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2025-03-21T14:48:31.023134
| 2020-05-19T19:07:41 |
360810
|
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"authors": [
"LSpice",
"Ville Salo",
"YCor",
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|
Stack Exchange
|
QI-closure of $\mathrm{NA}\times\mathrm{NA}$
Suppose we know the following about a class of groups $\mathcal{G}$.
If $G$, $H$ are f.g. r.p. nonamenable groups, then $G \times H \in \mathcal{G}$.
If $G \in \mathcal{G}$, $G$ is f.p., and $G$ is quasi-isometric to $H$, then $H \in \mathcal{G}$.
Here, f.g. means finitely generated, f.p. means finitely presented, r.p. means recursively presented.
What interesting/surprising/non-trivial corollaries can we deduce?
The motivation for this question is that I have a class for which I believe I can prove 1 and 2, so I'd like to know what interesting things my class contains just based on that. I already know a very interesting corollary, namely that my class contains a simple group Burger-Mozes. Unbelievable! Are there other surprises?
If it simplifies things, don't get too hung up on the r.p. and f.p. details. For instance if you can say something about the title, i.e. QI-closure of $\mathrm{NA}\times\mathrm{NA}$.
There appear to be at least three (1 2 3) "Burger–Mozes"s. Which one is the relevant one here?
Axiom 2 sounds weirdly stated, since $H$ QI to $G$ fp implies $G$ fp. So you have an isomorphism-closed class $\mathcal{G}$ of groups such that its subclass of fp elements is QI-closed, and contains all direct products of 2 fg rp non-amenable groups. The smallest class satisfying thus consists of (a) the fp groups QI to such a fp product (b) the rp fg direct products of two non-amenable groups. Probably the last formulation of the question is better
@LSpice in this case it probably means simple f.p. groups that occur as acting geometrically on a product of two trees. Other interesting groups QI to a product of non-amenable groups (and not virtually directly decomposable) are various irreducible lattices in products of Lie or non-archimedean groups. Rataggi also produced groups QI to a product of two bushy trees, that have no proper finite index subgroups, but have many normal subgroups of infinite index.
BTW "interesting" obviously depends on how your class is defined... I somewhat have the guess that your class is that of groups that admit a strongly aperiodic subshift, or something of this spirit, but this is hard to guess a priori...
I added the correct BM reference, the deduction is on page 760 of Drutu-Kapovich.
@YCor does your class of interesting examples include some where the group is QI to a product of two nonamenables, but you use something other than two trees?
Regarding last comment of @YCor, I think knowing the exact class doesn't affect the sort of things I'd consider interesting, and does not help suggesting groups in the class. You guess the spirit correctly (though I nitpick that all countable groups admit a SA subshifts, and not all products of f.p. nonamenable groups admit SA SFTs).
@YCor: On second thought, of course you are right, if some group in this class is one for which strongly aperiodic SFTs are not previously known, that's interesting, since indeed my condition implies that. Mainly non-RF Baumslag-Solitar groups spring to mind, are they QI to products of NA? I know they are a single QI-class, but I don't know what else is there.
Ok, I found the understandable classification of those (QI to $\mathrm{BS}(2,3)$ groups). I'll look at this subquestion. The main question was more or less successful: I have not missed anything completely obvious.
@VilleSalo a Baumslag-Solitar group $BS(m,n)$ with $|m|\neq |n|$ has a 1-dimensional asymptotic cone. Hence a direct product of two infinite f.g. groups cannot embed QI into it. In particular it cannot be QI to such a direct product.
Am I correct that even by repeatedly passing to subgroups and taking QI groups you cannot find such a product?
I.e. can distorted subgroups cause a problem
(I don't really understand your claim about the asymptotic cone, I suggested QI+subgroups cannot give you products due to some Bass-Serre theory arguments I spotted in the literature (and which I don't fully understand either).)
|
2025-03-21T14:48:31.023418
| 2020-05-19T20:01:01 |
360814
|
{
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"Jdoe",
"anon",
"https://mathoverflow.net/users/138629",
"https://mathoverflow.net/users/149169"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/360814"
}
|
Stack Exchange
|
The Fontaine Mazur conjecture for $\text{GL}_1$ over $\mathbf{Q}$
The Fontaine-Mazur cojectures says that if $\chi : \text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q}) \to \mathbf{Q}_p^\times$ is a character which is unramified almost everywhere and de Rham at $p$ then it appears as sub-quotient of the étale cohomology of some algebraic variety over $\mathbf{Q}$.
My question is : Can I explicitely find a variety $X$ and integer $r$ such that $\chi$ is a subquotient of $H^i_{\text{ét}}(X_{\overline{\mathbf{Q}}},\mathbf{Q}_p(r))$ ?
From what I understand $X$ will be of dimension $0$ and thus $i=0$ too. I have found these notes of Fargues : https://webusers.imj-prg.fr/~laurent.fargues/Motifs_abeliens.pdf
They are focused on the much more complicated case where $\mathbf{Q}$ is replaced by any number field and I can't seem to extract the answer to my question. In these notes it says that the Hecke character associated to $\chi$ by class field theory will be of the form $\eta*N^k$ where $\eta$ has finite order and $N$ is the norm (this I am fine with) and this implies that $\chi$ is the $p$-adic étale realization of $M(k)$ where $M$ is the simple Artin motive associated to $\eta$. Unfortunately I don't really know what this last part means and I would really like and explicit variety and understand how the $p$-adic étale cohomology gives rise to our original character.
Does the section on Artin Motives in Chapter 6 of Deligne and Milne, Tannakian Categories, answer your question?
Thanks Anon ! It does help but here is one thing I don't understand : given a character of finite order of $\text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ over $\mathbf{C}$ how do I get the finite dimensional representation of $\text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ over $\mathbf{Q}$ which then gives me the desired Artin motive. I think this is linked to viewing $\chi$ as a character over some number field and the notes of Fargues emphasize that there is some care to be taken with the coefficients (going between coefficients over $\mathbf{Q}_p$, $\mathbf{C}$ and number fields).
In my question I mentioned the fact that one can associate an Hecke character to $\chi$ but this requires going from $\mathbf{Q}_p$ to $\mathbf{C}$ and already this seems highly non canonical (choose an isomorphism between $\mathbf{C}$ and $\overline{\mathbf{Q}_p})$.
I found the following notes which are very helpful : http://math.stanford.edu/~conrad/modseminar/pdf/L11.pdf. What I understand now is the following : any continuous character $\chi : \text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q}) \to \mathbf{Q}_p^{\times}$ which is de Rham at $p$ is of the form $\eta \epsilon^n$ where $\eta$ is a finite order character and $\epsilon$ is the cyclotomic character. Thus what I am left to understand is what variety over $\mathbf{Q}$ has $p$-adic étale cohomology isomorphic to $\eta$.
This should clearly have an easy answer using "Artin motives" but what I am stumling on is the fact that $\eta$ has values in $\mathbf{Q}_p^\times$ and thus is not a finite dimensional representation over $\mathbf{Q}$. I think this can be solved by finding a model over $\mathbf{Q}$ but I haven't yet managed to do so. Any ideas ?
I was being a bit silly. Of course our finite order character $\eta : \text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q}) \to \mathbf{Q}_p^\times$ can be seen as a character to a $\mathbf{C}^\times$ and thus to a number field since it takes values in the roots of unity of $\mathbf{Q}_p$. Viewing this number field as a finite dimensional vector space over $\mathbf{Q}$ gives us the desired "Artin motive".
One last question (hopefully) is : do we really need class field theory for this ? The main point of the whole thing is the description of de Rham characters of the Galois group but I don't know if this relies on class field theory ?
|
2025-03-21T14:48:31.023671
| 2020-05-19T20:11:27 |
360815
|
{
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"authors": [
"Joseph O'Rourke",
"SMA.D",
"https://mathoverflow.net/users/6094",
"https://mathoverflow.net/users/97167"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/360815"
}
|
Stack Exchange
|
Properties of shapes defined by locus of points with a function of distances
Different shapes such as hyperbola and ellipse can be defined as a locus of points. For example, if we denote distance to points $P_1$ and $P_2$ from any arbitrary point as $d_1(x,y)$ and $d_2(x,y)$. The locus of points for which
$$d_1+d_2=\text{const}$$
will be an ellipse.
Are the properties of 2-dimensional shapes defined by the locus of points which is only a function of distances from several other points, well-studied? Does anyone know useful references on this subject?
More precisely, let by $P_1,P_2,\ldots,P_n$ be several fixed points in $\mathbb{R}^2$. Denote the distances of an arbitrary point $C=(x,y)$ from these fixed points by $d_1(x,y),d_2(x,y),\ldots, d_n(x,y)$. I'm interested in studying the properties of the curves defined as
$$ f(d_1,d_2,\ldots,d_n) = \text{const}.$$
As suggested by user @Blue, for the particular case where $f$ computes the sum of the distances, the resulting curve for k points is called a k-ellipse.
You may be interested in Cayley ovals, related to Cassini ovals.
@JosephO'Rourke Thank you for your comment. The curves were very interesting.
|
2025-03-21T14:48:31.023770
| 2020-05-19T22:41:21 |
360819
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/360819"
}
|
Stack Exchange
|
Closed-form expression for Riemannian exponential maps on symmetric spaces
Besides the Poincaré model for the hyperbolic disc, $S^n$, and Euclidean space, what are known instances of a symmetric space $M$ (finite-dimensional) for which the exponential map is known in closed-form?
|
2025-03-21T14:48:31.023817
| 2020-05-19T23:38:22 |
360822
|
{
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/360822"
}
|
Stack Exchange
|
Proof of parametrisation of $\hat{\mathfrak{g}}$-intertwiners of induced modules of affine Lie algebras
I am not 100% sure that this question belongs here, since I think the topic does but the specific problem I have might not. Feel free to tell me so if this is the case. It concerns the proof of Theorem 3.1.1. in the book "Lectures on Representation Theory and Knizhnik-Zamolodchikov Equations by" Etingof, Frenkel and Kirillov, Jr. A small part of it can be viewed here: https://books.google.nl/books?id=wNrxBwAAQBAJ&printsec=frontcover&hl=nl#v=onepage&q&f=false.
Let $\mathfrak{g}$ be a simple Lie algebra over $\mathbb{C}$, let $\mathfrak{h}$ be a Cartan subalgebra, let $L(\hat{\mathfrak{g}})=\mathfrak{g}\otimes\mathbb{C}[t,t^{-1}]$ be the loop algebra, let $\hat{\mathfrak{g}}=\mathfrak{g}\oplus\mathbb{C}\{c\}$ be the unique nontrivial central extension of $L(\mathfrak{g})$, and let $\tilde{\mathfrak{g}}=\hat{\mathfrak{g}}\oplus\mathbb{C}\{D\}$ be the affine Lie algebra, where $D$ is the derivation of $\hat{\mathfrak{g}}$ that acts as $\mathrm{id}_{\mathfrak{g}}\otimes t\frac{\mathrm{d}}{\mathrm{d}t}$ on $L(\mathfrak{g})$ and as zero on $\mathbb{C}\{c\}$. Let $\tilde{\mathfrak{g}}^+=\mathfrak{g}\otimes\mathbb{C}[t]\oplus\mathbb{C}\{c\}\oplus\mathbb{C}\{D\}$.
Let $\lambda\in\mathfrak{h}^\ast$, let $k,\Delta_\lambda\in\mathbb{C}$, and let $L_\lambda$ be the irreducible quotient of the Verma module $M_\lambda$ of $\mathfrak{g}$. Denote by $V_{\lambda,k}$ the induced module $\mathrm{Ind}^{\tilde{\mathfrak{g}}}_{\tilde{\mathfrak{g}}^+}L_\lambda=\mathcal{U}(\tilde{\mathfrak{g}})\otimes_{\mathcal{U}(\tilde{\mathfrak{g}}^+)}L_\lambda$, where $L_\lambda$ has been turned into a $\tilde{\mathfrak{g}}^+$-module by letting $\mathfrak{g}\otimes t\mathbb{C}[t]$ act as zero and $c$ and $D$ by constants $k$ and $-\Delta_\lambda$. We are assuming that $\lambda$ is generic with respect to $k$, meaning that $V_{\lambda,k}$ is an irreducible $\tilde{\mathfrak{g}}$-module. There is a gradation induced by $D$, i.e.
$$
V_{\lambda,k}=\bigoplus^\infty_{n=0}V_{\lambda,k}[n]
$$
where $V_{\lambda,k}[n]$ is the eigenspace of $D$ with eigenvalue $-n-\Delta$. The degree 0 component is isomorphic $L_\lambda$.
The theorem is now the statement that for any $\mathfrak{g}$-intertwiner $g:L_\lambda\to L_\mu\otimes V$, with $\mu\in\mathfrak{h}^\ast$ and $V$ a weight module of $\mathfrak{g}$ (so it has a weight decomposition with respect to $\mathfrak{h}$, with finite-dimensional weight spaces), there exists a unique $\hat{\mathfrak{g}}$-intertwiner $\Phi^g(z):V_{\lambda,k}\to V_{\mu,k}\hat{\otimes}V(z)$ such that on the 0th level this map is just $g$. Here $z$ is a nonzero complex number and the hat denotes the completed tensor product with respect to the gradation induced by $D$, and $V(z)$ is the $\mathfrak{g}$-module where $c$ acts as zero and $(x\otimes t^n)v=z^nxv$.
Two conditions characterise this intertwiner uniquely, the first one being the fact that $\Phi^g(z)w\in(V_{\lambda,k}\hat{\otimes}V(z))^{\mathfrak{g}\otimes t\mathbb{C}[t]}$ and the second that it equals $g$ on the 0th level.
The argument proceeds as follows: First of all, $\Phi^g(z)w\in(V_{\lambda,k}\hat{\otimes}V(z))^{\mathfrak{g}\otimes t\mathbb{C}[t]}=\mathrm{Hom}_{\mathfrak{g}\otimes t\mathbb{C}[t]}(V^\ast_{\lambda,k},V(z))$. Here $V^\ast_{\lambda,k}$ is the restricted dual but not with the action of $\hat{\mathfrak{g}}$ twisted by a Chevalley involuton, so $V^\ast_{\lambda,k}=\bigoplus_\Lambda V_{\lambda,k}[\Lambda]^\ast$, where $\Lambda$ ranges over the weights of $V_{\lambda,k}$. Furthermore, the fact that $V^\ast_{\lambda,k}$ is freely generated over $L^\ast_\lambda=(V_{\lambda,k}[0])^\ast$ implies that the restriction map $\mathrm{Hom}_{\mathfrak{g}\otimes t\mathbb{C}[t]}(V^\ast_{\lambda,k},V(z))\to\mathrm{Hom}_{\mathbb{C}}(L^\ast_\lambda,V)$ is an isomorphism (of what? $\mathfrak{g}$-modules? I hope so.). Finally, we have that $\mathrm{Hom}_{\mathbb{C}}(L^\ast_\lambda,V)=L_\lambda\otimes V$.
My problem lies with actually all of these equalities/isomorphisms, and a couple of other things.
To start with, why is $\mathrm{Hom}_{\mathbb{C}}(L^\ast_\lambda,V)=L_\lambda\otimes V$? There is some infinite-dimensional stuff swept under the rug here it feels like, and my tries at breaking both modules up in their weight spaces has so far been unsuccessful.
Same thing for $(V_{\lambda,k}\hat{\otimes}V(z))^{\mathfrak{g}\otimes t\mathbb{C}[t]}=\mathrm{Hom}_{\mathfrak{g}\otimes t\mathbb{C}[t]}(V^\ast_{\lambda,k},V(z))$, which I would say should be an iso instead of an equality anyways.
Why is $V^\ast_{\lambda,k}$ freely generated over $\mathfrak{g}\otimes t\mathbb{C}[t]$ by $L^\ast_\lambda=V_{\lambda,k}[0]^\ast$? Why are the weight spaces of $V_{\lambda,k}$ finite-dimensional?
Thanks so much for any help. I have gotten stuck in these details while I want to move on to KZ equations and the bigger story here :)
Extra thoughts
Intuitively, I thought that the iso $L_\lambda\otimes V\to(V_{\lambda,k}\hat{\otimes}V(z))^{\mathfrak{g}\otimes t\mathbb{C}[t]}$ should be just the inclusion, but this can't be it since $x\otimes t^n$ acts as zero on $L_\lambda$ but not on $V$., in general.
|
2025-03-21T14:48:31.024224
| 2020-05-19T23:45:36 |
360823
|
{
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],
"authors": [
"Alapan Das",
"Koko Nanahji",
"Max Alekseyev",
"https://mathoverflow.net/users/156029",
"https://mathoverflow.net/users/158321",
"https://mathoverflow.net/users/41291",
"https://mathoverflow.net/users/7076",
"მამუკა ჯიბლაძე"
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"url": "https://mathoverflow.net/questions/360823"
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|
Stack Exchange
|
Solving recursion of a complex function
I am trying to find a closed form formula for the following recursive function:
$$f_n(h)= \sum_{i=1}^{n-1} \binom{n-2}{i-1} \cdot (0.5)^{n-2} \cdot [ (f_{n-i}(h-1)\cdot \sum_{j=0}^{h-1}f_i(j)) + (f_{i}(h-1)\cdot \sum_{j=0}^{h-2} f_{n-i}(j))] $$
The base cases are the following:
$$ f_1(h)= \begin{cases}
1 & h=0 \\
0 & otherwise
\end{cases} \\
f_2(h)= \begin{cases}
1 & h=1\\
0 & otherwise
\end{cases}
$$
I have been trying to use the generating functions technique, but I have been unsuccessful so far and I was wondering if anyone has suggestions into how to solve this problem.
Thank you for your help in advance
Edit:
I added the base cases
If we start with $f_i(0)=1=g(0)$ we get all $f_i(h)=g(h)$. Then we get a simpler recurrence relation $$g(h+1)=2g(h)[\sum_{i=0}^{h} g(i)]$$. From this we can make a differential equation $\frac{\text{d}^2\text{ln}(g(x))}{\text{d}x^2}=2g(x)$.
Sorry, I forgot to add the base cases when I posted the question.
Define $g_k(m) := \sum_{j=0}^m f_k(j)$. Then the given recurrence becomes
\begin{split}
g_n(h)-g_n(h-1) &= 0.5^{n-2} \sum_{i=1}^{n-1}\binom{n-2}{i-1} [(g_{n-i}(h-1)-g_{n-i}(h-2))g_i(h-1)+(g_{i}(h-1)-g_{i}(h-2))g_{n-i}(h-2)] \\
&=0.5^{n-2} \sum_{i=1}^{n-1}\binom{n-2}{i-1} [(g_{n-i}(h-1)g_i(h-1)-g_{i}(h-2)g_{n-i}(h-2)].
\end{split}
Consider the generating function
$$G_h(x) := \sum_{n\geq 1} g_n(h) \frac{x^{n-1}}{(n-1)!}.$$
The initial conditions imply that $G_1(x)=1+x$ and $G_2(x)=1+x+\frac{x^2}2+\frac{x^3}{12}$.
Then the recurrence takes form:
$$G_h'(x) - G_{h-1}'(x) = G_{h-1}(x/2)^2 - G_{h-2}(x/2)^2$$
or
$$G_h'(x) - G_{h-1}(x/2)^2 = G_{h-1}'(x) - G_{h-2}(x/2)^2.$$
Unrolling the last recurrence, we get that for any $h\geq 2$
$$G_h'(x) - G_{h-1}(x/2)^2 = G_{2}'(x) - G_{1}(x/2)^2=0.$$
That is,
$$G_h'(x) = G_{h-1}(x/2)^2.$$
It seems that there is no simple expression for the solution to this recurrence, although we may notice that $\lim_{h\to\infty} G_h(x)=e^x$.
P.S. For a fixed $h$, the generating function for $f_n(h)$ can be expressed as
$$\sum_{n\geq 1} f_n(h) \frac{x^{n-1}}{(n-1)!} = G_h(x)-G_{h-1}(x).$$
Thank you for your time and clear explanation. I just have a question regarding $$G_h'(x) - G_{h-1}'(x) = G_{h-1}(x/2)^2 - G_{h-2}(x/2)^2.$$ From the lhs I get $$ \sum_{n\geq 1} (g_n(h)-g_n(h-1) ) \frac{x^{n-2}}{(n-2)!} $$ assuming $$ G_h'(x) = \frac{d(G_h(x))}{dx}$$ and I am not too sure how this equals to $$G_{h-1}(x/2)^2 - G_{h-2}(x/2)^2$$.
Not sure if this is usable, but for the generating function ${\mathcal G}(x,z):=\sum_hG_h(x)z^h$ from a formula for Hadamard product your relations give$$\frac\partial{\partial x}{\mathcal G}(x,z)=z\int_0^1{\mathcal G}(\frac x2,\sqrt ze^{2\pi it}){\mathcal G}(\frac x2,\sqrt ze^{-2\pi it})dt$$
@KokoNanahji: This is just an application of the formula for the product of two exponential generating functions.
Ok, sounds good. Thank you very much for your help
@მამუკაჯიბლაძე: Good point, thanks!
|
2025-03-21T14:48:31.024410
| 2020-05-20T00:00:28 |
360826
|
{
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"LSpice",
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|
Stack Exchange
|
Explanation of a proof of an embedding lemma of Bollobas and Thomason
I do not understand the proof of Bollobas and Thomason of an embedding lemma. There is a lot of notation to present first, then the statement of the lemma, then the precise question about the proof.
Let $U$ and $W$ be non-empty sets of vertices of a graph $G$. Let $e(U,W)$ be the number of edges with one endpoint in $U$ and the other in $W$. Let $d(U,W)=\frac{e(U,W)}{\lvert U\rvert\lvert W\rvert}$. The pair $(U,W)$ is $\eta$-uniform if $\lvert d(U,W)-d(U',W')\rvert<\eta$ whenever $\emptyset\ne U'\subseteq U$ and $\emptyset\ne W'\subseteq W$ and $\lvert U'\rvert>\eta\lvert U\rvert$ and $\lvert W'\rvert>\eta\lvert W\rvert$.
Lemma 3 (from Bollobas and Thomason, "Hereditary and Monotone Properties of Graphs" in The Mathematics of Paul Erdős II). Let $H$ be a graph with vertex set $\{x_1,\dots,x_k\}$. Let $0<\lambda,\eta<1$ satisfy $k\eta\le\lambda^{k-1}$. Let $G$ be a graph with vertex set $\bigcup_{i=1}^k V_i$ where the $V_i$ are disjoint sets each of order $u\ge 1$. Suppose that each pair $(V_i,V_j)$ is $\eta$-uniform and that $d(V_i,V_j)\le 1-\lambda$ if $x_i x_j\notin E(H)$ and that $d(V_i,V_j)\ge \lambda$ if $x_i x_j\in E(H)$.
Then there exist vertices $v_i\in V_i$ such that the map $x_i\mapsto v_i$ gives an isomorphism between $H$ and the subgraph of $G$ spanned by $\{v_1,\dots,v_k\}$.
In the proof, they say they may assume $H$ is a complete graph. Why?
I have a follow-up question that I will put in a new post.
By the way, I mention because I didn't know for the longest time: it is not Erdös but Erdős. I have edited accordingly.
Thanks. (For that matter, his name isn't "Paul.") Alert Springer Verlag, because I literally copied the title from the Springer Verlag site. On the cover of the book they used the "two dots" umlaut as well---a book co-edited by Ron Graham. https://link.springer.com/book/10.1007/978-3-642-60406-5 https://media.springernature.com/w306/springer-static/cover-hires/book/978-3-642-60406-5 (There is a different cover where they use the other mark.) So here's a question for MLA: if a book misspells a word, and one refers to the title of the book, should one correct the misspelling?
The reduction here is: if $x_ix_j$ is not an edge of $H$, then swap edges and non-edges between $V_i$ and $V_j$. Looking for a complete graph in the result, with vertices in specified parts, is the same as looking for a copy of $H$ in the original graph, and the uniformity is preserved.
|
2025-03-21T14:48:31.024604
| 2020-05-20T00:04:38 |
360827
|
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"Max Alekseyev",
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|
Stack Exchange
|
Chinese remaindering to solve solvable diophantine equations
Given a diophantine equation $$f(x_1,\dots,x_z)=0$$ where $f(x_1,\dots,x_z)\in\mathbb Z[x_1,\dots,x_z]$ is of total degree $d$ and each variable degree $d_i$ where $i\in\{1,\dots,z\}$ there is no method to solve it unless we are looking within a bound $$\|(x_1,\dots,x_z)\|_\infty<B.$$
Suppose we know $B$ is upper bound for solution size then we know $$f(x_1,\dots,x_z)=0$$ is solvable over $\mathbb Z^z$ iff $$f(x_1,\dots,x_z)=0\bmod p$$ is solvable for a suitable prime $p$ at least of size $$f_{max,B}=\max_{\substack{(x_1,\dots,x_z)\in\mathbb Z^z\\\|(x_1,\dots,x_z)\|_\infty<B}}\big|f_+(x_1,\dots,x_z)+f_-(x_1,\dots,x_z)\big|$$ where $$f(x_1,\dots,x_z)=f_+(x_1,\dots,x_z)-f_-(x_1,\dots,x_z)$$ where each coefficient of $f_+(x_1,\dots,x_z)$ and $f_-(x_1,\dots,x_z)$ is non-negative (in the iff only part I meant to say only look at $x_i:|X_i|<B$ even when considering $\bmod p$).
Is this the best we can do?
Can we reduce solving for $(x_1,\dots,x_z)\in\mathbb Z^z$ with $$\|(x_1,\dots,x_z)\|_\infty<B$$ to solving $$f(x_1,\dots,x_z)=0\bmod q_i$$ for $i\in\{1,\dots,t\}$ such that $\prod_{i=1}^tq_i>f_{max,B}$ holds or is there an obstruction to getting such a 'Diophantine' Chinese Remainder Theorem?
If above is too optimistic there perhaps are there other approaches?
Reason to believe anything canonical will not easily work. Take Diophantine equation $$f(x_1,x_2)=x_1x_2-PQ=0$$ where $P,Q$ are primes where $B=\lceil\sqrt{2PQ}\rceil$. Clearly if there is a neat 'Diophantine' Chinese Remainder theorem we can break $\mathsf{RSA}$.
I do not understand the "iff" part. Solutions modulo $p$ may not necessarily satisfy $|(x_1,\dots,x_z)|_\infty<B$, and thus they may not correspond to solutions over $\mathbb Z$. That is, solubility modulo $p$ does not seem to imply solubility over integers.
@MaxAlekseyev $|x_i|<B\ll p$ and so I meant to say solvable with $|x_i|<B$ even with mod $p$. For example $XY-PQ\equiv0\bmod p$ where $p\gg PQ$ and $|X|,|Y|<\lceil\sqrt{2PQ}\rceil$ should still solve factoring. The $\bmod p$ is not relevant nor analogous to Hasse principle. The problem is more of combinatorial flavor than arithmetic. I am just wondering if the obstructions that occur can be made explicit so that one might look for work around in applicable situations.
|
2025-03-21T14:48:31.024756
| 2020-05-20T02:03:22 |
360831
|
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|
Stack Exchange
|
Book on Rigorous Renormalization
Many years ago I came across Salmhofer's Renormalization book and I studied its first chapter for a while. At the time, a professor told me the aim of the book was to develop a perturbative fermionic field theory at positive temperature and (I don't remember exactly why) I decided I should focus on something else and quit the book. It's not like I completely forgot of this book until now, but I was recently doing a little search on RG books, specially those intended to apply these techniques to statistical mechanics. I understand RG is not a theory per se, in the sense that it has many ways to be implemented, so I was searching those books which covered more or less the topics I'd like to learn. This beautiful answer helped me a lot, and again Salmhofer's showed to be a great book. My point here is what follows. I know that the aim of the book is to study fermionic field theories, as I pointed out, and this is specially done in chapter 4. However, the first three chapters seem to me a great general exposition, which are not only applicable to fermionic systems but, rather, to a vast range of systems in statistical mechanics. For instance, the book discusses topics such as Gaussian integrals, polymer expansions, Feynman graphs and so on. My question is whether I'm misunderstanding something or these techniques are really meant to be much more general indeed. In other words, is anything deeply hidden behind the techniques developed in the first three chapters of the book so that it does not apply to non-fermionic systems (or need to be adapted) or am I reading it correctly and these techniques are worth understanding even if I'm not interested in fermionic systems?
For a fermionic system you would do Gaussian integrals with Grassman variables, while in a more general context you would also study complex or real variables. So yes, it makes sense to take the more general perspective, before focusing on the fermionic case.
which "beautiful answer" is that? you linked to the question not a particular answer.
@AbdelmalekAbdesselam Ops! It's your answer. Well, all the other answers are useful too but your answer was really detailed and enlightening.
@CarloBeenakker but that is the point. The book has an appendix on Grassmann variables and its associated calculus, but the exposure in the first few chapters states the results in real/complex variables. It really seems like a general approach since the discussion considers spin systems.
right, and since the difference real/complex/grassman is really not essential, you want to first introduce the topic with real or complex numbers, before adding the complexity of grassman calculus; the techniques are basically the same.
|
2025-03-21T14:48:31.024979
| 2020-05-20T02:40:07 |
360832
|
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"Gerhard Paseman",
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"მამუკა ჯიბლაძე"
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|
Stack Exchange
|
Indecomposable weirdos (cnt.)
This post is a continuation of Weirdos but algebraic.
Logically, the quoted post could follow the present one rather than precede it.
Question Does there exist an indecomposable weirdo which is neither an Abelian group nor is based on a module over certain left and right rings, with a distinguished invertible element for each of these rings?
These weirdos were described in my previous post. Also, indecomposable means indecomposable into a direct (Cartesian) product of two nontrivial weirdos.
=====================================
Examples of indecomposable weirdos:
every weirdo such that the number of its elements is a prime number is indecomposable;
every cyclic Abelian group of finite order $\ p^n,\ $ where $\ p\ $ is a prime, is indecomposable;
the additive Abelian group $\ \Bbb Z\ $ is indecomposable;
the set of ring $\ X:=\Bbb Z[\frac 12]\ $ is an indecomposable weirdo with respect to:
$$ \forall_{x\ y\in X}\quad \sigma(x\ y)\ :=\ \frac{x+y}2. $$
==============================================
EDIT
Please. check my "EDIT" from my previous post.
==============================================
==============================================
PS. Within a day or two, I will leave MO, this time for good.
MO had great potential, I saw it for all these years. It was not fulfilled at all (just the opposite). So be it.
What about the free one generated weirdo? Gerhard "It's Weird Enough For Me" Paseman, 2020.05.19.
A good question!
Actually an answer to this question was suggested in my previous answer: take a cancelative magma with the quasimedial property, in which the right multiplications are not all surjective, and modify arbitrary $\lambda$ outside the image of $(x,y)\mapsto (xy,y)$ (and similarly $\mu$ outside the image of $(x,y)\mapsto (x,xy)$). The answer here just plays this game in the case of the additive law on $\mathbf{N}$.
@YCor, you could write this in an explicit and complete way right here (? I'd welcome it).
@WlodAA It's even simpler. In both examples you mention (abelian, bimodules+ pair of invertible), the left and right multiplications are bijective (which is automatic for finite E-structures). So any indecomposable W-structure that for which it's not true yields an example. So even the monoid $\mathbf{N}$ works itself.
Actually, if $X$ is a cancelative magma, and $u,v$ are permutations of $X$, the new magma with law $(x,y)\mapsto u(x)v(y)$ is cancelative as well, defining an E-structure. If $X$ is commutative and $u$ is a single permutation, then $(x,y)\mapsto u(x)u(y)$ defines a W-structure. Even if $X$ is associative, it's not associative in general. So the most obvious examples of W-structures are those for which there is a commutative semigroup $X$ and a permutation $u$ of $X$ such that the law is given by $(x,y)\mapsto u(x)u(y)$. This covers all the given examples.
Not sure if the following example counts as being different
from those listed in the problem, but consider the weirdo
with universe $\mathbb N = \{0,1,2,\ldots\}$ and
$\sigma(x,y)=x+y$
$\lambda(x,y)=\rho(y,x) = x\ominus y$,
where $x\ominus y = x-y$ if $x\geq y$
while $x\ominus y = 0$ if $x<y$.
Comments.
Everybody knows that a quasigroup is a structure $\langle Q; *, /,\backslash \rangle$
satisfying the laws
(Axioms 1) $(x* y)/y = x$, $x\backslash(x* y) = y$, and
(Axioms 2) $(x/y)* y = x$, $x*(x\backslash y) = y$.
These say that the operation table of $x* y$ is a Latin square
on $Q$, or equivalently that the maps $x\mapsto a* x$
and $x\mapsto x* b$ are permutations of $Q$ with inverses
$x\mapsto a\backslash x$ and $x\mapsto x/b$.
If we take $\sigma(x,y)=x* y$, $\lambda(x,y) = x/y$,
and $\rho(x,y)=x\backslash y$ and only impose (Axioms 1), then
we get what the OP calls an ``eccentric''.
(Axioms 1) guarantee that multiplication is is cancellative
in each variable, but do not seem to force multiplication
to be a permutation in each variable.
The medial law for multiplication
is $(x* y)* (u* v) = (x* u)* (y* v)$.
The OP's weirdos are medial groupoids satisfying (Axioms 1).
Weirdos satisfying (Axioms 2) are medial quasigroups.
The following theorem from the 1940's is relevant to the question here:
Theorem.
(Bruck (1944), Murdoch (1941), Toyoda (1941))
The following are equivalent for a quasigroup $Q$.
(1) $Q$ satisfies the medial law for multiplication.
(2) There is an abelian group structure on $Q$
and two commuting abelian group automorphisms $\alpha, \beta: Q\to Q$
such that $x* y = \alpha(x)+\beta(y) + c$ for some $c\in Q$.
Thus, one does get an affine representation for
weirdos satisfying (Axioms 2).
If we delete (Axioms 2) we are only guaranteed that
multiplication is cancellative.
But the following is a consequence of Corollary 1.2 of
my paper
A quasi-affine representation.
Internat. J. Algebra Comput. 5 (1995), 673--702.
Consequence.
If $\langle Q; * \rangle$ is a medial, cancellative groupoid, then
$Q$ has a quasi-affine representation iff
$Q$ is abelian in the sense of commutator theory.
Thus abelian weirdos will have quasi-affine representations.
(Saying that $Q$ has a quasi-affine representation means
that there is an $R$-module $M$ containing
$Q$ as a subset and the multiplication is representable
as $x*y = \alpha(x)+\beta(y) + c$ for some
not-necessarily-invertible
ring elements $\alpha, \beta\in R$ and some $c\in Q$.
You will see from the first example above, involving $\mathbb N$,
why we need to talk about subsets of modules rather than full modules.)
I suppose it is not known whether there are nonabelian weirdos?
@მამუკა ჯიბლაძე, In my earlier post I gave an example, in full detail, of a non-abelian weirdo, the one which lives in $\ \Bbb Z[\frac 16].$
@WlodAA Sorry I should clarify that I mean abelian in the sense of commutator theory, as in this answer. Your example should be such, since it is affine (with $\alpha=2/3$, $\beta=1/3$, $c=0$), no?
Abelian (in the sense of commutator theory) is not the same as commutative. An algebra $A$ is abelian in the sense of commutator theory if the diagonal of $A\times A$ is a congruence class.
Keith, thank you for your answer. Also my thanks to @YCor. This is my last OM action (sorry that I don't have the energy to continue or even to complete the present threads. There are also at least equally interesting questions about field-like structures, both algebraic and topological). Good luck.
|
2025-03-21T14:48:31.025400
| 2020-05-20T04:42:28 |
360834
|
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|
Stack Exchange
|
Why is this "the first elliptic curve in nature"?
The LMFDB describes the elliptic curve 11a3 (or 11.a3) as "The first elliptic curve in nature". It has minimal Weierstraß equation
$$
y^2 + y = x^3 - x^2.
$$
My guess is that there is some problem in Diophantus' Arithmetica, or perhaps some other ancient geometry problem, that is equivalent to finding a rational point on this curve. What might it be?
Edit: Here's some extra info that I dug up and only mentioned in the comments. Alexandre Eremenko also mentions this in an answer below. The earliest-known example of an elliptic curve is one implicitly considered by Diophantus, in book IV of Arithmetica, problem 24 (Heath's translation): "To divide a given number into two numbers such that their product is cube minus its side". Actually this is a family of curves over the affine line, namely $y(a-y)= x^3-x$, though Diophantus, in his usual way, only provides a single rational point for the single curve corresponding to $a=6$. This curve is 8732.b1 in the L-functions and modular forms database (the Cremona label is 8732a1). So presumably the comment about 11a3 is not meant to mean "historically first".
h/t to Anton Hilado for bringing this to my attention on Twitter!
This text seems to have been written by Nicolas Mascot, if I have interpreted LMFDB correctly; so he might be a good person to ask. https://www.lmfdb.org/knowledge/show/ec.q.11.a3.top
Surely it's because it has smallest conductor. I'm guessing the congruent number curve (for 1, i.e. $y^2=x^3-x$) is the first elliptic curve in history.
product of two consecutive numbers equals product of three consecutive numbers?
As far as I know, this is an informal name was coined by John Coates (and, possibly, never in writing). It is one of the three curves of smallest conductor, and has the simplest equation among those.
@DrorSpeiser Well, Diophantus solved $6y - y^2 = x^3 - x$, as as example of the family $Ay - y^2 = x^3 - x$, or rather "To divide a given number [A] into two numbers such that their product is cube minus its side" (Problem IV-24; thanks to John Baez for pointing me to this)
This article Elliptic Curves from Mordell to Diophantus and Back by Brown and Myers in Am. Math. Monthly (doi:10.1080/00029890.2002.11919894) starts with a picture of the curve in my previous comment, labelled "The first elliptic curve".
FWIW, here is the curve considered by Diophantus: https://www.lmfdb.org/EllipticCurve/Q/8732/b/1
It is the minimal curve in the isogeny class with the minimal conductor. Though http://people.math.harvard.edu/~elkies/nature.html , which is a very nice source for elliptic curves in nature, puts it as the second and so did Cremona's tables, but it was reordered when it was taken into lmfdb.
The elliptic curve 11a3 is the curve with smallest Faltings height, which means in basic terms that the period lattice associated to the Néron differential has largest area. I don't know if this is recorded somewhere in the literature, but it can be checked numerically at least.
@FrançoisBrunault yep that is what I meant by "minimal". I believe that John's new way of enumerating curves in isogeny classes is such that the minimal one has label .a1, but I am not 100% sure.
@DrorSpeiser Diophantus treated many more elliptic curves as well (many of his so-called 'double equations' were curves of genus 1 with a natural choice of a rational point). It would be hard to pinpoint any one of them as somehow prior to the others.
@RP_ oh, interesting!
If you're interested, I wrote up some things about Diophantus' treatment of (some special cases of) intersections of two quadrics in P^3 here: https://arxiv.org/abs/1509.06138 (section 4). Most of it is directly based on the brilliant book of Thomas Heath, only he expressed his findings purely in the language of elementary algebra.
@liuyao : what you describe seems to be rather the curve https://www.lmfdb.org/EllipticCurve/Q/37/a/1, in my opinion
I asked Kevin Buzzard to ask John Coates directly, and it's basically as people have surmised: the moniker is due to the fact the curve appears first in Cremona's book as it has the smallest possible conductor, and it has the smallest coefficients. It is not due to historical priority, as Coates knows of 8th/9th century Arabic manuscripts that discuss $y^2 = x^3 - x$, whereas the first occurrence of the "first curve in nature" is apparently a book of Fricke on elliptic functions (I think from 1922, but I'm not sure).
I actually only wrote the part that says that this curve is a model for $X_1(11)$, not the first part, which I think was written by John Cremona.
It is standard to order elliptic curves by conductor (e.g. for statistics), and 11 is the smallest possible conductor. However, there are 3 curves with conductor 11, and no canonical way to order them as far as I know (though @François Brunault has an interesting point); for instance LMFDB labels do not order these 3 curves in the same way as Cremona labels.
This curve being the first one could maybe also be understood in terms of modular degree, although this is also ambiguous: if we order them by degree of parametrisation by $X_1(N)$, then this curve, being a model of $X_1(11)$, comes first, but if we order in terms of degree of parametrisation by $X_0(N)$, then 11.a2 comes first since it is a model for $X_0(11)$.
From (nearly) the horse's mouth! Unless Cremona lets us know why, this might be as good as we get (if you can get him to explain, I would be curious to know).
For each isogeny class of elliptic curves over $\mathbb{Q}$, one can define a graph (which I believe is canonical) where the vertices are the elliptic curves and the arrows are the étale isogenies of prime degree. Étale means, for example, that the pull-back preserves the Néron differentials. It turns out that this graph is a tree, and the root is the $X_1(N)$-minimal curve (the smallest degree of $X_1(N) \to E$). So there is a unique curve with smallest Faltings height. This last property can fail over number fields.
PS. I think what I wrote is conditional on Stevens's conjecture on the Manin constant of $X_1(N)$-parametrisations.
I can only echo Tim D's explanation: from Coates via Vlad to me. I did not know about it having minimal Faltings height.
But the lmfdb ordering is now deterministic, so there is an answer to why $X_1(11)$ is now the first. I believe to remember that you explained me your criteria for ordering and that it implies that the minimal curve (Falting height, or lattice or ... ) is always the first one. Or am I wrong here?
Hmm, the chain of whispers stretches back... I wonder if Coates could be prevailed on to explain? :-)
The closest thing I found in Diophantus is problem IV(24) which is solving the system
$$X_1+X_2=a,\quad X_1X_2=Y^3-Y.$$
Diophantus sets $X_1=x$ and eliminates $X_2$ obtaining
$$x(a-x)=Y^3-Y.$$
This seems to be the first elliptic curve encountered in the book of Diophantus; before that he only considers rational curves and surfaces.
Diophantus choses $a=6$ and obtains a solution $x=26/27,\; Y=17/19$.
(This little research is based on a Russian translation of Diophantus with
comprehensive comments by I. G. Bashmakova, published in Moscow in 1974.)
Yes, I mentioned this curve in the comments to the question; I was going to edit them into the main body tomorrow. I guess (I'm not an algebraic geometer) that your comment is showing that after base change to $\mathbb{Q}[i]$ one can get the Arithmetica IV(24) curve and 11a3 as fibres of the same family of curves over $\mathbb{A}^1$.
|
2025-03-21T14:48:31.025906
| 2020-05-20T05:08:31 |
360836
|
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|
Stack Exchange
|
Current status of axiomatic quantum field theory research
Axiomatic quantum field theory (e.g. the wightman formalism and constructive quantum field theory) is an important subject. When I look into textbooks and papers, I mostly find that the basic constructions involve functional analysis and operator algebra. Now, modern developments in the field of quantum field theory involve subjects such as algebraic geometry, topology, and knots. Mostly modern developments deal with supersymmetric quantum field theory.
My question is: What is the status of current research on the mathematical foundations of the dynamics of "non topological" quantum field theory? Are the older approaches to the subjects, such as the ones in Glimm and Jaffe's and Wightman's books, obsolete?
Paging @Urs Schreiber...
The answer to the second question "Are the older approaches to the subjects, such as the ones in Glimm and Jaffe's and Wightman's books, obsolete?" is: No.
The Axiomatic Quantum Field Theory from the 1950's has been rebranded as Algebraic QFT (keeping the same abbreviation), and the emphasis has shifted somewhat from quantum fields to local observables. The Wightman axioms for fields are then replaced by the Haag–Kastler axioms for the algebra of observables. For two relatively recent overviews see Current trends in axiomatic quantum field theory (1998) and Algebraic quantum field theory (2006). A concise summary is at nLab. An alternative approach to AQFT is FQFT (Functorial QFT), which focuses on states rather than observables.
Thanks. But algebraic QFT fails to capture topological and geometric phenomena in QFT, doesn't it? For example, it does not address the mathematical nature of quantum dualities, such as S-duality.
@Physicsstudent000 : AQFT -> blob homology -> causally local net of observables -> S-matrix.
@Physicsstudent000 You might possibly find this article interesting: https://arxiv.org/pdf/1511.00316.pdf Maybe it addresses your objections?
Sorry if I ask a stupid question. I understand that Euclidean field theory usually uses path integral as its way of quantizing fields. So why the finding of Osterwalder (focusing on the Euclidean formalism of QFT) would be classified into AQFT? From nLab, I understand that it is FQFT which is the formalization of path integral quantization.
|
2025-03-21T14:48:31.026222
| 2020-05-20T06:42:14 |
360843
|
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|
Stack Exchange
|
How is class of composition of two quadratic fields is related class numbers of quadratic field?
Let $K_1=\Bbb Q(\sqrt{d_1})$ , $K_2=\Bbb Q(\sqrt{d_2})$ and $K=\Bbb Q(\sqrt{d_1},\sqrt{d_2})$.Suppose $h_1,h_2,h$ be class number of $K_1,K_2,K$ respectively.
(i) Can we express $h$ in terms of $h_1,h_2$?
(ii) Knowing the divisibility properties of $h_1,h_2$, I want help with concluding about the divisibility of $h_1,h_2.$
You did not define $d$ in terms of $d_1$ and $d_2$. Also as is written, $K_1=K_2$.
I assume you mean $K_1=\mathbb Q(\sqrt{d_1})$.
1) If by $K$ you mean $\mathbb Q(\sqrt{d_1d_2})$, then there is no simple relation
between $h$ and $h_1$ and $h_2$.
2) If by $K$ you mean the quartic biquadratic field $\mathbb Q(\sqrt{d_1},\sqrt{d_2})$, a theorem of Herglotz says that $h=h_1h_2h_3/2^j$, where
$h_3$ is the class number of $\mathbb Q(\sqrt{d_1d_2})$ and $j=0,1,2$ which can
be computed in terms of the units of $K$.
Sorry I meant K to be the composition K_1 and K_2. Can you please provide some reference for theorem of Herglotz
https://link.springer.com/article/10.1007/BF01482079
|
2025-03-21T14:48:31.026323
| 2020-05-20T06:46:29 |
360845
|
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|
Stack Exchange
|
Is the module of Kähler differentials a coend?
Let $\phi\colon R\to S$ be a ring map. The module of Kähler differentials $\Omega_{S/R}$ of $\phi$ can be constructed as the following coequaliser:
$$\left(\bigoplus_{(a, b)\in S^2} S[(a, b)]\right) \oplus\left( \bigoplus_{(f, g)\in S^2} S[(f, g)]\right)\oplus\left( \bigoplus_{r\in R} S[r] \right)\underset{0}{\rightrightarrows}\bigoplus_{a\in S} S[a],$$
where the above map is defined by
$$
\begin{align*}
[(a,b)] &\mapsto [a+b]-[a]-[b],\\
[(f,g)] &\mapsto [fg] -f[g]-g[f],\\
[r] &\mapsto [\phi(r)].
\end{align*}
$$
This looks strikingly similar to the colimit formula for computing enriched co/ends, which given a $\mathsf{Mod}_R$-functor $P\colon\mathcal{C}^\mathsf{op}\times\mathcal{C}\to\mathsf{Mod}_R$, takes the form
$$
\int^{M\in\mathsf{Mod}_R}P(M,M)=\mathrm{Coeq}\left(\bigoplus_{\phi\colon M\to N\in\mathsf{Mod}_R}\mathrm{Hom}_{\mathsf{Mod}_R}(M,N)\otimes_RP(N,M)\rightrightarrows\bigoplus_{M\in\mathsf{Mod}_R}P(M,M)\right).$$
Question. Can we write $\Omega_{S/R}$ as a coend in a "nice"* way?
*That is, without directly using the definition of $\Omega_{S/R}$ (e.g. using the functor $\mathrm{Der}_R(S,-)$ to build the desired coend would certainly be "nice").
Here's a failed short attempt, for what it's worth. Using the Density Theorem and the isomorphism $\mathrm{Hom}_{\mathsf{Mod}_R}(R,M)\cong M$, one can write
$$\Omega_{S/R}\cong\int^{M\in\mathsf{Mod}_R}(-)\otimes_R\mathrm{Hom}_{\mathsf{Mod}_R}(-,\Omega_{S/R}),$$
but it now seems difficult to give a "nice" description of $\mathrm{Hom}_{\mathsf{Mod}_R}(-,\Omega_{S/R})$.
|
2025-03-21T14:48:31.026437
| 2020-05-20T07:52:47 |
360847
|
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"Matvey Tizovsky",
"abx",
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|
Stack Exchange
|
A Thom isomorphism for sheaves
Let say $\mathcal{F}$ is a locally free sheaf of abelian groups over $X$, where $X$ is an algebraic variety over $\mathbb{C}$ (or a field $k$) with analytic (or étale) topology and $Z$ is a closed subvariety of $X$ of codimension $d$. I want to ask if there exists, or under which condition there exists, an isomorphism
$$H^{2d}_{Z}(X, \mathcal{F}) \xrightarrow{\sim} H^0(Z,\mathcal{F}_{|Z}).$$
More specifically, can one deduce such an isomorphism from a suitable spectral sequence?
A sheaf of what? Over what kind of field, for what topology?
I have just edited
|
2025-03-21T14:48:31.026524
| 2020-05-20T07:53:00 |
360848
|
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"Brendan McKay",
"James Martin",
"Matthieu Latapy",
"gmvh",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/360848"
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|
Stack Exchange
|
Probability that a random multigraph is simple
Question.
Consider a given sequence of $n$ integers $d_1$, $d_2$, $\cdots$, $d_n$ with $\sum_i d_i$ even and $d_i\le n$ for all $i$. One may sample a random multi-graph having this degree sequence using the (classical) configuration model.
How to compute or precisely estimate the probability to obtain a simple graph (no loop, no multi-edge)?
For instance, if the sequence is $1, 1, 2, 2, 3, 3$ (thus $n=6$), then the probability experimentally seems close to $0.016$.
Remark.
Classical results (Erdös-Gallai and Havel-Hakimi) tell if a given sequence may lead to a simple graph (graphic sequences). Therefore, they tell (in linear time) if the probability under concern is non-zero.
Related work.
There are beautiful asymptotic results for the probability under concern, see in particular Janson papers and van der Hofstad book.
There are also beautiful results based on properties of the sequence, in particular for regular graphs (all $d_i$ are equal) and bounded sequences (the maximal $d_i$ is lower than a given bound). See in particular McKay and Wormald works.
These results hold for families of graphs and consider limits when $n$ grows to infinity, and/or families of sequences of which properties can be specified (like maximum degree bounded by a power of $n$).
Warning.
This is not the situation I face: I do have one specific sequence and want to estimate the probability that the configuration model gives a simple graph for this specific sequence.
Related questions.
Is there a way to compute this probability in linear time and space, with respect to $n$?
If, instead of just one sequence, I have two sequences of integers (of same sum) and sample bipartite multi-graphs with the configuration model, what is the probability to obtain a simple bipartite graph (no multi-edge)?
I wonder whether the probability is really well-defined without some knowledge as to the properties of your specific sequence?
@gmvh I may better define it as the fraction of multi-graphs with this degree sequence that are simple.
I think that @gmvh is asking about realizable degrees sequences. For example a degree sequence 1,2 does not corresponds to any graph. So the fraction is not well-defined.
Hi Matthieu! I'm not quite clear what you need. You say you want to estimate the probability, and you say $n$ is huge, and as you say there are asymptotic results - so presumably you are in some regime where you think these asymptotic results don't give a good estimate? Is your probability close to 0, or close to 1, or somewhere moderate in between? If the expected number of self-edges and multiple edges is small, a Poisson approximation is typically good. But if it's high, and you want a better estimate than "near 0", then different methods are needed. We may need more details....
@Sergey, thank you. I think I just need that $\sum_i d_i$ is odd, in order to make sure that there is at least one multi-graph with this sequence. Then, if there is no simple graph with this sequence (it is not realizable, or not graphic), the probability is just 0, right?
Hi @James! Yes, I do think (or at least, fear) that available asymptotic results do not give good enough estimates for my use. My probability is >0, since I know the sequence is realizable. It is <1 since I know not all graphs with this sequence are simple. I think it is "close to 0", but I need to know how close. For instance, $10^{-9}$ and $10^{-12}$ make a big difference to me.
Maybe I misunderstood something, but it seems that there are some degree sequences with no corresponding graphs at all, nor simple, nor multi. For instance, if we have a degree sequence "1, 2" then, at the best, we will have two connected nodes and a dangling link attached to the one of them. Thus it makes a fraction 0/0 with undefined value... Would your case imply the existence of at least one graph realizing the given degree sequence ?
Yes @Sergey, but if the sum of degrees is even then there is always a possible multi-graph. Sorry, I wrote 'odd' instead of 'even' in my previous reply :/
That already helps narrow it down. But maybe not enough? - I doubt there is anything off-the-shelf in general, which covers all the cases you might be talking about. I imagine the case of $n$ vertices each of degree $\log(n)$ needs a different approach to that of $1$ vertex of degree $n^{3/4}$ and $n-1$ of degree $2$. So probably we would need to engage more with the specifics of your problem! (But I may be too pessimistic; if so, others may be able to jump in....)
Sure @James, but since I have only a few and quite different empirical degree sequences, I do not have any model of them, so saying if the maximal degree is $\log(n)$, $\sqrt{n}$, or a constant seems difficult. Still, the number of edges is small compared to $n^2$ and the maximum degree is "large", although significantly smaller than $n$...
A random matching looks more or less like a bunch of random edges of the right density (I.e. edges selected independently) without 2-paths. In the latter model it’s easy to calculate the probability of no loop or multiple edge, since everything is independent. The models will give about the same answer (up to smallish constants) provided degrees grow with n but the sum of squared degrees is not too fast growing. Is your sequence like that?
Since the number of configurations is easy to compute, and each simple graph (with given degrees) corresponds to the same number of configurations, what you are asking for is just the number of simple graphs. I'm not aware of a fast way to bound that number rigorously.
It seems that counting labeled simple regular graphs is already not trivial. There is a bit of discussion on MO : https://mathoverflow.net/questions/135293/number-of-labeled-regular-graphs-on-n-vertices
@BrendanMcKay, the problem is indeed equivalent to computing the number of simple graphs, and I do fear too that there is no satisfactory solution known. I thought that maybe having the exact degree sequence and being open to a somewhat complex algorithmic computation may help, but, as Sergey points out, there is no simple solution even for regular graphs :/
Matthieu, you have probably already considered this approach. But or the sake of completeness, let me indicate a possible "empirical argument" that consists in sampling several graphs with the specific degree distribution, see how many among them are simple and compare this estimate with Poisson approximation, for example one from Corollary 3.7
of https://www.ndsu.edu/pubweb/~novozhil/Teaching/767%20Data/47_pdfsam_notes.pdf
You are right, @Sergey, and this is what I did for the small example in the question above. But for large instances, this is intractable, in particular if the case where the probability is very small. For small instances that I tested, unsurprisingly, the empirical observation is quite different from the asymptotic estimate.
Does Thm 5.2 of http://users.cecs.anu.edu.au/~bdm/papers/nickcount.pdf match your experiments better?
Thank you @BrendanMcKay for this interesting reference that I missed. The asymptotic formula of Theorem 5.2 did not help, but I can make experiments on very small examples only, and for rather simple degree sequences. Therefore, I do not know what sense they make. I am more and more thinking that an appropriate approach would be to seek upper and lower bounds, that may be empirically "tight enough".
A related question was posted recently here: https://mathoverflow.net/questions/379153/simple-graphs-with-prescribed-degrees-as-disjoint-union-of-simple-subgraphs-with/379427
|
2025-03-21T14:48:31.027023
| 2020-05-20T08:24:52 |
360851
|
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|
Stack Exchange
|
Do there exist graphs whose adjacency matrix is positive semi-definite?
If so, could you provide examples and specify the conditions under which this occurs? Thank you in advance
Hi Dante, this website is for research questions in mathematics, see MSE for general questions in mathematics. The adjacency matrix is (most of the time) a symmetric matrix, so its eigenvalues are all real. Note that the trace of a matrix is the sum of its eigenvalues. Since (unless you take loops into account) the trace is zero, and positive semi-definite means the eigenvalues are positive, this will [almost] never happen. I would advise you make the setup of your question clearer.
If you don't allow self-loops in the graph, then the trace is $0$. If the adjacency matrix is PSD then $0$ is the only eigenvalue. Since your adjacency matrix is symmetric, it must equal to $0$. So you essentially have the empty graph.
If you allow self-loops then you can also get a PSD adjacency matrix by adding some diagonal matrix. This corresponds to adding some self-loops in your original graph.
No problem. I suspect that MO might not be the right place for this question. Next time you can try math stack exchange first.
|
2025-03-21T14:48:31.027146
| 2020-05-20T08:34:33 |
360852
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/360852"
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|
Stack Exchange
|
maximal k-partite subgraph in a complete multipartite graph
What is the maximum number of edges a $k$-partite subgraph of a complete $s$-partite graph can have?
Bests,
Josefran
|
2025-03-21T14:48:31.027195
| 2020-05-20T09:53:07 |
360856
|
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"Otis Chodosh",
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"shurtados"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/360856"
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|
Stack Exchange
|
Length and curvature for closed curves in negatively curved spaces
In the Euclidean plane, for a closed smooth curve of length $\ell$ whose curvature is bounded above by $\epsilon$ we have the inequality
$$ \ell \ge 2\pi \epsilon^{-1} $$
which follows from the fact that the total curvature is $2|k|\pi$ with $k \not= 0$ the winding number.
Is there a known generalisation of this to closed curves in a CAT(0) (simply connected with negative sectional curvature) Riemannian manifold? That is given such a manifold $X$, is there a function $f : [0, 1] \to [0, +\infty[$ (say) so that for any closed smooth curve in $X$ of length $\ell$ and curvature bounded above by $\epsilon$ we have $\ell \ge f(\epsilon)$? (of course we want $\lim_{\epsilon\to 0} f(\epsilon) = 0$ and for pinched negative curvature I would expect it to go faster than $1/\epsilon$).
We can also ask a purely metric version of this question: given a CAT(0)-space $X$, is there a function $f : [0, 1] \to [0, +\infty[$ (say) so that for any closed curve in $X$ of length $\ell$ which is a local $(1+\epsilon)$-quasi-geodesic we have $\ell \ge f(\epsilon)$?
(By "local $(1+\epsilon)$-quasi-geodesic" I mean a curve $\gamma$ such that for any two points $x, y \in X$ such that $d(x, y) \le C$ and $x, y$ lie on $\gamma$, if $a$ is the length of the (shortest) arc between $x$ and $y$ on $\gamma$ then $a - d(x, y) \le \epsilon$. Here $C$ is a constant depending on $X$.)
Edit following comments: in Gromov-hyperbolic spaces it seems that the condition of being a "local-quasi-geodesic" implies (for sufficiently small $\epsilon$) that the curve is a global quasi-geodesic, in particular it cannot close. I think the proof of Theorem 1.13, p. 405 of Bridson--Haefliger can be immediately adapted to do this (and for this particular problem points (1) and (2) of the theorem are sufficient).
So we can take $f = +\infty$ in a neighbourhood of 0 (depending on the hyperbolicity constant). The comment by shurtados shows that for the hyperbolic plane we can take the neighbourhood to be $[0, 1[$.
As noted in Ycor's comment the "purely metric" version is not optimal for CAT(0) spaces (as opposed to hyperbolic) and for these a version that would include non-Riemannian $X$ and singular curves should probably involve some sort of "curvature measure" on the curve whose integral would be computable. This question seems interesting even for euclidean space.
Probably the right version should involve instead a punctual curvature condition given by some measure, so that for instance, it would encompass the fact that in the plane, the sum of $|\pi-\alpha_i|$ for $\alpha_i$ the angles of a closed polygonal path, is at least $2\pi$. The curvature is naturally bounded by some "measure", which in this case is supported by the vertices, while it it non-atomic for a $C^1$-curve. I'm not sure how to define this properly, but say for a curve parameterized with speed one I'd copy the curvature definition and take the derivative in distribution sense.
For the hyperbolic plane, a horosphere has constant geodesic curvature 1 and a circle of radius $r$ has constant geodesic curvature $\frac{1}{\tanh{r}}$, so I guess that for $X = \mathbb{H^2}$ we have $f(\epsilon) = \infty$ if $\epsilon < 1$ and $f(\frac{1}{\tanh{r}}) = \text{length}_{\mathbb{H}^2}(\partial B_r) = 2\pi \sinh(r)$. See https://math.stackexchange.com/questions/2430495/curvature-of-curves-on-the-hyperbolic-plane.
I think this paper answers some versions of the question https://mathscinet.ams.org/mathscinet-getitem?mr=3415662 (not that $K \leq -1$ can be replaced by $K\leq 0$ if one makes the appropriate changes to the proof/statement).
The Reshetnyak majorization theorem (see 9.56) states that any closed rectifiable curve $\alpha$ in a CAT(0) length space $U$ can be majorized by a convex plane figure $F$; that is, there is short (= 1-Lipschitz) map (= majorization) $m\colon F\to U$ such that $m|_{\partial F}$ is the arc-length parametrization of $\alpha$.
Note that majorization does not decrease the curvature; that is, curvature of $\alpha$ cannot be smaller than curvature of $\partial F$ at the corresponding point.
Therefore, the inequality
$$ \ell \ge 2\cdot \pi\cdot \varepsilon^{-1}$$
holds in CAT(0) spaces as well.
|
2025-03-21T14:48:31.027736
| 2020-05-20T10:29:45 |
360858
|
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|
Stack Exchange
|
Surjectivity in Deligne-Serre
Let $f$ be a newform of weight $k$ and level $N$ with integer coefficients. Deligne-Serre theorem theorem says there exist a nice associated representation $\rho_{f}^{(\ell)}:\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\to \text{GL}_2(\mathbb{Q}_{\ell}),$ for any prime $\ell$ not dividing $N$. In particular, this induces a representation $\rho_{f,\ell}:\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\to \text{GL}_2(\mathbb{F}_{\ell}).$ Due to Serre's open image theorem, both adelic and mod $\ell$ representations are surjective for all but finitely many $\ell$, when $f$ has weight $2$ and comes from an elliptic curve. For the general case (especially when the weight is more than $2$ ?) is the mod $\ell$ representation coincides with the matrices having determinant a $(k-1)^{th}$ power in $\mathbb{F}_{\ell}^*$, for all but finitely many $\ell$ ? If not, what's stopping it to be ? and what we can say about the images ?
In the setting that $f$ has integer Hecke eigenvalues, then the result is true, so long as $f$ has weight $k\ge 2$ and does not have complex multiplication. See Theorem 3.1 of this paper by Ribet: in this setting, $R=\mathbb Z$.
In general, the Hecke eigenvalues of $f$ will all lie in a finite extension $E$ of $\mathbb Q$, and $\rho_f$ will be valued in $\mathrm{GL}_2(E_{\lambda})$ for a prime $\lambda$ above $\ell$. Let $\mathbb F_\lambda$ be the residue field of $E_\lambda$. If $k\ge 2$ and $f$ does not have CM, then, for all but finitely many primes, the image of the mod $\lambda$ representation $\overline{\rho}_f$ in $\mathrm{GL}_2(\mathbb F_\lambda)$ will contain a subgroup conjugate to $\mathrm{SL}_2(\mathbb F_\ell)$ (but not necessarily $\mathrm{SL}_2(\mathbb F_\lambda)$).
To say anything further, we have to be a lot more careful: I'd suggest looking at the answers to this question.
Yes, I am happy when eigenvalues are integer.s In the Ribet's paper, are you talking about (2) at page 186 ? but that's only said for level 1, no ?
@dragoboy I'm referring to Theorem 3.1 on p191. You're right that the result on p186 only applies to level $1$ forms.
Great! thanks...
|
2025-03-21T14:48:31.027912
| 2020-05-20T10:33:30 |
360859
|
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|
Stack Exchange
|
fractional Laplacian estimates
Suppose $(-\Delta)^s u=f \geq 0$ in a ball $B_2$ and $u=0$ in $ R^N \setminus B_2.$ Also suppose $u$ is $C^{s}$ non-negative and $(-\Delta)^s u=0$ in $B_2 \setminus B_1$ and $u\leq a$ on $\partial B_1$ where $B_1, B_2$ is a ball of radius $1$ and $2$ and $a$ is a positive constant. Can one claim that $u\leq a$ in $B_2 \setminus B_1$ or can any estimate be obtained on the upper bound of $u$. In the case $s=1$, this is maximum principle.
Yes.
We have
$$ u(x) = \int_{B_1} G_{B_2}(x,y) f(y) dy , $$
where
$$ G_{B_2}(x,y) = C_{N,s} \frac{1}{|x - y|^{N - 2s}} \int_0^{T(x,y)} \frac{t^{s-1}}{(t+1)^{N/2}} dt , \\ T(x, y) = \frac{(4-|x|^2)(4-|y|^2)}{4|x-y|^2} $$
is the corresponding Green function. Fortunately, $1/|x - y|^{N-2s}$ and $T(x,y)$ are radially decreasing on $B_2 \setminus B_1$, so indeed the maximal value of $u$ over $B_2 \setminus B_1$ is taken somewhere over $\partial B_1$.
Can anything be deduced if $B_2$ is replaced by a smooth bounded domain $\Omega.$ Thank you.
I guess it is not: if $B_2$ is replaced by $\Omega = B_2 \setminus (\overline{B}_{1+\varepsilon} \setminus B_1)$ for a sufficiently small $\varepsilon > 0$, then the result is no longer true. I bet it could possibly be true if, say, $\Omega$ is convex, but these kind of problems are usually very difficult.
I think I meant something like $\Omega = B_3 \setminus (\overline{B}2 \setminus B{1+\varepsilon})$, sorry.
If $\Omega$ is smooth bounded domain and $(-\Delta)^s$ denotes the spectral fractional laplacian, do the result hold. That is $(-\Delta)^s u=0$ in $\Omega \setminus B$ and $u\leq a$ on $\partial B$ and $u=0$ on $\partial \Omega$ where $B$ is a ball of radius $1$ and $a$ is a positive constant. Can one claim that $u\leq a$ in $\Omega \setminus B.$
@Spal: No idea! Sounds much more plausible, but on the other hand I would not be surprised by a smart counter-example. I do not think such questions have been studied much, and my feeling is that they can easily get very difficult if true.
|
2025-03-21T14:48:31.028065
| 2020-05-20T10:45:28 |
360860
|
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"Leyli Jafari",
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|
Stack Exchange
|
Is the solvable radical of a finite perfect group contained in the Schur multiplier of the quotient of the group modulo the solvable radical?
Let $G$ be a finite perfect group, and let $N$ be the solvable radical
of $G$. If $G/N$ is a non-abelian simple group, then is it true that $N$
is contained in the Schur multiplier of $G/N$?
If this is not true in general, then does it hold at least in case
$G/N$ is either of type ${\rm PSL}(2,2^p)$ ($p$ prime) or isomorphic
to one of ${\rm PSL}(2,7)$ or ${\rm Sz}(8)$?
Furthermore, can the finite perfect groups whose quotient modulo their
solvable radical is isomorphic to one of the simple groups mentioned in
the previous paragraph reasonably be classified?
"Is contained", you mean "is isomorphic to a subgroup of"? Anyway, in many cases $N$ is non-abelian, so this sounds hopeless. Even with $N$ abelian, you have plenty of perfect semidirect products $G=N\rtimes S$ with $S$ simple non-abelian (for given $S$ you have such $G$ with $N$ arbitrary large). Classifying them for $N$ abelian of prime exponent is essentially classifying reps of $S$ in arbitrary finite fields. Classifying them for $N$ nilpotent is even much harder. So this sounds hopeless.
Thanks a lot! -- And can you perhaps also tell what would be the answer if the assume $N = {\rm Z}(G)$?
The answer is positive if $N=Z(G)$, but this is somewhat immediate from the definition. I think that the question belongs on MathSE rather than here.
The answer to the question as asked is definitely "no" in general. For any value of $n>1 $ and any odd prime $p$, we may take a perfect group $G$ which is a semidirect product of the form $E.{\rm Sp}(2n,p),$ where $E$ is extra special of order $p^{2n+1}$ and the action of the given symplectic group on $E$ is the natural one. Then $E$ is the solvable radical of $G$, and is non-Abelian, so is not (isomorphic to) a subgroup of any Schur multiplier.
|
2025-03-21T14:48:31.028221
| 2020-05-20T13:27:31 |
360866
|
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|
Stack Exchange
|
Quotient distance on $\mathbb{C}P^n$ is equivalent to distance induced by the Fubini-Study metric
Let $n$ be an even, positive integer. Then, is the metric (in the metric-space sense) on $\mathbb{C}P^n$ induced by the Fubini-Study (Riemannian) metric equivalent to the quotient (pseudo?)-metric
$$ d_Q([x],[y]) = \inf\{d(p_1,q_1)+d(p_2,q_2)+\dotsb+d(p_{n},q_{n})\}
> , $$ where the $\inf$ is taken over all finite sequences
$(p_1, p_2, \dots, p_n)$ and $(q_1, q_2, \dots,q_n)$
with $[p_1]=[x]$, $[q_n]=[y]$,
$[q_i]=[p_{i+1}]$, for $i=1,2,\dots, n-1$
and where $d$ is the great circle distance on $S^{2n+1}$, where we identify $\mathbb{C}P^n \cong S^{2n+1}/U(1)$. If this is indeed true, does someone have a reference?
I am not following what $p_i$ and $q_i$ are.
Sorry, I clarified the notation.
It is the quotient metric, so I think the answer is affirmative, but I have to think about your sequences.
Exactly, it's the quotient metric.
do you mean "equivalent" or "equal"?
equivalent, as in the identity map is Bi-Lipschitz.
This is a partial answer: we show that Fubini-Study metric does not exceed the quotient metric(and some ideas for other direction).
Let $(X,d)$ and $(Y,h)$ be metric spaces and let $q:X\to Y$ be a bijection. This map generates an equivalence relation on $X$: $x\sim z\Leftrightarrow q(x)=q(z)$. Moreover, we can view $Y=X/ \sim$.
Let $d_q$ be the quotient metric in the sense of the definition given in the question and the equivalence relation above. Note that $q$ generates one more semi-metric on $Y$: $$d'_q(y,w)=\inf \{d(x,z),~q(x)=y,~ q(z)=w\}.$$ The triangle inequality does not always hold. In fact, $d_q$ is the largest pseudo-metric such that $d_q\le d'_q$.
Proposition. $q$ is $\alpha$-Lipschitz iff $h\le \alpha d_q$.
If $h\le \alpha d_q$, then for any $x,z \in X$ we have $h(q(x), q(z))\le \alpha d_q(x,z)\le \alpha d(x,z)$.
If $q$ is $\alpha$-Lipschitz, then for $y,w\in Y$, any chain $y=y_0,...,y_n=w$, we have $$h(y,w)\le h(y_0,y_1)+...+h(y_{n-1},y_n)\le \alpha(d(x_1,z_1)+...+d(x_n,z_n)),$$ where $q(x_k)=y_{k-1}$ and $q(z_k)=y_k$. Since the chain was chosen arbitrarily, we conclude $h\le \alpha d_q$. $\square$
Now let $X=S^{2n+1}$ with the length distance $d$ induced by the Euclidean metric $\left<\cdot,\cdot\right>$, and let $Y=\mathbb{C}P^n$ with the distance $h$ induced by Fubini-Study metric. Let $q$ be the standard quotient map.
Note that the pull-back $\sigma$ of the Fubini-Study (Hermitean) metric with respect to $q$ is $\sigma_{w}(u,v)=\left<u,v\right>-\left<u,w\right>\left<w,v\right>$, where $w\in S^{2n+1}$ and $u,v\in \mathbb{C}^{n+1}$. Hence, $\sqrt{\sigma_{w}(u,u)}\le\|u\|$, but if we take $u\perp w$, then $\sqrt{\sigma_{w}(u,u)}=\|u\|$. Therefore, the norm of the tangent map $Tq_w:T_wX\to T_{q(w)}Y$ is $1$. Since this is true for all $w$, it follows that $q$ is $1$-Lipschitz with respect to $d$ and $h$, from where $h\le d_q$.
I don't have a complete proof of the other direction, but I think it should be possible to show that $d'_q$ is dominated by $h$ (we may need to switch to the Euclidean distance on $X$ instead of the length distance, but they are equivalent).
For that it may be useful to consider the following metric on $X=S^{2n+1}$: $$\delta(x,z)=\inf_{s,t\in\mathbb{R}}\|e^{is}x-e^{it}z\|=\inf_{s,t\in\mathbb{R}}\sqrt{2-2 Re~ e^{i(s-t)}\left<x,z\right>}=\sqrt{2-2 |\left<x,z\right>|}.$$
Then, $d'_q(q(x),q(z))=\delta(x,z)$. It is possible to show that if $\gamma$ is a smooth curve in $X$, then $$\lim_{t\to 0} \frac{\delta(\gamma(t),w)}{|t|}=\sqrt{\sigma_{w}(u,u)},$$
where $w=\gamma(0)$ and $u=\gamma'(0)$. This means that the length induced by $\sigma$ and $\delta$ coincide on $X$. The problem is that this is for the curve in $X$, not in $Y$, and I don't think any curve in $Y$ can be lifted to $X$.
|
2025-03-21T14:48:31.028583
| 2020-05-20T14:04:00 |
360868
|
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|
Stack Exchange
|
Representations in Archimedean quadratic modules
Let $\mathbb R [X] = \mathbb R [X_1,\dots,X_n]$ and $\Sigma[X] = \big\{ \, f \in \mathbb R[X] \mid \exists r \in \mathbb N, \ g_i \in \mathbb R[X] \colon f = g_1^2 + \dots + g_r^2 \,\big\}$ denote the Sum of Squares polynomials. $M \subset \mathbb R [X]$ is a quadratic module if $1\in M$, $\Sigma [X]\cdot M\subset M$ and $M+M\subset M$.
Let $M \subset \mathbb R [X_1, X_2]$ be the quadratic module generated by $X_1$, $X_2$ and $1-X_1-X_2$. The associated semialgebraic set is $S= \{ (x,y)\in \mathbb R^2 \mid x \ge 0, \ y\ge 0, \ 1-x-y \ge 0 \}$, which is compact. Since $M$ is generated by linear polynomials then $M$ is also Archimedean (Prestel Delzell, Positive Polynomials, Cor. 6.3.5).
Consider the polynomial $f = X_1 X_2 + 1$. Since $f > 0$ on $S$, then $f \in M$ by Putinar Positivstellensatz, i.e. $f = s_0 + s_1 X_1 + s_2 X_2 + s_3 (1 - X_1 - X_2)$ for some $s_i \in \Sigma [X]$.
I am interested in finding the explicit Sum of Squares coefficients in such a representation. It is well know that this is a semidefinite programming problem, but I could not find an easy access tool/software/library to compute explicitly the representation.
Then I am asking for references to this kind of software, and/or to explicit expressions for $s_0$, $s_1$, $s_2$ and $s_3$.
More generally, I am interested in the SoS coefficients in the representation $\prod X_i + 1 \in M(X_1,\dots, X_n, 1- \sum X_i)$.
It seems that this problem has been investigated recently in the literature. See this paper, section 3
|
2025-03-21T14:48:31.028705
| 2020-05-20T14:57:51 |
360876
|
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|
Stack Exchange
|
Degrees of syzygies of points in $\mathbb P^2$
Let $X$ be a collection of points in $\mathbb P^2$ over the complex numbers. Let $I_X$ be the defining ideal. I am interested in knowing when:
The syzygies of $I_X$ contains no linear forms. Since we are in $\mathbb P^2$, this just says that the Hilbert-Burch matrix contains no (non-zero) linear entries. $(*)$
One obvious case when this happens is when we take two general curves $F,G$ of degrees $a,b\geq 2$ and let $X=V(F)\cap V(G)$. Then the syzygies is just the Kozsul relation and has degrees $a,b$. I don't know further examples and would like to know if there are interesting geometric conditions that would imply $(*)$.
One obvious necessary condition is that the generators of $I_X$ have degree at least $2n-2$, where $n$ is the number of generators.
Also, perhaps if you fix the degree $d$ of $X$, then $(*)$ defines a closed subscheme (? I am not sure about this) of the Hilbert scheme $\mathbb P^{2[d]}$. If so, then knowing it's dimension would be nice.
If you take sufficiently many points in general position, shouldn't the condition be satisfied?
@Angelo: ideal of general points tend to have linear entries in the H-B matrix, but it's not clear when.
I see. This seems quite subtle.
I asked David Eisenbud and was told about Exercises 12 and 13 of Chapter 3 in his book "Geometry of Syzygies". Putting together, they show that for a generic set of $n$ points $X$ in $\mathbb P^2$, the syzygies of $I_X$ has no linear forms if and only if $n= \binom{2s+1}{2}+s$, for some positive integer $s$. It is unclear if a complete characterization can be found.
|
2025-03-21T14:48:31.028840
| 2020-05-20T16:07:22 |
360878
|
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|
Stack Exchange
|
SIR model constraint
During these past months, I've heard a lot about some pandemic modelling techniques, specially the so-called SIR model. Before I begin, I'd like to stress that my interest and question are just a matter of curiosity and I don't work with anything even remotely related to this topic. So, I became curious with it and decided to understand it at least superficially. Here are the three differential equations of the model:
$$\frac{dS}{dt} = -\beta I S $$
$$\frac{dI}{dt} = \beta IS - \gamma I$$
$$\frac{dR}{dt} = \gamma I$$
Here, $S, I$ and $R$ stand for the number of susceptible, infected and recovered individuals, respectivelly and $\beta, \gamma$ are two parameters of the model. The point that got me curious was the following. I've read that a constraint to this model is $S+ I+R = N$, where $N$ is the total number of people of, say, a given community. Well, this formula surely makes sense to me, but I don't understand why isn't $N$ also a function of time, $N= N(t)$. It seems reasonable, since the pandemic causes a certain number of deaths. It seems that, when you set $N$ to be fixed, you are modelling a pandemic that causes no deaths. Is that a limitation of the model or is it just me misunderstanding something?
@MattF. thanks for your answer! If we include mortalities in the recovered population, then the formula $S+I+R = N$ totally makes sense but I wonder why not to consider $N = N(t)$ instead. I imagine it is because the equations would be much more difficult to solve, but it is just a guess. Anyway, woudn't we expect the model to be, idk, a bit more accurate if we consider $N$ as a time-dependent parameter?
Add the three equations and see what happens.
$\frac{d}{dt}(S+I+R) = 0$. But this is just a consequence of $N$ being a constant, right?
in the socalled "SIR model with vital dynamics" birth and death processes are added, so $N$ becomes time dependent.
|
2025-03-21T14:48:31.029002
| 2020-05-20T16:45:03 |
360880
|
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|
Stack Exchange
|
Digitalized version of "Cours de topologie algébrique professé en captivité"
It is historically known that Jean Leray gave a course on algebraic topology while captive in the Officer's detention camp XVI in Edelbach, Austria during WW2. (References to this topic include an article by Sigmund, Michor and Sigmund in the mathematical Intelligencer, 27/2 (2005), 41-50.)
Is there a digitalized copy of the course material somewhere?
The course has been published in the Journal de Mathématiques Pures et Appliquées, volume 24 (1945), and can be found here: first part and second part and third part.
A quote from Leray's obituary:
The prisoners in the camp were mostly educated men, career or reserve
officers, many of them still students. As in several other camps, a
"university" was created and Leray became its rector. Classes were
taught, exams were given, and degrees granted, with some degree of
recognition by French authorities of the time. As for research, to
fight the feeling that he might be losing the best productive years of
his life, Leray wanted to resume his work. But he was confronted with
a dilemma. If he continued working in fluid mechanics, he might be
forced to collaborate with the German war effort. Instead, he decided
to pursue some ideas in algebraic topology that he had foreseen during
his collaboration with Schauder.
pages 95-167
pages 169-199
pages 201-248
|
2025-03-21T14:48:31.029152
| 2020-05-20T16:50:30 |
360882
|
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|
Stack Exchange
|
Maximally symmetric hyperbolic 3-manifolds with finite volume
In physics, standard cosmology is build with simple maximally symmetric 3-manifolds (spacelike time-slices of constant curvature, e.g. $S^3$ or less popular the hyperbolic space $H^3$). Since $S^3$ has a finite volume it seems natural to ask whether there also exists a maximally symmetric hyperbolic counterpart which also has a finite volume? (An answer in layman's terms would be fine, if possible.)
You probably want to read about Thurston geometries. Also about the notion of "locally symmetric space". That $H^3$ is "less popular"? well, hyperbolic 3-dimensional geometry has been a major theme in geometric topology in the last 40 years!
Sorry, but less popular in standard cosmology, because it is generally believed that space-time in general relativity is not negatively curved.
In any case: finite volume manifolds with curvature $-1$ usually have a very small amount of symmetries, whence the notion of locally symmetric space.
In GR we sometimes say "locally maximally symmetric" to mean: for every point on the manifold there exists an open neighbourhood of said point which is isometrically diffeomorphic to an open set on a "maximally symmetric" space. So for the purposes of this question said maximally symmetric space would be $H^3$. There should be lots of examples* which are not $H^3$ itself, but whether any have finite volume, I do not know. (*In the Lorentzian case the BTZ black hole geometry is some quotient of maximally symmetric $AdS_3$ but I do not remember the details now.)
Equivalently, such locally maximally symmetric spaces admit the maximum number of Killing vector fields on some neighbourhood of every point.
@AlexArvanitakis: there are many such examples with finite volume. Hyperbolic manifolds are a huge topic of research in low-dimensional topology. The simplest 3-dimensional example with finite volume is perhaps the complement of the figure-eight knot in the 3-sphere (see: https://en.wikipedia.org/wiki/Figure-eight_knot_(mathematics)). The simplest compact example is perhaps Seifert--Weber dodecahedral space (see https://en.wikipedia.org/wiki/Seifert%E2%80%93Weber_space).
Kojima has proved that every finite group occurs as the symmetry group of some compact hyperbolic 3-manifold. Hence there is nothing like a hyperbolic 3-manifold of maximal symmetry.
The same result is true in higher dimensions by Belolipetsky-Lubotzky.
Sorry, but I need some help to understand your answer. As far as I know there is a hyperbolic 3-manifold which satisfies homogenity and isotropy (space of constant sectional curvature -1), but it has an infinite volume. Do I understand right that there is no such manifold with finite volume?
Yes, there is no such manifold. A finite-volume complete hyperbolic manifold has only finitely many symmetries.
@Bruno Martelli thank you for the annotation.
|
2025-03-21T14:48:31.029348
| 2020-05-20T17:09:42 |
360884
|
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|
Stack Exchange
|
Is there any nontrivial characterization of weakly differentiable functions?
When $f\in L_\text{loc}^1$, it's distributional derivative can be defined as $D_{f'}\in\mathfrak{D}'$, such that $D_{f'}(\varphi)=-\int f\varphi'$ for all $\varphi\in\mathfrak{D}$, where $\mathfrak{D}$ is the space of test functions. Then from what I understood, $f$ is said weakly differentiable, if there exists $f'\in L_\text{loc}^1$ such that $D_{f'}(\varphi)=\int f'\varphi$ for all $\varphi\in\mathfrak{D}$. It is also well-known fact that not all locally integrable functions are weakly differentiable, because $L_\text{loc}^1$ is proper subset (in sense of isomorphism) of $\mathfrak{D}'$.
I tried to picture these definitions into my head in visual ways (which is not necessary), but having hard time to fully characterize the definition of weak differentiability. What I can say is $f$ must not make any jumps. I also tried to understand the space on the space of distributions : if we let $D(L_\text{loc}^1) \subset \mathfrak{D}'$ as the set of all distributions such that there exists it's representation $f\in L_\text{loc}^1$, and $d(L_\text{loc}^1)\subset\mathfrak{D}'$ a set of all distributional derivatives, then the space of weakly differentiable distributions (which means it's representation in $L_\text{loc}^1$ is weakly differentiable) will be $W(L_\text{loc}^1):=D(L_\text{loc}^1)\cap d(L_\text{loc}^1)$. But I don't have any idea to characterize this space in other ways.
Here are the questions : For any given $f\in L_\text{loc}^1$,
Is there any sufficient condition (which is not too trivial, for
example, $f$ is differentiable in classical sense) for $f$ to be
weakly differentiable?
Is there any necessary condition (again, not trivial one) for $f$ to be weakly differentiable?
In other words, is there any nontrivial characterization of $W(L_\text{loc}^1)$?
Thank you in advance.
In one dimension, $f'$ (distributional derivative) being locally integrable is equivalent to $f$ being absolutely continuous. The result is discussed in my distribution lecture notes here: https://math.ou.edu/~cremling/teaching/ln.html
The Sobolev space $W^{1,1}$ of $f\in L^1$ such that the distribution derivative $f'$ belongs also to $L^1$ is a nice space (by the way included in $L^{\frac{d}{d-1}}$). The space $BV$ of $f\in L^1$ such that the distribution derivative $f'$ is a Radon measure is larger (and nice as well).
Definition.
If $U\subset\mathbb{R}$ is open, we say that $u\in {AC}(U)$
if $u$ is absolutely continuous on every compact interval in
$U$. Let $\Omega\subset\mathbb{R}^n$. We say that
$u$ is absolutely continuous on lines, $u\in {ACL}(\Omega)$,
if the function $u$ is Borel measurable and for almost every line
$\ell$ parallel to one of the coordinate axes, $u|_\ell\in AC(\Omega\cap\ell)$.
Since absolutely continuous functions in dimension one are differentiable a.e.,
$u\in {ACL}(\Omega)$ has partial derivatives a.e.
Theorem. $f\in L^1_{\rm loc}(\Omega)$ has weak derivative $\nabla f\in L^1_{\rm loc}(\Omega)$ if and only if $f\in ACL(\Omega)$ and the classical partial derivatives (which exist a.e.) are in $L^1_{\rm loc}(\Omega)$. Moreover the classical partial derivatives of $f$ which exists a.e., equal to the weak partial derivatives.
Remark. Functions that are equal a.e. are identified so by writing $f\in ACL(\Omega)$ we mean that $f$ equals a.e. to a function that belongs to $ACL(\Omega)$.
The above result is Theorem 4.21 in [EG] or Theorem 2.23 in [H] or Theorem 1 p. 4 and Theorem 2 p. 6 in [M].
[EG] L. C. Evans, R. F. Gariepy, Measure theory and fine properties of functions. Revised edition.
Textbooks in Mathematics. CRC Press, Boca Raton, FL, 2015.
[H] P. Hajłasz Non-linear elliptic partial differential equations., 2010.
[M] V. Maz'ya, Sobolev spaces with applications to elliptic partial differential equations. Second, revised and augmented edition. Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], 342. Springer, Heidelberg, 2011.
Do we have local integrability of the gradient from ACL?
@NikitaEvseev Hi Nikita, you are right. I think one has to assume local integrability of partial derivatives. Thank you.
|
2025-03-21T14:48:31.029620
| 2020-05-20T17:23:47 |
360886
|
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|
Stack Exchange
|
Uniqueness of solution of the wave equation
Consider the wave equation
$$\frac{\partial^2 u}{\partial t^2}-\sum_{i=1}^n\frac{\partial^2 u}{\partial x_i^2}=0$$
with initial conditions
$$u|_{t=0}=\frac{\partial u}{\partial t}|_{t=0}=0$$
Does it follow that $u\equiv 0$? If not, are there sufficient extra conditions which guarantee that?
Remarks. (1) For $n=1$ the above uniqueness statement holds. I am particularly interested in $n=3$.
(2) Another version of initial condition which might be of interest for me is $u(x,t)=0$ for any $t\leq 0$ and any $x\in \mathbb{R}^n$.
Sorry if this question is too elementary for this site.
Version 2 follows from Version 1: In version 2 at $t = -1$ your assumptions imply that $u(x,-1)$ and $\partial_t u(x,-1)$ both vanish identically.
Since I haven't been able to track down Selberg's lecture notes since he moved to Bergen, and since the proof of the result I mentioned in this comment is super-short anyway, let me just include a proof here.
Theorem Let $\psi\in \mathscr{D}'(\mathbb{R}^{n+1})$ be a distributional solution of $\Box \psi = 0$. Suppose further that $\mathrm{supp}(\psi) \cap \{ t \leq 0\} = \emptyset$. Then $\psi \equiv 0$.
Proof: It suffices to show that for every $f\in C^\infty_0(\mathbb{R}^{n+1})$ that $\langle \psi, f\rangle = 0$. Let $T$ be sufficiently large such that $f \equiv 0$ for $t \geq T$. Solve the Cauchy problem $\Box u = f$ with "initial" data $u(T,x) = \partial_t u(T,x) = 0$; this can be done using, e.g. the representation formula. Notice that the representation formula gives $u \equiv 0$ for $t \geq T$. Additionally, the representation formula (finite speed of propagation) shows that for any $S$, on $\{t \geq S\}$, there exists some $R$ such that $u(t,x) \equiv 0$ when $x \geq R$. Let $S < 0$ and choose a smooth cut-off $\chi$ such that $\chi(t) \equiv 1 $ when $t \geq 0$ and $\chi(t) \equiv 0$ when $t \leq S$.
The function $\chi(t) u(t,x) \in C^\infty_0(\mathbb{R}^{n+1})$, and hence
$$ 0 = \langle \Box \psi, \chi u\rangle = \langle \psi, \Box(\chi u) \rangle = \langle \psi, f \rangle + \langle \psi, - u \chi'' - 2 \chi' \partial_t u \rangle.$$
Both $\chi'$ and $\chi''$ are supported on $[S,0]$. So $-u \chi'' - 2 \chi' \partial_t u$ is $C^\infty_0(\mathbb{R}^{1+n})$ with support in $\{ t \leq 0\}$, and hence its pairing against $\psi$ vanishes by assumption.
The same argument fails for the heat equation due to the failure of finite speed of propagation. Solutions of $\partial_t u - \Delta u = f$ when $f$ has compact support may have spatial tails of size $\exp(-|x|^2)$. You can compensate this by not allowing arbitrary distributions but only distributions "growing no faster than $\exp(|x|^2)$" (interpreted suitably), and get a version of Tychonoff's uniqueness theorem.
Nicely written!
+1 Very nice proof. It seems to me that the technique you used in the proof can be almost verbatim extended to the case of hyperbolic equations of $C^\infty$-smooth coefficients: while my answer relies on an explicit representation which cannot be easily obtained for variable coefficient equations (Hadamard docet), you need only to know that a compactly supported initial datum will propagate with a finite speed and thus remain compactly supported.
@DanieleTampieri: I believe you are correct.
The Wayback Machine has a copy of Selberg's old NTNU webpage. Maybe these are the notes you are referring to?
@HansLundmark: that's probably the right set (though it is best to ask Piero since he was the one who mentioned its existence in the other comment thread). It is funny to look back and think how some of the "modern" results that I was learning as a grad student have now become "classical".
The basic customary (and hidden) assumption I'm aware of which allows to conclude that the Cauchy problem
$$
\begin{cases}
\dfrac{\partial^2 u}{\partial t^2}-\displaystyle\sum_{i=1}^n\dfrac{\partial^2 u}{\partial x_i^2}\equiv\square u(x,t)=0\\
\\
u|_{t=0}=\left.\dfrac{\partial u}{\partial t}\right|_{t=0}=0
\end{cases}\label{cpwe}\tag{CPWE}
$$
has the unique null solution is that $u(x,t)$ is Fourier transformable in the sense of distributions, i.e. it is a slowly increasing (or temperate or Schwartz) distribution (see for example references [1], §2.8.1 pp. 148-150 or [2], §5.1-§5.2, pp. 74-78 for the relevant definitions)
$$
u(t,x)\in\mathscr{S}^\prime(\Bbb R^n\times\Bbb R)
$$
This is a consequence the fact that if we can apply the (partial) Fourier transform $\mathscr{F}_{x\to\xi}:\mathscr{S}^\prime(\Bbb R^n)\to\mathscr{S}^\prime(\Bbb R^n)$ respect to the spatial variable $x\in\Bbb R^n$ to the Cauchy problem \eqref{cpwe}, it becomes the following second order ODE Cauchy problem
$$
\begin{cases}
\dfrac{\partial^2 \hat{u}_p(\xi,t)}{\partial t^2} + |\xi|^2\hat{u}_p(\xi,t)=0
\\
\hat{u}_p|_{t=0}=\left.\dfrac{\partial\hat{u}_p}{\partial t}\right|_{t=0}=0
\end{cases}\label{1}\tag{1}
$$
which admits only (even when considered in $\mathscr{S}^\prime(\Bbb R)$), the (unique) the classical solution $\hat{u}_p(\xi,t)\equiv 0$. The only solution of \eqref{1} induce the uniqueness of the solution ${u}(x,t)\equiv 0$ to the problem \eqref{cpwe} by the isomorphism properties of the Fourier transform (see for example [2], §6.2, pp. 90-92).
Notes
Tempered distributions are, roughly speaking, distributions "that increase at infinity not faster than a polynomial": this perhaps includes the classes of solutions of the wave equations needed in your research. However, the Fourier transform can be extended to more general classes of generalized functions.
Strictly speaking, the hypothesis $u(t,x)\in\mathscr{S}^\prime(\Bbb R^n\times\Bbb R)$ is weaker than the strongest one allowed by this method: we could even assume that $u(x,t)\in\mathscr{S}^\prime(\Bbb R^n)\times\mathscr{D}^\prime(\Bbb R)$ as \eqref{1} can be solved in $\mathscr{D}^\prime$ and has the same and unique classical solution $\hat{u}_p(\xi,t)\equiv 0$: see for example [1], §1.5.2, pp. 25-26.
This method, with the same hypotheses, holds also for the solution of the non homogeneous equation, perhaps answering a question you asked yesterday.
References
[1] Shilov, G. E., Generalized functions and partial differential equations, Mathematics and Its Applications. Vol. 7, (in English)
New York-London-Paris: Gordon and Breach Science Publishers, XII, 345 p. (1968), MR0230129, Zbl 0177.36302.
[2] V. S. Vladimirov (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, Vol. 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR2012831, Zbl 1078.46029.
Thank you very much. Is there a counter-example without these conditions?
@MKO I am not aware of any counterexample. The references (even in the updated version of the answer) deal not specifically to the theory of hyperbolic PDEs. Also ,to my memory, while counterexample based on the growth of the solutions are known for parabolic PDEs even to non specialist, this not so for hyperbolic PDEs.
Uniqueness holds under minimal assumption, i.e., any solution $u\in C^1([0,T];D'(R^n))$ which satisfies the Cauchy problem in the sense of distributions of $D'(R^{n+1})$ with zero initial data, vanishes identically. The proof was written in some detail in a set of lecture notes by Sigmund Selberg on Nonlinear Wave Equations, it should be possible to find them around (or ask him directly)
Thus the behaviour at spatial infinity of the solution has no influence whatsoever on uniqueness
@PieroD'Ancona : Thanks very much. Sounds very interesting. For my current purposes $C^2$ regularity would be enough (but I would prefer to avoid the use of growth at infinity as much as possible). In this generality, is the uniqueness result more available?
If $u$ is a $C^2$ solution of the wave equation with zero initial data then it vanishes identically. No additional conditions are required. This is proved by classical local energy estimates
@PieroD'Ancona : A reference would be really helpful. Thank you.
Well, Evans, as stated before. Section 2.4.
Yes, as for any strictly hyperbolic equation you do have uniqueness and even much better, well-posedness: you can control the Sobolev norm of $u(t)$ by the Sobolev norms of the initial data $u(0), \dot u(0)$. For instance Theorem 23.2.2 in Hörmander's ALPDO III, gives you that result, but you can also look at Evans' book in the part about the wave equation. If you want a simple practical method, just multiply the equation by $\partial_t u$ and integrate by parts.
Thanks. It seems here you assume that some integrals converge (e.g. Sobolev norm is finite). This is interesting, but I am not sure this condition is satisfied in my situation.
@MKO: the result also holds for distributional solutions. It is convenient to state it using your second version: if $\psi$ is a distribution on $\mathbb{R}^{1+n}$ that solves the linear wave equation in the distribution sense, and if the support of $\psi$ is disjoint from ${t \leq 0}$, then in fact $\psi$ is identically zero.
@MKO: furthermore, the energy estimates can also be spatially localized due to finite speed of propagation; so if your worries are "growth at spatial infinity", you don't need to. See e.g. section 3 of my notes.
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2025-03-21T14:48:31.030202
| 2020-05-20T17:28:39 |
360887
|
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|
Stack Exchange
|
The first part of the Hilbert sixteenth problem for elliptic polynomials
A polynomial $P(x,y)\in \mathbb{R}[x,y]$ is called an elliptic polynomial if its highest homogeneous part does not vanish on $\mathbb{R}^2\setminus\{0\}$.
Inspired by the first part of the Hilbert 16th problem we ask that:
Is there an elliptic polynomial $P(x,y)\in \mathbb{R}[x,y]$ of degree $n$ which has a level set $P^{-1}(c)$ with more than $n$ connected components?
Can elliptic polynomials produce $M$-curves?
what do you mean by "last homogeneous part"?
@user347489 for example $x^2+y^2$ is the last homogenous part of $x^2+y^2+ax+by+c$.
Every polynomial is a sum if homogenous polynomials.the last homogenous part is the homogenous part with highest degree.
@user347489 please see the definition of an elliptic PDE
@user347489 I revised the word "last" to "highest"
What have you tried? Is it true for $n=2$? $n=4$? Also, possibly a more intrinsic way to describe your condition is in terms of projective space. Let $\overline P(x,y,z)\in\mathbb R[x,y,z]$ be the homogenization of your $P$. Then your condition is that $\overline P=0$ has no points on the line $z=0$ "at infinity" in $\mathbb P^2(\mathbb R)$.
@JoeSilverman thank you very much for your comment and very helpful suggestion on the projectivization of the question. I did not try the problem for any n but I think for n=2 each level set is a circle or an ellips.
What's an "$M$-curve" in this context?
@NoamD.Elkies An M curve is described here https://en.wikipedia.org/wiki/Harnack%27s_curve_theorem
An elliptic polynomial of degree $n=2m$
can have at least $m^2=n^2/4$ real components.
In particular the number of components can exceed $n$ once $n \geq 6$.
For example, here is a Sage plot of the nine-component sextic curves
$(x^3-x)^2 + (y^3-y)^2 = \epsilon$ for $\epsilon = .02, .07, .14$
in the square $|x|,|y| < 1.25$:
In general if $P,Q$ are polynomials of degree $m=n/2$, each with
$m$ distinct real roots, then the degree-$n$ curve $P(x)^2 + Q(y)^2 = \epsilon$
has $m^2$ components for $\epsilon>0$ sufficiently small;
as $\epsilon \to 0$ the components approximate ellipses (or circles)
centered at the $m^2$ points $(x,y)$ with $P(x)=Q(y)=0$.
Thank you very much for your very interesting answer.
You can get all the way up to $\binom{n-1}{2}+1 = g+1$, where $g$ is the genus. This is the maximal number of connected components a real curve of genus $g$ can have, so this is optimal.
To do this, I will use Viro's patchworking method. Itenberg and Viro already give an example of how to use patchworking to build a plane curve with $g+1$ connected components, so I will just show how to tweak it to use an elliptic polynomial.
I can't write a better explanation of patchworking than the one I just linked, so I'll assume you read it.
Take the triangulation from Figure 7. .
Along each edge on the outside of the figure, there are $2n$ small triangles. Pair them into $n$ pairs, and merge each pair with its common neighbor to make $n$ trapezoids. Leave the rest of the figure as before. It is easy to check that this subdivision is still coherent. The resulting plane curve has the same topology as the original figure, but the big loop which crossed the line at infinity $2n$ times before is now disjoint from it.
Poking around Viro's website, I came across slides from a talk on Hilbert's 16th problem. On page 48, he says that Hilbert, in 1891, found a construction of a curve with $g+1$ real components by perturbing a union of two conics. .
The slides don't give a detailed explanation of Hilbert's construction, but it looks like it would produce an elliptic polynomial.
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| 2020-05-20T17:55:56 |
360888
|
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Stack Exchange
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Notions of Lie 2-groupoids
The term Lie $2$-groupoid is used in the literature in more than one context. Some examples are given below:
Ginot and Stiénon's paper $G$-gerbes, principal $2$-group bundles and characteristic classes defines a Lie $2$-groupoid to be a double Lie groupoid in the sense of Brown - Determination of a double Lie groupoid by its core diagram satisfying certain conditions.
Rajan Amit Mehta and Xiang Tang's paper From double Lie groupoids to local Lie $2$-groupoids says that "The simplicial approach to $n$-groupoids goes back to Duskin [Higher-dimensional torsors and the cohomology of topoi: the abelian theory] in the discrete case, and the smooth analogue appeared in [Henriques - Integrating L-infinity algebras]." They define Lie $2$-groupoid to be a simplicial manifold $X=(X_k)$ with some condition on the horn maps.
Matias del Hoyo and Davide Stefani's paper The general linear $2$-groupoid defines a Lie $2$-groupoid to be a Lie $2$-category and a $2$-category satisfying certain conditions.
The n-lab page on Lie $2$-groupoids says "a Lie 2-groupoid is a 2-truncated $\infty$-Lie groupoid". No further discussion is given there. Clicking the link $\infty$-Lie groupoid takes you to the page Lie $n$-groupoid, which does not contain many details.
Questions :
Are these notions genuinely different notions introduced for different purposes or are they all the same candidate wearing different outfits?
Are there other notions of Lie $2$-groupoids in the literature?
Any comment/reference that helps to understand the above notions are welcome as answers...
Please do not use $\infty$-category theory language. I really do not know anything about it at present.
Sorry for so many typos. It was around 1 AM when I typed this. I misremembered Ginot and Stienon’s paper as some paper by Ginot alone. Thanks for correcting typos @LSpice
For your first question:
They are essentially all the same thing: some globular, some simplicial (taking the nerve goes from the former to the latter). The only subtlety is perhaps in the requirement on the maps $d_{2,0}, d_{2,2}$ to be surjective submersions in the del Hoyo–Stefani paper. This is not unusual for the simplicial approach to n-groupoids in non-finitely complete sites: the Kan condition (which is what these maps are about) shouldn't be just encoded by a certain map being an epimorphism, but being a cover of some sort. This is also the content of Definition 1.2 in Henriques' paper (note that you have linked to the arXiv version 2 of the paper, and that's what I'm referring to. Version 1 has slightly different material and you should also check it out). I don't recall offhand if this is automatic for the globular definition as in Ginot–Stiénon's paper.
So I would say with some confidence that 2.–4. are the same, apart from dealing with a 2-groupoid vs the nerve of a 2-groupoid, and 1. might be very slightly more general, though only in a small technical hypothesis that a certain small class of surjective maps are not required to be submersions a priori. I didn't dig into the techicalities of Brown–Mackenzie (cited by Ginot–Stiénon) to see if one gets submersions automatically. Ideally the nLab page on Lie 2-groupoids would get updated by some friendly soul who wants to put in the work to explain the more elementary viewpoint.
On a historical note, apart from the technical hypothesis, these definitions essentially go back to Charles Ehresmann, who defined (strict) 2-categories, double categories, internal categories (including Lie categories) and so on. Focusing specifically on the case 2-groupoids/double groupoids/Lie groupoids arose slowly, partly driven by Brown, Mackenzie, Pradines, Haefliger, .... and so on.
As far as different notion of 2-groupoids go, one can consider the composition functor on hom-groupoids to be a map of their associated stacks, hence an anafunctor as opposed to an internal functor. This viewpoint is implicit in
Christian Blohmann, Stacky Lie groups, Int. Mat. Res. Not. (2008) Vol. 2008: article ID rnn082, 51 pages, doi:10.1093/imrn/rnn082, arXiv:math/0702399
where the one-object case is considered (the general stacky notion is discussed by Breen in Bitorseurs et cohomologie non abélienne. In The Grothendieck Festschrift, Vol. I, Progr. Math., vol. 86, pp. 401–476 (1990)). More explicitly this notion is considered in
Chenchang Zhu, Lie $n$-groupoids and stacky Lie groupoids
arXiv:math/0609420
(with a more general theory in her followup paper 0801.2057) using the language of stacks. In general, what this means is that if you take the viewpoint on internal 2-groupoids as in Definition 2.1 of
A bigroupoid's topology (or, Topologising the homotopy bigroupoid of a space), Journal of Homotopy and Related Structures 11 Issue 4 (2016) pp 923–942, doi:10.1007/s40062-016-0160-0, arXiv:1302.7019.
(there given for topological bigroupoids, but one could repeat the definition mutatis mutandis for Lie 2-groupoids, taking the natural transformations $a,r,l,e,i$ to identities), then the hom-groupoid is a Lie groupoid over the square of the manifold of objects, with some properties, and then composition is an anafunctor between Lie groupoids over this manifold. And so on. This provides and extra layer of weakness to the structure. This viewpoint (in the special case of Lie 2-groups) was also used by Chris Schommer-Pries. One should view this as some kind of internal enrichment of a Lie groupoid in the (cartesian monoidal) category of differentiable stacks, much like ordinary 2-groupoids are groupoids enriched in the category of groupoids.
Ok. Thanks for the answer.. There are so many things to digest here :) I will go one by one and ask if I have any specific question...
I do not think the notions of Lie 2-groupoids in (2) and (3) are the same in the smooth case. The main difference lies in remark 1.3 in Henriques' paper. Here, he states that he does not assume the horn projections to have smooth sections. A choice of multiplication and inversion on 1-arrows corresponds to such a smooth section. In (3), they assume a multiplication operation for 1-arrows. So, there is a subtle difference here.
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2025-03-21T14:48:31.031042
| 2020-05-20T18:02:12 |
360889
|
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Stack Exchange
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Can the place of publication be harmful to one's reputation?
What can be said about publishing mathematical papers on e.g. viXra if the motivation is its low barriers and lack of experience with publishing papers and the idea behind it is to build up a reputation, provided the content of the publication suffices that purpose.
Can that way of getting a foot into the door of publishing be recommended or would it be better to resort to polishing door knobs at arXiv to get an endorsement?
Personal experience or that of someone you know would of course also be interesting to me.
Few if any serious mathematician looks at anything on Vixra.
Neither the ArXiv nor viXra, nor for that matter, posting on your personal website, count as a mathematical publication that enhance or detract from one's reputation. Publication in a reputable peer reviewed mathematics journal counts. (Accepted for publication also counts, given the long lag time between acceptance and publication. Accepted pending revision is almost as good. Under review does not count.) A paper posted on the ArXiv may be studied by others in your field, so that can help build reputation. Posting papers on viXra and putting them on your CV could be harmful.
viXra (just like arXiv) is a preprint server, and does not facilitate the formal publication process. What really counts is refereed articles in serious mathematical journals.
I'm not a professional mathematician and I have no experience with viXra or arXiv, thus my comment is unrelated to your question. I would like to add that it seems to me a good idea (if it is legitimate) that this site had a math journal associated to the site. I mean a mathematical journal associated to this site MathOverflow (it could even send periodically physical posters/letters printed on paper to university faculties showing a choice of the most relevant or new posts). I think that this would have many benefits, for example to recruit professors or PhD students, or retired professors.
arXiv and viXra shouldn't really be mentioned in the same sentence. The arXiv has a system so that not just anyone can submit, and what is submitted should look like a serious research paper (not checked for accuracy, but should at least superficially look professional). Anyone can submit to viXra and the standards are much, much lower. As in, it accepts stuff that looks like a bad, fail-grade school project.
If your work is arXivable, and you should be able to tell by comparing it to other papers in your field that appear there, you should really try and hunt down someone to endorse you. Once the paper is on the arXiv, you then have a better chance of getting other mathematicians to look at it if you email them. Getting over the minimal barrier that the arXiv has means that they can give more of a benefit of the doubt to an unannounced email referring to an unpublished paper from a relatively unknown author. Saying "read my attached paper" or "look at this viXra link" is a bit of a red flag; ...
...whereas saying "you might be interested in my recent arXiv preprint 20xx.?????" means they only have to overcome their own lack of time.
@DavidRoberts : I have come to realize that the arXiv can throw up more than a "minimal barrier." It isn't so easy for someone not in academia, or even in academia but publishing in a new area, to obtain an endorsement. A friend of mine, a math major, graduated with highest honors from Princeton University and then became a lawyer. He wrote an excellent article on an elementary topic but we couldn't find an endorser. People were surprisingly (to me) averse to endorsing. Many seem unaware of the barriers to the arXiv; e.g., see this MO question.
Yes, publishing in a journal which is known to be a predatory journal will harm your reputation. Vixra is a joke and contains many articles which are openly jokes or spoofs.
Yes, the place of publication can absolutely hurt your reputation. Specifically, I can tell you from having served on many hiring committees (and from conversations with professors at other universities about their hiring committees and tenure processes), that publications in predatory journals can hurt you. I'm talking specifically about journals whose model is to get the author to pay them, and whose peer review standards are a joke. Publications in journals like that can be interpreted as an author trying to side-step the normal process, or unethically inflate their numbers.
It may be hard to break into the absolute top journals with your first few papers (unless you have a famous advisor/coauthor or went to a prestigious school). But there are plenty of good journals around and after a track record of publishing in good journals you will have less difficulty publishing in top journals (of course, it'll always be extremely hard to publish in the Annals and other super elite journals). For people starting out, I recommend at least checking Beall's list of predatory publishers to be sure you don't end up publishing somewhere that might be frowned upon later in your career. Also, don't let fear paralyze you from trying. Lots of editors and referees will go gently on new PhDs. I wish this was even more common, rather than pushing young people out of academia.
"unless you have a famous advisor/coauthor or went to a prestigious school" <--- yes. I wonder why...
@DavidRoberts maybe because that's the next best replacement for the former criterion of being of noble descent to be admiited to a gentlemen's club.
"unless you have a famous advisor/coauthor or went to a prestigious school" > shouldn't this be "unless you have produced outstanding scientific results"...?
@Earthliŋ : On an ideal planet, yes.
If you want to build a reputation as a mathematician, post your preprint on the arXiv (this is a preprint server, not counted as publication, the posts are not refereed, and there are essentially no "barriers").
Then send your paper to a mainstream mathematical journal. Avoid those journals which are not reviewed in Mathscinet.
(Recently, many journals proliferated which do not really referee papers, but charge the authors for publication. Avoid them, if you want to build a reputation. The main criterion of a mainstream journal is that all its papers are reviewed in Mathscinet.) Mathscinet also publishes a rating of journals according to their citation rates. Publication in a highly rated journal (say of the first 100) will probably be good for your reputation,
but the "barriers" there are also high.
Still I believe that one's reputation is based more on the quality of papers, rather then the rating of the journals where they are published.
For example, the reputation of G. Perelman is mainly based on a few preprints
which he posted on the arXiv. He did not care to send them to journals.
This is an extreme example, but it is not unique. My 3-d highest cited paper is published in a volume of conference proceedings which is not even a journal (arXiv did not exist yet). My two most cited papers are in the journals which do not enter the list of 100 top journals according to MSN rating (though I also published in top journals). I conclude that there is very little correlation between the rating of an individual paper and the rating of the journal.
Pre-2002, "the reputation of G. Perelman [was] mainly based" on his proof of the Soul Conjecture, published in the Journal of Differential Geometry, a rather exclusive outlet, and work on spaces whose curvature is bounded below (which he spoke on at the 1994 Zurich ICM), published in places like JAMS.
At home, I cannot search MathSciNet to see if a given journal is indexed, and I cannot access the MCQ tables (my uni subscribes, so I know what all this is and how to get them). Recommending such resources may not be as helpful to the OP as you hope.
Sure, but he could only get away with that because he had a strong reputation to start with (ICM speaker!), and he then managed to solve a once-in-a-century problem. This is not a good example for someone who is asking about basic facts of the informal academic reputation system.
@David Roberts: I deleted my comments.
In complement to David Roberts' comment: I remember that when Perelman posted his papers on arXiv (Nov. 2002 to July 2003), I was told it was taken very seriously, based on his reputation. And obviously at that time, it meant his reputation prior to the arXiv preprints, which was based on published papers (and probably further mathematician activity, but published papers is the easiest/laziest clue to detect now).
regarding "there are essentially no barriers": [Tsang, Sandro. (2020). Re: How to get endorsement in arXiv?] (https://www.researchgate.net/post/How_to_get_endorsement_in_arXiv/5e456e08a7cbaf880b7965d4/citation/download) tells a different story for people that are in my situation, namely not affiliated with a research institution and no network of supportive arXiv endorsers
"I conclude that there is very little correlation between the rating of an individual paper and the rating of the journal." this is definitely true, and supported by hard data in certain science fields, where the studies have been done (biology, I think).
I agree with Manfred Weis's comment about barriers. See also my comment under the question itself, and this MO question.
|
2025-03-21T14:48:31.031696
| 2020-05-20T18:54:07 |
360891
|
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|
Stack Exchange
|
Distribution of prime ideals by norm
I was wondering if there exist results in arithmetic statistics which count the number of prime ideals of bounded degree in a given number field, for instance a (real or imaginary) quadratic number field? Concretly, I wonder if we have asymptotics of
$$\{\mathfrak{p}\in Spec(\mathcal{O}_K)\mid Nm(\mathfrak{p})\leq L\}$$
as $L\rightarrow \infty$? All the articles I've found give asymptotics for the number of number fields of boundd discriminant.
Yes, look for the prime ideal theorem.
This is the prime number theorem for $\zeta_K(s)$, the proof is the same as the one for $\zeta(s)$
|
2025-03-21T14:48:31.031805
| 2020-05-20T19:07:18 |
360892
|
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|
Stack Exchange
|
Theta-function in the lower half-plane
Standard theta function
$$\vartheta(q)=\sum_{n=-\infty}^\infty q^{n^2} \qquad\qquad(1)$$
has a natural boundary of analyticity at $|q|=1$. This means that it can not be used to regularize expressions of the type $\sum_{n=-\infty}^\infty q^{n^2}$ with $|q|>1$. I however encounter this type of sums in some heuristic computations and would like to have some way to make sense of it (there is a finite answer available for comparison). In a similar context heuristic gives $\sum_{n} n^k$ and the correct answer agrees with the zeta-function regularization $\zeta(-k)$. Is there any similar go-to way to think about (1) but with $|q|>1$?
I am not sure about $\theta$ itself, but your question is intimately linked to the notion of quantum modular form introduced by Don Zagier, which extends domain of validitiy from the upper half-plane by "going through" the roots of unity in the $q$ domain, equivalently through the rationals in the $\tau$ domain.
I show here the following, so i'd say it's simialir
https://math.stackexchange.com/questions/700299/a-ramanujan-like-summation-is-it-correct-is-it-extensible/3859675#3859675
$$\sum_{x=1}^{\infty} (c)^{-x^2}=\frac{\sqrt{\pi}}{2 \sqrt{ln(c)}}-1/2+\frac{ \sqrt{\pi}}{\sqrt{\ln(c)}}\sum_{x=1}^{\infty}e^{\frac{-\pi^2 x^2}{\ln(c)}}$$
$$\sum_{x=-\infty}^{\infty} (c)^{-x^2}=\frac{\sqrt{\pi}}{\sqrt{ln(c)}}+\frac{2 \sqrt{\pi}}{\sqrt{\ln(c)}}\sum_{x=1}^{\infty}e^{\frac{-\pi^2 x^2}{\ln(c)}}$$
If you are still active, I would be very interested to see the finite values you obtain in a physical context. For now, however, I will present two equal approaches we could take to regularize the sum.
Alternating Case
Ultimately, many series diverge because summation leaves some type of $x^N$ artifact remaining. The goal is to remove that artifact so that we end up with a sum that converges everywhere.
As an example, in the case of the geometric series the partial sums are $\sum_{n=0}^N x^n = \frac{1 - x^{N+1}}{1-x}$. If we let $S(N) = \sum_{n=0}^N x^n$, and define $S$ by the process $S(N+1) = S(N) + x^{N+1}$, then we realize we can take the sum both forwards or backwards. To take the sum backwards, write $S(N-1) = S(N) - x^N$, which gives rise to the sum $-\sum_{n=-1}^{-\infty} x^n = -\sum_{n=1}^\infty x^{-n} = \frac{1}{1-x}$ for $|x|>1$. This trick simply allows us to switch the artifact from $x^N$ to $x^{-N}$, which switches the convergent range from $|x|<1$ to $|x|>1$. Generally, we can use this trick to find representations for sums with only a single $x^N$ artifact. For instance $$\sum_{n=0}^{\infty}\sin\left(n+5\right)x^{n} \to -\sum_{n=1}^{\infty}\sin\left(-n+5\right)x^{-n}$$
or
$$\sum_{n=0}^{\infty}ne^{n}\cos\left(n-2\right)x^{n} \to -\sum_{n=1}^{\infty}\left(-n\right)e^{-n}\cos\left(-n-2\right)x^{-n}$$
The example is not too important, the main point is to say that finding a way to remove that $x^N$ artifact allows us to continue/regularize a function.
If we have an alternating function there's a nice approach to make this happen. We take $\sum_{n=0}^N (-1)^n f(n) x^n$ and transform it into
$$\sum_{n=0}^{N/2} f(2n) x^{2n} - \sum_{n=0}^{\frac{N-1}{2}} f(2n+1) x^{2n+1}$$
Next, we approximate $f(n)$ by a taylor series, so that we can define the sum for non-integer values. In the case of the sum $\sum_{n=0}^\infty (-1)^n q^{n^2}$ we get
$$\sum_{k=0}^{N/2} (-1)^n x^n q^{4n^2} = \sum_{n=0}^{N/2} \sum_{k=0}^\infty (-1)^n x^n \frac{(4\ln(q) n^2)^{k}}{k!} = \sum_{k=0}^\infty \frac{(4\ln(q))^{k}}{k!}\sum_{n=0}^{N/2} n^{2k} x^n (-1)^n = \sum_{k=0}^\infty \frac{(4\ln(q))^{k}}{k!} \left[\left(x \frac{d}{dx}\right)^{2k} \frac{1-x^{\frac{N}{2}+1}}{1-x}\right] $$
So, now we take $N=1$ and get the final sum as
$$\sum_{k=0}^\infty \frac{(4\ln(q))^{k}}{k!} \left[\left(x \frac{d}{dx}\right)^{2k} \frac{1-x^{\frac{1}{2}+1}}{1-x}\right] - f(1) x$$
This doesn't converge for $x=1$, and unfortunately becomes numerically unstable at around $|x-1| \approx .05$. However, I like this approach over some of the other ways of regularizing series because it converges even when $|q|<1$, which guarantees that this method will continuously extend all derivatives. When $x=1/2$, the graph looks like this: where blue is $\sum q^{n^2} (-1/2)^n$, and orange is the continuation.
To extend this to the general complex case, we instead look at the roots of unity. If we are looking at a ray cast in the direction $\theta = \frac{p}{q} \tau$, then $e^{ i \theta}$ is periodic with period q. This leads us to the expansion:
$$\sum_{n=0}^N e^{\frac{p}{q} \tau i n } f(n) = \sum_{k=0}^{q-1} \sum_{n=0}^{(N-k)/q} e^{\frac{p}{q} \tau i k} f(qn+k)$$
Integral Approach
I originally saw this approach from Jorge Zuniga here, and I recommend you read his answer first. However, there are a few things I would add. The general way I view the approach is transforming
$\sum_{n=0}^\infty f(n)$ into $\sum_{n=0}^\infty \mathcal{M}\left\{\mathcal{M}^{-1}\left\{f(n)\right\}\right\} = \mathcal{M}\left\{\sum_{n=0}^\infty \mathcal{M}^{-1}\left\{f(n)\right\}\right\}$. Borel's method can be viewed as doing almost this same transformation, except with the Laplace instead of the mellin transformation. In particular, if we look at $\sum \mathcal{L}\left\{\mathcal{L}^{-1}\{f(n)(1/x)^n\}\right\}$ we get
$$\sum \mathcal{L}^{-1}\{f(n)(1/x)^n\} = \sum \frac{f(n)x^n}{n!}$$
and then
$$ \sum \mathcal{L}\left\{\frac{f(n)x^n}{n!} \right\} = \sum \int_0^\infty e^{-t} \frac{f(n)x^n}{n!} dt = \int_0^\infty e^{-t} \sum \frac{f(n)x^n}{n!} dt$$
It makes sense to apply the inverse Mellin transform to the function $q^{n^2} = e^{\log(q) n^2}$ because the inverse Mellin transform takes a path that goes up/down the imaginary direction. While $e^{\log(q) n^2}$ grows quickly on the real line, traveling up the imaginary axis it converges to 0 very fast.
Anyway, once you have applied the method, you obtain
$$\sum_{n=0}^\infty q^{n^2} =\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}\frac{1}{t\left(1+xt\right)}\cdot\frac{1}{q^{\frac{1}{4}\log_{q}\left(t\right)^{2}}\ln\left(q\right)^{\frac{1}{2}}}dt$$
which converges only when $q>1$ (if $q<1$, you could obtain an integral representation by applying the mellin transform first, and then the inverse mellin transform, instead of the other way around).
The graph gives exactly the same result as the other approach. The integral approach is shown as black dots: .
This approach makes some interesting properties easier to see. For convenience, write
$$\theta(w,x) = \sum_{n=0}^\infty w^{n^2} x^n$$
Then the theta function satisfies the following equation inside the circle
$$\frac{\theta(w,x)-1}{x} = w\theta(w,w^2 x)$$
And it also formally holds this relationship outside the circle. Moreover, the partial sums asymptotically hold this relationship for $w>1$. We could show that the integral extension satisfies this functional equation.
Update
I was randomly thinking back on this question, and it occured to me to try to follow your approach using residues. The residue approach is well suited for this problem, because $e^{z^2}$ goes to zero very rapidly on the imaginary direction, so we can get a contour that converges even when the sum diverges. Thus, we can write
$$\vartheta(w,x) = \sum_{n=0}^\infty w^{n^2} (-1)^n x^n = \frac{1}{2 \pi i} \int_{-1/2 - i \infty}^{1/2 + i \infty} w^{n^2} x^t \Gamma(-t)\Gamma(t+1) dt$$
This has the benefit of now converging also when plugging in complex values for both $w$ and $x$. Moreover, previously, I had an issue with uniqueness because I only had the one functional equation $-\frac{\vartheta(w,x)-1}{x} = w\vartheta(w,w^2x)$. However, I have recently discovered the differential equation that is supposed to pair with that one, which is
$$\vartheta_w(w,x) = x^4 w^{11} \vartheta_{xx}(w, w^4 x) + 4 x^3 w^7 \vartheta_x(w, w^4 x) - x^2 w^2 \vartheta_x(w,w^2 x) + 2 x^2 w^3 \vartheta(w,w^4 x) - x \vartheta(w,w^2 x)$$
These two equations together with some initial conditions should uniquely define $\vartheta$, since the first equation allows you to extend the definition of $\vartheta(w,x)$ for any fixed $w$ into all values of $x$. The second equation allows you to use knowledge of $\vartheta(w,x)$ at all values of $x$ to get $\vartheta(w,x)$ at different values of $w$. These two functional equations actually naturally lead to the natural boundary that emerges at $|w|=1$, since values where $|w|<1$ will always depend on values of $\vartheta$ where $|w|<1$, and similarly points where $|w|>1$ will always depend only on values of $\vartheta$ where $|w|>1$. So I imagine the initial conditions would have to specify the behaviour on the boundary (probably in a distributional sense) to uniquely define the function.
Hi, thanks for an elaborate answer! Unfortunately I can not recall my motivation vividly enough, but I think I was playing around with expressing integrals as sums over residues. I noticed that sometimes the contour can not be closed (because the function does not decay where it would need to) and the sum of residues diverges, but regularized sum gives the right answer for original finite integral. So the sum in OP is can probably interpreted as a formal sum of residues for some converging integral.
The answer to the question being asked for, on Generalized Analytical Continuation (GAC) of $\theta_i, i=1,2,3,4$ functions can be found in Addenda 2 of this MO discussion that extends Borel´s summation (by means of Nachbin's Theorem) when nome $q$ holds $|q|>1$.
If you look such addenda with statistical eyes you realize that such formulation is not unique. In fact, the Integral for $s(\zeta,q)$ and their derivatives comes from an infinite sum of moments of a log-normal distribution that, as it is known, it cannot be uniquely determined by their moments.
So different measures can produce the same $s(\zeta,q)$ series. Thus a unique base $\theta_i$ function for $|q|<1$ can be analytically continued for $|q|>1$ to multiple (a truly infinite family of) integral functions. I rather prefer to talk about exo-analyticity since there is no "continuation" beyond the natural boundaries (the unit-$q$ disk in this case).
To illustrate this we use a classical example from Stieltjes
Stieltjes, T. -J., Investigations on continued fractions, Toulouse Ann. VIII, J1-J122 (1894); ibid. IX, A1-A47 (1895); C. R. CXVIII, 1401-1403 (1894). ZBL25.0326.01.
fitted to our notation with $\zeta\in\mathbb{C}\backslash(-\infty,0),\space q\in\mathbb{C}\wedge|q|>1$ and $|p|<1$,$$s(\zeta,q)=\sum_{n=0}^{\infty}(-\zeta)^n q^{n^2}=\frac{1}{2\sqrt{\pi\log q}}\int_{0}^{\infty}\exp(-\frac{1}{4}\frac{\log^2t}{\log q})\cdot\frac{1+p\sin(\pi\log_qt)}{1+t\zeta}\cdot\frac{dt}{t}$$
$p$ is an additional dummy parameter that does not contribute to the sum since all moments with the $\sin$ term vanish. Anyway the measure used in the mentioned MO discussion is a simple one (it corresponds to an unimodal distribution with $p=0$) that comes from basic Mellin Transforms. It allows to define a single equivalent exo-function for every $\theta_i$ Fourier $q$-series.
I leave a couple of references here
López-García, M., Characterization of solutions to the log-normal moment problem, Theory Probab. Appl. 55, No. 2, 303-307 (2011); translation from Teor. Veroyatn. Primen. 55, No. 2, 387-391 (2010). ZBL1225.60024.
Leipnik, R., The lognormal distribution and strong non-uniqueness of the moment problem, Theory Probab. Appl. 26, 850-851 (1982). ZBL0488.60024.
Sorry if this is too tangential, but do if any measures stand out in terms of their properties related to differentiation? I noticed when looking at the theta function in relation to the heat equation that the continuation from the unimodal distribution you provided seemed to have some nice properties such as satisfying the PDE and having a simple effect on the boundary. Do you know if different choices of measures have any known characterization in terms of their relationship to PDE's? I'm interested to read more on what you have said here--if any sources do exist
@Caleb_Briggs, I do not have much information, but I left some references. This topic deserves more research. Some ultra fast convergent series with super-factorial speed can build a natural boundary. Behind this boundary series is wildly divergent with terms that grow beyond factorial. The sequence of terms of these series can be represented by the sequence of moments of some distributions and these are Mellin transforms evaluated at integer values. These super growing divergent terms correspond to moments of some distributions. The growing rate defines if these distribution are unique.
|
2025-03-21T14:48:31.032685
| 2020-05-20T19:26:39 |
360894
|
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|
Stack Exchange
|
Cores in the tensor-train decomposition
Let $d_i\in\mathbb N$, $I_i:=\{1,\ldots,d_i\}$ and $u\in\mathbb R^{d_1}\otimes\mathbb R^{d_2}\otimes\mathbb R^{d_3}$. It's somehow clear to me that we may regard $u$ as a three-dimensional array (see Corollary 3 below) and hence consider its entry $u_{i_1i_2i_3}$ at the index $(i_1,i_2,i_3)\in I_1\times I_2\times I_3$. By fixing the middle index $i_2\in I_2$, we may define $$v^{(i_2)}_{i_1i_3}:=u_{i_1i_2i_3}\;\;\;\text{for }(i_1,i_3)\in I_1\times I_3$$ and I guess we can again consider $v^{(i_2)}$ as being an element of $\mathbb R^{d_1}\otimes\mathbb R^{d_3}$.
However, I'm really struggling to see that all the identifications involved are really legitimate. I'm used to think of tensor products in a more abstract fashion (see my definition of the tensor product below$^1$).
Is the process of "fixing the middle index", which is effectively a (surjective?) transformation from $\mathbb R^{d_1}\otimes\mathbb R^{d_2}\otimes\mathbb R^{d_3}$ to $\mathbb R^{d_1}\otimes\mathbb R^{d_3}$, a special case of a more general result of somehow "folding" a tensor product space $E_1\otimes E_2\otimes E_3$ to $E_1\otimes E_3$?
Remark: Tensors $u$ of order 3 like this are so-called cores in the tensor-train decomposition which I'm trying to understand.
$^1$ If $E_i$ is a $\mathbb R$-vector space, I'm defining $$(x_1\otimes x_2)(B):=B(x_1,x_2)\;\;\;\text{for }B\in\mathcal B(E_1\times E_2)\text{ and }x_i\in E_i,$$ where $\mathcal B(E_1\times E_2)$ is the space of bilinear forms on $E_1\times E_2$, and $$E_1\otimes E_2:=\operatorname{span}\{x_1\otimes x_2:E_i\in E_i\}\subseteq{\mathcal B(E_1\times E_2)}^\ast.$$
Lemma 1: If $I$ is a finite nonempty set, $(e_i)_{i\in I}$ is the standard basis of $\mathbb R^I$ and $E$ is a $\mathbb R$-vector space, then the linear extension $\iota_1$ of $$a\otimes x\mapsto(a_ix)_{i\in I}\tag1$$ is an isomorphism between $\mathbb R^I\otimes E$ and $E^I$.
Lemma 2: If $I_i$ is a finite nonempty set, then $$\iota_2:\left(\mathbb R^{I_2}\right)^{I_1}\to\mathbb R^{I_1\times I_2}\;,\;\;\;a\mapsto\left(a(i_1)(i_2)\right)_{(i_1,\:i_2)\in I_1\times I_2}$$ is an isomorphism.
Corollary 3: If $k\in\mathbb N$ and $d_1,\ldots,d_k\in\mathbb N$, then $$\bigotimes_{i=1}^k\mathbb R^{d_i}\cong\mathbb R^{d_1\times\cdots\times d_k}\tag2.$$
Proof: This follows by induction from Lemma 1 and Lemma 2.
Yes, mapping $A \otimes B \otimes C$ to $A \otimes C$ by choosing a coordinate of $B$ (equivalently, basis element of dual space $B^*$) is indeed a special case of choosing any element of $B^*$, considering it as a map $B \to k$ (where $k$ is the field), and then extending that map to $A \otimes B \otimes C \to A \otimes k \otimes C \cong A \otimes C$.
Your $v^{(i_2)}$ is sometimes called a slice, or more specifically a $2$-slice, since the index in the $2$nd position was fixed. This map, that sends a tensor to its $i_2$'th $2$-slice, is surjective: any matrix arises as the $2$-slice of a tensor given by sticking that matrix into the appropriate entries, and filling the rest with any old thing (zeros, whatever).
Thank you for your answer. The isomorphism is $(a\otimes b\otimes c)(b^\ast):=b^\ast(b)(a\otimes b)$, right? Is there an established notation which lets me denote the evaluation of $(a\otimes b\otimes c)$ at a basis functional in a more convenient way? The introduction of the notation $v^{(i_2)}_{i_1i_3}$ is quite annoying.
I'm not sure of any really great notation. Sometimes it's called a "contraction" (contracting by $b^$) and they use a symbol like $\neg$ ($b^ \neg a \otimes b \otimes c = b^(b)(a \otimes c)$), but I suspect that's far from universal. I think there are different conventions and different notations in geometry, physics, engineering, representation theory... :-( One problem with saying "oh we'll just write $b^$ for the map contracting tensors by $b^$ in the 2nd entry" is, what if $A=B=C$, so $b^$ could act in any position? Ugh. Maybe someone else has a good idea, but I don't know, sorry.
The notion of contraction that I'm aware of is also called the vector-valued trace. It is the mapping $$(A\otimes B^\ast)\otimes(B\otimes C)\to A\otimes C;,;;;(a\otimes b^\ast)\otimes(b\otimes c)\mapsto b^\ast(b)a\otimes c.$$ Is this what you mean?
Yeah, pretty much that. I guess I was thinking primarily of the situation where you have a fixed element $b^* \in B^$, yielding a map $A \otimes B \otimes C \to A \otimes C$; but that's just a restriction or specialization of what you wrote. I suppose the identification behind this is $\operatorname{Hom}(B^, \operatorname{Hom}(A \otimes B \otimes C, A \otimes C)) \cong \operatorname{Hom}((A \otimes B^*) \otimes (B \otimes C), A \otimes C)$. Your vector-valued trace is an element of the right-hand Hom. My version of contraction is a particular map in the left-hand Hom.
Okay, denoting the linearization of $$(E_1⊗ E_2^\ast)\times(E_2⊗ E_3)\to E_1⊗ E_3;,;;;(x_1⊗φ_2)\times(x_2⊗ x_3)\mapsto\langleφ_2,x_2\rangle_{E_2}x_1⊗ x_3$$ by $\operatorname{tr}{E_2}$, we may note that for any $φ_2\in E_2^\ast$, the linear extension $\iota{φ_2}$ of $$x_1⊗ x_2⊗ x_3\to x_1⊗φ_2⊗ x_2⊗ x_3$$ is an embedding of $E_1⊗ E_2⊗ E_3$ into $(E_1⊗ E_2^\ast)⊗(E_2⊗ E_3)$ and hence we could define $\operatorname{slice}{φ_2}:=\operatorname{tr}{E_2}\circ:\iota_{φ_2}$. By choosing $φ_2$ to be a basis functional, we can express $v^{(i_2)}_{i_1i_3}$ in terms of this slice operator.
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2025-03-21T14:48:31.033355
| 2020-05-20T21:28:02 |
360902
|
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|
Stack Exchange
|
An integral Jacobson-Morozov theorem?
$\DeclareMathOperator\SL{SL}$I want to ask if there exists a version of the Jacobson–Morozov theorem for integer matrices. A first approximation would ask: given an integral unipotent matrix $m \in \SL_n(\mathbb{Z})$ does there exist a homomorphism $\SL_2(\mathbb{Z}) \to \SL_n(\mathbb{Z})$ sending $$\begin{pmatrix} 1&1\\0&1\end{pmatrix} \to m?$$
Of course this quickly fails, if we take $n=2$ and $$m = \begin{pmatrix} 1&2\\0&1\end{pmatrix}$$ it's not hard to see that the map from $\SL_2(\mathbb{C}) \to \SL_2(\mathbb{C})$ we get from the Jacobson–Morozov theorem does not send $\SL_2(\mathbb{Z})$ to itself.
To remedy this instead of looking for a true homomorphism $\SL_2(\mathbb{Z}) \to \SL_n(\mathbb{Z})$ we can instead look for a "virtual homomorphism" by passing to a principal congruence subgroup $\Gamma_2(k) \subseteq \SL_2(\mathbb{Z})$ for some $k$, and looking for a map $\Gamma_2(k) \to \SL_n(\mathbb{Z})$ sending $$\begin{pmatrix} 1&k\\0&1\end{pmatrix} \to m$$ and ask if such a map always exists. In the previous example, we can take $k=2$ and just the identity map on $\Gamma_2(k)$ is such a homomorphism. When $n=2$ there is a classification of unipotent conjugacy classes, and indeed any unipotent matrix in $\SL_2(\mathbb{Z})$ virtually extends to a homomorphism in this way. My question is: does an integer unipotent matrix in $SL_n(\mathbb{Z})$ always extend to virtual copy of $SL_2$ in this sense?
This is a good question. Why not ask it for integer Chevalley groups in general? (I don't know the answer even for $\operatorname{SL}_n(\mathbb Z)$.) By the way, I think one shouldn't speak of the map coming from Jacobson–Morozov; it is well defined only up to conjugation by $\operatorname C_G(m)$.
Yes, I'd certainly be interested in the question for other Chevalley groups. I'd also be interested in replacing $\mathbb{Z}$ by a ring of integers $\mathcal{O}$, replacing the single matrix $m$ by a unipotent embedding of the additive group with some constraints. You are right that I shouldn't really say "the" map in general, but in the context I used it above I believe it's unique.
$\DeclareMathOperator\Int{Int}\DeclareMathOperator\SL{SL}$Even for $\SL_2$ it's not unique; you can use any $\Int\begin{pmatrix} 2 & b \ 0 & 1 \end{pmatrix}$. (I agree that none of them preserves $\SL_2(\mathbb Z)$, though.)
As far as I remember the group $\Gamma_2(2)$ is freely generated by $\left(\begin{array}{cc}1&2\0&1\end{array}\right)$ and its transpose, so the required homomorphism is easy to construct. Or do I miss something?
Hmm... good point. I think perhaps what I actually want to know is if there exists an embedding say of $SL_2(\mathbb{C})$ into $SL_n(\mathbb{C})$, such that the restriction to the congruence subgroup behaves as specified.
The answer is yes if you replace $m$ by a power of $m$. Let $\rho: SL_2 \rightarrow SL_n$ be a homomorphism defined over $\mathbb Q$ such that the unipotent element $\left(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}\right)$ maps to the fixed unipotent matrix $m$. This means that there exists a congruence subgroup $\Gamma$ os $SL_2(\mathbb Z)$ which maps into $SL_n(\mathbb Z)$ under $\rho$ and so that a generator of the upper triangular unipotent matrices in $\Gamma$ goes into a power of $m$.
|
2025-03-21T14:48:31.033600
| 2020-05-20T22:08:05 |
360903
|
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}
|
Stack Exchange
|
Conditional expectation values defined by expectation values
I asked this question a couple of days ago on Math.SE but without any echo (no upvotes, although I offered a bounty). But because I did it for oversight from a reputable/professional source I now think meta.MO might be the more suitable place to ask. I hope you don't mind the cross-posting.
Assume ${\bf x},{\bf z} \in \mathbb{R}^n$ denote real-valued and bounded random variables with continuous probability density $p({\bf x},{\bf z})$ and $f({\bf x})$ and $g({\bf x})$ are real-valued bounded scalar functions. Furthermore, $P_k({\bf z})$ denote all monomials for ${\bf z}$ indexed by $k\in\mathbb{N}$ (so that the expectation values of $E[P_k]$ are the corresponding mixed moments of $\bf z$).
Then, does the following implication hold?
For all $k$: $ E\left[f({\bf x})P_k({\bf z})\right] = E\left[g({\bf x})P_k({\bf z})\right]~~~ \implies ~~~E\left.\left[f({\bf x})\right|{\bf z}\right] = E\left.\left[g({\bf x})\right|{\bf z}\right]$
Note: All moments and expectation values exist because the random variables and functions are bounded.
I tried the following. (I'm not so sure with the exchange of the limits and the uniform convergence.)
Because the premise holds for all polynomials of $\bf z$ I can represent for each $\epsilon$ the $\eta_\varepsilon({\bf z})$ (Gaussian bump representation) in the Dirac-Delta function in $\mathbb{R}^n$
$\delta({\bf z})=\lim_{\varepsilon\to 0^+} \eta_\varepsilon({\bf z})$
as the powerseries of the exponential function in the Gaussian bump representation with coefficients $c_k$. This powerseries converges uniformly to the exponential function and, thus, the arguments of the expectation values converge uniformly to $f({\bf x})\eta_\varepsilon({\bf z})$ and $g({\bf x})\eta_\varepsilon({\bf z})$ (for arbitrary bounded $f$ and $g$). Then, I can exchange the integral of the expectation with the infinite sum and obtain
$$ \sum_k c_k E\left[f({\bf x})P_k({\bf z})\right] = \sum_k c_k E\left[g({\bf x})P_k({\bf z})\right] \\
E\left[f({\bf x})\sum_k c_k P_k({\bf z})\right] = E\left[g({\bf x})\sum_k c_kP_k({\bf z})\right]\\
E\left[f({\bf x})\eta_\varepsilon({\bf z} -{\bf z}_0)\right] = E\left[g({\bf x})\eta_\varepsilon({\bf z} -{\bf z}_0)\right]\\
\lim_{\varepsilon\to 0^+}E\left[f({\bf x})\eta_\varepsilon({\bf z} -{\bf z}_0)\right] = \lim_{\varepsilon\to 0^+}E\left[g({\bf x})\eta_\varepsilon({\bf z} -{\bf z}_0)\right]\\
E\left[f({\bf x})\delta({\bf z} -{\bf z}_0)\right] = E\left[g({\bf x})\delta({\bf z} -{\bf z}_0)\right]\\
E\left[f({\bf x})|{\bf z}_0\right] = E\left[g({\bf x})|{\bf z}_0\right]$$
Is this correct?
In general, I am interested if the statement holds without the assumption of a continuous probability density $p({\bf x},{\bf z})$ for arbitrary probability distributions, but I don't see any reason why it should. Any ideas?
A useful tool for proving results of this kind is the multiplicative system theorem.
Your desired conclusion does hold without the assumption of a probability density of $\bf x$ and $\bf z$, provided that the functions $f$ and $g$ are assumed to be Borel measurable.
Indeed, let $X:=f(\bf x)-g(\bf x)$ and $Z:=\bf z$, so that $X$ and $Z$ are bounded random variables with values in $\mathbb R$ and $\mathbb R^n$, respectively, such that
$$EXP(Z)=0$$
for all polynomials $P$ on $\mathbb R^n$. Since the set of all such polynomials is dense in $L^1(K)$ for any compact $K\subset\mathbb R^n$, whereas $X$ and $Z$ are bounded, we conclude that $EXh(Z)=0$ for any $h\in L^1(K)$. So, $EX\,1(Z\in B)=0$ for any Borel set $B\subseteq\mathbb R^n$, which means $E(X|Z)=0$, as desired.
Great, thanks. Yes, my $f$ and $g$ are continuous functions and thus Borel measurable. Do I get it right that your proof holds for any probability distribution with compact support (+ the constraints for $f$ and $g$)?
I've seen that you also have a math.SE account. If you are interested in a math.SE 100 points bounty you could repost your answer there and I'll assign it to you.
Could you please help me with your statement "It is unclear how you define the expressions $E\left[f({\bf x})\delta({\bf z} -{\bf z}_0)\right]$ and $E\left[f({\bf x})|{\bf z}_0\right]$". $E\left[f({\bf x})|{\bf z}0\right]$ should exist since $E[f({\bf x})]$ exists (by Radon–Nikodym theorem) and is almost surely unique. And $E[f({\bf x})\delta({\bf z} -{\bf z}0)] := \lim{\varepsilon\to 0^+}\int f({\bf x})\eta\varepsilon({\bf z} -{\bf z}_0) p({\bf x},{\bf z})d{\bf x}d{\bf z} = \int f({\bf x})p({\bf x},{\bf z}_0)d{\bf x}=E[f({\bf x})|{\bf z}_0]$ for density $p({\bf x},{\bf z})$, isn't it?
@Its_me : Concerning your first comment: Yes, in $\mathbb R^n$ compact support and bounded support mean the same, because the support set is closed, and a subset of $\mathbb R^n$ is compact if and only if it is bounded and closed.
Concerning your second comment: I have posted this answer on math SE. Thank you for this suggestion. Concerning your third comment: Sorry, I did not notice that you are assuming the existence of a joint pdf of $\bf x$ and $\bf z$. I have edited my answer accordingly.
I went again through the proof and afaik it applies also to Borel measurable functions that depend on both r.v., i.e. $f({\bf x},{\bf z})$ and $g({\bf x},{\bf z})$, doesn't it?
@Its_me : The proof is actually applicable in the even more general setting, when $f(\bf x)$ and $g(\bf x)$ are replaced by any bounded random variables whatsoever. In the proof, we only used the conditions that $X$ is bounded and $EXP(Z)=0$ for all polynomials $P$.
Great, thanks for your time and expertise.
|
2025-03-21T14:48:31.033937
| 2020-05-20T22:44:42 |
360904
|
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}
|
Stack Exchange
|
Image of the mapping class group of surfaces into automorphism group?
Let $S_{g,p}^n$ be a compact oriented surface of genus $g$ with $p$ punctures and $n$ boundary components, and $\operatorname{Mod}(S)$ and $\operatorname{PMod}(S)$ be the mapping class group and the pure mapping class group of $S$, respectively.
Suppose $n >0$ and fix a base point on a boundary.
I know that there is a natural monomorphism
$$\operatorname{Mod}(S_{g,p}^1)\to \operatorname{Aut}(\pi_1(S_{g,p}^1)),$$
and the image of $\operatorname{Mod}(S_{0,p}^1)$, which is isomorphic to the braid group, $\operatorname{PMod}(S_{0,p}^1)$, which is isomorphic to the pure braid group, and $\operatorname{Mod}(S_{g,0}^1)=\operatorname{PMod}(S_{g,0}^1)$ are known: respectively $$\{\phi \mid \phi(x_i)\textrm{ is a conjugate of }x_{\sigma(i)},\;\sigma\in \mathfrak{S}_p, \; \phi \textrm{ fixes the boundary}\},$$ $$\{\phi \mid \phi(x_i)\textrm{ is a conjugate of }x_i,\; \phi \textrm{ fixes the boundary}\},$$ $$\{\phi \mid \phi \textrm{ fixes the boundary}\},$$ where $x_i$ correspond the punctures, $\mathfrak{S}_p$ is the symmetric group.
How about the image of $\operatorname{PMod}(S_{g,p}^1)$, which is isomorphic to $\operatorname{Mod}(S_{g,0}^{p+1})/\langle\textrm{boundary Dehn twists}\rangle$), and $\operatorname{Mod}(S_{g,p}^1)$?
(Is the former
$$\{\phi \mid \phi(x_i)\textrm{ is a conjugate of }x_i,\; \phi \textrm{ fixes the boundary}\}?)$$
At least, I want to know whether they are known or open.
I asked a similar question here and Allen Hatcher gave an enlightening answer: https://mathoverflow.net/a/308391/8103
@MarkGrant I could not find any answer to my question. (Your question on the comment of Hatcher's reply is almost similar to my question, but there's no answer. Have you found the answer?)
We never found a reference, and so ended up changing the parameters of our problem. My question led to some email correspondence with another MO user, I'll email him to see if he made any progress.
@MarkGrant You mean that the problem is(was) unsolved, right?
I believe that there is no statement of Dehn-Nielsen-Baer for suraces with boundary and punctures in the literature. I also feel like someone who works a lot with mapping class groups should be able to figure out the answer, if they were sufficiently motivated. As I said, we shifted to looking at a slightly different problem, and the motiviation went away. (BTW, I'm talking about a project which led to this paper: https://arxiv.org/abs/1903.01916)
|
2025-03-21T14:48:31.034107
| 2020-05-20T23:08:38 |
360905
|
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"David Holmes",
"Joseph",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/360905"
}
|
Stack Exchange
|
Pullback of boundary divisors under forgetful maps
Let $\overline{\mathbf{M}}_{0,n}$ be the moduli space of stable $n-$pointed smooth rational curve of genus zero and $\overline{\mathbf{U}}_{0,n}$ the universal family described by $\pi_n:\overline{\mathbf{U}}_{0,n}\longrightarrow\overline{\mathbf{M}}_{0,n}$ with the n disjoints sections $\sigma_i: \overline{\mathbf{M}}_{0,n}\longrightarrow \overline{\mathbf{U}}_{0,n}.$ Joachim Kock in "An invitatation to quantum cohomology" Example 1.5.11 gave the relationship between a boundary $\mathbf{F}_n$ of $\overline{\mathbf{M}}_{0,n}$ with the boundary $\mathbf{F}_{n+1}$ of $\overline{\mathbf{M}}_{0,n+1}$ by the following formula
$$ \mathbf{F}_{n+1} = \varepsilon^*\mathbf{F}_n + \sum_{i}\sigma_i$$
where $\varepsilon: \overline{\mathbf{M}}_{0,n+1}\longrightarrow \overline{\mathbf{M}}_{0,n}$ is the forgetful maps.
Can someone explain me more how to get the formula?
You write `stable n-pointed smooth'; I think you want to omit the word smooth. As far as answering the question, have you worked out some examples with small n? I think if you work out the cases $n=3$ and $n=4$ by hand you will see what is going on. I think this is more useful than me writing out an argument, but feel free to disagree :-).
Yeah you are right, I should remove the word "smooth"
In fact I just listed all the divisors for n=3,4,5 and 6 of $\overline{\mathbf{M}}_{0,n}.$
|
2025-03-21T14:48:31.034234
| 2020-05-20T23:26:16 |
360906
|
{
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"Abdelmalek Abdesselam",
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"user42355"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/360906"
}
|
Stack Exchange
|
Moments of Plücker coordinates on complex Grassmannian
Consider the Grassmannian $Gr(k,N)\simeq U(N)/(U(k)\times U(N-k))$ which parametrizes $k$-dimensional subspaces of $\mathbb{C}^N$. Let us put on it the $U(N)$-invariant probability measure. Let $\Delta_{I}$ be one of the Plücker coordinates. My question is:
Are there explicit formulas for the moments $m_p=\mathbb{E}\ |\Delta_I|^{2p}$, for $p$ a nonnegative integer, of the random variable $|\Delta_I|^2$?
Equivalently, if $U$ is a Haar distributed unitary matrix in $U(N)$, what are the moments of $|\Delta|^2$ where $\Delta$ is your favorite $k\times k$ minor?
For $k=1$, which corresponds to projective space, we have
$$
m_p=\frac{p!}{N(N+1)\cdots(N+p-1)}\ .
$$
Edit: I did a rather brutal calculation for $k=2$. The result is
$$
m_p=\frac{(p+1)}{\binom{N+p-1}{p}\binom{N+p-2}{p}}\ .
$$
I think I know how to do the general case now, but I would be surprised if this computation has not been done before, so any reference would be appreciated.
Aug 2023 Update:
Please see my answer below for a generalization of the problem and more context.
There is still a few hours for the second bounty, but I need references on the following points, in particular.
Who first proved the formula $M_{\lambda}(N)=\frac{1}{{\rm dim}_{\mathbb{C}}(\mathbb{S}^{\lambda}(\mathbb{C}^N))}$? Are there standard textbooks which explain the proof?
Is there a name for functions like $|\Delta_1|^{2(\lambda_1-\lambda_2)}\cdots|\Delta_{\ell-1}|^{2(\lambda_{\ell-1}-\lambda_{\ell})}|\Delta_{\ell}|^{2\lambda_{\ell}}$ ? References, generalizations to general Lie groups?
Who first proved Theorem 5.3.1 in the book by Faraut about Haar measures in terms of decompositions $G=PQ$? Good references about this?
Does it make sense to talk about the 'value' of a projective coordinate?
I am presenting the variety as $U(N)/(U(k)\times U(N-k))$ instead of using $GL$ over a parabolic subgroup. So once you put the projective coordinate inside complex modulus square, it does have an honest value.
I assume that the orthogonal and unitary cases are similar. In the orthogonal case it seems to me that we are essentially interested in the angles of two random subspaces; more precisely, in the singular values of $X^T Y$ if $X$ and $Y$ are random $n \times k$ matrices with orthogonal columns. (If we fix $Y_{i,j} = \delta_{i,j}$, then we get back the original setting.) I've found this paper: https://math.mit.edu/~plamen/files/AEK.pdf, and then the book Aspects of Multivariate Statistical Theory, by Muirhead. The Wishart distribution, and Theorems 3.3.1 and 3.3.3 from the book seem relevant.
I added the automorphic form tag, because I am sure these Haar integrals are well known to folks in the area.
Notation: $\mathcal{CN}(0,1)$ denotes the complex standard normal distribution: if $Z \sim \mathcal{CN}(0,1)$, then $\sqrt{2} \operatorname{Re}(Z), \sqrt{2} \operatorname{Im}(Z)$ are independent real standard normal random variables.
Claim: Let $X = (X_{i,j})_{1 \le i,j \le N}$ be a random matrix with independent $X_{i,j} \sim \mathcal{CN}(0,1)$ (this is called the Ginibre ensemble).
Then $X$ is almost surely invertible, so its QR decomposition $X = QR$, with $Q \in U(N)$ and $R \in \mathbb{C}^{N \times N}$ upper triangular with positive diagonal entries, is uniquely defined almost surely.
Then $Q$ and $R$ are independent, $Q$ has the distribution given by Haar probability measure on $U(N)$, and $(R_{i,j})_{i \le j}$ are independent, with $R_{i,j} \sim \mathcal{CN}(0,1)$ for $i < j$, and $2 R_{i,i}^2 \sim \chi^2_{2(N-i+1)}$ for $i \in \{1, \dotsc, N\}$.
Using this Claim, the calculation of the moments becomes easy.
Take $k \le N$, and let $X'$, $Q'$, $R'$ denote the top-left $k \times k$ submatrices of $X$, $Q$, $R$.
Since $R$ is upper triangular, we get $X' = Q' R'$.
So $\det(X') = \det(Q') \prod_{i=1}^k R_{i,i}$, and here all factors are independent.
So
$$
E[|\det(X')|^{2p}] = E[|\det(Q')|^{2p}] \prod_{i=1}^k E[R_{i,i}^{2p}].
$$
Recall that if $Z \sim \chi^2_n$, then $E[Z^p] = n (n+2) \dotsm (n+2p-2)$.
So
$$
E[R_{i,i}^{2p}] = \prod_{j=1}^p (N-i+j) = p! {N-i+p \choose p}.
$$
Let $X' = \tilde{Q} \tilde{R}$ be the QR decomposition of $X'$, then using the Claim again, we get $|\det(X')| = \det(\tilde{R}) = \prod_{i=1}^k \tilde{R}_{i,i}$, where the $\tilde{R}_{i,i}$ are independent, and $2 \tilde{R}_{i,i}^2 \sim \chi^2_{2(k-i+1)}$.
So $E[|\det(X')|^{2p}] = \prod_{i=1}^k E[\tilde{R}_{i,i}^{2p}] = \prod_{i=1}^k (p! {k-i+p \choose p})$.
So
$$
E[|\Delta|^{2p}] = E[|\det(Q')|^{2p}] = \prod_{i=1}^k \frac{{k-i+p \choose p}}{{N-i+p \choose p}}.
$$
The only reference I have found for the above Claim is in the notes https://www.dam.brown.edu/people/menon/publications/notes/intro-rmt.pdf by Menon and Trogdon, see the section titled Other random matrix ensembles (warning: there $\sqrt{2} X$ is called the Ginibre ensemble, not $X$).
The orthogonal analogue of the Claim can be found e.g., in Theorem 2.3.18 of the book Matrix Variate Distributions, by Gupta, Nagar.
Very elegant proof! I was not clear, but my preferred answer would have been a quotable reference, although it is possible (unlikely?) it might not exist. So +1, and bounty bonus, but no accept, for now. I will start a new bounty.
Since the bounty is about to expire, I am posting my findings so far.
Using the method mentioned by user42355, one can prove the following more general result.
Let $\lambda=(\lambda_1,\ldots,\lambda_{\ell})$ be an integer partition with at most $\ell$ parts, i.e., we assume the $\lambda_i$'s are integers such that $\lambda_1\ge\cdots\lambda_{\ell}\ge 0$. Define, for $N\ge \ell$,
$$
M_{\lambda}(N):=\mathbb{E}_{U(N)}\left[
|\Delta_1(U)|^{2(\lambda_1-\lambda_2)}
|\Delta_2(U)|^{2(\lambda_2-\lambda_3)}\cdots
|\Delta_{\ell-1}(U)|^{2(\lambda_{\ell-1}-\lambda_{\ell})}
|\Delta_{\ell}(U)|^{2\lambda_{\ell}}
\right]\ .
$$
Here $\Delta_j(U)$ denotes the $j\times j$ principal minor determinant in the top left corner of the matrix $U$, and the latter is random and Haar distributed in the unitary group $U(N)$.
Let $G_{\lambda}(N)$ denote the expectation of the same integrand but where the matrix $X$ is now sampled according to the Ginibre ensemble, and let $T_{\lambda}(N)$ be the the similar expectation with now a random matrix $T$ which is upper triangular with positive diagonal entries, sampled as follows.
The entries are independent. For $i<j$, $T_{ij}$ is such that $\sqrt{2}{\rm Re}(T_{ij})$ and $\sqrt{2}{\rm Im}(T_{ij})$ are independent standard $N(0,1)$ real Gaussian random variables. For $1\le i\le N$, $2T_{ii}^2$ is a chi-squared random variable with $2(N-i+1)$ degrees of freedom.
As in the answer by user42355, the key fact one needs is the Gram (or QR for numerical analysts) decomposition $X=UT$, as an equality in distribution, with $U$ and $T$ independent and distributed as above. We then immediately have
$$
G_{\lambda}(N)=M_{\lambda}(N) T_{\lambda}(N)\ .
$$
Now since all the entries of the Ginibre matrix $X$ are independent, we see that $G_{\lambda}(N)=:G_{\lambda}$ is independent of $N$. As a result,
$$
M_{\lambda}(N)=\frac{T_{\lambda}(\ell)}{T_{\lambda}(N)}M_{\lambda}(\ell)
$$
Since $|\Delta_{\ell}(U)|=1$ for $U$ unitary of size $\ell$, we get the recursion relation
$$
M_{(\lambda_1,\ldots,\lambda_{\ell})}(\ell)=
M_{(\lambda_1-\lambda_{\ell},\lambda_2-\lambda_{\ell},
\ldots,\lambda_{\ell-1}-\lambda_{\ell})}(\ell)
$$
which reduces the length of the partition seen as a sequence of nonnegative integers.
Using the formula for the moments of the chi-squared or Gamma distributions, we easily get
$$
T_{\lambda}(N)=(N)_{\lambda_1}(N-1)_{\lambda_2}\cdots(N-\ell+1)_{\lambda_{\ell}}
$$
where I used the (rising) Pochhammer symbol $(x)_n=x(x+1)\cdots(x+n-1)$.
Putting everything together, and solving the recursion on $\ell$, we arrive at
$$
M_{\lambda}(N)=\frac{(\ell)_{\lambda_1}(\ell-1)_{\lambda_2}\cdots(1)_{\lambda_{\ell}}}{(N)_{\lambda_1}(N-1)_{\lambda_2}\cdots(N-\ell+1)_{\lambda_{\ell}}}\times C_{\lambda}
$$
where
$$
C_{\lambda}=\prod_{j=2}^{\ell}\left(
\prod_{i=1}^{j-1}
\frac{(j-i)_{\lambda_i-\lambda_j}}{(j-i+1)_{\lambda_i-\lambda_j}}
\right)
$$
$$
=\prod_{j=2}^{\ell}\left(
\prod_{i=1}^{j-1}
\frac{(j-i)}{(j-i+\lambda_i-\lambda_j)}
\right)\ .
$$
In other words, we end up with the beautiful and most certainly known formula,
$$
M_{\lambda}(N)=\frac{h_{\lambda}}{\prod_{\Box\in\lambda}(N+c(\Box))}\ .
$$
Here $\Box$ denotes a box in the Ferrers diagram of $\lambda$, drawn the English (rather than French) way. The content of $\Box$ is $c(\Box)=j-i$ if the box lies in the $i$-th row (from top to bottom) and the $j$-th column (from left to right).
Finally, $h_{\lambda}$ is the product of hook lengths for the partition $\lambda$, which is also given by
$$
h_{\lambda}=\frac{(\lambda_1+\ell-1)!\ (\lambda_2+\ell-2)!\cdots \lambda_{\ell}!}{\prod_{1\le i<j\le\ell}\ (j-i+\lambda_i-\lambda_j)}\ .
$$
It is easy to see that the moment in the question corresponds to the rectangular partition $\lambda=(p,\ldots,p)$, where $p$ is repeated $k$ times.
In a previous version of the post for my question, I suspected that the moment should be one over the dimension of a representation of $U(N)$, but then I deleted that because I dismissed it as too good to be true. Yet, it is true! Namely,
$$
M_{\lambda}(N)=\frac{1}{{\rm dim}_{\mathbb{C}}(\mathbb{S}^{\lambda}(\mathbb{C}^N))}
$$
where $\mathbb{S}^{\lambda}$ is the Schur functor associated to the partition $\lambda$.
This follows from the hook content formula for ${\rm dim}_{\mathbb{C}}(\mathbb{S}^{\lambda}(\mathbb{C}^N))$ which is the number of semistandard Young tableaux of shape $\lambda$ filled with numbers from $1$ to $N$.
As far as references, for the key probabilistic Gram or QR decomposition, I looked at the notes by Menon and Trogdon mentioned by user42355, as well as other random matrix references from the areas of statistics (e.g., the book by Eaton), probability, mathematical physics (e.g., related to quantum transport), but I did not like the presentation therein (just my personal taste). This is because one has to go through rather tedious independence or conditional independence statements about partitioned Wishart matrices. The most illuminating presentation I found was the book "Analysis on Lie Groups, An Introduction" by Jacques Faraut. The needed Gram decomposition is done in Exercise 5 of Chapter 5, as a corollary of Theorem 5.3.1 which shows how to build a left Haar measure on a locally compact group $G$ homeomorphically decomposed as a product $G=PQ$ of two closed subgroups, using
a left Haar measure on $P$, a right Haar measure on $Q$, and the module function of $G$.
After finishing the computation of $M_{\lambda}(N)$, and contemplating the beautiful final formula, I realized one can prove this in a completely different way using the Peter-Weyl type formula for orthogonality of matrix elements of irreducible unitary representations of compact groups. Indeed the integrand of $M_{\lambda}$ is essentially of the form
$$
|\langle v,\pi_{\lambda}(U)v\rangle|^2
$$
for a suitable highest weight vector $v$.
Finally, note that there is related formula in my previous post:
Integration of a function over 7-sphere
The integrand is essentially the same, but the difference is about taking the expectation of a Ginibre versus a complex Wishart random matrix.
|
2025-03-21T14:48:31.035012
| 2020-05-21T00:53:38 |
360911
|
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|
Stack Exchange
|
Critical probability for Erdos-Renyi digraphs to be strongly connected
Given $p \in [0,1]$, an Erdos-Renyi graph ${ER}(n,p)$ on $n$ vertices is constructed by defining, for each unordered couple of distinct vertices ${i,j}$ an edge between $i$ and $j$ with probability $p$.
Similarly we define an Erdos-Renyi digraph, which we shall denote as ${ER}_d(n,p)$, on $n$ vertices by defining, for each ordered pair of distinct vertices $(i,j)$ an edge $i \to j$ with probability $p$.
Erdos and Renyi proved that $p=\frac{\ln(n)}{n}$ is a sharp threshold for the connectedness of an (undirected) E-R graph, that is :
If ${\displaystyle p<{\tfrac {(1-\varepsilon )\ln n}{n}}}$, then ${ER}(n,p)$ will almost surely contain isolated vertices, and thus be disconnected.
If ${\displaystyle p>{\tfrac {(1+\varepsilon )\ln n}{n}}}$, then a graph ${ER}(n,p)$ will almost surely be connected.
Is there a similar sharp bound for the strong connectedness of ${ER}_d(n,p)$?
I collected statistics on ${ER}_d(n,p)$ for $n \le 300$ and a resolution of $0.02$ for $p$, with $200$ generations for each $(n,p)$. I measure a critical probability for (almost sure) strong connectedness asymptotically close to $p=\frac{2\ln(n)}{n}$.
https://onlinelibrary.wiley.com/doi/full/10.1002/rsa.20416?casa_token=kW4SC9LUA1MAAAAA%3A7gygvIaAjPVjJMvieLh6PJ2IiexXIMnpga-XFZjAXUVb6VESdq3uELU3c44PzE0X-Q-HsQTVu6LQD2s includes mention of sources that answer this question.
@Brendan McKay: That resource is blocked for me, what is the answer they give?
http://www.math.unl.edu/~xperezgimenez2/papers/dicores.pdf
It’s the same threshold. As in the undirected caseIt’s more or less immediate to see that this condition is what you need not to have components of size 1.
@AnthonyQuas Agreed if by "component" you mean "strong component".
This is basically the main result of a paper by Ilona Palásti:
On the strong connectedness of directed random graphs. Studia Sci.
Math. Hungar. 1 (1966), 205–214.
Here's the MathSciNet summary of the paper: "Let $G_{n,N}$ be a random directed graph having $n$ vertices and $N$ directed edges, the edges being chosen from the ${n \choose 2}$ possible edges so that each of the ${n^2\choose N}$ possible choices is equiprobable (i.e., loops are allowed). Let $P(n,N)$ denote the probability that $G_{n,N}$ is strongly connected. Then if $N_c"=[n\log n+cn]$, where $c$ is an arbitrary fixed number and $[x]$ denotes the integral part of $x$, it is proved that $\lim_{n \to \infty} P(n,Nc)=\exp(−2e−c)$.
The result is extended from the directed model with a fixed number of edges to the binomial directed random graph, which I think is what you're asking about, in the paper "A note on thresholds and connectivity in random directed graphs", by Alasdair J. Graham and David A. Pike; you can find that here.
well, the result in the paper summary is not what i measured. The critical probability for digraph follows from the argument by Erdos in the second paper you linked, however.
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2025-03-21T14:48:31.035232
| 2020-05-21T04:51:32 |
360920
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Stack Exchange
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Looking for a paper by Landau and one by Watson
For the purposes of a project, I've been looking for the following two papers referred to in Serre's "Divisibilité de certaines fonctions arithmétiques":
Landau (E.), - Über die Eitenlung der positiven ganzen Zahlen in vier Klassen nach der Mindestzahl der zu ihrer additiven Zusammensetzung erforderlichen Quadrate, Arch. der Math. und Phys.(3) 13, 1908, p. 305-312
Watson: (G.N.), - Über Ramanujansche Kongruenzeigenschaften der Zerfällungsanzahlen (I), Math. Z., t. 39, 1935, p. 712-731.
Despite my best attempts, I have been unable to find the papers themselves or the relevant issues of these journals either. I would be really grateful if someone to provide me a link to either or tell me how I might find them.
(Perhaps I'm asking for too much, but are there English translations of these available somewhere?)
Here's the Landau paper at the Internet Archive: https://archive.org/details/archivdermathem37unkngoog/page/n324/mode/1up
Here's the Watson paper at the EuDML: https://eudml.org/doc/168581, which links to the GDZ for a scan.
To answer your more general question of finding old stuff, EuDML has very good coverage of old European journals, especially German ones, and the Internet Archive is worth always a search for 19th century or early 20th century stuff.
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2025-03-21T14:48:31.035348
| 2020-05-21T05:17:18 |
360921
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How are the vector-valued trace and the unique linearization of $\mathfrak L(X,Y)\:\hat\otimes_π\:X→Y$ of $\mathfrak L(X,Y)×X→Y,\;(L,x)↦Lx$ related?
Let $X$ be a $\mathbb R$-Banach space and $X'\:\hat\otimes_\pi\:X$ denote the completion of the tensor product of $X'$ and $X$ with respect to the projective norm. The trace functional $\operatorname{tr}:X'\:\hat\otimes_\pi\:X\to\mathbb R$ is the unique linearization of the bounded bilinear form $$X'\times X\to\mathbb R\;,\;\;\;(\varphi,x)\mapsto\varphi(x)\tag1.$$ Now, I've seen that there are two related notions:
If $E$, $F$ and $G$ are $\mathbb R$-vector spaces, $$C_G:(E\otimes G')\otimes(G\otimes F)\to E\otimes F\;,\;\;\;(x\otimes u')\otimes(u\otimes y)\mapsto\langle u',u\rangle_{G',\:G}x\otimes y\tag2$$ is called tensor contraction or vector-valued trace
If $H$ and $K$ are separable $\mathbb R$-Hilbert spaces and $T$ is a trace-class operator on the Hilbert-Schmidt tensor product $H\:\hat\otimes_2\:K$, then there is a unique trace-class operator $\operatorname{tr}_K(T)$ on $H$, called partial trace, with $$\operatorname{tr}\left(\operatorname{tr}_K(T)B\right)=\operatorname{tr}\left(T\left(B\otimes_2\operatorname{id}_H\right)\right)\tag3$$ for all $B\in\mathfrak L(H)$
(I guess that in 2. there is a similar construction possible for the projective tensor product; but I haven't found any reference for that.)
Now, if I'm not missing anything, there should be a third thing we can construct: If $Y$ is another $\mathbb R$-Banach space, we could define $\operatorname{tr}_{X,\:Y}:\mathfrak L(X,Y)\:\hat\otimes_\pi\:X\to Y$ as the unique linearization of the bounded bilinear operator $$\mathfrak L(X,Y)\times X\to Y\;,\;\;\;(L,x)\mapsto Lx\tag4.$$ How are these notions (especially the tensor contraction and $\operatorname{tr}_{X,\:Y}$) related?
I think 2) can be understood as a version of 1) using the projective tensor product of operator spaces, but I have not sat down to check the details
admits a version for the projective tensor product of Banach spaces, which then looks very like your (4), but I admit that right now I don't see a direct deduction of (4) from the Banach-space version of 1)
The basic principle is that composition of two operators can be linearized to a contractive linear map ${\mathcal L}(E,F) \hat{\otimes}_\pi {\mathcal L}(F,G) \to {\mathcal L}(E,G)$. This seems to imply both 1) and 4)
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2025-03-21T14:48:31.035525
| 2020-05-21T06:22:07 |
360923
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Question about Berezin line bundle of odd cotangent bundle of supermanifold $\text{Ber}(\Pi T^*\mathcal N)$
The followings are from Mnev's paper about BV formalism.
Example 4.15 (Definition of split supermanifold)
Let $E \to M$ be a rank $m$ vector bundle over $n$-manifold $M$, then there exists a split $(n|m)$-supermanifold $\Pi E$ with body $M$ and structure sheaf $\mathcal O_{\Pi E} = \Gamma(M ,\bigwedge^{\bullet}E^*)$.
Section 4.2.4 (Definition of Berezin line bundle of split supermanifold)
Let $\mathcal M = \Pi E$ be the split $(n|m)$-supermanifold of vector bundle $E \to M$.
Berezin line bundle of $\mathcal M$ is $\text{Ber}(\mathcal M) := \bigwedge^{n} T^*M\otimes \bigwedge^m E$ over $M$.
For $T^*M \to M$, we have $\text{Ber}(\Pi T^*M) \cong (\bigwedge^{n} T^*M)^{\otimes 2}$. This is (161).
Then in (162), it writes
Similarly, for $\mathcal N$ a supermanifold, one has $\text{Ber}(\Pi T^*\mathcal N)|_{\mathcal N} \cong\text{Ber}(\mathcal N)^{\otimes 2}$.
Here we understand $\text{Ber}(\mathcal N)$ as a line bundle over $\mathcal N$ and the l.h.s. is a pullback of a line bundle over $\Pi T^* \mathcal N$ to $\mathcal N$.
I have difficulty in understanding the above sentence.
Suppose $\mathcal N$ is a $(k|n-k)$-supermanifold of body $N$, then $T^*\mathcal N$ has dimension $(2k|n-2k)$ and $\Pi T^*\mathcal N$ has dimension $(n|n)$.
Question 1:
What is the expression of $\text{Ber}(\Pi T^*\mathcal N)$ for supermanifold $\mathcal N$?
Question 2:
From Batchelor's theorem, we can assume $\mathcal N = \Pi{B}$ for some bundle $B \to N$.
By definition, $\text{Ber}(\mathcal N) = \bigwedge^{k}T^*N \otimes \bigwedge^{n-k} B$, so how can we show $$\text{Ber}(\Pi T^*\mathcal N)|_{\mathcal N} \cong \text{Ber}(\mathcal N)^{\otimes 2} ?$$
Update: My senior told me that this is a conclusion of adjunction formula for supermanifold, though I'm not very familiar with this and I can't understand the followings. Here's the proof:
$\text{Ber}$ can be defined on any locally free sheaf. Especially, when we say $\text{Ber}(X)$ we actually mean $\text{Ber}(T^*_X)$, here $T^*_X$ means cotangent sheaf on $X$.
$X \hookrightarrow Y$ is closed embedding of smooth supermanifold, ideal of $X$ is $\mathcal I$, then $$0 \to \mathcal I/\mathcal I^2 \to \Omega^{1}_{Y}|_{X} \to \Omega^{1}_{X} \to 0$$ is exact.
This implies $\text{Ber}(Y)|_{X} \cong \text{Ber}(X) \otimes \text{Ber}(\mathcal I / \mathcal I^2)$.
Take $Y = V(\Pi T_{X})$ (here $V(E)$ means $E^*$ bundle and $T_X$ means tangent sheaf), then $\mathcal I/\mathcal I^2 = \Pi T_X$ and $$\text{Ber}(\Pi T_X) = (\text{Ber}(T_X))^{-1} = \text{Ber}(X)$$
here $(\text{Ber}(T_X))^{-1}$ means dual bundle of $\text{Ber}(T_X)$.
Or is there any reference for this?
I really can't find too much information of Berezin bundle.
Thanks for your time and effort.
The Berezinian of a vector bundle $E$ arises via an associated bundle construction: it is reasonably easy to check that
$$
\begin{pmatrix}A&B\\C&D\end{pmatrix}\mapsto |\operatorname{det}(A)|\operatorname{det} ^{-1}(D- CA^{-1}B)
$$
defines a super Lie group homomorphism $\operatorname{Ber}:GL(\mathbb R^{m|n})\to\mathbb R^\times$. (Here $A\in GL(\mathbb R^n),B\in\mathbb R^{m\times n},C\in\mathbb R^{n\times m},D\in GL(\mathbb R^n)$ are the even-to-even etc. blocks of an element of a matrix in $GL(\mathbb R^{m|n})$). It arises by integration of the supertrace, which is a super Lie algebra homomorphism $\mathfrak{gl}(\mathbb R^{m|n})\to \mathbb R$. Then the Berezinian of $E$, thought of as a locally free sheaf over the base supermanifold, can be defined as the sheafification of the presheaf which assigns to an open subsupermanifold $U$ the $\operatorname{sMan}(U,GL(\mathbb R^{m|n}))$-equivariant functions from trivializations of $E|_U$ to $\mathcal O(U)$. In words, a trivialization of $E$ induces a trivialization of $\operatorname{Ber}(E)$, and a different trivialization of $E$ changes this trivialization by the Berezinian of the matrix-valued "gauge transformation" between them.
Your advisor's statement is that a short exact sequence of vector bundles $0\to F_1\to E\to F_2$ gives rise to a natural isomorphism of line bundles $\operatorname{Ber}(E)\cong \operatorname{Ber}(F_1)\otimes \operatorname{Ber}(F_2)$. In terms of the representation, such a short exact sequence is the same as a reduction of structure group from $GL(\mathbb R^{m|n})$ to the group of upper triangular matrices $\big(GL(\mathbb R^{m_1|n_1})\times GL(\mathbb R^{m_2|n_2})\big)\ltimes \mathbb R^{m_1|n_1\times m_2|n_2}$. Restricted to this subgroup, the Berezinian equals the product of the Berezinians of the two blocks, which under the associated bundle construction corresponds to this isomorphism. There are some fermion sign issues when switching the $F_i$; these become the standard Koszul signs if we stipulate that the parity of the Berezinian super line equals that of the odd dimension of the superspace.
Given any vector bundle $E\to X$ over a super manifold, we can form the total space which is also donated by $E$ and comes with a canonical map $p:E\to X$. Then there is a short exact sequence
$$
0\to p^* E\to TE\xrightarrow{p_*} p^* TX
$$
which gives rise to an isomorphism $\operatorname{Ber}(TE)\cong p^*\big(\operatorname{Ber}(E)\otimes \operatorname{Ber}(TX)\big)$; note that this isomorphism is only invariant under diffeomorphisms of the total space $E$ which preserve this exact sequence. This gives the formula for the Berezinian of a split supermanifold which you give in your question 2. However, when we pull back this exact sequence to $X$ along the zero section $s$, it is canonically split. For your question about the shifted cotangent bundle, this gives $\operatorname{Ber}(TX)\otimes \operatorname{Ber}(\Pi T^*X)$. Now it is straightforward to check at the level of representations that both changing the parity and dualizing replace the Berezinian by its dual, so that $\operatorname{Ber}(\Pi T^*X)\cong \operatorname{Ber}(TX)$, which then gives the isomorphism in (162). In general, since Batchelor's theorem does not give a natural split form for a supermanifold, it is usually not straightforward to give a natural construction using it, so I don't have an answer to your question 2.
Finally, there is a good reason why the Berezinian of the shifted cotangent bundle is the square of a line bundle: it carries a canonical odd symplectic structure, i.e. a reduction of structure group from $GL(\mathbb R^{n|n})$ to the group
$$
Sp(\mathbb R^{n|n}) = \left\{\begin{pmatrix}A&B\\ C&D\end{pmatrix}\mid A^TC=C^TA, D^TB=-B^TD, A^TD - C^TB = I\right\}
$$
It's straightforward to show that $(A^{-1})^T = D - CA^{-1}B$, so that $\operatorname{Ber}\begin{pmatrix}A&B\\ C&D\end{pmatrix}=\operatorname{det}^2(A)$ has a canonical square root. Under the associated bundle construction, this gives rise to the half-Berezinian line bundle, whose restriction to any Lagrangian submanifold is canonically identified with the Berezinian of that submanifold. A reference for the last part is Section 4.6 of Pavel Mnev's book "Quantum Field Theory: Batalin–Vilkovisky Formalism and Its Applications".
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2025-03-21T14:48:31.036098
| 2020-05-21T06:40:49 |
360924
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Stack Exchange
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Proofs where higher dimension or cardinality actually enabled much simpler proof?
I am very interested in proofs that become shorter and simpler by going to higher dimension in $\mathbb R^n$, or higher cardinality. By "higher" I mean that the proof is using higher dimension or cardinality than the actual theorem.
Specific examples for that:
The proof of the 2-dimensional Brouwer Fixed Point Theorem given by by Aigner and Ziegler in "Proofs from the BOOK" (based on the Lemma of Sperner). The striking feature is that the main proof argument is set up and run in $\mathbb R^3$, and this 3-dimensional set-up turns the proof particularly short and simple.
The proof about natural number Goodstein sequences that uses ordinal numbers to bound from above.
The proof of the Finite Ramsey Theorem using the Infinite Ramsey Theorem.
In fact, I would also be interested in an example where the theorem is e.g. about curves, lattice grids, or planar graphs $-$ and where the proof becomes strikingly simple when the object is embedded e.g. in a torus, sphere, or any other manifold.
Are you aware of proofs that use such techniques?
Do you count Buffon's noodle as an example of this?
https://en.wikipedia.org/wiki/Buffon%27s_noodle
Desargue’s Theorem is a famous example. There are two senses in which a move to something larger gives an easier proof for $\mathbb R^2$: moving to $\mathbb R^3$ and adding points at infinity.
Perhaps also: Showing Borel determinacy using a measurable cardinal (which implies analytic determinacy by a simpler argument).
I‘m not sure if this is exactly what you are asking for, but in the area of mass partitions there are a number of proofs that use some lifting into higher dimensions. For example to prove that an (open) necklace with beads of k colors can be fairly distributed among 2 robbers using k cuts, you can just lift the necklace to the moment curve in k-dimensional space and apply the Ham-Sandwich Theorem.
Here is a nice questions: given 3 disjoint circles of different radii in the plane, show that the 3 meeting points of the 3 pairs of bi-tangent lines to each pair of circles are co-linear.
@UriBader this is an interesting one! Are you aware of a proof taking the "higher order" approach? Maybe some kind of projection from $\mathbb R^3$ into the plane?
When proving existence of an algebraic closure of a field $K$, one issue which prevents naive applications of Zorn's Lemma is the lack of an ambient structure which would contain all algebraic extensions. An elegant way to get around that is to fix some set $S$ containing $K$ and of cardinality larger than $K$, and consider all algebraic extensions contained in $S$. Then an application of Zorn's Lemma gives an algebraic closure.
@Claus, yes, of course, this is why I put it here. This is not research level, hence it is just a comment . I will not provide the solution, not to take all the fun out, but the main hint is already given by the context.
There is the nice way of calculating $\int_\mathbb{R} e^{-x^2}dx$ by going planar and using polar coordinates...
@UriBader thanks a lot for your nice examples! Very much appreciated
@amakelov, thanks a lot for your comment! Just to confirm my understanding, you mean taking the noodle as the limit of piecewise linear needles? It is actually an interesting view, I think I should specify better in my question
@Claus - with the Buffon noodle, we take the original problem referring to a unit-length, straight needle, and we generalize in two directions: 1) to arbitrary length, and 2) to a curved needle ("noodle"). Neither of these is formally generalizing to higher dimension or higher cardinality - we just allow the needle to be longer (so, a finer notion than cardinality is being generalized here) and, if you choose to look at it this way, we give it "one more dimension" of wiggle room. I was just wondering what your point of view on this example is - does it fit your intended criteria? :)
@amakelov thanks a lot, this makes it clear. Thanks to your comment, I have now updated my question, as in this post I am most interested in the case where the proof uses higher dimension or cardinality than the actual theorem. I should have specified this earlier, my question was not in the direction of generalizing the theorem itself. Thanks again amakelov for helping me clarify!
Somewhat related question: https://mathoverflow.net/questions/336008/legendary-extra-parameters-to-simplify-a-counting-problem
Another simple example may be reducing higher order differential equations over $\mathbb{R}$ to first order differential equations over $\mathbb{R}^n$ by the standard trick of introducing extra variables, each begin the derivative of the previous one
Another somewhat related question: https://mathoverflow.net/questions/106705/2d-problems-which-are-easier-to-solve-in-3d/
@SamHopkins thank you for providing this link - great. Very much appreciated!
@OliverNash thanks a lot for your link! Your own example in this thread is great. Thanks again!
I believe (I'm not a professor) that an example could be A Generalization which is Easier to Prove than a Special Case, by R. A. Rosenbaum, The American Mathematical Monthly, Vol. 70, No. 4 (April, 1963), p. 427. I can to read it from my account of JSTOR. Isn't required a response for my comment, good day.
Another somewhat related question: https://mathoverflow.net/questions/271608/17-camels-trick/272143
Numberphile has a good video that explains something that only applies to a circle by using higher dimensions: https://www.youtube.com/watch?v=6_yU9eJ0NxA
@SamHopkins Sam, great link, thanks a lot
This video is a relatively simple, but nice example.
My professor in Model Theory said that none of the proofs work for Finite Model Theory because you always run out of elements to choose from.
There are entire compendia of real integrals performed using the residue theorem ...
Whitney's theorem is an example of this. To prove the weak version (i.e. embedding a manifold $M^n$ in $\mathbb{R}^{2n +1}$), you start by using a partition of unity to embed $M^n$ into
$\mathbb{R}^{N}$ where $N$ is very large. This is relatively easy to do when $M^n$ is compact and takes a little bit of thought otherwise, but is significantly easier than trying to get an embedding in a lower dimension from scratch. You can then use transversality arguments to show that a generic projection map preserves the embedding of $M^n$ to cut down $N$ until you get to $\mathbb{R}^{2n +1}$.
To get the strong version of the theorem (embedding $M^n$ in $\mathbb{R}^{2n}$), there is another insight needed, which is using Whitney's trick to get rid of double points. As such, it's really the weak version where the high-dimensionality approach is used.
this is a great example! Thank you very much for providing it!
Tarski's plank theorem (1932).
A plank of width $w$ in ${\bf R}^n$ is the closed region between
two parallel hyperplanes at distance $w$ from each other.
Q: Can a unit disc in ${\bf R}^2$ be covered with
a sequence of planks of total width less than $2$?
Note that a single plank of width $2$ suffices, and can be divided into
any number of parallel planks of total width $2$.
But it seems conceivable that one might reduce the total width using
planks that are not parallel, even if they must overlap.
We show that this is impossible
by going from a circle in ${\bf R}^2$ to a sphere in ${\bf R}^3$.
A: No. If the total width is $W$ then
a unit ball in ${\bf R}^3$ is covered by planks
of total width $W$. But by a classical theorem of Archimedes,
a plank of width $w$ in ${\bf R}^3$ meets the unit sphere $S$
in a subset of area at most $2\pi w$, with equality if and only if
both of the plank's bounding planes intersect $S$.
Therefore the planks cover at most $2\pi W$ of the area of $S$.
Since $S$ has area $4\pi$, we deduce that $W \geq (4\pi) / (2\pi) = 2$. QED
When I was a grad student, I learned this proof from Paul Chernoff. His formulation of the argument began something like "Imagine that the disk is a manhole and that a bubble of noxious gases is emerging from it."
@NoamD.Elkies this is a beautiful example, and striking. Thanks a lot for sharing. AndreasBlass, nice comment,
A rather obvious comment is that the analogue of Archimedes theorem does not hold in $\mathbb{R}^2$, so moving to $\mathbb{R}^3$ seems necessary here.
This argument also shows that if the planks cover each point of the disk $m$ times, the sum of their widths is at least $2m$. This generalization is no longer true for general Bang's theorem (which claims that if planks cover a convex compact set, the sum of their widths is not less than the width of the set).
@Fedor Petrov I see: an equilateral triangle of width $w$
is doubly covered by three planks of width $w/2$ ! Neat.
Is that kind of thing possible also for a convex compact set that's also
centrally symmetric, in ${\bf R}^2$ or some higher dimension?
(Uri Bader points this out in comments, but it should really be an answer.)
A classic example is the calculation of the one-dimensional integral $\int_{-\infty}^\infty e^{-x^2}dx$ by squaring it, considering that as a two-dimensional integral, and changing to polar coordinates.
thanks a lot! Great to have it in the answers now
Hindman's theorem comes to mind: If $\mathbb N$ is partitioned into finitely many pieces, then there is an infinite set $A$ such that not only $A$, but also all sums of finite subsets of $A$ are contained in the same piece of the partition. (This is a slightly simplified statement: see the link for the full version.)
The statement of the theorem does not mention any uncountable sets.
The theorem can be proved in a "purely combinatorial" way, without mentioning any uncountable sets or appealing to topology or algebra, and in fact this was how the original proof by Hindman went. But this proof is very complicated and hard to follow. (In Hindman's own words, "That proof is really only good for punishing graduate students.")
There is a much nicer proof that uses topological dynamics, specifically the shift map on the Cantor space (which has size $2^{\aleph_0}$). This proof can be found in the last chapter of the Graham-Rothschild-Spencer book Ramsey Theory.
But the nicest proof of Hindman's theorem uses an algebraic structure on the compact Hausdorff space $\beta \mathbb N$ (which has size $2^{2^{\aleph_0}}$) to construct a special kind of non-principal ultrafilter (which has size $2^{\aleph_0}$), and then uses the existence of this special ultrafilter to prove the theorem in a few lines.
thanks a lot for this cool example, great. It is completely striking that you go to such a high cardinality to get a simpler proof for a "countable result". And by the way cool quote from Hindman.
And Hindman's theorem was preceded by Folkman's theorem wis the same except that $A$ instead of being an infinite set is an arbitrarily large finite set. I don't know the original proof of Folkman's theorem, so I'm asking: is the easiest way to prove Folkman's nowadays to prove the stronger Hindman's theorem?
@bof the other contender is to prove the more general finite statement, Rado's theorem. Which is easiest probably depends on the reader's background in Ramsey theory or topology.
The interesting thing is that the progressively nicer proofs require stronger axioms: Hindman's original proof could be formulated in a weak fragment of second-order arithmetic, whereas the elegant ultrafilter proof requires the ultrafilter lemma (which can be proved in ZFC but not in ZF).
Borel determinacy is a good example. First, it is a fact about reals that actually requires using much larger sets to prove. This was shown by Harvey Friedman, and there is a recorded recent online talk of Menachem Magidor explaining Friedman’s argument. But this is an example where the use of larger cardinalities doesn’t just give an easier proof, but a proof at all. A nice presentation of the proof of Borel determinacy in ZF can be found in the textbook by Kechris.
However, Tony Martin’s argument for Borel determinacy was preceded by his argument for analytic determinacy (analytic being a larger class than Borel), using a measurable cardinal. The argument is simpler than the one for Borel determinacy in ZFC. Analytic determinacy is actually equivalent to the existence of $x^\sharp$ for every real $x$, which I would describe as countable “shadows” of a measurable.
thanks a lot for a great example. The list of examples is getting very rich
The Max-Cut problem for a graph asks for a subset $S$ of vertices such that the number of edges between $S$ and the complement of $S$ is maximized. This problem is NP-hard. In fact, Håstad showed that even getting within 5.8% of optimal is NP-hard.
However, Goemans and Williamson showed that you can get within 12.2% of optimal in polynomial time by using high-dimensional optimization. We replace the edges with repulsive springs, confine the vertices to a unit sphere, and cut across a random hyperplane. For the unit sphere in 1 dimension (i.e., two points), this process just yields a restatement of the original problem. However, when the dimension of ambient space is equal to the number of vertices, then the quadratic program becomes semidefinite, and relaxation can be done quickly.
thanks a lot for this example and the links, it looks like a great one and I will follow the links to get a good understanding!
There is one example from algebra which I'm pretty fond of, namely a proof that every field $K$ (which I assume to be infinite for simplicity) has an algebraic closure.
One issue which prevents naive applications of Zorn's Lemma is the lack of a natural ambient structure which would contain all algebraic extensions. An elegant way to get around that is to fix some set $S$ containing $K$ and of cardinality larger than $K$, and consider all algebraic extensions contained in $S$. Since under the axiom of choice all algebraic extensions of $K$ have the same cardinality, any maximal such extension has to be algebraically closed, so an application of Zorn's Lemma gives existence of an algebraic closure.
thanks for this great algebra example. Very much appreciated!
To avoid putting field structures on abstract sets, you could prove this theorem by using a large polynomial ring over $K$. The use of Zorn’s Lemma is then hidden in the appeal to a suitable maximal ideal in that polynomial ring. See https://kconrad.math.uconn.edu/blurbs/galoistheory/algclosureshorter.pdf.
Here is an example from projective geometry. Desargues's theorem states that for two triangles, if the lines connecting their corresponding vertices are concurrent, then the intersection of each pair of corresponding edges are collinear. (Wikipedia states if and only if, but IIRC the converse is an easy corollary.)
In a non-degenerate 3D case (where the triangles are not in the same plane), if we call the triangles $ABC$ and $abc$, then the proof goes something like this: $Aa$ and $Bb$ intersect, so $AaBb$ are on the same plane, and thus $AB$ and $ab$ intersect. Similarly, $BC$ and $bc$ intersect, and $CA$ and $ca$ intersect. These intersections must lie of the intersection of the planes containing $ABC$ and $abc$, which is a line.
A 2D case, which is just a degenerate 3D case, can be viewed as projecting a 3D case onto the plane. Proofs without using 3D geometry often rely on calculations.
great example! Thanks for adding this
Also mentioned by Monroe Eskew in a comment.
It is not a coincidence that proofs that do not rely on the 3D structure rely on calculations. One can give a combinatorial definition of a projective plane as two sets $P$ of points and $L$ of lines and a relation $R \subset P \times L$ that denotes which points lie on which lines, satisfying suitable axioms. A central question in the theory is whether such a structure is the projective plane over a skewfield $K$. One can try to costruct $K$ from the geometry: after all, $K$ should be in bijective correspondence with a line minus one point, and one can try to define the operations on $K$...
...using intersections. It turns out that this strategy works (hence the projective plane really is $\mathbb{P}^2(K)$) if and only if the Desargues theorem holds. Since there are nontrivial examples of combinatorial projective planes, Desargues theorem is equivalent to being able to introduce coordinates. This is why you see proofs using computations. The interesting thing is that one can define also a combinatorial $n$-space, but, due to your proof, once $n \geq 3$ you only get projective spaces over a skewfield. Hence the only combinatorially interesting case is $n = 2$.
@AndreaFerretti Andrea this is giving great extra perspective, thanks for that!
Liapunov's convexity theorem (1940). Let $\mu_1,\dots,\mu_n$ be finite, atomless, signed measures on a sigma-algebra $\mathcal F$. Then the set
$$\big\{\big(\mu_1(A),\mu_2(A),\dots,\mu_n(A)\big) \in \mathbb R^n: A \in \mathcal F\big\}$$ is closed and convex.
Liapounoff, A., Sur les fonctions-vecteurs complètement additives., Bull. Acad. Sci. URSS, Sér. math. 4, 465-478 (1940). ZBL66.0219.02.
In 1966, Lindenstrauss provided a shorter proof. This proof goes into an infinite-dimensional Banach space $X$, then uses the fact that a linear map $X \to \mathbb R^n$ cannot be injective.
Lindenstrauss, Joram, A short proof of Liapounoff’s convexity theorem, J. Math. Mech. 15, 971-972 (1966). ZBL0152.24403.
thanks a lot for a great example. This is exactly the kind of simpler proof I am interested in (and I was not aware of this result before)
I will gladly expand my above comment into an answer. I know this example from Matouseks great book „Using the Borsuk-Ulam theorem“.
Consider the Necklace Splitting Problem: two thieves have stolen a necklace (which is open) with beads made from k different gems. They want to divide the necklace in a fair way between them, i.e. in such a way that each thieve gets half of the stones of each type of gem. They further want to do so cutting the necklace as few times as possible.
The Necklace theorem now asserts that they can divide the necklace between them using at most k cuts. A possible proof is the following: place the necklace onto the moment curve in k-dimensional space. By the Ham Sandwich theorem there is a hyperplane splitting each type of gem in half. It can be shown that any hyperplane cuts the moment curve in at most k places, so the Ham Sandwich cut in k dimensions translates to a solution to the original one-dimensional problem.
As a side remark, a combination of lifting and Ham Sandwich cuts can be used to show a number of results about mass partitions. For example, lifting masses in 2D to the unit paraboloid in 3D, the Ham Sandwich theorem shows that there is always a circle (where a line is also a circle, just with infinite radius) which simultaneously bisects 3 masses. Another lifting can be used to show the existence of bisections by algebraic curves of fixed degree, which is the so called „Polynomial Ham Sandwich Theorem“.
As a second side remark, the lifting to the unit paraboloid in 3D can also be used to show that the Lawson flip algorithm to find a Delaunay triangulation terminates, see page 86 of these lecture notes. There is also a neat argument involving lifting in another chapter which is about counting embracing triangles (the lifting comes in at page 152, the fourth page of the chapter).
thanks a lot! It is a great example and your remarks are super interesting
The following is copied from an answer to another question on this site.
Here's an example in planar euclidean geometry. Consider an equilateral triangle of side $a$ and a general point in the plane distant $b$, $c$, and $d$ from the respective vertices. Then
$3(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2$.
This is an an awful slog to get by planar trigonometry. Even harder to do by trig in three dimensions is the corresponding result for the regular tetrahedron. However, it's easy to get the $(n - 1)$-dimensional result for a regular $(n - 1)$-dimensional simplex of side $d_0$, with vertex distances $d_1 ,..., d_n$ :
$n(d_0^4 +\cdots+ d_n^4) = (d_0^2 + \cdots + d_n^2)^2$.
You can do this by embedding the euclidean $(n - 1)$-dimensional space as the hyperplane of points $(x_1 ,..., x_n)$ in euclidean $n$-space such that $x_1 + \cdots+x_n = d_0/\surd2$. The vertices of the simplex can then be represented as the points $(d_0/\surd2)(1, 0 ,..., 0), ... , (d_0/\surd2)(0 ,..., 0, 1)$ in the hyperplane, and the result drops out in a few lines.
thank you very much for a great example. I can see how cumbersome it must be to prove it in the plane
Dear John Bentin, I am very interested in this theorem of yours as well as your two articles in The Mathematical Gazette? May I get your email?
@TranQuangHung : I'm wary of publishing my email address too directly online. However, if you search for my name on the Project Euclid website, you can find two open-access papers, and it won't be hard to locate my address in either of them.
This one comes to my mind (feel free to edit as I'm missing references, and might be wrong in details)
Let $n$ be a nonnegative integer. Let $F,F'$ be homeomorphic compact subsets of $\mathbf{R}^n$. Then $\mathbf{R}^n-F$ and $\mathbf{R}^n-F'$ have the same number of connected components [to be safe, say same finite number, or both $=\infty$].
This typically applies to Jordan's theorem on closed loops in the plane, and more generally to a topological $(n-1)$-sphere in $\mathbf{R}^n$: the complement has 2 connected components.
The proof, as I remember, consists in proving that a homeomorphism $F\to F'$ can be extended to a self-homeomorphism of $\mathbf{R}^{2n}$ ($\mathbf{R}^n$ being embedded in the standard way as the first $n$ coordinates). And then relating $H^0(\mathbf{R}^n-F)$ (whose dimension, finite or $\infty$, is the number of connected components) to the de Rham cohomology $H^n(\mathbf{R}^{2n}-F)$.
This beautiful trick is due to Victor Klee (1955). You can find references in https://mathoverflow.net/q/271846/6451
@BS. Great link! Thanks a lot
@YCor I wanted to thank you for adding a great example. Just noticed that I erroneously deleted my previous comment
The same argument also shows that the other Betti numbers of the complement match as well.
Perhaps I am just partially blind and someone has already said this, but the thing that springs to mind for me is to show that there no percolation (ie no infinite component) at criticality for $\mathbb Z^d$ with $d \ge 3$.
To learn more about this, there is a survey Sixty Years of Percolation by Hugo Duminil-Copin from IHÉS. He is one of the top people in the field. (In fact, he's one of the top young mathematicians in the world -- if he wins a Fields Medal, you heard the prediction here first!) He can the talk at ICM 2018. The final sentence of the abstract reads as follows:
This review is not intended to probabilists...: the target audience is mathematicians of all kinds.
Regarding the history, it is outlined in (the second half of) Section 1.2.
Let me summarise a little -- the complete history is not given there.
All references which I mention below can be found in the text of Duminil-Copin linked above.
It was originally proved by Hara and Slade for $d \ge 19$ using lace expansion. To quote the reference, "Each few years, more delicate uses of the lace-expansion enable [us] to reduce the dimension".
the The current best is $d \ge 11$, due to Fitzner and van der Hofstad.
(I think along the way vdH brought $d \ge 19$ down to something like $d \ge 14$.)
For some details on why higher dimensions may be easier, see Section 3.2. Roughly, it is to do with crossing probabilities of simple random walks in $\mathbb Z^d$.
It is well-known that $d \ge 3$ implies transience (so return to the origin finitely many times). However, I think you need $d \ge 5$ in order to say that two independent walks intersect only finitely many times. I forget the exact details. For very large $d$, a SRW on $\mathbb Z^d$ looks, at first, a little bit like a SRW on a $d$-regular tree (up until time $o(\sqrt d)$?).
It's an interesting history, showing how these tools originally really on work for sufficiently high dimensions. Unfortunately, $d \ge 3$ is still rather out of reach with current techniques...
Barsky, Grimmett and Newman showed, in 1991, that the analogous claim is true (for all $d \ge 3$) not for $\mathbb Z^d$ but for $\mathbb N \times \mathbb Z^{d-1}$. One would surely think the main conjecture is within touching distance given this. Amazingly, ~30 years later, basically no improvement for small $d$ has been obtained!
I found a nice reference: I have included that, and added some of my own comments :)
this is a great link, and the text you added gives great insights. Thanks a lot!
Glad to help! It was interesting for me reading up on it again -- I do discrete probability, so have some idea of the history, but I don't do percolation really. Was good to refresh the memory! (I've also added a note at the very end just now)
thanks for a great add, and a comment, because you make this good remark that 30 yrs later still no progress. In a similar direction, same thing for the Collatz Conjecture. This is even harder to believe
I was asked this in an interview one time! After a few preliminary questions on the algorithm, I was asked if it terminates. Needless to say, I couldn't give a proof! The interviewer quickly said that it's a famous open question
The Cauchy problem for the wave equation
$$\partial_t^2u=c^2\Delta_xu$$
is not too difficult to solve explicitly in $3$ space dimensions, by the method of spherical means. This yields a close formula for the fundamental solution.
It is much more difficult, if not impossible, to carry out the calculation directly in space dimension $2$. Actually, the explicit solution of the Cauchy problem and the fundamental solution are obtained by extending the initial data to ${\mathbb R}^3$ by $u_j(x_1,x_2)\mapsto v_j(x_1,x_2,x_3):=u_j(x_1,x_2)$ (here $j=0,1$ for the data of $u$ and $\partial_tu$ at initial time). This is called the descent method.
thanks a lot - great example
Not a single theorem per se, but in dynamical systems, it is often very useful to translate questions about properties of a continuous system $\dot{x}=f(x)$ or discrete-time system $x_{k+1}=f(x_k)$, where $x\in\mathbb{R}^n$, to questions about probability distributions or densities over $\mathbb{R}^n$. This is done by studying the associated Perron-Frobenius/Transfer operator that describes the density time evolution.
Arguably, questions such as existence and properties of invariant sets of $f$ are better handled in this infinite dimensional setting. The key point is that the infinite dimensional operators are linear, even if $f$ itself is nonlinear. This brings the spectral theory of linear operators into play.
great example! Thanks a lot for providing it and thanks also for a good outline
Differential equations of order $n$ in $\mathbb{R}$, like $\frac{d^n}{dt^n}x(t) = F\left(t, x(t), \frac{d}{dt}x(t), \dots, \frac{d^{n-1}}{dt^{n-1}}x(t)\right)$ can be transformed into first order differential equations in $\mathbb{R}^n$.
This is done by defining $x_1(t) = x(t), x_2(t) = x_1'(t), \dots, x_n(t) = x_{n-1}'(t)$ and observing that the equation reduces to $x_n'(t) = F(t, x_1(t), \dots, x_n(t))$, which - together with the relations defining $x_i(t)$ - is a differential equations of the first order in the vector $(x_1(t), \dots, x_n(t))$.
thanks again for a great example from analysis/differential equations. Great add, thanks a lot!
In 3D-graphics, 3D-points are translated into 4D-points using a technique refereed to as "homogeneous coordinates". Then 3D-perspective transformations and coordinate translations (which are non-linear in 3D), become linear in 4D. This allows one to concatenate all the successive transformations into a single linear transformation. This really enables the lightening fast 3D graphics you see today although it was discovered and used quite early on. I remember being amazed when I learned it.
There are correspondingly a host of perspective geometry theorems that are enabled as a result of this linearity in 4D, like the ability to clip at the end of the pipeline to get the same results of clipping at the beginning, etc.
thanks a lot, striking! I have to look it up. Do you have a good link to that?
@Claus: It's very simple, and there is totally no need for "homogenous coordinates". Put the 3d plane in 4d space with one of the coordinates being 1. A translation in that 3d plane can be done via a linear shear in 4d. So everything transform is represented by a matrix operator on 4d vectors. This means that any sequence of transforms is a product of matrices. Using standard knowledge from algorithms and data structures, joining or splitting transform sequences can be done in $O(\log(n))$ time where $n$ is their total size.
Not a theorem but a cool result regardless: Given a convex $n$ sided polygon in 2D, give an algorithm to find the largest circle that can fit inside it.
I am not aware of any non messy or particularly efficient approaches in 2D but if you consider the planes in 3D that contain each side length of the polygon and forming a 45 degree angle with the plane of the polygon, then the problem can be solved by finding the point with the largest third coordinate that is underneath all these planes. This can be done very efficiently with a linear program.
it is a great example!
The symmetry of the octahedron projects to
Simpson's Rule.
Recall that Simpson's Rule is the approximation
$$
\int_a^b f(x) \, dx \approx \frac{b-a}{6}
\Bigl(f(a) + 4 f\bigl(\frac{a+b}{2}\bigr) + f(b) \Bigr) \,,
$$
which is exact for $f$ polynomial with $\deg f \leq 3$, and thus true within
$2\epsilon(b-a)$ for a function that is approximated within $\epsilon$ by
a cubic polynomial. All intervals $[a,b]$ are equivalent under
affine-linear change of variable, and these changes of variable
preserve Simpson's Rule, so it is enough to consider the special case of
the interval $|x| \leq 1$, which is
$$
\int_{-1}^1 f(x) \, dx \approx \frac13 \bigl(f(-1) + 4 f(0) + f(1) \bigr).
$$
Now let $V = \{ \pm e_1, \pm e_2, \pm e_3 \}$ be the set of six vertices of
the standard octahedron inscribed in the sphere $S^2 \subset {\bf R}^3$;
and let $G$ be the group of symmetries of $V$, which are the $2^3 3! = 48$
signed permutation matrices of order $3$. Any polynomial function
$F: {\bf R}^3 \to {\bf R}$ that is invariant under $G$ and has
$\deg F \leq 3$ is a linear combination of $1$ and $x^2+y^2+z^2$.
It follows the average of $F$ over $S^2$ equals the average
$\frac16 \sum_{i=1}^3 (F(e_i) + F(-e_i))$ of $F$ over $V$.
That is, $V$ is a
spherical 3-design.
Now apply this to $F$ of the form $F(x_1,x_2,x_3) = f(x_1)$
with $f \in {\bf R}[x]$ of degree at most $3$. For any function
$G: S^2 \to {\bf R}$ of the form $G(x_1,x_2,x_3) = g(x_1)$,
the average of $G$ over $S^2$ equals the average of $g$ over $[-1,1]$,
essentially by the same theorem of Archimedes that I cited in an
earlier answer
to the same MO question (Tarski's plank theorem of 1932).
Of the six points in $V$, four have $x_1=0$ and one each has $x_1 = 1$ or $-1$,
so we have recovered Simpson's Rule.
I learned this from
Greg Kuperberg;
see his article
Numerical cubature from Archimedes' hat-box theorem, SIAM J. Numer. Anal. 44 (2006), 908-935 (arXiv:math/0405366).
"Cubature" is quadrature in higher dimensions (NB "quadrature" = "squaring"
as in "squaring the circle"). The article gives many other quadrature and
cubature formulas that can be obtained this way by projection from
symmetrical designs in higher dimension. For starters,
rotating $V$ so two faces are perpendicular to the $x$-axis,
or replacing $V$ by the cube $(\pm 1, \pm 1, \pm 1)/\sqrt 3$
(which has the same symmetries, and is thus also a $3$-design),
gives the quadrature rule
$$
\int_{-1}^1 f(x) \, dx \approx f(-1/\sqrt3) + f(1/\sqrt3)
$$
which is again exact for polynomials of degree at most $3$.
Remark: This is arguably a better example than the Tarski plank theorem,
not just because we must raise the dimension by $2$ but also
because here we make essential use of symmetries of ${\bf R}^3$:
for Tarski we could have integrated $dx \, dy \left/ \sqrt{1-x^2-y^2} \right.$
instead of invoking the third dimension.
On my first reading of this, I was confused by "Any polynomial function $F$ ... that is invariant under $G$ ....Now apply this to $F$ of the form $F(x_1,x_2,x_3)=f(x_1)$ ..." since $f(x_1)$ isn't invariant under $G$. Fortunately, it's explained on the first page of the Kuperberg's cited paper, and once it's explained it's obvious. OK, it should have been obvious anyway, but maybe mention it for the sake of other readers who are as sleepy as I am.
Yes I should rewrite that -- all the more so since there are two very different $G$'s ! But I need to go to sleep now too . . .
@NoamD.Elkies thank you for providing another great example here. It fits my question perfectly and great to have an example from numerics as well
The aperiodic Penrose tiling can be generated as a cross section of a regular tiling in 5 dimensions, which is periodic! See this answer for more details.
thanks for a great answer. This is truly a fascinating example as well. Thank you for the link with the details.
How do we need five dimensions? The quasilattice on which the nodes are based becomes periodic in only four.
The Toepliz problem, also known as "inscribed square problem" or "square peg problem", asks whether every Jordan curve in the plane contains the vertices of a square.
Vaughan's proof of the rectangular peg problem embeds the plane (and the curve) in $\mathbb{R}^3$ and works towards a contradiction. The proof is really beautiful, and there's a 3blue1brown video that fleshes it out.
Vaughan's idea has then been furthered developed by Hugelmeyer, who embeds the plane in $\mathbb{R}^4$ instead. His proof is really clever and works for smooth curves. Just last week Greene and Lobb posted a symplectic refinement of Hugelmeyer's ideas, leading to a much stronger statement about aspect ratios. Let me also advertise Matschke's work and his survey on the Toepliz problem.
Thanks a lot for this beautiful example! I like this proof a lot and the video is great as well. Thanks for the other links as well including the update from last week. Great example!
An example in the spirit (in fact a generalization) of the answer by Sam T is the triviality of the $\phi^4$ quantum field theories from lattice approximations.
In dimension 5 or more this was done a long time ago by Aizenman and Fröhlich. In dimension 4 this is a brand new result by Aizenman and Duminil-Copin. The reason I said generalization above is that this is part of a general phenomenon in statistical mechanics due to the notion of upper critical dimension. See this review by Gordon Slade for a general mathematical introduction.
thanks a lot for adding this example!
The famous result by Bang is that if a convex compact set $K\subset \mathbb{R}^n$ is covered by a finite number of open planks, then the sum of their widths is greater than a width of $K$. [The closed, correspondingly open, plank with normal $\theta$ of width $h\geqslant 0$ is the set of points lying between, correspondingly strictly between, two planes at distance $h$, both orthogonal to a unit vector $\theta$. The width $w(K)$ of $K$ is defined as the minimum of widths of closed planks containing $K$.]
If $n=2$ and $K$ is a unit disc, there is a short proof using the lifting to the third dimension, also mentioned in the answer by Noam Elkies : considering $K$ as a section of the unit ball in $\mathbb{R}^3$, for any plank $S$ of width $h$ its lifting $S\times \mathbb{R}=\{(s,x)\in \mathbb{R}^3: s\in S, x\in \mathbb{R}\}$ intersects the unit sphere by a set of area (at most) $2\pi h$ (this fact belongs to Archimedes himself). Since the whole unit sphere, which has area $4\pi$, must be covered by the liftings of our planks, we immediately get the desired lower bound 2 for the sum of their widths, it is strict for open planks.
Now about the general case, we again use the lifting but differently.
We use the following
Lemma. If $K\subset \mathbb{R}^n$ is a convex compact set and $f\in \mathbb{R}^n$, $\|f\|\leqslant w(K)=:h$, then
a) $K\cap (K+f)\ne \emptyset$;
b) $w(K\cap (K+f)) \geqslant h-\|f\|$.
Proof. a) Assume the contrary. Then by Hahn -- Banach $K$ and $K+f$ may be separated by a plane $\langle x,\theta\rangle=c$. That is, $\langle x,\theta\rangle< c<\langle x+f,\theta\rangle$ for any $x\in K$. Thus $K$ may be covered by an open plank of width $\langle f,\theta\rangle \leqslant \|f\|\leqslant h$, a contradiction.
b) Denote $g=f\cdot \frac{h}{\|f\|}$ (if $f\ne 0$, the case $f=0$ is trivial). Then $\|g\|=h$ and by a) there exists a point $a\in K\cap (K+g)$. We have by convexity $$\frac{h-\|f\|}h(K-a)\subset K-a,\\ \frac{h-\|f\|}h(K+g-a)\subset K+g-a,$$
that is equivalent to $a+\frac{h-\|f\|}h(K-a)\subset K\cap (K+f)$.
Therefore $w(K\cap (K+f))\geqslant w(a+\frac{h-\|f\|}h(K-a))=h-\|f\|$.
Now assume that $\sum h_i\leqslant h=w(K)$ and the open planks $S_i=\{x:|\langle x-x_0,\theta_i\rangle|< \frac{h_i}2 \}$, $i=1,\ldots,N$, cover $K$. In other words, we assume that there exists a point, called $x_0$, which belongs to all the middle planes of the planks ($x_0$ may belong to $K$ or not).
The $2^N$ sets
$K\pm \frac{h_1}2 \theta_1 \pm \frac{h_2}2 \theta_2\pm \ldots \pm \frac{h_N}2 \theta_N$ have a non-empty intersection: this follows from applying Lemma $N$ times (we start with $w((K-\frac{h_1}2\theta_1)\cap (K+ \frac{h_1}2\theta_1))=w(K\cap (K+h_1\theta_1))\geqslant h- h_1$ and proceed naturally, using the obvious inclusions like $(A\cap B)+x\subset (A+x)\cap (B+x)$.)
So, for certain $p\in \mathbb{R}^n$, the set $\Omega=\{p\pm \frac{h_1}2 \theta_1 \pm \frac{h_2}2 \theta_2\pm \ldots \pm \frac{h_N}2 \theta_N\}$ is contained in $K$. Choose the point $q\in \Omega$ on the maximal distance from $x_0$. We should have $|\langle q-x_0,\theta_i\rangle| <h_i/2$ for some $i$, and this implies (easily seen from the picture) that both points $q+h_i\theta_i$, $q-h_i\theta_i$ are further from $x_0$ than $q$. But one of these two points belongs to $\Omega$, a contradiction.
Now a general case. Assume that $K$ is covered by $N$ planks. If the normals of our planks are linearly independent, there middle planes have a common point and we are done. If $N\leqslant n$, we may move our planks a bit so that their normals become linearly independent and they still cover $K$. Finally, if $N>n$, we lift $K$ to a cylinder $C:=K\cdot [0,M]^{N-n}\subset \mathbb{R}^N$ (where $M$ is so large that $w(C)=w(K)$, $M=h$ is enough) and lift the planks $S_i$ to $S_i\times \mathbb{R}^{N-n}$. The problem is reduced to the case which is already done.
thanks a lot for providing! And great that you give the proof for general $n$ to it here!
Oh yes it is. I searched for Bang before posting, but not for Tarski, my bad.
Thanks @FedorPetrov. It seems that this argument has inspired other results around here as well, like this one by fedja: https://mathoverflow.net/questions/352720/reference-to-a-conjecture-on-unit-vectors-in-euclidean-space
@SandeepSilwal earlier it inspired fedja also to solve the coefficient problem for Fourier series (“The Bang solution of the coefficient problem”, St. Petersburg Math. J., 9:2 (1998), 407-419)
Amazing, I always learn something from fedja.
Given that you asked about planar graph : In graph theory, there is the Heawood conjecture proven in 1968 by Ringel and Youngs:
If a graph $G$ has genius $g>0$ then
$$ \chi(G)\leq \left\lfloor \frac{7+\sqrt{1+48g}}{2}\right\rfloor$$
Note that the case $g=0$ (not included in this theorem) would be the four color theorem for planar graph! It is a pretty surprising result to get a relatively simple theorem for any genius $g>0$ but not $g=0$.
thanks a lot for this striking example
For a graph $G$, its multi-dimensional characteristic polynomials $\Phi_G=\det(I_x-A)$ where $A$ is adjacency and $I_x=diag\{x_1,...,x_n\}$. It's definition depends on the labeling of the vertices but it is multi-affine real stable. One has $\Phi_{G-v_j}=\frac{\partial \Phi_G}{\partial x_j}$ which gives an intuitive reason why its contraction $\phi_{G-v_j}(x)$ interlaces $\phi_G(x),$ since $f'$ always interlace real rooted $f$.
Also the derivativ formula
$\frac{d}{dt} \Phi_G(x_1(t),...,x_n(t))=\sum_{j=1}^n\frac{\partial{\Phi_G}}{{\partial x_j}} \frac{d x_j}{dt},$ implies the formula
$ \frac{ d \phi_G(x)}{dx}=\sum_{j=1}^n \phi_{G-v_j} (x),$ if we identify $x_j=x=t$.
thanks a lot for your example! Can i ask you, please let me know, what's the name of this theorem?
This continued to hold if one replace determinant by permanent, in fact by the immanent attached to any irreducible character of $S_n$. If $\lambda$ is a partition of $n$ and $\Phi_G^\lambda:=imma_\lambda(I_x-A)$, then $ \frac{\partial \Phi^\lambda_G}{\partial x_n}=\sum_{\lambda_1' < \lambda} \Phi^{\lambda_1}{G-v_n}(x_1,...,x{n-1}),$, where $\lambda_1$ range over all partition of $n-1$ whose Young diagram is derive from that of $\lambda$ by removing one border square which remains a valid Young diagram.
This is a pure algebraic identity and it holds for all complex matrix $A$ not necessarily adjacency of a graph or Hermitian.
There are several examples related to the mathematics of quantum field theory in which using higher-dimensional
thinking in physics led to mathematical theorems answering questions which might not have even been asked without
said higher-dimensional thinking. (Those proofs don't necessarily use the same higher-dimensional methods, so maybe
this isn't exactly what you're looking for. If so, I'm happy to remove this answer.)
One common way to study quantum field theories is "compactification" from a higher-dimensional theory, by saying,
e.g., that your $n$-dimensional QFT on a manifold $M$ is the same thing as an $(n+2)$-dimensional QFT on $M\times
T^2$. (It doesn't have to be $T^2$). This often explains mysterious properties of the $n$-dimensional QFT in terms
of clearer information coming from the $(n+2)$-dimensional QFT (in one example, an $\mathrm{SL}_2(\mathbb
Z)$-symmetry on the original theory arising from the mapping class group of that $T^2$ in the $(n+2)$-dimensional
theory). Generally, these QFTs aren't mathematically well-defined, but their study still leads to mathematically
rigorous questions, and this perspective can help answer them.
Mirror symmetry is a great example. One of its avatars is a collection of conjectures (some of which are now
theorems) about six-dimensional Calabi-Yau manifolds, associating to such a manifold $X$ a "mirror" $X^\vee$,
another Calabi-Yau $6$-manifold, and equating certain data on $X$ with other data on $X^\vee$. These conjectures
arose in physics, where physicists suspected an equivalence between one kind of string theory on $\mathbb
R^{1,3}\times X$ and another kind of string theory on $\mathbb R^{1,3}\times X^\vee$. Without that insight, it is
very unlikely anyone would've thought to ask the questions leading to mirror symmetry, let alone answer them.
(There's no shortage of other examples, such as the study of anomalous QFTs as boundary theories of invertible
theories in one dimension higher, or the use of Theory $\mathfrak X$ to study lower-dimensional mathematical
objects…)
thanks a lot for your contribution based on quantum field theory! I think it is great to have such an example. My thinking, it is not a case where the proof gets simple because it goes "higher" dimension than the theorem, but it is still a great example in that the higher-dim case informs lower-dim results
This answer is quite different in spirit from my other answer, so I've factored it out.
The $n$th bordism group $\Omega_n$ is the abelian monoid of diffeomorphism classes of closed
smooth $n$-manifolds under disjoint union, modulo those which bound compact $(n+1)$-manifolds (this is in fact a
finitely generated abelian group). There are many variations on this, such as requiring everything to be oriented,
or have spin structures, or so on.
One can imagine computing these groups directly using topological or geometric methods, and this works up to
dimension 3 or so (e.g. this MO question and its answers, or
this paper of Stipsicz), but eventually these methods
aren't powerful enough.
Thom and Pontrjagin discovered a very different approach which requires higher-dimensional methods: use the Whitney
theorem to embed your manifold $M$ in $S^N$ for some $N$ large enough. The normal bundle of $M$ is classified by a
map from the normal bundle to the universal rank-$(N-n)$ vector bundle $V_{N-n}\to B\mathrm O_{N-n}$. One can
extend this to a map from $S^N$ to something called the Thom space $T_{N-n}$ of $V_{N-n}\to B\mathrm O_{N-n}$,
constructed by adding a basepoint at infinity in a suitable sense. One checks that homotopy classes of maps $S^N\to
T_{N-n}$ are in bijection with $\Omega_n$, and now computing bordism groups amounts to computing homotopy groups of
this Thom space.
Computing homotopy groups is not easy, but this method scales in $n$ much more nicely than more direct approaches,
and Thom completely solved this problem for all $n$. (Many variants of this problem are also completely solved,
thanks to work of Wall, Anderson-Brown-Peterson, Milnor, and many more.)
thanks a lot I think this example is a great fit.
Let $P$ be a convex polytope in $\mathbb{R}^d$ with vertices $v_1,\dots,v_n\in \mathbb{Z}^d$. A nice trick that helps to visualize, understand and prove that the number of lattice points in dilations $tP$ $(t\in\mathbb{N})$ is a polynomial in $t$, called the Ehrhart polynomial of $P$, is to add one dimension a consider the cone over $P$:
$$\mathrm{cone}(P)=\{r_1(v_1,1)+\cdots+r_n(v_n,1)\mid r_1,\cdots,r_n\ge0\}\subset\mathbb{R}^{d+1}.$$
Then the dilated polytope $tP\subset\mathbb{R}^d$ corresponds to the intersection of $\mathrm{cone}(P)$ with the hyperplane $\{(x_1,\dots,x_{d+1})\in\mathbb{R}^{d+1}\mid x_{d+1}=t\}$. This allows to work with some generating functions associated to polyhedra which simplify for cones.
super example! And very visual. Thanks a lot!
Hopefully, you will agree that infinite dimension
is higher dimension.
A fruitful approach to solve a nonlinear problem
on a finite dimensional space is to convert it into
a linear problem on an infinite dimensional space.
There are literally hundreds of examples in that vein.
Let me give just five of them.
Hurwitz's proof of the isoperimetric inequality in the plane using Fourier series. Convert a geometrical problem in the plane into a problem about complex valued functions.
More generally the resolution of the standard pdes (wave, heat) using Fourier series. See a function of two variables with finite dimensional range as a function of one variable with value in an infinite dimensional function space.
The theory of Schwartz's distributions. Functions as linear functionals on functions. The fact that every functions become differentiable simplify a lot of computations in mathematical physics.
The book of Laurent Schwartz
"Mathematics For The Physical Sciences"
is full of examples.
Koopmanism in dynamical systems. Replace the action
of a transformation $T : X \rightarrow X$ on a finite
dimensional manifold by the action of the linear operator
$f \rightarrow f \circ T$ on a well chose functional space,
for example $L^2(X,\mu)$ if $T$ preserves some measure $\mu$.
Von Neumann used that method to prove what is now known
as the Von Neumann ergodic theorem.
Another application of Koopmanism to the isomorphism problem
in dynamical systems : show that two rotations
on the circle are conjugate by a measurable transformation
preserving the Lebesgue measure if and only if their
angles are equal or opposite. This is easily done by
looking at the spectrum of the Koopman operator, which
is an invariant of measurable isomorphism.
Thanks a lot for a great spectrum of examples!
|
2025-03-21T14:48:31.040001
| 2020-05-21T07:11:40 |
360926
|
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"url": "https://mathoverflow.net/questions/360926"
}
|
Stack Exchange
|
Terminology for exact symplectomorphism
Let $(M,\omega = d\alpha)$ be an exact symplectic manifold. Then a symplectomorphism $\varphi \colon M \to M$ is said to be exact, iff $\varphi^*\alpha - \alpha$ is exact. Is there a terminology for the special case when $\varphi^*\alpha = \alpha$? I thought of something like the symplectomorphism preserves the exact symplectic form.
In his classic book on classical mechanics Whittaker calls these transformations Mathieu transformations. The term appears in Wikipedia.
|
2025-03-21T14:48:31.040088
| 2020-05-21T08:13:00 |
360930
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/360930"
}
|
Stack Exchange
|
Are p-boxes for discrete sample spaces meaningful?
The Wikipedia article about p-boxes only talks about cumulative probability density functions, which are meaningful for continuous sample spaces. https://en.wikipedia.org/wiki/Probability_box
Just out of curiosity, is it possible, and meaningful, to define p-boxes for discrete sample spaces? What axioms should they fulfil? Some quick google searches for "discrete p-box" and variants thereof did not turn up any results.
Take a three-element sample space $\{0,1,2\}$ for example. We can define a probability mass function (pmf) on it, like $\{0 \mapsto 0.5, 1 \mapsto 0.5, 2 \mapsto 0\}$, where 0 has probability 0.5, 1 has probability 0.5, and 2 has probability 0. Another one would be $\{0 \mapsto 0, 1 \mapsto 0.5, 2 \mapsto 0.5\}$.
My attempt at defining p-boxes on this would be to allow intervals for the probabilities, with the intention that it specifies all pmf's whose values lie in these intervals.
For example a p-box on the three-element set could be $\{0 \mapsto [0,0.5], 1 \mapsto [0,0.5], 2 \mapsto [0,0.5]\}$. This specifies all pmf's whose values lie between 0 and 0.5, like the two concrete ones above.
Not all of those would be meaningful, for example $\{0 \mapsto [0,0.2], 1 \mapsto [0,0.2], 2,[0 \mapsto 0.2]\}$ does not include any valid pmf, because no possible sum adds up to 1.
Another one that is not meaningful would be $\{0 \mapsto [0,\frac{1}{3}], 1 \mapsto [0,\frac{1}{3}], 2 \mapsto [0, \frac{1}{3}]\}$, because the lower bounds could never be used.
It seems that if one were to specify p-boxes in this way, there is some constraint between the upper and lower bounds.
Are p-boxes on discrete sample spaces meaningful? Are they described in some area of probability theory?
There is no such thing as a "cumulative probability density function". You seem to get confused between the notions of (i) the (cumulative) probability distribution function, (ii) the probability density function, and (iii) the probability mass function. I suggest you read an introductory textbook on probability or statistics, or just the corresponding Wikipedia articles.
Concerning now p-boxes specifically: The Wikipedia article linked in your post defines a p-box in terms of a cumulative probability distribution function, which in turn is defined for arbitrary real-valued random variables -- discrete, continuous, and neither discrete nor continuous. So, there is no reason to worry about p-boxes in the discrete case.
As for the usefulness/meaningfulness of p-boxes, even the Wikipedia article linked in your post has some criticisms of the notion of a p-box. I have never before encountered this notion, and I don't believe it could be really useful.
|
2025-03-21T14:48:31.040284
| 2020-05-21T09:10:39 |
360935
|
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|
Stack Exchange
|
Cayley-Bacharach property of points is linearly general position
Suppose $P$ be a zero dimensional subscheme of length $m$ in $\mathbb{P}^3$ which is in linearly general position. Further assume that there is a degree $d$ form passing through $P$. Can $P$ satisfy Cayley-Bacharach property for $\mathcal{O}(d)$
I got the answer of the question. In fact it can satisfy Cayley-Bacharach. For example if we choose 8 points in rational cubic curve in linearly general position then it always satisfies C-B for $\mathcal{O}(2)$.
why if $P$ is contained a unique curve of degree $d$, then it satisfies Cayley-Bacharach property for $\mathcal{O}(d)$ ? Is it possible to give some lower bound on number of independent conditions imposed by $m$ points in linearly general position on sections of $\mathcal{O}(d)$ ?
We say $P$ satisfy CB for $\mathcal{O}(d)$ if any section of $\mathcal{O}(d)$ which vanishes at a co-length $1$ subscheme then it vanishes on $P$.
Ah OK, sorry, I had something different in mind. I delete my previous comments.
In any case can we give a a lower bound on the number of independent conditions imposed on sections of $\mathcal{O}(d)$ by $m$ points in general position ?
If $m\leq \frac{1}{2} (d+1)(d+2)=\dim H^0(\mathscr{O}_{\mathbb{P}}(d))$, $m$ general points impose $m$ independent conditions.
$m$ points in linearly general position need not be general points. There is a lower bound that $m$ points in linearly general position imposes at least $\text{min}{3d +1, m}$ conditions. Can we say something more in some special cases? For example suppose we are given $14$ points in linearly general position not contained in rational normal curve, Can there be a quadric hypersurface passing through these $m$ points ?
|
2025-03-21T14:48:31.040427
| 2020-05-21T10:08:07 |
360938
|
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"Jochen Glueck",
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|
Stack Exchange
|
Equality in spectral inclusion theorem
I asked this question on Math SE but didn't receive any response.
Let $(T_t)$ be a $C_0$-semigroup on a Banach space $X$ with generator $A.$ If $\lambda_0\in \mathbb{C}$ is such that $e^{\lambda_0 t}$ is a pole of $R(\cdot,T(t)),$ then $\lambda_0$ is a pole of $R(\cdot,A)$ and $$k(e^{\lambda_0 t},T(t))\geq k(\lambda_0,A)\tag{1}$$ where $k(\cdot,\cdot)$ denotes the order of the corresponding pole. This is proved for example in Theorem IV.3.6 in the book by Engel and Nagel.
My question: Suppose we a priori know that $\lambda_0$ is a pole of $R(\cdot,A)$ and $e^{\lambda_0 t}$ is a pole of $R(\cdot,T(t))$ for all $t\geq 0.$ Are there any known conditions which guarantee an equality in $(1)$ for at least some $t>0?$
I would suspect that eventual compactness of the semigroup suffices to get equality for some times $t$ by a functional calaculus argument, but one would have to check the details. For more general semigroups, things might get more involved...
@JochenGlueck Do you know of a reference where similar arguments are used? Say for an eventually compact semigroup, how would one go about proving this?
Unfortunately, I don't know a reference, but the main line of argument should be as follows: Choose a right half plane $H$ that contains $\lambda_0$, then split off the part of the spectrum of $A$ that is contained in $H$ by means of a spectral projection. This reduces the problem to a finite dimensional situation, so one can use matrix analysis to prove the claim.
@JochenGlueck Reducing to finite dimensions make sense but how would one go about applying this to $R(\cdot,T(t))?$ Would using $R(\lambda,A)=R(e^{\lambda t},T(t)) \int_0^t e^{\lambda(t-s)}T(s) \ ds$ help?
In finite dimensions, the problem is essentially about the size of Jordan blocks. If I find time, I'll post the details in an answer this weekend.
@JochenGlueck Yes one needs to show that size of the largest Jordan block is at most $k(\lambda_0,A).$
Exactly -- and this follows immediately if you choose $t$ such that $\lambda \mapsto e^{t \lambda}$ maps all finitely many spectral values of $A$ on your finite dimensional space to distinct numbers. Does this answer your question, or should I add some more details?
@JochenGlueck I'd be grateful if you could add more details in an answer. I don't see how to achieve this.
Alright. About what point precisely are you unsure?
@JochenGlueck Mainly how to use eventual compactness of semigroup. Also is this technique of restricting to the finite-dimensional situations and using Jordan blocks common? It would be great if I could read similar arguments to get more understanding.
Reducing problems to the finite dimensional case is a quite common approach in spectral theory whenever sufficient compactness if present. (In my answer I've included a reference to a Corollary in the book of Engel and Nagel where you can see this in detail). Using Jordan blocks is probably the simplest way to understand the functional calculus in finite dimensions, so it is very common in the study of the matrix exponential function.
Here is a positive solution if the semigroup is eventually compact:
Consider, say, the open right halfplane
$$
H := \{\lambda \in \mathbb{C}: \, \operatorname{Re}\lambda > \operatorname{Re}\lambda_0 - 1\}.
$$
Then $H$ contains only finitely many spectral values of $A$, and all these spectral values are poles of the resolvent with finite-rank spectral projections. So if we denote the sum of these finitely many spectral projections by $P$, then the range $PX$ is a finite dimensional space, and the restriction of the semigroup to the complementary subspace $\ker P$ only contains spectral values with real part $\le \operatorname{Re}\lambda_0 - 1$.
All these properties follow from Corollary V.3.2 in the semigroup book of Engel and Nagel (2000).
The restriction of the semigroup to $\ker P$ is also eventually compact, so the spectral mapping theorem holds for it [op. cit., Corollary IV.3.12(i)]. Hence, $e^{t\lambda_0}$ is not in the spectrum of $T_t|_{\ker P}$ for any $t$.
So we only need to deal with the restriction of the semigroup to the finite dimensional space $PX$. Hence, we may asume from now on that $X$ itself is finite dimensional and that $A$ is a matrix. After a coordinate transformation, $A$ is in Jordan normal form. The matrix $A$ has distinct eigenvalues $\lambda_0, \dots, \lambda_m$, and for each $k \in \{0, \dots, m\}$ there are Jordan blocks $J_{k,1}, \dots, J_{k,\ell_k}$.
Since $A$ is the direct sum of all these Jordan blocks, we can compute $e^{tA}$ by computing $e^{tJ_{k,i}}$ for all Jordan blocks separately.
Now, if $t \in [0,\infty)$ is such that the numbers $e^{t\lambda_0}, \dots, e^{t\lambda_m}$ are all distinct, then out of all direct summands
$$
e^{tJ_{0,1}} , \dots, e^{tJ_{0,\ell_0}} , \quad \dots \quad , e^{tJ_{m,1}}, \dots, e^{tJ_{m,\ell_m}}
$$
of $e^{tA}$, only the matrices $e^{tJ_{0,1}} , \dots, e^{tJ_{0,\ell_0}}$ have the eigenvalue $e^{t\lambda_0}$. Hence, the dimension of largest Jordan block in $e^{tA}$ that belongs to the eigenvalue $e^{t\lambda_0}$ (i.e., the order of the pole $e^{t\lambda_0}$ of $R(\cdot,e^{tA})$) cannot be larger than the largest dimension of the matrices $e^{tJ_{0,1}} , \dots, e^{tJ_{0,\ell_0}}$ - which is the same as the largest dimension of the matrices $J_{0,1}, \dots, J_{0,\ell_0}$ (and this is, in turn, the order the pole $\lambda_0$ of $R(\cdot,A)$).
This solves the question for all those times $t$ for which the numbers $e^{t\lambda_0}, \dots, e^{t\lambda_m}$ are distinct.
Thank you for your answer. I have one question. Firstly, it seems to me that what you really used was SMT and not eventual compactness directly. So your proof technically works for all semigroups for which SMT holds? In particular, it holds for eventually norm continuous semigroups?
@Mark: Well, I used that there are only finitely many spectral values in the half plane $H$ and that these are poles of the resolvent with finite dimensional spectral spaces. But you are probably right that we can do better than that: If the semigroup is merely eventually norm continuous, and the spectral projection $P_0$ associated to the pole $\lambda_0$ has, say, finite rank, then I think it should work to merely split off $P_0$: if $t > 0$ is sufficiently small, than $T_t|_{\ker P_0}$ won't have $e^{t\lambda_0}$ in its spectrum (due to the spectral mapping theorem), and we're fine.
@Mark: If the spectral space $P_0X$ is not finite dimensional, though, then the reduction to matrix analysis does no longer work, of course. I'd be optimistic that the same result about the pole order still holds, but one would have to think about the details in the infinite dimensional case.
|
2025-03-21T14:48:31.040877
| 2020-05-21T10:57:53 |
360940
|
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|
Stack Exchange
|
Center-stable manifold theorem on manifold with boundary
I would like to see if there is a Center-stable manifold theorem on the phase space that is a manifold with boundary.
Suppose $f:M\rightarrow M$ is a diffeomorphism, according to Theorem III.7 in "Global stability of dynamical systems, M.Shub", at a fixed point $x^*\in M$, the tangent space $T_xM$ can be decomposed to $E^s\oplus E^c\oplus E^u$, based on that the norm of eigenvalues of $Df(x^*)$ are less than one, equal to one, greater than one). Then there is en embedded disc $W^{cs}$ tangent to $E^s$ at $x^*$ that is invariant under the iteration of $f$.
Question: Is there any results generalizing this theorem or giving counter example in the case when $M$ is a manifold with boundary?
|
2025-03-21T14:48:31.040997
| 2020-05-21T11:21:14 |
360941
|
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|
Stack Exchange
|
What is a good definition of a mathematical structure?
At the moment I am writing a textbook in Foundations of Mathematics for students and trying to give a precise definition of a mathematical structure, which is the principal notion of structuralist approach to mathematics, formed by Bourbaki. Intuitively (and on many examples) the notion of a mathematical structure is clear: this is a pair $(X,S)$ consisting of a set $X$, endowed with a structure $S$, which is a set somehow related to $X$. This relation of $S$ to $X$ is well-defined in universal algebras or first-order theories. What about the general case?
I arrived to the following definition and would like to ask some terminological questions.
The main idea is that a mathematical structure is determined by a list $\mathcal A$ of axioms. By an axiom I understand a formula $\varphi(x,s,c_1,\dots,c_n)$ in the language of Set Theory with free variables $x,s$ and parameters which are some fixed sets $c_1,\dots,c_n$.
Definition. A mathematical structure of type $\mathcal A$ is any ordered pair of sets $\langle X,S\rangle$ such that for any axiom $\varphi$ in the list $\mathcal A$, the formula $\varphi(X,S,c_1,\dots,c_n)$ is true.
The set $X$ is called the underlying set of the mathematical structure $\langle X,S\rangle$ and the set $S$ is called its structure.
In the list $\mathcal A$ of axioms we can encode all desired properties of the structure $S$, for example that it is an indexed family of some operations or some relations on $X$ that have some desirable properties.
The question is how to call the list of axioms $\mathcal A$ determining a type of a mathematical structure? Which properties of the list $\mathcal A$ guarantee that mathematical structures of type $\mathcal A$ form a category (for some natural notion of a morphism between mathematical stuctures of type $\mathcal A$)?
I have a strong feeling that such questions has been already studied (and some standard terminology has been elaborated), but cannot find (simple) answers browsing the Internet. I would appreciate any comments on these foundational questions.
Isn’t this just a specific kind of first-order two-sorted semantics?
But in this way higher-order properties can be encoded as well. For example, that $S$ is a subset of some $n$-th power-set of $X$. I assume that we can automatically use the axioms of set theory (for example, ZFC).
Right but nothing in the semantics forces $S$ to have some relation to the real powerset, right?
I understand a mathematical structure as a subset of the set-theoretic universe in which the real power-sets are well-defined. The formulas determining a mathematical structure are formulas of the language of Set Theory and use some fixed (for each formula) sets as parameters (but also can have no parameters at all). So, if necessary, I can use the true membership relation inside of the formulas. Then I obtain something related to the second-order logic.
What I want is to give a simple definition of a mathematical structure, which would include all examples of mathematical structures that arise in mathematical practice. I admit that the definition can be too general, but it should exist. Because it is not good to speak about mathematical structures as the main subject of study of mathematics (according to Bourbaki) and avoid providing a definition. If we accept the Set-Theoretic apprach, a unique undefined notion should be a set (or a class in NBG) and an element. Everything else should have a definition (desirably short and understandable).
I often see Bourbaki's structures quoted, but it never includes the definition (e.g., the "mathematical structure" Wikipedia page says essentially nothing about them— what they say suggest that the writer had little idea about what they are). I see little point in quoting them if one completely ignores their formalism, which is precisely defined.
@YCor By the way, what is the definition of a mathematical structure according to Bourbaki?
I once open the book at the right pages and it would take a while to read and understand, and I didn't so far. But it's your own question: quoting the "structuralist approach due to Bourbaki" let assume that you have an idea about what is it.
@YCor What is totally strange is that in French Wikipedia there is no page "Mathematical Structures" at all! But this should be the first language where this Bourbakist notion had to be explained! Moreover that there are whole books explaining the importance of mathematical structures of Bourbaki to the modern development of mathematics. Isn't it strange?
France or French Wikipedia has particular responsibility to advertise Bourbaki's foundations. I'm not sure what you mean with "whole books", but certainly I see no point in quoting work when one has no idea about what it is.
@YCor On one hand, every mathematician has a (I suspect vague) idea what is a mathematical structure via examples. The preceise definition is somewaht escaping as well as the precise definition of the subject of mathematics. What does it study? We should somehow to answer this question teaching students the couse in Foundations of Mathematics. After thinking a bit I arrived to the definition I suggested in my question: a mathematical structure is a pair of sets (or better classes) $(X,S)$ which are related somehow, i.e., via precisely defined formulas of Set Theory.
@YCor Concerning "whole books" I have just searched by Google and found some articles (not books, sorry): https://www.tau.ac.il/~corry/publications/articles/pdf/bourbaki-structures-synthese.pdf https://www.tau.ac.il/~corry/publications/articles/pdf/images-structures.pdf But there is also a book: https://link.springer.com/book/10.1007/978-3-0348-7917-0
All I understood is that Bourbaki's structures include topological spaces (which I didn't expect), so it's much more general than just algebraic systems. The amount of arid formalism discouraged me to go seriously into it so far. I don't know if it's considered as interesting now, or if it has been completely superseded, and hopefully some people know more about this.
@YCor So, this is my question to know the opinion of experts. As I understand, the Category Theory somehow overshaded the notion of a mathematical structure since often mathematical structures of given type singnature (of how it is called?) form a category and then we can treat them individually without inventing any strange definitions.
Interesting paper by Leo Corry. But it's really a paper about the history of mathematics around Bourbaki, and not a paper about foundations and structures, which would give credit to Bourbaki. So I rather see it as considering (when it was written, namely 1992) Bourbaki's structures as important at a historical level.
@YCor The formal definition of a Bourbaki's mathematical structure is described by Leo Corry on page 323 of his paper. Bourbaki had an aim to define a mathematical structure in a way that allows to define morphisms simultaneously, but I am not sure that this approach will give all possible mathematical structures because for example, the structure of power-set has two natural definitions of a morphism: using images or preimages. So, the mathematical structure $(X,P(X))$ has two different extensions to a category.
@YCor Having in mind the example of the structure $(X,P(X))$ (which are objects of two different categories), we can conclude that the connection between mathematical structures and Category Theory is not straitforward. Some mathematical structures form categories, some not, other form categories in many naturals ways. So, mathematical structures and categories are two different areas of mathematics with only a partial intersection. Mathematical structures are maninly related to the "object" part of categories. Right?
I can't say yes or no, not knowing precisely what a "mathematical structure" is. If I stick to algebraic systems, which is reasonably broad, but is at least a precise framework whose definition I understand [set endowed with a family of finitary relations and laws], I agree that they yield subcategories of the category of sets in sometimes several natural ways (even in basic cases such as totally ordered sets, or graphs).
Reading the thread, I made a bad typo and of course meant "France or French Wikipedia has NO particular responsibility to advertise Bourbaki's foundations" Having a look at the MathSciNet citations of the Set theory treaty by Bourbaki: there are quite few (10 times less than Lie, 4 times less than General topology), not particularly French-centered, and quite few seem to be related to structures (but I haven't looked into details).
Your highlighted question in the text seems different from the question title. Are you asking what would be a good definition of "mathematical structure" that would form categories? Or have you already settled on your proposed definition and are asking what properties of your axioms would ensure that you get a category?
@MikeShulman Initially I was asking about a precise definition of a mathematical structure but eventually I have found some reasonable answer myself and this answer is formulated in the question after some edit. Does this my definition agree with common understanding of a mathematical structure? I do not know. Mathematical structures in my understanding form a category whose morphisms are arbitrary functions between the underlying sets.
https://artofproblemsolving.com/community/c7h3068095
I doubt that there is any generally accepted definition of "structured set" in mathematics that includes a notion of morphism and does not already use the technology of category theory. (For a "behavioral" definition that does use category theory, see for instance here.) As has been noted in the comments, very few mathematicians have even ever seen Bourbaki's actual definition, and it probably had some issues.
The definition you propose seems too broad. Allowing arbitrary formulas of set theory enables axioms like "$x=\{\emptyset\}$", so you would have a type of structure such that $\{\emptyset\}$ admits that structure but $\{\{\emptyset\}\}$ does not. This is contrary to the general understanding of structuralism that a "structure" should be transportable across any bijection.
Probably the best-known general notion of "structured set" that forms a category (and is isomorphism-invariant) would be the models of a first-order theory. One can expand the class of models here by considering infinitary languages. However, this doesn't include examples such as topological spaces, which are still intuitively "structured sets".
The obvious way to remedy this difficulty is to use higher-order logic. The problem is that there is no obvious "correct" way to define non-invertible morphisms between models of a higher-order theory. How do you make continuous maps of topological spaces fall out of a general notion of morphism, given the contravariant character of continuity on open sets?
There are at least partial solutions to this problem, although I don't think any of them is standard or well-known. For instance, the double powerset functor is covariantly functorial in a canonical way (induced from the contravariant functoriality of the single powerset), so if we restrict our higher-order signatures to contain only relations between elements of iterated powersets $P^n(x)$ where $n$ is even, then there is a straightforward definition of morphism of structures. One can then represent and axiomatize topological spaces with such a signature having a single predicate on $P(P(x))$ that picks out the supersets of the topology, and the induced morphisms will be continuous maps. (We discovered this as part of our work on the higher structure identity principle.)
It's less clear that this approach can also represent morphisms between structures that should be covariant on subsets, but it seems to to be possible in at least some cases, such as suplattices. One could also try to augment a higher-order signature with explicit "variance information" that would determine the morphisms. Unfortunately, it's hard to make (let alone prove) a general claim that any such approach "always works" without any existing general notion of "structure" (with attendant notion of morphism) to compare it to!
Defining invertible morphisms between structures, on the other hand, is entirely straightforward. So if all you want is a groupoid of structures, then higher-order logic should do the job. This is one of the arguments for the "more foundational" nature of groupoids over categories: the groupoid of topological spaces (for instance) is uniquely and canonically determined by the notion of "topological space" (expressed, for instance, as a higher-order theory), but the same can't really be said for the usual category of topological spaces (from a purely abstract point of view, what privileges continuous maps over, say, open maps?).
So if your goal is just to have a definition with which to "speak about mathematical structures as the main subject of study of mathematics", I would say that higher-order logic is probably the best answer. If you also want to use this as a lead-in to introduce category theory, then my suggestion would probably be to discuss particular examples, then general morphisms of models of first-order theories, then isomorphisms of models of higher-order theories, then mention that defining a correct general notion of noninvertible morphism in terms of a higher-order theory is tricky, and finally use that difficulty as a motivation to refocus attention not on the notion of structure (i.e. the objects of the category) but on the entire category itself as an object of study.
Thank you for your answer from which I have learned many interesting things. My initial goal concerning mathematical structures was to give a desirably simple formal definition of a mathematical structure, which would include all know examples, confirming the known claim of Bourbaki that mathematical structures are main objects of study of mathematics. This definition had to be included into a textbook for undergraduates (that do not specialize in Logic of Set Theory). So, I eventually arrived to some solution, see the textbook at https://www.researchgate.net/publication/340899536
My main comment is this: I would do it differently.
My training tends to look at (set-based) determinations of structures as an arrangement or system of sorts which is a (often finite) tuple of sets, and then separately a language which has a tuple of symbols and rules for what characterizes a well formed sentence, and then some correspondence which allows one to interpret or apply sentences to the tuple of sorts to see if indeed the sentence is true or holds in the system of sets. Your initial attempt tries to rewrap these two notions into one, but I see that as causing problems later when you want to apply the sorts (or variations on them) to other languages (or variations on the first language): in your scheme you may have to throw out script A and rebuild from scratch script B in order to consider these changes.
This set based version will also not apply to those systems that emphasize a notion of relation over membership. It's taking me a long time to learn to use category theory because I am loathe to give up the habits developed with using membership, and it is hard for me to switch to manipulating objects and arrows without trying to interpret them as domains and functions. Yet a lot of new programming languages and structures in computer science benefit from adopting different perspectives on structures, especially viewing objects by their properties and not by their (membership-based) constituents.
You might try a goal-oriented approach. First determine what you want to do, and then try to organize structures to accomplish your goal. If a lot of your work depends on establishing a form of equality or containment, then go with set based. If much has to do with relations or elegant expression of procedures, consider a notation that captures the fundamental of the relational or procedural work you will do.
I recommend for inspiration George Bergman's invitation to universal constructions (his 245a class text of similar title), followed by chapter 3 of Algebras, Lattices,Varieties by McKenzie McNulty and Taylor, which has two formulations of category theory. Then try some books on Haskell or other functional programming language. Hilbert and Ackermann's classic text on higher order logics, and Hans Hermes book on computation (title escapes me) consider other systems like Fitch's minimal calculus. If your formulation(s) does not consider all of these for mathematical structures, I think you are setting the bar too low.
Gerhard "Ask Me About System Design" Paseman, 2020.05.21.
To address your question directly, the language part would be called an axiomatization or a theory. In computer science or software development it might be called a specification. I would call it a set of goals or requirements. Gerhard "Different Strokes For Different Folks" Paseman, 2020.05.21.
Thank you for the advise. I look at Bergman's book. It seems to be a more-or-less standard couse in Universal Algebra and this is the first examples that should be covered by a right definition of a mathematical structure. This indeed, can be covered by my approach: just form a list $\mathcal A$ of formulas saying that the mathematical structure $S$ is an indexed family of operations of a given signature and postulate their desirable properties (e.g. associativity commutativity, distibutivity etc). Everything like this can be easily done selecting specific formulas.
As I wrote in comments to comments of YCor, it seems that mathematical structures can (and should) be used only for definiting objects of suitable categories. Defining morphisms is a different task which is closure to Category Theory. A typical example is the mathematical structure $(X,P(X))$ containing a set $X$ and its power-set, this is the same as discrete topological spaces. This structure can be completed to a category in two ways: defining morphisms as images of preimages.
Indeed. Bergman' text is used for Universal Algebra, but my training says Category Theory is a separate and not always compatible (with UA) theory. (I never took Bergman's 245 class formally.) Bergman's text impressed me as an attempt to use both disciplines to do mathematics in a more or less harmonious fashion. Perhaps your formulation will strike me as even more harmonious. Gerhard "Have A Good Theme Song" Paseman, 2020.05.21.
I ask you to consider the following possibility. Don't settle on one formalism for structure. It is like trying to write a programming language to handle all tasks. You can do it, but then optimizing to do some tasks more efficiently than others becomes awkward and cumbersome, and does not lead to quick understanding. Instead, consider different formalisms and their relative advantages to one another. You can even try to arrange them in a formalist hierarchy if needed. Gerhard "That's How Mathematics Is Understood" Paseman, 2020.05.21.
In fact, my aim is not very ambitious: writing a textbook for student (not very deep), I want to mention that according to Bourbaki, the Mathematics studies mathematical structures, then give a general (but precise and correct) definition of a mathematical structure as a pair $(X,S)$ of a set $X$ and structure $S$ satisfying something (?? - that was my question), then give examples of some simple mathematical structures (semigroups, monoids, groups, rings, fields, graphs, orders, topological spaces), define their homomorphisms and switch smoothly to basic Category Theory.
By the way, the von Neumann-Bernays-Godel Set Theory (with classes) allows to give a precise set-theoretic definition of categories, functors, natural transformations, etc. The problem with set-theoretic foundations of Category Theory arised at the level of the category of functors between two non-small categories. Functors between such categories are classes which in NBG cannot be elements (or objects) of other classes.
I think for this text then, you want to use the phrases "list of requirements" and "mathematical theory", and then invent your own abbreviation for them. In other contexts they will find the synonyms "axiomatization" and "basis". Gerhard "Less Technical Is More Helpful" Paseman, 2020.05.21.
Thank you for the suggestion. I will write very simply: a mathematical structure is a pair of classes $(X,S)$ satisfying certain list of axioms $\mathcal A$ determining the type of a mathematical structure. By an axiom I understand a formula $\varphi(x,s,C_1,\dots,C_n)$ with free variables $x,s$ and parameters $C_1,\dots,C_n$ which are some fixed classes. I use the von Neumann-Bernays-Gedel Set Theory, so use classes, not sets. This allows me to include to the list of mathematical structures, the Convey's surreal line $\mathbf{No}$, see https://en.wikipedia.org/wiki/Surreal_number
The Convey's surreal line $\mathbf{No}$ has the structure of an ordered field, i.e., is a pair of classes $(\mathbf{No},S)$ where the structure $S$ is a 5-tuple of classes $(+,\cdot,0,1,<)$ consisting of two binary operations, two constants and a partial order. All (except of $0$ and $1$) are proper classes satisfying the axioms of an ordered fields. So, everything work even in such a strange "proper" context.
I offer proofreading services for not much more than a few cups of coffee. Having some familiarity with structures, I may give you even more value. Gerhard "Is Worth A Few Drinks" Paseman, 2020.05.21.
Thank you very much for the proposal of proofreading. Of course I would gladly drink a few cups of coffee with you at the earliest occasion (for example, you can visit my city Lviv at ScottishBook Fest: http://www.math.lviv.ua/SBF/). If the proofreading concerns the mentioned book, its unfinished version (without structures and categories, but with the surreal line) is here: https://www.researchgate.net/publication/340899536_CLASSICAL_SET_THEORY_THEORY_OF_SETS_AND_CLASSES
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2025-03-21T14:48:31.042561
| 2020-05-21T11:38:24 |
360942
|
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|
Stack Exchange
|
Generalization of multinomial theorem for powers of multinomial coefficients
I am trying to estimate the following expectation value in the multinomial probability distribution:
\begin{equation}
\mathbb{E}_{P}\left[ \left( \frac{n!}{x_1!..x_k!}\right)^{\alpha - 1} \right]
\end{equation}
where $P$ the usual multinomial distribution, $\theta_i \in [0,1] \forall i$, $\sum_{i=1}^{k} \theta_i = 1$, $\alpha \in (0,+\infty)\setminus \{1\}$ and $\sum_{i=1}^k x_i = n$.
1) Thus, I am wondering if there exists an extension of the multinomial theorem for the following:
\begin{equation}
\sum_{x_1 + .. + x_k = n} \left( \frac{n!}{x_1!..x_k!}\right)^{\alpha} \theta_1^{x_1}..\theta_k^{x_k}
\end{equation}
2) If not, would there be a way to estimate this w/o using Stirling on the $x_i$?
I found asymptotic estimates of the sum of powers of multinomial coefficients in my search (https://arxiv.org/abs/0807.5028), however this does not seem very useful here.
Any help would be greatly appreciated.
Thank you in advance
I doubt there exists an closed-form expression for
\begin{equation}
S_\alpha := \sum_{x_1 + .. + x_k = n} \left( \frac{n!}{x_1!..x_k!}\right)^{\alpha} \theta_1^{x_1}..\theta_k^{x_k}.
\end{equation}
Using the power mean inequality, it can be bounded as follows:
$$S_\alpha \lessgtr \binom{n+k-1}{k-1}^{1-\alpha}(\theta_1^{1/\alpha}+\cdots+ \theta_k^{1/\alpha})^{n\alpha},$$
where the sign is $\geq$ when $\alpha>1$, and $\leq$ otherwise.
Dear Max, Thanks for your reply. Could you maybe elaborate a bit on how you found this bound using the power mean inequality?
Just interpret $\left(\frac{S_\alpha}{\binom{n+k-1}{k-1}}\right)^{1/\alpha}$ as the $\alpha$-th power mean of the terms $\frac{n!}{x_1!\cdots x_k!} \theta_1^{x_1/\alpha}\cdots \theta_k^{x_k/\alpha}$, and compare it to the 1-st power mean of these terms.
Okay I got it. Then you use the multinomial theorem on the 1-st power means and resolve for $S_{\alpha}$. Thank you very much for your help
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2025-03-21T14:48:31.042717
| 2020-05-21T12:43:34 |
360944
|
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|
Stack Exchange
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Weak topology of Gaussian measures
Let us consider a space of Dirac measures $\delta_{x}$ on a Tychonoff space $X$. I know that this space is homeomorphic to $X$. A space of Gaussian measures (weak topology) on some loсally convex space is wider than the space of Dirac measures.
My question: Is the space of Gaussian measures on locally convex space $X$ homeomorphic to $X$? I'm interested about a case of an infinite-dimensional space $X$.
This seems already false when $X=\mathbb{R}$. For instance, I think that removing one point from the space of Gaussian measures on $\mathbb{R}$ will not disconnect it.
Or, consider a box $B = [a,b] \times [c,d] \subset \mathbb{R}^2$ with $c>0$, and consider the map that takes $(\mu, \sigma) \in B$ to the Gaussian measure on $\mathbb{R}$ with the corresponding mean and standard deviation. It seems clear that this map is injective and continuous, so the space of Gaussian measures contains a homeomorphic copy of the unit square, which of course $\mathbb{R}$ does not. I think a similar argument would work in any finite dimensional space.
And what about a case of an infinite dimensional space?
It may possibly be true abstractly in some cases, because there are just not that many homeomorphism types of infinite-dimensional spaces (e.g. all infinite-dimensional separable Banach spaces are homeomorphic). But I don't think you would get a homeomorphism that was "natural" in any way, and so I'm not sure how the mere existence of a homeomorphism would be useful even if it is true.
Thank you. I think it is false in any finite and infinite dimensional spaces. But I don't know how to prove it
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2025-03-21T14:48:31.042850
| 2020-05-21T12:52:42 |
360945
|
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|
Stack Exchange
|
Reducible polynomial among sequence of polynomials
Let $a_1$ and $a_2$ be two elements of a finte field $\mathbb{F}_{2^m}$ of even characteristic and $a_1^2\neq a_2$. Is it true that there always exists an element $a\in\{a_1,a_2,a_3,\ldots,a_{2^m}|a_{k+2}=a_1\cdot a_{k+1}+a_{k},k=1,2,\ldots\,2^m-2\}$ such that the polynomial $x^2+ax+1$ is reducible over $\mathbb{F}_{2^m}$?
Any help or comment is highly appreciated.
Have you tried calculating $a_4$ and $a_5$?
Yes, I tried. My computer calculations show that for $GF(2^5)$ it will be enough ${a_1,\ldots,a_7}$ elements, but not less. If field is growing then the number of elements that you need to take is also growing.
If that's true, then you have made a typo in the problem; as currently written, $a_4 = a_3 + a_2 = (a_2 + a_1) + a_2 = a_1$. I suspect the recurrence relation should be $a_{k + 2} = a_{k + 1} + a_k^2$.
Yes, I've made a typo, thanks. It should be $a_{k+2}=a_1\cdot a_{k+1}+a_k$.
In computer calculations is there any apparent relationship between the order of $a$ in the multiplicative group and the reducibility of $x^2+ax+1$?
I did not notice any relationship between the order of $a$ and the reducibility of $x^2+ax+1$.
|
2025-03-21T14:48:31.042960
| 2020-05-21T13:44:04 |
360952
|
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|
Stack Exchange
|
When is the quotient of a manifold by a discrete group of diffeomorphisms a diffeological covering space?
I was reading An Introduction to Diffeology by Patrick Iglesias-Zemmour and he defines a diffeological covering space as a diffeological fiber bundle with discrete fiber.
My question: Consider a manifold $M$ and a discrete subgroup $G$ of $\mathrm{Diff}(M)$ acting on $M$ freely. When is the quotient morphism $M \to M/G$ a diffeological covering space?
I have the impression the answer is "For every $x\in M$, the orbit of $x$ must be discrete in $M$", but I'm not sure.
In order to prove the result I believe the strategy is to prove the map
$$G \times M \to M \times M
$$
given by
$$(g,x) \mapsto (gx,x)$$
is a diffeomorphism to its image.
This map is injective because the action is free. Furthermore, it's smooth because the map $\mathrm{Diff}(M)\times M \to M$, given by $(g,x) \mapsto gx$, is smooth.
Observation: Every time I used the word discrete, I meant it in the diffeological sense.
By definition, diffeological covering spaces are nothing more than diffeological fiber bundles with discrete fibers. So I think your strategy is right for a partial answer, as you can see this in [Patrick Iglesias-Zemmour, Diffeology, art. 8.11], especially that your diffeological covering space would arise from group action. However, this works for any diffeological spaces, not only for manifolds.
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2025-03-21T14:48:31.043088
| 2020-05-21T13:50:01 |
360953
|
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|
Stack Exchange
|
Math software for 3d-visualisation
What are good and user-friendly math-softwares for 3d-visualization as in
https://www.youtube.com/watch?v=x7d13SgqUXg (sphere-eversion process).
https://www.youtube.com/watch?v=rB83DpBJQsE (divergence curl)
https://www.youtube.com/watch?v=r18Gi8lSkfM (Fourier Transform)
Given (2), this seems like a question for MESE; but anyway it does not seem to be a question about research-level mathematics, which means it probably doesn't belong on MO.
I ask this question to explain my research to graduate students
Right—to explain your research, which is in your capacity as a math educator. That is why I think it belongs on MESE.
If MO community can help with questions on highly-mathematical object visulaizations as in (1) https://mathoverflow.net/questions/60512/non-algebraic-curve-visualisation, (2) https://mathoverflow.net/questions/282100/any-visualization-software-for-the-intrinsic-metric-of-a-convex-polyhedron, (3) https://mathoverflow.net/questions/116333/software-reference-request-three-dimensional-staircase-visualizers-for-monomia.
Then I believe they can help with mine too, for in a visualization is many times intended to deliver concepts.
In principle, good visualization software could also be used to discover new things in the first place. (Probably it would make a good question to ask for real-life cases in which this has happened.)
True, especially if the visualization software(s) is user-friendly to play and experiment with. Visualizations of-course helps in more detailed observations that can in turn lead to new discoveries.
I use POV ray -- helps for both research and teaching.
Though I don't know if POV ray is a practical tool to model sphere eversion.
|
2025-03-21T14:48:31.043231
| 2020-05-21T13:57:05 |
360954
|
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|
Stack Exchange
|
Minimize a function through its upper approximation
I have some function $f(x) : \mathbb{R}^n \to \mathbb{R}$ (n is about several thousands, say $1000 \leq n \leq 10000$) to minimize over some constraints $g(x) \leq 0$ (by the way, $g$ is quadratic over $x$).
$f(x)$ is convex, differentiable (but not twice differentiable), nonnegative for all $x \in \mathbb{R}^n$, but its formula is too complicated (from the computational point of view). Thus I'm not able to write down its derivatives and pass them to the solver (or I can but have a strong opinion that it will not work).
But: there is an upper approximation $F(x)$ such as $F(x) \geq f(x) \geq 0$ for all $x \in \mathbb{R}^n$ which is somewhat "close" to $f(x)$ (though it is not yet proved how much close) and its formula is much better and it is much more convenient to optimize, and it has derivatives of all degrees.
For which condition on the quality of approximation ($|F(x) - f(x)|$) the minimization of $F(x)$ will imply the minimization of $f(x)$?
Sorry if this seems too trivial! But I cannot easily figure out in what direction to dig this problem but at the same time have a strong feeling that it should be some commonly known problem.
UPDATE:
Both $f(x)$ and $F(x)$ are convex.
You could try to do line search where the descend direction is found using $F$ and the step size is found using $f$. Not sure however what the conditions on $F$ is for this to work and what the performance will be...
If the derivatives are not known, you can use any constrained optimizer where the derivatives are approximated. Otherwise, I can recommended stochastic/blackbox/global optimization methods. The derivatives are not estimated and only function evaluations are needed.
But because your function is convex and has such a relatively high domain space, I would discourage stochastic optimization methods.
If it is not known how close $F(x)$ is to $f(x)$ for all $x \in \mathbb{R}^n$, then you cannot imply that $x^* = argmin F(x) = argminf(x)$.
To answer your question: the minimization of $F(x)$ will imply the minimization of $f(x)$ if $|F(x)-f(x)| = 0$ $\forall x \in \mathbb{R}^n$.
$F(x)$ can be quasiconvex, meaning it can have multiple local minima and the global minimum is not necessarily where the minima of $f(x)$ lies.
Thanks for the answer! I have forgotten to say that $F(x)$ is also convex. Can it help somehow?
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2025-03-21T14:48:31.043424
| 2020-05-21T14:43:45 |
360960
|
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|
Stack Exchange
|
Show that a tensor-train is contained in a recursive sequence of subspaces
Let
$p\in\mathbb N$;
$n_k\in\mathbb N$ and $\left(e^{(k)}_1,\ldots,e^{(k)}_{n_k}\right)$ denote the standard basis of $\mathbb R^{n_k}$ for $k\in\{1,\ldots,p\}$;
$u\in\bigotimes_{k=1}^p\mathbb R^{n_k}\cong\mathbb R^{\times_{k=1}^pn_k}$;
$r_0,\ldots,r_p\in\mathbb N$ with $r_0=r_p=1$;
$u^{(k)}\in\mathbb R^{r_{k-1}}\otimes\mathbb R^{n_k}\otimes\mathbb R^{r_k}$ for $k\in\{1,\ldots,p\}$ with $$u=\sum_{j_0=1}^{r_0}\cdots\sum_{j_p=1}^{r_p}u^{(1)}_{j_0,\::\:,\:j_1}\otimes\cdots\otimes u^{(p)}_{j_{p-1},\::\:,\:j_p}\tag1,$$ where $u^{(k)}_{j_{k-1},\::\:,\:j_j}$ denotes the vector in $\mathbb R^{n_k}$ obtained from $u^{(k)}$ by fixing the first and last index $j_{k-1}$ and $j_k$, respectively.
How can we show that there is a subspace $E_1$ of $\mathbb R^{n_1}$ with $\dim E_1=r_1$ and subspaces $E_k$ of $E_{k-1}\otimes\mathbb R^{n_k}$ with $\dim E_k=r_k$ for all $k\in\{2,\ldots,p\}$ such that $u\in E_p$?
I really struggle to see why this claim is true. If $p=1$, then $u=u^{(1)}_{1,\::\:,\:1}\in\mathbb R\otimes\mathbb R^{n_1}\otimes\mathbb R\cong\mathbb R^{n_1}$ and hence we may definie $E_1:=\mathbb Ru$.
Now, for $p>1$, I guess we need to use induction, but I'm not sure what precisely the induction hypotheses should be.
To keep things as simple as possible, just consider $p=2$. By a basis expansion of the second core, we can write $$u=\sum_{j_1=1}^{r_1}u^{(1)}_{1,\::\:,\:j_1}\otimes u^{(2)}_{j_1,\::\:,\:1}=\sum_{j_1=1}^{r_1}\sum_{i_2=1}^{d_2}u^{(2)}_{j_1i_21}u^{(1)}_{1,\::\:,\:j_1}\otimes e^{(2)}_{i_2}\tag2.$$ Now, the crucial observation seems to be that, for fixed $j_1\in\{1,\ldots,r_1\}$, $u^{(1)}_{1,\::\:,\:j_1}$ is just another instance of $(1)$ with $p=1$. So, some kind of induction hypotheses could be used to conclude. However, wouldn't this only yield the existence of a space $E_1^{(j_1)}\subseteq\mathbb R^{n_1}$ with $\dim E_1^{(j_1)}=1$ which depends on $j_1$? This seems to be problematic. My intention would be to define $E_1:=E_1^{(1)}+\cdots+E_1^{(r_1)}$, but all I know from $E_1$ is that $\dim E_1\le r_1$ ...
|
2025-03-21T14:48:31.043558
| 2020-05-21T14:58:20 |
360962
|
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|
Stack Exchange
|
Operational quantities characterizing upper semi-Fredholm operators
An operator $T:X\rightarrow Y$ is said to be upper semi-Fredholm if its range is closed and its kernel is finite-dimensional. M. Schechter (1972) introduced a quantity $$\nu(T):=\sup_{\operatorname{codim} M<\infty}\inf_{x\in M, \,\|x\|=1}\|Tx\|$$ and proved that $T$ is upper semi-Fredholm if and only if $\nu(T)>0$.
For a bounded subset $A$ of a Banach space $X$, let $\chi(A):=\inf\{\epsilon>0: A$ has a finite $\epsilon$-net in $X\}$. Then $A$ is relatively norm-compact if and only if $\chi(A)=0$. For an operator $T:X\rightarrow Y$, let $$h_{\mathrm{cb}}(T):=\inf\{\chi(TD):\chi(D)=1\},$$ where the infimum is taken over all countable bounded subsets $D$ of $X$ with $\chi(D)=1$.
M. González and A. Martinón (1995) noted that these two quantities are equivalent:
$$\frac{1}{2}h_{\mathrm{cb}}\leq \nu\leq 2h_{\mathrm{cb}}.$$
But they did not provide the proof. I need a detailed proof of this result.
|
2025-03-21T14:48:31.043670
| 2020-05-21T15:18:29 |
360964
|
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Stack Exchange
|
How do you check that your mathematical research topic is original?
Sorry if this question is not well-suited here, but I thought research in mathematics can be identified from other science field, so I wanted to ask to mathematicians.
I am just starting graduate study in mathematics (and my bachelor was in other field) so I have no research experience in mathematics. Recently I came up with a problem by myself, and thought it was interesting so devoted some time to draw the results. Now I have some results, but I am not sure this is already studied somewhere by someone. I tried to lookup some possible keywords in Google Scholar, but have nothing.
I am sure for more mature mathematicians they already know their fields of study and recent trends of the research, so it won't be a difficult problem to know this is original or not. But if you come up with some idea that are not seems to belong to any field, how do you know your result is original or not trivial result?
Thank you in advance!
Since you are starting a graduate study, I would recommend you to choose an advisor and to ask him/her, whether your problem is interesting and whether your results are original.
@MikhailBorovoi Yes, of course that's what I'll do. But I wonder what independent mathematicians will do in general.
The first question is whether your problem is interesting. A (good) professional knows what is interesting. Here the advisor can help.
Usually you know the history of your subject well enough to understand up to what point your problem is interesting and/or new, but there remains the possibility that the same problem has been tackled in a different form and different terminology in another area. It is still interesting to make the connection even if it takes away from your claims to originality.
The second question is whether your results are new. Try Google! However, it may happen that your problem and results are not new, but you don't know the standard terms, and therefore your Google search gives nothing. Again, here the advisor can help.
@MikhailBorovoi Yes, I strongly agree to that it's more likely that I don't know the standard terms :) Just one more thing, do you think this kind of situations (lack of knowledge on the standard terms) won't happen for professional mathematicians? For example, a mathematician who delve into the other topic from his/her previous areas. I guess in that case, he would talk with his/her colleagues who is expert on the field? Thank you for your precious comments!
@JingeonAn "do you think this kind of situations (lack of knowledge on the standard terms) won't happen for professional mathematicians?": in my case, yes, frequently. Sometimes converging to the right keywords takes time. In one case I was looking for reference to a result which sounded "too basic to be unknown", I spent hours without success, tried again weeks later with new ideas of keywords, found something which led me to try to ask some researcher (which I don't know, not in my department) and he confirmed me that it's standard, pointed out the right refs, the used terminology...
*"do you think this kind of situations (lack of knowledge on the standard terms) won't happen for professional mathematicians?" In my case, yes. Once I learned about previous work on the topic of my paper from the referee report. Then I had to revise the paper significantly, in particular, to change the title.
You've mention Google Scholar. For mathematics in particular one very valuable resource is MathSciNet which indexes and gives reviews for most published mathematical research articles. In most cases you can look up which articles have cited a given article etc. Hopefully your institution has a subscription to this.
"Some engineer out there has solved P=NP and it's locked up in an electric eggbeater calibration routine. For every 0x5f375a86 we learn about, there are thousands we never see." -xkcd.
I'm surprised no one has suggested MathOverflow, which is actually a great resource for this kind of thing.
(1) It depends a lot on the field. In fields that rely on specialized techniques discovered relatively recently or known only to a few, or fields where the questions involve recently-introduced objects, it's much easier to keep abreast of current research.
On the other hand, in fields with elementary questions that could have been studied a hundred years ago, sometimes even senior mathematicians discover that their work was studied a hundred years ago.
Of course, working in a trendy field carries its own risk, that someone else could be working on the same thing at the same time, but not much can be done about that.
(2) If you're working in a specialized field, as other have said, the best thing is to ask your advisor. If you have an advisor in a specialized field and have ideas in a different field, the best thing would be to ask someone in that field. As a grad student you probably want to start with fellow grad students, but a senior mathematician would probably asks someone on their own level.
If you have an idea that is more elementary, you should still ask your advisor, but there are certain mathematicians who know a lot of elementary and classical mathematics you could potentially ask.
(3) With regards to literature review, one trick that helps a bit when keyword searches fail is to use citations. If your idea generalizes work of Paper X, or answers a question from Paper X, or uses in a fundamental way the results of Paper X, anyone else who had the same idea would likely cite Paper X. You can produce a list of papers citing Paper X on both Google and MathSciNet.
(4) As a starting graduate student, even if your idea is completely new and original, it is likely that the greatest value it provides to you will be as practice for your future work. (I mean if you're good enough to do groundbreaking work right off the bat, you will probably do even more groundbreaking work once you get some experience under your belt.)
So don't feel bad at all if you find out something was already well-known - the experience of formulating and solving your own problem makes you well-placed to do original research once you learn a bit more, as compared to someone who knows a lot but hasn't done this.
A lot of my work has been contemporaneous, or nearly so, with other mathematicians. However, there is value in adding a new perspective or application. Don't worry about originality. Do worry about doing a literature search. You aren't expected to find the hundred year old paper quickly, but you are expected to keep looking. You are also expected to familiarize yourself with the basics. Gerhard "Advisor Shouldn't Do Your Job" Paseman, 2020.05.21.
Good advises, thank you!
As was echoed in the comments by YCor and Mikhail Borovoi, the question of originality (especially for results that could have been stated a long time ago) is one that is relevant to all mathematicians. An interesting recent example that comes to mind is this story on Terence Tao's blog.
So how can you guard yourself from this? I think that there are two tools you can use:
Experience. When you're around in an area for a few years, you should develop a pretty good sense of the literature. This is where an advisor can be really helpful when you're young, because by definition you don't have much experience at that point.
Experience also means knowing the correct search tools, like looking in papers that cite a given one. But sometimes you just don't know the correct words to search for, so then there is:
Asking other people. This is related to 1: one of the most useful skills is knowing whom to ask, and this again comes from experience and connections developed over a long period of time.
In both of these your advisor has an advantage because they've been around longer. But as you go on you will find that this becomes easier to do yourself.
Of course research output is not the only type of mathematical productivity, and you can still learn a lot by rediscovering an existing result. But it's incredibly frustrating to find out something you proved is already known (especially when you don't have so many papers yet), because it changes the meaning of the time investment you made.
If you just started your graduate study, you probably have an adviser. If you don't have one yet, try to find one as soon as possible. Anyway, in most universities that I now an adviser is required to defend a PhD.
Then show your result to your adviser. Even if it is from a different area from his/her scientific interests. Adviser will probably make a judgement, or if necessary
will know whom to ask and where to look.
It's not that hard nowadays to do a fairly exhaustive literature search to make sure something hasn't been done before. This is a skill which can take a bit of time to develop but will help and saves a lot of time in the long run, especially if you do not have someone you can show the result to. This will also help you to be independent, something which is starting to become increasingly important in academia (especially in mathematics).
Now with Google, Arxiv and many other tools which people in the past did not have access to it is much easier to check if something has already been done, although these checks are usually not foolproof. After a while you will probably also develop intuition for what is already a 'reasonably obvious' idea.
For example, a few weeks ago it occurred me that it would be nice to construct a quantum mechanical system whose Berry phase was a BPS t'Hooft-Polyakov monopole. It wasn't too hard to do some digging around and find that this had already been done: I encourage you to learn how to do this effectively if you do not already know how.
Edit: See this paper by Tao et al. where they investigate in great detail whether or not a result they have found in linear algebra is new or not and if not, exactly what is the history of the discovery. In Figure 1, they provide a very detailed tree of all the citations and previous mentions/studies of the result. Tao does not work directly in the field of linear algebra and his 3 co-authors are neutrino physicists, so how could they possibly have found this out except by doing some digging and deep searching around? OK, the graph might have been crowdsourced, but I doubt everyone contributing to it was exactly an expert.
Yes, I am willing to develop the skill, and I agree it is not as hard as before. Then this may be a trivial question : how do you develop the skill? I guess the answer is by doing it. Maybe I just want some tips from more experienced..
It's not that hard nowadays to do a fairly exhaustive literature search to make sure something hasn't been done before. - I often find it very hard to determine whether something I'm wondering about (in my field or not) is known. A big issue is that the way you formulate something often won't match the way other people do. Also, papers can be vague/unclear.
I don't really agree with this. This scenario is common: you come up with an idea in connection with problem A, it has previously been studied in connection with problem B, the terminology used is natural if you are thinking about B but you would never guess it if thinking about A.
I'm not saying it's foolproof to do a literature search, but it can be extremely helpful.
At times my experience of speaking with my supervisor is that sometimes they will not know either if a particular thing has been done and so they might ask you to look into it just in case, so I regard this as a useful skill. No-one's knowledge is boundless, you can't expect your supervisor to always know whether something is completely original, even Atiyah would often stumble across things which he thought were knew but which had already been done by previous mathematicians.
Google Scholar for example allows you to search for papers where a concept was first referred to (for example, one can find the paper where the Tits alternative was first referred to as such) and it allows you to find the most cited papers which themselves cite the paper which you are reading. These things didn't always exist and are very useful.
|
2025-03-21T14:48:31.044590
| 2020-05-21T15:24:04 |
360965
|
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|
Stack Exchange
|
Lattice structure in the root poset
Let $W$ be a Coxeter group with simple generators $s_1$, $s_2$, ..., $s_r$. Let $\Phi^+$ be the corresponding positive root system, with $\alpha_i$ the positive root corresponding to $s_i$. Bjorner and Brenti, Combinatorics of Coxeter Groups, Chapter Four define the root poset to be a partial order on $\Phi$ as follows:
If $\beta \in \Phi$ and $s_i \beta - \beta \in \mathbb{R}_{>0} \alpha_i$, then $\beta < s_i \beta$. The root poset is then the transitive closure of this relation.
Bjorner and Brenti, Exercise 4.15 asks:
Is the positive root poset $(\Phi^+, \leq)$, with a bottom element appended, a meet-semilattice?
I can't find the answer to this exercise. Can someone help?
What I would actually like to know is:
Are intervals $[\beta, \gamma]$ in the root poset lattices?
Below, some bibliographic notes:
I cheated slightly above: Bjorner and Brenti, as well as the sources below, actually only order $\Phi^+$, not $\Phi$. But I see no reason not to extend the order to the negative roots.
This poset was introduced in Henrik Eriksson's PhD thesis and, independently, by Brink and Howlett:
Brink, Brigitte; Howlett, Robert B., A finiteness property and an automatic structure for Coxeter groups, Math. Ann. 296, No. 1, 179-190 (1993). ZBL0793.20036.
This poset is not the same as defining $\beta \leq \gamma$ if $\gamma - \beta$ is in the positive span of the $\alpha_i$; a condition which is also sometimes called the root poset.
Let $\beta$ be a positive root and $t$ the corresponding reflection. Then $s_i \beta - \beta \in \mathbb{R}_{>0} \alpha_i$ if and only if $s_i$ is an inversion of $s_i t s_i$. Thus, we can define this relation in a purely Coxeter theoretic way, without mentioning root systems.
What's an easy example where the two orders on the root system differ?
$B_2$. I'll take my simple roots to be $(-1,1)$ and $(1,0)$ (so the Euclidean form is the standard one). The partial order I want is two chains: $(-1,-1) < (1,-1) < (-1,1) < (1,1)$ and $(0,-1) < (-1,0) < (1,0) < (0,1)$. Just asking that $\gamma -\beta$ be in the span of the positive roots gives $(-1,-1) < (0,-1) < [ (1,-1),\ (-1,0) ] < [ (-1,1),\ (1,0) ] < (0,1) < (1,1)$ where the square brackets surround incomparable pairs. Each incomparable pair is comparable to every element not in the pair. (Doing posets without diagrams is tough!) @LSpice
This order seems related to the order defining the existence of non-zero homomorphisms between Verma modules; see https://en.wikipedia.org/wiki/Verma_module#Homomorphisms_of_Verma_modules and also this previous MO question: https://mathoverflow.net/questions/338226/partial-ordering-on-mathfrakh-and-bruhat-ordering/338230.
It's worth pointing out that Björner and Brenti's definition of the "root poset" is probably not as widespread as the other definition (mentioned near the bottom of David's question). The other definition is the one that shows up in "antichains in the root poset" (AKA "nonnesting partitions") in Coxeter-Catalan combinatorics.
Another point related to @NathanReading's remark, and also the question about when these two orders differ, is that for say an irreducible crystallographic root system, the 'other' (more usual) root poset $\Phi^+$ is connected; whereas for the one David Speyer is considering, in general there will be two connected components corresponding to the long and short roots (which lie in different Weyl group orbits).
The root poset for $\tilde{A_2}$ is shown in Figure 4.5 in the same reference and copied below. One can check that the elements labelled $112$ and $221$ have both $100$ and $010$ as maximal common lower bounds, so meets don't exist in general. If you take an isomorphic example higher up in this same poset you can get an interval which is not a lattice.
$\tilde{A_2}$.">
John Stembridge recently pointed me towards his very nice paper Quasi-Minuscule Quotients and Reduced Words for Reflections, which gives a lot of insight into the root poset.
Here is what I understand from his paper.
Let $t$ be a reflection and let $\beta$ be the corresponding root.
Then $\phi: w \mapsto - w \beta$ is a surjective poset map from the weak order interval $[e,t]$ to the interval $[-\beta, \beta]$ in the root poset.
Of course, $[e,t]$ is a lattice, so, if $\phi$ is a poset isomorphism, then $[-\beta,\beta]$ is a lattice as well.
It turns out that $\phi$ is an isomorphism if and only if it is a bijection. Theorem 2.6 in Stembridge gives a necessary and sufficient condition for $\phi$ to be a bijection, but one important thing to note is that this always occurs if the Dynkin diagram is a forest (in particular, in all finite types and all affine types other than $\tilde{A}$).
I'm not sure if anyone but me cares about this old question, but I worked through the first example of a non-lattice in $\tilde{A}_2$ from this perspective. Let $t = (s_1 s_2 s_3)^2 s_1 (s_1 s_2 s_3)^{-2}$. I have drawn the interval $[e,t]$ in the image below: $[e,t]$ are the labels on the triangles inside the diamond region. In the bottom of the diamond, I have labeled each triangle with a reduced word; in the top, the words got too long so I just put black dots. The edges of the Hasse diagram are dual to the triangular tiling; moving down the page is going down in the lattice.
The root corresponding to $t$ is $\beta:=4 \alpha_1 + 3 \alpha_2 + 3 \alpha_3$. The map from $[e,t]$ to $[-\beta, \beta]$ identifies $(s_1 s_2 s_3)^2$ with $(s_1 s_3 s_2)^2$ (both indicated with light shading) and identifies $(s_1 s_2 s_3)^2 s_1$ with $(s_1 s_3 s_2)^2 s_1$ (dark shading). If we had chosen a larger reflection, we would have gotten a larger diamond $[e,t]$, and then $[-\beta, \beta]$ would be obtained by quotienting this diamond by a one dimensional group of translations so that the poset again had width $3$.
Now, look at the elements $x$ and $y$ (in red). The meet $x \wedge y$ in $\tilde{A}_2$ is $s_1 s_2 s_3 s_1 s_2 s_1$ (marked in black). However, $x$ dominates one of the dark shaded regions and $y$ dominates the other one so, in the quotient $[-\beta, \beta]$, the dark shaded element is a second, incomparable, lower bound for the images of $x$ and $y$.
|
2025-03-21T14:48:31.045272
| 2020-05-21T15:40:17 |
360966
|
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|
Stack Exchange
|
Examples of connection preserving maps in differential geometry
In synthetic differential geometry and tangent categories, linear connections on the tangent bundle are treated as a sort of algebraic gadgets that incorporate the tangent bundle. Like any other algebraic gadget, it seems natural to consider morphisms $f:M \to N$ that preserve a chosen linear connection $(M, \nabla), (N, \nabla')$. I've tried searching through the differential geometry literature, and connection-preserving morphisms doesn't seem to be discussed very much (except the case where an isometry between Riemannian manifolds induces a connection-preserving morphism between their Levi-Civita connections).
Have connection-preserving maps been considered by differential geometers, and if so, can someone point me in that direction?
I've been trying to get the term 'connectomorphism' going for some time.
For a prequantum bundle of a symplectic manifold, the group of connection-preserving automorphisms is known as the quantomorphism group https://ncatlab.org/nlab/show/quantomorphism+group. In the context of general principal bundles, the group of gauge transformations that preserve a given connection is finite dimensional and isomorphic to the centralizer of the holonomy group. This observation can be used to classify all possible groups that arise as connection-preserving gauge transformations. See eg https://arxiv.org/abs/hep-th/0203027
In chapter 6 of Kobayashi-Nomizu "Foundations of Differential Geometry vol. 1" these are covered under the keyword "affine mapping"
Affine geometry is maybe a topic of interest, see https://projecteuclid.org/euclid.aspm/1543086326
|
2025-03-21T14:48:31.045417
| 2020-05-21T16:49:08 |
360971
|
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|
Stack Exchange
|
Does there always exists a locally free resolution of quasi-coherent sheaves on quasi-projective noetherian scheme?
We consider a quasi-projective noetherian scheme. It is well known that for a coherent sheaf we can construct a sheaf resolution of locally free of finite rank. It is introduced in Hartshorne chapter III.6 for computing Ext and $\mathscr{Ext}$ for example.
In terms of quasi-coherent sheaves, does there always exist a locally free resolution of quasi-coherent sheaves on a quasi-projective noetherian scheme? If exists, what will happen if we do a similar cohomological argument using locally free resolution of infinite rank? If not, how do people talk about this kind of bounded above resolutions in the category of quasi-coherent sheaves? Will we suffer from infinite direct sums?
I know that in general if we remove the condition "quasi-coherent or coherent", we can construct flat resolutions. However, it may be useless in concrete computation for quasi-coherent sheaves.
Yes, you can find such resolutions: write your quasi-coherent sheaf as an increasing union of coherent sheaves, and then take a gigantic direct sum of locally free sheaves, one hitting each term of the direct system, as the first term of your resolution. This is however unlikely to be very useful unless you can also control what kind of twists show up in the direct sum.
Thank you. There is one thing I am afraid of. Infinite direct sum of quasi-coherent sheaves may not be quasi-coherent.
Infinite direct sums are fine, it's products that cause problems.
|
2025-03-21T14:48:31.045558
| 2020-05-21T16:53:21 |
360972
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/360972"
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|
Stack Exchange
|
Theory of extensions of non-archimedian local fields
I'm searching for a recommendable reference dealing with theory of
non-Archimedean local fields where I can find proofs of the following claims about
finite extensions $L/K$ of non-Archimedean local fields with finite
residue fields $l / k$. I'm pretty sure that this request might not have a research level but up to now I haven't obtained a satisfying answer asking the same at MSE.
Firstly, we use notations: We denote by $q \in O_K$ a uniformizer of $O_K$
and $\pi \in O_L$ a uniformizer of $O_L$. Then $n= [L:K]=e \cdot f$ with
$f= [l:k]$ and $(q)O_L= (\pi)^eO_L$.
Now I'm looking for rigorous proofs of the following statements:
(i) Case $L/K$ is Galois & unramified ($\Rightarrow$ $e=1$ & $[L:K]= [l:k]$)
and $p$ the characteristic of finite field $k$:
The claim is $G= \operatorname{Gal}(L/K)= \operatorname{Gal}(l/k)=g$. I'm looking for a proof making an
explicit construction of the "lift" $\operatorname{Gal}(l/k) \to \operatorname{Gal}(L/K)$, i.e. if we have
a $k$-automorphism of $l= k(a)= k[X]/ (X^f-1)$, how can it be uniquely lifted
to a $K$-automorphism of $L$?
(ii) If $L/K$ is unramified and $K= \mathbb{Q}_p$, then $L$ is cyclotomic:
$L= \mathbb{Q}_p(\zeta_n)$, for $\zeta_n$ an appropriate root of unity.
(iii)(1) Case $L/K$ totally ramified with $n =e$ coprime to $p=\operatorname{char}(k)$: there exists $b \in K$ with $L= K(\sqrt[e]{b})$.
How can this $b$ be
constructed? Can the $b$ be chosen as a uniformizer: i.e. $b= uq$ with $u \in O_K^\times$?
(2) If moreover $p \mid \lvert k \rvert -1$, why is $L/K$ cyclic & Galois extension?
See Fesenko and Vostokov - Local fields and their extensions. (i) is Proposition 3.3(2). (ii) is Proposition 3.2(1). (iii)(1) is Proposition 3.5(1) (and, yes, $b$ may be chosen as a uniformiser). For (iii)(2), I think you meant $e \mid \lvert k\rvert - 1$, not $p \mid \lvert k\rvert - 1$ (which is impossible). Then ($k$, hence) $K$ contains an $e$th root of unity, so $L = K(\sqrt[e]q)$ is (Galois and) cyclic over $K$.
Hi, thank you for your answer. A couple remarks: (you are right, of course, in (iii) I wanted to require $e \mid \lvert k\rvert - 1$). This requirement implies by finiteness of $k$ that $k$ contains an $e$-th root of unity. Why this imply that $K$ has also one? Do you use at this point the Hensel's lemma? Or is it a more more elementary result? Futhermore, you continue, that this imply that $L = K(\sqrt[e]q)$ (<= here we use already (iii)(1) ) is cyclic? How do you deduce this?
Indeed, I use Hensel's lemma to deduce that $K$ contains a (I forgot to say primitive) $e$th root of unity because $k$ does. It is a general fact in Galois theory (nothing to do with valued fields) that a radical extension is cyclic if and only if the ground field contains an appropriate root of unity: https://en.wikipedia.org/wiki/Radical_extension#Solvability_by_radicals . In this case, $X^e - q$ is irreducible by Eisenstein's criterion.
Great, that was exactly what I was looking for. Thank you a lot!
A nitpick: The proof in Proposition 3.3(2) that gives an answer two my (i) is indeed quite understandable (verify $Gal(L/K) \to Gal(l/k)$ surjective + $\vert Gal(L/K) \vert =\vert Gal(l/k) \vert$, but I'm quite curious if it also possible to build expliitely a lift $\operatorname{Gal}(l/k) \to \operatorname{Gal}(L/K)$, since I want to understand what this lift does "concretly" on $L$. Unfortunately, the proof 3.3(2) from the book not gives this "inside look". Do you know if it also possible to argue via these "lifts"? I mean every $\alpha \in Gal(l/k)$ is a power of
Frobenius $a \mapsto a^d$ with $d= \vert k \vert$. That is $\operatorname{Gal}(L/K)$ has also to be generated by a lift of this Frobenius. But I have no idea how this lift of the Frobenius in $Gal(L/K)$ shold act. My first idea was to do Teichmüller lifts of $\zeta_{d-1} \in l$ to $L$ and define to lift of the Frobenius on only on these Teichmüller lifts. Problem: do the Teichmüller lifts generate $L$ as $K$-algebra? Do you probably see another possibly more conventional approach to lift the Frobenius and thus to obtain the desired map $\operatorname{Gal}(l/k) \to \operatorname{Gal}(L/K)$?
Put $\mathfrak g = \operatorname{Gal}(l/k)$. For a concrete approach, I think you have to consider $\operatorname{char}(K)$. If it's $0$, then $\mathcal O_L$ is isomorphic as an $\mathcal O_K$-algebra to $W(l)$, and so you can define an action of $\mathfrak g$ on $\mathcal O_L$, hence on $L$. If it's $p$, then $L$ is isomorphic to $l((t))$, and you just let $\mathfrak g$ act trivially on $t$.
(Basically it's what you say, that the Teichmüller lifts generate $L$ as a topological $K$-algebra. If you just want to say what the action is, knowing it exists, then that's good enough; but to prove that this actually gives a well defined action would probably be harder than the proof in FV.)
|
2025-03-21T14:48:31.045914
| 2020-05-21T19:25:10 |
360980
|
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|
Stack Exchange
|
Minimal generation of simple groups and Ore's conjecture
The well known Ore's conjecture (now established) states that every element of a finite non-abelian simple group $G$ is a commutator of a pair of elements. Also we know that $G$ is $2$-generated.
I am trying to find out what is known about: given any $1 \neq x \in G$, can it be a commutator of two generating elements, i.e., $x = [a,b]$ so that $G$ is generated by $a, b$ as well.
If the answer is negative, are there known restrictions on the conjugacy class of $x$ for which this happens?
The question is motivated from the action of $G$ on Riemann surfaces that yield orbit genus $1$ corresponding to minimal signatures for the group.
This has been discussed elsewhere, and there are lots of counterexamples, the smallest of which is $A_5$. In $A_5$, if $[a,b]$ is an element of order $2$, then $a$ and $b$ both stabilize the same point and generate a subgroup $A_4$.
Thanks for pointing it out. My prime concern was while experimenting on action of $PSL(2,p)$ on Riemann surfaces with orbit genus $1$, where $p$ is a prime. It now seems like "almost often" (except the class $A_n$ and a few other low order cases), other than order $2$ element, rest of them are commutators with its components generating the group $G$. I need some time to close the question.
This question was answered on another forum, so I will just repeat the answer from there.
It is true for 'most' finite simple groups, but there are lots of exceptions, including ${\rm PSL}(2,2^n)$ for all $n$, ${\rm PSL}(3,3)$, ${\rm PSU}(3,3)$, $A_8$, ${\rm PSp}(4,3)$, and $M_{11}$. It is not true in general in ${\rm SL}(n,q)$ and ${\rm Sp}(2n,q)$, so they are also exceptions whenever they have trivial centre.
In particular, $A_5$ is an exception. If $a,b \in A_5$ with $[a,b]$ of order $2$, then $\langle a,b \rangle \cong A_4$.
Here is a character-theoretic argument which shows that for each $n > 1$, whenever an involution $t \in {\rm SL}(2,2^{n})= G$ has $t = [a,b]$, then $a,b \in B = N_{G}(S)$, where $S$ is the unique Sylow $2$-subgroup of $G$ containing $t$.
In general, when $G$ is a finite group and $x \in G$, then the number of ordered pairs $(a,b) \in G \times G$ with $x = [a,b]$ is expressible as $\sum_{ \chi \in {\rm Irr}(G)} \frac{|G|\chi(x)}{\chi(1)},$ where ${\rm Irr}(G)$ is the set of complex irreducible characters of $G$. This formula was probably known to W. Burnside (in fact, the fact that $x$ is a commutator if and only if the sum is positive appears in Burnside's book, and was later important in the proof of the Ore conjecture).
Letting $T,S, B, G$ be as above, we note that $B$ is a Frobenius group of order $2^{n}(2^{n}-1)$ , and has $2^{n}-1$ irreducible characters of degree $1$ (each with $t$ in their kernel), and one irreducible character of degree $2^{n}-1$ taking value $-1$ at $t$. Hence the number of order pairs $(a,b) \in B \times B$ with $t = [a,b] $ is
$(2^{n}(2^{n}-1) [ (2^{n}-1) - \frac{1}{2^{n}-1}] = 2^{3n}-2^{2n+1}$.
On the other hand, $G$ has one irreducible character of degree $1$, the trivial character, one irreducible character of degree $2^{n}$ (which vanishes at $t$), $2^{n-1}$ irreducible characters of degree $2^{n}-1$, all taking value $-1$ at $t$, and $2^{n-1}-1$ irreducible characters of degree $2^{n}+1$, all taking value $1$ at $t$.
Hence the number of ordered pairs $(a,b) \in G \times G$ with $t = [a,b]$ is
$2^{n}(2^{2n}-1) [ 1 - \frac{2^{n-1}}{(2^{n}-1)} + \frac{2^{n-1}-1}{2^{n}+1}],$ which also turns out to be $2^{3n} - 2^{2n+1}$.
Hence all ordered pairs $(a,b) \in G \times G$ with $t = [a,b]$ actually lie in $B \times B$, so it is not possible to express $t = [a,b]$ where $\langle a,b \rangle = G.$
|
2025-03-21T14:48:31.046286
| 2020-05-21T20:04:19 |
360983
|
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|
Stack Exchange
|
What is the standard definition of dual of disconnected planar graph when underlying graph derives 'product structure' over connected graphs?
Dual graph of a plane graph has a standard definition https://en.wikipedia.org/wiki/Dual_graph and an edgeless graph on $n$ vertices is planar. What is the standard dual graph of such a graph?
Update from comments:
It seems like $n$ vertex planar edgeless graph could be interpreted as 'some' (I do not know which would be appropriate) 'product' of $1$ vertex graph where the 'product' might have the interpretation that if the underlying graph is connected then the 'product' is connected. Perhaps for such 'product' graphs one might give special meaning to connectedness where usual connectedness fails thus salvaging some definition of duality?
The dual of the edgeless graph on 1 vertex is itself. As that Wikipedia article mentions, duals are usually only defined for connected planar graphs (at least, this is required for $(G^{})^{} = G$).
So there is no working definition for edgeless graph on itself? Well take a graph $G_1$ with vertices $u$ and $v$ and an edge and thus its dual has one vertex. So $(G_1^)^\neq G_1$ it seems. Since dual of any star graph is also one vertex $(G^)^=G$ seems restrictive. I think I miss details.
I don't understand your question. The most reasonable definition of the dual of the edgeless graph on $n$ vertices would be the edgeless graph on $1$ vertex (but then the dual of the dual will not be the primal graph).
The dual of a star graph (or indeed any tree) is a single vertex with a loop on it for each edge of the original graph. The way in which the loops are arranged in the plane allows the original tree to be recovered. If you don't allow loops in your graphs then planar duality really only makes sense for 2-edge-connected graphs.
Building off lambda's comment, the dual of any tree on $n$ vertices will be a graph which has one vertex and $n-1$ loops: what differs from tree to tree is how these loops are embedded in the plane. Indeed, duality makes the most sense for (connected) plane graphs, i.e., graphs with a fixed embedding in the plane. (But, as that Wiki page also mentions, by a theorem of Whitney, if your graph is 3-connected then the embedding, and hence the dual graph, is unique.)
It seems like $n$ vertex planar edgeless graph could be interpreted as 'some' (I do not know which would be appropriate) 'product' of $1$ vertex graph where the 'product' might have the interpretation that if the underlying graph is connected then the 'product' is connected. Perhaps for such 'product' graphs one might give special meaning to connectedness where usual connectedness fails thus salvaging some definition of duality?
|
2025-03-21T14:48:31.046490
| 2020-05-21T20:42:05 |
360987
|
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|
Stack Exchange
|
Large gaps in the norm of a subgroup and its centraliser
Take an infinite finitely generated group $G$ with an infinite subgroup $N$ which has an infinite centraliser $Z = Z_G(N)$.
Let $S$ be some [symmetric] generating set of $G$ and for $g \in G$,
denote by $\mathrm{norm}(g)$ the combinatorial distance between $g$ and the identity in the Cayley graph w.r.t. $S$.
Then $A= \mathrm{norm}(N)$ and $B=\mathrm{norm}(Z)$ are some subsets of the ($\geq 0$) integers.
Question: is it possible that $A$ and $B$ are asymptotically separated by large gaps,
that is for every $C>1$ there is a $n_0$ such that
$\forall a \in A \cap [n_0,\infty[$, $B \cap [a/C,aC] = \emptyset$
$\forall b \in B \cap [n_0,\infty[$, $A \cap [b/C,bC] = \emptyset$
Note that if either $Z$ or $N$ contain an element $x$ of infinite order,
then the answer is no (since $\mathrm{norm}(x^n) \to \infty$ and the triangle inequality implies $\mathrm{norm}(x^n) - \mathrm{norm}(x^{n-1}) \leq \mathrm{norm}(x)$,
so the largest gap has length $< \mathrm{norm}(x)$).
This question is partially motivated by the fact that it is possible for the centre of an [infinite finitely generated] group $C = Z_G(G)$ to be so that $\mathrm{norm}(C)$ [is infinite] and has large gaps (in the sense that forall $C>1$ there is a $n > 2$ such that $\mathrm{norm}(C) \cap ]n/C,Cn[ = \emptyset$)
These conditions are trivially fulfilled if $N$ or $Z$ is finite. Also your argument that this is "no" if $Z$ or $N$ is not torsion also carries over the case when $Z$ or $N$ is locally finite. That is, your conditions imply that $Z$ and $N$ are locally finite.
You are right: I forgot to add the adjective "infinite" before $N$ and $Z_G(N)$.
Your remark on locally finiteness is essentially an "upgrade" on my remark on the fact that no element (of $N$ or $Z$) may have infinite order... or am I missing an important point?
I said "your argument... carries over". It's essentially the same argument but it yields a much stronger conclusion.
Thanks (i just was not sure if there was nonetheless a new ingredient required). So for eventual readers (which are as slow as I am):
(1) If a subgroup $S$ is infinite, $\mathrm{norm}(S)$ is unbounded (because $G$ is fin.gen.)
(2) In a fin. gen. subgroup, the norm of the generating elements bound the size of the gap (again by the triangle inequality)
(3) Hence the $N$ and $Z$ [in the question] may not have infinite finitely generated subgroups (which is to say that they are locally finite)
For your question: first, I know finitely generated groups with a pair of infinite locally finite subgroups with your property (one can find such a pair in the lamplighter). But it's not a subgroup and its centralizer. I suspect an example how your require exists anyway, playing suitably with HNN extensions.
Thanks, I am quite sure such an example should exist, but I'm don't know how to build one (the example I mention at the end with $C= Z_G(G)$ having large gaps is quite convoluted). I'll try to look at HNN-extensions... (I assume the lamplighter subgroups are some parts of the lamp group? By that I mean $\oplus_{i_n} \mathbb{Z}2$ (as a subgroup of $\oplus{\mathbb{Z}} \mathbb{Z}_2$ with a well-chose sequence ${i_n} \subset \mathbb{Z}$).
Yes, in the lamplighter these are subgroups like this.
|
2025-03-21T14:48:31.046760
| 2020-05-21T20:48:16 |
360988
|
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|
Stack Exchange
|
Integrality certification for product of two matrices $A B^{-1}$
Let's consider two non-singular integer matrices $A,B \in\mathbb{Z}^{n\times n}$. I want a test to check if $A\times B^{-1}$ is integral (or no denominators). I am referring the unimodular certification paper by Storjohann. The inverse of a matrix $M$ can be written as a $x$-adic expansion: $$M^{-1}=c_0+c_1x+ c_2x^2+\cdots $$
If the matrix $M$ is unimodular, the $x$-adic expansion is finite and Storjohann provides a fast algorithm to check this. (Here $x\in \mathbb{Z}_{>2}$ is relatively prime to determinant of $M$).
My approach was: I computed few $x$-adic terms of $B^{-1}$. Say: $B^{-1}=b_0+b_1x+\cdots$. I can write matrix $A$ as a finite $x$-adic expansion. Let $A=a_0+a_1x+\cdots +a_m+x^m$. Similarly, if $A\times B^{-1}$ is integral, I should be able to write it as a finite $x$-adic expansion. Hence, I checked if $a_0 \times b_i$ (or $a_j\times b_i$) becomes zero at some point. I test for some examples (for matrices over Number fields), but it didn't work. Is there any problem with the theory? or Can someone suggest me a way to certify integrality of the product $A\times B^{-1}$.
This question is likelier to get an answer if you get more specific to the paper you are referring (which section? which criterion?) and explain what $x$ is (a big prime?).
My first attempt is to try for linear lifting section 2.1. Linear lifting algorithm is similar to the Dixon algorithm in paper :Exact Solution of Linear Equations §https://link.springer.com/article/10.1007/BF01459082§ for linear systems solving: $Ax=I$ taking RHS matrix $I$ to be the identity matrix. In this case $x$ is a prime or anything which does not divide Determinant of $A$. When it comes to the fast algorithm Double-plus-one Section 3, $x$ should be large enough. $x>max{10000, 3.61 n^2[A]} $, maximum entry =[A]
In "the inverse of a matrix $M$ can be written as a[n] $x$-adic expansion", what is $x$?
I considered $x$-adic expansions for $B^{-1}$ and $A$ separately and try to check whether product of terms becomes zero. It is not possible to test integrality of the product $AB^{-1}$ in this way as the series ($x$-adic expansion) grows (diverge) when B is not unimodular. Hence, I applied Dixon's algorithm to solve the linear system $yB=A$. This will compute the $x$-adic expansion of $AB^{-1}$ directly. Hence, I can check if this expansion becomes finite. Now, the next step is to modify "fast-Double-plus-One algorithm" to achieve this certification.
|
2025-03-21T14:48:31.046940
| 2020-05-21T21:10:01 |
360990
|
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|
Stack Exchange
|
How we can characterize all positive integers, multiples of 4, that cannot be expressed as $(p_1-1)(p_2-1),\;\;p_1,\,p_2$ distinct primes
I ask how we can characterize all positive integers
multiples of 4 that cannot be expressed as
$(p_1-1)(p_2-1),\;\;p_1,\,p_2$ distinct primes
The first multiples of 4 that cannot be expressed as above are:
$68, 76, 124, 128, 152, 188, 236, 244, 248, 284, ...$
Integers of the following forms cannot be expressed:
$4(2+15m),\;\;m\ge 1$
$4(21m-4),\;\;m\ge 1$
$4(1+30m),\;\;m\ge 1$
Also asked in math.stackexchange.com.
Many thanks.
It is unlikely that there's some nice characterization of these numbers.
Or in other words, multiples $n$ of $4$ such that for every divisor $d$ of $n$ with $d^2 \ne n$, $d+1$ or $n/d+1$ is composite. But that's just a restatement of the condition. The sequence is not in OEIS.
Its trivial that all numbers of the form $4p$, $p$ is a prime are in the sequence. Also it's trivial that the numbers which has less divisors are more likely to be in this sequence.
|
2025-03-21T14:48:31.047036
| 2020-05-21T21:13:09 |
360991
|
{
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"Joseph O'Rourke",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/360991"
}
|
Stack Exchange
|
Counting simple closed curves
I'm currently trying to understand how to count simple closed curves. I've been reading Alex Wright's survey (https://arxiv.org/pdf/1905.01753.pdf). However, I don't feel like I'm getting the big picture. All the surveys I can find on the subject try to show things in an explicit way, but I'd rather see an abstration. I feel the way dynamicist tacle this problem post Mirzakhani is by giving a flow on $M_{g,n}$, the Teichmüller geodesic flow. By the work of Mirzakhani, this is a ergodic flow. My question is how does this help us understand the function
$$f_L:M_{g,n}\rightarrow \mathbb{R}, \quad X\mapsto s_X(L):=\{ \gamma \mid \ell(\gamma)\leq L\}?$$
I feel that estimates of this function should be related to the Ergodic Theorem, but I can not find anything on that.
PS: I'm sorry for the tags, but I'm uncertain which tags I should use.
It may be that the 2-page Introduction of this paper could help? Cahn, Patricia, Federica Fanoni, and Bram Petri. "Mapping class group orbits of curves with self-intersections." Israel Journal of Mathematics 223, no. 1 (2018): 53-74.
arXiv abstract.
|
2025-03-21T14:48:31.047190
| 2020-05-21T21:57:22 |
360993
|
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"Paata Ivanishvili",
"Sandeep Silwal",
"Tanya Vladi",
"fedja",
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"https://mathoverflow.net/users/21907",
"https://mathoverflow.net/users/36721",
"https://mathoverflow.net/users/37436",
"https://mathoverflow.net/users/44191",
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"user44191"
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|
Stack Exchange
|
Fourier transform of $f_a(x)= a^{-2}\exp(-|x|^a)$, $a \in (0,2)$, is decreasing in $a$
Can one show that Fourier transform of
$$ f_a(x) = a^{-2} \exp(-|x|^a), \qquad a \in (0,2)$$
is decreasing in $a$?
I have a solution for $a \in (0,1]$ which cannot be used for $a\in (1,2)$.
@user44191: Doesn't look like a homework to me.
First of all it is not a homework. If you think it is a trivial question -please just refer me to a textbook. There is no closed form expression for Fourier for exp(-|x|^a) for any a in (0,2], just for a=1 and a=2. Context I am researching positive definite radial functions. I have solved the problem for a \in (0,1) and x\in R^d by using results from theory of k-times monotone functions.
@TanyaApanasovich: There must be a typo in your question: $f_a(0) = a^2$ is increasing in $a$, and hence the integral of the Fourier transform of $f_a$ is increasing in $a$. In fact, the Fourier transform of $f_a$ seems to be increasing in $a$ near the origin, but of course not in a neighbourhood of infinity (because tails become lighter as $a$ grows).
@MateuszKwaśnicki, yes, thank you, it should be exp(-||x||^a)a^{-\delta}, x \inR^d, \delta depends on d, for d=1, it is 2
I apologize; I misread the initial problem, and have retracted my close vote. That said, I do think the command in the original question ("Show that") contributed to the misreading and made me think it was taken from an exercise. For future reference, it's often better to phrase it as a question, especially because if you don't have a proof, then the conclusion is still in doubt.
@user44191 it was badly stated, I should know better. How I approached the problem- I took a derivative with respect to a, given that I can exchange integration and derivative, I need to show that Fourier transform of \exp(-|x|^a)(|x|^a log(|x|^a)+2) is positive
@AlapanDas: It is $F_a(z) = 2 a^{-2} \int_0^\infty e^{-x^a} \cos(2\pi x z) dx$, and as I understand, $z \in \mathbb{R}$ is arbitrary and fixed.
@AlapanDas you made a mistake when taking a derivative with respect to a, \partial (x^a)/\partial a - it is x^a log(x). Also Fourier of exp(-x^a) is not monotone in a
@TanyaApanasovich : Can share your solution for $a\le1$ (maybe by a link)? Also, plotting suggests your desired result holds even with $a^{-1}$ (but not with $a^{-0}$) in place of $a^{-2}$.
@AlapanDas I do not know how you conclude the integration will be positive, cos is an oscillating function (positive and negative). Can you please elaborate?
@IosifPinelis I study positive definite functions. Positive definite function has a non-negarive Fourier transform. There is a sufficient condition, called multiple monotonicity. I can exchange integral and derivative - so I need to show that f(x)=\exp(-x^a)(x^a log(x) +2/a) is positive definite. Using multiple monotonicity, for d=1, x \in R^1 , I just need to show that f(x) is positive, non-increasing and convex. I take two derivatives and show that. This sufficient condition does not work for a>1.
@Tanya Apanasovich Obviously that isn't necessary. But due to $e^{-x^a}$ function I just assumed that the value of integration will be positive. That's just an assumption based on data. That's why I said I am not sure whether it will be positive or negative, when $z$ is around $\frac{1}{4}$.
Maybe Bernstein's theorem on monotone functions will help? It implies that f_a is a nonnegative mixture of gaussians.
@SandeepSilwal Not really, at least I cannot see how I can focus on $a.$ The nice closed form expression for general $a$ does not exist
@AlapanDas why z around 1/4
@SandeepSilwal: Have you tried that? I remember that I checked that the derivative of $f_a$ is not of the form $\exp(-\phi(x^2))$ for a complete Bernstein function $\phi$. However, I do not remember now if I tried to check whether it is of the form $\psi(x^2)$ for a completely monotone $\psi$.
@MateuszKwaśnicki I just checked, its not of that form I think.
@MateuszKwaśnicki the derivative is not complete monotone, for sure. For $a\le 1$ the derivative is k-times monotone for various choices of power $\delta$. When $a>1$ the derivative cannot be k - times monotone for any delta.
try the explicit formula in terms of Bessel functions of the first kind together with the power series expansion for it. See the article by Blumenthal and Getoor https://www.ams.org/journals/tran/1960-095-02/S0002-9947-1960-0119247-6/
@AbdelmalekAbdesselam, thank you, I have tried it, but since the Bessel function of first kind can be positive and negative- did not help a bit, unless I am missing something. Taking a derivative with respect to $\alpha$ brings me to a more complicated problem to show that a function $\exp(-x^\alpha){x^\alpha\log(x)+2}$ is positive definite
yes but did you try the series expansion of the Bessel function plus commuting sum and integral and computing the integrals?
@AbdelmalekAbdesselam, I am not sure what you are referring to, I have tried using the Maclaurin series as in http://www.cl.eps.manchester.ac.uk/medialand/maths/pdfs/research/statistics-reports/psrr18-2009.pdf, I am interested in \alpha \in (1,2)
yes I meant the Taylor series. dead end then.
@AbdelmalekAbdesselam. I am not sure whether it is a dead end in general, for me it is as the series looks like $\sum_{j=0}a_j(\alpha) (-t^2)^j$, $a_j(\alpha)>0$, but the term $(-1)^j$ is uncomfortable given that $t^{2j}$ increases with $j$ when $|t|>1$ and decreases when $|t|<1$, but series is not my expertise, may be I am missing something
very long shot but how about taking derivative in alpha term by term and perhaps in some cases show the result is positive by alternating series test lower bound? this is not expertise but just superficial brainstroming
@AbdelmalekAbdesselam it is a power series and when $t>1$, $a_j$s increase, so alternative series test is not applicable
@TanyaVladi, there is a "subordination formula" $e^{-x^{a}} = \int_{0}^{\infty} e^{-x^{2} \tau} p_{a}(\tau) d\tau$ (true for any $a \in (0,2)$) holds for some probability density function $p_{a}(\tau) d\tau$. If you use it then the Fourier transform becomes $ \int_{0}^{\infty} e^{-s^{2}/4} a^{-2}p_{a}(s^{2})ds$. Have you tried to follow this path? Looks like algebra will be different, at least now $p_{a}(s^{2})$ is nonnegative.
@PaataIvanishvili, yes it is the first thing I have tried. $p_\alpha(\tau)$ does not have a nice analytic expression. All presented expressions include some sort of integral and the dependency on $\alpha$ is very convoluted. Let $U_p=({\sin(\pi x\alpha/2)\over \cos(\pi x/2)})^{\alpha/(1-\alpha)}{\cos(\pi x(\alpha-1)/2)\over \cos(\pi x/2)}$, the expression for $p_\alpha(t)$ is $C_\alpha |t|^{1/(\alpha-1)} \int_{0}^1U_\alpha(x)\exp{-|t|^{\alpha/(\alpha-1)}U_\alpha(x)}dx $, where $C_\alpha$ is a function of $\alpha$ only, also non-trivial as it inclues power and Gamma
@TanyaVladi I believe I can prove it when the test point is sufficiently far from the origin (distances around 20 and up should be fine). Would you be interested in such partial result or you need "all or nothing"?
Not sure why none of the comments mention that exp(-|x|^a) is the characteristic function of a symmetric stable law whose corresponding density is given by the so-called Bergstom-Feller expansions.
@AndreyF I did not mention because I do not know how that helps
Too long for a comment.
We have
$
\widehat{f_a}(\xi)=\int_{\mathbb R} a^{-2} e^{-\vert x\vert ^a-ix \xi} dx=
2\int_0^{+\infty} a^{-2} e^{-x^a} \cos(x\vert\xi\vert)dx,
$
so that
$$
J_a(\xi)=-\frac{a^3}2\partial_a\widehat{f_a}(\xi)=
\int_0^{+\infty} \bigl( 2 + x^a \ln (x^a)\bigr)e^{-x^a}\cos(x\vert\xi\vert)dx,\quad\text{and}
$$
$$
aJ_a(\xi)=\int_0^{+\infty} t^{\frac 1a-1}\bigl( 2 + t \ln t\bigr)e^{-t}\cos(t^{1/a}\vert\xi\vert)dt.
$$
We note that
$
\frac d{dt}(t\ln t)=\ln t+1
$
which is positive iff $t>1/e$ so that
$
2+t\ln t\ge 2-\frac{1}{e}>0,
$
proving that $J_a(0)>0$.
Remark. Positive values of $\xi$ remain to be checked. Note that for $\chi_0$ smooth equal to 1 near 0 and vanishing outside of $[0,1]$, we have
$$
t^{\frac 1a-1}\bigl( 2 + t \ln t\bigr)e^{-t}=t^{\frac 1a-1}\bigl( 2 + t \ln t\bigr)e^{-t}\chi_0(t) +\psi(t),
$$
where $\psi$ belongs to the Schwartz space, thus as well as its cosine transform. Somehow the main part of the integral is located near $t=0$ and
$$
\int_0^{+\infty} \cos(2π s\vert\xi\vert) ds 2πa=π a\delta_0(\xi),
$$
which is indeed positive for $a>0$.
I am struggling to understand why Fourier transform of $\psi$ is positive. Can you please give me a reference to that decomposition for $\chi$ and $\psi$ you have used in the proof. Thank you
"ln+1 which is positive iff >1/ " . and it seems the only condition for positivity of that Fourier transform of the derivative . So following your arguments I can replace the power in $1/\alpha^{2}$ from 2 to a smaller number as soon as it is bigger than $\exp(-1)$. However a simple numerical analysis shows that if the power is $0.4$ and $\alpha=1.2$, the Fourier transform can be negative for some $\omega$
@Tanya Vladi I did not write that the Fourier transform of $\psi$ is positive. I said that the $\partial_a\hat{f_a}(\xi=0)<0$. On the other hand it is indeed tempting to consider the positive function $\mathbb R_+\ni t\mapsto t^{1/a-1}(2+t\ln t)e^{-t}$ which is in $L^1$ and somewhat concentrated (in fact singular) at 0. I guess that for $\phi$ positive decreasing (and smooth), it is possible to decide the sign of the cosine transform, if $\phi(0)$ is large enough, but this case is indeed a bit different.
About $\chi_0$ and $\psi$, I simply claim that, given a positive $T$, I can find two smooth functions on $[0,+\infty)$ such that $\chi_0(t)=1$ on $[0,T]$ and $\chi_0(t)=0$ on $[2T,+\infty)$ so that with $\psi=1-\chi_0$, you get that $\psi$ vanishes near 0.
I guess I do not understand your proof at all. However a simple example, Fourier transform of $-{\partial\over \partial \alpha }[\exp(-x^{\alpha})\alpha^{-0.4}]$, for $\alpha$ can be negative for some $\omega$ (just from plotting using MATLAB), but according to your arguments it should be nonnegative
The only thing that I claimed is that $\partial_a\hat{f_a}(\xi=0)<0$ which does not preclude what happens for $\xi >0$.
but it is obvious as Fourier of $-{\partial\over \partial a} [\exp(-x^a)a^{-2}]$ is a positive definite function
I can calculate all the derivatives at $0$ of Fourier transform of the derivative with respect to $\alpha$,as I can express it as an analytic function $\sum_{m=0}^{\infty} { (-t^2)^m a_m \over m! }$ how does it help me?
Following this : https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S0004972700047511 , esp., the section "Complex Detours". I examined numerically the second derivative of $J_{\alpha}$ with a slightly complex $x$ (imaginary part ca 0.1 i ). The real part of the second derivatve seemed to fullfil the requirements indicated in the paper for $J_{\alpha}$ to be positive for $x>0$, at least for $1 < \alpha \le 1.5$. Maybe one could look in that direction, although the proof might be tedious.
@JohannesTrost, how do you choose $p$? I have tried to see if it works numerically, It seems $p$ should be small. However I failed to find $p$ for $\alpha>1.5$. Like you said $\alpha<1.5$ seems to be working
@JohannesTrost, the imaginary part is not decreasing for $\alpha>1$. the problem is around $0$
@Tanya Vladi For $a > 1.5$ (sorry for changing the name of the exponent to $\alpha$) there seems to be no imaginary shift, $p$, where the real part of th second derivative fulfils the requirement. I have seen the same.
@Tanya Vladi The imaginary part is only increasing for very small values around 0. If you look at the asymptotic expansion of the integral for very large $|x|$ (using the methods of Bleisteins book "Asymptotic Expansion of Integrals"), one can find a lower bound to $x$ above which the claim concerning $a$ is true. The wavelength of the sine under the integral is then so large that it will not "probe" the very small $t$ values around the origin, where the imaginary part increases. This is more a physicists handwaving argument, though. To prove this rigorously is quite tedious, I am afraid.
@JohannesTrost, thank you. The fact that my imaginary part increasing even at small value around 0 bothers me. I need the result for all values of cosine transform argument, I have looked into asymptotics and can prove the tail of transform is positive, which would correspond to small value of original function. , I guess I just cannot put all the pieces together. Not my area of expertise even though the intuition is there.
|
2025-03-21T14:48:31.047979
| 2020-05-21T22:23:50 |
360995
|
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|
Stack Exchange
|
Maximal $\mathbb{Q}$-split torus in center of Weil restriction of $\text{GL}_n$ over a number ring
I'm a topologist writing a paper where I have to do a bit of work with algebraic groups, and I've managed to confuse myself about something very basic.
Let $K$ be an algebraic number field. Regard $\text{GL}_n$ as an algebraic group over $K$, and let $G = R_{K/\mathbb{Q}}(\text{GL}_n)$ be the restriction of scalars of $\text{GL}_n$ to $\mathbb{Q}$. We thus have that $G(\mathbb{R}) = \text{GL}_n(K \otimes_{\mathbb{Q}} \mathbb{R})$, a product of copies of $\text{GL}_n(\mathbb{R})$ and $\text{GL}_n(\mathbb{C})$.
Consider the center $Z(G)$ of the $\mathbb{Q}$-algebraic group $G$. The real points $Z(G)(\mathbb{R})$ are thus a product of copies of $\mathbb{R}^{\times}$ (the center of $\text{GL}_n(\mathbb{R})$) and $\mathbb{C}^{\times}$ (the center of $\text{GL}_n(\mathbb{C})$).
Let $A$ be a maximal $\mathbb{Q}$-split torus in $Z(G)$.
Question: What are the real points $A(\mathbb{R})$? My first though was that this should be something like $\mathbb{R}^{\times}$, embedded as the diagonal in $Z(G)(\mathbb{R})$. But what if there are only complex embeddings?
EDIT: Let me explain where my "first thought" came from, which might help me pinpoint my confusion. We have that
$$Z(G)(\mathbb{Q}) = Z(\text{GL}_n(K)) = K^{\times}.$$
I was thinking that for the maximal $\mathbb{Q}$-split torus $A$, we should probably have
$$A(\mathbb{Q}) = \mathbb{Q}^{\times} < K^{\times} = Z(G)(\mathbb{Q}).$$
Anything bigger, and it would not split over $\mathbb{Q}$, right? Going up to $\mathbb{R}$, and it looks like we have
$$A(\mathbb{R}) = \mathbb{R}^{\times},$$
which is embedded diagonally as I said above. But I find this confusing, so I might be missing some simple point.
I guess that $A$ has dimension $d$, the number of factors in the algebra $K\otimes_\mathbf{Q}\mathbf{R}$ (product of copies of $\mathbf{R}$ and $\mathbf{C}$).
@YCor: I added some more information, which might help you figure out where I am going wrong.
$\newcommand\Q{\mathbb Q}\DeclareMathOperator\GL{GL}\DeclareMathOperator\Res{Res}$Notice that the reference to $\GL_n$ is a bit of a red herring; you're really dealing with $Z(G) = Z(\Res_{K/\Q} \GL_n) = \Res_{K/\Q} Z(\GL_n) = \Res_{K/\Q} \GL_1$. As you say, this has $\Q$-split rank 1. The $\mathbb R$-rational points of the maximal $\Q$-split torus are thus isomorphic to $\mathbb R^\times$, and they embed diagonally, including in the complex embeddings. (Maybe it helps to picture $K = \Q(i)$?)
|
2025-03-21T14:48:31.048290
| 2020-05-21T22:53:33 |
360997
|
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|
Stack Exchange
|
Is $[X, \_]$ a homology theory?
Let $X$ be a CW-spectrum. It is well-known that $[\_ ,X]$ is a generalized cohomology theory and, by Brown's representability theorem, every generalized theory is $H$ represented by a spectrum (namely, $H$ has this form).
What about $[X, \_]$? Is it a homology theory? (I do not claim every homology is corepresented by a spectrum.)
@LSpice thank you.
No worries. You can also rollback an edit, rather than making the opposite edit.
This holds only for compact objects (i.e. finite CW spectra), since it is easy to see that additivity fails otherwise (the other axioms of homology theories are satisfied). The usual way to obtain a homology theory from a spectrum $X$ is to consider $\pi_*(X\otimes -)$, note that for compact $X$ your $[X,-]$ is of this form as well, since
$$
[X,-] = \pi_*(DX\otimes -)
$$
by Spanier-Whitehead duality.
I also suspected that some conditions on X were necessary, but these notes left me puzzled (see page 4), thank you!.
|
2025-03-21T14:48:31.048396
| 2020-05-21T22:58:21 |
360999
|
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|
Stack Exchange
|
Maximal number of commuting functions of a finite set
Let $S$ be a finite set with $n$ elements and let $F_S$ denote the set of functions from $S$ to $S$. I wonder whether anything is known about the maximal cardinality of a commuting subset of $F_S$? A simple construction gives $2^{n-1}$ commuting elements, but I have no idea if this could be close to optimal. Is there any relevant literature?
You could send half the base set to a distinguished element and send the other half to the first half. This would give (n/2)^(n/2) elements. Gerhard "And Maybe There Are More" Paseman, 2020.05.21.
Very true... Thanks for this observation. I guess if you split the space differently, you could get $(n/\log n)^{(1-1/\log n)n}$, which is something like $e^{n\log n-n\log\log n}$.
$2^{n-1}$ is known to be the largest size of a set of commuting idempotent maps. I would Google maximal commutative subsemigroups of full transformation semigroups. I didn't find a precise reference that way, just a to appear that never appeared
According to the open questions at the end of https://www.sciencedirect.com/science/article/pii/S0195669810001265 this is open.
Gerhard's construction is building a null subsemigroup (the product of any two elements is the same). I guess one should be able to find the maximum size of a subsemigroup of this sort fromhttps://www.ams.org/journals/tran/1976-219-00/S0002-9947-1976-0404502-4/S0002-9947-1976-0404502-4.pdf but it might be annoying to extract
@BenjaminSteinberg: Thanks very much for this info! Also glad to hear that my guess, while far from the mark, is the answer to something!
Presumably you are associating to each subset of {1,...n-1} the mapping which fixes that subset and send the complement to n.
@BenjaminSteinberg: yes, exactly.
|
2025-03-21T14:48:31.048546
| 2020-05-21T23:11:36 |
361001
|
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|
Stack Exchange
|
Need reference for ideals and representations of $C_0(X,A)$
Let $A$ be $C^{\ast}$- Algebra and $X$ be a locally compact Hausdorff space and $C_{0}(X,A)$ be the set of all continuous functions from $X$ to $A$ vanishing at infinity. Define $f^{\ast}(t)={f(t)}^{\ast}$ (for $t\in X$). It is well known that $C_0(X,A)$ is $C^{\ast}-$ Algebra.
What’s known about ideals and representations of $C_0(X,A)$?
My guess is that it must be related with ideals and representations of $A$. Can someone give a reference or some ideas?
P.S: The same question was first posted on MSE but unfortunately I dint not get any answer so I am posting it here.
For each $x \in X$ let $I_x$ be a closed ideal of $A$. Then the set of $f \in C_0(X,A)$ satisfying $f(x) \in I_x$ for all $x$ is clearly an ideal of $C_0(X,A)$, and it shouldn't be too hard to show that every closed ideal has this form.
I assume this is "well known" but I don't have a reference.
As for representations, use the fact that $C_0(X,A)$ is $*$-isomorphic to the C*-algebra tensor product $C_0(X) \otimes A$. (Since $C_0(X)$ is abelian there is only one tensor product.) Thus the representations of $C_0(X,A)$ correspond to pairs of representations of $C_0(X)$ and $A$ on the same Hilbert space and whose ranges commute.
Thank you. Do you know about representations too?
I'm not sure what you want in regards to representations ...
I am mainly interested in seeing connection between representations of $A$ and of $C0(X,A)$.
Well, if $\pi$ is a representation of $A$ and $x \in X$ then $f \mapsto \pi(f(x))$ is a representation of $C_0(X,A)$. What else are you looking for?
What about the converse i.e.all representations of $C0(X,A)$ are of the form you mentioned?
@MathLover have you tried to work this out for yourself using the connections between ideals, homomorphisms and representations? It seems that Nik has already given you most of what you need to proceed
@YemonChoi I added that part of my answer after Math Lover asked the question in their comment.
|
2025-03-21T14:48:31.048705
| 2020-05-21T23:53:30 |
361003
|
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|
Stack Exchange
|
Conley Zehnder index for Floer homology of a symplectomorphism
I'm trying to get some intuition for the Conley-Zehnder index in the setting of Floer homology of a symplectomorphism $\phi : (M,\omega) \to (M,\omega)$. Let's assume that $\phi$ only has non-degenerate fixed points.
According to the paper of Dostouglu and Salamon "Self-Dual instants and Holomorphic curves" (DOI link pointing to publisher), the Maslov class determines a map $$ \mu : \text{Fix}(\phi) \to \Bbb Z_{2N}$$ (defined up to an additive constant) which satisfies $$\mu(u) = \mu_{CZ}(x_0) -\mu(x_1) \pmod{2N}$$ where $u$ is a Floer trajectory connecting $x_0$ to $x_1$ with Fredholm index $\mu(x)$.
I believe that they use this relation to define $\mu_{CZ}(x_1)$ say as $$\mu_{CZ}(x_1)=\mu_{CZ}(x_0) + \mu(u) \pmod{2N}$$ where the indeterminacy modulo $2N$ is given by the fact that by attaching a $J$-holomorphic sphere to the trajectory I can increase its Maslov index by $2N$ while preserving the fact that it's a solution of the Cauchy-Riemann equations. The indeterminacy up to translation comes from the fact that we are implicitly setting a value for index for $\mu_{CZ}(x_0)$, and we retrieve all the others starting from it.
Is this correct?
Why are these the only indeterminacies? I.e. why is $\mu_{CZ}(x_1)$ a well-defined number modulo $2N$ and not modulo some (smaller) divisors of $2N$? Equivalently, how can I see that the Fredholm index of a Floer trajectory going from $x_0$ to itself must be a multiple of $2N$?
If we were to prove that, as in the Hamiltonian case, the CZ-index of a critical point depends only depends on a choice of its lift to the Novikov cover of the twisted loop space of $\phi$ and that the Deck transformation group of such cover acts by attaching spheres like in the Hamiltonian case then we would be done I believe. But I can't see it stated anywhere and I'm not exactly sure it's the case. In fact if we consider the torus $T^2$, if the Deck transformation group of the Novikov cover acts by attaching spheres, then it would act trivially since $\pi_2(T^2)=0$. But in many papers dealing with surface automorphisms, the authors assume that the genus is at least $2$ to have an exact action functional (See my question here for example ). Hence I must confused somewhere along the way.
|
2025-03-21T14:48:31.048878
| 2020-05-22T00:09:23 |
361005
|
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"Rohil Prasad",
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|
Stack Exchange
|
Transverse invariant measures to vector fields
Given a smooth, non-vanishing vector field on a compact manifold, when does the 1-dimensional foliation given by its integral curves admit a transverse invariant measure?
I've seen examples of higher-dimensional foliations not admitting transverse invariant measures, but I'd imagine the same question is much easier to address one way of the other in the one-dimensional case.
Something that may work but I'm not sure about it enough to write an answer: Construct a Birkhoff section (likely disconnected) with a measure on it. Then the measure of a transversal is just defined to be the area of the parts of the Birkhoff section that the transversal first hits when we run it along the flow of the vector field.
I think that the best reference for this question is still the (relatively) old paper by Plante Foliations with measure preserving holonomy Ann. of Math. (2) 102 (1975), no. 2, 327–361, although it is a bit of an overkill for one-dimensional foliations. For instance, by Theorem 4.1 holonomy invariant measures exist for any foliation with a subexponential leaf. There are many other ways, of course.
Thanks for the reference!
As R W says, the answer is in Plante's paper, but the special case of $1$-dimensional foliations is actually simple, and well-known: let $M$ be the manifold, $*\in M$ be a basepoint, $(\phi^t)$ be the flow (assumed tangential to the boundary of $M$, if any). For every integer $n\ge 0$, let $\mu_n$ be the image in $M$ of the probability measure $dt/(2n)$ on the interval $[-n,+n]$ under the map
$$F:R\to M:t\mapsto\phi^t(*)$$
Since the space of Borelian probability measures on $M$ is compact for the weak topology, the sequence $(\mu_n)$ has a subsequence weakly converging to a Borelian probability measure $\mu$ on $M$.
Claim: $\mu$ is invariant by every $\phi^t$. Indeed, for a fixed $t$, the sequence of measures $$\phi^t_*\mu_n-\mu_n$$ goes to $0$ in the norm topology, since for any continuous real function $f$ on $M$, one has:
$$\vert(\phi^t_*\mu_n-\mu_n)f\vert=\frac{1}{2n}\vert\int_{-n+t}^{n+t}f(\phi^s(*))ds-\int_{-n}^{+n}f(\phi^s(*))ds\vert\le\frac{t}{n}\Vert f\Vert_\infty$$
Finally, the $R$-invariant measure $\mu$ amounts to a transverse invariant measure $\nu$ such that locally (in every local flow box $B\cong D^{n-1}\times I$) one has $\mu=\nu\otimes dt$. The measure $\nu(D)$ of every small transverse disk $D$ is the limit of some subsequence of the sequence
$$\frac{1}{2n}{\sharp(F^{-1}(D)\cap[-n,+n])}$$
In this construction, we have used the choice axiom (hidden in the compacity of the measures space); in particular, the construction of the subsequence is non-constructive.
I understand the construction, but I might need this spelled out a bit more explicitly for me. This is certainly an invariant measure, but how does it induce a transverse invariant measure? i.e. given some small transversal $\tau$, what would the total measure of $\tau$ be? Would it just be the limit as $n \to \infty$ of $\frac{1}{2n+1}$ times the number of intersection points of the trajectory with $\tau$?
|
2025-03-21T14:48:31.049099
| 2020-05-22T00:32:24 |
361006
|
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|
Stack Exchange
|
q-series identity related to Jackson-Slater, proof required
The question:
I have been trying to prove the following $q$-series identity for quite some time now:
$$
\sum_{k \geq 0} \frac{q^{2k^2}}{(q)_{2k}} = \sum_{m,k \geq 0} \frac{q^{m^2 + 3k m + 4k^2}}{(q)_k(q)_m} \left( 1 - q^{2m + 4k +1} \right) \tag{*}
$$
Where for a non-negative number $k$ we let $(q)_k = (1-q)...(1-q^k)$. I have verified this conjecture with SageMath to order $q^{2000}$.
I am no expert on the topic of $q$-series and partition identities so I have no intuition to decide if this should be an easy consequence of some well known $q$-hypergeometric summation formula or not. I am hoping that if that is the case, some expert here will point me in the right direction or share a proof.
I can prove that for every $n \geq 0$ the coefficient of $q^n$ on the RHS of (*) is bigger or equal to the same coefficient on the LHS, so only the other inequality is required.
Some background:
For a history of this identity please refer to the article we just posted with this conjecture [7]. I will collect below many things I know about this identity.
Here is a reason (besides pride and the list of techniques below that I have tried) that led me to believe that it may not be a trivial identity:
Sums like either of the two terms of the RHS of (*) are known as Nahm sums. In this particular case this is a Nahm sum associated to the matrix $(\begin{smallmatrix}8 & 3 \\ 3 & 2\end{smallmatrix})$. There is a famous conjecture [4] regarding modularity of these sums. This particular matrix is known to have the right asymptotic at $q \rightarrow 1$ for it to be modular. The LHS of (*) is the character of the Ising model vertex algebra which is known to be modular. Zagier studied this particular type of Nahm sums in [5] and did not find a modular sum with only one term and the matrix as in the RHS of (*). What this identity is saying is that by allowing linear combinations of Nahm sums of a fixed matrix that satisfies the right assymptotics, one may find modular sums.
Here are some techniques that I have tried so far:
The theory of Bailey pairs and Bailey chains. None of the Bailey pairs with parameters found in Sills recent article [1] works.
Polynomial approximations. Most of the RR type identities admit proofs along the lines of finding a polynomial identity that in certain limit gives the wanted identity. There are well known polynomial approximations to the LHS of this identity. For example one can look at the Santos Polynomials [2]. The RHS admits several polynomial approximations, for example one could take
$$
P_L(q) = \sum_{m=0}^L q^{m^2} \binom{L}{m}_q \sum_{k = 0}^m q^{2k^2 +km} \binom{m}{k}_q
$$
where the $q$-binomials in this setting is defined as $\binom{m}{n}_q = \frac{(m)_q}{(n)_q(m-n)_q}$.
Two variable generalizations. Most of the identities in Slater's list [3] admit two variable generalizations. In Sills list [1] mentioned above for example, there is a two variable generalization of Slater's identity (39) (which goes by the name in the title of this post), which has the LHS of (*) as one of the members. The RHS of (*) also comes with a natural two-variable generalization (see please [7] for this generating function). However I could not find a natural way of matching this two-variable generalization to any of the known expressions equal to the LHS of (*).
Functional equations: the generating series (or rather its two variable generalization alluded above) satisfies a functional equation of order $11$ so in principle if one were to guess the two variable version of the LHS a computer could decide if it satisfies this functional equation. This equation can be written as a system of $5$ (or even 2) variables and of order $2$ as written in our article.
Combinatorial description: The LHS of (*) counts naturally partitions satisfying some difference conditions due to a theorem of Hirschorn. There are many more families of partitions that are counted by the same generating function, a list below to the equivalent forms of this side should give you an idea. The RHS of (*) also is the generating function of a family of partitions. It arises by studying a Gröbner basis for an ideal sheaf in an affine arc space. As is typically the case in these Gröbner bases computations, the partitions involved are just horrible. It is somewhat of a miracle that we could find the generating series.
Here are some things that I have not tried:
The character of the Ising model satisfies an explicit modular differential equation. I do not know how to check that the RHS of (*) satisfies this equation or not. I do not even know how to compute $q \frac{d}{dq}$ of it.
Relations with other similar conjectures: if you look into our paper you will find similar conjectural expressions for the characters of other modules for the Ising vertex algebra. Since these are known to satisfy certain algebraic relations, perhaps these can be exploited to prove all of these identities at once.
Here are some equivalent forms for both sides of the identity:
The LHS of (*) is very well known and admits the following well known forms:
$$
\begin{aligned}
%
\sum_{k \geq 0} \frac{q^{2k^2}}{(q)_{2k}} &= \prod_{n=1}^\infty \frac{1}{1-q^n} \sum_{m \in \mathbb{Z}} \left( q^{12m^2+m} - q^{12m^2+7m+1} \right) \\
%
&= \frac{1}{2} \left( \prod_{m=1}^\infty \left( 1+q^{m-1/2} \right) + \prod_{m=1}^\infty \left( 1-q^{m-1/2} \right) \right) \\
%
&= \prod_{k=1}^\infty \frac{(1+q^{8k - 5})(1+q^{8k-3})(1-q^{8k})}{1-q^{2k}} \\
&= \sum_{k = (k_1, k_2, \ldots, k_8) \in \mathbb{Z}^8_{\geq 0}} \frac{q^{k^T C_{E_8}^{-1} k}}{(q)_{k_1} \dots (q)_{k_8}}
\end{aligned}
$$
This last form is striking, here $C_{E_8}$ is the Cartan matrix for the simple Lie algebra $E_8$.
The RHS can be easily written as
$$
\sum_{m \geq 0} \frac{q^{m^2}}{(q)_m} \sum_{k =0}^m q^{2k^2 + km} \binom{m}{k}_q (1 - q^k + q^m)
$$
Perhaps a better approach to try to use the trick of Section 5 of [6], the RHS (*) is also equivalent to
$$
\sum_{m \geq 0} \frac{q^{m^2}}{(q)_m} \left( 1 - q^{2m +1} \right) \sum_{0 \leq 2k \leq m} \frac{(q^{-m}, q^{1-m}; q^2)_k}{(q;q)_k} q^{(m+1)k}
$$
Finally, to show a not-so-straightforward equivalent form, by using Ismail-Garret-Statton generalization of the Rogers-Ramanujan identity one can sum first $m$ in (*) to obtain the equivalent form:
$$
\sum_{k \geq 0} (-1)^k \frac{q^{\frac{k(k+1)}{2}}}{(q)_k} \left( \frac{q^{2k}a_{3k} - a_{3k+2}}{(q,q^4;q^5)_{\infty}} - \frac{q^{2k} b_{3k} -b_{3k+2}}{(q^2,q^3;q^5)_{\infty}}\right)
$$
where $a_k$, $b_k$ are the Schur polynomials defined by the recursion $x_{k+2} = x_{k+1} + q^k x_k$ and the initial conditions
$$
a_0 = b_1 = 1, \qquad a_1 = b_0 = 0.
$$
References:
[1] A. Sills Identities of the Rogers-Ramanujan-Slater Type. International Journal of Number Theory. Vol. 03, No. 02, pp. 293-323 (2007) ]
[2] G. E. Andrews and J. P. Santos. Rogers-Ramanujan type identities for partitions with attached odd parts. The Ramanujan Journal, 1:91–99, 1997.
[3] L. J. Slater. Further identities of the rogers-ramanujan type. Proc. London Math. Soc., 54(2):147–167, 1952.
[4] W. Nahm. Conformal field theory and torsion elements of the Bloch group. In Frontiers in number theory, physics, and geometry. II, pages 67–132. Springer, Berlin, 2007.
[5] D. Zagier. The dilogarithm function. In Frontiers in number theory, physics, and geometry. II, pages 3–65. Springer, Berlin, 2007.
[6] G. Andrews, K. Bringmann, K. Mahlburg Double series representations for Schur's partition function and related identities Journal of Combinatorial Theory, Series A Volume 132, May 2015, Pages 102-119
[7] J. van Ekeren, R. Heluani The singular support of the Ising model
The following proof is due to George E. Andrews (who is now a coauthor on the mentioned paper).
Consider the following families of polynomials indexed by $n$
$$
S_n = \sum_{k \geq 0} q^{2k^2} \binom{n-k}{2k}_q
$$
$$
T_n = \sum_{m,k\geq 0} q^{m^2+3k m+4k^2} \left( \binom{n-3k-m}{k}_q
\binom{n-4k-m}{m}_q - q^k \binom{n-3k-m-1}{k}_q\binom{n-4k-m-1}{m-1}_q \right).
$$
The limit as $n \rightarrow \infty$ of $S_n$ is the LHS of the desired identity while the limit as $n \rightarrow \infty$ of $T_n$ is the RHS of the desired identity.
For every $n \geq 0$ one can check that $S_n = T_n$. We used the Mathematica package qMultiSum which was provided by RISC to check that both sides satisfy the order 8 recursion
$$
q^{4n+15}S_n + q^{2n+11}(1+q)(S_{n+3} - S_{n+4}) - q^3 S_{n+5} +
(1+q+q^2)(qS_{n+6} - S_{n+7}) + S_{n+8} = 0.
$$
In fact the same technique works for the three conjectures of our paper so now we have a PBW basis for the Ising model and three new quasi-particle sum expressions for the three characters of the minimal model at central charge $c=1/2$
I'll leave this question open since it still uses a computer to check the recursion. Perhaps someone can find a proof by pen and paper.
|
2025-03-21T14:48:31.049651
| 2020-05-22T00:52:09 |
361008
|
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"Stanley Yao Xiao",
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|
Stack Exchange
|
Prime factors of $p-1$
It is one of the consequences of Sieve theory is that number of primes $p\leq x$ such that all prime divisors of $p-1$ are greater than $p^{\varepsilon},$ is $\gg \frac{x}{\log^2x}.$ In particular, all of these $p-1$ have a bounded number of prime divisors. Due to Halberstam we know, number of distinct prime factors of $p-1$ is $\log \log p$ on an average. Given a constant $c$, is it known how many primes are there up-to $x$ such that all number of distinct prime factors of $p-1$ is bounded by $c$ ? I feel this actually same as answering the $p^{\varepsilon}$ thing, is it ?
No, not quite. The lower bound that you produce from the sieve has the wrong order of magnitude. You should look for papers on the Titchmarsh divisor problem. Also, the first sentence is wrong: for almost all primes $p$ the smallest prime divisor of $p - 1$ is $2$.
You are right, yes I should have said odd prime factors. But what's the correct order ? is it like $c\frac{x}{\log x}$ for some $0<c<1$ ?
|
2025-03-21T14:48:31.049757
| 2020-05-23T08:50:46 |
361138
|
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|
Stack Exchange
|
Determining if a morphism is a blowup along a given subvariety
Let $X,\tilde{X}$ be two smooth projective varieties over $\mathbb{C}$, and let $\pi:\tilde{X}\rightarrow X$ be a projective morphism. Let us moreover assume that there exists a smooth closed subvariety $Y\subset X$, such that $\pi$ is isomorphism outside $Y$, and $\pi^{-1}(Y)\rightarrow Y$ is a projective bundle of rank = codim($Y,X$).
I want to show that $\tilde{X}$ is indeed the blowup of $X$ along $Y$. I have seen smoething along this line mentioined in a paper. Of course, there will be a map from $\tilde{X}$ to the blowup by the universal perty of blowups, but I can't show it to be an isomorphism. I know that any projective birational morphism is a blowup along some closed subscheme, but it's not clear from the proof that the closed subscheme is indeed $Y$.
Any help would be appreciated.
The induced morphism $\tilde{X}\to Bl_Y(X)$ is a birational morphism of smooth projective varieties with the same Picard number (equal to the Picard number of $X$ plus one). This implies that it is an isomorphism (the equality of Picard numbers implies that there are no exceptional divisors).
@OlivierBenoist : Thank you for your prompt response. Could you please explain how you found the picard number of $\tilde{X}$? Also, why does same picard number imply isomorphism? Is there a reference to this?
The isomorphism statement follows from this post.
@R.vanDobbendeBruyn : In our case, isn't the exceptional subvariety for the map $\tilde{X}\rightarrow Bl_Y(X)$of codimension 1? Could you please explain how you are concluding from the post you have mentioned?
If the Picard ranks are the same there cannot be a codimension $1$ exceptional locus, because the exceptional divisor would necessary be linearly independent in Néron–Severi.
|
2025-03-21T14:48:31.049898
| 2020-05-23T08:50:59 |
361139
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/361139"
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|
Stack Exchange
|
Degree 2 plane curves on a cubic surface singular along a line in $\mathbb{P}^3$
Suppse $X$ is an irreducible cubic surface in $\mathbb{P}^3$ singular along a line $l_1$. Then clearly there is a plane $H$ containing $l$ such that $X \cap H = l_1^2l_2$.
My question is: can $X$ contain a degree $2$ plane curve other than curves of the form $l_1l$, where $l$ is some other line ?
Yes. $X$ is the projection of a cubic scroll $S\subset \mathbb{P}^4$ from a point $p\in \mathbb{P}^4\smallsetminus S$. This $S$ can be viewed as the image of the map $\mathbb{P}^2\rightarrow \mathbb{P}^4$ given by the linear system of conics passing through a fixed point. A general line in $\mathbb{P}^2$ maps to a conic in $\mathbb{P}^4$, which projects isomorphically to a conic in $X$.
|
2025-03-21T14:48:31.049986
| 2020-05-23T12:02:26 |
361145
|
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Stack Exchange
|
Action of fundamental group on homotopy fiber
For a Serre fibration of pointed topological spaces $f:X \to B$, there is an action of $\pi_1\left(B,b_0\right)$ on the fiber $F$. The construction of this action I'm familiar with uses a lift $F\times I \to X$ of the map $F \times I \xrightarrow{\pi} I \xrightarrow{\gamma} B$ for any $\left[\gamma \right] \in \pi_1 \left(B,b_0 \right)$.
Now for a general map between two $\infty$-groupoids $f:X \to B$, we can use some version of the Grothendieck construction to construct a map $\phi_f : B \to \operatorname{Grp}_\infty$, and then for an element $\left[\gamma \right] \in \pi_1 \left(B,b_0 \right)$, $\phi_f \left(\gamma \right)$ is an automorphism of $\phi_f \left(b_0 \right)$, which, I guess, generalizes the previous definition (please correct me if that is already false). Is there an explicit description of this automorphism for a specific $\gamma$? (i.e. in terms of pullbacks / pushouts / sections of the maps $f,\gamma$ etc.) I'm particularly interested in writing down obstructions for triviality of this action.
(This answer is written in a model-independent fashion -- translate to your favourite formalism).
For every path $\gamma:[0,1]\to B$ you get an isomorphism in the homotopy category $X_{\gamma0}\xrightarrow{\sim} X_{\gamma1}$ (where with $X_b$ I denote the homotopy fiber over $b\in B$). Probably the easiest and most geometric way of constructing it is to consider the space of lifts.
Let $\operatorname{Sec}_\gamma(f)$ be the space whose objects are sections up to homotopy over $\gamma$
$$\operatorname{Sec}_\gamma(f)=\{(\tilde\gamma,H)\mid \tilde\gamma:[0,1]\to X,\ H:f\tilde\gamma\sim \gamma\}$$
(this is nothing else but the homotopy fiber over $\gamma$ of the map $X^{[0,1]}\to B^{[0,1]}$). Then you have a zig zag of maps
$$ X_{\gamma0}\xleftarrow{ev_0} \operatorname{Sec}_\gamma(f)\xrightarrow{ev_1} X_{\gamma1}$$
where the two maps are evaluation at 0 and 1 respectively. Both maps are homotopy equivalences (this requires some proof, but it's not terribily hard: they're both trivial fibrations when $f$ is a fibration), and so you can define the map in the homotopy category as $ev_1 \circ ev_0^{-1}$.
In order to prove that the action of a loop is trivial, you'll have to prove that $ev_0$ and $ev_1$ are homotopic. I'm not aware of a general way of attacking this problem, but of course studying the behaviour of the two maps on various algebraic invariants can often provide obstructions.
Let $f:E\to B$ be a map of based spaces, and let $F$ be the homotopy fiber. Here is another way of constructing the action of $\Omega B$ on $F$. By definition, there is a homotopy pullback square
$$\require{AMScd}
\begin{CD}
F @>>> \ast\\
@VVV @VVV \\
E @>>> B.\\
\end{CD}$$
Taking homotopy pullbacks along the inclusion $\ast\to B$ produces a map to the above homotopy pullback square from the following one:
$$\require{AMScd}
\begin{CD}
\Omega B\times F @>>> \Omega B\\
@VVV @VVV \\
F @>>> \ast.\\
\end{CD}$$
The two morphisms in this square are the projections. The action of $\Omega B$ on $F$ is just the map between the top left corners of these squares; let's call this map $\mu$. This action is not just the projection: this construction shows that there is a homotopy pullback square
$$\require{AMScd}
\begin{CD}
\Omega B\times F @>{\mathrm{pr}}>> F\\
@V{\mu}VV @VVV \\
F @>>> E;\\
\end{CD}$$
if $\mu$ was just projection onto $F$, then the space $E$ in the bottom right corner would have to be replaced by $F\times B$.
(Note that this diagram shows that the composite $\Omega B \times F\to F\to E$ is trivial on $\Omega B$. You can also see this by the explicit model of this map in spaces: this composite just sends a pair $(\gamma, [e\in E, p:\ast\to f(e)])$ to $e$.) I don't know of general methods to show that the action is trivial. (Remark: one potential advantage of phrasing the construction in this way is that it works in any ($\infty$-)category with finite homotopy limits.)
Do you mean to have $\Omega B$ on the top right corner of your second square?
Whoops, yes, thanks for spotting the typo @DenisNardin.
|
2025-03-21T14:48:31.050387
| 2020-05-23T12:31:01 |
361148
|
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|
Stack Exchange
|
Explicit constant for Carbery–Wright inequality
The Carbery–Wright inequality is a seminal result about the anti-concentration of polynomials of Gaussian random variables.
See e.g. Meka, Nguyen, and Vu - Anti-concentration for polynomials of independent random variables, Theorem 1.4, for the precise statement.
However, I cannot find any reference where an explicit estimate on the constant $B$ on the r.h.s. of the inequality is given. Knowing this constant is crucial for the application I have in mind. Are there known estimates on it?
I should also say that the polynomial I have in mind is of the form
$p(g_1,g_2,\dotsc,g_k)=\langle g_1\otimes g_2 \dotsb\otimes g_k,A g_1\otimes g_2 \dotsb\otimes g_k\rangle$, where $A\in\mathbb{R}^{d^k\times d^k}$, $\operatorname{tr}(A)=0$, and the $g_i\in \mathbb{R}^d$ are random vectors with i.i.d. gaussian entries. Maybe this special structure helps to obtain better anti-concentration estimates.
Any help would be appreciated!
Have you looked at the original paper by A. Carbery and J. Wright, Distributional and $L^q$ norm inequalities for polynomials over convex bodies in $\mathbb R^n$? Theorem 8 page 244 is the famous inequality with a sharp constant.
what do you mean by sharp constant? I do not see an explicit constant in that Theorem. You mean this article: https://pdfs.semanticscholar.org/e997/358a17bdd75bd7ffe7126693da12baea325b.pdf ,right?
It only says "There exists an absolute constant C". Do we know what that constant is?
The way it is stated suggests that C is independent of the polynomial. The constant C is numerical.
Oh, now I understand the issue. Sure, the constant C is numerical, but my application requires bounding this constant.
Note that the statement of the inequality only becomes nontrivial for $\alpha\leq C^{-2d}$, where $d$ is the degree of the polynomial. In the application I have in mind, the degree of the polynomial grows, so the inequality is nontrivial only for exponentially small $\alpha$, with the exponent depending on $C$. That is why I actually want to know $C$ in the inequality.
Then, I guess you should be able to verify its order of magnitude numerically by testing different polynomials in dimension 1 for the Gaussian since it does not depend on the dimension, neither on the degree, neither on the polynomial neither on the measure...
I am not sure this is correct. The statement only tells us that there is a universal lower bound. But there is no guarantee that polynomials in dimension 1 are "representative". I.e., it could be the case that a much better constant $C$ holds for dimension $1$ compared to, say 59. Of course, this gives a lower bound on the constant, but not a universal upper bound.
Ok! I thought you were interested in a non-trivial lower bound.
|
2025-03-21T14:48:31.050607
| 2020-05-23T12:45:32 |
361149
|
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|
Stack Exchange
|
Does a compact Lie group have finitely many conjugacy classes of maximal Abelian Lie subgroups?
Let $G$ be a compact Lie group. An Abelian Lie subgroup $A \leq G$ is a maximal Abelian Lie subgroup if, for any Abelian Lie subgroup $A'$ such that $A \leq A' \leq G$, then $A' = A$.
Of course any maximal torus of $G$ (there is only one, up to conjugacy classes) is a maximal Abelian Lie subgroup, but there are other ones too, for example the Klein 4-group in $\mathrm{SO}(3)$.
What I'm wondering is if the number of conjugacy classes of maximal Abelian Lie subgroups of any compact Lie group $G$ is always finite?
This may be related, but I think your specific question just barely doesn't fit into any of the answers (or lemmas in them): https://mathoverflow.net/questions/265722/countability-of-conjugacy-classes-of-closed-subgroups
Yes it's true. It essentially follows from a lemma quoted in the linked answer by user Qayum Khan, which I quote verbatim:
Neighboring subgroups theorem [1942]
Any compact subgroup $H$ of an arbitrary Lie group $G$ admits a neighborhood $O$ in $G$ such that any subgroup contained in $O$ is $G$-conjugate to a subgroup of $H$.
Now let by contradiction $(A_n)$ be a sequence of pairwise non-conjugate closed maximal abelian subgroups in a compact Lie group. Let $A$ be a limit point in the Hausdorff topology; this is a compact abelian subgroup. By the above result, for $n$ large enough $A_n$ is conjugate to a subgroup of $A$. By maximality, this means that for $n$ large enough, $A_n$ is conjugate to $A$. This contradicts the non-conjugation.
Another approach: any abelian (or even nilpotent) subgroup of $G$ is contained in the normalizer of a maximal torus.
@abx I didn't know this result, which seems to provide an alternating approach reducing the problem to the case when $G^0$ is abelian.
Bourbaki's Lie IX, §5, Cor. 4 of Théorème 1.
@abx thanks. Anyway some further argument is needed to deal with $G^0$ abelian.
Just to emphasize a possible difficulty: while every abelian subgroup is contained in a maximal abelian one (obvious and true in every group), the same is false for nilpotent/maximal nilpotent subgroups in the compact Lie group $(\mathbf{R}/\mathbf{Z})\rtimes_\pm(\mathbf{Z}/2\mathbf{Z})$.
@abx, I asked a question on MSE a while ago to which this is an (perhaps the!) answer. I have posted an answer before I forget again, but I would prefer to give you the credit. If you would be willing to post your comment as an answer, then I will delete mine and accept yours.
@LSpice: Thank you, but that's fine as it is — it's just a reference.
|
2025-03-21T14:48:31.051063
| 2020-05-23T12:55:51 |
361150
|
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|
Stack Exchange
|
What is known about this generalization of planar dual?
So it is well known that given a planar graph, $G$, embedded in the plane (without edge crossing, so a planar embedding). One can construct the planar dual, $G^*$. What is perhaps slightly less well-known, and surprising, is that the resulting planar dual depends on the embedding. Not just the way in which the dual is embedded in the plane, but it's very graph isomorphism type can change with a different planar embedding of $G$. The easiest example to show this is a graph that looks like a rectangle with 2 triangles attached to opposite sides. This can alternatively be embedded with one of those triangles 'folded inside' the rectangle. Producing a different, non-isomorphic planar dual. This motivated me to ask: Is there a 'larger' abstract object which contains both of these (and more)? It turns out that I came up with just such an object.
Definition:
Given a 2-edge connected graph, $G$, the duplex graph, $G^{\triangle}$, of $G$ is constructed as follows:
1. For each cycle, $C_k$ in $G$ (by which, I mean a set of vertices and edges which form a subgraph isomorphic to a cyclic graph with at least 3 vertices), create a vertex, $v(C_k)$. 2. For each pair of distinct cycles, $C_j,C_k$, If the intersection of their edge sets, $E(C_j)\cap E(C_k)$, is nonempty, then for each edge, $e_d$, in the intersection, $E(C_j)\cap E(C_k)$, create an edge $e_d^{\triangle}$, between $v(C_j)$ and $v(C_k)$.
The result is a unique, embedding independent, graph $G^{\triangle}$ for any suitable graph $G$. In investigating this object and some close relatives. I discovered deep connections between these objects and the general embedding problem on 2-manifolds for graphs (essentially, the topological genus question for graphs and related ideas). Some digging revealed this question is generally attacked via the theory of rotation systems (several papers on this that I found citations for were from the early 1980s). What I think I have here is an entirely different way of looking at this question that seems very interesting.
My question is this: Is this idea (of producing a more general object related to the planar dual) present in the literature? (I have not encountered it, but perhaps someone else has?) If so, was any connection to graph embeddings (especially other than planar) investigated with it?
Matroids: see https://en.wikipedia.org/wiki/Matroid.
@RichardStanley I am unfamiliar with Matroid theory, but from skimming the article (and some related ones) it seems that it is definitely connected to the idea I've presented to some extent, but I don't think it quite captures what I have come up with here. Still, thank you for pointing this out, I suppose I should look into learning some Matroid theory then (recommendations welcome).
|
2025-03-21T14:48:31.051286
| 2020-05-23T15:00:42 |
361154
|
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|
Stack Exchange
|
Understanding the definition of weak solution for the eigenvalue problem of the Colding and Minicozzi's operator
I am trying to understand the corollary $5.15$ on page $23$ of the paper GENERIC MEAN CURVATURE FLOW I; GENERIC SINGULARITIES by Colding and Minicozzi. Specifically, I would like to understand why they stated the eigenvalue problem for the operator $L$, which is defined at the bottom of the page $21$, with weak solutions in the weighted $L^2$ space instead of the weighted $W^{1,2}$ space, which sounds more natural to me. I believe that the key for the answer is that $L$ is self-adjoint with respect the inner product of the weighted $L^2$ space by the theorem $5.2$ on page $22$, but I can not see how this can justify that the solutions live in the weighted $L^2$ space instead of the weighted $W^{1,2}$ space and I do not think that these weighted spaces are equal once that the respective spaces without the weight are not equal.
Have you read the corresponding section in Evans (Colding--Minicozzi's footnote 11)?
Yes, I have read, but I could not adapt the proof of the theorems in Evans' book for the Colding and Minicozzi's operator, then I followed the section $3$ of this lecture notes: http://www.math.mcgill.ca/gantumur/math580f18/laplacenotes.pdf and I forgot to come back to Evans' book when my doubt arised. Anyway, coming back to Evans' book, it seems clear now, the solutions live in the weighted $W^{1,2}$ space, but the spectral theorem states that the solutions are an orthonormal basis to the weighted $L^2$ space. Is this, isn't it?
In fact the solutions are completely smooth (by elliptic regularity)! But they are a orthonormal basis of $L^2$. Of course this seems a bit weird, but you should remember, for example that polynomials are dense in $L^2[0,1]$, for example. So it is not so strange that you can find an orthonormal basis of $L^2$ that consists of smooth functions.
|
2025-03-21T14:48:31.051514
| 2020-05-23T15:10:26 |
361156
|
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|
Stack Exchange
|
What is the nearest Ford circle for any point in $\mathbb R^2$
I want to draw Ford circles within a "distance Estimated system" (ray marching). Therefore, given a point $(x,y)$ from $\mathbb R^2$, I need the shortest distance to any circle with center $(p/q,1/2q^2)$ in $\mathbb R^2$ with radius $(1/2q^2)$ for coprime integers $p$, $q$.
It should be somehow possible to calculate (or at least approximate) with … maybe a continued fraction approximation of $x$?
What is a distance estimated system? Googling yields only distance estimated fractals; is that related? Also, is it clear that there is a unique closest Ford circle to every point?
yea, that link explains it. For sure there is almost anywhere a closest circle. however if tehre are more then one then at least the distance will be unique :-)
When p/q is the continued fraction of x which is approximated up to q>1/y then i always find the biggest circle below (x,y). Now i still need to find the bigger neighbour circles - which are also neigbours of p/q in the Farey Sequence F_q
|
2025-03-21T14:48:31.051659
| 2020-05-23T15:17:49 |
361157
|
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"Goldstern",
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|
Stack Exchange
|
Smallest ring whose field of fractions includes all the reals (subring of omnific integers?)
The surreal numbers have a subring, the ring of "omnific integers" or $\mathbf{Oz}$, which have the property that every surreal number is a quotient of two omnific integers. That is, the field of fractions of the omnific integers is the entire surreal number field, which in particular includes all the reals.
A ring whose field of fractions includes all the reals seems like a useful thing. The omnific integers would seem to be much larger than necessary if that is what we want. So we can ask for simpler examples.
Of course, $\Bbb R$ is a trivial example of a ring whose field of fractions includes all of $\Bbb R$. So, to be precise, I am interested in rings which do not already have all the reals, but whose field of fractions does have all the reals.
In particular, I have the following questions:
Does there exist some (ordered) ring $R$, which is not a superset of $\Bbb R$ but whose field of fractions is a superset of $\Bbb R$, that is "smallest" in the sense that $R$ is isomorphic to a (ordered) subring of any other (ordered) ring with this property?
Does the ring of omnific integers have any smallest subring with the above property?
Would any ring with the above property embed into the omnific integers anyway, making these criteria all equivalent?
To add some detail to the above:
In the omnific integers, for any real number $r$, we have that $r \omega$ is an omnific integer. So, to start, we can look at the following fragment of the omnific integers, with all elements of the form
$$z + r_1\omega + r_2\omega^2 + ... + r_n\omega^n$$
where $z$ is an integer, and the $r_n$ are all real numbers. I will notate this ring as
$$\Bbb Z \: \tilde \oplus \: \Bbb R \: \tilde \oplus \: \Bbb R \: \tilde \oplus \: ...$$
where the $\tilde \oplus$ is a kind of modified direct sum in which the polynomial coefficient multiplication is used rather than pointwise multiplication, which is always possible as long as each ring is a subring of the ring after it.
It is easy to see that the above ring has $\Bbb R$ in its field of fractions. It is also easy to see that this is true for any ring of the form
$$\Bbb Z \: \tilde \oplus \: \Bbb Z \: \tilde \oplus \: ... \tilde \oplus \: \Bbb Z \: \tilde \oplus \: \Bbb R \: \tilde \oplus \: \Bbb R \: \tilde \oplus \: ...$$
which the first ring embeds into, and which also embeds into the first ring.
So, one thing we can do is ask if there exists a countable sequence of rings $R_n \neq \Bbb R$ such that
$$ R_1 \subset R_2 \subset R_3 \subset ...$$
and
$$\Bbb R \subset \text{Quot}(R_1 \: \tilde \oplus \: R_2 \: \tilde \oplus \: R_3 \: \tilde \oplus \: ...)$$
One way this could be is if there is a countable sequence of $R_n$ such that the union of all the $R_n$ is $\Bbb R$, so that each real number appears at some point in the $R_n$.
If your $R$ has field of fractions $K$ containing $\mathbb R$, then it makes sense to embed both $R$ and $\mathbb R$ into $K$, and so to speak of $R \cap \mathbb R$. Could it happen that the field of fractions of $R \cap \mathbb R$ is smaller than $\mathbb R$? (Also, in your (1), do you want some kind of universality—e.g., a unique isomorphism (satisfying some extra conditions, I guess—or really just abstract isomorphism?)
As I understand it, Oz is a proper class, and you certainly do not need all of Oz to find the real numbers as a quotient field. If you take a sufficiently large and sufficiently closed ordinal $\delta$, then the set $Oz_\delta$ of all surreal integers born before day $\delta$ will be a ring, and its quotient field will include all the reals. I think that $\delta:=(2^{\aleph_0})^+$ will do, but perhaps much smaller smaller ordinals are enough, such as $\omega^ \omega$? (Clearly you need the ring to have at least continuum many elements.)
Algebraically, if we let $K = \overline{\mathbb{Q}} \cap \mathbb{R}$, then $\mathbb{R}$ is isomorphic to the real closure of $K$ adjoin uncountably many variables. So I imagine a "smallest" such ring would be some version of taking "integers" here -- e.g. taking the ring of real algebraic integers, adjoining infinitely many variables, and then adding a root of all monic odd-degree polynomials successively. Of course, this is totally non-canonical and you get many distinct subrings of $\mathbb{R}$ depending on which transcendence basis you pick.
Related: https://math.stackexchange.com/questions/1396872/is-there-any-proper-subring-of-mathbbr-with-field-of-fractions-equal-to-m
Assume for contradiction that such a ring $R$ exists.
Consider the ring $R_{\omega}:=\mathbb{Z} \oplus \omega\mathbb{R}[\omega]$ which is contained in $\mathbf{Oz}$. We have $\mathbb{R} \not\subseteq R_{\omega}$ and $\operatorname{Frac}(R_{\omega})\supseteq \mathbb{R}$. Since $R_{\omega}$ is discretely ordered, so must be $R$.
Consider a ring $R_0$ given by the answer linked by Kevin Casto. Since $R_0$ is archimedean, so must be $R$.
But $\mathbb{Z}$ is the only discretely ordered archimedean ordered ring. So we must have $R=\mathbb{Z}$, which cannot be.
Thanks, somehow missed this many years ago. Note also for those interested the answer to this: https://math.stackexchange.com/questions/1396872/is-there-any-proper-subring-of-mathbbr-with-field-of-fractions-equal-to-m
|
2025-03-21T14:48:31.052044
| 2020-05-23T15:29:34 |
361159
|
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|
Stack Exchange
|
Eigenspace of Gaussian Markov operator
Consider the (one-dimensional) Gaussian distribution $Q := N(\nu,\tau^2)$ and the (Gaussian) Markov operator
\begin{equation*}
\begin{array}{rccc}
R : & L_1(\mathbb{R},\mathcal{B}(\mathbb{R}),Q) & \to & L_1(\mathbb{R},\mathcal{B}(\mathbb{R}),Q) \\
& f & \mapsto & \int f(x)\, N(\cdot,\sigma^2)(\mathrm{d}x).
\end{array}
\end{equation*}
I am interested in the eigenspace $E_1 := \mathrm{kernel(I-R)},$ in particular in the dimension of $E_1.$
Obviously, the indicator function $\mathbb{1}_{\mathbb{R}}: x \mapsto 1$ and the identity $\mathrm{id}_{\mathbb{R}}: x \mapsto x$ are both eigenfunctions to the eigenvalue $1,$ that is, $\mathbb{1}_{\mathbb{R}},\ \mathrm{id}_{\mathbb{R}} \in E_1.$
Are there more linearly independent eigenfunctions?
If I understand correctly, your operator $R$ is the convolution operator with the Gauss–Weierstrass kernel. This is a Fourier multiplier with symbol $\lambda(\xi) = \exp(-\tfrac{1}{2} \sigma^2 |\xi|^2)$:
$$ \widehat{R f}(\xi) = \lambda(\xi) \hat f(\xi). $$
If $f$ is a tempered distribution, then $R f = f$ if and only if $$(\exp(-\tfrac{1}{2} \sigma^2 |\xi|^2) - 1) \hat{f}(\xi) = 0 ,$$ and this is equivalent to $\hat{f}$ being supported in $\{0\}$. This, in turn, implies that $f$ is a polynomial. By inspection, in dimension $1$ the eigenspace is indeed spanned by $f(x) = 1$ and $f(x) = x$. In higher dimensions, however, any harmonic polynomial (i.e. a polynomial $f$ such that $\Delta f = 0$) will do.
If one goes beyond tempered distributions, there are more solutions, even in dimension one. For example, $f(x) = e^{z x}$ is an eigenfunction corresponding to eigenvalue $$\lambda(z) = \exp(-\tfrac{1}{2} \sigma^2 z^2),$$ where $z$ is an arbitrary complex number. Choosing, for example, $$z = \sqrt{2 \pi} \sigma^{-1} (1 + i),$$ we get $\lambda(z) = \exp(-2 \pi i) = 1$, as desired.
If one insists on real-valued solutions, then $f(x) = \Re e^{z x} = e^{x \Re z} \cos(\Im z)$ works (as long as $\lambda(z)$ is real). Thus, to give a specific real-valued example,
$$ f(x) = e^{\sqrt{2 \pi} x / \sigma} \cos(\sqrt{2 \pi} x / \sigma) $$
is another eigenfunction with eigenvalue $1$.
Thank you! Since I am not trained and experienced in distribution theory (and Fourier transforms of distributions), I cannot see that the Fourier transform of $f(x)=x$ for $x \in \mathbb{R},$ that is, essentially $\delta'$ (correct?) is supported in ${0}.$ What is the difference to polynomials of higher order? Do you know any good reference for these basics?
The Fourier transform of $x$ is indeed $i \delta_0'$, and more generally, the Fourier transform of a polynomial $P(x)$ is $P(-i\partial_x) \delta_0$. I am not sure I have a good reference; Vladimirov's Methods of the Theory of Generalized Functions is one of the standard references, I think.
|
2025-03-21T14:48:31.052247
| 2020-05-23T16:21:32 |
361164
|
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|
Stack Exchange
|
An elementary proof of Davies' inequality
In the paper Lipschitz continuity of functions of operators in the Schatten classes, Davies proved the following matrix inequality.
Let $a_i,b_i>0$ for $1\leq i\leq n$ and $A$ be an $n\times n$ matrix, and
$B_{ij}=\frac{a_i-b_j}{a_i+b_j}A_{ij}$.
Then it holds that $\|B\|_p\leq \gamma_p\|A\|_p$ for $1<p<\infty$ and
$\gamma_p=\begin{cases}cp \quad\quad\quad\quad\quad\mbox{if $2\leq p<\infty$}\\ cp/(p-1) \quad\quad\mbox{if $1<p\leq 2$}\end{cases}$,
where $c\geq 1$ is an absolute constant and $\|\cdot \|_p$ is the Schattern $p$-norm.
The proof crucially relies on a fact about Volterra operators in Hilbert space in an old book "Theory and Applications of Volterra Operators in Hilbert space".
Is there a more elementary proof about this inequality? "An elementary proof" means a matrix-analytic proof, which only involves operators in finite dimensional space, but not the operators in infinite dimensional Hilbert space, like Volterra operators.
|
2025-03-21T14:48:31.052351
| 2020-05-23T16:25:26 |
361165
|
{
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"Brian Hopkins",
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|
Stack Exchange
|
Is there a 4-polytope without 3-gonal and 4-gonal faces, other than the 120-cell?
The question is in the title:
Question: Is there any 4-dimensional polytope without 3-gonal and 4-gonal faces (of dimension two), other than the 120-cell?
I consider only convex polytopes (convex hull of finitely many points) that are full-dimensional (not contained in a proper subspace).
And I consider a polytope to be distinct from the 120-cell if it has a non-isomorphic face-lattice.
It is known that any 4-polytope must have a 3-gonal, 4-gonal or 5-gonal face of dimension two.
The 120-cell has only 5-gonal faces of dimension two.
"other than" means: not isomorphic as polyhedral complex? (this is a reasonable isomorphism notion; an a priori stronger one would be being isotopic, i.e., have a continuous deformation from one to another)
@YCor Yes, thanks. I edited that into the question.
Did you find who proved that the only possible faces are 3-, 4-, or 5-gonal? (An earlier version requested a citation for that result.)
@BrianHopkins I still don't have a source, but I realized the following: one can show (via a standard double counting arguments) that a planar graph has a vertex of degree 5 or smaller, or equivalently (considering its dual), a 3-gonal, 4-gonal or 5-gonal face. Since the edge-graph of a (3-dimensional) polyhedron is a planar graph, this proves it in dimension three. This then carries over to higher dimensions by considering the 3-faces of the polytopes.
I found this highly relevant question with an equivalently relevant answer.
There are other polytopes. To construct one let's do the following. Remember first that in the hyperbolic $4$-space there exists a regular compact right-angled 120-cell. Here, right-angled means that any two adjacent faces intersect under angle $\frac{\pi}{2}$. Regular means, that all the faces are isomeric, and the polytope has the same group of self-isometries as the Euclidean 120-cell. This polytope is discussed, for example, in
https://pdfs.semanticscholar.org/a0eb/ccbed0687d966a9aaaac2f370bc930a556be.pdf
at the bottom of page 65. The references to more classical articles are given there.
Now, if we double it in one face then we get a new convex polytope, and it is not hard to see, that it doesn't have 2-faces that are triangles and quadrilaterals. But any convex hyperbolic polytope is also combinatorially equivalent to a Euclidean one.
More generally, you can take any compact right-angled hyperbolic polytope in $\mathbb H^4$. Since it is hyperbolic and right-angled, it can not have $2$-faces that are triangles of quadrilaterals. And there is a infinite number of such polytopes in dimension 4. Each of them gives a Euclidean one as well.
Thank you for your answer. I am not familiar with some terminology: 1. what is this "regular compact right-angled 120-cell" in hyperbolic 4-space. Is it like a tiling of 4-space? I wasn't able to find out what a "convex hyperbolic polytope" is via google. 2. What does it mean to "double it on one face"?
I added a reference. Yes, starting with such a polytope you get a tiling of $\mathbb H^4$, in the same way as cubes tile $\mathbb R^n$. And what I say is that you need to take the union of two tiles that share a common hyperface to get a new polytope as you want. This can be continued as far as you want.
Dmitri's answer is definitely correct. I just want to add my geometric intuition, and a generalization, which, in hindsight, is quite obvious.
All in all, we can have the following:
If $P\subset\Bbb R^d$ is a polytope with $n$ facets, each of which is combinatorially (or projectively) equivalent to $Q\subset\smash{\Bbb R^{d-1}}\!$, then for each $k\ge 1$ there also exists a polytope $P_k\subset\Bbb R^d$ with $k(n-2)+2$ facets, all of which are combinatorially (or projectively) equivalent to $Q$.
With this, it should be clear that there are many 4-polytopes with only 5-gonal 2-faces.
The main idea is visualized below.
Construction:
Fix a face $\sigma\subset P$.
Let $P'$ be the polytope obtained from $P$ by applying a certain projective transformation that a) fixes $\sigma$, and b) moves all vertices of $P$ "beyond" $\sigma$ (see the image). This construction is related to the idea behind the Schlegel diagram, in particular, this transformations always exists.
Glue $P'$ and $P$ on their common face isomorphic to $\sigma$ (if we have chosen the correct transformation in 2., then this is a convex polytope).
Repeat this to obtain as many $Q$-facets as you like.
Still, it might be interesting to determine the atomic $Q$-facetted polytopes, i.e. those, which are not "stacked" in the sense above.
|
2025-03-21T14:48:31.052703
| 2020-05-23T19:09:19 |
361175
|
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|
Stack Exchange
|
Generalized tensor-train decomposition
If $U\in\bigotimes_{k=1}^p\mathbb R^{n_k}$ and $U^{(k)}\in\mathbb R^{r_{k-1}}\otimes\mathbb R^{n_k}\otimes\mathbb R^{r_k}$ with$^1$ $$U_{i_1,\:\ldots\:,i_p}=\sum_{j_0=1}^{r_0}\cdots\sum_{j_p=1}^{r_p}U^{(1)}_{j_0i_1j_1}\cdots U^{(p)}_{j_{p-1}i_pj_p}\tag1,$$ then $(1)$ is called a tensor-train decomposition of $U$. $(1)$ has to be understood with respect to the isomorphisms $\bigotimes_{k=1}^p\mathbb R^{n_k}\cong\mathbb R^{n_1\times\cdots\times n_p}$ and $\mathbb R^{r_{k-1}}\otimes\mathbb R^{n_k}\otimes\mathbb R^{r_k}\cong\mathbb R^{r_{k-1}\times n_k\times r_k}$.
Is there a generalization of this decomposition type to general $\mathbb R$-vector spaces (or Banach/Hilbert spaces) $E_i,F_j$? Something like $U\in\bigotimes_{k=1}^p E_k$, $U^{(k)}\in F_{k-1}\times E_k\times F_k$, $$U=\iota\bigotimes_{k=1}^pU^{(k)}\tag2,$$ where some identification/transformation $\iota$ is applied to the right-hand side?
I'm quite sure that it should be possible to write $(1)$ in the form of $(2)$ using a suitable tensor contraction$^2$. I'm strongly interested in a way to express $(1)$ without the identification with matrices.
EDIT 1: Clearly, we have the canonical isomorphisms, \begin{equation}\begin{split}\Xi_k:\mathbb R^{r_{k-1}\times n_k\times r_k}&\to\mathbb R^{r_{k-1}}\otimes\mathbb R^{n_k}\otimes\mathbb R^{r_k},\\\left(a_{j_{k-1}i_kj_k}\right)&\mapsto\sum_{j_{k-1}=1}^{r_{k-1}}\sum_{i_k=1}^{n_k}\sum_{j_k=1}^{r_k}a_{j_{k-1}i_kj_k}\left(f^{(k-1)}_{r_{k-1}}\otimes e^{(k)}_{i_k}\otimes f^{(k)}_{j_k}\right),\end{split}\end{equation} where $\left(e^{(k)}_1,\ldots,e^{(k)}_{n_k}\right)$ and $\left(f^{(k)}_1,\ldots,f^{(k)}_{r_k}\right)$ denote the standard bases of $\mathbb R^{n_k}$ and $\mathbb R^{r_k}$, respectively, and $$\Lambda_k:\mathcal L(\mathbb R^{n_k},\mathbb R^{r_{k-1}}\otimes\mathbb R^{r_k})\to\mathbb R^{r_{k-1}}\otimes\mathbb R^{n_k}\otimes\mathbb R^{r_k}\;,\;\;\;A\mapsto\sum_{i_k=1}^{n_k}e^{(k)}_i\otimes Ae^{(k)}_i.$$ Now, it's easy to see that $$\Lambda_k^{-1}\left(U^{(k)}\right)=\sum_{i_k=1}^{n_k}\left\langle\;\cdot\;,e^{(k)}_{i_k}\right\rangle\operatorname{tr}_{21}\left(U^{(k)}\otimes e^{(k)}_{i_k}\right)\tag3.$$ So, $$\Lambda_k^{-1}\left(U^{(k)}\right)(x_k)=\operatorname{tr}_{21}\left(U^{(k)}\otimes x_k\right)\;\;\;\text{for all }x_k\in\mathbb R^{n_k}.\tag4$$ Does this allow us to rewrite $(1)$?
EDIT 2: I guess the natural generalization is the following: Let $E_i$ be a $\mathbb R$-vector space, $I$ be a finite nonempty set and $U\in\bigotimes_{i\in I}E_i$. Let $\langle\;\cdot\;,\;\cdot\;\rangle_{\bigotimes_{i\in I}E_i,\:\bigotimes_{i\in I}E_i^\ast}$ denote the canonical duality pairing between $\bigotimes_{i\in I}E_i$ and $\bigotimes_{i\in I}E_i^\ast$. Then we may look for a factorization of $$\left\langle U,\bigotimes_{i\in I}\varphi_i\right\rangle_{\bigotimes_{i\in I}E_i,\:\bigotimes_{i\in I}E_i^\ast},\tag5$$ where $\varphi_i\in E_i^\ast$ for $i\in I$. In the case of $E_i=\mathbb R^{n_i}$ we obtain the matrix entry $U_{i_1,\:\ldots\:,i_p}$ by inserting for $\varphi_{i_k}$ the basis functional corresponding to the $i_k$th standard basis vector of $\mathbb R^{n_k}$. Now, I guess the cores are naturally assumed to be some $U^{(i)}\in F_{i-1}^\ast\otimes E_i\otimes F_i$, where $F_i$ is another $\mathbb R$-vector space. I guess we need to build the trace $$\operatorname{tr}\left(\bigotimes_{i\in I}\varphi_i\otimes U^{(i)}\right)\in\bigotimes_{i\in I}F_{i-1}^\ast\otimes F_i\subseteq\bigotimes_{i\in I}\mathcal L(F_{i-1},F_i).$$ Now I think the operators induced by the embedded on the right-hand side only need to be concatenated (which might be express in terms of the tensor constraction as well; see my related question Tensor contraction (vector-valued trace) on $\bigotimes_{i=1}^k\mathcal L(E_{i-1},E_i)$) and we need to assume that $\dim F_0=\dim F_p=1$.
It would be great if someone could put these pieces together to formulate a generalization.
$^1$ $r_0=r_p=1$.
$^2$ Let $I,J\subseteq\mathbb N$ be finite and nonempty and $E_i,F_j$ be $\mathbb R$-vector spaces for $(i,j)\in I\times J$. If $I_0\subseteq I$ and $J_0\subseteq J$ with $|I_0|=|J_0|\ge1$, $\phi:I_0\to J_0$ denotes the canonical bijection and $$E_i=F_{\phi(i)}^\ast\;\;\;\text{for all }i\in I_0\tag6,$$ then the tensor contraction (or vector-valued trace) with respect to $(I_0,J_0)$ is given by \begin{equation}\begin{split}\operatorname{tr}_{I_0J_0}:&\bigotimes_{k\in I}E_k\otimes\bigotimes_{l\in J}F_l\to\bigotimes_{k\in I\setminus I_0}E_k\otimes\bigotimes_{l\in J\setminus J_0}F_l,\\&\bigotimes_{k\in I}x_k\otimes\bigotimes_{l\in J}y_l\mapsto\prod_{i\in I_0}\left\langle x_i,y_{\phi(i)}\right\rangle_{F_{\phi(i)}}\bigotimes_{k\in I\setminus I_0}x_k\otimes\bigotimes_{l\in J\setminus J_0}y_l.\end{split}\end{equation} If $(i,j)\in I\times J$, then $\operatorname{tr}_{ij}:=\operatorname{tr}_{\{i\}\{j\}}$.
Sure. One way is to define the cores in the tensor train as matrix valued functions. The term to search for is "functional tensor train"
@NickAlger Thank you for the search term. Do you have an explicit reference at hand which is particularly suitable for my purpose?
@NickAlger Please note that I've edited the question once again.
@0xbadf00d: your question is close to unreadable. To remedy this you need to use a graphical/diagrammatic notation as discussed in this MO question https://mathoverflow.net/questions/168888/who-invented-diagrammatic-algebra It's OK to write by hand, scan and insert the picture in the MO question.
I'm not sure about your approach. The conventional way to extend tensor trains to Hilbert spaces (or more generally) is as follows. One may identify tensors with multilinear functions,
$T:H_1 \times H_2 \times \dots H_p \rightarrow \mathbb{R}$, where $H_i$ are Hilbert spaces. A tensor train is then a sequence of functions
$$
G_i : H_i \rightarrow \mathbb{R}^{r_i \times r^{i+1}}, \quad i=1,\dots,p
$$
where $r_1=1$ and $r_p=1$ such that
$$
T(u_1, u_2, \dots, u_p) = G(u_1)G(u_2) \dots G(u_p)
$$
where the $G_i(u_i)$ are multiplied with usual matrix multiplication.
Here is a paper:
Gorodetsky, Alex, Sertac Karaman, and Youssef Marzouk. "A continuous
analogue of the tensor-train decomposition." Computer Methods in
Applied Mechanics and Engineering 347 (2019): 59-84. https://arxiv.org/pdf/1510.09088.pdf
So the cores $G_i$ are infinite in the externally facing mode, but finite in the internally facing modes. If you want to allow the internal modes of the cores to be infinite, it would be pretty straightforward to replace the matrices $G_i(u_i)$ with linear operators between some suitable spaces.
|
2025-03-21T14:48:31.053206
| 2020-05-23T21:01:38 |
361180
|
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|
Stack Exchange
|
Any comparison between the category of cubes and its opposite?
To model topological spaces combinatorially, one can use simplicial sets -- or cubical sets. Simplicial sets are defined as presheaves on the simplex category $\Delta$, the category of non-empty finite ordered sets and order-preserving maps. In general $\Delta^{op}$ and $\Delta$ are quite different, but there is the following observation. Let $\Delta_{alg}$ be the category of all ordered finite sets (i.e. including the empty one). Then $\Delta_{alg}^{op}$ can be seen as a non-full subcategory in $\Delta^{alg}$ -- of non-empty sets with maps that preserve endpoints.
Now consider $\square$, the category of cubes (in any of its version, with connections or without). Is there any similar comparison between $\square$ and $\square^{op}$?
|
2025-03-21T14:48:31.053290
| 2020-05-23T21:16:00 |
361182
|
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|
Stack Exchange
|
Classes of Banach algebras (that aren't operator algebras) whose bidual comes from a "universal representation"
Are there any classes of (Arens regular) Banach algebras that are not operator algebras whose bidual comes from a “universal representation”, as in the case of C*-algebras?
This sort of depends on what you mean by "universal representation". For $C^*$-algebras, I think the statement is usually the following: Given a $C^*$-algebra $A$ and a representation $\pi:A\rightarrow B(H)$, let $M(\pi) = \pi(A)''$ denote the von Neumann algebra generated by $\pi(A)$. There is a unique surjective normal $*$-homomorphism $\tilde\pi:A^{**}\rightarrow M(\pi)$. (Here I followed Takesaki's book.)
For Banach algebras, we need to decide what the analogue of "von Neumann algebra" is, and perhaps what "normal" should mean.
One solution is to focus on dual Banach algebras, that is, Banach algebras $A$ for which there is some predual $A_*$ making the multiplication separately weak$^*$-continuous. Then "normal" means weak$^*$-continuous. Such algebras have long been studied, with Runde undertaking a lot of study. In my paper Dual Banach algebras: representations and injectivity arXiv:math/0604372 I study representations of dual Banach algebras on reflexive Banach spaces (using ideas of Young). Typically I fail to make a clean statement, but we have:
If $E$ is a reflexive Banach space then $B(E)$ is a dual Banach algebra with predual $E\widehat\otimes E^*$, and so any weak$^*$-closed subalgebra of $B(E)$ is a dual Banach algebra.
By Corollary 3.8 every dual Banach algebra is isometrically weak$^*$-weak$^*$-continuously isomorphic to a weak$^*$-closed subalgebra of $B(E)$ for some reflexive $E$.
There is a closed submodule $\newcommand{\wap}{\operatorname{wap}}\wap(A^*)$ of $A^*$ which is maximal so that the quotient map $A^{**}\rightarrow\wap(A^*)^*$ induces a single product on $\wap(A^*)^*$ for either Arens product. In particular, $A$ is Arens regular if and only if $\wap(A^*)=A^*$.
$\wap(A^*)^*$ is hence a dual Banach algebra, and it satisfies the universal property we hope for: see Proposition 2.9 (due to Runde). If $B$ is any dual Banach algebra, a homomorphism $A\rightarrow B$ admits a unique extension a weak$^*$-continuous homomorphism $\wap(A^*)^*\rightarrow B$.
So let $A$ be Arens regular, and consider a representation $\pi:A\rightarrow B(E)$ for some reflexive $E$. Then we obtain a unique extension $\tilde\pi:\wap(A^*)^* = A^{**} \rightarrow \overline{\pi(A)}^{w^*}$. If $\pi$ induces a bounded below map $A/\ker\pi\rightarrow B(E)$ then $\tilde\pi$ will be surjective (from Hahn-Banach). In general, I don't see that $\tilde\pi$ will be surjective.
To better answer the original question: "yes". If we take $\pi:A\rightarrow B(E)$ to be a "maximal" representation on a reflexive $E$ (e.g. the sum of all cyclic representations, "cyclic" chosen to obtain a set not a proper class) then $\pi$ is an isometry (follows from Theorem 3.6, also from Young's work) and $\tilde\pi:A^{**}\rightarrow B(E)$ is an isometry, so $A^{**}$ is the weak$^*$-closure of $\pi(A)$.
(In the last few years, Thiel and Gardella have studied similar things, but I don't have an exact reference.)
A second approach is to take a more permissive approach to what a "representation" is: just any bounded homomorphism $\pi:A\rightarrow B(E)$ for any Banach space $E$. This is the same notation as a (bounded) left $A$-module, and I will sometimes use the module language. Any $A$ has an isometric representation: let $E=A\oplus_1\mathbb C$ be the unitisation, and $\pi$ the left-regular representation, so $a\cdot (b,\alpha) = (ab+\alpha a, 0)$ for $a,b\in A, \alpha\in\mathbb C$. But $B(E)$ carries no (natural) weak$^*$-topology.
Instead, we could use $E=A^{**}\oplus_1\mathbb C$ with the predual $A^*\oplus_\infty\mathbb C$, and the left-regular representation. Then the predual becomes a right module for the action $(a^*,\beta)\cdot a = (a^*\cdot a, \langle a^*,a \rangle)$ with the usual action of $A$ on $A^*$. Then $B(E)$ does not become a dual Banach algebra, but it does become a one-sided dual Banach algebra (see work of Spain, and Thiel and Gardella, for more on this notion). Furthermore, there is a standard way to extend the action of $A$ on $E$ to an action of $A^{**}$ which turns $E$ into a left $(A^{**},\Box)$ module, where $\Box$ is the first Arens product. A check shows that this agrees with the weak$^*$-extension of the action of $A$, and the image in $B(E)$ is $A^{**}$. Thus again $A^{**}$ arises as the weak$^*$-closure of a "universal" representation of $A$.
The difference here is that this only works with the first Arens product (a similar construction uses the second Arens product) but we don't need Arens regularity.
Thanks for the detailed answer. How do you get that $\tilde{\pi} : A^{**} \to B(E)$ is an isometry? The usual proof in the C*-algebra case uses the Kaplansky Density Theorem, and the simplest alternative proof I can think of uses the polar decomposition of normal functionals.
Does Lemma 3.4 in your Dual Banach Algebras paper suffice as a replacement for polar decomposition?
Yes, exactly that! Lemma 3.4 gives you the norm control you need.
|
2025-03-21T14:48:31.053603
| 2020-05-24T00:58:33 |
361187
|
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Stack Exchange
|
Categorical Kähler differentials and the Leibniz rule
From nlab, the module of Kähler differentials over some category $\mathcal{C}$ is the free functor:
$$\Omega: \mathcal{C} \to \mathsf{Mod_{\mathcal{C}}}$$
left-adjoint to the (forgetful) embedding:
$$u: \mathsf{Mod}_{\mathcal{C}} \cong \mathsf{Ab}(\mathcal{C}) \hookrightarrow \mathcal{C}$$
with $\mathsf{Ab}(\mathcal{C})$ denoting abelian group objects of $\mathcal{C}$, and $\mathsf{Mod}_{\mathcal{C}}$ denoting the category of modules of $\mathcal{C}$. By this definition, we automatically get a bijection of hom-sets:
$$\mathsf{Mod_{\mathcal{C}}}(\Omega(R), M) \cong \mathcal{C}(R, u(M))$$
Also, according to the same nlab page as above, $R$-derivations taking values in some $R$-module $M$ (with $R$ some object of $\mathcal{C}$) are morphisms:
$$d: \Omega(R) \to M$$
in $\mathsf{Mod}_R \cong \mathsf{Mod}_{R/\mathcal{C}}$. Thus, they could be identified by $\mathcal{C}$-morphisms:
$$X: R \to u(M)$$
because of the adjunction $\Omega \dashv u$.
At this point, I have two questions:
When $\mathcal{C} = \mathsf{CRing}$, the category of commutative and unital rings, do we automatically get the Leibniz/product rule ? Why or why not ?
If we do automatically get the Leibniz rule, then is it also the case in categories more general than $\mathsf{CRing}$ ?
Thank you.
1. The Leibniz rule follows immediately from the last description
of derivations as morphisms of commutative rings X:R→u(M).
Indeed, u(M) is the square-zero extension of some R-module M'
(in the traditional sense), i.e., u(M)=R⊕M'.
Now a morphism of commutative rings f:R→R⊕M' in the slice category C/R
(not in C, as is claimed in the main post) necessarily has the form
r↦(r,φ(r)), for some linear map φ.
Since f is a homomorphism, we have f(1)=(1,φ(1))=(1,0), so φ(1)=0.
Also $$(rr',φ(rm'+r'm))=f(rr',rm'+r'm)=f((r,m)(r',m'))=f(r,m)f(r',m')=(r,φ(m))(r',φ(m'))=(rr',rφ(m')+r'φ(m)),$$
so $$φ(rm'+r'm)=rφ(m')+r'φ(m).$$
This is precisely the Leibniz rule.
2. Yes, for instance, this is true for algebras over Fermat theories.
See Carchedi and Roytenberg's Homological Algebra for Superalgebras of Differentiable Functions.
Why do we know that $u(M)$ is the square-zero extension of some $R$-module $M'$ ?
@DatMinhHa: This is explained in the link to the nLab that you yourself cited in the main post: https://ncatlab.org/nlab/show/K%C3%A4hler+differential#the_correct_definition_of_the_notion_of_module_
|
2025-03-21T14:48:31.053798
| 2020-05-24T01:36:29 |
361190
|
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|
Stack Exchange
|
Conjugacy of elements in a parabolic subgroup
Let $G$ be a complex connected reductive group, and let $P \subseteq G$ be a parabolic subgroup. My question is the following: if $g$ and $h$ are elements of $P$ which are conjugate as elements of $G$, are they necessarily also conjugate in $P$?
Edit: In order to rule out the cases pointed out in the comments, I'm refining my question.
Let $U \subset P$ be the unipotent radical, and $L = P/U$ the reductive quotient. Let $g$ and $h$ be elements of $P$, which project to the same element in $L$, and which are conjugate as elements of $G$. Are they then necessarily also conjugate in $P$?
Further Edit:
If $g$ and $h$ are elements of $P$ which project to the same element of $L$, and if $g$ and $h$ both lie in some (but not necessarily the same) Levi subgroup, then they are conjugate in $P$. This is because all Levi subgroups are conjugate by an element of $U$, and so we can conjugate $g$ and $h$ into the same Levi (without changing their projection), and then they must be equal. This implies that without loss of generality, we can assume that $g$ and $h$ have the same semisimple component.
Now there are counter-examples if we do not require $h$ and $g$ to lie in some Levi. For example, let $G = GL(3,\mathbb{C})$, and $P$ to be the subgroup of upper triangular matrices. Then the following two elements are conjugate in $G$ but not in $P$, and both project to the same element of the Levi:
$\begin{pmatrix}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \ \ \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$.
To rule out this case, we should then also require that $h$ lies in some Levi. The question is then whether the conjugate element $g$ must also lie in some (possibly different) Levi. By the above discussion, this would imply that $h$ and $g$ are conjugate in $P$.
No: if $P$ is Borel and $T$ a maximal torus, two elements of $T$ are conjugate in $P$ iff they're equal. But could be conjugate in $G$ (Weyl group action).
@YCor, that is certainly true for regular elements of $T$, but is it true for any two elements?
@LSpice in a Borel $B$, $T$ is a retract, so two elements of $T$ are conjugate in $B$ iff they're conjugate in $T$. Since $T$ is abelian, this means being equal.
@YCor okay what if we also assume that g and h project to the same element in the reductive quotient of P.
@unknownymous it might be a reasonable; please edit your question accordingly? if you know some particular cases (e.g., when $g,h$ belong to the unipotent radical of $P$) could you mention it too?
You're writing "if $g$ and $h$ are elements of $P$ which project to the same element of $L$, and if $g$ and $h$ both lie in some Levi subgroup, then they are conjugate in $P$". I agree, but it seems it even implies that that $g=h$. Indeed if $L$ is a Levi and $L'=sLs^{-1}$ is another Levi, and $p_L$ is the projection to the Levi $L$, then $p_{L'}(x)=sp_L(x)s^{-1}$ for all $x$. Hence the condition $p_L(x)=p_L(y)$ does not depend on $L$. In particular it holds and if in addition $x,y\in L'$, this forces $x=y$.
@YCor the Levi subgroups are conjugate by elements of the unipotent radical, and this does not change the projection to the quotient. So for example, take $h$ unipotent and lying in some Levi. Then for appropriately chosen element $u$ from the unipotent radical, $g = uhu^{-1}$ is not equal to $h$, but projects to the same element.
@unknownymous well, in my comment I was viewing the projection $p_L$ to $L$ as a map onto $L$, so $p_L\neq p_{L'}$ if $L\neq L'$.
@YCor ohh I think I wrote the statement in a confusing way. I meant to say that $g$ and $h$ lie in possibly different Levi subgroups.
I think your question has a negative answer.
For reductive $G$, there are only finitely many conjugacy classes of unipotent elements in $G$. However, for a parabolic subgroup $P < G$ with unipotent radical $U$, usually the action of $P$ on $U$ by conjugation has an infinite number of orbits.
Here is an explicit example from [1]. Let $G = \operatorname{GL}_6(K)$ with $K$ an infinite field and let $P$ be the parabolic (Borel) subgroup formed by the lower triangular matrices. (This also gives you examples for $G = \operatorname{GL}_n(K)$ for any $n > 6$). For $\alpha \in K$ define $$u_{\alpha} = \begin{pmatrix}
1 & & & & & \\
1 & 1 & & & & \\
0 & 0& 1 & & & \\
0 & 1 & 1 & 1 & & \\
0 & \alpha & 0 & 0 & 1 & \\
0 & 0 & 0 & 1 & 1 & 1 \end{pmatrix}$$ A calculation shows that $u_{\alpha}$ and $u_{\beta}$ are $P$-conjugate if and only if $\alpha = \beta$. On the other hand if $\alpha, \beta \not\in \{ 0,-1 \}$, then $u_{\alpha}$ and $u_{\beta}$ are $G$-conjugate.
For the question of when a parabolic subgroup $P$ has finitely many orbits on $U$, see for example [2], [3], [4] below and references therein.
[1] Djoković, Dragimir Ž.; Malzan, J. Orbits of nilpotent matrices. Linear Algebra Appl. 32 (1980), 157–158.
[2] Hille, L.; Röhrle, G. A classification of parabolic subgroups of classical groups with a finite number of orbits on the unipotent radical. Transform. Groups 4 (1999), no. 1, 35–52.
[3] Jürgens, U.; Röhrle, G. MOP—algorithmic modality analysis for parabolic group actions. Experiment. Math. 11 (2002), no. 1, 57–67.
[4] Goodwin, S.; Röhrle, G. Finite orbit modules for parabolic subgroups of exceptional groups. Indag. Math. (N.S.) 15 (2004), no. 2, 189–207.
Very interesting example! I give another counter example in my question. But I've refined my question once again to rule out these counter examples.
|
2025-03-21T14:48:31.054141
| 2020-05-24T01:52:06 |
361191
|
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|
Stack Exchange
|
Applications of model categories
I was wondering if someone could explain some of the concrete applications of model categories. My possibly naive understanding of the motivation is that one wants to mimic the category of topological spaces in some sense or to define a homotopy theory for a category.
For example, more concretely on the Wikipedia page it is stated that the category of chain complexes of $R$-modules for some commutative ring $R$ is a model category and that homology can be viewed as a type of homotopy which allows generalisation of homology to objects such as groups and $R$-algebras. Is there some reference where there is explained a bit more? How do model categories allow one to generalise homology?
Could you focus the question? "[What are] concrete applications of model categories" and "How do model categories allow one to generalise homology" seem to go in different directions...
Arguably, the primary concrete application is to model all sorts of computations with the underlying (∞,1)-category by performing computations with the underlying 1-category. This includes deriving functors of all sorts, in particular, limits and colimits, monoidal product and internal hom, etc. This is explained in detail here: https://ncatlab.org/nlab/show/homotopy+theory+FAQ
"one wants to mimic the category of topological spaces in some sense or to define a homotopy theory for a category": I am not sure what the mimicking process refers to. One does not need model categories to define homotopy categories, though, since merely having weak equivalences is already sufficient.
Hi Dmitri, yes this looks like what I had in mind, thanks for this.
There are many references where model categories, and their connection to homology, are described more. See this MO question for a list. For the example of $Ch(R)$, there are several model structures. Those that have quasi-isomorphisms as weak equivalences capture homological algebra (e.g., a morphism $f_*: C_* \to D_*$ is a weak equivalence if the map on homology $H_*(f)$ is an isomorphism). Similarly, there are model structures on the category of topological spaces, either for classical homotopy theory (where $f$ is a weak equivalence if all $\pi_*(f)$ are isomorphisms) or for homology (where $f$ is a weak equivalence if all $H_*(f)$ are isomorphisms). Indeed, this can be done for generalized homology theories, and the machinery of Bousfield localization allows you to change your model structure to focus on an enlarged class of weak equivalences in this way.
When the wikipedia page mentioned groups and $R$-algebras, it probably had in mind Quillen's early work producing model structures for simplicial groups and simplicial $R$-algebras. In these cases, the homotopy theory is lifted from the category of simplicial sets (a model for topological spaces), and the Dold-Kan correspondence gives an equivalence $Ch_{\geq 0}(\mathcal{A}) \simeq Fun(\Delta^{op},\mathcal{A})$ for an Abelian category $\mathcal{A}$. This is very clearly described in a note by Akhil Mathew. The homotopy theory of homological algebra on one side is equivalent to the homotopy theory coming from spaces on the other side, justifying the sentence on wikipedia.
There are also model structures on the category of groups (e.g., the trivial model structures), but they are boring. There are model structures on $R$-algebras for entirely different homotopy theories related to representation theory, e.g., you can start with the model structure on $R$-modules whose homotopy category is the stable module category (described in Hovey's book among other places), and can lift it to a model structure on $R$-algebras, as I wrote about here. Homotopy in these cases is about having the same representation theory in the stable module category (i.e., after quotienting out boring representations). This is probably not what wikipedia had in mind, but provides lots of applications of model categories.
There's far too much to possibly fit in a single MO answer, so I encourage you to read the references linked above, and ask more targeted questions as you go. Model categories have lots of applications, going far beyond "just" modeling computations in the underlying $(\infty,1)$-category (though, surely, this is an important application of the theory). For instance, when Javier Gutierrez and I proved the Blumberg-Hill conjecture using model categories, this really required the cofibrations, rather than only the weak equivalences. I don't know how to carry out that proof on the level of $(\infty,1)$-categories. The same is true of my recent work with Michael Batanin proving a generalized form of the Baez-Dolan Stabilization Hypothesis.
Hi David, thanks for this. I'm wondering for example, one can have homological representations of certain types of group (for example, braid groups act on the homology of topological spaces obtained from punctured discs via functor constructions), might it be possible to use model categories to generalise these constructions and obtain new representations for those groups?
That's an interesting question. I would need to think carefully about it. My first instinct was "no: model categories are for organizing information, not obtaining new representations" but actually I can imagine a framework where you could create new representations via categorical procedures. I'm curious to think about encoding braid group actions on homology in an abstract framework. I don't have enough time now to think about this (textbook deadline soon), but maybe in a few weeks.
In your previous comment you mentioned categorical procedures, do you mean categorical methods as compared to model categories?
I guess both, but of course you only need the model structure for the latter. Like, you could create new representations from old via pushouts or other colimits (categorical) or via cofibrant replacement (model category technique). All of this is speculative since we don't actually have a specific model structure in mind.
|
2025-03-21T14:48:31.054515
| 2020-05-24T02:22:45 |
361194
|
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|
Stack Exchange
|
Smoothening pseudo-Anosov flows
A topological Anosov flow on a closed 3-manifold can be replaced by a smooth Anosov flow using an argument of Fried: use Markov partitions to find a surface of section, put in other terms, one can blow up some closed orbits so that the flow is a suspension of a pseudo-Anosov map on a surface with boundary. Then take a smooth representative of this pseudo-Anosov map within its isotopy class, and blow down the orbits to get back the original manifold.
Is this statement (or some analogue of it) still true for pseudo-Anosov flows? The above argument doesn't work anymore since the singular orbits cannot lie in the interior of a Markov rectangle, so one cannot find a surface of section near these singular orbits in the same way. Is an alternate way of thinking about all of this that generalizes more easily to pseudo-Anosov flows? Any help is appreciated!
In fact, the argument by Fried that you describe is incomplete. It is unclear that such a scheme can be made to work and there is evidence that it may not work (why should the blow down be smooth, and if it were, why would it be a smooth Anosov flow rather than a smooth topological Anosov flow).
However, the result has been clarified and rigourously proved recently in Mario Shannon's thesis http://www.theses.fr/2020UBFCK035#
It is likely that his techniques also respond to your problem and are the technique you seem to be looking for.
|
2025-03-21T14:48:31.054772
| 2020-05-24T04:07:53 |
361195
|
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|
Stack Exchange
|
Boundedness of total current in electrical network
Consider the following symmetric matrix (adjacency matrix):
$$A=(a_{ij})_{1\leq i,j\leq n}$$
such that $a_{ij}=a_{ji}, a_{ii}=0$ and $a_{ij}=0$ for $|i-j|\geq k$ where $k\geq3$. We also have $1\leq a_{ij}\leq2, 0<|i-j|<k$. This just means that the adjacency matrix $A$ has a banded structure, i.e. entries far away from the diagonal are 0.
Consider the solution of the following linear system:
$$\begin{cases}
(\sum_{j\neq1}a_{1j})x_1-\sum_{j\neq1}a_{1j}x_j=1\\
(\sum_{j\neq2}a_{2j})x_2-\sum_{j\neq2}a_{2j}x_j=-1\\
(\sum_{j\neq3}a_{3j})x_3-\sum_{j\neq3}a_{3j}x_j=0\\
(\sum_{j\neq4}a_{4j})x_4-\sum_{j\neq4}a_{4j}x_j=0\\
\vdots\\
(\sum_{j\neq n}a_{nj})x_n-\sum_{j\neq n}a_{nj}x_j=0
\end{cases}$$
In matrix notation, it is just $Lx=e_1-e_2$ where $L=D-A, D=diag\{d_1,...,d_n\}, d_i=\sum_{j\neq i}a_{ij}$. $L$ is just the graph Laplacian. I conjecture that there exists constant $C>0$ independent of $n$ such that
$$\sum_{i,j}a_{ij}|x_i-x_j|\leq C$$
The physical meaning of the conjecture is that, if we flow 1 unit of current from node 1 to node 2, the sum of current of each edge in the given electrical network is bounded.
Simulation result indicates that this is indeed the case. However, I somehow can only prove bound involving $n$. I think this may be related to the bandwidth of a graph?
This is also related to my previous question:
Exponential decay of voltage potential difference
I have an intuitive idea of why this is the case. Since the total current flows out of node 1 is 1, then this 1 unit of current gets split into at least $k$ parts to the neighbors of 1. For each neighbor of 1, when the current flows out, it again gets split into at least $k$ parts and this somehow forms a geometric series that is summable.
Another idea is to use induction. The physical intuition is to examine how will the total current change when we gradually add new nodes and connections to the existing network. Adding new connections will decrease the effective resistance between 1 and 2 but new current will flow in new edges, they can somehow be balanced making the total current bounded.
Edit: By request, I have added some explanations at the end. The first bullet may be helpful (it introduces a little notation). I also misread the question, and used a constant $k=3$ (instead of $k\ge 3$). This is now fixed, but $k$ must be fixed; for now the resulting bound depends on it...
Edit 2: I have added an idea on how to eliminate this issue, and (in comments) how to sharpen another bound. But the result here still depends on $k$.
Let's think of the problem as occuring on a multigraph (some edges are doubled) and interpret the Laplacian using hitting probabilities of a random walk.
The graph is more or less a long line segment, with edges between some nearby pairs of points. The vertex set is $\{1,\ldots,n\}$. Your solution $x$ is, up to a multiplicative constant $\alpha$ in $[0,1]$, the unique function $h:\{1,\ldots,n\}\rightarrow \mathbb{R}_+$ which is harmonic except at $\{1,2\}$ (by which I mean $(Lh)(k)=0$ for $k\notin \{1,2\}$) and satisfies $h(1)=0,h(2)=1$. This $h$ is nonnegative, taking values in the unit interval.
In fact, $h(i)$ is the probability that a random walk on the graph starting from $i$ will reach $2$ before $1$. You wish to prove $|h(i)-h(j)|$ decays at least exponentially in $\min(i,j)$ at a rate independent of $n$. This can be seen as follows: the probability that a random walk starting from $i$ and one starting from $j$ pass through a common vertex tends exponentially to $1$ in $\min(i,j)$, and if they pass through a common vertex with probability $\ge p$ then $|h(i)-h(j)| \le 2-2p$.
To see the last point, let's denote the random walk starting at $i$ and at $j$ by $W_i$ and $W_j$ respectively. Note that: $$|h(i) - \sum_{u=3}^n P(W_i \text{ intersects $W_j$ at vertex u and at no vertex $\ell > u$})\cdot h(k)| < P(W_i,W_j\text{ do not meet})$$ since the events "intersecting at vertex $k$ and no higher-index vertex" are mutually disjoint and each value of $h$ is in $[0,1]$ (then apply the law of total probability and the fact $h$ is a hitting probability). The same holds for $W_j$ and $h(j)$.
To see that the probability of intersecting does in fact approach $1$ exponentially, suppose $j>i$. Consider the random walk starting from $j$, and look at the first time it reaches a vertex $\ell\le i$. Then $i-\ell <k$ (where $k$ is a constant we fixed ahead of time). Now, if $i=\ell$ they do intersect; otherwise look at the next step of the random walk from $i$. It has a probability of at least $\frac{1}{4k-1}$ of landing at $\ell$ after one step. Observe this walk until the first time it reaches a vertex $i'$ smaller than $\ell$; if they do not intersect until then, then again $\ell - i' < k$, and the process continues... Thus there are at least $\min(i,j)/k$ opportunities for the random walks to intersect, each with probability at least $\frac{1}{4k-1}$ (a gross underestimate, but nevermind). This goes to $1$ exponentially in $\min(i,j)$, since the complement is at most $$\left(\frac{4k-2}{4k-1}\right)^{(\min(i,j)/k)}.$$
The case of $k$ large - an approach
First note that the multiplicative constant $\alpha \rightarrow 0$ when $k\rightarrow \infty$ - $\alpha = O(1/k)$.
Second, consider a vertex among $\{3,\ldots,k\}$. Its probability of landing at $1$ or at $2$ within one step is bounded by $4/k$. So the expected number of steps a random walk makes in $\{3,\ldots,k\}$ is bounded below by something like $k/4$, and the expected number of distinct vertices it hits is $O(k)$, with the probability of just $tk$ being hit being at most $O(t)$ for $t\in(0,\frac{1}{10})$.
Consider two random walks: let $S$ be the set of vertices the first one encounters among $\{3,\ldots,k\}$. Suppose there are $tk$ of these. Then the probability the second does not intersect it is at most $\approx(1-t/3)^s$, where $s$ is the second walk's number of steps among $\{3,\ldots,k\}$ (the $1/3$ appears because the odds of encountering a vertex among $\{3,\ldots,k\}$ need not be uniform, they depend on the matrix entries).
One can use the law of total probability to condition on $t,c$. Carrying this out to the end should show the non-intersection probability is bounded by a function of $k$ which approaches $0$ as $k\rightarrow\infty$, I guess it is something like $\frac{\log(k)}{k}$.
The next natural step is to sharpen the bound on $|h(i)-h(j)|$ given that random walks starting at $i,j$ are likely to intersect. (See comments.) It is hard to tell exactly how good the obtained bounds are without carrying out the computations, but it seems reasonable that this solves at least the cases in which $k$ is allowed to grow, but $n/k$ is bounded (a regime in which $k$ is assumed to be "large"), along with some further cases.
Some more details:
We regard the graph laplacian $L$ of a graph $(V,E)$ as an operator $V^\mathbb{R}\rightarrow V^\mathbb{R}$. It takes a function $h:V\rightarrow\mathbb{R}$ to the function $Lh$ defined by $$(Lh)(v)=\sum_{w\text{ a neighbor of $v$}}(h(v)-h(w)).$$
This is consistent with the possibly more familiar definition as a matrix $D-A$, where the $v$-th row of $L$ is given by the vector
$$\deg(v)e_v - \sum_{w\text{ a neighbor of $v$}}e_w.$$
As in the question, some entries of the adjacency matrix may be $2$ and not just $1$. In this case, imagine two edges between the appropriate vertices, and sum over the corresponding term $(h(v)-h(w))$ with the appropriate multiplicity. This generalizes to arbitrary weights, though for a probabilistic interpretation we want them nonnegative...
If $h:V\rightarrow\mathbb{R}$ is given by $$h(v) = \text{the probability of reaching the vertex 2 before 1 in a random walk},$$ then $h$ is harmonic at every vertex except the vertices $1,2$ by the law of total probability (at the vertices $1$ and $2$, it takes the values $0$ and $1$ respectively regardless of the values of the neighbors). Indeed, denote by $X_{v,w}$ the event that the first step of a random walk starting at $v$ is a step to $w$, and $p_{v,w}=P(X_{v,w}).$ Then using the definition of $h$ we see:
$$h(v)=\sum_\text{$w$ a neighbor of $v$}P(\text{the random walk from $w$ reaches 2 before 1}\vert X_{v,w})\cdot P(X_{v,w})$$
$$ = \sum_\text{$w$ a neighbor of $v$}h(w)\cdot p_{v,w},$$
and $p_{v,w}$ is just $$\frac{a_{v,w}}{\sum_\text{$u$ a neighbor of $v$}a_{v,u}}$$ in the language of the question. Hence $h(v)=(Ah)(v)/\deg(v)$, and $(Lh)(v)=0$.
Given $h$ defined as above, $h$ assumes only values between $0$ and $1$ (they are all probabilities by definition). Using $h(1)=0$ and $h(2)=1$, we have
$$(Lh)(1) = -h(2) - \sum_\text{$x$ another neighbor of $1$} h(x) \le -1,$$
and similarly
$$(Lh)(2) = h(2)-0 + \sum_\text{$x$ another neighbor of $2$}(h(2)-h(x)),$$
and each of the $h(x)$ is at most $1=h(2)$. So $(Lh)(2)\ge 1$.
Now, the image of $L$ is always orthogonal to the constant function (each column of the matrix sums to $0$,) and $(Lh)(v)=0$ unless $v\in\{1,2\}$. Therefore $(Lh)(1)=-(Lh)(2)$.
The inequality $(Lh)(1)\le -1$, together with $(Lh)(1)=-(Lh)(2)$, implies there is some $\alpha \in [0,1]$ such that $f=\alpha\cdot Lh$ satisfies $$f(1)=-1,f(2)=1,$$ so it is the desired solution $x$.
How does the multiplicative constant depend on $n$, though?
If I understand the question correctly, it doesn't matter: the maximal sum considered in the question occurs when it is 1.
ah yes, you are right, that's because the conductivity is bounded below by $1$.
Thanks! I'm not so familiar with random walks. Could you elaborate more on why is the solution $x$ up to some multiplicative constant the function $h$?
@neverevernever - done, see at the end. I have erased my comments so as not to crowd this space, since they are now contained in the answer.
@gyashfe Thanks that's very helpful! Another question is about bounding the probability that $W_i$ and $W_j$ do not meet. I do not quite follow. Why the first time $j$ reaches a node $\leq i$ can only be $i$ or $i-1$? $i-2,...,i-k$ should still be possible.
@neverevernever -
You only have edges between a node $k$ and the nodes $k-2,k-1,k+1,k+2$. Now, $j>i$. Consider a path from $j$, and the first time it reaches a node $k\le i$. Then at the previous step, it was at some $\ell > i$. Thus ${k,\ell}$ is an edge of the graph, and $|k-\ell| \le 2$, so $k\le i<\ell = k+1$ or $k\le i<\ell = k+2$...
In other words this follows from your assumption that $a_{i,j}=0$ for $|i-j|>2$.
I now see I may have misread your question... So this is a solution if the $k$ you fix at the beginning is $3$ (shall we call it the number of bands?) I will try to generalize.
@neverevernever - done, but for now the resulting bound depends on $k$.
@gyashfe I see. So the argument does not work for large $k$, for example, when the graph is fully connected. However, when the graph is fully connected, simulation result still confirms the boundedness of the total current.
@neverevernever When $k=n$ or is linear in $n$, I believe the argument should become simpler. One can give direct bounds on the differences $|h(v)-h(w)|$. I wonder what happens if $k$ is around $\log(n)$ or $n^s$ for some $s\in(0,1)$.
@neverevernever - note also that there could be significant differences between an adversarially-designed $A$ and a random one. In the latter case I have no doubt it's true in general.
@gyashfe Your argument except bounding the meeting probability of $W_i$ and $W_j$ seems sharp to me. To consider growing $k$, do you think it is enough if we can bound the meeting probability sharper?
@neverevernever Yes, and I am beginning to think it should work (the bound should be independent). As $k\rightarrow\infty$, the probability for two random walks to intersect somewhere within the vertices ${3,\ldots,k}$ should approach $1$.
@gyashfe BTW, the multiplicative constant that transform $x$ to $h$ is at the order of $1/k$, which is just the resistance distance between node 1 and 2. So we only need to prove that $\sum_{i,j}a_{ij}|h(i)-h(j)|$ does not grow faster than $k$.
@neverevernever - I have added a sketch of how to finish using this and another simple idea. Since it's late here I will stop for today, and perhaps fill things in during the weekend. But more likely you don't need my help for this: filling in the sketch seems routine.
@gyashfe Thanks! It looks promising. But how will $h(i)-h(j)$ decay in this case? It seems that the larger $k$ is, the slower the decay, but the meeting probability within ${3,...,k}$ will be higher. It seems that there is some sort of balance which will make $\sum_{i,j}a_{ij}|x_i-x_j|$ constant.
@neverevernever - You are right, it is incomplete. For now I have several remarks:
The lemma which bounds |h(i)-h(j)| using intersection probability can be sharpened. The RHS should be replaced with a much smaller quantity, depending on the distribution of h(x) over x after one step given that the walks do not intersect. This is a sort of bootstrapping procedure - use bounds found before to apply a slightly tighter inequality. This does make the sketch work in the regime where $n/k$ is bounded above.
We only considered intersection probabilities within ${3,...,k}$ for now.
...
There are more tools we can try to apply. But I am not convinced that the conjecture is true: it isn't clear what happens in some kind of "middle range" where $k$ is large and $n$ is larger, but not too much - e.g., if $n \approx k^\alpha$ for some specific $1<\alpha$, and $k$ grows, it is not clear to me what happens.
Random matrices $A$ could behave differently from well-chosen ones: for example you can try to arrange that $2$ subsets of the vertices have little cross-interaction by setting $a_{i,j}=2$ for each $i,j$ within one set and $a_{i,j}=1$ for every $i,j$ intersecting both.
@neverevernever - By the way, thinking about it in bed, last night I came to the conclusion that the non-intersecting probability within ${3,\ldots,k}$ is probably something like $\log(k)/k$. So the sharpened bound from above may be necessary even for $n\approx c\cdot k$.
@gyashfe How does the $\log k/k$ come from? I though it should be something exponential in $k$, since $(1-t/3)$ is roughly constant and $s$ is roughly linear in $k$.
@neverevernever - Consider just the probability coming from the part of the sample space in which each of the two walks reaches ${1,2}$ in at most $k/10$ steps. Ignoring the possibility that the first walk ever steps on a vertex twice (this only lowers the intersection probability,) the probability is then (by estimating the probability of ending within $i$ steps $\approx \frac{1}{k}$, which is OK):
$\approx\frac{1}{k^{2}}\sum_{i}^{k/10}\sum_{j=1}^{k/10}\left(1-\frac{i}{k}\right)^{j}\approx\frac{\log\left(k/10\right)-c}{k}$.
(The expression in the middle can be computed directly).
@gyashfe I'm thinking about partition the graph into several layers, each layer has the same shortest path to one of the absorbing state. Then, maybe we can look at distinct sites a random walk visits at each layer. Since we only need to look at $i,j$ such that they are connected. They will start at adjacent layers. They will have some probability of meeting at each layer.
@gyashfe Although I'm still struggling with the how to deal with large $k$, I'm gonna give you the bounty. Your answer is very helpful! Thanks!
@neverevernever Thanks, I appreciate it. I have become too busy to come back to this in the last few days, but might still do so later. In any case it was fun to think about.
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