parent_url
stringlengths 37
41
| parent_score
stringlengths 1
3
| parent_body
stringlengths 19
30.2k
| parent_user
stringlengths 32
37
| parent_title
stringlengths 15
248
| body
stringlengths 8
29.9k
| score
stringlengths 1
3
| user
stringlengths 32
37
| answer_id
stringlengths 2
6
| __index_level_0__
int64 1
182k
|
|---|---|---|---|---|---|---|---|---|---|
https://mathoverflow.net/questions/449188 | 1 | I come across an interesting question.
>
> Let $ B\_r=\{x\in\mathbb{R}^3:|x|\leq r\} $ be the ball in $ \mathbb{R}^3 $ with radius $ r $. Assume that $ u \in C(\mathbb{R}^3\setminus B\_1) $ satisfies
> $$
> \Delta u\leq -u^3,\,\,u\geq 0,\,\,\forall |x|\geq 1.
> $$
>
> where $ \Delta u $ is defined in the sense of distribution. Show that $ u=0 $ in $ \mathbb{R}^3\setminus B\_1 $.
>
>
>
I guess I need to use maximal principle but cannot go on, can you give me some hints or references?
| https://mathoverflow.net/users/241460 | How to show that $ u $ is vanishing in $ \mathbb{R}^3\setminus B_1 $? | A proof that $u=0$ is as follows. Assume that such $u \geq 0 $ is not identically zero. Since $\Delta u \leq 0$, $u$ can never vanish, otherwise it would have an interior minimum. Taking averages on the unit sphere of $\mathbb R^3$ we may assume also that $u$ is radial (the inequality in the differential equation is preserved by Jensen inequality). Then $u''+\frac 2r u' \leq -u^3$ or $(r^2u')' \leq -r^2u^3 \leq 0$, in particular $r^2 u'$ is decreasing.
There exists $r\_0$ such that $u'(r\_0)<0$. If not, $u$ would be increasing and $u \geq c>0$ gives $\Delta u \leq -c^2 u$ which has no positive solutions in $\mathbb R^3 \setminus B\_R$ (I assume this to be known but a proof follows with the arguments below). Then $r^2 u'(r) \leq -\alpha <0$ for $r \geq r\_0$ and then $u$ is decreasing and $\lim\_{r \to \infty}u(r)=0$ (by the same argument, if the limit is $\ell>0$ we have the inequality $\Delta u \leq -\ell^2 u$).
For $r \geq r\_0$ we have $u \geq \frac c r$ ($c=r\_0 u(r\_0))$ by subharmonic comparison, since both function tend to zero at infinity.
The inequality $(r^2u')' \leq -r^2 u^3 \leq - c^3/r$ gives $u' \leq -\alpha /r^2 -c^3/r^2 \log (r/r\_0)$. Integrating this inequality between $r$ and $\rho$ and letting $\rho \to \infty$ we obtain $u(r) \geq a/r+(b/r) \log r$ for suitable $b>0$ and then $u(r) \geq k/r$ for $r \geq M$ and $k \geq 1$.
Form this we obtain $u''+\frac 2r u' + \frac {k^2}{r^2} u \leq 0$ for $r \geq M$ and, setting $u=w/r$, $w''+\frac{k^2}{r^2} w \leq 0$. However this last equation does not have positive solutions. In fact every solution of the equation $v''+\frac{k^2}{r^2} v=0$ oscillates (it has power like solutions) and if $w>0$ satisfies the above inequality, setting $p=w''/w$ it follows that $w''-pw=0$ with $-p \geq k^2/r^2$ contradicitng the Sturm oscillation theorem.
| 2 | https://mathoverflow.net/users/150653 | 449212 | 180,808 |
https://mathoverflow.net/questions/449206 | 2 | Let $\{V\_n\}\_{n=1}^\infty$ be a collection of finite dimensional vector subspaces of $L^2[0,1]$ such that $V\_n \subset V\_{n+1}$ and $\bigcup\_{n=1}^\infty V\_n$ is dense in $L^2[0,1]$. Suppose further that each $V\_n$ consists of smooth functions.
Consider a collection of invertible linear mappings $\{T\_n\}\_{n=1}^\infty$ defined on each $V\_n$ such that $T\_n(A) \subset T\_{n+1}(A)$ for all $A \subset V\_n$.
Now, if we fix some $m \in \mathbb{N}$, then the image $T\_n(V\_m)$ has the same dimension as $V\_m$ for all $n \geq m$. Now, I wonder what the vector space
\begin{equation}
\bigcup\_{n=1}^\infty T\_n(V\_m)
\end{equation}
looks like. It certainly looks like finite-dimensional, but is it possible to say more about this space? For example, does it still consist of smooth functions?
Also, if $v \in V\_m$, in what sense does $T\_n(v)$ converge in $\bigcup\_{n=1}^\infty T\_n(V\_m)$ as $n \to \infty$?
I think LF or LB spaces are relevant topics, but cannot find some concrete information to apply to this case. Could anyone please help me?
Edit : OK, I need to change the condition $T\_n(A) \subset T\_{n+1}(A)$ by the one that
"there exists a projection $P\_{n,n'} : T\_n(V\_m) \to T\_{n'}(V\_m)$ for all $n \geq n' \geq m$ such that $P\_{n', k} \circ P\_{n,n'}= P\_{n,k}$ for $n \geq n' \geq k$ and $P\_{n,n}=Id$." Also, I further asssume that the image of each $T\_n$ consists of smooth functions as well. In this case, Is the above union a finite-dimensional space consisting of smooth functions, so that I can use "any" $L^p$ norm on it?
| https://mathoverflow.net/users/56524 | LF or LB space that happens to be finite dimensional | The expression
\begin{equation}
\tilde W\_m:=\bigcup\_{n=1}^\infty T\_n(V\_m)
\end{equation}
is undefined in general for $m\ge2$, because $T\_n$ is defined (and is invertible) only on $V\_n$ and hence $T\_n(V\_m)$ may be undefined for $n<m$.
So, instead of $\tilde W\_m$, one may want to consider
\begin{equation}
W\_m:=\bigcup\_{n=m}^\infty T\_n(V\_m).
\end{equation}
However, the condition that $T\_n(A)\subset T\_{n+1}(A)$ for all $A\subset V\_n$ implies that $T\_m(V\_m)\subset T\_n(V\_m)$ for all $n\ge m$. It is also given that $T\_n(V\_m)$ has the same dimension as $V\_m$ for all $n\ge m$. So, $T\_n(V\_m)=T\_m(V\_m)$ for all $n\ge m$, and hence we simply have
\begin{equation}
W\_m=T\_m(V\_m).
\end{equation}
As to whether $W\_m$ will still consist of smooth functions, the answer is: not in general. E.g., let $V\_n$ be the space of all polynomials of degree $\le n$. Let $(h\_n)\_{n=0}^\infty$ be an enumeration of (say) the [Haar basis](https://en.wikipedia.org/wiki/Haar_wavelet).
For a polynomial $p\in V\_n$ such that $p(t)=a\_0+a\_1 t+\cdots+a\_n t^n$ for some scalars $a\_0,\dots,a\_n$ and all $t\in[0,1]$, let
\begin{equation}
T\_n(p):=a\_0 h\_0+a\_1 h\_1+\cdots+a\_n h\_n.
\end{equation}
Then all your conditions on the $V\_n$'s and the $T\_n$'s hold, whereas $W\_m$ contains the (discontinuous) Haar functions $h\_0,\dots,h\_m$.
| 4 | https://mathoverflow.net/users/36721 | 449214 | 180,809 |
https://mathoverflow.net/questions/449213 | 1 | I am writing up some notes and the following occurred to me and I would like to see if there are **a variety of ways to prove it**. Just for reference, the identity pops out of equality between constant term evaluations of [Laurent polynomials from my earlier quest](https://mathoverflow.net/questions/449007/have-you-seen-this-constant-term-extraction). There should be easier ways as well, I think.
>
> **QUESTION.** Can you provide a proof for
> $$\prod\_{j=0}^n\frac{(2j)!^2}{(n+j)!\,j!}=2^n.$$
>
>
>
| https://mathoverflow.net/users/66131 | Product/quotient of factorials beget dyadic powers | This is easy. Start with $$n!.2^n=(2n)(2n-2)\dots 2.$$ This implies that $$((2n-1)!(2n-3)!\dots 1!).n!.2^n=(2n)!(2n-2)!\dots 2!.$$ Then $$((2n)!(2n-1)!\dots 1!).n!.2^n=((2n)!(2n-2)!\dots 2!)^2.$$ Thus $$\left(\prod\_{j=0}^n (n+j)!j!\right).2^n=\prod\_{j=0}^n (2j)!^2. $$ Finally, divide both sides by the product on the left.
| 3 | https://mathoverflow.net/users/460592 | 449215 | 180,810 |
https://mathoverflow.net/questions/449217 | 2 | Let $S$ be a Noetherian scheme, $f : X\to S$ a smooth quasi-projective morphism, $g : X\to Y$ a morphism of finite type, and $h : Y\to S$ a smooth projective morphism with $h\circ g =f$.
Can we say anything about the scheme-theoretic image of $g$?
For example, let $Z$ be the scheme theoretic image of $g$, with $i : Z\to Y$ the closed immersion, and $p : Z\to S$ the structure morphism $h\circ i$.
>
> Is $p$ an lci morphism?
>
>
>
| https://mathoverflow.net/users/nan | Images of smooth schemes under lci morphisms | It's very hard to say anything. In fact, if $S = \operatorname{Spec} \mathbf C$, then *any* (integral) projective $k$-variety $Z$ arises in this way. Indeed, since $Z$ is projective, there exists a closed immersion $i \colon Z \hookrightarrow \mathbf P^n$ for some $n$, so we may take $Y = \mathbf P^n$. Then we may choose a resolution of singularities $X \twoheadrightarrow Z$, so that the composition $X \twoheadrightarrow Z \hookrightarrow \mathbf P^n$ is a morphism of smooth projective $S$-schemes whose scheme-theoretic image is $Z$.
For a concrete example to see that $Z$ need not be a local complete intersection, even without using resolution of singularities, we can choose $Y = \mathbf P^2 \times \mathbf P^2$ and $X = \mathbf P^2 \amalg \mathbf P^2$, mapping the first component to $\mathbf P^2 \times p$ and the second to $p \times \mathbf P^2$ for some $p \in \mathbf P^2(\mathbf C)$. The scheme-theoretic image is the union of two planes meeting in a point, which is not Cohen–Macaulay by Hartshorne's connectedness theorem (see e.g. Eisenbud, Thm. 18.12 and Figure 18.2); in particular it cannot be a local complete intersection.
(Even worse, the map $X \to Z$ need not be surjective, e.g. we could even take the same $Z = (\mathbf P^2 \times p) \cup (p \times \mathbf P^2)$ above and take $X = Z \setminus (p,p)$ which is smooth.)
| 6 | https://mathoverflow.net/users/82179 | 449223 | 180,814 |
https://mathoverflow.net/questions/448915 | 4 | *Disclaimer*: This question [was initially](https://math.stackexchange.com/q/4718589) asked yesterday in Mathematics Stack Exchange but left unanswered there.
---
I am interested in learning about differential graph theory or differential operators on graphs, something related to what [E. Bautista](https://mathoverflow.net/users/173461/e-bautista) introduced in his answer [here](https://mathoverflow.net/questions/392391/differentiation-of-functions-over-graphs). Can one suggest a textbook in which such topics are discussed? (The quoted answer refers to some papers, but I prefer a book covering a comprehensive treatment of the topic.)
| https://mathoverflow.net/users/106458 | Reference request for differential graph theory | Well, here's a recent monograph on different flavours of graph Laplacians by Kostenko and Nicolussi: [https://www.mat.univie.ac.at/~kostenko/list/GraphLaplInf.pdf](https://www.mat.univie.ac.at/%7Ekostenko/list/GraphLaplInf.pdf)
But I'd say that the topic is too vast to be sufficiently covered in any single coherent book; you have such different appearances of operators related to edgewise derivative! To list just a few:
* random walks on finitely generated groups (there are thousands of papers on this!)
* expander graphs, and other isoperimetric properties related to amenability (Cheeger constants, cost of a group)
* geodesic flow on hyperbolic groups
* spectral theory of random graphs
* Selberg-like trace formulas
* Gromov's proof of Stallings splitting theorem via harmonic functions
* *...here someone more knowledgeable may insert ten times more topics, because I do not know much of it outside of applications to group theory; I would be interested myself to see an operator which is* not *a variant of Laplacian used to do something meaningful*.
| 1 | https://mathoverflow.net/users/81055 | 449232 | 180,815 |
https://mathoverflow.net/questions/449234 | 5 | Let $f: \mathbb R \to \mathbb R$ be a locally integrable measurable function.
We say $f$ satisfies the *intermediate value property* if given any $a, b\in \mathbb R$ with $a < b$, whenever $u \in \mathbb R$ is such that $\min(f(a), f(b)) \leq u \leq \max(f(a), f(b))$, there exists some $x \in [a, b]$ such that $f(x) = u$.
**Question:** Suppose every point of $\mathbb R$ is a Lebesgue point of $f$. Does it follow that $f$ satisfies the intermediate value property?
*Note: We use the “strong” definition of Lebesgue point, given [here](https://en.m.wikipedia.org/wiki/Lebesgue_point).*
| https://mathoverflow.net/users/173490 | If every point is a Lebesgue point of $f$, does $f$ satisfy the intermediate value property? | If $F(x)=\int\_a^x f(x)dx$ (for some fixed $a$), then $x$ being a Lebesgue point of $f$ yields $F'(x)=f(x)$; and the derivatives enjoy the intermediate value property by Darboux theorem.
| 6 | https://mathoverflow.net/users/4312 | 449239 | 180,817 |
https://mathoverflow.net/questions/449238 | 3 | Let $P(n)$ denote the largest prime factor of $n$. I wanted to know if an analogue of Baker-Harman estimate for smooth shifted primes in arithmetic progressions with fixed moduli is there in literature.
Baker-Harman proved that there exist infinitely many primes $p$ such that $P(p-1)<p^{0.2961}$ and the exponent was recently improved to $0.2844$ by Lichtman [recently](https://arxiv.org/abs/2211.09641). Has someone considered the problem of obtaining an inifnitude of such primes which are of form $1\pmod{a}$ for some fixed modulus $a$?
Thanks in advance for any help.
| https://mathoverflow.net/users/160943 | Infinitude of smooth shifted primes in arithmetic progression with fixed moduli | The only relevant result I found is [a paper](https://projecteuclid.org/journals/michigan-mathematical-journal/volume-52/issue-3/Smooth-values-of-shifted-primes-in-arithmetic-progressions/10.1307/mmj/1100623415.full) by Banks, Harcharras, and Shparlinski. They did not establish existence of such smooth numbers but instead gave an upper bound.
Let
$$
\pi\_h(x,y;q,a)=\#\{p\le x:P(p-h)<y\wedge p\equiv a\pmod q\}.
$$
Then uniformly for $x\ge y\ge2$ and $q\ge1$, they proved that
$$
\pi\_h(x,y;q,a)\ll{u\rho(u)\over\varphi(q)}\prod\_{\substack{p|h\\p>2}}{p-1\over p-2}\pi(x).
$$
| 1 | https://mathoverflow.net/users/449628 | 449262 | 180,822 |
https://mathoverflow.net/questions/449277 | 2 | Let $R$ be a regular local ring, $X$ and $Y$ smooth $R$-schemes, $T\to Y$ a regular closed immersion over $R$ with $T$ smooth over $R$, and $f: X\to Y$ an $R$-morphism.
>
> Is the fiber product scheme $X\_T:=X\times\_YT$ Cohen-Macaulay?
>
>
>
What conditions on $f$ would ensure that $X\_T$ is Cohen-Macaulay?
| https://mathoverflow.net/users/501361 | Cohen-Macaulay fiber products | Ok, $T \hookrightarrow Y$ is locally a complete intersection, right? (Closed regular subscheme of a regular scheme). So you'd want that regular sequence (locally) to become a regular sequence on $X$. That's not always true, but if $f : X \to Y$ is *flat* it is ok (since a weakly regular sequence will stay weakly regular...).
Without the flatness though I don't see how to guarantee it. Explicitly, let $R = k$ be a field, let $Y = \mathbb{A}^3\_k = \mathrm{Spec}\, k[x,y,z]$ and let $f : X \to Y$ be the blowup of the origin with exceptional divisor $E \cong \mathbb{P}^2$. Let $T = V(x,y) \subseteq Y$, note $T \cong \mathbb{A}^1\_k$. Then $X\_T$ is not equidimensional (it's a union of $E$ with the strict tranform of $T$), and hence it can't be Cohen-Macaulay.
Similar constructions work over other regular $R$s too (ie, $R$ need not be a field).
| 3 | https://mathoverflow.net/users/3521 | 449279 | 180,827 |
https://mathoverflow.net/questions/449265 | 5 | Fix a positive integer $n$ and consider the $2$-Wasserstein space $\mathcal{P}\_2(\mathbb{R}^n)$. Let $X$ be the cone of $n\times n$ symmetric positive semidefinite matrices with Frobenius norm and consider the map $f:\mathbb{R}^n\times X \rightarrow \mathcal{P}\_2(\mathbb{R}^n)$ given by
$$
f(a,B) = N(a,B)
$$
where $N(a,B)$ is the $n$-dimensional Gaussian measure with mean $a$ and covariance $B$.
Clearly the map $f$ is locally-Lipschitz and injective. However, is its inverse on its image in $\mathcal{P}\_2(\mathbb{R}^n)$ also locally-Lipschitz?
| https://mathoverflow.net/users/496781 | Local Lipschitzness of parameterization of Gaussians in Wasserstein space | $\newcommand{\R}{\mathbb R}\newcommand{\tr}{\operatorname{tr}}$The answer is yes.
Indeed, it is easy to see (cf. e.g. [Proposition 7](https://projecteuclid.org/journals/michigan-mathematical-journal/volume-31/issue-2/A-class-of-Wasserstein-metrics-for-probability-distributions/10.1307/mmj/1029003026.full) or the beginning of its proof) that the Wasserstein distance between $N(a,A)$ and $N(b,B)$ is
\begin{equation\*}
W\_2(N(a,A),N(b,B))=\sqrt{\|a-b\|^2+W\_2(N(0,A),N(0,B))^2}, \tag{10}\label{10}
\end{equation\*}
where $\|\cdot\|$ is the Euclidean norm. So,
\begin{equation\*}
\|a-b\|\le W\_2(N(a,A),N(b,B)). \tag{20}\label{20}
\end{equation\*}
Let now $X\sim N(0,A)$ and $Y\sim N(0,B)$. Then for any unit vector $u\in\R^n=\R^{n\times1}$
\begin{equation\*}
E\|X-Y\|^2\ge E(u^\top X-u^\top Y)^2\ge(\sqrt{u^\top A u}-\sqrt{u^\top B u})^2,
\end{equation\*}
where the last inequality holds (say, by mentioned [Proposition 7](https://projecteuclid.org/journals/michigan-mathematical-journal/volume-31/issue-2/A-class-of-Wasserstein-metrics-for-probability-distributions/10.1307/mmj/1029003026.full)) because $u^\top X$ and $u^\top Y$ are real-valued Gaussian zero-mean random variables with variances $u^\top A u$ and $u^\top B u$.
So, by \eqref{10} and the [definition of the $W\_2$-distance](https://en.wikipedia.org/wiki/Wasserstein_metric#Definition), for any unit vector $u\in\R^n=\R^{n\times1}$,
\begin{equation\*}
\begin{aligned}
W\_2(N(a,A),N(b,B))&\ge W\_2(N(0,A),N(0,B)) \\
&\ge|\sqrt{u^\top A u}-\sqrt{u^\top B u}| \\
&=\frac{|u^\top A u-u^\top B u|}{\sqrt{u^\top A u}+\sqrt{u^\top B u}} \\
&\ge\frac{|u^\top(A-B)u|}{\sqrt{\|A\|}+\sqrt{\|B\|}} \\
&=\frac{\|A-B\|}{\sqrt{\|A\|}+\sqrt{\|B\|}}
\end{aligned}
\end{equation\*}
for some unit vector $u\in\R^n=\R^{n\times1}$, where $\|M\|$ is the spectral norm of a matrix $M$.
So,
\begin{equation\*}
\|A-B\|\le(\sqrt{\|A\|}+\sqrt{\|B\|\,})\, W\_2(N(a,A),N(b,B)). \tag{30}\label{30}
\end{equation\*}
Thus, by \eqref{20} and \eqref{30}, we have the desired local Lipschitz property. $\quad\Box$
*Remark:* Inequality \eqref{20} is "exact in the limit", when $A$ and $B$ are each (close to) the zero matrix.
Inequality \eqref{30} turns into the equality when $n=1$ to, more generally, "exact in the limit" when $A$ and $B$ are (close to) commuting matrices of rank $1$ each. $\quad\Box$
---
[User ABIM wrote in a comment](https://mathoverflow.net/questions/449265/local-lipschitzness-of-parameterization-of-gaussians-in-wasserstein-space/449287#comment1161164_449287): "@Justin\_other\_PhD OP claimed that $f$
is bi-Lipschitz but why is the upper-bound true?"
Let me answer this question as well. First of all, it is not true that $f$ is Lipschitz (and the OP did not claim that). Indeed, even when $n=1$, we have $W\_2(N(0,A),N(0,B))=|\sqrt A-\sqrt B|$, so that there is no real $L>0$ such that $W\_2(N(0,A),N(0,B))\le L|A-B|$ for all real $A,B>0$.
What the OP said, and what is true, is that $f$ is **locally** Lipschitz. Indeed,
let $X\sim N(0,A)$ and $Y:=B^{1/2}A^{-1/2}X$. Then $Y\sim N(0,B)$ and hence
\begin{equation\*}
\begin{aligned}
W\_2(N(0,A),N(0,B))^2&\le E\|X-Y\|^2 \\
&=\tr(A+B-B^{1/2}A^{1/2}-A^{1/2}B^{1/2}) \\
&=\tr[(A^{1/2}-B^{1/2})^2]=\|A^{1/2}-B^{1/2}\|\_F^2 \\
&\le n\,\|A^{1/2}-B^{1/2}\|^2,
\end{aligned}
\tag{40}\label{40}
\end{equation\*}
where $\tr$ denotes the trace and $\|\cdot\|\_F$ is the Frobenius norm.
Next, for real $a\ge0$,
\begin{equation\*}
a^{1/2}=\frac1\pi\int\_0^\infty dt\,t^{-1/2}\Big(1-\frac t{t+a}\Big)
\end{equation\*}
and hence the value of the derivative of the function $M\mapsto M^{1/2}$ at $A$ at a symmetric matrix $H$ is
\begin{equation\*}
\begin{aligned}
(A^{1/2})'(H)&=\frac1\pi\int\_0^\infty dt\,t^{-1/2}(tI+A)^{-1}H(tI+A)^{-1} \\
& \le\frac1\pi\int\_0^\infty dt\,t^{-1/2}(t+c)^{-2}|H|
=\frac1{2c^{1/2}}\,|H|
\end{aligned}
\end{equation\*}
provided that $A\ge cI$ for some $c\in(0,\infty)$ (where $I$ is the identity matrix), so that the operator norm of $(A^{1/2})'$ is $\le\dfrac1{2c^{1/2}}$.
So, by \eqref{10} and \eqref{40},
\begin{equation\*}
W\_2(N(a,A),N(b,B))\le\|a-b\|+W\_2(N(0,A),N(0,B)) \\
\le \|a-b\|+\dfrac{n^{1/2}}{2c^{1/2}}\,\|A-B\|
\end{equation\*}
provided that $A\ge cI$ and $B\ge cI$ for some $c\in(0,\infty)$. Thus, $f[=N]$ is locally Lipschitz. $\quad\Box$
| 5 | https://mathoverflow.net/users/36721 | 449287 | 180,828 |
https://mathoverflow.net/questions/449285 | 3 | I am wondering if there is an analog of the following theorem by Morgan and White:
Suppose $g\_E$ is the Euclidean metric on $\mathbf R^3$. If $\gamma$ is a closed $C^{k,\alpha}$ curve in $(\mathbf R^3,g\_E)$ that bounds a unique area minimizing surface, then every closed curve sufficiently closed to $\gamma$ in $C^{k,\alpha}$ also bounds a unique area minimizing surface.
Specifically, I am wondering what happens if we fix a curve, say a circle in $\mathbf R^3$, and let the metric change. Intuitively, when the metric is closed to $g\_E$, one expects that $\gamma$ still bounds a unique minimal surface. Is there any theorem that settled this question? More generally, if I consider the set of all circles on $\mathbf R^3$, do I expect that when the metric $g$ is closed to $g\_E$, all of them bound a unique area minimizing surface?
Any insight or reference would be very appreciated.
| https://mathoverflow.net/users/175594 | Minimal surface on $R^3$ with with non Euclidean metric | Below I sketch the proof of the following theorem:
>
> Theorem: Suppose that $\Sigma^n\subset (M^{n+1},g\_0)$ is smooth and uniquely area-minimizing relative to it's boundary $\Gamma : = \partial\Sigma$ and is strictly stable (no nontrivial Jacobi fields with Dirichlet boundary conditions). Then if $g$ is a sufficiently small $C^{k,\alpha}$ perturbation of $g\_0$, there's a unique area minimizer with boundary $\Gamma$ for the metric $g$ (and it's still smooth and strictly stable).
>
>
>
---
Remarks:
1. I think $k =1, \alpha \in (0,1)$ should suffice but I didn't think through each step very carefully to verify this.)
2. Note that this theorem actually proves the Euclidean one since if $\gamma,\gamma'$ are nearby boundaries, then you can find $\Phi : \mathbb{R}^n\to \mathbb{R}^n$ diffeomorphism (small) taking $\gamma'$ to $\gamma$. Then the $\gamma'$ solution (for $g\_E$) is a $\gamma$ solution for $\Phi^\*g\_E$.
---
Proof: Suppose that $g\to g\_0$ and $\Sigma\_1,\Sigma\_2$ are distinct $g$-minimizers with boundary $\Gamma$ (singular is fine). Since $\Sigma$ is smooth then Allard's theorem (interior and boundary) imply that $\Sigma\_1,\Sigma\_2$ converge smoothly to $\Sigma$ all the way up to the boundary. In other words, there's $u\_1,u\_2$ on $\Sigma$ so that $\textrm{graph}(u\_i) = \Sigma\_i$ and $u\_i |\_{\partial\Sigma} = 0$ (where the graph is taken using the normal exponential map of $g\_0$) and $u\_i\to 0$ in some $C^{k',\alpha}$ space (which you have to work out depending on $k$). Write $H(u,g)$ for the mean curvature of the graph (as above) of $u$ with respect to $g$. Note that $H(u\_i,g) = 0$ by assumption. Thus,
$$
0 = H(u\_2,g)-H(u\_1,g) = D\_1H(u\_1,g) (u\_2-u\_1) + O(||u\_2-u\_1||\_{C^2}^2)
$$
(we used Taylor's theorem). The error is uniform with respect to the metric.
Now, divide by $||u\_2-u\_1||\_{C^2}$ to get that $w:=(u\_2-u\_1)/||u\_2-u\_1||\_{C^2}$ satisfies
$$
0 = D\_1H(u\_1,g) w + o(1)
$$
Using Schauder estimates we can get that $||w||\_{C^{2,\alpha}} = O(1)$ (you have to check that the $o(1)$ term is bounded in $C^\alpha$ and that $D\_1(u\_1,g)w$ is a uniformly elliptic operator. Note that as $g\to g\_0$ and $u\_1\to 0$, $D\_1H(u\_1,g)$ limits to the Jacobi operator on $\Sigma$ with respect to $g\_0$ (this is the definition of the Jacobi operator $J=D\_1H(0,g\_0)$. Moreover, by the $C^{2,\alpha}$ bounds for $w$, we get that the limit $w\_0$ is $C^2$ and nonzero and solves $Jw\_0 = 0$ (with Dirichlet boundary conditions).
This is a contradiction, proving uniqueness for $g$ minimizers, $g$ close to $g\_0$. Strict stability follows from the fact that the first Dirichlet eigenvalue of $J$ is continuous as the metric and minimizer varies.
---
For your second question about the result being uniform for all disks, this cannot be true. Here is a strange proof (probably there is a simpler way):
Construct a non-flat $g$ on $\mathbb{R}^3$ rotationally symmetric with scalar curvature $> 0$ and so that $g=g\_E + O(r^{-1})$ at infinity (along with derivatives). This is called an asymptotically flat metric. It turns out that you can do this with $||g-g\_E||\_{C^k}$ as small as you like. (You can also do this while arranging that there are no closed minimal surfaces in $(\mathbb{R}^3,g)$). This can all be accomplished by looking at the scalar curvature of $dr^2 + \varphi(r)^2 g\_{S^2}$ and playing around with the resulting differential inequality.)
Thus, it suffices to prove that for this metric $g$ fixed, for $R$ sufficiently large, the circle of radius $R$ in the $xy$-plane has non-unique minimizers. Suppose the minimizer was unique. Then, by symmetry it must be the disk in the $xy$-plane. (The symmetry of the manifold already shows this is a minimal surface, but it might not be the minimizer; if not the minimizer then $z\mapsto -z$ takes the minimizer to a second one!).
If each disk of radius $R$ was a minimizer then the $xy$-plane will be a minimizer (on compact sets). However, the [proof](https://mathscinet.ams.org/mathscinet/article?mr=0526976) of the positive mass theorem shows that the scalar curvature would vanish along the plane. (Alternatively, see [here](https://arxiv.org/pdf/1510.07406.pdf) and the references contained within.)
---
For your question in the comment, I think the following is true:
>
> Theorem: Fix $R\_0>0$ and suppose that $||g-g\_E||\_{C^{k,\alpha}} \leq \epsilon=\epsilon(R\_0)$. Then there's a unique $g$-minimizer for any circle of radius $R\leq R\_0$.
>
>
>
Proof: If not there's $g\to g\_E$ in $C^{k,\alpha}$ and circles $C$ with radius $R\leq R\_0$ where things fail. You can translate $C$ to the origin and rescale so that they have radius $1$. The rescaled metrics $g'$ still converge in $C^{k,\alpha}$ (this is where we used $R\leq R\_0$). Thus, the previous proof applies.
| 4 | https://mathoverflow.net/users/1540 | 449291 | 180,829 |
https://mathoverflow.net/questions/449302 | 2 | I'm looking for an example of a function $u \in H\_2$ such that $u \notin H\_\infty$, where $H\_p$ is the Hardy space on the right-half plane. Since this notation is perhaps not standard, here is a complete definition.
Let $\mathbb{C}^+ := \{s\in \mathbb{C} \mid \text{Re}(s) > 0 \}$ and $\bar{\mathbb{C}}^+ := \{s\in \mathbb{C} \mid \text{Re}(s) \geq 0 \}$ denotes the open and closed right-half plane, respectively.
Let $u:\bar{\mathbb{C}}^+\to \mathbb{C}$ be a function. Consider the following four conditions:
1. $u$ is analytic in the open right-half plane $\mathbb{C}^+$.
2. For almost all $\omega\in\mathbb{R}$, we have $\displaystyle \lim\_{\sigma \to 0^+} u(\sigma+i\omega) = u(i\omega)$
3. $\displaystyle \sup\_{\sigma \geq 0} \int\_{-\infty}^\infty |u(\sigma+j\omega)|^2\,\mathrm{d}\omega < \infty$
4. $\displaystyle \text{ess}\sup\_{s \in \bar{\mathbb{C}}^+} |u(s)| < \infty\;$ (this is the *essential supremum*, so ignoring any sets of measure zero)
We say that $u\in H\_2$ if conditions 1,2,3 hold. We say that $u\in H\_\infty$ if conditions 1,2,4 hold. Note that when conditions 1 and 2 hold, the suprema in conditions 3 and 4 will occur on the imaginary axis.
| https://mathoverflow.net/users/7667 | Hardy space inclusion in the right-half plane | Here is a suggestion (but I have not worked through the details). What follows is given for the upper-half plane (UHP for short) but of course a trivial rotation will convert the example to one on the right-half plane as you requested.
There is a weighted composition operator giving an isometry of Hilbert spaces from $H^2$ of the UHP to $H^2$ of the open unit disc. One can find an explicit formula on the [Wikipedia page for Hardy spaces](https://en.wikipedia.org/wiki/Hardy_space#Hardy_spaces_for_the_upper-half-plane). If $M: H^2({\rm UHP})\to H^2({\bf D})$ is the particular isometry described at that link, then inspecting the formula there shows that for each $f\in (H^2\cap H^\infty)({\rm UHP})$, the function $M(f)$ belongs to $H^2({\bf D})$ and is essentially bounded as $z\to -1$.
It therefore suffices to pick any $g\in H^2({\bf D})$ which is **not** essentially bounded as $z\to -1$, and then $M^{-1}(g)$ will be a function in $H^2({\rm UHP})$ that is not essentially bounded.
One such $g$ is given by
$$
\displaystyle g(z)=\sum\_{n=1}^\infty \frac{(-1)^{n-1}}{n}z^n=\log(1+z).
$$
| 3 | https://mathoverflow.net/users/763 | 449303 | 180,834 |
https://mathoverflow.net/questions/449308 | 3 | Let $p$ be a prime and let $w$ be a primitive $p$-th root of unity in $\mathbb{C}$. There is an element $e$ of order $p$ in $G=\operatorname{PGL}\_n(\mathbb{C})$ where $n=pk$ and
$$e=\left[\left(\begin{matrix}I\_k&&&&\\&{wI}\_k&&&\\&&{w^2I}\_k&&\\&&&\ddots&\\&&&&{w^{p-1}I}\_k\\\end{matrix}\right)\right].$$
My focus is on the elementary abelian $p$-subgroups of rank 2 of $G$. Let $E$ be such a group.
Computation shows if all the nontrivial elements of $E$ are conjugate to $e$, then $N\_{G}(E) /C\_{G}(E) \cong \operatorname{GL}\_2(p)$.
I can't see it clearly. I observe that if $E=\langle u,v \rangle = \langle uv,v \rangle = \langle u,uv \rangle$, then a change of basis matrix will be
$$\left(\begin{matrix}1&0\\1&1\\\end{matrix}\right) \, \text{or} \, \left(\begin{matrix}1&1\\0&1\\\end{matrix}\right).$$
These two matrices generate $\operatorname{SL}\_2(p)$. Am I arguing along the right lines?
| https://mathoverflow.net/users/488802 | normalizer quotient is $\operatorname{GL}_2(p)$ | Let $p$ be an odd prime. Look at the group $P$ generated by the following elements of $GL\_p(\mathbb{C})$: $\left(\begin{smallmatrix} 1&&&\\&w&&\\&&\ddots&\\&&&w^{p-1}\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix}&1&&\\&&1&\\&&&\ddots\\1&&&&\end{smallmatrix}\right)$. Then $P$ is an extraspecial group of order $p^3$ and exponent $p$. Its image in $PGL\_p(\mathbb{C})$ is elementary abelian $E$ of order $p^2$, and the normaliser modulo centraliser is $SL\_2(p)$, acting as the group of all automorphisms of determinant one of $E$. There are $p-1$ representations like this, permuted by $GL\_2(p)$; in other words, related by automorphisms of $E$. You can distinguish between these $p-1$ representations by the scalar representing a non-identity central element.
You can't mix different ones of these representations if the central element of $P$ has to act as a scalar on the entire space. So there are only two possibilities: Either your representation consists of $k$ copies of one of the $p-1$ representations above, in which case the normaliser modulo centraliser is $SL\_2(p)$; or $k$ is divisible by $p$, and your representation is a direct sum of $p/k$ copies of the regular representation of the elementary abelian $E$, in which case the normaliser modulo centraliser is $GL\_2(p)$.
**Note** I have edited this answer several times, for the sake of both correctness and readability. Apologies for any confusion caused.
| 3 | https://mathoverflow.net/users/460592 | 449310 | 180,836 |
https://mathoverflow.net/questions/393787 | 7 | From Alexandrov's work we know that any metric on the sphere with lower curvature bound $\kappa$ (in the sense of Alexandrov) can be realized as a closed convex surface (i.e. boundary of a compact convex domain) in the $3$-space form of constant curvature $\kappa$.
For $\kappa=0$, the surface is unique up to isometry of the surrounding space $\mathbb{R}^3$ due to Pogorelov. Is the same true for the elliptic and hyperbolic case? In other words, are two closed convex surfaces in these spaces congruent when they are isometric with respect to their inner metrics?
I asked this question [here](https://math.stackexchange.com/questions/4148093/rigidity-for-convex-surfaces-in-elliptic-hyperbolic-space) on Math SE, but did not get an answer.
| https://mathoverflow.net/users/111820 | Rigidity for convex surfaces in elliptic/hyperbolic space | Yes, closed convex surfaces in spaces of constant curvature are rigid. See Chapter V of Pogorelov's book [Extrinsic Geometry of Convex Surfaces](https://www.ams.org/books/mmono/035/mmono035-endmatter.pdf).
As described at the beginning of the chapter on p. 270-271, the main idea is to reduce the problem to the Euclidean case. Pogorelov says that this reduction is similar for elliptic and hyperbolic cases, and only considers the elliptic case, which is established in Thm. 1 on p. 321.
| 3 | https://mathoverflow.net/users/68969 | 449321 | 180,839 |
https://mathoverflow.net/questions/449329 | 8 | Let me preface this by saying that I have next to no background in representation theory. I come from geometry but the following representations showed up naturally in my work.
We let $ V = C^\infty \left( \mathbb{R}^2 \setminus \left\{ 0 \right\} \right) $ be equipped with its standard Frechet space topology. Then, $ V $ is a continuous representation of $ GL \left( 2, \mathbb{R} \right) $ via the action $$ gf \left( x \right) = f \left( g^{-1} x \right). $$ One easily sees that for any complex number $ w \in \mathbb{C} $ and any parity $ \varepsilon \in \left\{+,-\right\} $, the space $ V^\varepsilon\_w \subset V $ of smooth, $ w $-homogeneous and even (if $ \varepsilon = + $) resp. odd (if $ \varepsilon = - $) functions forms a closed subrepresentation of $ V $. My question now is: Is the representation $ V\_0^- $ irreducible? If someone knows the answer for general values of $ w $ and $ \varepsilon $, this would of course be even better.
My guess is that this question should somehow be connected to the so called principal series representations as these are also indexed by a complex number and a parity. I have looked this up in Knapp's book "Representation Theory of Semisimple Groups". However, the representations there look very different (at least to the untrained eye?!). Moreover, he only considers them as representations over $ SL(2,\mathbb{R}) $ while I am interested in the bigger group $ GL(2, \mathbb{R}) $.
EDIT: Instead of homogeneous of degree $ w $ I should have more precisely said positively homogeneous of degree $ w $, that is $ f (ax) = a^w f(x) $ for all $ a>0 $ and $ x \in \mathbb{R}^2 \setminus \left\{ 0 \right\} $. By even ($ \varepsilon = + $) resp. odd ($ \varepsilon = - $), I mean that $ f (-x) = \varepsilon f(x) $ for all $ x $.
| https://mathoverflow.net/users/507338 | Is the GL(2,R)-representation of smooth, odd and 0-homogeneous functions on the punctured plane irreducible? | Here's the general set up, as I understand it. Recall that a character of $\mathbb{R}^{\times}$ is a continuous homomorphism from $\mathbb{R}^{\times}$ to $\mathbb{C}^{\times}$. Every character of $\mathbb{R}^{\times}$ is of the form $\operatorname{sgn}(x)^{\kappa} |x|^s$ for some $\kappa \in \{0,1\}$ and $s \in \mathbb{C}$.
Associated to a pair of characters $\operatorname{sgn}^{\kappa\_1} |\cdot|^{s\_1},\operatorname{sgn}^{\kappa\_2} |\cdot|^{s\_2}$ of $\mathbb{R}^{\times}$ is a principal series representation of $\mathrm{GL}\_2(\mathbb{R})$. The vector space of this representation is called the induced model: it consists of smooth functions $f : \mathrm{GL}\_2(\mathbb{R}) \to \mathbb{C}$ satisfying
$$f\left(\begin{pmatrix} a & b \\ 0 & d \end{pmatrix} g\right) = \operatorname{sgn}(a)^{\kappa\_1} |a|^{s\_1} \operatorname{sgn}(d)^{\kappa\_2} |d|^{s\_2} \left|\frac{a}{d}\right|^{1/2} f(g)$$
for all $a,d \in \mathbb{R}^{\times}$, $b \in \mathbb{R}$, and $g \in \mathrm{GL}\_2(\mathbb{R})$. The group $\mathrm{GL}\_2(\mathbb{R})$ acts on such functions via right-translation: $(h \cdot f)(g) = f(gh)$. In this way, we obtain a representation of $\mathrm{GL}\_2(\mathbb{R})$, namely the principal series representation associated to this pair of characters. (It is the representation obtained by normalised parabolic induction from these characters.)
---
Such a principal series representation is usually, but not always, irreducible. Reducibility occurs only in two situations. The first is if $\operatorname{sgn}(x)^{\kappa\_1} |x|^{s\_1} \operatorname{sgn}(x)^{\kappa\_2} |x|^{-s\_2} = \operatorname{sgn}(x)^k |x|^{k - 1}$ for some positive integer $k \geq 2$ (i.e. $\kappa\_1 + \kappa\_2 \equiv k \pmod{2}$ and $s\_1 - s\_2 = k - 1$): there is an infinite-dimensional irreducible subrepresentation and a finite-dimensional irreducible subquotient. The second is if $\operatorname{sgn}(x)^{\kappa\_1} |x|^{s\_1} \operatorname{sgn}(x)^{\kappa\_2} |x|^{-s\_2} = \operatorname{sgn}(x)^k |x|^{1 - k}$ for some $k \geq 2$: there is an infinite-dimensional subquotient and finite-dimensional subrepresentation. In both cases, the irreducible subrepresentation or subquotient is a discrete series representation of weight $k$ twisted by a power of $\left|\det\right|$. (The twist is by $\left|\det\right|^{s\_1 - \frac{k - 1}{2}} = \left|\det\right|^{s\_2 + \frac{k - 1}{2}}$ for the former case and by $\left|\det\right|^{s\_1 + \frac{k - 1}{2}} = \left|\det\right|^{s\_2 - \frac{k - 1}{2}}$ for the latter case.)
---
Given a function $f$ in the induced model of such a principal series representation, let $F : \mathbb{R}^2 \setminus \{(0,0)\} \to \mathbb{C}$ be the smooth function given by
$$F(x,y) = (-1)^{\kappa\_1 + \kappa\_2} \left|x^2 + y^2\right|^{-\frac{s\_1 - s\_2 + 1}{2}} f\begin{pmatrix} \frac{x}{\sqrt{x^2 + y^2}} & \frac{y}{\sqrt{x^2 + y^2}} \\ -\frac{y}{\sqrt{x^2 + y^2}} & \frac{x}{\sqrt{x^2 + y^2}}\end{pmatrix},$$
so that if $y \neq 0$,
$$F(x,y) = f\left(\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} y & -x \\ 0 & \frac{1}{y} \end{pmatrix}\right),$$
while
$$F(x,0) = f\begin{pmatrix} \frac{1}{x} & 0 \\ 0 & x \end{pmatrix}.$$
This function satisfies
$$F(ax,ay) = \operatorname{sgn}(a)^{\kappa\_1 + \kappa\_2} |a|^{s\_2 - s\_1 - 1} F(x,y)$$
for $a \in \mathbb{R}^{\times}$. In particular, it has parity $\varepsilon = (-1)^{\kappa\_1 + \kappa\_2}$ and is $w$-homogeneous for $w = s\_2 - s\_1 - 1$.
Conversely, any smooth function $F : \mathbb{R}^2 \setminus \{(0,0)\} \to \mathbb{C}$ satisfying this property gives rise to a smooth function $f : \mathrm{GL}\_2(\mathbb{R}) \to \mathbb{C}$ in the induced model: by the Iwasawa decomposition, every $g \in \mathrm{GL}\_2(\mathbb{R})$ is of the form
$$g = \begin{pmatrix} a & b \\ 0 & d \end{pmatrix} \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}$$
for some $a,d \in \mathbb{R}$, $b,\theta \in \mathbb{R}$, in which case we define
$$f(g) = \operatorname{sgn}(a)^{\kappa\_1} |a|^{s\_1} \operatorname{sgn}(d)^{\kappa\_2} |d|^{s\_2} \left|\frac{a}{d}\right|^{1/2} (-1)^{\kappa\_1 + \kappa\_2} F(\cos \theta,\sin \theta).$$
We call the vector space of such smooth functions $F$ the plane model of this principal series representation.
---
It remains to define an appropriate action of $\mathrm{GL}\_2(\mathbb{R})$ on smooth functions $F : \mathbb{R}^2 \setminus \{(0,0)\} \to \mathbb{C}$ satisfying
$$F(ax,ay) = \operatorname{sgn}(a)^{\kappa\_1 + \kappa\_2} |a|^{s\_2 - s\_1 - 1} F(x,y)$$
that intertwines between these two models. This action is given by
$$(h \cdot F)(x,y) = \operatorname{sgn}(\det h)^{\kappa\_2} \left|\det h\right|^{s\_2 - \frac{1}{2}} F((x,y) \, {}^t h^{-1}).$$
---
You are interested in two special cases of the above, where the action is $(h \cdot F)(x,y) = F((x,y) \, {}^t h^{-1})$.
The first is the case $\kappa\_1 = 0$, $\kappa\_2 = 0$, $s\_1 = -w - 1/2$, and $s\_2 = 1/2$, so that the action of $\mathrm{GL}\_2(\mathbb{R})$ on elements $F$ of the plane model is $(h \cdot F)(x,y) = F((x,y) \, {}^t h^{-1})$, while elements of the plane model are even and $w$-homogeneous, so that they satisfy $F(ax,ay) = |a|^w F(x,y)$ (i.e. the plane model is simply the vector space $V\_w^1$). The associated principal series representation is irreducible so long as $-w - 1 \neq \pm (k - 1)$ for some $k \geq 2$ with $k \equiv 0 \pmod{2}$ (i.e. $w$ is not an even integer).
The second is $\kappa\_1 = 1$, $\kappa\_2 = 0$, $s\_1 = -w - 1/2$, and $s\_2 = 1/2$. The action of $\mathrm{GL}\_2(\mathbb{R})$ on elements $F$ of the plane model is again $(h \cdot F)(x,y) = F((x,y) \, {}^t h^{-1})$, while elements of the plane model are instead odd and $w$-homogeneous, so that they satisfy $F(ax,ay) = \operatorname{sgn}(a) |a|^w F(x,y)$ (i.e. the plane model is $V\_w^{-1}$). The associated principal series representation is irreducible so long as $-w - 1 \neq \pm (k - 1)$ for some $k \geq 2$ with $k \equiv 1 \pmod{2}$ (i.e. $w$ is not an odd integer except possibly $w = -1$).
| 6 | https://mathoverflow.net/users/3803 | 449340 | 180,842 |
https://mathoverflow.net/questions/449337 | 6 | **Setting:** Consider sampling two orthonormal vectors $\mathbf{u},\mathbf{v}$ in $\mathbb{R}^p$ (where $p\ge2$) from a "uniform" distribution over the $p$-dimensional sphere (alternatively, sample $\mathbf{u},\mathbf{v}\sim\mathcal{S}^{p-1}$ and then orthogonalize and normalize $\mathbf{v}$).
Let $\lceil p/2\rceil\le m\le p$.
**Question:** Can we upper bound the following expectation?
$$ \mathbb{E}\_{\substack{\mathbf{u},\mathbf{v}\sim\mathcal{S}^{p-1}:\\\mathbf{u}\perp\mathbf{v}}}
\left(\mathbf{u}^\top\left[\begin{array}{cc}
\mathbf{I}\_{m}\\
& \mathbf{0}\_{p-m}
\end{array}\right]\mathbf{v}\right)^2
=
\mathbb{E}\_{\substack{\mathbf{u},\mathbf{v}\sim\mathcal{S}^{p-1}:\\\mathbf{u}\perp\mathbf{v}}}
\left(\sum\_{i=1}^{m}u\_iv\_i\right)^2
$$
**Example:** when $m$ is very close to $p$, the expectation should be very small, since the vectors are nearly orthogonal.
**Direction:** We thought about approximating this with just two independent Gaussian vectors and look for the concentration bounds when $p\to\infty$, but this seems perhaps too loose?
**Observation:** Since the vectors are orthogonal, we have $
\left(\mathbf{u}^\top\left[\begin{array}{cc}
\mathbf{I}\_{m}\\
& \mathbf{0}\_{p-m}
\end{array}\right]\mathbf{v}\right)^2
=
\left(\mathbf{u}^\top\mathbf{v}-\mathbf{u}^\top\left[\begin{array}{cc}
\mathbf{I}\_{m}\\
& \mathbf{0}\_{p-m}
\end{array}\right]\mathbf{v}\right)^2
=
\left(\mathbf{u}^\top\left[\begin{array}{cc}
\mathbf{0}\_{m}\\
& \mathbf{I}\_{p-m}
\end{array}\right]\mathbf{v}\right)^2$, and thus we can equivalently focus on $0\le m\le \lfloor p/2\rfloor$.
Any help and ideas would be greatly appreciated!
| https://mathoverflow.net/users/100796 | Expectation of the inner product of a subset of two random orthonormal vectors | Denote $\alpha=\mathbb{E} u\_1^2v\_1^2$, $\beta=\mathbb{E} u\_1v\_1u\_2v\_2$. Then by the symmetry and linearity of expectation we have $$f(m):=\mathbb{E} (u\_1v\_1+\ldots+u\_mv\_m)^2=m\alpha+m(m-1)\beta.$$
We have $f(p)=0$, thus $\beta=-\alpha/(p-1)$, and $f(m)=\alpha m(p-m)/(p-1)$.
It remains to bound $\alpha$. Choose a vector $w=(w\_1,\ldots,w\_p)\in \mathcal{S}^{p-1}$ independent of $u,v$, then $\alpha=\mathbb{E} \langle u,w\rangle^2\cdot \langle v,w\rangle^2$, as the conditional expectation clearly does not depend on $w$ (here $\langle \cdot,\cdot\rangle$ stands for the inner product). On the other hand, the conditional expectation does not depend on the pair $u,v$ of orthogonal vectors $u,v$, thus we may take $u=(1,0,\ldots,0)$, $v=(0,1,\ldots,0)$, and $\alpha=\mathbb{E} w\_1^2w\_2^2$.
Next, we have
$$1=\mathbb{E} \left(\sum\_{i=1}^p w\_i^2\right)^2=\alpha\cdot p(p-1)+p\cdot \mathbb{E} w\_1^4,$$
thus $$\alpha=\frac{1-p\cdot \mathbb{E} w\_1^4}{p(p-1)}.$$
To find $\mathbb{E} w\_1^4$, we may think that $w=(w\_1,\ldots,w\_p)=(\xi\_1,\ldots,\xi\_p)/\sqrt{\sum \xi\_i^2}$ where $\xi\_i$ are i.i.d. standard normal. Then
$$
\mathbb{E} w\_1^4=\mathbb{E} \frac{\xi\_1^4}{(\sum \xi\_i^2)^2}=:\Theta
$$
By some form of law of large numbers like Chernoff bound, the probability that $\sum \xi\_i^2<p/2$ is exponentially small in $p$, that gives exponentially small contribution to the expectation $\Theta$. If $\sum \xi\_i^2>p/2$, then $$\frac{\xi\_1^4}{(\sum \xi\_i^2)^2}<\frac{4}{p^2}\xi\_1^4,$$
and since $\mathbb{E} \xi\_1^4$ is a finite constant, we conclude that $\Theta=O(1/p^2)$. Thus $\alpha=1/p^2+O(1/p^3)$.
| 3 | https://mathoverflow.net/users/4312 | 449362 | 180,849 |
https://mathoverflow.net/questions/449324 | 2 | For reference, the linked paper is Composite parameterization and Haar measure for all unitary and special unitary groups by Christoph Spengler, Marcus Huber and Beatrix C. Hiesmayr (J. Math. Phys. 53, 013501 (2012)).
I am implementing an algorithm that "scrambles" a Hamiltonian by a random (w.r.t. to the Haar measure) special unitary operator:
$$H\_{\text{scram}} = UH\_0U^\dagger.$$
I would like these unitaries to be parameterized as the algorithm minimizes a quantity by varying $U$. I found that [this](https://arxiv.org/pdf/1103.3408.pdf) paper provides precisely what I need as it gives a parameterization of $\mathrm{SU}(N)$ and the Haar measure in terms of the same parameters.
It would not be too bad to implement the results of the linked paper from scratch, but I was wondering if a Python package already exists which samples random special unitaries and provides a parameterization of said unitaries, so that I can scramble a Hamiltonian as defined above and then perform an optimization procedure w.r.t. the parameters parameterizing $\mathrm{SU}(N)$.
| https://mathoverflow.net/users/505917 | Does there exist a Python package that samples random special unitary matrices such that the matrices are parameterized | I would just use Euler angles to parameterise the unitary, and then vary the angles according to the Haar measure. Below I copy the relevant equations from [Zyczkowski and Kus](http://yaroslavvb.com/papers/zyczkwoski-random.pdf). I do not have a Python code (when I used this method we were still using Fortran...), but it should be only a few lines. (For $SU(N)$ instead of $U(N)$ you would omit the angle $\alpha$.)
There is an alternative method that is computationally more intensive, but can be explained in one line: Parameterize a general complex symmetric matrix $H$ with normally distributed elements and compute the unitary matrix $U$ of eigenvectors.
---
| 2 | https://mathoverflow.net/users/11260 | 449365 | 180,851 |
https://mathoverflow.net/questions/449358 | 3 | Let $a>0$ be a fixed number and consider the Hermite operator (or harmonic oscillator) defined by
\begin{equation}
Hf(x)=x^2f(x)-f''(x)
\end{equation}
for any smooth function $f$ compactly supported on the interval $(-a,a)$. Then, we may extend $H$ as an unbounded operator on $L^2(-a,a)$.
Extending from the Wikipedia link : <https://en.wikipedia.org/wiki/Dirichlet_eigenvalue> I would like to compute the first Dirichlet eigenvalue $\lambda\_a$ of the above harmonic operator $H$ on the interval $(-a,a)$.
When working on whole $\mathbb{R}$, it is well-known that the first eigenvalue of $H$ is $1$. What I suspect from connection with the standard Gaussian measure is the following formula:
\begin{equation}
\lambda\_a=\frac{\sqrt{2\pi}}{\int\_{-a}^a e^{-\frac{x^2}{2}}dx}
\end{equation}
However, I cannot find a way to justify my guess. I tried to look for a relevant reference but all seem to deal with the Laplacian only, let alone the explicit formula for the first Dirichlet eigenvalue..
Could anyone please help me?
| https://mathoverflow.net/users/56524 | First Dirichlet eigenvalue of the harmonic oscillator on a bounded interval $(-a,a)$? | Your formula for the first Dirichlet eigenvalue cannot be correct: for $a=1/\sqrt{2}$, the first eigenvalue is $5$.
Indeed, the eigenfunction
$$y(x)=e^{-x^2/2}(2x^2-1)$$
is positive on $(-1/\sqrt{2},1/\sqrt{2})$, zero at the ends,
and satisfies
$$x^2y-y''=5y,$$
therefore $5$ is the smallest eigenvalue.
Your formula gives $2.071...$, which is not close.
For arbitrary $a$, eigenvalues and eigenfunctions can be expressed in terms of [Weber functions](https://en.wikipedia.org/wiki/Parabolic_cylinder_function).
| 5 | https://mathoverflow.net/users/25510 | 449368 | 180,852 |
https://mathoverflow.net/questions/448818 | 4 | This question is "take 2" of [this older one](https://mathoverflow.net/questions/420048/comparing-bornologies-for-domination-escaping), following a suggestion of Francois Dorais. Consider the following [bornologies](https://en.wikipedia.org/wiki/Bornology) $\mathbb{D},\mathbb{E}$ on the set $\mathcal{N}$ of all functions from $\mathbb{N}$ to $\mathbb{N}$:
* $\mathbb{D}=\{A: \exists f\in\mathcal{N}\forall g\in A\exists m\in\mathbb{N}\forall n>m(f(n)>g(n))\}$. (**Dominatable** sets)
* $\mathbb{E}=\{A: \exists f\in\mathcal{N}\forall g\in A\forall m\in\mathbb{N}\exists n>m(f(n)>g(n))\}$. (**Escapable** sets)
Clearly we must have $\mathfrak{b}=\mathfrak{d}$ in order for these to yield bornomorphic structures on $\mathcal{N}$, and Francois Dorais showed at the above-linked question that the converse holds as well. However, the "tame maps" version of the question is still open:
>
> Suppose $\mathfrak{b}=\mathfrak{d}$. Must there be a **Borel** $i:\mathcal{N}\rightarrow\mathcal{N}$ such that the $i$-image of each set in $\mathbb{D}$ is in $\mathbb{E}$ and the $i$-preimage of each set in $\mathbb{E}$ is in $\mathbb{D}$?
>
>
>
It's not even clear to me (per James Hanson's comment below) what the answer is assuming $\mathsf{V=L}$.
| https://mathoverflow.net/users/8133 | Comparing bornologies for cardinal characteristics via Borel maps | A bornomorphic map $i\colon\mathcal{N}\to\mathcal{N}$ for these bornologies cannot be Borel.
First, I shall use $f<^\infty g$ to mean that $(\forall m\in\mathbb{N})(\exists n>m)f(n)<g(n)$, and $f<^\*g$ to mean that $(\exists m\in\mathbb{N})(\forall n>m)f(n)<g(n)$.
For $f\in\mathcal{N}$, let $\mathbb{D}\_f=\{g\in\mathcal{N}\mid g<^\* f\}$, and let $\mathbb{E}\_f=\{g\in\mathcal{N}\mid g<^\infty f\}$. Since the $\mathbb{D}\_f$ and $\mathbb{E}\_f$ generate their respective ideals, $i$ is a bornomorphism if and only if for all $f\in\mathcal{N}$ there is $a(f)\in\mathcal{N}$ such that $i``\mathbb{D}\_f\subseteq\mathbb{E}\_{a(f)}$ and $u(f)\in\mathcal{N}$ such that $i^{-1}(\mathbb{E}\_f)\subseteq\mathbb{D}\_{u(f)}$.
We shall focus on the $u$ case. Another way of saying this is for all $g,f\in\mathcal{N}$, if $i(g)<^\infty f$ then $g<^\*u(f)$.
Adjoin to the universe a Cohen generic function $c\colon\mathbb{N}\to\mathbb{N}$, and let $d=\tilde{i}(c)$, where $\tilde{i}$ is the Borel function with the same code as $i$. Since ground model elements of $\mathcal{N}$ are a $<^\infty$-dominating family in the extension, there must be $f\in\mathcal{N}$ in the ground such that $d<^\infty f$. Let $h=u(f)$. Then, by our assumption, in the ground model we get
$$(\forall x\in\mathcal{N})(i(x)<^\infty f\implies x<^\*h)$$
This is a $\Pi^1\_1$-statement involving only $f$, $h$, and (a code for) $i$, so it remains true in our extension. In particular, generic cohen real $c$ is satisfies $c<^\*g$. However, Cohen generic functions $\mathbb{N}\to\mathbb{N}$ are never dominated by ground model reals.
---
This proof is a small adjustment of Andreas Blass's Theorem 4.15 in *Combinatorial Cardinal Characteristics of the Continuum*.
*Blass, Andreas*, [**Combinatorial cardinal characteristics of the continuum**](https://doi.org/10.1007/978-1-4020-5764-9_7), Foreman, Matthew (ed.) et al., Handbook of set theory. In 3 volumes. Dordrecht: Springer (ISBN 978-1-4020-4843-2/hbk; 978-1-4020-5764-9/ebook). 395-489 (2010). [ZBL1198.03058](https://zbmath.org/?q=an:1198.03058).
| 6 | https://mathoverflow.net/users/478588 | 449381 | 180,854 |
https://mathoverflow.net/questions/449375 | 7 | The [replica trick](https://en.wikipedia.org/wiki/Replica_trick) attempts to calculate the expectation of the logarithm $X=\log(Z)$ of a random variable $Z$. The wikipedia article describes the logarithm as the limit
$$
\log(Z) = \lim\_{n\to 0}\frac{Z^n -1}{n}, \quad\text{or}\quad \log(Z) = \lim\_{n\to 0} \frac{\partial Z^n}{\partial n}
$$
But I think it is much more natural to calculate the moment generating function of $X$ (and not just the expectation $\mathbb{E}[X]$):
$$
M(\lambda) = \mathbb{E}[\exp(\lambda X)] = \mathbb{E}[Z^\lambda]
$$
Now for $\lambda = n\in \mathbb{N}$ this is simply calculating the expectation of $n$ "replicas" of $Z$, which is where the name comes from.
The expectation can easily be obtained as $\mathbb{E}[X]=M'(0)$. For this we need to let $\lambda\to 0$
>
> The crux of the replica trick is that while the disorder averaging is done assuming $n$ to be an integer, to recover the disorder-averaged logarithm one must send $n$ continuously to zero. This apparent contradiction at the heart of the replica trick has never been formally resolved, however in all cases where the replica method can be compared with other exact solutions, the methods lead to the same results. - [Wikipedia](https://en.wikipedia.org/wiki/Replica_trick)
>
>
>
So what is happening is: By calculating $M(n)$ we hope to obtain a function $f$ which coincides with $M$ on the natural numbers and hope that it coincides with $M$ completely, i.e.
$$
f: \mathbb{C} \to \mathbb{C},\quad M(n) = f(n)\quad \forall n\in \mathbb{N}\overset?\implies M\equiv f
$$
Were we view $f$ and $M$ as complex functions because this Wikipedia entry already contains the suggestion to use Carlson's Theorem
>
> To prove that the replica trick works, one would have to prove that Carlson's theorem holds, that is, that the ratio $\frac{Z^n − 1}{n}$ exponential type less than pi.
>
>
>
Now I am surprised about this phrasing, since that requirement
of Carlson's Theorem is actually straightforward to prove.
So here are the requirements of [Carlson's Theorem](https://en.wikipedia.org/wiki/Carlson%27s_theorem) which we want to apply to $g:= f- M$:
1. $g$ is an entire function of exponential type
2. There exists $C\in \mathbb{R}$ and $c<\pi$ such that
$$
|g(iy)|\le Ce^{c|y|} \quad \forall y\in \mathbb{R}
$$
3. $g(n)=0\quad \forall n\in\mathbb{N}$
which implies $g\equiv 0$ and therefore $M\equiv f$. The third requirement is the heart of the replica trick, the second requirmenent seems to be the one referred to as the problematic one in the [replica trick](https://en.wikipedia.org/wiki/Replica_trick) wikipedia entry. Now we can either prove this requirement for $g$, or for $f$ and $M$ which is sufficient (by the triangle inequality and replacing $C$ with $2C$). But this property is trivial for $M$, since
$$
|M(iy)| = |\mathbb{E}[\exp(iyX)]| \le \mathbb{E}[|\exp(iyX)|] = 1
$$
So $C=1$ and $c=0<\pi$ does the job. Since $f$ is the function we guess in the replica trick, we would simply need to check this requirement for the guessed function too.
So it seems like the much more difficult problem is proving that $g$ is an entire function of (arbitrary) exponential type (i.e. the first requirement).
In [this paper](https://arxiv.org/pdf/2108.12846.pdf) the first requirement of Carlson's theorem is further relaxed to be:
>
> $g$ needs only be holomorphic in the closed right half plane (not even of exponential type)
>
>
>
My questions now are:
* am I overlooking something important?
* What are sufficient requirements on $Z$ such that the moment generating function $M$ of $X=\log(Z)$ is holomorphic (assuming that $M$ holomorphic and $f$ holomorphic implies $g$ is holomorphic - holomorphic functions are a vector space right?)
To prove the replica trick works, practicioners would then only need to check this condition on $Z$ and that their guess of $f$ is holomorphic, and satisfies the second requirement of Carlson's theorem (it can't possibly be a moment generating functions if it doesn't anyway). Additionally they get the entire moment generating function $M$ and not just the expectation. This feels like it is too good to be true.
| https://mathoverflow.net/users/122659 | Proving the Replica Trick works | **Q:** *Am I overlooking something important?*
I think you are ignoring the role played by the thermodynamic limit.
There are two interplaying limits here, the replica limit $n\rightarrow 0$ and the thermodynamic limit $N\rightarrow \infty$, where $N$ quantifies the system size. The usual practice in the replica method is to take the limit $N\rightarrow\infty$ first, for integer $n$, and then send $n\rightarrow 0$ by analytic continuation (relying on Carlson's theorem). The correct procedure takes the limit $n\rightarrow 0$ first, and then sends $N\rightarrow\infty$. The two answers will differ if the analytic continuation for finite $N$ exhibits a singularity in the $N\rightarrow\infty$ limit.
An example of such a breakdown of analyticity is given in [Exact analytic continuation with respect to the replica number in the discrete random energy model of finite system size.](https://arxiv.org/abs/cond-mat/0310490)
| 10 | https://mathoverflow.net/users/11260 | 449387 | 180,858 |
https://mathoverflow.net/questions/449355 | 0 | The $\|\cdot\|\_{\infty}$-norm on $\mathbb{R}^n$ for $n\in \mathbb{Z}^+$ is not a smooth function. However, I came across [this post](https://stats.stackexchange.com/questions/298849/soft-version-of-the-maximum-function) which essentially says that a pointwise approximation to the maximum function, and therefore $\|\cdot\|\_{\infty}$ is
$$
m\_{\lambda}(x) = \sum\_{i=1}^n\,\frac{e^{\lambda\, x\_i}x\_i}{\sum\_{j=1}^n\,e^{\lambda\, x\_j}}
$$
where $x=(x\_i)\_{i=1}^n\in \mathbb{R}^n$ is arbitrary.
Fix $\lambda>0$ and consider the map $d\_{\lambda}:\mathbb{R}^n\times \mathbb{R}^n\rightarrow [0,\infty)$ defined by
$$
d\_{\lambda}(x,y) = m\_{\lambda}(|x-y|)
$$
where $|z|:=(|z\_i|)\_{i=1}^n$ for any $z\in \mathbb{R}^n$.
Does $d\_{\lambda}$ define a quasi-metric on $\mathbb{R}^n$?
It certainly satisfies positivity, symmetry and $d\_{\lambda}(x,y)=0$ if and only if $x=y$ by its not obvious that it should satisfy a relaxed triangle inequality: ie
$$
d\_{\lambda}(x,y) \le C\big(d\_{\lambda}(x,z)+d\_{\lambda}(z,y)\big)
$$
for every $x,y,z\in \mathbb{R}^n$ and some $C\ge 1$ independent of $x,y,z$.
| https://mathoverflow.net/users/496781 | Is this a smooth approximation to the $\ell$-infinity distance actually a quasi-metric? | $\newcommand\R{\mathbb R}$By rescaling, without loss of generality $\lambda=1$.
So, the question becomes the following: is there a real $C$ not depending on $u=(u\_1,\dots,u\_n)\in\R\_+^n$, $v=(v\_1,\dots,v\_n)\in\R\_+^n$, and $w=(w\_1,\dots,w\_n)\in\R\_+^n$ such that for all such $u,v,w$
$$w\le u+v\implies m(w)\le C(m(u)+m(v)) \tag{1}\label{1},$$
where $w\le u+v$ means $w\_i\le u\_i+v\_i$ for all $i\in[n]:=\{1,\dots,n\}$ and $m(u):=\dfrac{\sum\_{i\in[n]}e^{u\_i}u\_i}{\sum\_{i\in[n]}e^{u\_i}}$?
Clearly, $m(u)\le\max u:=\max\_{i\in[n]}u\_i$ for $u\in\R\_+^n$. Also, by (say) [Chebyshev's sum inequality](https://en.wikipedia.org/wiki/Chebyshev%27s_sum_inequality), $m(u)\ge\bar u:=\frac1n\sum\_{i\in[n]}u\_i$ for $u\in\R\_+^n$. So, for all $u,v,w$ in $\R\_+^n$ such that $w\le u+v$,
$$m(w)\le\max w\le\max u+\max v\le n\bar u+n\bar v\le n m(u)+n m(v),$$
so that \eqref{1} holds with $C=n$.
| 1 | https://mathoverflow.net/users/36721 | 449395 | 180,860 |
https://mathoverflow.net/questions/449389 | 3 | For a polynomial $f \in \mathbb{R}[x\_1, \cdots, x\_n]$, we say that $f$ is *coercive* (see my earlier question: [Real polynomials that go to infinity in all directions: how fast do they grow?](https://mathoverflow.net/questions/444925/real-polynomials-that-go-to-infinity-in-all-directions-how-fast-do-they-grow)) if
$$\displaystyle \lim\_{\lVert \mathbf{x} \rVert} f(\mathbf{x}) = \infty.$$
More precisely, $f$ is coercive if for all $c \in \mathbb{R}$ there exists $M(c)$ such that whenever $\lVert \mathbf{x} \rVert \geq M(c)$ we have $f(\mathbf{x}) \geq c$. Further, it is known that there is always a positive *order of coercivity* $q(f)$ satisfying the property that for all $c > 0$ there exists $M(c)$ satisfying
$$\displaystyle \lVert \mathbf{x} \rVert \geq M(c) \Rightarrow f(\mathbf{x}) \geq c \lVert \mathbf{x} \rVert^{q(f)}.$$
It is known that $q(f)$ can be arbitrarily small (see: [How fast do coercive polynomials grow?](http://num.math.uni-goettingen.de/preprints/files/2017-1.pdf)).
My question is: does the following statement hold?
Let $f \in \mathbb{R}[x\_1, \cdots, x\_n]$ be coercive. Then there exists a positive number $c\_1$ and a real number $c\_2$ such that
$$\displaystyle f(x\_1, \cdots, x\_n) \geq c\_1 \min\{|x\_1|, \cdots, |x\_n|\}^2 - c\_2.$$
| https://mathoverflow.net/users/10898 | Lower bound for coercive polynomials | A counterexample is given by
$$
f(x,y)=(x^4-y^3)^2+y .
$$
Clearly, $f(R, R^{4/3})=R^{4/3}\ll R^2 = \left( \min \{ R, R^{4/3} \} \right)^2$, so $f$ doesn't satisfy your condition.
However, $f$ is coercive: If $y\le 0$, then $f(x,y)\ge x^8+y^6 +y$ and now either $|y|$ is not large and there are no problems or if $|y|\gg 1$, then $y^6+y\ge (1/2)y^6$, say.
If $y\ge 0$, then $f(x,y)\ge \max\{ (x^4-y^3)^2, y\} \to\infty$ as $|(x,y)|\to\infty$, $y\ge 0$.
| 3 | https://mathoverflow.net/users/48839 | 449411 | 180,864 |
https://mathoverflow.net/questions/449415 | 1 | Reading on Infinitary languages I'm only seeing first order infinitary languages $\mathcal L\_{\kappa, \lambda}$, i.e. in all of these languages no quantification over predicate and function symbols is allowed.
>
> Are there second (or even higher) order infinitary languages? I mean something like $\mathcal L^2\_{\kappa, \lambda}$ (more generally $\mathcal L^n\_{\kappa, \lambda}$)rendering of the infinitary first order language $\mathcal L\_{\kappa, \lambda}$. Or, is it the case that all of those are reducible to first order infinitary languages, and so dispense with all of them?
>
>
>
>
> If there are, what are the recommended sources on those?
>
>
>
| https://mathoverflow.net/users/95347 | Are there second (or higher) order infinitary logic languages? References? | Sure there are. They even come up in practice from time to time; e.g. to show that assuming Vopenka's Principle the modal analogue of *second-order* logic (a la [Hamkins/Woloszyn](https://arxiv.org/abs/2009.09394)) has definable-in-$V$ semantics, the only argument I'm aware of goes through $\mathcal{L}\_{\theta,\theta}^2$ where $\theta$ is the smallest limit of extendible cardinals. *(This is basically a Los-Tarski-type argument, see [here](https://mathoverflow.net/questions/428896/how-strong-is-this-modal-definability-principle).)*
That said, even finitary second-order logic is incredibly complicated. In particular, it doesn't have any good metatheorems (compare forcing absoluteness for $\mathcal{L}\_{\infty,\omega}$ or [complicated analogues of Barwise compactness](https://www.sciencedirect.com/science/article/pii/0168007284900125?via%3Dihub) for $\mathcal{L}\_{\infty,\omega\_1}$). So I'm not aware of any source treating infinitary extensions of $\mathsf{SOL}$ or higher-order logics in detail; it's more something that is treated as it shows up.
It's also worth noting that there's a finitary second-order sentence not equivalent even to any $\mathcal{L}\_{\infty,\infty}$ sentence:
>
> "The cardinality of the universe is a successor cardinal."
>
>
>
| 2 | https://mathoverflow.net/users/8133 | 449416 | 180,867 |
https://mathoverflow.net/questions/448607 | 5 | Let $\mathfrak S\_w(x\_1,\ldots,x\_n)$ be a Schubert polynomial. It's known that if we pick an index $i$, there are nonnegative integer coefficients $c\_{w'}^w(i,j)$ such that
$$\mathfrak S\_w(x\_1,\ldots,x\_n)=\sum\_{w'\in S\_\infty,j}c\_{w'}^w(i,j)x\_i^j\mathfrak S\_{w'}(x\_1,\ldots,x\_{i-1},x\_{i+1},\ldots,x\_n)$$
When $i=1$, Bergeron and Billey found a positive formula for $c\_{w'}^w(i,j)$. I know a positive formula for all $i$, but I'm wondering if anyone else does. Does anyone have a reference?
I'm writing about it in a paper, and it would actually make my paper more interesting if someone else found a formula for it, especially if it was found before $1995$.
| https://mathoverflow.net/users/62135 | Pulling out a variable from a Schubert polynomial | This is in a Section five of a paper of mine with Bergeron in the Transactions of the AMS, which identifies it.
| 2 | https://mathoverflow.net/users/507423 | 449425 | 180,870 |
https://mathoverflow.net/questions/448489 | 3 | Let $X$ be a K3 over $\overline{\mathbb{F}\_p}$. The (crystalline version's) Tate conjecture predicts:
$c\_1: Pic(X)\otimes\mathbb{Q}\_p\rightarrow H^2\_{crys}(X/W)^{\Phi=p}\otimes\mathbb{Q}\_p$
is an isomorphism.
If $\lambda=1-\frac{1}{h}$ is the smallest slope of $H^2\_{crys}(X/W)$, then $H^2\_{crys}(X/W)$ must be isogenous to $H\_{\lambda}\oplus H\_1^{22-2h} \oplus H\_{2-\lambda}$, where $H\_\alpha$ is the F-crystal that $\Phi^t=p^s$ for $\alpha=s/t$. The $\Phi=p$ parts of the latter F-crystal apparently has rank $22-2h$, so the $\Phi=p$ part of $H^2\_{crys}(X/W)$ must also has rank $22-2h$? Which means the above conjecture is equivalent to $\rho=22-2h$ always hold true? But the later one is not true in general, as a general K3 in char $p$ should have Picard number $2$, which is usually less than $22-2h$.
Where was I wrong?
| https://mathoverflow.net/users/177957 | (crystalline cohomology version's) Tate's conjecture for K3 surfaces | This is related to a phenomenon called 'hypersymmetry' (or the lack of it). The term comes from Chai and Oort, who explored this for abelian varieties. The essential point is that the category of F-isocrystals over the algebraic closure of a finite field is *not* the filtered colimit of the corresponding categories over its finite subfields.
For the simplest example, take a product of two non-isogenous ordinary elliptic curves $A = E\_1\times E\_2$ over $\overline{\mathbb{F}}\_p$. You can get such a pair by choosing two Weil $q$-numbers of weight $1$ generating distinct quadratic extensions of $\mathbb{Q}$. Then we have (with the Tate twist on the left hand side)
$H^2\_{crys}(A/W(\overline{\mathbb{F}}\_p)(1) = \mathrm{Hom}(H^1\_{crys}(E\_1/W(\overline{\mathbb{F}}\_p),H^1\_{crys}(E\_2/W(\overline{\mathbb{F}}\_p))\oplus H^2\_{crys}(E\_1/W(\overline{\mathbb{F}}\_p))(1) \oplus H^2\_{crys}(E\_2/W(\overline{\mathbb{F}}\_p) )(1)$
Since both elliptic curves are ordinary, they have isogenous (as $F$-crystals) crystalline cohomology groups over $\overline{\mathbb{F}}\_p$. In particular, if you take $\varphi$-invariants, you will get a 4-dimensional $\mathbb{Q}\_p$-vector space.
However, if you repeat this over any finite field containing the $j$-invariants of $E\_1$ and $E\_2$, then the crystalline Tate conjecture says that the $\varphi$-invariants will only be 2-dimensional over $\mathbb{Q}\_p$, since the first Hom space should have no $\varphi$-invariant elements.
If you take the Kummer K3 surface associated with $A$, you will find a K3 surface that has the same feature: The $\varphi$-invariants in $H^2\_{crys}$ over $W(\overline{\mathbb{F}}\_p)$ will have two extra ‘transcendental’ dimensions that are not visible over any $W(\mathbb{F}\_q)$.
| 3 | https://mathoverflow.net/users/7868 | 449426 | 180,871 |
https://mathoverflow.net/questions/449282 | 0 | I'm trying to construct lists with elements of type $A$ as the initial algebra over a base endofunctor in $\mathsf{Set}/\mathcal{P}(A)$, such that the list is indexed by the set of its elements.
My idea is to model `Cons` using a bifunctor $C\colon\ \mathsf{Set}/\mathcal{P}(A)\times\mathsf{Set}/\mathcal{P}(A)\to \mathsf{Set}/\mathcal{P}(A)$.
My definition of $C$ would be $C\_0 (X,f) (Y,g) := (X\times Y, \lambda(a,b).\,f(a)\cup g(b))$.
My List base functor would then be $L\_0(R,r) := (1,\lambda\\_.\,\emptyset ) + C\_0\left(\left(A,\lambda x.\,\{x\}\right),\left(R,r\right)\right)$.
My question is: Is this kind of bifunctor $C$ a known construction? I find what it does reminiscent of the $\mu$ (join) of the [Action Monad](https://ncatlab.org/nlab/show/action+monad) (aka Writer monad), but can't quite put my finger on it.
| https://mathoverflow.net/users/502814 | Is this kind of functor $\mathsf{Set}/M×\mathsf{Set}/M\to \mathsf{Set}/M$, with $M$ a monoid, a known construction? | In general, for any monoidal category $(\mathcal M, \otimes, I)$ and a monoid $(m, \*, i) \in \mathcal M$, the [slice category](https://ncatlab.org/nlab/show/over+category#examples) $\mathcal M/m$ inherits the monoidal structure pointwise, i.e. the tensor product is given by taking morphisms $f : x \to m$ and $g : y \to m$ to the morphism $x \otimes y \xrightarrow{f \otimes g} m \otimes m \xrightarrow{\*} m$, and the unit is given by $i : I \to m$.
In your example, we take $(\mathbf{Set}, \times, 1)$ to be the category of sets equipped with [cartesian monoidal structure](https://ncatlab.org/nlab/show/cartesian+monoidal+category) and $(m, \*, i)$ to be the power set monoid $(\mathcal P A, \cup, \varnothing)$.
Abstractly, we can see resulting tensor product on $\mathbf{Set}/\mathcal{P}A$ to be given by the following composite
$$(\mathbf{Set}/\mathcal{P}A)^2 \simeq (\mathbf{Set}^{\mathcal{P}A})^2 \xrightarrow{\langle-,-\rangle} (\mathbf{Set}^2)^{({\mathcal{P}A}^2)} \xrightarrow{\times^{({\mathcal{P}A}^2)}} \mathbf{Set}^{({\mathcal{P}A}^2)} \simeq \mathbf{Set}/({\mathcal{P}A}^2) \xrightarrow{\mathbf{Set}/\cup} \mathbf{Set}/{\mathcal{P}A}$$
where the equivalences are given by the correspondence between indexed sets and $\mathbf{Set}$-functors from discrete categories.
The unit is given similarly by
$$1 \xrightarrow{1} \mathbf{Set} \simeq \mathbf{Set}/1 \xrightarrow{\mathbf{Set}/i} \mathbf{Set}/{\mathcal{P}A}$$
| 5 | https://mathoverflow.net/users/152679 | 449441 | 180,879 |
https://mathoverflow.net/questions/449451 | 8 | Let $k$ be an algebraically closed field, and $X$ a projective variety over $k$. Let $i : X\subset \mathbf{P}^d\_k$ be a closed immersion into a projective space of high enough dimension.
>
> Is there a smooth projective variety $X'$ that is a global complete intersection in some projective space, together with an alteration $X'\to X$?
>
>
>
In other words, can one alter projective varieties into smooth global complete intersections? Presumably the answer is "no" (for instance when $X$ itself is smooth) and there's some cohomological obstruction that I'm unaware of.
| https://mathoverflow.net/users/501361 | Alterations and smooth complete intersections | As you guessed, there are cohomological obstructions. Indeed, if $f \colon Y \twoheadrightarrow X$ is a surjective morphism of smooth projective varieties and $H$ is a Weil cohomology theory (with coefficients containing $\mathbf Q$), then the pullback $f \colon H^i(X) \to H^i(Y)$ is *injective*; see for instance [Kleiman, Prop. 1.2.4].
So in particular, if $\dim X > 1$ and $H^1(X) \neq 0$, then this is impossible since a smooth complete intersection $Y$ satisfies
$$h^i(Y) = \begin{cases}1, & i \text{ even} \\ 0, & i \text{ odd}\end{cases}$$
for $i < \dim Y$. (Since $H^1(Y)$ is a birational invariant of smooth projective varieties, replacing $Y \twoheadrightarrow X$ by a dominant rational map $Y \dashrightarrow X$ doesn't help.)
---
**References.**
[Kleiman] S. L. Kleiman, *Algebraic cycles and the Weil conjectures*. In: Dix exposés sur la cohomologie des schémas, Adv. Stud. Pure Math. **3**, p. 359-386 (1968). [ZBL0198.25902](https://zbmath.org/?q=an:0198.25902).
| 11 | https://mathoverflow.net/users/82179 | 449452 | 180,881 |
https://mathoverflow.net/questions/449445 | 3 | Let $X$ be a smooth and proper scheme over $\mathbb{Q}$ and choose integers $n,i$ such that $n>\frac{i}{2}+1$. Then we have
$$ ord\_{s=i+1-n}L(H^i(X),s)=\dim H^{i+1}\_{\mathcal{D}}(X\_\mathbb{R},\mathbb{R}(n))=\text{Ext}^1\_{\mathbb{R}-MHS}(\mathbb{R},H^i(X(\mathbb{C}),\mathbb{R}(n))$$
(see for instance 3.1.4 in [this survey](https://www.dpmms.cam.ac.uk/%7Eajs1005/preprints/d-s.pdf)). However I'm confused by the indexing: if we assume that $X$ is an elliptic curve then we hope/conjecture that $ord\_{s=1} L(H^1(X),s)$ is equal to $\dim X(\mathbb{Q})\otimes \mathbb{Q}\cong \text{Ext}\_{\mathcal{MM}}^1(1,H^1(X)(1))$. Then it would actually be natural to have $i=1,n=1$ to have
$$ord\_{s=1+1-1=1} L(H^1(X),s)=\dim \text{Ext}^1\_{\mathbb{R}-MHS}(\mathbb{R},H^1(X(\mathbb{C}),\mathbb{R}(1))\\
\overset{?}{=}\text{Ext}\_{\mathcal{MM}}^1(1,H^1(X)(1))\cong \text{rank } X(\mathbb{Q})$$
that is the Beilinson conjecture $(?)$ implies (the rank part of) the Birch/Swinnerton-Dyer conjecture (which makes sense since Beilinson's conjecture are supposed to generalize BSD). However this does not hold up as $1 \not>\frac{1}{2}+1$. Presumably my indexing is off by $1$ somewhere or this reasoning doesn't work and my question is which is it? Thanks!
| https://mathoverflow.net/users/152554 | Order of vanishing of $L$-function and mixed Hodge-structures | There is a good reason why *this particular form* of the Beilinson conjecture cannot possibly be valid for the particular $i$ and $n$ you mention.
The real $\mathbb{R}$-MHS $H^1(X(\mathbb{C}), \mathbb{R})$ is the same for all elliptic curves $X$: it's always a 2-dimensional pure Hodge structure of weight 1, with $(0, 1)$ and $(1, 0)$ parts 1-dimensional and interchanged by complex conjugation. So the dimension of $\operatorname{Ext}^1\_{\mathbb{R}-MHS}(\mathbb{R}, H^1(X(\mathbb{C}), \mathbb{R}(1)))$ is the same for all elliptic curves (it is always zero). However not all elliptic curves have the same rank!
A way of thinking about this is the following. Since $X$ is smooth proper, $L(H^i(X), s)$ is given by a *convergent* Euler product for $Re(s) > \tfrac{i}{2} + 1$. So for $n > \tfrac{i}{2} + 1$, the order of vanishing at $s = n$ is always 0, and hence the order of vanishing at $s = i + 1 - n$ can be read off from the shape of the functional equation (which depends only on the Hodge structure over $\mathbb{R}$). This is why it is reasonable to expect Ext groups of Hodge structures to fully capture the order of vanishing.
On the other hand, if $s = \tfrac{i + 1}{2}$ is the central value, then the order of vanishing of the L-function is a really deep, mysterious global invariant and in particular can't be read off just from knowing local information over $\mathbb{R}$. So this is why you should expect the formulation of the conjecture to be more difficult (and more interesting) in this central-value setting.
(The near-central value $s = \tfrac{i}{2} + 1$ is delicate for a different reason, because of vanishing of local Euler factors at finite primes.)
EDIT. I see you have $Ext^1\_{\mathcal{MM}}$ in your answer. There is always a map from $Ext^1\_{\mathcal{MM}}$ to $Ext^1\_{\mathbb{R}-MHS}$, but it isn't always an isomorphism. In the elliptic curve case we should always have $ord\_{s = 1}\, L(H^1(X), s) = dim\, Ext^1\_{\mathcal{MM}}$, but (if the rank is $> 0$) the map from $Ext^1\_{\mathcal{MM}}$ to $Ext^1\_{\mathbb{R}-MHS}$ will have a kernel.
| 4 | https://mathoverflow.net/users/2481 | 449453 | 180,882 |
https://mathoverflow.net/questions/449455 | 2 | I am looking for the english translation of the paper in russian [Variations of knotted graphs, geometric technique of n-equivalence, St. Petersburg Math. J. 12-4 (2001)](https://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=aa&paperid=1116&option_lang=eng) by Gusarov.
There is a .ps file on [Dror Bar-Natan website](https://www.math.toronto.edu/%7Edrorbn/Goussarov/) but I am unable to read the file. I already tried reading it with a postscript reader, or converting it to a pdf file.
Would someone have the english translation of that paper, that they could send to me? Or would someone know of a place where I could find it?
Thank you for your help!
| https://mathoverflow.net/users/504366 | English version of a paper by Gusarov | Try this [link](https://drive.google.com/file/d/14uqJjQ8a8Cckza5ctofDJF6c4k_Zxs0O/view?usp=sharing). I put the paper into pdf format; it looks ok to me. I didn't have any trouble reading the original postscript file with ghostscript on Ubuntu. If you can't read the pdf or the postscript file, maybe your computer is misbehaving.
| 4 | https://mathoverflow.net/users/13268 | 449456 | 180,883 |
https://mathoverflow.net/questions/449463 | 1 | Suppose $n\in\mathbb{N}$ and set $A\subseteq\mathbb{R}^{n}$.
If we define a sequence of sets $\left(F\_r\right)\_{r\in\mathbb{N}}$ with a [set theoretic limit](https://en.wikipedia.org/wiki/Set-theoretic_limit) of $A$; how do we define the rate *at* which $\left(F\_r\right)\_{r\in\mathbb{N}}$ converges to $A$?
| https://mathoverflow.net/users/87856 | Convergence rate of a sequence of sets to a set-theoretic limit? | $\newcommand\R{\mathbb R}\newcommand\si\sigma\newcommand\Si\Sigma$If $\mu$ is a finite measure on a $\si$-algebra $\Si$ over $\R^n$ and $(F\_r)$ is sequence in $\Si$ such that $\lim\_r F\_r=A$, then ($A\in\Si$ and) $\lim\_r \mu(F\_r)=\mu(A)$.
So then, one may define the rate of convergence of $(F\_r)$ to $A$ as the rate of convergence of $(\mu(F\_r))$ to $\mu(A)$.
Alternatively, again if $\mu$ is a finite measure on a $\si$-algebra $\Si$ over $\R^n$ and $(F\_r)$ is sequence in $\Si$ such that $\lim\_r F\_r=A$, then $\lim\_r (F\_r+A)=\emptyset$ and hence $\lim\_r \mu(F\_r+A)=0$, where $+$ denotes the symmetric difference.
So then, one may define the rate of convergence of $(F\_r)$ to $A$ as the rate of convergence of $(\mu(F\_r+A))$ to $0$. (Cf. e.g. [this paper](https://link.springer.com/article/10.1007/s11117-017-0507-8) or [its arXiv version](https://arxiv.org/abs/1702.01142).)
| 2 | https://mathoverflow.net/users/36721 | 449464 | 180,885 |
https://mathoverflow.net/questions/449361 | 25 | Let $a(n)$ be [A301897](https://oeis.org/A301897), i.e., number of permutations $b$ of length $n$ that satisfy the Diaconis-Graham inequality $I\_n(b) + EX\_n(b) \leqslant D\_n(b)$ with equality. Here
$$a(n)=\frac{1}{n+1}\binom{2n}{n}+\sum\limits\_{k=1}^{n-2}\sum\limits\_{j=1}^{n-k-1}\binom{n}{k-1}\binom{n-1}{k+j}\binom{n-k+j-1}{j-1}\frac{1}{j}$$
Let
$$R(n,q)=\sum\limits\_{j=0}^{q+q\operatorname{mod}3+1}R(n-1,j),$$ $$R(0,q)=1$$
I conjecture that
$$R(n,0)=a(n+1)$$
Here is the PARI/GP prog to check it numerically:
```
R2_upto(n)=my(v1, v2, v3); v1=vector(3*n+1, i, 1); v2=v1; v3=vector(n+1, i, 0); v3[1]=1; for(i=1, n, for(q=0, 3*(n-i), v2[q+1]=sum(j=0, q+q%3+1, v1[j+1])); v1=v2; v3[i+1]=v1[1];); v3
a(n)=binomial(2*n,n)/(n+1)+sum(k=1,n-2,sum(j=1,n-k-1,binomial(n,k-1)*binomial(n-1,k+j)*binomial(n-k+j-1,j-1)*(1/j)))
test(n)=R2_upto(n)==vector(n+1,i,a(i))
```
Is there a way to prove it?
| https://mathoverflow.net/users/231922 | Elegant recursion for A301897 | Here is an expanded version of the generating function argument I sketched in a comment.
For $i=1,2,3$, define the generating functions $F\_i(x,y) := \sum\_{n=0}^\infty \sum\_{q=0}^\infty R(n,3q+i) x^n y^q$, which are well defined for $x,y$ small. If one starts with the recursive identities
\begin{align\*} R(n,3q) &= \sum\_{0 \leq r \leq q} R(n-1,3r) + \sum\_{0 \leq r \leq q} R(n-1,3r+1) \\
&\quad + \sum\_{0 \leq r \leq q-1} R(n-1,3r+2)\\
R(n,3q+1) &= \sum\_{0 \leq r \leq q+1} R(n-1,3r) + \sum\_{0 \leq r \leq q} R(n-1,3r+1) \\
&\quad + \sum\_{0 \leq r \leq q} R(n-1,3r+2)\\
R(n,3q+2) &= \sum\_{0 \leq r \leq q+1} R(n-1,3r) + \sum\_{0 \leq r \leq q+1} R(n-1,3r+1) \\
&\quad + \sum\_{0 \leq r \leq q+1} R(n-1,3r+2)
\end{align\*}
for $n \geq 1$, multiplies by $x^n y^q$, and then sums using the geometric series formula and the initial condition $R(0,3q+i)=1$, one obtains after some calculation the equations
\begin{align\*}
F\_0(x,y) &= \frac{1}{1-y} + \frac{x}{1-y}(F\_0(x,y) + F\_1(x,y) + y F\_2(x,y))\\
F\_1(x,y) &= \frac{1}{1-y} + \frac{x}{1-y}\left(\frac{F\_0(x,y)}{y} + F\_1(x,y) + F\_2(x,y)\right) - \frac{x\alpha(x)}{y}\\
F\_2(x,y) &= \frac{1}{1-y} + \frac{x}{1-y}\left(\frac{F\_0(x,y)}{y} + \frac{F\_1(x,y)}{y} + \frac{F\_2(x,y)}{y}\right) - \frac{x\beta(x)}{y}
\end{align\*}
for almost all small $x,y$, where
$$ \alpha(x) := \sum\_{n=0}^\infty R(n,0) x^n$$
and
$$ \beta(x) := \sum\_{n=0}^\infty (R(n,0)+R(n,1)+R(n,2)) x^n.$$
Note that it is $\alpha$ that we want to understand. So the strategy will be to eliminate the other unknowns $F\_0,F\_1,F\_2,\beta$ to isolate a formula for $\alpha$.
We have a linear system of three equations in three unknowns $F\_0,F\_1,F\_2$. Solving this system using a standard symbolic algebra package, one can eliminate these unknowns, obtaining for instance
$$ F\_1(x,y) = \frac{(x^3-yx^2 - y^2 x + yx)\alpha(x) +yx^2 \beta(x) - y^2+x^2}{P(x,y)}$$
for almost all small $x,y$, where $P$ is the (irreducible) cubic
$$ P(x,y) := y^3 - (1-2x) y^2 + xy - x^3;$$
there are similar formulae for $F\_0$ and $F\_2$ that we shall discard (they give equivalent constraints to the one (1) we will end up using).
Since $F\_1$ is analytic at the origin, we conclude the constraint
$$ (x^3-yx^2 - y^2 x + yx)\alpha(x) +yx^2 \beta(x) - y^2+x^2 = 0 \quad (1)$$
whenever $x,y$ are small and $P(x,y)=0$. So now the main challenge is to use this relation (1) to eliminate $\beta$.
When $x=0$, the equation $P(x,y)$ has a double zero at $y=0$. Thus for small $x$, there are two small solutions $y\_1,y\_2$ to $P(x,y)=0$ and one large solution $y\_3$ (which is near $y=1$, since $P(0,1)=0$). Since (1) holds for $y=y\_1$ and $y=y\_2$, we may eliminate $\beta(x)$ to conclude after some algebra that
$$ \alpha(x) = -\frac{x^2 + y\_1 y\_2}{x (x^2 + y\_1 y\_2 - x)}.$$
However, as $y\_1,y\_2,y\_3$ are the roots of $P(x,y)=0$ we have $y\_1 y\_2 y\_3 = x^3$, so we can simplify to
$$ \alpha(x) = \frac{y\_3+x}{y\_3 - xy\_3 - x^2}.$$
From the implicit function theorem $y\_3$ is an analytic function of $x$ for $x$ small, so this in fact describes $\alpha$ completely as an element of ${\bf Q}(x,y\_3) \equiv {\bf Q}(x,y)/(P(x,y))$, which is a cubic extension of ${\bf Q}(x)$ and should therefore obey a cubic equation with coefficients in ${\bf Q}(x)$. Indeed, using a symbolic algebra package, one can verify the identity
$$ x^3 \alpha(x)^3 + (4x^2-3x+1) \alpha(x)^2 + (5x-3) \alpha(x) + 2 = \frac{x P(x,y\_3)}{(y\_3 - xy\_3 - x^2)^3};$$
but $P(x,y\_3)$ vanishes, hence
$$ x^3 \alpha(x)^3 + (4x^2-3x+1) \alpha(x)^2 + (5x-3) \alpha(x) + 2 = 0.$$
Writing $A(x) := x\alpha(x)$, we then get the generating function identity
$$ x^2 A(x)^3 + (4x^2-3x+1) A(x)^2 + (5x^2-3x) A(x) + 2x^2 = 0$$
for the sequence $a(n)$ in [A301897](https://oeis.org/A301897). This uniquely specifies $A$ if we enforce the asymptotics $A(x) = x + O(x^2)$, so we obtain $R(n,0)=a(n+1)$ as claimed. With a similar effort one could obtain explicit formulae for $\beta, F\_0, F\_1, F\_2$ which would lead eventually to some combinatorial formula for $R(n,q)$; one could also analyze the singularities of these generating functions to obtain asymptotics for these sequences using standard analytic combinatorics methods (e.g., the residue theorem).
EDIT: I have some further commentary regarding the means I arrived at this answer [here](https://mathstodon.xyz/@tao/110601051375142142).
| 36 | https://mathoverflow.net/users/766 | 449471 | 180,886 |
https://mathoverflow.net/questions/449477 | 2 | Do there exist some non-zero rational numbers $x, y$ such that $x \neq \pm y$ and
$$4y^p = x^2 + 3 \tag{1}$$
for some odd prime $p$?
If one lets $y=a/b$ and $x=c/d$ where $a, b, c, d$ are non-zero integers and $\gcd(a, b)=1=\gcd(c, d)$, you obtain $\frac{a^p}{b^p} = \frac{c^2 + 3d^2}{4d^2}$ thus both $c^2 + 3d^2$ and $2d$ must be integral $p-th$ powers. That is, $d=u^p/2$ and
$$v^p = c^2 + 3d^2 = c^2 + 3(u^p /2)^2 \tag{2}$$
for some coprime integers $u, v$. Now, Euler famously demonstrated that (2) is false for $p=3$. Does his argument also work for $p>3$ ?
| https://mathoverflow.net/users/492235 | On the equation $4y^p= x^2 + 3$ | Note that $(y,p,x)=(7,3,37)$ is a solution since $4(7^3) = 37^2 + 3 = 1372$.
You got that
$$\frac{a^p}{b^p} = \frac{c^2 + 3d^2}{4d^2}$$
Since $\gcd(c,d) = 1$, if $d$ is even, then $\gcd(4d^2, c^2 + 3d^2) = 1$, so the rest of your analysis would then be correct.
However, if $d$ is odd (e.g., $d = 1$ in my example solution), then $c$ can't be even (if it was, then the numerator would be odd, with just $2$ factors of $2$ in the denominator, which isn't possible for $p \gt 2$), so $c$ must also be odd. In that case, we instead get that $c^2+3d^2$ is a multiple of $4$, with $\gcd\left(\frac{c^2+3d^2}{4},d^2\right) = 1$. Thus, $\frac{c^2+3d^2}{4}$ and $d^2$ are both integral $p$-th powers so, since $p$ is odd, then $d$ is also a $p$-th power. With $d = u^p$, your $(2)$ would then instead be
$$v^p = \frac{c^2+3d^2}{4} = \left(\frac{c}{2}\right)^2 + 3\left(\frac{u^p}{2}\right)^2$$
| 4 | https://mathoverflow.net/users/129887 | 449479 | 180,889 |
https://mathoverflow.net/questions/449421 | 6 | $\DeclareMathOperator\SL{SL}\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\PSp{PSp}$The braid groups $ B\_1=1 $ and $ B\_2\cong \mathbb{Z} $ have no perfect quotients.
$ B\_3 $ has quotients $ \SL(2,\mathbb{Z}) $ and $ \PSL(2,\mathbb{Z})\cong B\_3/Z(B\_3) $. As a result, $ B\_3 $ has infinitely many perfect quotients $ \SL(2,p) $ and $ \PSL(2,p) $ (of course $ p \neq 2,3 $ for it to be perfect). Indeed since $ B\_3/Z(B\_3) \cong \PSL(2,\mathbb{Z}) \cong C\_2 \* C\_3 $ then every $ (2,3) $ generated group is a quotient of $ B\_3 $. This includes all except for a few of the infinite families of finite simple groups see [Is every finite simple group a quotient of a braid group?](https://math.stackexchange.com/questions/4726854/is-every-finite-simple-group-a-quotient-of-a-braid-group) and
[On $(2,3)$-generation of finite simple classical groups](https://mathoverflow.net/questions/365374/on-2-3-generation-of-finite-simple-classical-groups?_gl=)
Does this hold for all $ n\geq 3 $? In other words, do all braid groups $ B\_n, n \geq 3 $ have infinitely many non-isomorphic (finite) perfect quotients?
Update: Here is an update based on the accepted answer from Ian Agol. Considered the reduced Burau representation <https://en.wikipedia.org/wiki/Burau_representation> evaluated at $ t=-1 $
$$
\rho: B\_n \to U(n-1)
$$
Note that this representation is not continuous i.e. the image is not a closed subgroup of $ U(n-1) $. The image of $ \rho $ turns out to be a finite index subgroup of $ \Sp(2m,\mathbb{Z}) $ by some topological/geometric argument I don't quite understand about preserving a symplectic form on the homology. Also the way that $ m $ is related to $ n $ is somehow topological. I guess $ m $ is the rank of the first homology $ H\_1 $ of some hyperelliptic curve. Again not exactly sure what's going on with the topology/geometry/algebraic curve stuff.
An example of this phenomenon where there is a quotient of $ B\_n $ which is a (finite index subgroup of) $ \Sp(2m,\mathbb{Z}) $ can already be seen for the first non-trivial case $ n=3 $. For $ B\_3 $, there is a central quotient of $ B\_3 $ isomorphic to $ \SL(2,\mathbb{Z}) \cong \Sp(2,\mathbb{Z}) $.
So for $ n=3, B\_3 $ we have $ m=1 $ and the index of $ \rho(B\_n) $ in $ \Sp(2,\mathbb{Z}) $ is just $ 1 $.
For higher values of $ n $ the values of $ m $ increase strictly monotonically with $ n $ (at least I think so, again there is some topology I don't quite understand here). And for higher values of $ n $ the index $ [\Sp(2m,\mathbb{Z}): \rho(B\_n)] $ also form a strictly increasing sequence of integers, an explicit form for the sequence is given in a corollary to theorem 1 in the paper "Tresses, monodromie et le groupe symplectique."
Now that you have a quotient of $ B\_n $ which is (finite index) subgroup of $ \Sp(2m,\mathbb{Z}) $ you can take congruence subgroups modulo $ p $ prime. Since $ p $ is prime then the image of $ \rho(B\_n) $ in the simple group $ \Sp(2m,p) $ must be either everything or trivial. Direct inspection of the matrices for the reduced Burau representation <https://en.wikipedia.org/wiki/Burau_representation> reveals that they are not congruent to the identity mod any $ p $ (there is a $ 1 $ off the diagonal). Thus the image is all of $ \Sp(2m,p) $.
| https://mathoverflow.net/users/387190 | Perfect quotients of braid groups | A’Campo showed that the finite symplectic group $Sp(2m,p)$, $p>2$ prime, is a quotient of the braid group $B\_n$ for some $m$ depending on $n$. Hence the finite groups $PSp(2m,p)$ are quotients of $B\_n$. [These groups are simple](https://groupprops.subwiki.org/wiki/Projective_symplectic_group_is_simple) non-abelian hence perfect for most $m, p$.
*A’Campo, Norbert*, [**Tresses, monodromie et le groupe symplectique**](https://doi.org/10.1007/BF02566275), Comment. Math. Helv. 54, 318-327 (1979). [ZBL0441.32004](https://zbmath.org/?q=an:0441.32004) [MR0535062](https://mathscinet.ams.org/mathscinet/article?mr=0535062)
| 4 | https://mathoverflow.net/users/1345 | 449481 | 180,890 |
https://mathoverflow.net/questions/449293 | 3 | Heuristically, I want to know, given a smooth, projective morphism from a scheme to a discrete valuation ring, if the generic fiber can be 'covered' by a family of geometrically integral curves, is it true that the specialization of a general member of this family is also geometrically integral. More precisely,
Let $\pi: \mathcal{X} \to \mathrm{Spec}(R)$ be a smooth, projective morphism of relative dimension at least $2$ and $R$ is a discrete valuation ring with residue field isomorphic to $\mathbb{C}$.
Let $B$ be an irreducible variety mapping surjectively to $\mathrm{Spec}(R)$.
Consider a family of curves $\mathcal{C} \subset \mathcal{X} \times\_R B$ parameterized by $B$ i.e., $\mathcal{C}$ is a closed subscheme in $\mathcal{X} \times\_R B$ and the natural morphism from $\mathcal{C}$ to $B$ is flat and projective. Suppose that the natural morphism from $\mathcal{C}$ to $\mathcal{X}$ is surjective.
Denote by $K$ the fraction field of $R$ and $\mathcal{C}\_K:= \mathcal{C} \times\_R \mathrm{Spec}(K)$ the family of curves parameterized by the generic fiber $B \times\_R \mathrm{Spec}(K)$.
Similarly, denote by $\mathcal{C}\_k:= \mathcal{C} \times\_R \mathrm{Spec}(k)$ the family of curves parameterized by the special fiber $B \times\_R \mathrm{Spec}(k)$. Note that, there are natural flat, projective morphisms from $\mathcal{C}\_K$ (resp. $\mathcal{C}\_k$) to $B\_K$ (resp. $B\_k$).
Is it true that if a general fiber of the natural morphism from $\mathcal{C}\_K$ to $B\_K$ is geometrically integral, then a general fiber for the morphism from $\mathcal{C}\_k$ to $B\_k$ is also geometrically integral?
Any hint/reference will be most welcome.
| https://mathoverflow.net/users/45397 | Degeneration of curves in smooth families | I am just writing my comments as one answer. Without further hypotheses, there are counterexamples. Even without a specific example of $\mathcal{X}$, there are plenty of examples of a $K$-scheme $B\_K$ and a family of smooth, projective, geometrically connected relative curves $\mathcal{C}\_K\to B\_K$ such that for every fppf $R$-scheme $B\_R$ that has $B\_K$ as its generic fiber, for every flat, proper relative curve $\mathcal{C}\_R\to B\_R$ that extends the $K$-family, the geometric fibers over $B\_k$ are not integral.
Indeed, for every proper, flat $R$-scheme $B'\_R$ whose geometric fibers are integral, for every proper, flat morphism $\mathcal{C}'\_R\to B'\_R$ whose fibers are at-worst-nodal, connected curves with ample dualizing sheaf, for every "modification" of the family over a proper, flat $R$-scheme $B\_R$ whose geometric fibers are integral,
$\mathcal{C}\_R \to B\_R$, the "stabilization" of the geometric generic fiber of $\mathcal{C}\_k\to B\_k$ equals the geometric generic fiber of $\mathcal{C}'\_k\to B'\_k$: this is part of the uniqueness in "stable reduction". So if the geometric generic fiber of $\mathcal{C}'\_k\to B'\_k$ is reducible, the same is true for the geometric generic fiber of $\mathcal{C}\_k\to B\_k$.
Thus, to get a positive answer, you need to add the hypothesis that for the family $\mathcal{C}\_K\to B\_K$ whose geometric fibers are assumed to be integral, there exists an extension $\mathcal{C}\_R\to B\_R$ over an fppf $R$-scheme $B\_R$ whose geometric generic fiber over $B\_k$ is integral (just as a proper, flat family of abstract curves, with no morphism to $\mathcal{X}\_R$ specified).
Even with this hypothesis, there are still counterexamples, e.g., the example in my comment where $\mathcal{X}\_K$ is a Hirzebruch surface $\mathbb{P}^1\times \mathbb{P}^1$, yet $\mathcal{X}\_k$ is a different Hirzebruch surface, e.g., the minimal (crepant) resolution of a singular quadric cone in $\mathbb{P}^3$. The usual way to deal with this is to change the question slightly: does there exist an extension $\mathcal{C}\_R \to B\_R\times\_{\text{Spec}(R)} \mathcal{X}\_R$ and an irreducible component $\mathcal{C}\_{k,i}$ of $\mathcal{C}\_k$ satisfying all of your conditions. If you begin with a $K$-family of curves $\mathcal{C}\_K\to B\_K\times\_{\text{Spec}(K)} \mathcal{X}\_K$ that is constructed in a "sufficiently general" way, e.g., the family of all complete intersection curves of sufficiently very ample divisors in given linear equivalence classes on $\mathcal{X}\_K$, then this new question has a positive answer if and only if there exists an irreducible component $\mathcal{X}\_{k,i}$ of $\mathcal{X}\_k$ that is geometrically irreducible.
| 2 | https://mathoverflow.net/users/13265 | 449486 | 180,891 |
https://mathoverflow.net/questions/449462 | 3 | Recall that two elements $h\_1,h\_2$ of a finite group $G$ are called *conjugate* when $h\_2 = gh\_1 g^{-1}$ for some $g \in G$, and *algebraic-conjugate* when $h\_2 = gh\_1^a g^{-1}$ for some $a \in (\mathbb{Z}/\mathrm{order}(g))^\times$; equivalently when the cyclic subgroups $\langle h\_1\rangle$ and $\langle h\_2 \rangle$ are conjugate. In general, algebraic-conjugacy is a much coarser relationship than conjugacy. But sometimes they are equivalent: in the symmetric group, for example, both conjugacy and algebraic-conjugacy are simply the cycle structure of the element.
>
> How common is it for algebraic-conjugacy to equal conjugacy?
>
>
>
One reason why you might ask this is the following. Suppose you have a representation $V$ of $G$. Then its character is, of course, a *class function*: $\mathrm{tr}\_V(g)$ depends only on the conjugacy class of $g \in G$. But if $V$ happens to be defined over $\mathbb{Q}$, then $\mathrm{tr}\_V(g)$ in fact depends only on the algebraic conjugacy class of $g$. In particular, if every representation of $G$ is defined over $\mathbb{Q}$, then by character theory, algebraic-conjugate elements are automatically conjugate.
An intermediate relation between conjugacy and algebraic conjugacy is what I'll call, based on its character-theory interpretation, *real conjugacy*: I'll say that $h\_1$ and $h\_2$ are *real-conjugate* if $h\_2 = g h\_1^{\pm 1}g^{-1}$ for some $g$ and some sign.
>
> How common is it for algebraic-conjugacy to equal real-conjugacy?
>
>
>
| https://mathoverflow.net/users/78 | How often does algebraic-conjugacy imply conjugacy? | It is reasonably standard to call a finite group $G$ a rational group if all its complex irreducible characters are rational-valued ( equivalently, if $g \in G,$ then $g$ is conjugate within $G$ to all generators of $\langle g \rangle )$. Thanks to work of Feit-Seitz, and of J.G. Thompson,( see Feit-Seitz "On rational finite groups and related topics", Illinois Journal of Mathematics, 33,1 Spring 1988 and J.G. Thompson "Composition factors of rational finite groups", Journal of Algebra, 319,2,Jan 2008,558-594 ( Feit memorial issue)) it is known that every Abelian composition factor of a rational finite group has order at most $11$, and that there are only five possibilities for isomorphism types of non-Abelian composition factors for $G$ (other than alternating composition factors). The five possible non-alternating non-Abelian composition factors are :
${\rm PSp}(4,3),{\rm Sp}(6,2),{\rm O}^{+}\_{8}(2)′, {\rm PSL}(3,4)$ and ${\rm PSU}(4,3)$.
Also, R. Gow proved that solvable finite rational groups are $\{2,3,5 \}$-groups, and P.Hegedus refined this somewhat. In this sense, the rational finite groups are somewhat rare.
On the other hand, if we call an individual element $g$ of a finite group $G$ rational if $g$ is conjugate (within $g$) to all generators of $\langle g \rangle,$ then there may be many rational elements within a non-rational group $G$. For example, all unipotent elements in ${\rm GL}(n,q)$ are rational, so determination of which elements of ${\rm GL}(n,q)$ reduces to determining which semisimple elements are rational, and there are many other instances of similar behaviour among classical groups and other finite groups of Lie type.
| 11 | https://mathoverflow.net/users/14450 | 449488 | 180,892 |
https://mathoverflow.net/questions/449501 | 0 | Let $(a\_n)\_n$ be an increasing real sequence with $a\_n=O(\sqrt n)$.
Is it true that there exists an increasing function $\phi:\mathbb N\to\mathbb N$ such that $$\lim \left|\sum\limits\_{k=1}^{\phi(n)}\cos(a\_k)\right|+\left|\sum\limits\_{k=1}^{\phi(n)}\sin(a\_k)\right|=\infty?$$
<https://artofproblemsolving.com/community/c7h3080371_majoration_of_cosinus_and_sinus_addition>
| https://mathoverflow.net/users/110301 | Series involving sine and cosine | Yes, it suffices that $a\_n=o(n)$.
Denote $h(n)=|\sum\_{j=1}^n \cos a\_j|+|\sum\_{j=1}^n \sin a\_j|$.
For any fixed integer $M>0$ there exists arbitrarily large $n$ for which $a\_{n+M}-a\_n<1/M^2$. Thus by triangle inequality and 1-Lipschitz property of functions $\cos$ and $\sin$ we have
$$h(n+M)+h(n)\geqslant |\cos a\_{n+1}+\ldots+\cos a\_{n+M}|+|\sin a\_{n+1}+\ldots+\sin a\_{n+M}|\\\geqslant M|\cos a\_{n+1}|-M/M^2+M|\sin a\_{n+1}|-M/M^2\geqslant 2M-2$$
that yields that $h(n+M)$ or $h(n)$ is at least $M-1$. Your claim follows immediately.
| 2 | https://mathoverflow.net/users/4312 | 449505 | 180,896 |
https://mathoverflow.net/questions/449137 | 0 | Consider the following line of reasoning that shows certain simplicial complexes (of arbitrary dimension) are completely determined by corresponding graphs:
1. The facet complex of any [simplicial polytope](https://en.wikipedia.org/wiki/Simplicial_polytope) is a simplicial complex.
2. The facet complex completely determines (and is completely determined by) the original simplicial polytope.
3. The simplicial polytope is dual to a [***simple*** polytope](https://en.wikipedia.org/wiki/Simple_polytope).
4. The dual ***simple*** polytope completely determines (and is completely determined by) the original simplicial polytope.
5. By a Theorem of Whitney, cf. the references [a] and [b] below, the dual ***simple*** polytope is completely determined by (and completely determines) its 1-skeleton.
>
> **Question:** For which simplicial complexes does the above chain of reasoning generalize?
>
>
> In other words, what is a necessary and sufficient characterization of simplicial complexes with the property that their (polyhedrally) dual polyhedral complexes are completely determined by the 1-skeleton of the dual?
>
>
> Equivalently, which polyhedral complexes generalize ***simple*** polytopes in having the two properties that (1) they are (polyhedrally) dual to a simplicial complex, and (2) they are completely determined by their 1-skeleton?
>
>
>
**Note:** Feel free to assume all involved simplicial complexes, cell complexes, and graphs have finitely many vertices.
**Note:** This is a [cross-post of an unanswered question](https://math.stackexchange.com/questions/4715194/which-simplicial-complexes-are-completely-determined-by-the-1-skeleton-of-their) on math.stackexchange. I'm not sure whether this is a research-level question, but it might be because it's asking for references about generalizations of results from the late '80's, in a field (combinatorial topology) that seems to be still actively researched. I am an outsider to this field myself, however, so I'm uncertain how basic this question is. (Likewise I wasn't sure which tags to add.)
**References**
**[a]** Blind, Roswitha; Mani-Levitska, Peter (1987), *"[Puzzles and polytope isomorphisms](https://link.springer.com/article/10.1007/BF01830678)"*, Aequationes Mathematicae, 34 (2–3): 287–297, doi:10.1007/BF01830678, MR 0921106
**[b]** Kalai, Gil (1988), *"[A simple way to tell a simple polytope from its graph](https://www.sciencedirect.com/science/article/pii/0097316588900647?via%3Dihub)"*, Journal of Combinatorial Theory, Series A, 49 (2): 381–383, doi:10.1016/0097-3165(88)90064-7, MR 0964396
**Related, but distinct, questions on MathOverflow:** Unlike this related question [Cyclic polytopes whose boundary is a flag complex](https://mathoverflow.net/questions/266009/cyclic-polytopes-whose-boundary-is-a-flag-complex) I am not asking about whether the simplicial complex is determined by the simplicial complex's 1-skeleton, I am asking about whether its dual polyhedral complex is determined by its dual polyhedral complex's 1-skeleton.
This other question ([Visualizing polyhedra from their 1-skeletons](https://mathoverflow.net/questions/119455/visualizing-polyhedra-from-their-1-skeletons)) also appears to be (indirectly related), but I am asking specifically about the level of generality of arbitrary polyhedral complexes, not only inidividual polyhedra (for which the answer is already known and given in the two references below).
This question ([Can I build infinitely many polytopes from only finitely many prescribed facets?](https://mathoverflow.net/questions/351241/can-i-build-infinitely-many-polytopes-from-only-finitely-many-prescribed-facets)) is asking about a case where the maximal polyhedral cells of a (single polytope considered as a pure) polyhedral complex are restricted to belonging to a certain family of polyhedra, but does not seem to require that the *facet complex* itself be specified a priori (the latter of which is what I am asking). So it also seems to be a different question.
(Also technically I'm not certain that the facet complex would uniquely specify the original polytope, the converse is trivial of course, but I also don't see how it wouldn't.)
| https://mathoverflow.net/users/142698 | Which simplicial complexes are completely determined by the 1-skeleton of their dual polyhedral complexes? | A simplicial polytope is determined by the 1-skeleton of its dual *as
a polytope*, not as a simplicial complex. For instance, take a
sufficiently large simplicial polytope of dimension at least three,
and identify two far away vertices (so that it remains a simplicial
complex). The dual polyhedral complexes of these two simplicial
complexes have the same 1-skeleton.
As suggested by Sam Hopkins, we can ask for families of simplicial
complexes whose members *within the family* are determined by their
1-skeleton. Possible families, in increasing order of generality, are
triangulations of spheres, Gorenstein\* complexes, and doubly
Cohen-Macaulay complexes. On the other hand, the family of
Cohen-Macaulay complexes is too general. There are two nonisomorphic
triangulations of a 2-dimensional ball with four facets (triangles)
whose 1-skeletons are both paths of length three.
| 2 | https://mathoverflow.net/users/2807 | 449509 | 180,897 |
https://mathoverflow.net/questions/449517 | -1 | Let V and W be any two subspaces of $(\mathbb{C}^d)^{\otimes n}$ such that there exists two irreducible and non-isomorphic representations $\rho\_V: G \to GL(V)$ and $\rho\_W: G \to GL(W)$. Does this imply: V and W are orthogonal subspaces?
I am new to representation theory, so apologies if this is basic knowledge. I could not find a good proof/counterexample for this fact, but something along the lines is used in Proposition 1 in [Stembridge - Orthogonal sets of Young symmetrizers](https://doi.org/10.1016/j.aam.2009.08.004).
In this link, it is shown that Young symmetrizers corresponding to two different shapes are orthogonal by using the fact that they project onto irreducible and non-isomorphic representations.
| https://mathoverflow.net/users/173056 | Orthogonality of irreducible and non-isomorphic representations | I wasn't sure what the question meant; to make sense of it, $V$ and $W$ would have had to be subspaces of a common space with some sort of form with respect to which we could measure orthogonality, and two abstract representations need not be presented in this way.
Your reference to [Stembridge - Orthogonal sets of Young symmetrizers](https://doi.org/10.1016/j.aam.2009.08.004) suggested that you were thinking of, say, complex representations of a finite group $G$. In this setting, the group algebra $\mathbb C[G]$ has a $V$-isotypic subspace (consisting of the matrix coefficients $g \mapsto \langle v^\*, g\cdot v\rangle$ for $v \in V$ and $v^\* \in V^\*$) and a $W$-isotypic subspace (analogously defined). These are orthogonal for the usual ($G$-equivariant) inner product on $\mathbb C[G]$, by the [Schur orthogonality relations](https://en.wikipedia.org/wiki/Schur_orthogonality_relations).
EDIT: You have updated the question to indicate that you just want $V$ and $W$ to be two subspaces of $(\mathbb C^d)^{\otimes n}$, presumably with the usual inner product inherited from its natural identification with $\mathbb C^{d n}$. (I'm not sure what the purpose of the separate $d$ and the $n$ parameters is.) In this generality, the answer can be ‘no’ in all but the most trivial cases: consider, for example, the cyclic group of order $2$ acting trivially on a line $V$ in $(\mathbb C^2)^{\otimes1}$, and non-trivially on a non-orthogonal line $W$ in $(\mathbb C^2)^{\otimes1}$. (This can even be extended, as you probably meant, to an action of the cyclic group on $(\mathbb C^2)^{\otimes1}$ that makes its actions on $V$ and $W$ subrepresentations.)
If you added, as you probably meant to do, the requirement that $G$ actually act on $(\mathbb C^d)^{\otimes n}$, preserving some inner product $\langle\cdot, \cdot\rangle$ there (not necessarily the standard one!), and that the representations of $G$ on $V$ and $W$ make them subrepresentations of this big $d n$-dimensional representation (as is the case for finite groups $G$, again, by the [Schur orthogonality relations](https://en.wikipedia.org/wiki/Schur_orthogonality_relations), when we embed abstract representations $G$-equivariantly in $\mathbb C[G] \cong (\mathbb C^{\lvert G\rvert})^{\otimes 1}$, equipped with its standard ‘$L^2$’ inner product), *then* the answer is once again ‘yes’, that the spaces of $V$ and $W$ are orthogonal *with respect to this $G$-fixed inner product*. (I emphasise the need for consistency in the choice of inner product because *every* action of a finite group on a complex vector space preserves *some* inner product there. For example, if we chose the lines in the example of the previous paragraph to be $V = \operatorname{Span}(1, 1)$ and $W = \operatorname{Span}(1, 0)$, then the inner product with squared norm $(x, y) \mapsto 2\lvert x\rvert^2 - 4\operatorname{Re}(x\overline y) + 5\lvert y\rvert^2$ is preserved by the cyclic group of order $2$, and, indeed, $V$ and $W$ are orthogonal with respect to this inner product even though they are not with respect to the ‘standard’ pairing.)
I will now prove the modified statement. For every $v \in V$, the [Riesz representation theorem](https://en.wikipedia.org/wiki/Riesz_representation_theorem) produces a unique element $T(v) \in W$ such that $\langle T(v), w\rangle$ equals $\langle v, w\rangle$ for all $w \in W$. We have that $T : V \to W$ is $G$-equivariant (because $G$ preserves the inner product), so, since $V$ and $W$ are irreducible and non-isomorphic, $T$ must be identically $0$. Then $\langle v, w\rangle = \langle T(v), w\rangle$ equals $0$ for all $v \in V$ and $w \in W$.
| 4 | https://mathoverflow.net/users/2383 | 449519 | 180,899 |
https://mathoverflow.net/questions/449305 | 2 | Let $\omega\_1,\cdots,\omega\_n$ be $n$ elements of $\overline{\mathbb F\_q(T)}$ that are $\mathbb F\_q(T)$ linearly independant. Denote by $\Lambda$ the lattice $\Lambda=\mathbb F\_q[T]\omega\_1+\cdots+\mathbb F\_q[T]\omega\_n$. Can one estimate (with at least two significant terms) the number of elements of $\Lambda$ whose degree ($\overline{\mathbb F\_q(T)}$ is considered embeded in the completion of $\overline{\mathbb F\_q\left(\left(\frac1T\right)\right)}$) is less than $r$ ($r$ is a positive integer)?
Surely Riemann-Roch theorem (or Riemann hypothesis for curves) could help, but I do not see how.
| https://mathoverflow.net/users/33128 | Numbers of points in lattice | There are two possibilities. If $\omega\_1,\dots, \omega\_n$ are not $\mathbb F\_q((\frac{1}{T}))$-linearly independent, then the count is infinite for all $r$ sufficiently large. Indeed, if $\sum\_{i=1}^n a\_i \omega\_i$ is a relation, then for $f$ in $\mathbb F\_q[T]$ any polynomial, if we take the power series $fa\_1,\dots, f a\_n$ and remove all terms with negative powers of $T$ to obtain polynomials $g\_1,\dots, g\_n$ then $\sum\_{i=1}^n g\_i \omega\_i$ will equal $\sum\_{i=1}^n (g\_i -f a\_i) \omega\_i$ and thus will have degree less than the maximum degree of the $\omega\_i$. This gives infinitely many elements of $\Lambda$ with bounded degree.
Otherwise, $\omega\_1,\dots, \omega\_n$ generate a $\mathbb F\_q((\frac{1}{t}))$-submodule of $\overline{\mathbb F\_q((\frac{1}{t}))}$ of rank $n$. Consider the intersection of this submodule with the ring of integers, which is a $\mathbb F\_q[[\frac{1}{T}]]$-module, necessarily free of rank $n$, and choose a generating set $\alpha\_1,\dots, \alpha\_n$. We can write $\alpha\_i =\sum\_{j=1}^n c\_{ij} \omega\_j$ for a unique $c\_{ij} \in \mathbb F\_q((\frac{1}{T}))$. Let $C$ be the $n\times n$ matrix with entries $c\_{ij}$.
It is easy to see from Riemann-Roch that the number of elements of $\Lambda$ with degree $\leq r$ is equal to $q^{ (r+1) n + \deg \det C}$ for all $r$ sufficiently large.
Indeed, this is equal to the number of vectors $a\_1,\dots, a\_n \in \mathbb F\_q[T]^n$ such that $C^{-1} \begin{pmatrix} a\_1 \\ \dots \\ a\_n \end{pmatrix} $ is a tuple of vectors of degree $\leq r$. We construct a vector bundle $V$ on $\mathbb P^1$ by gluing $\mathcal O\_{\operatorname{Spec} \mathbb F\_q[T]}^n$ to $\mathcal O\_{\operatorname{Spec} \mathbb F\_q[[ \frac{1}{T} ]] }^n$ using the gluing data $C$ on their intersection $\operatorname{Spec} \mathbb F\_q(( \frac{1}{T} ))$. The desired count is equal to the number of sections of $H^0(\mathbb P^1, V(r) )$, which by Riemann-Roch for vector bundles is
$$ q^{n + \deg V(r)} = q^{n + nr + \deg V }= q^{n + nr + \deg \det V} =q^{ n + nr + \deg \det C}$$ as soon as $H^1(\mathbb P^1, V(r))=0$, which happens for all $r$ sufficiently large.
I am not sure what you mean by at least two significant terms.
| 1 | https://mathoverflow.net/users/18060 | 449523 | 180,901 |
https://mathoverflow.net/questions/449504 | 5 | *Crossposted from <https://math.stackexchange.com/questions/4717613>*
---
An $\omega$-cover $\mathscr U$ of a space $X$ is a collection of open sets so that $X \not\in\mathscr U$ and every finite subset of $X$ is contained in a member of $\mathscr U$. Similarly, a $k$-cover $\mathscr U$ of a space $X$ is a collection of open sets so that $X \not\in\mathscr U$ and every compact subset of $X$ is contained in a member of $\mathscr U$.
A space is an $\varepsilon$-space (aka $\omega$-Lindelöf) if every $\omega$-cover has a countable subset which is an $\omega$-cover. Likewise, a space is $k$-Lindelöf if every $k$-cover has a countable subset which is a $k$-cover.
It can be shown that $k$-Lindelöf $\implies$ $\varepsilon$-space $\implies$ Lindelöf. The Sorgenfrey line is an example of a space which is Lindelöf but not an $\varepsilon$-space. This leads us to the following question.
>
> Is there an $\varepsilon$-space which is not $k$-Lindelöf? Can the example be Hausdorff? Metrizable?
>
>
>
If it can be of any help, it would suffice to find a non-compact $\varepsilon$-space $X$ so that $\mathbb K(X)$, the space of compact subsets of $X$ with the Vietoris topology, is not Lindelöf by Cor. 4.16 of <https://doi.org/10.1016/j.topol.2021.107772>.
No such example exists in the realm of separable metrizable spaces.
>
> **Claim.** Every separable metrizable space is $k$-Lindelöf.
>
>
>
*Proof.*
If $X$ is separable metrizable, then $\mathbb K(X)$ is also separable metrizable; hence, Lindelöf. If we let $[U] = \{ K \in \mathbb K(X) : K \subseteq U \}$, then, for a given $k$-cover $\mathscr U$ of $X$, $\mathscr W := \{ [U] : U \in \mathscr U \}$ is an open cover of $\mathbb K(X)$. Then a countable subcover of $\mathscr W$ corresponds to a countable subset of $\mathscr U$ which is a $k$-cover of $X$. $\square$
| https://mathoverflow.net/users/57800 | Is there an $\varepsilon$-space which is not $k$-Lindelöf? | (MA) There exists a countable, hereditarily (strongly) paracompact space
X, with ω<kL(X) (Example 6.4 in "Tightness, character and related properties
of hyperspace topologies" Topology and its Applications 142 (2004) 245–292).
| 0 | https://mathoverflow.net/users/112417 | 449529 | 180,903 |
https://mathoverflow.net/questions/449490 | 6 | Let $H\_n$ denote the set of $n \times n$ nilpotent matrices with complex entries. The set $H\_n$ may be regarded as an algebraic variety. Indeed, consider the polynomial ring $\mathbb{C}[A\_{i,j} : 1 \leq i,j \leq n]$ in $n^2$ variables, and for $m \geq 0$, define the polynomial
\begin{equation\*}
\phi\_{m;i,j} := \sum\_{ 1 \leq k\_1,\ldots,k\_{m-1} \leq n} A\_{i,k\_1}A\_{k\_1,k\_2}\ldots A\_{k\_{m-1},j},
\end{equation\*}
with the understanding that $\phi\_{1;i,j} = A\_{i,j}$, and $\phi\_{0;i,j} = \delta\_{i,j}$.
Then $H\_n$ is the zero set $V(I)$ of the ideal generated by the $n^2$ polynomials $\{\phi\_{n;i,j} : 1 \leq i,j \leq n \}$.
Alternatively, for $0 \leq k \leq n$ define the polynomials
\begin{equation}
p\_k := \sum\_{S \subset [n], |S|=k} \det\_{i,j \in S} A\_{i,j},
\end{equation}
with $p\_0=1$.
Then the characteristic polynomial of a matrix takes the form
\begin{align\*}
\Delta\_A(t) := \det(tI - A) = \sum\_{k=0}^n t^{n-k}(-1)^k p\_k.
\end{align\*}
Since $A$ is nilpotent if and only if $\det(tI-A) = t^n$, it follows than $H\_n$ may alternatively be characterised as the $V(J)$ of the ideal $J$ generated by the $n$ polynomials $\{p\_1,\ldots,p\_n\}$.
In particular, by Hilbert's Nullstellensatz, we have the equality of radicals, $\sqrt{I} = \sqrt{J}$. Clearly $J$ is not a subset of $I$, since $J$ contains polynomials of degree $1$, whereas $I$ contains only polynomials of degree $n$ and higher.
Conversely however, we can show that $I \subset J$. Indeed, by the Cayley–Hamilton theorem, $A$ is a root of its own characteristic polynomial, so that for each $1 \leq i,j \leq n$ we have the identity in $\mathbb{C}[A\_{i,j} : 1 \leq i,j \leq n ]$
\begin{equation}
0 = \sum\_{k=0}^n (-1)^k \phi\_{n-k;i,j} p\_k.
\end{equation}
Thus, since $p\_0=1$, $\phi\_{n;i,j}$ can be written as a linear combination of $p\_k$, and hence $\phi\_{n;i,j} \in J$.
Question:
Have the combinatorics of the nullstellensatz for $H\_n$ been studied? Here are some concrete examples of questions I'm interested in:
1. By the Nullstellensatz, for each $k$, there is some integer $r \geq 1$ such that $p\_k^r$ can be written as a polynomial linear combination of the $\{\phi\_{n;i,j} : 1 \leq i,j \leq n\}$. Can this linear combination be found explicitly?
2. If an $n \times n$ matrix $A$ satisfies $A^m = 0$ for some $m \geq n$, then it is guaranteed to satisfy $A^n = 0$. Thus, if we consider the ideal $I\_m := \{ \phi\_{m;i,j} : 1 \leq i,j \leq n \}$, then for $m \geq n$, $I\_m$ and $I=I\_n$ have the same zero set. It's easily verified $I\_{m+1} \subset I\_m$. However, by the nullstellensatz, for any $m' \geq m$, each $\phi\_{m';i,j}$ has a power that can be written as a polynomial linear combination of the $\{ \phi\_{m;i,j} : 1 \leq i,j \leq n \}$. Can this linear combination be found explicitly?
3. In the previous two questions, do we have a concept of how the exponent $r$ behaves as a function of $n,m,m'$?
Have this circle of questions been studied anywhere? There are some sort of related questions on this site and stackexchange, though without explicit reference to the combinatorics:
[Here](https://math.stackexchange.com/questions/4681296/is-the-ideal-generated-by-coefficients-of-characteristic-polynomial-of-a-matrix "Is the ideal generated by coefficients of characteristic polynomial of a matrix prime?") someone asks if the ideal $J$ above is prime. See also [here](https://math.stackexchange.com/questions/473266/ideal-defining-the-nilpotent-cone-of-mathfrakgl-nk "Ideal defining the nilpotent cone of \mathfrak{gl}_n(k)").
[Here](https://mathoverflow.net/questions/327340/has-the-geometry-of-the-variety-of-nilpotent-matrices-over-mathbbc-been-stu "Has the geometry of the variety of nilpotent matrices over \mathbb{C} been studied?") someone asks if the geometry of $H\_n$ has been studied. There are references to Jantzen's Lie Theory and Chriss-Ginzburg, as well as papers by Ness and by Richardson.
[Here](https://math.stackexchange.com/questions/405291/variety-of-nilpotent-matrices "Variety of Nilpotent Matrices") someone show that $H\_n$ has dimensions $n^2-n$.
I should also mention the combinatorial proofs of the Cayley–Hamilton theorem: see [Straubing - A combinatorial proof of the Cayley-Hamilton theorem](https://doi.org/10.1016/0012-365X(83)90164-4).
Thank you for reading this far, and my apologies if I have used any unusual notation or terminology — I'm a beginner in algebraic geometry!
| https://mathoverflow.net/users/380543 | The combinatorics of the Nullstellensatz for the variety of nilpotent matrices | Thanks everyone for your replies! As Darij suggested in his answer, it appears the clever trick used in [Gert Almkist's generalisation of a mistake of Bourbaki](https://sites.math.rutgers.edu/%7Ezeilberg/mamarim/mamarimhtml/gert.html) can be generalised to tackle the problem. Thanks Darij for the suggestion and for the reference.
If I'm not mistaken, here are the details. In the originally setting of Almkist's trick, we show that $(\sum\_{i=1}T\_{ii})^{n(n-1)+1}$ is a linear combination of the $\phi\_{n;i,j}$ as follows:
Note that if $e\_1,\ldots,e\_n$ is the standard basis of $\mathbb{C}^n$, then
\begin{equation}
(\sum\_{i=1}^nT\_{ii})\det(e\_1,\ldots,e\_n) = \sum\_{k=1}^n \det(e\_1,\ldots,Te\_k,\ldots,e\_n).
\end{equation}
By multilinearity, for any vectors $b\_1,\ldots,b\_n$ we have
\begin{equation}
(\sum\_{i=1}^nT\_{ii})\det(b\_1,\ldots,b\_n) = \sum\_{k=1}^n \det(b\_1,\ldots,Tb\_k,\ldots,b\_n).
\end{equation}
In particular, for any natural number $N$ we have
\begin{equation}
(\sum\_{i=1}^nT\_{ii})^N = \sum\_{k\_1 + \ldots + k\_n=N} \frac{N!}{k\_1!\ldots k\_n!} \det(T^{k\_1}e\_1,\ldots,T^{k\_n}e\_n),
\end{equation}
where the sum is over nonnegative integers $k\_1,\ldots,k\_n$ adding to $N$. When $N \geq n(m-1)+1$, it is guaranteed that at least one of the $k\_i$ will be $\geq m$, so that the quantity on the right-hand-side is a polynomial linear combination of $\phi\_{m;i,j}$.
To generalise this method to the polynomials $p\_k$, we begin by noting
\begin{equation}
p\_k \det(e\_1,\ldots,e\_n) = \sum\_{1 \leq i\_1 < \ldots < i\_k \leq n} \det(e\_1,\ldots,Te\_{i\_1},\ldots,Te\_{i\_2},\ldots,Te\_{i\_k},\ldots,e\_n),
\end{equation}
where we understand that between the dots we have $e\_i$s. In particular, iterating like before, for any $N$ we have
\begin{equation}
p\_k^N = \sum\_{k\_1+\ldots+k\_N = kN} C\_{N,k}(k\_1,\ldots,k\_n) \det(T^{k\_1}e\_1,\ldots,T^{k\_n}e\_n),
\end{equation}
where $C\_{N,k}(k\_1,\ldots,k\_n)$ is the number of ways of putting balls into $n$ bins in $N$ rounds with $k\_j$ balls in bin $j$ at time $N$, with the rule that on each round, we put $k$ balls into $k$ different bins. In any case, provided $kN \geq n(m-1)+1$, we are guaranteed to have some $k\_i \geq m$ for each summand, and hence for such $N$, this formula expresses $p\_k^N$ as a polynomial linear combination of $\{\phi\_{n=m;i,j} \}$.
Note that as $k$ increases, we need only take smaller powers of $p\_k$ (i.e. the $N\_k^{\text{th}}$ power with $N\_k$ the smallest integer greater than $(n(m-1)+1)/k$) to express $p\_k^{N\_k}$ in terms of $\phi\_{m;i,j}$. This makes sense, as $p\_k$ is a higher degree polynomial. In fact, $p\_k^{N\_k}$ has degree $n(m-1)+1+r(k)$, where $0 \leq r(k) < k$.
This method also gives us a (probably quite inefficient) way of expressing a power of $\phi\_{m;i,j}$ in terms of the $\phi\_{m';i',j'}$. Write $\phi\_{m;i,j} = \sum\_k \lambda\_kp\_k$, where $\lambda\_k$ are polynomials. Then $\phi\_{m;i,j}^{n(N-1)+1}$ is a polynomial in powers of $p\_k$, which again by this pigeonhole style principle, has every monomial containing a power of a $p\_k$ of at least $N$. Provided $kN \geq n(m'-1)+1$, this power can then be written as a polynomial linear combination of the $\phi\_{m';i',j'}$.
| 5 | https://mathoverflow.net/users/380543 | 449531 | 180,905 |
https://mathoverflow.net/questions/449525 | 5 | Let $D$ be a possibly unbounded domain in $\mathbb{R}^d$, $d \ge 2.$
We say that $D$ has the property P if there exists $C>0$ such that such that any pair of points $x,y \in D$ can be joined by a curve $\gamma$ in $D$ satisfying $\text{length}(\gamma) \le C|x-y|.$ Here, $\text{length}(\gamma)$ denotes the length of $\gamma$.
The property P appears in the Whitney extension theorem (see Theorem in [W](https://www.jstor.org/stable/pdf/1968745.pdf?refreqid=excelsior%3Ab0aedd84cb53f3564899ca902fad3d0a&ab_segments=&origin=&initiator=&acceptTC=1) for details), and is sometimes referred to the quasi-convexity.
It is known that bounded Lipschitz domains are quasi-convex.
Can we show that unbounded Lipschitz domains are also quasi-convex?
| https://mathoverflow.net/users/68463 | On the property P in the Whitney extension theorem | No. The complement $M$ in $\mathbb R^2$ of the union of the closed unit ball and the *half strip* $\{(x,y)\in\mathbb R^2: x\ge 0, |y|\le 1\}$ has $C^1$-boundary but curves in $M$ joining $(n,2)$ and $(n,-2)$ (which have distance $4$) have length $>2n$.
| 6 | https://mathoverflow.net/users/21051 | 449535 | 180,906 |
https://mathoverflow.net/questions/449493 | 4 | Let $X = \mathbb{R}^2$ with $(x, y)\in X$ for $y\neq 0$ isolated and $(x, 0)$ having neighbourhood basis of the form $$U\_n(x) = \{(x, y) : y\in (-1/n, 1/n)\}\cup \{(x+y+1, y) : 0 < y < 1/n\}\cup \{(x+\sqrt{2}+y, -y) : 0 < y < 1/n\}.$$ The space $X$ is called Mysior plane, and it's an example of a space which is a union of two closed subspaces, $X\_+ = \mathbb{R}\times [0, \infty)$ and $X\_- = \mathbb{R}\times (-\infty, 0]$, which are realcompact, but isn't itself realcompact.
Since the map $f:X\_+\sqcup X\_-\to X$ is perfect, this shows realcompactness is not invariant under perfect mappings.
The space $X$ is Tychonoff: Say $(x, y), (w, t)\in X$. If $y\neq 0$ or $t\neq 0$ this is quite obvious since one of the points is isolated. If $y = t = 0$, and $x < w$, then by splitting into three cases $x < w \leq x+1, x+1 < w \leq x+\sqrt{2}$ and $w\geq x+\sqrt{2}$ one can easily see that we have disjoint neighbourhoods. Thus $X$ is Hausdorff. If $(x, y)\notin F$ and $F$ is closed, of course the only interesting case is when $y = 0$ for otherwise $\{(x, y)\}$ is clopen. In that case we can find $n$ with $U\_n(x)\subseteq F^c$. By defining $f(w, t) = 1$ for $(w, t)\notin U\_n(x)$ and $f(w, t) = nt$ for $(w, t)\in U\_n(x)$, we see that $f$ is continuous and so $X$ is Tychonoff.
To see that $X\_+$ and $X\_-$ are realcompact, lets focus on $X\_+$. If $\mathcal{F}$ is a real z-ultrafilter on $X\_+$, then since $\mathbb{R}\times [0, a]$ for $a > 0$ is clopen, we can see that if $\mathbb{R}\times (a, \infty)\in\mathcal{F}$, then $\mathcal{F}\restriction\_{\mathbb{R}\times (a, \infty)}$ is a real z-ultrafilter on the realcompact space $\mathbb{R}\times (a, \infty)$, and as such is fixed, so that $\mathcal{F}$ is fixed. Otherwise, if for all $a > 0$, $\mathbb{R}\times [0, a]\in \mathcal{F}$, then $\mathbb{R}\times \{0\}\in\mathcal{F}$ since $\mathcal{F}$ is closed under countable intersections. One can observe that $([a, b]+\mathbb{Z})\times \{0\}$ for $a < b$ are zero sets of $X\_+$ by construction appropriate functions (similar to the one in the proof that $X$ is Tychonoff). Thus $([0, 1/2]+\mathbb{Z})\times \{0\}\in \mathcal{F}$ or $([1/2, 1]+\mathbb{Z})\times\{0\}\in \mathcal{F}$, for example the former case holds. Take $a = \sup \{x\in [0, 1/2] : ([x, 1/2]+\mathbb{Z})\times\{0\}\in \mathcal{F}\}$ and $b = \inf\{x\in [a, 1/2] : ([a, x]+\mathbb{Z})\times\{0\}\in\mathcal{F}\}$. Once again one would show that $[a, b]+\mathbb{Z}\in \mathcal{F}$ from closure under countable intersections, and if $a\neq b$ we could take $a < c < b$ and by considering $([a, c]+\mathbb{Z})\times\{0\}$ and $([c, b]+\mathbb{Z})\times\{0\}$ obtain a contradiction with choice of $a, b$. Thus $(a+\mathbb{Z})\times\{0\}\in\mathcal{F}$ for some $a\in \mathbb{R}$. However, since $(a+\mathbb{Z})\times\{0\} = \bigcup\_{n\in \mathbb{Z}}\{(a+n, 0)\}$ is a countable union of zero sets, and since $\mathcal{F}$ is a real z-ultrafilter, we must have $\{(a+n, 0)\}\in\mathcal{F}$ for some $n$, that is $\mathcal{F}$ is a fixed z-ultrafilter. This proves $X\_+$ is realcompact, the proof for $X\_-$ is the same.
Could anyone help me on how to show that $X$ is not realcompact?
**Edit:** A z-ultrafilter is an ultrafilter on the lattice of zero sets (as to distinguish them from ultrafilters on the lattice of sets), and a real z-ultrafilter is a z-ultrafilter closed under countable intersections.
[Crossposted with math.stackexchange](https://math.stackexchange.com/questions/4718866/mysior-plane-is-not-realcompact).
**Edit 2:** The reason why I reposted my question here, is that I think only small amount of people on math.stackexchange know what realcompactness is. I want to gain exposure to knowledgeable people that have a higher chance to help me and maybe even know the example I'm talking about. The original paper this comes from is "A union of realcompact spaces" by A. Mysior. All books and articles only briefly mention this example (e.g. Engelking and plethora articles about realcompactness and various alternatives). My guess is that the $1$ and $\sqrt{2}$ are there for a reason, and I tried to consider what would happen if e.g. $f\in C(X)$ were $0$ on $[0, 1]\times \{0\}$, because it seems like maybe I could prove that a zero set containing this will contain $[1, 2]\times \{0\}$ and $[\sqrt{2}, 1+\sqrt{2}]\times\{0\}$ etc., but I feel like while $f$ on $U\_n(x)$ for $x\in [0, 1]$ must approach $0$ as $n$ gets larger, they can do so at distinct rates for all $x$, hence my difficulties. Overall, my first intuition would be to figure out the zero sets contained in $\mathbb{R}\times\{0\}$.
| https://mathoverflow.net/users/150060 | Mysior plane is not realcompact | The proof that I can think of applies a Baire-category argument to the normal topology of the real line.
As in your argument for realcompactness you need to look at zero-sets that are subsets of the $x$-axis. Let $Z\_f$ be such a zero-set, and look at $Z=\{x:f(x,0)=0\}$. There are two cases: (1) every nonempty open interval contains a nonempty open interval in which $Z$ is of first category (union of countably many nowhere dense sets), or (2) there is an open interval $(a,b)$ such that for no subinterval $(p,q)$ of $(a,b)$ the intersection $Z\cap(p,q)$ is of first category in $(p,q)$.
In the first case you can show that $Z$ is of first category in $\mathbb{R}$.
Consider the second case; we show that $(a+1,b+1)\setminus Z$ and $(a+\sqrt2,b+\sqrt2)\setminus Z$ are both of first-category.
Let $\varepsilon>0$. For every $x\in(a,b)\cap Z$ there is an $n\_x$ such that $f[U\_{n\_x}(x)]\subseteq(-\varepsilon,\varepsilon)$.
Let $(p,q)$ be a subinterval of $(a,b)$, by our assumption there is an $n$ such that $F=\{x\in(p,q)\cap Z:n\_x=n\}$ is not nowhere dense, that is there is a further subinterval $(r,s)$ of $(p,q)$ that is contained in the closure (in the normal topology of $\mathbb{R}$) of $F$.
Let $x\in(r,s)$ and take an increasing sequence $\langle t\_k\rangle$ in $F$ that converges to $x$. Now check that $U\_m(x+1)$ interesects $U\_n(t\_k)$ for infinitely many $k$; this implies that $|f(x+1,0)|\le\varepsilon$; similarly you'll find that $|f(x+\sqrt2,0)|\le\varepsilon$.
This shows that the sets $O\_1(\varepsilon)=\{x\in(a,b):|f(x+1, 0)|\le\varepsilon\}$ and $O\_2(\varepsilon)=\{x\in(a,b):|f(x+\sqrt2, 0)|\le\varepsilon\}$ both contain dense open subsets of $(a,b)$.
Now let $D=\bigcap\_m\bigl(O\_1(2^{-m})\cap O\_2(2^{-m})\bigr)$ then $(a,b)\setminus D$ is of first-category and for all $x\in D$ we have $f(x+1,0)=f(x+\sqrt2,0)=0$.
Using simlar methods you can show that $(a-1,b-1)\setminus Z$ and $(a-\sqrt2,b-\sqrt2)\setminus Z$ are both of first-category.
Now use the fact that $\{a+m+n\sqrt2:m,n\in\mathbb{Z}\}$ is dense in $\mathbb{R}$ to see that, in fact, the complement of $Z$ in $\mathbb{R}$ is of first category.
So a zero-set contained in the $x$-axis either is of first-category in $\mathbb{R}$ or its complement is. The latter form a zero-set ultrafilter with the countable intersection property but with an empty intersection.
| 1 | https://mathoverflow.net/users/5903 | 449540 | 180,907 |
https://mathoverflow.net/questions/449377 | 9 | Crossposted from [MSE.](https://math.stackexchange.com/questions/4721350/changing-variables-in-discrete-calculus)
In discrete calculus one defines the $h$-difference operator $$\Delta\_h[f(x)] = f(x+h) - f(x)$$
and we often define $\Delta = \Delta\_1.$ We can similarly define the [indefinite sums](https://en.wikipedia.org/wiki/Indefinite_sum) $\Delta\_h^{-1}$ and set $\Delta^{-1} = \Delta\_1^{-1}.$
Most books on discrete calculus only include results for $h=1.$
For example, the above link gives that
$$\Delta^{-1}\sin rx = \frac{-\cos(r[x-\frac{1}{2}])}{2\sin\frac{r}{2}}$$
Through an ugly computation I managed to work out that $$\Delta\_h^{-1}[\sin rx] = \frac{-\cos(r[x-\frac{h}{2}])}{2\sin\frac{rh}{2}}$$
What I'd like is a uniform procedure for recovering the "$h$-version" from such results. How could I have derived (or even just guessed) the second formula from the first? Is there something like the change of variable formula in calculus? Maybe some version of the 'dimensional analysis' heuristics in ordinary calculus that tells you where to insert the $h$'s? As a test, given a formula at random from the previous link like $\Delta[\log x] = \log(1 + \frac{1}{x})$ I'd like to be able to see immediately that $\Delta\_h[\log x] = \log(1 + \frac{h}{x})$ without carrying out the actual calculation.
A search trough the standard references didn't turn up much, so any help or references would be welcome.
| https://mathoverflow.net/users/506963 | Change of variable formulas in discrete calculus? | In terms of the differential operator $\partial\_x\equiv d/dx$ one has $f(x+h)=e^{h\partial\_x}f(x)$, hence
$$\Delta\_h=e^{h\partial\_x}-1.$$
Upon Fourier transformation, $\hat{f}(k)=\int\_{-\infty}^\infty e^{ikx}f(x)\,dx$, one has $\partial\_x\mapsto -ik$, hence
$$\hat{\Delta}\_h=e^{-ihk}-1\Leftrightarrow \hat{\Delta}\_h^{-1}=\frac{e^{ihk}}{1-e^{ihk}}.$$
In this way you can calculate, for any function $f(x)$ with Fourier transform $\hat{f}(k)$:
$$\Delta\_h^{-1}f(x)=\int\_{-\infty}^\infty\frac{e^{i(h-x)k}}{1-e^{ihk}}\hat{f}(k)\,\frac{dk}{2\pi}=\int\_{-\infty}^\infty F\_h(x'-x)f(x')\,dx'$$
with [Dirac-delta-function](https://www.google.com/search?client=safari&rls=en&q=dirac+delta+function&ie=UTF-8&oe=UTF-8) kernel
$$F\_h(x)=\int\_{-\infty}^\infty\frac{e^{i(h+x)k}}{1-e^{ihk}}\,\frac{dk}{2\pi}=\sum\_{n=1}^\infty\delta(x+nh).$$
The relation $F\_h(x)=(1/h)F\_1(x/h)$ gives the scaling with $h$ requested by the OP:
>
> $\Delta\_h^{-1}f(x)$ follows from $\Delta\_1^{-1}f(rx)$ by rescaling
> $r\mapsto rh$ and $x\mapsto x/h$.
>
>
>
---
The "test cases" mentioned by the OP include: $\Delta\_1^{-1}\log rx=x\log r+\log \Gamma(x)$, hence
$$\Delta\_h^{-1}\log rx=(x/h)\log(rh)+\log\Gamma(x/h).$$
Similarly, $\Delta\_1^{-1}a^{rx}=a^{rx}/(a^r-1)$, hence
$$\Delta\_h^{-1} a^{rx}=a^{rx}/(a^{rh}-1),$$
and $\Delta\_1^{-1}(rx)^a=r^aB\_{a+1}(x)/(a+1)$, hence
$$\Delta\_h^{-1}(rx)^a=(rh)^aB\_{a+1}(x/h)/(a+1).$$
| 5 | https://mathoverflow.net/users/11260 | 449547 | 180,908 |
https://mathoverflow.net/questions/449560 | 2 | This question was originally posted in ME: <https://math.stackexchange.com/questions/4725157/what-is-an-explicit-subset-of-mathbbz3-that-makes-bigl-sinn-cdot-x>
but more and more I think about it, this problem looks nontrivial. So, I ask for help here.
Basically I would like to find an explicit orthonormal basis of $L^2([0,1]^3, \mathbb{R})$ as eigenfunctions of $-\Delta$. Here, we assume periodic boundary conditions on $[0,1]^3$.
Let $n \in \mathbb{Z}^3$ and $x \in [0,1]^3$. Then, it is clear that the collection
\begin{equation}
\Bigl( \sqrt{2}\sin(2\pi n \cdot x),\sqrt{2}\cos(2\pi n \cdot x) \Bigr)\_{n \in \mathbb{Z}^3}
\end{equation}
are eigenfunctions of $-\Delta$ on $[0,1]^3$ with the eigenvalues $4\pi^2 \lvert n \rvert^2$. Moreover this collection clearly generates the Hilbert space $L^2([0,1]^3, \mathbb{R})$ and each element is of unit norm.
However, I would like to extract some "orthonormal basis" from this generating set. For example, the sine and cosine functions corresponding to $n=(1,1,1)$ and $n=(-1,-1,-1)$ are NOT linearly independent.
If I regard $L^2([0,1]^3, \mathbb{R})$ as $3$-fold tensor product of $L^2([0,1], \mathbb{R})$ and construct its orthonormal basis as tensor products of sines and cosines such as $\sin(n\_1 x\_1) \cos(n\_2x\_2) \sin(n\_3 x\_3)$ then, it would suffice to set $n\_1, n\_2, n\_3 \geq 0$.
On the other hand, the situation becomes much more complicated if I seek an orthonormal basis of the form mentioned above..
Could anyone please help me?
| https://mathoverflow.net/users/56524 | What is a subset of $\mathbb{Z}^3$ making $\Bigl( \sin(n \cdot x),\cos(n \cdot x) \Bigr)_{n \in \mathbb{Z}^3}$ linearly independent? | One can use, for example, $n \in \mathbb{Z}^3$ with the following restrictions (edited upon comment by Alexei Kulikov to be careful about cases with $n\_i =0 $): $n\_1 >0 \ \ \lor \ (n\_1 =0 \land n\_2 > 0) \ \lor \ (n\_1 = n\_2 =0 \land n\_3 \geq 0) $.
To abbreviate things, introduce the notation
$$
sss = \sin 2\pi |n\_1 | x\_1 \sin 2\pi |n\_2 | x\_2 \sin 2\pi |n\_3 | x\_3 \\
ssc = \sin 2\pi |n\_1 | x\_1 \sin 2\pi |n\_2 | x\_2 \cos 2\pi |n\_3 | x\_3 \\
\ldots \hspace{8.2cm}
$$
Then, for given $n$ with $n\_1 , n\_2 , n\_3 \geq 0$, your preferred functions are
$$
\sin (2\pi n\cdot x) = scc-sss+csc+ccs \\
\cos (2\pi n\cdot x) = ccc-css-ssc-scs
$$
Note that the right hand sides contain all 8 functions that would (as you note) constitute a basis, namely, $sss,css,scs,ssc,ccs,csc,scc,ccc$ (reduced in number in cases where a $n\_i $ vanishes, stipulating that the zero function is not a basis vector and shall simply be discarded whenever it arises); however, of those 8 functions (or a reduced number), we only have 2 linear combinations.
Let's start by considering $n\_1 > 0$. To disentangle the 8 individual functions (or a reduced number), it is sufficient to supplement with $n\_2 < 0$, yielding the combinations
$$
scc+sss-csc+ccs \\
ccc+css+ssc-scs \ ;
$$
with $n\_3 < 0$, yielding the combinations
$$
scc+sss+csc-ccs \\
ccc+css-ssc+scs \ ;
$$
and with $n\_2 , n\_3 < 0$, yielding the combinations
$$
scc-sss-csc-ccs \\
ccc-css+ssc+scs \ ;
$$
Adding and subtracting the first two sets yields the combinations
$$
scc+ccs \\
sss-csc \\
ccc-scs \\
css+ssc
$$
and adding and subtracting the last two sets yields the combinations
$$
scc-ccs \\
sss+csc \\
ccc+scs \\
css-ssc
$$
Finally, adding and subtracting these new sets of 4 functions each fully disentangles the 8 functions (or a reduced number) that we know to be part of a basis.
Now, in the case $n\_1 =0$, consider to begin with $n\_2 > 0$. In this case, we need only the 4 functions $ccc,csc,ccs,css$ (or a reduced number if $n\_3 =0$), of which we already have 2 linear combinations; to disentangle the functions, we only need to supplement with $n\_3 <0$, i.e., we again have disentangled the functions that we know to be part of a basis.
Finally, for $n\_1 =n\_2 =0$, we only need the 2 functions $ccc,ccs$, which are already disentangled, i.e., we are left with the restriction $n\_1 =n\_2 =0$ and $n\_3 \geq 0$.
Altogether, we thus have a one-to-one mapping between the described cases and a known basis; therefore, the described set of your preferred functions forms a basis (again, the zero function is discarded whenever it arises).
| 3 | https://mathoverflow.net/users/134299 | 449564 | 180,916 |
https://mathoverflow.net/questions/449572 | 11 | Let $X$ be a smooth, projective (complex) variety of dimension $n$ and $Z \subset X$ be a subvariety of codimension $k$ (if necessary assume $Z$ is non-singular). We know the cohomology class $[Z]$ of $Z$ is an element in $H^{2k}(X,\mathbb{C})$. Denote by $\phi\_Z \in H^{2n-2k}(X,\mathbb{C})^{\vee}$ the element corresponding to $[Z]$ under the Poincaré duality $H^{2k}(X,\mathbb{C}) \cong H^{2n-2k}(X,\mathbb{C})^{\vee}$. Is it true that $\phi\_Z$ is a linear combination of integrations of the form $\int\_{W\_i}$ where $W\_i$ is a subvariety in $X$ of codimension $n-k$? Of course, if the Hodge conjecture is true then this is true as well. I wanted to know if this statement holds without the Hodge conjecture? Any hint / reference will be most welcome.
**EDIT** Assume $2k > 2n-2k$.
| https://mathoverflow.net/users/45397 | Does Poincaré duality preserve algebraic cycles? | A positive answer to your question\* would imply one of Grothendieck's standard conjectures (conjecture A) for complex varieties. See his note: [Standard conjectures on algebraic cycles](https://web.archive.org/web/20210822125536/http://www.math.tifr.res.in/~publ/studies/SM_04-Algebraic-Geometry.pdf). It's safe to say it's open!
\* As interpreted it. See Will Sawin's answer for clarifying remarks.
Also as I mentioned in the comments. The standard conjectures over $\mathbb{C}$ are in some sense weaker than the Hodge conjecture. They are known in many more cases (e.g. abelian varieties, Hilbert schemes of points on a surface, and certain hyper Kähler varieties).
| 11 | https://mathoverflow.net/users/4144 | 449574 | 180,920 |
https://mathoverflow.net/questions/449380 | 3 | Is it consistent with $\sf ZFC + \text{ countable models of } ZFC \text { exist}$, that every countable model of $\sf ZFC$ is a subset of some parameter free definable pointwise-definable model of $\sf ZFC$?
| https://mathoverflow.net/users/95347 | Is every countable model of ZFC a subset of some parameter free definable pointwise-definable model of ZFC? |
>
> The answer is in the positive. Since by the completeness theorem the existence of a countable model of ZFC is equivalent to Con(ZFC), the argument below works with the assumption that ZFC + Con(ZFC) is consistent.
>
>
>
Wojowu's answer already points out that the proof of the following fact (S) can be found in the paper [Pointwise Definable Models of Set Theory](https://arxiv.org/abs/1105.4597) (JSL 2013) by Hamkins, Linetsky, and Reitz.
**(S)** Every countable model of ZFC has a (class) generic extension to model of ZFC that is pointwise definable.
[*Historical note:* (S) was asserted with a proof outline in my paper [Models of set theory with definable ordinals](https://www.researchgate.net/publication/220592523_Models_of_set_theory_with_definable_ordinals), which includes a detailed proof of the weaker statement: *every countable model of ZFC has a generic extension in which all the ordinals are parameter-free definable*. The paper of Hamkins-Linetsky-Reitz presents a detailed proof of not only (S), but also of a corresponding result for countable models of GBC].
Note that (S) is a theorem of ZFC (indeed, it is provable in much weaker systems , but that is a different story). It is easy to see that if the theory ZFC + Con(ZFC) is consistent, then so is $T$ = ZFC + Con(ZFC) + V = L. The fact that $T$ includes the statement V = L implies that $T$ has a global well-ordering of its ambient universe and therefore $T$ has definable Skolem functions. This immediately implies that $T$ has a pointwise definable model $M$. Since (S) is a theorem of ZFC, it holds in $M$, and moreover EVERYTHING in $M$ is parameter-free definable. Thus $M$ is a model in which every countable $N$ model of ZFC has a (class) generic extension $N[G]$ such that (1) $N[G]$ is pointwise definable, and (2) $N[G]$ is parameter-free definable in $M$.
| 6 | https://mathoverflow.net/users/9269 | 449576 | 180,921 |
https://mathoverflow.net/questions/449211 | 7 | Consider the finite Boolean lattice $B\_n$ of subsets of $[n]:=\lbrace 1,\dots,n\rbrace$ ordered by inclusion, let $1\leq j,k\leq n$ and consider the poset:
$$A\_{j,k}=\lbrace\emptyset\neq U\in B\_n\mid (1,\dots,j)\nsubseteq U, (n-k,\dots,n)\nsubseteq U \rbrace$$
I want to compute the geometric realization for every $j,k$. Using the ideas of ([Homotopy type of the geometric realization of a poset](https://mathoverflow.net/questions/448347/homotopy-type-of-the-geometric-realization-of-a-poset?noredirect=1#comment1160121_448347)) I was able to prove the following:
If $j\leq n-k-1$ then:
$$|A\_{j,k}|=|((B\_{j}\setminus\lbrace\hat{1}\rbrace)\times B\_{n-j-k-1}\times (B\_{k+1}\setminus\lbrace\hat{1}\rbrace))\setminus\lbrace\hat{0}\rbrace|=$$$$=|B\_{j}\setminus\lbrace\hat{1},\hat{0}\rbrace|\ast| B\_{n-j-k-1}\setminus\lbrace\hat{0}\rbrace|\ast |B\_{k+1}\setminus\lbrace\hat{1},\hat{0}\rbrace|=$$$$=\mathbb{S}^{j-2}\ast | B\_{n-j-k-1}\setminus\lbrace\hat{0}\rbrace|\ast \mathbb{S}^{k-1}=\mathbb{S}^{j+k-2}\ast | B\_{n-j-k-1}\setminus\lbrace\hat{0}\rbrace|$$
Then, if $n=j+k+1~|A\_{j,k}|=\mathbb{S}^{j+k-2}=\mathbb{S}^{n-3}$ and if $n\neq j+k+1~|A\_{j,k}|$ is contractible.
But I don't know how we can prove it for $j\geq n-k$, since in this case the two sets $(1,\dots,j)$ and $(n-k,\dots,n)$ intersect, so I can't split $A\_{j,k}$ into a direct product.
Any hint or help will be thanked.
Edit: Using the hints of @ChristopheLeuridan and @DavidWhite I did the following for the case $j\geq n-k$:
First of all, we can decompose $A\_{j,k}=A\_{j,k}^1\cup A\_{j,k}^2$ where:
$$A\_{j,k}^1=\lbrace\emptyset\neq U\in B\_n\mid (n-k,\dots,j)\nsubseteq U\rbrace$$
$$A\_{j,k}^2=\lbrace\emptyset\neq U\in B\_n\mid (n-k,\dots,j)\subseteq U\text{ and }(1,\dots,n-k-1)\nsubseteq U,(j+1,\dots,n)\nsubseteq U\rbrace$$
We can compute the geometric realization of those two subposets:
$$|A\_{j,k}^1|=|((B\_{j-n-k+1}\setminus\lbrace\hat{1}\rbrace)\times B\_{2n-j+k-1})\setminus\lbrace\hat{0}\rbrace|=$$
$$=|B\_{j-n-k+1}\setminus\lbrace\hat{0},\hat{1}\rbrace|\ast|B\_{2n-j+k-1}\setminus\lbrace\hat{0}\rbrace|=\mathbb{S}^{j-n+k-3}\ast|B\_{2n-j+k-1}\setminus\lbrace\hat{0}\rbrace|$$
Hence, if $2n-j+k-1=0~|A\_{j,k}^1|=\mathbb{S}^{j-n+k-3}=\mathbb{S}^{n+2k-4}$ and if $2n-j+k-1\neq0~|A\_{j,k}^1|$ is contractible, but it is easy to see that we can't have $2n-j+k-1=0$. On the other hand:
$$|A\_{j,k}^2|=|((B\_{n-k-1}\setminus\lbrace\hat{1}\rbrace)\times(B\_{n-j}\setminus\lbrace\hat{1}\rbrace))\setminus\lbrace \hat{0}\rbrace|=$$
$$=|B\_{n-k-1}\setminus\lbrace\hat{0},\hat{1}\rbrace|\ast|B\_{n-j}\setminus\lbrace\hat{0},\hat{1}\rbrace|=\mathbb{S}^{n-k-3}\ast\mathbb{S}^{n-j-2}=\mathbb{S}^{2n-k-j-4}$$
Hence, since those two posets are disjoint we have:
$$|A\_{j,k}|=|A\_{j,k}^1|\cup|A\_{j,k}^2|=\mathbb{S}^{2n-k-j-4}$$
| https://mathoverflow.net/users/482329 | Geometric realization of a poset | Put $P\_0=\{1,\dotsc,n-k-1\}$ and $P\_1=\{n-k,\dotsc,j-1\}$ and $P\_2=\{j,\dotsc,n\}$, so $[n]=P\_0\amalg P\_1\amalg P\_2$ with each $P\_i$ nonempty. Any subset $U$ can be decomposed as $\coprod\_{i=0}^2U\_i$ with $U\_i\subseteq P\_i$. You are looking at the space
$$ A = \{(U\_0,U\_1,U\_2): U\_0\cup U\_1\cup U\_2\neq\emptyset, U\_0\cup U\_1\neq P\_0\cup P\_1, U\_1\cup U\_2\neq P\_1\cup P\_2\}. $$
This can be written as $B\cup C$, where
\begin{align\*}
B &= \{(U\_0,U\_1,U\_2): U\_0\cup U\_1\cup U\_2\neq\emptyset,U\_1\neq P\_1\} \\
C &= \{(U\_0,U\_1,U\_2): U\_0\cup U\_1\cup U\_2\neq\emptyset,U\_0\neq P\_0,U\_2\neq P\_2\} \\
D &= B\cap C = \{(U\_0,U\_1,U\_2): U\_0\cup U\_1\cup U\_2\neq\emptyset,U\_0\neq P\_0,U\_1\neq P\_1,U\_2\neq P\_2\}
\end{align\*}
We can define $f\_0,f\_1,f\_2\colon B\to B$ by $f\_0=\text{id}$ and $f\_1(U\_0,U\_1,U\_2)=(P\_0,U\_1,P\_2)$ and $f\_2(U\_0,U\_1,U\_2)=(P\_0,\emptyset,P\_2)$. We have $f\_0\leq f\_1\geq f\_2$, so the identity map $|f\_0|\colon|B|\to|B|$ is homotopic to the constant map $|f\_2|$, so $B$ is contractible. Similarly, we can define $g\_0,g\_1,g\_2\colon C\to C$ by $g\_0=\text{id}$ and $g\_1(U\_0,U\_1,U\_2)=(U\_0,P\_1,U\_2)$ and $g\_2(U\_0,U\_1,U\_2)=(\emptyset,P\_1,\emptyset)$, and use these to see that $|C|$ is contractible. We also have a pushout square of subcomplex inclusions
$\require{AMScd}$
\begin{CD}
|D| @>>> |B|\simeq \* \\
@VVV @VVV\\
|C|\simeq \* @>>> |A|
\end{CD}
which gives $|A|\simeq\Sigma|D|$. Now put $D\_i=\{U\_i : \emptyset\subset U\_i\subset P\_i\}$, so $|D\_i|\simeq S^{|P\_i|-2}$. It is easy to see that $|D|$ is the same as the join $|D\_0|\*|D\_1|\*|D\_2|$, and $S^r\*S^s\simeq S^{r+s+1}$, so $|D|\simeq S^{|P\_0|+|P\_1|+|P\_2|-4}=S^{n-4}$ and $|A|\simeq\Sigma S^{n-4}=S^{n-3}$.
| 2 | https://mathoverflow.net/users/10366 | 449596 | 180,924 |
https://mathoverflow.net/questions/449251 | 1 | I would like to see an example of a fiber bundle $\mathbb{T}^2 \to X \xrightarrow{\pi} B$ whose fibers are tori $\mathbb{T}^2:=(S^1)^2$ and whose vertical tangent bundle $\ker(d\pi) \to X$ is not decomposable? Which homotopy groups does one need to look at to construct such an example?
(This problem has been edited using inputs from the comments.)
| https://mathoverflow.net/users/15197 | A torus bundle whose vertical tangent bundle is indecomposable | Put $T=\{(z\_0,z\_1,z\_2)\in(S^1)^3:z\_0z\_1z\_2=1\}$, so $T$ is homeomorphic to $S^1\times S^1$ and has an obvious action of the symmetric group $\Sigma\_3$. The action of $\Sigma\_3$ on $H\_1(T;\mathbb{R})$ is indecomposable, and this homology group can also be identified with the tangent space to $T$ at the identity element.
Now put $F=\{(u\_0,u\_1,u\_2)\in\mathbb{C}^3:u\_i\neq u\_j\text{ for } i\neq j\}$. This also has a natural action of $\Sigma\_3$, which is free. Put $E=(T\times F)/\Sigma\_3$ and $B=F/\Sigma\_3$, so we have a fibre bundle $T\to E\to B$. This will have the required indecomposability. Note that the identity element of $T$ gives a section of the projection $E\to B$.
The cohomology of $F$ is a well-known example. For $i\neq j$ we can define $\pi\_{ij}\colon F\to\mathbb{C}\setminus\{0\}$ by $\pi\_{ij}(z)=z\_j-z\_i$, and by pulling back the generator of $H^1(\mathbb{C}\setminus\{0\})$ we get an element $a\_{ij}\in H^1(F)$. These satisfy $a\_{ij}=a\_{ji}$ and $a\_{ij}^2=0$ and the Arnol'd relation $a\_{ij}a\_{jk}+a\_{jk}a\_{ki}+a\_{ki}a\_{ij}=0$. It follows that $\{1,a\_{01},a\_{02},a\_{12},a\_{01}a\_{02},a\_{01}a\_{12}\}$ is a basis for $H^\*(F)$. From this description you can calculate $H^\*(B)$ and $H^\*(E)$ using the Serre spectral sequence. With rational coefficients this just degenerates to $H^\*(B;\mathbb{Q})=H^\*(F;\mathbb{Q})^{\Sigma\_3}=\mathbb{Q}$ and $H^\*(E;\mathbb{Q})=H^\*(T\times F;\mathbb{Q})^{\Sigma\_3}$. It can't be too hard to calculate that last ring explicitly, but I have not done so.
If you prefer to work with homotopy groups, it is standard that $\pi\_1(B)$ is the braid group $Br\_3$. This has a standard surjective homomorphism to $\Sigma\_3$, whose kernel is the pure braid group $PBr\_3$, which can be identified with $\pi\_1(F)$. It is also known that $\pi\_i(F)=\pi\_i(B)=\pi\_i(T)=0$ for $i>1$. The map $PBr\_3\to\Sigma\_3$ gives an action of $PBr\_3$ on $\pi\_1(T)\simeq\mathbb{Z}^2$, using which we can form the semidirect product $PBr\_3\ltimes\mathbb{Z}^2$. Using the split fibre bundle $T\to E\to B$ we can identify $\pi\_1(E)$ with this semidirect product, and also see that $\pi\_i(E)=0$ for $i>1$.
| 2 | https://mathoverflow.net/users/10366 | 449601 | 180,925 |
https://mathoverflow.net/questions/449578 | 1 | **Imprecise Question**: Suppose I have a function defined on non-codimension-zero strata of a smooth manifold with a stratification, and I know the function is smooth when restricted to each of these strata. Is there any setting in which the function extends smoothly to the whole smooth manifold?
Here is the baby case in which I am interested (where corners also come into play).
Fix $n \geq 0$ and let $\mathbb{R}\_{\geq 0}^n$ denote standard Euclidean octant -- vectors whose coordinates are non-negative.
For any subset $P \subset \{1,\ldots,n\}$, we let $C\_P \subset \mathbb{R}\_{\geq 0}^n $ denote the locus of points whose $i$th coordinate vanishes for all $i \in P$. Note we know abstractly what it means for a function $f\_P : C\_P \to \mathbb{R}$ to be smooth, treating each $C\_P$ as a smooth manifold with corners in the standard way.
Suppose that for every non-empty $P$, we are given a smooth function $f\_P$, and we also know that $f\_P|\_{C\_Q} = f\_{P \cup Q}$ for all pairs $P,Q$ with $P \cap Q \neq \emptyset$.
**Precise Question:** Then is there a globally defined function $f = f\_{\emptyset}: \mathbb{R}\_{\geq 0}^n \to \mathbb{R}$ which is smooth, and for which $f|\_{C\_P} = f\_P$?
A colleague remarked this is a consequence of Whitney's Extension Theorem, but I was not able to see why even after looking up the theorem.
| https://mathoverflow.net/users/507571 | Smooth extension of piecewise smooth function on a corner | You have a function defined on the boundary of $\mathbb R^n\_{\ge 0}$. Its restriction to each face is smooth. You can extend it to all of $\mathbb R^n\_{\ge 0}$ by writing
$$
f(x)=\sum\_P (-1)^{|P|-1}f\_P(x\_P),
$$
where $x\mapsto x\_P$ is the projection $\mathbb R^n\_{\ge 0}\to C\_P$.
| 2 | https://mathoverflow.net/users/6666 | 449607 | 180,927 |
https://mathoverflow.net/questions/449580 | 2 | The following question arised in my research, and I was unable to settle it after playing with it for sometime. Let $\{a^k\_i\}\_{i\geq 1}$ (for $k\in \{1,2,3,4\}$) be four sequences of real numbers. For each $k\in \{1,2,3,4\}$, consider the function $f\_k:]-r\_k,r\_k[\rightarrow \mathbb{R}$ given by $x\mapsto\sum\_{i=1}^{\infty}a\_i^kx^i$ (Where $r\_k$ is defined as $\frac{1}{\limsup\_{i\rightarrow \infty}|a\_i^k|^{\frac{1}{i}}}$).
It is given that for all $x$ sufficiently close to $0$, we have $f\_2(f\_1(x))=f\_4(f\_3(x))$. It is also given that for some $x\_0\in\mathbb{R}$, we have that both $f\_2(f\_1(x\_0)), f\_4(f\_3(x\_0))$ exist. Must $f\_2(f\_1(x\_0))= f\_4(f\_3(x\_0))$ ? I would be very grateful if you can help, thank you.
**Edit:** "$f\_2(f\_1(x\_0))$ exists" means that $x\_0$ belongs to domain of $f\_1$ and that $f\_1(x\_0)$belongs to domain of $f\_2$. Similarly, "$f\_4(f\_3(x\_0))$ exists" means that $x\_0$ belongs to domain of $f\_3$ and that $f\_3(x\_0)$belongs to domain of $f\_4$
I know that the answer is yes in some special cases like $f\_1$ equals identity map, or $f\_1,f\_2,f\_3,f\_4$ are rational functions (Which raises the question what if the $f\_k$(s) are algebraic functions ?). I also considered extending the domain of the $f\_k$(s) to an open subset of the complex plane and possibly open the route for analytic continuations and complex analysis machinery but was not successful with that.
| https://mathoverflow.net/users/32135 | Local equality of functions implies global equality? | Consider $$\eqalign{f\_1(x) &= (x+1)^2-1\cr f\_2(x) &= \sqrt{x+1}-1\cr f\_3(x) &= f\_4(x) = x
}$$
$f\_1, f\_3$ and $f\_4$ being polynomials, their radius of convergence is $\infty$, while $f\_2(x)$ has a Maclaurin series with radius of convergence $1$. $f\_2(f\_1(x)) = f\_4(f\_3(x)) = x$ for $x \in (-1,1)$. But since $f\_1(x) = f\_1(-x-2)$, $f\_2(f\_1(x))$ is also defined for $x \in (-3,-1)$, with $f\_2(f\_1(x)) = -2-x$ there while $f\_4(f\_3(x)) = x$.
| 4 | https://mathoverflow.net/users/13650 | 449608 | 180,928 |
https://mathoverflow.net/questions/449598 | 0 | Assume I have the following function
$f\left(n\right)=\frac{-2\sqrt{n}}{A}e^{-A\left(\frac{k}{\sqrt{n}}\right)}\left(e^{-\frac{A}{\sqrt{n}}}-1\right)$
Where $n-k\gg1$ and $k\gg\sqrt{n}$ and $A=2$ is a constant.
And $n\rightarrow\infty$.
My thoughts is that $f\left(n\right)\rightarrow 0$ (I have ran some examples in walfram alpha and desmos)
But I am not quite sure how to approach this problem and I would like some help.
| https://mathoverflow.net/users/491400 | asymptotic behavior of function $f\left(n\right)=\frac{-2\sqrt{n}}{A}e^{-A\left(\frac{k}{\sqrt{n}}\right)}\left(e^{-\frac{A}{\sqrt{n}}}-1\right)$ | You have
$$f(n)=\sqrt{n} (1-e^{-2/\sqrt{n}})
e^{-2k/\sqrt{n}}\sim2e^{-2k/\sqrt{n}}\to0,
$$
since $k>>\sqrt{n}$. So, $f(n)\to0$.
| 1 | https://mathoverflow.net/users/36721 | 449612 | 180,930 |
https://mathoverflow.net/questions/449628 | 1 | $\sf V=HOD$ is stated as:
$\forall X \, \exists \theta \, \exists \alpha < \theta \, \exists \varphi : X=\{y \in V\_\theta\mid V\_\theta\models\varphi(y,\alpha)\}$
This use two ordinal parameters (other than the code for $\varphi$) $``\theta; \alpha"$.
Can we do with just ONE parameter (other than the code for $\varphi$)? that is:
$\forall X \, \exists \alpha \, \exists \varphi: X= \{y \in V\_\alpha \mid V\_\alpha \models \varphi(y)\}$
| https://mathoverflow.net/users/95347 | Can we state $\sf V=HOD$ using a single ordinal parameter(other than the formula code)? | Yes, because there is a definable ordinal pairing function.
Specifically, if you want to get the set $\{y\in V\_\theta\mid V\_\theta\models\varphi(y,\alpha)\}$, then let $\beta=\langle\theta,\alpha\rangle$ be the ordinal coding the pair, and then look at $V\_{\beta+1}$. Inside this structure, we have $\beta$ as the largest ordinal, and the model can decode $\beta$ as the pair $\langle\theta,\alpha\rangle$, and then get the set $\{y\in V\_\theta\mid V\_\theta\models\varphi(y,\alpha)\}$. So this set is definable inside $V\_{\beta+1}$ without any extra parameter.
Ultimately one should view $\varphi$ also as a parameter (and formulas can be coded by finite numbers, hence by ordinals, a fact that is important when proving that HOD as defined internally is the same as the external concept of definable-from-ordinal parameters, since the internal definition might use nonstandard $\varphi$, but this is still OK since the code of a nonstandard formula is still a number, hence an ordinal). But this is fine, since we can consider $\beta=\langle\theta,\alpha,\ulcorner\varphi\urcorner\rangle$ and then proceed as I explained.
| 6 | https://mathoverflow.net/users/1946 | 449629 | 180,933 |
https://mathoverflow.net/questions/449638 | 3 | I'm writing a Master's thesis on knot invariants and I'm trying to chase down the original source that introduced the unknotting number and perhaps proved that it is a knot invariant. The texts I've consulted that reference the unknotting number already discuss it in terms of being an established invariant. But I can't find what source has first introduced it as a concept and proved that it is a knot invariant. I think it's obvious that Reidemeister moves don't affect the unknotting number (they preserve the number of crossings), but what is it that makes this number well-defined?
I'm a bit wary of ChatGPT, but it's pointed me in the direction of (Gordon & Luecke, 1989). I scanned through the paper but must admit found it too dense to read. Equally, the unknotting number was already an established concept by 1989 (it pops up in (Cochran & Lickorish, 1986)), so I'm curious if there isn't a different, earlier proof.
| https://mathoverflow.net/users/137916 | Straightforward reference on the unknotting number being a knot invariant | I agree with Tom Goodwillie that it is immediate from the definitions that the unknotting number is well-defined. I think that what might be tripping you up is being taught that “knot invariants” are really diagram invariants that are invariant under the Reidemeister moves. That makes invariants like this one that cannot be computed from a given diagram in a straightforward way confusing.
Anyway, on to its origins. The earliest reference I know about is
[Wendt, H. Die gordische Auflösung von Knoten. Math. Z. 42, 680–696 (1937)](https://link.springer.com/article/10.1007/BF01160103)
It's been a very long time since I read this paper (and my German is mediocre, so it would take some effort for me to re-read it today), so I can't remember if it references anything else.
As evidence that this is the original source, Reidemeister's classic [book](https://www.maths.ed.ac.uk/%7Ev1ranick/papers/reidemeister_eng.pdf) on the subject references Wendt's paper when it introduces the unknotting number. See the beginning of Chapter 2.
| 6 | https://mathoverflow.net/users/317 | 449643 | 180,936 |
https://mathoverflow.net/questions/449553 | 8 | I am looking for examples of results about large cardinals, large cardinal axioms, or other objects of high (or seemingly high) consistency strength that are almost inconsistencies. I am looking for large cardinal axioms that only remain consistent because they "got off on a technicality".
In particular, I am looking for axioms $C$ along with theorems $T$ proven using $C$ such that the theorem $T$ calls into question the consistency of $C$ but which does not establish a contradiction under $C$.
These examples should fall under the standard large cardinal hierarchy. For example, the partition relation $\kappa\rightarrow(\aleph\_0)^{\aleph\_0}\_2$ (this says that for each function $f:[\kappa]^{\aleph\_0}\rightarrow\{0,1\}$, there is an infinite $A$ where $f|\_{[A]^{\aleph\_0}}$ is constant) is an inconsistent strengthening of the notion of an $\omega$-Erdos cardinal, but I would not consider this axiom since it is not implied by the existence of a non-trivial rank-into-rank embedding (unless non-trivial rank-into-rank embeddings are inconsistent).
Any near inconsistency of an axiom $C$ where $\text{Con}(\text{ZFC}+\text{I0})\rightarrow\text{Con}(C)$ and where $\text{Con}(C)\rightarrow\text{Con}(\text{ZFC})$ should be fine. I am not currently interested in near inconsistencies that are reformulations of the Kunen inconsistency nor am I currently interested in exceedingly strong axioms such as Berkeley cardinals.
When an axiom which we shall call Axiom $C$ (Axiom $C$ could be the existence of a Vopenka or a rank-into-rank cardinal) is originally formulated and it is not immediately obvious that the consistency of Axiom $C$ is implied by stronger large cardinals, then it is natural to be skeptical about the consistency of Axiom $C$. But I am generally not currently interested in these apparent near inconsistencies because the doubt of the consistency of Axiom $C$ seems to wane over time as the theory around Axiom $C$ or about stronger axioms is developed. I am instead looking for a case where after Axiom $C$ is formulated and the theory of Axiom $C$ is developed, a result $T$ derived from Axiom $C$ causes people to doubt the consistency of Axiom $C$.
Perhaps Jack Silver's concerns about the consistency of ZFC or of measurable cardinals should count as a near inconsistency.
**My own example of a near inconsistency**
It is easy to prove that if $j:V\_\lambda\rightarrow V\_\lambda$ is an elementary embedding and $\alpha<\lambda$, then $(j\*j)(\alpha)\leq j(\alpha)$, but there are many finite algebraic structures that appear similar to algebras of elementary embeddings but which violate this inequality.
An algebra $(X,\*,1)$ that satisfies the identities $x\*(y\*z)=(x\*y)\*(x\*z)$ and $x\*1=1,1\*x=x$ is said to be nilpotent if for all $x\in X$, there is some $n$ with $x^{[n]}=1$ where we define $x^{[n]}$ recursively by letting $x^{[1]}=x,x^{[n+1]}=x\*x^{[n]}$. Define $x\_{[1]}=x,x\_{[n+1]}=x\_{[n]}\*x$. Then there are 6854 nilpotent self-distributive algebras $(X,\*,1,a,b)$ generated by $a,b$ up to a critical equivalence preserving $a,b$ with $a\_{[64]}=1,b\*b=1$ and where $\text{crit}[X]$ is linearly ordered but only 1145 of these (16.7056%) satisfy the inequality $\text{crit}((x\*x)\*y)\leq\text{crit}(x\*y)$. This looks like a near inconsistency since it seemed feasible to obtain one of those 5709 algebras from rank-into-rank embeddings.
| https://mathoverflow.net/users/22277 | Large cardinal near inconsistencies | I would argue that a "restricting versions" of large cardinals are such.
Starting from the top down, we have the inconsistent Berkeley cardinals:
* $κ$ is Berkeley if for every transitive $M\ni\kappa$ and $ν<κ$ we have an elementary embedding on $M$ with critical point between $\nu$ and $κ$.
A way to make this cardinal consistent with $\sf AC$ is by restricting the sets $M$ can be:
* $κ$ is $\sf HOD$-Berkeley if the above condition holds for $M\in \sf HOD$
This large cardinal is very important, the consistency of $\sf ZFC+HOD$-Berkeley cardinal is implied by the consistency of $\sf ZF$+Berkeley cardinal (±some other large cardinals), and if $\sf ZFC+HOD$-Berkeley is consistent it means that $\sf HOD$ is very far from $V$ (again, ±some other LCA).
Similarly to Berkeley, we can weaken Reinhardt cardinals to smaller classes:
* $κ$ is $\sf HOD$-Reinhardt if there exists an elementary embedding on $\sf HOD$ with critical point $κ$.
Similarly to the Berkeley case, the consistency of $\sf ZF$+Reinhardt implies the consistency of $\sf ZFC+HOD$-Reinhardt (±LCA), and the existence of (sufficiently large) $\sf HOD$-Reinhardt implies that $\sf HOD$ is far from $V$ (±LCA).
The $\sf HOD$-Berkeley/Reinhardt cardinals are $\sf HOD$-analog to $0^\sharp$ of $L$, which brings us to:
* a non-trivial elementary embedding on $L$.
* a non-trivial elementary embedding from $V$ to some transitive class
I would argue that the only reason this is less concerning to be consistent is that we know that $V$ can be *very* far from $L$, and because we know the theory of ultrapowers.
| 7 | https://mathoverflow.net/users/113405 | 449648 | 180,940 |
https://mathoverflow.net/questions/449610 | 5 | Let $f,g\colon K\to \mathcal{C}$ be diagrams in a nice $\infty$-category $\mathcal{C}$. I have two general questions:
1. If I have a natural transformation $\eta\colon f\Rightarrow g$ which is a monomorphism in $\mathcal{C}$ at every component, is it a monomorphism in $Fun(K,\mathcal{C})$?
2. If every 1-simplex in the image of $f$ is a monomorphism, is it true that the maps $lim(f)\to f(\sigma)$ are monomorphisms for all 0-simplices $\sigma\in K$?
I don't know if these are going to be true at this level of generality. Ultimately I'm interested in the very special case of $\mathcal{C}=Gpd\_\infty$ and $K$ being the subcategory of $\Delta$ spanned by injections (i.e. semicosimplicial objects), so if there's a simpler answer in that case I'd be very happy to know it.
| https://mathoverflow.net/users/11546 | Monomorphisms of diagrams in an $\infty$-category | For completeness, and because I cannot figure out the general case (cf. my comment below Daniel's answer), let me prove the following: if $f:\Delta\to C$ is a functor which, when restricted to $\Delta\_{inj}$ takes values in monomorphisms, then $\lim\_\Delta f\to f([0])$ is a monomorphism.
This is simply the following observation : $f([0])\simeq \lim\_\Delta f(-\*[0])$ (as $f(-\*[0])$ is a split-augmented simplicial object), and along thsi identification, the projection map corresponds to $\lim\_\Delta (f\to f(-\*[0]))$. Now each $[n]\to [n]\*[0]$ is an injection, and so this map is $\lim\_\Delta$ of monomorphisms, and hence is a monomophism itself.
This seems to use something specific about the combinatorics of $\Delta$.
When $K$ is general, and say has a final object $\infty$, then one can argue similarly: one has a composite $\lim\_K f\to f(k) \to f(\infty)$, and thus by some version of 2-out-of-3 it suffices to prove that $\lim\_K f\to f(\infty)$ is a monomorphism, but this follows from it simply being $\lim\_K (f\to f(\infty))$. My attempt for a general $K$ with $|K|\simeq \*$ was to try and prove that the inclusion $K\to K^\triangleright$, where we add a final object, preserves the property of every map being a mono, i.e. that for every $k$, $f(k)\to \mathrm{colim}\_K f$ would be a monomorphism - this step is where one would use $|K|\simeq \*$, but I didn't manage to prove it.
There is a somewhat easy-to-prove version which is when $K$ is weakly contractible via a zig-zag of functors : if there is a zigzag $id\_K \leftarrow h\_1 \to h\_2 \leftarrow \dots \to h\_n$ where $h\_n$ is a constant functor, then one can do a similar argument to the one for $\Delta$ and prove the result. I don't remember exactly how often we can expect this: is it the case that any weakly contractible $K$ is a filtered colimit of $K$'s admitting such a zigzag ?
| 4 | https://mathoverflow.net/users/102343 | 449649 | 180,941 |
https://mathoverflow.net/questions/449656 | 6 | $\DeclareMathOperator\Vect{Vect}\DeclareMathOperator\Mat{Mat}$This question was originally asked on [MSE](https://math.stackexchange.com/questions/4723838/are-algebroids-just-matrices) but may be better here.
[Algebroids](https://ncatlab.org/nlab/show/algebroid) are particularly interesting structures: they are basically categories enriched over $\Vect\_K$ for some field $K$. This means the hom-sets are all vector spaces, and composition is "bilinear" in a certain sense. (Let's just focus on when $K$ is a field for now rather than an arbitrary ring.)
Most examples I can think of are basically equivalent to some subset of matrices with the usual addition and multiplication rules, as long as we are willing to be creative and allow "infinite matrices" of various cardinalities to exist. In general, for any $n$ and $m$, the set of $n \times m$ matrices forms an algebroid, as long as we also allow the $n \times n$ and $m \times m$ identity matrices to exist. The union of any such sets also forms an algebroid, as long as matrix compositions exist when expected (meaning if we have $5 \times 4$ and $4 \times 3$ matrices, we must also have $5 \times 3$ matrices, as well as the relevant identity matrices). This is also true if $n$ and $m$ are arbitrary infinite cardinals, with the caveat that only finitely many elements of each column of the matrix can be nonzero.
So, we can ask if this is basically "what algebroids are." It is a little bit difficult to even figure out how to formalize the question, but we can think about it in the following sense:
Let's say that the category $\Mat^+\_K$ is basically an extension of the usual $\Mat\_K$, but with the objects as all possible *cardinals* rather than only natural numbers. For objects $\kappa$, $\lambda$, the morphisms $\kappa \to \lambda$ are (possibly infinite) matrices of size $\kappa \times \lambda$ (treating these cardinals as initial ordinals), with finitely many nonzero coefficients in each column, taking values in $K$.
As a sort of first pass, we would then like to ask the following "pre-question":
**Naive Pre-Question**: is every possible $K$-algebroid equivalent to a subcategory of $\Mat\_K^+$?
The main problem with this is that we can take a disjoint union of two $\Mat\_K^+$'s. This will be a $K$-algebroid, but as we haven't drawn explicit isomorphism arrows between the two copies of each object, I don't think the two categories will be equivalent.
But, we can salvage the spirit of the question by noting that $\Mat\_K^+$ is a skeleton category of $\Vect\_K$, which has infinitely many copies of the things in $\Mat\_K^+$ with all possible linear transformations (and thus isomorphisms, when they exist) between them. So, we can thus ask our real question:
**Real Question**: is every possible $K$-algebroid equivalent to a subcategory of $\Vect\_K$?
---
EDIT: It has been pointed out in the [comment](https://mathoverflow.net/questions/449656/are-algebroids-just-matrices#comment1161885_449656) that this can fail due to size issues, e.g. you can have an algebroid so large that it isn't concretizable. I guess that technically answers the way I formulated the "real question" above, as I hadn't thought of that snag, but still looking to get clarity on the algebraic relationship of these things beyond that sort of thing, so would be happy to see it addressed for small algebroids.
| https://mathoverflow.net/users/24611 | Are algebroids "just matrices"? | If we take the large but locally small category $\mathcal{C}$ described by Isbell in Example 2.4 of
* *Two set-theoretical theorems in categories*, Fundamenta Mathematicae **53** Issue 1 (1964) pp 43-49, ([EuDML](https://eudml.org/doc/213746)),
namely, with class of objects $(\{1,2\}\times\mathrm{ORD}) \sqcup \{X,Y\}$, and generated by maps $f\_{\alpha,X}\colon (1,\alpha) \to X$, $f\_{\alpha,Y}\colon (1,\alpha) \to Y$, $g\_{X,\alpha}\colon X\to (2,\alpha)$, $g\_{Y,\alpha}\colon Y\to (2,\alpha)$, for each ordinal $\alpha \in \mathrm{ORD}$, with the relation that $\forall \alpha$ forces $g\_{X,\alpha}\circ f\_{\alpha,X} = g\_{Y,\alpha}\circ f\_{\alpha,Y}$, and such that we also keep the two composites $g\_{X,\beta}\circ f\_{\alpha,X}$ and $g\_{Y,\beta}\circ f\_{\alpha,Y}$ distinct for $\alpha\neq \beta$.
The free $K$-algebroid $K[\mathcal{C}]$ on this category, namely take the free vector space on the hom-sets over any field $K$, and extend composition bilinearly, cannot admit a faithful functor to $K\text{-}\mathrm{Vect}$, since otherwise there would be a faithful functor $\mathcal{C} \to K[\mathcal{C}] \to K\text{-}\mathrm{Vect} \to \mathrm{Set}$, which Isbell proved cannot happen.
EDIT: For small algebroids, there is no issue: for every small algebroid $\mathcal{A}$ we can send an object $a$ to the vector space that is the *direct sum* over all $\mathcal{A}(x,a)$ as $x$ varies over all objects. This gives a faithful functor to vector spaces.
It is only for large algebroids where the issues come in. There are large algebroids where the construction in the previous paragraph still works and gives a functor to the category of vector spaces. For example, a large algebroid $\mathcal{A}$ such that for each object $a$, only a set's worth of hom-spaces $\mathcal{A}(x,a)$ are nontrivial vector spaces. For example the free $K$-algebroid on the opposite of $\mathrm{ORD}$ (arrows point to smaller ordinals), so that hom-spaces are either $K$ or $0$, and only set-many are $K$.
| 8 | https://mathoverflow.net/users/4177 | 449660 | 180,944 |
https://mathoverflow.net/questions/449657 | 5 | The statement in the title seems to be generally accepted as true, but I have not seen proof. They are?
The strict formulation I have in mind is the following. By an algebraic category we mean the category of algebras of some monad on $\mathrm{Set}^S$. In particular, this includes all the usual finitary algebraic categories: groups, modules, etc [1]. Also note that the product of every set of algebraic categories is an algebraic category (this way we can combine every set of invariant into one).
**Question**. Is there an essentially injective functor $F: \mathrm{Hmpt} \to A \times B^{\mathrm{op}}$, where $\mathrm{Hmpt}$ is the category of homotopy types (which can be equivalently defined as the homotopy category of cw-complexes, delta-generated spaces, simplicial sets, or infinity-groupoids), and $A, B$ are algebraic categories (so we can use both covariant and contravariant invariant). Essentially injectivity means that $F(X) \cong F(Y)$ implies that $X \cong Y$.
For example, $F$ can take all of the following data
* all homotopy groups with their Whitehead product (in particular, the action of $\pi\_1$ on higher homotopy groups is stored)
* some set generalized homology and cohomology theories (we can't take all of them because they form a proper class) with all cohomological operations
**Sub-question**. Is this $F$ essentially injective?
The closest thing I can think of is that by [Freyd's theorem](http://www.tac.mta.ca/tac/reprints/articles/6/tr6.pdf) there can't be such a faithful functor, because $A \times B^{\mathrm{op}}$ is [concrete](https://ncatlab.org/nlab/show/concrete+category) over $\mathrm{Set}$. But I don't see how that would help. At the same time, it seems to me really curious to know whether there can be some set of algebraic invariants, by comparing which, we can conclude that the spaces are homotopically equivalent.
[1]: In addition to them, we have categories where operations in algebras can have infinite arity, as well as categories where the arity of operations is unlimited - the latter gives, for example, compact Hausdorff spaces. This doesn't bother me at all, because firstly these categories still behave quite similarly to algebraic ones (and seem to still be quite efficient invariants), and secondly, I think you can replace $A \times B^{\mathrm{op}}$ to any concrete category and the answer should still be no.
| https://mathoverflow.net/users/148161 | Can we prove that any number of algebraic data cannot be a complete invariant of a homotopy type? | I'll answer the corresponding question for the homotopy category $\mathcal{S}$ of spectra. I doubt that this makes much difference, but I have not checked the details. We can choose a list $X\_0,X\_1,\dotsc$ containing one representative of every homotopy equivalence class of finite spectra, and then put $X=\bigvee\_iX\_i$ and $R=[X,X]$. This has an idempotent $e\_i$ for each $i\in\mathbb{N}$, corresponding to the obvious map $X\to X\_i\to X$. Define $F\colon\mathcal{S}\to\text{Mod}\_R$ by $F(E)=E\_0R=\pi\_0(E\wedge R)$. Given $F(E)$ we can use the idempotents $e\_i$ to determine the groups $E\_0X\_i$, and $R$ contains $e\_jRe\_i=[X\_i,X\_j]$ so we can determine the full homology theory on finite spectra represented by $E$. It is a standard consequence of Brown representability theory that this determines $E$ up to a weak equivalence (which is itself unique up to phantoms).
Of course there is no hope of understanding this $R$ explicitly, so this answer may be considered unsatisfying, but it does answer the question as asked.
Here are some similar questions where positive answers are known.
* The category of rational spectra is equivalent to the category of graded rational vector spaces. More generally, let $G$ be a finite group, let $H\_1,\dotsc,H\_r$ be a list containing one subgroup from each conjugacy class, and let $W\_i=N\_G(H\_i)/H\_i$ be the associated Weyl group, and put $R=\prod\_i\mathbb{Q}[W\_i]$. Then the category of rational $G$-spectra is equivalent to the category of graded $R$-modules.
* Fix a prime $p$, and let $\mathcal{F}\_p$ be the category of spectra that can be expressed as the $p$-completion of a finite spectrum. We then have a functor $\pi\_\*\colon\mathcal{F}\_p\to\text{Mod}\_{\pi\_\*(S\_p)}$. Freyd conjectured (the ``Generating Hypothesis'' or GH) that this is faithful. It is known that many consequences would follow if GH were true; in particular $\pi\_\*(X)$ would be an injective module over $\pi\_\*(S\_p)$ for all $X\in\mathcal{F}\_p$, and the functor $\pi\_\*\colon\mathcal{F}\_p\to\text{Mod}\_{\pi\_\*(S\_p)}$ would in fact be a full and faithful embedding, so $\mathcal{F}\_p$ would be an algebraic category.
* Let $\mathcal{M}$ be the category of Moore spectra, i.e. spectra $X$ with $\pi\_i(X)=0$ for $i<0$ and $H\_i(X)=0$ for $i\neq 0$. Define a *Moore diagram* to be a diagram $A\xrightarrow{\phi}B\xrightarrow{\psi}A$ with $\psi\phi=0$ and $\phi\psi=2.1\_B$. We say that such a diagram is *exact* if the induced sequence $A/2\to B\to\text{ann}(2,A)$ is short exact. It can be shown (<https://arxiv.org/abs/1205.2247>) that the category of Moore spectra is equivalent to the category of exact Moore diagrams. However, even though this is a rather simple example, there is a surprisingly large amount of work in the proof, which is not a promising sign if you want to generalise further.
* You could relax your requirements and ask for good functors from a category of spectra to the derived category of an abelian category, rather than to the abelian category itself. The most important example is as follows. We can fix a prime $p$ and an integer $n>0$ and consider the category $\mathcal{L}(p,n)$ of spectra that are Bousfield-local with respect to the Johnson-Wilson spectrum $E(p,n)$. This category is a central player in the chromatic approach to stable homotopy theory. It is known that when $p$ is large relative to $n$, the category $\mathcal{L}(p,n)$ is equivalent to a kind of derived category of differential graded comodules. The first results in this direction were due to Franke, but the paper <https://arxiv.org/abs/1903.10003> (by Barthel, Schlank and Stapleton) is probably the best place to look to understand the current status.
| 10 | https://mathoverflow.net/users/10366 | 449669 | 180,947 |
https://mathoverflow.net/questions/449670 | 2 | I'm reading about the Weierstrass zeta function. In this context,
$\phi(z)=\zeta(z)-\pi\bar{z}$
is periodic over the lattice
$$\mathcal{L}=\{a+bi\mid a,b\in\mathbb{Z}\}.$$
If we take $w\in\mathcal{L}\setminus\{\mathbf{0}\}$, then
$$\phi(z+w)-\phi(z)=w\sum\_{q\in\mathcal{L}\setminus\{0\}}q^{-2}-\pi\bar{w}.$$ Since $\phi$ is periodic, we get
$$\frac{1}{\pi}\sum\_{q\in\mathcal{L}\setminus\{0\}}q^{-2}=\frac{\bar{w}}{w}.$$
At this point, I wonder the value of this summation. If someone knows about this, please let me know.
**Question.** What is the value of
$$\sum\_{q\in\mathcal{L}\setminus\{0\}}q^{-2}?$$
| https://mathoverflow.net/users/482837 | The sum of $q^{-2}$ over nonzero Gaussian integers | The series
$$\sum\_{q\in\mathcal{L}\setminus\{0\}}q^{-2}$$
is not absolutely convergent. If we arrange the terms in groups of four of shape $\{\pm q,\pm iq\}$, then the series converges to zero since
$$q^{-2}+(iq)^{-2}+(-q)^{-2}+(-iq)^{-2}=0,\qquad q\in\mathcal{L}\setminus\{0\}.$$
For more general sums of similar shape, see [Eisenstein series](https://en.wikipedia.org/wiki/Eisenstein_series).
| 6 | https://mathoverflow.net/users/11919 | 449673 | 180,949 |
https://mathoverflow.net/questions/449676 | 1 | Let $X,Y,Z$ be Banach spaces and let $m\,:\,X\times Y\to Z$ be a bilinear map such that $\|m(x,y)\|\leq C \|x\|\|y\|$ for some fixed constant $C$. Moreover, let $\mu$ be a Borell vector measure on $\mathbb{R}^d$ valued in $Y$ with finite total variation and let $f\,:\,\mathbb{R}^d\to X$ be bounded and measurable (in the sense that an inverse image of a Borel set is Borel). Can one make sense of the following integral
$$
\int\_{\mathbb{R}^d} m(f(u),\mu(du)) \in Z
$$
(possibly under some extra conditions like the condition that the Banach spaces are separable).
| https://mathoverflow.net/users/47256 | Integration of vector function against vector measure | $\newcommand\R{\mathbb R}$Suppose that $X$ is separable. Let
$$\int\_{\R^d}m(x\,1\_B,d\mu):=m(x,\mu(B))$$
for all $x\in X$ and Borel $B\subseteq\R^d$. Then, for simple functions
$$s\_n:=\sum\_{j=1}^n x\_j\,1\_{B\_j},$$
with $x\_j\in X$ and Borel $B\_j\subseteq\R^d$ for all $j$,
let
$$\int\_{\R^d}m\big(s\_n,d\mu\big):=\sum\_{j=1}^k m(x\_j,\mu(B\_j)).$$
Then, by approximation, extend this notion of integral to bounded measurable functions $f\colon\R^d\to X$. Here, as in the theory of the Bochner integral, we can use the Pettis theorem implying that a measurable function with values in a separable Banach space is a $\|\mu\|$-almost everywhere limit of a sequence of measurable simple functions, where $\|\mu\|$ is the variation measure for $\mu$ -- cf. e.g. Sections V.4 and V.3 of K. Yosida's Functional Analysis, Sixth Edition.
| 1 | https://mathoverflow.net/users/36721 | 449681 | 180,950 |
https://mathoverflow.net/questions/449679 | 3 | Is the following a theorem of $\sf ZF+[V=HOD]$?
If $Q$ is a property definable in a parameter free manner, then: $$\operatorname {Con}(\sf ZF + [V=HQD]) \implies V=HQD$$
where $\sf V=HQD$ means:
$$\forall X \exists v\_0 \exists v\_1: Q(v\_0) \land Q(v\_1) \land \rho(v\_0) > \rho(v\_1) \land \exists \varphi:\\ X=\{y \in V\_{\rho(v\_0)} \mid V\_{\rho(v\_0)} \models \varphi(y,v\_1) \}$$
Where $\rho$ is the known rank function.
| https://mathoverflow.net/users/95347 | Does V=HOD prove all kinds of consistent universal hereditary definability? | The answer is no. Indeed, one can rarely move from consistency to truth in this way.
For a counterexample, let $Q(v)$ be the property "CH holds and $v$ is an ordinal."
If CH holds, then $Q$ expresses the property of being an ordinal, but if CH fails, then $Q$ never holds. Consider a model of ZF+V=HOD in which CH fails, but Con(ZF) holds. So V=HQD is false in this model, since Q is never true. But from Con(ZF) we can prove Con(ZF+V=HQD), since it can build a model of V=HOD in which CH holds, and that will be a model of V=HQD.
The point is that the consistency statement is true in this countermodel, because it is consistent that CH holds with V=HOD, but the statement itself is not true, because CH fails.
| 6 | https://mathoverflow.net/users/1946 | 449682 | 180,951 |
https://mathoverflow.net/questions/449613 | 3 | I already asked this on [Math.SE](https://math.stackexchange.com/questions/4722773/how-to-determine-the-type-of-a-divisor-d-on-a-product-of-elliptic-curves), but didn't receive an answer yet.
---
Say $E\_1, \dotsc, E\_n$ are elliptic curves (everything over $\mathbb C$), and $D \subset E\_1 \times \dotsc \times E\_n$ is an effective divisor. How can I determine the type $(d\_1, \dotsc, d\_n)$ of the line bundle $\mathcal O(D)$?
If $D = \sum\_i \operatorname{pr}\_i^\* D\_i$ for divisors $D\_i \subset E\_i$, then $d\_i = \deg D\_i = \deg (D \cdot \iota\_i(E\_i))$, where $\iota\_i: E\_i \to E\_1 \times \dotsc \times E\_n$ embeds $E\_i$ by $z \mapsto (0, \dotsc, z, \dotsc, 0)$. Is it true in general that one gets $d\_i$ as the degree of the intersection of $D$ with $\iota\_i(E\_i)$?
---
Concretely, I'm interested in the following two cases:
1. Let $D\_0 \subset E \times E$ be the union of the diagonal and the anti-diagonal,i.e.
$$D\_0 = \{(z,z) : z \in E\} \cup \{(z,-z):z\in E\}.$$
2. Take the quotient $A = E \times E / \langle (\frac 1 2,\frac 1 2)\rangle$, by the $2$-torsion point $(\frac 1 2, \frac 1 2)$, and divisor $D\_1 = \overline{D\_0}$. Using the isomorphism $$E \times E \to E \times E, (z\_1, z\_2) \mapsto (z\_1, z\_1 - z\_2)$$ one sees that $A$ is isomorphic to $E / \langle \frac 1 2 \rangle \times E$, so again a product of ellpitic curves.
Any help would be appreciated :)
| https://mathoverflow.net/users/111897 | How to determine the type of a divisor on a product of elliptic curves? | I managed to calculate my examples. In the first one, $D\_0$ has indeed polarization type $(2,2)$. To see this, let $E = \mathbb C / (\mathbb Z + \tau \mathbb Z)$ be an elliptic curve, and consider the isogeny
$$\varphi: E \times E \to E \times E, (z\_1, z\_2) \mapsto (z\_1 + z\_2, z\_1 - z\_2).$$
Then $\varphi^2(z\_1, z\_2) = (2z\_1, -2z\_2)$, so $\deg \varphi^2 = 16$ and $\deg \varphi = 4$.
Note that $D\_0 = \varphi^\* \tilde D$, where $\tilde D$ is the divisor
$$\tilde D = E \times \{0\} \cup \{0\} \times E.$$
Clearly $\tilde D$ has type $(1,1)$, and the corresponding hermitian form on $\mathbb C^2$ is
$$H\left(\left(\begin{smallmatrix}u\_1 \\ u\_2 \end{smallmatrix}\right),\left(\begin{smallmatrix}v\_1 \\ v\_2 \end{smallmatrix}\right)\right) = \frac 1 {\operatorname{Im}\tau} (u\_1 \bar v\_1 + u\_2 \bar v\_2).$$
If $\phi: \mathbb C^2 \to \mathbb C^2$ is the analytic representation of $\varphi$, an easy calculation shows $\phi^\* H = 2H$. Hence $D$ has type $(2,2)$.
I think my second example actually serves as a counterexample to the claim that the $d\_i$ can be computed as $E\_i \cdot D$. The projection $E \times E \to A$ has degree $2$, and the above isogeny $\varphi$ factors over $A$, so $\tilde D$ has to pull-back to a polarization of type $(1,2)$ on $A$ (for combinatorial reasons). But under the identification $A \cong F \times E$, with $F = E / \langle \frac 1 2 \rangle$, $D\_1$ is given by
$$D\_1 = (F \times 0) \cup \{(z,2z)\} \subset F \times E.$$
Then $0 \times E$ intersects both components in the point $(0,0)$; $F \times 0$ doesn't intersect itself, but intersects the second component in the points $(0,0)$ and $(\frac \tau 2, 0)$. All intersections are transversal, so this would give a polarization type $(2,2)$.↯
| 1 | https://mathoverflow.net/users/111897 | 449683 | 180,952 |
https://mathoverflow.net/questions/449634 | 7 | I’m wondering what is known about the cohomology of $\operatorname{GL}\_3(\mathbb{Z}\_2)$, the general linear group of $3\times3$ matrices over the finite field $\mathbb{Z}\_2$. There are results in Chapter I of Knudson’s “[Homology of Linear Groups](https://doi.org/10.1007/978-3-0348-8338-2)” as well as Chapter VII of Adem and Milgram’s “[Cohomology of Finite Groups](https://doi.org/10.1007/978-3-662-06280-7)” that consider the cohomology of $\operatorname{GL}\_n(\mathbb{F}\_p)$, but not this specific case.
Update: It seems that Adem and Milgram does the trick once we identify $\operatorname{GL}(3,2) =\operatorname{SL}(3,2)$. The relevant lemma says that
$$H^\*(\operatorname{SL}(3,2)) \oplus H^\*(D\_8) \cong H^\*(S\_4)\oplus H^\*(S\_4). $$
| https://mathoverflow.net/users/503849 | Cohomology of $\operatorname{GL}_3(\mathbb{F}_2)$ | As you mention in your update, you have a general answer, but if you want a concrete answer for the low-dimensional integral cohomology of $G = \operatorname{GL}(3,2)$ (or any other finite group!), you can use [GAP](https://www.gap-system.org/index.html), specifically the `hap` package. The following code computes $H^i(G, \mathbf{Z})$ for $i \leq 10$:
```
LoadPackage("hap");
G = GL(3,2);
List([1..10], x->GroupCohomology(G, x));
```
This will output:
```
[ [ ], [ ], [ 2 ], [ 4, 3 ], [ ], [ 2, 2, 7 ], [ 2 ], [ 4, 3 ], [ 2, 2 ], [ 2, 2 ] ]
```
which is a list of the abelian invariants of the cohomology groups $H^i(G, \mathbf{Z})$ for $1 \leq i \leq 10$, i.e. the first ten integral cohomology groups are $1, 1, C\_2, C\_4 \times C\_3, 1, C\_2 \times C\_2 \times C\_7, C\_2, C\_4 \times C\_3, C\_2 \times C\_2$ and $C\_2 \times C\_2$ (for readability I here use $C\_n = \mathbf{Z}/n\mathbf{Z}$).
If you want to compute the mod $p$ cohomology, and don't feel like using the above with the universal coefficient theorem by hand, you can compute this just as easily by using `GroupCohomology(G, x, p)`. When $p=2$, the sequence of exponents of the first ten cohomology groups (together with $H^0(G, \mathbf{Z}/2\mathbf{Z})$) is
$$
1, 0, 1, 2, 1, 2, 3, 2, 3, 4, 3
$$
e.g., $H^3(G, \mathbf{Z}/2\mathbf{Z}) \cong (\mathbf{Z}/2\mathbf{Z})^2$. This sequence (of course) agrees with the generating function in @DaveBenson's answer, which expands as:
$$
\frac{1+t^3}{(1-t^2)(1-t^3)} = 1 + \quad + t^2 + 2 t^3 + t^4 + 2 t^5 + 3 t^6 + 2 t^7 + 3 t^8 + 4 t^9 + 3 t^{10} + ...
$$
| 11 | https://mathoverflow.net/users/120914 | 449684 | 180,953 |
https://mathoverflow.net/questions/439203 | 0 | The Crust algorithm by Amenta, Bern, and Eppstein computes a polygonal reconstruction of a smooth curve $C$ without boundary from a discrete set of sample points $S$. It is known that if $S$ is an a $\epsilon$-sample of $C$ with $\epsilon < \frac{1}{5}$, then the Crust algorithm computes the correct polygonal reconstruction, in the sense that it correctly connects all pairs of sample points that are consecutive along $C$. (Here, $\epsilon$-sample is defined with respect to the medial axis of $C$, as in Tamal Dey, "Curve and Surface Reconstruction", [http://www.csun.edu/~ctoth/Handbook/chap35.pdf](http://www.csun.edu/%7Ectoth/Handbook/chap35.pdf) .)
In practice, the Crust algorithm often computes correct polygonal reconstructions from samples that are less dense than a $\frac{1}{5}$-sample. What is the smallest $\epsilon > 0$ for which there exists a smooth curve $C$ without boundary and an $\epsilon$-sample $S$ of $C$ such that the Crust algorithm computes an *incorrect* polygonal reconstruction of $C$? Clearly such $\epsilon$ must be at least $\frac{1}{5}$.
| https://mathoverflow.net/users/48162 | What is the most dense sample for which the Crust algorithm returns an incorrect polygonal reconstruction? | In 2022, Håvard Bakke Bjerkevik gave a 0.72-sample of a particular curve for which reconstruction is not possible. Moreover, Bjerkevik showed that curve reconstruction is always possible from a 0.66-sample. This work appears in "Tighter Bounds for Reconstruction from ϵ-samples," <https://doi.org/10.4230/LIPIcs.SoCG.2022.9>.
It seems the question of curve reconstruction from an ϵ-sample is still open for ϵ-samples with 0.66 < ϵ < 0.72.
| 0 | https://mathoverflow.net/users/48162 | 449691 | 180,955 |
https://mathoverflow.net/questions/449687 | 4 | This question is somewhat similar to [Minimizing the L1 norm of odd-term trigonometric polynomial](https://mathoverflow.net/questions/260077/minimizing-the-l1-norm-of-odd-term-trigonometric-polynomial). The context of the question is based on the paper [Hardy's Inequality and the $L^1$ norm of Exponential Sums](https://www.jstor.org/stable/2007000). In that paper, they prove the inequality that given any increasing sequence $n\_1, \dots, n\_k$ of integers,
$$
\left\|\sum a\_j e^{2\pi i n\_j x}\right\|\_1 \geq C \sum \frac{|a\_j|}{j}
$$ for some absolute constant $C$.
A key idea of the paper is that they are able to construct a function $F$ such that
1. $\|F\|\_{\infty} \leq 1$.
2. $\hat F(n\_j) = \Theta(1/j)$ (This is the big-theta notation)
Can one extend this idea to give the same/worse bound for functions of the form $\sum a\_j \cos(2\pi n\_j x)$ where the $n\_j$'s to be positive integers? Or perhaps is there a simpler way by saying something like
$$
\left\|\sum a\_j e^{2\pi i n\_j x}\right\|\_1 \leq C(k) \left\|\sum a\_j \cos(2 \pi i n\_j)\right\|\_1
$$ for some constant $C(k)$ that's not too big?
Concretely, can one prove:
$$
\left\|\sum a\_j \cos(2 \pi n\_j x)\right\|\_1 \geq C \sum \frac{|a\_j|}{j}
R$$ for some absolute constant $C$? Or maybe a worser bound?
| https://mathoverflow.net/users/482554 | A lower bound for the $L^1$ norm of real trigonometric polynomials | This is not true.
Consider the [Fejer Kernel](https://en.wikipedia.org/wiki/Fej%C3%A9r_kernel) $F\_n$. From the explicit formula for $F\_n$ we see that $$\|F\_n\|\_{L^1} \leq C $$ independent of $n$.
On the other hand since $ F\_n = 2 \sum\_{0\leq k \leq n-1} \left(1 - \frac{|k|}{n} \right) \cos(k x)$ we have that $$ \sum\_{0\leq k \leq n-1} \frac{1}{k} \left(1 - \frac{|k|}{n} \right) \sim \log n.$$
| 4 | https://mathoverflow.net/users/630 | 449699 | 180,956 |
https://mathoverflow.net/questions/449697 | 3 | $\sf ZF + Def$ is the theory that extends $\mathcal L(=,\in)\_{\omega\_1,\omega}$ with axioms of $\sf ZF$ (written in $\mathcal L(=,\in)\_{\omega, \omega}$) and the axiom of definability:-
$\textbf{Define: } Dx \iff \bigvee x= \{ y \mid \Phi \}$
where $\Phi$ range over all formulas in $\mathcal L(=,\in)\_{\omega, \omega}$ in which only the symbol "$y$" occurs free, and the symbol "$y$" never occurs bound.
**Axiom of definability:** $\forall x Dx$
This theory has its models being exactly the pointwise-definable models of $\sf ZF$ [[Hamkins](https://mathoverflow.net/a/448434/95347)]
Now, if one wants to confine matters to $\mathcal L(=,\in)\_{\omega,\omega}$, i.e. the usual $\textbf{FOL}(=,\in)$, then one can add this rule to $\sf ZFC$:
$\textbf{Definability: }$ if $\phi\_1,\phi\_2, \phi\_3,...$ are all formulas in which only symbol "$y$" occurs free, and "$y$" never occur bound, and that doesn't use the symbol "$x$", and $\psi$ is a formula in which only symbol "$x$" occurs free, and "$x$" never occur bound; then:
$\underline {[i=1,2,3,...; \\ \forall x \, (x=\{y \mid \phi\_i\} \to \psi)]} \\ \forall x: \psi$
In English: if a parameter free formula holds for all parameter free definable sets, then it holds for all sets.
This was proved by [Hamkins](https://mathoverflow.net/a/448132/95347) to be equivalent over $\sf ZFC$ to the set theoretic axiom $\sf V=HOD$.
The finitary version of definability doesn't manage to confine all of its models to be the pointwise-definable models of $\sf ZFC$. [see [here](https://mathoverflow.net/a/448132/95347)]. But, it can be argued to be the finitary parallel of the infinitary version.
My question here is:
>
> Are there finitary sentences (i.e. in $\mathcal L(=,\in)\_{\omega,\omega}$) that are theorems of $\sf ZF + Def$ yet not provable in $\sf ZFC + \text { Definability rule}$ (or equivalently in $\sf ZFC+[V=HOD])$? Or is the former a conservative extension of the latter?
>
>
>
>
> If there are such sentences, are there clear examples?
>
>
>
| https://mathoverflow.net/users/95347 | Is ZF + Def a conservative extension of ZFC+HOD? If not, what are counter-examples? | Any model of ${\sf ZFC}+V=\sf HOD$ has an elementary equivalent pointwise definable model.
If $M$ models $V=\sf HOD$, it has a parameter free definable well ordering, for each formula $φ$ consider the Skolem function that gives the least witness using that well ordering, the class $M\_0$ of definable (without parameters) elements of $M$ will be closed under those Skolem functions, so $M\succ M\_0$, because $M\_0$ is elementary substructure of $M$, the definable elements of $M\_0$ are exactly equal to those of $M$, so by construction $M\_0$ is pointwise definable model that is elementary equivalent to $M$.
In other words, $\sf ZFC$+"being pointwise definable" does not prove anything that ${\sf ZFC}+V=\sf HOD$ does not prove in the language of FOST
| 6 | https://mathoverflow.net/users/113405 | 449700 | 180,957 |
https://mathoverflow.net/questions/449611 | 6 | A finite group $G$ is called *rational* if every element $g \in G$ is conjugate to all of its primitive powers $g^a, a \in (\mathbb{Z}/\operatorname{order}(g))^\times$.
Analogously, I'll call $G$ *real* if every $g$ is conjugate to its inverse. The ratio of these notions is something I'll call *rational-relative-to-real* (RRR): for every $g$, every primitive power $g^a$ should be conjugate to either $g$ or $g^{-1}$ (or both).
These conditions arise naturally in character theory. Given a field $\mathbb{k}$, let me write $\mathrm{R}\_{\mathbb{k}}(G)$ for the ring of $\mathbb{k}$-linear representations of $G$. So $\mathrm{R}\_{\mathbb{C}}(G) = \mathrm{RU}(G) = \mathrm{R}(G) $ and $\mathrm{R}\_{\mathbb{R}}(G) = \mathrm{RO}(G)$. Then $G$ is rational when $$\mathrm{R}\_{\mathbb{Q}}(G) \otimes \mathbb{Q} \to \mathrm{R}\_{\mathbb{C}}(G) \otimes \mathbb{Q}$$
is an isomorphism; $G$ is real when
$$\mathrm{R}\_{\mathbb{R}}(G) \otimes \mathbb{Q} \to \mathrm{R}\_{\mathbb{C}}(G) \otimes \mathbb{Q}$$
is an isomorphism; and $G$ is RRR when
$$\mathrm{R}\_{\mathbb{Q}}(G) \otimes \mathbb{Q} \to \mathrm{R}\_{\mathbb{R}}(G) \otimes \mathbb{Q}$$
is an isomorphism.
According to [Geoff Robinson's answer to a previous question](https://mathoverflow.net/a/449488/78), finite groups are almost never rational: any abelian composition factor has order at most 11; other than alternating groups, there are only five nonabelian simple groups that can appear.
The question I most care about it:
>
> Which finite simple groups are RRR?
>
>
>
For example, which sporadic groups are RRR?
| https://mathoverflow.net/users/78 | Which finite simple groups are rational-relative-real? | Rational-relative-to-real groups are frequently called inverse semi-rational groups (or cut groups). Inverse semi-rational groups are a subclass of semi-rational groups. A finite group $G$ is *semi-rational* if for every $g \in G$ there exists a positive integer $m\_0$ such that every primitive power $g^a$ ($a \in (\mathbb{Z}/\operatorname{order}(g))^\times$) is conjugate to $g$ or $g^{m\_0}$. A semi-rational group is *inverse semi-rational* if we can take $m\_0 = -1$ for every $g \in G$ in the above definition. The terms semi-rational groups and inverse semi-rational groups were introduced in the article [Chillag–Dolfi, [Semi-rational solvable groups](https://doi.org/10.1515/JGT.2010.004), *Journal of Group Theory* **13**(4), 2010, 535-548]. In [Alavi–Daneshkhah, [On semi-rational finite simple groups](https://doi.org/10.1007/s00605-016-0964-3), *Monatshefte für Mathematik* **184**(2), 2017, 175-184] the authors determined the finite simple semi-rational groups.
Note that, in the statement of Theorem 1.1 of Alavi–Daneshkhah, the group $G\_2(4)$ is missing. The Tits group, the Suzuki groups, the Ree groups and the Steinberg triality groups are discarded in the proof, so should actually be removed from the list in Theorem 1.1.
The simple inverse semi-rational groups were derived in Theorem 5.1 of [Bächle-Caicedo-Jespers-Maheshwary, [Global and local properties of finite groups with only finitely many central units in their integral group ring](https://doi.org/10.1515/jgth-2020-0165), *Journal of Group Theory* **24**, 2021, 1163-1188].
| 6 | https://mathoverflow.net/users/153043 | 449702 | 180,958 |
https://mathoverflow.net/questions/449677 | 7 | I am looking for the analytic continuation of
\begin{align\*}
& f\_m(v,w) := \sum\limits\_{k,l=0}^\infty v^k w^l {k+l+m \choose k} {k+l+m \choose l} \ ,
\end{align\*}
where $m \in \{1,2,...\}$ is fixed. The sums converge for small enough $|v|$ and $|w|$ and I have already made a few observations:
We can write
\begin{align\*}
& f\_m(v,w) = \sum\limits\_{k=0}^\infty v^k \frac{k!}{(k+m)!} \partial\_w^m [\phantom{|}\_2F\_1(k+1,k+1;1;w)]
\end{align\*}
and the hypergeometric function $\phantom{|}\_2F\_1(k+1,k+1;1;w)$ has the analytic continuation $(1-w)^{-k-1} P\_k(\frac{1+w}{1-w})$, where $P\_k$ is the $k$-th Legendre Polynomial. I think the property $\sum\limits\_{k=0}^\infty t^k P\_k(x) = \frac{1}{\sqrt{1-2xt+t^2}}$ or maybe the analogon for Jacobi-Polynomials (see [here](https://en.wikipedia.org/wiki/Jacobi_polynomials#Generating_function)) might be helpful.
Another observation is that the Chu-Vandermonde identity allows us to split up the binomial coefficients ${k+l+m \choose k}$ and ${k+l+m \choose l}$ into sums, which leads to the (not quite recursive) porperty
\begin{align\*}
& f\_m(v,w) = \sum\limits\_{i,j=0}^m w^i v^j {m \choose i} {m \choose j} f\_{i+j}(v,w) \ .
\end{align\*}
Finally the special case $v=w$ by Chu-Vandermonde identity yields
\begin{align\*}
& f\_m(v,v) = \sum\limits\_{a=0}^\infty v^a \sum\limits\_{k=0}^a {a+m \choose k} {a+m \choose k+m} = \sum\limits\_{a=0}^\infty v^a {2(a+m) \choose a} = \phantom{|}\_2{F}\_{1}\Big( m+\frac{1}{2} , m+1 ; 2m+1 ; 4v \Big) \ .
\end{align\*}
Any help in finding the analytic continuation is greatly appreciated!
| https://mathoverflow.net/users/409412 | Help finding an analytic continuation | Your $f\_m(v,w)$ is a special case of [Appell's $F\_4$ hypergeometric function](https://mathworld.wolfram.com/AppellHypergeometricFunction.html),
$$f\_m(v,w) = F\_4(m+1;m+1;m+1,m+1;v,w).$$
Some information about analytic continuation of $F\_4$ can be found in <https://arxiv.org/abs/2005.07170> and the references cited there.
| 12 | https://mathoverflow.net/users/10744 | 449703 | 180,959 |
https://mathoverflow.net/questions/449696 | 3 | Let $ f : [0,1] \to \mathbb{R} $ be a function satisfying: 1.) $ |f(x)| \leqslant a $ for some $ a < 1 $, and 2.) $ \int\_0^1 f(x) {\mathrm d}x = 0 $. I would like to know whether the following inequality holds:
$$
\int\_0^1 \log(1+f(x)) \,{\mathrm d}x \; +\; \int\_0^1 (1-f(x)) \log(1-f(x)) \, {\mathrm d} x \;\stackrel{?}{\geqslant}\; \log(1-a^3)
$$
Note that $ 1 + f(x) $ and $ 1 - f(x) $ are probability densities on $ [0,1] $, so the stated inequality can also be written in the form:
$$
D\_{KL}(1 \parallel 1+f) - D\_{KL}(1-f \parallel 1) \;\stackrel{?}{\leqslant}\; \log\frac{1}{1-a^3}
$$
where $ D\_{KL}(p \parallel q) = \int p(x) \log \frac{p(x)}{q(x)} {\mathrm d}x $ is the Kullback-Leibler divergence (relative entropy).
The inequality comes up in an information-theoretic setting and it "should" be true, but I failed to prove it formally. I did confirm it numerically in some special cases, e.g., when $ f(x) = a \cos(2\pi n x) $ for various $ a, n $, when $ f(x) = \pm a $ (each on one half of the domain), etc.
| https://mathoverflow.net/users/83189 | Bound on an integral representing a difference of two relative entropies | Yes, the inequality is true. Indeed, the inequality in question can be rewritten (or, if you prefer, generalized) as
\begin{equation}
Eg(Y)\ge\ln(1-a^3), \tag{10}\label{10}
\end{equation}
where
$$g(t):=\ln(1+t)+(1-t)\ln(1-t)$$
and $Y$ is a random variable (r.v.) such that
\begin{equation}
P(|Y|\le a)=1\quad\text{and}\quad EY=0. \tag{20}\label{20}
\end{equation}
(One can take $Y:=f(X)$, where $X$ is a r.v. uniformly distributed on $[0,1]$.)
By well-known results (see e.g. [this paper](https://projecteuclid.org/journals/annals-of-mathematical-statistics/volume-26/issue-2/The-Extrema-of-the-Expected-Value-of-a-Function-of/10.1214/aoms/1177728543.full) or [this paper](https://www.jstor.org/stable/3689412) or [Corollary 13](https://link.springer.com/article/10.1007/s00186-015-0530-0) or [formula (2.13)](https://projecteuclid.org/journals/electronic-journal-of-probability/volume-14/issue-none/Optimal-two-value-zero-mean-disintegration-of-zero-mean-random/10.1214/EJP.v14-633.full)), without loss of generality (wlog) the distribution of the r.v. $Y$ is supported on a set $\{u,v\}\subset[-a,a]$ of cardinality $\le2$. Given such $u$ and $v$, if $a$ is now replaced by the smallest possible value of $b$ such that $\{u,v\}\subseteq[-b,b]$, the left-hand side of \eqref{10} will not change whereas the right-hand side of \eqref{10} will not decrease. So, wlog one of the values $u$ or $v$ coincides with one of the endpoints of the interval $[-a,a]$; that is, wlog either $v=-a$ or $v=a$. To cover both of these cases at once, assume wlog that the distribution of the r.v. $Y$ is supported on a set of the form $\{-a,u,a\}$ with $u\in[-a,a]$.
So then, it remains to show that
\begin{equation\*}
L:=pg(-a)+qg(u)+rg(a)-\ln(1-a^3)\ge0 \tag{30}\label{30}
\end{equation\*}
given the conditions
\begin{equation\*}
-a\le u\le a,\quad p,q,r\ge0,\quad \\
p+q+r=1,\quad p(-a)+qu+sa=0. \tag{40}\label{40}
\end{equation\*}
Solving the latter two equations in \eqref{40} for $p$ and $r$ and substituting the resulting expressions for $p$ and $r$ into \eqref{30}, we get
\begin{equation\*}
L=M(q,a,u):=P(a)+qR(a,u),
\end{equation\*}
where
\begin{equation\*}
\begin{aligned}
P(a)&:=a \tanh ^{-1}(a)+\ln \left(1-a^2\right)-\ln \left(1-a^3\right) \\
Q(a,u)&:=g(u)-(1-u/2) \ln \left(1-a^2\right)-a \tanh ^{-1}(a).
\end{aligned}
\end{equation\*}
Note that $M(q,a,u)$ is affine in $q$.
So, to prove \eqref{30} given \eqref{40}, it suffices to show that
\begin{equation\*}
M(0,a,u)\ge0\quad\text{and}\quad M(1,a,u)\ge0 \tag{50}\label{50}
\end{equation\*}
for $u\in[-a,a]$ and $a\in(0,1)$.
We have $M(0,a,u)=P(a)$,
$P(0)=0=P'(0)$ and
\begin{equation\*}
P''(a)=H(a):=\dfrac{a \left(6+10 a+2 a^2-3 a^3+2 a^4+a^5\right)}{(1-a)^2 (1+a)^2
\left(1+a+a^2\right)^2}, \tag{60}\label{60}
\end{equation\*}
which is obviously $>0$ for $a\in(0,1)$. So, $M(0,a,u)=P(a)>0$ for $a\in(0,1)$.
Next,
\begin{equation\*}
M(1,a,u)=F(a):=F(a,u):=g(u)-\ln \left(1-a^3\right) \\
+\frac u2\, \ln \left(1-a^2\right),
\end{equation\*}
\begin{equation\*}
F'(a)\frac{(1 - a) (1 + a) (1 + a + a^2)}a
=3 a (1 + a) - (1+ a +a^2) u \\
\ge3 a (1 + a) - (1+ a +a^2) a
=a (2 + 2 a - a^2)>0.
\end{equation\*}
So, $F$ is increasing in $a$ and hence $M(1,a,u)=F(a,u)\ge F(|u|,u)$, because $|u|\le a$. If now $u\in[0,1)$, then $F(|u|,u)=F(u,u)=P(u)\ge0$, by what was shown in the previous paragraph.
Finally, if $u\in(-1,0)$, then $F(|u|,u)=F(-u,u)=h(u):=g(u)+\frac u2\, \ln(1 - u^2) - \ln(1 + u^3)$, $h(0)=0=h'(0)$, $h''(u)=H(|u|)>0$ -- cf. \eqref{60}; so, $M(1,a,u)=F(a,u)\ge F(|u|,u)=F(-u,u)=h(u)\ge0$.
Thus, \eqref{50} is proved. $\quad\Box$
| 3 | https://mathoverflow.net/users/36721 | 449706 | 180,960 |
https://mathoverflow.net/questions/449704 | 5 | My question stems from the following result about holomorphic functions on the unit disc:
>
> "A function, continuous on the closed unit disc, holomorphic inside, and vanishing on an open subset of the boundary, vanishes identically. A simple proof of this well-known proposition is obtained by considering its Cauchy integral representation." (*Generalized Analytic Functions* by Richard Arens and I.M.Singer)
>
>
>
There are other more direct proofs too, including the one by P.R. Chernoff, using a product of finitely many rotations of $f: \Bbb D \to \Bbb C$ and the identity theorem. See details [here](https://math.stackexchange.com/questions/443406/holomorphic-function-zeros-on-the-circle), for example.
In P.R. Chernoff's article, I found the statement of a stronger result - the set of zeros of $f: \Bbb D \to \Bbb C$ on $∂$ cannot have positive Lebesgue measure unless $f$ vanishes identically. The proof is more involved and can be found in Chapter 4 of *Banach Spaces Of Analytic Functions* by Kenneth Hoffman. See the Corollary on Pg. 52 of a Theorem on Pg. 51.
**Question:** It is natural to ask what happens if $\Bbb D$ is replaced by an arbitrary open (connected?) subset $\Omega \subset \Bbb C$. Does the result continue to hold, i.e., is it true that the set of zeros of a holomorphic function $f:\Omega\to \Bbb C$ on $\partial \Omega$ cannot have positive Lebesgue measure unless $f$ vanishes identically on $\Omega$? Does a weaker result hold? One may require $\Omega$ to be bounded.
Thanks a lot!
| https://mathoverflow.net/users/157422 | Boundary zeros of a holomorphic function $f: \Omega \to \Bbb C$ | It is not clear in your question what "Lebesgue measure on $\partial\Omega$" really means.
Let us begin with the unit disk. In the unit disk, every bounded holomorphic function which is zero on a set $E$ of positive measure, in the sense that
$$\limsup\_{r\to 1}|f(re^{i\theta})|=0,\quad e^{i\theta}\in E,$$ vanishes. This is the direct consequence of Jensen's inequality
$$\log|f(0)|\leq\frac{1}{2\pi}\int\_{-\pi}^\pi\log|f(re^{i\theta})|\,d\theta, \quad r\in(0,1).$$
(If $f\neq 0$, you can find an automorphism $\phi$ of the unit disk
such that $g=f\circ\phi(0)\neq 0$. Applying Jensen's inequality to this $g$ and letting $r\to 1$ gives a contradiction.)
Now for other regions. If $\Omega$ is simply connected, you can use a conformal map of the unit disk onto $\Omega$. But then the question is what is the correct "measure" on $\partial \Omega$. It is called the harmonic measure. For a simply connected $\Omega$ it can be defined as the image of
the Lebesgue measure on the unit circle under the conformal map. But it can be defined for arbitrary region possessing
a Green function (in particular for any bounded region, or more generally for any region for which $\partial D$ contains a continuum). Let $G$ be the Green function, with
the pole at some point $z\_0\in\Omega$. Extend it to the whole plane by setting $G(z)=0,z\not\in\Omega$. The resulting function is subharmonic in ${\mathbf{C}}\backslash\{z\_0\}$,
and harmonic measure is the Riesz measure, $(2\pi)^{-1}\Delta G$ of this subharmonic function. Here $\Delta$ is in the sense of Schwarz distributions.
Now your question has the following general answer for arbitrary region possessing a Green function: if $f$ is bounded, and tends to zero as $z\to E$, where $E$ is a set
of positive harmonic measure then $f=0$.
The question remains how to characterize the sets of positive or zero harmonic measure in terms of some usual metric properties,
for example Hausdorff measures, in particular the $1$-dimensional measure when $\partial G$ is rectifiable etc.
The answer depends on the class of domains $\Omega$ considered,
and this is a subject of a very large body of literature,
of which I can recommend the modern book:
MR2150803 Garnett, John B.; Marshall, Donald E. Harmonic measure, Cambridge, 2005.
If $\Omega$ is simply connected, and $\partial\Omega$ is a rectifiable Jordan curve, then, according to a theorem of F. and M. Riesz, the sets of zero harmonic measure are the same as those of zero 1-Hausdorff measure.
F. and M. Riesz (1916), "Über die Randwerte einer analytischen Funktion", Quatrième Congrès des Mathématiciens Scandinaves, Stockholm, pp. 27–44.
| 5 | https://mathoverflow.net/users/25510 | 449718 | 180,964 |
https://mathoverflow.net/questions/449729 | 3 | Let $A,B$ be $C^\*$-algebras and $E$ be a right $A$-Hilbert $C^\*$-module. We can form the Hilbert $A\otimes B$ (minimal tensor product) module $E \otimes B$. If $\omega \in B^\*$, there is a unique bounded map
$$\iota \otimes \omega: E \otimes B \to E$$
extending $\iota \odot \omega$.
Now, let $F$ be a closed submodule of $E$, and assume that $z\in E \otimes B$ satisfies
$$(\iota \otimes \omega)(z)\in F$$
for all $\omega \in B^\*$. Is it true that $z\in F \otimes B?$
I'm not even sure this is true if $E= A$ is a $C^\*$-algebra!
| https://mathoverflow.net/users/216007 | Property that follows from conditions involving slice maps on Hilbert module | Theorem 8 from the paper "A Pathology in the Ideal Space of $L(H)\otimes L(H)$" by Simon Wassermann states that there exists an element $x \in B(H)\otimes B(H)$ such that all slices belong to the algebra of compact operators, but $x \notin K(H) \otimes B(H)$, so in general it does not hold.
The property you are after is much more common in the setting of von Neumann algebras.
| 5 | https://mathoverflow.net/users/24953 | 449740 | 180,967 |
https://mathoverflow.net/questions/449723 | 2 | Consider a projective morphism of Noetherian schemes $p:X\to \mathrm{Spec}(A)$. Let $\mathcal{E},\mathcal{F}$ be coherent $\mathcal{O}\_X$-modules flat over $A$. For every (Noetherian) ring map $A\to B$, denote $X\_B:=X\times\_{\mathrm{Spec}(A)}\mathrm{Spec}(B)$, $\mathcal{E}\_B:=(X\_B\to X)^\*\mathcal{E}$, and $p\_B:X\_B\to\mathrm{Spec}(B)$.
Choose $i\in\mathbb{Z}$. We say $\mathrm{Ext}^i\_p(\mathcal{E},\mathcal{F})$ commutes with base change if for all $A\to B$, the base change map is an isomorphism
$$\tau^i:\mathrm{Ext}^i\_p(\mathcal{E},\mathcal{F})\otimes\_AB\xrightarrow{\cong} \mathrm{Ext}^i\_{p\_B}(\mathcal{E}\_B,\mathcal{F}\_B). $$
I have the following conjecture.
>
> If $\mathrm{Ext}^i\_p(\mathcal{E},\mathcal{F})$ commutes with base change, then there exists a finitely generated $A$-module $Q$ such that $\mathrm{Ext}^{i+1}\_p(\mathcal{E}\_B,\mathcal{F}\_B)\cong \mathrm{Hom}\_A(Q,B)$ functorial for $A\to B$.
>
>
>
Is this conjecture ture? If yes, then can we always get a module $Q$ and a natural transformation
$$\mathrm{Hom}\_A(\mathcal{Q},B)\to \mathrm{Ext}\_p^{i+1}(\mathcal{E}\_B,\mathcal{F}\_B)$$
without the base change assumption on $\mathrm{Ext}^i$.
Similar results hold true for sheaf cohomologies, derived functors of $\Gamma(X,-)$, which can be found in Section III.12 in *Algebraic Geometry* by Hartshorne. However, it is not clear to me whether the method can deform to one argument for $\mathrm{Ext}^i\_p$.
| https://mathoverflow.net/users/105537 | If $\mathrm{Ext}^i(E,F)$ commutes with base change, then is $\mathrm{Ext}^{i+1}(E,F)$ representable? | **Edit.** I started this answer earlier, but then I had to do something else.
**Positive answer.**
Without any further hypothesis, the answer is positive if $i$ equals $0$. This is one of the theorems the follows from the "Exchange Property" in Section 7.7 of EGA III\_2. If $i$ equals $1$, at least there is a well-defined functor: because $\mathcal{E}$ is $A$-flat, for every short exact sequence in the Yoneda-Ext group $\text{Ext}^1\_{\mathcal{O}\_{X}}(\mathcal{E},\mathcal{F})$, the base change to $B$ is still a short exact sequence.
When $X$ is $A$-flat, there are well-defined functors for every $i$. Moreover, there is a bounded below complex whose terms are finitely generated, locally free $A$-modules whose homology modules compute the Ext groups compatibly with arbitrary base change.
Since $p$ is projective, there exists a $p$-ample invertible sheaf $\mathcal{O}\_X(1)$. Thus, there exists a bounded above resolution of $\mathcal{E}$ by $\mathcal{O}\_X$-modules that are finite direct sums of twists $\mathcal{O}\_X(-d)$ with $d$ so positive that $\mathcal{F}(d)$ has vanishing higher direct image sheaves for $p$. Thus, we can compute $R\textit{Hom}\_{\mathcal{O}\_X}(\mathcal{E},\mathcal{F})$ by forming the usual sheaf Hom of this resolution into $\mathcal{F}$. By construction, this new bounded below complex has terms that are finite direct sums of coherent sheaves $\mathcal{F}(d)$ having vanishing higher direct image sheaves. Thus, $Rp\_\*$ is just $p\_\*$. Altogether, this gives a bounded below complex of finite free $A$-modules whose homologies compute the Ext groups, compatible with arbitrary base change (since everything, including the sheaves $\mathcal{O}\_X(-d)$, is $A$-flat).
**Negative answer.**
For $i\geq 2$, the conjecture is false without further hypotheses, e.g., flatness of $p$. If you assume that $p$ is flat, then I believe the conjecture is true (I will try to return to this soon).
Consider the case where $A$ equals $k[s]$, where $X$ is the effective Cartier divisor of $\mathbb{P}^3\_A = \text{Proj}\ A[T,U,V,W]$ with defining equation $s(T^3+U^3+V^3+W^3)$, where $\mathcal{E}$ equals the structure sheaf of the closed subscheme $C$ of $X$ with defining ideal $\langle T+U,V+W \rangle$, and where $\mathcal{F}$ equals the pushforward to $X$ of the dualizing sheaf of $C$. For $i=0$, the formation of $\text{Hom}\_{\mathcal{O}\_X}(\mathcal{E},\mathcal{F}) = H^0(C,\omega\_C)$ is compatible with arbitrary base change: it is always zero since $C$ is just $\mathbb{P}^1\_A$ and $\omega\_C$ is just $\mathcal{O}\_{\mathbb{P}^1}(-2)$.
The flattening stratification of $X$ is the locally closed partition of $\text{Spec}(A)$ into two subsets: the basic open subset $D(s) = \text{Spec}(A[1/s])$ and the closed point $\text{Zero}(s) = \text{Spec}(A/sA)$. Over each of these locally closed subschemes, the restriction of $X$ is smooth: either a smooth cubic surface over $D(s)$ or all of $\mathbb{P}^2$ over $\text{Zero}(s)$. Moreover, the curve $C$ is smooth over $\text{Spec}(A)$; just a line. Thus, over each of these locally closed subschemes, the curve $C$ is a "regular embedding" in $X$ with locally free normal sheaf $N\_{C/X}$ (the sheaf Hom into $\mathcal{O}\_C$ of the ideal sheaf of $C$ in $X$). Over $D(s)$, the locally free normal sheaf $N\_{C/X}$ is just $\mathcal{O}\_{\mathbb{P}^1}(-1)$. Over $\text{Zero}(s)$, the locally free normal sheaf $N\_{C/X}$ is just $\mathcal{O}\_{\mathbb{P}^1}(+1)^{\oplus 2}$.
In fact, in this example $C$ is a transverse intersection of divisors: it is a divisor in $X$ over $D(s)$, and it is an intersection of two hyperplanes in $X$ over $\text{Zero}(s)$. This means that the object $R\textit{Hom}\_{\mathcal{O}\_X}(\mathcal{O}\_C,\mathcal{F})$ is particularly simple when restricted over $D(s)$ and over $\text{Zero}(s)$: it is quasi-isomorphic to a complex that is simply the direct sum of its homology sheaves (i.e., the differentials are all zero), and these homology sheaves are just $\mathcal{H}^q = \bigwedge^q\_{\mathcal{O}\_C}(N\_{C/X})\otimes\_{\mathcal{O}\_C} \mathcal{F}$. Thus, $Rp\_\*$ (over each stratum) is also just the direct sum of the complexes $Rp\_\*(\bigwedge^q\_{\mathcal{O}\_C}(N\_{C/X})\otimes\_{\mathcal{O}\_C} \mathcal{F})[-q]$. That makes it particularly easy to compute the ranks of the Ext groups over each stratum.
More precisely, over $D(s)$, the complex $R\textit{Hom}\_{\mathcal{O}\_X}(\mathcal{E},\mathcal{F})$ is just $\mathcal{O}\_{\mathbb{P}^1}(-2)[0]\oplus \mathcal{O}\_{\mathbb{P}^1}(-3)[-1]$. Thus $Rp\_\*R\textit{Hom}\_{\mathcal{O}\_X}(\mathcal{E},\mathcal{F})$ is just $H^1(\mathbb{P}^1,\mathcal{O}\_{\mathbb{P}^1}(-2))[-1]\oplus H^1(\mathbb{P}^1,\mathcal{O}\_{\mathbb{P}^1}(-3))[-2]$.
Similarly, over $\text{Zero}(s)$, the complex $R\textit{Hom}\_{\mathcal{O}\_X}(\mathcal{E},\mathcal{F})$ is just $\mathcal{O}\_{\mathbb{P}^1}(-2)[0]\oplus \mathcal{O}\_{\mathbb{P}^1}(-1)^{\oplus 2}[-1]\oplus \mathcal{O}\_{\mathbb{P}^1}(0)[-2]$. Thus $Rp\_\*R\textit{Hom}\_{\mathcal{O}\_X}(\mathcal{E},\mathcal{F})$ is just $H^1(\mathbb{P}^1,\mathcal{O}\_{\mathbb{P}^1}(-2))[-1]\oplus H^0(\mathbb{P}^1,\mathcal{O}\_{\mathbb{P}^1})[-2]$.
Now consider the functor $\text{Ext}^1\_{p\_B}(\mathcal{E},\mathcal{F})$. For the base change to $A[1/s]$, it is representable by a locally free sheaf $Q\_{1/s}$ of rank $1$. Also, for the base change to $A/sA$, it is representable by a locally free sheaf of rank $1$. So if your conjecture is correct, then the functor is representable by a locally free sheaf $Q$ of rank $1$.
But then the same hypotheses apply for the functor $\text{Ext}^2\_{p\_B}(\mathcal{E},\mathcal{F})$. However, now the base change to $A[1/s]$ is representable by a locally free sheaf of rank $2$, whereas the base change to $A/sA$ is representable by a locally free sheaf of rank $1$. There is no finitely generated $A$-module $R$ such that $R\otimes\_A A[1/s]$ is free of rank $2$ yet $R/sR$ is free of rank $1$.
| 3 | https://mathoverflow.net/users/13265 | 449741 | 180,968 |
https://mathoverflow.net/questions/449341 | 5 | Suppose $X\_1, X\_2, X\_3 \sim N(0, 1)$ are three independent standard normal random variables. I am interested in showing that:
$$\text{Var}[X\_2\mid X\_2 \geq X\_1 - a, X\_1 \leq X\_3 + b] < 1,$$
where the inequality is strict. I've run some simulations and this seems to be true for various choices of $a$ and $b$, and it's also intuitively reasonable, since for any fixed value of $X\_1$, this is just the variance of a normal random variable truncated to $(-\infty, X\_1 - a)$. I've tried both using the law of total variance (to no avail) and also reparameterizing in terms of $[X\_2, X\_1 - X\_2, X\_1 - X\_3]$ and using the closed form for the [variance of a truncated multivariate normal](https://arxiv.org/pdf/1206.5387.pdf) (Eq. 16), also to no avail.
I'm happy with a loose bound as long as I can show it's strictly less than one!
| https://mathoverflow.net/users/128729 | Bounding the variance of a truncated Gaussian random variable | The inequality is a special case of the following claim.
Claim:
If $X = (X\_1, \dotsc, X\_d) \sim N(\mu, \Sigma)$ is an $\mathbb{R}^d$-valued normal random variable with invertible covariance matrix $\Sigma$, $L : \mathbb{R}^d \to \mathbb{R}$ is linear, and $K \subseteq \mathbb{R}^d$ is convex and has a nonempty interior, then $\operatorname{Var}(L(X) \mid X \in K) \le \operatorname{Var}(L(X))$, with equality if and only if $L(X)$ and $I(X \in K)$ are independent.
(Here $I(X \in K)$ is the indicator function of the event $X \in K$.)
Proof:
We may assume that $L \neq 0$.
After a linear change of coordinates, we may assume that $X$ is standard normal and $L(X) = X\_1$.
Then the PDF of $X\_1$ is $\frac{1}{\sqrt{2 \pi}} e^{-\frac{x^2}{2}}$, and the PDF of $Z = (X\_1 \mid X \in K)$ is $\varphi(x) = C e^{-\frac{x^2}{2}} P(X' \in K\_x)$, where $X' = (X\_2, \dotsc, X\_d)$,
$$
K\_x = \{(x\_2, \dotsc, x\_d) \in \mathbb{R}^{d-1}; \, (x, x\_2, \dotsc, x\_d) \in K\}
$$
and $C > 0$.
The function $\psi \colon \mathbb{R} \to [0,1]$, $x \mapsto P(X' \in K\_x)$ is log-concave:
if $x, y \in \mathbb{R}$ and $\lambda \in (0,1)$, then
$$
\psi((1-\lambda) x + \lambda y) \ge P(X' \in (1-\lambda) K\_x + \lambda K\_y) \ge \psi(x)^{1-\lambda} \psi(y)^{\lambda},
$$
because normal distributions are log-concave (see <https://en.wikipedia.org/wiki/Logarithmically_concave_measure>).
Note that $X\_1$ and $I(X \in K)$ are independent if and only if $\psi$ is a constant function.
The Claim follows from the following lemma.
Lemma: If $Z$ is a real random variable with PDF $\varphi(x)$ such that $e^{\frac{x^2}{2}} \varphi(x)$ is log-concave, then $\operatorname{Var}(Z) \le 1$, with equality if and only if $Z \sim N(E[Z],1)$.
Proof:
We may add a constant to $Z$, so we may assume that $\varphi$ is maximal at $0$.
Then $\varphi$ is decreasing in $\mathbb{R}\_{\ge 0}$, increasing in $\mathbb{R}\_{\le 0}$, and $\varphi(x) \le \varphi(0) e^{-\frac{x^2}{2}}$ for every $x \in \mathbb{R}$.
We have
\begin{align}
& 1 - \operatorname{Var}(Z) \ge \int\_{\mathbb{R}} (1-x^2) \varphi(x) \, dx \\[6pt]
= {} & \int\_0^\infty (1-x^2) \varphi(x) \, dx + \int\_0^\infty (1-x^2) \varphi(-x) \, dx.
\end{align}
We prove that $\int\_0^\infty (1-x^2) \varphi(x) \, dx \ge 0$.
This is clear if $\varphi(1) = 0$, so let $\varphi(1) > 0$.
There is a $c \ge 0$ such that $\tilde{\varphi}(1) = \varphi(1)$, where $\tilde{\varphi}(x) = \varphi(0) e^{-c x - \frac{x^2}{2}}$.
Then $\varphi(x) \ge \tilde{\varphi}(x)$ for $x \in [0,1]$, and $\varphi(x) \le \tilde{\varphi}(x)$ for $x \in [1,\infty)$, so $\int\_0^\infty (1-x^2) \varphi(x) \, dx \ge \int\_0^\infty (1-x^2) \tilde{\varphi}(x) \, dx = \varphi(0) \int\_0^\infty (1-x^2) e^{-c x - \frac{x^2}{2}} \, dx$.
We can do the same for $\int\_0^\infty (1-x^2) \varphi(-x) \, dx$.
So all we need now is that
$$
\int\_0^\infty (1-x^2) e^{-c x - \frac{x^2}{2}} \, dx > 0
$$
for $c > 0$.
We have $\int\_0^\infty e^{-c x - \frac{x^2}{2}} \, dx = \frac{\sqrt{\pi}}{\sqrt{2}} e^{\frac{c^2}{2}} \operatorname{erfc}(\frac{c}{\sqrt{2}})$, and differentiating twice in $c$ we get $\int\_0^\infty x^2 e^{-c x - \frac{x^2}{2}} \, dx$.
In the end, the inequality boils down to $c e^{\frac{c^2}{2}} \operatorname{erfc}(\frac{c}{\sqrt{2}}) < \frac{\sqrt{2}}{\sqrt{\pi}}$.
Equivalently, $\frac{\sqrt{\pi}}{2} \operatorname{erfc}(x) < x^{-1} e^{-x^2}$ for $x > 0$.
This follows from $(\frac{\sqrt{\pi}}{2} \operatorname{erfc}(x) - x^{-1} e^{-x^2})' = (1+x^{-2}) e^{-x^2} > 0$, since the limits at $x \to \infty$ are $0$.
| 4 | https://mathoverflow.net/users/42355 | 449743 | 180,969 |
https://mathoverflow.net/questions/449739 | 2 | Let $X\subseteq \mathbb{P}^n$ be a closed irreducible subvariety, with vanishing ideal $I(X)\subseteq k[x\_0,\ldots,x\_n]$, where $k$ is the ground field, assumed to be algebraically closed. Let $F\in k[x\_0,\ldots,x\_n]$ be a homogeneous polynomial. If $F$ belongs to the ideal $I(X)^2$, then $F$ is singular along $X$ (the singular locus of $F$ contains $X$). Is the converse true?
I am mostly interested in the case where $n=3$ and $X$ is a curve, of small degree (say $\le 9$). But the general case seems interesting too.
| https://mathoverflow.net/users/23758 | What is the ideal of hypersurfaces singular at a given irreducible variety? | If $X=\mathbb{V}(I)$ is given by the ideal $I$, then the $m$th symbolic power $I^{[m]}$ consists of all those functions vanishing to multiplicity $m$ at the generic point of $X$. Thus a hypersurface $\mathbb{V}(F)$ will be singular along $X$ if and only if $F\in I^{[2]}$. (Thanks to Zach Teitler for pointing this out in the comments below.) To give a counterexample it is therefore enough to find an $X$ for which $I^2\subsetneq I^{[2]}$.
A (non-irreducible, affine) example is $X=\mathbb{V}(xy,yz,zx)$, the union of the three coordinate axes in $\mathbb{A}^3$ and $F=xyz\notin I^2$. Then the hypersurface $\mathbb{V}(xyz)$ has multiplicity $2$ on every component of $X$. In general if $X\subset \mathbb{P}^3$ is an irreducible space curve which is smooth apart from a triple point (i.e. a point locally analytically isomorphic to the example above) then there should also be function $F\in I^{[2]}\setminus I^2$ as above.
(P.S. If you want to compute $I^{[m]}$ explicitly (e.g. in the case you are interested in, of a curve $X$ of small degree) then you can ask your favourite computer algebra package for the primary decomposition of $I^m$. The unique primary component supported on the whole of $X$ will be $I^{[m]}$.)
| 9 | https://mathoverflow.net/users/104695 | 449744 | 180,970 |
https://mathoverflow.net/questions/449290 | 0 | Consider prime constellations $p,p+2s$ where both $p,p+2s$ are prime.
For instance for $s=1$ we get the twin primes.
We define the counting function $\pi\_{2s}(n)$ to count the number of such pairs $p,p+2s$ below or equal to $n$.
Does this conjecture hold ?
$$ \pi\_{2a}(n+(6a+4)^3)+(6a+4)^3 > \pi\_{4a}(n)$$
for all positive integer $n>2$ and all positive integer $a>0$ ?
I know that according to Hardy-Littlewood this probably is false, but has any counterexample actually been found ?
I know the skewes numbers for (cousin primes) $\pi\_4 >$ the skewes number for (prime twins) $\pi\_2$ and I know the skewes number for sexy primes is unknown if it even exists.
($1369391$ for twins vs $5206837$ for cousins )
see [Wikipedia entry on Skewes numbers](https://en.wikipedia.org/wiki/Skewes%27s_number) (for prime k-tuples).
| https://mathoverflow.net/users/80790 | Prime gap conjecture $ \pi_{2a}(n+(6a+4)^3)+(6a+4)^3 > \pi_{4a}(n)$ counterexamples? | It is false, try $a=1, n=250\,003\,639$.
| 2 | https://mathoverflow.net/users/6043 | 449747 | 180,971 |
https://mathoverflow.net/questions/371232 | 3 | If $S$ and $T$ are monads on a category $C$, and $\lambda:S\to T$ is a morphism of monads, it is well-known that there is a functor $\lambda^\*:C^T\to C^S$ which assigns to the $T$-algebra $(A,a:TA\to A)$ the $S$-algebra $(A,a\circ\lambda:SA\to A)$.
Does a similar phenomenon happen for pseudomonads and their pseudoalgebras (as defined for example [here](https://www.sciencedirect.com/science/article/pii/S0001870899918819))?
A reference would be welcome too.
| https://mathoverflow.net/users/30366 | Morphism of pseudomonads induces pullback functors between pseudoalgebras | Yes, Theorem 3.4 of Gambino–Lobbia's [On the formal theory of pseudomonads and pseudodistributive laws](https://arxiv.org/abs/0907.1359) establishes that pseudomonad morphisms are in correspondence with liftings to pseudoalgebras. They work more generally in the setting of pseudomonads in a Gray-category, and with pseudomonads on different objects, rather than pseudomonads on a fixed object, but the theorem specialises to the result you are looking for.
| 1 | https://mathoverflow.net/users/152679 | 449755 | 180,972 |
https://mathoverflow.net/questions/449735 | 0 | Assume that $ \Omega $ is a smooth bounded domain in $ \mathbb{R}^n $. Consider a functional
$$
\mathcal{F}(u)=\int\_\Omega(|\nabla u|^2+h^{-1}|u-u\_0|^2) \, dx
$$
where $ h>0 $ is a parameter and $ u\_0\in H\_0^1(\Omega) $. For the minimizing problem
$$
\min\_{u\in H\_0^1(\Omega)}\mathcal{F}(u),
$$
it is easy to get by standard arguments that there is a unique minimizer $ u\_h\in H\_0^1(\Omega) $, i.e. $ u\_h=\operatorname{argmin}\_{u\in H\_0^1(\Omega)} \mathcal{F}(u) $. A natural observation is that $ \|u\_h-u\_0\|\_{L^2(\Omega)}\to 0 $. I want to ask the following question.
>
> Let $ h\_0>0 $. Assume that $ h\to h\_0 $. If there is the result $ \|u\_{h}-u\_{h\_0}\|\_{L^2(\Omega)}\to 0 $?
>
>
>
Such question is natural and intuitively true. However I cannot solve it, can you give me some hints or references?
| https://mathoverflow.net/users/241460 | Limit of minimizers of a class of functionals | By simple calculations, we can obtain that $ u\_{h} $ and $ u\_{h\_0} $ satisfy the following Euler-Lagrange equation
\begin{align}
\frac{u\_h-u\_0}{h}-\Delta u\_{h}=0,\\
\frac{u\_{h\_0}-u\_0}{h}-\Delta u\_{h\_0}=0.
\end{align}
Then it can be got that
$$
\begin{aligned}
0&=\frac{u\_h-u\_0}{h}-\frac{u\_{h\_0}-u\_0}{h\_0}-\Delta(u\_{h}-u\_{h\_0})\\
&=\frac{u\_h-u\_0}{h}-\frac{u\_h-u\_0}{h\_0}+\frac{u\_h-u\_0}{h\_0}-\frac{u\_{h\_0}-u\_0}{h\_0}-\Delta(u\_{h}-u\_{h\_0})\\
&=(u\_h-u\_0)\frac{h\_0-h}{hh\_0}+\frac{u\_h-u\_{h\_0}}{h\_0}-\Delta(u\_h-u\_{h\_0}).
\end{aligned}
$$
Testing the above equation by the function $ u\_h-u\_{h\_0} $, we obtain that
$$
\int\_{\Omega}|\nabla(u\_h-u\_{h\_0})|^2dx+\int\_{\Omega}\frac{(u\_{h}-u\_{h\_0})^2}{h\_0}dx\leq\int\_{\Omega}\frac{h-h\_0}{hh\_0}(u\_h-u\_0)(u\_h-u\_{h\_0})dx.
$$
By Cauchy inequality, we have
$$
\frac{1}{2}\int\_{\Omega}(u\_h-u\_{h\_0})^2dx\leq\frac{(h-h\_0)^2}{h^2}\int\_{\Omega}(u\_h-u\_0)^2dx.
$$
From this, we can complete the proof.
| 1 | https://mathoverflow.net/users/241460 | 449766 | 180,975 |
https://mathoverflow.net/questions/449719 | 1 | Are there any examples where a usual algebraic 1-stack $X$ and the corresponding derived stack enhancement $\mathbb{R}X$ coincide?
Let me take an example from notes of Bertrand Toen, page 41 of <https://arxiv.org/pdf/math/0604504.pdf>.
Let $C$ be a smooth projective curve and consider the functor $i: \mathrm{St}(k) \to \mathrm{dSt}(k)$. Take $X$ to be the mapping stack
$$
X = \mathrm{Map}(C,BG) = \mathrm{Bun}\_G(C).
$$
The derived enhancement is
$$
\mathbb{R}X = \mathbb{R}\operatorname{Map}(i(C), i(BG)),
$$
and $X$ is called a *truncation* of $\mathbb{R}X$. I am therefore requesting examples where this truncation procedure is trivial!
My naive hope is that this is the case since $\dim(C) = 1$ and so there are no higher cohomologies/homotopies above degree 1.
| https://mathoverflow.net/users/465579 | Examples when algebraic 1-stack = derived enhancement? | If $S$ is a smooth projective variety of dimension $d$, then the derived stack $X=Bun\_G(S)$ has cotangent complex perfect of amplitude $[-(d-1), 1]$. If $S=C$ is a curve then it is in $[0,1]$, in particular $H^\*(\mathbb{L}\_X)=0$ for $\*<0$. For stacks, this condition (+ finite presentation) is exactly smoothness.
Smooth morphisms $X \to Y$ are flat, which in the derived context means flatness on classical truncations but also $\pi\_i(O\_Y) \otimes\_{\pi\_0(O\_Y)} \pi\_0(O\_X) \simeq \pi\_i(O\_X)$ on higher homotopy. In particular, saying that a derived stack is smooth over the base field $k$ automatically implies that its derived structure is trivial. I would refer you to the original references on DAG, such as [HAG II, 2.2.2].
For $X = Higgs\_G(S)$ we typically have one extra obstruction $H^{-d}(\mathbb{L}\_X) \ne 0$ compared to $Bun\_G(S)$. For $S=C$ a curve this means that $X$ is only quasi-smooth in general. The extra obstruction $H^{-1}$ does vanish in certain special cases, e.g. if $G$ is semisimple, see Beilinson-Drinfeld "Quantization of Hitchin..." 2.1.2 and also Halpern-Leistner's paper on Verlinde for Higgs, Prop 3.14.
| 3 | https://mathoverflow.net/users/85136 | 449769 | 180,976 |
https://mathoverflow.net/questions/448614 | 2 | Let $(X,d)$ be a metric space. If set $A\subseteq X$, let $H^{\alpha}$ be the $\alpha$-dimensional Hausdorff measure on $A$, where $\alpha\in[0,2]$ and $\text{dim}\_{\text{H}}(A)$ is the Hausdorff dimension of set $A$.
**Motivation:** If the set $A\subseteq[0,1]\times[0,1]$, I would like to measure set $A$'s deviation from a uniform subset $A^{\prime}$ of $[0,1]\times[0,1]$ with Hausdorff dimension $\alpha$, where:
>
> If set $A^{\prime}\subseteq[0,1]\times[0,1]$ has Hausdorff dimension $\alpha$ less than $2$; for all real $x\_1,x\_2\in J\_n$ *where* sequence of sets $(J\_n)\_{n\in\mathbb{N}}=(\left\{1/n,2/n,3/n,\cdot\cdot\cdot,1\right\})\_{n\in\mathbb{N}}$, if $0\le x\_1<x\_2\le 1$ and $\mu$ is the [*counting measure*](https://en.wikipedia.org/wiki/Counting_measure) *where*:
>
>
> $$\mu^{\star}(A)=\begin{cases}
> 1/\mu(A) & \mu(A)<+\infty\\
> 0 & \mu(A)=+\infty
> \end{cases}$$
>
>
> with $\mu^{\star}(A)< x\_2-x\_1\le 1$; we *get* $A^{\prime}$ is **uniform** in $[0,1]\times[0,1]$, when for $j\in\mathbb{N}$ and $j< n$, the number of squares $[x\_1,x\_2]\times[x\_1,x\_2]$ with area $j^2/n^2$ satisfying:
>
>
> $$H^{\alpha}(([x\_1,x\_2]\times[x\_1,x\_2])\cap A^{\prime})=H^{\alpha}(\text{dom}(A^{\prime})){H}^{\alpha}(\text{range}(A^{\prime}))(x\_2-x\_1)^2$$
> divided by the total # of squares with area $j^2/n^2$ *approaches* $1$ *as* $n,j\to\infty$ and $j/n\to 0$.
>
>
>
Moreover, I want to define a measure of uniformity to be *between* (and including) zero and one (or zero and infinity) such that the larger the measure, the smaller the *non-uniformity*.
**Question:** How do we define such a measure?
| https://mathoverflow.net/users/87856 | Defining a measure of uniformity for measurable subsets of $[0,1]^2$ w.r.t dimension $\alpha\in[0,2]$ | Here is another possible approach, perhaps closer to what the OP had in mind.
Let $S:=[0,1]^2$ be the unit square. "Partition" $S$ naturally into four congruent squares $S\_{1,j}$ (with side length $1/2$ each), where $j=1,\dots,4$; the quotation marks are used here because the $S\_{1,j}$'s will have some common boundary points. Next, "partition" each $S\_{1,j}$ naturally into four congruent squares (with side length $1/2^2$ each), so that we get $4^2$ squares $S\_{2,j}$ for $j=1,\dots,4^2$. Continue doing so, so that at the $k$th step we get $4^k$ squares $S\_{k,j}$ for $j=1,\dots,4^k$, for each $k=1,2,\dots$.
Take any subset $A$ of $S$. For each $k=1,2,\dots$ and each $j=1,\dots,4^k$, let
$$A\_{k,j}:=(A\cap S\_{k,j})-s\_{k,j},$$
where $s\_{k,j}$ is the southwest vertex of the square $S\_{k,j}$, so that $A\_{k,j}\subseteq S\_k:=2^{-k}S$.
Suppose that for each $k$ we have a "measure" $D\_k$ of dissimilarity for subsets of $S\_k$, so that for any two subsets $B$ and $C$ of $S\_k$ we have a nonnegative real number $D\_k(B,C)$, which is the greater the more "dissimilar" $B$ and $C$ are (and is $0$ if $B=C$); here the term "measure" is used in the general sense, not necessarily in the sense of measure theory. For instance, $D\_k(B,C)$ may depend on the [Hausdorff distance](https://en.wikipedia.org/wiki/Hausdorff_distance#Definition) between $B$ and $C$ or on some "measure" of the symmetric difference of the sets $B$ and $C$ or on some combination thereof.
Then the distance of the set $A$ from uniformity can be defined by the formula
$$D(A):=\sum\_{k=1}^\infty\frac1{L^k}\sum\_{j=1}^{4^k}\sum\_{m=1}^{4^k}
\frac{D\_k(A\_{k,j},A\_{k,m})}{1+D\_k(A\_{k,j},A\_{k,m})},$$
where $L$ is a real number $>16$ (to ensure the convergence of the series). Then $D(A)$ will be small if, for "most" levels $k$ of "zooming", "most" of the intersections of the set $A$ with all the "$k$-level" small squares $S\_{k,j}$ "look similar" to one another.
(Of course, $D(A)$ will depend on the choices of $L$ and the dissimilarity "measures" $D\_k$.)
For instance, for any $L$ and any $D\_k$'s we have $D(S)=0$ -- of course, the unit square $S$ is at distance $0$ from uniformity (in itself).
As another example, for the uniform grid $G\_n$ (defined in the [previous answer](https://mathoverflow.net/a/449757/36721)) with $n=2^K$ for a natural $K$, any real $L>16$, and any $D\_k$'s we have
$$D(G\_n)\le \sum\_{k=K+1}^\infty\frac1{L^k}\,16^k
=Cn^{-p}\to0$$
as $n=2^K\to\infty$, where $C:=\dfrac{16}{L-16}$ and $p:=\log\_2\dfrac L{16}$.
So, we see that $G\_n$ is close to uniformity for large $n$, even though the Hausdorff dimension of $G\_n$ is $0$ for all $n$ (in big contrast with the Hausdorff dimension of $S$, which is $2$). Thus, again we see that the Hausdorff dimension can hardly have anything to do with the idea of uniformity.
| 2 | https://mathoverflow.net/users/36721 | 449772 | 180,977 |
https://mathoverflow.net/questions/449776 | 13 | Does the following infinite series have a closed form?
$$
\sum\_{n=1}^{\infty} {(-1)^n \frac{\Gamma(\frac{1}{3}+\frac{n}{3})}{\Gamma(1+\frac{n}{3})} \sin(\frac{2\pi n}{3})}
$$
| https://mathoverflow.net/users/105725 | Closed form of an infinite series | **Q:** Does the following infinite series have a closed form?
It does, according to Mathematica:
$$\sum\_{n=1}^{\infty} {(-1)^n \frac{\Gamma(\frac{1}{3}+\frac{n}{3})}{\Gamma(1+\frac{n}{3})} \sin\left(\frac{2\pi n}{3}\right)}=-2^{-10/3} \Gamma \left(\tfrac{1}{3}\right)\Big[2 \sqrt{3}+9\pi^{-1} \, \_2F\_1\left(\tfrac{2}{3},\tfrac{2}{3};\tfrac{5}{3};-1\right)\hspace{-1mm}\Big].$$
| 11 | https://mathoverflow.net/users/11260 | 449778 | 180,979 |
https://mathoverflow.net/questions/449313 | 0 | If I have a compound Poisson process whose characteristic function is known, is there a way to calculate the **joint** characteristic function of this process and its quadratic variation process?
If not possible in general, are there specific examples of compound Poisson processes for which this function is known in closed form?
| https://mathoverflow.net/users/109513 | Characteristic function of quadratic variation of compound Poisson process | Suppose that $N=\{N(t), t\in\mathbb{R}\_+\}$ is a Poisson process with rate $\lambda $, and $\{D\_j,j\ge 1\} $ are i.i.d. random variables, with distrbution function $F$, which are also independent of $N$. Let
\begin{equation\*}
X(t) = \sum\_{j=1}^{N(t)}D\_j, \quad t\in\mathbb{R}\_+, \tag{1}
\end{equation\*}
then the continuous-time stochastic process $X=\{X(t), t\in\mathbb{R}\_+\}$ is a
compound Poisson process. $X$ is Lévy process and also step process(all its trajectories are cádlág step funtion having at most a finite number of jumps in every finite interval). The quadratic variation process of $X$ is
\begin{equation\*}
Y(t) \stackrel{\triangle}{=} [X,X]\_t =\sum\_{s\le t}\Delta X^2\_s
= \sum\_{j=1}^{N(t)}D\_j^2, \tag{2}
\end{equation\*}
hence $Y=\{Y(t), t\in\mathbb{R}\_+\}$ is also a compound Poisson process. Now the characteristic function of $(X,Y)$ is
\begin{align\*}
&\varphi\_{(X(t),Y(t))}(u,v)\\
&\quad=\mathsf{E}[\exp\{\mathrm{i}(uX(t)+vY(t))\}]\\
&\quad=\mathsf{E}[\mathsf{E}[\exp\{\mathrm{i}(uX(t)+vY(t))\}|N(t)]]\\
&\quad=\sum\_{k=0}^{\infty}\Big[\mathsf{E}\Big[\exp\Big\{\sum\_{j=1}^{N(t)} \mathrm{i}(uD\_j+vD\_j^2)\Big\}\Bigm|N(t)=k\Big]\mathsf{P}(N(t)=k)\Big]\\
&\quad=\sum\_{k=0}^{\infty}\Big[\int\_{\mathbb{R}}e^{\mathrm{i}(ux+vx^2)}\, \mathrm{d}F(x) \Big]^k \frac{(\lambda t)^k}{k!} \mathrm{e}^{-\lambda t} \\
&\quad=\exp\Big[\lambda t\int\_{\mathbb{R}}(e^{\mathrm{i}(ux+vx^2)}-1)\, \mathrm{d}F(x)\Big].
\end{align\*}
| 1 | https://mathoverflow.net/users/103256 | 449787 | 180,983 |
https://mathoverflow.net/questions/449764 | 1 | I am reading the following two papers:
* Pappas, *On the arithmetic moduli schemes of PEL Shimura varieties*, 1999 (it seems to be difficult to find online nowadays - only a .ps file remains available),
* Krämer, *Local models for ramified unitary groups*, 2003.
Let $n\geq 1$ and consider integers $r,s\geq 0$ such that $r+s = n$. Let $K\_0$ be a $p$-adic field with $p\not = 2$, and let $K$ be a ramified quadratic extension of $K\_0$. The (naive) local model of signature $(r,s)$ associated to a $K/K\_0$-hermitian space of dimension $n$ is a projective scheme $M\_{r,s}$ over $\mathcal O\_K$, defined by an explicit moduli problem of linear algebra. In general, eg. when $|r-s|> 1$, it is not flat. To rectify this, Pappas introduces the local model $M\_{r,s}^{\mathrm{loc}}$ by a slight modification of the moduli problem. It comes along with a closed immersion $M\_{r,s}^{\mathrm{loc}}\hookrightarrow M\_{r,s}$. In the special case $(r,s) = (n-1,1)$, Pappas proves that $M\_{n-1,1}^{\mathrm{loc}}$ is normal, Cohen-Macaulay, flat over $\mathrm{Spec}(\mathcal O\_K)$ and smooth outside of a single closed point $y\in M\_{n-1,1}^{\mathrm{loc}}$ of the special fiber.
In both papers, the authors consider the setting as above. They then build a resolution of the singularity of $M\_{n-1,1}^{\mathrm{loc}}$, but a priori in two different ways:
* Pappas considers the blow-up $\mathrm{Bl}(M\_{n-1,1}^{\mathrm{loc}})$ of $M\_{n-1,1}^{\mathrm{loc}}$ at $y$.
* Krämer builds a new local model $\mathcal M\_{n-1,1}$ by adding extra datum to the moduli problem defining $M\_{n-1,1}^{\mathrm{loc}}$.
Both constructions, $\mathrm{Bl}(M\_{n-1,1}^{\mathrm{loc}})$ and $\mathcal M\_{n-1,1}$, give rise to projective regular schemes over $\mathcal O\_K$ with semistable reduction. They both come with a natural map to the local model $M\_{n-1,1}^{\mathrm{loc}}$, inducing an isomorphism outside of the fiber over $y$.
It seems very likely that both constructions are actually isomorphic, ie. $\mathcal M\_{n-1,1} \simeq \mathrm{Bl}(M\_{n-1,1}^{\mathrm{loc}})$. However, this does not seem to be stated anywhere in the literature, as far as I can tell. Besides, Krämer refers to Pappas' work by saying that
>
> First we have to resolve the singularities of the local model. In [Pappas], this was done by blowing up the singular locus. Our approach is different: We define a resolution $\mathcal M\_{n-1,1} \to M\_{n-1,1}^{\mathrm{loc}}$ by posing a moduli problem analogous to the Demazure-resolution of a Schubert variety in the Grassmannian.
>
>
>
In other words, Krämer seems to emphasize that this construction *a priori* differs from Pappas' one. If they really did give isomorphic resolutions, one would expect a mention of this fact at least in the introduction...
Thus, I'd like to ask anyone who would be familiar with local models. Are the two above constructions known to be isomorphic or not?
| https://mathoverflow.net/users/125617 | Is Krämer's local model for ramified unitary groups isomorphic to the blow-up of Pappas' flat model at the singular point? | In Yousheng Shi's [paper](https://arxiv.org/abs/2004.07158):
**Proposition 2.2.** $N^{\text{Kra}}$ is the blow-up of $N^{\text{Pap}}$ along its singular locus Sing.
Also see the words before the Proposition: "The following fact should be well-known to experts. However due to the lack of
a precise reference, we prove it in Appendix A"
| 1 | https://mathoverflow.net/users/486528 | 449792 | 180,985 |
https://mathoverflow.net/questions/449705 | 1 | Let $X$ be a $\mathbb R^d$ valued continuous stochastic process. I am interested in bounding $$P(\|X\|\_\gamma>R).$$
The standard technique to do so, is to apply Markov inequality and then Garsia-Rodemich-Rumsey inequality. Recall that GRR inequality says that
>
> For any $\alpha>1$, $\delta>1/\alpha$ there is some constant $C:=C(\alpha, \delta)$ so that for all continuous $f\in C([0,1],\mathbb R^d)$ and $s,t\in [0,1]$ we have $$|f(t)-f(s)|^\alpha\leq C |t-s|^{\delta\alpha-1}\int\_s^t \int\_s^t \frac{|f(u)-f(v)|^\alpha}{|u-v|^{\delta\alpha + 1}}du\,dv.$$
>
>
>
If we have the Kolmogorov type bound $E[|X(t)-X(s)|^\alpha]\leq K|t-s|^{1+\beta}$ and $\gamma<\beta/\alpha$ we may follow the answer [here](https://mathoverflow.net/questions/278977/kolmogorov-continuity-theorem-and-holder-norm/279085#279085), and apply Markov to get that
\begin{align\*}
P(\|X\|\_\gamma>R)&=P(\|X\|\_\gamma^\alpha>R^\alpha)\\
&\leq \frac{E[\|X\|\_\gamma^\alpha]}{R^\alpha}\\
&\le \frac{1}{R^\alpha}C(\delta,\alpha) \int\_0^T\int\_0^T \frac{\mathrm{E} \left[|X(u)-X(v)|^\alpha\right]}{|u-v|^{\delta\alpha + 1}}du\,dv \\
&\le \frac{1}{R^\alpha}C(\delta,\alpha) K\int\_0^T\int\_0^T |u-v|^{\beta-\delta\alpha}du\,dv \\
&= \frac{1}{R^\alpha} C(\delta,\alpha,\beta,T)K.
\end{align\*}
However, Markov in general is a pretty crude bound. Furthermore, $X$ may not have too many moments. I am wondering if we only have tail bounds of the form
$$P\left(\frac{|X(t)-X(s)|}{|t-s|^p}>R\right)\leq \Phi(R)$$
for some decreasing function $\Phi$, what can we say? Assume $\Phi$ is continuous if necessary.
That is, can we express
$$P\left(\int\_0^T\int\_0^T \frac{|X(u)-X(v)|^\alpha}{|u-v|^{\delta\alpha + 1}}du\,dv>R^\alpha\right)$$
in terms of $\Phi$?
| https://mathoverflow.net/users/479223 | Garsia-Rodemich-Rumsey without Markov | Apologies for the earlier error. Upon examining what went wrong, I think I have discovered that no such bound can hold - there is in general no relationship between $\Phi$ and the Holder norm of $X$.
The idea is to define $X$ to be a small bump placed uniformly at random. Changing the parameters of this bump gives us arbitrarily nice functions $\Phi$ but arbitrarily bad Holder norms simultaneously.
More precisely, for any fixed positive integer $n \geq 22$, let $X^n$ be defined as follows - uniformly at random pick a point $p$ in, say $[\frac{1}{4}, \frac{3}{4}]$, and define
$$
X^n :=
\begin{cases}
0 & \text{on } [0, p], \\
n(t - p)^\gamma & \text{on } (p, p + \frac{1}{n^{2/\gamma}}], \\
\frac{1}{n} & \text{on } (p + \frac{1}{n^{2/\gamma}}, 1]. \\
\end{cases}
$$
Then it can be computed that for every fixed $ R$,
$$\sup\_{t \neq s} \mathbb P \left(\frac{|X^n(t)-X^n (s)|}{|t-s|^\gamma}>R\right) \to 0$$
as $n \to \infty$, but $\| X^n \|\_{\gamma} = n$ almost surely, which of course tends to $\infty$.
| 2 | https://mathoverflow.net/users/173490 | 449797 | 180,986 |
https://mathoverflow.net/questions/449798 | 0 | Suppose we have a parameter $\theta \in R^{n}$ that defines some noisy observation $z=\mu(\theta)+\eta, z\in R^{m}$ where the noise follows a Gaussian distribution whose covariance is a function of the parameter, i.e. $\eta\sim N(0,\Sigma(\theta))$. Then the observation random variable is $z \sim N(\mu(\theta),\Sigma(\theta))$. Assume observations of $z$ are iid and $\Sigma=diag(\sigma\_1^{2}(\theta), \sigma\_2^{2}(\theta)...\sigma\_m^{2}(\theta))$, where $\sigma\_i^{2}\ne\sigma\_j^{2}$.
Then, the log-likelihood function would be:
\begin{align}
L\_z\left(z;\theta\right)&=log\left(\prod^m\_{j=1}\frac{1}{\sqrt{2 \pi \sigma\_j^2\left(\theta\right)}} e^{-\frac{\left(z\_j-\mu\_j(\theta)\right)^2}{2 \sigma\_j^2\left(\theta\right)}}\right)\\
&=\sum\_{j=1}^{m} -\frac{1}{2}log(2\pi)-\frac{1}{2} log(\sigma\_{j}^{2}(\theta))-\frac{(z\_j-\mu\_j(\theta))^{2}}{2\sigma\_j^{2}(\theta)}\\
\end{align}
The first derivative of the log-likelihood function w.r.t the parameters would be:
\begin{aligned}
\frac{\partial L}{\partial \theta}= \sum\_{j=1}^{m}-\frac{1}{2}\left(\sigma\_j^2(\theta)\right)^{-1} \frac{\partial \sigma\_j^2}{\partial \theta}-(z\_j-\mu\_j(\theta)) \frac{\partial \mu\_j}{\partial \theta}\left(\sigma\_j^2(\theta)\right)^{-1}-\frac{1}{2}\left(\sigma\_j^2(\theta)\right)^{-2} (z\_j-\mu\_j(\theta))^2 \frac{\partial \sigma\_j^2(\theta)}{\partial \theta}
\end{aligned}
What I'm getting confused about is: how would the log-likelihood function derivative be zero at the parameter value, when the first term of the derivative does not depend on $(z\_j-\mu\_j(\theta))$?
| https://mathoverflow.net/users/506618 | Derivative of log-likelihood function for Gaussian distribution with parameterized variance | $\newcommand\th\theta\newcommand\si\sigma\newcommand\p\partial\newcommand\ol\overline$There is no reason to get confused here. Indeed, that "the first term of the derivative does not depend on $(z\_j-\mu\_j(\theta))$" does not at all prevent the derivative from taking the zero value.
If e.g. $\mu(\th)=\th$ and $\si\_i(\th)=\th$ for all real $\th>0$ and all $i=1,\dots,m$, then
$$\frac{\p L}{\p\th}=-\frac n{\th^3}\,(-\ol{z^2}+\th\ol z+\th^2),$$
where $\ol z:=\frac1m\,\sum\_{i=1}^m z\_i$ and $\ol{z^2}:=\frac1m\,\sum\_{i=1}^m z\_i^2$, so that $\frac{\p L}{\p\th}=0$ at a real $\th>0$ if and only if
$$\th=\hat\th:=\hat\th\_m:=\frac{-\ol z+\sqrt{\ol z^2+4\ol{z^2}}}2.$$
Moreover, here $\hat\th$ is the maximum likelihood estimator of $\th$. Using (say) the law of large numbers, one can easily check that $\hat\th\_m$ is consistent: $\hat\th\_m\to\th$ in probability as $m\to\infty$ (assuming that $\th$ is the true value of the parameter).
| 2 | https://mathoverflow.net/users/36721 | 449803 | 180,989 |
https://mathoverflow.net/questions/449785 | 5 | Let $H$ be a Hilbert space, and let $B(H)$ be the set of bounded linear operators $t \colon H \to H$. Recall that we say $t\_i \to t$ in the *strong operator topology* if $t\_i \xi \to t \xi$ for every $\xi \in H$.
**Question:** Let $B := (B(H)\_1, SOT)$ be the unit ball of $B(H)$ equipped with the strong operator topology. Is $B$ a Baire space, that is, is it the case that any intersection of countably many open dense sets is still dense?
The answer is positive if $H$ is separable. Indeed, see [here](https://math.stackexchange.com/questions/2515140/metrizability-of-the-strong-operator-topology-on-the-set-of-unitary-operators-of) for a proof of the fact that $B$ is then completely metrizable, and therefore a Baire space by the Baire Category Theorem. I'm hence mostly interested in the case $H$ is not separable, which, in turn, implies that $B(H)\_1$ is much larger, and hence (roughly speaking) it is easier for a set to be nowhere dense.
| https://mathoverflow.net/users/147609 | Is the unit ball of $B(H)$ a Baire space (with the SOT)? | I would say it is. Below is a sketch to more-or-less reduce to the separable case.
Given a finite set $V$ of unit vectors and some $\epsilon>0$, let
$$
U\_{V,\epsilon}=\{t\in B(H)\_1\mid \|tv\|\leq\epsilon \text{ for every }v\in V\}.
$$
The sets $\bar t + U\_{V,\epsilon}$ form a basis neighborhood for the strong operator topology.
Let now $\mathcal U\_n$ be a sequence of dense open sets in $(B(H)\_1, SOT)$ and $U\_0$ any open. We wish to show that $U\_0\cap \bigcap\_{n\in\mathbb N}\mathcal U\_n$ is not empty. We start the argument as for the usual proof of the Baire theorem. Since $\mathcal U\_1$ is dense, we may find $t\in B(H)\_1$, $V\_1$ and $\epsilon\_1$ so that the open set $U\_1= t\_1+U\_{V\_1,\epsilon\_1}$ has closure contained in $U\_0\cap \mathcal U\_1$. In turn, $U\_1$ intersects $\mathcal U\_2$ etc... This way we construct a sequence of nested open sets $U\_n= t\_n+ U\_{V\_n,\epsilon\_n}$. Moreover, we may also assume that $V\_n\subseteq V\_{n+1}$, and that $\epsilon\_{n+1} \ll \epsilon\_n$, if necessary.
Since each $V\_n$ is finite, their union $V=\bigcup\_{n\in\mathbb N}V\_n$ is countable. Let
$$
H'=\overline{<t\_n(V)\mid n\in\mathbb N>}^{\|\cdot\|}
$$
be the norm closure of the span of $V$ and its image under any of the $t\_n$. Observe that $H'$ now is separable and hence completely metrizeable.
Let $p$ the projection onto $H'$ and observe that for every $n\in\mathbb N$ the operator $pt\_np$ still belongs to $U\_n$.
It follows that the restrictions of $U\_n$ to $B(H')\_1$ (i.e. the sets $U\_{n}'=\{t\in B(H')\_1\mid \|(t-pt\_np)(v)\|\leq\epsilon\_n\ \forall v\in V\_n \in U\_{n}\}$) still are non-empty nested open sets, each containing the closure of all the following ones. Since they shrink very fast by construction, it should now be easy to deduce that they contain a Cauchy sequence in $B(H')\_1$. Extending the resulting limit operator by $0$ on the orthogonal complement of $H'$ yields the desired element in $B(H)\_1$.
(The sketch is inspired by the proof that arbitrary products of complete metric spaces are Baire. See Exercise 17 in Bourbaki's General Topology part 2, pp 254.)
| 3 | https://mathoverflow.net/users/54309 | 449809 | 180,990 |
https://mathoverflow.net/questions/449811 | -1 | I'm reading Titchmarsh's "The theory of the Riemann zeta function", and on p.81 it is claimed that
$$ \int\_{x}^{x+i\infty} (\cot(\pi z)+i)z^{-s} \, \mathrm{d}z \ll \frac{x^{-\sigma}}{2(n+1)\pi-|t|/x} \tag{1}$$
where $\sigma=\Re(s), t=\Im(s), x$ is a half-positive integer and $n$ is an arbitrary positive integer.
Titchmarsh proves (1) by invoking the identity
$$-\cot(\pi z) - i =2i\sum\_{v=1}^{n} e^{2\pi i vz} + \frac{2ie^{2(n+1)\pi i z}}{1-e^{2\pi iz}},$$
but doesn't show the steps. That's what I'm asking for.
| https://mathoverflow.net/users/507786 | On the bound for $\int_{x}^{x+i\infty} (\cot(\pi z)+ i)z^{-s} \, \mathrm{d}z$ | Your inequality does not make sense, since the RHS has $n$ in it, while the LHS does not. What Titchmarsh claims is the bound
$$\int\_{x}^{x+i\infty} (\cot(\pi z)+i)z^{-s} \mathrm{d}z \ll \frac{x^{-\sigma}}{2\pi-|t|/x}.$$
And he provides a detailed proof stretching 5 lines: "In the second integral we put $z=x+ir$ etc." The identity you quote is not needed for this proof. It is mentioned after the half-sentence "and the theorem follows".
| 0 | https://mathoverflow.net/users/11919 | 449820 | 180,994 |
https://mathoverflow.net/questions/449770 | 2 | In Higher Algebra 4.2.8.19, Lurie shows that the symmetric monoidal structure on spectra is uniquely defined (on the $\infty$-category level) by the following properties:
1. The sphere spectrum is the unit object
2. The tensor product bifunctor Sp x Sp -> Sp preserves colimits in each variable
Let R be an $E\_\infty$ ring. It seems clear to me like there should be a unique symmetric monoidal structure on $R$-modules such that
1. R is the unit object
2. The corresponding bifunctor preserves colimits in each variable.
but I am struggling to find a reference for this statement. Is there a standard one? Does it follow from something else in Higher Algebra?
(Even more generally, what if I have a small set of generators of a stable infinity-category such that Maps(X,Y) = 0 for any pair of distinct generators? Does a symmetric monoidal structure on the generators uniquely give a colimit-preserving symmetric monoidal structure on the entire $\infty$-category?)
| https://mathoverflow.net/users/131360 | Is the symmetric monoidal product on the $\infty$-category of $R$-modules unique? | If by "$R$ is the unit object" you mean : 1- the object $R$ in $Mod\_R$ is the unit **and** 2- The induced commutative algebra structure on $R = map(1,1)$ is the given commutative algebra structure on $R$, then the answer is yes.
Without the "and", namely if you only have 1-, then there are as many such symmetric monoidal structures as there are commutative algebra structures on $R$ (typically more than one).
This follows from the symmetric monoidal adjunction in *Higher Algebra*, 4.8.5.11 and 4.8.5.16 (4.8.5.11 provides the adjunction and 4.8.5.16. provides a symmetric monoidal structure on the left adjoint "$\Theta$" - $\Theta$ is the functor $R\mapsto (Mod\_R,R)$).
Together, 4.8.5.11 and 4.8.5.16 imply that the data of a colimit-preserving symmetric monoidal structure on $(Mod\_R,R)$ is equivalent to the data of an $E\_\infty$-structure on $R$, which explains my 2- above.
With regards to your parenthetical question: the property you describe provides a splitting of your category as a product of module categories (here I'm assuming you mean *compact* generators). You can then povide each factor with a symmetric monoidal structure at your leisure and get one on the product, certainly (although there might be other ones that aren't products, such as $Sp[\mathbb N]$ or something).
| 2 | https://mathoverflow.net/users/102343 | 449829 | 180,997 |
https://mathoverflow.net/questions/449808 | 7 | I've seen it stated in the $\infty$-categorical literature (without proof or reference) that every object in the $\infty$-category $\operatorname{Fun}(\Delta^1, \mathcal{C})$ of morphisms in a stable $\infty$-category $\mathcal{C}$ is a geometric realization of arrows of the form $A \rightarrow A \oplus B$. I may be missing some assumptions, but those should be "mild". Why is this the case?
| https://mathoverflow.net/users/507783 | Are morphisms in a stable $\infty$-category generated by split injections? | Any map $f:A\to B$ fits in a cofiber sequence of arrows $(0\to A)\to (A\to A\oplus B)\to (A\to B)$
In other words, any map is a cofiber of split inclusions. But now cofibers (as any colimit, by the Bousfield-Kan formula) can be rewritten as geometric realizations of coproducts of the involved terms, and thus we can conclude (coproducts of split inclusions are clearly split inclusions).
| 7 | https://mathoverflow.net/users/102343 | 449831 | 180,999 |
https://mathoverflow.net/questions/449722 | 0 | * Let $f(n)$ be an arbitrary function such that $f(n)\in\mathbb{Z}$.
* Let
$$
F(x)=\sum\limits\_{m\geqslant 0}f(m)x^m
$$
* Define the operator $\operatorname{SR}$, which is associated with the [series reversion](https://mathworld.wolfram.com/SeriesReversion.html).
* Let $a(n)$ be an integer sequence with generating function $A(x)$ where
$$
A(x)=\frac{1}{x}\operatorname{SR}(xF(x))
$$
* Let $b(n)$ be an integer sequence with generating function $\frac{1}{G(0)}$ where
$$
G(j)=F\left(\frac{x}{G(j+1)}\right)
$$
Here we have
$$
G(0)=F\left(\frac{x}{G(1)}\right)=F\left(\frac{x}{F\left(\frac{x}{G(2)}\right)}\right)=F\left(\frac{x}{F\left(\frac{x}{F\left(\frac{x}{G(3)}\right)}\right)}\right)
$$
and so on.
I conjecture that $$a(n)=b(n).$$
Is there a way to prove it?
| https://mathoverflow.net/users/231922 | Series reversion using something like continued fraction | We assume $F(0) \neq 0$, since otherwise we don't satisfy the assumptions for the series reversion. Let $G = G(0)$ be the fixpoint of the recurrence given:
$$G(x) = F\left(\frac{x}{G(x)}\right)$$
Multiply both sides by $\frac{x}{G(x)}$:
$$x = \frac{x}{G(x)} F\left(\frac{x}{G(x)}\right)$$
By inspection of the structure we have $\frac{x}{G(x)} = \operatorname{SR}(xF(x))$, or $$\frac{1}{G(x)} = \frac{1}{x} \operatorname{SR}(xF(x))$$ as desired.
| 1 | https://mathoverflow.net/users/46140 | 449836 | 181,001 |
https://mathoverflow.net/questions/449841 | 5 | Let $A \subset [0, 1]$ be a measurable set, and $\mathbf 1\_A$ its indicator function, viewed as a function on $\mathbb R$. Define for each $\delta > 0$, the function $f\_{A, \varepsilon}: \mathbb R \to [0, 1]$ given by
$$f\_{A, \varepsilon} (x) = \sup\_{r \leq \varepsilon} \frac{1}{2r} \int\_{x-r}^{x+r} |\mathbf 1\_A (y) - \mathbf 1\_A (x)| \, dy.$$
By the [Lebesgue density theorem](https://en.wikipedia.org/wiki/Lebesgue%27s_density_theorem), $f\_{A, \varepsilon}$ converges pointwise a.e. to the zero function as $\varepsilon \to 0$, and hence in $L^1$ by the dominated convergence theorem.
**Question:** Do there exist some absolute constants $C, \delta > 0$ such that for any measurable subset $A$ of $[0, 1]$, and all $\varepsilon < \delta$, we have
$$\|f\_{A, \varepsilon}\|\_{L^1} < C\varepsilon?$$
**Remark:** The role of $\delta$ is not quite necessary, but it saves us the trouble of having to deal with the trivial case of large $\varepsilon$.
| https://mathoverflow.net/users/173490 | Quantitative Lebesgue density theorem | Split $[0,1]$ into $n$-dyadic intervals and consider the set of alternating intervals:
$$A\_{n}\triangleq\Big[0,\frac{1}{2^{n}}\Big]\cup \Big[\frac{2}{2^{n}},\frac{3}{2^{n}}\Big]\cup\cdots\cup \Big[1-\frac{2}{2^{n}},1-\frac{1}{2^{n}}\Big]$$
and the function $f\_{n}(x)=1\_{A\_{n}}(x)$. Then we have that for $x\in A\_{n}$ and $r>2^{-n}$
$$
\begin{split}
\frac{1}{2r} &\int\_{x-r}^{x+r} |\mathbf 1\_A (y) - \mathbf 1\_A (x)| \, dy\\
& = 1-\frac{|A\_n \cap (x-r,x+r)|}{2r}=\frac{1}{2}.
\end{split}
$$
So for the $L^1$-norm and $\epsilon>2^{-n}$ we have the lower bound
$$\int\_{A\_{n}}f\_{A,\epsilon}(x)dx=|A\_{n}|\frac{1}{2}=\frac{1}{4}.$$
This example is from [T.Tao's notes](https://terrytao.wordpress.com/2007/06/18/the-lebesgue-differentiation-theorem-and-the-szemeredi-regularity-lemma/). In the same notes, he does provide some interesting rate of convergence theorems.
| 6 | https://mathoverflow.net/users/99863 | 449844 | 181,004 |
https://mathoverflow.net/questions/449838 | 10 | It is known that a one-dimensional Noetherian local ring is a discrete valuation ring if it is integrally closed.
Then, even if it is not Noetherian, would a one-dimensional local ring become a valuation ring if it is integrally closed?
| https://mathoverflow.net/users/500859 | Is it a valuation ring? | This is (essentially) a conjecture of Krull; a counter-example was given by P. Ribenboim: [*Sur une note de Nagata relative à un problème de Krull*](https://link.springer.com/article/10.1007/BF01166564), Math. Zeit. 64, 159-168 (1956). Note that "primaire" means "local of dimension 1", and that "complètement intégralement clos" is stronger than "integrally closed".
| 17 | https://mathoverflow.net/users/40297 | 449845 | 181,005 |
https://mathoverflow.net/questions/449789 | 3 | Given $k \in \mathbb N$, we define $f\_k: \mathbb N \longrightarrow \mathbb N$ by
$$ f\_k(x) = \begin{cases} \,\quad\dfrac{x}2 &\text{ if } x \text{ is even} \\\\ \dfrac{3x+3^k}{2} & \text{ if } x \text{ is odd} \end{cases} $$
For $k=0$, we have the function of the infamous Collatz conjecture.
*Does there exist for some $k\geq 0$ a strictly positive natural integer $N\_k$ whose iterates under $f\_k$ do not end up in the
trivial cycle $\lbrace 3^k,2\cdot 3^k\rbrace$ of $f\_k$?*
| https://mathoverflow.net/users/4556 | A mutation of the Collatz disease | Maybe it is worth to write my comments into an answer:
1. The question is equivalent to asking whether all elements of the form $n/3^k$ end up in the collatz cycle.
2. It is still an open question whether there is an integer $n$ where the collatz iterates go to infinity
3. It is still an open question whether there are other integral positive Collatz cycles
4. Iterating with starting value $n/3^k$ for some $k$ and $n\neq 0$, we always end up with an integer after finitely many steps. Each $x\mapsto 3x+1$ decreases the exponent of $3$ in the denominator by one. So the question is equivalent to asking whether the conjectures 2,3 hold.
| 7 | https://mathoverflow.net/users/3969 | 449849 | 181,007 |
https://mathoverflow.net/questions/449834 | 2 | Let $X$ be any topological space and denote by $\tau\_X$ the topology on $C\_b(X;\mathbb{R})$ that is induced by the family of seminorms $(\|\cdot\|\_\psi\mid\psi\in B\_0(X))$ with $\|f\|\_\psi:=\sup\_{x\in X}|f(x)\psi(x)|$ and
$$\tag{1}B\_0(X):=\{\psi:X\rightarrow\mathbb{R} \text{ bounded}\mid \forall\,\varepsilon>0\, : \,\exists\, K\subset X \text{ compact} \ : \ \sup\nolimits\_{x\in X\setminus K}|\psi(x)|<\varepsilon\}.$$
A result of [[Giles](https://www.ams.org/journals/tran/1971-161-00/S0002-9947-1971-0282206-4/S0002-9947-1971-0282206-4.pdf), Thm. 3.1] asserts that if $\mathfrak{A}$ is a point-separating and pointwise non-vanishing subalgebra of $C\_b(X;\mathbb{R})$, then $\mathfrak{A}$ is dense in $C\_b(X;\mathbb{R})$ wrt. $\tau\_X$.
**Question:** Given $c : X\rightarrow \mathbb{R}$ continuous, can we infer that also
$$\tag{2}\mathfrak{A}\_c:=\{p\in\mathfrak{A} \mid p\leq c\} \quad\text{is $\tau\_X$-dense in} \quad C\_{b\,|\,c}(X):=\{f\in C\_b(X;\mathbb{R})\mid f\leq c\} \ ?$$
(The above inequalities are understood to hold pointwise on $X$.) More regularity on $c$ can be assumed if necessary.
| https://mathoverflow.net/users/472548 | Does global boundedness ruin Stone-Weierstrass denseness? | If $c\notin\mathfrak{A}$ in general it is not true: e.g.: consider the case where $X$ is the real line , $\mathfrak{A}$ is the algebra of the polynomials, and $c$ is the function $-e^x$. Then $\mathfrak{A}\_c$ is empty.
If $\mathfrak{A}$ is also a lattice, it is true:
Note that for any $u,v\in C\_b(X;\mathbb{R})$ one has $|u\wedge c-v\wedge c|\le |u-v|$ point-wise. Therefore $\|u\wedge c-v\wedge c\|\_\psi\le \|u-v\|\_\psi
$ for any weight $\psi$, and $F:u\mapsto u\wedge c$ is a continuous self-map on $\big(C\_b(X;\mathbb{R}), \tau\_X\big)$. So $ C\_{b|c}(X) =F(C\_b(X;\mathbb{R}))=F(\overline{\mathfrak{A}}) \subset \overline{F (\mathfrak{A})}= \overline{ \mathfrak{A}\_c}$.
| 2 | https://mathoverflow.net/users/6101 | 449853 | 181,008 |
https://mathoverflow.net/questions/436711 | 5 | $\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}\DeclareMathOperator\SL{SL}$A closed subgroup $ \Gamma $ of a Lie group $ G $ is called Lie primitive if it is not contained in any proper positive dimensional closed subgroup.
Let $ G $ be a compact group. Then a Lie primitive subgroup $ \Gamma $ of $ G $ must be finite.
Must a compact simple Lie group $ G $ have only finitely many isomorphism types of Lie primitive subgroups?
Context:
This is true for $ \SO\_3(\mathbb{R}) $ which has only $ 3 $ Lie primitive subgroups up to conjugacy: the exceptional rotation groups of the platonic solids $ A\_4$, $S\_4$, $A\_5 $.
This is also true for $ \SU\_3 $ and $ \SU\_4 $ due to work by Blichfeldt. For an explicit list for $ \SU\_4 $ for example see [Hanany and He - A Monograph on the Classification of the Discrete Subgroups of SU(4)](https://arxiv.org/abs/hep-th/9905212).
Since it is true for $ \SU\_4 $ then it must be true for $ \SO\_6 $ by the double cover $ \SU\_4 \to \SO\_6 $.
Also this seems relevant: [Primitive subgroup of $ SU\_n $ contained in maximal finite subgroup?](https://math.stackexchange.com/questions/4468936/primitive-subgroup-of-su-n-contained-in-maximal-finite-subgroup)
Some comments about the case when $ G $ is not compact:
A closed subgroup not contained in any positive dimensional subgroup must be dimension $0$ i.e. it must be discrete. Peter McNamara [points out](https://mathoverflow.net/questions/436711/compact-lie-group-has-finitely-many-lie-primitive-subgroups#comment1132858_436711) that the only proper positive dimensional closed subgroups of $ \SL\_2(R) $ are solvable and thus can't contain a free group. Since a free group on any number of letters is a subgroup of $ \SL\_2(R) $ that means there are infinitely many non isomorphic Lie primitive subgroups of $ \SL\_2(R) $.
| https://mathoverflow.net/users/387190 | Compact Lie group has finitely many Lie primitive subgroups | Yes, it is true. Let us show that, in a compact Lie group $G$ with $G^0$ nonabelian, finite subgroups that are not contained in any proper subgroup of positive dimension have bounded cardinal.
By contradiction, let finite subgroups $G\_n$ have cardinal tending to infinity, and not contained in any proper subgroup of positive dimension.
Let $B$ be a neighborhood of 0 in the Lie algebra of $G$ such that $\exp|\_B$ is a homeomorphism onto its image, with inverse denoted by $\log$ (we will let $B$ vary, but can assume it is contained in a single such neighborhood $B\_0$ once and for all, hence the function $\log$ will not depend on $B$).
Define $T\_n^B=\exp\big(\mathrm{span}\big(\log(G\_n\cap\exp(B))\big)\big)$.
The first point is that for each $B\subset B\_0$ and for $n$ large enough, $G\_n\cap\exp(B)$ consists of commuting elements. This is close to Jordan's theorem and is a particular case of Zassenhaus' theorem (see for instance [Theorem 1.3 here](https://arxiv.org/abs/1402.0962), with in mind that connected nilpotent subgroups of compact groups are abelian). We can extract once and for all to suppose that this holds for all $n$.
The next point is that $T\_n^B$ is a torus. For an arbitrary commutative subset, it might have been dense in a torus. But the point is that elements of finite subgroups are well-behaved. More precisely: for this point, we can suppose that $G$ is embedded in the unitary group $\mathrm{U}(d)$. Every element of $G$ can be diagonalized with eigenvalues $e^{i\pi k\_1/m},\dots,e^{i\pi k\_d/m}$. We can arrange $B\_0$ so that it avoids elements with eigenvalues $-1$ and define the log accordingly, so assume all $k\_i$ to be in $]-m,m[$. Hence the log of such an element will be, in the same diagonal basis, the matrix with eigenvalues $i\pi k\_1/m,\dots i\pi k\_d/m$, and the exp of its span will be the 1-dimensional torus of diagonal matrices with eigenvalues $\theta^{k\_1},\dots,\theta^{k\_d}$ for $\theta$ in the unit circle. Now when we consider all $G\_n\cap \exp(B)$, we just consider the torus generated by several such (commuting) tori.
Now consider $\liminf\_n \dim(T\_n^B)$. This is positive (because $G\_n$ is infinite, so that $G\_n\cap\exp(B)$ is never reduced to $\{1\}$. This liming decreases when $B$ decreases, and hence we can find $B\_1$ for which its value, say $c>0$, is minimal. After extracting once again, we an suppose that $\lim\_n\dim(T\_n^{B\_1})=c$. Define $T\_n=T\_n^{B\_1}$. Hence, for every neighborhood $B\subset B\_1$, there exists $n\_0$ (depending on $B$) such that for all $n\ge n\_0$ we have $T\_n^B=T\_n$ (because $T\_n^B\subset T\_n$ and these are tori of the same dimension).
This observation allows to deduce that $T\_n$ is, for large $n$, normalized by $G\_n$. Indeed, for $g\_n\in G\_n$, we have $$g\_nT\_n^Bg\_n^{-1}=\exp(\mathrm{span}\log(G\_n\cap \exp(g\_nBg\_n^{-1})));$$
since $1$ has a basis of conjugation-invariant neighborhoods, we can choose $B$ such that $\bigcup\_g gBg^{-1}$ is contained in $B\_1$. We deduce that $g\_nT\_ng\_n^{-1}\subseteq T\_n$ for large $n$, hence this is an equality.
Then $G\_nT\_n$ is a positive-dimensional closed subgroup containing $G\_n$. By assumption, it equals $G$, and this is a contradiction since $G^0$ is non-abelian.
| 2 | https://mathoverflow.net/users/14094 | 449855 | 181,009 |
https://mathoverflow.net/questions/449602 | 5 | I apologize as I am certain this is not research-level, but several days have gone by without an answer on stackexchange (<https://math.stackexchange.com/questions/4724245/jacobian-criterion-for-zariski-cotangent-space-over-arbitrary-field>).
Section 7.2 of the notes on etale cohomology (<https://people.dm.unipi.it/tdn/CoomologiaEtale/Note.pdf>) by Lombardo and Maffei states the following (up to a small typo):
Let $k$ be any field, let $A = \frac{k[x\_1,...,x\_n]}{\langle f\_1,...,f\_m \rangle}$, and let $Jf = [\frac{\partial f\_j}{\partial x\_i}]$ be the Jacobian matrix of the $f\_j$. Let $\mathfrak{m}$ be a maximal ideal of $A$, and $k(\mathfrak{m}) = \frac{A}{\mathfrak{m}}$. Then $\dim\_{k(\mathfrak{m})} \frac{\mathfrak{m}}{\mathfrak{m}^2} \le n - $ rank $Jf(\mathfrak{m})$, where $Jf(\mathfrak{m})$ is obtained from $Jf$ by reducing all entries mod $\mathfrak{m}$.
The proof is omitted in the notes. Can anyone please supply a proof or a reference?
I emphasize that there is no further assumption on $k$ or the extension $k(\mathfrak{m})/k$. If $k$ is algebraically closed this is standard and if the extension is separable it is easy to prove from the notes.
| https://mathoverflow.net/users/91041 | Jacobian criterion for Zariski cotangent space over arbitrary field (X-post from SE) | Here's an attempt at solution.
Let $I=(f\_i)$ be the ideal defining the scheme $Spec(A)$ and $\mathtt{n} \subset k[\underline{x}\_i]$ be the maximal ideal s.t. $\mathtt{n}/I=\mathtt{m}$. Then we have the following commutative diagrams
$\require{AMScd}$
\begin{CD}
I/(I \cap \mathtt{n}^2) @>\alpha>> \mathtt{n}/\mathtt{n}^2 @>>>\mathtt{m}/\mathtt{m}^2 @>>> 0\\
@AAA @V \theta VV @VV V @.\\
k(\mathtt{m}) \otimes I/I^2=I/\mathtt{n}I @>\delta>>\oplus^{n}\_{i=1} k(\mathtt{m}) dx\_i @>>> \Omega^1\_{A/k}(\mathtt{m}) @>>> 0
\end{CD}
where the first vertical map is the obvious surjection and the next two vertical maps are maps defined naturally(as in Prop'n 6.13 of the notes alluded to in the question). The top row is a short exact sequence and the bottom row is exact. The bottom is also obtained as in Prop'n 6.13 cited before, but only after tensoring by $k(\mathtt{m})$.
By commutativity, $\theta(im(\alpha))=im(\delta)$. Since $dim\_{k(\mathtt{m})}im(\delta)=$ rank$Jf(\mathtt{m})$, so $dim\_{k(\mathtt{m})}im(\alpha) \geq rankJf(\mathtt{m})$ or equivalently $dim\_{k(\mathtt{m})}\mathtt{m}/\mathtt{m}^2 \leq n-rankJf(\mathtt{m})$(since $k[\underline{x}\_i]$ is regular $dim\_{k(\mathtt{m})}\mathtt{n}/\mathtt{n}^2=n$, noting that $k[\underline{x}\_i]/\mathtt{n}=A/\mathtt{m}$).
| 2 | https://mathoverflow.net/users/495875 | 449877 | 181,013 |
https://mathoverflow.net/questions/449860 | 0 | I have scouted books on statistics, probability, and consulted with mathematicians whom I know, and I have not succeeded in getting a satisfactory answer to this basic question:
**Question:** Exactly what does it mean for a sequence of points to be sampled from a given probability measure?
I am looking for the proper, abstract framework in which to formulate this concept, along with a formal definition in said context. Assume full knowledge of topology, measure theory, functional analysis, etc.
I have in mind statements such as «let the sequence $(x\_k)$ be sampled with density $f$», where $f \ge 0$ is a Borel function on $\mathbb{R}^n$ such that $\int f = 1$, or for such $f$ defined on appropriate submanifolds of $\mathbb{R}^n$.
My intuition is something along the lines that this should hold iff for every open set $U \subseteq \mathbb{R}^n$,
$$\lim\_m \, (\text{density of points } x\_k \text{ for } k \le m \text{ in } U ) = \int\_U f,$$
but it is not obvious to me how this notion of density should be defined or if this is sound. Certainly the Euclidean structure has nothing to do with the phenomenon, and the concept should be expressible in suitable topological spaces (with a Borel measure) or even in general probability spaces.
| https://mathoverflow.net/users/507776 | Definition of sequence sampled from a measure | What you are asking about is called **equidistribution** with respect to a given probability measure $\mu$ on a topological space $X$. Namely, a sequence $(x\_n)$ is $\mu$-equidistributed if the empirical distributions $(\delta\_{x\_1} + \dots + \delta\_{x\_n})$ weakly converge to $\mu$. Whether $X=\mathbb R^n$, or whether $\mu$ has a density (either with respect to the Lebesgue measure, or with respect to any given reference measure in the general case) is completely irrelevant.
If you are interested in the "higher order" equidistribution as well (this question arises in the theory of pseudorandom number generators), then one should also require, for any $d>1$, the convergence of order $d$ empirical distributions (i.e., those of the "vectors" $(x\_n,x\_{n+1},\dots,x\_{d+d-1})$) to the product of $\mu$ by itself $d$ times.
In the case when $X$ is finite, and $\mu$ is the uniform distribution on $X$, these are precisely the definitions of, respectively, the "simply normal" and the "normal" real numbers (identified with their base $|X|$ expansions).
However, as pointed out by Dieter Kadelka, one should realize that usually, when talking about a point or a sequence (which is a point in the space of sequences) sampled from a certain distribution $\mu$, one just means a random point whose distribution is the measure $\mu$.
| 0 | https://mathoverflow.net/users/8588 | 449891 | 181,019 |
https://mathoverflow.net/questions/449904 | 3 | Say I have a reductive complex algebraic group $G$ with maximal torus $T$ and associated Weyl group $W$. I would like to be able to say that the characters of $G$ are in bijection with the $W$-invariant characters of $T$, and the bijection is given by restriction. Is this correct? If so, how could it be proven?
| https://mathoverflow.net/users/507634 | Describing characters of a reductive group in terms of characters of a maximal torus | I thought at first that you meant "trace character of a finite-dimensional, irreducible representation of $G$" by "character of $G$". Then this is false; for example, if $G$ is $\operatorname{SL}\_2$, then $G$ certainly has non-trivial, finite-dimensional, irreducible representations (e.g., the defining representation), but the only $W(G, T)$-fixed character of a maximal torus $T$ in $G$ is the trivial one.
If you mean literally "homomorphism $G \to \mathbb C^\times$", then certainly such a character gives a $W(G, T)$-fixed character of $T$. Conversely, a $W(G, T)$-invariant character of $T$ is trivial on $[W(G, T), T] = T \cap [G, G]$ (where the bracket notation means the group generated by commutators, not the set of commutators), and $T/[W(G, T), T] \to G/[G, G]$ is an isomorphism, so we obtain a map $G/[G, G] \to \mathbb C^\times$, as desired.
In case you're curious about that equality and isomorphism, they use the structure of reductive groups:
* Since $N\_G(T) \to W(G, T)$ is surjective, certainly $[W(G, T), T]$ is contained in $T \cap [G, G]$. Conversely, $T \cap [G, G]$ is the smallest subtorus of $T$ containing the image of $\alpha^\vee$ for every root $\alpha$ of $T$ in $G$, but $[t, w\_\alpha]$ equals $\alpha^\vee(\alpha(t))$ for every such root and every $t \in T$.
* We have just shown that $T/[W(G, T), T] \to G/[G, G]$ is an embedding. Since $G$ is the almost direct product of its maximal central torus $Z$ with $[G, G]$, already $Z \to G/[G, G]$ is surjective; but $Z$ (being a central torus) is contained in $T$, so $T/[W(G, T), G] \to G/[G, G]$ is also surjective.
Incidentally, I also used the surjectivity of $N\_G(T) \to W(G, T)$ to assert that the restriction of a homomorphism $G \to \mathbb C^\times$ to $T$ is $W(G, T)$-fixed.
| 5 | https://mathoverflow.net/users/2383 | 449906 | 181,024 |
https://mathoverflow.net/questions/435265 | 10 | I'm currently learning about the Kubota–Leopoldt $p$-adic $L$-function and I'm noticing that many people view the Kubota–Leopoldt $p$-adic $L$-function as a *measure* as opposed to a $p$-adic analytic function or as a power series.
**My question is:** What is the motivation behind viewing a $p$-adic $L$-function as a measures? It seems a bit unnatural, at least from a beginner's perspective. Why is it natural to view $p$-adic $L$-functions as measures (as opposed to, say, as a $p$-adic analytic function or as a power series), and what is the advantage of viewing them as measures?
| https://mathoverflow.net/users/394740 | Why $p$-adic measures? | This is extremely late, but hopefully it's still of some use/interest to you.
p-adic L-functions are usually described as 'p-adic interpolations of special values of L-functions'. For Kubota--Leopoldt's p-adic interpolation of the Riemann zeta function, the relevant special values are all the values $\zeta(1-k)$, where $k$ is a positive integer.
Riemann zeta is a complex meromorphic function $\zeta : \mathbb{C} \to \mathbb{C}$. Probably the most naive hope is that you can construct a p-adic L-function as an $p$-adic meromorphic function $\zeta\_p : \mathbb{Z}\_p \to \mathbb{C}\_p$ with the appropriate interpolation property, namely that $\zeta\_p(1-k) = (\*) \zeta(1-k)$ for all $k > 0$ and some correction factor $(\*)$. Unfortunately this doesn't quite work: there is no single such function that captures all the special values. In particular, $\mathbb{Z}\_p$ is the 'wrong' domain for the p-adic L-function.
**From complex analytic functions to complex-valued measures:** To fix this, as Chris mentioned in his comment, we use Tate's measure-theoretic interpretation of L-functions. You can identify $\mathbb{C}$ with the space of characters $\mathbb{R}\_{>0} \to \mathbb{C}$, where $s$ corresponds to the character $x \mapsto x^s$. Then we view an $L$-function as a function on $\mathrm{Hom}\_{\mathrm{cts}}(\mathbb{R}\_{>0},\mathbb{C})$, which looks like a $\mathbb{C}$-valued measure on $\mathbb{R}\_{>0}$. You can expand this by adding in the finite places to obtain a measure-theoretic description on the ideles $\mathbb{A}^\times$, capturing not only the original $L$-function but all of its twists by Dirichlet characters.
**p-adic measures:** In the p-adic world, we might instead consider $\mathbb{C}\_p$-valued measures on the ideles. For topological reasons this essentially reduces (up to finite-order twist) to considering measures on $\mathbb{Z}\_p^\times$, and - hey presto! - you get the measure-theoretic language in which you see statements about p-adic L-functions. Now, there is a single (pseudo-)measure $\mu$ on $\mathbb{Z}\_p^\times$ that does interpolate *all* the special values $\zeta(1-k)$. (Here pseudo-measure = 'measure with simple poles'). In particular, we have
$$\int\_{\mathbb{Z}\_p^\times} x^k \cdot \mu = (1-p^{k-1})\zeta(1-k)$$
for all $k > 0$, a very clean interpolation.
**p-adic measures to p-adic analytic functions:** From this, one can indeed recover a description as an analytic function. There is a correspondence (Fourier or Amice transform) between $\mathbb{C}\_p$-valued measures on $\mathbb{Z}\_p^\times$ and bounded rigid analytic functions on $\mathrm{Hom}\_{\mathrm{cts}}(\mathbb{Z}\_p^\times,\mathbb{C}\_p)$. The latter is the $\mathbb{C}\_p$-points of the weight space $\mathcal{W}$, that shows up in the theory of p-adic families of modular forms/Hida theory/the eigencurve. The p-adic L-function, then, is naturally a p-adic meromorphic function on $\mathcal{W}(\mathbb{C}\_p)$, with at worst simple poles.
Now, what about that original naive guess? There is a theory in this direction; it just isn't as conceptually nice. Note $\mathcal{W}(\mathbb{C}\_p)$ breaks up into $p-1$ connected components (if $p$ is odd), each indexed by a character $\psi$ of $(\mathbb{Z}/p)^\times$. The possible characters are $\omega^i$, where $\omega$ is the Teichmuller character and $0 \leq i \leq k-2$. On the connected component corresponding to a fixed $\omega^i$, we are then reduced to considering characters of $1+p\mathbb{Z}\_p$. This naturally contains $\mathbb{Z}\_p$, where $s \in \mathbb{Z}\_p$ corresponds to the character $x \mapsto x^s := \mathrm{exp}(s \ \mathrm{log}(x))$. (Note we're 'reversing' the process we adopted over $\mathbb{C}$!) Define, then,
$$\zeta\_{p,i} : \mathbb{Z}\_p \to \mathbb{C}\_p, \ \ \ s \mapsto \int\_{\mathbb{Z}\_p^\times} \omega^i(x) \langle x \rangle^{1-s} \cdot \mu,$$
where $\langle - \rangle : \mathbb{Z}\_p^\times \to 1 + p\mathbb{Z}\_p$ is the projection. We get $p-1$ p-adic meromorphic functions $\zeta\_{p,0},...,\zeta\_{p,k-2}$ on $\mathbb{Z}\_p$. Apart from a simple pole for $i=0$, each $\zeta\_{p,i}$ is a power series, as you might hope. The function $\zeta\_{p,i}$ has the following interpolation:
$$ \zeta\_{p,i}(1-k) = (1-p^{k-1})\zeta(1-k) \ \ \ \ \forall k \equiv i \mod p-1.$$
In particular, each $\zeta\_{p,i}$ only sees *some* of the special L-values, and you need to consider the collection of all $p-1$ of them -- or, in other words, the measure/analytic function on $\mathcal{W}(\mathbb{C}\_p)$ -- to see *all* of them.
**Twists by Dirichlet characters:** Here's another benefit, that to me is an absolute clincher for the measure-theoretic language. Let $\chi$ be any Dirichlet character of $p$-power conductor. Then the measure-theoretic p-adic L-function also satisfies the following magic interpolation:
$$\int\_{\mathbb{Z}\_p^\times} \chi(x)x^k \cdot \mu = (1- \chi(p)p^{k-1}) L(\chi,1-k),$$
for any $k > 0$. In other words, the measure-theoretic object sees *so many more L-values* than the corresponding analytic function on $\mathbb{Z}\_p$. (You can write down twisted analytic functions on $\mathbb{Z}\_p$ that each sees some of these $L$-values; but somehow the single measure-theoretic object is parcelling together an infinite number of interesting analytic functions on $\mathbb{Z}\_p$ into one package!) This is very much in the spirit of Tate's original observations, in which an L-function and its twists get parceled into the same measure-theoretic object in complex analytic terms.
Everything I've written here is explained in more detail in the notes I wrote with Joaquin Rodrigues on Kubota--Leopoldt: <https://da380198-735d-4021-b7f2-6247f0586806.filesusr.com/ugd/946d8a_60256058271940f7b9d2c14dc721bf91.pdf>
**Generalisations:** If you weren't already convinced, here's one final observation. If you want to construct p-adic L-functions for motives/automorphic forms over more general base fields, then the shape of the theory changes. Here one can observe that $\mathbb{Z}\_p^\times$ is either the ray class group $\mathrm{Cl}\_{\mathbb{Q}}^+(p^\infty)$, or the Galois group of the maximal abelian extension of $\mathbb{Q}$ unramified outside $p\infty$. For a number field $F$, measures on $\mathbb{Z}\_p^\times$ generalise to measures/distributions on $\mathrm{Cl}\_F^+(p^\infty)$, or on the Galois group of the maximal abelian extension of $F$ unramified outside $p\infty$. This is a very clean generalisation; for example, if $F$ is imaginary quadratic, then you get two-variable measures, with cyclotomic and anticyclotomic variation. This admits a clean description in terms of analytic functions on the corresponding weight space. The analogous generalisation of analytic functions on $\mathbb{Z}\_p$ is a mess that doesn't seem natural to me at all.
| 8 | https://mathoverflow.net/users/61424 | 449907 | 181,025 |
https://mathoverflow.net/questions/449889 | 3 | $\DeclareMathOperator\Fl{Fl}$It is known that $H^\*(\Fl(m)) \cong R^{\mathbb Z}(m)$, where $\Fl(m)$ denotes the variety of complete flags in $\mathbb C^m$, and $R^{\mathbb Z}(m)$ is the coinvariant algebra, which may be defined by $$\frac{\mathbb Z[x\_1,\dotsc,x\_m]}{e\_1(x\_1,\dotsc,x\_m),e\_2(x\_1,\dotsc,x\_m),\dots,e\_m(x\_1,\dotsc,x\_m)}$$
More generally the coinvariant algebra is just the quotient $\mathbb C[x\_1,\dotsc,x\_m]/I\_m$ (the above just replaces $\mathbb Z$ with $\mathbb C$), where $I\_m$ is the ideal generated by the non-constant, homogeneous $S\_m$-invariants, where $S\_m$ acts on $\mathbb C[x\_1,\dotsc,x\_m]$ in the natural way. The above definition presented is where we choose the elementary symmetric polynomials $e\_i$ as a basis of $I\_n$. The isomorphism in that case is realized by sending the generator $x\_i$ of $R^{\mathbb Z}(m)$ to $-c\_1(U\_i/U\_{i-1})$, where $U\_i$'s are the standard bundles over $\Fl(m)$. These successive quotients filter the trivial bundle, $\mathbb C^m \times \Fl(m)$, by line bundles, thus the $i$th chern class of the trivial bundle is given by $e\_i(x\_1,\dotsc,x\_m)$ and must vanish hence the isomorphism. I would say this constitutes a geometric understanding of why the relations $e\_i$ appear.
If we chose a different basis for $I\_m$, the power sum symmetric polynomials, there is of course another isomorphism $H^\*(\Fl(m)) \cong \mathbb Z[x\_1,\dotsc,x\_m]/(p\_1(m),p\_2(m),\dotsc,p\_m(m))$. What is the geometric significance of the $p\_i(m)$'s, if any? Can they be realized in the same way as above but corresponding to different bundles?
The power sum symmetric polynomials are defined by
$$
p\_k(n) = x^k\_1 + x^k\_2 + \dotsb + x^k\_n.
$$
| https://mathoverflow.net/users/131046 | Alternative bases of symmetric polynomials in cohomology ring of flag varieties and coinvariant algebras | I assume we compute cohomology with $\mathbb{Q}$ coefficients. Let $x\_i$ map to $c\_1(U\_i/U\_{i-1})$, like you suggest. Then the polynomial $p\_k(n)$ gets sent to $k!$ times the $k$-th graded piece of the Chern character of the trivial bundle.
| 1 | https://mathoverflow.net/users/470175 | 449908 | 181,026 |
https://mathoverflow.net/questions/449884 | 3 | Suppose I have a global field $K$ and a finite Galois extension $L/K$ of Galois group $G$. It is often written without proofs (it seems that this is a very common statement) that for every $G$-module $M$ of permutation (that is a free $\mathbf{Z}$-module that admits a basis permuted by the action of $G$), the Tate-Shafarevich group of degree $2$
$$ \mathrm{Sha}^2(G,M)=\mathrm{Ker}\left(H^2(G,M)\longrightarrow\prod\_{v\in\Omega\_K}H^2(G\_v,M)\right) $$
where $\Omega\_K$ is the set of places of $K$ and $G\_v=\mathrm{Gal}(L\_w/K\_v)$ for $w$ a place of $L$ that extends $v$, is trivial. Do you have any reference for this ? or a proof (if it's not too long) ?
| https://mathoverflow.net/users/170999 | The second Tate-Shafarevich group of a permutation module is trivial | We write $G\_w={\rm Gal}(L\_w/K\_v)$.
>
> **Definition.** For $n\ge 1$, we denote
> $$Ш\_\omega^n(G,M)=\ker\Big(H^n(G,M)\to\prod\_C H^n(C,M)\Big)$$
> where $C$ runs over the cyclic subgroups of $G$.
>
>
>
>
> **Remark.** $Ш^2(L/K,M)\subseteq Ш\_\omega^2(G, M)$. Indeed,
> by the Chebotarev density theorem, for any cyclic subgroup $C\subseteq G$
> there exist $v$ and $w$ such that $G\_w=C$.
>
>
>
>
> **Definition.** A permutation $G$-module is a torsion free $G$-module admitting a $G$-invariant (as a subset) ${\mathbb Z}$-basis $B$.
>
>
>
>
> **Theorem.** If $M$ is a permutation $G$-module, then $Ш\_\omega^2(G,M)=0$, and therefore $Ш^2(L/K,M)=0$.
>
>
>
*Sketch of proof.*
*Step 1.*
We have
$$ Ш\_\omega^2(G,M\_1\oplus M\_2)=Ш\_\omega^2(G,M\_1) \oplusШ\_\omega^2(G, M\_2),$$
which reduces the theorem to the case when $G$ acts on the basis $B$ transitively.
Then using Shapiro's lemma we reduce the theorem to the case $M={\mathbb Z}$ (with trivial $G$-action).
*Step 2.* We have a short exact sequence of $G$-modules (with trivial $G$-action)
$$0\to {\mathbb Z}\to{\mathbb Q}\to {\mathbb Q}/{\mathbb Z}\to 0,$$
which gives rise to cohomology exact sequences
\begin{align\*}
&0= H^1(G,{\mathbb Q})\to H^1(G,{\mathbb Q}/{\mathbb Z}) \to H^2(G,{\mathbb Z})\to H^2(G,{\mathbb Q})=0\\
&0= H^1(C,{\mathbb Q})\to H^1(C,{\mathbb Q}/{\mathbb Z}) \to H^2(C,{\mathbb Z})\to H^2(C,{\mathbb Q})=0.
\end{align\*}
Therefore, we may identify
$$Ш\_\omega^2(G,{\mathbb Z})=Ш\_\omega^1(G,{\mathbb Q}/{\mathbb Z}),$$
and it suffices to show that $Ш\_\omega^1(G,{\mathbb Q}/{\mathbb Z})=0$.
By definition,
$H^1(G,{\mathbb Q}/{\mathbb Z})={\rm Hom}(G,{\mathbb Q}/{\mathbb Z})$ and $H^1(C,{\mathbb Q}/{\mathbb Z})={\rm Hom}(C,{\mathbb Q}/{\mathbb Z})$.
It remain to show that
$$\ker\Big({\rm Hom}(G,{\mathbb Q}/{\mathbb Z})\to \prod\_C{\rm Hom}(C,{\mathbb Q}/{\mathbb Z})\Big)=0,$$
but this is obvious because $G$ is the union of its cyclic subgroups $C$.
*Reference:* Lemma 1.9 of J.-J. Sansuc, Groupe de Brauer et arithmétique des groupes algébriques linéaires sur un corps de nombres, J. Reine Angew. Math. 327 (1981), 12–80.
| 6 | https://mathoverflow.net/users/4149 | 449921 | 181,029 |
https://mathoverflow.net/questions/449927 | 0 | Let
* $H$ be a infinite dimensional, separable, complex Hilbert space,
* $\{v\_{1\_n}\}\_{n \in \mathbb{N}}$ be a sequence in $H$,
* $V\_1=\operatorname{span}\{v\_{1\_n}\}\_{n \in \mathbb{N}}$
* $U\_1=\overline{V\_1}$.
We know that $\{v\_{1\_n}\}\_{n \in \mathbb{N}}$ is a Schauder basis of $U\_1$
Let $\{v2\_n\}\_{n \in \mathbb{N}}$ be a sequence in $H$. Furthermore let,
* $V\_2=\operatorname{span}\{v\_{1\_n}\}\_{n \in \mathbb{N}}$,
* $U\_2=\overline{V\_2}$.
We know that $\{v\_{2\_n}\}\_{n \in \mathbb{N}}$ is a Schauder basis of $U\_2$ and furthermore, we know that $V\_1 \cap V\_2 = \{0\}$.
**The question**: Is it true that $U\_1 \cap U\_2 = \{0\}$.
| https://mathoverflow.net/users/108867 | Intersection of Hilbert spaces with Schauder basis | No, the intersection of the $U$'s might even be full. Note that it suffices to find $V$'s of countable dimension, since we can then use the Gram-Schmidt algorithm to find an orthonormal basis, which will be automatically Schauder.
Set $H=L^2([0,1])$, $V\_1$ the space generated by the $\mathbb1\_{[k/n,(k+1)/n)}$, and $V\_2$ that of the polynomials. They are dense, so the corresponding $U$'s are equal (they are precisely $H$). The intersection of the $V$'s is precisely the constants, but you can get rid of them by considering $H'\subset H$ the functions of zero mean, and $V'\_i=V\_i$.
For an explicit example of sequences, consider the basis of Haar wavelets in $H'$ and that of the non-constant Legendre polynomials.
| 4 | https://mathoverflow.net/users/129074 | 449929 | 181,031 |
Subsets and Splits