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https://mathoverflow.net/questions/439855 | 2 | According to the standard definition, $\mathcal{S}\_0^{\beta}(\mathbb{R})$ is a subspace of smooth functions on $\mathbb{R}$ with the property that
\begin{equation}
\lvert x^k f^{(q)}(x) \rvert \leq CA^kB^q q^{\beta q}
\end{equation}
for some constants A, B, C that depend on each $f$. I am aware that $f$ must be compactly supported and $\beta>1$ is required for a nontrivial $f$ to exist.
Now, if I tensor-product $n$ copies of $f$, that is, define
\begin{equation}
F(x\_1, \cdots, x\_n):= f(x\_1) \times \cdots \times f(x\_n)
\end{equation}
it is clear that $F$ belongs to $\mathcal{S}\_0^{\beta}(\mathbb{R}^n)$, with the same $A,B,C$ for each coordinate.
Now, let us think of any invertible $n \times n$ matrix $\mathfrak{A}$ and consider the function
\begin{equation}
G(x\_1, \cdots, x\_n):= F\bigl( \mathfrak{A}(x\_1, \cdots, x\_n) \bigr).
\end{equation}
Then, I believe that $G$ must be in $\mathcal{S}\_0^{\beta}(\mathbb{R}^n)$ as well. However, I cannot figure out further information for the bounds on $G(x\_1, \cdots, x\_n)$. For example, is it possible to express the constants $A\_G^1, B^1\_G, C^1\_G$ such that
\begin{equation}
\lvert (x\_1)^k (\partial\_1)^q G(x\_1, \cdots, x\_n) \rvert \leq C\_G^1 (A\_G^1)^k (B\_G^1)^q q^{\beta q}
\end{equation}
in terms of $A, B, C,$ of the orignal function $f$ and some quantities related to $\mathfrak{A}$, like its operator norm?
This kind of detailed estimate seems quite delicate for me.. Could anyone please help me?
| https://mathoverflow.net/users/56524 | Specific estimation of the norm for a linearly transformed function in $\mathcal{S}_0^{\beta}(\mathbb{R}^n)$ | $\newcommand\be\beta\newcommand\A{\mathfrak A}\newcommand\R{\mathbb R}\newcommand\N{\mathbb N}$Suppose that
\begin{equation}
|x\_j^k F^{(q)}(x)| \le CA^kB^{|q|} q^{\beta q}
\end{equation}
for some positive real $A,B,C,\be$, all $k\in\N\_0:=\{0,1,\dots\}$, all $j\in[n]:=\{1,\dots,n\}$, all $x=[x\_1,\dots,x\_n]^\top\in\R^n$, and all $q=[q\_1,\dots,q\_n]^\top\in\N\_0^n$, where $|q|:=q\_1+\dots+q\_n$ and $q^{\be q}:=\prod\_{j\in[n]}q\_j^{\be q\_j}$.
Let
$$G(x):=F(\A x)$$
for some invertible matrix $\A=[A\_{i,j}\colon(i,j)\in[n]^2]$ and all $x\in\R^n$.
Then for any $j\in[n]$ and all $x\in\R^n$
$$D\_j G(x)=\sum\_{i\in[n]}A\_{i,j}D\_iF(\A x),$$
where $D\_i$ is the operator of the partial differentiation with respect to $x\_i$.
So, for any $r\in\N\_0$, any $j\_1,\dots,j\_r$ in $[n]$, and all $x\in\R^n$ one has
$$D\_{j\_r}\cdots D\_{j\_1}G(x)
=\sum\_{i\_r\in[n]}A\_{i\_r,j\_r}\cdots \sum\_{i\_1\in[n]}A\_{i\_1,j\_1}
D\_{i\_r}\cdots D\_{i\_1}F(\A x).$$
So,
$$|G^{(q)}(x)|\le m(\A)^{|q|}|F^{(q)}(\A x)|$$
for all $x\in\R^n$ and $q\in\N\_0^n$, where $$m(\A):=\max\_{j\in[n]}\sum\_{i\in[n]}|A\_{i,j}|=\|\A\|\_{1,1},$$
the [$(\ell^1,\ell^1)$ operator norm](https://en.wikipedia.org/wiki/Matrix_norm#Matrix_norms_induced_by_vector_p-norms) of the matrix $\A$.
Also, $\|x\|\_\infty:=\max\_{1\le j\le n}|x\_j|\le\|\A^{-1}\|\_{\infty,\infty}\|\A x\|\_\infty$ for all $x\in\R^n$, where $\|\cdot\|\_{\infty,\infty}$ is the $(\ell^\infty,\ell^\infty)$ operator norm.
Thus, as desired,
\begin{equation}
|x\_j^k G^{(q)}(x)| \le C(A\|\A^{-1}\|\_{\infty,\infty})^k
(B\|\A\|\_{1,1})^{|q|} q^{\beta q}
\end{equation}
for all $k\in\N\_0$, $j\in[n]$, $x\in\R^n$, and $q\in\N\_0^n$.
| 2 | https://mathoverflow.net/users/36721 | 439877 | 177,627 |
https://mathoverflow.net/questions/421682 | 2 | Let $G$ be a compact matrix Lie group under the Killing form metric $\langle \xi, \eta \rangle\_g = -\frac{1}{2}\text{tr}((g^{-1}\xi)^T(g^{-1}\eta))$ for $g \in G$ and $\xi,\eta \in T\_gG$. Let $C \subset G$ be a geodesically convex set. Pick finitely many $g\_1,...,g\_N \in C$ and define $\Omega$ to be the smallest closed convex set containing those points. What does the boundary of $\Omega$ look like? Is it like the smallest geodesic polygon that fits those points, like in Euclidean space? Or is it something more complicated?
| https://mathoverflow.net/users/141449 | What does the boundary of convex hulls look like in matrix Lie groups? | I guess you wanted to say *smallest geodesic polytope* (not *polygon*).
It is unclear what is polytope in a the matrix Lie group, but it seems to require geodesic hypersurfaces.
They do not exist in most Riemannian manifold starting from dimension 3 and matrix Lie groups are not exceptional.
BTW, if you are interested in convex sets in general Riemannian manifolds,
then check our paper ["About every convex set..."](https://arxiv.org/abs/2103.15189).
| 1 | https://mathoverflow.net/users/1441 | 439889 | 177,635 |
https://mathoverflow.net/questions/3347 | 38 | The definition in the title probably needs explaining. I should say that the question itself was an idea I had for someone else's undergraduate research project, but we decided early on it would be better for him to try adjacent and less technical questions. So it's not of importance for my own work per se, but I'd be interested to know if it easily reduces to a known conjecture/fact/counterexample in number theory.
Apologies if the question is too technical/localized/unappealing/bereft of schemes.
Given a subset $X$ of the natural numbers $N$, and given $n \in N$, we write $X-n$ for the backward translate of $X$, i.e. the set $\{x-n : x\in X\}$.
We say that $X$ is *translation-finite* if it has the following property: for every strictly increasing sequence $n\_1 < n\_2 < \dotsb$ in $N$, there exists $k$ (possibly depending on the sequence) such that
$$(X-n\_1) \cap (X-n\_2) \cap \dotsb\cap (X-n\_k)$$
is finite or empty.
Thus every finite set is trivially translation-finite: and if the elements of $X$ form a sequence in which the difference between successive terms tends to infinity, then $X$ is translation-finite *and* we can always take $k = 2$. Moreover:
* if $X$ contains an infinite arithmetic progression, or if it has positive (upper) Banach density, then it is NOT translation finite;
* there exist translation-finite sets which, when enumerated as strictly increasing sequences, grow more slowly than any faster-than-linear function.
* there exist translation-finite sets containing arbitrarily long arithmetic progressions.
These resultlets suggest the question in the title, but I don't know enough about number theory to know if it's a reasonable question. Note that if, in the definition, we were to fix $k$ first (i.e. there exists $k$ such that for any sequence $(n\_j)$…) then we would get something related to Hardy–Littlewood conjectures; but I was hoping that this might not be necessary to resolve the present question.
**EDIT (2nd Nov)** It's been pointed out below that the question reduces in some sense to a pair of known, hard, open problems. More precisely: if the answer to the question is yes, then we disprove the Hardy–Littlewood $k$-tuples conjecture; if the answer is no, then there are infinitely many prime gaps bounded by some absolute constant, and this is thought to be beyond current techniques unless one assumes the Eliott–Halberstam conjecture.
**Added in 2013:** Stefan Kohl points out that the latter is [Yitang Zhang's famous recent result](https://www.simonsfoundation.org/features/science-news/unheralded-mathematician-bridges-the-prime-gap/). However, as Will Sawin points out in [comments](https://mathoverflow.net/questions/3347/is-the-set-of-primes-translation-finite#comment349721_3347), a negative answer to the main question would imply there are 3-tuple configurations occurring infinitely often in the primes, and (see the link in Will's [comment](https://mathoverflow.net/questions/3347/is-the-set-of-primes-translation-finite#comment349721_3347)) this is thought to be out of reach even if we assume the EH conjecture holds.
| https://mathoverflow.net/users/763 | Is the set of primes "translation-finite"? | As pointed out to me by the OP, Theorem 1.5 of [my recent paper with Tamar Ziegler](https://arxiv.org/abs/2301.10303) answers this question in the negative. (The situation has changed since the 2009 and 2013 comments due to the breakthrough [work of Zhang on prime gaps](https://mathscinet.ams.org/mathscinet-getitem?mr=3171761) in 2014, followed by [the stronger results of Maynard](https://mathscinet.ams.org/mathscinet-getitem?mr=3272929) in 2015; our own paper relies heavily on the Maynard sieve.)
| 18 | https://mathoverflow.net/users/766 | 439890 | 177,636 |
https://mathoverflow.net/questions/439893 | 2 | In "Number: The Language of Science" (1930), Tobias Dantzig refers to what we call the base case of mathematical induction as "the induction step" (and refers to what we call the induction step as "the recurrence step" or "the proof of the hereditary property"). Was this standard terminology a century ago, or was Dantzig confused?
Note that he wrote this way back when mathematical induction was commonly called complete induction as opposed to Baconian or incomplete induction. Since verification of a single base case could be viewed as a minimalist version of Baconian induction, Dantzig's terminology does not seem totally illogical to me. Perhaps his use of the phrase "induction step" was standard a century ago, and over time its meaning shifted so that it now has the "opposite" meaning (that is, it now refers to the other component of proof by mathematical induction).
I'd be grateful for comments by those who know more history of mathematics than I do, as well as those who can bring a multicultural perspective to this question (what sort of terminology for mathematical induction is used in other languages?).
| https://mathoverflow.net/users/3621 | Terminology associated with mathematical induction | Not really an answer (too long for a comment) but I hope that this will be helpful:
In Jeff Miller's "Earliest Uses of Some Words of Mathematics" (<https://mathshistory.st-andrews.ac.uk/Miller/mathword/>) we find the following:
The term MATHEMATICAL INDUCTION was introduced by Augustus de Morgan (1806-1871) in 1838 in the article Induction (Mathematics) which he wrote for the Penny Cyclopedia. De Morgan had suggested the name successive induction in the same article and only used the term mathematical induction incidentally. The expression complete induction attained popularity in Germany after Dedekind used it in a paper of 1887 (Burton, page 440; Boyer, page 404).
and
COMPLETE INDUCTION (vollständige Induktion) was the term employed by Dedekind in his Was sind und Was sollen die Zahlen? (1887) for what is nowadays called "mathematical induction", and whose "scientific basis" ("wissenschaftliche grundlage") he claimed to have established with his "Theorem of complete induction" (§59). Dedekind also used occasionally the phrase "inference from n to n + 1", but nowhere in his booklet did he try to justify the adjective "complete".
In Concerning the axiom of infinity and mathematical induction (Bull. Amer. Math. Soc. 1903, pp. 424-434) C. J. Keyser referred to "complete induction" as
a form of procedure unknown to the Aristotelian system, for this latter allows apodictic certainty in case of deduction only, while it is just characteristic of complete induction that it yields such certainty by the reverse process, a movement from the particular to the general, from the finite to the infinite.
| 2 | https://mathoverflow.net/users/10744 | 439896 | 177,638 |
https://mathoverflow.net/questions/439917 | 3 | Let $X$ be a smooth projective curve over a field $k$ and $K\_X$ be its canonical line bundle. By the Serre duality, $\text{H}^1(X,K\_X)$ is a one-dimensional $k$-vector space. On the other hand, $\text{H}^1(X,K\_X)=\text{Ext}^1(O\_X,K\_X)$ so this vector space corresponds to the set of extensions of $O\_X$ by $K\_X$. Let $V$ be such an extension that corresponds to a nonzero vector in $\text{H}^1(X,K\_X)$. This is a $2$-dimensional vector bundle on $X$.
**Question.** Is there exists a simple/concrete description of $V$?
| https://mathoverflow.net/users/95601 | Extension of the trivial bundle by the canonical bundle on a curve | If $X$ is a smooth curve, the canonical bundle $\omega\_X$ is nothing but the bundle of differentials $\Omega\_X^1$, and the corresponding extension defines the **jet bundle**
$$
0 \to \Omega\_X^1 \to J\_X \to \mathcal{O}\_X \to 0.
$$
| 4 | https://mathoverflow.net/users/4428 | 439921 | 177,648 |
https://mathoverflow.net/questions/304119 | 9 |
>
> Given a bi-Lipschitz homeomorphism
> $\Phi:\mathbb{B}^n(0,1)\to\mathbb{R}^n$, (that is a bi-Lipschitz map onto the image), can one find a bi-Lipschitz homeomorphism $\Psi:\mathbb{R}^n\to\mathbb{R}^n$ such that
> $$
> \Psi|\_{\mathbb{B}^n(0,\frac{1}{2})}=\Phi|\_{\mathbb{B}^n(0,\frac{1}{2})}\ \ ?
> $$
>
>
>
It follows from the planar bi-Lipschitz Schoenflies theorem **[T]** that
when $n=2$, the answer is yes since you can actually extend $\Phi$ from $\mathbb{B}^2(0,1)$ to $\mathbb{R}^2$ in a bi-Lipschitz manner, but in higher dimensions it is not possible to extend from $\mathbb{B}^n(0,1)$, and this is why I am asking about extension from a smaller ball.
Thickening of the Fox-Artin arc creates a domain $\Omega$ in $\mathbb{R}^3$ that is bi-Lipschitz homeomorphic to $\mathbb{B}^3(0,1)$, but no bi-Lipschitz homeomorphism $\Phi:\mathbb{B}^3\to\Omega$ can be extended to a homeomorphism of $\mathbb{R}^3$. This was observed by Fred Gehring in 1968, but details were not provided. Details are provided in Theorem 3.7 in **[M]**.
**[M] Martin, G.:** Quasiconformal and bi-Lipschitz homeomorphisms, uniform domains and the quasihyperbolic metric. *Trans. Amer. Math. Soc.* 292 (1985), 169–191.
**[T] Tukia, P.:** The planar Schönflies theorem for Lipschitz maps. *Ann. Acad. Sci. Fenn. Ser. A I Math.* 5 (1980), no. 1, 49–72.
| https://mathoverflow.net/users/121665 | Bi-Lipschitz extension | The following result answers the question by providing a locally bi-Lipschitz extension.
>
> **Theorem.** Given a bi-Lipschitz homeomorphism
> $\Phi:\mathbb{B}^n(0,1)\to\mathbb{R}^n$, (that is a bi-Lipschitz map onto the image), one can find a locally bi-Lipschitz homeomorphism $\Psi:\mathbb{R}^n\to\mathbb{R}^n$ such that
> $$
> \Psi|\_{\mathbb{B}^n(0,\frac{1}{2})}=\Phi|\_{\mathbb{B}^n(0,\frac{1}{2})}.
> $$
>
>
>
This result is a straightforward consequence of the Theorem 5.10 in [TV]. If we regard $\mathbb{B}^n(0,1)$ as a subset of $\mathbb{S}^n$ (stereographic projection + one point compactification of $\mathbb{R}^n$), then $\Phi$ can be extended from $\mathbb{B}^n(0,\frac{1}{2})$ to a bi-Lipschitz map of $\mathbb{S}^n$. Such a map gives a locally bi-Lipschitz map of $\mathbb{R}^n$.
Theorem 5.10 is a version of the generalized Schoenflies theorem for bi-Lipschitz maps.
**Remark.** I think it is possible to find not only a locally bi-Lipschitz, but a bi-Lipschitz extension. Argument would be the same as above, with the only difference that we would place at the north poles (from which we have stereographic projections) points there the extension and its inverse are differentiable. But, I have to think about this argument a bit more.
**[TV] Tukia, P.; Väisälä, J.:** Lipschitz and quasiconformal approximation and extension. *Ann. Acad. Sci. Fenn. Ser. A I Math.* 6 (1981), no. 2, 303–342 (1982).
| 1 | https://mathoverflow.net/users/121665 | 439924 | 177,650 |
https://mathoverflow.net/questions/439951 | 3 | I'm studing C. Petit's work "Faster algorithms for isogeny problems using torsion point images" ([link](https://eprint.iacr.org/2017/571.pdf)) and he talks about meet-in-the-middle approach/strategy for solve some isogenies problems.
Well, what does Petit mean with meet-in-the-middle approach/strategy?
I've read that it's an approach that allows you to reduce the complexity of a problem with the aim of using brute force to solve it, but what does it mean in mathematical language?
Any help will be greatly appreciated.
| https://mathoverflow.net/users/497497 | What is meant by a meet-in-the-middle approach? | The "meet in the middle approach", also known as "bidirectional search", is a method to find shortest paths in graphs. It was proposed by Pohl [1] and first used by Galbraith [2] to construct isogenies between elliptic curves $E$ and $E'$. One builds two trees of isogenies from both sides of $E$ and $E'$, and finds a collision between the two trees to obtain the shortest path from $E$ to $E'$.
[1] [Bi-directional and heuristic search in path problems](https://www.semanticscholar.org/paper/Bi-directional-and-heuristic-search-in-path-Pohl/e609ab97186834bd97d3fbc0c97a19744d752c23), I Pohl (1969).
[2] [Constructing isogenies between elliptic curves over finite fields](https://www.math.auckland.ac.nz/~sgal018/iso.pdf), S.D. Galbraith (2011).
| 3 | https://mathoverflow.net/users/11260 | 439953 | 177,659 |
https://mathoverflow.net/questions/439864 | 1 | I recently encountered a particular delay PDE in my work, the solution of which corresponds to the Laplace transform of some probability distribution. I'm having trouble to solve this equation. The solution $f(q,s)$ defined on $q\geq 0, s>1$ should verify $f(0,s) = 1 \forall s$ and $f(q, s\to \infty) = \delta\_{q,0}$ (the Kronecker delta). It should also the verify the following PDE
$$\partial\_q f(q,s-1) = -\sqrt{s(2+s)}f(q,s) $$
It looks a bit like a "deformed" Hurwitz zeta function relation (which is $\partial\_q \zeta(q,s) = -s\zeta(q,s+1)$). I looked at some simple ansätze but couldn't come up with anything useful. I also am not sure if the boundary conditions I specified above are enough to obtain a unique solution...
I would be glad if you have any suggestions.
Thanks a lot !
| https://mathoverflow.net/users/498690 | Solving a particular delay PDE $\partial_q f(q,s-1) = -\sqrt{s(2+s)}f(q,s)$ | Inspired by your comparison with the Hurwitz zeta function, which has the integral representation
$$ \zeta (s,a)={\frac {1}{\Gamma (s)}}\int \_{0}^{\infty }{\frac {x^{s-1}e^{-ax}}{1-e^{-x}}}\mathrm{d}x \:, $$
I tried to come up with an integral representation for your function $f$. The factor $s$ that comes out when taking the $q$ derivative of $\zeta(s,q)$ arises from the $\Gamma(s)$ in the denominator. To get your factor $\sqrt{s(2+s)}$, we need to change this expression into something of the form
$$ f(q,s) = \frac{1}{\sqrt{\Gamma(s+1)\Gamma(s+3)}} \int\_0^\infty x^{s-1} e^{-q x} G(x) \mathrm{d} x \:, $$
with an unknown function $G$. This expression satisfies your PDE.
The function $G$ can be determined from your condition $f(0,s) = 1$, which actually gives the Mellin transform of $G$,
$$ \int\_0^\infty x^{s-1} G(x) \mathrm{d}x = \sqrt{\Gamma(s+1)\Gamma(s+3)}
= \Gamma(s) s\sqrt{(s+2)(s+1)} \:. $$
In this form, you can obtain $G$ as a power series via Ramanujan's master theorem, which gives
$$ G(x) = -\sum\_{k=3}^\infty \frac{(-x)^k}{(k-1)!}\sqrt{(k-1)(k-2)} \:. $$
| 2 | https://mathoverflow.net/users/498819 | 439964 | 177,665 |
https://mathoverflow.net/questions/439918 | 7 | We know about volume: The $L\_{\infty}$ ball of radius one-half, i.e. the hypercube, has volume $1$ in all dimensions. On the other hand, I believe that for every $1 \leq p < \infty$, the volume of the inscribed $L\_p$ ball $\{x : \|x\|\_p \leq \tfrac{1}{2}\}$ goes to zero as the dimension $d \to \infty$. (For example, a simple generalization of [this argument](https://mathoverflow.net/a/8270/29697).)
My question is the analogous one for surface area. The surface area of the $L\_2$ ball goes to zero as the dimension diverges. But the surface area of the $L\_{\infty}$ ball is $2d \to \infty$. So, we might wonder if there is a "happy medium" $p \in (2,\infty)$ where the surface area is constant in all dimensions. But I doubt it. **Does the surface area of the radius-$\tfrac{1}{2}$ $L\_p$ ball go to zero for all $p < \infty$**?
Asymptotics about the surface area of Lp balls are [apparently hard to pin down](https://mathoverflow.net/questions/234314/surface-area-of-an-ell-p-unit-ball). Tools from [this mathoverflow Q&A](https://mathoverflow.net/questions/293840/volume-ratio-of-general-ell-p-balls-and-surfaces) and linked paper might be useful.
| https://mathoverflow.net/users/29697 | Does the surface area of the unit Lp ball go to zero for all $p < \infty$? | The surface area of the unit $L^p$ ball (of radius $1$) goes to $0$ as $n$ goes to infinity, if $p<\infty$. Below I show it for $p>2$, but this is enough because if we have convex sets $A\subseteq B$, then the surface area of $B$ is bigger than that of $A$, which follows from the proof in [this MSE answer](https://math.stackexchange.com/a/58556/807670) to [Surface area of a convex set less than that of its enclosing sphere?](https://math.stackexchange.com/questions/58456/surface-area-of-a-convex-set-less-than-that-of-its-enclosing-sphere).
First note that the minimum value of $\sum\_{i=1}^n|x\_i|^p$ in the sphere $\mathbb{S}^{n-1}\subseteq\mathbb{R}^n$ is achieved when $x\_1=\dots=x\_n$. This can be deduced by setting $y\_i=x\_i^2$ and then $\sum y\_i^{p/2}$ is convex, so for any $y\_1,\dots,y\_n\geq0$, if $\sum y\_i=1$ then $\sum y\_i^{p/2}\geq n\left(\frac{1}{n}\right)^{p/2}$, with equality when $y\_1=\dots=y\_n$.
This implies that the ball $B$ of radius $n^\frac{p-2}{2p}$, which is tangent to the $L^p$ ball at the points $(\pm\frac{1}{n^{1/p}},\dots,\pm\frac{1}{n^{1/p}})$, contains the $L^p$ ball. So the $n-1$-volume of $\partial B$ is bigger than that of the $L^p$-ball, and we just have to prove that the area of $B$ goes to $0$ when $n\to\infty$.
Now, the area of $\mathbb{S}^{n-1}$ is $\frac{2\pi^{n/2}}{\Gamma(\frac{n}{2})}$, so the area of $\partial B$ is $A=\frac{2\pi^{n/2}}{\Gamma(\frac{n}{2})}\left(n^\frac{p-2}{2p}\right)^{n-1}$. So
$$\ln(A)=-\ln\left(\Gamma\left(\frac{n}{2}\right)\right)+n\frac{p-2}{2p}\ln(n)+O(n),$$
which by [Stirling's approximation](https://en.wikipedia.org/wiki/Stirling%27s_approximation#Stirling%27s_formula_for_the_gamma_function) is
$$-\frac{n}{2}\ln\left(\frac{n}{2}\right)+n\ln(n)\frac{p-2}{2p}+O(n)=-\frac{n}{2}\ln(n)+n\ln(n)\frac{p-2}{2p}+O(n),$$
which goes to $-\infty$ as $n$ tends to $\infty$, concluding the proof.
| 5 | https://mathoverflow.net/users/172802 | 439968 | 177,666 |
https://mathoverflow.net/questions/439279 | 9 | From discussions [1](https://mathoverflow.net/questions/377032/more-identities-for-the-lambert-w-function), [2](https://mathoverflow.net/questions/377012/an-identity-for-the-lambert-w-function), @HenriCohen wrote a paper on [Lambert $W$-Function Branch Identities](https://arxiv.org/pdf/2012.11698.pdf) which includes identities such as $$\sum\_\limits{k
\in\Bbb Z}\frac1{(W\_k(x)+1)^2}=\sum\_\limits{k
\in\Bbb Z}\frac1{(W\_k(x)+1)^3}=\frac1{xe+1}$$ by repeated differentiation of $\prod\_{k\in\Bbb Z}(1-t/W\_k(x))=e^{-t/2}-te^{t/2}/x$.
>
> Naturally, we ask: is $(m,n)=(2,3)$ the only solution to $$\sum\_\limits{k
> \in\Bbb Z}\frac1{(W\_k(x)+1)^m}=\sum\_\limits{k
> \in\Bbb Z}\frac1{(W\_k(x)+1)^n}$$ for positive integers $m<n$ and for all $x>-1/e$?
>
>
>
E.g. When $m>2$ and $n=m+1$, it suffices to show that equality does not hold when $x=2$ for instance, and it then suffices to show that $$\frac{d^m}{ds^m}\left(2m\log(e^{-2s}-s)+\frac{1+2s}{e^{-2s}-s}\right)\bigg\vert\_{s=-1/2}\ne0.$$
| https://mathoverflow.net/users/113397 | Is $(m,n)=(2,3)$ the only solution to $\sum_\limits{k \in\Bbb Z}\frac1{(W_k(x)+1)^m}=\sum_\limits{k \in\Bbb Z}\frac1{(W_k(x)+1)^n}$? | The answer is "yes":
Define the generating function
$$
G(x,t) = \sum\_{k \in \Bbb Z}\log\left[1- \frac{t}{W\_k(x)}\right].\tag{1}
$$
Then,
$$
\sum\_{k \in \Bbb Z} \frac 1{(W\_k(x)+1)^m}
= -\frac{1}{\Gamma(m)} \, G^{(0,m)}(x,-1),\tag{2}
$$
where $G^{(0,m)}(x,-1)$ denotes the $m$-th derivative w.r.t. $t$ at $t=-1$.
From the cited paper, (1) equals
$$
G(x,t) = \log(e^{-t/2} - t x^{-1} e^{t/2}),\tag{3}
$$
such that we need to calculate the $m$-th derivative of (3) at $t=-1$.
While this generates complicated rational polynomials, it is sufficient to expand them to first order around $x=0$ and look at the linear term, note that $G^{(0,m)}(0,-1)/\Gamma(m)=-1$ for $m>1$. We find
\begin{align}
\frac{1}{e\,\Gamma(m)}\,G^{(1,m)}(0,-1)
&= \frac{!m}{\Gamma(m)} = \frac{\Gamma(m+1,-1)}{e\,\Gamma(m)}\tag{4a}\\
&= 0, 1, 1, \tfrac{3}{2}, \tfrac{11}{6}, \tfrac{53}{24},
\tfrac{103}{40}, \tfrac{2119}{720}, \ldots,\tag{4b}
\end{align}
where $!m$ denotes the [subfactorial](https://mathworld.wolfram.com/Subfactorial.html) of $m$, and $\Gamma$ is the incomplete gamma function.
Note that
$$
\frac{!m}{m!}=\sum\_{k=0}^{m}\frac{(-1)^k}{k!}.\tag{5}
$$
So, only $m=2,3$ have the same constant and linear order near $x=0$.
| 5 | https://mathoverflow.net/users/90413 | 439969 | 177,667 |
https://mathoverflow.net/questions/439973 | 5 | For (hopefully) simplicity, let a premouse be defined coarsely as in Martin and Steel's 1994 paper, *Iteration Trees*.
Is (or is it consistent that) there is a premouse that is $\omega\_1$-iterable but not $(\omega\_1+1)$-iterable?
Note: if adding some extra conditions will get partial answers, that would also be very helpful.
| https://mathoverflow.net/users/9324 | Are there premice that are $\omega_1$-iterable but not $(\omega_1+1)$-iterable? | It is consistent (relative to large cardinals). There is an example given in Example 3.6 [here](https://arxiv.org/abs/2012.07185). For a brief summary: the model is the minimal proper class mouse $S$ such that $\mathbb{R}^S$ is closed under the $M\_1^\#$-operator. In particular, $M\_1^\#$ is a proper segment of $S$. And $S$ satisfies "$(M\_1,\delta^{M\_1})$ is a premouse and $(M\_1,\delta^{M\_1})$ is $\omega\_1$-iterable but not $(\omega\_1+1)$-iterable" (with these definitions in the sense of "Iteration trees"). In $S$, there is a tree $\mathcal{T}$ on $M\_1$, formed via a variant of self-genericity iteration, which has length $\omega\_1$ and no cofinal branch in $S$. (The usual sort of self-genericity iteration would be formed by iterating to make some initial segment of $S$ generic for the $\delta$-generator extender algebra. In the variant used, one also interweaves countable linear iterations at appropriate stages of the process, each of which "moves past the next instance of the $M\_1^\#$-operator on the $S$-sequence"; this ensures that $\delta(\mathcal{T})$ is a limit of $M\_1^\#$-operator instances on the $S$-sequence, which in turn ensures that $\delta(\mathcal{T})=\mathrm{lh}(\mathcal{T})=\omega\_1^S$.)
| 8 | https://mathoverflow.net/users/160347 | 439979 | 177,670 |
https://mathoverflow.net/questions/439976 | 11 | For a finite graph $X$, let $A\_X$ denote the associated right-angled Artin group. Thus $A\_X$ is generated by the vertices of $X$ subject to the relations $[v,w]=1$ whenever vertices $v$ and $w$ are connected by an edge.
I have seen references to the following theorem in several places, but I can neither figure out a proof myself not find a reference that proves it:
Let $X$ be a finite graph. Assume that $A\_X \cong G \times H$, where $G$ and $H$ are nontrivial groups. Then $X$ decomposes as a nontrivial join, i.e., we can partition the vertices of $X$ into two nonempty sets $V\_1$ and $V\_2$ such that each vertex in $V\_1$ is connected by an edge to each vertex in $V\_2$ and also each vertex in $V\_2$ is connected by an edge to each vertex in $V\_1$.
Can anyone provide me a proof or a reference for this?
| https://mathoverflow.net/users/498832 | Right-angled Artin groups that split as direct products | The place I've seen this is in Koberda's RAAG notes [here](https://users.math.yale.edu/users/koberda/raagcourse.pdf), see Corollary 2.15. This relies on the description of centralizers in Proposition 2.14, which is also proved in Behrstock and Charney's paper [here](https://link.springer.com/article/10.1007/s00208-011-0641-8) (if you want an officially-published reference). Given Proposition 2.14, Corollary 2.15 is immediate because if $A\_X\cong G\times H$ for non-trivial $G$ and $H$ then every non-trivial element $g$ has non-cyclic centralizer, so picking some $g$ that "really uses" every vertex of $X$, Proposition 2.14 implies $X$ is a non-trivial join.
| 8 | https://mathoverflow.net/users/164670 | 439992 | 177,673 |
https://mathoverflow.net/questions/439966 | 2 | In Beauville's "Variétés de Prym et jacobiennes intermédiaires", Proposition 3.5, it is claimed that $\textrm{Corr}(T)$ is torsion-free for a smooth projective variety $T$. Here $$\textrm{Corr}(T) := \textrm{Pic}(T \times T)/ (\pi\_1^\*\textrm{Pic}(T) \oplus\pi\_2^\*\textrm{Pic}(T))$$ Why is this true? A reference would suffice.
I can see why this is true for an Abelian variety $A = T$. It would also be enough to show that there is no $2$-torsion.
| https://mathoverflow.net/users/160814 | Reference for torsion-freeness of the group of correspondences on a smooth projective variety | It's not just torsion-free, we can actually compute it in terms of the Picard and Albanese variety of $T$, by the following classical result:
**Lemma.** *If $X$ and $Y$ are smooth projective varieties over a field $k$, then there is a canonical short exact sequence of group schemes*
$$0 \to \mathbf{Pic}\_X \times \mathbf{Pic}\_Y \to \mathbf{Pic}\_{X \times Y} \to \mathbf{Hom}\big(\mathbf{Alb}^1\_X,(\mathbf{Pic}^0\_Y)^\text{red}\big) \to 0.$$
Here, $(\mathbf{Pic}^0\_X)^\text{red}$ is an abelian variety; see for instance [FGA Explained, Thm. 9.5.4]. Its dual is denoted $\mathbf{Alb}^0\_X$, and without choosing a base point this gives a torsor $\mathbf{Alb}^1\_X$ under $\mathbf{Alb}^0\_X$ together with a morphism $X \to \mathbf{Alb}^1\_X$ that is initial in the category of maps from $X$ to torsors under abelian varieties. Finally, $\mathbf{Hom}$ denotes the (ind-étale) group scheme of morphisms of abelian varieties (or probably morphisms of pairs of an abelian variety with a torsor under it, up to translation ― once you choose a base point on $X$, this should be the same thing).
Torsion-freeness then follows since $\mathbf{Hom}$ schemes between abelian varieties are torsion-free.
For a proof of the lemma, see for instance my dissertation [vDdB, §4.4], although I claim no originality here! (That said, I also didn't find this precise statement in the literature...)
A different argument goes by choosing base points $x \in X(k)$ and $y \in Y(k)$ (first enlarge the field if necessary and worry about descent later). These give a section $\mathbf{Pic}\_{X \times Y} \to \mathbf{Pic}\_X \times \mathbf{Pic}\_Y$. To compute the kernel, consider a line bundle $\mathscr L$ on $X \times Y \times T$ for some $k$-scheme $T$ that is trivial along $x \times Y \times T$ and $X \times y \times T$. The second trivialisation gives a morphism $X \times T \to \mathbf{Pic}\_Y$ via the rigidified Picard functor, and the first says that the image lands in $\mathbf{Pic}\_Y^0$. This gives a morphism of $T$-schemes $X \times T \to \mathbf{Pic}\_Y^0 \times T$, which by the universal property of $\mathbf{Alb}^1\_X$ corresponds to a $T$-point of $\mathbf{Hom}(\mathbf{Alb}^1\_X,(\mathbf{Pic}^0\_Y)^\text{red})$. Now check that this doesn't depend on the choice of section and hence descends down to $k$.
---
**References.**
[FGA Explained] B. Fantechi, L. Göttsche, L. Illusie, S. L. Kleiman, N. Nitsure, A. Vistoli, [*Fundamental algebraic geometry: Grothendieck’s FGA explained*](https://bookstore.ams.org/surv-123-s). Mathematical Surveys and Monographs **123**. American Mathematical Society, 2005. [ZBL1085.14001](https://zbmath.org/?q=an:1085.14001).
For Kleiman's chapter, see also arXiv:[math/0504020](https://arxiv.org/abs/math/0504020)
[vDdB] R. van Dobben de Bruyn, [*Dominating varieties by liftables ones*](https://webspace.science.uu.nl/%7Edobbe012/doc/Dominating%20varieties%20by%20liftable%20ones.pdf). PhD thesis, Columbia University, 2018.
| 2 | https://mathoverflow.net/users/82179 | 439994 | 177,674 |
https://mathoverflow.net/questions/439972 | 11 | I am interested in seeing examples of research problems which fall into one of the two following categories:
1. A problem which is solved in the case of primes (or prime powers), but which remains open in the case of composite integers.
2. A problem which historically was first solved for primes, and then significant additional work was needed to prove the result for all integers.
This is a rather broad question, so I'm mainly interested in combinatorial or algebraic problems that appear to be easy over primes because of the existence of certain structures which exist for primes, but not for all integers (e.g., for any prime power $q$, there is a finite field $\mathbb{F}\_q$ of size $q$, but there is no finite field of size $6$).
An example of the sort of problem I'm looking for in the first category is *tesselations of integers*. We say a finite set $A$ of integers is a "tile" if there exists an infinite set of integers $X$ such that for every integer $n$, the equation $n = a +x$ has exactly one solution $(a,x)\in A\times X$. In the 70s, [Newman](https://www.sciencedirect.com/science/article/pii/0022314X77900543?via%3Dihub) showed that for any prime power $q$, there is a simple characterization for the set of tiles $A$ of size $q$. [Additional](https://arxiv.org/abs/math/9802122) [work](https://arxiv.org/abs/math/0109127) has characterized the possible tile sets $A$ when $|A|$ has very few distinct prime factors, for example, but in general characterizing the possible tile sets of size $n$ for an arbitrary integer $n$ remains open.
| https://mathoverflow.net/users/138628 | What are examples of problems we know how to solve for primes (or prime powers), but not for composites? | There is a projective plane of order $N$ for every prime power $N$. The existence of projective planes of other orders is an open question; in particular, it is not known whether there is a projective plane of order $12$. See, e.g., <https://en.wikipedia.org/wiki/Projective_plane#Finite_projective_planes>
| 14 | https://mathoverflow.net/users/3684 | 439999 | 177,676 |
https://mathoverflow.net/questions/439983 | 1 | After the GNS representation for $C^{\*}$-algebras is presented in Thirring's book [Quantum mathematical physics](https://doi.org/10.1007/978-3-662-05008-8), the author states the following theorem.
>
> **The Spectral Theorem:** For any given Hermitian (self-adjoint) element $a$ of a $C^{\*}$-algebra $A$, every representation of $A$ is equivalent to a representation $\mathscr{H} = \bigoplus\_{i}\mathscr{H}\_{i}$, for which $\mathscr{H}\_{i} = L^{2}(\sigma(a),d\mu\_{i})$ and $\pi(a)\vert\_{\mathscr{H}\_{i}}: \varphi(\alpha) \mapsto \alpha \varphi(\alpha)$. In this representation, $a$ acts as a multiplication operator.
>
>
>
I want to understand this theorem, but I did not follow Thirring's arguments which led to its proof. The argument uses the GNS construction: $A$ is a $C^{\*}$-algebra with unit and $\omega$ a state, there exists a representation $\pi\_{\omega}: A \to \mathscr{B}(\mathscr{H})$, where $\mathscr{H}$ is just the completion of $A/J$, $J$ being the left ideal defined by the set of $a \in A$ such that $\omega(a^{\*}a) = 0$. In his notation, $\pi\_{\omega}(a): b \mapsto ab$.
>
> By the axiom of choice, we can choose $b\_{i} \in \mathscr{H}\_{i} \equiv $ the completion of the sets of linear combinations of $a^{n}b\_{i}$, $n=0,1,\dotsc$ spans all of $\mathscr{H}$. Each $\mathscr{H}\_{i}$ provides a representation of the (Abelian) $C^{\*}$-algebra generated by $a$ and has $b\_{i}$ as a cyclic vector.
>
>
>
I really don't follow the argument. $\mathscr{H}\_{i}$ spans $\mathscr{H}$ in which sense? Is it always possible to find $b\_{i}$ such that $\mathscr{H}\_{i}$ spans $\mathscr{H}$? Is the index $i$ countable, uncountable? $\mathscr{H}\_{i}$ is a Hilbert space, so does a representation between $A$ and $\mathscr{H}\_{i}$ always exist?
Can someone help with these arguments? Maybe giving more details at each step? Or maybe providing a reference in which the theorem is proved more carefully?
| https://mathoverflow.net/users/152094 | GNS Representation — A theorem from Thirring’s book | The C${}^\*$-algebra $A$ is a red herring here. All the result is really saying is that if $T$ is a self-adjoint operator on a Hilbert space $H$ then we can find a family of measures $\mu\_i$ on $\sigma(T)$ and an isomorphism $H \cong \bigoplus L^2(\sigma(T), d\mu\_i)$ which takes $T$ to the operator of multiplication by $x$ on each $L^2(\sigma(T), d\mu\_i)$. Not my favorite version of the spectral theorem on aesthetic grounds, but it is easy to use.
To prove this, let $v$ be any nonzero vector in $H$ and let $A\_0$ be the C${}^\*$-algebra generated by $T$. Then $A\_0$ is abelian and by Gelfand's theorem it is isomorphic to $C(\sigma(T))$. Thus the map $\omega: S \mapsto \langle Sv, v\rangle$ is a positive linear functional on $A\_0$ of norm at most $\|v\|^2$, so it is given by $\omega(f(T)) = \int\_{\sigma(T)} f\, d\mu$ for some positive measure $\mu$ on $\sigma(T)$. Now $A\_0v = \{f(T)v: f \in C(\sigma(T))\}$ is a (not necessarily closed) subspace of $H$ and one can check that the map $f(T)v \mapsto f$ is an isometry from $\overline{A\_0v}$ onto $L^2(\sigma(T), d\mu)$. Also the restriction of $T$ to $\overline{A\_0v}$ is taken to the operator of multiplication by $x$ on $L^2(\sigma(T), d\mu)$.
Now $\overline{A\_0v}$ is invariant for $T$, and since $T$ is self-adjoint so is its orthocomplement $H \ominus \overline{A\_0v}$. So we can now choose another nonzero vector $w$ in the orthocomplement of $\overline{A\_0v}$, and keep going until there is nothing left in the orthocomplement. This argument can be made rigorous using Zorn's lemma, or just a simple transfinite induction.
If $H$ is nonseparable this is going to require uncountably many summands, since each summand is separable. If $H$ is separable then obviously there will only be countably many summands.
If I recall correctly, this version of the spectral theorem is given a simple proof, probably essentially the same as the one I just gave, in Reed and Simon vol. 1.
| 2 | https://mathoverflow.net/users/23141 | 440000 | 177,677 |
https://mathoverflow.net/questions/440010 | 3 | All manifolds will be assumed to be closed, oriented, and connected.
Let $f\colon M\to M$ be a map of degree $\pm 1$. It is not hard to show that $\pi\_1(f)$ is surjective.
>
> What is an example of a *non* $\pi\_1$-injective, degree one, self-map of a three-manifold?
>
>
>
If $\dim(M) = 2$, then $\pi\_1(f)$ is injective because the fundamental group of any surface is residually finite, and any finitely generated residually finite group is Hopfian.
If $\dim(M) = 3$ and if $M$ is either a spherical, or hyperbolic, or Haken manifold, then again $\pi\_1(M)$ is Hopfian. Thus $\pi\_1(f)$ is injective.
A related conjecture is the [Hopf conjecture](https://en.wikipedia.org/wiki/Hopf_conjecture#Self-maps_of_degree_1).
| https://mathoverflow.net/users/363264 | Example of a non $\pi_1$-injective, degree one, self-map of a three-manifold | In the book [Three-manifold groups](https://arxiv.org/abs/1205.0202) we find a giant flow chart showing what is known to follow from the assumption that $N$ is an irreducible, compact, orientable three-manifold with empty or toroidal boundary (such that $\pi\_1(N)$ is neither finite nor solvable). As one particular consequence, such manifolds are Hopfian.
Manifolds with finite fundamental group are Hopfian, so those are dealt with. Three-manifolds with solvable fundamental group are toroidal, thus Haken, and so dealt with. Three-manifolds which are reducible have fundamental group being a free product. However, residually finiteness passes is preserved under taking free products. So reducible manifolds are dealt with. (See, for example, Hempel's paper [Residual finiteness for three-manifold groups](https://www.degruyter.com/document/doi/10.1515/9781400882083-018/html).)
Thus all closed, oriented, and connected three-manifolds have fundamental group being Hopfian.
---
Further references: [a mathoverflow question](https://mathoverflow.net/questions/61661)
| 6 | https://mathoverflow.net/users/1650 | 440015 | 177,680 |
https://mathoverflow.net/questions/440012 | 3 | In Squier's short, yet influential, paper about the Burau representation, he made two conjectures that might have provided a proof for the faithfulness of the Burau representation (which we now know to be false by work of Bigelow, for instance).
**Intro.** The (reduced) Burau representation $\beta$ sends the Artin generators $\sigma\_i$ for the braid group $B\_n$ to the $(n-1)\times(n-1)$ matrix over the Laurent polynomials $\mathbb Z[t^\pm]$ given by
$$\beta(\sigma\_i) = \operatorname{Id}\_{i-2} \oplus \begin{pmatrix} 1 & 0 & 0 \\ t & -t & 1 \\ 0 & 0 & 1 \end{pmatrix} \oplus \operatorname{Id}\_{n-i-2}$$
with a slight, easy modification in the $i=1$ and $i=n-1$ cases when this formula doesn't make sense. The powers of $\sigma\_i$ have a nice formula that ends up being a bit cleaner to write if we substitute $t=-q$. We get
$$\beta(\sigma\_i^d) = \operatorname{Id}\_{i-2} \oplus \begin{pmatrix} 1 & 0 & 0 \\ -q(1+q+\dotsb+q^{d-1}) & q^d & (1+q+\dotsb+q^{d-1}) \\ 0 & 0 & 1 \end{pmatrix} \oplus \operatorname{Id}\_{n-i-2}$$
**Squier's conjecture.** In case $q$ "specialized" to be a primitive $d$-th root of unity, you can see that $\beta(\sigma\_i)$ has order $d$. Squier conjectured the following:
>
> (C1) The kernel of the composite map $\beta\_{-q}:B\_n \xrightarrow{\beta} \operatorname{GL}\_{n-1}(\mathbb Z[t^\pm]) \xrightarrow{t \mapsto -q} \operatorname{GL}\_{n-1}(\mathbb C)$, when $q$ is a primitive $d$-th root of unity, is exactly equal to the normal subgroup $\langle\langle \sigma\_i^d \rangle\rangle$.
>
>
>
**My issue.** Consider the full twist braid $\Delta^2=(\sigma\_1\dotsb\sigma\_{n-1})^n$ in $B\_n$. You can show that $\beta(\Delta^2)=t^n\cdot \operatorname{Id}\_{n-1}$. What power of $\Delta^2$ lies in $\ker \beta\_{-q}$? Consider the case $n=3$, $d=6$. Then $\beta(\Delta^2) = t^3 \mapsto -q^3 = 1$ under the specialization at $-q$, so $\Delta^2 \in \ker \beta\_{-q}$. But $\Delta^2 \notin \langle\langle \sigma\_i^6 \rangle\rangle$! (For instance, forgetting one strand sends eleemnts of $\langle\langle \sigma\_i^6 \rangle\rangle$ to 6-th powers in $B\_2$, but $\Delta^2$ maps to $\sigma\_i^2$.) It shouldn't be hard to find lots of examples like this, powers of $\Delta^2$ that easily lie in $\ker \beta\_{-q}$ but don't lie in $\langle\langle \sigma\_i^d \rangle\rangle$.
**Question.** Did Squier just overlook this seemingly simple counterexample to his conjecture? Does the conjecture seem more reasonable if we replace $\langle\langle \sigma\_i^d \rangle\rangle$ with $\langle \Delta^{2k} \rangle\langle\langle \sigma\_i^d \rangle\rangle$ for some power $k$ depending on $d$?
*Squier, Craig C.*, [**The Burau representation is unitary**](http://dx.doi.org/10.2307/2045338), Proc. Am. Math. Soc. 90, 199-202 (1984). [ZBL0542.20022](https://zbmath.org/?q=an:0542.20022).
| https://mathoverflow.net/users/151664 | Squier's conjecture on Burau at roots of unity | Presumably, this conjecture appears from the observation that the suitable powers of the standard generator $\sigma\_1$ lies in the kernel, although it looks that Squier missed other obvious elements of kernels as you mentioned.
I mention the paper by Funar and Kohno
On Burau’s representations at roots of unity.
<https://link.springer.com/article/10.1007/s10711-013-9847-0>
where they proved a strengthened form of another Squier's conjecture (C2) that appeared in the same Squier's paper. The paper also contains several discussions and results on Squire's conjectures.
It seems that a more suitable formulation of Squier's conjecture is as follows.
For $N=(n\_1,\ldots,n\_{k-1})$, let $B\_k\{N\}$ be the subgroup generated by
$\sigma\_1^{2n\_1}, (\sigma\_1\sigma\_2)^{3n\_2}, \ldots, (\sigma\_1\sigma\_2\cdots \sigma\_{k-1})^{kn\_{k-1}}$.
Geometrically, the group $B\_k\{n\}$ is the group generated by $n\_i$-th powers of the Dehn twists along simple closed curves in $n$-punctured disk $D\_n$
enclosing $i+1$ puncture points.
When $N=\{n,n,n,\ldots\}$, we simply write $B\_k\{N\}$ by $B\_k\{n\}$.
**Modified Squier's conjecture, weak version**
If $q$ is a primitive root of unity then the kernel of Burau representation $\beta\_{-q}$ is equal to $B\_k\{N\}$ for suitable $N$.
**Modified Squier's conjecture, strong version**
If $q$ is a primitive root of unity then the kernel of Burau representation $\beta\_{-q}$ is equal to $B\_k\{n\}$ for some $n$.
If the Modified conjecture (strong version) is true for infinitely many distinct primitive root of unity, then the Burau representation is faithful.
This follows from Theorem 2.1 of Funar-Kohno's paper that asserts that the intersection of $B\_k\{n\}$ over an infinite set of integers $n$ is trivial.
Thus for $k>4$, the Modified conjecture (strong version) can be true for only finitely many primitive root of unities.
On the other hand, Corollary 3.2 of Funar-Kohno's paper shows that if $q$ is a primitive $2n$-th root of unity for odd $n$ and $2n\geq7$, then modified conjecture (strong version) is true:
Corollary 3.2 says that the kernel $K$ of the Burau representation of 3-braids at $-q$ is normally generated by $\sigma\_1^{2n}, \sigma\_2^{2n} ,(\sigma\_1^{2}\sigma\_2^{2})^{n}$. Since
$ (\sigma\_1\sigma\_2)^{3n}=(\sigma\_1^{2}(\sigma\_2\sigma\_1^{2}\sigma\_2))^{n} = \sigma\_1^{2n} (\sigma\_2 \sigma\_1^{2}\sigma\_2)^{n}$
and $(\sigma\_2 \sigma\_1^{2}\sigma\_2)^{n}$ is conjugate to $(\sigma\_1^{2}\sigma\_2^{2})^{n}$, the kernel $K$ coincides with $B\_3\{n\}$.
These results say that the modified conjecture (weak or strong version) is meaningful and subtle for $k=4$ case. I feel it reasonable to investigate the modified conjectures as a possible approach of the faithfulness problem of Burau representation of 4-braids.
| 5 | https://mathoverflow.net/users/193957 | 440017 | 177,682 |
https://mathoverflow.net/questions/439932 | 4 | The question in the title seems a natural one to ask, and I suspect that it is already considered, even completely solved, somewhere in the literature. Although I prefer some explanation of the idea(s), relevant references are also appreciated.
| https://mathoverflow.net/users/40789 | Is every Riemannian metric conformally equivalent to one that is geodesically complete? | This is not hard to prove. The idea of a proof is as follows. Take any metric $g$ on a manifold $M$. Define $d:M\to \mathbb{R}\_+$ by saying that $d(x)$ is the infimum of all lengths of curves $\gamma:[0,b)\to M$ with $\gamma(0)=x$, such that $\gamma$ is proper (In particular: For every compact subset K\subset M, there is a $t\in [0,b)$ with $\gamma(t)\not\in K$). Choose a smooth function $f:M\to (0,1]$ with $0<f<d$. Then $\bar g:= f^{-2}g$ is complete.
There is much more possible, namely bounded geometry. This was published in
Müller, Olaf; Nardmann, Marc;
Every conformal class contains a metric of bounded geometry.
Math. Ann. 363 (2015), no. 1-2, 143–174.
| 6 | https://mathoverflow.net/users/110127 | 440031 | 177,688 |
https://mathoverflow.net/questions/440035 | 6 | Let $A$ be an $n \times n$ invertible complex matrix. Let $Gr(k)=Gr(k,\mathbb{C}^n)$ be the complex $k$-Grassmannian, $1\leq k \leq n$. Since $A$ is invertible, it maps a $k$-dimensional subspace to a $k$-dimensional subspace, so it gives a function (which I'll call $A\_k$) on $Gr(k)$. The fact that matrices have eigenvalues lets us deduce that $A$ has a $k$-dimensional invariant subspace - i.e. this map $A\_k$ has a fixed point.
My question is this: is there a fixed point theorem on $Gr(k)$ (which ideally does not in any way depend on the existence of eigenvalues) that we can invoke to arrive at the same conclusion?
| https://mathoverflow.net/users/322473 | Invariant subspaces for matrices via fixed points on Grassmannians | I think the Lefschetz fixed point theorem still applies. If a self-map M→M of a compact orientable manifold M has no fixed points than the Euler characteristic of M is zero. But if M is a complex Grassmannian then its odd Betti numbers vanish so the Euler characteristic is positive (since b\_0=1).
| 7 | https://mathoverflow.net/users/14830 | 440036 | 177,689 |
https://mathoverflow.net/questions/439996 | 7 | This is a curiosity question that came out of teaching abstract algebra.
Let $F$ be a field, and $n>1$ an integer.
Let $F^{n \leq n}$ be the $F$-algebra of all upper-triangular $n\times n$-matrices $\begin{pmatrix} a\_{1,1} & a\_{1,2} & \cdots & a\_{1,n} \\ 0 & a\_{2,2} & \cdots & a\_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a\_{n,n} \end{pmatrix}$ with entries in $F$.
Let $J$ be the subset of $F^{n \leq n}$ that consists of all such matrices that satisfy $a\_{i,j} = 0$ for all $\left(i,j\right) \neq \left(1,n\right)$. In other words, $J$ is the set of all $n\times n$-matrices whose only nonzero entry (if any) is in the northeasternmost corner. It is easy to see that $J$ is an ideal of $F^{n \leq n}$. Thus, a quotient $F$-algebra $F^{n \leq n} / J$ is defined (and its elements can be thought of as upper-triangular matrices whose northeasternmost entry is undetermined).
>
> **Question.** What is the smallest $m$ such that there is an injective $F$-algebra homomorphism from $F^{n\leq n} / J$ to the matrix ring $F^{m\times m}$? In other words, what is the smallest dimension of a faithful representation of the $F$-algebra $F^{n \times n} / J$?
>
>
>
My suspicion is that it is $2n-2$. Indeed, it is certainly $\leq 2n-2$, since there is an injective $F$-algebra homomorphism $F^{n\leq n} / J \to F^{\left(2n-2\right)\leq \left(2n-2\right)}$ that sends the residue class of a matrix $\begin{pmatrix} a\_{1,1} & a\_{1,2} & \cdots & a\_{1,n} \\ 0 & a\_{2,2} & \cdots & a\_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a\_{n,n} \end{pmatrix}$ to $\begin{pmatrix} a\_{1,1} & a\_{1,2} & \cdots & a\_{1,n-1} & 0 & 0 & \cdots & 0 \\ 0 & a\_{2,2} & \cdots & a\_{2,n-1} & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a\_{n-1,n-1} & 0 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 & a\_{2,2} & a\_{2,3} & \cdots & a\_{2,n} \\ 0 & 0 & \cdots & 0 & 0 & a\_{3,3} & \cdots & a\_{3,n} \\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & a\_{n,n} \end{pmatrix}$.
For $n = 3$, this is an embedding $F^{3 \leq 3} / J \to F^{4 \leq 4}$, and I think I have convinced myself by a long and messy argument that no embedding $F^{3 \leq 3} / J \to F^{3 \leq 3}$ exists, but this doesn't rule out an embedding $F^{3 \leq 3} / J \to F^{3 \times 3}$ into arbitrary (rather than triangular) matrices.
The whole thing originated in [my attempts to illustrate the variety (common-sense meaning) of quotient rings](https://www.cip.ifi.lmu.de/%7Egrinberg/t/23wa/23wa.pdf). Specifically, I was trying to show that not every quotient ring is just a subring in disguise. The easiest example for this is $\mathbb{Z} / n\mathbb{Z}$, but I was looking for something less obvious.
| https://mathoverflow.net/users/2530 | Smallest faithful representation of an upper-triangular matrix quotient | Here's an elementary proof that $2n-2$ is a lower bound.
Suppose that
$$V\_1\xrightarrow{\alpha\_1}V\_2\xrightarrow{\alpha\_2}\dots\xrightarrow{\alpha\_{n-2}}V\_{n-1}\xrightarrow{\alpha\_{n-1}}V\_n$$
is a representation of the linearly ordered $A\_n$ quiver $Q$ that is a representation of $FQ/I$, where $I$ is the one-dimensional ideal of the path algebra $FQ$ generated by the longest path, so that $\alpha\_{n-1}\dots\alpha\_1=0$, and that it is faithful as a representation of $FQ/I$.
By faithfulness, $\alpha\_{n-2}\dots\alpha\_1(v\_1)\neq0$ for some $v\_1\in V\_1$. For $1<i<n$ let $v\_i=\alpha\_{i-1}\dots\alpha\_1(v\_1)\in V\_i$.
Also by faithfulness, $\alpha\_{n-1}\dots\alpha\_2(w\_2)\neq0$ for some $w\_2\in V\_2$. For $2<i\leq n$ let $w\_i=\alpha\_{i-1}\dots\alpha\_2(w\_2)\in V\_i$.
For $1<i<n$, $v\_i$ and $w\_i$ are linearly independent (consider their images in $V\_{n-1}$ and $V\_n$), so $v\_1,\dots,v\_{n-1},w\_2,\dots,w\_n$ span a ($2n-2$)-dimensional subrepresentation (which is isomorphic to the representation described in the OP).
| 4 | https://mathoverflow.net/users/22989 | 440046 | 177,695 |
https://mathoverflow.net/questions/439995 | 1 | Let $k$ be a field, $m$ be a positive integer and $R$ be the subring $k[x,xy,xy^2,…,xy^m]$ of the polynomial ring $k[x,y]$. Let $B$ be the quotient ring $R/xR$. Then $B$ is the finitely generated $k$-algebra $k[xy,xy^2,…,xy^m]$.
How do I compute the associate reduced ring for the $k$-algebra $B$? Is this reduced ring always isomorphic to the polynomial ring $k[t]$?
I have verified that this associated reduced ring for $B$ is isomorphic to the polynomial ring $k[t]$ for the cases $m=1,2$.
What I did was to find all the nilpotent elements in the $k$-algebra generators for $B$ and take the ideal in $B$ generated by these elements as the nilradical of $B$. Then I took the quotient ring of $B$ by its nilradical to compute the reduced ring associated to $B$.
But I am not sure about the general case. It seems to me that in the general case, there are many relations to be considered when the quotient of $B$ by its nilradical is taken. This makes the determination of the isomorphism type of the associated reduced ring of $B$ challenging.
Could someone please help me with this problem? Thank you so much for your kind help.
| https://mathoverflow.net/users/477848 | How to compute the associated reduced ring for this finitely generated algebra? | In $R/xR$, all $xy^l, l<m$ are nilpotent. To see this, notice that $(xy^l)^m= x^{m-l}(xy^m)^l$. Thus, modulo nilpotents, $R/xR$ is generated by the single element $xy^m$ over $k$. The rest is clear.
| 1 | https://mathoverflow.net/users/9502 | 440053 | 177,698 |
https://mathoverflow.net/questions/440048 | 0 | Consider complex smooth hypersurfaces $X\subset\mathbb{P}^n$ and $Y\subset\mathbb{P}^m$ for $m,n\geq 4$ and a morphism $f\colon X\rightarrow Y$ which satisfies one of the properties
1. $f\_\*\mathcal{O}\_X\cong\mathcal{O}\_Y$ and $f\_\*\mathcal{O}\_X(1)\cong\mathcal{O}\_Y(1)$
2. $f\_\*\mathcal{O}\_X(i)\cong\mathcal{O}\_Y(i)$ for $i=0,1,\dots,b$ for some $b>1$
3. $f\_\*\mathcal{O}\_X(i)\cong\mathcal{O}\_Y(i)$ for all $i\in\mathbb{Z}$
Here $\mathcal{O}\_X(1)$ and $\mathcal{O}\_Y(1)$ are the natural polarizations induced from $\mathbb{P}^n$ and $\mathbb{P}^m$ respectively.
Do we have (1) $\Rightarrow$ (2) $\Rightarrow$ (3)? In the case (3), can we say that $f\_\*:D^b(X)\rightarrow D^b(Y)$ induces an isomorphism of the full subcategory of line bundles?
| https://mathoverflow.net/users/nan | morphisms between smooth hypersurfaces that preserve many line bundles | Your assumption about $f$ is very strong, it implies that $X \cong Y$ and $f$ is an isomorphism.
Indeed, let $Z \subset X$ be a general fiber of $f$. Then the assumption that $f\_\*\mathcal{O}\_X(1)$ is a line bundle implies that
$$
\dim H^0(Z, \mathcal{O}\_X(1)\vert\_Z) = 1.
$$
But $\mathcal{O}\_X(1)\vert\_Z$ is a very ample line bundle on $Z$, hence $Z$ is a single point. Thus, $f$ is birational. If it is not an isomorphism, the relative Picard group must be nontrivial, but since
$$
\mathrm{Pic}(X) = \mathrm{Pic}(Y) = \mathbb{Z},
$$
this is impossible, hence $f$ is an isomorphism.
| 4 | https://mathoverflow.net/users/4428 | 440054 | 177,699 |
https://mathoverflow.net/questions/440033 | 3 | Given a partition of the edges of $K\_{n,n}$ into $n$ colours, where each colour appears exactly $n$ times, prove that there exists a vertex incident to at least $\sqrt{n}$ colours.
| https://mathoverflow.net/users/497926 | For every partition of $E(K_{n,n})$ into $n$ colour classes of size $n$, there is a vertex incident to at least $\sqrt{n}$ colours | Here is a short proof. Thanks to David Speyer for simplifying an earlier proof of mine (see the comments below).
For each colour $i \in [n]$, let $a\_i$ be the number of vertices incident to an edge of colour $i$. Observe that $a\_i \geq 2\sqrt{n}$ for all $i$, where equality is obtained if and only if the edges of colour $i$ induce a $K\_{\sqrt{n}, \sqrt{n}}$. Thus, $\sum\_{i \in [n]} a\_i \geq 2n \sqrt{n}$. On the other hand, $\sum\_{i \in [n]} a\_i$ is the number of ordered pairs $(v,i)$, where $v$ is a vertex, and $i$ is a colour incident to $v$. Therefore, since there are only $2n$ vertices, some vertex must be incident to at least $\sqrt{n}$ colours.
| 4 | https://mathoverflow.net/users/2233 | 440058 | 177,702 |
https://mathoverflow.net/questions/440069 | 1 | Let $X$ be a closed Riemannian manifold and consider the function $f\_n : X \times \cdots \times X \to \mathbb{R}$ where the domain of $f\_n$ is the $n$-fold cartesian product of $X$ and where $f\_n(p\_1,...,p\_n) = \sum\_{i \neq j} d(p\_i, p\_j)$ where $d$ is the distance function on $X$. The function $f\_n$ is invariant under the action of the isometry group $G = \text{Isom}(X)$ on the $n$-fold product $X^n$ as well as the action of the symmetric group on $n$ letters $S\_n$ acting on $X^n$ by permuting the factors
>
> Are the maxima of $f\_n$ on $X^n / (G \times S\_n)$ isolated?
>
>
>
As an example, this is the case for $X = S^1$ where the maxima is unique and achieved by the regular $n$-gon. My initial inspiration was that I was considering this question for $S^2$ with the round metric (where I have a feeling the answer is well known and I suspect the maximum is unique, but I do not know and would appreciate a reference if anyone knows -- [here](https://www.jstor.org/stable/2006132) as an older paper of Berman and Hanes with some estimates of the maximum value in this case which have subsequently been improved upon).
| https://mathoverflow.net/users/419791 | Isolated maxima for sum of distances of points on a manifold | Not necessarily. Consider the sphere $(\mathbb{S}^2,g)$ with its usual metric and give it a new metric $hg$, where $h\leq1$, $h$ has three local minima $h(p\_1)=0.7,h(p\_2)=0.8,h(p\_3)=0.9$ (where $p,q,r$ are close to one another) and $h=1$ outside very small neighborhoods of $p\_1,p\_2,p\_3$. Then Isom$(\mathbb{S}^2,hg)=\{\text{Id}\}$, but $f\_2$ has the same maximum as in the usual metric, and this maximum is achieved in uncountably many points of $(\mathbb{S}^2)^2$, so they are not isolated.
| 1 | https://mathoverflow.net/users/172802 | 440070 | 177,705 |
https://mathoverflow.net/questions/440081 | 3 | Let $S$ be a separable irreducible Noetherian scheme and let $X$ be a projective smooth curve over $S$. Let $\mathcal F$ be a coherent sheaf on $X$ which is flat over $S$. Suppose the restriction $\mathcal F\mid\_{X\_s}$ of $\mathcal F$ on the fiber $X\_s$ is locally free for some point $s\in S$.
**Question:** is $\mathcal F$ also locally free?
| https://mathoverflow.net/users/11750 | Is a coherent and flat sheaf locally free? | As pointed out in the comments, being a vector bundle at a point is (by definition) an open property and for coherent sheaves flat over a base, being a vector bundle at a point can be checked on fibers. This means,
**Prop** Let $f : X \to Y$ be a flat map of schemes and $\mathcal{F}$ a coherent $\mathcal{O}\_{X}$-module flat over $Y$. Suppose that $\mathcal{F}|\_{X\_y}$ is a vector bundle on $X\_y$ for some $y$. Then there is an open $U \subset X$ containing $X\_y$ such that $\mathcal{F}|\_U$ is a vector bundle.
If $f : X \to Y$ is flat and proper and $\mathcal{F}$ is coherent and flat over $Y$, this implies there is actually an open of the base over which $\mathcal{F}$ is a vector bundle.
Now, in your question I take that $X \to S$ is a *relative curve*. Then there exist degenerations of a vector bundle to a non-locally-free coherent sheaf on a family $X \to S$ which is as nice as possible.
Indeed consider
Let $\pi\_1 : X = \mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^1 = S$ be the projection. Let $x = X$ be a point and $\mathcal{I} \subset \mathcal{O}\_{X}$ the ideal sheaf of $x = (0,0) \in X$. For each fiber $X\_t$ with $t \neq 0$ we have $\mathcal{I}|\_{X\_t} = \mathcal{O}\_{X\_t}$ is a vector bundle. However, $\mathcal{I}$ is not a vector bundle. I claim that $\mathcal{I}$ is $\pi\_1$-flat. This is clear on $X \setminus \{ x \}$ so I we consider the local structure around $x$. On a dense open we have the following algebra problem,
$$ A = k[x]\_{(x)} \to k[x,y]\_{(x,y)} = B \quad \text{ with the ideal } \quad I = \mathfrak{m}\_B = (x,y) \subset k[x,y]\_{(x,y)} $$
I claim that $I$ is flat over $A$. There is an exact sequence,
$$ 0\to B \xrightarrow{(y \; -x)} B^2 \xrightarrow{(x \; y)} I \to 0 $$
Then applying <https://stacks.math.columbia.edu/tag/00MK> we just need to show that $B/\mathfrak{m}\_A B \to (B / \mathfrak{m}\_A B)^2$ is injective which is true because $y$ is a non zero-divisor on $B / \mathfrak{m}\_A B$. Thus $I$ is $A$-flat. Furthermore, we get the local structure,
$$ I / \mathfrak{m}\_A I \cong k \oplus k[y]\_{(y)} $$
but its image in $B / \mathfrak{m}\_A B$ is just $(y)$ which is locally free. This we see that $\mathcal{I}|\_{X\_0} \cong \mathcal{O}\_{X\_0}(-1) \oplus \iota\_\* k$ which has degree zero as it must because $\mathcal{I}|\_{X\_t} \cong \mathcal{O}\_{X\_t}$ for $t \neq 0$ and degree is constant in flat families.
Another example is given by considering the relative cotangent bundle of the flat projective family,
$$ f : X = \mathrm{Proj}{(k[t][X,Y,Z]/(XY - t Z^2))} \to \mathrm{Spec}{(k[t])} = S $$
| 2 | https://mathoverflow.net/users/154157 | 440094 | 177,715 |
https://mathoverflow.net/questions/440102 | 4 | Fix a group $G$ and fix a presentation of $G$ as $\langle X\mid R\rangle$. A *presentationally finite* extension of $G$ is any group that can be presented as $H=\langle X\cup X'\mid R\cup R'\rangle$, where $X',R'$ are finite. (Mind, the natural homomorphism $G\to H$ may not be injective. If necessary, we can restrict ourselves to cases where it is.)
For example, the presentationally finite extensions of the trivial group are the finitely presentable groups.
>
> **Question**: Is there an existing name for this kind of group homomorphism/extension?
>
>
>
Unless I've missed something, it's not very hard to see that this does not depend on the choice of presentation of $G$ (sketch: adding extraneous generators to $X$ and the corresponding generation relations to $R$ does not change the isomorphism type (over $G$) of $\langle X\cup X'\mid R\cup R'\rangle$, neither does adding any relations that follow from the existing ones; thus, we can just add all elements of $G$ plus any number of their copies we want, and all relations true in $G$ about them and then remove any redundant ones we do not want to obtain all possible presentations).
Thus, this is really a property of the map $G\to H$ (or the extension $G\leq H$ if the map is injective) and not of a particular presentation.
I'm mostly interested in the case of finitely generated groups, but I think the idea works just as well for arbitrary groups, of arbitrary cardinality.
(Another way to describe this is simply as the groups of the form $H=(G\*F)/N$, where $F$ is a free group of finite rank and $N$ is finitely generated as a normal subgroup of $G\*F$, such that $N\cap G$ is trivial if we demand that the map $G\to H$ actually be injective.)
| https://mathoverflow.net/users/54415 | Presentationally finite group "extensions" | The kind of presentation you are giving is called a relative presentation and has been studied extensively by Steve Pride and his coauthors. So relatively finitely presented over $G$ is probably the best name (I didn't search his papers to see if he uses this term.)
| 6 | https://mathoverflow.net/users/15934 | 440105 | 177,721 |
https://mathoverflow.net/questions/440040 | 8 | The categories of modules over commutative rings are especially notable Abelian categories. Wanting to extend this class a bit, I thought of this question:
Let $R$ be a commutative ring with $1$. Does there exist a Grothendieck topos $E$ such that $\mathrm{Ab}(E) \simeq R\text{-}\mathrm{Mod}$ (equivalence of categories)?
See the related question I was able to find: [Is every Grothendieck category with a generator a category of sheaves?](https://mathoverflow.net/questions/333955/is-every-grothendieck-category-with-a-generator-a-category-of-sheaves)
| https://mathoverflow.net/users/148161 | Is the category of modules over a commutative ring the category of abelian objects in a topos? | **No — in particular, not if $R$ has any non-integer rationals.**
Briefly: From $\newcommand{\Z}{\mathbb{Z}}\newcommand{\Ab}{\mathrm{Ab}}\newcommand{\RMod}{{R\text{-}\mathrm{Mod}}}\RMod$, we can recover $R$ as the ring of endomorphisms of the identity functor, $\newcommand{\End}{\mathrm{End}}\newcommand{\id}{\mathrm{id}}R \cong \End(\id\_\RMod)$. But for a topos $\newcommand{\E}{\mathcal{E}}\E$, $\End(\id\_{\Ab(\E)})$ retracts onto $\Gamma(\Z\_\E)$ (global sections of the integers of $\E$); and this will never have any non-trivial rationals.
Unwinding this more elementarily: Suppose $n \in \Z$ is invertible in $R$, with $n \neq \pm 1$. Then multiplication by $\frac{1}{n}$ gives a natural endomorphism $\mu\_{\frac{1}{n}}$ of the identity functor on $\RMod$, satisfying $n \cdot \mu\_{\frac{1}{n}} = 1$ (where $1$ is the identity endomorphism of the identity functor).
But the $\mathbf{Ab}$-enrichment of any Abelian category is determined by the category structure ([using the biproducts](https://ncatlab.org/nlab/show/biproduct#BiproductsImplyEnrichment)). So an equivalence $\RMod \cong \Ab(\E)$ would transfer $\mu\_{\frac{1}{n}}$ to an endomorphism $\mu'$ of $\id\_{\Ab(\E)}$ with $n \cdot \mu' = 1$; then applying this at $\Z\_\E$, and taking global sections, would give an inverse for $n$ in $\Gamma(\Z\_\E)$. But this can never exist unless $\E$ is the trivial topos, since as $n$ is non-unital you can prove constructively “$n$ is not invertible in $\Z$”, so this holds in the internal language of $\E$. So then $\Ab(\E)$ is also trivial, and the equivalence $\RMod \cong \Ab(\E)$ implies $R \cong 1$.
So we’ve shown: **if an integer $n \neq \pm 1$ is invertible in $R$, and $\RMod \cong \Ab(\E)$, then $R$ is the zero ring.**
| 7 | https://mathoverflow.net/users/2273 | 440110 | 177,724 |
https://mathoverflow.net/questions/440099 | 7 | Let us define a pure vector state of a quantum system as a vector $\psi$ in a Hilbert space $\mathscr{H}$ with norm $\|\psi\| = 1$. Let $\mathscr{B}(\mathscr{H})$ be the Banach space of bounded linear operators on $\mathscr{H}$.
In the $C^{\*}$-algebra formulation of quantum mechanics, one defines observables as the $C^{\*}$-algebra $\mathscr{B}(\mathscr{H})$ of bounded linear operators on $\mathscr{H}$, and a state as a norm one positive functional $\omega$ on this $C^{\*}$-algebra. Strocchi gives a nice interpretation in his book on why these states should be interpreted as average values of observables. In addition, if $A \in \mathscr{B}(\mathscr{H})$ is fixed, with spectrum $\sigma(A)$ and $C^{\*}(A)$ is the unit $C^{\*}$-algebra generated by $A$, then by Riesz-Markov Representation Theorem, to each state $\omega$ there exists a measure $\mu\_{\omega,A}$ such that:
$$\omega(A) = \int\_{\sigma(A)}\tilde{A}(\lambda)d\mu\_{\omega,A}(\lambda)$$
where $\tilde{A}$ is the Gelfand transform of $A$. If the state is $\omega(A) = \langle \psi, A\psi\rangle$, there is a natural interpretation of $\langle \psi, A\psi\rangle$ as the average value of $A$.
In non-algebraic approaches, one usually postulates that given an observable $A$ and a pure vector state $\psi$, the probability that $A$ will take any value inside a Borel set $E \subset \sigma(A)$ is the spectral measure:
$$p\_{\psi,A}(E) = \langle \psi, \chi\_{E}(A)\psi\rangle$$
where $\chi\_{E}$ is the characteristic function of the set $E$.
My question is: is there a nice interpretation or justification of this definition? I know that, a posteriori, this leads to the various formulas that physicists use, but I wanted something a priori. Why is this natural? Putting it another way, I understand the spectral measure is a probability measure on $\sigma(A)$, but why should I use it as the probability of obtaining $E$ in the state $\psi$? Notice that, in this case, $A$ can be unbounded too.
| https://mathoverflow.net/users/152094 | Interpretation of spectral measures in quantum mechanics | (Small correction: We can take the observables to be the *self-adjoint* elements of $B(H)$, or any C${}^\*$-algebra, and in your whole discussion $A$ should be assumed self-adjoint.)
This can be reduced to the following more fundamental principle. Let $v$ be a unit vector in some Hilbert space $H$ and let $E$ be a closed subspace of $H$. Write $H = E \oplus E^\perp$ and decompose $v = v\_1 + v\_2$ accordingly; then the probability the state represented by $v$ will be measured to belong to $E$ is $\|v\_1\|^2$, the probability it will be measured to belong to $E^\perp$ is $\|v\_2\|^2$.
Now if you have an observable (self-adjoint operator) $A$ with finite spectrum, in particular if $H$ is finite dimensional, then we can decompose $H$ as the orthogonal sum of eigenspaces of $A$, $H = E\_1 \oplus \cdots \oplus E\_n$. If the corresponding eigenvalues are $\lambda\_1, \ldots, \lambda\_n$, then the probability a state represented by a unit vector $v$ will be measured to be in $E\_i$ is $\|v\_i\|^2$ where $v\_i$ is the component of $v$ in $E\_i$, and so the expected value of $A$ for $v$ is $\lambda\_1\|v\_1\|^2 + \cdots + \lambda\_n\|v\_n\|^2 = \langle Av,v\rangle$.
That is, the expected value of $A$ for $v$ is the sum of (value of $A$ on $E\_i$)(probability $v$ belongs to $E\_i$). Make sense?
In the general (infinite dimensional) case you can norm approximate any self-adjoint operator with a self-adjoint operator with finite range, so we can argue by approximation that $\langle Av,v\rangle$ is still the expected value.
| 7 | https://mathoverflow.net/users/23141 | 440113 | 177,726 |
https://mathoverflow.net/questions/440131 | 0 | Let $\mu$ be a Radon measure on $[0, 1]$, and $f: [0, 1] \to \mathbb R$ a Borel measurable function.
**Question:** Is it true that for $\mu$ almost every $x \in [0, 1]$, we have
$$f(x) \leq \mu\text{-esssup}\_{[0, x]} \, f?$$
Here the esssup is taken with respect to $\mu$.
| https://mathoverflow.net/users/173490 | An inequality involving the essential supremum | Yes. It suffices to prove that for every rationals $p<q$ the set $A$ of those $x$ for which simultaneosly $\mu\text{-esssup}\_{[0,x]} f<p$ and $q<f(x)$ satisfies $\mu(A)=0$. Note that if $x\in A$, then $\mu(A\cap [0,x])=0$, otherwise we would get $\mu\text{-esssup}\_{[0,x]} f\geqslant q$. It remains to note that $A$ is at most countable union of the sets of form $A\cap [0,x]$ with $x\in A$.
| 3 | https://mathoverflow.net/users/4312 | 440132 | 177,731 |
https://mathoverflow.net/questions/440120 | 2 | I am looking for a proof for the following fact: for $U>0,\beta>0,$ there exists $C>0,\epsilon>0$ such that $$\forall u\in [0,U],\alpha\in\left]0,1\right],\int\_0^u\int\_{[-1,1]^2}\int\_{[-1,1]^2} \frac{1}{r}e^{-\alpha^2|x-y|^2/r} \, dx\,dy\,dr\leq Cu^{\epsilon}\alpha^{-2\beta},$$ where $|\cdot|$ is the euclidean norm.
| https://mathoverflow.net/users/138491 | $\int_0^u\int_{[-1,1]^2}\int_{[-1,1]^2}\frac{1}{r}e^{-\alpha^2|x-y|^2/r} \, dx\,dy\,dr\leq Cu^{\epsilon}\alpha^{-2\beta}$ | $\newcommand{\al}{\alpha}\newcommand{\be}{\beta}\newcommand\ep\epsilon$The integral in question is
\begin{equation\*}
I(u,\al):=\int\_0^u \frac{dr}r \int\_{[-1,1]^2} dy \int\_{[-1,1]^2} dx\,e^{-\al^2|x-y|^2/r}.
\end{equation\*}
Using the substitution $r=\al^2 s$ (suggested by Giorgio Metafune), we get
\begin{equation\*}
I(u,\al)=J(u/\al^2),
\end{equation\*}
where
\begin{equation\*}
J(v):=\int\_0^v \frac{ds}s \int\_{[-1,1]^2} dy \int\_{[-1,1]^2} dx\,e^{-|x-y|^2/s}
=\int\_0^v \frac{ds}s\, K(s)^2,
\end{equation\*}
where
\begin{equation\*}
K(s):=\int\_{-1}^1 da\int\_{-1}^1 db\,e^{-(a-b)^2/s}
\le\int\_{-1}^1 da\int\_{-\infty}^\infty db\,e^{-(a-b)^2/s}
=\int\_{-1}^1 da\, \frac c2\,\sqrt s=c\sqrt s;
\end{equation\*}
here and in what follows $c$ denotes universal positive real constants (possibly different even within the same formula), and $s$ is any positive real number. Also,
\begin{equation\*}
K(s)\le\int\_{-1}^1 da\int\_{-1}^1 db=4.
\end{equation\*}
So,
\begin{equation\*}
K(s)^2\le c\min(1,s).
\end{equation\*}
So,
\begin{equation\*}
J(v)\le c\int\_0^v \frac{ds}s\,\min(1,s)=c(\min(1,v)+\ln\max(1,v))\le c\ln(1+ v);
\end{equation\*}
the latter inequality follows because (say) $\min(1,v)+\ln\max(1,v)$ and $\ln(1+ v)$ are (i) positive and continuous in $v\in(0,\infty)$ and (ii) asymptotically equivalent to each other as $v\downarrow0$ and as $v\to\infty$.
So,
\begin{equation\*}
I(u,\al)\le c\ln\Big(1+\frac u{\al^2}\Big).
\end{equation\*}
So, upon the substitution $u=\al^2 v$, the inequality in question reduces to the inequality
\begin{equation\*}
\ln(1+v)\le C v^\ep\al^{2\ep-2\be} \tag{1}\label{1}
\end{equation\*}
for some real $C>0$, some real $\ep>0$, all real $v>0$, and all $\al\in(0,1]$.
If we now take $\ep=\min(1,\be)$, then \eqref{1} will hold for some real $C>0$ depending only on $\be$. $\quad\Box$
| 2 | https://mathoverflow.net/users/36721 | 440145 | 177,737 |
https://mathoverflow.net/questions/440152 | 5 | I am an undergrad senior math major taking a gap year looking to become an actuary. However, I still want to continue learning pure math. I've been looking for a relatively high level text to self study for the next few months. I'm already a good chunk through the first chapter of Lang's "Algebra" and everything is flowing nicely. However, I have not done any of the exercises. Also, there are a lot of topological examples and things not purely algebraic which I would need to review (I have access to Munkres). In the end, I want to learn a lot of math (which could possibly help with my thesis) and solve a lot of problems. For background, my algebra class sophomore year used Artin's Algebra and I have also taken a seminar on algebraic combinatorics. Are there other texts (within or outside abstract algebra) better-suited for what I'm looking for? And finally, is the Companion to Lang's a supplement/fleshing out of the material, or more of a guide to getting at the solutions?
| https://mathoverflow.net/users/498786 | Lang's "Algebra" as a self-study book | In my opinion, Lang's *Algebra* has an excellent choice of topics for someone who wants to do further work in algebraic number theory or algebraic geometry. I'm not sure whether I'd recommend it for self-study, however. My feeling is that the exposition, and the exercises, are rather uneven. If you're studying from it on your own, I think you could easily get stuck at various points where it's Lang's fault and not yours. But that's just my experience, and your mileage may vary.
Since you mentioned algebraic combinatorics, I feel obliged to mention Richard Stanley's *Enumerative Combinatorics*. This contains a ton of excellent exercises with estimated difficulty ratings and solutions, so I think it's perfect for self-study. Algebraic combinatorics is my field, so I'm biased, but I'd recommend Stanley over Lang for your purposes.
| 13 | https://mathoverflow.net/users/3106 | 440154 | 177,741 |
https://mathoverflow.net/questions/440088 | 5 | Background
----------
For a finite graph $G$, let $\tilde{G}$ denote the [universal cover](https://en.wikipedia.org/wiki/Covering_graph#Universal_cover) of $G$. For a vertex $v$, let $p\_{2n}(v)$ denote the number of paths of length $2n$ that start and end at $v$. The spectral radius of $\tilde{G}$, denoted $\rho(\tilde{G})$, is equal to
$$
\rho(\tilde{G})=\lim\_{n\rightarrow\infty} p\_{2n}(v)^{\frac{1}{2n}}
$$
for any vertex $v\in\tilde{G}$.
**Known Examples:**
For some graphs, the combinatorics of counting pathes simplifies greatly. For $d$-regular graphs, the cover is the $d$-regular tree, and the spectral radius of the universal cover, the $d$-regular tree, equals $2\sqrt{d-1}$.
For [biregular graphs](https://en.wikipedia.org/wiki/Biregular_graph), with degrees $d\_1,d\_2$, calculating the number of paths can be similarly done, and the universal cover has radius $\sqrt{d\_1-1}+\sqrt{d\_2-1}$.
Question
--------
Can we explicitly calculate the radius of the universal cover for other irregular graphs? I couldn't find any explicit examples in the literature other than the two mentioned above.
I would specifically like to know how to calculate it for the following two cases:
1. Let $G\_1$ denote $K\_4$ minus an edge. What is $\rho(\tilde{G}\_1)$? Numerically, it is around $~2.508\dots $
2. Let $G\_2$ denote $K\_5$ minus an edge. What is $\rho(\tilde{G}\_2)$? Numerically, it is around $~3.262\dots $
The problem I run into when trying to calculate the number of paths in $\tilde{G}\_1$ is that backtracking changes the distribution of vertices, potentially in a complicated way with multiple backtracks, but I might be missing something.
| https://mathoverflow.net/users/12176 | Can we calculate the spectral radius of the universal cover for specific graphs? | For the complete graph minus an edge $K\_n-e$, the spectral radius is the largest zero of
\begin{align\*}&x^{14}+(30-10 n) x^{12}+(2 n^{3}+21 n^{2}-202 n +357) x^{10}\\
&+(-10 n^{4}+26 n^{3}+456 n^{2}-2288 n +2888) x^{8} \\
&+(n^{6}-4 n^{5}+76 n^{4}-1520 n^{3}+9320 n^{2}-23056 n +20360) x^{6}\\
&+(-4 n^{7}+48 n^{6}-272 n^{5}+2184 n^{4}-16472 n^{3}+66016 n^{2}-126496 n +93120) x^{4}\\
& +(4 n^{8}-80 n^{7}+720 n^{6}-4432 n^{5}+22800 n^{4}-91712 n^{3}+242704 n^{2}-358944 n +222672) x^{2}\\
& -16 (n -4)^{3} (n^{2}-6 n +10)^{3}.
\end{align\*}
Values for $n=4,5,6,7,8$: 2.50828679, 3.26287647, 3.85572357, 4.36125025, 4.80992000. Experimentally, it is very close to $2\sqrt{n-2}$.
I confess to not tying up every last bit of the proof, but at least the method is valid. Label $K\_n-e$ as $0,1,\ldots,n-1$ where $\{0,1\}$ is the missing edge.
Apply the labelling to the cover.
For a vertex $v\_i$ labelled $i$ in the cover, and neighbour $v\_j$ labelled $j$, let $w\_{ij}=w\_{ij}(x)$ be the ordinary generating function for closed walks that start at $v\_i$, step to $v\_j$, then do anything they like except that they only return to $v\_i$ in the last step. Because $K\_n-e$ is so symmetrical, there are only 3 values for this generating function, represented by $w\_{02},w\_{20},w\_{23}$. Also let $w\_0$ be the ogf for all closed walks starting at $v\_0$. We can write relations between them:
$$ w\_{02} = \frac{x^2}{1-w\_{20}-(n-3)w\_{23}},\quad
w\_{20} = \frac{x^2}{1-(n-3)w\_{02}}, $$
$$ w\_{23} = \frac{x^2}{1-2w\_{20}-(n-4)w\_{23}}
\quad w\_0 = \frac{1}{1-(n-2)w\_{02}}. $$
These equations can be solved for $w\_0$, though I didn't find any way to write the solution in a pretty form.
It satisfies a quartic polynomial with coefficients that are polynomial in $n,x$.
The final step is to identify the smallest singularity (which is on the real line since all the coefficients are non-negative). The reciprocal of that is the spectral radius.
| 5 | https://mathoverflow.net/users/9025 | 440155 | 177,742 |
https://mathoverflow.net/questions/439306 | 3 | We assume ZFC+U.
A *category* is an ordered pair $(\operatorname{Ob} \mathcal{C},\operatorname{Mor} \mathcal{C},\operatorname{dom},\operatorname{codom},e,∘)$ of sets (not classes) and maps satifying some conditions.
Let $\mathbb{U}$ be a Grothendieck universe.
An element of $\mathbb{U}$ is called a *$\mathbb{U}$-set*.
A set is called *$\mathbb{U}$-small* if it is isomorphic to a $\mathbb{U}$-set.
In the following, we suppose that $\mathbb{N} \in \mathbb{U}$.
In SGA4, a category $\mathcal{C}$ is called *$\mathbb{U}$-small*
if $(\operatorname{Ob} \mathcal{C},\operatorname{Mor} \mathcal{C},\operatorname{dom},\operatorname{codom},e,∘)$ is $\mathbb{U}$-small
as a set (if my understanding is correct).
However, I don't see this definition working well.
For any set $a$ and $b$,
an ordered pair $(a,b)$ is always $\mathbb{U}$-small
since $(a,b)=\{\{a\},\{a,b\} \}$ is a set consisting of exactly two elements,
which is isomophic to $2:=\{\emptyset,\{\emptyset\}\} \in \mathbb{U}$.
Thus, $\mathbb{U}$-smallness imposes nothing on categories.
In particular, it is not equivalent to $\operatorname{Ob} \mathcal{C}$ and $\operatorname{Mor} \mathcal{C}$ are $\mathbb{U}$-small.
I think I am mistaken somewhere, where is it?
| https://mathoverflow.net/users/137654 | On the definition of small categories in SGA4 | You’re correct: read literally, those definitions are mismatched, for the reasons you give. The solution is to fix the definition of “$U$-small category” to say that “$\newcommand{\C}{\mathcal{C}}\newcommand{\mor}{\mathrm{mor}}\mor(\C)$ is $\newcommand{\ob}{\mathrm{ob}}U$-small” — this is what the authors of SGA4 clearly intended (as clarified in the footnote on p2 of [this modern edition](http://www.normalesup.org/%7Eforgogozo/SGA4/tomes/SGA4.pdf), pointed out [by @abx in comments](https://mathoverflow.net/questions/439306/on-the-definition-of-small-categories-in-sga4#comment1133094_439306)), but expressed in a way that doesn’t depend on the precise encoding of the definition of categories or of ordered pairs.
There are two general points here:
* Outside of explicit investigations of set-theoretically foundational issues (and usually even within such contexts), *nothing* should ever depend on the specific set-theoretic implementation of ordered pairs. Anything that seems to depend on it can very safely be assumed to be a misunderstanding, an abuse of notation, or a mismatch of definitions.
* More generally, mathematics is usually written in “implementation-independent” ways as far as possible. Of course, there are often lapses from this in practice, and that sometimes leads to mismatches, as here. When such mismatches happen, the right fix is to rewrite the later definitions in more implementation-independent ways, *not* to tweak the implementation of the earlier definitions so that the implementation-dependent later definitions work. In ordinary human-practiced mathematics, this usually isn’t a problem, because it’s clear what people meant. But in computer-formalised mathematics (and programming more generally), this is a serious concern: if you go back and change the implementation of the earlier definition to make one later definition work, then that may break *anything else* that was written in an implementation-specific way. Implementation-dependent definitions are inherently fragile — so fix *them*, don’t take them as god-given and twist other things around to try to work with them.
| 6 | https://mathoverflow.net/users/2273 | 440169 | 177,746 |
https://mathoverflow.net/questions/440181 | 28 | I was thinking about the idea that succession, addition, multiplication, exponentiation, tetration and so on form a sequence of operations where each is defined as a repeated self application of the previous one.
And then it struck me that the first 2 operations in this sequence are commutative but this breaks at exponentiation.
What exactly breaks? When repeated self application of a commutative operation is itself commutative and when it's not?
That is, for an operation:
$$f: \mathcal{N} \times \mathcal{N} \to \mathcal{N}$$
If I define:
$$g(m, 2) = f(m, m)$$
$$g(m, n) = f(m, g(m, n-1))$$
for any $n \geq 2$.
What conditions $f$ must satisfy for $g$ to be commutative? That is, for:
$$g(a,b) = g(b,a)$$
for all $a$ and $b$?
| https://mathoverflow.net/users/757 | Addition and multiplication are commutative but exponentiation and tetration are not. Do we know why? | Not really an answer, but too long for a comment: it's worth noting that if we assume that
* $f$ is associative,
* $g$ is associative,
* $g$ is cancellative for at least one $a$, meaning that $g(a,u)=g(a,v)$ implies $u=v$ for this particular $a$,
then $f$ and $g$ *must* be addition and multiplication.
Indeed, let $a,p,q$ be natural numbers, with $a$ such that $g(a,—)$ is cancellative. Then $g(a,pq)$ is the application of $f$ to $pq$ terms all equal to $a$. By associativity of $f$ we can group this as $q$ terms all of which are the application of $f$ to $p$ terms, i.e., $g(a,p)$, so that $g(a,pq) = g(g(a,p),q)$. By associativity of $g$, we can rewrite this as $g(a,g(p,q))$. By the cancellativity assumption, we get $pq = g(p,q)$. We then have $g(1,n) = n$, and since $g(1,m+n)=f(g(1,m),g(1,n))$ (again, by associativity of $f$) we get $m+n=f(m,n)$, as claimed. ∎
**Update:** similarly, if we assume that
* $f$ is associative,
* $g$ has a unit element $e$, meaning that $g(e,n) = n$ for all $n$,
then the same conclusion holds.
The proof is pretty much the same: as above, $g(e,pq) = g(g(e,p),q)$ so the fact that $e$ is a unit for $g$ means $pq = g(p,q)$, and the rest of the proof is identical. ∎
(Note that for all this I'm assuming that $g(c,1) = c$, which is logical if $g(c,n)$ means “$n$-fold application of $f$ to $c$”, but you didn't actually make this part of your definition. I suppose it was an oversight.)
| 34 | https://mathoverflow.net/users/17064 | 440184 | 177,750 |
https://mathoverflow.net/questions/440179 | 9 |
>
> Find all continuous and bounded functions $g$
> with :
> $$\forall x \in \mathbb R, 4g(x)=g(x+1)+g(x-1)+g(x+\pi)+g(x-\pi).$$
>
>
>
I have posted this question [here](https://math.stackexchange.com/questions/4630194/x-ens-functional-equation), but received no answer.
| https://mathoverflow.net/users/110301 | How may I find all continuous and bounded functions g with the following property? | $\newcommand\de\delta$Considering $g$ a distribution (in the generalized-function sense), let $\hat g$ be the Fourier transform of $g$. Then your functional equation yields
$$4\hat g(t)=e^{it}\hat g(t)+e^{-it}\hat g(t)+e^{i\pi t}\hat g(t)+e^{-i\pi t}\hat g(t),$$
or
$$(\cos t+\cos\pi t-2)\hat g(t)=0,$$
for real $t$.
The equality $\cos t+\cos\pi t-2=0$ for real $t$ implies $\cos t=1=\cos\pi t$ and hence $t=0$ (because $\pi$ is irrational). So,
the support of $\hat g$ is $\{0\}$. So (see e.g. "For every compact subset $K\subseteq U$ there exist constants
$C\_{K}>0$ and $N\_{K}\in \mathbb {N}$ such that for all $f\in C\_{c}^{\infty }(U)$ with support contained in $K$ [...]" [here](https://en.wikipedia.org/wiki/Distribution_(mathematics)#Definitions_of_test_functions_and_distributions)), we have $\hat g=\sum\_{j=0}^n c\_j\de^{(j)}$ for some $n\in\{0,1,\dots\}$ and some complex $c\_j$'s, where $\de^{(j)}$ is the $j$th derivative of the delta function. So, $g$ is a polynomial. Since $g$ is bounded, it is constant. $\quad\Box$
| 18 | https://mathoverflow.net/users/36721 | 440186 | 177,751 |
https://mathoverflow.net/questions/440150 | 12 | In the paper [The Continuumproblem](https://www.pnas.org/doi/abs/10.1073/pnas.24.2.101), Felix Bernstein introduces a new axiom and uses it to conclude the continuum hypothesis.
(1) As the paper is relatively old and the writing style is somehow informal, I am wondering if there is a more exact and concrete proof of the result? (or may please give a better presentation of the argument!)
(2) Is the system introduced by Bernstein consistent? (he does not discuss this in the paper)
| https://mathoverflow.net/users/11115 | Bernstein's proof of the continuum hypothesis | This is really a long comment. This paper has been reviewed twice by zbMATH: [one by H. B. Curry](https://zbmath.org/0019.00903), which is not informative; [another by W. Ackermann](https://zbmath.org/64.0035.02), which is in German. The following is the (manipulated) translation of part of his review by DeepL.
>
> ... However, a certain sense cannot be connected with this axiom, since it is not said what is to be understood by the existence of an element. If one tries to find this sense in the application of the axiom, one is also disappointed. The point of the proof is that a subset of power $\aleph\_1$ is extracted from a certain set of the power of the continuum by multiple application of the axiom of choice, and it is claimed that the selection can be made so arbitrarily that it is impossible to determine whether the subset is a real subset or not without giving any reason for it. (Indeed, in the presence of a proof for $2^{\aleph\_0}\neq\aleph\_1$ this statement would be possible). In any case, the validity of Cantor's hypothesis remains just as indeterminate after the present remarks as before.
>
>
>
---
Edit: It seems, in [this paper](https://zbmath.org/?q=ti%3ADie%20Theorie%20der%20reellen%20Zahlen%20ai%3Abernstein.felix) (1905), Bernstein has sketched a proof of the Continuum Hypothesis for a submission to Math. Ann.
| 11 | https://mathoverflow.net/users/38866 | 440190 | 177,755 |
https://mathoverflow.net/questions/440185 | 6 | The objects of the desired category are epimorphisms of sets $E \to B$ (in what follows, the notation $E/B$ will be used instead of the arrow). Is it possible to naturally define morphisms such that:
1. $E/B \cong (E \times A) / (B \times A)$ for each inhabited $A$ and there are no other non-trivial isomorphisms (non-trivial isomorphism is what is true in addition $\mathrm{Mor~Set}$ isomorphism)
2. The subcategory $E/1$ is isomorphic to $\mathrm{Set}$ by canonical functors
3. The category is a Grothendieck topos.
Perhaps the last condition is too restrictive, then any weakening of it would be interesting. The more good categorical properties that can be preserved, the better. In fact, I don't yet see how to define such a category at all.
P.S. Of course, the condition 1 means that $f \cong f \times \mathrm{id}\_A$, if you write more carefully.
| https://mathoverflow.net/users/148161 | Is there a topos of quotients of sets? | I'll argue Requirement (1) and (2) together are impossible - at least not without making some highly unnatural construction. To be honest the main problem is with (1) alone.
Informally, the idea is that being of the form $E \times A \to B \times A$ is a structure, there are many way of being of this form, and each such way should corresponds to a different isomorphism with $E \to B$. And they should be different, because the corresponds to different way identifying the fibers of $E \to B$ with the fibers of $E \times A \to B \times A$. But the set of isomorphisms between $E \times A \to B \times A$ and $E \to B$ needs to be in bijection with (be a torsor under to be precise) the set of automorphisms of $ E\to B$, so there can't be that many of those. especially if we fix $E \to B$ and let $A$ vary.
Let's try to make a more formal argument - I can't completely formalize it because your requirement are a bit vague, so I'll make some assumption that we want the construction we are building to satisfies some fairly weak naturality condition.
First, I'm interpreting (1) as meaning that if you have isomorphisms $E \to E'$ and $B \to B'$ that makes the square commute then this gives an isomorphism (what you refer to as a trivial isomorphism) in your category, and that this construction should be functorial. I'm doing this because this is strongly suggested in the OP, but also because this requirement encode a minimum of requirement of naturality of the construction on sets: it just means that one does not distinguish between isomorphic sets.
So, I'm assuming you'll have a group homomorphism $TAut(E/B) \to Aut(E/B)$ where $TAut$ denotes the group of these "trivial isomorphisms" and $Aut$ is the group of automorphism in your category.
Pick $E \to B$ to be of the form $X \times B \to B$, so that it should isomorphic to $X \to 1 $ in your category. In particular $Aut(E \to B) \simeq Aut(X)$ by your requirement (2). It follows that we have a group homomorphism
$TAut(X \times B \to B) \to Aut(X)$
If I now look at the "trivial" automorphisms of $X \times B \to B$ that only acts on the $B$ component and are the identity on $X$ component, I have a subgroup of the trivial isomorphisms group isomorphic to $Aut(B)$, so such construction would produce for any sets $E$ and $B$ a choice of a group homomorphism $Aut(B) \to Aut(X)$ for any set $B$ and $X$.
The only reasonable choice here, if you want any kind of naturality for an operation doing this is that the isomorphism is trivial (And even if you want to look at non-natural things, one can talk about it, but if you take $B$ and $X$ finite with at $B$ having at least 5 elements and $X$ smaller than $B$, then the only two possibilities are trivial or factoring two the signature and an element of order $2$ of $Aut(X)$, and I'm relatively confident that one can reach a contradiction in this second case).
But then, one can simply observe that $TAut(X \times B \to B)$ identifies $Aut(X)^B \rtimes Aut(B)$, so we obtain a morphism
$Aut(X)^B \rtimes Aut(B) \to Aut(X)$
that is trivial on elements of $Aut(B)$. That is, we have a group homomorphism $Aut(X)^B \to Aut(X)$ which is invariant under permutation of the component of the products.
Here again, that operation would have to exists for any set $B$ and $X$, and it would have at the minimum to be the identity when $B = \{\*\}$. If you want any kind of naturality, this obviously cannot exists - but I'm also relatively confident that if look at it for some finite set $X$ and $B$ one can show by group-theoretic argument that no such non-trivial operation can exists.
Also, a very natural requirement in you setting would be that in the isomorphism $(E \times A \to B \times A) \simeq (E \to B)$ the trivial automorphisms of $E \to B$ corresponds to their "diagonal" action on $(E \times A \to B \times A)$ which would corresponds to the requirement that the map $Aut(X)^B \to Aut(X)$ above sends $(f,f,f,...f)$ to $f$, which is also definitely something impossible.
| 8 | https://mathoverflow.net/users/22131 | 440191 | 177,756 |
https://mathoverflow.net/questions/431429 | 14 | It was shown by [Hamilton](https://www.degruyter.com/document/doi/10.1515/9781400882571-013/html#:%7E:text=AN%20ISOPERIMETRIC%20ESTIMATE%20FORTHE%20RICCI%20FLOWON%20THE%20TWO-SPHERE,%E2%80%94%20RQij%20where%20R%20is%20the%20scalar%20curvature.) in the 1990s that the isoperimetric ratio $C\_H$ on the $2$-sphere improves along the Ricci flow.
A way to prove this is to use the fact that if $(M^2, g(t))$ is a solution of the Ricci flow on a topological $2$-sphere, then the isoperimetric ratios $C\_H (\rho , t)$ of parallel loops $\gamma\_{\rho}$ measured with respect to the metric $g(t)$ satisfy a heat-type equation
$\frac{\partial }{\partial t} (\log C\_H) = \frac{\partial^2}{\partial {\rho}^2 } (\log C\_H) + \frac{\Gamma}{L} \frac{\partial}{\partial \rho} (\log C\_H)+ \frac{4 \pi - C\_H}{A} \bigg( \frac{A+}{A\_-} + \frac{A\_-}{A\_+} \bigg) $,
where $\Gamma$ is an integral of the signed curvature and more definitions can be found in the textbook of Chow and Knopf.
The isoperimetric ratio $C\_H$ is similar to, but distinct from the [Cheeger isoperimetric constant](https://en.wikipedia.org/wiki/Cheeger_constant).
1.) Is it known if the Cheeger constant $h$ also satisfies a heat-type equation for a solution of the Ricci flow on a topological $2$-sphere?
2.) For a topological $2$-sphere, is it the case that
$C\_H (M) \leq h (M)?$
| https://mathoverflow.net/users/119114 | Does the Cheeger constant satisfy a heat-type equation? | Answer to (2) is negative. For a counterexample, consider a $2$-sphere and blow up the radius. The isoperimetric ratio $C\_H$ remains the same, but the Cheeger constant $h$ shrinks, essentially because $C\_H$ is a dimensionless quantity which is independent of the scaling, whereas $h$ is not.
Actually in Lemma 5.85 of the textbook of Chow and Knopf, it is shown that if $(M^2, g)$ is a closed orientable Riemannian surface, then
$C\_H (M) \leq 4 \pi,$
which is obviously not true for the Cheeger constant.
I did a few calculations and looks as if answer to (1) is positive.
| 3 | https://mathoverflow.net/users/119114 | 440193 | 177,758 |
https://mathoverflow.net/questions/440196 | 12 | **Introduction**
This question is about Picard spectra for the symmetric monoidal $\infty$-category of spectra. We say that a spectrum $X$ is invertible if there is another spectrum $Y$ such that $X\wedge Y\simeq S^0$. It is well-known that any such $X$ is equivalent to $S^n$ for some $n\in\mathbb{Z}$, and that the space of endomorphisms of $S^n$ is $QS^0=\lim\_{\to k}\Omega^kS^k$. This space has $\pi\_0(QS^0)=\mathbb{Z}$, and we write $Q\_{\pm 1}S^0$ for the union of the two components corresponding to $1$ and $-1$, which is the space of self-homotopy equivalences of $S^n$. We write $\text{pic}(S)$ for the $K$-theory spectrum of the symmetric monoidal category of invertible spectra, so $\text{pic}(S)$ is $(-1)$-connected with $\pi\_0(\text{pic}(S))=\mathbb{Z}$ and $\Omega^{\infty + 1}(\text{pic}(S))=Q\_{\pm 1}S$.
Given an invertible spectrum $S^{2n}$, we have a naively homotopical commutative ring spectrum $R(n)=\bigvee\_{k\in\mathbb{Z}}S^{2nk}$ with ${\pi\_{\ast}}(R)={\pi\_{\ast}}(S)[x,x^{-1}]$ where $|x|=2n$. One might like to build a strictly commutative (or $E\_\infty$) version of $R(n)$, but this is not obviously possible.
Various people have studied the strict Picard spectrum $\text{spic}(S)$, which is the $(-1)$-connected cover of $F(H\mathbb{Z},\text{pic}(S))$; in particular there is the paper *[On the Strict Picard Spectrum of Commutative Ring Spectra](https://arxiv.org/abs/2208.03073)* by Carmeli. There it is shown (amongst many other things) that $\pi\_0(\text{spic}(S))$ maps trivially to $\pi\_0(\text{pic}(S))=\mathbb{Z}$, and that the strict Picard spectrum is related to the realisation problem mentioned above, so that $R(n)$ cannot be made $E\_\infty$ unless $n=0$. However, we only get to this conclusion after developing an extensive theory.
**The question**
Is there a simple direct proof that $R(n)$ cannot be made $E\_\infty$ for $n\neq 0$?
| https://mathoverflow.net/users/10366 | Why are ordinary spheres not strictly invertible? | An $E\_{\infty}$ structure extending the $E\_1$ structure on $R(n)$ in particular yields maps $(\mathbb{S}^{2n})^{\otimes p}\_{hC\_p} \to \mathbb{S}^{2pn}$ splitting the inclusion of the bottom cell. The left hand spectrum can be viewed as homotopy orbits of the $C\_p$ action on the representation sphere $\mathbb{S}^{2n \rho}$, which can be interpreted as Thom spectrum on $BC\_p$. The existence of a map as above is precluded by the action of the Steenrod algebra on $\mathbb{F}\_p$ cohomology, which can be computed through the Thom isomorphism.
| 14 | https://mathoverflow.net/users/39747 | 440199 | 177,760 |
https://mathoverflow.net/questions/440210 | 4 | For a semisimple complex Lie algebra $\frak{g}$ it is well known that irreducible finite-dimensional representation are not characterised by their dimension.
More formally, let us define an equivalence relation on dominant weights by $\lambda ~ \mu$, for $\lambda, \mu \in \mathcal{P}^+$, is it holds that
$$
\mathrm{dim}(V\_{\lambda}) = \mathrm{dim}(V\_{\mu}).
$$
As just mentioned, classes can in general have more than one element. Is there an upper bound on the number of elements a class can have, or can one find classes with arbitrarily many elements?
| https://mathoverflow.net/users/378228 | Number of representations of a semisimple Lie algebra of any given dimension | For $\mathfrak{sl}\_2\times \mathfrak{sl}\_2$, the number of irreps of dimension $n$ is the number of factorizations $n=n\_1n\_2$ (you tensor the irreps of the two $\mathfrak{sl}\_2$'s), so there's no upper bound.
For $\mathfrak{sl}\_3$, the Weyl dimension formula says that these dimensions are $\frac{1}{2}(n\_1+1)(n\_2+1)(n\_1+n\_2+2)$. For a fixed dimension, $\frac{1}{2}(n\_1+1)(n\_2+1)(n\_1+n\_2+2)=d$.
Completing the square in one variable, this is $(n\_1+1+\frac{1}{2}(n\_2+1))^2=\frac{2d}{n\_2+1}+(\frac{1}{2}(n\_2+1))^2$
which implies that $n\_1+1=-\frac{1}{2}(n\_2+1)\pm\frac{1}{n\_2+1}\sqrt{\frac{2d}{n\_2+1}+(\frac{1}{2}(n\_2+1))^2}$. Seems pretty unlikely that one can find a $d$ that gets you a bunch of integral RHSs for n\_2 integral, but maybe I'm missing something.
| 6 | https://mathoverflow.net/users/66 | 440220 | 177,765 |
https://mathoverflow.net/questions/440223 | 6 | I checked some relations between primes, here $1<n<10^5$ and $p\_n$ is the $n$th prime.
$a) p\_n^{1/3} - p\_{n-1}^{1/3}<1/2$
$b) p\_n^{1/n} - p\_{n-1}^{1/n}<1/n $
$c) (\log p\_n)^{1/2} - (\log p\_{n-1})^{1/2} < 1/4$
$d) (\log p\_n)^{1/n} - (\log p\_{n-1})^{1/n} < 1/n^X, n\geq7,X=2$
In $d)$ I tried to find a larger $X$, but I failed.
Maybe some will fail for a larger $n$. Are any of these know to be true? Also, what can be deduced from they?
| https://mathoverflow.net/users/126334 | Some conjectures about prime gaps | The first three statements are true for $n$ sufficiently large. The fourth statement is equivalent to a very strong bound on prime gaps (which has a chance to hold but is right on the edge and hopeless to prove or disprove at present), while any $X>2$ produces a false bound for all $n>n\_0(X)$.
Regarding a), a classical result of Ingham (1937) shows that
$$p\_n^{1/3} - p\_{n-1}^{1/3}<\frac{p\_n-p\_{n-1}}{3p\_{n-1}^{2/3}}=o(1).$$
Regarding b), the Prime Number Theorem gives that
$$p\_n^{1/n} - p\_{n-1}^{1/n}\sim\frac{p\_n-p\_{n-1}}{n p\_{n-1}}=o\left(\frac{1}{n}\right).$$
Regarding c), Bertrand's postulate shows for $p\_n>7$ that
$$(\log p\_n)^{1/2}-(\log p\_{n-1})^{1/2}<\frac{\log p\_n-\log p\_{n-1}}{2(\log p\_{n-1})^{1/2}}<\frac{\log 2}{2(\log p\_{n-1})^{1/2}}<\frac{1}{4}.$$
In fact the left-hand side tends to zero (as $n\to\infty$).
Regarding d), the Prime Number Theorem gives that
$$(\log p\_n)^{1/n}-(\log p\_{n-1})^{1/n}\sim\frac{p\_n-p\_{n-1}}{n p\_n\log p\_n}\sim\frac{p\_n-p\_{n-1}}{n^2\log^2 n}.$$
Under some standard (but bold) conjectures the right-hand side stays below $2/n^2$, and perhaps it is even $o(1/n^2)$. So statement d) restricted to large $n$ is equivalent to some expected but hopelessly difficult upper bound on prime gaps. On the other hand, the above calculation shows that d) fails for any $X>2$ and $n>n\_0(X)$.
| 9 | https://mathoverflow.net/users/11919 | 440227 | 177,767 |
https://mathoverflow.net/questions/440114 | 9 | Consider a system of $n$ linear equations with $n$ unknowns, all of whose coefficients and right hand sides are nonnegative integers, with a unique solution consisting of nonnegative rational numbers. Is it always possible to solve the system using restricted subtraction-moves that only let us subtract one equation from another if the coefficients and right hand side of the latter respectively dominate the coefficients and right hand side of the former? In addition to restricted subtraction, we’re also allowed to add equations without restriction, or to multiply an equation by a positive rational number.
Example (with $n=2$): To solve $2x+y=7$, $x+2y=8$, we may not subtract either equation from the other (since that would lead to a negative coefficient), but we can double the former, obtaining $4x+2y=14$, subtract $x+2y=8$ to get $3x=6$, and then multiply by $1/3$ to get $x=2$, and we can solve for $y$ in a similar way.
I do not require that at each stage one retain only $n$ equations (though if my question has an affirmative answer, that will be my next question, either here or in a separate post).
| https://mathoverflow.net/users/3621 | Solving systems of linear equations without introducing negative numbers | Here is a proof that it is always possible by keeping at most $n+1$ equations throughout.
Suppose the system $Ax=b$ has a unique solution $c \in \mathbb{Q}\_{\geq 0}^n$, where $A \in \mathbb{Z}\_{\geq 0}^{n \times n}$ and $b \in \mathbb{Z}\_{\geq 0}^n$. Note that the equation $x\_1=c\_1$ can be written as a linear combination of equations which appear in $Ax=b$. Thus, there exist $q\_1, \dots, q\_n \in \mathbb{Q}$ such that $\sum\_{i=1}^n q\_iA\_i=e\_1$ and $\sum\_{i=1}^n q\_i b\_i=c\_1$, where $A\_i$ is the $i$th row of $A$ and $e\_1$ is the first standard basis vector. Let $I$ be the set of indices $i$ such that $q\_i \geq 0$. The coefficients of $\sum\_{i \in I} q\_iA\_i$ dominate the coefficients of $\sum\_{i \notin I} -q\_iA\_i$, and $\sum\_{i \in I} q\_ib\_i \geq \sum\_{i \notin I} -q\_ib\_i$. Thus, we may subtract $\sum\_{i \notin I} -q\_iA\_i$ from $\sum\_{i \in I} q\_iA\_i$ to derive $x\_1=c\_1$. Then just repeat for each variable.
| 4 | https://mathoverflow.net/users/2233 | 440239 | 177,770 |
https://mathoverflow.net/questions/435153 | 1 | If I sample three points independently, uniformly at random on an $n$-dimensional sphere of radius $R$, what is the probability density function of their [polar sine](https://en.wikipedia.org/wiki/Polar_sine)?
More generally, for $k<n$ if I sample $k$ points independently uniformly at random on an $n$-dimensional sphere of radius $R$, what is the probability density function of their polar sine?
An asymptotic form for large $n$ can be deduced in the case $k=2$ from Lemma 4 of Appendix A of Laarhoven's [Sieving for shortest vectors in lattices using angular locality-sensitive hashing](https://eprint.iacr.org/2014/744.pdf), but my attempts to generalise to $k=3$ do not match empirical distributions for $n=10$.
| https://mathoverflow.net/users/489157 | Probability density function for the polar sine of uniformly distributed points on the sphere | For $k=3$, $0\le s\le 1$ it is
$$c\_ns^{n-3}\left(\frac\pi2-\arcsin(s)\right)$$
where
$$c\_n=\begin{cases}\frac2\pi(n-2)\frac{2^{n-2}\left(\frac{n-2}2\right)!}{(n-2)!}&\text{ for $n$ even}\\
\frac{(n-2)!}{2^{n-3}\left(\frac{n-3}2\right)!}&\text{ for $n$ odd}.\end{cases}$$
Using polyspherical coordinates, the cumulative density function for fixed a given sine $s\_0$ between the first two vectors is
$$\frac{A\_1A\_{n-3}}{\frac12B(1,(n-1)/2)A\_{n-1}}\int\_0^u\cos\theta\sin^{n-3}\theta$$
where
$$u=\begin{cases}\arcsin\left(\frac s{s\_0}\right)&\text{ for $s<s\_0$}\\
\frac\pi2&\text{ for $s\ge s\_0$}\end{cases}.$$
Integrating this over $s\_0=\sin\theta$ gives a cumulative distribution function for $s$ whose derivative is the pdf.
| 0 | https://mathoverflow.net/users/489157 | 440241 | 177,771 |
https://mathoverflow.net/questions/440235 | 3 | I am trying to figure out whether or not the following property is first-order expressible in the language of groups.
$$\text{$G$ has a subgroup $H$ with which it forms a Frobenius pair $(H,G)$.}$$
My search online yielded no answers. Even if there is a positive or negative result, it seems to be not very well-known.
| https://mathoverflow.net/users/498986 | Is having a Frobenius pair first-order expressible in the language of groups? | I'll answer here positively the natural variant of the question:
>
> does there exist a 1st order sentence such that for every finite group $G$, the group $G$ is Frobenius over some subgroup iff the given sentence holds in $G$.
>
>
>
In a sense, this sound to me like a more natural question, since Frobenius pairs are really motivated by finite group theory, and such a result allows to extend the notion to pseudofinite groups.
(Concerning infinite groups, my bet would be that the result is false although I can't prove it at the moment.)
In a group $G$ and $g\in G$, write $C\_g$ the centralizer of $g$, and $L\_g=\{h:[g,hgh^{-1}]=g\}$. Let $A\_G$ be the set of elements of $G$ which commute with all their conjugates (i.e., are contained in some abelian normal subgroup). Let $B\_G$ be the set of $g\in G$ such that $C\_g$ is a normal subgroup, and $B'\_G$ the set of $g\in B\_G$ such that every element of $G\smallsetminus C\_g$ acts by conjugation on $C\_g$ with no nontrivial fixed point. Let $D\_G$ be the set of $g\in G$ such that $L\_g$ is a subgroup.
**Proposition.** A group is Frobenius iff there exists $z\in B'\_G\smallsetminus\{1\}$ and $h\in D\_G\smallsetminus\{1\}$ such that $G=C\_z\rtimes L\_h$.
*Proof:* For basic facts on Frobenius pairs see the [Wikipedia page](https://en.wikipedia.org/wiki/Frobenius_group).
Suppose $G=K\rtimes H$ is Frobenius. So $K$ is nilpotent (Thompson) and nontrivial, choose any central $z\in G$: then $K=C\_z$ and $z\in B'\_G$. Also, $H$ is a Frobenius complement, and hence has a nontrivial solvable radical (follows from results of Zassenhaus: it is indeed either solvable and nontrivial, or has a normal copy of $\mathrm{SL}\_2(\mathbf{F}\_5)$, which has a nontrivial center), hence has a nontrivial abelian normal subgroup; choose $h$ therein. Then $L\_h=H$ follows easily: choose $g\in L\_h$ since if $[h,ghg^{-1}]=1$ and the centralizer of $h$ is contained in $H$ it follows that $ghg^{-1}\in H$ and in turn $g\in H$ follows.
Conversely, if $G$ has these properties, then it is straightforward that $L\_h$ is malnormal (i.e. $L\_h\smallsetminus\{1\}$ is disjoint each from its conjugates by elements not in $L\_h$) and this implies that $G$ is Frobenius.$\Box$
Now the property of the proposition is obviously expressible by a first-order sentence $F\_0$ in $G$. Hence for every finite group $G$, $G$ satisfies $F\_0$ iff $G$ is Frobenius.
In particular, we can define a pseudofinite group to be Frobenius iff it satisfies $F\_0$. Note that this does not depend on $F\_0$. This can be defined directly as: a group $G$ is pseudofinite Frobenius if for every 1st order sentence $F$ satisfied by $G$ there exists a finite Frobenius group satisfying $F$.
While the latter could be defined in any case, the existence of $F\_0$ above ensures the following: if a group $G$ is pseudofinite Frobenius, then there exists a 1st order formula $F$ such that every finite group $G'$ satisfying $F$ is Frobenius. (More informally: it means that a pseudofinite group cannot be both pseudo-Frobenius and pseudo-(non-Frobenius).)
[For context, there also exists a first-order formula that characterizes, among finite groups, the solvable ones (Wilson). But there is none that characterizes, among finite groups, the nilpotent ones (YC-Wilson).]
| 4 | https://mathoverflow.net/users/14094 | 440254 | 177,775 |
https://mathoverflow.net/questions/440250 | 0 | Let $\lambda \in \mathcal{P}^+$ be a dominant weight for $\frak{sl}(n,\mathbb{C})$. When does it hold that the zero weight space, of the associated finite-dimensional $L(\lambda)$, is non-trivial?
What about the same question for the other seires?
| https://mathoverflow.net/users/378228 | Non-trivial weight spaces of finite-dimensional irreducible $\frak{g}$-modules | I'm not quite sure this question rises to the level of MathOverflow, which is why I initially posted only a comment, but at the request of the question-asker I am converting my comment to an answer.
For any semisimple Lie algebra $\mathfrak{g}$ and dominant weights $\mu,\lambda \in P^+$, the condition for the $\mu$-weight space of the irreducible finite dimensional $\mathfrak{g}$-representation $L(\lambda)$ to be nonzero is that $\lambda - \mu$ is a nonnegative sum of positive roots (equivalently, simple roots). This is normally just denoted $\mu \leq \lambda$ (or $\mu \preceq \lambda$). This is the most common partial order on (dominant) weights. See, e.g., Humphreys' "Introduction to Lie Algebras and Representation Theory," Chapter VI.
There is an equivalent, but more geometric, criterion too: we have $\mu \leq \lambda$ if and only if $\mu$ and $\lambda$ lie in the same coset of $P/Q$ (the weight lattice modulo the root lattice) and the convex hull of $W\mu$ is contained in the convex hull of $W\lambda$. (These convex hulls are sometimes called $W$-permutohedra.) This condition is discussed in Stembridge's "The Partial Order of Dominant Weights" (<https://doi.org/10.1006/aima.1998.1736>), and as he mentions there they probably also appear in Bourbaki's book on Lie groups and Lie algebras.
| 2 | https://mathoverflow.net/users/25028 | 440260 | 177,776 |
https://mathoverflow.net/questions/440236 | 1 | [Originally posted at [math.stackexchange](https://math.stackexchange.com/q/4626210/573047) without answer]
Consider a family $\mathcal{F}$ of $n=|\mathcal{F}|$ sets, $\emptyset \not\in \mathcal{F}$ and an universe $U(\mathcal{F})$ of $q=|U(\mathcal{F})|$ elements.
It is known that at least a fraction $r\binom{n}2$, $0 \lt r \lt 1$, of the unordered couples of sets of $\mathcal{F}$, have at least one element in common, i.e. $|\{\{A\_1,A\_2\}: A\_1,A\_2 \in \mathcal{F} \land A\_1 \not= A\_2 \land A\_1 \cap A\_2 \not= \emptyset \}| \ge r\binom{n}2$.
If we find the lowest $m$ such that:
$$\binom{m}2 \ge \frac{r\binom{n}2}q$$
we can then say that there exists an element belonging to at least $m$ sets of $\mathcal{F}$.
However since those $m$ sets cannot be made of only one element, I think the bound can be improved, i.e. we can say that there is an element in at least $m'$ sets:
$$m' = f(r,n,q) \gt m$$
Any idea for doing that?
| https://mathoverflow.net/users/136218 | Improving a lower bound for the minimum of the maximum frequency of an element in a family of sets | I do not think the bound can be improved without further assumptions. For example, consider the family $\mathcal{F}$ consisting of all singleton subsets of $\{1, \dots, n-1\}$ together with $\{1, \dots, n-1\}$. Here $q=n-1$, and $r\binom{n}{2}=n-1$. Thus, the smallest $m$ for which $$\binom{m}{2} \geq \frac{r\binom{n}{2}}{q}$$ is $m=2$. However, there is no element which is contained in $3$ sets of $\mathcal{F}$, so the bound is tight.
| 2 | https://mathoverflow.net/users/2233 | 440263 | 177,777 |
https://mathoverflow.net/questions/440237 | 7 | Has Pontryagin duality been extended to condensed abelian groups? The obvious approach being to define $\hat M$ as the internal hom to the circle group. Is it true that $\hat{\hat M}=M$ with this definition?
| https://mathoverflow.net/users/473423 | Condensed Pontryagin duality | This cannot be true for all condensed abelian groups. Indeed, in [this answer](https://mathoverflow.net/a/356261/82179) to [Are there (enough) injectives in condensed abelian groups?](https://mathoverflow.net/questions/352448/are-there-enough-injectives-in-condensed-abelian-groups), Scholze explains that there are no nonzero injective condensed abelian groups. So it suffices to prove the following:
**Lemma.** *Let $\mathscr A$ be an abelian category and let $\mathbf D \colon \mathscr A^{\text{op}} \to \mathscr A$ be a left exact functor such that $\mathbf D^2$ is faithful and preserves monomorphisms (e.g. $\mathbf D^2 \cong \mathbf 1\_{\mathscr A}$). Then $\mathbf D$ is exact.*
Indeed, this would prove that $\mathbf T = \mathbf R/\mathbf Z$ is injective if $\mathbf D = \mathbf{Hom}(-,\mathbf T)$ is an auto-duality.
*Proof of Lemma.* If $0 \to A \to B \to C \to 0$ is an exact sequence in $\mathscr A$, then
$$0 \to \mathbf D(C) \to \mathbf D(B) \to \mathbf D(A)$$
is exact by hypothesis. Let $M$ and $N$ be the image and cokernel of $\mathbf D(B) \to \mathbf D(A)$ respectively, so that we get exact sequences
$$\begin{array}{ccccccccc}
0 & \to & \mathbf D(C) & \to & \mathbf D(B) & \to & M & \to & 0,\\
0 & \to & M & \to & \mathbf D(A) & \to & N & \to & 0.
\end{array}$$
Dualising again gives exact sequences
$$\begin{array}{ccccccc}
0 & \to & \mathbf D(M) & \to & \mathbf D^2(B) & \to & \mathbf D^2(C),\\
0 & \to & \mathbf D(N) & \to & \mathbf D^2(A) & \to & \mathbf D(M).
\end{array}$$
The composition $\mathbf D^2(A) \to \mathbf D(M) \to \mathbf D^2(B)$ is injective by hypothesis, so the second sequence gives $\mathbf D(N) = 0$. Faithfulness of $\mathbf D^2$ gives $N = 0$, so $\mathbf D(B) \to \mathbf D(A)$ was surjective to begin with. $\square$
Presumably there is a direct example showing that $\mathbf T$ is not injective. If you want to work with $\mathbf{Cond}\_\kappa$ instead of $\mathbf{Cond}$, then the same argument doesn't work (because $\mathbf{Cond}\_\kappa$ does have enough injectives), and I don't know if $\mathbf T$ is injective in $\mathbf{Cond}\_\kappa$ for any strong limit cardinal $\kappa$. (Surely not, right?)
| 7 | https://mathoverflow.net/users/82179 | 440272 | 177,781 |
https://mathoverflow.net/questions/438912 | 2 | Let $\mathbb{N}\_+$ denote the set of positive integers and let $\mathbb{N}\_0 = \mathbb{N}\_+\cup\{0\}$. Fix $\alpha\in[0,1]\setminus \mathbb{Q}$. For $n\in\mathbb{N}\_+$ we let the *approximation radius* of $n$ be $$\text{rad}\_\alpha(n) = \min\Big\{\Big|\alpha-\frac{x}{n}\Big|:x\in\mathbb{N}\_0\text{ and } x \leq n\Big\}.$$
Inductively define the strictly increasing *approximation radius sequence* $(\text{appr}\_\alpha)\_{n\in\mathbb{N}\_+}$ by
>
> $\text{appr}\_\alpha(1) = 1$ and $\text{appr}\_\alpha(n+1) = \min\{x\in\mathbb{N}\_+: x>n \text{ and }\text{rad}\_\alpha(x) < \text{rad}\_\alpha(n)\}$ for all $n\in\mathbb{N}\_+$.
>
>
>
For $f,g: \mathbb{N}\_+\to \mathbb{N}\_+$ we say that $f\leq^\*g$ if there is $N\in\mathbb{N}\_+$ such that for all $x\in\mathbb{N}\_+$ with $x\geq N$ we have $f(x) \leq g(x)$.
**Question.** Given any function $f: \mathbb{N}\_+\to \mathbb{N}\_+$, is there $\alpha\in[0,1]\setminus\mathbb{Q}$ such that $f\leq^\*\text{appr}\_\alpha$?
| https://mathoverflow.net/users/8628 | Measuring how "close" $\alpha\in[0,1]\setminus\mathbb{Q}$ is to being rational | I think the answer is yes.
We will assume that $\text{appr}\_{\alpha}$ is defined for $\alpha\in\mathbb{Q}$ also but $\text{appr}\_{\alpha}(n) = \infty$ for all large $n$. Consider another function $\text{Appr}\_{\alpha}(n)\colon \mathbb{N}\to\mathbb{N}\_+^2$ which will return the pair $(x, \text{appr}\_{\alpha}(n))$, where $x$ is the optimal numerator from the definition of $\text{rad}\_{\alpha}$. For positive integers $n,x, N$, define the set
$$
A\_{n,x,N} = \left\{\alpha\mid \text{Appr}\_{\alpha}(n) = (x,N)\right\}.
$$
**Lemma.** Let $n, x, N, N'$ be positive integers and $I = [x/N,r]$ be a segment such that $I\subset A\_{n,x,N}$. Then there exists $M > N'$ such that for some $y$ and $r' > y/M$ we have $[y/M, r']\subset (I\cap A\_{n + 1,y,M})$.
First let us derive the statement from the lemma. Let $I\_0 = [1/2, 3/5]\subset A\_{2, 1,2}$ be some starting segment. We have $n\_0 = 2, x\_0 = 1, N\_0 = 2$. On each step apply lemma to $n = n\_k, x = x\_k, x = N\_k$ and $N' = f(n\_k + 1)$ and define $n\_{k + 1} = n\_k + 1, x\_{k + 1} = y, N\_{k + 1} = M$ then we have the nested sequence of the segments $I\_0, I\_1, \ldots$ such that $I\_k = [x\_k/N\_k, r\_k]$ and $I\_k\subset A\_{n\_k,x\_k,N\_k}$. There exists $\alpha\in \cap I\_k$. By definition we have $\text{appr}\_{\alpha}(k) = N\_k \ge f(k)$.
*Proof of the lemma*. There exists $r\_1$ such that for all $\alpha\in[x/N, r\_1]$ we have $\text{appr}\_{\alpha}(n + 1) > N'$ (this is all points such that $x/N$ is closer to them then any other fraction with denominator $\le N'$). Take an arbitrary rational number $y/M\in (x/N, r\_1)$. Again we can choose $r'$ such that $y/M$ is the closest rational number with denominator $\le M$ for all point in $[y/M, r\_1]$. The lemma follows.
| 1 | https://mathoverflow.net/users/498423 | 440277 | 177,784 |
https://mathoverflow.net/questions/400602 | 9 | In the known paper *On the reconstruction of topological spaces from their group of homeomorphisms* by Matatyahu Rubin several deep reconstruction theorems of the form "if $X$ and $Y$ are topological spaces in a broad class of spaces $K$ and there is an isomorphism between $\mathrm{Homeo}(X)$ and $\mathrm{Homeo}(Y)$, then $X$ and $Y$ are homeomorphic" are proved. Moreover the following result is claimed
>
> Assume $V=L$. If $X$ and $Y$ are second countable connected Euclidean manifolds and $\mathrm{Homeo}(X)$ is elementary equivalent to $\mathrm{Homeo}(Y)$, then $X$ and $Y$ are homeomorphic.
>
>
>
to appear in *Second countable connected manifolds with elementarily equivalent homeomorphism groups are homeomorphic in the constructible universe*. Unfortunately I cannot find any information on a paper with this title online. Has a proof of this theorem been published by Rubin? What is known about this result in $\mathsf{ZFC}$ without extra set theoretic assumptions?
| https://mathoverflow.net/users/49381 | On a result by Rubin on elementary equivalence of homeomorphism groups and homeomorphisms of the underlying spaces | There's a new paper that proves something even stronger for compact manifolds: <https://arxiv.org/abs/2302.01481>. I don't know about noncompact manifolds, though, and it seems neither do the authors.
| 4 | https://mathoverflow.net/users/499018 | 440283 | 177,787 |
https://mathoverflow.net/questions/440264 | 0 | I am learning local cohomology from Hartshorne’s Local Cohomology book.
My question is about the notion of essentially zero inverse system of abelian groups, which is defined to be an inverse system of abelian groups $(M\_{m})$ indexed by the non-negative integers, such that for every $m$, there is some integer $m’\geq m$, such that $M\_{m’}\rightarrow M\_{m}$ is the zero map.
Hartshorne’s book has the following remark, which I have trouble verifying:
If we have an exact sequence of inverse systems
$$0\rightarrow (M’\_{m})\rightarrow (M\_{m})\rightarrow (M''\_{m})\rightarrow 0,$$
then the middle one is essentially zero if and only if the two outside ones are essentially zero.
I could only show that if the middle inverse system is essentially zero, then so are the two outside inverse systems.
The idea is to use a commutative diagram of the two short exact sequences induced by the zero map $M\_{m’}\rightarrow M\_{m}$.
So my question is, how to show that if the two outside inverse systems are essentially zero, then so is the middle one?
I have tried to apply the same idea as above, but did not get a proof.
Thank you so much for your kind help.
| https://mathoverflow.net/users/477848 | Essentially zero inverse system of abelian groups | I don't like indexing with primes, so let's consider an exact sequence of inverse systems:
$$ 0 \to (A\_m) \to (B\_m) \to (C\_m) \to 0 $$
Now we are assuming that $(A\_m)$ and $(C\_m)$ are essentially zero inverse systems, which means that for a given $m$ we can find a larger $m'$ so that both $A\_{m'} \to A\_m$ and $C\_{m'} \to C\_m$ are the zero map.
At this point we get a map of short exact sequences:
$\require{AMScd}$
\begin{CD}
0 @>>> A\_{m'} @>i\_{m'}>> B\_{m'} @>>> C\_{m'} @>>> 0 \\
@. @VV 0 V @VVf\_{m', m} V @VV 0 V @. \\
0 @>>> A\_{m} @>>i\_m> B\_{m} @>>> C\_{m} @>>> 0 \\\end{CD}
Now the tricky point is that it does not follow that $f\_{m', m}$ is the zero map. There can certainly be maps of short exact sequences where the outside maps are zero, but the central map is non-zero. However we can record two facts.
First, by the commutivity of the diagram and the fact that the rows are exact we see that the image $f\_{m', m}(B\_{m'})$ is contained in the image $i\_m(A\_m)$. Second, by the commutativity of the diagram, the image of $i\_{m'}(A\_{m'})$ lies in the kernel of $f\_{m', m}$.
Summarizing, the sequence $(B\_m)$ satisfies the property that for each $m$ there exists an $m' \geq m$ such that
1. $f\_{m', m}(B\_{m'}) \subseteq i\_m(A\_m)$
2. $i\_{m'}(A\_{m'}) \subseteq \ker f\_{m', m}$
Now given $m$ we apply this property twice, first to $m$ to get $m'$ and then to $m'$ to get $m''$.
Then the image of $B\_{m''}$ under $f\_{m'', m'}$ lies in the image of $A\_{m'}$, which in turn lies in the kernel of $f\_{m', m}$. Thus the map
$$f\_{m'', m} = f\_{m', m} \circ f\_{m'', m'}: B\_{m''} \to B\_m$$
is zero.
| 3 | https://mathoverflow.net/users/184 | 440284 | 177,788 |
https://mathoverflow.net/questions/440176 | 3 | $\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Gal{Gal}$Recall that $n$-simplices of the classifying space (simplicial set) $BG$ of a group are orbits of the diagonal action by shifts
on $n+1$-tuples of its elements. But what if we take elements from some other set the group acts on ?
That is, the same definition can be made for an arbitrary action of a group on a set,
e.g. the Galois action on $\Bbb{\bar Q}$, or a symmetric group acting by permutations. Note that for the trivial action
of a group on a singleton I get the trivial simplicial object, and thus this is not the Borel construction.
Where can I read about this ? What is the right terminology ?
In notation: let $G$ be a group acting on a set $X$, i.e. we have $\tau:G\rightarrow \Aut X$.
Recall that $B(G)\_n = G^{n+1}/G$, and I define
$$B'(G \xrightarrow{\tau} \Aut X):=X^{n+1}/G$$
Note that for $X=G$ and $\tau:G\rightarrow \Aut G$ the action by shifts,
$B(G)=B'(\tau)$, and for the trivial action of a group on a singleton,
$B'(G\rightarrow \Aut(\{\mathrm{pt}\}))$ is trivial as well.
What what can be said about $B'(\tau)$ and where can I read about it ?
In particular, for $G=\Gal( \Bbb{\bar Q/ Q})$ action on $\mathbb{\bar Q}$, or perhaps
$\Gal(K/\Bbb{Q})$ action on $K$ for a number field ? Has this been considered in number theory ?
Is this construction trivial for some reason ?
For the permutation representation of an infinite symmetric group one gets a simplicial set representing equivalence relations;
for the finite symmetric group $S\_n$ it classifies equivalence relations with at most $n$ equivalence classes.
I am also interested in the following "ordered" modification of the construction for the group $G=\Aut(\Bbb{Q}^\leq)$
of automorphisms of a dense linear order $\Bbb{ Q}$: an $n$-simplex is an orbit of the action on *non-decreasing* $n+1$-tuples.
This simplicial set has the property that in each dimension there is a unique non-degenerate simplex, and, moreover,
all its faces are non-degenerate, and is weakly contractible, according to [answers here](https://mathoverflow.net/questions/439687/the-simplicial-set-with-a-unique-non-degenerate-simplex-in-each-dimension).If you truncate the simplicial set, for odd dimensions you get a sphere up to homotopy, and for even
dimensions something weakly contractible, e.g. for $n=2$ the dunce hat.
| https://mathoverflow.net/users/494312 | Defining the classifying space of a group acting on a set | As in Tom Goodwillie's comment, if you take the construction that you discuss for a $G$-set $X$, the equivariant homotopy type of the space obtained before you quotient out by the action of $G$ is called a classifying space for a family of subgroups of $G$; the family being the subgroups that fix some point of $X$.
There is a survey by Wolfgang Lueck "Survey on Classifying Spaces for Families of Subgroups", available on arXiv at <https://arxiv.org/abs/math/0312378> or as a chapter of a volume published by Birkhauser <https://link.springer.com/chapter/10.1007/3-7643-7447-0_7>. The survey considers topological groups, but as he says in the introduction, if you just want to consider discrete groups you should just ignore Sections 2 and 3.
The cases when stabilizers in $X$ contain exactly the finite subgroups and exactly the virtually cyclic subgroups have attracted a lot of interest, partly due to their appearance in "assembly conjectures" as discussed in Section 7 of Lueck's survey.
| 4 | https://mathoverflow.net/users/124004 | 440296 | 177,794 |
https://mathoverflow.net/questions/440271 | 5 | Let $u \in W^{1,2}(\mathbb{R}^2)$ be a given function satisfying $$\frac{1}{|B\_r|}\int\_{B\_r(x)} |\nabla u|^2 dy \leq \frac{1}{r^{\delta}}$$ for all $r \leq 1$, $x \in \mathbb{R}^2$ and some fixed $\delta \in (0,1)$.
Hueristically, this looks like that the gradient cannot go to infinity too quickly. In this sense, does it rigorously imply that $\nabla u \in L^{2+\epsilon}\_{loc}$ for some $\epsilon >0$?
| https://mathoverflow.net/users/100801 | Higher integrability for Sobolev functions | Since the [OP asked for a discussion of features](https://mathoverflow.net/questions/440271/higher-integrability-for-sobolev-functions#comment1135653_440271), I provide one by way of an explanation of Christian Remling's counterexample:
Holder's inequality states that
$$ | \int fg | \leq \| f\|\_p \|g\|\_q $$
if $p^{-1} + q^{-1} = 1$, and the dual characterization of $L^p$ states that
$$ f \in L^p \iff \exists M \forall g\in L^q, \int fg \leq M \|g\|\_q $$
Your question can be interpreted as "does the dual characterization of $L^p$ work if, instead of all $g$ in $L^q$, we only look at those $g$s which are characteristic functions of balls (here we use that for $g$ the characteristic function of $B(x,r)$ in $\mathbb{R}^n$, its $L^q$ norm is $\approx r^{n/q}$).
While one may expect that the set of characteristic functions of balls form a nice "basis" for $L^q$, the problem however is that we don't see some convexity properties: knowing that the bounds hold for $g\_1$ and $g\_2$ only gives
$$ \int f (g\_1 + g\_2) \leq M (\|g\_1\|\_q + \|g\_2 \|\_q) $$
the right hand side of which is not $\leq \|g\_1 + g\_2\|\_q$. In Christian Remling's answer his $f$ is built out of functions
$$ \chi\_{B(0, 2^{-k} + 2^{-k^2 - k})} - \chi\_{B(0,2^{-k})} $$
We have for this function, in $\mathbb{R}^n$
$$ \|\chi\_{B(0, 2^{-k} + 2^{-k^2 - k})} - \chi\_{B(0,2^{-k})}\|\_q \approx 2^{-nk/q - k^2/q} \ll 2^{-nk/q} \approx \|\chi\_{B(0, 2^{-k} + 2^{-k^2 - k})}\|\_q + \|\chi\_{B(0,2^{-k})}\|\_q $$
These functions are, in some sense, "orthogonal" to the characteristic functions of balls, and so their largeness are not detectable by integration against balls.
---
**Edit**: since there are some questions about the actual construction of the counterexample, here's an outline of how one can construct one.
1. We will work in $\mathbb{R}$ for simplicity. You can do the same on $\mathbb{R}^2$ using balls instead of intervals.
2. We will study $f = |\nabla u|^2$ to simplify notation. Throughout $f$ will be non-negative. The requirements are now
$$ \int\_{I} f \leq r^{1-\delta} \tag{req}$$
for all $r\in (0,1]$, and $I$ an interval of width $r$.
3. Given an interval $I$ with with $|I| \leq 1$. Let $J$ be the interval with the same center as $I$, but width $|J| = \frac1{3^{1/(1-\delta)}} |I|^{1/(1-\delta)} \leq\frac13 |I|$. Observe that as $I$ shrinks in length, the interval $J$ shrinks faster. Denote by
$$ f\_I(x) = \begin{cases} \frac{1}{|J|^\delta} & x\in J \\ 0 & x\not\in J \end{cases} $$
One can check that $f\_I$ satisfies the requirement (req), as given any interval $I'$ we have $$\int\_{I'} f\_I \leq |J|^{-\delta} \min(|I'|,|J|) \leq |I'|^{1-\delta}$$
4. Now take $\mathscr{I}$ a collection of disjoint intervals, then $f = \sum\_{I\in \mathscr{I}} f\_I$ is a function that satisfies (req): Given an arbitrary interval $I'$, if it only intersects one of the $J$s from our construction, then by the previous part we have that the inequality holds. When $I'$ spans multiple $J$s, let $\mathscr{I}'$ be the set of the corresponding $I$s, we have
$$ \int\_{I'} f \leq \sum\_{I\in \mathscr{I}'} |J|^{1-\delta} = \sum\_{I\in \mathscr{I}'} \frac13 |I| $$
But the length of $I'$ must be at least $\frac13 \sum\_{I\in \mathscr{I}'} |I|$ as the $I$ in $\mathscr{I}'$ are disjoint intervals, and the $J$s are in their centers.
5. The $L^1$ norm of $f\_I$ is exactly $\frac13 |I|$. For $p > 1$, the $L^p$ norm of $f\_I$ is
$$ \| f\_I\|\_{p}^p = \left(\frac13|I|\right)^{\frac{1-\delta p }{1-\delta}} $$
Note that the exponent is $< 1$.
6. Finally, take $w\_n$ a sequence of numbers, each $\leq \frac13$, such that $\sum w\_n$ converges but $\sum (w\_n)^{\alpha}$ diverges for every $\alpha < 1$ (so something like $w\_n = ( n \ln(n)^2)^{-1}$). Choose a disjoint family of intervals $I\_n$ such that $|I\_n| = w\_n$, and setting $f = \sum f\_{I\_n}$. This gives an example of an $L^1$ function satisfying condition (req) that is not in any $L^p$ for $p > 1$.
| 7 | https://mathoverflow.net/users/3948 | 440301 | 177,797 |
https://mathoverflow.net/questions/440319 | 3 | Let $C$ be a smooth projective geometrically connected curve of genus 2 defined over a number field $k$. Here are some definitions:
The *index* $I$ of a curve $C$ is the greatest common divisor of all effective divisors $D \in \mathrm{Div}(C)$. Equivalently, it is the greatest common divisor of the degrees $[L:k]$, where $[L:k]$ ranges over algebraic extensions such that $C(L) \neq \emptyset$. The *period* $P$ is the smallest positive degree of rational divisor classes, i.e., those given by divisors linearly equivalent to their Galois conjugates.
Let us list 2 lemmas:
>
> For a hyperelliptic curve (which includes all genus 2 curves), we have $I=P$ (see Lemma 5 of *Brauer groups of local elliptic and hyperelliptic curves and central division algebras over their function fields* by Yanchevski and Margolin).
>
>
>
>
> Let $C$ be a smooth projective curve of genus $g \geq 2$. Then $C$ has a closed point of degree at most $2g-2$.
>
>
>
So here is the issue. If $C(k)= \emptyset$, then all closed points are of degree 2. In other words, for any closed point $P$, we can find a degree 2 extension $L/k$ such that $P \in C(L)$. By the definition of the index of a curve, we must have $I = 2$ since for any finite extension $K/k$ such that $C(K) \neq \emptyset$, $K$ must contain a degree 2 extension and thus $[K:k]$ must be even $\implies I = 2$. Then $P = I = 2$ and so the smallest positive degree of $k$-rational divisor classes is 2.
However, in the paper *The Hasse Principle and the Brauer-Manin obstruction for curves* by Flynn, Corollary 4 states that for $k = \mathbb{Q}$, there are genus 2 curves with no rational points but has a degree 1 rational divisor class. What is going on?
| https://mathoverflow.net/users/172132 | If a genus 2 curve has no $k$-rational points, can it have a $k$-rational divisor class of degree 1? | Rummaging a bit through the LMFDB turns up the curve
<https://www.lmfdb.org/Genus2Curve/Q/129600/b/129600/1>
with equation $y^2 = -(2x^3+3x-2)(2x^3+4x^2+x-2)$
with no rational points (indeed trivial Mordell-Weil group)
but a degree-3 divisor $2x^3+3x-2 = y = 0$.
| 7 | https://mathoverflow.net/users/14830 | 440321 | 177,803 |
https://mathoverflow.net/questions/440304 | 15 | The triangle inequality seems much stronger than necessary for a lot of analysis. So I will define a "loose metric" on a set $X$ to be a function $d \colon X \times X \to [0,\infty)$ with the following properties:
* $d(x,y)=d(y,x)\ $ for all $x,y \in X;$
* every $\, x,y \in X\ $ has $\ d(x,y)=0\ $ if and only if $\ x=y$;
* there exists a function $\rho \colon [0,\infty) \to [0,\infty]$, with $\rho(t) \to 0$ as $t \to 0$, such that for all $x,y,z \in X$,
$$ | d(x,z) - d(y,z) | \,\leq\, \rho(\, d(x,y) \, )\text{.} $$
>
> Is the topology generated by a loose metric necessarily metrisable?
>
>
>
Obviously, by "the topology generated by a loose metric $d$ on $X$", I mean the collection of all $d$-open subsets of $X$, where a set $U \subset X$ is called $d$-open if for all $x \in U$ there exists $\delta>0$ such that $\{y \in X : d(x,y) < \delta\} \subset U$. I believe it is easy to show that this topology is a regular Hausdorff topology, and so the Nagata-Smirnov metrisation theorem would give that it is sufficient to show that the topology has a $\sigma$-locally finite basis. However, I suspect that there will be a more direct way of constructing a classical metric on $X$ that is topologically equivalent to a given loose metric $d$.
| https://mathoverflow.net/users/15570 | Is the topology generated by this weaker notion of a metric necessarily metrisable? | For a loose metric $d$ as above, we can consider the function
$$d\_1(x,y):=\sup\{|d(x,z)-d(y,z)|;z\in X\}.$$
It is easy to verify that $d\_1$ is a metric, and $d(x,y)\leq d\_1(x,y)\leq\rho(d(x,y))$ for all $x,y$, so $d\_1$ and $d$ generate the same topology.
Edit: As mentioned in the comments, we can let $d\_2=\min(d\_1,1)$ if we prefer metrics which don't take the value $\infty$.
| 17 | https://mathoverflow.net/users/172802 | 440322 | 177,804 |
https://mathoverflow.net/questions/440312 | 0 | Let $g: \mathbb R \to \mathbb R$ be a function of locally bounded variation, and $f$ a locally integrable function with respect to $dg$, the Lebesgue–Stieltjes measure associated with $g$.
Let $\eta$ be a smooth, compactly supported function. Define
$$F(x) := \int\_0^x f(s) \, dg(s).$$
**Question:** Is it true that
$$(F \ast \eta)(x) = \int\_0^x f \ast \eta (s) \, dg(s)?$$
*Remark: This is true for ordinary integrals, i.e. when $g$ is the identity.*
| https://mathoverflow.net/users/173490 | Does convolution commute with Lebesgue–Stieltjes integration? | If (say) $g$ is absolutely continuous (with an almost-everywhere derivative $g'$), then the left-hand side of your identity is
$$L(\eta):=(F\*\eta)(x)=\int dy\,\eta(x-y)\int\_0^y ds\,g'(s) f(s)$$
and its right-hand side is
$$R(\eta):=\int\_0^x dg(s)\,(f\*\eta)(s)=\int\_0^x ds\,g'(s)\,\int dt\,f(t)\eta(s-t).$$
We see that the identity cannot hold in general, because the arguments of $g'$ and $f$ are the same on the left but different on the right.
---
More specifically, suppose that for all natural $n$ we have $L(\eta\_n)=R(\eta\_n)$, where $f(s)=g'(s)=e^{-s^2}\,1(s>0)$ for real $s$, and the $\eta\_n$'s are smooth compactly supported positive functions such that $\eta\_n\to1$ monotonically pointwise. Then for all real $x>0$ we have $\infty=L(1)=R(1)<\infty$, a contradiction. $\quad\Box$
| 3 | https://mathoverflow.net/users/36721 | 440327 | 177,806 |
https://mathoverflow.net/questions/440307 | 0 | Consider $p(u,x)=(4\pi u)^{-d/2}e^{-\frac{|x|^2}{4u}},u>0,x\in \mathbb{R}^d.$
Prove that there exists $C>0$ such that for all $0<u\leq v,r\in[0,u[,$ $$\int\_{\mathbb{R}^d}|p(v-r,x)-p(u-r,x)|\, dx \leq C\frac{v-u}{u-r}$$
Any ideas how to prove it?
| https://mathoverflow.net/users/138491 | $\int_{\mathbb{R}}|p(v-r,x)-p(u-r,x)|\,dx \leq C\frac{v-u}{u-r}$ | $\newcommand\R{\mathbb R}$Letting $s:=u-r$ and $t:=v-r$, rewrite the inequality in question as
\begin{equation\*}
\int\_{\R^d}dx\,|p(t,x)-p(s,x)| \le C\Big(\frac ts-1\Big) \tag{0}\label{0}
\end{equation\*}
given that $0<s\le t<\infty$.
Note that
\begin{equation\*}
|p(t,x)-p(s,x)|\le\int\_s^t dw\,|D\_w p(w,x)|,
\end{equation\*}
where $D\_w$ is the operator of partial differentiation with respect to $w$. So,
\begin{equation\*}
\int\_{\R^d}dx\,|p(t,x)-p(s,x)| \le\int\_s^t dw\, \int\_{\R^d}dx\,|D\_w p(w,x)|. \tag{1}\label{1}
\end{equation\*}
Next,
\begin{equation\*}
\int\_{\R^d}dx\,D\_w p(w,x)=D\_w\int\_{\R^d}dx\, p(w,x)=D\_w1=0
\end{equation\*}
and
\begin{equation\*}
D\_w p(w,x)=p(w,x)\Big(\frac{|x|^2}{4w^2}-\frac d{2w}\Big).
\end{equation\*}
So, with $z\_+:=\max(0,z)$ for real $z$,
\begin{equation\*}
\int\_{\R^d}dx\,|D\_w p(w,x)|
=2\int\_{\R^d}dx\,p(w,x)\Big(\frac d{2w}-\frac{|x|^2}{4w^2}\Big)\_+
\le2\int\_{\R^d}dx\,p(w,x)\frac d{2w}=\frac dw.
\end{equation\*}
Thus, by \eqref{1},
\begin{equation\*}
\int\_{\R^d}dx\,|p(t,x)-p(s,x)| \le d\,\ln\frac ts,
\end{equation\*}
whence \eqref{0} follows, with $C=d$.
| 1 | https://mathoverflow.net/users/36721 | 440334 | 177,810 |
https://mathoverflow.net/questions/440345 | 1 | Imagine to have a set of random variables $\{ X\_i \}\_{i=1}^{n}$ independent (**Non** identically distributed). In these scenario, if the Lindeberg's condition hold we can extend the result of the CLT, *i.e.*, calling $\mu\_i = \mathbb{E}(X\_i)$ , $s\_n^2 = \sum\_i \left(\mathbb{E}(X\_i^2) - \mathbb{E}(X\_i)^2\right)$ :
$$
\frac{\sum\_{i=1}^n X\_i-\mu\_i}{s\_n} \xrightarrow{n\rightarrow\infty} \mathcal{N}(0,1)
$$
(for more details see for example <http://shannon.cm.nctu.edu.tw/prob/c27s08.pdf> or <https://drive.google.com/file/d/1beeTLVbuXEQabqUHReI3K0eJiZ7xfek3/view>).
Now even if the set of random variable is not identically distributed, assume that the distribution of $X\_i$ respect the CLT hypothesis $\forall i$ (**Hp 1**).
Is this a sufficient hypothesis for the Lindeberg's condition to hold and get the CLT for the set $\{ X\_i \}\_{i=1}^{n}$? If yes how could I show it formally, *i.e.* that (**Hp 1**) implies the Lindeberg's condition?
**Follow up question**
Thanks to *Iosif Pinelis* for the nice and clear answer.
Since Lindeberg's condition intuitively guarantees that fluctuations of the random variables from the mean stays "small", I was wondering if assuming finite variances that doesn't grow with $n \rightarrow \infty$, *i.e.* $\textrm{Var} \left( X\_n \right) = \mathcal{O} \left( 1 \right)$ (**Hp 2**) would be a sufficient condition to satisfy the Lindeberg condition.
In the example that you proposed (**Hp 2**) is not satisfied since
$$
\textrm{Var} \left( Y\_n \right) = (n!)^2 2^{-n} \xrightarrow{n\rightarrow\infty} (2 \pi n) \left( \frac{n^2}{e^2 2} \right)^n = \mathcal{O} \left( n^{2n+1} \right)
$$
| https://mathoverflow.net/users/174176 | Hypothesis to guarantee Lindeberg's condition | $\newcommand\ep\varepsilon$No, of course the Lindeberg condition is not necessary for the CLT.
E.g., for each natural $i$ let
\begin{equation\*}
X\_i:=Z\_i+Y\_i,
\end{equation\*}
where $Z\_i\sim N(0,1)$, $P(Y\_i=i!)=2^{-i-1}=P(Y\_i=-i!)$, $P(Y\_i=0)=1-2^{-i}$, and the random variables (r.v.'s) $Z\_1,Y\_1,Z\_2,Y\_2,\dots$ are independent.
Then
\begin{equation\*}
\frac1{\sqrt n}\,\sum\_{i=1}^n X\_i=\frac{S\_n}{\sqrt n}+\frac{T\_n}{\sqrt n},
\end{equation\*}
where $S\_n:=\sum\_{i=1}^n Z\_i$ and $T\_n:=\sum\_{i=1}^n Y\_i$.
For any real $\ep>0$, any natural $k$, and any natural $n\ge k$,
\begin{equation\*}
P\Big(\Big|\frac{T\_n}{\sqrt n}\Big|>2\ep\Big)\le P\_{1n}+P\_{2n},
\end{equation\*}
where
\begin{equation\*}
P\_{1n}:=P\Big(\Big|\frac{T\_k}{\sqrt n}\Big|>\ep\Big),\quad
P\_{2n}:=P\Big(\Big|\frac{T\_n-T\_k}{\sqrt n}\Big|>\ep\Big).
\end{equation\*}
Clearly, $P\_{1n}\to0$ for each $k$ as $n\to\infty$. Also,
\begin{equation\*}
P\_{2n}\le\sum\_{i>k}P(Y\_i\ne0)=\sum\_{i>k}2^{-i}=2^{-k}\to0
\end{equation\*}
as $k\to\infty$. So, $P\big(\big|\frac{T\_n}{\sqrt n}\big|>2\ep\big)\to0$ as $n\to\infty$ for each real $\ep>0$. That is, $\frac{T\_n}{\sqrt n}\to0$ in probability and hence in distribution. Also, $\frac{S\_n}{\sqrt n}\sim N(0,1)$. Thus, the CLT for the $X\_i$'s holds.
On the other hand,
\begin{equation\*}
s\_n^2=\sum\_{i=1}^n EX\_i^2=\sum\_{i=1}^n EZ\_i^2+\sum\_{i=1}^n EY\_i^2
=n+\sum\_{i=1}^n (i!)^2 2^{-i}\sim (n!)^2 2^{-n}\to\infty \tag{1}\label{1}
\end{equation\*}
(as $n\to\infty$), whereas for each real $\ep>0$ and all large enough $n$ (such that $n!\ge2\ep s\_n$) we have
\begin{equation\*}
\begin{aligned}
s\_{n,\ep}^2&:=\sum\_{i=1}^n EX\_i^2\,1(|X\_i|\ge\ep s\_n) \\
&\ge EX\_n^2\,1(|X\_n|\ge\ep s\_n) \\
& \ge E(Z\_n+Y\_n)^2\,1(|Y\_n|\ge2\ep s\_n,|Z\_n|\le\ep s\_n) \\
& = E(Z\_n^2+Y\_n^2)\,1(|Y\_n|\ge2\ep s\_n)\,1(|Z\_n|\le\ep s\_n) \\
& \ge EY\_n^2\,1(|Y\_n|\ge2\ep s\_n)\,1(|Z\_n|\le\ep s\_n) \\
& = EY\_n^2\,1(|Y\_n|\ge2\ep s\_n)\,P(|Z\_n|\le\ep s\_n) \\
& = (n!)^2 2^{-n}\,P(|Z\_n|\le\ep s\_n) \\
& = (n!)^2 2^{-n}\,P(|Z\_1|\le\ep s\_n) \\
& \sim (n!)^2 2^{-n}\sim s\_n^2,
\end{aligned}
\end{equation\*}
by \eqref{1}. Thus, the Lindeberg condition $s\_{n,\ep}^2/s\_n^2\to0$ fails to hold. $\quad\Box$
---
Then OP additionally asked if there is a counterexample with $s\_n^2$ bounded. Here is a simple one: Let $X\_1\sim N(0,1)$ and $X\_2=\cdots=X\_n=0$. Then $\sum\_{i=1}^n X\_i\sim N(0,1)$, whereas $s\_n^2=1$
\begin{equation\*}
s\_{n,\ep}^2=EX\_1^2\,1(|X\_1|\ge\ep s\_n)\not\to0
\end{equation\*}
for any real $\ep>0$. So, the Lindeberg condition $s\_{n,\ep}^2/s\_n^2\to0$ fails to hold. $\quad\Box$
---
Now, the positive news. Suppose that we have a triangular array $(X\_{nk}\colon n=1,2,\dots,\ k=1,\dots,k\_n)$ of independent zero-mean r.v.'s satisfying the normalization condition
\begin{equation}
s\_n^2:=\sum\_k EX\_{nk}^2\to1, \tag{$\*$}\label{0}
\end{equation}
the infinite smallness condition
\begin{equation\*}
\max\_{1\le k\le k\_n}P(|X\_{nk}|\ge\ep)\to0
\end{equation\*}
for each real $\ep>0$, and the CLT condition
\begin{equation\*}
S\_n:=\sum\_k X\_{nk}\to Z \tag{2}\label{2}
\end{equation\*}
in distribution, where $Z\sim N(0,1)$.
Then, by Theorem 15 of Chapter IV by [Petrov](https://books.google.com/books?id=zSDqCAAAQBAJ&printsec=frontcover#v=onepage&q&f=false), for every real $\ep>0$
\begin{equation\*}
\sum\_k P(|X\_{nk}|\ge\ep)\to0 \tag{3}\label{3}
\end{equation\*}
and
\begin{equation\*}
\sum\_k(EY\_{nk}^2-E^2Y\_{nk})\to1, \tag{4}\label{4}
\end{equation\*}
where $Y\_{nk}:=X\_{nk}\,1(|X\_{nk}|<\ep)$.
Since the $X\_{nk}$'s are zero-mean, for any real $\ep>0$ and any real $M>\ep$ we have
\begin{equation\*}
-EY\_{nk}=EX\_{nk}\,1(|X\_{nk}|\ge\ep)=EX\_{nk}\,1(|X\_{nk}|\in[\ep,M))+EX\_{nk}\,1(|X\_{nk}|\ge M),
\end{equation\*}
\begin{equation\*}
\sum\_k|EX\_{nk}\,1(|X\_{nk}|\in[\ep,M))|\le \sum\_k MP(|X\_{nk}|\ge\ep)\to0
\end{equation\*}
by \eqref{3},
\begin{equation\*}
\sum\_k|EX\_{nk}\,1(|X\_{nk}|\ge M)|\le\frac1M\,\sum\_k EX\_{nk}^2\to0
\end{equation\*}
by \eqref{0} as $M\to\infty$. So,
\begin{equation\*}
\sum\_k|EY\_{nk}|\to0.
\end{equation\*}
Also, $|Y\_{nk}|\le\ep$ and hence $E^2Y\_{nk}\le\ep|EY\_{nk}|$. So,
\begin{equation\*}
\sum\_k E^2Y\_{nk}\to0.
\end{equation\*}
So, by \eqref{4},
\begin{equation\*}
\sum\_k EY\_{nk}^2\to1. \tag{5}\label{5}
\end{equation\*}
So, for
\begin{equation\*}
s\_{n,\ep}^2:=\sum\_k EX\_{nk}^2\,1(|X\_{nk}|\ge\ep)
\end{equation\*}
we get
\begin{equation\*}
\frac{s\_{n,\ep}^2}{s\_n^2}\sim s\_{n,\ep}^2=\sum\_k EX\_{nk}^2-\sum\_k EX\_{nk}^2\,1(|X\_{nk}|<\ep)
=\sum\_k EX\_{nk}^2-\sum\_k EY\_{nk}^2\to1-1=0,
\end{equation\*}
so that the Lindeberg condition $s\_{n,\ep}^2/s\_n^2\to0$ holds.
| 2 | https://mathoverflow.net/users/36721 | 440362 | 177,823 |
https://mathoverflow.net/questions/440326 | 3 | I apologize in advance if the answer to this question is well-known to experts.
So let $F$ be a $p$-adic field, and $G$ a reductive group over $F$. Let $P$ be an $F$-parabolic subgroup of $G$ and $M$ the Levi of $P$. (For convenience, I just use the same notations for an algebraic group and a group of rational points.) We know that roughly, irreducible admissible (smooth) representations of $G$ come from (normalised) parabolic induction: if $\pi$ is a non-cuspidal irreducible representation of $G$, then $\pi$ can be embedded into some $I\_Q^G(\sigma)$ for some representation $\sigma$ of Levi of some proper parabolic $Q$.
Meanwhile, we know what is local Langlands correspondence. Let $WD(F):=W(F)\times\operatorname{SL}\_2(\mathbb{C})$ be the Weil-Deligne group, then an $L$-parameter of $G$ is a conjugacy class of homomorphism
$$\phi:WD(F)\rightarrow {^LG}.$$
Suppose now we have an irreducible representation $\sigma$ of $M$ (setting as above), and then we consider $I\_P^G(\sigma)$, and take some irreducible composition factor $\pi$, which is an irreducible representation of $G$. Suppose $\phi\_\sigma:WD(F)\rightarrow {^LM}$ is the $L$-parameter of $\sigma$. We know that $M\subset G$ provides us with ${^LM}\subset {^LG}$, so we get a parameter of $G$: $WD(F)\rightarrow {^LM}\subset {^LG}$.
**Question:** some expert told me that this new parameter is just the parameter for the **Langlands quotient** of $I\_P^G(\sigma)$. This is related to the **Langlands classification** of $p$-adic groups (for which I'm a total beginner). I want to ask where can I find the proof of this fact in the literature? And I also want to ask: in general, is there any systematic relation between the parameter of $\sigma$ and the parameter of $\pi$ (a composition factor of the parabolic induction)? How to characterise tempered/discrete series/cuspidal representations via parameters? (I know such facts about $\operatorname{GL}\_n$ from Kudla's famous survey article, but I'm curious about whether there's a general result for general reductive groups, which so far I didn't find in any paper, but seems to be folklore to experts...)
Again, I want to apologize if this question is standard for veterans/experts. Thanks a lot in advance for any explanation!
| https://mathoverflow.net/users/149922 | $L$-parameters and parabolic induction | I think what you say is a part of the local Langlands Conjecture. See Conjecture 4.1 (7)(8)(10)in Kaletha and Taibi's Lecture notes on LLC for IHES 2022. The local Langlands conjectures for different groups should be compatible with parabolic inductions. As a special case, you can consider spherical representations and calculate their Satake parameters.
For your other problem, here is my understanding (these can also be found in the lecture mentioned above):
* A Langlands parameter is tempered if its image under the projection to $\widehat{G}(\mathbb{C})$ is bounded （Thanks to Kenta's comment, here "the image of the Weil group is bounded" should be more precise）. By the LLC for quasi-split groups, there should be a surjective map from the set of isomorphism classes of tempered irreducible admissible representations to the set of equivalence classes of tempered parameters.
* We should have: the image of a tempered Langlands parameter is not contained in any proper parabolic subgroup of the L-group (equivalently, its centralizer in $\widehat{G}(\mathbb{C})$ is finite modulo $Z(\widehat{G})^{G\_{F}}$) if and only if one element (equivalently, every element) in its L-packet is essentially square-integrable. We call this kind of parameter *essentially discrete*.
* Here we just assume the coefficient field $C=\mathbb{C}$ or $\mathbb{Q}\_{l},l\neq p$ so that we can view cuspidal and supercuspidal as the same thing. We should have: a Langlands parameter is essentially discrete and trivial on Deligne's $\mathrm{SL}\_{2}(\mathbb{C})$ if and only if every member in its L-packet is supercuspidal. However, even for $\mathrm{GL}\_{n}$ it can happen that for an essentially discrete Langlands parameter nontrivial on $\mathrm{SL}\_{2}(\mathbb{C})$ its L-packet contains a supercuspidal representation. Of course you can try to describe the correspondence in terms of *enhanced Langlands parameters*.
* For discrete series, it should correspond to some enhanced Langlands parameter consisting of an essentially discrete Langlands parameter and a character on its component group satisfying some extra condition. Notice that here we need the *refined LLC*.
I am also not so familiar with the representation theory of $p$-adic groups so my answer may contain some errors and comments are welcome!
| 3 | https://mathoverflow.net/users/168680 | 440368 | 177,824 |
https://mathoverflow.net/questions/440353 | 3 | Let $S$ be the set of real matrices with at least one real logarithm. For some couple of its elements, for example those with at least (one real logarithm each with submultiplicative norm smaller than $\ln 2/4$), the product of the couple can be proven to lie in $S$ via the Baker-Campbell-Hausdorff formula.
My question is more general- is S closed under multiplication as a whole? My guess is that one can use Lindelöf summation, which would converge on the Mittag-Leffler star(in this case, $S$ itself), on the BCH series, but I'm not sure if it works on matrix functions as well.
---
Simple proof: Apply Sylvester's formula to each term of the Lindelöf limit of the BCS series and pull the limit out(legal since Sylvester's formula is a finite series).
| https://mathoverflow.net/users/113020 | Is the set of real matrices with at least one real logarithm closed under multiplication? | This is already not true for $2$-by-$2$ matrices: Consider
$$
A = \begin{pmatrix}2 & 0 \\0 &\frac12\end{pmatrix}\quad
\text{and}\quad
B = \begin{pmatrix}-1 & 0 \\0 &-1\end{pmatrix}.
$$
$A = \exp\bigl(\ln(2) K\bigr)$ and $B = \exp\bigl(\pi J\bigr)$ for
$$
K= \begin{pmatrix}1 & 0 \\0 &-1\end{pmatrix}\quad\text{and}\quad J = \begin{pmatrix}0 & 1 \\-1 &0\end{pmatrix}.
$$
However, $AB$ is not $\exp(X)$ for any real $2$-by-$2$ matrix $X$ because the trace of $AB$ is less than $-2$.
| 13 | https://mathoverflow.net/users/13972 | 440373 | 177,828 |
https://mathoverflow.net/questions/440318 | 6 | Suppose $A,G,H$ are finitely generated groups and $A$ is quasi-isometrically embedded into $G$ and $H$. Does it follow that the natural embeddings of $G$ and $H$ into $G\*\_AH$ are quasi-isometric?
I can more or less see that this is true if $A$ is finite (in particular, for the free product), or, more generally, if $G$ and $H$ have finite generating sets which are invariant under conjugation by $A$, but I can't quite solve the general case.
This seems like something that should be well-known (one way or the other), but I could not find any information on this in the literature (not even for the case of the free product). If this is true, I would appreciate a reference. (If necessary, I don't mind assuming that $A,G,H$ are hyperbolic, but this seems like something that should be true in general.)
Edit: I am also happy to assume that the embedding of $H$ into $G$ and $H$ is conjugate-separated i.e. that $A\cap A^g$ is finite for $g\in (G\cup H)\setminus A$.
| https://mathoverflow.net/users/54415 | Are the canonical embeddings into $G*_AH$ quasi-isometric? | In the setting when $G, H$ are hyperbolic groups and $A$ is almost malnormal in $G, H$ (the one you are actually interested in, per your comments), the positive answer is given in
*Kapovich, Ilya*, [**The combination theorem and quasiconvexity**](http://dx.doi.org/10.1142/S0218196701000553), Int. J. Algebra Comput. 11, No. 2, 185-216 (2001). [ZBL1025.20028](https://zbmath.org/?q=an:1025.20028).
On the other hand, if you only assume hyperbolicity of $G, H$ and $G\*\_AH$, the subgroups $G, H$ need not be quasiconvex in $G\*\_AH$.
Examples that I know come from topology. Suppose that $S$ is a closed nonorientable hyperbolic surface, $F\to S$ is its orientation double-cover, $C(F\to S)$ the mapping cylinder. The latter is a 3-dimensional manifold with boundary (homeomorphic to $F$). Take two copies $M\_1, M\_2$ of $C(F\to S)$ and a pseudo-Anosov homeomorphism $h: \partial M\_1\to \partial M\_2$ (this makes sense due to canonical hoomeomorpisms of these boundaries to $F$). Glue $M\_1, M\_2$ via $h$. The result is a closed hyperbolic 3-manifold $M$. Of course,
$$
\pi\_1(M)=\pi\_1(M\_1)\*\_{\pi\_1(F)} \pi\_1(M\_2)
$$
and amalgam of two hyperbolic groups over a hyperbolic subgroup of index $2$. At the same time, $M$ admits a 2-fold covering $p: N\to M$ such that $N$ is fibered over the circle, such that the fiber corresponds to a connected component of $p^{-1}(F)$. The fundamental group of the fiber is not quasiconvex in $\pi\_1(N)$, hence, $\pi\_1(F)$ is not quasiconvex in $\pi\_1(M)$.
I am not quite sure what happens if you drop the hyperbolicity assumption but keep (almost) malnormality. I think the answer is still positive.
| 6 | https://mathoverflow.net/users/39654 | 440374 | 177,829 |
https://mathoverflow.net/questions/440377 | 1 | I'm stuck with the following problem:
In Petit's work "[Faster Algorithms for Isogeny Problems using Torsion Point Images](https://eprint.iacr.org/2017/571)", p. 8, he says that we can deduce $\ker \psi\_{N\_2}$ knowing the action of $\psi = \psi\_{N\_1'}\circ\psi\_{N\_2}$ on $E[N\_2],$ where $\psi\_{N\_1'}$ and $\psi\_{N\_2}$ are isogenies of degree respectively $N\_1'$ and $N\_2$ and $\psi = \psi\_{N\_1'}\circ\psi\_{N\_2}$ an endomorphism on the elliptic curve $E$.
Well, I don't understand this statement. I've tried writing $[N\_2] = \psi\_{N\_2}\circ \hat{\psi}\_{N\_2}$ and then fiddling with the expressions, but to no avail.
I have a feeling this problem is easy, but I'm stuck and would appreciate any help/ideas.
| https://mathoverflow.net/users/497497 | Deduce kernel of isogeny from action on torsion points | The key details here are
1. $N\_2$ is "smooth by assumption", so solving discrete logarithms in $E[N\_2]$ is supposed to be easy (using Pohlig-Hellman; how easy this is depends on how smooth $N\_2$ is).
2. $\gcd(N\_1,N\_2) = 1$, so the kernel of $\psi\_2$ is equal to the kernel of the restriction of $\psi$ to $E[N\_2]$.
Now, if you evaluate $\psi$ on a basis of $E[N\_2]$, then by solving
discrete logarithms you can view the restriction of $\psi$ to
$E[N\_2]$ as a matrix over $\mathbb{Z}/N\_2\mathbb{Z}$, and then
computing the kernel (and hence the kernel of $\psi\_2$) is a matter of linear algebra.
| 2 | https://mathoverflow.net/users/156215 | 440386 | 177,833 |
https://mathoverflow.net/questions/440379 | 2 | I was wondering whether anything is known on the following: Let
$h\_k (x)= (-1)^k e^{x^2/2} \frac {d^k}{dx^k} \, e^{-x^2/2}$, $k \geq 0$, be the classical
Hermite polynomials ($h\_0(x) = 1$, $h\_1(x) = x$,
$h\_2(x) = x^2 -1$, $h\_3(x) = x^3 - 3x$, ...).
Is there anything known on the asymptotic distributional behavior (if any), as $k \to \infty$,
of the sequence
of random variables $X\_k$, $k \geq 0$, such that the law of $X\_k$ has a density
proportional to $h\_k^2 \, e^{-x^2/2}$, $x \in \mathbb{R}$, with respect to the
Lebesgue measure?
| https://mathoverflow.net/users/499080 | Random variables with density distributions given by squared Hermite polynomials | This is related to the probability density of the [quantum harmonic oscillator.](https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator) (The Hermite polynomials are the eigenstates of that problem.) For large $k$ the density
$$P\_k(x)=\frac{1}{\sqrt{2\pi}k!} [h\_k(x)]^2 \, e^{-x^2/2},$$
normalized to unity, has the approximation
$$p\_k(x)=\frac{1}{\pi}(4k-x^2)^{-1/2},\;\;|x|<2\sqrt{k}.$$
The boundaries $\pm 2\sqrt{k}$ are the classical turning points of the harmonic oscillator. In that context $E=4k$ is the energy of the particle and $V(x)=x^2$ is the harmonic potential in which it moves. The large-$k$ approximation is known as the [WKB approximation](https://en.wikipedia.org/wiki/WKB_approximation). More generally, for an arbitrary potential $V(x)$ the density varies as $[E-V(x)]^{-1/2}$.
Alternatively, one can use the asymptotic expansion of the [Hermite polynomials,](https://en.wikipedia.org/wiki/Hermite_polynomials)
$$e^{-x^2/4}h\_k(x) \propto \cos \left(x \sqrt{ k}- \pi k/2 \right)\left(4k-x^2\right)^{-1/4}.$$
Squaring and averaging over the rapidly oscillating $\cos^2$ gives $p\_k(x)$.
---
The plot compares $P\_k$ and $p\_k$ for $k=100$. The function $p\_k$ is a smoothed version of $P\_k$, suitable for averaging quantities that vary little on the scale of $1/\sqrt{k}$. For example, one can check by explicit computation that the average of $x^n$, $n\in\mathbb{N}$, agrees to leading order in $k$,
$$\lim\_{k\rightarrow\infty} k^{-n}\int P\_k(x)x^{2n}\,dx=\frac{2^{2n}\,\Gamma(n+1/2)}{\sqrt{\pi}\,\Gamma(n+1)}=k^{-n}\int p\_k(x) x^{2n}\,dx.$$
| 0 | https://mathoverflow.net/users/11260 | 440397 | 177,836 |
https://mathoverflow.net/questions/440365 | 4 | Let $A$ be a commutative ring, and $M$ be an $A$-module, and $M^\*$ be $\mathrm{Hom}\_A(M,A)$. Let $f$ be the map from $M \otimes\_A M^\*$ to $\mathrm{Hom}\_A(M,M)$, such that, for all $x=\sum\_i a\_i \otimes b\_i \in M \otimes\_A M^\*$, $f(x)$ is the homomorphism $y \in M \mapsto \sum\_ib\_i(y)a\_i \in M$.
Is it true that $f$ is always a monomorphism ?
If not, is there a necessary and sufficient condition on $M$ for $f$ to be a monomorphism ?
| https://mathoverflow.net/users/456131 | Tensor product and homomorphism | Here is how I would start.
Let $S$ be the class of all pairs of $A$-modules $(M,N)$ such that the canonical map
$$M \otimes N^\* \to \hom(N,M), ~ m \otimes \omega \mapsto (n \mapsto \omega(n) \cdot m)$$
is a monomorphism. Both sides are additive functors in both variables. It follows formally that $S$ is closed under finite direct sums (in both variables) as well as under direct summands (in both variables).
Clearly, $(A,N) \in S$ and $(M,A) \in S$. It follows that $(M,N) \in S$ whenever $M$ and $N$ are finitely generated and projective (in which case the map is actually an isomorphism).
We can also prove that $S$ is closed under infinite direct sums in the *first* variable: Let $I$ be any index set and $(M\_i,N) \in S$ for $i \in I$, then $(\bigoplus\_{i \in I} M\_i,N) \in S$: This is because the map
$$\textstyle (\bigoplus\_{i \in I} M\_i) \otimes N \to \hom(N,\bigoplus\_{i \in I} M\_i)$$ is mono iff the composition
$$\textstyle (\bigoplus\_{i \in I} M\_i) \otimes N \to \hom(N,\bigoplus\_{i \in I} M\_i) \hookrightarrow \hom(N,\prod\_{i \in I} M\_i)$$
is mono, which identifies with the mono
$$\textstyle \bigoplus\_i (M\_i \otimes N) \hookrightarrow \bigoplus\_{i \in I} \hom(N,M\_i) \hookrightarrow \prod\_{i \in I} \hom(N,M\_i).$$
In particular $(M,N) \in S$ whenever $M$ is a free module, and hence also when $M$ is a projective module. (This special case has been mentioned in the comments already, but I wanted to show a conceptual proof.)
Remark: $S$ is not closed under infinite direct sums in the second variable. In fact, it can happen that $M \otimes A^I \to M^I$ is not a mono, so that $(M,\bigoplus\_{i \in I} A) \notin S$.
The next case would be to look at two cyclic modules, say
$$M = A/I,~N = A/J$$
for two ideals $I,J$. Then
$$M \otimes N^\* \cong A/I \otimes \mathrm{Ann}(J) \cong \mathrm{Ann}(J)/(I \cdot \mathrm{Ann}(J))$$ and
$$\hom(N,M) \cong \{[a] \in A/I : aJ \subseteq I\} = (I:J)/I.$$
The canonical map then identifies with $[a] \mapsto [a]$. This is clearly not a monomorphism in general, even for $I=J$ in which case we just get the map $$\mathrm{Ann}(I) \to A/I,~a \mapsto [a]$$
with kernel $I \cap \mathrm{Ann}(I)$. This kernel can be non-trivial: take $A = k[X]/\langle X^2 \rangle$ and $I = \langle X \rangle$.
| 8 | https://mathoverflow.net/users/2841 | 440399 | 177,838 |
https://mathoverflow.net/questions/440394 | 5 | Consider the space $\mathcal{S}'$ of functions $\mathbb{R}^n\to\mathbb C$ that are (real-)analytic and with exponential decay at infinity. This is an analogue of [Schwartz space](https://en.wikipedia.org/wiki/Schwartz_space), but real-analytic rather than smooth. (Exponential decay [suffices](https://mathoverflow.net/q/23679/) for the Fourier transform being analytic.)
Consider the commutative unital ring $\mathcal{D}$ of [pseudodifferential operators](https://en.wikipedia.org/wiki/Pseudo-differential_operator) $D$ of the form
$$(Df)(x)=\frac1{(2\pi)^n}\int\mathrm{d}^n\xi\,\exp(\mathrm ix\cdot\xi)p\_D(\xi)\,\hat f(\xi)$$
where $p\_D$ is an analytic function on $\mathbb R^n$ that is slowly growing (i.e. is a [tempered distribution](https://en.wikipedia.org/wiki/Distribution_(mathematics)#Tempered_distribution)). In particular, it includes the ring $\mathbb R[\partial\_1,\dotsc,\partial\_n]$ of differential operators (with constant coefficients) in the special case where $p\_D$ is a polynomial.
Now, I’d like the following to hold:
1. $\mathcal{S}'$ is closed under pointwise products. (That is, it forms a non-unital associative commutative algebra over $\mathbb R$.)
2. $\mathcal{S}'$ is closed under the action of $\mathcal{D}$. (That is, it forms a module over $\mathcal{D}$.)
3. An analytic [Leibniz rule](https://en.wikipedia.org/wiki/Product_rule) holds in the sense that, for $f,g\in\mathcal{S}'$ and $D=\sum\_{I\in\mathbb N^n}c\_I\partial\_I$,
$$D(fg)=\sum\_{I\in\mathbb N^n}\sum\_{\substack{I',I''\in\mathbb N^n\\I'+I''=I}}c\_I\binom I{I'}(\partial\_{I'}f)(\partial\_{I''}g),$$
where $I$ is a [multi-index](https://en.wikipedia.org/wiki/Multi-index_notation) and where convergence is taken in some suitable topology.
4. We have $\mathcal{S}'\otimes\_{\mathcal{D}}\mathcal S'=\mathcal{S}'$. (That is, the evident inclusion $\mathcal{S}'\otimes\_{\mathcal{D}}\mathcal S'\subset\mathcal{S}'$ is surjective.)
I'd like to know:
1. Do the above desired properties hold? If not, is there some variation of the definitions of $\mathcal{D}$ and $\mathcal{S}'$ that satisfies them?
2. Is the space $\mathcal{S}'$ or some variation thereof discussed anywhere in the literature?
3. $\mathcal{D}$ fails to be a Hopf algebra (unlike the ring of true differential operators $\mathbb R[\partial\_1,\dotsc,\partial\_n]\subset\mathcal{D}$) because the Leibniz rule yields infinite (rather than finite) sums. But can it be regarded as some sort of topological Hopf algebra, in which the coproduct $\Delta\colon\mathcal{D}\to\mathcal{D}\mathbin{\hat\otimes}\mathcal{D}$ takes values in some completed tensor product $\mathcal{D}\mathbin{\hat\otimes}\mathcal{D}$?
Apologies if I’ve made any mistakes. Analysis is not my usual alley.
*Edit*: I had originally asked the above about $\mathcal S$ being real-analytic functions whose Fourier transforms are real-analytic, but @reuns has quickly supplied the counterexample of $f=\sin(\exp(x^2))\in\mathcal S$ where $f^2\not\in\mathcal S$.
| https://mathoverflow.net/users/499084 | Real-analytic analogue of Schwartz functions | I'll assume that $n=1$.
I'd say that the natural choice is to consider the space $A$ of functions $f:\Bbb{R\to C}$ such that for some $c>0$, $f$ extends analytically to the strip $|\Im(z)|< c$ and $f(x+iy)=O(e^{-c|x|})$ in that strip.
Due to the Cauchy integral theorem this implies that the same holds for $\hat{f}$: for
$|v|<c,|y|<c$
$$\hat{f}(u+iv)=\int\_\Bbb{R} f(x)e^{-i x(u+iv)}dx =
\int\_\Bbb{R} f(x+iy)e^{-i (x+iy)(u+iv)}dx $$
so that $\hat{f}(u+iv)$ is holomorphic on $|v|<c$ and it is $O(e^{-uy})$.
Then it is natural to consider the space $P$ of functions $p: \Bbb{R\to C}$ that extend analytically to some strip $|\Im(z) |< d$ and such that $\forall \epsilon >0, p(x+iy)=O(e^{\epsilon |x|})$ in that strip. This way $A P=A$.
Also $AA = A$ as $u=\frac1{e^{\epsilon z}+ e^{-\epsilon z}+i}$ is in $A$ and $\forall f\in A$ taking $\epsilon $ small enough then $f/u\in A$.
| 4 | https://mathoverflow.net/users/84768 | 440403 | 177,839 |
https://mathoverflow.net/questions/440404 | 0 | I have problems to understand a proof in [this paper](https://www.pierrickdartois.fr/homepage/wp-content/uploads/2022/04/seminar_report.pdf) by Pierrick Dartois on Abelian varieties:
**Theorem 1.13 (rigidity lemma).** Let $ \varphi: X \times\_k Y \to Z$ be a morphism of $k$-schemes. Assume that $X$ is proper and geometrically integral, that $ X \times\_k Y $ is geometrically irreducible, that $ Z$ is separated and that there exist $ k$-valued points $x\_0 ∈ X(k), y\_0 ∈ Y(k) $ and $z\_0 ∈ Z(k)$ such that: $ \varphi(X \times \{y\_0\}) = \{z\_0\} = \varphi({x\_0} \times Y)$ Then, $ \varphi(X \times Y) = \{ z\_0 \}$.
The proof starts with:
**Proof.** One can easily see that the properties of the theorem are invariant under base change, so we can assume that $k$ is algebraically closed. Let $U\_0$ be an open affine neighborhood of $z\_0$ ...
*Question:* Why $ k$ can be assumed to be algebraically closed? Since all properties of the schemes in the assumptions are geometrically, the conditions not change if we replace the schemes by fibre products $ X \times \text{Spec} (\overline{k})$ with algebraic closure of $k$. But having proved the lemma for them, how can we derive the statement for schemes over not algebraically closed $k$?
*Remark:* I noticed that the proof strategy is rather similar with lemma 1.12 in [these notes](http://van-der-geer.nl/%7Egerard/AV.pdf) by by Moonen & Geer with one exception that there is not assumed the existence of a $k$-point $x\_0 ∈ X(k)$. This can be only assured after base change to algebraic closure. But the problem remains the same: how to retain the claim after hsving proved it for algebraic closed base field?
| https://mathoverflow.net/users/108274 | Proof of rigidity lemma | If $\varphi\_{\bar{k}}(X\_{\bar{k}} \times\_{\bar{k}} Y\_{\bar{k}})=\{z\_0\}$ then since $\varphi(X\times Y) \subset \varphi\_{\bar{k}}(X\_{\bar{k}} \times\_{\bar{k}} Y\_{\bar{k}})$ we also have $\varphi(X\times Y)\subset\{z\_0\}$.
| 2 | https://mathoverflow.net/users/327 | 440408 | 177,841 |
https://mathoverflow.net/questions/440251 | 2 | $\DeclareMathOperator\codim{codim}$Let $X=V(I)$, $Y=V(J)$ be two affine varieties. I'd like to know possibile strategies to understand when their schematic intersection, i.e. $X\cap Y=V(I+J)$, is reduced.
I am aware that reduceness is equivalent to Serre's conditions $R\_0$+$S\_1$ (at least under Noetherian hypothesis), but I don't really know how to use them efficiently, at least in my specific case.
More precisely, the case I am interested in is the following:
Let $V=\mathbb A^{2n}$ be an even dimensional vector space, let $X$ be an half dimensional vector subspace and let $Y=\bigcup\_i Y\_i \cup \bigcup\_j Z\_j$, where each $Y\_i$, $Z\_j$ is again a half dimensional vector subspace of $V$ (the reason for using different notations for them will become clear).
In general, $X\cap Y$ is not reduced: for example it's easy to see that, if $X=V(y-x)$ is the diagonal and $Y=V(xy)$ is the cross inside $\mathbb A^2$, then $X\cap Y=V(x^2, y-x)$ is not reduced…so one certainly needs to put more assumptions on $X,Y$.
I'd like to know if the following two assumptions are actually enough to ensure reduceness for $X\cap Y$:
(*i*) The $X\cap Y\_i$'s are all distinct subspaces, of codimension 1 in $X$.
(*ii*) For each $Z\_j$, there exists at least one $Y\_i$ such that $X\cap Z\_j \subsetneq X\cap Y\_i$ (in particular, $\codim\_X X\cap Z\_j\geq 2$).
My intuition is that such an intersection should be at least generically reduced, namely reduced outside the closed locus $\bigcup\_j X\cap Z\_j$, and I hope these conditions are enough to ensure that nothing bad happens even inside such a locus, but I don't know how to prove that (basically I don't know how to exclude embedded primes).
P.S. If this turns out to be actually false, anyway I'd be interested to know which sort of other ‘reasonable’ conditions on the $Y\_i$'s and $Z\_j$'s could ensure reducedness of such an intersection.
| https://mathoverflow.net/users/95513 | How to show that the intersection of two certain affine varieties is reduced? | It seems to me that if there are no $Z$'s, then this works. Indeed, both $X$ and the $Y\_i$ are linear spaces, so the intersection $X\cap Y$ is just a union of pairwise different linear spaces, so it is reduced. I think one could write down this with equations.
However, adding the $Z\_i$'s is problematic. I think the following example works:
Let $Y=Z(xz,xt,yz,yt)\subseteq \mathbb A^4$. Then $Y=Y\_1\cup Z\_1$, where $Y\_1=Z(z,t)$ and $Z\_1=Z(x,y)$. Further let $X=Z(y-z,t)$. Then $X\cap Y\_1=Z(y,z,t)$, a hyperplane in $X$ and $X\cap Z\_1=Z(x,y,z,t)$, which is contained, but not equal to $X\cap Y\_1$. Finally, $X\cap Y=Z(xy,y^2,y-z,t)$ is not reduced.
Here is also a theoretical reason for this example. (This had been the motivation behind my previous attempts at giving an example. In fact, this is essentially the first example I gave, but it didn't work, because I tried to do it without the Z's). Anyway the point is, that $Y$ is a simple example of a union of linear spaces that is not $S\_2$. A hyperplane section of something that's not $S\_2$ is not $S\_1$, so cannot be reduced. The only trick is to figure out how to choose $X$, so its intersection with $Y$ is actually a hyperplane cut of $Y$.
Finally, a little nit picking: $\mathbb A^n$ is **not** a vectorspace, nor are the $X$, the $Y\_i$'s and the $Z\_i$. They are affine spaces. If you want to say that they all share a common point, then say that.
| 1 | https://mathoverflow.net/users/10076 | 440426 | 177,844 |
https://mathoverflow.net/questions/440289 | 4 | $C/ \Bbb{Q}: 3X^3 + 4Y^3 + 5Z^3 = 0$ is known to be a nontrivial element of the Tate–Shafarevich group of the elliptic curve $E/\Bbb{Q}:X^3 + Y^3 + 60Z^3 = 0$. It is also an example of an abelian variety for which finiteness of Sha is known. In fact, $|\mathrm{III}(E/\Bbb{Q})| = 3^2$.
But I have never seen the proof of $|\mathrm{III}(E/\Bbb{Q})| = 3^2$.I don't either know the proof of $[C]$ has order $3$ in $\mathrm{III}(E/\Bbb{Q})$.
Maybe this is first proved by Rubin, but I cannot find article which discusses this example.
Could you give me an reference for the proof?
| https://mathoverflow.net/users/144623 | Tate–Shafarevich group of Jacobian of Selmer curve $3X^3 + 4Y^3 + 5Z^3 = 0$ | $\DeclareMathOperator{\sha}{Ш}$
I am not sure that the proof that Sha has order 9 is anywhere spelled out in full. Here the ideas how to do it.
First, that the order of $C$ is three in the $\sha$ is just saying that it has a point over a field of degree 3 (index=period), which is obvious, and none of degree 1, which was first proven by Selmer. See Cassel's lectures, chapter 18.
A $2$-descent, proves that $E(\mathbb{Q})$ has rank $0$ and that $\sha[2]$ is trivial.
Calculating the torsion subgroup of $E$ (trivial), the Tamagawa numbers (all trivial and the value of $L(E,1)$, either using cm-theory or modular symbols, reveals that the Birch and Swinnerton-Dyer conjecture is equivalent to $\sha$ having $9$ elements.
Rubin's work on the main conjecture of elliptic curves with CM, as in "Tate-Shafarevich groups and $L$-functions of elliptic curves with complex multiplication" for instance, shows that the order of the $p$-primary part of $\sha$ is correctly predicted by BSD for all primes not dividing the order of the units in $\mathbb{Q}(\sqrt{-3})$. For our curve this means $\sha[p]=0$ for all primes $p>3$. But this does not cover the very bad prime $3$; though maybe has done this since, I don't know.
Selmer's original work proved that $\sha[3]$ contains $9$ elements, by doing a second descent $E\to E'$ via an isogeny of degree $3$. This exact example appears in magma's documentation of its ThreeDescent function: <https://magma.maths.usyd.edu.au/magma/handbook/text/1515#17622> .
More interestingly this calculation also shows that the $3$-Selmer group of the curves $E'$ at the other end of the $3$-isogeny is trivial. Hence the $3$-part of BSD is true for that curve and since this is invariant under isogeny, it is also true for ours. Therefore $\sha$ has order $9$ with the structure being $\mathbb{Z}/3\mathbb{Z}\times \mathbb{Z}/3\mathbb{Z}$.
This last step should also be possible using Heegner points with $D=-17$. Sage uses this in "prove\_BSD" for the curve $E'$.
| 6 | https://mathoverflow.net/users/5015 | 440441 | 177,846 |
https://mathoverflow.net/questions/440348 | 1 | Let $\frak{g}$ be a complex semisimple Lie algebra with root lattice $Q$ and positive weight space $P^+$. Let $\lambda, \mu \in Q \cap P^+$, with corresponding respective fin-dim irreducible representations $V\_{\lambda}$ and $V\_{\mu}$. Form the tensor product $V\_{\mu} \otimes V\_{\lambda}$ and then decompose it into irreducibles
$$
V\_{\mu} \otimes V\_{\lambda} \simeq V\_{\gamma\_1} \oplus \cdots \oplus V\_{\gamma\_m}.
$$
Will the dominant weights $\gamma\_i$ still be contained in $Q \cap P^+$ or will they in general only be contained in $P^+$? If they are only contained in $P^+$, then what is a simple example that illustrates this?
Finally, if in general there exist such $\gamma$'s that live outside $Q \cap P^+$, can we categorize/classify the subset of all such dominant weights?
| https://mathoverflow.net/users/378228 | Tensoring irreducible representations corresponding to root lattice elements | Just to summarize what was mentioned in the comments ([1](https://mathoverflow.net/questions/440348/tensoring-irreducible-representations-corresponding-to-positive-root-lattice-ele#comment1136048_440348) [2](https://mathoverflow.net/questions/440348/tensoring-irreducible-representations-corresponding-to-positive-root-lattice-ele#comment1136051_440348)) and to have this question marked as answered, what follows is the explanation of why the answer is *yes*, the irreps that appear in the decomposition of $V\_{\mu}\otimes V\_{\lambda}$ will be indexed by root lattice elements.
First notice that every weight that appears in both $V\_{\mu}$ and $V\_{\lambda}$ must also belong to the root lattice (because we can get to every weight from the highest weight by going down along simple roots: this is a well-known fact about the structure of irreps for semisimple Lie algebras). Then note that $V\_{\mu}\otimes V\_{\lambda}$ is spanned by $u\otimes v$ where $u$ is a weight basis element of $V\_{\mu}$ and $v$ a weight basis element of $V\_{\lambda}$. But the weight of $u\otimes v$ is the sum of the weights of $u$ and of $v$. So all the $u\otimes v$ have weights that belongs to the root lattice. In particular, all highest weight vectors do too, hence all irreps that appear are indexed by root lattice elements.
| 3 | https://mathoverflow.net/users/25028 | 440449 | 177,851 |
https://mathoverflow.net/questions/440439 | 1 | I was working more on the topic on my previous question when I have to know whether the following statement is true to circumvent the "exception" caused by division by singular matrices; again, long story short, the statement follows:
If two singular matrices $A, B$ exist s.t. the determinant of $EA-B$ is identically zero for all real matrices $E$, then either $A=YB$ or $B=ZA$, $Y$ and $Z$ being undetermined matrices.
Is it true (vacuously or not) in general?
| https://mathoverflow.net/users/113020 | Singularity of matrix pencil-like expression | No.
The first condition is satisfied if (and only if) there is some vector in the kernel of $A$ that is also in the kernel of $B$.
The second condition is satisfied (if and) only if the kernel of $A$ is contained in the kernel of $B$ or the kernel of $B$ is contained in the kernel of $A$.
To make a counterexample, we choose any two subspaces with nonempty intersection but neither contained in the other, and choose matrices with those kernels. This first occurs in dimension 3, with two 2-dimensional subspaces, giving [Federico Poloni's answer](https://mathoverflow.net/a/440447/18060).
| 2 | https://mathoverflow.net/users/18060 | 440450 | 177,852 |
https://mathoverflow.net/questions/440412 | 0 | Let $k$ be an algebraically closed field. Let $X\subset \mathbb{A}^n\_{k}$ be a conical closed subvariety. In other words,
$\mathcal{O}(X)=k[x\_1,\cdots, x\_n]/I$, where $I$ is generated by homogeneous polynomials. Assume also that $X$ is a normal variety. In this general setting, is there anything known about the Picard group of $X$? For example, is it always true that
the Picard group of $X$ is trivial? (This is my naive hope.)
| https://mathoverflow.net/users/481692 | Picard group of a normal conical affine variety | Let $X$ be the affine cone over a normal projective variety $Y$.
Let
$$
\pi \colon \tilde{X} \to X
$$
be the blowup of $X$ at the vertex. Then $\tilde{X}$ comes with a projection
$$
p \colon \tilde{X} \to Y
$$
that identifies $\tilde{X}$ with the total space of a line bundle on $Y$ (the restriction to $Y$ of the tautological bundle from the ambient projective space). Moreover, the zero section of this line bundle coincides with the exceptional divisor $E \subset \tilde{X}$ of the blowup; in particular, $p \colon E \to Y$ is an isomorphism.
Now, the pullback map
$$
\pi^\* \colon \mathrm{Pic}(X) \to \mathrm{Pic}(\tilde{X})
$$
is injective and its image is contained in the subgroup
$$
\{ L \in \mathrm{Pic}(\tilde{X}) \mid L\vert\_E \cong \mathcal{O}\_E \}.
\tag{\*}
$$
On the other hand, the map
$$
p^\* \colon \mathrm{Pic}(Y) \to \mathrm{Pic}(\tilde{X})
$$
is an isomorphism, and since the composition of $p^\*$ and the restriction to $E$ is an isomorphism, it follows that the subgroup $(\*)$ is zero,
hence $\mathrm{Pic}(X) = 0$.
| 2 | https://mathoverflow.net/users/4428 | 440453 | 177,853 |
https://mathoverflow.net/questions/440443 | -1 | Let $(X,D)$ be a pair and $f:Y\rightarrow X$ a log resolution. Write
$$
K\_Y + \widetilde{D} = f^{\*}(K\_X) + \sum\_{i}a\_iE\_i
$$
where $\widetilde{D}$ is the strict transform of $D$. I found the following definition:
the pair $(X,D)$ is $(t,c)$ is $X$ is terminal and $(X,D)$ is canonical meaning that $a\_i\geq 0$ for all $i$.
Now, assume that $X$ is smooth and $D = D\_1 + \dots + D\_r$, where the $D\_i$ are prime divisors, is simple normal crossing. Then the identity is a log resolution and hence $(X,D)$ is $(t,c)$. Is this correct? Or must one interpret the absence of exceptional divisors as the possibility of having arbitrarily negative discrepancies?
I am asking since this does not seem to match the arguments in a paper that I am reading. Are there different definitions of $(t,c)$ pair?
For instance is $\mathbb{P}^2$ with the divisor $D = \{xyz=0\}$ a canonical pair?
Thanks a lot.
| https://mathoverflow.net/users/14514 | Definition of canonical pair | You are using the wrong equation to compute discrepancies. It should be
$$ K\_Y = f^\*(K\_X + D) + \sum a\_E(X,D) E $$
where the $E$ are not all necessarily exceptional.
For example if $(X,D)$ is already smooth and simple normal crossing then
$$ K\_X = \operatorname{id}\_X^\*(K\_X + D) - D $$
and so every irreducible component $D\_i\subset D$ has discrepancy $a\_{D\_i}(X,D)=-1$. Thus the simple normal crossing pair $(X,D)$ is *log canonical*, and not canonical.
| 2 | https://mathoverflow.net/users/104695 | 440456 | 177,854 |
https://mathoverflow.net/questions/440491 | 6 | The motivation for this question is the startling fact that there is an order-preserving injective map (embedding) from $\mathbb{R}$ into ${\cal P}(\omega)$. (Think [Dedekind cuts](https://en.wikipedia.org/wiki/Dedekind_cut).) I am wondering how "large" a linear order can become and still be embeddable in ${\cal P}(\omega)$. To be able to formalize this, I would like to focus on well-orders.
**Question.** What is the smallest ordinal $\mu$ such that there is no injective order-preserving map $\varphi:\mu\to{\cal P}(\omega)$?
| https://mathoverflow.net/users/8628 | Smallest ordinal $\mu$ not embeddable in ${\cal P}(\omega)$ | This is just $\omega\_1$. The naive argument ("pick the least new thing") that no uncountable linear order embeds into $\mathcal{P}(\omega)$ actually establishes that no uncountable *well*-order embeds into $\mathcal{P}(\omega)$: if $f$ is an injection of an ordinal $\theta$ into $\mathcal{P}(\omega)$, the map $$\hat{f}:\theta\rightarrow \omega: \alpha\mapsto\min(f(\alpha+1)\setminus f(\alpha))$$ is an injection of $\theta$ into $\omega$. This doesn't contradict the injectibility of $\mathbb{R}$, since there is no injection from $\omega\_1$ into $\mathbb{R}$ either.
(Interestingly, if we order $\mathcal{P}(\omega)$ by **mod-finite** containment instead of genuine containment, we can indeed inject $\omega\_1$.)
---
EDIT: As bof pointed out below, we can improve the above drastically: there is an order preserving (but not *reflecting* of course) injection $\mathcal{P}(\omega)\rightarrow\mathbb{R}$, so the linear orders embeddable in $\mathcal{P}(\omega)$ are exactly those embeddable in $\mathbb{R}$. Of course that still leaves plenty of open questions (e.g. are any two $\aleph\_1$-dense suborders of $\mathbb{R}$ order-isomorphic?) but I think it lets us stop thinking about $\mathcal{P}(\omega)$ as such.
| 12 | https://mathoverflow.net/users/8133 | 440492 | 177,863 |
https://mathoverflow.net/questions/440480 | 4 | Let $a\_n$ be a sequence of strictly positive real numbers such that $\lim\_{n \to \infty}a\_n=0$. Find all functions $f: \mathbb{R} \to \mathbb{R}$ that admit primitives (i.e. there exists a function $F:\mathbb{R} \to \mathbb{R}$ such that $\frac{dF(x)}{dx}=f(x), \forall x \in \mathbb{R}$) and satisfy the following equality $$2f(x)=f(x+a\_n)+f(x-a\_n), \forall x \in \mathbb{R}, \forall n \in \mathbb{N}$$
I have already posted this question [here](https://math.stackexchange.com/questions/4634456/trying-to-solve-a-functional-equation) but I got no answer.
| https://mathoverflow.net/users/480453 | Solving the functional equation $2f(x)=f(x+a_n)+f(x-a_n)$ | $\newcommand{\De}{\Delta}$This problem can be solved by using the Fourier transform -- cf. [this previous answer](https://mathoverflow.net/a/440186/36721).
Let us present here an elementary solution:
Letting $G\_n(x):=F(x+a\_n)+F(x-a\_n)-2F(x)$, we get $G'\_n(x)=f(x+a\_n)+f(x-a\_n)-2f(x)=0$ for all $x$. So,
\begin{equation\*}
c\_n:=G\_n(x) \tag{1}\label{1}
\end{equation\*}
does not depend on $x$.
Take any real $A$ and $B$ such that $A<B$. Let $k\_n:=\lfloor\frac{B-A}{a\_n}\rfloor$, so that
\begin{equation\*}
k\_n\sim \frac{B-A}{a\_n},
\end{equation\*}
$A+k\_n a\_n\to B$, and, by \eqref{1}, $\De\_x(a\_n):=F(x+a\_n)-F(x)=\De\_{x-a\_n}(a\_n)+c\_n$ for all $x$, so that
\begin{equation\*}
F(A+k\_n a\_n)-F(A)=k\_n \De\_A(a\_n)+\frac{k\_n(k\_n-1)}2\,c\_n \\
=(1+o(1))(B-A)\frac{\De\_A(a\_n)}{a\_n}+\frac{1+o(1)}2\,(B-A)^2\frac{c\_n}{a\_n^2},
\end{equation\*}
whence
\begin{equation\*}
\frac{\De\_A(a\_n)}{a\_n}
=(1+o(1))\frac{F(A+k\_n a\_n)-F(A)}{B-A}-\frac{1+o(1)}2\,(B-A)\frac{c\_n}{a\_n^2}. \tag{2}\label{2}
\end{equation\*}
Similarly, for any real $C>B$,
\begin{equation\*}
\frac{\De\_A(a\_n)}{a\_n}
=(1+o(1))\frac{F(A+m\_n a\_n)-F(A)}{C-A}-\frac{1+o(1)}2\,(C-A)\frac{c\_n}{a\_n^2}, \tag{3}\label{3}
\end{equation\*}
where $m\_n:=\lfloor\frac{C-A}{a\_n}\rfloor$.
The function $F$ is differentiable and hence continuous. So, $F(A+k\_n a\_n)\to F(B)$ and $F(A+m\_n a\_n)\to F(C)$. Subtracting now \eqref{2} from \eqref{3}, we get
\begin{equation}
\frac12\,\frac{c\_n}{a\_n^2}\to\frac{\De\_A(C-A)-\De\_A(B-A)}{C-B}.
\end{equation}
It follows now by \eqref{2} that $\frac{\De\_A(a\_n)}{a\_n}$ converges to a finite limit as well, and thus
\begin{equation}
F(B)-F(A)=K\_1(B-A)+\frac{K\_2}2\,(B-A)^2
\end{equation}
for some real $K\_1,K\_2$ and all real $A$ and $B$ such that $A<B$.
We conclude that $F$ is a quadratic polynomial and hence $f=F'$ is an affine function. (Vice versa, any affine function $f$ satisfies your system of functional equations.)
---
The above proof can be simplified a bit by noting that the limit of $\frac{\De\_A(a\_n)}{a\_n}$ exists (and equals $F'(0)$). However, the advantage of the above proof is that it shows that the conclusion that the only solutions to the system \eqref{1} of functional equations are quadratic polynomials can be reached assuming a priori only the continuity of $F$.
| 3 | https://mathoverflow.net/users/36721 | 440497 | 177,865 |
https://mathoverflow.net/questions/440544 | 3 | Let $s\_1, s\_2, \dotsc$ be a real sequence and define
$$\sigma\_n = \frac{s\_1 + s\_2 + \dotsb + s\_n}{n}.$$ The inequality
$$\operatorname{lim sup}\sigma\_n \leq \operatorname{lim sup} s\_n$$
is well known and trivially proved.
Consider a real valued continuous function $f(x)$ defined on the positive real line and oscillating in sign infinitely often.
The integral analog of the Cesàro mean for a sequence is the Cesàro mean of the integral
$$ s(T) := \int\_0^T f(x) \ dx, $$
which is defined to be
$$ \sigma(T) := \int\_0^T f(x) (1-\frac{x}{T}) \ dx. $$
1. Does the following inequality hold?
$$\operatorname{lim sup}\sigma(T) \leq \operatorname{lim sup} s(T), \quad T \uparrow \infty.$$
2. Suppose $\sigma(T)$ diverges to $\infty$ as $T \uparrow \infty$. Does it follow that $s(T)$ diverges to $\infty$ too?
One expects that the answer to (1) is Yes and the answer to (2) is No but I do not have a proof of the one and a counterexample for the other.
| https://mathoverflow.net/users/156678 | Integral analog of an inequality for the Cesàro mean of a sequence | $\newcommand\si\sigma$Note that
$$\si(T)=\frac1T\,\int\_0^T dx f(x)\,\int\_x^T dt
=\frac1T\,\int\_0^T dt\,\int\_0^t dx\,f(x)
=\frac1T\,\int\_0^T dt\,s(t).$$
Take any real $L>\limsup\_{T\to\infty} s(T)$ (if such $L$ exists, that is, if $\limsup\_{T\to\infty} s(T)<\infty$) and then take any real $A>0$ such that $s(t)\le L$ for all real $t>A$. Then
$$\limsup\_{T\to\infty}\si(T)\le\limsup\_{T\to\infty}\frac1T\,\int\_0^A dt\,s(t)+\limsup\_{T\to\infty}\frac1T\,\int\_A^T dt\,s(t)\le0+L=L,$$
for any real $L>\limsup\_{T\to\infty} s(T)$.
So, the answer to your first question is yes.
---
The answer to your second question is no. E.g., suppose that $s(t)=t\sin^2 t$ for real $t\ge0$. Then $\si(T)=\frac{1}{8} \left(2 T^2-2 T \sin (2 T)-\cos (2 T)+1\right)\to\infty$ but $s(T)\not\to\infty$ as $T\to\infty$.
| 7 | https://mathoverflow.net/users/36721 | 440546 | 177,875 |
https://mathoverflow.net/questions/440549 | 0 | If $p \in \mathbb Z[x]$ has non-negative coefficients $\le n$ and if $q$ is a proper divisor of $p$, are the absolute values of the (integer) coefficients of $q$ bounded by some function of $n$; if so, what is a good bound for the case $n=1$?
| https://mathoverflow.net/users/499203 | Factorising single variable polynomials with non-negative integer coefficients | Cyclotomic polynomials divide $p=1+x+\cdots+x^m$ but the (absolute values of) coefficients of cyclotomic polynomials grow unboundedly: see e.g. “ON THE SIZE OF THE COEFFICIENTS OF THE CYCLOTOMIC POLYNOMIAL” by Bateman, available online at <https://www.jstor.org/stable/44165422>
So there is no bound even in the case $n=1$.
| 2 | https://mathoverflow.net/users/25028 | 440550 | 177,877 |
https://mathoverflow.net/questions/440475 | 2 | Here exponentially localized can be thought in a non-rigorous manner as a measure that is mostly supported on a sparse number of nodes.
Some intuition can gained by thinking about a diffusion process, e.g., we know that in presence of a "confining potential" e.g., $V(x)\approx x^2/2$, the Fokker-Planck equation in 1D will have an exponentially localized stationary distribution (Boltzmann distribution).
$\dfrac{\partial P}{\partial t}-\partial\_x(P\partial\_x V)=\sigma\partial\_{xx}P$
Is there corresponding Markov Chain literature that also studies such localization properties, e.g. providing conditions on the Markov chain transition matrix that lead to such localization ?
| https://mathoverflow.net/users/30684 | When is a stationary measure of a Markov chain "exponentially localized"? | I will give some answers in terms of diffusion processes, since the examples are easiest for me to describe in that context. There are more general examples which follow the same pattern, but typically require additional care to state rigorously.
Suppose that you are interested in the Markov process
$$\mathrm{d} X = b(X) \, \mathrm{d} t + \mathrm{d} W,$$
with $b$ some vector field. If $b$ is contractive, in the sense that for some $K > 0$, the inequality
$$\langle b(X) - b(Y), X - Y\rangle \leqslant - K \| X - Y \|\_2^2$$
holds globally, then it follows that the invariant measure of the process is unique and has sub-Gaussian concentration with a constant depending on $K$, see e.g. [here](https://doi.org/10.1007/978-3-319-11970-0_9) for details.
Actually, suppose that it only holds that
$$\langle b(X) - b(Y), X - Y\rangle \leqslant - K ( \| X - Y \| ) \cdot \| X - Y \|\_2^2$$
for some regular function $K$ which may be negative, but satisfies $\lim \inf\_{r \to \infty} K(r) > 0$, i.e. 'distant contractivity'. In this case, sub-Gaussian concentration again holds for the invariant measure (under a few bonus regularity assumptions, and with a slightly harder-to-derive constant) as a consequence of the results in [this work](https://doi.org/10.1007/s00440-015-0673-1).
In another direction, suppose that the same diffusion satisfies a drift condition, i.e. there is some positive, coercive function $V$ with reasonable sub-level sets such that
$$LV \leqslant -\gamma V + C$$
for some $\gamma, C > 0$, where $L$ is the infinitesimal generator of the diffusion. Then, results along the lines of [this work](https://doi.org/10.1214/ECP.v13-1352) give that the invariant measure satisfies a Poincaré inequality, and hence admits sub-exponential concentration. Under stronger drift conditions, a subset of the same authors are able to obtain various stronger concentration properties; see [this chapter](https://doi.org/10.1017/CBO9781107297296.012) for a collection of examples to this effect.
[ Note: I realised after initial posting that this latter example requires the reversibility of the process, which is often difficult to check without knowing the invariant measure of the process ]
| 2 | https://mathoverflow.net/users/121692 | 440566 | 177,881 |
https://mathoverflow.net/questions/440500 | 1 | Suppose I have an extension of fields $L/K$, a group scheme $G\_K$ over $\operatorname {Spec} K$. Let $G\_L$ denote the pullback of $G\_K$ to $\operatorname{Spec} L$. Then, under what conditions on the extension $L/K$ can one say that we have an equality of $L$-algebras of the form $\operatorname{Lie} (G\_L) = \operatorname{Lie} (G\_K)\otimes \_K L$?
| https://mathoverflow.net/users/499148 | Lie algebras and pulled back group schemes | As the link in Erica's [comment](https://mathoverflow.net/questions/440500/lie-algebras-and-pulled-back-group-schemes#comment1136246_440500) shows, you can find this in SGA3 Exp. 2, but it is not so easy to extract from the very general language. Here is a rough guide: From Definition 3.9.0, the Lie algebra of $G$ over $K$ is the pullback of the total tangent space $T\_{G/K} \to G$ along the unit section $e: \operatorname{Spec} K \to G$. Base-changing this whole diagram along $\operatorname{Spec} L \to \operatorname{Spec} K$ yields the Lie algebra of $G\_L$ over $L$, once we can identify $(T\_{G/K})\_L$ with $T\_{G\_L/L}$, but this follows from Proposition 3.4. It comes down to the fact that the Hom functor (in this case, applied to the spectrum of dual numbers) commutes with arbitrary base change.
| 3 | https://mathoverflow.net/users/121 | 440575 | 177,885 |
https://mathoverflow.net/questions/440560 | 2 | $\DeclareMathOperator\Coh{Coh}\DeclareMathOperator\ev{ev}\DeclareMathOperator\cone{cone}\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Ext{Ext}$This is a specific question concerning a statement in [Symplectic structures on moduli spaces of sheaves via the Atiyah class](https://www.sciencedirect.com/science/article/pii/S0393044009000503).
Let $Y\subset\mathbb{P}^5$ be a smooth cubic 4-fold, and $\mathcal{I}\_\ell$ be the ideal sheaf of a line $\ell$. Consider the functor
$$\mathbb{L}\colon D^b(\Coh(Y))\rightarrow D^b(\Coh(Y)),\quad E\mapsto \cone(\ev:\mathcal{O}\_Y\otimes\Hom^{\bullet}(\mathcal{O}\_Y,E)\rightarrow E)$$
then one has $\mathbb{L}(\mathcal{I}\_\ell (1))\cong\mathcal{F}\_\ell[1]$ in $D^b(\Coh(Y))$ where $\mathcal{F}\_\ell$ is defined by
$$0\rightarrow\mathcal{F}\_\ell\rightarrow\mathcal{O}\_Y^{\oplus 4}\rightarrow\mathcal{I}\_\ell(1)\rightarrow0$$
as $\mathcal{I}\_\ell(1)$ is generated by global sections. Applying $\Hom(\mathcal{I}\_\ell(1),-)$ one has a long exact sequence
$$\cdots\rightarrow0\rightarrow\Ext^1(\mathcal{I}\_\ell(1),\mathcal{I}\_\ell(1))\stackrel{\alpha}{\rightarrow}\Ext^2(\mathcal{I}\_\ell(1),\mathcal{F}\_\ell)\rightarrow\cdots$$
It states in [[1](https://www.sciencedirect.com/science/article/pii/S0393044009000503), Proposition 5.4] that the map
$$\mathbb{L}\colon\Ext^1(\mathcal{I}\_\ell(1),\mathcal{I}\_\ell(1))\rightarrow\Ext^1(\mathcal{F}\_\ell,\mathcal{F}\_\ell)$$
coincides with the composition
$$\Ext^1(\mathcal{I}\_\ell(1),\mathcal{I}\_\ell(1))\stackrel{\alpha}{\rightarrow}\Ext^2(\mathcal{I}\_\ell(1),\mathcal{F}\_\ell)\cong\Ext^1(\mathcal{F}\_\ell,\mathcal{F}\_\ell).$$
However, I can not see why it is true. Why they should coincide?
[1] [Kuznetsov and Markushevich - Symplectic structures on moduli spaces of sheaves via the Atiyah class](https://doi.org/10.1016/j.geomphys.2009.03.008).
| https://mathoverflow.net/users/nan | Homomorphism between Ext induced by the left mutation functor | To say that mutation is a functor, we must define its action on morphisms. It is defined as follows: any morphism $f \colon F\_1 \to F\_2$ in the derived category induces $H^\bullet(f) \colon H^\bullet(F\_1) \to H^\bullet(F\_2)$ and there is a unique morphism $\mathbb{L}(f)$ that fits into a commutative diagram
$$
\require{AMScd}
\begin{CD}
H^\bullet(F\_1) \otimes \mathcal{O} @>>>
F\_1 @>>>
\mathbb{L}(F\_1)
\\
@VH^\bullet(f)VV @VfVV @V\mathbb{L}(f)VV
\\
H^\bullet(F\_2) \otimes \mathcal{O} @>>>
F\_2 @>>>
\mathbb{L}(F\_2)
\end{CD}
$$
Applying this to $F\_1 = I\_\ell(1)$ and $F\_2 = I\_\ell(1)[1]$ gives a commutative square
$$
\require{AMScd}
\begin{CD}
I\_\ell(1) @>>> F\_\ell[1]
\\
@VfVV @V\mathbb{L}(f)VV
\\
I\_\ell(1)[1] @>>> F\_\ell[2].
\end{CD}
$$
It remains to note that the map $\alpha$ is given by composition with the bottom arrow, and the isomorphism in the question is given by composition with the top arrow.
| 0 | https://mathoverflow.net/users/4428 | 440580 | 177,887 |
https://mathoverflow.net/questions/435640 | 1 | Let $(X, \mathcal A, \mu)$ be a $\sigma$-finite measure space and $(E, |\cdot|)$ a Banach space. Here we use the [Bochner integral](https://math.stackexchange.com/questions/4298588/dominated-convergence-theorem-for-banach-space). Let $p \in [1, \infty)$ and $q \in (1, \infty]$ such that $p^{-1}+q^{-1}=1$. In an attempt to formalize the ideas in [this comment](https://math.stackexchange.com/questions/4577980/let-h-h-n-in-l-p-such-that-h-n-f-hf1-x-n-for-all-f-in-l-p-a#comment9636279_4577980), I have come across below questions, i.e.,
1. Let $\psi$ be an isometric automorphism of $L\_{p}(X,\mu, E)$. Let $f, f\_n \in L\_{p}(X,\mu, E)$ such that $f\_n \to f$ pointwise $\mu$-a.e. Does $\psi(f\_n) \to \psi(f)$ pointwise $\mu$-a.e.?
2. Let $X^{\*}$ have the [Radon-Nikodým property](https://math.stackexchange.com/questions/4546633/construct-conditional-expectation-on-banach-spaces-with-radon-nikod%c3%bdm-property) with respect to $\mu$. Then there is a canonical isometric isomorphism $\varphi:L\_{p}(X,\mu, E)^\* \to L\_{q} (X,\mu, E^\*)$ such that
$$
K (f) = \int\_\Omega \langle \varphi(K), f \rangle \mathrm d \mu \quad \forall K \in L\_{p}(X,\mu, E)^\*, \forall f \in L\_{p}(X,\mu, E)
$$
Let $H,H\_n \in L\_{p}(X,\mu, E)^\*$ such that $H\_n \to H$ pointwise. Does $\varphi(H\_n) \to \varphi(H)$ pointwise $\mu$-a.e.?
I feel that isometric isomorphisms do not preserve pointwise convergence. However, I could not come up with a counter-example. Could you elaborate on my questions?
---
I posted [this question](https://math.stackexchange.com/questions/4580930/does-an-isometric-automorphism-of-l-p-mu-x-preserve-pointwise-convergence) on MSE, but have not received any satisfying answer. So I post it here.
| https://mathoverflow.net/users/99469 | Does an isometric automorphism of $L_p (X,\mu, E)$ preserve pointwise convergence? | Point 1) does not hold for the Fourier transform on $L^2({\bf R}, {\bf C})$, which is an isometry for a well chosen normalization.
Consider the sequence $$f\_n(x) = n {\bf 1}\_{[-1/n,1/n]}(x)$$ which converges to 0 for all $x\neq 0$. The Fourier transform of $f\_n$ is proportional to $${sin(x/n) \over {x/n}}$$ which converges to $1$ as $n$ goes to infinity for all $x \neq 0$. And the constant function 1 is not the Fourier transform of 0.
| 2 | https://mathoverflow.net/users/6129 | 440589 | 177,890 |
https://mathoverflow.net/questions/440525 | 4 | Is there any known example of a one-ended finitely presented group with exponential growth that does not contain a quasi-isometric copy of the hyperbolic plane?
This question is motivated by the following question of Papasoglu mentioned in the paper *'Quasi-hyperbolic planes in hyperbolic group'* by Bonk–Kleiner which asks whether every one-ended finitely presented group $G$ contains a quasi plane- the image of a uniform embedding $P\rightarrow G$ where $P$ is a complete Riemannian plane with bounded geometry.
Thank you for reading this.
| https://mathoverflow.net/users/429294 | Groups that don't contain quasi-hyperbolic plane | It is a result of Buyalo and Schroeder [BS, Corollary 1.2] that for every $n\ge 2$ there is no QI embedding of the hyperbolic space $\mathbb{H}^n\_\mathbf{R}$ into any product of $n-1$ trees with a Euclidean space.
In particular, the 1-ended group $F\_2\times\mathbf{Z}$ (as well as $F\_2\times\mathbf{Z}^d$ for arbitrary $d\ge 1$) contains no QI-copy of the hyperbolic plane, thus answering your question.
[BS] S. Buyalo and V. Schroeder. The hyperbolic dimension of metric spaces. Algebra i Analiz, 19(1):93–108, 2007. [ArXiv link](https://arxiv.org/abs/math/0404525)
| 7 | https://mathoverflow.net/users/14094 | 440590 | 177,891 |
https://mathoverflow.net/questions/440594 | 1 | Suppose that $(X,d)$ is a locally compact metric space and $\mu$ is a $\sigma$-finite Radon measure on the Borel sigma-algebra of this space. I am aware that if $(X,d)$ is separable and $\mu$ has full support then $L^2(X,\mu)$ is separable i.e. it admits a dense countable subset.
My question is the following: does there exist a *countable* subset $G \subset L^2(X,\mu) \cap C\_c(X)$ such that
>
> for every $f \in C\_c(X) \cap L^2(X,\mu) \cap L^\infty(X,\mu), f \geq 0$ a.s., there exists an *increasing* (i.e. $f\_n \geq f\_m$ for all $n \geq m$) sequence of functions $f\_n \in G$ such that $f\_n \uparrow f$ pointwise?
>
>
>
If this occurs, then by the monotone convergence theorem, we will have $f\_n \to f$ in $L^2$ as well, which is great.
Essentially, given the existence of such a countable $G$ allows me to do the following : given $f \in C\_c(X)$ and such a sequence, I can obtain an inequality of the form $L(f\_i-f\_j)(x) \leq C\|f\_i-f\_j\|\_{2}^2$, however this can only be obtained when $f\_i-f\_j \geq 0$ a.e. This becomes an inequality of the form $L(f\_i)(x)-L(f\_j)(x) \leq C\|f\_i-f\_j\|^2\_2$ as $L$ is linear, and because the RHS converges to $0$ as $i,j \to \infty$, $L(f\_i)(x) \to Lf(x)$ (the argument goes on from here, but that's basically why non-negativity is required).
Note : I asked a slightly similar question over at [Mathematics Stack Exchange](https://math.stackexchange.com/questions/4636432/density-of-functions-in-increasing-pointwise-sense) as well, it's a simplification of this question that can be attempted before this one.
| https://mathoverflow.net/users/327983 | On the existence of a countable dense family in "increasing" pointwise convergence | Yes, this is true. First note that if there is a sequence $f\_n$ that converges to $f$ from below, $f\_n \leq f$, then there is an increasing sequence converging to $f$ given by $\tilde{f}\_n = max(f\_1,...,f\_n)$.
You can add such functions to your dense countable set without changing its cardinal.
So we are left to show that we can find a sequence converging from below.
Let me give the argument when $X = {\bf R}$ for simplicity. You can find a simple function less than $f$ whose integral is $\varepsilon$-close to the integral of $f$, from a set of countable simple functions, the one with rational values and associated to intervals with rational endpoints. This is actually true for any Riemann-integrable functions. This simple function can be approximated from below by an affine function, here again from a countable subset of the sets of all affine functions, e.g. the one with rational slope and rational turning points.
The same process work for general separable locally compact metric spaces, you just need to check how the approximating functions from the countable set mentioned in your question are built.
| 1 | https://mathoverflow.net/users/6129 | 440600 | 177,898 |
https://mathoverflow.net/questions/440588 | 6 | Let $S\_g$ be a closed orientable surface of genus $g>1$.
*How can one prove that its mapping class group $\mathrm{Mod}(S\_g)$
is not generated by two Dehn twists?*
A pair of simple closed curves in $S\_g$ may be very complicated
(e.g. there are *filling* pairs of curves, such that they are in minimal position
and they divide $S\_g$ into discs).
Maybe the question is still open, is it?
---
I also know several results in this direction, such as:
* $\mathrm{Mod}(S\_g)$ can be generated by two elements -- a product of $2$ Dehn twists and a product of $2g$ Dehn twists
[*Wajnryb, Mapping class group of a surface is generated by two elements, 1996*]
* $\mathrm{Mod}(S\_g)$ can be generated by two elements, one of which is a Dehn twist
[*Korkmaz. Generating the surface mapping class group by two elements, 2004*]
* $\mathrm{Mod}(S\_g)$ can be generated by two torsion elements, of order $g$ if $g\ge6$
[*Yildiz. Generating mapping class group by two torsion elements, 2020*]
Maybe you can share some similar interesting results?
| https://mathoverflow.net/users/76500 | Generate $\mathrm{Mod}(S_g)$ by two Dehn twists | In
Humphries, Stephen P.
Generators for the mapping class group. Topology of low-dimensional manifolds (Proc. Second Sussex Conf., Chelwood Gate, 1977), pp. 44–47,
Lecture Notes in Math., 722, Springer, Berlin, 1979.
Humphries proves that no collection of less than or equal to 2g Dehn twists generates the mapping class group of a closed genus g surface, but that a specific collection of 2g+1 twists does generate it.
By the way, it is easy to see that no collection of at most 2g-1 twists generates the mapping class group since such a collection of twists will always fix some nonzero vector in first homology (the point being that the first homology group has rank 2g, and a Dehn twists acts trivially on a codimension one subgroup of homology).
| 8 | https://mathoverflow.net/users/317 | 440605 | 177,900 |
https://mathoverflow.net/questions/440597 | 2 | Consider the analytic function $f : (0,\infty) \to (0,\infty)$ given by
$$
f(x) = \bigg( \sum\_{i=1}^n a\_i b\_i^{1/x} \bigg)^x
$$
where $n\in\mathbb N$, $a\_i>0$ and $0<b\_1<\ldots<b\_n<1$. I am interested in analytic continuations of $f$ to the right half plane $\mathbb H\_+ = \{z\in\mathbb C : \mathrm{Re}(z)>0\}$.
My guess is that that such an analytic continuation to all of $\mathbb H\_+$ exists only in the trivial case $n=1$ (where the continuation is $f(z) = b\cdot a^{z}$). Even the $n=2$ case seems to be quite hard.
Any suggestions and ideas would be much appreciated!
| https://mathoverflow.net/users/485160 | Analytic continuation of a function on the half line | Partial answer:
$g(z)=\sum\_{k=1}^n a\_kb\_k^{1/z}$ is analytic on the punctured plane and clearly $f(z)=e^{z\log g(z)}$ defines $f$ analytic outside the zeroes of $g$ - as usual locally and then we can stitch together a global $f$ with appropriate cuts.
Now $g >0$ on the real axis outside the origin by definition, so it comes down to figuring out if $g$ has zeroes only in the left-hand plane.
Since $1/z$ preserves the right hand and left hand planes, we need to look at the zeroes of $\sum a\_kb\_k^w=\sum a\_ke^{c\_k w}, c\_k=\log b\_k$ which are known as exponential or Ritt polynomials and there is a large literature about.
Edit later - actually since in the case $n=2$ the zeroes of $h(w)=a\_1e^{c\_1w}+a\_2e^{c\_2w}, c\_k=\log b\_k, c\_1<c\_2<0$ are clearly periodic with period $\frac{2k \pi i}{c\_2- c\_1}$, so all have the same real part, the function $f$ will be analytic in the right half plane precisely when one, hence all, are in the left half plane as then the zeroes of $g(z)=h(1/z)$ are like that too as noted.
This happens when with $a=\frac{a\_1}{a\_2}$ so one such zero is $w\_0=\frac{\log (-a)}{c\_2-c\_1}$ we have $\Re w\_0 \le 0$ and that happens precisely when $|a| \le 1$ or $a\_1 \le a\_2$, so we can actually completely solve the case $n=2$
Edit later per comments: Since $f(x)$ seems to be defined the usual way for $x>0$ where we have a unique positive $1/x$ root for the positive number $g(x)$, $\log g(x)$ is defined uniquely and continuously (real analytically too) by its usual real value and we can extend it (uniquely) to an analytic $\log g$ on a small (simply connected) neighborhood of the positive real axis in $\Re z>0$ - note that it may not be a strip if zeroes of $g$ accumulate towards the real axis. So already $f$ has unique analytic continuation to a complex open set in $\Re z >0$ and the only question is how much more we can go and still be analytic in $\Re z >0$ and that is precisely until we hit zeroes of $g$ where we have to start making cuts. If there are no zeroes, then $\log g$ and hence $f$ have analytic continuation to the whole half plane.
If $n=1$ we have $g(x)=a^xb, x>0$ so $\log g(x)=x \log a +b, x>0$ and that clearly extends to $\Re z>0$ so $f(z)=b a^z$ is the analytic continuation to $\Re z >0$ and actually to all $z$ finite as the singularity at $0$ is removable as we see.
| 3 | https://mathoverflow.net/users/133811 | 440607 | 177,901 |
https://mathoverflow.net/questions/440583 | 4 | **Question.** Is there an entire function $F$ satisfying first two or all three of the following assertions:
* $F(z)\neq 0$ for all $z\in \mathbb{C}$;
* $1/F - 1\in H^2(\mathbb{C}\_+)$ -- the classical Hardy space in the upper half-plane;
* $F$ is bounded in every horizontal half-plane $\{z\colon \text{Im}(z) > \delta\}$?
**Thoughts**. Let $G= 1/F$. Then we have $G(z) = 1 + \int\_0^{\infty} h(x)e^{izx}\, dx$ for some $h\in L^2[0,\infty)$ and all $z\in \mathbb{C}\_+$. For nice functions $h$ (e.g., for super-exponentially decreasing) this integral representation can be extended to the whole complex plane and probably the example can be constructed in terms of $h$. However, I don't know if it is possible to find $h$ such that $G$ is non-zero for every $z$.
| https://mathoverflow.net/users/498423 | Existence of nonzero entire function with restrictions of growth | There is a zero-free entire function bounded in every left half-plane, and such that $f-1$ is in $H^2$ in every left half-plane.
Let $\gamma$ be the boundary of the region $$D=\left\{ x+iy: |y|<2\pi/3, x>0\right\}
.$$ Consider the function
$$g(z)=\int\_\gamma \frac{\exp e^\zeta}{\zeta-z}d\zeta,\quad z\in {\mathbf{C}}\backslash D.$$ The integral evidently converges and $g(z)=O(1/z)$ in ${\mathbf{C}}\backslash D.$ Now, $g$ has an analytic continuation to an entire function: deforming the contour to $\partial D\_t$,
where $D\_t=\{ x+iy:|y|<2\pi/3, x>t\},\; t>0$ does not change $g$ in $D$,
and shows that $g$ has an analytic continuation to ${\mathbf{C}}\backslash D\_t$, and this is for every $t>0$, so $g$ is entire. Now $f(z)=e^{g(z)}$ is the desired function. If you want upper half-planes take $f(iz)$.
Remark. You can improve the estimate $g(z)=O(1/z)$.
Evidently, $g$ has infinitely many zeros $z\_1,z\_2,\ldots$. Then $g\_k(z)=g(z)/((z-z\_1)\ldots(z-z\_k))$
satisfies $g\_k(z)=O(z^{-k-1})$ as $z\to\infty$ outside $D\_t$.
Remark 2. This construction is standard in the theory of entire functions, see, for example, [Entire function bounded at every line](https://mathoverflow.net/questions/190837/entire-function-bounded-at-every-line/190874#190874)
Sometimes this $g$ is called the Mittag-Leffler function.
| 5 | https://mathoverflow.net/users/25510 | 440610 | 177,902 |
https://mathoverflow.net/questions/440427 | 1 | Given only an arc of a circle, we can easily reconstruct it fully without any use of analytic geometry - indeed using only compass and straightedge. Note that by "given only an arc", we mean that "from a circle that was originally drawn on the plane, everything except an arc got erased".
**Question:** Given an arc of an ellipse, can we reconstruct it without using analytic or algebraic geometry but *synthetically* - using objects not restricted to compass and straightedge (for example, one could use threads)?
**Sub-questions:** Does it matter if the arc is more than half of the full ellipse? What if more than one arc, say 2 arcs, of the ellipse are given (here, we mean the arcs given only as broken off fragments of the wireframe boundary of the ellipse and their relative position is not known)?
Note: analogous questions could be asked about the other conic sections.
| https://mathoverflow.net/users/142600 | Reconstructing an ellipse from an arc, synthetically | Five points in general position lie on a unique conic. With the help of Pascal's theorem one can construct arbitrarily many points on the same conic.
EDIT. Answering the question about construction of the foci: yes, it is possible. If 5 points on a conic are given, then the coefficients of the corresponding quadratic polynomial are constructible: solution of a linear system requires only rational functions of the coefficients. The center and the principal axes are constructible: rational functions plus square root extraction when finding the eigenvectors. Ellipse vertices are constructible: intersection of a line with a conic requires square roots and arithmetic operations only. The foci are constructible: for the focal distance arithmetic operations and square roots only are needed.
The above results in a terribly long algorithm. I am sure that the descriptive geometers knew a shorter construction in the 19th century already.
| 2 | https://mathoverflow.net/users/98590 | 440611 | 177,903 |
https://mathoverflow.net/questions/440606 | 4 | I have been working on my master's thesis which is about the equivalence of the Hardy-Littlewood conjecture on primes represented by quadratic polynomials and the Lang-Trotter conjecture for CM elliptic curves. Although I haven't yet worked out all the details but based on some recent papers by Wan/Xi and H. Qin it looks like it is quite possible to complete this work and establish such equivalence for all CM elliptic curves which means the difficult Lang-Trotter conjecture can be interpreted in terms of a relatively simple conjecture (only in the case of CM elliptic curves).
But I couldn't find much relevant work on this conjecture by Hardy and Littlewood. So could somebody refer to me any recent work on this conjecture or any promising work on this conjecture?
| https://mathoverflow.net/users/483436 | Recent works on the Hardy-Littlewood conjecture on primes represented by quadratic polynomials | For the convenience of the reader, I have written Hardy and Littlewood's conjecture from their paper [linked in the comments above](https://link.springer.com/article/10.1007/BF02403921):
Suppose that $a,b,c$ are integers and $a$ is positive; that $\gcd(a,b,c) = 1$; that $a+b$ and $c$ are not both even; and that $D = b^2 - 4ac$ is not a square. Then there are infinitely many primes of the form $am^2 + bm + c$. The number $P(n)$ of such primes less than $n$ is given asymptotically by
$$\displaystyle P(n) \sim \frac{\varepsilon C}{\sqrt{a}} \frac{\sqrt{n}}{\log n} \prod\_{p | \gcd(a,b)} \left(\frac{p-1}{p} \right) $$
where $\varepsilon$ is $1$ if $a + b$ is odd and $2$ if $a+b$ is even. The constant $C$ is given by
$$\displaystyle C = \prod\_{w \geq 3, w \nmid a} \left(1 - \frac{1}{w-1} \left(\frac{D}{w} \right)\right).$$
To answer the question, there has not been any recent work on this problem for some time due to its perceived difficulty. In fact it is a long-standing debate between prominent analytic number theorists whether this conjecture is harder or easier than the twin prime conjecture. Despite the significant progress made by Zhang and Maynard about ten years ago on the latter, we are still very far away from twin primes themselves.
The best results towards this conjecture are due to Iwaniec ([Almost primes represented by quadratic polynomials](https://eudml.org/doc/142575)), who proved that the polynomial $x^2 + 1$ takes on infinitely many integer values with at most two prime divisors (including multiplicity) and that the density of such values is $\gg \sqrt{n}/\log(n)$. His arguments can be readily generalized to arbitrary quadratic polynomials satisfying the conditions imposed by Hardy and Littlewood above. On the other hand, Iwaniec's arguments have no chance of being improved to give actual primes without very significant new ideas, due to the parity barrier.
| 9 | https://mathoverflow.net/users/10898 | 440612 | 177,904 |
https://mathoverflow.net/questions/387536 | 3 | Let $n\geq 3$ be an integer and $0<\alpha\_1, \dots ,\alpha\_{n-2}<1$. Let's say a tuple of positive numbers $(e\_1,\dots, e\_n)$ is *nice* if there is a convex $n$-gon $A\_1\dots A\_n$ such that $\hat A\_i=\pi\alpha\_i$ and edge lengths $\overline{A\_iA\_{i+1}}=e\_i$.
(The convexity condition probably will make things rather more complicated, so I will be happy to drop that)
Pythagorean theorem says that with $n=3$ and $\alpha\_1=1/2$, the nice tuples are precisely the intersection of the variety $e\_2^2=e\_1^2+e\_3^2$ with the positive orthant. Of course similar result holds for any $\alpha\_1$ and $n=3$.
Question: what about $n>3$? My guess is the set of nice tuples is still the intersection of an algebraic scheme with the positive orthant (or some polytope in it). If so, what is known about the equations defining such schemes (obviously they are homogenous, but what about things like degrees)? Any reference for such kind of problems?
(Update Apr 2021): I asked Hendrik Lenstra and he confirmed that the statement mentioned by Henri Cohen below appeared as a Proposition in his thesis (which unfortunately he has no copies left). Also according to Lenstra, it might have appeared in [Nieuw Archief voor Wiskunde](http://www.nieuwarchief.nl/), but quite probably in Dutch, and I have not been able to track it down.
| https://mathoverflow.net/users/2083 | Generalizing Pythagorean Theorem: equations defining edges of a (convex) $n$-gon with $n-2$ prescribed angles? | One has $$\overrightarrow{A\_nA\_1} + \cdots + \overrightarrow{A\_{n-2}A\_{n-1}} = \overrightarrow{A\_nA\_{n-1}}.$$
Taking the squared norm of both sides one arrives at
$$\sum\_{i=1}^{n-1} e\_{i-1}^2 + 2 \sum\_{1\le i < j\le n-1}e\_{i-1}e\_{j-1}\cos\sum\_{k=i}^{j-1}(\pi-\alpha\_i) = e\_{n-1}^2,$$
which is the equation we are looking for.
For $n=4$ I studied this equation some years ago from a different viewpoint: given the side lengths of a quadrilateral, how are two of its adjacent angles related? The above equation answers this question. It turns out that it can be rewritten as follows:
$$a\_1^2a\_2^2x\_{22} + a\_1^2x\_{20} + a\_2^2x\_{02} + 2a\_1a\_2x\_{11} + x\_{00} = 0,$$ where
\begin{gather\*}
a\_1=\cot\frac{\alpha\_1}2, \quad a\_2 = \cot\frac{\alpha\_2}2,\\
x\_{22} = (e\_1-e\_2-e\_3-e\_4)(e\_1-e\_2+e\_3-e\_4),\\
x\_{20} = (e\_1+e\_2+e\_3-e\_4)(e\_1+e\_2-e\_3-e\_4),\\
x\_{02} = (e\_1-e\_2+e\_3-e\_4)(e\_1-e\_2-e\_3+e\_4),\\
x\_{11} = -4e\_2e\_4,\\
x\_{00} = (e\_1+e\_2-e\_3+e\_4)(e\_1+e\_2+e\_3+e\_4).
\end{gather\*}
| 2 | https://mathoverflow.net/users/98590 | 440617 | 177,905 |
https://mathoverflow.net/questions/440618 | 5 | By $MRDP$ resolution of Hilbert's tenth, we infer, counting number of solutions to Diophantine equations is undecidable.
Is parity of number of solutions to Diophantine equations undecidable?
| https://mathoverflow.net/users/10035 | Parity of number of solutions to Diophantine equations | While it’s unclear what you mean by “parity” when the number of solutions is infinite, all such questions have been answered by M. Davis, *On the number of solutions of Diophantine equations*, Proc. AMS 35 (1972), 552–554, [doi link](https://doi.org/10.2307/2037646):
>
> **Theorem:** Fix $\varnothing\subsetneq A\subsetneq\omega\cup\{\aleph\_0\}$. Then it is undecidable whether the number of solutions of a given Diophantine equation belongs to $A$.
>
>
>
| 16 | https://mathoverflow.net/users/12705 | 440621 | 177,907 |
https://mathoverflow.net/questions/394390 | 2 | Given $4$ distinct points on the Riemann sphere, thought of as the sphere at infinity of hyperbolic $3$-space $H^3$, one may define the cross-ratio in the usual way. Note that the cross-ratio is the ratio of products of $2$ pairwise differences (using an affine coordinate on the Riemann sphere), so in this sense it has degree $2$.
I came across an expression, depending on $4$ distinct points in $H^3$, say $p\_1$, $p\_2$, $p\_3$ and $p\_4$, more precisely depending on ideal points lying on hyperbolic lines joining pairs of points in that configuration, and such that when the $p\_i$ all lie on the sphere at infinity, for $i = 1, \ldots, 4$, then one gets the usual cross-ratio. Interestingly, this extended cross-ratio has degree $3$, not $2$.
Let $p\_1$, $p\_2$, $p\_3$ and $p\_4$ be points in $H^3$ such that
$$ p\_1 \neq p\_2 \neq p\_3 \neq p\_4.$$
Let $t\_{12}$ denote the ideal point on the Riemann sphere which is the limiting point of the geodesic ray starting at $p\_1$ and passing through $p\_2$, and so on. I will use a complex affine coordinate, so that the $t\_{ab}$ are actually complex numbers (possibly $\infty$). Consider
$$ C(p\_1,p\_2,p\_3,p\_4) = \frac{(t\_{12}-t\_{23})(t\_{21}-t\_{34})(t\_{32}-t\_{43})}{(t\_{12}-t\_{21})(t\_{23}-t\_{32})(t\_{34}-t\_{43})}.$$
In the limiting case where all the points $p\_i$ lie on the sphere at infinity, and let us denote them by $z\_1$, $z\_2$, $z\_3$ and $z\_4$ in this case, we then get
$$ C(z\_1,z\_2,z\_3,z\_4) = \frac{(z\_{2}-z\_{3})(z\_{1}-z\_{4})(z\_{2}-z\_{3})}{(z\_{2}-z\_{1})(z\_{3}-z\_{2})(z\_{4}-z\_{3})}.$$
Simplifying one gets
$$ C(z\_1,z\_2,z\_3,z\_4) = - \frac{(z\_{1}-z\_{4})(z\_{2}-z\_{3})}{(z\_{1}-z\_{2})(z\_{3}-z\_{4})},$$
as claimed.
I wonder if it is new, though I doubt it, as these things are quite classical. But I have only seen the cross-ratio defined for $4$ points on the Riemann sphere, but not for $4$ points in the "bulk", meaning in hyperbolic $3$-space.
Since this extended cross-ratio is invariant under the group of orientation preserving hyperbolic isometries, and gets complex conjugated under orientation reversing hyperbolic isometries, I suspect its real part to be expressible in terms of pairwise hyperbolic distances, and its imaginary part maybe to be expressible in terms of hyperbolic volume (and pairwise hyperbolic distances possibly).
I thus pose the problem of expressing this extended cross-ratio in terms of invariants of hyperbolic geometry (hyperbolic pairwise distances, hyperbolic areas/volumes etc.).
Edit 1: it is known that the volume of an ideal hyperbolic tetrahedron is the Bloch-Wigner function $D\_2$ applied to their cross-ratio. Note that "my" (?) extended cross-ratio $C(p\_1,p\_2,p\_3,p\_4)$ depends real analytically on its arguments, provided $p\_1 \neq p\_2 \neq p\_3 \neq p\_4$. So for instance, $D\_2 \circ C$ is a real analytic extension of the signed hyperbolic volume of an ideal tetrahedron. Could it possibly be the signed hyperbolic volume of the tetrahedron with vertices $p\_i$ ($1 \leq i \leq 4$)? Maybe it does not have the right symmetries... But still, isn't it plausible that $D\_2$ of some analogous real analytic extension of the usual cross-ratio, defined in a way similar to how $C$ is defined, may give the hyperbolic volume of the tetrahedron with vertices $p\_i$?
| https://mathoverflow.net/users/81645 | How to express this hyperbolic extension of the cross-ratio in terms of hyperbolic distances and volumes? | The quantity $C(p\_1,p\_2,p\_3,p\_4)$ does not change if $p\_1$ is replaced by any other point on the ray $p\_2p\_1$. It also does not change if $p\_4$ is replaced by any other point on the ray $p\_3p\_4$. Thus this is an invariant of a triple of oriented lines $\ell\_1 = p\_1p\_2, \ell\_2 = p\_2p\_3, \ell\_3 = p\_3p\_4$. It is thus possible to express $C(p\_1,p\_2,p\_3,p\_4)$ in terms of the following four quantities: angle between $\ell\_1$ and $\ell\_2$, angle between $\ell\_2$ and $\ell\_3$, angle between the planes $\ell\_1\ell\_2$ and $\ell\_2\ell\_3$, distance between $p\_2$ and $p\_3$ (i. e. between the intersection points $\ell\_1 \cap \ell\_2$ and $\ell\_2 \cap \ell\_3$).
For computation let us use Poincare half-space model and send $t\_{23}$ to $0$ and $t\_{32}$ to $\infty \in \mathbb{C}\mathrm{P}^1$. One then has
$$C(p\_1,p\_2,p\_3,p\_4) = \frac{t\_{12}(t\_{21}-t\_{34})}{(t\_{21}-t\_{12})(t\_{34}-t\_{43})}.$$
The points $t\_{12}$ and $t\_{21}$ lie on a line through $0$, on the opposite sides from it, the line $\ell\_1$ is a vertical half-circle with diameter $t\_{12}t\_{21}$. Similarly for $t\_{34}$ and $t\_{43}$. The quotient $\frac{t\_{12}}{t\_{21}-t\_{12}}$ equals $\frac{1-\cos\alpha}{2\cos\alpha}$, where $\alpha$ is the angle between $\ell\_1$ and $\ell\_2$. It is more complicated to decypher $\frac{t\_{21}-t\_{34}}{t\_{34}-t\_{43}}$, but in the end one gets some trigonometric functions of the angles between the lines, and a hyperbolic trigonometric function of the distance $p\_2p\_3$, plugged into some rational function. That is, nothing related to the volume of a simplex.
One more remark: $C(p\_1,p\_2,p\_3,p\_4)$ can be rewritten as a product of certain cross-ratios of points $t\_{ij}$. Thus it has a meaning also for arbitrary lines $\ell\_1, \ell\_2, \ell\_3$.
| 1 | https://mathoverflow.net/users/98590 | 440624 | 177,908 |
https://mathoverflow.net/questions/440625 | 0 | I am wondering why a standard notation for open sets is $G$ and that for closed sets is $F$. I mean, $F$ precedes $G$ in the alphabet, whereas open sets are usually introduced before closed ones.
| https://mathoverflow.net/users/36721 | Notations for open and closed sets | I see that @LSpice has already provided an answer in their comment (I see that Emil has added a clarification, so to speak). Mine will compliment the comment by @LSpice a little.
Historically, closed sets were before the open sets (I believe so). Kazimierz Kuratowski defined topology (of general $T\_1$-spaces} via the closure operation hence the closed sets came before open sets. We are talking here about the years 1921 and 1933.
I seem to remember that also Wacław Sierpiński was introducing the closed sets before open sets.
Earlier, I believe, that topologists/mathematicians were talking about **neighborhoods** rather than open sets. Perhaps(?) people started to talk about open sets seriously only after the results/papers by Paul S. Urysohn (I am not a historian -- please, do double-check my vague reminiscences).
Finally, the notation F (for closed sets) comes from the French "Fermé"; then open G followed. In the old days, French and German were ruling mathematics before English.
---
---
**SOURCES:**
* Hausdorff definition of topological spaces via neighborhoods:
$\qquad$https://mathworld.wolfram.com/HausdorffAxioms.html
See also Bourbaki foundational text on General Topology, it presents the Hausdorff neighborhood axioms soon after the today standard definition via open sets.
Hausdorff gutsy topological axioms absorbed Hilbert's axioms on 2d-geometric spaces.
| 2 | https://mathoverflow.net/users/110389 | 440626 | 177,909 |
https://mathoverflow.net/questions/435445 | 12 | $\DeclareMathOperator\SO{SO}$I'm not asking for a proof but for some hint that might be helpful to understand this "anomaly" in 4 dimensions. I'm aware of the parallelism with the $A\_4$ finite group case as shown in the [answer](https://mathoverflow.net/a/145066) by @Benoit Kloeckner which I would also like to see elaborated on as I'm not sure I understood how the non-simplicity arises from the 3 decompositions into pairs of orthogonal planes in [The non-simplicity of $SO(4)$ and $A\_4$](https://mathoverflow.net/questions/145026/the-non-simplicity-of-so4-and-a-4), but I'm more centered on the particularity of 4 dimensions.
| https://mathoverflow.net/users/118787 | Why is $\operatorname{SO}(4)$ not a simple Lie group? | Here's a different approach than the ones that have been suggested so far. For any $n, k$ we can consider the action of $SO(n)$ on the exterior power $\Lambda^k(\mathbb{R}^n)$. These representations are almost all irreducible, *except* when $n = 2k$ and $k \ge 2$ is even; in this case the [Hodge star](https://en.wikipedia.org/wiki/Hodge_star_operator) $\star : \Lambda^k(\mathbb{R}^n) \to \Lambda^{n-k}(\mathbb{R}^n)$ restricts to an involutive automorphism (as an $SO(n)$-representation) of $\Lambda^k(\mathbb{R}^{2k})$ and hence splits it up into a direct sum of its $+1$-eigenspace and its $-1$-eigenspace, each of which is irreducible (I think). (When $k$ is odd the Hodge star squares to $-1$ so it no longer has real eigenvalues.)
These representations have dimension $\frac{1}{2} {2k \choose k}$, and the special feature of $SO(4)$ is that this dimension is smaller than the dimension $n = 2k$ of the defining representation iff $k = 2$. In the $k = 2$ case the exterior square $\Lambda^2(\mathbb{R}^4)$ is $6$-dimensional and the eigenspaces of the Hodge star are $3$-dimensional. This exhibits the double cover $SO(4) \to SO(3) \times SO(3)$ which shows that $SO(4)$ is not simple, and there is no need to know anything about the quaternions, spin groups, or spin representations.
More explicitly, writing $e\_1, e\_2, e\_3, e\_4$ for the standard basis of $\mathbb{R}^4$, and choosing the orientation $\omega = e\_1 \wedge e\_2 \wedge e\_3 \wedge e\_4$, the Hodge star $\star : \Lambda^2(\mathbb{R}^4) \to \Lambda^2(\mathbb{R}^4)$ by definition satisfies
$$\alpha \wedge (\star \beta) = \langle \alpha, \beta \rangle \omega$$
where $\langle \alpha, \beta \rangle$ is the Gram determinant inner product as described on Wikipedia; it is completely specified by saying that the wedge products $e\_i \wedge e\_j, i < j$ form an orthonormal basis. This lets us compute the action of $\star$ explicitly:
$$\star (e\_1 \wedge e\_2) = e\_3 \wedge e\_4$$
$$\star (e\_1 \wedge e\_3) = - e\_2 \wedge e\_4$$
$$\star (e\_1 \wedge e\_4) = e\_2 \wedge e\_3$$
$$\star (e\_2 \wedge e\_3) = e\_1 \wedge e\_4$$
$$\star (e\_2 \wedge e\_4) = - e\_1 \wedge e\_3$$
$$\star (e\_3 \wedge e\_4) = e\_1 \wedge e\_2.$$
The $+1$-eigenspace is spanned by $e\_1 \wedge e\_2 + e\_3 \wedge e\_4$, $e\_1 \wedge e\_4 + e\_2 \wedge e\_3$, and $e\_1 \wedge e\_3 - e\_2 \wedge e\_4$, while the $-1$-eigenspace is spanned by $e\_1 \wedge e\_2 - e\_3 \wedge e\_4$, $e\_1 \wedge e\_4 - e\_2 \wedge e\_3$, and $e\_1 \wedge e\_3 + e\_2 \wedge e\_4$. This corresponds to $3$ decompositions of $\mathbb{R}^4$ into pairs of orthogonal planes which means it presumably has something to do with the answer you quoted from the other MO question but I'm not sure what exactly.
| 7 | https://mathoverflow.net/users/290 | 440634 | 177,911 |
https://mathoverflow.net/questions/440638 | 5 | As far as I know, there are two versions of Waldspurger's formula (classical and adelic), which can be vaguely stated as follows
* (Classical version) Let $f$ be a half-integral weight modular form of weight $k/2$ (where $k$ is odd), level $N$ and character $\chi$ and $F = \operatorname{Sh}(f)$ be its Shimura correspondent in $S\_{k-1}(N/2, \chi^2)$. Then $|a\_d(f)|^2$, (square of) Fourier coefficients of $f$ for square-free $d$, can be represented as $L(1/2, F \otimes \chi\_d)$, the special $L$-value of quadratic twist of $F$.
* (Adelic version, brought from [Horawa's notes on Zydor's lectures on periods of automorphic forms](https://people.maths.ox.ac.uk/horawa/math_678_periods.pdf)) Let $\pi$ be an automorphic cuspidal representation of $\mathrm{GL}\_2(\mathbb{A}\_F)$ and $\phi = \bigotimes\_v \phi\_v \in\pi$ be an automorphic form. Let $T \leq \mathrm{GL}\_2$ be a non-split torus. Then the square of the toric period is equal to product of special $L$-values and local periods:
$$
|\mathcal{P}\_T(\phi)|^2 = \bigg|\int\_{T(\mathbb{A}\_F) / T(F)} \phi(t) dt\bigg|^2 = c\cdot L(1/2, \pi) \cdot L(1/2, \pi \otimes \eta) \cdot \prod\_v \mathcal{L}\_v(\phi\_v).
$$
I wonder how the adelic version can be thought as a generalization of the classical version. To be precise, I can't find a connection with toric periods and Fourier coefficients of half-integral weight modular forms (there's even nothing about Shimura correspondence in the adelic statement. Note that Waldspurger mention about the classical result in the introduction of his original paper). Is it possible to somewhat *adelize* the classical statement as follows: for an automorphic cuspidal representation $\pi'$ of $\operatorname{Mp}\_2(\mathbb{A}\_F)$ (or maybe 2-cover of $\operatorname{GL}\_2(\mathbb{A}\_F)$), and a corresponding automorphic representation $\pi$ of $\mathrm{GL}\_2(\mathbb{A}\_F)$ via Shimura correspondence, can we related a certain kind of period of $\pi'$ with special $L$-values of quadratic twists of $\pi$?
In the above note by Zydor, it does not mention about half-integral weight modular forms, instead it consider sum over Heegner points. Maybe this is the right classical statement that corresponds to the above adelic statement, but I still don't get why this corresponds to the toric periods.
---
*Edit*: Based on [Kimball's answer](https://mathoverflow.net/a/440640) and [Peter Humphrey's comment](https://mathoverflow.net/questions/440638/waldspurgers-formula-and-toric-periods-classical-and-adelic-versions#comment1136548_440638), they are separate theorems and the second is NOT an adelic version of the first one. For completion, here are the list of relevant papers from Waldspurger (found on mathscinet)
* Correspondance de Shimura ((J. Math. Pures. Appl. 1980): *Adelic* version of Shimura correspondence.
* Sur les coefficients de Fourier des formes modulaires de poids demi-entier (J. Math. Pures. Appl. 1981): *Classical* Waldspurger's formula for half-integral weight modular forms and $L$-values.
* Sur les valeurs de certaines fonctions $L$ automorphes en leur centre de symetrie (Compositio, 1985): Above *adelic* results on toric periods and $L$-values.
* Correspondences de Shimura et quaternions (Forum math, 1991): Shimura correspondence between $\mathrm{Mp}\_2$ and $D^\times$, viewed as a composition of the original Shimura correspondence and Jacquet-Langlands correspondence?
| https://mathoverflow.net/users/95471 | Waldspurger's formula and toric periods — classical and adelic versions | These are two separate theorems, proved in different papers of Waldspurger (I think in 1980/1981 and 1985, respectively), so you shouldn't conflate them. The first theorem can be viewed as an "$L$-value correspondence between automorphic representations of GL(2) and Mp(2) (Shimura correspondence), and the second between automorphic representations of GL(2) and $B^\times$ where $B$ is a quaternion algebra (Jacquet-Langlands correspondence).
So really Waldspurger proved a triangle of relations between three things: $L$-values of quadratic twists, Fourier coefficients of half-integral weight forms, and toric periods.
| 4 | https://mathoverflow.net/users/6518 | 440640 | 177,913 |
https://mathoverflow.net/questions/440609 | 0 | Recently I got interested in the following property of topological spaces:
$(X,\mathcal{T})$ satisfies (P) if the following holds:
For any nonempty closed subsets $F$ and $G$ with $F\ne G$, there are closed subsets $F'\subseteq F$ and $G'\subseteq G$ satisfying the following conditions:
* $F'$ has nonempty interior in $F$,
* $G'$ has nonempty interior in $G$,
* $F'\cap G'=\varnothing$.
It is not difficult to show that regularity implies property (P). Moreover, for $T\_0$ spaces, property (P) implies $T\_1$ (this is also easily seen by letting $x\ne y$ such that $\overline{\{x\}}\subseteq \overline{\{y\}}$, $\overline{\{y\}}\not\subseteq \overline{\{x\}}$, choosing $F=\overline{\{y\}}$ and $G=\overline{\{x\}}$, and deriving a contradiction.).
In view of this situation, I was wondering if this property appears in the literature or if anyone knows any relation to other well-known properties.
| https://mathoverflow.net/users/499248 | Property stronger than $T_1$ and weaker than regularity | 1. Hausdorff spaces need not have this property. Consider [Bing's Countable connected Hausdorff space](https://doi.org/10.1090/S0002-9939-1953-0060806-9) (Example 75 in Steen and Seebach's *Counterexamples in Topology*); it has the property that for every pair of nonempty open sets $U$ and $V$ the closures $\overline U$ and $\overline V$ intersect. Let $U$ and $V$ be nonempty, open, and disjoint, and let $F=\overline U$ and $G=\overline V$. Let $F'$ and $G'$ be closed subsets of $F$ and $G$ respectively with non-empty relative open interior, say $F'\supseteq U'\cap F\neq\emptyset$ and $G'\supseteq G\cap V'\neq\emptyset$ for some open sets $U'$ and $V'$. Then also $U'\cap U$ and $V'\cap V$ are nonempty and $F'$ and $G'$ contain their closures, so $F'\cap G'\neq\emptyset$.
2. Urysohn spaces do have this property. In Urysohn spaces distinct points have disjoint *closed* neighbourhoods. So, assume $F\neq G$ and, wlog, take $x\in F\setminus G$ and $y\in G$. Let $U$ and $V$ be open sets with $x\in U$, $y\in V$ and $\overline U\cap\overline V=\emptyset$. Let $F'$ be the closure of $U\cap F$ and let $G'$ be the closure of $V\cap G$. These are as required.
| 1 | https://mathoverflow.net/users/5903 | 440642 | 177,914 |
https://mathoverflow.net/questions/440582 | 2 | Assume that $\boldsymbol{v}\_{\boldsymbol{1}}, \ldots, \boldsymbol{v}\_{\boldsymbol{n}} \in \mathbb{R}^n$ satisfy $\forall i, j \in[n],i \neq j,\left\langle\boldsymbol{v}\_{\boldsymbol{i}}, \boldsymbol{v}\_{\boldsymbol{j}}\right\rangle=0,\left\|\boldsymbol{v}\_{\boldsymbol{i}}\right\|=1$. Let $\mathcal{L}=\left[\boldsymbol{v}\_{\boldsymbol{1}}, \ldots, \boldsymbol{v}\_{\boldsymbol{n}}\right] \mathbb{Z}^n$.
The problem is the following: for any polynomial $p$ given $\boldsymbol{B}$ (a basis of $\mathcal{L}$ ) and $\boldsymbol{x}\_1, \ldots, \boldsymbol{x}\_{p(n)}$, each $\boldsymbol{x}\_{\boldsymbol{i}}=t\_{i 1} \boldsymbol{v}\_1+\ldots+t\_{i n} \boldsymbol{v}\_{\boldsymbol{n}}$, where $t\_{i j} \in\{-1,+1\}$ are chosen uniformly at random. The goal is the output $v\_1, \ldots, v\_n$.
| https://mathoverflow.net/users/482299 | A problem about matrix | Nice question!
First of all, notice that the most we can hope for is to get $v\_i$ up to a permutation and $\pm 1$. Besides these obvious obstructions, the answer is yes.
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The $x\_i$ are independent and identically distributed, and thus all the information about the $x\_i$ comes from taking functions $f : \mathbb{R}^n \to \mathbb{R}$ and computing the expected value of $f(x\_i)$ by taking the average over many $x\_i$. So, our strategy will be to take certain functions $f$ and compute this average by sampling the $x\_i$. Ideally, $f$ would behave nicely with respect to addition because our $x\_i$'s are given as sums of many variables. Thus, it makes sense that $f(x) = g(\langle x, v \rangle)$ for some vector $v$ and some function $g : \mathbb{R} \to \mathbb{R}$ which behaves nicely with respect to addition.
The most obvious choice would be a linear function, which for obvious reasons does not give any information.
Another logical choice would be $g(x) = c^x$. This might be doable, but there is a technical problem here that the variance is very large, so it is hard to get a reliable estimate by sampling. Thus, our next choice is polynomials..
Let $g(x) = x^k$. Then, $g(\langle x\_i, v \rangle) = \left( \sum\_j t\_{i1} \langle v\_i, v \rangle \right)^k$. Expanding out, we see that $g$ consists of a sum of monomials in $\langle v\_i, v \rangle$ with coefficients with the corresponding $t\_{i1}$'s. Taking the expected value, we see that the only terms that remain are those where each $t\_{i1}$ appears to an even power. In particular, there is no reason to look at $k$ odd.
For $k = 2$ the expected value of $g$ is $\sum\_j \langle v\_j, v \rangle^2 = \langle v, v \rangle$, so we get no new information. Thus, we need to take $k = 4$.
For $k = 4$ we basically get $\sum\_j \langle v\_j, v \rangle^4$. Note that both the expected value and the variance are $n^{\Theta (1)}$ and thus with a polynomial number of samples we can compute the expected value to high accuracy. Replacing $v$ with $v + t u$, and taking $5$ values of $t$ we see that we can compute $\sum\_j \langle v\_j, v \rangle^s \langle v\_j, u \rangle^{4 - s}$ for $0 \leq s \leq 4$. In particular taking $s = 1$ we can compute $\sum\_j \langle v\_j, v \rangle \langle v\_j, u \rangle^3 = \langle \sum\_j \langle v\_j, u \rangle^3 v\_j, v \rangle$.
In particular, letting $v$ run over the standard basis vectors we see that we can compute $\sum\_j \langle v\_j, u \rangle^3 v\_j$. Iterating this process $r$ times, we get the vector $\sum\_j \langle v\_j, u \rangle^{3^r} v\_j$. Notice that this vector very quickly becomes proportional to $v\_j$, where $j$ is such that $\lvert \langle v\_j, u \rangle \rvert$ is largest, and thus we get a good approximation for $v\_j$.
Do this $n \log n$ times with different random starting vectors $u$ until (by the coupon collector's problem) we will end up with good guesses for all $v\_j$. From this it is easy to find the $v\_j$ exactly, because for each $x\_i$ we can find out what $t\_{ij}$ is by taking the scalar product with our guess for $v\_j$ and checking the sign, and then we just get a problem in linear algebra.
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As I've mentioned above, each average can be computed to high accuracy with a polynomial number of samples, and we compute this things a polynomial number of times, thus this whole process requires a polynomial number of vectors $x\_i$. With a small amount of effort this can be made explicit and I think the constants involved should be quite small.
| 3 | https://mathoverflow.net/users/88679 | 440644 | 177,916 |
https://mathoverflow.net/questions/440602 | 2 | It is a well-known result that there is a bijective correspondence between harmonic sections and cohomology classes of an elliptic complex in a Riemannian/Hermitian manifold. Now consider Riemannian/Hermitian metrics on the vector bundles of a differential complex, that might be non-elliptic. Is this still true if we discard the ellipticity?
I am interested in particular in knowing:
* If it is false, true or unknown the bijective correspondence for general differential complexes on compact manifolds.
* If it is false, a counterexample would be appreciated.
* If it is true, a reference to the proof would be appreciated.
I have on my mind the following basic example in which this seems to work. Consider any manifold $X$ and any complex of vector bundles where the differential morphisms $d\_i$ are all 0. In this case it is trivial that this holds. In this case we do not need compactness either.
For a less trivial example we can consider the torus $T=S^1\times S^1$. Let $d$ be the exterior derivative on $S^1$. Then define a differential operator $d':C^\infty(T)\to \Omega^1(T)$ given by $d'(f)(x,y)={\pi\_1}^\* d(f|\_{S^1\times\{y\}} )$ where ${\pi\_1} :T \to S^1$ is the projection on the first coordinate. The adjoint of $d'$ is $d^\* (fdx+gdy)=\frac{\partial f}{\partial x}$ where $dx={\pi\_1}^\* (ds)$, $dy={\pi\_2}^\* (ds)$, $\frac{\partial}{\partial x}$ is the vector field given by $dx(\frac{\partial}{\partial x})=1$ and $ds$ is any differential form on $S^1$ that does not vanish. In this case we would have Laplacians $\Delta\_0=d^\* d'$ and $\Delta\_1=d'd^\* $. It is clear that $\Delta\_0(f)=\frac{\partial^2 f}{\partial x^2}$ and $\Delta\_1(fdx+gdy)=\frac{\partial^2 f}{\partial x^2}dx$. Thus, the kernel of $\Delta\_0$ equals the kernel of $d'$ and the kernel of $\Delta\_1$ is also $\Omega^1(T)/ d'(C^\infty(T))$.
| https://mathoverflow.net/users/494777 | Hodge decomposition for non-elliptic complexes | This is false. Equip $T^2 = S^1 \times S^1$ with the Lorentzian metric $g = -ds^2 + dt^2$. For concreteness, regard $S^1 = \mathbb{R} / \mathbb{Z}$. Consider the de Rham complex. The Hodge Laplacian on functions is now the wave operator $\Box = -\frac{\partial^2}{\partial s^2} + \frac{\partial^2}{\partial t^2}$. Its kernel is infinite-dimensional (e.g. $\cos(2\pi ks)\cos(2\pi kt)$ is in the kernel for all nonnegative integers $k$), so is very far from being isomorphic to $H^0(T^2) = \mathbb{R}$.
There are situations where one does have a Hodge isomorphism theorem for a non-elliptic complex. For example, this is true when the complex is only subelliptic, such as in the [Rumin complex](https://dx.doi.org/10.4310/jdg/1214454873). More strikingly, it is also [true](https://www.jstor.org/stable/j.ctt1bd6jn8) for the Kohn—Rossi complex, even though some of the operators in this complex are not even hypoelliptic.
| 5 | https://mathoverflow.net/users/121820 | 440646 | 177,918 |
https://mathoverflow.net/questions/439894 | 8 | For $n\in\omega+1$ let $\mathsf{ZFC}\_n$ be $\mathsf{ZC}$ + $\{\Sigma\_k$-$\mathsf{Rep}: k<n\}$. Let $\widehat{\mathsf{ZFC}}$ be the strongest *consistent* theory $\mathsf{ZFC}\_n$ (so if $\mathsf{ZFC}$ is consistent then $\widehat{\mathsf{ZFC}}=\mathsf{ZFC}$). Note that this definition is entirely "internal" - for example, for every standard natural number $n$ every model of $\mathsf{ZFC}$ thinks that $\widehat{\mathsf{ZFC}}$ has a model and that $\widehat{\mathsf{ZFC}}$ strictly extends $\mathsf{ZFC}\_n$ (see [here](https://mathoverflow.net/a/51786/8133)).
I'm interested in the "downward-looking" modal logic this gives rise to. Ignoring set/class issues for simplicity, let our class of worlds $\mathbb{W}$ be the class of models of $\mathsf{ZFC}$, with the accessibility relation $\mathcal{M}\rightarrow\mathcal{N}$ holding whenever $\mathcal{N}$ is a structure in $\mathcal{M}$ such that $\mathcal{M}\models(\mathcal{N}\models\widehat{\mathsf{ZFC}})$. Intuitively $\Diamond \varphi$ expresses that $\varphi$ is "as consistent as possible" with $\mathsf{ZFC}$.
Finally, and as with for example the modal logic of forcing (see [Hamkins](https://arxiv.org/abs/math/0509616)), we'll restrict attention to "first-order expressible" valuations $\nu$; these are valuations such that for each propositional atom $p$ there is some sentence in first-order set theory $\varphi$ with $\nu(p)=\{\mathcal{M}\in\mathbb{W}: \mathcal{M}\models\varphi\}.$ Basically, we're not just treating $(\mathbb{W},\rightarrow)$ as a pure Kripke frame, we're also restricting attention to "model-theoretically meaningful" ways of assigning truth values to sentences at worlds.
>
> **Question 1**: What is the corresponding modal logic? That is, what propositional modal sentences are made true at every world with respect to every valuation satisfying the above expressibility criterion?
>
>
>
Since this is rather broad, let me localize things a bit. Note that we can make sense of the definition of the modal logic above - call it "$\mathfrak{D}$" - within the language of set theory itself, so we can ask what a given model of $\mathsf{ZFC}$ thinks about $\mathfrak{D}$. Semantically, this amounts to restricting attention to worlds within a given model of $\mathsf{ZFC}$.
>
> **Question 2**: Is there a (standard) propositional modal sentence $\varphi$ such that it is independent of $\mathsf{ZFC}$ whether $\varphi\in\mathfrak{D}$?
>
>
>
A positive answer to question 2 would suggest that question 1 does not have a simple answer.
---
My primary interest is actually in more complicated languages based on the above picture - in particular, when $\mathcal{M}\rightarrow\mathcal{N}$, the starting structure $\mathcal{M}$ thinks that some of the elements of $\mathcal{N}$ are standard and so we get a partial counterpart relation and can start playing with object quantifiers - but the "pure sentential" side of things seems like a good starting point already.
| https://mathoverflow.net/users/8133 | Modal logic of "mostly-satisfiability" | First point is that this logic is simply the provability logic of formalized $\widehat{\mathsf{ZFC}}$-provability, i.e. the provability logic of the provability predicate:
$$\mathsf{Prv}\_{\widehat{\mathsf{ZFC}}}(x)\colon\;\;\;\;\; \exists y(\mathsf{Con}(\mathsf{ZFC}\_y)\land \mathsf{Prv}\_{\mathsf{ZFC}\_y}(x)).$$
This, of course. simply is a variant of Feferman's provability (see [1]) for the case of $\mathsf{ZFC}$. The fact that this provability logic coincides with the one based on the frame of models from the question, could be shown in the fashion of the paper [2] of Paula Henk, whom considered the case of $\mathsf{PA}$ with usual provability and the frame built from $\mathsf{PA}$ models. Basically we simply show by induction on the modal depth of modal formulas $\varphi(\vec{x})$ that for any first-order valuation $[\vec{\psi}/\vec{x}]$ and model $\mathcal{M}\models \mathsf{ZFC}$ we have
$$\mathcal{M}\Vdash \varphi(\vec{x})[\vec{\psi}/\vec{x}]\iff \mathcal{M}\models (\varphi(\vec{x})[\vec{\psi}/\vec{x}])^{\star},$$
where $\cdot^\star$ is the provability interpretation mapping $\Box$ to $\mathsf{Prv}\_{\widehat{\mathsf{ZFC}}}$.
Hence the question is basically what is the provability logic of this version of Feferman provability. The provability logic of Feferman provability for $\mathsf{PA}$ was studied by Vladimir Shavrukov [3]. Namely his main focus was on the provability predicate
$$\exists y (\mathsf{Con}(\mathsf{I}\Sigma\_{y})\land \mathsf{Prv}\_{\mathsf{I}\Sigma\_y}(x))$$
Although, I haven't carefully checked this, I am almost sure, that the result would transfer to $\mathsf{Prv}\_{\widehat{\mathsf{ZFC}}}$ without any troubles. For the step of induction we simply use that each $\mathcal{M}\models \mathsf{ZFC}$ satisfies completeness theorem.
So assuming that Shavrukov's result indeed transfers, then the joint provability logic of $\mathsf{Prv}\_{\mathsf{ZFC}}$ ($\Box$) and $\mathsf{Prv}\_{\widehat{\mathsf{ZFC}}}$ ($\triangle$) is
1. $\mathsf{GL}$ for $\Box$
2. $\mathsf{K}$ for $\triangle$
3. $\Box\varphi\to \Box\triangle \varphi$
4. $\Box\varphi\to \triangle\Box\varphi$
5. $\Box\varphi\mathrel{\leftrightarrow} (\triangle \varphi\lor \Box\bot)$
6. $\triangledown \top$
7. $\triangle \varphi \to \triangle((\triangle \psi\to\psi)\lor\triangle \varphi)$
The answer to the original Question 1 is the $\triangle$-fragment of this logic.
[1] S. Feferman. Arithmetization of metamathematics in a general setting. Fundamenta mathematicae, vol. 49 no. 1 (1960), pp. 35–92.
[2] Henk, Paula. "Kripke models built from models of arithmetic." Logic, Language, and Computation: 10th International Tbilisi Symposium on Logic, Language, and Computation, TbiLLC 2013, Gudauri, Georgia, September 23-27, 2013. Revised Selected Papers. Berlin, Heidelberg: Springer Berlin Heidelberg, 2015. <https://eprints.illc.uva.nl/id/eprint/489/1/PP-2014-01.text.pdf>
[3] Shavrukov, V. Yu. "A smart child of Peano's." Notre Dame Journal of Formal Logic 35.2 (1994): 161-185. <https://projecteuclid.org/journalArticle/Download?urlId=10.1305%2Fndjfl%2F1094061859>
| 3 | https://mathoverflow.net/users/36385 | 440659 | 177,922 |
https://mathoverflow.net/questions/440639 | 6 | **Deep Choice**:
$\forall X \ [\forall a,b \in X \, ( a \neq \emptyset \land (a \neq b \to a \cap b = \emptyset)) \to \\ \exists Y \exists f \,(f: X \rightarrowtail Y \land \forall x \in X \,( f(x) \in \operatorname {tc}(x)))]$
In English: for every family $X$ of pairwise disjoint nonempty sets, there is an injective function that sends each element $x$ of $X$ to an element of the transitive closure of $x$.
Denote this principle “$\mathsf{DpC}$”.
Is the following true?
>
> $(\mathsf{ZF} + \mathsf{DpC}) \to \mathsf{AC}$.
>
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| https://mathoverflow.net/users/95347 | Can Deep Choice entail Axiom of Choice? | It proves AC. For this, recall it's enough to see that for every ordinal $\alpha$, $\mathcal{P}(\alpha)$ is wellorderable, and for that it's enough to see that $\mathcal{P}(\mathcal{P}(\alpha))\backslash\{\emptyset\}$ has a choice function. Supposing this fails for some $\alpha$, let $\alpha$ be the least such; then $\alpha$ is an infinite cardinal. Let $\pi:3\times\alpha\to\alpha$ be a bijection. Let $\pi^+:\mathcal{P}(\alpha)^3\to\mathcal{P}(\alpha)$ be the induced bijection, i.e. $$\pi^+(x,y,z)=\pi[x]\cup\pi[\alpha+y]\cup\pi[2\times\alpha+z],$$ where $\alpha+y=\{\alpha+\beta\bigm|\beta\in y\}$ etc. Now for $x\in\mathcal{P}(\alpha)$ and $X\in\mathcal{P}(\mathcal{P}(\alpha))$ and $z\in\mathcal{P}(\alpha)$ define $(x,X,z)^\*\in\mathcal{P}(\mathcal{P}(\alpha))$ by $$ (x,X,z)^\*=\{\pi^+(x,y,z)\bigm|y\in X\}.$$
Clearly $(x,X,z)\mapsto(x,X,z)^\*$ is injective in all 3 arguments $x,X,z$. For $x\in\mathcal{P}(\alpha)$ and $X\in\mathcal{P}(\mathcal{P}(\alpha))$ let
$$(x,X)^{\*\*}=\{(x,X,z)^\*\bigm|z\in X\}.$$
Note that if $X,Y\in\mathcal{P}(\mathcal{P}(\alpha))$ and $x,y\in\mathcal{P}(\alpha)$ with $(x,X)\neq(y,Y)$ then $(x,X)^{\*\*}\cap(y,Y)^{\*\*}=\emptyset$.
Also if $X\in\mathcal{P}(\mathcal{P}(\alpha))\backslash\{\emptyset\}$ then $(x,X)^{\*\*}\neq\emptyset$.
Now let $$P=\Big\{(x,X)^{\*\*}\Bigm|x\in\mathcal{P}(\alpha)\wedge X\in\mathcal{P}(\mathcal{P}(\alpha))\backslash\{\emptyset\}\Big\}.$$
So by the remarks above, we can apply Deep Choice to the family $P$. Let $f:P\to V$ be an injection witnessing this.
I claim that for some $x\in\mathcal{P}(\alpha)$, we have $f((x,X)^{\*\*})\notin\alpha$ for all $X\in\mathcal{P}(\mathcal{P}(\alpha))\backslash\{\emptyset\}$. Otherwise, let $\beta\_x$ be the least $\beta$ of form $f((x,X)^{\*\*})$ for some such $X$; then $x\mapsto\beta\_x$ is injective with domain $\mathcal{P}(\alpha)$, so $\mathcal{P}(\alpha)$ is wellorderable, contradiction.
Fix $x$ witnessing the claim. Then note that for each $X\in\mathcal{P}(\mathcal{P}(\alpha))\backslash\{\emptyset\}$, either
* (i) $f((x,X)^{\*\*})=(x,X,z)^\*$ for some $z\in X$, or
* (ii) $f((x,X)^{\*\*})=\pi^+(x,y,z)$ for some $y,z\in X$
(this is because the only other options are further down in the transitive closure of $(x,X)^{\*\*}$, but such objects are in the transitive closure of some $\pi^+(x,y,z)$, but these are only ordinals ${<\alpha}$, which have been ruled out already).
But now we can define a choice function $c$ for $\mathcal{P}(\mathcal{P}(\alpha))\backslash\{\emptyset\}$: given $X\in\mathcal{P}(\mathcal{P}(\alpha))\backslash\{\emptyset\}$, let $c(X)=$ the unique $z\in X$ witnessing (i), if (i) holds; or if (i) is false, then let $c(X)=$ the unique $z\in X$ witnessing (ii). The existence of $c$ contradicts our hypothesis.
| 8 | https://mathoverflow.net/users/160347 | 440668 | 177,925 |
https://mathoverflow.net/questions/440655 | 2 | I asked a question a few days ago and got a response
But my follow-up question was not answered (maybe my email was not sent successfully)
[A question about computability and Turing machines](https://mathoverflow.net/questions/440338/a-question-about-computability-and-turing-machines)
My quesion is:
1. If $E$ is not well-based, should the range of $F$ also exist ?
2. For every $n<w$, what is the set {$m | (w,E)$ satisfies $mEn$} ?
| https://mathoverflow.net/users/499050 | A question about computability and Turing machines Part 2 | I believe you refer to "well founded" not "well based".
You are asking about a relation $E$ on $\omega$ and a function $F$ for which
$$F(n)=\{F(m)\mid m\mathrel{E} n\}.$$
Such a function $F$ would have the property that
$$m\mathrel{E} n\implies F(m)\in F(n)$$
And from this it follows that $E$ must be well-founded, since $\in$ is well founded. Every nonempty subset of $\omega$ would map to a set in the range of $F$, and that set would have a minimal element, whose preimage would be $E$-minimal.
If $E$ is not well-founded, therefore, there is no such function $F$ obeying the requirement.
These kinds of functions are especially useful when $E$ is an extensional well-founded relation, and the function $F$ is known as the [Mostowski collapse](https://en.wikipedia.org/wiki/Mostowski_collapse_lemma). The basic fact is that every well-founded extensional relation $E$ on a set $X$ is isomorphic to a unique transitive set $M$. The function is simply
$$F(x)=\{F(y)\mid y\mathrel{E} x\}.$$
This definition is made by transfinite recursion of $E$, which is legitimate since $E$ is well-founded. Since $E$ is extensional, one proves that $F$ is one-to-one, and since
$$y\mathrel{E}x \iff F(y)\in F(x)$$
it follows that $F$ is an isomorphism between $\langle X,E\rangle$ and $\langle\text{ran}(F),\in\rangle$.
Regarding question 2, the set
$$\{m\mid m\mathrel{E} n\}$$
is the set of objects in the structure $\langle\omega,E\rangle$ that this structure thinks are "elements" of the object $n$. The $E$ relation is simply a coded version of $\in$, but on the natural numbers instead of sets.
When one is working with Turing machines or infinite time Turing machines, the machines do not compute directly with sets, but the sets can be represented in terms of their codes, and that is the reason for introducing these relations $E$, which allow us to represent sets in a way that is suitable for computation.
| 4 | https://mathoverflow.net/users/1946 | 440669 | 177,926 |
https://mathoverflow.net/questions/440670 | 26 | Let ${\mathcal A}$ be the family of subsets $A$ of the natural numbers ${\mathbf N}$ which are co-infinite (i.e., their complement is infinite). We partially order this family by set inclusion. A [cofinal](https://en.wikipedia.org/wiki/Cofinal_(mathematics)) subset of ${\mathcal A}$ is a subcollection ${\mathcal A}'$ such that every $A$ in ${\mathcal A}$ is contained in some $A' \in {\mathcal A}'$. The [cofinality](https://en.wikipedia.org/wiki/Cofinality) of ${\mathcal A}$ is the minimum cardinality of a cofinal subset ${\mathcal A}'$ of ${\mathcal A}$. ${}{}{}$
An easy application of the Cantor diagonal argument shows that the cofinality of ${\mathcal A}$ is uncountable, and the cofinality is of course dominated by the cardinality of the continuum; thus on the continuum hypothesis the cofinality is equal to the cardinality of the continuum. In general, what are the possible values of this cofinality?
Many years ago I [asked a similar question](https://mathoverflow.net/questions/29624/how-many-orders-of-infinity-are-there) about $\omega^{\omega}$, but the poset ${\mathcal A}$ seems to have a rather different structure (it is not closed under joins, for instance), and so the answer to this question may be rather different.
| https://mathoverflow.net/users/766 | What is the cofinality of the co-infinite subsets of ${\bf N}$? | Every such cofinal family $\mathcal{A}'$ must have size continuum. The reason is that there is an almost disjoint family $\mathcal{D}$ of size continuum, a family of infinite co-infinite sets $A\subseteq\mathbb{N}$ for which any two have finite intersection. To construct such an almost disjoint family, label the nodes of the infinite binary tree with distinct natural numbers, and take the sets of labels arising along any branch of the tree. This tree has continuum many branches, and any two of them have finite intersection in the tree, so the family of label sets will be almost disjoint.
Now, if we have an almost disjoint family $\mathcal{D}$ of size continuum, your dominating family $\mathcal{A}'$ will have to contain covers of the complements of these sets, that is, covering $\mathbb{N}-A$ for every $A\in \mathcal{D}$. But if a set $X$ covers the complement of $A$, then the complement of $X$ is contained in $A$. Therefore no coinfinite set $X$ can cover the complements of two different $A,B\in\mathcal{D}$, since the complement of $X$ would be contained in $A\cap B$, which is finite. So you will need a different covering set for every $\mathbb{N}-A$ for $A\in \mathcal{D}$. Thus, you will need continuum many sets in $\mathcal{A}'$.
| 36 | https://mathoverflow.net/users/1946 | 440671 | 177,927 |
https://mathoverflow.net/questions/440664 | 1 | Let $n \in \mathbb N^\* , \alpha \in \mathbb R$, and $w\_i \in \mathbb R$ for all $i=1, \ldots, n$. Consider the map $f:\mathbb R\_{\ge 0} \to \mathbb R$ defined by
$$
f(x) := \sum\_{i=1}^n w\_i x^{\color{red}{1/i}}.
$$
I'm interested in the solutions of $f(x)=\alpha$. I would like to search on Google Scholar, but I don't know the terminology for this kind of equation.
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> Could you please elaborate on the references of this kind of equation?
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| https://mathoverflow.net/users/477203 | The solutions of $\sum_{i=1}^n w_i x^{1/i} = \alpha$ | Setting $x=e^t$ we obtain an exponential sum. Exponential sums were much studied, from various points of view. One reference is Pólya–Szegő, Problems and theorems in Analysis, vol. 2, part V, Chap 1, section 6, where they are studied in the real domain. Also notice that exponential sums are solutions of linear differential equations with constant coefficients.
| 4 | https://mathoverflow.net/users/25510 | 440672 | 177,928 |
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