $G^{[1]}=G$, $G^{[2]}=[G,G]G^2$, $G^3=G,G^{[2]}^2$, then consider the quotient $G/G^{[3]}$. It's the largest quotient of $G$ that is central extension where both the kernel and quotient have exponent dividing 2. Is there a manual or book of the software Magma where I can look to compute these things? Yes, YCor's definition is correct. It is called the lower exponent $p$ central series, and you will get plenty of hits if you search for that. It is very well known in computational circles, because there is an efficient algorithm (due to Mike Newman and George Havas) to compute it, and it has been used to compute orders of restricted Burnside groups, including $R(2,7)$ of order $7^{20416}$. There are also nice theoretical properties arising from the fact that the $p$-covering group of a group (as opposed to ordinary Schur cover) is unique. I will show you my Magma calcualtion shortly, but it you are more familiar with GAP then you could easily do the same calculation there. All of the time is taken up by typing in the presentation! @DerekHolt excuse me,sir. I've been experimenting with the Magma calc (very helpful), but I have a doubt. When I input (for example as in your calculation) "P1 := pQuotient(G1,2,3 : Print:=1)"; in the Ycor's notation, it gives me back $G/G[3]$ or just $G[3]$? – @AleTolcachier As the name suggests, it is calculating a quotient! The quotient groups that it calculates are always finite $p$-groups, but the terms $G^{[i]}$ themselves are often infinite groups, and they are all infinite in this example. In YCor's notation it is actually calculating $G/G^{[4]}$. Claim. The groups $G_A$ and $G_B$ are not isomorphic. We will use the following lemmas. Lemma 1. Let $A \in \text{GL}_n(\mathbb{Z})$ and let $G_A \Doteq \mathbb{Z} \ltimes_A \mathbb{Z}^n$. Then the following hold: The center $Z(G_A)$ of $G_A$ is generated by $\{0\} \times \ker(A - 1_n)$ and $(\omega, (0, \dots, 0))$ where $1_n$ is the $n \times n$ identity matrix and $\omega$ is the order of $A$ in $\text{GL}_n(\mathbb{Z})$ if $A$ has finite order, zero otherwise. The derived subgroup $[G_A, G_A]$ of $G_A$ is $\{0\} \times (A - 1_n)\mathbb{Z}^n$. More generally, setting $\gamma_{i + 1}(G_A) \Doteq [\gamma_i(G_A), G_A]$ with $\gamma_1(G_A) \Doteq G_A$, we have $\gamma_{i + 1}(G_A) = \{0\} \times (A - 1_n)^i \mathbb{Z}^n$. Proof. Straightforward. For $A$ and $B$ as in OP's question, we have thus $$Z(G_A) = 4\mathbb{Z} \times \ker(A - 1_5), \, Z(G_B) = 4\mathbb{Z} \times \ker(B - 1_5)$$ with $\ker(A - 1_5) = \ker(B - 1_5) = \mathbb{Z} \times \{ (0, 0, 0, 0) \} \subset \mathbb{Z}^5$. Lemma 2. Let $A$ and $B$ as in OP's question and set $\Gamma_A \Doteq G_A / Z(G_A)$ and $\Gamma_B \Doteq G_B / Z(G_B)$. Then we have $\Gamma_A/ [\Gamma_A, \Gamma_A] \simeq (\mathbb{Z}/ 2 \mathbb{Z})^3 \times \mathbb{Z}/ 4 \mathbb{Z}$ and $\Gamma_B/ [\Gamma_B, \Gamma_B] \simeq \mathbb{Z}/ 2 \mathbb{Z} \times (\mathbb{Z}/ 4 \mathbb{Z})^2$. Proof. Write $\Gamma_A = \mathbb{Z} / 4 \mathbb{Z} \ltimes_{A'} \mathbb{Z}^4$ and $\Gamma_B = \mathbb{Z} / 4 \mathbb{Z} \ltimes_{B'} \mathbb{Z}^4$ where $A', B' \in \text{GL}_4(\mathbb{Z})$ are obtained from $A$ and $B$ by removing the first row and the first column. Use then the description of the derived subgroup of Lemma 1 which still applies to $\Gamma_A$ and $\Gamma_B$ if we replace $A$ by $A'$ and $B$ by $B'$. Proof of the claim. If $G_A$ and $G_B$ are isomorphic, then so are $\Gamma_A$ and $\Gamma_B$. This is impossible since the two latter groups have non-isomorphic abelianizations by Lemma 2. Addendum. Let $C_A$ be the cyclic subgroup of $G_A$ generated by $a \Doteq (1, (0, \dots, 0))$ and $K_A$ the $\mathbb{Z}[C_A]$-module defined as in Johannes Hahn's answer (and subsequently mine) to this MO question. Let $\omega(A)$ be the order of $A$ in $\text{GL}_n(\mathbb{Z})$, that we assume to be finite, and set $e_0 \Doteq (\omega(A), (0, \dots, 0)) \in G_A$. Let us denote by $(e_1, \dots, e_n)$ the canonical basis of $\mathbb{Z}^n \triangleleft G_A$. It has been established that the pair $\{K_A, K_{A^{-1}}\}$ of $\mathbb{Z}[C]$-modules is an isomorphism invariant of $G_A$, where $C = C_A \simeq C_{A^{-1}}$ with the identification $a \mapsto (1, (0, \dots,0)) \in G_{A^{-1}}$. For the instances of this MO question, straightforward computations show that $$\left\langle e_0, e_2, e_3, e_5 \, \vert \, (a - 1)e_0 = (a + 1)e_2 = (a + 1)e_3 = (a^3 -a^2 + a - 1)e_5 = 0\right\rangle$$ is a presentation of both $K_A$ and $K_{A^{-1}}$ and $$\left\langle e_0, e_1, e_2, e_3, e_5 \, \vert \, (a - 1)e_0 = (a -1)e_1 = (a + 1)e_2 = (a + 1)e_3 = (a^2 + 1)e_5 + e_1 + e_2 = 0\right\rangle$$ is a presentation of $K_B$. From the above presentations, we easily infer the following isomorphisms of Abelian groups: $K_A/(a + 1)K_A \simeq \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/4 \mathbb{Z} \times \mathbb{Z}^2$ and $K_B/(a + 1)K_B \simeq (\mathbb{Z}/2\mathbb{Z})^2 \times \mathbb{Z}^2$. As result, the groups $G_A$ and $G_B$ are not isomorphic. Woah, excellent! I tried to compute the invariant $G/Z(G)$ but I didn't think of the abelianization of that space. Very very helpful your answer indeed, sir. Thanks Excuse me, I've been doing some more computations in other similar examples. I wanted to ask, what was your idea to compute the abelianization of $G_A/Z(G_A)$?. Because I 've a similar example and the groups $\Gamma_A$ and $\Gamma_B$ are isomorphic, and I want to look to some kind of invariant like you've proposed. Dear Ale, what I had initially in mind was to inspect the nilpotent quotients $\Gamma_i(G) = G/ \gamma_i(G)$ of $G$ for $G \in {G_A, \Gamma_A}$. Here $\gamma_i$ is defined as in my answer and by means of "inspection" I would start looking at the center or the derived subgroup. It turned out that $\Gamma_1(G_A)$ was just enough. I had also in mind to study $M_A = [G_A, G_A]$ as a module over $\mathbb{Z}[A] \subset M_n(\mathbb{Z})$, in particular the computation of the Fitting invariants of $M_A$ (this resembles Alexander's polynomial of a knot group). Note that if $A$ and $B$ are $GL_n(\mathbb{Q})$-conjugate, then $\mathbb{Z}[A] \simeq \mathbb{Z}[B]$ and an isomorphism between $G_A$ and $G_B$ induces an isomorphism between $M_A$ and $M_B$. Eventually, looking for non-isomorphic finite quotients of $G_A$ and $G_B$ by characteristic subgroups is a very common strategy; Derek Holt gave you excellent hints. In my above reply, one should read "$\Gamma_2(\Gamma_A)$ was just enough" instead of "$\Gamma_1(G_A)$ was just enough". I've just read about Fitting Invariants. They seem such an interesting invariant. Are they easy to compute in the case of $M_A$? @AleTolcachier The Fitting invariants of a module $M$ over $R$ are ideals of $R$ generated by the minors of fixed size of a matrix defining a presentation of $M$ ($M$ must be isomorphic to the cokernel of this matrix). Given a finite presentation, they are easy to compute (of course the number of generators and relators do matter). However, deciding whether two ideals are equal in a ring can be a difficult problem. In the case of $M_A$, they are easy to compute. Here is my Magma calculation - I did the $2$-quotient calculations to class 3. Please check that I have entered the group presentations correctly. Note that $(a,b)$ is Magma's notation for the commutator $a^{-1}b^{-1}ab$, and $a^t$ means $t^{-1}at$. > G1 := Group<a,b,c,d,e,t | (a,b), (a,c), (a,d), (a,e), (b,c), (b,d), (b,e), > (c,d), (c,e), (d,e), a^t=a, b^t=b^-1, c^t=c^-1, d^t=e*a, e^t=d^-1 >; > > G2 := Group<a,b,c,d,e,t | (a,b), (a,c), (a,d), (a,e), (b,c), (b,d), (b,e), > (c,d), (c,e), (d,e), a^t=a, b^t=b^-1, c^t=c^-1, d^t=e*a*b, e^t=d^-1 >; > P1 := pQuotient(G1,2,3 : Print:=1); Lower exponent-2 central series for G1 Group: G1 to lower exponent-2 central class 1 has order 2^4 Group: G1 to lower exponent-2 central class 2 has order 2^9 Group: G1 to lower exponent-2 central class 3 has order 2^14 > P2 := pQuotient(G2,2,3 : Print:=1); Lower exponent-2 central series for G2 Group: G2 to lower exponent-2 central class 1 has order 2^4 Group: G2 to lower exponent-2 central class 2 has order 2^8 Group: G2 to lower exponent-2 central class 3 has order 2^13 Thanks for showing this. Unfortunately, I have some technical complications. I've tried to download Magma but it seems that it isn't a free Software. I didn't find how to do it with GAP either :/ You can do it on GAP by using the Package "anupq" You should be able to do it with the Magma calculator you have to copy and paste all of the input into the wondow, and then click on "submit". Here is Derek Holt's computation done in GAP: gap> LoadPackage("anupq"); gap> F := FreeGroup("a","b","c","d","e","t");; gap> AssignGeneratorVariables(F); gap> comms := List(Combinations(GeneratorsOfGroup(F){[1..5]},2),Comm);; gap> G1 := F/Concatenation(comms, > [Comm(a,t),b^t*b,c^t*c,d^t*a^-1*e^-1, e^t*d]);; gap> G2 := F/Concatenation(comms, > [Comm(a,t),b^t*b,c^t*c,d^t*b^-1*a^-1*e^-1,e^t*d]);; gap> Pq(G1:Prime:=2,ClassBound:=2); <pc group of size 512 with 9 generators> gap> StructureDescription(last); "(C4 x C4 x C4 x C2) : C4" gap> Pq(G2:Prime:=2,ClassBound:=2); <pc group of size 256 with 8 generators> gap> StructureDescription(last); "C2 x ((C4 x C4 x C2) : C4)" Thanks! I'll keep it.
2025-03-21T14:48:31.781119
2020-08-15T04:26:13
369215
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Andy Putman", "C.F.G", "Christos", "Deane Yang", "Fedor Petrov", "Mohammad Ghomi", "Robert Bryant", "https://mathoverflow.net/users/13972", "https://mathoverflow.net/users/317", "https://mathoverflow.net/users/4312", "https://mathoverflow.net/users/613", "https://mathoverflow.net/users/68969", "https://mathoverflow.net/users/85181", "https://mathoverflow.net/users/90655" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632110", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369215" }
Stack Exchange
Alternate proofs that hyperbolic plane can’t be isometrically immersed in $\mathbb{R}^3$ A famous theorem of Hilbert says that there is no smooth immersion of the hyperbolic plane in 3-dimensional Euclidean space. The expositions of this that I know of (in eg do Carmo’s book on curves/surfaces, and in Spivak vol 3) are very analytic and non-geometric, with lots of delicate formulas. However, in Thurston’s book on 3d geometry and topology, he suggests that this is really a geometric fact that should be proved by looking at the lines of curvature and how they twist around as you go off to infinity. Question: does anyone know any alternate proof of this (or at least novel expositions of the usual ones) that emphasize the geometry and minimize the pages and pages of formulas? I don’t care about minimizing smoothness assumptions (I would rather have analytic simplicity!). If you don't like the construction of Tchebychev coordinates that is the heart of Hilbert's proof (and accounts for most of the formulae), then perhaps you should take a look at Efimov's theorem that there is no isometric immersion of a complete surface with curvature $K\le-1$ into $\mathbb{E}^3$. Since the surface is not assumed to satisfy an equation, maybe Efimov's proof (which I have never looked at, but the standard reference is Sov. Math. (Doklady) 4 (1963), 843–846, so it appears to be at most 4 pages) is more along the lines you want. Also, cool is the fact that there is a higher dimensional version studied by Tenrnblat and Terng, and it’s natural to ask whether the theorem extends to that case. @RobertBryant: While I have not tried to read it myself, my understanding is that like many Russian papers from that era Efimov's paper contains only the briefest indication of the proof. A complete account can be found in Milnor, Tilla Klotz, Efimov's theorem about complete immersed surfaces of negative curvature. Advances in Math. 8 (1972), 474-543. @AndyPutman: Thanks for the reference to T. Milnor's exposition of Efimov's theorem. I've never looked at either one, but I would not be surprised to find that Efimov's paper is telegraphic. https://www.sciencedirect.com/science/article/pii/0001870872900072 Here is Efimov's paper http://mi.mathnet.ru/dan28206 it is indeed telegraphic And idea: You could try and show that the length of the second fundamental form does not have exponential growth. (see here https://link.springer.com/article/10.1007%2Fs00013-020-01565-x). However, I have to admit that I don't know if such a proof exists. It might be possible though to find an alternative proof this way without using the Tchebychev net. @RobertBryant, the Effimov paper which you cite is only the announcement of the result. The full proof is presented in Mat. Sb. (N.S.) 64(106) (1964), 286–320, which is 35 pages. There's a nice discussion of local isometric immersions of hyperbolic planes in Robert McLachlan's expository article: McLachlan, Robert, A gallery of constant-negative-curvature surfaces, Math. Intell. 16, No. 4, 31-37 (1994). ZBL0812.53003. See also his accompanying website gallery and Richard Palais' gallery of pseudospherical surfaces. Ghys has a nice survey too in french, but I didn't see how to fill in the sketch of proof that he gives of Hilbert's theorem. I'm not sure what Thurston had in mind referring to the twisting of the lines of curvature. Following Hilbert's proof, the asymptotic curves give natural coordinates on an isometrically immersed hyperbolic surface which form a Chebyshev net. In these coordinates, the metric (first fundamental form) has the form $$\mathrm{I} = \left(\begin{array}{cc}1 & \cos(\omega) \\\cos(\omega) & 1\end{array}\right), $$ where $\omega$ is the angle between the asymptotic curves. Note that the area form is $dA=\sqrt{det(\mathrm{I})}dxdy=\sin(\omega)dxdy$. A nice proof of this is given in the Proposition on p. 192 of Ghys' paper (he says, translated: "Let us try to demonstrate it by avoiding too many calculations. . ."). He shows that the shape operator acts on vectors in the two asymptotic directions by rotation by $\pi/2$ and $-\pi/2$. Then giving a local parameterization by coordinates parallel to the asymptotic directions, he uses this to show that the mixed second partial derivative of this parameterization is orthogonal to the surface, which is equivalent to the Chebyshev net condition. I'm not sure if there's an intuitive way to understand this. One thing to note is that the asymptotic curves have constant torsion ($\pm 1$), which I think is related to the result about the action of the shape operator. McLachlan and Ghys describe a Chebyshev net as a piece of clothing with fibers going in two directions, where the fiber lengths and their crossings remain rigid, but the angles between them can vary. Then any such piece of clothing can be draped over a piece of a surface by varying the fiber angles, also like a fish net or fishnet stocking (indeed, Chebyshev considered the fitting of clothes in the presentation he made in 1878). See also Terng and Uhlenbeck "The Geometry of Solitons". Then the second fundamental form has the formula $$\mathrm{I\!I} = \left(\begin{array}{cc}0 & \sin(\omega) \\\sin(\omega) & 0\end{array}\right). $$ The $0$ on the diagonal are equivalent to taking asymptotic curves for the coordinates. One checks that $\frac{det(\mathrm{I\!I})}{det(\mathrm{I})}+-1 = 0,$ agreeing with Gauss' Theorema Egregrium. Now McLachlan notes that the Gauss-Codazzi-Mainardi equations imply that the angle $\omega$ satisfies the sine-Gordon equation and has $0 < \omega <\pi $, since the angle between the coordinates must be non-zero. $$\frac{\partial^2\omega}{\partial x\partial y}= \sin(\omega).$$ On any rectangle in the coordinate chart, the sine-Gordon equation implies that the area is bounded by $2\pi$ ("Hazzidakis' Formula"). Let the rectangle $X=[x_1,x_2]\times[y_1,y_2]$, then $$\omega(x_1,y_1)-\omega(x_1,y_2)-\omega(x_2,y_1)+\omega(x_2,y_2) = \iint_X \frac{\partial^2 \omega}{dxdy} dx dy \\ = \iint_X \sin(\omega) dx dy = \iint_{\psi(X)} dA = Area(X).$$ Since $0<\omega < \pi$, the area is bounded by $2\pi$. Now suppose that a global isometric immersion exists $\varphi:\mathbb{H}^2\to \mathbb{E}^3$. Take a point $p$ in the hyperbolic plane, then we have canonical Chebyshev coordinates about $p$, since the asymptotic curves are complete (first follow one asymptotic line through $p$ a distance $x$, then follow the transverse asymptotic curve a distance $y$ to get to the coordinates $(x,y)$ to get $\psi:\mathbb{R}^2\to \mathbb{H}^2$). If we pull back the metric to this chart, then we cannot get a complete metric because the area is bounded by $2\pi$, but the area of the hyperbolic plane is infinite. Hence by Hopf-Rinow there must be some geodesic line $\gamma:\mathbb{R}\to \mathbb{R}^2$ in this chart that is finite length, so that $\underset{t\to ∞}{\lim} \psi(\gamma(t))=q\in \mathbb{H}^2$. But along this curve then we must have $\omega$ getting arbitrarily close to $0$ or $\pi$, since if $\omega$ remains bounded away from these, then the length of the tangent vector is bounded away from $0$, so the curve would have infinite length. The principal curvatures are $\cot(\omega)\pm \csc(\omega)=\frac{\cos(\omega)\pm 1}{\sin(\omega)}$, and thus as $\omega\to 0$ or $\omega \to \pi$, the principal curvatures get arbitrarily close to $0$ and $\infty$. But by completeness of the immersion, the limit point $\underset{t\to ∞}{\lim} \varphi(\psi(\gamma(t)))=\varphi(q)$ would have to exist with finite principal curvatures, a contradiction.
2025-03-21T14:48:31.781755
2020-08-15T05:05:42
369217
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Asvin", "Joël", "https://mathoverflow.net/users/350297", "https://mathoverflow.net/users/58001", "https://mathoverflow.net/users/9317", "mhahthhh" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632111", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369217" }
Stack Exchange
Regular representations of Galois groups Suppose $\mathcal{G}_k$ is the absolute Galois group of a number field $k$. $\mathcal{G}_k$ is a topological group, with profinite topology. How does the theory of harmonic analysis of regular representations of locally compact groups apply to it? Which function spaces on $\mathcal{G}_k$ is it meaningful to consider; how do (left or right) regular representations of $\mathcal{G}_k$ on them decompose into irreducibles; which irreducibles occur; and what is the analog of the Plancherel measure? The absolute Galois group (of any field) is not only locally compact, it is compact. This makes the harmonic analysis of it rather trivial. For instance, the regular representation is the Hilbert direct sum of each irreducible representation, which all have finite dimension, each of them with multiplicity equal to its dimension... And if it is true that any finite group is a quotient of the absolute galois group, then there is really no hope of saying anything useful about which irreps can appear. The absolute Galois group (of any field) is not only locally compact, it is compact. This makes the harmonic analysis of it completely solved by Peter-Well theory. In particular, the regular representation is the Hilbert direct sum of each irreducible representation, which all have finite dimension, each of them with multiplicity equal to its dimension. What about the Weil group associated to the absolute Galois group? Since the Weil group is not compact, is there any hope on doing harmonic analysis on the Weil group?
2025-03-21T14:48:31.781893
2020-08-15T07:32:26
369223
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Jon Pridham", "Walter field", "https://mathoverflow.net/users/103678", "https://mathoverflow.net/users/148068" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632112", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369223" }
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Two definitions of cotangent complex I have reading a paper by Professor Pridham(https://arxiv.org/abs/0905.4044v4). Page 47-48 contains a comparison of the two definitions of the cotangent complex, but there is a part I don't understand. In detail, Prop 7.21 on page 48 says $\mathbb{L}^{X/Y}_{\mathscr{X}/\mathscr{Y}}$ is the sheafification of $\mathbb{L}^{X/Y}$. Question1 What is the definition of “sheafification” written in the previous sentence. And, I have one more question. Question2 Does the last sentence on page 47 means the same as the statement of the proposition ? I would appreciate if anyone who knows can tell me. I am also very grateful for the comments. Thank you ! Sheafification of that presheaf in the etale topos $X_{\bullet,et}$ from section 4 of the Olsson paper cited there, i.e. taking the sheaf associated to that presheaf. No - if you look at the first line of the proof, you will see that the expression there is not the same as the formula at the end of p47, so the purpose of the proof is to show that they are quasi-isomorphic. Thank you for your reply. (1): to me, $\mathbb{L}^{X/Y}$ seems a chain complex, but how should I sheafify $\mathbb{L}^{X/Y}$ ? (2):So what does “ sheafification ” in the last sentence of page 47 mean? In terms of the terminology on p29 of Olsson's paper, I meant any (quasi-coherent) complex on $X_{\bullet,et}$ whose restriction to $X_n$ is quasi-isomorphic $(\mathbb{L}^{X/Y})^n$. That means $\mathbb{L}^{\mathcal{X}/\mathcal{Y}}_{X/Y}$ is quasi-isomorphic to $\mathbb{L}^{X/Y} $? Yes, though they live in slightly different categories iirc. What you said is that $\mathbb{L}_{\mathcal{X}/\mathcal{Y}}$ in the Olsson paper is in $\mathcal{D}'(\mathcal{X})$, but the associated complex of $\mathbb{L}^{X/Y}$ is in $\mathcal{D}(\mathcal{X})$ ? Thank you for your reply many times. But, to me, it seems that $u^N^s\Omega(^{X}/Y)$ in $\mathbb{L}^{\mathcal{X}/\mathcal{Y}}_{X/Y}$ and $u^ \Omega(^{X}/Y)$ in $\mathbb{L}^{X/Y} $ are different although I thought $\mathbb{L}^{\mathcal{X}/\mathcal{Y}}_{X/Y}$ is quasi-isomorphic to $\mathbb{L}^{X/Y} $. Something's not rendering in your comment, but the answer might be that normalised and unnormalised chains are quasi-isomorphic. See for instance the simplicial sections of Weibel's homological algebra. I'm sorry for the typo. Thank you for giving me the reference. And, what is the definition of Tot ? It seems that the complexes are not double complexes. When Tot's applied it is to a double complex, the two directions coming from the simplicial and cosimplicial structures. Thank you very much. I understood. After all, are the definitions of cotangent complex by Olsson and by Toën-Vezzosi same ? To me, it seems that the former is in $\mathcal{D}(\mathcal{X})’$ while the latter is in $\mathcal{D}(\mathcal{X})$. What is the definition of “sheafification” in the last sentence on p47 ? The quasi-coherent complex of sheaves associated to that $O(X)$-module. That means $\mathbb{L}^{X/Y}{\mathcal{X}/\mathcal{Y}}$ in Olsson paper(the formula (8.2.3)) is different from $\mathbb{L}^{X/Y}{\mathcal{X}/\mathcal{Y}}$ in the paper in general? I think the issue's just $O(X)$-modules versus $\mathcal{O}_{X,et}$-modules, so they're the same for practical purposes. Thank you very much. It seems to me that “sheafification” of Prop 7.21 does not make sense if $\mathbb{L}^{X/Y}_{\mathcal{X}/\mathcal{Y}}$ and $\mathbb{L}^{X/Y}$ are $N^sO(X)$-modules. I am a little confused. In the prop, is $ \mathbb{L}^{X/Y}{\mathcal{X}/\mathcal{Y}}$ regarded as a $\mathcal{O}{X,et}$-module ? that's my recollection Thank you very much. So, does “sheafification ” in the prop also mean the chain complexes of $\mathcal{O}_{X, et}$-modules associated to $N^sO(X)$-modules ? I am sorry for asking the same questions.
2025-03-21T14:48:31.782174
2020-08-15T08:49:24
369225
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Christian Remling", "Lao", "Pietro Majer", "https://mathoverflow.net/users/143757", "https://mathoverflow.net/users/48839", "https://mathoverflow.net/users/6101" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632113", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369225" }
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What is the weak limit of $f_n \ \mathrm{sign}(f_n - 1)$ if $f_n \to f$ weakly in $L^p([0,1])$? Let $f_n: [0,1] \to \mathbb R$ be a uniformly bounded sequence in $L^p$. Then there exists a subsequence such that $f_{n_k} \to f$ weakly in $L^p([0,1])$. What is the weak limit of the sequence of functions $$g_{n_k} = f_{n_k} \ \mathrm{sign}(f_{n_k} - 1),$$ where sign is the signum function? Can we write it in terms of $f$? Note that $g_{n_k}$ is also uniformly bounded in $L^p$, hence it has a weak limit $g$ (up to subsequences). What is the relationship between $g$ and $f$? It depends. For example, if $f_n\to 1$, then we can obviously have $g_n\to 1$ as well, but also $g_n\to 0$ if $f_n-1$ changes sign frequently. The question may be reformulated e.g. as: what is the closure of the graph of the map $f\mapsto f\mathrm{sign} (f-1)$ from $L^\infty[0,1]$ to itself, wrto a given topology @ChristianRemling I see: for example, what assumptions do I need to add to obtain $f_{n_k}\mathrm{sign}(f_{n_k}-1) \rightharpoonup f \chi$, where $$\chi \in \begin{cases} 1 \text{ if } f > 1 \ [-1,1] \text{ if } f = 1 \ -1 \text{ if } f < 1 \end{cases},$$ weakly in $L^p$?
2025-03-21T14:48:31.782279
2020-08-15T09:02:46
369226
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632114", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369226" }
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Weibel's H-book, Milnor's exact sequence for spectral sequence of filtered complex, Theorem 5.5.5 This is a question which I asked on StackExchange first, but might be more suited here. I got stuck on the proof of Theorem 5.5.5 in Weibel's book. Not only that, but I also could not even find the statement of said theorem in any other source, so I am completely at a loss how to proceed. The result is called the Eilenberg-Moore Filtration Sequence for complete complexes. Interestingly, I was unable to find this result under this name anywhere else. Suppose the filtered chain complex $C$ is complete with respect to it's filtration $F_*C$, i.e $C=\lim\limits_{\leftarrow} C/F_pC.$ In this case the exact sequence $$0\longrightarrow\mathbin{\lim\limits_{\longleftarrow}}^1H_{n+1}(C/F_pC)\longrightarrow H_n(C)\longrightarrow\mathbin{\lim\limits_{\longleftarrow}} H_n(C/F_pC)\longrightarrow0$$ takes the form $$0\longrightarrow\bigcap F_pH_n(C)\longrightarrow H_n(C)\longrightarrow H_n(C)/\bigcap F_pH_n(C)\longrightarrow0$$ and $${\lim\limits_{\longleftarrow}} H_n(C/F_pC)\cong {\lim\limits_{\longleftarrow}} H_n(C)/F_pH_n(C). $$ The first exact sequence is Milnor's $\mathbin{\lim\limits_{\longleftarrow}}^1$ exact sequence (Theorem 3.5.8): Let $C_i$ be a tower of chain complexes satisfying the Mittag-Leffler condition, and set $C=\mathbin{\lim\limits_{\longleftarrow}} C_i$. Then there is an exact sequence for each n: $$0\longrightarrow\mathbin{\lim\limits_{\longleftarrow}}^1H_{n+1}(C_i)\longrightarrow H_n(C)\longrightarrow\mathbin{\lim\limits_{\longleftarrow}} H_n(C_i)\longrightarrow0$$ The proof then proceeds as follows: Taking the limit of the exact sequence of towers $$0\longrightarrow F_pH_*(C)\longrightarrow H_*(C)\longrightarrow H_*(C)/F_pH_*(C)\longrightarrow0$$ we find that there is a monomorphism $H_*(C)/\bigcap F_pH_*(C)\to {\lim\limits_{\longleftarrow}} H_*(C)/F_pH_*(C)$. Also, there is an exact sequnce $$0\longrightarrow H_*(C)/F_pH_*(C)\to H_*(C/F_pC),$$ which also gives a monomorphism ${\lim\limits_{\longleftarrow}} H_*(C)/F_pH_*(C)\to \lim\limits_{\longleftarrow} H_*(C/F_pC)$. This combines into a monomorphism $H_*(C)/\bigcap F_pH_*(C)\to\lim\limits_{\longleftarrow} H_*(C/F_pC)$. Then Weibel says to "combine this with the $\mathbin{\lim\limits_{\longleftarrow}}^1$ sequence of 3.5.8". And I don't get this last step at all. It seems we have not yet connected $\mathbin{\lim\limits_{\longleftarrow}}^1$ and $\bigcap F_pH_n(C)$. This result is used quite a bit in later sections, so I would really like to understand it. I would be thankful for any help here. You have constructed, up to this point, a monomorphism $H_n C / \bigcap F_p H_n C \to \lim H_n(C/F_p C)$, and this is compatible with the map from $H_n C$. This allows you to construct the right-hand square, and then assemble all of, the following map of exact sequences: $$ \require{AMScd} \begin{CD} 0 @>>> \bigcap F_pH_n(C) @>>> H_n(C) @>>> H_n(C)/\bigcap F_pH_n(C) @>>> 0\\ @. @VVV @| @VVV @.\\ 0 @>>> \mathbin{\lim\limits_{\longleftarrow}}^1H_{n+1}(C/F_pC) @>>> H_n(C) @>>> \mathbin{\lim\limits_{\longleftarrow}} H_n(C/F_pC) @>>>0. \end{CD}$$ The middle map is an isomorphism and you have shown that right-hand vertical map is injective. Applying the snake lemma, we find that all the vertical maps are isomorphisms.
2025-03-21T14:48:31.782724
2020-08-15T09:09:27
369227
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632115", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369227" }
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Primality test for specific class of natural numbers using factors of Lucas polynomials This question is related to my previous question. Can you prove or disprove the following claim: Let $N=2n+1$ where $n$ is an odd natural number greater than one , let $L_m(x)$ be the mth Lucas polynomial and let $F_m(x)$ denote an irreducible factor of degree $\varphi(m)$ of $L_m(x)$ . If there exists an integer $a$ such that $F_{n}(a) \equiv 0 \pmod{N} $ then $N$ is a prime. You can run this test here. I have verified this claim only for small values of $N$ , that is $N \in [7,1000]$ with $a \in [1,100]$ , because my PARI/GP implementation of the test is too slow. EDIT For values ​​of $n$ that are odd prime numbers this test runs in polynomial time (PARI/GP implementation) . List of Sophie Germain primes can be found here. This claim can be proved in essentially the same way as the previous one. We have $$F_n(x)=\prod_{\substack{|m|<n/2\\(m,n)=1}}(x+\zeta^m-\zeta^{-m}),$$ where $\zeta\in\mathbb{C}$ is a primitive $2n$-th root of unity. The splitting field of $F_n(x)$ is the $n$-th cyclotomic field. Assume that $q\nmid n$ is a prime number such that the reduction of $F_n(x)$ mod $q$ has a root in $\mathbb{F}_q$. The roots of $F_n(x)$ in $\overline{\mathbb{F}_q}$ are of the form $\xi^m-\xi^{-m}$, where $\xi\in\overline{\mathbb{F}_q}$ is a primitive $2n$-th root of unity. By assumption, the Frobenius automorphism $t\mapsto t^q$ fixes one of these roots, which is only possible when $q\equiv 1\pmod{2n}$. It follows that, for any $a\in\mathbb{Z}$, the prime factors of $F_n(a)$ coprime to $n$ are congruent to $1$ modulo $2n$. In particular, if $2n+1$ divides $F_n(a)$, then the only prime factor of $2n+1$ can be itself, i.e., $2n+1$ is prime.
2025-03-21T14:48:31.782877
2020-08-15T10:42:42
369233
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Eli Bartlett", "Yuhang Liu", "https://mathoverflow.net/users/474723", "https://mathoverflow.net/users/70120" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632116", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369233" }
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Analysis of solutions to a system of nonlinear ODEs arising from differential geometry Consider the system of ODEs: \begin{equation} \varphi''\varphi'^{q-1}\psi'^{p-2}=\varphi^{p-1}\psi^{q-1}, \end{equation} \begin{equation} \varphi'^2+\psi'^2=1, \end{equation} where $\varphi>0$, $\psi'>0$ and $p\geq2,\ q\geq1$ are integers. This problem has a geometric background: consider the linear $SO(p)\times SO(q)$-action on $\mathbb{R}^{p+q}$, and hypersurfaces with constant Gauss-Kronecker curvature $\pm1$ in $\mathbb{R}^{p+q}$ invariant under this action. By choosing appropriate parametrization $(\varphi(t),\psi(t))$ of the generating curve, we obtain the above equations. The sign of the curvature is not important here, as both choices of the sign lead to similar equations. We only discuss one case here and the other case with reverse sigh should be similar. A special case where $p=3,\ q=1$ has been discussed in the previous question:Analysis of solutions to a nonlinear ODE, and Robert Bryant managed to solve the equation explicitly by separation of variables. In the general situation, we do not expect to solve the equations completely, but we still hope to have certain ways of analyzing the properties of the solution. If the general case is too hard, any attempts on other special cases are also helpful. (e.g. we are trying the case of $p=q=2$ now.) Are you still interested in this question? I've got a solution for the general case of $q=1$. Thank you. We also solved the general case of $q=1$. We are interested in the cases of $p\geq 2,\ q\geq 2$.
2025-03-21T14:48:31.783011
2020-08-15T11:21:26
369235
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Louis D", "Sam Hopkins", "https://mathoverflow.net/users/12481", "https://mathoverflow.net/users/17798", "https://mathoverflow.net/users/25028", "joro" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632117", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369235" }
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Infinitely many counterexamples to Nash-Williams's conjecture about hamiltonicity? Question from 2013 gives one counterexample to Nash-Williams's conjecture about hamiltonicity of dense digraphs. Later, we found tens of counterexamples on more than 30 vertices and believe there are infinitely many counterexamples. Define $K_{x_1,x_2,...x_n}$ to the complete multipartite digraph with partitions $x_i$ and every edge is oriented in both directions. Let $L=\max x_i$. Conjecture 1: as $n,L$ vary, there are infinitely many counterexamples Q1 Does this give infinitely many counterexamples? sagemath code for $K_{1,1,2,5}$: G1=graphs.CompleteMultipartiteGraph((1,1,2,5)).to_directed() sage: print G1.edges(False) [(0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (0, 8), (1, 0), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (2, 0), (2, 1), (2, 4), (2, 5), (2, 6), (2, 7), (2, 8), (3, 0), (3, 1), (3, 4), (3, 5), (3, 6), (3, 7), (3, 8), (4, 0), (4, 1), (4, 2), (4, 3), (5, 0), (5, 1), (5, 2), (5, 3), (6, 0), (6, 1), (6, 2), (6, 3), (7, 0), (7, 1), (7, 2), (7, 3), (8, 0), (8, 1), (8, 2), (8, 3)] For counterexample on 15 vertices take $x_i=(1, 1, 1, 2, 2, 8)$. Added The suggested counterexamples are wrong and were the result of a program bug. The suggested counterexamples are wrong and were the result of a program bug. These examples are symmetric digraphs, i.e. graphs. For graphs, the Nash-Williams conjecture just becomes Chvatal's theorem (If $G$ is a graph on $n\geq 3$ vertices with degree sequence $d_1\leq d_2\leq \dots\leq d_n$ and for all $1\leq i<n/2$, $d_i\geq i+1$ or $d_{n-i}\geq n-i$, then $G$ has a Hamiltonian cycle). In other words, these examples can't be counterexamples to Nash-Williams conjecture. Of course there is no Hamiltonian cycle in these examples since there is an independent set larger than $n/2$, but Nash-Williams condition is not met. Look at the example $K_{1,1,1,1,5}$ for instance; both degree sequences are $[4,4,4,4,4,8,8,8,8]$, but $d_4=4$ and $d_{9-4}=d_5=4$. Thanks. I had typo in the first graph. According to my computations, the following non-hamiltonian digraphs satisfy nash's hypothesis, would you please verify them: (1, 1, 1, 3, 7) and (1, 1, 1, 2, 2, 8) Again, my answer explains that these can’t be counterexamples. All of your examples are subgraphs of the non-Hamiltonian graph $K_{1,1,\dots,1,(n+1)/2}$. In this graph $d_{(n-1)/2}=(n-1)/2$ and $d_{(n+1)/2}=(n-1)/2$ so Nash-Williams doesn’t hold. Since you aren’t providing the degree sequence I can’t tell why you think it does. Thanks again! I might have a bug in my program, trying to debug it and write then. This question is entirely wrong because of program bug (very likely in upstream sage). Sorry about wasting your time. Probably will delete the question. @joro: IIRC you can't delete a question that has an answer with positive score (nor should you!).
2025-03-21T14:48:31.783216
2020-08-15T13:01:47
369239
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "LSpice", "Mikhail Borovoi", "https://mathoverflow.net/users/2383", "https://mathoverflow.net/users/4149" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632118", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369239" }
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Indecomposable integral representations of a group of order 2 "by hand" This question is a duplicate of that 2010 MO question. I am interested in classifying isomorphism classes of $n$-dimensional integral representations of the cyclic group $C_2$ of order $2$. Clearly, any integral representation of $C_2$ is a direct sum of indecomposable integral representations. The following result is well-known: Theorem. The group $C_2$ has exactly 3 isomorphism classes of indecomposable integral representations: (1) trivial; (2) the sign representation; (3) the 2-dimensional representation with matrix $\left(\begin{smallmatrix}0 & 1\\ 1 & 0\end{smallmatrix}\right).$ This result was stated in the answer of Victor Protsak. See also the answer of Todd Leason. In his comment Victor Protsak gives a reference. He writes: "Curtis and Reiner, Chapter 11. It's a special case of a theorem in Section 74 which classifies integral representations of cyclic groups of prime order. Naturally, this case is much easier and can be done by hand." Question. How to prove the above theorem "by hand", without reference to the book by Curtis and Reiner? Motivation: I am working now with algebraic $\mathbb R$-tori. They are classified by integral representations of the Galois group ${\rm Gal}({\mathbb C}/{\mathbb R})$, which is a group of order $2$. In order to understand the well-known classification of indecomposable $\mathbb R$-tori, I need to understand the well-known classification of indecomposable integral representations of ${\rm Gal}({\mathbb C}/{\mathbb R})$. I asked this seemingly elementary question on Mathematics StackExchange, but got no answers or comments, so I ask it here. Casselman has a nice write-up of this classification of indecomposable tori … somewhere, but I can't find it right now. Ah, here we go. Theorem 2, p. 3, of Casselman - Computing with real tori. Let me know if this reference suffices as an answer; I leave it as a comment for now in case you're hoping for a more MO-answer-sized proof. Excellent, thank you! Let us wait... @LSpice: If you post this reference to Bill Casselman as an answer, I will be happy to accept it. Many thanks! In Computing with real tori, Casselman has a nice write-up of this theorem from the point of view of not just proving that these are the only indecomposable tori, but, supposing you are given an explicit integral representation of $\operatorname C_2$, explicitly finding/computing its decomposition into these three representations. In fact, if you (you the general reader, not necessarily @MikhailBorovoi) aren't familiar with Bill Casselman's recent work, it's well worth checking out his page http://www.math.ubc.ca/~cass; he's been very interested for a while in doing actual computations, in the sense of things that can be fed into a computer, relating to algebraic groups. The above is one example; others can be found at http://www.math.ubc.ca/~cass/research/publications.html, including, for example, The computation of structure constants according to Jacques Tits—things that we all know can be done but that most of us (at least I!) would shrink from actually doing, here laid out in a way that demonstrates how to carry it out practically. (There's also some nice stuff on mathematical graphics!) The first link in your answer does not open. Huh, sorry; it opens for me. The target is https://www.math.ubc.ca/~cass/research/pdf/realtori.pdf . If you can figure out what needs to be done to make it open for you, then please feel free to edit. See Appendix A in M. Borovoi and D. A. Timashev, Galois cohomology and component group of a real reductive group.
2025-03-21T14:48:31.783463
2020-08-15T13:08:32
369240
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Seva", "https://mathoverflow.net/users/9924" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632119", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369240" }
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Ellenberg and Gijswijt's result on arithmetic progressions in subsets of $\mathbb{F}_q^n$ and a generalisation to sets of linear equations Ellenberg and Gijswijt showed that the largest subset of $\mathbb{F}_q^n$ with no three terms in arithmetic progression has size $c^n$ where $c<q$. Ellenberg and Gijswijt actually proved a generalisation: Let $x_1, x_2, x_3\in S \subset \mathbb{F}_q^n$ then $a_1x_1+a_2x_2+a_3x_3=0$ with $a_1+a_2+a_3=0$ has no solutions (excluding the trivial $x_1=x_2=x_3$) $\implies$ $|S|<c^n$ where $c<q$. (Choosing $(a_1,a_2,a_3)=(1,-2,1)$ gives the case of AP's) My question is: How do the above bounds for |S| improve if we add additional equations and ask that all of them individually have no solutions? Specifically consider a set of equations $$a_{i1}x_1+a_{i2}x_2+a_{i3}x_3=0,$$ $1 \leq i \leq m$ with $a_{i1}+a_{i2}+a_{i3}=0$, distinct (up to multiplication by a constant) and the $a_{ij}$'s all non-zero. If $x_i\in S \subset \mathbb{F}_q^n$. How large can $|S|$ be if none of the above equations have a (non-trivial) solution? Note that my condition is not that the system of equations as a whole has no solution but rather the much stronger assertion that none of the equations of the system individually can be satisfied non-trivially. Any guesses, heuristics, plausible conjectures are welcome. Related: https://mathoverflow.net/q/254958/9924
2025-03-21T14:48:31.783578
2020-08-15T15:31:56
369246
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Emil Jeřábek", "https://mathoverflow.net/users/12705" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632120", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369246" }
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Deduction theorem for the modal mu-calculus Does the modal mu-calculus have a deduction theorem? If yes, how is it stated? Does it have the 'classical' form (i.e. as in classical propositional logic) or is it more involved? Well it can’t possibly have the same form as classical propositional logic, as it conservatively extends modal logic, for which the classical form already fails: modal logic only has deduction theorem in the form that $\Gamma,\phi\vdash\psi$ iff there exists $n$ such that $\Gamma\vdash\bigwedge_{i=0}^n\underbrace{\Box\dots\Box}_i\phi\to\psi$ (this is for a single modality; there is an obvious extension to more types of boxes). I think it is plausible that this form of the deduction theorem also holds for the modal $\mu$-calculus if $\phi$ is a sentence (i.e., if all variables inside $\phi$ ... ... are bound by $\mu$ or $\nu$ operators), but otherwise it would have to be even more complicated.
2025-03-21T14:48:31.783829
2020-08-15T17:47:32
369253
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Igor Khavkine", "JustWannaKnow", "Konstantinos Kanakoglou", "Mirco A. Mannucci", "Nik Weaver", "Robert Furber", "https://mathoverflow.net/users/150264", "https://mathoverflow.net/users/15293", "https://mathoverflow.net/users/23141", "https://mathoverflow.net/users/2622", "https://mathoverflow.net/users/61785", "https://mathoverflow.net/users/85967" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632121", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369253" }
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Creation and annihilation operators in QFT As I said before, I'm not a QFT expert but I'm trying to understand the basics of its rigorous formulation. Let's take Dimock's book, where the foundation of QM and QFT is discussed. If we consider, say, two particles, one living in a Hilbert space $\mathcal{H}_{1}$ and the other in another Hilbert space $\mathcal{H}_{2}$, the description of the state of the two-particle system is given in terms of the tensor product $\mathcal{H}^{(2)}=\mathcal{H}_{1}\otimes \mathcal{H}_{2}$. Of course, we could go furhter and study a system $\mathcal{H}^{(N)}=\mathcal{H}_{1}\otimes \cdots \otimes \mathcal{H}_{n}$. If all the particles are identical, then $\mathcal{H}_{1}=\cdots = \mathcal{H}_{n} \equiv \mathcal{H}$ and we must take into account symmetric and anti-symmetric subspaces of $\mathcal{H}^{(N)}$, which correspond to the fact that particles may be either bosons or fermions, respectivelly. At this point, one defines symmetrization and anti-symetrization operators. The next step is to consider a system of an arbitrary number of particles. At this point, one defines Fock spaces $\mathcal{F}^{\pm}(\mathcal{H}) = \bigoplus_{n=0}^{\infty}\mathcal{H}_{n}^{\pm}$ for bosons and fermions. Also, one defines creation and annihilation operators $a(h)$ and $a^{\dagger}(h)$ on $\mathcal{F}^{\pm}(\mathcal{H})$. Now, as far as I understand, this is all quantum mechanics, not QFT. However, these ideas seem to find analogues in QFT, and this is the point where I get confused. On section I.5 of Feldman, Trubowitz and Knörrer's book there is a quick discussion on (fermionic) QFT and it is stated that, in this context, creation and annihilation operators are special families $\{\varphi^{\dagger}(x,\sigma):\hspace{0.1cm} x \in \mathbb{R}^{d}, \hspace{0.1cm} \sigma \in \mathcal{S}\}$ and $\{\varphi(x,\sigma):\hspace{0.1cm} x \in \mathbb{R}^{d}, \hspace{0.1cm} \sigma \in \mathcal{S}\}$ on a Hilbert space $\mathcal{H}$. This is very different than the creation and annihilation operators mentioned above. For instance, these are now families of operators indexed by $x$ and $\sigma$. I believe this is a reflection of the fact that we passed from QM to QFT. But I'm really lost here and I don't know what's the difference between these two constructions and definitions. Can anyone help me, please? I'm mainly interested in understanding the second approach, since the first one I believe I understand (at least sufficiently well). If, in addition, you could suggest some reference where these ideas of Feldman, Trubowitz and Knörrer are discussed in more details and with rigor, I'd appreciate! ADD: Based on Feldman, Trubowitz and Knörrer's book, it seems to me that the understanding of these objects (to be more precise, the objects they briefly describe in the first 2 pages of section I.5) is fundamental to understand the formulation of a bunch of QFT models (at least for fermions). Thus, I'd appreciate if someone could elaborate a little more on the structure behind these creation and annihilation operators and its connections to the quantum case that is needed to understand the rest of the discussion on FTK's book. In other words, I think I just need to better understand these first definitions (and how are they connected with the usual quantum case I (seem) to know) to be able to understand the rest of the text. Morally speaking, a linear functional on a function space is an integral. So $a(h) = \int a(x) h(x) dx$. Replace $a$ by $\varphi$ and make $h(x)$ multicomponent-valued to account for spin to get the FTK notation. If you shift the focus from rigor to motivation and historical context, you can't go wrong with Dirac's classic Principles of Quantum Mechanics (1930), §59–65. The general idea is that in QFT -unlike QM- the wave functions themselves are "upgraded" to operators. If you wish to understand the basics, i suggest you first focus on the radiation field, that is QED (quantum electrodynamics). Among all field theories this is the first to be formulated (historically) and maybe the only one which has a rigorous formulation. Mandl's book: https://archive.org/details/FranzMandlGrahamShawQuantumFieldTheoryWiley2010/page/n161/mode/2up (especially the first few chapters), could provide a descent starting point. @IgorKhavkine this is something that came to my mind, too. But this is just a different way to represent these operators, right? I mean, is it all? And, besides, why doing that? (Sorry if my questions are really basic, I'm still getting acquainted with all this). @KonstantinosKanakoglou thank you so much for the comments! I'm going to take a look at this book right now. @IamWill Yes, it is all. The notation $a$ stands for annihilation, $\varphi$ for $\varphi$ield and of course $h$ for whave-function (joke!). They are all justified in some sense, so to each (author) their own! @KonstantinosKanakoglou: two complaints. One, I've never heard that in QFT wave functions become operators. Hilbert space operators? This makes no sense to me. Two, QED (in 3 + 1 dimensions) emphatically does NOT have a rigorous formulation. @Nik Weaver, regarding your first complaint, to tell the truth i was really surprised when i read your comment but on a second thought i believe it possibly has to do with an abuse of terminology on my side (which however i think is very common in QFT courses and texts -at least in physics depts): Speaking of "wavefunctions" i was actually refering to fields, understood as wavefunctions of matter. In the frame of second quantization (multiparticle systems) or of the canonical quantization (QFTs) the quantized field (or the multiparticle system) is frequently described (or at least i think so) as a linear combination of creation/annihilation operators with coefficients single particle wavefunctions (or the other way around). These field (or wavefunction) operators act on a ground state and produce a space which is not exactly a Hilbert space (because it is generally not complete). The spaces produced so are dense subspaces of a Hilbert space (and i think they are refered to as preHilbert spaces in some textbooks and/or lecture notes). As to your second complaint: granted! The word "rigorous" was an overstatement :) I strongly believe (modulo my knowledge of course) that hardly anything is really rigorously formulated in QFT's.... Maybe i should have said "the one which leaves the least number of unanswered questions on a first study" ... In any case, i'd love to hear your thoughts, especially if you think that my comments are misleading. Ah. So the "wave functions", i.e., states, are obtained by applying field operators to the vacuum state. That makes more sense (though it is less meaningful in curved spacetimes where there is no preferred vacuum state). The connection can be seen by taking $H = L^2(\mathbb{R}^3)$ in the first explanation. This is the Hilbert space of a nonrelativistic, spinless, three-dimensional particle. By direct summing the symmetric (antisymmetric) tensor powers of $H$ we get the Hilbert space of an ensemble of noninteracting Bosonic (Fermionic) nonrelativistic, spinless, three-dimensional particles, known as Fock space. The $n$th tensor power represents the states in which $n$ particles are present. Now we have "creation" and "annihilation" operators which take states in the $n$th tensor power into the $(n \pm 1)$st tensor power. For each state $h$ in the original Hilbert space $H$ there is a creation operator which tensors with $h$ and symmetrizes (antisymmetrizes), taking the $n$th tensor power into the $(n+1)$st, and its adjoint which goes in the opposite direction and removes a tensor factor of $h$. In the physics literature one usually works with idealized creation/annihilation operators for which the state $h$ is a fictional Dirac delta function concentrated at some point of $\mathbb{R}^3$. This is what is described in your second explanation. As is usual in physics, the Hilbert space is unspecified, but in the case of free fields it corresponds to the Fock space in the first explanation. Fock space is inadequate to model interacting fields (indeed, here the mathematical issues become deep and fundamentally unresolved). However, it is not trivial; for instance, one can study free quantum fields against a curved spacetime background and derive Hawking radiation, the Unruh effect, etc. Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics by Wald is an excellent, mathematically rigorous explanation of this setting. In QFT the intuition is that one has a separate Hilbert space at each point of space, and one takes their tensor product to get the Hilbert space of the entire field. I indicated how, intuitively, the Fock space models a "measurable tensor product" of a family of harmonic oscillators (Bosonic case) or two-state systems (Fermionic case) indexed by all the points of space in my answer here. See Section 2.5 of my book Mathematical Quantization for a full explanation. very nice! Bravo Nik. And thanks for filling us in with a more accurate pciture Thank you Mirco! @NikWeaver thank you so much for the amazing answer. This clarifies the things to me. Let me ask you something. You said that my second representation arises when the state $h$ is taken to be a delta. Dimock's book discusses many particle systems where $\mathcal{H}=L^{2}(\mathbb{R}^{3})$. In this case, $\mathcal{F}^{\pm}=\bigoplus_{n=0}^{\infty}\mathcal{H}^{\pm}$ , where $\mathcal{H}^{\pm} = L^{2}_{\pm}(\mathbb{R}^{3n})$ are, respectivelly, symmetric and anti-symmetric subspaces of $L^{2}(\mathbb{R}^{3n})$. continues... At some point, Dimock states: "If $\psi \in L^{2}{\pm}(\mathbb{R}^{3n})$ is a continuous function, then in $a(h)\psi$ we can take $h$ to be a $\delta$-function and define an operator $a(x)$ by $(a(x)\psi)(x{1},...,x_{n-1}) = \sqrt{n}\psi(x,x_{1},...,x_{n-1})$." Question: according to you answer, this is precisely what my $\varphi^{\dagger}(x,\sigma)$ are, right? The only difference is that, in this case, the spins are not being taking into account, so that $\sigma$ is omitted. Yes, I believe that is what the $\phi^{\dagger}(x,\sigma)$ are. Just be careful about $\psi$ needing to be continuous --- this makes the operator $a(x)$ unbounded on each summand, whereas $a(h)$ for $h \in L^2(\mathbb{R}^3)$ would be bounded on each summand ... Disclaimer: I am not a mathematical physicist. Even with one Hilbert space, namely the quantum harmonic oscillator, you can define "creation-annihilation" operators, except that in this case they simply raise or downgrade the energy level of the single particle system. Now, you consider the Fock space $\mathcal{F}^{\pm}(\mathcal{H}) = \bigoplus_{n=0}^{\infty}\mathcal{H}_{n}^{\pm}$ the way you describe above: it is actually a functor, hence the infamous dictum that second quantization is a functor. Therein, you define again the two operators, but you re-interpret them as ladder operators which, from the ground state, create and destroy particles. Formally they behave very much as with the toy harmonic oscillator, and that analogy is far-reaching: basically it tells you that the quantum field described by the Fock functor can get "excited": particles are excitations of the void (in fact there are some beautiful pictures of quantum fields as (infinite) ensembles of (coupled ) harmonic oscillators, see here). What has this to do with the second definition? If the quantum field creates and annihilates particles, it can do it at each point of your ambient space. Hence the indexes... Regarding whether each quantum field can be represented using creation and annihilation operators on a Fock space -- the answer is no. Hello again Mirco! And thanks for the answer! You said that the Fock space formulation is already QFT, right? This is confusing to me. See, I know that, rigorously, a QFT theory is defined by Wightman axioms, right? But intuitively (since I'm not an expert), this seems really like a QM problem. What is you understanding on this? @Iamwill you see the beaty of MO? You try to answer a question, and sometimes you mak a fool out of yourself :) . Now, assuming that one can take it (and I can), it is GOOD. Why? because you learn what you do not know. My picture was an oversimplification, and was based on the erroneous assumption that every QFT can somehow be "represented" as a Fock space, in other words, that the Hilbert space which is assumed by the Wightman axioms can be realized as a Fock tensor product. Robert disabused me of this. So, here i my new understanding (perhaps also faulty): some QFT can be so realized, but actually they are in a way the silly ones. Now, the exact picture still escapes me @MircoA.Mannucci I think to recognize one's mistake is as important as learning from others. In my opinion, this is a very nice humble behavior of yours! This is well done science for sure!
2025-03-21T14:48:31.784733
2020-08-15T18:29:56
369254
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Stack Exchange
Iterated Gentzen: or, a Sith objection to the proof of consistency of PA $\DeclareMathOperator\PRA{PRA}\DeclareMathOperator\WF{WF}\DeclareMathOperator\Con{Con}\DeclareMathOperator\PA{PA}$Preamble: In the year … in a galaxy far far away, a nasty Sith named Darth Dubious (DD) asks a Jedi, Obi Wan Mathobi (OWM), about the consistency of PA: DD: How do you know that PA is consistent? OWM: Don't you know that many centuries ago a great Mathematician from Terra by the name of Gentzen proved its consistency? DD: Of course I do. He proved that $\PRA + \WF(\epsilon_0) \vdash \Con(\PA)$, but …. OWM: But what? Are you going to tell me the usual story that the proof is not finitistic enough? DD: I dare not, Master Mathobi. I happily concede that the proof is valid. Yet, I have still a doubt lingering in my brain: how do you know that $G(0) =\PRA + \WF(\epsilon_0)$ itself is not contradictory? As I said, the proof seems quite acceptable to me, but not any argument that $G(0)$ is consistent because it has a model in some very infinitistic theory such as ZF. Rather, let us say that we apply Gentzen's argument to $G(0)$, and that we can prove the following: $\PRA + \WF(\alpha_0) \vdash \Con( G(0))$. Now, I would surmise that $\alpha_0 > \epsilon_0$, right? OWM: It would seem so. I suspect (but I am not sure) that else $\PA$ would be able to prove the consistency of $G(0)$. Need to ask some other Jedis more skillful in Ordinal Analysis …. DD: Ok, waiting for them I tell you where I am going, although methinks you know it already: I am going to repeat my argument again, and create a chain of theories $G(i)$ such that each one Gentzen-proves the consistency of the previous one by an ever greater countable ordinal. If necessary, we can iterate beyond $\omega$. Now, is this series of countable recursive ordinals cofinal in the set of all countable recursive ordinals? If yes, I am afraid you ask for too much, because then I would have to accept induction all the way to $\omega^{CK}_1$. If, on the other hand, it does not, I would like to know which is the upper bound. THAT ordinal is the actual price to pay to secure $\PA$'s consistency. Question: What is wrong with DD's argument? Or, if it is sound, any clues on the upper bound ordinal which would secure the consistency of the entire chain of iterated Gentzen theories ADDENDUM I started with $\PA$ but, mutatis mutandis, you can begin from $\operatorname Q$, in which case instead of $\epsilon_0$ you use $\omega$. There's a step "Now, I would surmise that $\alpha_0 > \epsilon_0$, right?" which is not fully justified, and seems to be the obvious place to look for trouble. LSpice, thanks for your editing. My eye sight is not that good, and I am quite sloppy. As for your point, of course, that is the "critical point", but we need details, don't we? Some Jedi Masters here know how to dig it, I trust.... PS Even if the ordinal inequality is wrong at stage 0, to secure Gentzen we would need an argument that it is UNIFORMLY so, ie that it never goes up. If the argument already fails at step 0 then it has nowhere further to go, as $\operatorname{Con}(\epsilon_0) \vdash \operatorname{Con}(\alpha_0)$ in this case (right?). Anyway I am no logician, and agree that experts will chime in. I was just pointing out that there was an obvious point on which to focus. I do not think so. Probably, but not surely. It needs an argument that for each alpha (index of G_alpha), the ordinal does not get pushed up. And, of course, this argument must also be finitistic... In order to really formalize this question, you may have to be more careful in terms of formalizing what you mean by $\mathrm{WF}(\alpha)$. $\mathrm{PRA}$ is a first-order theory of arithmetic, so right off the bat full well-foundedness is not formalizable. You can only talk about well-foundedness relative to descending sequences definable with formulas of some given complexity (and, indeed, Gentzen only needs it for some relatively low complexity). The other issue is that really you have to talk about specific ordinal notations (as in specific computable copies of the ordinals). Over a sufficiently strong base theory these different forms of the assumption should be provably equivalent, but I don't know if this true for arbitrarily large computable ordinals, and I don't know if it's true for $\mathrm{PRA}$. @JamesHanson what you say is correct, and helped me correct a silly statement I made inside: if the sequence of ordinals were cofinal in the RECURSIVE ordinals, it would have as upper bound the Kleene ordinal. So, thank you. Having said that, though, I do not see your point: of course we are talking about ordinals and well-foundedness inside an arithmetical theory, and of course if you choose as a starting point a weak theory you must pay attention to how you encode the recursive relation, the ordinal notation, and whatelse. But the bottom line remains: it can be done, so that G(0) is a perfectly legitimate arithmetical theory, seemingly incomparable in strength with PA, and proving its consistency. So, Darth Dubious's argument applies: what is the ordinal strength needed to prove CON(G(0))? and also, you can define the same issue for a recursive list of ordinal numbers. Now, aside the little SF story, what is at stake is: perhaps the ordinal strength necessary to prove CON(G(0)) is not greater than epsilon_0, and the same applies to all successive iterations. In that case Master Mathobi wins: you only have to believe induction up to e_0 and bingo! you can sleep well at night. But I am a bit skeptical about that. Anyway thanks for your contribution anyway I don't really understand what you're asking here: specifically, the question "[What] ordinal is the actual price to pay to secure PA's consistency[?]" is confusing to me because I don't know what "secure PA's consistency" means. I don't understand why $\epsilon_0$ isn't enough - the fact that we can further question the consistency of $PRA+WF(\epsilon_0)$ (ignoring precise phrasing issues) doesn't seem relevant to me, since we don't need the additional hypothesis of that theory's consistency to conclude $Con(PA)$ within that theory. So what exactly do you mean by "secure the consistency of?" Ok @NoahSchweber: suppose for one second that G(0) were inconsistent. If G(0) were inconsistent, it would prove ANYTHING, ex falso sequitur quodlibet, right? So, what Gentzen proves in fact is: IF G(0) is consistent, THEN PA is consistent too. Now, how do I know that G(0) is consitent? Of course, I can resort to, say, ZF, but then why bother with G(0) in the first place? So, I say, very well, let us apply the same therapy to G(0). Assume that to use the same type of argument, I need another ordinal, say alpha_1. Now I repeat the chain. I have a sequence of recursive ordinals, @MircoA.Mannucci So what exactly are you looking for? To me, "secure the foundation of PA" means exactly "when granted as an assumption, it implies $Con(PA)$." So the issue you raise just isn't a problem for how I understand the question: if we grant that $PRA+WF(\epsilon_0)$ is true, then surely it's consistent. e_0 =alpha_0, alpha_1, ...... etc. Now I wanna know what is the sup of this sequence. If I know it, I know how much of ordinal induction is necessary to swallow, to ensure that PA is consitent, WITHOUT resorting to ZF or anything else for that matter. Makes sense now? No, because I don't understand what is satisfactory, only one thing you find unsatisfactory. For example, what does "etc" mean - does that sequence go on for $\omega$-many steps, or what? By the way, I hope you are one of the downgraders of this question. You know why? If you are, I like you. And I like you because you had the good taste of saying this question does not make sense to me, rather than saying nothing at all. Perhaps my question does not make sense, though alllow me to say that so far it does, but I need criticisms, objections, etc. (Yes, I did downvote - I don't downvote without saying why unless it's a question by a user I already know becomes very aggressive in comments. Those are, granted, extremely rare.) Good point. Let us start with that, let us continue omega steps. But even better, because perhaps I have more than omega times to say I do not trust the last step, let us continue till allowed, namely as long as I can go staying within recursive ordinals. Now, many things can happen: perhaps the series is stationary at epsilon_0, in which case I would be very happy, or perhaps is monotonic increasing, or perhaps oscillates. I hope someone will tell me which one BRAVO Noah. Actually, twice bravo, one for verbalizing your objections, and the other one to say clearly so. I have no ego here, it may well that my question was not as transparent as I hope it to be, or whatever. Thanks ! PS You said the magic word: TRUE. I know exactly what you mean, but to me it means nothing. There is nothing true anywhere in math, as far as I can see . There are PROOFS, and if by true you mean in N, then I tell you again: why bother? PA is true in N just as well. But point is, that is not what Gentzen had in mind.Gentzen wanted to show CON(PA) with what he understood to be finitistic arguments, and there ain't anything finitistic about N... I once asked somewhere how to find the ordinal strength of PA+CON(PA) and it turned out to be $\epsilon_0$, the same as that of PA itself. I don't remember the explanation of this. The main conclusion that ordinal strength wasn't as linear as it might sound. Maybe your $\alpha_0$ also turns out to be the same as $\epsilon_0$. Yes, it can be so, that is what Mathobi hopes, and perhaps it is indeed the case. But remember, that takes care of G(0), would be great that all G(i) had ordinal strength $\epsilon_0$. We shall see.... I was going to write this as an answer to the question supporting Obi Wan Mathobi's hope, but it's too incomplete to post as one. In "The Realm of Ordinal Analysis" (proposition 2.24, preprint) M. Rathjen writes that if we have a subsystem of $Z_2$ that includes all axioms of $PA$ (e.g. $PA$ itself), adding a semantically true $\Sigma_1^1$ statement doesn't change its proof-theoretic ordinal. So $\varepsilon_0=\vert PA\vert=\vert PA+Con(PA)\vert=\vert PA+Con(PA+Con(PA))\vert=\ldots$, including all finite iterations. But Rathjen's proposition uses a formulation of proof-theoretic ordinal called $\vert\cdot\vert_{sup}$, which is different from the formulation $PRA+WF(\alpha_i)\vdash Con(G(i))$ in the question. Also I'm not sure if the proof of Rathjen's proposition is finitistic enough for Obi Wan Mathobi. Mathobi is making a different argument in the comments on this post than in the question itself: in the post Mathobi is considering how far we need to justify transfinite induction to prove $G(i)$ consistent over PRA, while in the comments you mentioned Mathobi's hope would be supported if PA+Con(PA) had proof-theoretic ordinal $\varepsilon_0$. In favor of the position in the comments, the theories PA, PA+Con(PA), PA+Con(PA+Con(PA))), etc. (for finite iterations) all have the same strength, $\varepsilon_0$. However, we can show $G(1)$ is quite a bit stronger than PA+Con(PA), and each $G(i+1)$ is stronger than $G(i)$. But first, we need to distinguish a few concepts: there's more than one formulation of proof-theoretic ordinal, the most commonly used one is $\vert\, .\,\vert_{sup}$, and $\vert\, .\,\vert_{Con}$ is the rendering of Gentzen's result in your question, $\textrm{PRA+TI}(\alpha)\vdash\textrm{Con(T)}$. (These are described in more detail in Rathjen's "The Realm of Ordinal Analysis".) However $\vert\, .\,\vert_{Con}$ is only useful with respect to a "natural" ordinal representation system fixed in advance, so for the rest of this answer we fix a representation system such as one based on Veblen's functions. Let's also define theories $H(0)=\textrm{PA}$, and $H(i+1)=\textrm{PA+Con}(H(i))$. Here we get our first result comparing proof-theoretic ordinals of these theories, which is that for all $i<\omega$ we have $\vert H(i)\vert_{sup}=\varepsilon_0$. Since each $H(i)$ is a recursive theory, $\textrm{Con}(H(i))$ is a $\Pi_1^0$-sentence, and by proposition 2.24 from Rathjen's "The Realm of Ordinal Analysis", we have $\vert\textrm{PA+}\phi\vert_{sup}=\varepsilon_0$ where $\phi$ is any true arithmetical sentence. So we have $\vert\textrm{PA}\vert_{sup}=\vert \textrm{PA+Con}(H(i))\vert_{sup}$ for all $i<\omega$, and all those proof-theoretic ordinals are $\varepsilon_0$. This doesn't fully answer the original question though, which was framed in terms of $\vert G(i)\vert_{Con}$ instead of $\vert H(i)\vert_{sup}$. The next step to answering the main question is utilizing this result to compare $\vert H(i)\vert_{Con}$, which also gives us that $\vert H(i)\vert_{Con}=\varepsilon_0$ for all $i<\omega$. (I can provide proof of this fact.) However, failure to replicate this result for the hierarchy $G(i)$ shows the discrepancy between Mathobi's points: while for all $i<\omega$ we have $\vert H(i)\vert_{Con}=\vert H(i+1)\vert_{Con}$, in contrast we have $\vert G(i)\vert_{Con}<\vert G(i+1)\vert_{Con}$! (This is because otherwise $G(i+1)=\textrm{PRA+TI}(\vert G(i)\vert_{Con})=\textrm{PRA+TI}(\vert G(i+1)\vert_{Con})$ would prove its own consistency.) As a result $\textrm{PRA+TI}(\varepsilon_0)$ is quite a bit stronger than PA+Con(PA) at least in terms of $\vert\, .\,\vert_{Con}$. C7X, you got my vote. A for a comment, give me some time to process your answer, and then I will add my reaction as an addendum to my post
2025-03-21T14:48:31.785673
2020-08-15T18:48:12
369255
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632123", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369255" }
Stack Exchange
Ideal of the boundary of $G/U \subset \overline{G/U}$ Let $G$ be a semi simple algebraic group, $B \subset G$ is a Borel subgroup and $U \subset B$ is the unipotent radical of $B$. We can consider the variety $G/U$. Let us also denote $\overline{G/U}:=\operatorname{Spec}(\mathbb{C}[G/U])$. It is known that the natural morphism $G/U \rightarrow \overline{G/U}$ is an open embedding. Let $\partial{G/U}$ be the boundary of $G/U$ inside $\overline{G/U}$. Note now that $\mathbb{C}[G/U]=\bigoplus_{\mu} V(\mu)$, where the sum runs through dominant characters $\mu$ of $G$ (we fix some maximal torus $T \subset B$, here $V(\mu)$ is the irreducible representation of $G$ with highest weight $\mu$). Claim: the ideal of $\partial{G/U} \subset \overline{G/U}$ is generated by $V(\mu)$ with $\mu$ being regular (strictly dominant). How to prove this claim? Maybe there are any references? Here is one way to see it, via classifying $G$-invariant radical ideals. (This has the bonus that it implicitly describes the boundary.) Lemma: $G$-invariant ideals $I$ of $\mathbb{C}[G/U]$ are in bijection with sets of weights $S$ so that for $\lambda\in S$ and $\mu > \lambda$, $\mu\in S$. Such an ideal is radical iff for all $\lambda\notin S,$ we have $n\lambda\notin S$ for all positive integers $n$. To see this, note that $G$-invariance tells you that $I$ must split as a sum $$\displaystyle\bigoplus_{\lambda\in S}V(\lambda)$$ for some set $S$. Now if $\lambda\in S,$ the multiplication map $V(\mu-\lambda)\otimes V(\lambda)\rightarrow V(\mu)$ is surjective and hence $\mu > \lambda$ must also be in $S$. The statement about radical ideals follows similarly. From this statement, you can see that the minimal nonzero $G$-invariant radical ideal (which necessarily cuts out the boundary) corresponds to taking $S$ the set of all regular weights.
2025-03-21T14:48:31.785834
2020-08-15T19:29:39
369257
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Fedor Petrov", "Giorgio Metafune", "Keba", "LSpice", "Tony419", "https://mathoverflow.net/users/150653", "https://mathoverflow.net/users/157356", "https://mathoverflow.net/users/2383", "https://mathoverflow.net/users/4312", "https://mathoverflow.net/users/44919" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632124", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369257" }
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Example of a function with a curious property Denote by $L^1(0,1)$ the space of Lebesgue integrable functions on the interval $(0,1)$. $\textbf{Question:}$ Does there exist a function $F:(0,1)\rightarrow\mathbb{R}$ such that: $\frac{F(x)}{x}\in L^1(0,1)$, $\frac{F'(x)}{x}\in L^1(0,1)$, $\frac{F(x)}{x^2}\notin L^1(0,1)$? I'm guessing that the answer is positive and the point is to construct $F$ such that $F$ and $F'$ behave suitably near zero. It seems quite delicate. I checked that $F$ cannot be a polynomial or a power function (since then $F'\simeq \frac{F}x$, thus conditions 2 and 3 cannot hold simultaneously). I would appreciate any hints! Not if $F$ is nonnegative and increasing near $0$: Condition 1 implies $F(0) = 0$, hence by the mean value theorem $F(x) \le x F'(\xi) \le x F'(x_0)$ for some $\xi \in (0, x_0)$ and small $x_0 > 0$. My intuition tells me that then it should not be possible in general (for instance, multiplying with something like $\sin(1/x)$ worsens the problem). There is no such function. First of all, $|F(a)-F(b)|\leqslant \int_a^b |F'(x)|dx\to 0$ when $a,b\to 0$. So $F$ has a limit $c$ at point 0. If $c\ne 0$, then 1) fails. So $\lim_{x\to 0} F(x)=0$. Next, $$|F(a)|\leqslant \int_{0}^a|F'(x)|dx\leqslant a\int_{0}^a\frac{|F'(x)|}x dx=o(a),\quad\text{when}\quad a\to 0.\quad (1)$$ Now $$ \int_a^b \frac{F(x)}{x^2}dx=\frac{F(a)}a-\frac{F(b)}b+\int_a^b \frac{F'(x)}xdx. \quad(2) $$ Consider two cases: $F$ has fixed sign near 0. Then choosing $a,b$ close to 0 we conclude from (1) and (2) that $\int \frac{F(x)}{x^2}dx$ converges at 0, but this is equivalent to the convergence of $\int \frac{|F(x)|}{x^2}dx$ which we need. $F$ has infinitely many zeroes in any neighborhood of 0. Then choosing $(a_k,b_k)$ being inclusion-maximal intervals of the open set $\{x:F(x)\ne 0\}$ and applying (2) for $a=a_k,b=b_k$ we bound $\int_0^c \frac{|F(x)|}{x^2}dx$ via $\int_0^c \frac{|F'(x)|}{x}dx$. Here $c=b_1$, for example. Cool, thanks! :] You’re proving that the answer to the stated question (does there exist $F$ such that $F(x)/x \in L^1$, $F’(x)/x \in L^1$, $F(x)/x^2 \notin L^1$) is “no”, right? @LSpice yes, now added A similar argument: once $F(0)=0$ has been proved, then $F(x)=\int_0^x F'(t)dt$ and $$\int_0^1 \frac{|F(x)|}{x^2}dx\le \int_0^1 \frac{1}{x^2} dx\int_0^x |F'(t)|dt=\int_0^1 |F'(t)|dt \int_t^1 x^{-2}dx=\int_0^1 |F'(t)|\frac{1-t}{t}dt$$.
2025-03-21T14:48:31.786147
2020-08-15T19:32:36
369258
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Moishe Kohan", "YCor", "https://mathoverflow.net/users/14094", "https://mathoverflow.net/users/39654" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632125", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369258" }
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Positive genus Fuchsian groups Let $G$ be a lattice in $SL(2,\mathbb{R})$. Is it always true that there exists a finite index subgroup $F$ of $G$ such that the quotient surface $\mathbb{H}/F$ has positive genus? Is the statement true under some general enough set of assumptions? Please can you add a reference? Yes, this is true, but proving this is easier than finding a reference. Every finitely-generated matrix group (e.g. a lattice in $PSL(2, {\mathbb R})$ contains a torsion-free subgroup. The general result is due to Selberg, but for discrete subgroups of $PSL(2, {\mathbb R})$ it was surely known earlier. In view of 1, it suffices to prove that every surface $S$ homeomorphic to the 2-dimensional sphere with $n\ge 3$ punctures admits a finite covering $S'\to S$ such that $S'$ has positive genus. Suppose first that $n$ is odd. Surround punctures $p_i$ by small loops $c_i$. I will think of these as elements of $H_1(S, {\mathbb Z}_2)$. Now, consider the homomorphism $$ \alpha: \pi_1(S)\to H_1(S, {\mathbb Z}_2)\to {\mathbb Z}_2$$ where the first arrow is Hurewicz and the second sends $[c_1], [c_2]$ to $1$ and the rest of $[c_i]$'s to $0$. Take the 2-fold covering $S_1\to S$ corresponding to the kernel of $\alpha$. Then $S_1$ is $2+ 2(n-2)$-times punctured sphere. Thus, the problem is reduced to the case of spheres with even number of punctures. Let $S$ be $S^2$ with $n=2k\ge 4$ punctures. Similarly to (2), define the homomorphism $$ \beta: \pi_1(S)\to H_1(S, {\mathbb Z}_2)\to {\mathbb Z}_2 $$ where the second arrow sends all $[c]_i$'s to the nonzero element of ${\mathbb Z}_2$. Let $S'\to S$ denote the 2-fold covering corresponding to the kernel of $\beta$. Then $S'$ will have $2k$ punctures and genus $k-1>0$. (This is an exercise in topology of surfaces. The natural extension of $S'\to S$ to a ramified covering of compact surfaces is called a hyperelliptic covering map.) Edit. 1. If you want a reference, an optimal result is in Edmonds, Allan L.; Ewing, John H.; Kulkarni, Ravi S., Torsion free subgroups of Fuchsian groups and tessellations of surfaces, Invent. Math. 69, 331-346 (1982). ZBL0498.20033. It can be stated as: Suppose that $F_1, F_2$ are lattices in $G=PSL(2, {\mathbb R})$. Then $F_2$ embeds in $F_1$ (as an abstract group) with index $k$ if and only if the Riemann-Hurwitz condition is satisfied: $$ \chi(F_2)/\chi(F_1)=k. $$ Once you unravel the definitions, it implies the positive answer to the positive genus question. In order to apply their result, one needs to know (and they take it for granted) that every lattice in $G$ has the presentation $$ \langle a_1, b_1,...,a_p, b_p, c_1,...,c_r, d_1, ..., d_s| \prod_{i=1}^p [a_i, b_i] \prod_{j=1}^rc_i \prod_{k=1}^s d_k =1, c_1^{e_1}=...=c_s^{e_s}=1\rangle. $$ This presentation one can find in Poincare's papers on Fuchsian functions. Whether he actually had a proof is hard to tell (this applies to pretty much everything written by Poincare that I tried to read, but others might disagree), but he had a tool for proving the result, namely convex fundamental domains. A more solid proof is likely to be found in Dehn's papers (I did not try). The earliest solid reference I know for the existence of a finite generating set for lattices $\Gamma< G=PSL(2, {\mathbb R})$ is Siegel, Carl Ludwig, Some remarks on discontinuous groups, Ann. Math. (2) 46, 708-718 (1945). ZBL0061.04505. Unsurprisingly, Siegel uses fundamental polygons to prove the result: He proves the existence of a finitely-sided fundamental polygon and, as a consequence, concluded an explicit upper bound on the number of generators in terms of the area of the quotient ${\mathbb H}^2/\Gamma$. This finiteness theorem holds in much greater generality, for lattices in connected Lie groups, but this is another story (that also has complicated history to the point that it is unclear whom to credit with this, clearly fundamental, result). One thing, that I am not sure about is: While the existence of finite generating sets for lattices in connected Lie groups is known, I do not know a solid reference to an explicit upper bound on the number of generators in terms of the volume of the quotient (in the non-torsion free case). Regarding "Fenchel's Conjecture" that each lattice in $G=PSL(2, {\mathbb R})$ contains a torsion-free subgroup of finite index: The story is somewhat bizarre. When the conjecture was first stated is hard/impossible to tell. It is mentioned in Nielsen's paper J. Nielsen, Kommutatorgruppen for det frie produkt af cykliske grupper, Matematisk Tidsskrift. B (1948), pp. 49-56. Nielsen's paper, remarkably, contains no references whatsoever. However, by the time of appearance of Nielsen's paper, Fenchel's conjecture was already proven. The proof is mostly contained in: Mal’tsev, A. I., On the faithful representation of infinite groups by matrices, Am. Math. Soc., Transl., II. Ser. 45, 1-18 (1965); translation from Mat. Sb., N. Ser. 8 (50), 405-422 (1940). ZBL0158.02905. Now, each lattice $\Gamma< G=PSL(2, {\mathbb R})$ is finitely generated and contains only finitely many $\Gamma$-conjugacy classes of finite order elements. (This, at the very least, comes from Siegel's theorem on fundamental polygons which, as I said, was likely to have been known to Poincare.) Mal'tsev's theorem implies that if $\Gamma$ is a finitely-generated matrix group, then for every finite collection of nontrivial $\Gamma$-conjugacy classes $C_1,...,C_k$, there exists a finite-index subgroup $\Gamma'< \Gamma$ disjoint from $C_1,...,C_k$. By combining the two results, every lattice in $G=PSL(2, {\mathbb R})$ contains a torsion-free subgroup of finite index. A complete solution of Fenchel's conjecture was claimed by Fox in Fox, Ralph H., On Fenchel’s conjecture about (F)-groups, Mat. Tidsskr. B 1952, 61-65 (1952). ZBL0049.15404. who was clearly unaware of Mal'tsev's paper. Fox's solution turned out to be partially erroneous, with an error (in one of the cases) corrected in: Chau, T. C., A note concerning Fox’s paper on Fenchel’s conjecture, Proc. Am. Math. Soc. 88, 584-586 (1983). ZBL0497.20035. By that time (23 years earlier), Selberg proved an even more general result in: Selberg, Atle, On discontinuous groups in higher-dimensional symmetric spaces, Contrib. Function Theory, Int. Colloqu. Bombay, Jan. 1960, 147-164 (1960). ZBL0201.36603. Selberg proved that each finitely-generated matrix group contains a torsion-free subgroup of finite index. Selberg was also unaware of Mal'tsev's paper but, at least he was not reporoving something which was already there. The thing is that a finitely generated matrix group $\Gamma$ can have infinitely many $\Gamma$-conjugacy classes of finite subgroups, hence, one cannot simply apply Mal'tsev's result. I don't think the case of finitely generated subgroups of $(P)SL(2,\mathbf{R})$ is any easier than the general case, or even known before. However, for finitely generated discrete subgroups it's possibly an earlier result. Another issue is that it's not a trivial fact that lattices are finitely generated (it's trivial for cocompact lattices), but finite generation of arbitrary lattices is the main issue. @YCor: I meant discrete subgroups. As for finite generation of lattices in $PSL(2,R)$, it was known much earlier than the general case, the earliest reference I know is in Siegel's 1945 Annals paper "Some remarks on discontinuous groups." He even gives a bound on the number of generators in terms of area. Siegel's argument is by looking closely at fundamental polygons, it is likely that his proof goes back to Poincare. But probably the Poincaré/Siegel argument yields the whole result (not only finite generation)? @YCor: After some group-theoretic work, yes, but Siegel says nothing of sorts. It's possible that he was unaware of the result. A remark on Step (1) in Moishe Kohan's proof. This problem (of finding a finite index, torsion-free subgroup of a lattice in $\mathrm{PSL}(2, \mathbb{R})$) was called "Fenchel's Conjecture". It was resolved by Ralph H. Fox. See his paper: On Fenchel's Conjecture about F-Groups and later work (for other proofs, and for corrections to earlier work). Curious: Apparently, Fox's proof was partially incorrect, see T.C.Chau, A note concerning Fox's paper on Fenchel's conjecture. Proc. Amer. Math. Soc. 88 (1983), no. 4, 584–586. He was seemingly unaware of Selberg's 1960 paper which solved the problem in much greater generality. (And Selberg was unaware of Fenchel's conjecture.)
2025-03-21T14:48:31.786677
2020-08-15T20:26:33
369262
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Is the $\sqrt{{\rm time}}$ spread of a stochastic process about the global minima the ubiquitous phenomenon? Given a function $f$ with a global minima at $x^*$, consider a stochastic process given as, $x_{t+1} = x_t - \nabla f(x_t) + \xi$ where $\xi$ is a random variable. Now we want to understand the occurrence of concentration of the following form ($\lambda >0$ is an arbitrary parameter and $x_0$ is an arbitrary choice of the starting point), $$P \left [ \Vert x_T - x^* \Vert > {\cal O} \left ( \sqrt{ \lambda + T + \Vert x_0 - x^* \Vert^2 } \right ) \right ] \leq {\cal O} (T \cdot e^{-\frac{\lambda^2}{T}}) $$ I know of explicit examples of such phenomenon i.e of the process after time $t$ being spread out about the global minima of $f$ upto a distance of $\sqrt{{\rm t}}$. Is this kind of spreading the generic behaviour or the exception? Are there large classes of $f$ known for which this $\sqrt{{\rm time}}$ spreading about the global minima is known to hold?
2025-03-21T14:48:31.786774
2020-08-15T20:26:43
369263
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "R. van Dobben de Bruyn", "abx", "https://mathoverflow.net/users/40297", "https://mathoverflow.net/users/82179" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632127", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369263" }
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Localization of Chow groups and flat base change For any flat morphism $f:X\rightarrow Y$, we have a flat pullback of Chow groups $$Ch^i(Y)\rightarrow Ch^i(X).$$ A particular example of this is of course an open immersion $U\rightarrow X$. In that case we in fact know that the induced morphism on Chow groups is surjective (and we also know the kernel). Hence I'm curious if this localization property is also true for finer topologies, i.e. how fine can we make our topology such that we still have surjectivity of restriction? I think this fails already for the Nisnevich topology. Consider a scheme $X$ and the Nisnevich open $$X\coprod X\rightarrow X$$ then we don't get a surjection $$Ch^i(X)\rightarrow Ch^i(X\coprod X)\cong Ch^i(X)\times Ch^i(X), \quad [D]\mapsto ([D],[D]).$$ The main examples I would like to be able to apply this would be for a flat base change, i.e. if $X$ is defined over some field $k$, and we compare the chow groups $$Ch^i(X) \rightarrow Ch^i(X\times_{\text{Spec} k} \text{Spec } l)$$ for some (possibly infinite) field extension $l/k$. Note that $X \amalg X$ is already a covering of $X$ in the (big) Zariski site. Somehow the objects that appear in the site and the families that form coverings are separate (unrelated) notions. This question is only about objects and does not see the Grothendieck (pre)topology. For $X$ a smooth projective curve of genus $\geq 1$ defined over $\mathbb{Q}$, $CH^1(X)=\operatorname{Pic}(X) $ is finitely generated, while $CH^1(X\otimes _{\mathbb{Q}}\mathbb{C})$ is an extension of $\mathbb{Z} $ by a complex torus.
2025-03-21T14:48:31.786905
2020-08-15T20:30:46
369265
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Chris Ramsey", "Robert Israel", "Yemon Choi", "adenali", "https://mathoverflow.net/users/13650", "https://mathoverflow.net/users/151417", "https://mathoverflow.net/users/763", "https://mathoverflow.net/users/76593" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632128", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369265" }
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Necessary (and sufficient) conditions for the following matrix product to be symmetric positive definite? Fix some $n\times n$ symmetric positive definite matrix $A$. Consider the following matrix product, $$B = AC$$ where $C$ is an arbitrary $n\times n$ matrix. Given $A$, I would like to know if there are known necessary and sufficient conditions on all square matrices $C$ such that the resulting matrix $B$ is also symmetric positive definite? I am more interested in knowing (if possible) necessary conditions. Edit: I am only concerned with real matrices. If $C$ is positive and commutes with $A$ then $AC$ will be positive. Could you clarify two things: (1) are you only working with real matrices? (2) when you say positive definite, do you mean strictly positive definite, or positive semi-definite? @ChrisRamsey thanks for your answer. Any idea if something similar could be a necessary condition? @YemonChoi I edited my question to clarify that 1) yes, I am working with real matrices. For 2) I mean strictly positive definite not positive semi-definite. The obvious necessary and sufficient condition on $C$ is that $C = A^{-1} B$ for some symmetric positive definite $B$. @RobertIsrael Thank you but that does not help in understanding the required relationship between all possible $C$ matrices and a fixed $A$ that result in a positive definite product. If $C$ is a positive definite real matrix that commutes with $A$ then $AC = C^{1/2}AC^{1/2}$ which is positive definite. So this is certainly a sufficient condition. However, it is far from necessary. Consider that $$ \left[\begin{matrix}2 & 1 \\ 1 & 2\end{matrix}\right]\left[\begin{matrix}2 & 0 \\ 1 & 4\end{matrix}\right] = \left[\begin{matrix}5 & 4 \\ 4 & 8\end{matrix}\right]. $$ I am not convinced there is going to be a nice condition that completely describes such $C$. One necessary condition is that $$ AC = (AC)^T = C^TA \ \ \ \ \textrm{or} \ \ \ ACA^{-1} = C^T $$ If in addition $C$ is symmetric then it commutes with $A$ and then $A^{1/2}CA^{1/2} = AC > 0$ which implies that $C$ is positive definite since $A^{-1}$ is positive as well. Hardly a complete answer, but that's all I have for now. Thank you for your answer. I guess it's not easy to say anything beyond this, but it's still helpful :)
2025-03-21T14:48:31.787068
2020-08-15T21:29:11
369270
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Jiří Rosický", "Mike Shulman", "Simone Virili", "Tim Campion", "fosco", "https://mathoverflow.net/users/152679", "https://mathoverflow.net/users/2362", "https://mathoverflow.net/users/24891", "https://mathoverflow.net/users/49", "https://mathoverflow.net/users/73388", "https://mathoverflow.net/users/7952", "varkor" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632129", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369270" }
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Interlocking (weak) factorization systems I'm interested in instances of the following data: $C$ is a (possibly higher) category; $(L,M)$ is a weak factorization system (wfs) on $C$; $(M,R)$ is a unique factorization system (fs) on $C$. Definition: I'll call such data $(L,M,R)$ interlocking factorization systems on $C$ because the right half of the wfs coincides with the left half of the fs. Examples: On the 1-category $Set$ we have the $(Mono,Epi)$ wfs and the $(Epi,Mono)$ fs. On the $\infty$-category $Spaces$ we have (as discussed here) the $((n+1)\text{-skel_r}, n\text{-conn})$ wfs and the $(n\text{-conn},n\text{-trunc})$ fs [1], for any $n \in \mathbb Z_{\geq -2}$. On the 1-category $Ch_{\geq 0}(R\text{-Mod})$, we have the $(\text{acyclic mono with projective cokernel}, \text{epi})$ wfs and the $(\text{epi}, \text{mono})$ fs, for any ring $R$. Question 1: What are some other examples of the above data of "interlocking factorization systems"? Observations: If $(L,M,R)$ forms interlocking factorization systems on $C$, then $(M,R)$ is a modality (i.e. a fs with basechange-stable left class). Moreover, $M$ is closed under co-transfinite-composition. In fact, we might think of $(L,M,R)$ as being fundamentally a fs $(M,R)$ where $M$ satsifies some further closure conditions like these (though it's not clear that there's actually an equivalent formulation along these lines). Dually, we might think of an interlocking factorization system $(L,M,R)$ on $C$ as being fundamentally a wfs $(L,M)$ where $M$ satisfies some further closure conditions (namely, cobase-change and colimits in the arrow category). If $C$ is locally presentable, then modulo checking that $M$ is accessibly embedded, this is actually an equivalent formulation. Question 2: Given a class of morphisms $M$ in a ($\infty$-)category $C$ (perhaps assumed to be locally presentable), does knowing that $M$ is the left half of a fs $(M,R)$ in any way "simplify" the task of checking whether $M$ is the right half of a wfs $(L,M)$? dually, does knowing that $M$ is the right half of a wfs $(L,M)$ imply anything interesting about whether $M$ is also the left half of a fs $(M,R)$? Question 3: Are interlocking factorization systems just a curiosity, or is there anything special you can do with them? For instance, do they lead to some kind of obstruction theory? [1] Here, $(n+1)\text{-skel_r}$ denotes the retracts of relative $(n+1)$-dimensional CW complexes $n\text{-conn}$ denotes the $n$-connected maps (= maps with $n$-connected fibers, off by 1 from the most classical convention) $n\text{-trunc}$ denotes the $n$-truncated maps (= maps with $n$-truncated fibers, which again may be off by 1 from your favorite convention) One example worth understanding could be the following: it is known that t-structures on the homotopy category of a given $(\infty,1)$-category are in bijection with a suitable class of factorization systems (called normal torsion theories). It could be interesting to study which $t$-structures correspond to "normal interlocking torsion theories"... e.g., are they the TTFs studied in the PhD thesis of Pedro Nicolas (https://arxiv.org/pdf/0801.0507.pdf) or do we need to impose some more conditions to get a bijection? This isn't quite the same phenomenon, but ternary factorisation systems share similar structure in being overlapping factorisation systems in some sense, though they're stronger than the notion you're interested in (at least when $R_1 = L_2$). Long ago I asked what's up with those FS such that $(E,M)$ is a OFS and $(M,E)$ a WFS; this doesn't seem to be understood. (I would call these particular interlocking FS Frobenius, for obvious reasons). Fiorenza, Marchetti and I studied the notion of "stable n-ary OFS" to describe Postnikov towers in https://arxiv.org/abs/1501.04658: not that this was really unknown. It just come out better when done $\infty$. Hopefully you can adapt that technology? @Fosco Thanks! I just came across this question of yours related to the first question above, as well as this one which yields, following Joyal's Catlab, a class of examples generalizing the first and third above. Namely, if $C$ is any variety of algebras, then $({}^\square Surj, Surj, Inj)$ is an interlocking factorization system on $C$. Awesome! I suppose another class of examples would be the following. Let $C^{op}$ be a Grothendieck topos or a Grothendieck abelian category. Then the triple $((Mono^\square)^{op},Mono^{op},Epi^{op})$ is an interlocking factorization system on $C$. Tim Campion. Your last observation can be extended. Dual of an interlocking factorization system is $(R,M,L)$ where $(R,M)$ is a factorization system and $(M,L)$ is a weak factorization system. I met this situation in my arXiv:1702.08684 paper. Examples are on Boolean algebras, Banach spaces, or commutative $C^\ast$-algebras. The last example yields an interlocking factorization system on compact Hausdorff spaces. Trivial observation about question 2: at least it's automatically closed under retracts... (-:O @MikeShulman: It contains the isomorphisms and is closed under composition, too :)
2025-03-21T14:48:31.787403
2020-08-15T21:52:49
369272
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "https://mathoverflow.net/users/54415", "tomasz" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632130", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369272" }
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S-unital compact rings are profinite It is well-known that compact Hausdorff topological unital rings are profinite. The proof generalises to (left or right) s-unital rings (i.e. rings such that for all $r\in R$ we have $r\in Rr$ or for all $r\in R$ we have $r\in rR$). Is there a reference for this more general fact? Is there a further generalisation (i.e. an interesting class of rings, containing s-unital rings, for which compact Hausdorff implies profinite)? (Note that this is not true for all rings, as given any compact Hausdorff abelian group $A$, we can endow $A$ with zero multiplication, making it a compact Hausdorff topological ring.) This is essentially answered in one of the answers to Is every compact topological ring a profinite ring?. If a compact ring $R$ either admits no element $r\neq 0$ with $rR=0$ or the left-right dual condition then it is profinite. This is the condition that the multiplication map induces and embedding of $R$ into the endomorphisms of the Pontryagin dual of its additive group which is what you use to prove total disconnectedness. See Thm 3 of On Compact Topologica Rings. by Hirotada Anzai https://projecteuclid.org/euclid.pja/1195573244 Thanks, this is just what I was looking for.
2025-03-21T14:48:31.787512
2020-08-15T22:04:54
369273
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Daniele Tampieri", "Gary Moon", "https://mathoverflow.net/users/113756", "https://mathoverflow.net/users/114164", "https://mathoverflow.net/users/17773", "kodlu" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632131", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369273" }
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Looking for an electronic copy of Lebeau's paper I would like to know if anyone has an electronic copy of the paper "Gilles Lebeau - Contrôle De L'Équation De Schrödinger"? This article appeared in Journal de Mathématiques Pures et Appliquées, Vol. 71, (1992), no. 3, 267–291. The website of the Bibliothèque nationale de France - Gallica has all volumes all volumes published by this journal from 1836 to 1945 and the website of the journal, the volumes published from 1997 to the present. So I couldn’t find that article on the internet. I've checked in my institutional library. Most likely, it seems those volumes of the journal were not digitized. Plus the physical library is off-limits due to COVID-19, but they didn't have paper copies in any case. I think this will be difficult unless someone who has physical copies of the journal sees this and contacts you. The years from 1946 to 1996 have not been digitized yet (or they have been, but are not yet available on the web): however, I think that you could find a library that have not been blocked by the COVID-19, and ask them for a digitized scam of the paper. Otherwise you could ask him directly a paper copy. This is an old post, so you may no longer be interested in the paper. However, if you are, I have a copy. I don't have a link or anything, but I'd be happy to email it to you.
2025-03-21T14:48:31.787640
2020-08-15T22:58:36
369276
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Robert Furber", "Tim Campion", "https://mathoverflow.net/users/2362", "https://mathoverflow.net/users/24078", "https://mathoverflow.net/users/61785", "truebaran" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632132", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369276" }
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Alfsen Shultz theorem-the space of states of $C^*$-algebra depends only on Jordan structure According to the article on nLab the Alfsen Shultz theorem states that the space of states of a given $C^*$-algebra depends on somehow weaker structure namely on the so called Jordan algebra structure. This article gives reference for this theorem: however I've checked the cited paper and I don;t see how the main theorem (namely theorem 9.5 in this paper) implies the above statement. For convienience I quote this theorem below: Let $A$ be a JB algebra (see the Introduction of the paper). Then there is unique Jordan ideal $J$ such that $A/J$ has faithful, isometric Jordan representation as an JC algebra and every factor representation of $A$ not anihilating $J$ is onto the exceptional algebra $M_3^8$ . The fact that you can get back the Jordan multiplication follows because you can recover the squaring map because the state space of a unital C$^*$-algebra is strongly spectral, a notion defined here, although perhaps the place to look is the two books by Alfsen and Shultz called State Spaces of Operator Algebras and Geometry of State Spaces of Operator Algebras. This would give the one direction: that the state space is enough to recover the Jordan product, right? How about the converse implication? Wait -- isn't the fact that the state space depends only on the Jordan structure pretty much trivial? A state is a positive linear functional, and "positive" and "linear" depend only on the Jordan structure. It's the other direction -- recovering the Jordan algebra from the state space -- which is hard. @truebaran An element $x$ of a Jordan algebra is positive iff it is a square, i.e. there is a $y$ such that $x = y \circ y$. (This agrees with the fact that positive elements of C$^*$-algebras are the squares of self-adjoint elements.) The positive cone and unit allow you to define when a linear functional is a state, and get the state space. Thank you, you are both right
2025-03-21T14:48:31.787807
2020-08-16T00:46:05
369281
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Gordon Royle", "LeechLattice", "Per Alexandersson", "Richard Stanley", "https://mathoverflow.net/users/1056", "https://mathoverflow.net/users/125498", "https://mathoverflow.net/users/1492", "https://mathoverflow.net/users/161863", "https://mathoverflow.net/users/2807", "mtsecco" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632133", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369281" }
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On a limit involving a transform of the chromatic polynomial I was playing around with the chromatic polynomial (denoted here by $\chi_G(x)$) and I have made the following conjecture. Let $(G_n)_{n \ge 1}$ be a sequence of graphs with $v(G_n) \to \infty$ ($v(G_n)$ denotes the number of vertices of $G_n$) and $e(G_n) \to \infty$ ($e(G_n)$ denotes the number of edges of $G_n$). For each $x \neq 0$, let us define the following transform of the chromatic polynomial of $G_n$ $$ \psi_{G_n}(x) = \frac{x^{v(G_n)}}{e(G_n)^{v(G_n)}} \chi_{G_n}\left( \frac{e(G_n)}{x} \right). $$ The conjecture is that for each fixed real number $x \neq 0$, we have $\psi_{G_n}(x) \to \exp(-x)$ as $n$ goes to infinity. I have checked the conjecture for a few sequences of graphs: for example, $G_n$ being the complete graph $K_n$, for $G_n$ being a tree on $n$ vertices and for $G_n$ being a collection of $n$ independent edges (a matching on $2n$ vertices). Does anyone know if this is well-known? PS: I am not sure if the conditions on $v(G_n)$ and $e(G_n)$ are the right one. Any comments on this are welcome as well. There doesn’t seem to anything in this problem that says “chromatic polynomial” as opposed to much a broader class of polynomials. Actually there's a heuristic justifying the equation when $k=e(G)/x$ is a natural number: Color the graph randomly with $k$ colors. If $k$ is large enough, the probability of the coloring being proper is close to $((k-1)/k)^{e(G)}$, so $\psi(x)$ is close to $((k-1)/k)^{kx}$. There are two limitations of the heuristic: the first is that one cannot modify $k$ without dealing with $G_n$ (this can be overcome by effective probability bounds), and the second is that $e(G)/x$ may not be a natural number, so the counting interpretation does not work. I would like to know counting-like interpretations of $χ$ at non-natural values. @LeechLattice The only non-natural for which a counting interpretation of $\chi$ is known seems to be $-1$, but any interpretation for a non-integer would be a major breakthrough. @GordonRoyle Though irrelevant to the problem, a counting interpretation of $\chi$ is known for every integer. Yikes, my two comments both rapidly shown to be utterly wrong... oh well, there’s always other questions. @GordonRoyle well, take any derivative, evaluated at any negative integer, and there is an interpretation, see https://arxiv.org/abs/1904.01262 Here is a heuristic argument which perhaps someone can make rigorous. I write $v_n=v(G_n)$ and $e_n=e(G_n)$. Let $$ \chi_{G_n}(x) = x^{v_n}-c_{n,v_n-1} x^{v_n-1}+c_{n,v_n-2}x^{v_n-2}-\cdots. $$ I claim that for fixed $k\geq 0$, $$ \lim_{n\to\infty} \frac{c_{n,v_n-k}}{e_n^k} = \frac{1}{k!}. $$ One can prove this by noting that (by the Broken Circuit Theorem, for instance, which shows that $c_{n,v_n-k}$ increases as we add more edges to $G_n$) $c_{n,v_n-k}$ is bounded below by its value when $G_n$ is a tree, and is bounded above by its value when $G_n$ is a complete graph. The claimed result is easily verified for trees and complete graphs (in the latter case, using known asymptotics for the Stirling numbers of the first kind). Perhaps there is a more direct proof, but in any case, if we don't worry about justifying interchanging limits and sums, we get $$ \lim_{n\to\infty} \frac{x^{v_n}}{e_n^{v_n}}\chi_{G_n}\left( \frac{e_n}{x}\right) = \sum_{k\geq 0} \lim_{n\to\infty} \frac{(-1)^k c_{n,v_n-k}x^k}{e_n^k} $$ $$ \qquad = \sum_{k\geq 0} \frac{(-1)^k x^k}{k!} = \exp(-x). $$ That's exactly what I was thinking with a friend earlier today. Thank you!
2025-03-21T14:48:31.788171
2020-08-16T01:47:04
369284
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632134", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369284" }
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When is it possible to use the Parseval-Plancherel identity to solve an integral? The integral is of the form $\int_{-\infty}^\infty \sigma(x)\mu(x)\,\mathrm{d}x$. Where the Fourier transform of the $\sigma$ function is $\tilde \sigma(p)= e^{-iap}\frac{1}{1+e^{-c|p|}}$ and the function $\mu(x)$ is given by $\mu(x)=-2 \tan ^{-1}\left(\frac{2 x-2}{c}\right)$. The Fourier transform of $\mu(x)$ can be found quite easily $\tilde \mu(p)=\frac{e^{-i p} \left(2 i \pi e^{-\frac{c | p| }{2}}\right)}{p}$. The question is: Is it possible to use the the Parseval-Plancherel identity and write the above integral as $\frac{1}{2 \pi}\int_{-\infty}^\infty \tilde\sigma(p)\tilde \mu(p)\,\mathrm{d}p$? If so, the above integral becomes $\frac{i}{2}\int_{-\infty}^\infty dp \frac{ e^{-i (a+1) p} \text{sech}\left(\frac{c p}{2}\right)}{p}$ Which looks like a Fourier Transform of $\frac{sech(\frac{cp}{2})}{p}$ function. How is this Fourier transform computed? Recall the identity that Fourier transform of $K(x)=\text{sech}(x)$ is $\tilde K(p)=\pi \text{sech}\left(\frac{\pi p}{2}\right)$. Using this identity the Fourier transform of $\frac{\text{sech} {x}}{x}$ can be easily computed \begin{equation} \int_{-\infty}^{-\infty} e^{-i x p} \frac{\text{sech}{x}}{x} \, \mathrm{d}x= -i \int \pi \text{sech}\left(\frac{\pi p}{2}\right) \mathrm{d}p= -2 i \tan ^{-1}\left(\sinh \left(\frac{\pi p}{2}\right)\right) \label{ident} \end{equation} Using equation this relation, the given integral can be easily integrated \begin{equation} \frac{i}{2}\int_{-\infty}^\infty dp \frac{ e^{-i (a+1) p} \text{sech}\left(\frac{c p}{2}\right)}{p}= \tan ^{-1}\left(\sinh \left(\frac{\pi (\Lambda_h+1)}{|c|}\right)\right) \label{rest} \end{equation} Checking the answer numerically. Plot: Constant a Plot Constant c
2025-03-21T14:48:31.788542
2020-08-16T04:07:37
369287
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Fedor Petrov", "Paul B. Slater", "https://mathoverflow.net/users/4312", "https://mathoverflow.net/users/47134" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632135", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369287" }
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What are the John ellipsoids for a pair of (9- and 15-dimensional) convex sets of $4 \times 4$ positive-definite matrices? What are the John ellipsoids (JohnEllipsoid) for the 9- and 15-dimensional convex sets ($A,B$) of $4 \times 4$ positive-definite, trace-1 symmetric (Hermitian) matrices (in quantum-information parlance, the sets of “two-rebit” and “two-qubit” “density matrices” [DensityMatrices], respectively)? (Are these bodies "centrally-symmetric", in the sense of one aspect of the underlying theorem JohnTheorem?) Further, what is the relation (intersections, …) of these ellipsoids to the important convex subsets of $A$ and $B$ composed of those matrices that remain positive-definite under the (not completely positive) operation of partial transposition—by which the four $2 \times 2$ blocks of the $4 \times 4$ matrices are transposed in place? (It has been established [MasterLovasAndai] that the fractions of Euclidean volume occupied by these "PPT" [positive-partial-transpose/separable/nonentangled] convex subsets are $\frac{29}{64}$ for $A$ and $\frac{8}{33}$ for $B$.) Also, what is the further relation of these ellipsoids to the "inspheres" (the maximal balls inscribed in $A$ and $B$ [SBZ])? The inspheres also lie within the PPT sets. Might the John ellipsoids and inspheres simply coincide? Additionally, what might be the John ellipsoids themselves for these PPT sets? There is an interesting concept of a "steering ellipsoid", referred to in the following quotation p. 28 [SteeringEllipsoid]: For two-qubit states, the normalized conditional states Alice can steer Bob’s system to form an ellipsoid inside Bob’s Bloch sphere, referred to as the steering ellipsoid (Verstraete, 2002; Shi et al., 2011, 2012; Jevtic et al., 2014). However, the "Bloch sphere" is 3-dimensional, so the steering ellipsoid of a two-qubit state can not be the (15-dimensional) John ellipsoid requested above. Of course, the question what are the John ellipsoids can be asked for the convex sets of $m \times m$ symmetric and $n \times n$ Hermitian (positive-definite, trace 1) density matrices ($m,n \geq 2$). For $m,n=2$, the answers appear to be trivial, namely the convex sets themselves. For $m,n =3$, it seems possibly nontrivial. Only, however, for composite values of $m,n$, do we have subsidiary questions regarding the convex subsets of PPT-states. The Wikipedia article given by the first hyperlink above describes the "maximum volume inscribed ellipsoid as the inner Löwner–John ellipsoid". [DensityMatrices]: Slater - A concise formula for generalized two-qubit Hilbert–Schmidt separability probabilities [JohnTheorem]: Howard - The John ellipsoid theorem [MasterLovasAndai]: Slater - Master Lovas–Andai and equivalent formulas verifying the $\frac8{33}$ two-qubit Hilbert–Schmidt separability probability and companion rational-valued conjectures [SBZ]: Szarek, Bengtsson, and Życzkowski - On the structure of the body of states with positive partial transpose [SteeringEllipsoid]: Uola, Costa, Nguyen, and Gühne - Quantum steering Thanks--LSpice! They are not centrally symmetric. Let us begin with two apparently relevant formulas. The first is for the volume of a $k$-dimensional ellipsoid [Thm. 2.1, EllipsoidVolume], \begin{equation} vol_k=\frac{2 \pi ^{k/2} \prod _{i=1}^k a_i}{k \Gamma \left(\frac{k}{2}\right)}, \end{equation} where the $a_i$’s are the lengths of the semi-axes. The other is for the volume of the set of $m \times m$ symmetric, positive-definite matrices of trace 1 [(7.7), RebitVolume]. \begin{equation} Vol_m=\frac{2^{\frac{1}{4} (m-1) m+m} \sqrt{m} \pi ^{\frac{1}{4} (m-1) m-\frac{1}{2}} \Gamma \left(\frac{m+1}{2}\right) \prod _{l=1}^m \Gamma \left(\frac{l}{2}+1\right)}{m! \Gamma \left(\frac{1}{2} m (m+1)\right)}. \end{equation} For the (“two-rebit”) case $m=4$ ($k=9$) of immediate interest, the formula yields \begin{equation} \frac{\pi ^4}{60480} \approx 0.0016106. \end{equation} So, the question of particular interest to us is what proportion of this volume is occupied by the inner Lowner-John ellipsoid for the convex set of the indicated 9-dimensional set of $4 \times 4$ (density) matrices. Further, what is its magnitude in comparison with $\frac{29}{64}$, the fraction established by Lovas and Andai for the separability—equivalently, PPT—probability of the two-rebit states? Also, in comparison with the volume of the insphere (for which we have no immediate present computation). So, to approach these questions, we generated pairs of randomly-generated “two-rebit density matrices” (sec, 4, RandomDensityMatrices), using Ginibre-ensemble methods. Then, we took the absolute values of their differences and divided by 2. Nine independent entries (three diagonal ones, and the six upper off-diagonal ones) of the resultant matrix, were taken as the semi-axes. At this point in time, we have generated close to sixteen million such pairs. The pair of $4 \times 4$ density matrices for which we have found the associated maximum ellipsoid volume, $6.98613 \cdot 10^{-8}$ (only 0.0000432642 of $\frac{\pi ^4}{60480} \approx 0.0016106$), so far are \begin{equation} \left( \begin{array}{cccc} 0.424772 & -0.147161 & -0.3345 & -0.177458 \\ -0.147161 & 0.164668 & 0.146384 & 0.0925659 \\ -0.3345 & 0.146384 & 0.29387 & 0.157489 \\ -0.177458 & 0.0925659 & 0.157489 & 0.11669 \\ \end{array} \right) \end{equation} and \begin{equation} \left( \begin{array}{cccc} 0.135144 & 0.189631 & -0.03164 & 0.145386 \\ 0.189631 & 0.449171 & -0.180868 & 0.347037 \\ -0.03164 & -0.180868 & 0.126351 & -0.128246 \\ 0.145386 & 0.347037 & -0.128246 & 0.289334 \\ \end{array} \right). \end{equation} One-half of the absolute differences for these two matrices of the leading three diagonal entries and the upper six off-diagonal entries are used as the nine semi-axes in the first formula given above. Let us also point out that there is an alternative—but equivalent up to certain normalization factors—approach to computing the volumes of $m \times m$ density matrices (AndaiVolume). Andai, however, restricted attention to the $2 \times 2$ Hermitian case, and did not give an explicit alternative to the volume formula of Zyczkowski and Sommers presented above--so, at this point in time, we are not sure of what form it would take.
2025-03-21T14:48:31.788961
2020-08-16T04:18:52
369288
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "AWO", "Luc Guyot", "hans", "https://mathoverflow.net/users/138357", "https://mathoverflow.net/users/163822", "https://mathoverflow.net/users/84349" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632136", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369288" }
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Ideal norm in orders Let $\overline{T}$ be a Dedekind ring such that $\overline{T}/\overline{I}$ is finite for every nonzero ideal $\overline{I}$ of $\overline{T}$. Let $T$ be a subring of $\overline{T}$ with the same total ring of fractions (i.e. an order). Let $I$ be an ideal of $T$ and let $\overline{I} = I\overline{T}$. The norm $N_T(I)$ of $I$ is defined to be the cardinality of $T/I$. Question: Is there a formula relating $N_T(I)$ and $N_{\overline{T}}(\overline{I})$? For example, it seems plausible that the discrepancy is measured by some "tor" group. Remarks: If $I$ is projective then $N_T(I)$ and $N_{\overline{T}}(\overline{I})$ are equal. Localization reduces the problem to the case when $T$ is local (and $\overline{T}$ is semi-local), and both $I$ and the conductor of $T$ are proper ideals. (Thanks to Luc Guyot) If $T$ is a Bass ring ($\leftrightarrow$ every intermediate ring $T \subset R \subset \overline{T}$ is Gorenstein $\leftrightarrow$ every ideal is generated by two elements), and $T = \{a \in \overline{T} : a I \subset I \}$, then by [2, Proposition 5.8] $I$ is projective. It follows that $N_T(I)$ and $N_{\overline{T}}(\overline{I})$ are equal (by the first remark). (generalization of third remark) if $T$ is a Gorenstein integral domain and $T = \{a \in \overline{T} : a I \subset I \}$, then $I$ is projective. This follows from combining Theorem 6.2(4) with Proposition 7.2 of [1]. It follows that $N_T(I)$ and $N_{\overline{T}}(\overline{I})$ are equal (by the first remark). [1] H. Bass, "On the ubiquity of Gorenstein rings", 1963. [2] L. Levy and R. Wiegand, "Dedekind-like behavior of rings with 2-generated ideals", 1985. You might get something by looking at minimal ring extensions. There are some restrictions one what can happen on the "residue side" see Section 5.1 of https://arxiv.org/pdf/1909.10860.pdf and the references there. Regarding your first remark, can you prove it without further assumptions? It is clear in the number field setting and works as well in the more general context of Del Corso and Dvornicich paper that you cite below. But is it the same context as yours?. I'll begin with a general remark which will be illustrated by a computation in an arbitrary order of quadratic number field. If $\overline{I}$ contracts to $I$, i.e., if $\overline{I} \cap R = I$, then the inclusion $R \rightarrow \overline{R}$ induces an injective $R$-module homomorphism $R/I \rightarrow \overline{R}/\overline{I}$. As a result, $N_R(I)$ divides $N_{\overline{R}}(\overline{I})$ and in particular we have $N_R(I) \le N_{\overline{R}}(\overline{I})$. If for instance $I$ is a prime ideal, then $N_R(I)$ divides $N_{\overline{R}}(\overline{I})$. The underlying question that I fail to answer is: Question. Is it always true that $N_R(I)$ divides $N_{\overline{R}}(\overline{I})$, or at least that $N_R(I) \le N_{\overline{R}}(\overline{I})$? Edit. The OP answer contains a proof that $N_R(I) \le N_{\overline{R}}(\overline{I})$ holds true for every non-zero ideal of $R$. I will not address the above question. Instead, I'll introduce a condition on $R$ under which $N_R(I)$ divides $N_{\overline{R}}(\overline{I})$ for every non-zero ideal $I$ of $R$. Proposition. If a non-zero ideal $I$ of $R$ is projective over its ring of multipliers $\varrho(I) \Doteq \{ r \in \overline{R} \, \vert \, rI \subseteq I\}$, then we have $$ N_{\overline{R}}(\overline{I}) = N_R(I) \vert \varrho(I)/R \vert. $$ Side note. that $\varrho(I) = \{ r \in K \, \vert \, rI \subseteq I\}$ where $K$ denotes the field of fractions of $R$, since $R$ is Noetherian. Lemma 1 (OP's Claim). If $I$ is an invertible ideal of $R$ then $N_{\overline{R}}(\overline{I}) = N_R(I)$. Proof. First, prove the statement for a non-zero principal ideal $I$. Then decompose the $R$-module of finite length $\overline{R}/\overline{I}$ as a direct sum of its localizations with respect to the maximal ideals of $R$ [4, Theorem 2.13]. Do the same for $R/I$ and compare the cardinalities of the summands. Proof of the Proposition. By Lemma 1, we have $N_{\overline{R}}(\overline{I}) = N_{\varrho(I)}(I)$. Hence $N_{\overline{R}}(\overline{I}) = [\varrho(I) : R][R: I] = \vert \varrho(I)/R\vert N_R(I)$. Note that if $R$ is an order whose ideals are two-generated (e.g., an order in a quadratic field or an order whose discriminant is fourth-power free [2, Theorem 3.6]), then every non-zero ideal of $R$ satisfies the hypothesis of the above proposition, see e.g., [1], [2] and Theorem 4.1, Corollaries 4.3 and 4.4 of Keith Conrad's notes. The OP discusses similar results in his remarks and his answer. Let $m$ be a square-free rational integer. We set $K \Doteq \mathbb{Q}(\sqrt{m})$ and denote by $\mathcal{O}(K)$ the ring of integers of the quadratic field $K$. Loose Claim. Given an order $R$ of $K$ and an ideal $I \subseteq R$, we shall compute $N_{\mathcal{O}(K)}(I\mathcal{O}(K))$ as a function of $N_R(I)$ and of a binary quadratic form associated to $I$. To do so, we introduce some notation and definitions. Setting $$\omega = \left\{ \begin{array}{cc} \sqrt{m} & \text{ if } m \not\equiv 1 \mod 4, \\ \frac{1 + \sqrt{m}}{2} & \text{ if } m \equiv 1 \mod 4, \\ \end{array}\right. $$ we have $$\mathcal{O}(K) = \mathbb{Z} + \mathbb{Z} \omega$$ and any order of $K$ is of the form $\mathcal{O}_f(K) \Doteq \mathbb{Z} + \mathbb{Z} f \omega$ for some rational integer $f > 0$ [2, Lemma 6.1]. Moreover, the inclusion $\mathcal{O}_f(K) \subseteq \mathcal{O}_{f'}(K)$ holds true if and only if $f'$ divides $f$. If $I$ is an ideal of $\mathcal{O}_f(K)$, then its ring of multipliers $\varrho(I) \Doteq \{ r \in \mathcal{O}(K) \, \vert \, rI \subseteq I\}$ is the smallest order $\mathcal{O}$ of $K$ such that $I$ is projective, equivalently invertible, as an ideal of $\mathcal{O}$ [2, Proposition 5.8]. Let us fix $f > 0$ and set $$R \Doteq \mathcal{O}_f(K), \quad \overline{R} \Doteq \mathcal{O}(K).$$ An ideal $I$ of $R$ is said to be primitive if it cannot be written as $I = eJ$ some rational integer $e$ and some ideal $J$ of $R$. The main tool is the Standard Basis Lemma [5, Lemma 6.2 and its proof]. Lemma 2. Let $I$ be a non-zero ideal of $R$. Then there exist rational integers $a, e > 0$ and $d \ge 0$ such that $-a/2 \le d < a/2$, $e$ divides both $a$ and $d$ and we have $$ I = \mathbb{Z} a + \mathbb{Z}(d + e f \omega). $$ The integers $a, d$ and $e$ are uniquely determined by $I$. We have $\mathbb{Z}a = I \cap \mathbb{Z}$ and the integer $ae$ is equal to the norm $N_R(I) = \vert R /I \vert$ of $I$. The ideal $I$ is primitive if and only if $e = 1$. Note that, since $\mathbb{Z}a = I \cap \mathbb{Z}$, the rational integer $a$ divides $N_{K/\mathbb{Q}}(d + e f \omega)$. We call the generating pairs $(a, d + ef \omega)$ the standard basis of $I$. Let us associate to $I$ the binary quadratic form $q_I$ defined by $$q_I(x, y) = \frac{N_{K/\mathbb{Q}}(xa + y(d + ef\omega))}{N_R(I)}.$$ Then we have $$eq_I(x, y) = ax^2 + bxy + cy^2$$ with $$b = Tr_{K/\mathbb{Q}}(d + ef \omega) \text { and } c = \frac{N_{K/\mathbb{Q}}(d + ef \omega)}{a}.$$ We define the content $c(q_I)$ of $q_I$ as the greatest common divisor of its coefficients, that is $$c(q_I) \Doteq \frac{\gcd(a, b, c)}{e}.$$ Remark. We have $c(q_I) = \frac{\gcd(a, d, ef)}{e} = \frac{f}{f'} = \vert \varrho(I) / R \vert$ where $f'$ is the divisor of $f$ such that $\varrho(I) = \mathcal{O}_{f'}$. Claim. Let $I$ be a non-zero ideal of $R$. Then we have $$N_{\overline{R}}(\overline{I}) = N_R(I) \vert \varrho(I)/R \vert \text{ with } \vert \varrho(I)/R \vert = c(q_I).$$ Proof. Since $N_R(xI) = N_R(Rx) N_R(I)$ and $N_R(Rx) = N_{\overline{R}}(\overline{R}x) = \vert N_{K/\mathbb{Q}}(x) \vert$ for every $x \in R \setminus \{0\}$, we can assume, without loss of generality, that $I$ is primitive, i.e., $e = 1$. It follows immediately from the definitions that $$\overline{I} = \overline{R} I = \mathbb{Z}a + \mathbb{Z}a \omega + \mathbb{Z}(d + f \omega) + \mathbb{Z}v$$ where $$v = \left\{ \begin{array}{cc} f \omega^2 + d \omega & \text{ if } m \not\equiv 1 \mod 4, \\ f \frac{m - 1}{4} + (d + f) \omega & \text{ if } m \equiv 1 \mod 4. \\ \end{array}\right.$$ Now it suffices to compute the Smith Normal Form $\begin{pmatrix} d_1 & 0 \\ 0 & d_2 \\ 0 & 0 \\ 0 & 0 \end{pmatrix}$ of the matrix $A \Doteq \begin{pmatrix} a & 0 \\ 0 & a \\ d & f \\ v_1 & v_2 \end{pmatrix}$ where $(v_1, v_2)$ is the matrix of $v$ with respect to the $\mathbb{Z}$-basis $(1, \omega)$ of $\overline{R}$. The coefficient $d_1$ is the greatest common divisor of the coefficients of $A$ and is easily seen to be $\gcd(a, d, f) = \gcd(a, b, c)$. The coefficient $d_2$ is the greatest common divisor of the $2 \times 2$ minors of $A$ divided by $d_1$ and is easily seen to be $\frac{a \gcd(c(q_I), q_I(0, 1))}{d_1} = \frac{a c(q_I)}{d_1}$. Thus $N_{\overline{R}}(\overline{I}) = d_1 d_2$ has the desired form. [1] J. Sally and W. Vasconcelos, "Stable rings", 1974. [2] C. Greither, "On the two generator problem for the ideals of one-dimensional ring", 1982. [3] L. Levy and R. Wiegand, "Dedekind-like behavior of rings with $2$-generated ideals", 1985. [4] D. Eisenbud, "Commutative algreba with a view toward algebraic geometry", 1995. [5] T. Ibukiyama and M. Kaneko, "Quadratic Forms and Ideal Theory of Quadratic Fields", 2014. Thank you for a very nice and helpful answer. For the time being I will leave the question open in the hopes that someone may attempt to say something about the general case. I have used your reference to [1] to add a useful remark to the original question. For the application I have in mind, one may assume that $T$ is the full ring of multipliers of the ideal $I$, in which case the problem is even solved by [1, Proposition 5.8]. Do you know whether Proposition 5.8 might extend in some fashion to more general classes of orders? Thanks for the edit, I appreciate! And thanks for the feedback. Regarding [1, Proposition 5.8], I don't know of any generalisation at the moment ("Stable rings" by Sally and Vasconcelos, 1973, might be worthwhile reading), but I'll keep you posted if anything dawns on me (I like the question). I also hope that someone will come with something more general (bookmark on; I feel that my computation fails to capture some significant and general features of the problem at hand). It looks like $e$ divides $a$, $b$, and $c$, and $q_I(0,1) = c/e$, so it looks like we have $\gcd(c(q_I), q_I(0,1)) = c(q_I)$? It also looks like $(a,b,c) \mapsto (ea,eb,ec)$ under $I \mapsto eI$, so it seems like the formula simplifies to $N_{\overline T}(\overline I) = N_T(I) \gcd(a,b,c)$ even if $I$ is not primitive, or even $N_{\overline T}(\overline I) = N_T(I) f/f'$ by your remark. Thanks, indeed, it can be simplified. I was actually editing my answer while you where writing this comment. It seems that you made also some progress, very interesting. I am recording for the benefit of others what is to my knowledge the full extent of what is known about the general problem. Luc Guyot has provided a nice and explicit answer for the case of quadratic orders. I do not mark this post as "the answer" as the original question has not yet been answered. Let the discrepancy of a $T$-ideal $I$ be defined as $ds(I) = N_{\overline{T}}(\overline{I})/N_T(I)$ (non-standard definition). When does $ds(I) = 1$? The following theorem is the main tool of the paper [1]. The statement uses the module index notation of [2]. Theorem [1; Theorem 1]: $[\overline{T}:\overline{I}] \subset [T:I]$. $[\overline{T}:\overline{I^{-1}}] \subset [I:T]$. $[{T}:{I^{-1}}] \subset [\overline{I}:\overline{T}]$. Moreover, the following are equivalent: Any subset relation among (1), (2), (3) is an equality. All subset relations among (1), (2), (3) is an equality. $I$ is invertible. This theorem has the following corollaries for the "discrepancy". Recall that the different of $T$ is defined to be $\mathfrak D_{T} = (T^\vee)^{-1}$ where $T^\vee$ is the dual of $T$ for the trace form. Corollary: $ds(I) \geq 1$ with equality if and only if $I$ is invertible. Corollary: The following are equivalent: The discrepancy of $\mathfrak D_{T}$ is $1$. For every ideal $I$ of $T$, $ds(I) = 1$ if and only if $T = (I:I)$. $T$ is Gorenstein. Everything in these corollaries follows immediately from the theorem except the second point of the second corollary which follows from the well-known equivalence $T=(I:I) \iff I \text{ invertible}$ when $T$ is Gorenstein (cf. e.g. [3; Proposition 5.8] or [4; Proposition 2.7]). Quadratic case [Following the notation in Luc Guyot's answer] Using the above corollaries we revisit the quadratic case. The discrepancy is invariant under homotheties and so we may assume the ideal $I$ is primitive ($e = 1$). By [5; Lemma 6.5], the ideal $I$ satisfies $R = (I:I)$ if and only if $\gcd(a,b,c) = 1$. Indeed, the formula for the discrepancy in Luc Guyot's answer is precisely $\gcd(a,b,c)$. (By the remark in Luc Guyot's answer, we even have $ds(I) = f/f'$ where $f$ is the conductor of $T$ and $f'$ is the conductor of $(I:I)$.) Thus the formula $ds(I) = c(q_I)$ is consistent with the second corollary. Upper bound We will derive an upper bound for $ds(I)$ which is independent of $I$. I assume that $T$ is a domain for simplicity. We may suppose that $T \neq \overline{T}$ and set $S = \overline{T}$. Let $\mathfrak f$ denote the conductor of $T$. Upper bound: For any T-fractional ideal $I$, $ds(I) \leq |S/T||S/\mathfrak f|.$ Two $T$-fractional ideals are in the same genus if they are locally isomorphic; equivalently, there exists an invertible T-ideal which multiplies one ideal into the other. Claim: Any $T$-fractional ideal $I$ is in the same genus as a $T$-fractional ideal $J$ such that $\mathfrak f \subset J \subset S.$ Proof: Let $P$ be a prime ideal of $T$ and let $S_P$ denote the integral closure of $T$ (integral closure commutes with localization). It suffices to construct a $T_P$-fractional ideal which is isomorphic to $I_P$ such that $\mathfrak f_P \subset J_P \subset T_P$ where subscript denotes tensoring with $T_P$. $S_P$ is a finite product of local Dedekind rings so it is a PID. Hence $I_PS_P = \alpha S_P$ for some $\alpha$ in $Quot(T)$. Let $J_P = \alpha^{-1}I_P$. Then $J_P \subset S_P$, but also $$J_P \supset J_P \mathfrak f_P = J_P S_P \mathfrak f_P = \mathfrak f_P.$$ Claim: The discrepancy $ds(I)$ is constant on genera. Proof: This is proven by localizing and using that an invertible ideal of $T$ is locally principal (this latter fact follows from [5; Proposition 2.3]). Putting these claims together, we have that for $I$ any $T$-fractional ideal, $ds(I) = ds(J)$ for some $T$-fractional ideal $J$ such that $\mathfrak f \subset J \subset S$. From [1; Theorem 1], $|T/J| \leq |S/SJ|$. We also have $S\mathfrak f = \mathfrak f \subset SJ \subset S$, and so $|S/SJ| \leq |S/\mathfrak f|$. Write $M' = M/\mathfrak f$ for any module containing $\mathfrak f$. Putting the inequalities together we have $$ds(I) = |S/SJ|/|T/J| \leq |S/\mathfrak f|/ |T/J| = |S'|/(|T'|/|J'|) = |S/T| |J/\mathfrak f| .$$ The last term is bounded from above by $|S/T| |S/\mathfrak f|$. Conclusion The discrepancy function satisfies the inequality, $1 \leq ds(I) \leq |\overline{T}/T||\overline{T}/\mathfrak f|$, for any $T$-fractional ideal $I$, and admits an explicit and natural formula in terms of conductors in the quadratic case. However it appears to be unknown whether the discrepancy function can be given a "closed form" in general (e.g., an expression in terms of the conductor of $T$, the differents or discriminants of $T$ and $\overline{T}$, Ext or Tor groups over $T$ or $\overline{T}$). References: [1] I. Del Corso, R. Dvornicich, Relations among Discriminant, Different, and Conductor of an Order, 2000. [2] A. Fröhlich, Local fields, from J. W. S. Cassels and A. Fröhlich, Algebraic number theory, 1967. [3] L. Levy and R. Wiegand, Dedekind-like behavior of rings with 2-generated ideals, 1985. [4] J. Buchmann and H. W. Lenstra, Jr., Approximating rings of integers in number fields, 1994. [5] V. M. Galkin, $\zeta$-functions of some one-dimensional rings, 1973. In "Upper Bound" and "Conclusion", what are the assumptions made on $T$? Is $T$ any domain whose proper quotients are finite? Throughout the post, $T$ is assumed to be an order of a Dedekind domain $\overline T$ whose proper quotients are finite. In "Upper Bound" and "Conclusion", it is further supposed that $T$ is not integrally closed. Thanks for making this clear. I was confused by the sentence "I assume that $T$ is a domain for simplicity." since $T$ is a domain throughout. In the original post I intended for a "Dedekind ring" to mean a finite product of Dedekind domains. Otherwise I would have been bothered that the problem as stated did not strictly make sense in the local setting since the localization of an order may have zero divisors. Is the context of "Relations among Discriminant, Different, and Conductor of an Order" the same as your question, i.e., is it possible to identify R, K, F, S and A of the cited paper in your setting? Neither setting generalizes the other entirely. In my original question I stated it for "Dedekind rings", whereas their paper considers the case of a Dedekind domain. I also made the additional assumption that the nontrivial quotients of the ring were finite, whereas it seems "Relations among..." does not make this restriction. In considering the overlap of the settings, one should at least assume $\overline{T}$ is a Dedekind domain with finite residue fields, in which case one should take $F=\mathrm{Frac},\overline{T}$, $S=\overline{T}$, and $A=T$. [cont'd] [cont'd] I don't see how to identify $R$ and $K$ just in terms of the data of $(\overline{T},T)$, but in the situations I can imagine it is easy enough to identify what one should take. For instance, if $\overline{T}$ is the ring of integers of a number field then one would identify $R=\mathbb Z$, $K=\mathbb Q$. In the function-field setting, one would use $R = \mathbb F_q[t]$, $K = \mathbb F_q(t)$. If $\overline{T}$ is the ring of integers in a finite extension $F$ of $K=\mathbb Q_p$, one would use $R=\mathbb Z_p$.
2025-03-21T14:48:31.790049
2020-08-16T06:58:42
369291
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Fedor Petrov", "H A Helfgott", "HenrikRüping", "Mikael de la Salle", "Will Sawin", "https://mathoverflow.net/users/10265", "https://mathoverflow.net/users/18060", "https://mathoverflow.net/users/3969", "https://mathoverflow.net/users/398", "https://mathoverflow.net/users/4312" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632137", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369291" }
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Eigenvalue bounds and triple (and quadruple, etc.) products Very basic and somewhat open-ended question: Let $A$ be a symmetric operator on functions $f:X\to \mathbb{R}$, where $X$ is a finite set. Assume that the $L^2\to L^2$ norm of $A$ is $\leq \epsilon$, i.e., every eigenvalue $\lambda$ of $A$ has $|\lambda|\leq \epsilon$, or, what is the same, $|\langle f,Af\rangle|\leq \epsilon |f|_2^2$ for all $f:X\to \mathbb{R}$, where $\langle f,Af\rangle:= \sum_{x\in X} f(x) Af(x)$. Can we conclude anything about sums of the type $$\sum_{x\in X} f(x) Af(x) A^2f(x),\;\;\;\;\; \sum_{x\in X} f(x) Af(x) A^2f(x) A^3f(x),\;\;\;\;\text{etc.?}$$ Assume $|f|_2=1$ and $|f|_\infty$ bounded, if needed. (I'm interested both in counterexamples that show that one cannot conclude anything and in cases where one can say something, possibly given some auxiliary information.) You look for bounds which do not depend on $|X|$, yes? Well, probably, but I'm looking for anything, really. What sort of bound would depend on $|X|$? WEll maybe a bit too simple, but we have for pointwise multiplication of functions that $ |f\cdot g|_2 \le |f|_2\cdot |g|_2$, and thus we ould estimate the L_2-Norm of $f \cdot Af\cdot A^2f$ by $\varepsilon^3 |f|_2^3$. And similarly the L_2-norm of $f\cdot Af\cdot A^2 f\cdot A^3f$ by $\varepsilon^6 |f|_2^4$ etc. Then $|f|_1$ and $|f|_2$ differ at most by a factor of $\sqrt{|X|}$. Ah, well, yes, I was thinking that one might be able to afford factors of $\log X$ or $\log \log X$. A factor of $\sqrt{X}$ is too large. I would start with writing $A=\sum \lambda_i P_i$ for projectors $P_i$ and changing the order of summation. We get $\sum_{i,j} \lambda_i \lambda_j^2 \sum_x f(x) f_i(x)f_j(x)$ where $f_i(x)=(P_i f)(x)$. Do you see anything more clever than bounding this as $\varepsilon^3 \sum_{i,j}|\sum_x f(x)f_i(x)f_j(x)|$? @HenrikRüping Why use the $L^2$ - $L^1$ norm estimate and not just multiply in two batches and use Cauchy-Schwarz? If $f_1,f_2, f_3$ have $L_2$ norms at most $a_1,a_2,a_3$ then $\left| \sum_{x \in X} f_1(x)f_2(x) f_3(x) \right| \leq a_1 a_2 a_3$. (But maybe the $L^2$ norm is supposed to be normalized so $X$ has total mass one and not point mass one?) You can do a little better using the boundedness of $f_1$ to bound some higher norm. The worst case for these simple estimates is when the functions $f_1,f_2,f_3$ are both highly concentrated and highly correlated. I don't see any obstruction to this happening except the boundedness of $f_1$, which is mild. Take $A$ to be an $n\times n$ symmetric matrix where the first row and column are all $\epsilon/\sqrt{|X|}$ except for the first entry which is $\epsilon$, and all other entries are $0$. Take $f_1$ to be the all $1/\sqrt{X}$ vector (or all $1$s depending on normalization). In this setting $f_2,f_3,\dots$ all attain the maximal $L^2$ norm, or close to it, and a positive proportion of their $L^2$ mass is concentrated at the point $0$, so these types of bounds will basically be sharp. Just to be sure, when you write "$|f|\infty$ is bounded", do you mean that $\sup{x \in X} |f(x)| \leq C/\sqrt{|X|}$? Well, I meant to define the inner product with a factor of $1/X$, and then bounded just means bounded. So in this normalization the inequality is $||fg||2 \leq \sqrt{X} ||f||2 ||g||2$ and we can get a bound of $|\sum{x \in X} \prod{i=0}^{n-1} A^i f(x) | \leq X ||\prod{i=0}^{n-2} A^i f ||_2 ||A^{n-1} f ||2 \ll X ||\prod{i=1}^{n-2} A^i f ||_2 ||A^{n-1} f ||2 \leq X^{1 + \frac{n-3}{2}} \prod{i=1}^{n-1} || A^i f||_2 \leq X^{ \frac{n-1}{2}} \epsilon^{ n (n-1)/2}$ which, if I calculated correctly, my counterexample shows is sharp to within a constant factor. ... on the other hand, if you make the additional assumption that the $L^\infty \to L^\infty$ norm of $A$ is $\leq 1$, by interpolation and Hoelder you get the bound $|\frac{1}{X} \sum_x \prod_{i=0}^{n-1} A^i f(x) | \leq \epsilon^{2n-3}$. Same bound for $n=3$, but better for $n>3$ in the regime $\log X>>-\log \epsilon$.
2025-03-21T14:48:31.790455
2020-08-16T08:12:15
369295
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Donu Arapura", "IMeasy", "Jason Starr", "https://mathoverflow.net/users/13265", "https://mathoverflow.net/users/4096", "https://mathoverflow.net/users/4144" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632138", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369295" }
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Decomposition of a morphism with positive dimensional fibers It is well known that any birational morphism between projective varieties is a sequence of blow ups. Suppose now that I have a morphism $f:X \to Y$ with positive dimensional fibers, that is a projective bundle over an open subset of $Y$. We can even assume $Y$ smooth, even if I don't think it is necessary. Is it still true that $X$ is the blow-up of a projective bundle on $Y$? No that is not true. Consider the projective Abelian cone associated to a reflexive sheaf on $Y$ that is not locally free. For an explicit example, consider a general hypersurface of bidegree $(1,1)$ in $\mathbb{P}^2\times \mathbb{P}^3$ with its projection to the second factor. @IMeasy I'm guessing "projective bundle" = "projective space bundle". @Donu : yes, of course that’s what I meant I am posting my comment as an answer. This already fails for relative dimension $1$ when the base scheme has dimension $n$ at least $3$. Let $k$ be a field. Let $n\geq 3$ be an integer. Denote $\text{Proj}\ k[x_0,x_1,x_2, \dots,x_n]$ by $\mathbb{P}^n_k$. Denote $\text{Proj}\ k[y_0,y_1,y_2]$ by $\mathbb{P}^2_k$. Denote by $X$ the hypersurface in $\mathbb{P}^2_k\times_{\text{Spec}\ k}\mathbb{P}^n_k$ with bihomogeneous defining equation, $$f=x_0y_0 + x_1y_1 + x_2y_2.$$ The projection from $X$ to $\mathbb{P}^2_k$ is a Zariski-locally-trivial projective space bundle of relative dimension $n-1$. In particular, $X$ is a smooth $k$-scheme. By the Grothendieck-Lefschetz Theorem on Picard groups from SGA 2, the restriction homomorphism of Picard groups is an isomorphism, $$\text{res}:\text{Pic}(\mathbb{P}^2_k\times_{\text{Spec}\ k}\mathbb{P}^n_k) \xrightarrow{\cong} \text{Pic}(X).$$ Of course the first Picard group is $\mathbb{Z}\times \mathbb{Z}$. Moreover the nef cone in the first Picard group is $\mathbb{Z}_{\geq 0}\times \mathbb{Z}_{\geq 0}$. One way to see this is to consider restriction of ample invertible sheaves to linear rational curves ("lines") in fibers of each projection. Since the closed subscheme $X$ contains such lines as well, it follows that the restriction isomorphism also induces an isomorphism of nef cones. In particular, the ample cone of $X$ equals $\mathbb{Z}_{>0}\times \mathbb{Z}_{>0}$, so that the non-ample nef divisors are those in the "boundary" of the nef cone, i.e., $$\{(0,0)\}\sqcup \left(\mathbb{Z}_{>0}\times \{0\}\right) \sqcup\left( \{0\}\times \mathbb{Z}_{>0}\right). $$ The invertible sheaf in the first component of this partition of the boundary is just the structure sheaf, and the associated contraction of $X$ is the constant $k$-morphism to $\text{Spec}\ k$. The second component gives the projection to $\mathbb{P}^2_k$, and the third component gives the projection to $\mathbb{P}^n_k$. Since none of these contractions is birational, it follows that $X$ is not a blowing up of some projective scheme, except as a "blowing up" that is an isomorphism. Thus, the projection morphism from $X$ to $\mathbb{P}^n$ does not factor through a nontrivial blowing up. The restriction of this projection is flat over the closed subscheme $\text{Zero}(x_0,x_1,x_2)$, and the restriction is flat over the open complement of this closed subscheme. However, the fiber dimension over the closed subscheme is $2$, whereas the fiber dimension over the open subscheme is $1$. Thus, this projection is not a projective space bundle, although it is a projective space bundle of relative dimension $2$, resp. of relative dimension $1$, when restricted over the closed subscheme, resp. over the open subscheme.
2025-03-21T14:48:31.791146
2020-08-16T12:00:35
369301
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Iosif Pinelis", "dohmatob", "https://mathoverflow.net/users/36721", "https://mathoverflow.net/users/78539" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632139", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369301" }
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How tight is the bound $P(\|X\|^2 \ge t |\langle a,X\rangle|) \ge 1 - t\sqrt{\frac{2}{m-1}}$, where $X \sim N(0, I_m)$ and $\|a\| = 1$? Let $X$ be a random vector in $\mathbb R^m$ with iid $N(0,1)$ coordinates and let $a$ be a fixed unit vector in $\mathbb R^m$. In another post (SE link here https://math.stackexchange.com/a/3792730/168758), I'm able to show using anti-concentration results for quadratic forms in gaussians, that $$ P(\|X\|^2 \ge t |\langle a,X\rangle|) \ge 1 - t\sqrt{\frac{2}{m-1}}. $$ Question. Can the above bound be improved ? $\newcommand\Ga\Gamma$ $\newcommand{\R}{\mathbb R}$ This bound can be greatly improved, even using rather rough estimates. Indeed, in view of the spherical symmetry of the distribution of $X$, without loss of generality $a$ is the first vector of the standard basis of $\R^m$. So, your inequality can be rewritten as \begin{equation*} p_n(t):=P(W<t|Z_1|)\le t\sqrt{\frac2n}=:r_n(t), \label{1}\tag{1} \end{equation*} where $t>0$, $W:=Z_1^2+\dots+Z_{n+1}^2$, the $Z_i$'s are iid standard normal, and \begin{equation*} n:=m-1\ge1. \end{equation*} Note that for $t\ge\sqrt{n/2}$ inequality \eqref{1} is trivial. So, without loss of generality \begin{equation*} 0<t<\sqrt{n/2}. \label{2}\tag{2} \end{equation*} Let us give an upper bound on $p_n(t)$ which is much better than $r_n(t)$. Letting $Y:=Z_2^2+\dots+Z_{n+1}^2$, we have \begin{equation*} p_n(t)=P(Y<t|Z_1|-Z_1^2)=\int_0^\infty P(Y<y)P(t|Z_1|-Z_1^2\in dy). \end{equation*} Since $Y$ has the gamma distribution with parameters $n/2$ and $2$, for $y>0$ we have \begin{equation*} P(Y<y)=c_n\int_0^y u^{n/2-1}e^{-u}\,du\le c_n y^{n/2}, \end{equation*} where \begin{equation*} c_n:=\frac1{\Ga(n/2)2^{n/2}}. \end{equation*} Thus, \begin{equation*} p_n(t)\le c_ne_n(t), \end{equation*} where \begin{align*} e_n(t)&:=\int_0^\infty y^{n/2}P(t|Z_1|-Z_1^2\in dy) \\ &=E\max(0,t|Z_1|-Z_1^2)^{n/2} \\ &=\frac1{\sqrt{2\pi}}\int_{-t}^t (t|z|-z^2)^{n/2}e^{-z^2/2}\,dz \\ &=\frac2{\sqrt{2\pi}}\int_0^t (tz-z^2)^{n/2}e^{-z^2/2}\,dz \\ &\le\frac2{\sqrt{2\pi}}\int_0^t (tz-z^2)^{n/2}\,dz \\ &=\frac{2t^{n+1}}{\sqrt{2\pi}}\int_0^1 (u-u^2)^{n/2}\,du \\ &=\frac{2t^{n+1}}{\sqrt{2\pi}}\frac{\Ga(n/2+1)^2}{\Ga(n+2)}=:f_n(t). \end{align*} We conclude that \begin{equation*} p_n(t)\le q_n(t):=c_n f_n(t). \end{equation*} The ratio of the new bound, $q_n(t)$, to the old bound, $r_n(t)$, is \begin{equation*} R_n(t):=\frac{q_n(t)}{r_n(t)}=\frac{t^n\sqrt n}{\sqrt{\pi}}\frac{\Ga(n/2+1)^2}{\Ga(n+2)}\frac1{\Ga(n/2)2^{n/2}} =\frac{t^n n^{3/2}}{\sqrt{\pi}}\frac{\Ga(n/2+1)}{\Ga(n+2)}\frac1{2^{n/2+1}} \le R_n(\sqrt{n/2}), \end{equation*} in view of \eqref{2}. Also, by Stirling's formula, \begin{equation*} R_n(t)\le R_n(\sqrt{n/2})\sim\frac{\sqrt{2n/\pi}}4\,\Big(\frac e8\Big)^{n/2} \end{equation*} as $n\to\infty$. So, the new bound is exponentially smaller than the old bound uniformly over $t$ as in \eqref{2}, and the improvement is only much greater for $t=o(\sqrt n)$. Here are the graphs $\{(t,R_n(t))\colon0<t<\sqrt{n/2}\}$ of the ratios of the new upper bound, $q_n(t)$, to the old bound, $r_n(t)$, for $n=10$ (left) and $n=20$ (right). This is indeed an exponential improvement! Thanks. Question: Is it possible to get a bound which is valid for $t \gg \sqrt{n}$ (e.g $t=n$) ? Or even for $t=n/\sqrt{\log n}$. Should be possible to improve your bounds by using Jensen's inequality on the concave function $u \mapsto (u - u^2)^{m/2}$ and the distribution with density $\propto e^{-t^2u^2/2}du$ on $[0, 1]$. Will write it down asap. Here there are two players, $n$ and $t$, and good bounds on (or asymptotics) will depend on an (unknown to me) relation between $n$ and $t$, with possibly several distinct cases. The case considered here, of $t=O(\sqrt n)$, seems comparatively simple, and even here only rough estimates were used, as I said, to avoid an answer to long by MO customs. However, your use of Jensen's inequality will not work, since $(u-u^2)^{n/2}$ is not concave in $u\in(0,1)$ for $n>2$. OK, makes sense. Thanks (Indeed, my use of Jensen's inequality won't work since the function is only concave on an interval of length $\mathcal O(1/\sqrt{n})$ around $u=1/2$. I got misled by looking at a plot of the situation and overestimating the size of this interval.)
2025-03-21T14:48:31.791420
2020-08-16T12:48:45
369304
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Grisha Taroyan", "Steve Costenoble", "Tyler Lawson", "https://mathoverflow.net/users/143549", "https://mathoverflow.net/users/360", "https://mathoverflow.net/users/58888" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632140", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369304" }
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Bredon cohomology of a permutation action on $S^3$ I've seen a couple of similar questions asking to verify computations of Bredon cohomology here and here, so I will ask one such question myself. Let $\mathbb{Z}/2$ act on $S^3\subset \mathbb{C}^2$ by restriction of a permutation action on $\mathbb{C}^2.$ I wanted to compute Bredon cohomology $\mathcal{H}^*_{\mathbb{Z}/2}(S^3;\underline{\mathbb{Z}}).$ I have a cell decomposition based on a decomposition of complex $1$-dimensional disk into $3$ cells: $\mathbb{D}=D\sqcup T\sqcup *.$ Here $T\sqcup *=S^1=\partial \mathbb{D}$ and $D$ is the interior of $\mathbb{D}.$ Then we have a decomposition of $S^3=\mathbb{D}\times S^1 \cup S^1\times \mathbb{D}$ into cells compatible with the $\mathbb{Z}/2$ action. The fixed point set of an action is a circle given by $\{z_1=z_2\}\cap S^3\subset \mathbb{C}^2.$ Since the orbit category of $\mathbb{Z}/2$ consists of $*$ and $\mathbb{Z}/2$ there are the following equivariant chains: \begin{array}{|c|c|c|c|} \hline \operatorname{dim} &*& \mathbb{Z}/2 & \operatorname{cells corresponding to} \underline{C}_n(S^3)(\mathbb{Z}/2)\\ \hline 0 & \mathbb{Z} & \mathbb{Z} & * \times *\\ 1 & 0 & \mathbb{Z}\oplus\mathbb{Z},\quad \begin{pmatrix} 1 \\ 0\end{pmatrix}\xrightarrow{\overline{1}} \begin{pmatrix} 0 \\ 1\end{pmatrix} & T\times *,*\times T\\ 2 & 0 & \mathbb{Z}\oplus \mathbb{Z} \oplus \mathbb{Z},\quad \begin{pmatrix} 1 \\ 0\\0\end{pmatrix}\xrightarrow{\overline{1}}\begin{pmatrix} 0 \\ 1\\0\end{pmatrix};\;\begin{pmatrix} 0 \\ 0\\1\end{pmatrix}\xrightarrow{\overline{1}}\begin{pmatrix} 0 \\ 0\\-1\end{pmatrix} & D\times *, *\times D, T\times T\\ 3 & 0 & \mathbb{Z}\oplus \mathbb{Z},\quad \begin{pmatrix} 1 \\ 0\end{pmatrix}\xrightarrow{\overline{1}} \begin{pmatrix} 0 \\ 1\end{pmatrix} & D\times T, T\times D\\ \hline \end{array} So it seems that the cochains valued in $\underline{\mathbb{Z}}$ are: \begin{array}{|c|c|} \hline \operatorname{dim} & \\ \hline 0 & \mathbb{Z}\\ 1 & \mathbb{Z}\\ 2 & \mathbb{Z}\\ 3 & \mathbb{Z}\\ \hline \end{array} Since $(T\times T)^*=0$ in cochains, we have $\mathcal{H}^3_{\mathbb{Z}/2}(S^3;\underline{\mathbb{Z}})=\mathbb{Z}.$ Differential $d_1$ is an isomorphism since $\partial(D\times *)=T\times *.$ It seems that $\mathcal{H}^*_{\mathbb{Z}/2}(S^3;\underline{\mathbb{Z}})=H^*(S^3;\mathbb{Z}).$ It is a bit odd to me that the quotient is a homological sphere. Sure, the group $\mathcal{H}^3_{\mathbb{Z}/2}(S^3;\underline{\mathbb{Z}})=\mathbb{Z}$ since orientation is preserved, but maybe I've missed some $2$-torsion in lower degrees? I think you're correct. $\underline{\Bbb Z}$ coefficients have the property that they compute the integral cohomology of the orbit space $S^3 / (\Bbb Z/2)$, and by a slightly different cell decomposition I think this space is homotopy equivalent to $S^3$. Your final answer is correct, but the cell structure you're using isn't a $G$-CW structure: $T\times T$ can't be used as a cell in this way. I would approach it like this: The action of $G = {\mathbb Z}/2$ on $\mathbb{C}\times\mathbb{C}$ can be written as the representation $\mathbb{C}\oplus\mathbb{C}^\sigma$, where $G$ acts trivially on $\mathbb{C}$ and by negation on $\mathbb{C}^\sigma$. The sphere $S(\mathbb{C}\oplus\mathbb{C}^\sigma)$ is also the one-point compactification $S^{1+2\lambda}$, where $\lambda$ denotes the real line with $G$ acting by negation. This has a $G$-CW structure with one $G$-fixed 0-cell, one $G$-fixed 1-cell, one $G$-free 2-cell, and one $G$-free 3-cell, so that the skeleta are $*$, $S^1$, $S^{1+\lambda}$, and $S^{1+2\lambda}$. From here you can work out that the $\underline{\mathbb{Z}}$-cochain complex is $$ \mathbb{Z} \xrightarrow{0} \mathbb{Z} \xrightarrow{1} \mathbb{Z} \xrightarrow{0} \mathbb{Z}. $$ A way to check that the answer is correct is to write $$ H_G^n(S^{1+2\lambda}) \cong \tilde H_G^n(S^0) \oplus \tilde H_G^n(S^{1+2\lambda}) \cong \tilde H_G^n(S^0)\oplus \tilde H_G^{n-1-2\lambda}(S^0) $$ and then use the known calculation of the $RO(G)$-graded cohomology of a point (originally due to Stong (unpublished), since published in various places). It is not a G-CW structure because fixed points are not a G-CW subspace? All the cells in a $G$-CW complex have to have the form $G/H\times D^n$ where $G$ acts trivially on $D^n$. $T\times T$ doesn't have that form, it looks like the disc of a nontrivial representation of $G$. Is there a way to work with "cells" with nontrivial action of the stabilizer? I'm now aware that any cell with nontrivial action can be subdivided into "good" G-cells, but is there a way to avoid this step? Not easily, at least not if you want to calculate the integer-graded part of the cohomology. Any filtration gives rise to a spectral sequence that could, theoretically, be used for computation, but that's not going to be straightforward. There is a notion of $G$-CW($V$) complexes, using cells of the form $G/H\times D(V)$, but that calculates the cohomology in grading $V$. I think Stefan Waner first noticed this, Gaunce Lewis published an exposition, and then Stefan and I published a book using a very generalized version. One other situation where you can use arbitrary cells: At least for $G = \mathbb{Z}/p$, if you have a space built out of cells of the form $G/H\times D(V)$ where $V$ is even-dimensional, but can vary, then, with some additional assumptions, you can conclude that the $RO(G)$-graded cohomology is a free module over the cohomology of a point, though the generators may not be where you expect them to be. I think this was first proved by Ferland and Lewis. Thank you for great reference! I had another question about possibly non-trivial representation on the cells. If I restrict to the case of the coefficient system $\underline{\mathbb{Z}}$ I'm just computing the cohomology of the quotient space. Assume that cellular structure is preserved in a sense that the image of a cell under a group action is once again a cell. Then the spectral sequence of the filtration on the quotient space collapses at $E_2$ for dimensional reasons, just like in the case of cellular cochains. Am I getting it wrong? I'm not sure I understand exactly what you're suggesting, but watch out for the fact that the quotient of $(D(V),S(V))$ by $G$ may have cohomology in several degrees, both even and odd, so I don't see that the spectral sequence necessarily collapses.
2025-03-21T14:48:31.791802
2020-08-16T13:52:44
369307
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Bad English", "Paul Siegel", "geometricK", "https://mathoverflow.net/users/4362", "https://mathoverflow.net/users/78729", "https://mathoverflow.net/users/8906" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632141", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369307" }
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Does the Gysin map in $K$-theory respect bordism? Let $X_1$ and $X_2$ be two closed spin$^c$ manifolds that are bordant via a spin$^c$ manifold-with-boundary $W$. Let $Z$ be a closed spin$^c$ manifold with $\dim Z=\dim X_1$ mod $2$. Let $$f_1:X_1\to Z,\qquad f_2:X_2\to Z,\qquad F:W\to Z$$ be smooth maps such that $F|_{X_1}=f_1$ and $F|_{X_2}=f_2$. We can associate to $f_1$ and $f_2$ two wrong-way (or Gysin) maps in $K$-theory: $$f_{1!}:K^0(X_1)\to K^0(Z),$$ $$f_{2!}:K^0(X_2)\to K^0(Z).$$ Let $E_1\to X_1$ and $E_2\to X_2$ be two $\mathbb{C}$-vector bundles such that there exists a vector bundle $\Omega\to W$ satisfying $\Omega|_{X_1}\cong E_1$ and $\Omega|_{X_2}\cong E_2$. Let $[E_i]\in K^0(X_i)$ denote the $K$-theory classes defined by $E_i$. Question: Is it true that $f_{1!}[E_1]=f_{2!}[E_2]\in K^0(Z)$? Added after: I would be most interested in an approach not directly using Poincare duality for K-theory/K-homology. I know how to do this in K-homology, which gives me confidence that it also works for K-theory, but I haven't worked it out. In K-homology you can attach cylindrical ends $X_1 \times [0, \infty)$ and $X_2 \times [0, \infty)$ to $W$ and argue that the Gysin maps factor through the Mayer-Vietoris boundary maps on the new object $W'$. You then lift the K-homology class on W to a class in a certain K-homology-like group for W'; this requires care since W' is non-compact. (One way to do it involves C*-algebras and coarse geometry.) Functoriality of Mayer-Vietoris completes the proof. @PaulSiegel Could you elaborate how the Gysin maps factor through the Mayer-Vietoris boundary maps on $W'$ and why this implies the result in $K$-homology? I would like to adapt your argument to $K$-theory. Let $N^n=\partial M^{n+1}$, $E\in K^\bullet(M)$ and $f:M\to X$ Choose a smooth embedding $i:X\to \mathbb{R}^N,N>>1$, denote by $\chi$ the normal bundle of $X$ and by $\mu$ the normal bundle of $M$ after suitable small deformation of $i\circ f$. Let $\nu=\mu|_N$ and $\eta$ be the normal bundle of $N\subset M$ (which is trivial and one-dimensional) By considering tubular neighborhoods we get the natural map: $t:Th_\chi X\to Th_{\nu+\eta}N$, where $Th$ denotes a Thom space. After applying the Thom isomorphism $th$ on $K^\bullet$ we obtain the definition of a Gysin map (going in "right-way" on a $Th$'s). So for $f_!(E|_N)=0$ it's sufficient to prove that $t^* th_{\nu+\eta}(E|_N)=0$ Actually $t^*$ is passing through a connecting homomorphism. Namely, there is a commutative diagram: $\begin{matrix} Th_{\chi}X&\to& Th_{\mu}M/Th_\nu N&\\ \downarrow{t}&\swarrow{\sigma}&\downarrow{\Sigma}&\\ Th_{\nu+\eta}N&\xrightarrow{\sim}& \Sigma Th_{\nu}N&\\ \end{matrix}$ The top arrow comes from the tubular neighborhoods. The horizontal isomorphism comes from triviality of $\eta$, while suspension $\Sigma$ from Puppe cofiber sequence: $Th_\nu N\to Th_\mu M\to Th_\mu M/Th_\nu N\xrightarrow{\Sigma} \Sigma Th_{\nu}N$ The map $\sigma$ explains commutativity and is coming from: $Th_\mu M/Th_\nu N\sim Th_\mu M/Th_\mu (N\times [0,\varepsilon))\to$ $Th_\mu (N\times(-\varepsilon,\varepsilon))/Th_\mu (N\times [0,\varepsilon))\to Th_{\nu+\eta}N$ where $N\times [0,\varepsilon)\subset M$ is a collar of $N$. Finally, $\Sigma^*$ is the connecting homorphism and it follows that $\Sigma^* th_{\nu}(F)=0$ for all $F\in Im( K^\bullet(M)\to K^\bullet(N))$, so $t^* th_{\nu+\eta}(E|_N)=0$ I have a couple of questions. How is the map $t$ defined? Also, it seems to me that $\nu\oplus\eta$ is just the normal bundle of $N$ in $\mathbb{R}^N$ - is this correct? @geometricK, correct. Let $D$ and $S$ denote a small disk and sphere bundles respectively. I assume that $D_{\nu+\eta}N\subset D_\chi X$. This inclusion automatically gives a "wrong-way" map $t:D_\chi/S_\chi\to D_{\nu+\eta}/S_{\nu+\eta}$ The horizontal map is obtained in the same way Am I correct in understanding that your argument in fact shows a stronger result, namely that the Gysin map $i_!$ for the inclusion $i:\partial M\hookrightarrow M$ is always $0$? (I assume $i_!$ is the $K$-map induced by the composition $Th_{\mu}M\to Th_{\mu}M/Th_{\nu}N\xrightarrow{\Sigma}\Sigma Th_{\nu}N\sim Th_{\nu+\eta}N)$. Then the fact that $f_!$ factors through $i_!$ gives the result I was after originally. @geometricK, of course we can forget about $X$ and prove that $i_!(E|N)$ is zero. Always in a sense that we have to know that $\Sigma^(E_N)=0$, which is coming from $K^\bullet(M/N\xrightarrow{\Sigma} \Sigma N\to \Sigma M)$, then the same holds after applying $th$-isomorphism, so $\Sigma^ th\nu (E|N)=0$. I don't know how to prove that the two long sequences naturally isomorphic at $K^\bullet(M/N)$ and $K^\bullet(Th\mu(M)/Th_\mu(N))$, but for required vanishing it's not essential Thanks. There should be a relative version of the Thom isomorphism, which I'm trying to locate and which should give the isomorphism $K^(M/N)\to K^(Th_{\mu}M/Th_{\nu}N)$. As you say this isn't strictly necessary but would still be nice to have. The answer is yes, using general properties of orientations and fundamental classes. Let $X_1$ and $X_2$ be $n$--dimensional. Then $f_{!i}$ is the composite $$K^0(X_i) \xrightarrow[\sim]{\cap [X_i]} K_n(X_i) \xrightarrow{f_{i*}} K_n(Z) \xleftarrow[\sim]{\cap [Z]} K^0(Z).$$ Meanwhile Poincare duality for $W$ has the form $K^0(W) \xrightarrow{\cap [W]} K_{n+1}(W, X_1 \coprod X_2)$, and $d([W]) = [X_1]-[X_2]$. Thus $ d(\Omega \cap [W]) = (E_1 \cap [X_1], -E_2 \cap [X_2])$, and so $$ (f_{1*})(E_1 \cap [X_1]) - (f_{2*}(E_2 \cap [X_2]) = F_* i_* (d(\Omega \cap [W])) = 0,$$ since the composite $$K_{n+1}(W,X_1\coprod X_2) \xrightarrow{d} K_n(X_1 \coprod X_2) \xrightarrow{i_*} K_n(W)$$ is zero. Although I completely agree with your answer, I was secretly hoping for an approach without Poincare duality, as (although I didn't mention this) Poincare duality may not be available to me in the setting I'm actually working in.
2025-03-21T14:48:31.792278
2020-08-16T13:52:47
369308
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "LSpice", "Xiaosong Peng", "https://mathoverflow.net/users/131781", "https://mathoverflow.net/users/157954", "https://mathoverflow.net/users/2383", "https://mathoverflow.net/users/83554", "lefuneste", "user131781" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632142", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369308" }
Stack Exchange
Bi-annihilator of a subspace of the dual of an infinite-dimensional vector space Let $V$ be an infinite-dimensional vector space and $V^*$ its dual. For a linear subspace $W\subset V$ define $W^ \circ\subset V^*$ as the subspace of linear forms on $V$ vanishing on $W$. Dually, for $\Gamma\subset V^*$ define $\Gamma^\diamond \subset V$ as the set of vectors $v\in V$ such that $\gamma(v)=0$ for all linear forms $\gamma\in \Gamma$. It is slightly surprising but not too difficult to show that that we have for all subspaces $W\subset V$ the equality $(W^\circ) ^\diamond=W$. But is it true that for all $\Gamma\subset V^*$ we have $(\Gamma^\diamond)^\circ=\Gamma$ ? And is there a reference (article, book, lecture notes,...) where this problem is mentioned? A maybe-more accessible question that keeps it all on the dual side: since $((\Gamma^\diamond)^\circ)^\diamond$ equals $\Gamma^\diamond$, we can ask: if $\Gamma$ is contained in $\Lambda$ and $\Gamma^\diamond$ equals $\Lambda^\diamond$, then under what circumstances does $\Gamma$ equal $\Lambda$? In the infinite dimensional case, you can’t get by without topology, in this case the weak topologies $\sigma(V,V^\ast)$ and $\sigma(V^\ast,V)$. It is well known that the bipolars of subspaces of either of your spaces are precisely their closures for the corresponding topologies. So if they are not closed, then your claims will fail. @user131781 Of course we can get by without topology: already two answers show this, just 17 minutes after the question was posted! I think @user131781's point is less that the problem can't be stated, or even answered, without topology, and more that it is probably most useful to think of the problem in topological terms even if one is not required to do so (for example, this reveals 'heuristically' that the answer must be no, even if one is left to, as you and I did, construct a specific non-closed subspace). For example, your surprising true fact is the statement that all subspace are closed in the weak topology (with respect to the full algebraic dual). Dear LSpice, thank you for your explanations. I know nothing about topological vector spaces but I appreciate your point that if I did I would have immediately realized that the question is quite easy, whereas in reality I spent much time coming up with a solution. Maybe you can see Proposition 1.3.5 of the book "Hopf algebras" by David E. Radford No, $(\Gamma^\diamond)^\circ$ need not always equal $\Gamma$. Let $\mathcal B$ be a basis for $V$, and let $\Gamma$ be the span of the 'dual' set $\{e_b \mathrel: b \in \mathcal B\}$, so $e_b(c)$ is the Iverson bracket $[b = c]$ for all $b, c \in \mathcal B$. Then $\Gamma^\diamond$ is $0$, so $(\Gamma^\diamond)^\circ$ is all of $V^*$; but $\Gamma$ itself does not contain, for example, the element $\sum_{b \in \mathcal B} e_b$ of $V^*$. Your symbol $\sum_{b \in \mathcal B} e_b$ does not make sense because you cannot sum infinitely many non-zero vectors in a vector space. Sure it does: $\bigl(\sum_{b \in \mathcal B} e_b\bigr)(v) = \sum_{b \in \mathcal B} e_b(v)$ is a finitely supported sum for each $v \in V$. I think that, as @user131781 points out, this is a convergent sum in a weak topology; but anyway it can be defined without direct reference to topology. OK, I agree, what you write makes sense. However I don't know about weak topology and I thank you for your definition which doesn't make reference to topology. The topology @user131781 mentions is the weak* topology, weakest making all eval. fncls. continuous. Since $\lim_{F \subseteq \mathcal B} \bigl(\sum_{b \in F} e_b\bigr)(v)$ equals $\bigl(\sum_{b \in \mathcal B} e_b\bigr)(v)$ for all $v \in V$, this topology must declare that $\lim_{F \subseteq \mathcal B} \sum_{b \in F} e_b$ equals $\sum_{b \in \mathcal B} e_b$ (the limit taken over the finite subsets $F$ of $\mathcal B$—the so called "unordered sum"). Thanks again for your clear explanations, LSpice. I have upvoted your answer and your comments. The equality is false in general. Here is a counterexample: fix a basis $v_i, i\in I$ of $V$ and consider the set of coordinate linear forms $v^*_i, i\in I$. These forms are linearly independant but never form a basis since $V$ is infinite-dimensional. So complete these forms to a basis $(v^*_j), j\in J$ with $J\setminus I\neq\emptyset$. Choose $l\in J\setminus I$ and put $J'=J\setminus \{l\}$ If you define $\Gamma \subset V^*$ as the vector space generated by the $v_j^*, j\in J'$, then $\Gamma^\diamond =0$ (since already the subspace of $V^*$ generated by the $v_i^*, i\in I$ kill all vectors in $V$) so that $\Gamma\subsetneq (\Gamma^\diamond)^\circ=\{0\}^\circ=V^*$ yielding the required counterexample. This seems to be the same as my answer. Yes, it is the same idea, but I posted my answer without seeing yours. The technical details are different and I find your formulation indeed simpler. Also I wrote a comment to your post which you should take into account.
2025-03-21T14:48:31.792588
2020-08-16T15:36:12
369317
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Alexandre Eremenko", "Ariyan Javanpeykar", "Ben C", "Felipe Voloch", "François Brunault", "https://mathoverflow.net/users/154157", "https://mathoverflow.net/users/2290", "https://mathoverflow.net/users/25510", "https://mathoverflow.net/users/4333", "https://mathoverflow.net/users/6506", "https://mathoverflow.net/users/8726", "inkspot" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632143", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369317" }
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Is there an algorithm to determine if there exists a dominant map between two curves? Suppose I am given two smooth projective curves $C_1$ and $C_2$ over a field $k$ I want to know if there is an algorithm to decide whether there exists a nonconstant (and thus dominant) map $f : C_1 \to C_2$. To make this question more precise, suppose I am given an affine plane models $f_1(x, y) = 0$ and $f_2(x, y) = 0$ for some polynomials $f_1, f_2 \in k[x,y]$ birationally equivalent to the curves $C_1$ and $C_2$. Then is there an algorithm to determine when there exists a nonconstant map $f : C_1 \to C_2$ defined over $k$? Of course, it suffices to find a nonconstant rational map between these affine models so we may reduce the problem to asking if there exist two nonconstant polynomials $a(x,y), b(x,y) \in k[x,y]$ such that $f_2(a(x,y), b(x,y)) \in f_1(x,y) \cdot k[x,y]$. From Riemann-Hurwitz we can easily bound the degree of a possible morphism $f : C_1 \to C_2$ so it suffices to find an algorithm to compute if there exist morphisms $f : C_1 \to C_2$ of a fixed degree $n$. I am particularly interested in the case for morphisms of degree $2$ and $3$ and $k = \overline{\mathbb{F}_p}$ for my own applications. However, I am interested also in the general case out of curiosity if this problem is decidable. Thank you in advance for any suggestions. If there is an (efficient) algorithm known I would be very appreciative if somebody could point me to it in the literature or to an implementation if one exists. (1): You mean Hurwitz, not Hurewicz. They are different people. If you believe that this distinction is unimportant then ask yourself this: how soon do you want to be confused with another person? (2) The Tate conjecture (a theorem for several decades) says that two elliptic curves over a finite field are isogenous if their zeta functions are equal. However, this has, as far as I know, not been made effective. Thank you. Edited with apologies to Poland. Since you have the degree bound, this is a question about solvability of a polynomial system of equations. This general problem is decidable (in principle), and the algorithm is called "elimination theory". So the question can only be "how fast and effective this algorithm can be made for your specific problem". You might find some of the results and references in this paper https://arxiv.org/abs/1711.00125 useful Morphisms between elliptic curves may have arbitrarily large degree. Otherwise, Magma has some functionality with subfields of function fields https://magma.maths.usyd.edu.au/magma/handbook/text/452 but it doesn't seem to address your question directly (but you may check the references). G. Möhlmann, Einbettungen globaler Funktionenkörper Diplomarbeit TU Berlin 2008 @AlexandreEremenko But the answer to Hilbert's 10th problem is not known e.g. for Q. Elimination theory will work only over algebraically closed fields. @FrançoisBrunault ah yes, I overlooked this important case. However, I am most interested in the case $g \ge 2$ for both curves. Thank you for the reference. @FelipeVoloch this reference looks very relevant, do you know if anyone has implemented this algorithm? Unfortunately, my German is quite poor... @BenC There is something implemented in Magma, maybe just Hess's algorithm to test isomorphism. You should write to Hess. @Francois Brunault: But I thought that the field $\overline{F}_q$ in the question IS algebraically closed, or did I misunderstand the notation? @AlexandreEremenko Yes but he also asks for general k.
2025-03-21T14:48:31.792835
2020-08-16T18:31:04
369328
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "A.Skutin", "LSpice", "Mohan", "https://mathoverflow.net/users/100359", "https://mathoverflow.net/users/2383", "https://mathoverflow.net/users/9502" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632144", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369328" }
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Is it true that $g-t$ is divisible by $f$? Assume $f\in k[x_1,\ldots, x_n]$ is irreducible. Let for $g\in k[x_1,\ldots, x_n]$, $\partial(g)$ is divisible by $f$ for each derivation $\partial$ with $f\in\ker\partial$. Is it true that $g-t$ is divisible by $f$ for some $t\in k$? Your definition of slice does not refer to the derivation $D$. Why? Slice is $f$ with $D(f) = 1$. OK; I don't know the definition of a slice, so it is what you say it is. You said "being a slice is equivalent to $(f'{x_1}, \dotsc, f'{x_n}) = (1)$", where I guess $f'_x$ means $\partial_x f$, and that doesn't mention $D$. The conclusions of your Question and Question' also don't seem to depend on $D$, but maybe that is intentional. By a slice I mean $f$ for which exists $D$. And if the ideal is $1$ then such $D$ can be constructed. Consider the map $g:X\to\mathbb{A}^1$, where $X$ is defined by $f=0$ in $\mathbb{A}^n$. Your condition implies $dg=0$ and thus this map must be constant. This is what you wanted to prove. Is there a reference of the proof that if $dg = 0$ on the variety $X$ then $g$ is constant on $X$? Did you use that $(f'{x_1},\ldots, f'{x_n}) = (1)$ in the proof? @Algorithm Actually $dg=0$ implies $g$ is locally constant (just use the definition). So, my proof gives what you want only if $X$ is irreducible. I don't know a single slice which is not irreducible . $X$ is irreducible because $f$ is irreducible. Are you sure that $dg = 0$ is equivalent to $f|D(g)$ for each $D$ with $0=D(f)$, or it is just equivalent to $f|D(g)$ for each derivation $D$? @Algorithm I do not understand what you mean. For a smooth irreducible variety over complex numbers, the kernel of the map $d:O_X\to \Omega^1_X$ is just the constants.
2025-03-21T14:48:31.792973
2020-08-16T19:11:20
369329
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Goulifet", "Ryan Hendricks", "https://mathoverflow.net/users/39261", "https://mathoverflow.net/users/91878" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632145", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369329" }
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The space of periodizable tempered distribution The periodization operator $\mathrm{Per}$ is defined for a Schwartz function $\varphi \in \mathcal{S}(\mathbb{R})$ as \begin{equation} \mathrm{Per} \{ \varphi \} (x) = \sum_{n \in \mathbb{Z}} \varphi( x - n ), \quad \forall x \in \mathbb{R}. \tag{1} \end{equation} The sum in (1) is of course well-defined pointwise due to the rapid decay of $\varphi$ and we then have that $\mathrm{Per}\{\varphi\}$ is an infinitely smooth $1$-periodic function. More generally, it is possible to define the periodization operator $\mathrm{Per}$ over rapidly decaying distributions $\mathcal{O}_{C}'(\mathbb{R})$ (see for instance this paper for details). We then have \begin{equation} \mathrm{Per} : \mathcal{O}_{C}'(\mathbb{R}) \rightarrow \mathcal{S}'(\mathbb{T}) \tag{2} \end{equation} continuously, the latter space being the space of $1$-periodic distributions. Question: Can we define a proper subspace of $\mathcal{S}(\mathbb{R})$ that maximally extends the periodization in a precise sense? That is, a space on which the periodization is well-defined, with a natural topology that makes the periodization continuous, with good reasons for its "maximality"? Would this just be the set of functions ${f: \forall x\in\Bbb{R}, \sum\limits_{n\in\Bbb{Z}} f(x-n)<\infty}$? @fierydemon First I also want to include generalized functions with no pointwise definition (e.g., the Dirac comb $\sum_{n\in\mathbb{Z}}\delta(\cdot - n)$. But even with well-defined pointwise values, it is not obvious to me what is a good function space to deal with. In your suggestion, what is the meaning of the infinite sum? Pointwise convergence for each $x$? Where is the function $f$? These are typically the questions I am facing.
2025-03-21T14:48:31.793124
2020-08-16T20:10:09
369333
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Aaron Meyerowitz", "Asvin", "Joshua Holden", "https://mathoverflow.net/users/163849", "https://mathoverflow.net/users/58001", "https://mathoverflow.net/users/8008" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632146", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369333" }
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Is it possible to find a (nonsquare) integer which is a quadratic residues modulo a given infinite list of primes? I'm wondering if it's possible, given a prime p and an infinite list of primes $q_1$, $q_2$, ... to find an integer d which (1) is not a square mod p, but (2) is a square mod $q_i$ for all i. Always, sometimes, never? Probably sometimes --- what are some conditions? In the application I have in mind, the $q_i$ are all the prime divisors of the numbers $p^{2^n}-1$ as n ranges from 1 to infinity, but that's somewhat flexible. (The application, by the way, involves taking a p-adic interpolation of exponentiation of rational integers, and extending it to rings of integers in towers of number fields.) [ETA: I forgot to mention that d should also be a square mod 8 for the application, which rules out the answer of -1 given below.] For a finite list, d can be constructed using the Chinese Remainder Theorem, but that doesn't seem to help here. Given d, quadratic reciprocity gives an infinite set of primes for which d is a square, but I need the primes specified first. Grunwald-Wang says, if I understand it correctly, that condition (1) implies that d is not a square modulo $q$ for infinitely many primes $q$, but doesn't say anything about primes which d is a square for. The Chebotarov Density Theorem seems to imply that the set of possible d has density zero, but doesn't rule out (or imply) that one such d exists. Thanks for any help, sources, or advice! ----Josh Well, if $q$ were the set of all primes (Except p) or even just density more than half, then we know that $d$ is a square in the rationals. So we should assume the $q$ are sparse enough. In the other direction, if the sequence $q_i$ is such that for every (squarefree?) $d$, the $q_i$ fill up more than half the residue classes modulo $d$, then again the condition can't be satisfied. Both these observations follow from the fact that for fixed $d$, exactly half the primes (and half the residue classes modulo $d$) split in $\mathbb Q(\sqrt(d))$. Note that we can find a sequence of primes $q_i$ that fill up residue classes modulo $d$ for every $d$ but are very sparse among all primes: We just enumerate through each d and each residue class modulo $d$ and the choose the next prime to be in this residue class but very large otherwise. So these two conditions are really independent. It depends on the given list of primes. A simpler but necessary condition is that there be a $d$ so that all the primes of the list (greater than $d$) are concentrated in a few congruence classes $\bmod 4d.$ We can stick to odd prime divisors since everything is a quadratic residue $\bmod 2.$ If the list is all primes congruent to $1 \bmod 4$ then $-1$ is a common quadratic residue. That probably doesn't seem very exciting. If the list is all odd prime divisors of $3^{2^n}-1$ as $n$ ranges over the positive integers then $-1$ is again a common quadratic residue. That is the kind of thing you were mentioning. But the reason is that all those primes are $1 \bmod 4$ If I am not mistaken, and for the same reason, $-1$ is a common quadratic residue of of the prime divisors of $p^{2^n}-1$ as $n$ ranges over the integers starting at $2.$ For certain primes , such as $5,7,17,19,31,53,59$ we can expand the list to all prime divisors of $p^{2^n}-1$ with the exception of $3.$ In general it is sufficient to discard any divisors of $p^2-1$ which are $3 \bmod 4.$ The facts behind this are $p^{2^n}-1=(p-1)(p+1)(p^2+1)(p^4+1)\cdots(p^{2^{n-1}}+1)$ every odd factor of $p^{2^m}+1$ is of the form $2^{m+1}q+1$ $-1$ is a quadratic residue for primes which are $1 \bmod 4.$ Think first about this (easy) question. For fixed $d$ what are the odd primes $q$ such that $d$ is a quadratic residue $\bmod q?$ Call this set $G_d.$ We may assume that $d$ is squarefree. Then the members of $G_d$ are the prime divisors of $d$ along with those primes in a union of certain congruence classes $\bmod 4d.$ Half of the classes $(r \bmod 4d)$ with $\gcd(r,4d)=1$ In some cases ($d$ even or $d$ odd with all divisors $1 \bmod 4$) it suffices to consider congruence classes $\bmod 2d$. However what is written is still correct. I will ignore your $p$ on the assumption that the goal was to rule out $d$ being a square. Then the specific $d$ works for a particular instance of your problem, precisely if the chosen list is one of the uncountably many infinite subsets of $G_d.$ On the other hand, suppose it is given that the members of the list (other than the divisors of $d$ in the list, if any) are chosen from some $k \ll \phi(d)$ of the congruence classes $\bmod 4d$. Then, if the $k$ are chosen at random, the chance that $d$ will work is less than $2^{-k}$. So starting from a list $\mathbf{q}=q_1,q_2,\cdots$ the first question is "Is there some reason to suspect that there is an $M$ so that all the members of $\mathbf{q}$ (prime to $M$) are concentrated in a few of the congruence classes $\bmod M?$" If that does not happen, then there is no hope. If it does happen for a certain $M,$ then chances still may be low. So it very much depends on where $\mathbf{q}$ comes from. By the way, the problem of finding a $d$ which is a quadratic non-residue relative to all $q \in \mathbf{q},$ is equally difficult. Thanks very much, Aaron! In your example, I'm assuming that you took $p=3$ and $d=7$ just as convenient small numbers? In those now deleted lines I took $p=3$ since it was small and $7$ as a prime which is $1 \bmod 2p.$ I was probably confused and thinking of this fact: The prime factors of $2^p-1$ are all $1 \bmod 2p.$ So you might be able to get results for certain instances of "the list of all prime factors of $2^p-1$ where $p$ is a prime congruent to $u \bmod v.$" But, again, only because that restricts congruence classes $\bmod 2d.$ Ah, I see. Thanks for the update! I hadn't thought about the properties of generalized Fermat numbers before. It turns out -1 doesn't quite work for the application I wanted (see edit) but this gives me something I can work with! (In particular this shows how to explicitly construct a sequence of integers $d_n$ that work for larger and larger $n$, and that seems like the next best thing. I'm still wondering if there's a single $d$ that works but I admit now that it seems unlikely!)
2025-03-21T14:48:31.793571
2020-08-16T20:13:44
369334
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Ben C", "Emily", "Jason Starr", "Jürgen Böhm", "Qfwfq", "https://mathoverflow.net/users/130058", "https://mathoverflow.net/users/13265", "https://mathoverflow.net/users/154157", "https://mathoverflow.net/users/21940", "https://mathoverflow.net/users/4721", "https://mathoverflow.net/users/64302", "user2520938" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632147", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369334" }
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How to define jet bundles algebraically? Let $B\to X$ be a surjective submersion over the smooth integral scheme $X$ over $\mathbb{C}$. Associated to this we have in the $C^\infty$ world the notion of the $k$ jet-bundles $J_k(B)$, which are affine bundles over $X$. I wonder what the best way to define this notion in the setting of algebraic geometry is. If $B=TX$ there is an algebraic description on Wikipedia which makes sense, but I cannot figure out how to give a satisfactory definition in general. I am fine with restricting to the case where $B$ is the spectrum of a sheaf of $\mathcal{O}_X$ algebras. Edit: I am talking here about jet bundles in the sense of e.g. this wikipedia article. Please state the definition of "jet bundle" you use (there are multiple interpretations in algebraic geometry, e.g., truncated arc spaces and principal parts bundles). Is this related: https://ncatlab.org/nlab/show/arithmetic+jet+space? @JasonStarr please see the edit. @Emily Thanks. I will need to read up on some of things written there to be to judge to what extend that page suggests the answer to my question, but it definitely seems related.. These are a (relative) version of "truncated arc spaces". One analogue in algebraic geometry of the Cartan extension method mentioned in that wikipedia article is Chapter 3 of the PhD thesis of Jan Gutt: http://www.math.stonybrook.edu/alumni/2013-Jan-Gutt.pdf I found the following short manuscript from Ravi Vakil quite enlightening: http://math.stanford.edu/~vakil/files/jets.pdf Let $S$ be a scheme, e.g., $\text{Spec}\ \mathbb{C}$. Let $f:X\to S$ be a morphism that is separated and smooth. Denote the associated relative diagonal morphism by $$\Delta_{X/S}:X \to X\times_S X.$$ This is a closed immersion whose ideal sheaf $\mathcal{I}$ is everywhere locally generated by a regular sequence. For every integer $e\geq 0$, the ideal sheaf $\mathcal{I}^{e+1}$ is the defining ideal sheaf of a closed subscheme of $X\times_S X$. Denote this closed subscheme by, $$\Delta_{e,X/S}:X_e \to X\times_S X.$$ Because $f$ is smooth, each associated projection morphism to $X$ is a finite and flat morphism (in particular, it is proper), $$p_i: X_e \to X, \ \ i=1,2.$$ By representability of the functor from Part IV.4.c, p. 267 (p.20 of the NUMDAM edition) in Fondements de la Géometrie Algébrique, for every scheme $X$, for every flat and projective morphism, $$p:Y\to X,$$ for every finitely presented, quasi-projective morphism, $$q:Z\to Y,$$ there exists a universal pair, $$(r:\Pi_{Z/Y/X} \to X, \ s:\Pi_{Z/Y/X}\times_X Y \to Z),$$ of a morphism from a scheme $T$ to $X$ and a $Y$-morphism from $T\times_S Y$ to $Z$. In particular, for the flat and finite morphism $p_2$ from $X_e$ to $X$, for every smooth, quasi-projective morphism $q$ from a scheme $Z$ to $X_e$, there is such a pair, $$(r:\Pi_{Z/X_e/X} \to X, \ s:\Pi_{Z/X_e/X}\times_X X_e \to Z).$$ Finally, for every finitely presented, quasi-projective morphism, $$\pi:B\to X,$$ the base change morphism is also finitely presented and quasi-projective, $$B\times_{X,\text{pr}_1} (X\times_S X) \to X\times_S X.$$ Thus, the pullback of this morphism over the closed subscheme $X_e$ is also a finitely presented and quasi-projective morphism. Denote this pullback by $$\pi_e:B_e\to X_e.$$ The "relative truncated sections" parameter space is the universal pair, $$(r:\Pi_{B_e/X_e/X} \to X,\ s:\Pi_{B_e/X_e/X}\times_X X_e \to B_e).$$ If the morphism $q$ is smooth, then every "truncated section" parameterized by $\Pi_{B_e/X_e/X}$ extends to a formal section by Hensel's lemma. Thank you for this detailed answer. Can you provide one more detail by giving a more precise reference for this construction of Grothendieck? I realise that this might be common knowledge, but I do not know this. That is a good point. I added a page number in FGA for the construction. According to Mustață, Jet schemes of locally complete intersection canonical singularities, the $m$-th jet scheme of (the $\mathbb{C}$-scheme) $X$ is the scheme $J_m X$ over $X$ representing the functor $\mathrm{Sch} \to \mathrm{Set}$ given by $$ S \mapsto \mathrm{Hom}(\Delta^m\times S,X)$$ where $\Delta^m:=\mathrm{Spec}\frac{\mathbb{C}[t]}{\langle t^{m+1} \rangle}\;.$ So the closed points of $J_m X$ are the morhisms of schemes $\Delta^m\to X$. And the fiber of $J_m X\to X$ over $x\in X$ has as set of closed points the set $\mathrm{Hom}(\mathcal{O}_{X,x}\;,\mathbb{C}[t]/\langle t^{m+1} \rangle)$. Another way of seeing it is: let $\mathbf{F}:\mathrm{Sch}\to\mathrm{Sch}$ be the functor $\mathbf{F}(S)=\Delta^m\times S$, than it has a right adjoint given by $X\mapsto J_m X$. Remark. According to the above construction we have $J_1 X=TX$, the tangent bundle (or total space of tangent sheaf if $X$ isn't smooth). This is only a special case of the type of jet bundle that I am talking about, corresponding to the surjective submersion $X\times \Delta\to \Delta$. I was about to add that mine is a pocket version of Jason Starr's answer. But now that I read better, you're right, it actually doesn't answer your question, which is a request for the relative version. I'll still leave my answer there, as might be useful as reference on a version of jet spaces for future readers. To be honest, your planned addendum on how this relates to Jason Starr's answer would still be very helpful for me, since I am familiar with the jet spaces as described in your answer, while I'm having some trouble relating this this the constructino on Jason Starr. (The comment that I wrote wasn't correct, so I deleted it) I am having trouble seeing how this is a special case of Jason Starr's answer. Does anyone know?
2025-03-21T14:48:31.794351
2020-08-16T20:17:11
369335
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Cranium Clamp", "Ennio Mori cone", "LSpice", "https://mathoverflow.net/users/122729", "https://mathoverflow.net/users/152391", "https://mathoverflow.net/users/2383" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632148", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369335" }
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On the structure of Hilbert schemes While studying and solving some exercises on Hilbert schemes, I've come across many problems in Hartshorne's book on deformation theory which ask the reader to show certain properties such as irreducibility, openness or non-singularity. Below I list a few examples. (Assume $ k $ is an algebraically closed field.) (1) In $ \operatorname{Hilb}^{3t+1} ( \mathbb{P}^3_k ) $, the twisted cubic curves form a non-singular open subset of an irreducible component $ H_0 $ of dimension $ 12 $, also courtesy of Piene and Schlessinger's paper On the Hilbert scheme compactification of the space of twisted cubics. This Hilbert scheme has another irreducible component $ H'_0 $ corresponding to plane elliptic curves union a point not on the curve. (2) In $ \operatorname{Hilb}^8( \mathbb{P}^4_k ) $, there is an irreducible component of dimension $ 32 $ that has a non-singular open subset corresponding to $ 8 $-tuples of distinct points in $ \mathbb{P}^4_k $. (3) The complete intersection curves in $ \mathbb{P}^3_k $ obtained by homogeneous polynomials of degrees $ a,b $ form a non-singular open subset of an irreducible component of the corresponding Hilbert scheme. Question: How does one justify these properties, particularly the openness of certain subsets and irreducibility of components? Although these assertions seem intuitively clear, are there strategies in general that allow you to see these rigorously? For example, in the paper by Piene and Schlessinger, $ H'_0 $ is stated to be irreducible without proof and I'm not sure how to do this. I edited your question to include the name of the paper, and, while I was there, bolded the question to help it stand out. I hope that was all right. One can show that these spaces are irreducible by parameterizing a dense open subset. For instance, the Hilbert scheme $M$ of plane cubics in $P^3$ is parameterized by first choosing a plane, and then a cubic in that plane; so it is birational to a projective bundle over $Gr(2,4)$, and thus it is irreducible. Hence $H_0'$ is also irreducible, since it is birational to the product $M\times P^3$ (letting the extra point vary over the $P^3$). Such `set-theoretic' arguments are enough for openness and irreducibility, which are topological in nature. @LSpice, thank you for the edits. @wnx, can you elaborate on the projective bundle part? I think you made a typo, should be $ Gr(3,4) $. If I understand correctly, that has a canonical vector bundle $ \mathcal{E} $ of rank 3 and then taking $ \mathbb{P}( \operatorname{Sym}^3 \mathcal{E}^{*} ) $ gives you a cubic? Yes, indeed Gr(3,4). Also, your interpretation looks correct.
2025-03-21T14:48:31.794561
2020-08-16T20:40:03
369336
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Mathman", "https://mathoverflow.net/users/153090" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632149", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369336" }
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Bounds on variance of sum of dependent random variables Let $x_1, \ldots, x_n$ be possibly dependent random variables, each taking values $x_i \in \{0, 1, 2\}$. Suppose further that in every outcome the number of random variables that equal 2 is exactly 1. Now for each $i \in \{1, \ldots, n\}$ define $$ f_i = \begin{cases} \Pr[x_i = 2 \mid x_i \geq 1] & \text{if } x_i \geq 1\\ 0 & \text{if } x_i =0 \end{cases}, $$ and let $ f = \sum_i f_i. $ My question is how large can the variance of $f$ be? My conjecture is that we should be able to bound it by $O(1)$ but don't know how to prove this. Note: In case it helps, it is easy to prove that $E[f] = 1$: $$ E[f] = \sum_i E[f_i] = \sum_i \Pr[x_i \geq 1] \times \Pr[x_i = 2 \mid x_i \geq 1] = \sum_i \Pr[x_i = 2] = 1, $$ where the last equality comes from our initial assumption that in all outcomes exactly one of the $x_i$'s equals 2. $Var\,f$ can be on the order of $n$ (but not more than that). Indeed, let $U$ and $N$ be independent random variables such that $P(U=1)=:p=1-P(U=0)=:q$ and $P(N=i)=1/n$ for all $i\in[n]:=\{1,\dots,n\}$. Let $$x_i:=1(U=1,N\ne i)+2\times1(N=i). $$ Then with $p=1/n$ $$Var\,f\sim n/4\tag{1}$$ (as $n\to\infty$). On the other hand, $$Var\,f\le Ef^2=\sum_{i,j\in[n]}Ef_if_j\le\sum_{i,j\in[n]}Ef_i =n\sum_{i\in[n]}Ef_i=n.$$ Details on (1): We have $$Ef^2=\sum_{i,j\in[n]}Ef_if_j \\ =\sum_{i,j\in[n]}P(x_i=2|x_i\ge1)P(x_j=2|x_j\ge1) P(x_i\ge1,x_j\ge1),\tag{2}$$ $$P(x_i\ge1)=1-P(x_i=0)=1-P(U=0)P(N\ne i)=1-q(1-1/n)=p+q/n,$$ $$P(x_i=2)=P(N=i)=1/n,$$ $$P(x_i=2|x_i\ge1)=\frac{P(x_i=2)}{P(x_i\ge1)}=\frac{1/n}{p+q/n},$$ and $$P(x_i\ge1,x_j\ge1)=1-P(x_i=0\text{ or }x_j=0)=1-P(x_i=0)-P(x_j=0)+P(x_i=0,x_j=0) =1-2q(1-1/n)+q(1-2/n)=1-q=p$$ for $i\ne j$. Choosing now $p=1/n$, we have $$Ef^2\sim n/4.$$ Since $Ef=1$, (1) now follows. Looking back at (2), now the idea behind the construction should become transparent: We want to make $P(x_i\ge1,x_j\ge1)$ for $i\ne j$ much greater than $P(x_i\ge1)P(x_j\ge1)$ and at the same time not to make $P(x_i\ge1,x_j\ge1)$ too small. The choice $p=1/n$ is nearly optimal in this regard. Thank you for the prompt response and the excellent demonstration of the distribution Iosif! Now I understand why I wasn't able to prove my conjecture lol. Btw, I posted a related question (which was the main thing I was trying to prove via bounding the variance of $f$, which you refuted here). It would be great if you take a look at that one too if you have time. https://mathoverflow.net/questions/369355/independent-sampling-of-dependent-random-variables
2025-03-21T14:48:31.794741
2020-08-16T20:56:48
369337
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Avishek Dutta", "David Roberts", "Gabe K", "Mirco A. Mannucci", "Yemon Choi", "https://mathoverflow.net/users/125275", "https://mathoverflow.net/users/15293", "https://mathoverflow.net/users/163854", "https://mathoverflow.net/users/4177", "https://mathoverflow.net/users/763" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632150", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369337" }
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How to study to learn differential geometry for applying it to statistics Basically I want to learn information geometry or specifically the application of differential geometry in statistics to do a project. I am from a statistical background and have a knowledge about real analysis, several variable calculus, linear algebra. One of my professors told me that the first three chapters from Do Carmo's Differential geometry would be sufficient. Can someone assure me if that's enough or do I need to learn Riemannian geometry. And If I need to learn Riemannian geometry then what should be my path for learning. I don't want to learn rigorous mathematics. I just want to apply it to statistics. https://stats.stackexchange.com/questions/86734/does-differential-geometry-have-anything-to-do-with-statistics I have read that post..But the recommended books there require some mathematical maturity.Like I can't understand Amari's books without any knowledge of differential geometry. I have only glanced at the book of Murray and Rice, and that was almost 20 years ago, but my recollection is that it was relatively accessible and developed the necessary differential-geometric language as it went along Our department head, a statistician, apparently says this book is good: https://en.wikipedia.org/wiki/Calculus_on_Manifolds_(book) He is a very theoretically-minded person FWIW. This book might also be helpful: https://www.routledge.com/Differential-Geometry-and-Statistics/Murray-Rice/p/book/9780412398605 @DavidRoberts Indeed, the 2nd book you link to is the one mentioned in the link that Mirco gives, and which I mentioned in my previous comment :) @Yemon gah, I totally missed that, both times! :-/ Avishek, not easy to answer with the little context you provided. I would go first with what your prof said, and yes, Do Carmo is the place to go. There, you will learn all about surfaces in $R^n$, which is basically classical differential geometry. If, on the other hand, your project is at the research level (say master thesis or beyond), then download this article. That has to do with abstract information geometry, which in turn relies on modern differential geometry: manifolds, tensor calculus, etc. Basically, the chief difference between the first and the second is that in manifold theory you do not start from embedded manifold, rather you define the entire machinery intrinsically. If you do not know classic geometry of surfaces, you still have to spend a few days on Do Carmo. Then prepare for a lot of sweat, to get into the modern approach. Hope it helps I am going to do a project for my master's thesis.Now I have studied Do Carmo till regular surfaces.How much should I read. And after finishing it ,what book should I follow to study rest of the modern approach. I have no one to ask .As very few people whom I know are working on this .So please guide me through this How about all of it? At least a cursory look. If you go directly into manifold theory it will sound a bit exotic, but if you are steeped in the classical one, it will make sense (after a while). Then download this one: https://im0.p.lodz.pl/~kubarski/AnalizaIV/Wyklady/L-Tu-1441973990.pdf. Just found it, but looks as easy going as it comes. Should I read this book on manifold totally or should I read a part of it? My friend, how can I possibly know? I can tell you that in that book there is more than some basic math grad course in manifolds, so I would download it, show it to your prof, and ask. I would assume that you need only the fundamentals of differentiable manifolds, but ask him anyway I think Do Carmo is a good option. Personally, I'm a fan of John Lee's Introduction to Smooth Manifolds and its sequel Riemannian Manifolds. While these are written at a higher level, they really emphasize the geometric picture at work. I think the survey by Nielsen is a good article and I found it very helpful to get a broad overview of IG. However, I would not recommend using it to learn differential geometry. Most books about information geometry take a very idiosyncratic approach to geometry, which can give rise to various misunderstandings. These are not a big deal if you are already familiar with differential geometry but are more of a problem if you are trying to learn it. Both of these works are well worth reading if you are interested in IG, but I'll give an example of what I mean. Both Amari's book and the survey article by Nielsen state that the holonomy of a flat connection is trivial (although they don't use this language). In information geometry, the flat connections of interest are generally on exponential families (where this ends up being true). However, in general, the holonomy of a flat connection is not zero (it's induced by fundamental group). Furthermore, for this result, the connection must be both curvature- and torsion-free (not just curvature-free). Statistical manifolds are generally taken to have torsion-free connections, so this is not an issue in applications. These are relatively minor points if you are familiar with differential geometry, but would be misleading for someone learning it. If I want to read Amari's Information geometry and it's application totally,How much Differential geometry should I know.and from which book should I learn that as a beginner, That's a good question, and I don't really have a good answer. The issue is that Amari's books aren't trying to be fully rigorous and so have a lot of things that aren't exactly right. I suspect in applications that these issues don't matter and that I'm splitting hairs, but on the other hand I think differential geometry is a topic where it's easy to go astray. But if you want to learn differential geometry, I would highly recommend Lee's books (Spivak's textbooks are also really good) and to keep those as reference texts. As I mentioned in my answer, information geometry takes a somewhat idiosyncratic approach to differential geometry, which is rooted in affine differential geometry and considers embeddings quite a bit more than is traditionally done. [cont.] Stylistically, I think the DG textbooks that are most similar to Amari's approach are Foundations of Differential Geometry by Nomizu and Kobayashi, and Affine Differential Geometry: Geometry of Affine Immersions by Nomizu and Sasaki. However, these are written at a much more advanced level. They also emphasize a "structural point of view," which avoids any sort of pictures. To be honest, something like Do Carmo which focuses on surfaces in $\mathbb{R}^3$ is probably a better way to learn about embeddings for the first time. [cont.] Personally, the resource that was most helpful to me for learning information geometry was reading a series of blog posts by John Baez and working through some calculations involving the multinomial and Gaussian families. I would highly recommend working through a few examples in detail, because otherwise it might seem like there is a huge amount of machinery but not much to say about them. Once I considered a few common exponential families, everything became a lot more clear.
2025-03-21T14:48:31.795216
2020-08-16T21:32:19
369338
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Use of Flattening Stratification part 2 (Nitsure's construction of Hilbert and Quot schemes) I study Nitin Nitsure's paper Construction of Hilbert and Quot Schemes (arXiv:math/0504590) and have some problems with the content of imposed universal property (F) in the section "Use of Flattening Stratification" (p 26). Reading the imposed universal property (F) following parts confuse me: Question 1: About the composition map $h$. First of all I guess the author had $W_Y \otimes_{O_Y} \operatorname{Sym}^r V_Y$ in mind and not $W \otimes_{O_S} \operatorname{Sym}^r V$, or not? Let assume that. But why is $\pi_Y^* W_Y \otimes_{O_Y} \operatorname{Sym}^r V_Y = \pi_Y^* \pi_{Y*} E_Y$ and not $\pi_Y^* \pi_{Y*} E_Y(r)$? I think that this identification should arise in same way as in the construction of $\alpha:$ $ \frak{Quot} $ $^{\Phi, L}_{E/X/S} \to $ $\frak{Grass} $ $(W \otimes_{O_S} \operatorname{Sym}^r V, \Phi(r))$ from the section "Embedding Quot into Grassmannian" (the previous part on same page). Let introduce abbreviations for these functors: We set $G:=$ $\frak{Grass}$ $(W \otimes_{O_S} \operatorname{Sym}^r V, \Phi(r))$ and $Q:=$ $\frak{Quot}$ $^{\Phi, L}_{E/X/S}$. The explicit construction of $\alpha$ is given there. I assume that the author hasn't changed notations when passing to "Use of Flattening Stratification" and therefore it should still be $\pi_Y^* W_Y \otimes_{O_Y} \operatorname{Sym}^r V_Y = \pi_Y^* \pi_{Y*} E_Y(r)$? Question 2: Based on insights from Question 1 I think that the cokernel of $h$ should be $q: E_Y(r) \to F$ because $h$ should be $\pi_Y^* K_Y \to E_Y(r)$. Then I come to conclusion that the universal property (F) on $F$ should be read as follows: (UP) $F(-r)$ (ie twist of $F$ by $O_Y(-r)$ and not $F$ itself !) is flat over $Y$ with its Hilbert polynomial on all fibers equal to $\Phi$ if and only if $\phi : Y \to T$ factors via $T' \to T$. But then the proof shows an embedding of $ \frak{Quot} $ $^{\Psi, L}_{E(-r)/X/S} $ with $\Psi(X) := \Phi(X-r)$ and not $ \frak{Quot} $ $^{\Phi, L}_{E/X/S} $. That's a problem. The reason I think so is the definition of relative representability, see Tag 0023 from the Stacks Project. Reparaphrasing this we have to show for fixed $f \in G(Y) \cong \operatorname{Hom}(Y, \operatorname{Grass}(W \otimes_{O_S} \operatorname{Sym}^r V, \Phi(r)))$ that $h_T \times_{f, G, \alpha} Q(Y)$ is the set of pairs $(\phi, q) \in h_T(Y) \times Q(Y)$ where $q: E_Y \to D$ is the quotient with $G(\phi)(f)=\alpha(q)$. In this sense we have to show that this is representable. Now what does this condition mean? By construction $ G(\phi)(f)= f_Y: W_Y \otimes_{O_Y} \operatorname{Sym}^r V_Y \to \phi^* \mathcal{J}$ and for a quotient $q: E_Y \to D \in Q(Y)$ the image $\alpha(q)$ is by construction of $\alpha$ exactly $\pi_{Y*} E_Y(r) \to \pi_{Y*} D(r) $ and the requirement is that $\alpha(q)= f_Y$. Therefore I think that the correct formulation in the last sentence of universal (F) should be as in (UP). Is that true or do I make a thinking error and misunderstand the construction? Question 3: I not understand why if we take at the end $T:= \operatorname{Grass}(W \otimes_{O_S} \operatorname{Sym}^r, \Phi(r))$ then the corresponding locally closed subscheme $T' \subset T$ represents the functor $\frak{Quot}$ $^{\Phi, L}_{E/X/S}$ "by construction". Precisely, by the construction of what? Of $\alpha$?
2025-03-21T14:48:31.795443
2020-08-16T21:56:46
369340
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "AlexArvanitakis", "gradstudent", "https://mathoverflow.net/users/22757", "https://mathoverflow.net/users/89451" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632152", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369340" }
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Spectrum of the Witten Laplacian on compact Riemannian manifolds Below I have given what I am calling as the ${\rm Witten{-}Laplacian}_{s,p}$ on a Riemannian manifold $(M,g)$ for any constant $s >0$ and $p \in C^2(M,g)$ How generally is it true that this ${\rm Witten{-}Laplacian}_{s,p}$ is positive semi-definite? And when the above is true are there examples of compact manifolds and (preferably ``usual") functions on them $p$ s.t we know the exact spectrum of the corresponding ${\rm Witten{-}Laplacian}_{s,p}$? (or at least a lower bounds on its smallest non-zero eigenvalue) By "usual" functions I mean we do not assume that $f$ is Morse or that $f$ satisfies the confining or the Villani condition or such. And I happy to know if the above is known even for just spheres or AdS. For ``nice" real valued functions $f$ on $(M,g)$ we have for the square of the gradient of $f$, $\Vert {\nabla_g f} \Vert ^2 = g(\nabla_g f,\nabla_g f) =  \sum_{j=1}^n  \sum_{i=1}^n g^{ij} \partial_i f  \partial_j  f$ and the Laplacian of $f$ being, $\nabla_g^2 f := \frac{1}{\sqrt{\det(g)}} \sum_{i,j=1}^n \frac{\partial }{\partial x_i} \left (  \sqrt{\det(g)} g^{ij} \frac{f}{\partial x_j}\right )$ where we define the metric as $g = [g_{ij}] = g \left ( \partial_{x_i}, \partial_{x_j} \right )$ and $g^{-1} = [g^{ij}]$. Then the ``${\rm Witten{-}Laplacian}_{s,p}$" will be the operator mapping, $$ C^2(M,g) \ni h \mapsto \left ( -s^2 \nabla_g^2 + \Vert{\nabla_g p}\Vert ^2 -s \nabla_g^2 p \right ) h \in C^2(M,g)$$ Since your operator is of the form "Laplacian + perturbation" it will be positive-definite (possibly modulo finite numbers of zero modes and negative modes) exactly as often as the Laplacian itself is. You probably want to look at works by Gilkey. I don't remember an exact reference off the top of my head though. There is a list of them in the physics-oriented review arxiv.org/abs/hep-th/0306138. Well, we do not want any negative eigenvalues at all. One thing that I guess is true is that it is PSD over the span of the solutions ${ \psi_s }$ of the Fokker-Plank equation given as, $\frac{\partial \psi_s}{\partial t} = - \frac{1}{2s} \cdot {\rm Witten-Laplacian}_{s,p} ( \psi_s )$. So I believe it should be PSD on any function space in which the solutions of the Fokker-Plank are dense. But I dont know what is that function space. The Fokker-Planck equation (?) you mention is (also) the heat equation for this "Witten-Laplacian" operator (possibly modulo an overall factor involving $s$) which therefore has the same domain as the general class of operators in the review I referenced above. I cant see anything in that review which looks like the Witten-Laplacian! Am I missing something? (2.1) apparently. Okay! So you are setting $a$ there to $0$ and $b = \Vert{\nabla_g p}\Vert ^2 -s \nabla_g^2 p$ here. Wonder if these people can compute the eigenvalues for this! Though, even if one does this identification, how does one get to the smallest positive eigenvalue from the heat-kernel expansion?
2025-03-21T14:48:31.795667
2020-08-16T22:23:46
369343
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Greg Friedman", "LSpice", "TheGeekGreek", "https://mathoverflow.net/users/2383", "https://mathoverflow.net/users/6646", "https://mathoverflow.net/users/98139" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632153", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369343" }
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Connected manifold without connected regular level set admits exactly two connected components Let $M$ be a connected smooth manifold and $f \in C^\infty(M)$ such that $0$ is a regular value of $f$. Moreover, suppose that $f^{-1}(0)$ is connected. Is it true that $M \setminus f^{-1}(0)$ has exactly two connected components? As $0$ is a regular value of $f$, we know that $f^{-1}(-\infty,0]$ and $f^{-1}[0,+\infty)$ are both regular domains, but I am not sure if my assumptions are sufficient. The assumptions certainly aren't necessary (for example, take $M = \mathbb R$ and $f \colon x \mapsto x^2$), so I guess you just mean sufficient? @LSpice Yes, indeed. Good example. Another one would be the torus with the height function. Since $M$ is connected and a manifold, it is path-connected. Thus, any two points $x,y \in M$ such that $f(x), f(y) > 0$ can be joined by a path $\gamma$. Suppose $\gamma$ does not lie entirely in $f^{-1}(0,\infty)$. Then let $p = \text{inf}\{ t \in [0,1] : f(\gamma(t)) \leq 0\}$ and $q = \text{sup}\{ t \in [0,1] : f(\gamma(t)) \leq 0\}$. Then $p, q \in f^{-1}(0) = N$, and since $N$ is path-connected, we can join $p$ and $q$ by a path $\ell$ in $N$. As $f \circ \gamma$ takes positive values on $[0,p) \cup (q,1]$, first running through $\gamma_{[0,p]}$, then through $\ell$, and finally through $\gamma_{[q,1]}$ produces a path from $x$ to $y$ entirely contained in $f^{-1}[0,\infty)$. Using a tubular neighbourhood of $N$ in $M$, we can homotope this path so that it lies in $f^{-1}(0,\infty)$. Thus, in any case, $x,y$ can be joined by a path in $f^{-1}(0,\infty)$, so this subset of $M$ is path-connected. The same is true for $f^{-1}(-\infty,0)$. Why does using the tubular neighbourhood make it obvious that you can homotope into $f^{-1}(0, \infty)$? Is it because $f^{-1}(0)$ disconnects the neighbourhood, and you know you can always homotope into a chosen one of the components of the complement? @LSpice Sort of. In this case the tubular neighborhood will be diffeomorphic to $N\times \mathbb{R}$. In fact, since $0$ is a regular value, such a neighborhood will be given by the inverse image $f^{-1}(-\epsilon,\epsilon)$ for some $\epsilon>0$. Once you know that, you can think of the part of the path in $f^{-1}([0,\infty))$ as being a path in $N\times [0,\infty)$. Then using an isotopy of $\mathbb R$ supported on an appropriate compact set, you can push all of $N\times [0,\infty)$ diffeomorphically to, say, $N\times [a,\infty)$ for some $a>0$, keeping things far enough from $N$ fixed. Now compose the path with the non-identity end of that isotopy to get the desired path. I think in this case an easier homotoping argument can be given by simply using the flow of the normalised gradient vector field with respect to any Riemannian metric on $M$. No need to use tubular neighbourhoods.
2025-03-21T14:48:31.795866
2020-08-16T22:40:32
369344
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Free abelian group on a space and fibrations Let $X$ be a topological space. Endow the free abelian group on $X$, $\mathbb Z[X]$, the quotient topology coming from the surjection $\bigsqcup_n X^n \times \mathbb Z^n \to \mathbb Z[X]$. For $Y$ a sub-CW-complex of a CW-complex $X$, the quotient map $\mathbb Z[X] \to \mathbb Z[X]/\mathbb Z[Y]$ is a quasi-fibration (see e.g. Salvatore Lemma 6.1 https://arxiv.org/pdf/math/9907073.pdf). Is this quotient map a Serre fibration?
2025-03-21T14:48:31.795926
2020-08-16T23:06:22
369346
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Alapan Das", "Iosif Pinelis", "LSpice", "https://mathoverflow.net/users/156029", "https://mathoverflow.net/users/2383", "https://mathoverflow.net/users/36721" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632155", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369346" }
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Probability of getting two subsets with the same sum Let $A=\{1,...,n\}$. Two subsets of $A$, not necessarily distinct, chosen uniformly at random. What is the probability that both subsets have the same sum? Alternatively, is there a known upper bound? Just to be sure, "subsets … chosen uniformly at random" means that each subset has probability $2^{-n}$ of being chosen (so that the question is really how many pairs of subsets have the same sum), right? Assuming the two subsets are selected independently, the probability in question is $$p_n=a_n/2^{2n},$$ where $a_n$ the sum of the squares of the coefficients in the polynomial $$\prod_{k=1}^n (1+x^k).$$ The sequence $(a_n)$ is the sequence A047653, and its asymptotics, given on that page, is $$a_n\sim\frac{\sqrt{3/\pi}\,4^n}{n^{3/2}} \tag{1}$$ (as $n\to\infty$). So, $$p_n\sim\frac{\sqrt{3/\pi}}{n^{3/2}}.\tag{2}$$ For (1), A047653 refers to Vaclav Kotesovec, where I have been unable to find a link/reference to (1). However, the equivalent relation (2) was actually proved by van Lint. (I found a link to van Lint on the related OEIS page A000980.) $a_n = s_n$, right? @LSpice : Thank you for your comment. I am sorry for the confusion. I was hesitating between $s_n$ (for the sum of the squares) and $a_n$ (the notation used on the linked page). Now this typo is fixed. The asymptotic comes from the integral $$4^n\int_{0}^{1} \left(\prod_{k=1}^{n} \cos^2(\pi kt) \right)dt$$. But how to calculate the integral? @AlapanDas : I have added a response to your comment.
2025-03-21T14:48:31.796055
2020-08-16T23:55:20
369347
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Dominic van der Zypen", "https://mathoverflow.net/users/8628" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632156", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369347" }
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Weak $1$-factorizability A simple, undirected graph is said to be $1$-factorizable if there is a partition of the edge set $E$ such that every member of the partition is a perfect matching of $G$. Let us call $G$ weakly $1$-factorizable if there is a partition of $E$ into maximal (but not necessarily perfect) matchings. $K_3$ is weakly $1$-factorizable, but not $1$-factorizable for the trivial reason that every singleton edge set is a maximal matching and their union is the edge set of $K$, but it has an odd number of vertices and therefore no perfect matching The graph $(V,E)$ with $V = \{0,1,2,3\}$ and $E = \big\{\{k, k+1\}: k \in \{0,1,2\}\big\}$ has a perfect matching, but it is not regular (i.e. not all vertices have the same degree), therefore not $1$-factorizable, but it is weakly $1$-factorizable. Question. What is an example of a weakly $1$-factorizable but not $1$-factorizable regular graph $G=(V,E)$ such that $G$ has a perfect matching? Thank you - beautiful! The Petersen graph $G$ is $3$-regular and has a perfect matching but is not $1$-factorizable. To see that $G$ is weakly $1$-factorizable, regard it as the complement of the line graph of $K_5$. For each vertex $v$ of $K_5$, let $M_v$ be the set of all pairs $\{e,f\}$ where $e$ and $f$ are edges of $K_5$ which are not incident with $v$ and not adjacent to each other in $K_5$ (so they are adjacent vertices in $G$). Then $M_v$ is a maximal matching in $G$, and $\{M_v:v\in V(K_5)\}$ is a partition of $E(G)$ into $5$ maximal matchings. Let $V(K_5)=\{v,w,x,y,z\}$. Plainly $M_v=\{\{wx,yz\},\{wy,xz\},\{wz,xy\}\}$ is a matching in $G$. To see that $M_v$ is a maximal matching, consider an edge $e\in E(G)\setminus M_v$, say $e=\{vw,xy\}$, and observe that $e$ is adjacent to the edge $\{wz,xy\}$ which is in $M_v$. Alternatively, $E(G)$ can be partitioned into $4$ maximal matchings $M_1,M_2,M_3,M_4$ where $M_1=\{\{wx,yz\},\{wy,xz\},\{wz,xy\}\}$; $M_2=\{\{vw,xy\},\{vx,wz\},\{vy,xz\},\{vz,wx\}\}$; $M_3=\{\{vw,xz\},\{vx,yz\},\{vy,wz\},\{vz,wy\}\}$; $M_4=\{\{vw,yz\},\{vx,wy\},\{vy,wx\},\{vz,xy\}\}$. The observation that the Petersen graph is weakly $1$-factorizable can be generalized. Theorem. Let $G$ be an $n$-regular (finite or infinite) graph, $2\le n\lt\aleph_0$. (1) If $E(G)$ can be partitioned into $t$ maximal matchings, then $n\le t\le2n-1$. (2) $E(G)$ can be partitioned into $2n-1$ maximal matchings if and only if $G$ is a covering graph of the odd graph $O_n$ also known as the Kneser graph $K(2n-1,n-1)$. Corollary. For each $n\ge2$ the odd graph $O_n$ is weakly $1$-factorizable; in fact, the edges of $O_n$ can be partitioned into $2n-1$ maximal matchings. (Recall that $O_2=K_3$ and $O_3$ is the Petersen graph.) Remark. It's easy to construct a $3$-regular infinite connected graph $G$ such that $E(G)$ can not be partitioned into $3$ or $5$ maximal matchings; e.g., if $E(G)$ can be partitioned into $5$ maximal matchings, then $G$ must be triangle-free. But perhaps the edges of every $3$-regular infinite connected graph can be partitioned into $4$ maximal matchings?
2025-03-21T14:48:31.796359
2020-08-17T00:29:28
369348
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "https://mathoverflow.net/users/40120", "username" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632157", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369348" }
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Recommendation for books on boundary-value problems that include perturbed boundaries and many solved problems I am looking for a book or resource that contains applied math analytical methods and a lot of solved problems in Boundary-Value Problems for second-order PDEs, and if it could be related to wave-equation problems in 2D or 3D domains/structures, such as electromagnetics, it would be even better. I have found texts that have extensive treatment of perturbations (e.g. Bender and Orzag), but they mainly focused on initial value problems, not perturbations of the boundary geometry itself. I also like the book by Dudley, but it doesn't have much solved problems with perturbations. The classic tomes by Morse and Feshbach are also great, but have very few solved problems that help one to practice more. Ideally, something in the spirit of Schaum's outline series for solved problems would be perfect, but I couldn't find such a source yet for advanced boundary-value problems and/or perturbed boundaries. The aim is to practice more in this topic and develop my techniques in approximate analytical solutions for different boundary shapes and perturbations. Thanks for any advice! Non-Homogeneous Media and Vibratin Theory by Sanchez-Palencia has a number of examples, Here are 2 worked examples: Problems in PERTURBATION by Ali Hasan Nayfeh ISBN 0-471-82292-2, 15.36,15.41. Here is another non worked problem: Methods of Mathematical Physics, Jeffreys & Jeffreys, ISBN 0-521-09723-1.24.3. Here is another method of approximation: Calculus of Variations, Robert Weinstock, ISBN 0-486-63069-2.
2025-03-21T14:48:31.796483
2020-08-17T01:20:53
369350
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Dieter Kadelka", "Sina Baghal", "https://mathoverflow.net/users/100904", "https://mathoverflow.net/users/59151" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632158", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369350" }
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Does the following sequence $\{g_n\}$ converge? Consider a function sequence $\{f_n(t)\}$ ($n\in\mathbb{N}^+$) defined on the interval $(\frac{1}{2},1)$, where \begin{eqnarray}\label{eqn:constraint1} f_n(t)=\frac{\exp\left(n\left(\log R(h_t) - th_t\right)\right)}{h_t\sigma(h_t)\sqrt{2\pi n}}\left(1+o\left(1\right)\right) \end{eqnarray} as $n\to \infty$ uniformly in $t$ in any closed subinterval of the interval $(\frac{1}{2},1)$, \begin{eqnarray} R(h)=\frac{e^h-1}{h}, \end{eqnarray} \begin{eqnarray} m(h)=\frac{R'(h)}{R(h)}=\frac{e^h}{e^h-1}-\frac{1}{h}, \end{eqnarray} \begin{eqnarray} \sigma(h)=m'(h), \end{eqnarray} and $h_t$ is the positive real root of the equation $m(h)=t$. Then we study the following equation of $g_n$ \begin{eqnarray} c=g_n+\frac{\int_{g_n}^1f_n(t)~dt}{f_n(g_n)},~\frac{1}{2}< g_n<1, \end{eqnarray} where the constant number $\frac{1}{2}< c<1$. Try to prove: the convergence of the sequence $\{g_n\}$ and compute $A=\lim_{n\to \infty}g_n$; $\lim_{n\to \infty}n(g_n-A)=0$. did you try to code it up? for $o(1)$ just try different things. Can you tell where this problem comes from please. Sometimes a problem is more easily solved in its original form and then follows by applying an appropriate continuous map.
2025-03-21T14:48:31.796589
2020-08-17T01:22:21
369351
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Antoine Labelle", "Derek Holt", "Nourddine Snanou", "Qiaochu Yuan", "https://mathoverflow.net/users/123061", "https://mathoverflow.net/users/160416", "https://mathoverflow.net/users/290", "https://mathoverflow.net/users/35840" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632159", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369351" }
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On the number of structure of $F_p[G]$-modules Let $A$ be an abelian group and $G$ be a group. A short exact sequence of groups like $1\longrightarrow A\longrightarrow E\longrightarrow G\longrightarrow 1$ is called an extension. We say that $E$ is an extension of $A$ by $G$. This extension makes $A$ into a $G$-module. Assume that $A$ is an elementary abelian p-subgroup of rank $m$ and $G$ be an elementary abelian p-subgroup of rank $2$. Then $A$ is a $F_{p}[G]$ -module with $F_{p}$ is a finite field of $p$ elements. Obviously, this structure of $F_{p}[G]$-module is not unique. Thus, one can ask the following question: Question: What is the number of structure of $F_{p}[G]$-modules defined on $A$?. Any help would be appreciated so much. Thank you all. I don't see how the extension gives to $A$ a $G$-module structure. How do you define the action of $G$? @Antoine: the conjugation action of $E$ on $A$ factors through $G$ because $A$ is abelian. Dos this question have anything to do with group extensions? You seem to be just asking for the number of isomorphism classes of ${\mathbb F}_pG$-modules over ${\mathbb F}_p$, where $G$ is elementary abelian of order $p^2$. That is a question in representation theory. Yes, exactly. Thank you for your clarification. Can you explain why that is a question in representation theory. Are there similar questions in representation theory. @Nourddine Snanou Group representations of group $G$ over a field $K$ are the same as $K[G]$-modules and this is exaftly what representation theory studies. In your case $K$ is $\mathbb{F}_p$ and $G\cong (\mathbb{Z}/p\mathbb{Z})^2$. Since the characteristic of $K$ divides $|G|$, Maschke's theorem does not hold, so this is more precisely a question of modular representation theory. I don't know much about modular representation theory, though, so I can't help you more. There is a large number of distinct (up to module isomorphism) ${\mathbb F}_pG$-modules, even in small dimensions, and I don't think you can hope for any kind of reasonable classification. I did some quick computer calculations. For $p=2$, the number of isomorphism classes of dimensions 1,2, …, 7 is 1, 5, 11, 28, 53, 111, 199. For $p=3$ in dimensions 1,2,3,4,5, it is 1,6, 25, 78, 235. For $p=5$ in dimensions 1,2,3,4, it is 1,8,47,310. Of course, it is sufficient to classify the indecomposable modules, but ${\mathbb F}_pG$ is of so-called wild representation type, which means more or less that this cannot easily be done! For example, with $p=5$ in dimension 4, 242 of the 310 modules are indecomposable. On the other hand, I am by no means an expert in modular representation theory, and it is possible that more can be said.
2025-03-21T14:48:31.796782
2020-08-17T01:49:55
369355
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Mathman", "Ron P", "https://mathoverflow.net/users/153090", "https://mathoverflow.net/users/85550" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632160", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369355" }
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Independent sampling of dependent random variables Let $x_1, \ldots, x_n$ be possibly dependent random variables, each taking values $x_i \in \{0, 1, 2\}$. Suppose further that in every outcome the number of random variables that equal 2 is exactly 1. Now for each $i \in \{1, \ldots, n\}$ define $$ f_i = \begin{cases} \Pr[x_i = 2 \mid x_i \geq 1] & \text{if } x_i \geq 1\\ 0 & \text{if } x_i =0 \end{cases}, $$ and for each $i \in \{1, \ldots, n\}$ let $y_i$ be a Bernoulli random variable that is 1 independently with probability $f_i$ and 0 otherwise. Is the following conjecture correct or is there a distribution on $x_i$'s refuting it? Conjecture: There is a fixed $\epsilon > 0$ (i.e. $\epsilon$ being independent of $n$) such that with probability at least $\epsilon$, there is exactly one index $i$ where $y_i = 1$. Related question: Bounds on variance of sum of dependent random variables A clarifying question. Is the following description correct: first the $f_i$s are sampled, and then the $y_i$s are independent conditioned on the $f_i$s. @RonP Correct. The $f_i$'s are first drawn according to the distribution of $x_i$'s and are not independent. Then each $y_i$ is drawn independently according to probability $f_i$. The answer is "no" (if i understand the question correctly). Consider the following exchangeble joint distribution of the $x_i$s. In event $A$, which occur with probabiluty $1/\sqrt n$, all the $x_i$s are 1, exept for one 2. In the complement event $B$, all the $x_i$s are 0 exept for one 2. Under this distribution, $f_i$ is either 0 or $1/\sqrt n$. Let $Y=\sum y_i$. Since $E[ Y|A]=\sqrt n$, and $E[Y|B]=1/\sqrt n$, in either event it is too far from 1; therefore the probability that there is exaxtly one positive $b_i$ is vanishing. This is excellent Ron. Thank you!
2025-03-21T14:48:31.796933
2020-08-17T06:07:07
369363
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Christian Bueno", "Gerald Edgar", "Jochen Glueck", "Jochen Wengenroth", "Michael Greinecker", "https://mathoverflow.net/users/102946", "https://mathoverflow.net/users/12939", "https://mathoverflow.net/users/21051", "https://mathoverflow.net/users/35357", "https://mathoverflow.net/users/454" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632161", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369363" }
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Do distance functionals separate probability measures? Let $(\Omega,d)$ be a compact metric space and $\mathcal P(\Omega)$ its space of Borel probability measures. Let $D=\{ d_p\mid p\in\Omega\}$ where $d_p(x)=d(p,x)$ be the set of all "distance functionals". As usual, we can think of $D$ acting on $\mathcal P(\Omega)$ (or vice versa) via integration i.e. $\langle d_p,\mu\rangle = \int_\Omega d_p(x)\,\mathrm d\mu(x)$. Title Question Does $D$ acting on $\mathcal P(\Omega)$ via integration separate points? Or equivalently, If $\mu,\nu \in \mathcal P(\Omega)$ and $\langle d_p,\mu\rangle = \langle d_p,\nu\rangle$ for all $p\in \Omega$, then must $\mu=\nu$? Alternative Formulations There are a few other ways to frame the question as well. Probabilistic Formulation Rewriting all integrals as expectations the question becomes, If $\mathbb E_{X\sim\mu}[d_p(X)] = \mathbb E_{Y\sim\nu}[d_p(Y)]$ for all $p\in \Omega$, then must $\mu=\nu$? In other words, does knowing the expected distance to a point for all points determine the measure? Geometric Formulation Recall that the 1-Wasserstein distance is given by $W_1(\mu,\nu) = \inf_{\gamma\in\Gamma(\mu,\nu)} \int_{\Omega\times\Omega} d(x,y) \,\mathrm d\gamma(x,y)$ where $\Gamma(\mu,\nu)$ is the set of couplings between $\mu$ and $\nu$ i.e. Borel probability measures on $\Omega\times\Omega$ with marginals $\mu$ and $\nu$ respectively. Since the product measure $\delta_p\otimes\mu$ is the unique coupling between a Dirac delta measure $\delta_p$ and $\mu$, we have that $$W_1(\delta_p,\mu)=\int_{\Omega\times\Omega} d(x,y)\,\mathrm d(\delta_p\otimes\mu)(x,y)=\int_\Omega d(p,y)\,\mathrm d\mu(y)=\langle d_p,\mu\rangle$$ Now the question can be stated geometrically as If $W_1(\delta_p,\mu)=W_1(\delta_p,\nu)$ for all $p\in \Omega$, then must $\mu=\nu$? In other words, does knowing the $W_1$ distance to the extreme points of $\mathcal P(\Omega)$ completely determine the probability measure? Integral Transform Formulation Define the distance transform of $\mu\in\mathcal P(\Omega)$ as the function $\phi_\mu:\Omega\to\mathbb R$ given by $\phi_\mu(p) = \int_\Omega d(p,x)\,\mathrm d\mu(x)$. The question can now be restated as, Is the distance transform injective on $\mathcal P(\Omega)$? Moreover, by the geometric formulation we have $\phi_\mu(p) = W_1(\delta_p,\mu)$. We will use the weak-$*$ topology for $\mathcal P(\Omega)$ (which coincides with the $W_1$ topology). Since the map $p\mapsto \delta_p$ is an embedding of $\Omega$ into $\mathcal P(\Omega)$, it follows that $\phi_\mu:\Omega\to\mathbb R$ is continuous. Denote the distance transform by $\Phi(\mu)=\phi_\mu$. Since $\mathcal P(\Omega)$ is compact Hausdorff and $C(\Omega)$ is Hausdorff we can restate the question as If $\Phi:\mathcal P(\Omega)\to C(\Omega)$ is continuous, is it an embedding? Final Thoughts Are any of these equivalent statements true? I have unfortunately only been able to reformulate the question and have not identified any clear proof, though I wouldn't be surprised if there is an easy one I'm overlooking. The geometric formulation of the problem leads me to believe that $D$ does indeed separate points in $\mathcal P(\Omega)$. However, if the answer is affirmative then I feel the resulting nice properties of $\Phi$ would make it something that would be easy to look up. Any insight would be appreciated. Update: In light of George Lowther's elegant 4-point counter-example and Pietro Majer's affirmative answer for $\Omega=[0,1]$, it would be interesting to better understand what factors determine whether the underlying metric space yields an affirmative answer. George's counter-example can be extended to counter-examples where $\Omega$ is a sphere (with intrinsic metric). Thus, requiring $\Omega$ to be positive-dimensional, a manifold, connected, path-connected, simply-connected, etc, will not make the issue go away. On the other hand, Pietro suspects that the answer is again affirmative in the case when $\Omega$ is a compact convex subset of Euclidean space. I don't know the answer, but one can reduce the problem to the case that $\Omega$ is finite, since probability measures with finite support are weak*-dense. Isn't this a consequence of the Riesz representation theorem and the Stone-Weierstraß approximation theorem? But Stone-Weierstraß only implies that the algebra generated by the distance functionals is dense. One needs the weak$^*$ density of the generated subspace. @JochenWengenroth: You're right, of course. I confused a few arguments and thus thought this would not be a problem; but of course, it actually is a problem... See also this answer: https://math.stackexchange.com/a/90511/442 No. Suppose that $\Omega$ consists of four points arranged in a square, where adjacent points have distance 1 between them and opposite points have distance 2. Specifically, if the points are labeled A,B,C,D then \begin{align} & d(A,C)=d(B,D)=2,\\ & d(A,B)=d(B,C)=d(C,D)=d(D,A)=1. \end{align} For example, A,B,C,D could be equally spaced around a circle, using the internal circle metric. There are precisely two probability measures assigning probability 1/2 to each of two opposite points and probability zero to the remaining two points. \begin{align} & \mu(\{A\})=\mu(\{C\}) = 1/2,\ \mu(\{B,D\})=0,\\ & \nu(\{B\})=\nu(\{D\})=1/2,\ \nu(\{A,C\})=0. \end{align} You can check that these two measures give the same integral for all `distance functions'. The average distance from every point is equal to 1 under both of these. Elegant counter-example.This idea can be pushed a bit further as well though it was perhaps implicit: The empirical measure of a pair of antipodal points on the unit circle $\delta_p / 2+\delta_{-p} / 2$ produces an expected intrinsic distance of $\pi/2$ to any point on the circle. Thus, any two "empirical measure of antipodal points" cannot be separated by distance functionals. This shows the obstruction isn't simply due to finiteness of $\Omega$ and that the obstruction can't be avoided by requiring the underlying space to be nonzero dimensional, a manifold, connected, path-connected, etc. Actually, this idea using antipodal points works for spheres as well. Thus being simply-connected is not a sufficient condition to ensure an affirmative answer either. On the positive side, the answer is affirmative if $\Omega$ is the unit interval $[0,1]$ with its standard distance. In this case $\phi_\mu$ is a convex $1$-Lipschitz function (in fact, it is also defined for all $p\in\mathbb{R}$, with $\phi'(p)=\mathrm{sgn} p$ for $p\notin[0,1]$), with left and right derivatives $$\phi_-'(p)=\mu[0,p)-\mu[p,1]= 2\mu[0,p)-1$$ $$\phi_+'(p)=\mu[0,p] -\mu (p,1]= 2\mu[0,p] -1=1-2\mu(p,1]$$ so that $\mu$ is determined on all intervals, hence on all Borel subsets. Conversely, note that any convex function $\phi$ as above may be written in the form $\phi(p)=\int_{[0,1]}|t-p|dm(t)$ for some Borel probability measure $m$ on $[0,1]$. This because $g:= \frac{1}{2}\big(1-\phi_+'\big) $ is a nonnegative bounded cadlag function, so there is a Borel probability function $m$ such that $g(p)=m(p,1]$, whence $\phi(p)=\int_{[0,1]}|t-p|dm(t)$ follows easily from the above relations. I'd guess the answer is also affirmative for $\Omega$ a convex compact set of $\mathbb{R}^n$ with the Euclidean distance.
2025-03-21T14:48:31.797383
2020-08-17T07:43:30
369367
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632162", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369367" }
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Does strict convexity plus asymptotic affinity imply bounded mean? I am not sure if this is exactly research-level, but I am struggling to find a proof for the following claim: Let $F:[0,\infty) \to [0,\infty)$ be a $C^2$ strictly convex function. Let $\lambda_n \in [0,1],a_n\le c_0<b_n \in [0,\infty)$ satisfy $$ \lambda_n a_n +(1-\lambda_n)b_n=c_n $$ and suppose that $c_n \to c>c_0$. Set $D_n=\lambda_nF(a_n)+(1-\lambda_n)F(b_n)-F\big(c_n\big) $, and suppose that $\lim_{n \to \infty}D_n=0$ Question: Must $b_n$ be bounded? I have a quite simple proof (which I present below) for the special case where $a_n=a,c_n=c$ are constant sequences, but I am having trouble generalizing it. The proof for the simplified case: We have $ \lambda_n a +(1-\lambda_n)b_n=c$. Given $x \ge r$, let $\lambda(x) \in [0,1]$ be the unique number satisfying $$ \lambda(x) a +(1-\lambda(x))x=c. $$ We have $\lambda(b_n)=\lambda_n$. Define $$g(x) = \lambda(x) F(a) + (1-\lambda(x))F(x).$$ The strict convexity of $F$ implies that $g$ is a strictly increasing function of $x$. The assumption $D_n \to 0$ is equivalent to $g(b_n) \to F(c)$. Since $g(b_n) \ge F(c)$ (by convexity) and $g$ is strictly increasing, we conclude that $b_n$ must be bounded. Yes, $b_n$ must be bounded. Assume the contrary. Passing to a subsequence we may suppose that $a_n\to a$, $b_n\to \infty$. We have $$\lambda_n=\frac{b_n-c_n}{b_n-a_n}\to 1;\, 1-\lambda_n=\frac{c_n-a_n}{b_n-a_n}\sim (c-a)b_n^{-1},$$ and using $F(b_n)\geqslant F(c_n)+(b_n-c_n)F'(c_n)$ we get $$ D_n+F(c_n)=\lambda_n F(a_n)+(1-\lambda_n)F(b_n)\geqslant \lambda_n F(a_n)+(1-\lambda_n)F(c_n)+(1-\lambda_n)(b_n-c_n)F'(c_n)\\ \to F(a)+(c-a)F'(c)>F(c), $$ thus $\liminf D_n>0$.
2025-03-21T14:48:31.797509
2020-08-17T08:15:02
369372
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "InfiniteLooper", "JensVF", "LSpice", "MatthieuMeo", "https://mathoverflow.net/users/163866", "https://mathoverflow.net/users/163889", "https://mathoverflow.net/users/2383", "https://mathoverflow.net/users/86526" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632163", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369372" }
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When is the periodisation of a function continuous? Consider a function $f\in\mathcal{C}_0(\mathbb{R})$, where $\mathcal{C}_0(\mathbb{R})$ denotes the space of bounded continuous functions vanishing at infinity. I am interested in the $T$-periodisation of such a function, defined as: $$f_{T}(t)=\sum_{n\in\mathbb{Z}} f(t-nT),\quad \forall t\in \mathbb{R}.$$ As explained in Fischer - On the duality of discrete and periodic functions, $f_{T}$ is a $T$-periodic tempered distribution if $f$ is a rapidly decaying function —i.e. vanishing at infinity faster than any polynomial. My question concerns the regularity of $f_T$: For which functions $f\in\mathcal{C}_0(\mathbb{R})$ is the periodised generalised function $f_{T}$ defined above an ordinary, continuous function? In other words, what should be the assumptions on $f$ so that its periodisation is continuous? Any lead would be greatly appreciated. Thank you very much in advance! What would be the periodisation of a function equal for great $x$ to $\frac{1}{x}$ ? I do not believe it exists since this function is not rapidly decaying. You just need that $f$ decreases fast enough to make the series uniformly convergent on compact sets. E.g., it would be enough that $|x|^p |f(x)|$ is bounded for some $p>1$. Then you can estimate the terms of the series uniformly on a compact interval $[-a,a]$ for $nT>2a$ by $cn^{-p}$ with a constant $c$. Thank you for your answer. This seems indeed like a reasonable sufficient condition, but I am actually interested in a necessary and sufficient condition. I want to find the space of all functions whose periodisation is continuous. But this is probably too much to ask. Short answer: e.g. for Schwartz functions. Long answer: The Fourier transform of "periodic" is "discrete" and the Fourier transform of "discrete" is "periodic". This is a one-to-one mapping. It is explained in this Fischer - On the duality of discrete and periodic functions. Analogously, the Fourier transform of "regular" is "local" and the Fourier transform of "local" is "regular". It is another one-to-one mapping. It is explained in Fischer - On the duality of regular and local functions. The term "regular" refers to ordinary, infinitely differentiable functions which do not grow faster than polynomials. These (regular) functions are so-called multiplication operators for tempered distributions. Their multiplication product with any tempered distribution is again a tempered distribution. The term "local" refers to tempered distributions which are "local", i.e., they rapidly decay to zero (faster than polynomials). These (generalized) functions are so-called convolution operators for tempered distributions. Their convolution product with any tempered distribution is again a tempered distribution. The properties of "regular" and "local" fulfill a convolution theorem on tempered distributions. Now, properties of "periodic", "discrete", "regular" and "local" can be combined. For example, "local+regular" are Schwartz functions and the Fourier transform of Schwartz functions are, again, Schwartz functions ("local+regular"). Moreover, the Fourier transform of "discrete periodic" is again "discrete periodic". It yields the Discrete Fourier Transform (DFT). Now, the precondition for generalized functions which can be periodized is that they are "local" and the precondition for generalized functions which can be discretized is that they are "regular". So, back to the original question, in order to periodize an (ordinary or generalized) function, it must be "local" and in order to allow it to be an ordinary function it must be "regular". In other words, Schwartz functions fulfill these two requirements, they are "regular+local". This property of Schwartz functions of being "regular" and "local" simultaneously, explains their special role as test functions in distribution theory and in quantum physics. However, there is a difference of "being smooth" in the ordinary and in the generalized functions sense. One may recall, every generalized function is smooth (infinitely differentiable) and, hence, "continuous". To answer this question in the ordinary functions sense, embedded in generalized functions theory, there are more functions beside Schwartz functions. The rectangular function, for example, is smooth in the generalized functions sense but not smooth in the ordinary functions sense. Its periodization, however, yields the function that is constantly 1 for suitable T which is a smooth, ordinary function (in particular continuous). So, obviously, functions which are continuous on an interval [-T/2,+T/2] and such that f(-T/2)=f(+T/2) do also fulfill the requirement. Your first reference is to the paper that @MatthieuMeo mentions in their post. Hello, thank you very much for this very detailed and pedagogical answer. I agree with your statement that Schwartz functions are good candidates. However, as you point out yourself with the example of the rectangular function, there exists some functions much less regular whose periodisation is continuous. And this is precisely what I am interested in: I want to find the subspace of all functions whose periodisation is continuous. More precisely, I am interested in characterising the space: $$ \left{f\in\mathcal{C}_0(\mathbb{R}): \Delta\Delta\Delta_T f \in \mathcal{C}([0,T])\right}$$ Case 1: Let f be a function of compact support on [0,T]. Because f is already continuous, the only additional condition required is that f(T) = f(0). Case 2: Let f be a function of compact support but not on [0,T]. This case depends on "how often" the given interval "fits into" the interval [0,T] or vice versa. It's a matter of divisibility. In the worst case, the discontinuity is "sliding" such that only Schwartz functions fulfill the condition. Case 3: Infinite support: Only Schwartz functions fulfill the condition.
2025-03-21T14:48:31.797843
2020-08-17T08:19:37
369373
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Narutaka OZAWA", "https://mathoverflow.net/users/7591" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632164", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369373" }
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Sums over products over short paths in an expander graph Let $\Gamma=(V,E)$ be an undirected graph of degree $d$. (Say $d$ is a large constant and the number of vertices $n=|V|$ is much larger.) Let $W_0$ be the space of functions $f:V\to \mathbb{C}$ with average $0$. Assume $\Gamma$ is a strong expander graph, meaning that, for $A$ the adjacency operator $Af(w) = \sum_{\{w,v\}\in E} f(v)$ of $\Gamma$ restricted to $W_0$, all the eigenvalues of $A$ are considerably smaller than $d$. Say they are all $\leq 2 \sqrt{d}$, i.e., the graph is basically a Ramanujan graph. Then, by definition, for all $f\in W_0$ and $\sum_{v\in V} |f(v)|^2\leq n$, $$\left|\sum_{v_1,v_2\in V: \{v_1,v_2\}\in E} f(v_1) \overline{f(v_2)}\right| \leq 2\sqrt{d} \cdot n.$$ Is it possible to give a nontrivial upper bound on $$\left|\sum_{v_1,v_2,v_3\in V: \{v_1,v_2\},\{v_2,v_3\}\in E} f(v_1) f(v_2) f(v_3)\right|?$$ Assume that $f$ is real-valued and $|f|_\infty=1$, if it helps. (If yes: what about sums of longer products $f(v_1) f(v_2) \dotsc f(v_k)$, over $v_1,\dotsc,v_k\in V$ such that $\{v_1,v_2\},\dotsc,\{v_{k-1},v_k\}\in E$? Assume $k$ bounded. If no: what sort of auxiliary hypothesis might help?) Let's begin with the trivial upper bound $$|\sum f(v_1)f(v_2)f(v_3)|=|\sum_{v_2} (Af)(v_2)f(v_2)(Af)(v_2)| \le |f|_\infty|Af|2^2,$$ where $(Af)(v)=\sum{w: {w,v}\in E}f(v)$. I don't know your purpose, but here's some not-probably-sharp estimate that works for any $k$. Put $\rho=\frac{1}{d}\|A|_{({\mathbb C}1)^\perp}\|$ and $\gamma$ to be the positive root of $t^2-\rho t -\rho=0$. One has $\gamma<\sqrt{2\rho}<1$ when $\rho<\frac{1}{2}$. Then for any $f$ with $\sum f(v)=0$ and $\|f\|_\infty\le1$, one has $$\frac{1}{|V|\cdot d^{k-1}}\left|\sum_{v_1,v_2,\ldots,v_k : \{v_i,v_{i+1}\}\in E} f(v_1)\cdots f(v_k)\right| \le \gamma^k.$$ Proof. For $D:=\mathrm{diag}\,f \in B(\ell_2V)$ and $B:=\frac{1}{d}AD$, the LHS is $\frac{1}{|V|}|\langle B^{k-1}1_V,f\rangle|$. With respect to the orthogonal decomposition $\ell_2V={\mathbb C}1_V\oplus ({\mathbb C}1_V)^\perp$, one writes $B$ as an operator matrix $B=\left[\begin{smallmatrix} 0 & b \\ c & d \end{smallmatrix}\right]$, where $\| b\|\le 1$, $\|c\|\le\rho$, and $\|d\|\le\rho$. Hence for $C:=\left[\begin{smallmatrix} 0 & 1 \\ \rho & \rho \end{smallmatrix}\right] \in M_2({\mathbb R})$ with the eigenvalue $\gamma>0$ and the eigenvector $\left[\begin{smallmatrix} 1 \\ \gamma \end{smallmatrix}\right]$, one gets $$\frac{1}{|V|}|\langle B^{k-1}1_V,f\rangle| \le \left[\begin{smallmatrix} 0 & 1 \end{smallmatrix}\right] C^{k-1} \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right] \le \left[\begin{smallmatrix} 0 & 1 \end{smallmatrix}\right] C^{k-1} \left[\begin{smallmatrix} 1 \\ \gamma \end{smallmatrix}\right]=\gamma^k.$$ It's probably worth noting that the same proof shows $$\frac{1}{|V|}\sum_{v_1\in V}\left|\frac{1}{d^{k-1}}\sum_{v_2,\ldots,v_k : \{v_i,v_{i+1}\}\in E} f_1(v_1)\cdots f_k(v_k)\right|^2 \le 2\gamma^{2(k-1)}.$$
2025-03-21T14:48:31.798147
2020-08-17T09:34:43
369377
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "ABIM", "Dejan Govc", "https://mathoverflow.net/users/16447", "https://mathoverflow.net/users/36886" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632165", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369377" }
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Projecting Graph of a Function acted on by a homeomorphism Let $X,Y$ be compact, connected, simply-connected, and separable, metric spaces each with at-least $2$-points, and let $f,g:X\rightarrow Y$ be continuous functions. Does there always exist a homeomorphism $\Phi:X\times Y \rightarrow X\times Y$ such that $$ g(x) =\pi_Y\circ \Phi(x,f(x)) $$ for all $x \in X$, where $\pi_Y$ is the canonical projection of $X$ onto $Y$? Heuristically (but not exactly): "Can we always perturb the graph of a continuous function so that it becomes the graph of another function?" No. Take $I=[0,1]$, $Q=[0,1]\times[0,1]$ and $J=\{0\}\times[1,2]$. Let $X=I$ and $Y=Q\cup J$. Let $f:X\to Y$ be the constant function $f(x)=(0,\frac32)$ and let $g$ be the constant function $g(x)=(\frac12,\frac12)$. Then any homeomorphism $\Phi$ of $X\times Y$ will preserve the square $I\times J$, so $\pi_Y(\Phi(x,f(x)))\in J$ and therefore cannot be equal to $g(x)$. Added a bit later: The answer is negative even if you additionally assume that the two spaces are homogeneous. Take $X=Y=S^2$. Let $f:X\to Y$ be the identity map of $S^2$ and let $g:X\to Y$ be a constant map taking the value $c\in S^2$. Then the requirement $g(x)=\pi_Y(\Phi(x,f(x)))$ becomes $\pi_Y(\Phi(x,x))=c$. In other words, $\Phi(x,x)=(h(x),c)$ where $h:S^2\to S^2$ is a homeomorphism. But $(S^2\times S^2)\setminus\Delta$ (here $\Delta$ is the diagonal) and $(S^2\times S^2)\setminus (S^2\times\{c\})$ are not homeomorphic, so there is no such $\Phi$. What if $X=Y=\mathbb{R}^n$? Would things workout better? Yes, then you can take $\Phi(x,y)=(x,y-f(x)+g(x))$. You could also e.g. replace $\mathbb R^n$ by a Lie group in which case $\Phi(x,y)=(x,yf(x)^{-1}g(x))$ would do the job. Do we really need continuity of both $f$ and $g$ in this case.. I guess not? @James_T: Well, mainly to ensure $\Phi$ is continuous. But if the discontinuities of $f$ and $g$ "cancel out", this still works, I guess.
2025-03-21T14:48:31.798297
2020-08-17T09:41:43
369378
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Tom Copeland", "https://mathoverflow.net/users/12178" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632166", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369378" }
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More or less universal formula for regularization of divergent integrals? Is there a simple formula that would produce the regularized value for the most common divergent integrals? I know, there is a formula for Cesaro integration, but it is applicable only to Cesaro-summable integrals. Other formulas for regularization essentially convert the integral into a series and regularize that series. What I am looking for is an integral analog of Faulhaber's formula for Ramanujan's summation: $$\operatorname{reg} \sum _{n=0}^{\infty} f(n)= -\sum_{n=1}^{\infty} \frac{f^{(n-1)} (0)}{n!} B_n $$ Or even its more universal variant: $$\operatorname{reg} \sum _{n=0}^{\infty} f(n)=\frac{f(0)}{2}+i \int_0^{\infty } \frac{f(i t)-f(-i t)}{e^{2 \pi t}-1} \, dt$$ Yes, it is also not universally applicable, but it is universal enough. Is there an analog of this formula, but for integrals? UPDATE I came to the following formula: $$\operatorname{reg}\int_0^\infty f(x)\,dx=\frac{i}{2} \int_0^{\infty } \frac{f(i t)-f(-i t)}{1-e^{-2 \pi t}} \, dt-\frac{i}{2} \int_0^{\infty } \frac{f(i t)-f(-i t)}{e^{2 \pi t}-1} \, dt$$ Unfortunately, the first term tends not to converge. The formula works though for integral $\int_0^\infty e^x dx$ (giving $-1$ as expected). Have you looked at "New summation and transformation formulas of the Poisson, Müntz, Möbius and Voronoi type" by Semyon Yakubovich? https://arxiv.org/abs/1410.7934
2025-03-21T14:48:31.798416
2020-08-17T09:55:25
369379
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Alexandre Eremenko", "Francesco Polizzi", "Michael Albanese", "aglearner", "gaoqiang", "https://mathoverflow.net/users/13441", "https://mathoverflow.net/users/161514", "https://mathoverflow.net/users/21564", "https://mathoverflow.net/users/25510", "https://mathoverflow.net/users/7460" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632167", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369379" }
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Elliptic, parabolic and hyperbolic Riemann surfaces: classification? In the book of Kra and Farkas on Riemann surfaces the following (slightly unusual) definition is given: Definition IV.3.2 (Section IV.3). Let $M$ be a Riemann surface. We will call $M$ elliptic if and only if $M$ is compact. We will call $M$ parabolic if and only if $M$ is not compact and $M$ doesn't carry a non-negative subharmonic function. We will call $M$ hyperbolic if and only if $M$ does carry a negative non-constant subharmonic function. Question. Is there some geometric way to characterise parabolic and hyperbolic surfaces? For example, suppose $M$ is a compact Riemann surface and $x_1,\ldots, x_n$ are points on it. Is the surface $M\setminus \{x_1,\ldots, x_n\}$ parabolic? Strange definition, indeed. Usually, compact Riemann surfaces of genus $\geq 2$ are called hyperbolic, because they are uniformized by the upper half-space (or, equivalently, by the disk). Should the second occurrence of $M$ in each of the defining sentences be $\widetilde{M}$, the universal cover of $M$? No Michael, I've just copied the definition from the book. And I am certain that this is exactly what they meant. This is somewhat unusual terminology, but it is common in the theory of classification of open Riemann surfaces. The more standard notation is $P_G$ for "parabolic", and $O_G$ for "hyperbolic". The surface $M\backslash\{ x_1,\ldots,x_n\}$ is parabolic in this sense, by the "removable singularity theorem" (a subharmonic function which is bounded from above in a punctured neighborhood of the point extends to a subharmonic function in the full neighborhood). There are some criteria, especially, for surfaces of the form $M\backslash E$, where $M$ is compact and $E$ is a closed subset. But these criteria are not very geometric: they use capacity. Some results can be given in terms of Hausdorff measures of $E$ but they are not "necessary and sufficient". Classical results can be found in the books M. Tsuji, Potential theory in modern function theory, Maruzen, Tokyo, 1959 (there is an AMS reprint). Ahlfors, Sario, Riemann surfaces, Princeton UP, 1960. Can you give me some references on the "removable singularity theorem" for subharmonic functions? I can only find references on this theorem for harmonic functions. L. Carleson, Selected Problems on Exceptional Sets.
2025-03-21T14:48:31.798603
2020-08-17T10:50:55
369383
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Jens Reinhold", "Tyrone", "https://mathoverflow.net/users/14233", "https://mathoverflow.net/users/54788" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632168", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369383" }
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Hopf invariants of elements from spherical fibrations Let $G_n$ be the space of homotopy-equivalences of $S^{n-1}$. Evaluation produces a map $G_{n} \to S^{n-1}$. For $n = 2m+1$, I would like to understand the induced map on $\pi_{4m-1}$. More precisely, what I would really like to know are the Hopf invariants of elements in the image of $\pi_{4m-1} G_{2m+1} \to \pi_{4m-1} S^{2m}$. For $m = 1$, the classical Hopf fibration gives rise to a generator of $\pi_3 S^2 = \bf Z$, and this element even lies in the image of $\pi_3 SO(3) \to \pi_3S^2$ which factors through $\pi_3 G_3 \to \pi_3 S^2$. For $m = 2$, I do not know the answer, but I can prove that every map lying in the image of $\pi_7 SO(5) \to \pi_7S^4 \cong {\bf Z} \oplus {\bf Z}/12\bf Z$ has Hopf invariant divisible by $12$. Moreover, the question whether $1$ is attained might be equivalent to asking whether ${\bf H}P^3$ is homotopy equivalent to the total space of a fibration with fiber $S^4$ and base $S^8$. I do not know what happens for $m \geq 3$. The case $m = 4$, where $\pi_{15} S^8$ contains an element of Hopf invariant 1, would be especially interesting. Maybe there is also a relation to the non-existence of ${\bf O}P^3$? The fibre of the evaluation map $e:G_n\rightarrow S^n$ is $\Omega^nS^n$. The fibration connecting map $\partial:\Omega S^n\rightarrow\Omega^nS^n$ acts on homotopy groups to send $\alpha\in\pi_k\Omega S^n\cong\pi_{k+1}S^n$ to the Whitehead product $[\alpha,\iota_n]\in\pi_{n+k}S^n\cong\pi_{k}\Omega S^n$. This was worked out by G. Lang in The evaluation map and EHP sequences, Pacific J. Math. Volume 44, Number 1 (1973), 201-210. See the last section, where the interactions between the evaluation sequence and James's EHP sequence are discussed. This is very interesting, thanks! So to fully answer my question one needs to determine the order of $[e,\iota] \in \pi_{6m-2} S^{2m}$, where $e \in \pi_{2m}S^{2m}$ is a generator and $\iota \in \pi_{4m-1} S^{2m}$ generates $\pi_{4m-1} S^{2m} / \text{torsion} \cong \bf Z$, right? Tyrone, would you be interested in converting (and possibly expanding) your comment into an answer?
2025-03-21T14:48:31.798757
2020-08-17T14:21:40
369396
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632169", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369396" }
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Laplacian of Fourier-like function The question that follows has to do with the effects of a turbulent atmosphere on wave propagation. The structure function, $D(\vec{r})$, which is defined as, \begin{equation} D(\vec{r}) = \left\langle \left[ n(\vec{x} - \vec{d}) - n(\vec{x}) \right]^2\right\rangle \, , \end{equation} and is what our instruments observe, is related to the power spectrum, $\Phi(\vec{k})$, thought \begin{equation} D(\vec{r}) = 2 \int_0^{\infty} \int_0^{\infty} \int_0^{\infty} \Phi(\vec{k}) [1 - \cos(\vec{k} \cdot \vec{r})] d^3k \label{1}\tag{1} \end{equation} According to the textbook I am reading on the topic, I can obtain the inverse of \eqref{1} by operating with a Laplacian on it, which results in, \begin{equation} \Phi(\vec{k}) = \frac{1}{16 \pi^3 k^2} \int_0^{\infty} \int_0^{\infty} \int_0^{\infty} \cos(\vec{k} \cdot \vec{r}) \, \Delta D(\vec{r}) d^3r \label{2}\tag{2} \end{equation} where $\Delta$ is the Laplace operator. Under the assumption of a locally isotropic field, I can introduce spherical coordinates and integrate over the angular variables to obtain \begin{equation} \Phi(k) = \frac{1}{4 \pi^2 k^2} \int_0^{\infty} \frac{\sin(kr)}{kr} \frac{d}{dr} \left[ r^2 \frac{d}{dr} D(r) \right] dr \label{3}\tag{3} \end{equation} Objective: Derive the equivalent formula to Eq. \eqref{3} in the 2D case. Question: My calculus is a bit rusty so I am struggling to work out how to derive Eq. \eqref{2} from Eq. \eqref{1}. I assume Eq. \eqref{2} above is for a 3D field and the relation will be different in the 2D case (correct me if I am wrong on this). If I have Eq. \eqref{2} in 2D, I can easily derive Eq. \eqref{2} by introducing polar coordinates.
2025-03-21T14:48:31.798887
2020-08-17T15:34:00
369400
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Iosif Pinelis", "OmarR", "https://mathoverflow.net/users/148279", "https://mathoverflow.net/users/36721" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632170", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369400" }
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Number of duplicate pairs in multiple samplings My universe has M different items. I run m=10 independent samplings over M. In each sampling, n elements are picked without replacement (n<<M). What is the expected number of pair duplicates we shall get across the m independent samplings? I know that the universe has M(M-1)/2 distinct pairs, and in each sampling one combination among nCM is selected, still the exact probability formula does not seem straightforward. What do you mean by a pair duplicate? Any pair of items (i,j) that is found more than once across samplings. Could you please explain how the probability $9(n^2-n)/(M^2-M)$ is obtained because this formula does not make a distinction between sampling with and without replacement (I guess). @OmarR. : Does "any pair of items (i,j) that is found more than once across samplings" mean "any subset of cardinality $2$ of the set ${1,\dots,M}$ that is found in more than one sampling"? @losifPinelis. Exactly! Thanks for clarifying. The answer is $$\binom M2(1-m p q^{m-1}-q^m),$$ where $$p:=\frac{n(n-1)}{M(M-1)},\quad q:=1-p.\tag{0}$$ Details: Fix any "pair" -- that is, any subset $a\subseteq[M]:=\{1,\dots,M\}$ of cardinality $2$. For each $i\in[m]$, let $S_i$ denote the $i$th random sampling, that is, the $i$th random set of size $|S_i|=n$, selected uniformly at random from the set $[M]$. The random sets $S_1,\dots,S_m$ are independent and identically distributed (iid). Consider now the $m$ iid trials with the success in the $j$th trial defined as $a\subseteq S_j$, for $j\in[m]$, so that that the success probability in each trial is $p$, with $p$ as defined in (0). Then the event -- say $B(a)$ -- that $a$ is contained in more than one of the random sets $S_1,\dots,S_m$ is the event of $\ge2$ successes in the $m$ trials, and hence $$P(B(a))=P:=1-m p q^{m-1}-q^m,$$ with $q=1-p$, as defined in (0). It remains to note that the number of the "pair duplicates", that is, the random number of subsets of cardinality $2$ of the set $[M]$ that are contained in more than one of the sets $S_1,\dots,S_m$ is $$N:=\sum_{a\in\binom{[M]}2}1_{B(a)}$$ and hence $$EN=\binom M2 E1_{B(a)}=\binom M2 P,$$ as claimed.
2025-03-21T14:48:31.799044
2020-08-17T16:16:42
369403
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Ben Howard", "Jason Starr", "https://mathoverflow.net/users/13265", "https://mathoverflow.net/users/163886" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632171", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369403" }
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Extensions of (semi-)abelian schemes Let $S$ be a regular Noetherian scheme, and let $U\subset S$ be the complement of a divisor. If $A\to B$ is an isogeny of abelian schemes over $U$, and $A$ extends to a semi-abelian scheme over $S$, does $B$ also extend to a semi-abelian scheme over $S$? I'm happy to assume that the degree of the isogeny is invertible in $\mathcal{O}_S$. Welcome new contributor. The answer I wrote before was wrong. I will try to write again when I have more time . . . Under the hypothesis that the degree of the isogeny is invertible in $S$, the answer is yes; I attempted to sketch a proof below under slightly more general assumptions on the base. [On the other hand, if the degree of the isogeny is not invertible in $S$, the answer is no. There are counterexamples over regular rings of dimension $2$. You can look at page 192 of the book by Faltings-Chai for a counterexample in equal characteristic $p$; or to chapter 6 of De Jong-Oort's "On extending families of curves" for an example in mixed characteristic.] Statement: let $S$ be a normal noetherian integral scheme, $U\subset S$ a non-empty open, $f\colon A\to B$ an isogeny of $U$-abelian schemes, of degree $m$ invertible on $S$. Suppose $A$ extends to a semiabelian scheme $\mathcal A/S$; then $B$ extends to a semiabelian scheme $\mathcal B/S$, and $f$ extends to an isogeny $\mathcal A\to \mathcal B$ (i.e. a fibrewise finite surjective homomorphism). Proof: the $m$-torsion $\mathcal A[m]$ is a quasi-finite etale group scheme over $S$, and the inclusion $\mathcal A[m]\to \mathcal A$ is a closed immersion, as semiabelian schemes are separated. Let $K/U$ be the kernel of $f$, a finite etale group scheme. The closed immersion $K\to A[m]$ is etale hence an open immersion. We can therefore write $A[m]$ as a disjoint union of schemes $K\sqcup Z$. Claim: the schematic closure of $K$ inside $\mathcal A[m]$ (or $\mathcal A$), is etale over $S$. Proof of claim: Write $\mathcal A[m]$ as the disjoint union $Y_1\sqcup Y_2\sqcup \ldots Y_n$ of its connected components. As $S$ is geometrically unibranch, so is $\mathcal A[m]$, hence each $Y_i$ is also an irreducible component. In particular, the restrictions $Y_{1 ,U},\ldots,Y_{n,U}$ are the irreducible (and connected) components of $A[m]$, and by reordering the indexes we can write $K=Y_{1,U}\sqcup \ldots \sqcup Y_{k,U}$, $Z=Y_{k+1,U}\sqcup\ldots\sqcup Y_{n,U}$. It follows that $\mathcal A[m]$ is the disjoint union of $\overline K$ and $\overline Z$, the schematic closures of $K$ and $Z$. This proves the claim. It is clear that the multiplication map of $\mathcal A$ maps $\overline K\times \overline K$ inside $\overline K$, hence $\overline K$ is a subgroup of $\mathcal A$. The sheaf quotient $\mathcal B=\mathcal A/\overline K$ is a smooth group algebraic space extending $B$; it is separated because quotient of a separated group scheme by a closed subgroup; one argues easily that the fibres are semiabelian varieties. It remains to show that it is a scheme; here you can apply theorem XI 1.13 from Raynaud's "Faisceaux amples sur les schemas en groupes et les espaces homogenes" which shows that $\mathcal B/S$ is quasi-projective. It is here that the hypothesis that $S$ is normal, noetherian, integral is used. Thanks, G_Orecchia! I realized this argument myself a few days ago, but I didn't know about the counterexamples when the degree is not invertible. Very interesting.
2025-03-21T14:48:31.799622
2020-08-17T16:33:11
369405
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Manfred Weis", "Nate Eldredge", "https://mathoverflow.net/users/31310", "https://mathoverflow.net/users/4832" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632172", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369405" }
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Directed vertex-disjoint 3-cycle covers from Windy Postman tours? It is a result of Zaw Win (paywall) that optimal Windy Postman tours in eulerian digraphs can be calculated in polynomial time. Bodo Manthey has shown that directed 3-Cycle Covers are APX hard to approximate. It is a trivial observation that weighted digraphs can be interpreted as instances of the Windy Postman Problem by combining pairs of antiparallel arcs into an undirected edge, whose weight depends on the direction of traversal. The advent of undesirable 2-cyles can then be ruled out by taking the 3-Cycle Cover of the optimal Windy Postman tour instead of on the complete digraph with its antiprallel arcs as a compromise. For the purpose of finding good(?) directed 3-Cycle Covers efficiently any non-eulerian complete instance can be turned into an eulerian one via a variant of the 17 camels trick by adding $3$ vertices $x,y,z$ and setting the weights of all adjazent edges to $INF$ and the weight of each edge on the subtour $(x,y,z,x)$ to $0$ if the cover of minimum weight is sought. The so modified instance will replicate the original cycle cover plus $(x,y,z,x)$, so the original solution can be easily recovered. Question: what is can be said about the quality of 3-Cycle Covers obtained by removing antiparallel arcs via the polynomial-time eulerian Windy Postman algorithm of Zaw Win for arcweights retricted to $\lbrace 0,1\rbrace$ or drawn uniformly distributed in $[0,1]$ assuming $n=2k+1$? Every time I see a reference to this problem, my brain hears "windy" with a short i, and I imagine a postman who is either talkative, cowardly, or afflicted with digestive problems... When I first encountered the Chinese Postman Problem I thought it was about a problem that only postmen in China have.
2025-03-21T14:48:31.799779
2020-08-17T16:53:27
369407
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Antonio", "R. van Dobben de Bruyn", "https://mathoverflow.net/users/163887", "https://mathoverflow.net/users/82179" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632173", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369407" }
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When is there a path between two minimal prime ideals? Recall a path from a point $x$ to a point $y$ in a topological space $X$ is a continuous function $f$ from the unit interval $[0,1]$ to $X$ with $f(0) = x$ and $f(1) = y$. Now let $R$ be a commutative ring with $1$ and $Spec(R)$ be the set of all prime ideals of $R$ with the Zariski topology. It is well-known that if $P$ and $Q$ are two prime ideals of $R$ with $P\subseteq Q$, then there exists a path from $P$ to $Q$. Now let $\frak{p}_1$ and $\frak{p}_2$ be two minimal prime ideals of $R$. I am looking for equivalent conditions under which there exists a path between $\frak{p}_1$ and $\frak{p}_2$. Shouldn't this just be equivalent to the existence of a zig-zag of specialisations from one to the other? @R. van Dobben de Bruyn: as you stated, if there exists a finite zig-zag of specializations from one to the other we are done. But does the existence of a path imply a finite zig-zag of specializations from one to the other? Also, infinite zig-zag of specializations (especially uncountable) is also a problem.
2025-03-21T14:48:31.799874
2020-08-17T18:11:39
369409
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Qiaochu Yuan", "https://mathoverflow.net/users/290", "https://mathoverflow.net/users/40804", "mme" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632174", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369409" }
Stack Exchange
Riemannian version of topological $K$-theory Let $X$ be a compact Hausdorff space.Put $Vec(X)$, the space of all real (or complex) vector bundles over $X$.We put also $Vec_g(X)$, the space of all Riemannian vector bundles over $X$, that is the space of all vector bundles equiped with a continuous fiber wise inner product. By imitation of three equivalent relations defined on $Vec(X)$ in the theory of topological $K$-theory, we define a metric version of the same equivalent relations as follows: 1)Two Riemannian vector bundles $(E,g)$ and $(F,h)$ are equivalent if there is isometric bundle isomorphism between them. 2)Two Riemannian vector bundles $(E,g)$ and $(F,h)$ are equivalent if there is a trivial bundle $\epsilon_k$ with the obvious metric structure such that $E\oplus \epsilon_k$ is equivalent to $F \oplus \epsilon_k$ in the sense of 1) where each direct sum bundle is equipped with direct sum.metric 3)Two Riemannian vector bundles $(E,g)$ and $(F,h)$ are equivalent if there are two trivial bundles $\epsilon_m$ and $\epsilon_n$ with obvious metrics such that $E\oplus \epsilon_m$ is equivalent to $F \oplus \epsilon_n$ in the sense of 1) where each direct sum bundle is equipped with direct sum.metric. Does this idea produce a new kind of $K$_theory? Does it introduce a functor from the category of compact Hausdorff space to the category of Groups? Can we extend this functor from the category of commutative unital $C^*$ algebras to the category of non commutative unital $C^*$ algebras. Two Riemannian vector bundles are (1)-equivalent iff they are isomorphic as vector bundles, so you can get nothing new from this. One way of thinking of this is the fact that a Riemannian vector bundle is classified by a map to $BO(n)$, while a vector bundle is classified by a map to $BGL(n)$. Because the inclusion $O(n) \to GL(n)$ is a homotopy equivalence the induced map on classifying spaces is too. Said another way, the space of Riemannian metrics on a given vector bundle is convex, hence contractible. So the natural forgetful map from Riemannian vector bundles to vector bundles has contractible fibers.
2025-03-21T14:48:31.800010
2020-08-17T18:36:59
369410
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Giorgio Metafune", "https://mathoverflow.net/users/147835", "https://mathoverflow.net/users/150653", "rihani" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632175", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369410" }
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Spectrum of Laplace-Beltrami with piecewise constant coefficients By the Laplace-Beltrami with piecewise constant coefficients I means the operator $-div (f\, \nabla .)$ in the 2-sphere. Where $f$ is a piecewise constant function that takes two values $1$ and $a>0$. The spectrum of such operator is known to be discret. Except for the case $a=1$, i can't find any work related to the computation of the eigenvalues and their multiplicity. Is there any work in this direction? I doubt that there is something published, but one never knows. The problem is only apparently simple, since the geometry of the set $E={f=a}$ enetrs and I believe it is unsolvable in general. Possibly, some assumption on $E$ (it could be an interval with respect to an angular coordinate) could allow to do the computation by separation of variables. Sorry, but I do not know more. Yes, in my case of interest E is interval with respect to an angular coordinate. Did you try the case of $S^1$ first (with $E$ being an interval)? Could be a nice warm-up. Yes, the case of $S^1$ (with E being an interval) can be done. we solve the equation in E and $S^1\backslash E$ separably. In these expressions two parameters have to be found. using the continuity and the continuity of the flux we can find a relation between a and $\lambda$ ( the eigenvalue). we can show that all the eigenvalues,, except 0, are simple.
2025-03-21T14:48:31.800134
2020-08-17T20:16:05
369414
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "André Henriques", "Chris Ramsey", "Qiaochu Yuan", "https://mathoverflow.net/users/290", "https://mathoverflow.net/users/5690", "https://mathoverflow.net/users/76593" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632176", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369414" }
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Characterisation of finite dimensional C*-algebras? $\DeclareMathOperator\Spec{Spec}$Let $A$ be a finite dimensional $*$-algebra over $\mathbb C$. (Namely, an associate algebra equipped with an involution $*:A\to A$ satisfying $(ab)^*=b^*a^*$ and $(\lambda a)^*=\bar\lambda a^*$.) Assume that for $\forall a\in A$ we have $\Spec(a^*a)\subset\mathbb R_+$. Does it follow that $A$ is a C*-algebra? Here, the spectrum $\Spec(x)$ of an element $x$ is the set of scalars $\lambda\in \mathbb C$ such that $x-\lambda$ is not invertible. I suppose you mean "does there exist a norm such that $A$ is a C$^*$-algebra?". @Chris: the norm is completely determined by the underlying *-algebra (if it exists); the grammar of the question is fine. @QiaochuYuan Yes, I agree with you. I was making the point that the algebraic structure can be paired with many other norms which do not yield C$^*$-algebras. @Ruy Indeed, the algebra $\mathbb C[x]/x^2$ with conjugation $(a+bx)^*:=\bar a+\bar b x$ is a counterexample. You may post it as an answer and I will accept it. Let $V$ be a complex vector space equipped with an involutive anti-linear star operation (e.g. a C*-algebra whose multiplication has been forgotten). Equip $V$ with the identically zero multiplication, namely $xy=0$ for all $x$ and $y$ in $V$. Then the unitization of $V$ is a counter-example. In fact, every element $a$ of $V$ is nilpotent so $\text{spec}(a) = \{0\}$. Consequently the spectrum of any element of the form $a-\lambda$ is $\lambda$ from where one easily checks the required condition. However $a^*a=0$ for every $a$ in $V$, so $\tilde V$ cannot possibly be a C*-algebra.
2025-03-21T14:48:31.800279
2020-08-17T20:37:52
369416
{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Eoin", "Gjergji Zaimi", "TomGrubb", "Will Sawin", "https://mathoverflow.net/users/106264", "https://mathoverflow.net/users/18060", "https://mathoverflow.net/users/2384", "https://mathoverflow.net/users/65919" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632177", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369416" }
Stack Exchange
Why should the number of $\mathbb{F}_q$ points on degree $d$ curves $C\subset \mathbb{P}_{\mathbb{F}_q}^n$ decrease as $n$ increases? This question concerns some counterintuitive results (to me at least) regarding the number of points on a projective curve over a finite field. Namely, if one fixes the degree of the curve, but increases the dimension of the ambient projective space, one can get tighter bounds on the number of $\mathbb{F}_q$ points on the curve, despite there being a larger number of $\mathbb{F}_q$ points in the ambient space. Let me make this more precise with two examples. Let $C\subset \mathbb{P}^n_{\mathbb{F}_q}$ be a projective curve of degree $d$. Suppose $C$ is nondegenerate in the sense that it is not contained in any smaller projective space $\mathbb{P}^k_{\mathbb{F}_q}$, $k<n$. Work of Homma (extending work of Homma and Kim) has shown $$ \#C(\mathbb{F}_q)\leq (d-1)q+1, $$ with a single exception (up to isomorphism) over $\mathbb{F}_4$. This is the so called Sziklai bound, and is tight for $n=2$. This bound is not tight for $n>2$; recently Beelen and Montanucci show that if $C\subset \mathbb{P}^3_{\mathbb{F}_q}$ is nondegenerate then in fact $$ \#C(\mathbb{F}_q)\leq (d-2)q+1. $$ They further conjecture than if $C\subset \mathbb{P}^n_{\mathbb{F}_q}$, the general bound should be $$ \#C(\mathbb{F}_q)\leq (d-n+1)q+1. $$ This is reminiscent of a phenomenon from work of Bucur and Kedlaya. For example: a random smooth curve in $\mathbb{P}^2_{\mathbb{F}_q}$ is expected to have $$q+1$$ points over $\mathbb{F}_q$ as its degree grows to infinity. A random complete intersection of two smooth degree $d$ surfaces in $\mathbb{P}^3_{\mathbb{F}_q}$ is expected to have $$ q+1 - \frac{q^{-2}(1+q^{-1})}{1+q^{-2}-q^{-5}} < q+1 $$ points over $\mathbb{F}_q$, again as $d\to\infty$. These results are counterintuitive to me, as the number of points in the ambient projective space grows (exponentially) as $n$ does, so in particular it seems to me that it should be easier for curves to have $\mathbb{F}_q$ points when they are embedded in larger projective spaces. Does anyone have any intuition as to why the opposite should be true? References: Beelen and Montanucci: A bound for the number of points of space curves over finite fields Bucur and Kedlaya: The probability that a complete intersection is smooth Homma: A bound on the number of points of a curve in projective space over a finite field Intuitively, if you fix the degree and assume the curve is nondengerate in a projective space of some dimension, and then increase the dimension, I would assume that the genus of your curve decreases (this is related to the Castelnuovo bound). E.g., A cubic curve in $\Bbb{P}^2$? Elliptic curve. A cubic curve in $\Bbb{P}^3$? Twisted cubic. Of course, weird things happen over finite fields but if the field is large enough I think the intuition would hold. One way to get some intuition comes from looking at the (weaker) combinatorial bound. Suppose you had a nondegenerate curve $C$ in some projective space $\mathbb P^n$. Suppose that that $L$ is a subspace of codimension $2$ in $\mathbb P$ and that $|C\cap L|=m$. The higher the dimension $n$ gets, the higher value we are allowed to pick for $m$. Indeed we can always find at least $n-1$ points in $C$ that span a $\mathbb P^{n-2}$. Bezout tells you that for any hyperplane $H$ that contains $L$, the number of points of $C$ that lie in $H$ and don't lie in $L$ is at most $d-m$. Since the number of such hyperplanes is $q+1$, independent of the dimension, we get $|C|-m\le (q+1)(d-m)$ or equivalently from rearranging terms $$|C|\le (d-m)q+d.$$ For $m=n-1$ this gives the bound $|C|\le (d-n+1)q+d$ for all nondegenerate curves $C$. Of course this is weaker than the conjecture and theorems that you mention in the post, but (1) it holds true for all curves including the one that violates the Sziklai bound (2) it already exhibits the phenomenon "bound gets tighter as $n$ goes up". Thanks for your answer Gjergji. I'd like to leave the question up just a bit longer to see if it attracts any other discussion, but this does help. Is this really much weaker? It seems only a little bit weaker to me. @WillSawin I changed it to just say weaker. The papers above do some work to really refine this bound and settle the conjecture in dimensions $2,3$. So the problem gets harder the higher $n$ (and therefore $d$) gets, but that's where this bound starts getting worse.