Datasets:

euclaise

/

mathoverflow-accepted

like 3

https://mathoverflow.net/questions/210676

8

Consider a smooth, connected and complete Riemannian manifold $M$. It is well known that harmonic functions defined on some open subset of $M$ are $C^\infty$. I'm interested in knowing whether there are quantitative $C^1$ estimates for such harmonic functions which depend only on some bound from below on the curvature (either sectional or Ricci) and an upper bound on the dimension. To give an example of what I mean, consider the same question with "$C^1$" replaced by "Lipschitz". It is then a classical fact of geometric analysis that Lipschitz estimates follows from a lower bound on the Ricci and an upper bound on the dimension, the proof being an application of the Bochner inequality together with elliptic regularity theory. Yet, I don't know any result which improves such Lipschitz estimate to $C^1$. I'm interested both in positive statement (e.g. with a lower bound on the sectional there is an estimate) and in negative ones (e.g. a lower bound on the Ricci is not sufficient because along a certain sequence of manifolds the $C^1$ norm of properly defined harmonic functions blows-up). The question comes from the study of non-smooth spaces arising as (measured)-Gromov-Hausdorff limits of manifolds with prescribed curvature bounds. Basically, what I'm trying to understand is "how smooth" can be functions defined on them, so that it is natural to start looking at harmonic ones. Still, in fact I don't really know how smooth harmonic functions are on Riemannian manifolds, and given that any estimate on limit spaces is (more-or-less) also a quantitative estimate on the smooth ones, before trying to reinvent the wheel I'd like to know if there are known results available in the literature. Any help is appreciated, thanks.

https://mathoverflow.net/users/58975

$C^1$ regularity of harmonic functions on Riemannian manifolds

For the people interested in the solution to this problem, Gigli encouraged Nuñez Zimbrón and De Philippis to look for a solution to this problem, which now can be found here <https://arxiv.org/pdf/1909.05220.pdf>

3

https://mathoverflow.net/users/510475

452743

https://mathoverflow.net/questions/452708

3

Let $n$ be any integer greater than $2^{10^6}$. Given any $s\le (\log\_2 n)/1000$ integers $1=q\_1\le q\_2\le \cdots q\_{s-1}\le q\_s=n$. Prove that $$\min\_\ell\left(\sum\_{i=1}^\ell q\_i\right)\left(\sum\_{i=\ell+1}^s\frac{1}{q\_i}\right)\le \frac{1}{64}$$

https://mathoverflow.net/users/510447

Min problem on integers

Let us denote $$\sigma\_\ell:=\sum\_{i=1}^\ell q\_i\qquad\text{and}\qquad\tau\_\ell:=\sum\_{i=\ell+1}^s\frac{1}{q\_i}.$$ Then $$\prod\_{\ell=1}^{s-1}\left(\frac{q\_\ell}{q\_{\ell+1}}\cdot\frac{\sigma\_{\ell+1}}{\sigma\_\ell}\cdot\frac{\tau\_{\ell-1}}{\tau\_\ell}\right)=\frac{q\_1}{q\_s}\cdot\frac{\sigma\_{s}}{\sigma\_1}\cdot\frac{\tau\_0}{\tau\_{s-1}}=\sigma\_s\tau\_0>n,$$ hence there exists $\ell\in\{1,2,\dotsc,s-1\}$ such that $$\frac{q\_\ell}{q\_{\ell+1}}\cdot\frac{\sigma\_{\ell+1}}{\sigma\_\ell}\cdot\frac{\tau\_{\ell-1}}{\tau\_\ell}>n^{1/(s-1)}>n^{1/s}=2^{1000}.$$ On the other hand, the left-hand side equals $$\frac{q\_\ell}{q\_{\ell+1}}\left(1+\frac{q\_{\ell+1}}{\sigma\_\ell}\right)\left(1+\frac{1}{q\_\ell\tau\_\ell}\right) =\frac{q\_\ell}{q\_{\ell+1}}+\frac{q\_\ell}{\sigma\_\ell}+\frac{1}{q\_{\ell+1}\tau\_\ell}+\frac{1}{\sigma\_\ell\tau\_\ell},$$ so for the same $\ell$ we have $$\frac{q\_\ell}{q\_{\ell+1}}+\frac{q\_\ell}{\sigma\_\ell}+\frac{1}{q\_{\ell+1}\tau\_\ell}+\frac{1}{\sigma\_\ell\tau\_\ell}>2^{1000}.$$ The first three terms on the left-hand side do not exceed $1$, hence in fact $$\sigma\_\ell\tau\_\ell<\frac{1}{2^{1000}-3}.$$

9

https://mathoverflow.net/users/11919

452748

https://mathoverflow.net/questions/452696

1

We consider the minimal surface equation $$ (1+|\nabla u|^2) \, \Delta u=\sum\_{i,j=1}^n\partial\_iu \, \partial\_ju \, \partial\_{ij}u\quad\hbox{in $B\_1\subset\mathbb R^n.$} $$ If $u\in C^2(B\_1)$ is a positive solution of above equation, does the Harnack inequality hold in $B\_{1/2}$? That is, is there a constant $C>0$, which is only dependent on $n$, such that $$\sup\_{B\_{1/2}}u\leq C\inf\_{B\_{1/2}}u?$$ If above equation is uniformly elliptic, then Harnack inequality for the uniformly elliptic equation concludes that the constant $C$ may be dependent on the elliptic constants. However, can we derive the Harnack inequality for the minimal surface equation without the uniformly elliptic condition?

https://mathoverflow.net/users/105893

Harnack inequality for the minimal surface equation

The minimal surface equation is uniformly elliptic, at least in the sense that its linearization at any solution is uniformly elliptic. It will be convenient to rewrite the equation as $$\nabla \cdot \left(\frac{\nabla u}{\sqrt{1 + |\nabla u|^2}}\right) = 0$$ (this is the standard way that the equation is written on euclidean space, anyways). We now have the coefficient $A\_{ij} = (1 + |\nabla u|^2)^{-1/2} \delta\_{ij}$, and the PDE is $Pu := \nabla \cdot (A\nabla u) = 0$. By elliptic regularity, the solution $u$ is smooth, so there exists $\varepsilon > 0$ such that on $B\_{3/4}$ (say) we have $|\nabla u| < \varepsilon^{-1}$. On $\{|\nabla u| < 1\} \cap B\_{3/4}$ we have $A\_{ii} > 1/10$, and on $\{|\nabla u| \geq 1\} \cap B\_{3/4}$ we have $A\_{ii} > \varepsilon^{1/2}/10$. Therefore $P$ is a uniformly elliptic operator and we can apply the usual proof of Harnack's inequality. EDIT 2: Here's an explicit example. Recall that the catenoid is the surface of revolution corresponding to the catenary $$z = f(x) := \varepsilon \cosh\left(\frac{x}{\varepsilon}\right)$$ and it is minimal. Taking only half of the surface of revolution, we get a minimal graph $z = u(x, y)$ with $u > 0$. Consider the restriction of $u$ to the ball $B$ of radius $1/4$ centered on $(x, y) = (1, 0)$. Then the graph of $u$ contains the the catenary, so $\inf\_B u \leq f(0.75)$ and $\sup\_B u \geq f(1.25)$. Now we estimate the ratio $$\frac{\sup\_B u}{\inf\_B u} \geq \frac{f(1.25)}{f(0.75)} = \frac{\cosh(1.25/\varepsilon)}{\cosh(0.75/\varepsilon)}$$ which blows up as $\varepsilon \to 0$.

1

https://mathoverflow.net/users/109533

452752

https://mathoverflow.net/questions/452753

1

Let $X=\operatorname{Spec}(A)$ be an affine [Dedekind domain](https://en.m.wikipedia.org/wiki/Dedekind_domain#Alternative_definitions) with field of fractions $K$. Let $\widetilde{A}$ be the integral closure of $A$ in separable closure $ K^{\text{sep}}$. A closed point $x$ of $X$ is a nonzero prime ideal $\mathfrak{p}$ of $A$, and the choice of a prime ideal $\widetilde{p}$ of $\widetilde{A}$ lying over $\mathfrak{p}$ determines a geometric point $\overline{x}= \operatorname{Spec}k(x)^{\text{sep}} \to X$ a geometric point of $X$. The etale stalk with respect to $\overline{x}$ is given by $$ \mathcal{O}\_{X,\overline{x}}^{\text{et}}= \varinjlim\_{(A\_U,u) \text{ etale nbhds of } A \text{ over }\overline{x}} (A\_U,u)$$ **Question:** How to see that the field of fractions of the ring $\mathcal{O}\_{X,\overline{x}}^{\text{et}}$ equals to the fixed field $(K^{\text{sep}})^{I(\widetilde{\mathfrak{p}})}$ where the inertia group $I(\widetilde{\mathfrak{p}})$ of $\widetilde{\mathfrak{p}}$ is defined as $$I(\widetilde{\mathfrak{p}})=:\{ g \in G \ | \ g(\widetilde{\mathfrak{p}})=\widetilde{\mathfrak{p}} \text{ and } g | \_{\kappa(\widetilde{\mathfrak{p}})} = \text{id}\_{\kappa(\widetilde{\mathfrak{p}})} \} \subset \text{Gal}(K^{\text{sep}}/K)$$ Source: Milne's LEC script on Etale Cohomology (Example 12.4, page 82 last paragraph) Note that previous considerations in the linked script show that the field of fractions of the ring $\mathcal{O}\_{X,\overline{x}}^{\text{et}}$ is given by inductive limit of those finite separable field extensions $L/K$ ( contained in $ K^{\text{sep}}$) such that the normalization $X\_L$ of $ X$ in $L$ is unramified at some point lying over $x$, but I not see why this inductive limit equals in this case to $(K^{\text{sep}})^{I(\widetilde{\mathfrak{p}})}$ as claimed. Especially, how the requirement that normalization $X\_L$ of $ X$ in $L$ is unramified at some point lying over $x$ is related to the property of $L$ being fixed by the inertia group $I(\widetilde{\mathfrak{p}})$. Thematically the problem is closely related to [this](https://mathoverflow.net/questions/452621/calculate-stalk-of-etale-derived-pushforward-sheaf-milnes-lec) question I asked before. Ideas: Here a way to reduce the problem at least to finite extension case. There exist a finite extension $M/K$, such that the associated map $\overline{x}= \operatorname{Spec}k(x)^{\text{sep}} \to X$ to the considered geometric point factors scheme theoretically through $X\_M$, the Spec of integral closure of $A$ in $M$. So we can assume that all field extensions discussed about about to be finite and we can replace avove everywhere $ K^{\text{sep}}$ by finite extension $M$, letting the inertia group living inside a finite Galois group. But not see how to proceed from here.

https://mathoverflow.net/users/108274

Field of fractions of etale stalk of Dedekind domain (Example from Milne's LEC)

As Milne (sort of) explains, the following two data are exactly the same: 1. A connected étale neighbourhood $U \to X$ with a lift $\bar x \to U$ of $\bar x \to X$, up to shrinking Zariski-locally around the image of $\bar x$ in $U$ (I suppose you could call these 'Zariski germs of connected étale neighbourhoods'); 2. A finite separable field extension $K \to L$ together with a prime $\mathfrak q$ above $\mathfrak p$ in the integral closure $B$ of $A$ in $L$, together with a map $\kappa(\mathfrak q) \to \kappa(\bar x)$, such that the map $\operatorname{Spec} B \to \operatorname{Spec} A$ is étale at $\mathfrak q$. Indeed, to pass from (1) to (2), we saw in [my answer](https://mathoverflow.net/q/452621/82179) to your previous question that the generic fibre of $U \to X$ is a field if $U$ is connected, and $\operatorname{Spec} B \to \operatorname{Spec} A = X$ is étale at $\mathfrak q$ by assumption. Conversely, to go from (2) to (1), we may for instance take $U$ to be the (open) locus of $\operatorname{Spec} B$ where $\operatorname{Spec} B \to \operatorname{Spec} A$ is étale, which is nonempty since it contains the generic point (as $K \to L$ is separable). The choice of $\mathfrak q$ above $\mathfrak p$ together with the map $\kappa(\mathfrak q) \to \kappa(\bar x)$ can be taken care of by choosing a prime ideal $\widetilde{\mathfrak p}$ in $\widetilde A$ together with an isomorphism $\kappa\big(\widetilde{\mathfrak p}\big) \stackrel\sim\to \kappa(\bar x)$. So the only thing to think about is when $\operatorname{Spec} B \to \operatorname{Spec} A$ is étale at $\mathfrak q := \widetilde{\mathfrak p} \cap B$. Standard Galois ramification theory tells us that $(\bar K)^{I(\widetilde{\mathfrak p})}$ is the unique largest subfield of $\bar K$ such that for any finite subextension $K \to L \to (\bar K)^{I(\widetilde{\mathfrak p})}$, the map $\operatorname{Spec} B \to \operatorname{Spec} A$ is étale at $\mathfrak q := \widetilde{\mathfrak p} \cap B$ (where as usual $B$ is the integral closure of $A$ in $L$). A sample reference is Neukirch's *Algebraic Number Theory*, Ch. I, Prop. 9.6.

4

https://mathoverflow.net/users/82179

452758

https://mathoverflow.net/questions/452760

-3

In chapter 4 of Analysis I by Terence Tao, we have the following note about the set theoretic construction of the integers: > > In the language of set theory, what we are doing here is starting with the space N × N of ordered > pairs (a, b) of natural numbers. Then we place an equivalence relation ∼ on these pairs by declaring > (a, b) ∼ (c, d) iff a + d = c + b. The set-theoretic interpretation of the symbol a — b is that it is the > space of all pairs equivalent to (a, b): a — b := {(c, d) ∈ N × N : (a, b) ∼ (c, d)}; the existence of > the set Z = {a — b : (a, b) ∈ N × N} of integers then follows from two applications of the axiom > of replacement. However, this interpretation plays no role in how we manipulate the integers and > we will not refer to it again. A similar set-theoretic interpretation can be given to the construction > of the rational numbers later in this chapter, or the real numbers in the next chapter. > > > What caught my eye in this note is that Tao suggests we need to apply the Axiom of Replacement twice in order to justify the existence of Z, but Z may be constructed with a single application of replacement: for every (a, b) ∈ N × N, we replace it by a — b and thus get Z. This reasoning seems plausible, but something still feels off about it. Is this proof really correct, and hence a simplification of Tao's idea? Or is there some error in the logic behind it?

https://mathoverflow.net/users/510487

Analysis I, simpler proof of Tao's construction of the integers

In fact one doesn't need the replacement axiom at all in order to implement this set-theoretic construction of the integers. The entire construction can be undertaken in Zermelo set theory, which lacks the replacement axiom. Specifically, for a fixed copy of the natural numbers $\mathbb{N}$, one can form the set of pairs using applications of the power set axiom and mere separation, and then define the set of equivalence classes using another application of power set and separation. Indeed, by applying first a suitable number of power set operations, with instances of union, one can apply a single application of the separation axiom to arrive at the desired copy of $\mathbb{Z}$ realized as equivalence classes of the same-difference equivalence relation. The larger lesson is that the replacement axiom is needed in set theory generally only for associations mapping the elements of a set into the universe in such a way that there is no previously known upper bound of a set from which those associated individuals might arise. This reaching-upward aspect is what gives the power of ZFC set theory over Zermelo's strictly weaker theory Z without replacement. It is consistent with Zermelo's theory, for example, that $\aleph\_\omega$ does not exist, even when $\omega$ exists and we can define the map $n\mapsto\aleph\_n$. One needs replacement to show that the cardinals do not end prematurely in this way, as in Zermelo set theory there is no way to provide a bounding set for the $\aleph\_n$'s. But when one already has an upper bound for the witnessing objects, as is the case here since the equivalence classes are sets of pairs, and the set of all equivalence classes is thus a set of sets of pairs, then we can use merely the separation axiom to get at what we want.

13

https://mathoverflow.net/users/1946

452763

https://mathoverflow.net/questions/452730

1

Let $V > 0$ and let $\Phi(\cdot)$ be the standard normal CDF. Consider the infimum of $$f(x\_1, x\_2,x\_3, p\_1, p\_2, p\_3) := p\_1 \Phi(x\_1) + p\_2 \Phi(x\_2) + p\_3 \Phi(x\_3)$$ with respect to $x\_1, x\_2, x\_3, p\_1, p\_2$ and $p\_3$ satisfying $\begin{align} &p\_1 \geq 0,\\\\ &p\_2 \geq 0,\\\\ &p\_3 \geq 0,\\\\ &p\_1 + p\_2 + p\_3 = 1,\\\\ &p\_1 x\_1 + p\_2 x\_2 + p\_3 x\_3 = 0,\\\\ &p\_1 x\_1^2 + p\_2 x\_2^2 + p\_3 x\_3^2 \leq V. \end{align} $ The objective is a continuous function and the feasible set is a subset of $\mathbb{R}^6$. One way to show that the infimum is a minimum is to show that the feasible set is compact. But even with the constraint $p\_1 x\_1^2 + p\_2 x\_2^2 + p\_3 x\_3^2 \leq V$, it can happen that $p\_1$ is arbitrarily small and $x\_1$ is arbitrarily large. I think I need to use the fact that the objective function is bounded between $0$ and $1$, which together with the last constraint limits $x\_i$'s becoming very large.

https://mathoverflow.net/users/145647

Showing that the infimum is a minimum

$\newcommand{\R}{\mathbb R}\newcommand{\tx}{\tilde x}$Indeed, the infimum (say $f^\*$) in question is attained for each $V\in(0,\infty)$. To prove this, for each $j\in\{1,2,3\}$ let $(x\_j^n)\_{n=1}^\infty$ and $(p\_j^n)\_{n=1}^\infty$ be sequences such that for each $n$ all your conditions are satisfied with $x\_1^n,x\_2^n,x\_3^n,p\_1^n,p\_2^n,p\_3^n$ in place of $x\_1,x\_2,x\_3,p\_1,p\_2,p\_3$ and \begin{equation} f(x\_1^n,x\_2^n,x\_3^n,p\_1^n,p\_2^n,p\_3^n)\to f^\* \end{equation} (as $n\to\infty$). Passing to subsequences, without loss of generality assume that for each $j\in\{1,2,3\}$ \begin{equation} x\_j^n\to x\_j^\*\in[-\infty,\infty],\quad p\_j^n\to p\_j^\*\in[0,1]. \end{equation} Let \begin{equation} J:=\{j\in\{1,2,3\}\colon x\_j^\*\in\R\},\quad J^c:=\{1,2,3\}\setminus J. \end{equation} Since $p\_j^n (x\_j^n)^2\le V<\infty$, we see that for each $j\in J^c$ we have $p\_j^\*=0$ and $p\_j^n x\_j^n\to0$. So, letting $\tx\_j:=x\_j^\*\,1(j\in J)$ for $j\in\{1,2,3\}$, we have \begin{equation} p\_j^\*\ge0\ \forall j\in\{1,2,3\},\quad \sum\_{j=1}^3 p\_j^\*=1, \end{equation} \begin{equation} \sum\_{j=1}^3 p\_j^\*\tx\_j=\sum\_{j\in J} p\_j^\*x\_j^\*=\lim\_n \sum\_{j=1}^3 p\_j^n x\_j^n=\lim\_n 0=0, \end{equation} \begin{equation} \sum\_{j=1}^3 p\_j^\*\tx\_j^2=\sum\_{j\in J} p\_j^\*(x\_j^\*)^2=\lim\_n \sum\_{j\in J}p\_j^n (x\_j^n)^2 \le\limsup\_n \sum\_{j=1}^3 p\_j^n (x\_j^n)^2\le V, \end{equation} \begin{equation} \sum\_{j=1}^3 p\_j^\*\Phi(\tx\_j)=\sum\_{j\in J} p\_j^\*\Phi(x\_j^\*) =\lim\_n \sum\_{j\in J} p\_j^n \Phi(x\_j^n) =\lim\_n \sum\_{j=1}^3 p\_j^n \Phi(x\_j^n)=f^\*. \end{equation} So, all your conditions are satisfied with $\tx\_1,\tx\_2,\tx\_3,p\_1^\*,p\_2^\*,p\_3^\*$ in place of $x\_1,x\_2,x\_3,p\_1,p\_2,p\_3$ and the infimum $f^\*$ is attained at $(x\_1,x\_2,x\_3,p\_1,p\_2,p\_3)=(\tx\_1,\tx\_2,\tx\_3,p\_1^\*,p\_2^\*,p\_3^\*)$. $\quad\Box$

2

https://mathoverflow.net/users/36721

452766

https://mathoverflow.net/questions/452663

4

Suppose that $(X,\rho)$ is a compact [doubling metric space](https://en.wikipedia.org/wiki/Doubling_space). Does there necessarily exist an $\epsilon>0$ and a maximal $\epsilon$-net $\{x\_i\}\_{i=1}^n\subseteq X$ such that the map $$ \begin{aligned} \Phi:(X,\rho) & \rightarrow (\mathbb{R}^n,|\cdot|\_2) \\ x&\mapsto \big(\rho(x,x\_i)\big)\_{i=1}^n \end{aligned} $$ is bi-Lipschitz? (A trivial upper-Lipschitz bound of $\sqrt{n}$ is clear but the lower-Lipschitz bound is far from obvious for me). --- My question is rooted in the following observations. **Motivation/Intuition:** The motivation for my question is rooted in the following two observations. 1. The Assouad embedding theorem, see e.g. [this paper for a recent formulation](https://web.math.princeton.edu/%7Enaor/homepage%20files/assouad-N(K).pdf), shows that every doubling metric space admits a bi-Hölder embedding into a Euclidean space. Moreover, it is known that bi-Hölder is necessary, due to the *global* non-embeddability of the Heisenberg group, since the distortion of any closed ball diverges as the radius grows; [this paper](https://web.math.princeton.edu/%7Enaor/homepage%20files/UCHeis9.pdf). 2. As remarked in this [old MO post](https://mathoverflow.net/questions/417185/finite-approximations-to-the-kuratowski-fr%C3%A9chet-embedding), in [this paper of Katz and Katz](https://link.springer.com/article/10.1007/s10711-010-9497-4) (with un unpublished quantitative version [found here](https://arxiv.org/pdf/1305.1529.pdf)) shows we know that there is a bi-Lipschitz embedding of any closed and connected Riemannian manifold $(M,g)$ into some Euclidean space $(\mathbb{R}^n,|\cdot|\_2)$ given by $$ \varphi:\,M\ni x\mapsto \big(\rho\_g(x,x\_i)\big)\_{i=1}^n \in \mathbb{R}^n $$ where $\{x\_i\}\_{i=1}^n$ is any maximal $\epsilon$-net for some sufficiently small $\epsilon>0$ and $\rho\_g$ is the geodesic distance on $(M,g)$. Clearly, compactness is needed here, since it is well-known that the hyperbolic plane cannot be bi-Lipschitz embedded into any Euclidean space. 3. I comment that smoothness is not needed in (1) since the existence of $\Phi$ is obvious for any finite metric space. --- **Update:** **Claim:** Suppose that there exists some $0<\epsilon\le 1$ and a finite set $Y\subseteq X$ (depending on $\epsilon>0$) with the property that for every $x,z\in X$ there is some (possibly not unique) $y\_{x,z}\in Y$ satisfying $$ \epsilon \rho(x,z)\le |\rho(x,y\_{x,z})-\rho(z,y\_{x,z})|. $$ Then, the map $\Phi(x)\mapsto (\rho(x,y))\_{y\in Y}$ is a bi-Lipschitz embedding into the $|Y|$-dimensional Euclidean space. *Proof:* If this case, the finiteness of $Y$ implies that $$ \epsilon \rho(x,z)\le |\rho(x,y\_{x,z})-\rho(z,y\_{x,z})| \le \max\_{y\in Y}\, |\rho(x,y)-\rho(z,y)| \le \|\Phi(x)-\Phi(y)\|\_2 \le |Y|^{1/2}\,\max\_{y\in Y}\, |\rho(x,y)-\rho(z,y)| \le |Y|^{1/2}\rho(x,y) $$ where I used the fact that $|Y|<\infty$ and $x\mapsto \rho(x,y)$ is $1$-Lipschitz for every $y\in Y$. --- **Updated Question:** So doesn't every $\epsilon$-packing do the trick and shouldn't any packing exist since packing numbers are upper-bounded by covering numbers, and in a doubling space all covering numbers are $O((\operatorname{diam}(X,\rho)\epsilon)^{-d})$ where $d$ is the doubling dimension of $(X,\rho)$ and $\operatorname{diam}(X,\rho)$ is its diameter? In which case, why not take $\epsilon=1$? *(Besides the fact that the distortion may be sub-optimal).* --- **Thoughts:** I guess the main challenge is to control the distortion and to show that one can get it to converge to $1$ as $\epsilon\downarrow 0$ (and this $n\uparrow \infty$ in general).

https://mathoverflow.net/users/36886

Bi-Lipschitz embeddings of compact doubling spaces

No. There are many compact, doubling metric spaces with no bi-Lipschitz embedding in a Hilbert space. Some examples: * The closed unit ball in the (continuous) Heisenberg group with its Carnot-Carath'eodory metric. (See, e.g., Theorem 6.1 of "Differentiability of Lipschitz maps from metric spaces to Banach spaces" and observe that the differentiability argument is purely local.) * The example in Theorem 2.3 of "Bilipschitz embeddings of metric spaces into space forms" by Lang and Plaut. * The examples constructed by Laakso in "Ahlfors $Q$-regular spaces with arbitrary $Q>1$ admitting weak Poincar'e inequality" and (suitably compactified) "Plane with $A\_\infty$-weighted metric not bilipschitz embeddable to $\mathbb{R}^n$". See Proposition 1.23 of "Doubling conformal densities" by Bonk, Heinonen, and Rohde.

2

https://mathoverflow.net/users/510495

452767

https://mathoverflow.net/questions/452761

2

Let $M$ be a type $\rm{II}\_{1}$ factor with trace $\tau$, acting by the GNS representation on $B(L^{2}(M,\tau))$. Let $R\subset M \subset B(L^{2}(M,\tau))$ be a hyperfinite $\rm{II}\_{1}$ subfactor of $M$. > > **Question:** Is there a norm one Banach space projection $\Phi: B(L^{2}(M,\tau))\rightarrow R$ such that $\Phi|\_M=\mathbb{E}\_{R}$, > where $\mathbb{E}\_{R}$ is the $\tau$-preserving conditional > expectation of $M$ onto $R$? > > > (I actually expect this is never possible, but have been unable to prove this. I figured this should be well-known, so a reference would be a fine answer!)

https://mathoverflow.net/users/6269

Hyperexpectations from injective subfactors of a type $II_1$ factor

Such a projection $\Phi$ actually always exists. In the terminology of Definition 2.2 of [N. Ozawa and S. Popa, On a class of II$\_1$ factors with at most one Cartan subalgebra. Ann. of Math. 172 (2010), 713-749] the existence of $\Phi$ is saying that $R$ is amenable relative to $\mathbb{C} 1$ inside $M$ (see Theorem 2.1.(3) of the same paper). By Proposition 2.4.(2) of the same paper, this relative amenability holds automatically since $R$ is amenable. A direct argument can be distilled as follows. Denote by $\hat{x} \in L^2(M)$ the vector that corresponds to $x \in M$. Assume that we have a copy of $M\_n(\mathbb{C}) \subset R \subset M$. Denote by $(e\_{ij})$ the matrix units of $M\_n(\mathbb{C})$. Then $$\Phi\_0 : B(L^2(M)) \to M\_n(\mathbb{C}) : \Phi\_0(T) = \sum\_{i,j=1}^n n \, \langle T \hat{e\_{j1}},\hat{e\_{i1}}\rangle \, e\_{ij}$$ defines a conditional expectation of $B(L^2(M))$ onto $M\_n(\mathbb{C})$ whose restriction to $M$ equals the unique trace preserving conditional expectation. Generating $R$ by an increasing union of matrix algebras and taking a pointwise weak$^\*$ limit point of the corresponding $\Phi\_0$ would provide your $\Phi$.

3

https://mathoverflow.net/users/159170

452775

https://mathoverflow.net/questions/451346

2

Let $(X\_{jn})\_{1\leq j \leq n}$, $n\in \mathbb N$, be a triangular array of random vectors in $\mathbb R^d$ (the $X\_{jn}$ are understood to be independent in $j$ for fixed $n$.). We say that the triangular array is *null* if $X\_{jn} \overset{p}{\to} 0$, as $n \to \infty$, uniformly in $j$, i.e.,: \begin{equation} \lim\_{n \to \infty} \max\_{1\leq j\leq n} P( \, |X\_{jn}|> \epsilon \, )=0, \quad (\forall \, \epsilon>0). \end{equation} For any random vector $X$ with distribution $\mu$, we introduce an associated compound Poisson random vector $X^{\tilde{}}$ (Note the tilde, please) with characteristic measure $\mu$, i.e.: \begin{equation} \log \varphi\_{X^{\tilde{}}}(u) = \int\_{\mathbb R^d} (e^{iux}-1)\mu (dx), \quad u \in \mathbb R^d. \end{equation} For a triangular array $(X\_{jn})\_{1\leq j \leq n}$, the corresponding compound Poisson vectors $(X\_{jn}^{\tilde{}})\_{1\leq j \leq n}$ are again assumed to have row-wise independent entries. By $X\_n \overset{d}{\sim} Y\_n$ we mean that, if either side converges in distribution along a sub-sequence, then so does the other along the same sequence. We can show the proposition about compound Poisson approximation: Let $(X\_{jn})\_{1\leq j \leq n}$, $n\in \mathbb N$, be a null array of random vectors in $\mathbb R^d$. Fix $h > 0$, and define: \begin{equation}\label{k1}\tag{K1} b\_{jn}= E(X\_{jn}\,;\,|X\_{jn}|\leq h). \end{equation} Then \begin{equation}\label{k2}\tag{K2} \sum\_j X\_{jn} \overset{d}{\sim} \sum\_j \left\{ (X\_{jn} - b\_{jn})^{\tilde{}} +b\_{jn} \right\} \end{equation} (This the Proposition 7.11 from [Foundations of Modern Probability](https://link.springer.com/book/10.1007/978-3-030-61871-1), Third edition) **Update** Allow me to rephrase my question. We say that a Infinite Divisible r. vector has the levy kintchine representations if its characteristic function is given by: $$\varphi\_X(u)= \exp\left\{i u'b + \frac{1}{2}u'au + \int\_{\mathbb R^d} \left[e^{iu'x}-1 - i u'x c(x)\right] d\nu(x) \right\} $$ where $c(x)$ is a integrable function. We denote this as $X \sim (b\_c, a, \nu)\_c$. In the case above, we have that $X^{\tilde{}} \sim (0,0,\mu)\_0$. Now, we can change the truncation function $c(x)$ by other $h(x)$. If $X \sim (b\_c, a, \nu)\_c$, we have that $$X \sim (b\_h, a, \nu)\_h, \quad b\_h = b\_c + \int\_{\mathbb R^d}x [ h(x)- c(x)]d\nu(x)$$ **Question:** Given a **Null** triangular array $(X\_{jn})\_{1\leq j \leq n}$ with $X\_{jn} \sim \mu\_{jn}$. Suppose $E[X\_{jn}]=0$ and we also have that: \begin{equation}\label{I}\tag{I} \sum\_{j=1}^n \mathbf{v}(X\_{jn})\leq C< \infty,\quad \forall \, n \end{equation} where $X\_{jn}=(X\_{jn\_{1}},..., X\_{jn\_{d}})$ and: $$\mathbf{v}(X\_{jn}):=\hbox{trace} \left(E\left[X\_{jn}X\_{jn}^{\,\,'}\right]\right)= \sum\_{k=1}^d \hbox{var}(X\_{jn\_{k}})$$ Note that $X\_{jn\_{k}}$ is unidemsional, $E[X\_{jn\_{k}}]=0$ and $\hbox{var}(X\_{jn\_{k}})= E[(X\_{jn\_{k}})^2]$. Define $S\_n' \sim (0,0,\mu\_n)\_h$ with $h(x)\equiv 1$ and $$\mu\_n := \sum\_{j=1}^n \mu\_{jn}$$ So, how to show, using (\ref{k1}) and (\ref{k2}), that: $$S\_n := \sum\_j X\_{jn} \overset{d}{\sim} S\_n' $$

https://mathoverflow.net/users/479236

Convergence of the row sums in a triangular null array with zero mean

$\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}$Let us rephrase the question a bit. For each natural $n$, let $X\_{1,n},\dots,X\_{j\_n,n}$ be independent zero-mean random vectors in $\R^d$ such that (i) for each real $\ep>0$ \begin{equation\*} \max\_{j\in J\_n} P(\|X\_{j,n}\|>\ep)\to0 \tag{0}\label{0} \end{equation\*} (as $n\to\infty$) and (ii) for some real $C>0$ and all $n$ \begin{equation\*} \sum\_{j\in J\_n} E\|X\_{j,n}\|^2\le C, \tag{10}\label{10} \end{equation\*} where $j\_n$ is a positive integer, $J\_n:=\{1,\dots,j\_n\}$, and $\|\cdot\|$ denotes the Euclidean norm. Let \begin{equation\*} S\_n:=\sum\_{j\in J\_n} X\_{j,n}. \end{equation\*} For any random vector $Z$ in $\R^d$, let $f\_Z$ denote the characteristic function (ch. f.) of $Z$, so that $f\_Z(t)=Ee^{it\cdot Z}$ for $t\in\R^d$, where $\cdot$ denotes the dot product. For each natural $n$, let $Y\_{1,n},\dots,Y\_{j\_n,n}$ be independent random vectors in $\R^d$ such that for all $j\in J\_n$ and all $t\in\R^d$ \begin{equation\*} f\_{Y\_{j,n}}(t)=\exp(f\_{X\_{j,n}}(t)-1). \end{equation\*} Let \begin{equation\*} T\_n:=\sum\_{j\in J\_n} Y\_{j,n}. \end{equation\*} The problem is then to show that \begin{equation\*} \text{$S\_n$ converges in distribution iff $T\_n$ converges in distribution.} \tag{20}\label{20} \end{equation\*} Here is a proof. Take any $t\in\R^d$. Note that $|e^{iu}-1|\le\min(2,|u|)$ for all real $u$. So, for each real $\ep>0$, \begin{equation\*} |f\_{X\_{j,n}}(t)-1|\le E|e^{it\cdot X\_{j,n}}-1|\,1(\|X\_{j,n}\|\le\ep) +E|e^{it\cdot X\_{j,n}}-1|\,1(\|X\_{j,n}\|>\ep) \le\|t\|\ep+2P(\|X\_{j,n}\|>\ep) \end{equation\*} and hence, by \eqref{0}, \begin{equation\*} \limsup\_n\max\_{j\in J\_n}|f\_{X\_{j,n}}(t)-1|\le\|t\|\ep. \end{equation\*} So, \begin{equation\*} \max\_{j\in J\_n}|f\_{X\_{j,n}}(t)-1|\to0\text{ uniformly in $t$ in any bounded set}. \tag{30}\label{30} \end{equation\*} Note also that $|e^{iu}-1-iu|\le u^2/2\le u^2$ for all real $u$. So, \begin{equation\*} |f\_{X\_{j,n}}(t)-1|=|E(e^{it\cdot X\_{j,n}}-1-it\cdot X\_{j,n})|\le E(t\cdot X\_{j,n})^2 \le\|t\|^2 E\|X\_{j,n}\|^2. \end{equation\*} So, by \eqref{10}, \begin{equation\*} \sum\_{j\in J\_n}|f\_{X\_{j,n}}(t)-1|\le C\|t\|^2. \tag{40}\label{40} \end{equation\*} It follows from \eqref{30} that for all large enough $n$ (depending on $t$) and all $j\in J\_n$ the value of $\ln f\_{X\_{j,n}}(t)$ is defined and \begin{equation} |\ln f\_{X\_{j,n}}(t)-(f\_{X\_{j,n}}(t)-1)|\le|f\_{X\_{j,n}}(t)-1|^2. \end{equation} So, by \eqref{40} and \eqref{30}, \begin{equation} \Big|\sum\_{j\in J\_n}\ln f\_{X\_{j,n}}(t)-\sum\_{j\in J\_n}(f\_{X\_{j,n}}(t)-1)\Big| \le\sum\_{j\in J\_n}|f\_{X\_{j,n}}(t)-1|^2 \\ \le\max\_{j\in J\_n}|f\_{X\_{j,n}}(t)-1|\,\sum\_{j\in J\_n}|f\_{X\_{j,n}}(t)-1| \le C\|t\|^2\max\_{j\in J\_n}|f\_{X\_{j,n}}(t)-1|\to0. \tag{50}\label{50} \end{equation} Thus, in view of \eqref{50} and \eqref{30}, $S\_n$ converges in distribution iff $\prod\_{j\in J\_n}f\_{X\_{j,n}}$ converges pointwise to a function continuous at $0$ iff $\sum\_{j\in J\_n}\ln f\_{X\_{j,n}}$ converges pointwise to a function continuous at $0$ iff $\sum\_{j\in J\_n}(f\_{X\_{j,n}}-1)$ converges pointwise to a function continuous at $0$ iff $T\_n$ converges in distribution. So, we have \eqref{20}. $\quad\Box$

1

https://mathoverflow.net/users/36721

452788

https://mathoverflow.net/questions/452692

1

(Asking again in a new question because the [previous version](https://mathoverflow.net/q/452519/36721) had insufficient conditions, as pointed out in the [answer there](https://mathoverflow.net/a/452659/36721).) Define the densities: $$p(\phi;\theta,r) = \Big(f\big(r\cos(\phi-\theta)\big) - f\big(r\cos(\phi+\theta)\big)\Big)\hspace{0.5pt} \frac{\sin(2\phi)}{\sin(2\theta)}, \quad 0 \le \phi,\theta\le \pi/2,\,r>0$$ where $f(x)=g(x^2)$, and the following hold: $g$ is twice-differentiable, increasing, and convex or concave on $(0,\infty)$, $g''(x)$ is non-decreasing on $(0,\infty)$, and $f'(0^+) < \infty$. Remarkably, the area of these densities is independent of $\theta$ whenever $g$ is increasing and concave or convex (without requiring the monotonicity of $g''(x)$), which can be shown using an integral representation. For $f(x)=|x|$, we have $\int\_0^{\pi/2} p(\phi;\theta,r)d\phi=\frac{2}{3}r$. Show that for all $r$, $p(\phi;\theta,r)$ has a monotone likelihood ratio (decreasing for concave $g$, increasing for convex $g$). I.e., for $0\le\theta\_1 < \theta\_2\le\pi/2$: $$ h(\phi) = \frac{f\big(r\cos(\phi-\theta\_2)\big) - f\big(r\cos(\phi+\theta\_2)\big)}{f\big(r\cos(\phi-\theta\_1)\big) - f\big(r\cos(\phi+\theta\_1)\big)}$$ is monotonic on $[0,\pi/2]$. Examples of functions are $f(x) = |x|^p$, $1\le p<2$, or for $p>2$. (For $p=2$, $p(\phi;\theta) = \sin^2(2\phi)$.) And the function $f(x) = \log( \cosh(x))$, which is concave in $x^2$, i.e. $g(x) = \log(\cosh(\sqrt{x}))$ is concave on $[0,\infty)$, and twice differentiable at $x=0$, unlike $|x|^p$, $p<2$. An answer showing the this holds for either convex or concave $g$ is acceptable. This question can be considered as only asking for one of them. This result is important to prove uniqueness of stable optima in the unmixing and deconvolution of linear mixtures of independent random variables with strongly sub- and super-gaussian densities, using the Karlin-Rubin theorem. The result can be proved for $f(x)=x^4$ by simplifying the derivative expression. This corresponds to using kurtosis as the the cost function, and the uniqueness result is already known in this case. For $f(x) = |x|$, the likelihood ratio is non-increasing, constant around $\phi=0$ and $\phi=\pi/2$.

https://mathoverflow.net/users/510206

Monotone likelihood ratio of a family of densities with convexity property

$\newcommand{\ep}{\varepsilon}$The "convex" part of this conjecture is not true in general. Indeed, suppose it is true. Then (letting $x:=\phi$, $t:=\theta\_1$, and $\theta\_2\downarrow\theta\_1=t$) we see that for any strictly increasing convex smooth function $g$ with $g'''\ge0$ and all $x$ and $t$ in $(0,\pi/2)$ we would have $h\_2(g;x,t):=\partial\_x\partial\_t\,\ln(g(\cos^2(x-t))-g(\cos^2(x+t))\ge0$. (Note that for all $x$ and $t$ in $(0,\pi/2)$ we have $\cos^2(x-t)-\cos^2(x+t)=\sin2x\,\sin2t>0$, so that $h\_2(g;x,t)$ is well defined.) For $c$ and $c\_\*$ in $[0,\infty)$ and real $\ep>0$, let $g(c):=g\_{c\_\*,\ep}(c):=(\sqrt{(c-c\_\*)^2+\ep^2}+c-c\_\*)^2$. Then the function $g$ is strictly increasing, convex, and smooth on $\mathbb R$, and $g'''>0$. However, $h\_2(g;x,t)=-461586.955\ldots\not\ge0$ if $c\_\*=\frac{526}{1000}$, $\ep=\frac1{1000}$, $x=\frac{812}{1000}$, and $t=\frac{157}{100}$. So, the "convex" part of your conjecture is not true in general. $\quad\Box$

1

https://mathoverflow.net/users/36721

452792

https://mathoverflow.net/questions/452794

0

(Asking a final time in a new question because the [previous version](https://mathoverflow.net/q/452519/36721) had insufficient conditions, as pointed out in the [answer there](https://mathoverflow.net/a/452659/36721).) Define the densities: $$p(\phi;\theta,r) = \Big(f\big(r\cos(\phi-\theta)\big) - f\big(r\cos(\phi+\theta)\big)\Big)\hspace{0.5pt} \frac{\sin(2\phi)}{\sin(2\theta)}, \quad 0 \le \phi,\theta\le \pi/2,\,r>0$$ where $f(x)=g(x^2)$, with $g'(x)>0$, $g''(x)>0$, $g'''(x)> 0$, and $g'''(x)$ monotonic on $(0,\infty)$. Remarkably, the area of these densities is independent of $\theta$ whenever $g$ is increasing and convex, which can be shown using an integral representation. Show that for all $r$, $p(\phi;\theta,r)$ has a monotone likelihood ratio. I.e., for $0\le\theta\_1 < \theta\_2\le\pi/2$: $$ h(\phi) = \frac{f\big(r\cos(\phi-\theta\_2)\big) - f\big(r\cos(\phi+\theta\_2)\big)}{f\big(r\cos(\phi-\theta\_1)\big) - f\big(r\cos(\phi+\theta\_1)\big)}$$ is monotonic on $[0,\pi/2]$. Examples of functions are $f(x) = |x|^p$ for $p>2$ and $f(x)=\cosh(x)$. This result is important to prove uniqueness of stable optima in the unmixing and deconvolution of linear mixtures of independent random variables with strongly sub- and super-gaussian densities, using the Karlin-Rubin theorem. The result can be proved for $f(x)=x^4$ by simplifying the derivative expression. This corresponds to using kurtosis as the the cost function, and the uniqueness result is already known in this case. For $f(x) = |x|$, the likelihood ratio is non-increasing, constant around $\phi=0$ and $\phi=\pi/2$.

https://mathoverflow.net/users/510206

Sign Regularity of a Density Kernel with Convexity Properties

$\newcommand{\ep}{\varepsilon}$This conjecture is not true in general. Indeed, suppose it is true. Then (letting $x:=\phi$, $t:=\theta\_1$, and $\theta\_2\downarrow\theta\_1=t$) we see that for any strictly increasing convex smooth function $g$ with $g'''>0$ and $g''''>0$ and all $x$ and $t$ in $(0,\pi/2)$ we would have $h\_2(g;x,t):=\partial\_x\partial\_t\,\ln(g(\cos^2(x-t))-g(\cos^2(x+t))\ge0$. (Note that for all $x$ and $t$ in $(0,\pi/2)$ we have $\cos^2(x-t)-\cos^2(x+t)=\sin2x\,\sin2t>0$, so that $h\_2(g;x,t)$ is well defined.) For $c$ and $c\_\*$ in $[0,\infty)$ and real $\ep>0$, let $g(c):=g\_{c\_\*,\ep}(c):=(\sqrt{(c-c\_\*)^2+\ep^2}+c-c\_\*)^3$. Then the function $g$ is strictly increasing, convex, and smooth on $\mathbb R$, and $g'''>0$ and $g''''>0$. However, $h\_2(g;x,t)=-132194.575\ldots\not\ge0$ if $c\_\*=\frac{5}{10}$, $\ep=\frac1{1000}$, $x=\frac{118}{100}$, and $t=\frac{39}{100}$. So, your conjecture is not true in general. $\quad\Box$

1

https://mathoverflow.net/users/36721

452800

https://mathoverflow.net/questions/452803

1

I'm looking for references for two facts that are stated without proof in the paper: > > *Talagrand, M.*, Are all sets of > positive measure essentially convex?, Lindenstrauss, J. (ed.) et al., > Geometric aspects of functional analysis. Israel seminar (GAFA) > 1992-94. Basel: Birkhäuser. Oper. Theory, Adv. Appl. 77, 294-310 > (1995). [ZBL0835.60003](https://zbmath.org/?q=an:0835.60003). > > > Let $\gamma\_m(ds) := (2\pi)^{-m/2}e^{-||s||\_2^2/2}ds$ be the standard Gaussian measure on $\mathbb{R}^m$. Given a convex set $C$, define the guage $||x||\_C := \inf \{ \lambda : x \in \lambda C \}$. A convex set is balanced if $s \in \mathbb{C}$ implies $\lambda s \in \mathbb{C}$ for every $\lambda \in [-1,1]$. **Lemma 1.** For each $\varepsilon \in (0,1)$, there is a universal $R(\varepsilon)$ with the following property. For any $m$ and any balanced convex $C \subset \mathbb{R}^m$ satisfying $\gamma\_m(C) > \varepsilon$, we have $\int\_{\mathbb{R}^m}||s||\_C \gamma\_m(ds) \leq R(\varepsilon)$. We say that an $\mathbb{R}^m$-valued random variable $Y = (Y\_1,\ldots,Y\_m)$ is subgaussian if $\mathbb{E}[e^{ \sum\_{i=1}^m a\_i Y\_i } ] \leq e^{ \frac{1}{2} \sum\_{i=1}^m a\_i^2} $ for all $a\_1,\ldots,a\_m \in \mathbb{R}$. We have the following generalisation: **Lemma 2.** For each $\varepsilon \in (0,1)$, there is a universal $S(\varepsilon)$ with the following property. For any $m$ and any balanced convex set $C$ satisfying $\gamma\_m(C) > \varepsilon$, and any $Y$ subgaussian, we have $\mathbb{E}[||Y||\_C] \leq S(\varepsilon)$. Talagrand implies that the latter lemma follows from results in another paper of his: > > *Talagrand, Michel*, [**Regularity > of Gaussian processes**](https://doi.org/10.1007/BF02392556), Acta > Math. 159, No. 1-2, 99-149 (1987). > [ZBL0712.60044](https://zbmath.org/?q=an:0712.60044). > > > However, I wasn't able to find a statement like Lemma 2 in the latter. I'd be very grateful if someone could point me to a reference, or even a proof!

https://mathoverflow.net/users/380543

Reference request: Inequalities involving convex sets and Gaussian variables stated in a paper by Talagrand

$\newcommand{\R}{\mathbb R}\renewcommand{\th}{\theta}\newcommand{\ep}{\varepsilon} $This indeed follows immediately from [Theorem 15](https://doi.org/10.1007/BF02392556) of Talagrand, Regularity of Gaussian processes, Acta Math. 159, No. 1-2, 99-149 (1987). --- Details: The mentioned theorem by Talagrand states the following: > > Let $(X\_t)\_{t\in T}$ be a Gaussian process, and $(Y\_t)\_{t\in T}$ be any other centered > process indexed by the same set. Assume that for each $\th\in\R$, we have > \begin{equation} > E \exp \th(Y\_u-Y\_v)\le E \exp \th(X\_u-X\_v) = \exp(\th^2 d(u, v)^2/2). \quad (40) > \end{equation} > Then we have $E\sup\_{t\in T}Y\_t\le K E\sup\_{t\in T}X\_t$ where $K$ is a universal constant. > > > Let now $T$ be a set of vectors $t=(t\_1,\dots,t\_m)\in\R^m$ such that $\|x\|\_C=\sup\_{t\in T}t\cdot x$ for all $x\in\R^m$, where $\cdot$ denotes the dot product. For $t=(t\_1,\dots,t\_m)\in\R^m$, let $Y\_t:=t\cdot Y$ and $X\_t:=t\cdot X$, where $X$ is a standard normal random vector in $\R^m$. Then (40) will hold with $d(u,v)^2=(u-v)\cdot(u-v)$. So, we will have $E\|Y\|\_C\le K E\|X\|\_C$. So, one may take $S(\ep):=KR(\ep)$. $\quad\Box$

2

https://mathoverflow.net/users/36721

452804

https://mathoverflow.net/questions/452789

7

Let $A\_n$ be the coinvariant algebra of the symmetric group $S\_n$. This algebra has vector space dimension $n!$. $A\_n$ is the quotient algebra of the polynomial ring $K[x\_1,...,x\_n]$ by the elementary symmetric polynomials $e\_i$. Since $e\_1=x\_1+....+x\_n$, one can "eliminate" the variable $x\_n$ to obtain an algebra with admissible relations. > > Question: Is there a canonical monomial basis $b\_p$ (in the literature) for $A\_n$ parametrised by the elements of the symmetric group $p \in S\_n$ such that the variable $x\_n$ does not appear? > > > There is the Artin basis, see theorem 1.7 in <https://www.jpswanson.org/talks/2017_coinvariant_zoo.pdf> ,which can do this up to a choice of a bijection to the symmetric group elements and I wonder what bijection to the symmetric group elements one should use (probably there is a convention or a canonical way for this already). One idea might be to "set" $r\_i:=(i,i+1,...,n)$, then the symmetric group should be having also the elements $r\_i$ as a basis meaning all elements of the symmetric group have the unique form $\Pi r\_i^{a\_i}$ as in the Artin basis. Is this the canonical way that should be used?

https://mathoverflow.net/users/61949

Basis parametrized by the symmetric group elements for the coinvariant algebra

So the Artin basis already does this. There is a natural bijection of the symmetric group $S\_n$ with sequences $(a\_1,\ldots,a\_{n-1})$ of integers $a\_i$ with $0\leq a\_i\leq n-i$. Namely, let $w:[n]\to [n]$ be a permutation. Then define $$a\_i(w)=\#\{j>i\mid w(i)>w(j)\}$$ This is known as the Lehmer code. It's a little bit more work to get back to the permutation from the Lehmer code, but it's still not very hard. For an algorithm, start with a $0$-indexed sequence $A^0=(1,2,\ldots,n)$. To construct $w$ given $(a\_1,\ldots,a\_{n-1})$, at iteration $i$ set $w(i)=A^{i-1}\_{a\_i}$ and ``pop'' the element at index $a\_i$ to obtain a new $0$-indexed sequence $A^i$. For the last index $w(n)$, the value will be the only remaining element in $A^{n-1}$.

8

https://mathoverflow.net/users/62135

452809

https://mathoverflow.net/questions/452811

11

I am wondering if the category of small categories $\mathbf{Cat}$ is known to (not) have the Cantor–Schröder–Bernstein property? That is, for any two categories $\mathcal{C}$ and $\mathcal{D}$, does the statement that there exist embedding functors (monomorphisms in $\mathbf{Cat}$) $\mathcal{F}: \mathcal{C} \rightarrow \mathcal{D}$ and $\mathcal{G}: \mathcal{D} \rightarrow \mathcal{C}$ imply that $\mathcal{C} \cong \mathcal{D}$? If not, what can we say about the validity of the following weaker statement? For any two categories $\mathcal{C}$ and $\mathcal{D}$, the statement that there exist embedding functors $\mathcal{F}\_1: \mathcal{C} \rightarrow \mathcal{D}$ and $\mathcal{F}\_2: \mathcal{D} \rightarrow \mathcal{C}$ implies that there exists bimorphisms (epic embedding functors) in $\mathbf{Cat}$ $\mathcal{G}\_1: \mathcal{C} \rightarrow \mathcal{D}$ and $\mathcal{G}\_2: \mathcal{D} \rightarrow \mathcal{C}$. Thank you in advance for your help!

https://mathoverflow.net/users/510524

Does $\mathbf{Cat}$ have the Cantor–Schröder–Bernstein property?

One can take some of the standard violations of CSB with other kinds of mathematical structures and transfer them to categories. For example, with linear orders, we have the two linear orders $$\langle\mathbb{Q},\leq\rangle\qquad \langle\mathbb{Q}^{\geq 0},\leq\rangle,$$ which each order-embed into each other, but they are not isomorphic, since the latter has a minimal element and the former does not. But any linear order can be viewed as a category, where one takes the nodes of the order as objects, and whenever $x\leq y$ there is a unique morphism from $x$ to $y$, which is the identity morphism when $x=y$. The order embeddings give rise to embedding functors in each direction for the categories, but these two categories are not isomorphic, since the latter one has an initial object, but the former does not. This example answers directly the first part of the question. But actually it answers also the second part of the question, as explained in [Peter's comment](https://mathoverflow.net/questions/452811/does-mathbfcat-have-the-cantor-schroder-bernstein-property#comment1171201_452814).

21

https://mathoverflow.net/users/1946

452814

https://mathoverflow.net/questions/452580

2

I am reading "The Uncertainty Principle" by Fefferman (Bull. AMS, 1983) and have some issues following the arguments. In Lemma $C$ we have the following setting: > > Let $Q^0\subseteq \mathbb{R}^n$ be some dyadic cube, $V: \mathbb{R}^n \rightarrow (-\infty, 0]$ some function, $p>1$ and > $$ V^+\_Q(x):= \sup\_{\_{Q' \text{ cube } \ : \ x\in Q' \subseteq Q}} \left( \frac{1}{\vert Q' \vert} \int\_{Q'} \vert V(y) \vert^p dy \right)^{1/p}. $$ > and we assume that there exists a constant $\gamma>0$ such that for every subcube $Q\subseteq Q^0$ we have > $$ \left( \frac{1}{\vert Q\vert} \int\_Q \vert V\vert^p \right)^{1/p}\leq \gamma \text{diam}(Q)^{-2}.$$ > > > In the proof it is then claimed that there exists a constant $C\_p$ depending only on $p$ such that $$ \frac{1}{\vert Q \vert} \int\_Q V\_Q^+ \leq C\_p \left( \frac{1}{\vert Q \vert} \int\_Q \vert V \vert^p \right)^{1/p}. $$ I'd appreciate any pointers on how to prove this (or a reference where to find such a proof). This looks a bit like the boundedness of the maximal operator (restricted to our cube), but the powers look completely off. **Added:** This claim does not appear in the statement of Lemma $C$, but in its proof. Namely it says "the maximal theorem and the hypothesis imply $$\text{Av}\_Q V\_Q^+\leq C\_p (\text{Av}\_Q \vert V\vert^p)^{1/p} \leq C\_p \gamma (\text{diam}(Q)^{-2})."$$ Here $\text{Av}\_Q$ denotes the average over $Q$. It is the first inequality that is unclear to me.

https://mathoverflow.net/users/91098

Maximal function on small cubes

By Holder’s inequality $$ \frac{1}{\lvert Q\rvert}\int\_{Q}V^{+}dm \leq \frac{1}{\lvert Q\rvert} \lvert Q\rvert^{1/q}\cdot \left(\int\_{Q} \left(V^{+}\right)^{p}dm \right)^{1/p} = \left(\lvert Q\rvert^{-1}\int\_{Q} \left(V^{+}\right)^{p}dm \right)^{1/p}.$$ The RHS term is $\lVert V^{+}\rVert\_{p}$ (up to the $Q$-normalization). Applying the strong maximal inequality of Hardy-Littlewood gives $$ \left(\lvert Q\rvert^{-1}\int\_{Q} \left(V^{+}\right)^{p}dm \right)^{1/p} \leq C\_{p} Q^{-1/p}\left\lVert V\right\rVert\_{p},$$ where the integrals all happen inside $Q$, so the inequality follows as $Q^{-1/p}\lVert V\rVert\_{p} = \lvert Av\_{Q}(V)^{p}\rvert^{1/p}$.

2

https://mathoverflow.net/users/8857

452819

https://mathoverflow.net/questions/452805

0

Let $T>0$ and $p \in [1, \infty)$. Let $f \in L^p ([0, T] \times {\mathbb R}^d)$. By a theorem in [this](https://math.stackexchange.com/questions/4301414/if-f-n-is-cauchy-and-converges-a-e-to-f-there-is-a-cauchy-subsequence) thread, there is a Lebesgue null subset $N$ of $[0, T]$ such that $f(t, \cdot)$ is Lebesgue measurable and that $f(t, \cdot) \in L^p ({\mathbb R}^d)$ for all $t \in [0, T] \setminus N$. For $t \in N$, we don't know the measurability and thus the integrability of $f(t, \cdot)$. However, we can re-define $f$ such that $f(t, \cdot) \equiv 0$ for all $t \in N$. Then writing $f(t, \cdot) \in L^p ({\mathbb R}^d)$ is rigorously valid for *every* $t \in [0, T]$. > > Is there a modification of $f$ on a Lebesgue null subset of $[0, T]$ such that the map > $$ > F: [0, T] \to L^p ({\mathbb R}^d), t \mapsto f(t, \cdot) > $$ > is [Bochner measurable](https://www.wikiwand.com/en/Bochner_measurable_function)? > > > Intuitively, the answer is positive. But I don't know how to prove it...

https://mathoverflow.net/users/99469

Is there a modification of $f$ on a null set such that $F: [0, T] \to L^p ({\mathbb R}^d), t \mapsto f(t,\cdot)$ is Bochner measurable?

Let * $(X, \mathcal A, \mu)$ and $(Y, \mathcal B, \nu)$ be complete $\sigma$-finite measure spaces, and $(E, | \cdot |)$ a Banach space. * $S (X)$ the space of $\mu$-simple functions from $X$ to $E$. * $\mathcal C :=\mathcal A \otimes \mathcal B$ the product $\sigma$-algebra of $\mathcal A$ and $\mathcal B$. * $\lambda := \mu \otimes \nu$ the product measure of $\mu$ and $\nu$. Here we use Bochner integral. Then we have > > Let $p \in [1, \infty)$. > > > * [Lemma 1](https://math.stackexchange.com/questions/4737770/tensor-product-of-lp-spaces-is-dense-in-the-product-lp-space) Then > $$ > \overline{S(X) \otimes S(Y)} = L^p(X \times Y), > $$ > where the closure is with respect to the norm in $L^p(X \times Y)$. Here $S (X) \otimes S(Y)$ is identified with a subspace of $L^p(X \times Y)$ through the natural embedding > $$ > f \otimes g \mapsto \big( (x,y) \mapsto f(x) g(y) \big). > $$ > * [Lemma 2](https://math.stackexchange.com/questions/4301414/if-f-n-is-cauchy-and-converges-a-e-to-f-there-is-a-cauchy-subsequence) Let $(f\_n) \subset S (X \times Y)$ be a Cauchy sequence in $L^p(X \times Y)$ that converges $\lambda$-a.e. to $f$. Then there is a subsequence $\varphi$ of $\mathbb N$ such that for $\mu$-a.e. $x \in X$, we have $(f\_{\varphi (n)} (x, \cdot))\_n \subset S (Y)$ is a Cauchy sequence in $L^p (Y)$ and converges to $f(x, \cdot)$ both $\nu$-a.e. and in $L^p (Y)$. > > > Let $X := [0, T]$ and $Y := \mathbb R^d$. Let $\mathcal A, \mathcal B$ be the corresponding Lebesgue $\sigma$-algebras of $X$ and $Y$. Let $\mu, \nu$ be the corresponding Lebesgue measures on $X, Y$. By **Lemma 1**, there is a sequence of $(f\_n) \subset L^p(X \times Y)$ with $f\_n = g\_n h\_n$ for some $g\_n \in S (X)$ and $h\_n \in S (Y)$ such that $\|f\_n-f\|\_{L^p(X \times Y)} \to 0$. Clearly, $(f\_n)$ is a Cauchy sequence in $S(X \times Y)$. Also, each $f\_n$ is of the form $$ f\_n (t, x) = \sum\_{k=1}^{\varphi\_n} e\_{n, k} 1\_{A\_{n, k}} (t) 1\_{B\_{n, k}} (x) \quad \forall t\in X, x \in Y, $$ for some $e\_{n, k} \in \mathbb R$ and $A\_{n, k} \in \mathcal A, B\_{n, k} \in \mathcal B$ such that $\mu(A\_{n, k}) + \nu (B\_{n, k}) < \infty$. Then $f\_n (t, \cdot) \in L^p(Y)$ for all $t \in X$. Let $$ F\_n : X \to L^p (Y), t \mapsto f\_n (t, \cdot) \quad \forall n \in \mathbb N. $$ Clearly, $F\_n \in S(X, L^p(Y)) \subset L^p(X, L^p(Y))$ for all $n \in \mathbb N$. By Fubini's theorem, $$ \begin{align\*} \| F\_n - F\_m\|\_{L^p(X, L^p(Y))}^p &= \int\_0^T \mathrm d t \, \|F\_n (t) - F\_m (t)\|\_{L^p (Y)}^p \\ &= \|f\_n-f\_m\|\_{L^p(X \times Y)}^p. \end{align\*} $$ Then $(F\_n)$ is a Cauchy sequence in the Banach space $L^p(X, L^p(Y))$. Then there is $F \in L^p(X, L^p(Y))$ such that $\|F\_n-F\|\_{L^p(X, L^p(Y))} \to 0$. Convergence in $L^p$ implies a.e. convergence of a subsequence, so WLOG we assume $f\_n \to f$ $\lambda$ a.e. and $F\_n \to F$ $\mu$-a.e. By **Lemma 2**, there is a subsequence $\varphi$ of $\mathbb N$ su for $\mu$-a.e. $t \in X$ we have $$ \|f\_{\varphi (n)} (t, \cdot)-f(t, \cdot) \|\_{L^p (Y)} \xrightarrow{n \to \infty} 0. $$ On the other hand, for $\mu$-a.e. $t \in X$ we have $$ \|f\_n (t, \cdot) - F (t)\|\_{L^p(Y)} =\|F\_n (t) - F (t)\|\_{L^p(Y)} \xrightarrow{n \to \infty} 0. $$ It follows that for $\mu$-a.e. $t \in X$ we have $$ \|f (t, \cdot) - F(t) \|\_{L^p (Y)} = 0. $$ The claim then follows.

0

https://mathoverflow.net/users/99469

452821

https://mathoverflow.net/questions/452816

1

Let $X$ be an extremally disconnected compact Hausdorff space *with no open points*, and $f:X\to\mathbb{C}$ be a non-constant continuous function. Let $D\_f$ be the linear span of the functions of the form $$u\to \gamma(u) f(\sigma(u))$$ where $\sigma:X\to X$ is a homeomorphism, and $\gamma:X\to\mathbb{C}$ is continuous with $|\gamma|=1$. > > **Question:** What is an example of $X$ and $f$ so that $D\_f$ isn't dense in $C(X)$? > > >

https://mathoverflow.net/users/164350

Subspaces generated by the orbits of the group of isometries on $C(K)$

Take two extremally disconnected spaces $Y$ and $Z$ without open points and with different cardinality (say $Z$ is the one with larger cardinality). Define $X$ as the disjoint union of $Y$ and $Z$ and let $f$ be the indicator function of $Y$. For every sequence $(g\_n)$ in $D\_f$ the union of the supports $\{g\_n \not= 0\}$ is, for cardinality reasons, a proper subset of $X$. So $(g\_n)$ cannot approximate the constant $1$-function on $X$.

5

https://mathoverflow.net/users/102946

452822

https://mathoverflow.net/questions/452843

0

The real hyperbolic fixed points of $\mathrm{SL}\_2(\mathbb{Z})$ are the points $x\in\mathbb R\smallsetminus\mathbb{Q}$ with $$ \frac{ax+b}{cx+d}=x $$ for some $\left(\begin{array}{2} a&b\\ c&d\end{array}\right)\in\mathrm{SL}\_2(\mathbb{Z})$. These are special quadratic numbers and I wonder, what their properties are. Do they have a name? Arithmetic meaning?

https://mathoverflow.net/users/473423

Hyperbolic fixed points of SL(2,Z)

Am I missing anything? Any fixed point $x$ is a root of quadratic equation $px^2+qx+r=0$ with $p,q,r\in\mathbb Z$, where $D=q^2-4pr$ is not a square; we assume that $(p,q,r)=1$. This $x$ must also satisfy $cx^2+(d-a)x-b=0$, which yields $c=\nu p$, $b=-\nu r$, $d=\nu q+a$ for an integer $\nu$. So we need $$ 1=\left|\matrix{a& -\nu r\\\nu p& \nu q+a}\right|=a^2+\nu q a+\nu^2pr $$ to be solvable in integer $a$. This is so iff $(\nu q)^2-4(\nu^2pr-1)=\nu^2D+4$ is a square. But, since $D$ is not a square, a Pell's equation $n^2-D\nu^2=4$ has a non-trivial solution $(n,\nu)$. Starting at that, we get $$ a=\frac{n-\nu q}2, \quad b=-\nu r, \quad c=\nu p, \quad d=\nu q+a $$ which form a desired element in $\mathrm{SL}\_2(\mathbb Z)$. So *every* quadratic irrational (in $\mathbb R$ or in $\mathbb C$, depending of what you are transforming) is a fixed point, and this is a complete description.

8

https://mathoverflow.net/users/17581

452845

https://mathoverflow.net/questions/451815

1

Let $X\_t, Y\_t \in C^\infty(\mathbb{R}; \mathfrak{X}^\infty(M))$ be (smooth, or something else if it's necessary) time dependent vector fields. Is there some analogue of the following fact in finite dimensions (with no time dependence, identifying Lie group elements with exponentials of algebra elements could also be false when you involve time): $[A,B] = 0 \implies \operatorname{Ad}\_{e^A}B = e^A B e^{-A} = B$ Where now I want to say something along the lines of $(\phi^X\_t)\_\* Y\_t = Y\_t$ I know infinite dimensional Lie groups are very easy to get misled by intuition from the finite case, so I'm looking for a reference that takes care of the details.

https://mathoverflow.net/users/125989

Commuting time dependent vector fields and pullback invariance

Let $\phi^t$, $\psi^t$ be the respective flows of two vector fields $A$, $B$ at time $t$ on a manifold $M$ (assume to fix ideas that these flows do exist for every time). If $A, B$ do ***not*** depend on time, then of course one has an equivalence between the 4 properties i) $[A,B]=0$; ii) $A$ is invariant by every $\psi^t$; iii) $B$ is invariant by every $\phi^t$; iv) $\phi^s\circ\psi^t=\psi^t\circ\phi^s$ for every two times $s$, $t$. (see e.g. Spivak, "A Comprehensive Introduction To Differential Geometry" volume 1). For time-dependant vector fields, nothing of the like can hold. For example, on $M=R$ the real line, consider $A(t)=f(t)d/dx$ and $B(t)=g(t)xd/dx$ where $f:R\to R$ and $g:R\to R$ are two smooth functions such that $f(t)=0$ except for $t\in(0,1/2)$ and $g(t)=0$ except for $t\in(1/2,1)$ and $\int\_Rf=\int\_Rg=1$. Certainly, at every time $t$ one has $[A(t),B(t)]=0$ since at least one of the two vector fields is identically $0$ at this time! But their respective flows at time $1$ are $\phi^1:x\mapsto x+1$ and $\psi^1:x\mapsto ex$, which don't commute. Also, if say $g(3/4)\neq 0$, then clearly the translation $\phi^{3/4}=\phi^1$ does not preserve the linear vector field $B(3/4)$.

2

https://mathoverflow.net/users/105095

452864

https://mathoverflow.net/questions/452865

3

Is this true that finitely generated flat module over an integral domain is projective. If Yes, please provide a proof.

https://mathoverflow.net/users/510594

On Flat and Projective Modules over integral domain

Yes, this is true. Lemma 5 on page 249 of Cartier, "Questions de rationalité des diviseurs en géométrie algébrique" says that over an integral domain, whether or not it is Noetherian, if the localisation of a finitely generated module at every maximal ideal is free, then the module is projective. Over a local ring, every finitely generated flat module is free (see for example Matsumura, "Commutative Ring Theory" Theorem 7.10). Putting these together gives you a positive answer to your question.

5

https://mathoverflow.net/users/460592

452867

https://mathoverflow.net/questions/452866

2

$\newcommand{\std}{\mathrm{std}}\newcommand{\SL}{\mathrm{SL}}\newcommand{\mmod}{/\!\!/}$Fix the base field to be the complex numbers $\mathbf{C}$. Let $\std = \mathbf{A}^2$ denote the standard representation of $\SL\_2$, so that there is a natural action of $\SL\_2^{\times 3}$ on $\std^{\otimes 3}$. Let $Y$ denote the variety of pairs $(q\_1(x,y), q\_2(x,y))$ of binary quadratic forms with the same discriminant, so that it admits a natural action of $\SL\_2^{\times 2}$ (via the natural action of $\SL\_2$ on the space of binary quadratic forms). Some considerations from homotopy theory/relative Langlands suggest that there is a relationship between the stacks $\std^{\otimes 3}/\SL\_2^{\times 3}$ and $Y/\SL\_2^{\times 2}$. My basic question is whether there is some natural construction which relates these two stacks. For instance, are they derived equivalent? Do they even have the same coarse moduli spaces? It is not too hard to show that the GIT quotient $Y\mmod\SL\_2^{\times 2} \cong \mathbf{A}^1$ via the discriminant; is it also true that the GIT quotient $\std^{\otimes 3}\mmod\SL\_2^{\times 3} \cong \mathbf{A}^1$? If so, what is the resulting map $\std^{\otimes 3} \to \mathbf{A}^1$? Or, is there some natural construction from the theory of binary quadratic forms which takes as input an $\SL\_2^{\times 2}$-orbit of pairs $(q\_1(x,y), q\_2(x,y))$ with the same discriminant and produces an $\SL\_2^{\times 3}$-orbit of $\std^{\otimes 3}$? Apologies for the somewhat vague question, and thanks in advance!

https://mathoverflow.net/users/102390

Pairs of quadratic forms and $\mathbf{A}^8/\mathrm{SL}_2^{\times 3}$

$\newcommand{\std}{\mathrm{std}}\newcommand{\SL}{\mathrm{SL}}\newcommand{\mmod}{/\!\!/}$ A lowbrow answer to your question is given by Bhargava's theory of $2 \times 2 \times 2$ cubes explained in his [Higher Composition Laws Paper](https://annals.math.princeton.edu/wp-content/uploads/annals-v159-n1-p03.pdf). In particular, you're correct that $\std^{\otimes 3}\mmod\SL\_2^{\times 3} \cong \mathbf{A}^1$, with the invariant being the discriminant listed at the bottom of page 220 in the paper (page 4 in the pdf). For a higher-brow approach to this general topic you may be interested in Melanie Wood's paper [Gauss composition over an arbitrary base](https://arxiv.org/abs/1007.5285), though I don't think it talks about cubes directly.

6

https://mathoverflow.net/users/422

452870

https://mathoverflow.net/questions/452443

2

Let $F$ be an order unit Banach space with order unit $e$ and topological dual space $F^\*$ ordered by the dual cone. Let $E\subset F^\*$ be a closed subspace that separates points of $F$ and such that the dual cone intersected with $E$ is spanning for $E$. Consider the initial topology $\sigma(E,F)$ induced on $E$ by $F$. Given a net $\omega\_\lambda\in E$ converging to $0$ in $\sigma(E,F)$, do there exist nets $0\le \psi\_\lambda, \varphi\_\lambda$ of positive elements converging to $0$ in $\sigma(E,F)$ such that $\omega\_\lambda = \psi\_\lambda - \varphi\_\lambda$? If not in this generality, are there further conditions on $E$ under which one can say more, e.g. $E$ being a Banach lattice/AL space? For example if $E$ is base normed by the base $B = \{\omega\in E\ |\ \omega\ge 0, \omega(e) = 1\}$, the statement holds true if $\sigma(E,F)$ is replaced by the norm topology. That is, every net converging to $0$ in norm can be decomposed into positive nets converging to $0$ in norm. Is the weak version also true in this setting? More concretely, I suspect the weak version to be true at least for $F = \ell^\infty(\mathbb N, \mathbb R)$ and $E=\ell^1(\mathbb N, \mathbb R)$ with the usual order and decomposition into positive and negative part of a sequence. However, even in this case I don't have a proof nor a counterexample so far and would appreciate either. By **order unit Banach space** I mean a real Banach space that is ordered by a closed and pointed cone (that is the cone does not contain any non-trivial subspace). Furthermore there shall exist an element $e$ with the property that for every $x\in F$ there is $\lambda \ge 0$ with $\lambda e\ge x\ge -\lambda e$ and such that $\|x\|$ is the infimum over all such $\lambda$. The **dual cone** consists of the positive linear functionals on $F$. By **base normed by the base $B$** I mean that every positive element $\omega\ge 0$ of $E$ is contained in exactly one set of the form $\lambda B$ where $\lambda\ge0$ and with the property that $$\forall\_{\omega\in E}: \|\omega\| = \inf\_{\substack{\psi,\varphi\ge 0\\\omega = \psi - \varphi}} \psi(e) + \varphi(e).$$

https://mathoverflow.net/users/493899

Decomposition of weak* convergent nets into positive weak* convergent nets

Based on the comment by @BillJohnson: If $0\le \omega\in E$ then $\|\omega\| = \omega(e)$ implying that a net of positive elements in $E$ converges to $0$ in weak$^\*$ topology if and only if it converges to $0$ in norm. Thus, a positive answer to the question would imply that a net (not necessarily positive) converges to $0$ in norm if and only if it converges weak$^\*$ to $0$, which is clearly false. Explicitly, if $E$ is infinite-dimensional then the unit sphere of $E$ is weak$^\*$ dense in the unit ball of $E$. Hence, there exists a net $\omega\_\lambda$ with $\|\omega\_\lambda\| = 1$ but converging to $0$ weak$^\*$, so the question has a positive answer if and only if $E$ is finite-dimensional.

2

https://mathoverflow.net/users/493899

452874

https://mathoverflow.net/questions/452783

1

Oguiso writes[1] > > **Theorem 1.1** *Let $f: X \to \mathbf P^n$ be an abelian fibered HK [hyperkähler] manifold. Let $K = \mathbf C(\mathbf P^n)$ and let $A\_k$ be the generic fiber of $f$. Then, $\rho(A\_K)= 1$. Here $\rho(A\_K)$ is the Picard number of $A\_K$ over $K$.* > > > It can happen that $\rho(X\_t) \geq 2$ for all smooth closed fiber of $f$ ([...]). The statement is of arithmetical nature. Geometrically, it means that two horizontal divisors on $X$ are proportional in $NS(X)$ up to vertical divisors. > > > A divisor is called horizontal if it dominates $\mathbf P^n$, otherwise it is called vertical. I'm struggling with the last sentence of Oguiso. Is that correct? Given two line bundles $L\_1, L\_2$ on $X$, the theorem tells us that they are proportional in $NS(A\_K)$, so there exist $n,m \in \mathbb Z$ such that $$ n L\_1|\_{A\_K} + m L\_2|\_{A\_K} = 0 \in NS(A\_K).$$ I guess that Oguiso then alludes to an exact sequence of the kind $$Z^0(X \setminus U) \to NS(X) \to NS(U) \to 0,$$ where $U = f^{-1}(V)$ is the preimage of an open set $V \subset \mathbf P^n$, and $Z^0(X\setminus U)$ is the free group generated by the irreducible components of $X \setminus U$. But I don't see a reason why $nL\_1 + mL\_2$ should vanish in $NS(U)$, for appropriate $U$. [As far as I know](https://math.stackexchange.com/questions/4745392/line-bundles-which-are-fiber-wise-algebraically-trivial), a line bundle being trivial in the Neron-Severi group of all fibers does not mean that it is trivial in the total Neron-Severi group. Did I miss anything here? --- [1] Keiji Oguiso, *Picard number of the generic fiber of an abelian fibered hyperkähler manifold*, 2009, [arXiv:0803.1205](https://arxiv.org/abs/0803.1205)

https://mathoverflow.net/users/111897

Are horizontal divisors on abelian fibered hyperkähler manifolds proportional in $NS(X)$ up to vertical divisors?

This is simply false, already for K3 surfaces with an elliptic fibration. The quotient of $NS(X)$ by the subgroup generated by the vertical components + the zero section is the Mordell-Weil group of the generic fiber, it can very well be nontrivial. See for instance *Elliptic surfaces* by Schütt and Shioda, Adv.Stud. Pure Math. 60.

2

https://mathoverflow.net/users/40297

452878

https://mathoverflow.net/questions/452737

0

Let $\frak{g}$ be a semisimple Lie algebra and $U(\frak{g})$ its universal enveloping algebra. The adjoint action of $\frak{g}$ on itself extends to an action of $\frak{g}$ on $U(\frak{g})$. How does this action interact with the PBW basis of $U(\frak{g})$? More precisely: are the irreducible submodules of $U(\frak{g})$ spanned by the elements of the PBW basis?

https://mathoverflow.net/users/507923

Adjoint action on the universal enveloping algebra and the PBW theorem

Since the sum of all the trivial subrepresentations of $U(\mathfrak{g})$ with the adjoint action is its centre, if you could do this then the centre must have a monomial basis but it does not. For example even if $\mathfrak{g}=\mathfrak{sl}\_2$ with basis $e,h,f$ in that order then the centre consists of polynomials in $$C=ef+fe+\frac 1 2 h^2=2ef+\frac 1 2(h^2-h)$$ which does not live in the monomial basis.

6

https://mathoverflow.net/users/345

452889

https://mathoverflow.net/questions/452887

10

Is there a Bell number $B\_n$ of the form $2^k$ for some $k>1$? If there is, are there infinitely many?

https://mathoverflow.net/users/101817

Can a Bell number be a power of 2?

No. It's easy to see that $B\_n=4$ is impossible, and $B\_n$ is never divisible by 8. This follows from the fact that $B\_n$ is periodic modulo 8 with period 24. See, e.g., W. F. Lunnon, P. A. B. Pleasants, and N. M. Stephens, [Arithmetic properties of Bell numbers to a composite modulus I](http://matwbn.icm.edu.pl/ksiazki/aa/aa35/aa3511.pdf/), Acta Arithmetica 35 (1979), 1–16, Theorem 6.4. The Bell numbers $B\_n \bmod 8$ for $n$ from 0 to 23 are 1, 1, 2, 5, 7, 4, 3, 5, 4, 3, 7, 2, 5, 5, 2, 1, 3, 4, 7, 1, 4, 7, 3, 2. So $B\_n$ is never 0 or 6 modulo 8.

25

https://mathoverflow.net/users/10744

452891

https://mathoverflow.net/questions/452855

17

On Mathematics Stack Exchange, I asked the following question: [Why are infinite-dimensional vector spaces usually equipped with additional structure?](https://math.stackexchange.com/questions/4751895/why-are-infinite-dimensional-vector-spaces-usually-equipped-with-additional-stru) Although it received one good answer, I feel that there is more to be said, and a more technical explanation would be welcome. I thus ask a modified version of my question here. Finite-dimensional vector spaces have a range of applications in pure mathematics. Although infinite-dimensional vector spaces are also widely studied, say, in functional analysis, it seems that most of the time they appear "naturally", they have additional structure, such as an inner product, norm, or a topology. My question is why this phenomenon occurs. Is there a reason for why "pure" infinite-dimensional vector spaces are not more pervasive? (I also welcome answers that challenge the premise of the question. Perhaps finite-dimensional vector spaces are also most useful in applications when they are equipped with extra structure, or perhaps there are areas of mathematics which *do* make use of "pure" vector spaces, including infinite-dimensional ones.)

https://mathoverflow.net/users/144779

Why do infinite-dimensional vector spaces usually have additional structure?

Here is a supplement to the nice [answer](https://math.stackexchange.com/a/4751952/87579) that you got at MSE. Much of the theory of infinite dimensional vector spaces is motivated by solving concrete problems in **analysis**. To solve differential equations, it is often profitable to use vector spaces of functions, and it is for this purpose that the theory of Banach spaces and other areas of functional analysis were developed. It is no surprise that in an analytic context one is concerned with questions of distance/absolute value, defining infinite sums, different notions of convergence etc. On the other hand, as noted in Mikhail's [response](https://mathoverflow.net/a/452860), the algebraic theory of infinite dimensional vector spaces is not particularly interesting on its own. For the most part, one of the following two things happens 1. There is an entirely analogous theory to the finite dimensional case (e.g. there is one isomorphism class of vector space for each cardinality of set; a linear transformation is determined uniquely by its values on a basis; a linear transformation is invertible iff the kernel and cokernel vanish etc.). 2. There is no possible analogous theory to the finite dimensional case, without introducing some notion of convergence/topology (e.g. there aren't infinite dimensional determinants that you can use to detect invertibility). This makes the algebraic theory less interesting than the analytic one. In summary, the theory of infinite dimensional vector spaces has an analytic flavor because the historical motivations and applications are analytic, and most of the *new* nontrivial theory lies in an analytic direction. **Edit:** One final comment: so-called "pure" infinite dimensional vector spaces actually do appear in mathematical practice quite frequently (at least in algebra). But there typically aren't classes or books devoted specifically to them, for the reasons mentioned above.

15

https://mathoverflow.net/users/494541

452892

https://mathoverflow.net/questions/452895

9

What is known about the (asymptotic?) behaviour of the number of equivalence classes of functions $n\to n$, where two functions are considered equivalent if they differ by a permutation of $n$, i.e., $f\sim g$ if there is a permutation $\sigma$ of $n$ so that $f=\sigma g\sigma^{-1}$? I am guessing this a known or elementary thing that I am somehow unable to google. When restricted to bijections, this is the partition function, which has been famously estimated by Hardy and Ramanujan.

https://mathoverflow.net/users/2225

How many functions are there from a set to itself, up to isomorphism?

The OEIS page linked by Sam Hopkins also refers to page 308 of Finch, Mathematical Constants (which refers back to the Meir and Moon paper). Finch writes that the generating function $$ P(x)=\sum\_1^{\infty}P\_nx^n $$ satisfies $$ 1+P(x)=\prod\_1^{\infty}(1-T(x^k)))^{-1} $$ where $$ T(x)=\sum\_1^{\infty}T\_nx^n $$ is given on page 296 as the generating function for $T\_n$, the number of nonisomorphic rooted trees of order $n$. $T(x)$ satisfies $$ T(x)=x\exp\left(\sum\_1^{\infty}{T(x^k)\over k}\right),\quad T\_{n+1}={1\over n}\sum\_1^{\infty}\left(\sum\_{d\mid k}dT\_d\right)T\_{n-k+1} $$ Finch gives the asymptotic $$ P\_n\sim\eta\_Pn^{-1/2}\alpha^n $$ where $$ \eta\_P={1\over2\pi}\left({2\pi\over\beta}\right)^{1/3}\prod\_2^{\infty}\left(1-T(\alpha^{-i})\right)^{-1}=0.4428767697\dotsc=(1.2241663491\dotsc)(4\pi^2\beta)^{-1/3} $$ Here $\alpha=2.9557652856\dotsc=(0.3383218568\dotsc)^{-1}$ is the unique positive solution of the equation $T(x^{-1})=1$, and $$ \beta={1\over\sqrt{2\pi}}\left(1+\sum\_2^{\infty}\alpha^{-k}T'(\alpha^{-k})\right)^{3/2}=0.5349496061\dotsc $$ $T'$ being the derivative of $T$.

11

https://mathoverflow.net/users/3684

452901

https://mathoverflow.net/questions/452893

6

Let $f(x)$ be a real-valued Riemann integrable function supported in $[0,1]$ with range in $[0,1]$. Let $\alpha$ be irrational. Consider the weighted Riemann sum $$S\_N:=\frac{1}{N}\sum\_{k=1}^Nf\left(\frac{k}{N}\right)e^{2\pi i \alpha k}.$$ What can be said about $\lim\_{N\to\infty} S\_N$? It seems like a kind of Cesàro average.

https://mathoverflow.net/users/499631

Twisted Riemann sums

$\newcommand\abs[1]{\lvert#1\rvert}$As hinted at by [Achim Krause](https://mathoverflow.net/questions/452893/twisted-riemann-sums#comment1171477_452893), the limit is $0$ for all such functions $f$, however the approximation procedure is somewhat subtle. First let us show it is $0$ for all step functions $f$. By linearity, it suffices to prove it in the case where $f$ is the indicator function of an interval $[a, b]$. But we have \begin{gather\*} S\_N(f) = \frac{1}{N}\sum\_{k=1}^N 1\_{[a, b]}\left (\frac{k}{N} \right ) e^{2\pi i \alpha k} \\ = \frac{1}{N} \sum\_{k = \lceil Na \rceil}^{\lfloor Nb \rfloor} e^{2\pi i \alpha k}, \end{gather\*} where $\lceil \cdot \rceil$, $\lfloor \cdot \rfloor$ denote the ceiling and floor functions respectively. This is in turn less than or equal to in magnitude to \begin{gather\*} \frac{1}{\lfloor Nb \rfloor - \lceil Na \rceil} \sum\_{k = \lceil Na \rceil}^{\lfloor Nb \rfloor} e^{2\pi i \alpha k} \\ = e^{2 \pi i \alpha \lceil Na \rceil} \left ( \frac{1}{\lfloor Nb \rfloor - \lceil Na \rceil} \sum\_{k = 0}^{\lfloor Nb \rfloor - \lceil Na \rceil} e^{2\pi i \alpha k} \right ). \end{gather\*} Since irrational rotations are uniquely ergodic, by the pointwise ergodic theorem, the term in brackets converges to $0$. Taking absolute values and noting that $ \abs{ e^{2 \pi i \alpha \lceil Na \rceil} }$ is uniformly bounded by $1$ allows us to conclude. Now let $f$ be Riemann integrable on $[0, 1]$ with the given bound on the range. Actually, the assumption on the range is unnecessary since all Riemann integrable functions are uniformly bounded, and the proof works just as well for bounded functions. However, I will assume wlog that the range lies in $[0, 1]$; it should be clear how to adapt the proof in the general bounded case. Fix $\varepsilon > 0$, and let us choose a partition $0 = a\_0 < \dotsb < a\_{n-1} = 1$ such that the following holds. Write $P\_k = [a\_k, a\_{k+1})$, and denote by $\mathcal G$ the set of intervals $P\_k$ such that $\sup\_{x, y \in P\_k} \abs{f(x) - f(y)} < \varepsilon$. We choose the partition such that $\mu(\bigcup\_{P\_k \in \mathcal G} \, P\_k) > 1 - \varepsilon$. The existence of such a partition follows directly from the definition of the Darboux sums for the Riemann integral. Define the step function $g$ by $g(x) = \sum\_{k = 0}^{n-1} \mathbb 1\_{P\_k} (x) \inf\_{y \in P\_k} f(y)$. Clearly we have $\abs{f - g} \leq \varepsilon$ on $\mathcal G$, and by the bound on $f$ we have $\abs{f - g} \leq 1$ on the complement of $\mathcal G$. Now we write $$\abs{S\_N (f)} \leq \abs{S\_N (f - g)} + \abs{S\_N (g)}.$$ By the earlier discussion, the last term may be made arbitrarily small by choosing $N$ large enough. On the other hand, we have $$\abs{S\_N (f - g)} = \frac{1}{N}\sum\_{k, \frac{k}{N} \in \bigcup\_{P\_k \in \mathcal G} \, P\_k} \left ( f(\frac{k}{N}) - g(\frac{k}{N}) \right ) e^{2\pi i \alpha k} + \frac{1}{N}\sum\_{j, \frac{j}{N} \in [0, 1] \setminus \bigcup\_{P\_k \in \mathcal G}\, P\_k} \left ( f(\frac{j}{N}) - g(\frac{j}{N}) \right ) e^{2\pi i \alpha j}.$$ The summands in the first term are bounded in absolute value by $\varepsilon$, while those in the second are bounded by $1$. Additionally, the number of terms in the second sum is at most $\varepsilon N + n$, where we recall that $n$ is the number of terms in the partition. Thus $$\abs{S\_N (f - g)} \leq \varepsilon + \frac{1}{N} (\varepsilon N + n) \leq 3 \varepsilon$$ for all $N$ large enough. Since $\varepsilon$ was arbitrary, we conclude that $S\_N (f) \to 0$ as desired.

11

https://mathoverflow.net/users/173490

452902

https://mathoverflow.net/questions/452454

5

In Shelah's [paper](https://shelah.logic.at/papers/679/) [679](https://arxiv.org/abs/math/0003163), he proves primitive recursive bounds for the polynomial Hales-Jewett theorem and thus for the polynomial van der Waerden theorem. How about for the multidimensional polynomial HJ and multidimensional polynomial vdW theorems? In particular, for the "multidimensional polynomial van der Waerden theorem" -- see e.g. Theorem B of [Bergelson-Leibman](https://www.ams.org/journals/jams/1996-9-03/S0894-0347-96-00194-4/S0894-0347-96-00194-4.pdf) which does the density version (and the corresponding "multidimensional polynomial Hales-Jewett") -- can we obtain primitive recursive bounds? I do think that Shelah's method probably can be used or maybe does get a primitive recursive bound for all of these questions, but I am not very familiar with it. I use the multidimensional polynomial vdW theorem in [a recent paper](https://arxiv.org/abs/2307.08901) and I want to know if the bounds there can actually be made primitive recursive.

https://mathoverflow.net/users/119533

Primitive recursive bounds for multidimensional polynomial vdW / HJ

It is a standard fact that the (linear) Hales--Jewett theorem tensorizes to yield its multidimensional version. The same observation applies to the polynomial version. I will describe the idea for the two-dimensional case: assume that $n=2k$ is an even integer, $A$ is a finite alphabet and $c:A^{[n]^2}\to [r]$ is an $r$-coloring. Split the discrete interval $[n]$ into two successive subintervals $I\_0:=[k]$ and $I\_1:=\{k+1,\dots,2k\}$; this induces a partition of the index set $[n]^2$ into four squares $I\_0\times I\_0$, $I\_0\times I\_1$, $I\_1\times I\_0$ and $I\_1\times I\_1$ that are all isomorphic to $[k]^2$. Therefore, the $r$-coloring $c$ also induces an $r$-coloring of $(A^4)^{[k]^2}$, which I shall denote by $C$. Then observe that a monochromatic, with respect to $C$, polynomial combinatorial line of $(A^4)^{[k]^2}$ unfolds to a two-dimensional polynomial combinatorial subspace of $A^{[n]^2}$ that is monochromatic with respect to $c$. Note that this subspace is quite special: its wildcard sets are successive and have the same size. This argument yields the estimate $$ MPHJ(d,m,k,r) \leq PHJ(d,k^{m^d},r), $$ where * $PHJ(d,k,r)$ denotes the polynomial Hales--Jewett number, that is, the least positive integer $N$ such that for every integer $n\geq N$ and every alphabet $A$ with $k$ letters, an arbitrary $r$-coloring $c:A^{[n]^d}\to [r]$ has a monochromatic polynomial combinatorial line, and * $MPHJ(d,m,k,r)$ denotes the corresponding multidimensional polynomial Hales--Jewett number, that is, the least positive integer $N$ such that for every integer $n\geq N$ and every alphabet $A$ with $k$ letters, an arbitrary $r$-coloring $c:A^{[n]^d}\to [r]$ has a monochromatic $m$-dimensional polynomial combinatorial subspace. The paper of Shelah you cited has an ingenious proof that the shows that, for every fixed dimension $d$, the function $(k,r) \mapsto PHJ(d,k,r)$ is upper bounded by a primitive recursive function that belongs to the class $\mathcal{E}\_{7+3d}$ of Grzegorczyk’s hierarchy. (Here, the constants are most likely not optimal, but the linear dependence on the dimension $d$ is necessary.) The same paper also has a proof that claims that the polynomial Hales--Jewett numbers are upper bounded by a primitive recursive function (namely, we have a uniform control of their growth as the dimension $d$ increases) but that proof is problematic (see Fact 4.7 in the first preprint). Summing up, for every fixed dimension $d$, the function $(m,k,r)\mapsto MPHJ(d,m,k,r)$ is upper bounded by a primitive recursive function that belongs to the class $\mathcal{E}\_{7+3d}$ of Grzegorczyk’s hierarchy. From this, one can easily extract bounds for the multidimensional polynomial Van der Waerden theorem.

4

https://mathoverflow.net/users/510630

452922

https://mathoverflow.net/questions/452907

0

Prove that $p\mid\genfrac[]0{}{p^w}k$ where $p$ is an odd prime, $w \in \mathbb{N}$, $1<k<p^w$ and $k \neq p^v$ for some positive integer $v<w$. This has to be already done I just can't find where. Where in the literature has this been stated?

https://mathoverflow.net/users/265714

Prime divisibility of Stirling numbers of first kind

(The $w=1$ case is well-known, so we suppose $w\ge2$ in what follows; we also let $p$ denote any prime, which need not be odd.) See the remarks following the proof of Theorem 2.2 in Fredric T. Howard, [“Congruences for the Stirling numbers and associated Stirling numbers”](https://eudml.org/doc/206267), *Acta Arithmetica* **55** (1990), 29–41. In particular, it is stated that ${hp+m\brack k}\equiv0\pmod p$ if $m\in\{0,1,2,p-2,p-1\}$ and $h\ge1$ (with a few exceptions given); set $h=p^{w-1}$ and $m=0$, so that the only applicable exception is $${hp\brack h+(p-1)i}\equiv{h\choose i}(-1)^{h-i}\pmod p$$ for $0\le i\le h$. Then, since $p$ divides ${h\choose i}$ for $0<i<h$, we deduce that $${p^w\brack k}\equiv0\pmod p,$$ except for the two cases $${p^w\brack p^{w-1}}\equiv-1 \quad\text{and}\quad {p^w\brack p^w}\equiv1$$ (modulo $p$). Interestingly, an analogous result for Stirling numbers of the second kind is also given later in the paper (following Theorem 4.2): For $t>0$, we have $$\begin{align\*} {p^t\brace k}\equiv0\pmod p\qquad&\hbox{if $k\ne p^r$, $0\le r\le t$,}\\ {p^t\brace p^r}\equiv1\pmod p\qquad&(r=0,1,\dots,t). \end{align\*}$$

1

https://mathoverflow.net/users/118745

452925

https://mathoverflow.net/questions/452911

2

In [this post](https://mathoverflow.net/q/397649), an answer claims that the normal bundle of the Veronese decomposes into a filtration, such that the associated graded is $$\bigoplus\_{i=2}^d S^i T,$$ where $T$ is the tangent bundle. But I thought it was actually $$\bigoplus\_{i=2}^d S^i T \otimes \mathcal{O}(i-1)$$... I attempted to prove this in the following way: We know that the normal bundle sits in the exact sequence $$0 \to V \otimes \mathcal{O}(1) \to S^d V \otimes \mathcal{O}(d) \to \mathcal{N} \to 0.$$ Given the Euler exact sequence $$0 \to \mathcal{O} \to V \otimes \mathcal{O}(1) \to T \to 0,$$ symmetrizing yields the exact sequence $$0 \to S^{d-1}V \otimes \mathcal{O} \to S^dV \otimes \mathcal{O}(1) \to S^d T \to 0.$$ Iterating this and twisting, we obtain exact sequences $$0 \to S^{d-i}V \otimes \mathcal{O}(d-i) \to S^{d-i+1}V \otimes \mathcal{O}(d-i+1) \to S^{d-i+1}T \otimes \mathcal{O}(d-i) \to 0$$ Then we have a filtration of $S^d V \otimes \mathcal{O}(d)$ via $$V \otimes \mathcal{O}(1) \subset S^2V \otimes \mathcal{O}(2) \subset \cdots \subset S^{d-1}V \otimes \mathcal{O}(d-1) \subset S^d V \otimes \mathcal{O}(d)$$ such that quotienting out by $V \otimes \mathcal{O}(1)$ gives us the filtration $$S^2 T \otimes \mathcal{O}(1) \subset \cdots \subset \frac{S^{d-1}V \otimes \mathcal{O}(d-1)}{V \otimes \mathcal{O}(1)} \subset \frac{S^d V \otimes \mathcal{O}(d)}{V \otimes \mathcal{O}(1)} = \mathcal{N}$$. But then the associated graded is $$\bigoplus\_{i=2}^d S^i T \otimes \mathcal{O}(i-1).$$ Is this correct?

https://mathoverflow.net/users/510609

Normal bundle of veronese as iteration extension of symmetric powers

Your third exact sequence is incorrect --- the correct form is $$ 0 \to S^{d-1}V \otimes \mathcal{O}(d-1) \to S^{d}V \otimes \mathcal{O}(d) \to S^dT \to 0. $$

2

https://mathoverflow.net/users/4428

452926

https://mathoverflow.net/questions/452920

10

In Proposition 7.2.1.14 of *Higher Topos Theory*, Lurie asserts the following: > > Let $\mathcal{X}$ be an $\infty$-topos and let $\tau\_{\leq0}:\mathcal{X}\to\tau\_{\leq0}\mathcal{X}$ denote a left adjoint to the inclusion. A morphism $\phi:U\to X$ is an effective epimorphism if and only if $\tau\_{\leq 0}$ is an effective epimorphism in the ordinary topos $\operatorname{h}(\tau\_{\leq0}\mathcal{X}$). > > > His proof relies on Lemma 7.2.1.13, which in turn relies on Proposition 6.5.1.20. His proof of Proposition 6.5.1.20, in turn, relies on Proposition 7.2.1.14. **This is circular.** Hopefully, the proof of Proposition 6.5.1.20 only uses the following weaker version of Lemma 7.2.1.13: > > ($\ast$) Let $\mathcal{X}$ be an $\infty$-topos and let $f:X\to Y$ be a morphism of $\mathcal{X}$. Suppose that $X$ and $Y$ are $1$-connective. Then $f$ is an effective epimorphism. > > > So my question is: **Does anyone know how to prove ($\ast$) without using Proposition 7.2.1.14?** Thanks in advance.

https://mathoverflow.net/users/144250

Effective epimorphisms and 0-truncations (HTT, 7.2.1.14)

Here's an easy way to resolve the circularity. Proposition 7.2.1.13 is only used in the proof of 7.2.1.14 to establish the following statement: (1) If $f\colon V\to X$ is a monomorphism and is surjective on $\tau\_{\leq 0}$, then it is an isomorphism. This in turn follows from: (2) If $f\colon V\to X$ is a monomorphism, then it is pulled back from $\tau\_{\leq 0}f\colon \tau\_{\leq 0}V\to\tau\_{\leq 0}X$. Now (2) is obviously preserved by left exact localizations, and it holds in $\infty$-categories of presheaves since it holds in $\mathcal S$, so it holds in any $\infty$-topos. This gives a direct proof of 7.2.1.14 and in particular of (\*).

9

https://mathoverflow.net/users/20233

452929

https://mathoverflow.net/questions/452932

0

Below we use [Bochner measurability](https://www.wikiwand.com/en/Bochner_measurable_function) and [Bochner integral](https://www.wikiwand.com/en/Bochner_integral). Let * $(X, \mathcal A, \mu)$ and $(Y, \mathcal B, \nu)$ be complete $\sigma$-finite measure spaces, * $(E, | \cdot |)$ a Banach space, * $S (X)$ the space of $\mu$-simple functions from $X$ to $E$, * $L^0 (X)$ the space of $\mu$-measurable functions from $X$ to $E$, * $L^1 (X)$ the space of $\mu$-integrable functions from $X$ to $E$, * $\mathcal C :=\mathcal A \otimes \mathcal B$ the product $\sigma$-algebra of $\mathcal A$ and $\mathcal B$, * $\lambda := \mu \otimes \nu$ the product measure of $\mu$ and $\nu$. I'm trying to prove a claim in [this](https://mathoverflow.net/a/184921/99469) thread. > > **Theorem** Let $f \in L^0 (X \times Y)$. There is a sequence $(f\_n) \subset S(X \times Y)$ such that $f\_n \to f$ $\lambda$-a.e. and that each $f\_n$ is of the form > $$ > f\_n (x, y) = \sum\_{i,j=1}^{\varphi\_n} e\_{n, i,j} 1\_{A\_{n, i}} (x) 1\_{B\_{n, j}} (y), > \tag{$\*$} > $$ > where $e\_{n, k} \in E$ and $(A\_{n, i})\_{i=1}^{\varphi\_n}, (B\_{n, j})\_{j=1}^{\varphi\_n}$ are finite measurable partitions of $X,Y$ respectively. > > > Such $f\_n$ has a very nice form that I could not construct. Could you elaborate on how to get it? --- *My failed attempt* First, we assume $\mu (X) + \nu (Y) +\sup\_{(x, y) \in X \times Y} |f(x, y)| < \infty$. Then $\lambda(X \times Y) < \infty$ and $f \in L^1(X \times Y)$. There is a sequence $(f\_n) \subset S(X \times Y)$ such that $f\_n \to f$ $\lambda$-a.e. and in $L^1(X \times Y)$. Let $$ f\_n (x, y) = \sum\_{k=1}^{\varphi\_n} e\_{n, k} 1\_{C\_{n, k}} (x, y), $$ where $e\_{n, k} \in E$ and $C\_{n, k} \in \cal C$ such that $\lambda(C\_{n, k}) < \infty$.

https://mathoverflow.net/users/99469

How to construct this sequence that converges a.e. in product measure and that has a very particular form?

$\newcommand\C{\mathcal C}\newcommand\A{\mathcal A}\newcommand\B{\mathcal B}$Using the $\sigma$-finiteness condition, truncation of $f$, and your *attempt*, we see that without loss of generality $f=1\_C$ for some $C\in\C$. But in this case the result follows by (say) Theorem 1.4 of [this paper](https://link.springer.com/article/10.1007/s11117-017-0507-8) or [its arXiv version](https://arxiv.org/abs/1702.01142). Indeed, by that theorem, every $C\in\C$ can be approximated by sets in the algebra generated by product sets $A\times B$ such that $(A,B)\in\A\times\B$ with respect to the pseudometric $d$ given by the formula $d(E,F):=(\mu\otimes\nu)(E+F)$ for $E$ and $F$ in $\C$, where $+$ denotes the symmetric difference.

2

https://mathoverflow.net/users/36721

452934

https://mathoverflow.net/questions/452930

4

A 1950 result of Tur'an establishes an equivalence between any prime number theorem of the form $\operatorname{li}(x)-\pi(x)= O(xe^{-C(\log x)^\alpha}) \ (x \to \infty)$ and a certain class of zero-free regions of $\zeta(s)$. See: P. Tur'an, On the remainder-term of the prime-number formula, II, Acta Math. Acad. Sci. Hungar. 1 (1950) 155--166. Has this equivalence been extended to prime number theorems of the form $\operatorname{li}(x)-\pi(x)= O(xe^{-C(\log x)^\alpha(\log \log x)^\beta}) \ (x \to \infty)$? If so, what is the most general known equivalence of this type?

https://mathoverflow.net/users/17218

Zero-free regions of $\zeta(s)$ equivalent to prime number theorems with error bound

Yes. All of the results quoted in the answer are stated in Pintz's paper "[On the remainder term of the prime number formula. II. On a theorem of Ingham.](https://zbmath.org/0447.10038)" (Acta Arith. 37, 209-220 (1980)). Below, your $(\alpha,\beta)$ correspond to my $(1/(a+1),b/(a+1))$. 1. One direction is older than Turán and goes back to Ingham. He established that if $\zeta(s)$ does not vanish when $\Re s>1-\eta(|\Im s|)$, and $\eta >0$ is a continuously-differentiable decreasing function satisfying some mild conditions, then one has $$\pi(x)-\mathrm{Li}(x)\ll\_{\varepsilon} xe^{-\frac{1}{2}(1-\varepsilon)\omega(x)}$$ for $$\omega(x) := \min\_{t \ge 1} (\eta(t)\log x+\log t)=\min\_{v \ge 0} (\eta(e^v)\log x+ v).$$ In particular, this applies to $\eta(t) = c(\log t)^{-a}(\log \log t)^{b}$ ($a>0$), which gives $\omega(x) \sim d\_{c,a,b}(\log x)^{\frac{1}{a+1}} (\log \log x)^{\frac{b}{a+1}}$ by a short computation. This is Theorem 22 in Ingham's book "[The distribution of prime numbers](https://zbmath.org/0006.39701)" (1932). The formulation above is from Pintz's paper. This resolves one direction of your question. Pintz was interested in optimal exponents, and in Theorem 1 he sharpened Ingham's result, proving, in the same notation as above, that $$\pi(x)-\mathrm{Li}(x)\ll\_{\varepsilon} \frac{x}{\log x}e^{-(1-\varepsilon)\omega(x)}.$$ Moreover, he doesn't require $\eta$ to be differentiable, it can be any continuous decreasing function taking values in $(0,1/2]$. (In particular, the mild conditions of Ingham, which include $1/\eta(t) = O(\log t)$, are not needed, and any choice of $a>0$ and $b \in \mathbb{R}$ is allowed.) 2. [Turán](https://zbmath.org/0041.37102) and [Staś](https://zbmath.org/0104.27001) studied converses to Ingham's result. As you are already familiar with Turán's work, let me state Staś's as it appears in Pintz. If $\eta> 0$ is a continuous decreasing function satisfying some mild conditions, and $$|\psi(x)-x|\le c xe^{-\frac{1}{2}(1-\varepsilon)\omega(x)}$$ for the same $\omega$ as before, then $\zeta(s) \neq 0$ for $$\Re s > 1-\frac{1-\varepsilon}{1600}\eta(|\Im s|^{4/(1-\varepsilon)}), \qquad |\Im s| \gg\_{\varepsilon} 1.$$ Again, this applies to $\eta(t) = c(\log t)^{-a}(\log \log t)^{b}$, for which $\omega(x) \sim d\_{c,a,b}(\log x)^{\frac{1}{a+1}} (\log \log x)^{\frac{b}{a+1}}$. This shows that if $|\psi(x)-x| \ll x e^{-c'(\log x)^{\frac{1}{a+1}} (\log \log x)^{\frac{b}{a+1}}}$ then $\zeta(s) \neq 0$ for $\Re s > 1-c''(\log t)^{-a}(\log \log t)^{b}$. This gives the second direction of your question and might satisfy you already. Although you ask about $\psi(x)-x$ and not $\pi(x)-\mathrm{Li}(x)$, this difference is just a matter of partial summation. 3. A sharper version of Staś's result is Theorem 2 of Pintz (which has slightly different assumptions). The theorem says that if there is an infinite sequence of zeroes of $\zeta$ with $\Re s > 1-\eta(|\Im s|)$, where $\eta>0$ is a continuous decreasing function satisfying some mild conditions, then $$\pi(x)-\mathrm{Li}(x) = \Omega\_{\pm}(x e^{-(1+\varepsilon)\omega(x)}).$$ This might not be as useful as Staś's work due to the requirement of infinitely many zeros. However, see Theorem 2' of Pintz.

10

https://mathoverflow.net/users/31469

452936

https://mathoverflow.net/questions/452937

3

I mean specifically real-valued Schwartz distributions on the real line.  That is linear functionals  on $C^{\infty}\_c(\mathbb{R})$ continuous in the canonical LF topology.  My question is, what are all of these that have weak derivative 0?    Right now the only examples I know are distributions corresponding to constant functions.   Are there non-constant locally integrable functions with distributional derivative 0?  Are there distributions on the real line with weak derivative 0 but not corresponding to any locally integrable function? A locally integrable function with distributional derivative 0 must have no step discontinuity, and must have strong derivative 0 wherever it has a strong derivative.  But that is all I can prove about it now.  Question https://mathoverflow.net/posts/415151 deals with a holomorphic version of this question.

https://mathoverflow.net/users/38783

What real distributions solve $f'=0$?

You can take Fourier transforms $\widehat{f'}(t)=it\widehat{f}(t)=0$ to conclude that $\widehat{f}$ is supported by $\{0\}$, so $\widehat{f}=\sum\_{j=0}^n c\_j \delta^{(j)}$, so $f$ is a polynomial and then $f=c$. (If $f$ is locally integrable with distributional derivative in $L^1\_{\textrm{loc}}$, then $f$ is absolutely continuous, so can be recovered by integrating its derivative.) Or you can do it by hand, in the more general version for arbitrary (not necessarily tempered) distributions. See Theorem 2.6 of my lecture notes [here.](http://www2.math.ou.edu/%7Ecremling/teaching/lecturenotes/ln-dist.pdf)

4

https://mathoverflow.net/users/48839

452938

https://mathoverflow.net/questions/452279

0

Let $\hat{g}$ and $\bar{g}$ be two smooth Riemannian metrics defined, say, on $\mathbb{R}^n.$ Consider a smooth function $\xi$ that acts as an interpolation function between the two metrics above on an annulus $B\_{2R}(0)\backslash B\_R(0).$ If we define $g=\xi\hat{g} + (1-\xi)\bar{g},$ which is clearly a smooth Riemannian metric, can we find write $Rm(g)$ in terms of $Rm(\hat{g})$ and $Rm(\bar{g})$ plus an error? More importantly, what would be the structure of this error? I am sure someone did this calculation already, but I cannot find it anywhere. The calculation gets very messy very quickly, so I would appreciate a reference if anyone has one. Thank you.

https://mathoverflow.net/users/100597

Curvature tensor of interpolation of two metrics

After some Googling, the trick seems to be to consider the difference between the two metrics $ h = \tilde{g}-\bar{g}.$ Thus, $g= \bar{g} + \xi h.$ Using the notation above, we can write the Christoffel symbols of $g$ schematically as: $\Gamma(g) = \Gamma(\bar{g}) + \frac{1}{2}g \* \bar{\nabla}(\xi h),$ where $ h = \tilde{g}-\bar{g}.$ Thus, $Rm(g)= Rm(\bar{g}) + g^{-1} \*[ \bar{\nabla}^2\xi \*h + \bar{\nabla}\xi\*\bar{\nabla}h+ \xi^2\bar{\nabla}^2h] + g^{-2}[\*\bar{\nabla}\xi\*\bar{\nabla}\xi\*h\*h + \bar{\nabla}h\*\bar{\nabla}\xi\*\xi h +\xi^2\*\bar{\nabla}h\*\bar{\nabla}h],$ where $A\*B$ denotes any linear combination of terms involving contractions of the tensor product $A\otimes B.$ The details can be found in this paper by Gursky and Viaclovsky: <https://www.sciencedirect.com/science/article/pii/S0001870815301596?via%3Dihub>.

0

https://mathoverflow.net/users/100597

452940

https://mathoverflow.net/questions/452942

4

Let $E$ be an elliptic curve over a number field $K$ and $p$ a prime. Suppose that $E$ has a $K$-rational $p$-torsion, which gives the short exact sequence $0\to\mathbb{Z}/p\to E[p]\to\mu\_p\to0$ of Galois modules. Let us assume that this sequence splits, i.e. $E[p]\simeq\mathbb{Z}/p\times\mu\_p$ as Galois modules. Set $E':=E/\mu\_p$. Now, my questions is: > > Is the set of bad primes for the elliptic curve $E'$ the same as the set of bad primes for $E$? > > > I am also wondering whether there is a general theory in this line, for example, how to compute the set of bad rimes for the quotient elliptic curve $E/H$ for "Galois stable" subgroup $H$ of $E$.

https://mathoverflow.net/users/44005

Primes of bad reductions for quotients of elliptic curves

As has already been remarked, but more generally, if $K$ is a number field and $A/K$ and $B/K$ are abelian varieties that are isogenous over $K$, then the criterion of Neron-Ogg-Shafarevich has as a quick corollary that $A$ and $B$ have the same set of primes of bad reduction. The paper with the proof is by Serre and Tate. (*Serre, Jean-Pierre; Tate, J.*, [**Good reduction of abelian varieties**](https://doi.org/10.2307/1970722), Ann. Math. (2) 88, 492-517 (1968). [ZBL0172.46101](https://zbmath.org/?q=an:0172.46101).) If you want a proof just for elliptic curves, you could look at Section VII.7 in (*Silverman, Joseph H.*, [**The arithmetic of elliptic curves**](https://doi.org/10.1007/978-0-387-09494-6), Graduate Texts in Mathematics 106. New York, NY: Springer (ISBN 978-0-387-09493-9/hbk; 978-0-387-09494-6/ebook). xx, 513 p. (2009). [ZBL1194.11005](https://zbmath.org/?q=an:1194.11005)).

7

https://mathoverflow.net/users/11926

452944

https://mathoverflow.net/questions/452906

0

I have a question about affine Coxeter groups when reading Humphreys's book: Let $(W,S)$ be an irreducible affine Coxeter group, $M=(m\_{ij})$ be its Coxeter matrix, and $\{\alpha\_s\}\_{s\in S}\in V$ be the system of simple roots in the standard geometric realization, so the $\{\alpha\_s\}$ are linearly independent. The bilinear form $\bullet$ given by $$\alpha\_s\bullet\alpha\_s = \begin{cases}-\cos\frac{\pi}{m\_{s,t}}, & m\_{s,t}<\infty\\ -1, & m\_{s,t}=\infty\end{cases}$$ has signature $(n-1, 0)$. It's well-known that ${\rm Rad}(\bullet)$ is one-dimensional, and is spanned by $\delta=\sum\_{s}c\_s\alpha\_s$ where all $c\_s>0$, and $\bullet$ is positive-definite on the quotient $U=V/\mathbb{R}\delta$, which can be identified with the hyperplane in the dual space $U^\ast=\langle\cdot,\delta\rangle=0$. $W$ also acts on $U^\ast$ as a reflection group, this gives a homomorphism $W\to{\rm GL}(U^\ast)$, and let $W'$ be its kernel. Prove that $W'$ is non-trivial. Is there a direct proof of this? Update: Sorry for the confusion. By 'direct', I mean a general one that doesn't need to check the list of irreducible affine root systems.

https://mathoverflow.net/users/70248

A question on irreducible affine Coxeter groups

When you ask for a *direct* proof, I'm not sure what I'm allowed to use. But if I am allowed to know how to construct the affine root system from an appropriate finite root system, the fact that $W'$ is non-trivial becomes easy: If you use the standard construction of an affine root system for your Coxeter group, the root system is the set of vectors $\beta+k\delta$ such that $\beta$ is a root in the finite root system and $k$ is an integer. For two roots with the same $\beta$ and different $k$, the corresponding reflections have the same action on $U^\*$.

5

https://mathoverflow.net/users/5519

452954

https://mathoverflow.net/questions/336330

11

$\def\inv{\mathrm{inv}}\def\Acyc{\mathrm{Acyc}}$Let $G$ be a graph whose vertices are numbered $\{ 1,2, \ldots, n \}$. Given an orientation $\omega$ of $G$, define the inversions of $\omega$, written $\inv(\omega)$, to be the set of edges $(i,j)$ with $i<j$, which are oriented $i \leftarrow j$. Define one orientation $\omega\_1$ to be $\leq$ another orienation $\omega\_2$ iff $\inv(\omega\_1) \subseteq \inv(\omega\_2)$. Obviously, the set of all orientations of $G$ form a boolean lattice in this way. Let $\Acyc(G)$ be the set of acyclic orientations of $G$. Restricting the above partial order to $\Acyc(G)$ makes $\Acyc(G)$ into a poset. > > What is known about when $\Acyc(G)$ is a lattice? > > > Some thoughts below: --- $\bullet$ If $G$ is the complete graph $K\_n$, this is the weak order on $S\_n$, known to be a lattice. $\bullet$ We could ask more strongly when the obvious surjection $\Acyc(K\_n) \to \Acyc(G)$ is a map of lattices or, in other words, if $\Acyc(G)$ is a quotient of $\Acyc(K\_n)$. This can be studied using Nathan Reading's classification of quotients of weak orders (see [Reading, Section 4](https://arxiv.org/abs/1405.6904)). The answer is that, if $i<j<k$, and $(i,k)$ is an edge of $G$, then $(i,j)$ and $(j,k)$ must also be edges of $G$. However, this strong condition is not necessary to make $\Acyc(G)$ into a lattice. Note that $\Acyc(F)$ will be a lattice for any forest $F$, and a tree with $\geq 3$ edges will not obey the above condition. $\bullet$ $\Acyc(G)$ is the regions of the graphical hyperplane arrangement coming from $G$, see Björner, Edelman and Ziegler, "[Hyperplane arrangements with a lattice of regions](https://mathscinet.ams.org/mathscinet-getitem?mr=1036875)" and Reading's Chapter 9, "Lattice Theory of the Poset of Regions" in [Lattice Theory: Special Topics and Applications](https://link.springer.com/book/10.1007%2F978-3-319-44236-5) for relevant background. So we can phrase this questions as "when do the regions of a graphical hyperplane arrangement form a lattice"? $\bullet$ An example of a graph where this does NOT hold is the one with edge set $\{ (1,2), (2,4), (1,3), (3,4) \}$. The elements $1 \to 2 \to 4 \to 3 \leftarrow 1$ and $1 \to 2 \to 4 \leftarrow 3 \to 1$ have no join.

https://mathoverflow.net/users/297

When is the poset of acyclic orientations of a graph a lattice?

Vincent Pilaud's recent paper ["Acyclic reorientation graphs and their lattice quotients"](https://arxiv.org/abs/2111.12387) is a thorough answer to this question and every question like it. In particular, here is the answer to the particular question which is asked: Given a directed acyclic graph $G$, the **transitive reduction** of $G$ is defined to be the subgraph $G'$ obtained from $G$ by deleting those edges $i \to j$ for which there is a directed path $i \to \cdots \to j$ of length $\geq 2$. The directed acyclic graph $G$ is called **vertebrate** iff the transitive reduction of every induced subgraph is a forest. > > **Theorem (Pilaud)** $A(G)$ is a lattice iff and only if $G$ is vertebrate. > > > Pilaud also characterizes when $A(G)$ is distributive, semidistributive, congruence uniform, and many other properties. Check out the paper!

6

https://mathoverflow.net/users/297

452969

https://mathoverflow.net/questions/452971

4

*Note: We view the sphere $S^1$ as $[0,1]$ with the endpoints identified, and equip it with its usual addition structure, and Lebesgue measure.* **Question:** Does there exist an absolute constant $C > 0$ such that for all $L^1$ functions $f: S^1 \to \mathbb R$, $$\sup\_{t \in S^1} \int\_{S^1} |f(x + t) - f(x)| \, dx \geq C\int\_{S^1} \left|f(x) - \left(\int\_{S\_1} f(y) \, dy\right) \right| \,dx\text{?}$$

https://mathoverflow.net/users/173490

The maximal difference between a function and translates of itself

Using the condition $\int\_{S^1}dx=1$ and Jensen's inequality, we have $$\sup\_{t\in S^1} \int\_{S^1}dx\,|f(x + t)-f(x)| \ge\int\_{S^1}dt\, \int\_{S^1}dx\,|f(x + t)-f(x)| \\ =\int\_{S^1}dx\, \int\_{S^1}dt\,|f(x)-f(x + t)| \ge\int\_{S^1}dx\, \Big|f(x)-\int\_{S^1}dt\,f(x + t)\Big| \\ =\int\_{S^1}dx\, \Big|f(x)-\int\_{S^1}dy\,f(y)\Big|. $$ So, your conjectured inequality holds with $C=1$. $\quad\Box$

5

https://mathoverflow.net/users/36721

452977

https://mathoverflow.net/questions/452989

0

Suppose that $G=(V,E)$ is a simple, undirected graph. We say that $D\subseteq V$ is *dominating* if for all $v\in V\setminus D$ there is $d\in D$ such that $\{v,d\}\in E$. We say $D$ is *minimal dominating* if for all $d\in D$ we have that $D\setminus\{d\}$ is no longer dominating. There are graphs that [do not have minimal dominating sets](https://mathoverflow.net/questions/260931/minimal-dominating-subsets-in-infinite-graphs). We call $G=(V,E)$ *flat* if for all $v\in V$ the neighborhood $N\_G(v)=\{w\in V:\{v,w\}\in E\}$ is finite. If $G=(V,E)$ is a flat graph, does $G$ necessarily contain a minimal dominating set? Is it even the case that every dominating set contains a minimal dominating set? (Only answering the first question is sufficient for acceptance, but it would be lovely to have the answer to both questions.)

https://mathoverflow.net/users/8628

Minimal dominating sets in flat graphs

For flat graphs, dominating sets satisfy Zorn's condition: the intersection of a chain of dominating sets $D:=\cap D\_\alpha$ is dominating. Indeed, for any vertex $v$ the finite set $\{v\cup N(v)\}$ has non-empty intersection with every $D\_\alpha$, thus this finite non-empty intersection stabilizes and its limit is contained in $D$. So, the answer is positive by Zorn lemma.

2

https://mathoverflow.net/users/4312

452991

https://mathoverflow.net/questions/452970

5

I am interested in the hypersurface $X\subset\mathbb{A}^4\_{\mathbb{F}\_{5^n}}$ defined by $$ X = \{x^3 + 3xy^2 + z^3 + 3zw^2 + 1 = 0\} $$ over a finite field $\mathbb{F}\_{5^n}$ with $5^n$ elements. Via some computer experiment I have noticed that when $n$ is odd the number of points of $X$ is equal to the number of points of $\mathbb{A}^3\_{\mathbb{F}\_{5^n}}$. Is this just a coincidence or is there a theoretical reason for this? Thank you.

https://mathoverflow.net/users/510696

Points on affine hypersurface over finite field

Note that the defining equation for $X$ can be rewritten as $$(x+y)^3+(x-y)^3+(z+w)^3+(z-w)^3=-2.$$ As the linear transformation $(x,y,z,w)\mapsto(x+y,x-y,z+w,z-w)$ is invertible over any field of characteristic not equal to $2$, the number of $\mathbb{F}\_{5^n}$-points on $X$ is equal to the number of $\mathbb{F}\_{5^n}$-points on the hypersurface $Y$ with defining equation $$a^3+b^3+c^3+d^3=-2.$$ As $3$ does not divide $5^n-1$, every element of $\mathbb{F}\_{5^n}$ is a cube in a unique way, so the number of points on $Y$ are the same as the number of points on the hypersurface defined by $$e+f+g+h=-2,$$ but this hypersurface is isomorphic to $\mathbb{A}^3,$ as desired.

10

https://mathoverflow.net/users/51424

452992

https://mathoverflow.net/questions/452990

1

I have recently learned about relative Poincaré duality in the book *Weil conjectures, perverse sheaves and $\ell$-adic Fourier transform* by Kiehl and Weissauer (2001). The reference is section II.7. When trying to apply the statement to some toy examples however, I end up with an absurd conclusion. Thus, there must be something I'm doing wrong, but I can't find it out. Let me explain. Let $k$ be a finite field of characteristic $p>0$ (or an algebraic closure thereof) and let $\ell \not = p$ be a prime number. For a variety $X$ over $k$, we denote by $D\_c^b(X,\overline{\mathbb Q\_{\ell}})$ the bounded "derived" category of constructible $\overline{\mathbb Q\_{\ell}}$-sheaves. For a morphism $f:X\to S$ of varieties over $k$ and for $K \in D\_c^b(X,\overline{\mathbb Q\_{\ell}})$ and $L \in D\_c^b(S,\overline{\mathbb Q\_{\ell}})$, the relative Poincaré duality is a natural isomorphism $$Rf\_{\*}R\mathcal{Hom}(K,f^!L) \simeq R\mathcal{Hom}(Rf\_!K,L),$$ which is functorial in both $K$ and $L$. In particular, taking $S = \mathrm{Spec}(k)$ and $L := \overline{\mathbb Q\_{\ell}}$, the identity becomes $$Rf\_\*(D\_X(K)) \simeq D\_S(Rf\_!(K)),$$ where $D$ denote the dualizing functor. Taking cohomology gives $$H^i(X,D\_X(K)) \simeq H^{-i}\_c(X,K)^{\vee}.$$ If we further assume that $X$ is smooth of dimension $d$, we have $$D\_X(K) = R\mathcal{Hom}(K,\overline{\mathbb Q\_{\ell}}[2d](d)) = K^{\vee}[2d](d),$$ where $K^{\vee} := R\mathcal{Hom}(K,\overline{\mathbb Q\_{\ell}})$. Thus, we obtain \begin{equation} H^{i+2d}(X,K^{\vee})(d) \simeq H^{-i}\_c(X,K)^{\vee}, \tag{$\*$} \end{equation} which looks like the usual Poincaré duality except that $K$ is allowed to be any object of $D\_c^b(X,\overline{\mathbb Q\_{\ell}})$. To me, this sounds doubtful. I always thought that Poincaré duality requires the coefficient sheaf to be smooth (lisse), not just constructible. Eg. let us assume $X = \mathbb P^1$ and take $K = \mathcal F$ be just a skyscraper sheaf at a closed point of $X$ with value $\overline{\mathbb Q\_{\ell}}$ in degree $0$. Then the RHS of $(\*)$ is $1$-dimensional for $i=0$, and vanishes otherwise. On the other hand, $K^{\vee} \simeq K$ and the LHS is $1$-dimensional for $i = -2$, and $0$ otherwise. So clearly, something isn't right. Would anybody be able to point out where I wrote something wrong?

https://mathoverflow.net/users/125617

Confusion about relative Poincaré duality in the context of $\ell$-adic cohomology

In your last paragraph you don't have $K\simeq K^\vee$. Let $i$ be the inclusion of the closed point, so $K=i\_!\mathbf Q\_\ell$. Then by local Verdier duality $$K^\vee = RHom(i\_!\mathbf Q\_\ell,\mathbf Q\_\ell)=i\_\ast RHom(\mathbf Q\_\ell,i^!\mathbf Q\_\ell)$$which is a skyscraper sheaf concentrated in degree 2.

4

https://mathoverflow.net/users/1310

452994

https://mathoverflow.net/questions/234507

11

Let $A$ be a C\*-algebra. We identify $A$ with its canonical image in the bidual $A^{\*\*}$. Consider the following conditions: (1) $A$ is a von Neumann algebra. (2) There is a multiplicative conditional expectation from $A^{\*\*}$ onto $A$, that is, a map $\pi\colon A^{\*\*}\to A$ that is a \*-homomorphism and such that $\pi(a)=a$ for all $a\in A$. Then (1) implies (2): Consider the predual $A\_\*$ of $A$, and the canonical embedding $\kappa\colon A\_\*\to (A\_\*)^{\*\*}$. Then the dual of $\kappa$ has the desired properties. Note that $A^{\*\*}$ is always a von Neumann algebra. One can show that (2) implies that $A$ is a monotone complete AW\*-algebra. However, is it also a von Neumann algebra? > > Q1: Does (2) imply (1) ? > > > Assuming (2), let $J$ denote the kernel of $\pi$. Then $A$ is (\*-isomorphic to) the quotient $A^{\*\*}/J$. However, a quotient of a von Neumann algebra by a closed, two-sided ideal need not be a von Neumann algebra. (For example, the Calkin algebra is such a quotient and not a von Neumann algebra.) In the commutative case, we might also consider the following more general question: > > Q2: Is the quotient of a commutative von Neumann algebra by a closed, two-sided ideal again a von Neumann algebra? > > > Equivalently, is every closed subset of a hyperstonean space again hyperstonean?

https://mathoverflow.net/users/24916

von Neumann algebras as C*-algebras with multiplicative conditional expectation $A^{**}\to A$

This is a seven-year-old question, so I'm not sure if an answer is still meaningful at this point, but no, $A$ needs not be a vNa. There is even a commutative counterexample. In fact, fix any monotone complete commutative $C^\*$-algebra $A$ which is not a vNa. Then $A$ is injective in the category of unital commutative $C^\*$-algebras and $\*$-homomorphisms. (See, for example, <https://arxiv.org/abs/0706.2995> .) Hence, the identity map $A \rightarrow A$ can be extended to a multiplicative conditional expectation $\pi: A^{\*\*} \rightarrow A$.

5

https://mathoverflow.net/users/504602

452997

https://mathoverflow.net/questions/452993

2

Do there exist positive integers $A, B, C$ such that all seven numbers $$A, B, C, A+B, B+C, A+C, A+B+C$$ are perfect squares?

https://mathoverflow.net/users/4312

3-dimensional Boolean cube of Squares

If such $(A, B, C)$ exist, then $(\sqrt{A}, \sqrt{B}, \sqrt{C})$ are sides of a perfect cuboid (see [here](https://en.m.wikipedia.org/wiki/Euler_brick)). Such a cuboid has not yet been found.

4

https://mathoverflow.net/users/507773

452999

https://mathoverflow.net/questions/452557

5

I would like to find a reference that describes the semantics of constructive higher order logic *with function types* in toposes. In particular, it seems that if we are to take function types as primitive, then we need the **definite description operator** $\iota x. P(x)$, which produces a term that satisfies $\exists ! x. P(x) \implies P(\iota x. P(x))$. This is required to prove the principle of unique choice. And if we want the function type to actually be useful this principle is a must. For Isabelle/HOL and HOL Light, the even stronger *indefinite* description operator $\epsilon x. P(x)$ is used, validating choice. However, I can't work out the semantics of this operator myself. This operator would produce a term no matter whether $P(x)$ actually has a unique solution $x$. So this would produce a map $\iota : \mathcal P(X) \to X$ in any topos where $X$ is the interpretation of the type of $x$. In the classical case this is fine because we can just case split and assign a junk value. But this doesn't sound possible in an arbitrary elementary topos. In the literature: the Elephant did mention function types, but it did not talk about definite descriptions at all, and in its definition of the formal system ($\tau$-theory) it did not include function types because we already have powerset types and "we shall keep the logic as simple as possible by leaving them out". This avoids the problem because defining functions as functional relations $\psi : \mathcal P(A \times B)$ makes the principle of unique choice a theorem. In *Introduction to Higher Order Categorical Logic* by Lambek and Scott, there are also no function types. So, what is the right way of having function types, and ensuring the principle of unique choice? Do we not use definite description, and instead just add an axiom? Or is there another way out? Any reference on this matter is welcome.

https://mathoverflow.net/users/136535

Topos semantics of constructive higher order logic

Categorical logic texts such as Lambek & Scott's "Introduction to higher-order categorical logic" and Johstone's "Elephant" usually focus on the categorical side of things and are often a bit cavalier about the details of the internal language – enough so to require significant adjustments before one can actually think about implementing it. Regarding the definite description operator, you can look at Theorem 5.9 of Lambek & Scott, but that's in a section about the internal language, so the authors have the freedom to switch to the categorical side whenever needed, and they do. Regarding setting up the internal language of a topos qua cartesian-closed category with a subobject classifier, finite coproducts, and coequalizers, I recommend Wesley Phoa's [An introduction to fibrations, topos theory, the effective topos and modest sets](http://www.lfcs.inf.ed.ac.uk/reports/92/ECS-LFCS-92-208/). The precision of detail in the text surpasses other sources that I am familiar with. In particular, the appendix spells out both the details of the formal system, and its interpretation in a topos. The definite description operator is treated on page 61, and in the appendix on pages 134 and 142. As you will see there, the categorical construction of the semantics of the definite description is the most complicated part of the semantics of the internal language of a topos. Good luck converting a Postscript file to a readable PDF!

6

https://mathoverflow.net/users/1176

453003

https://mathoverflow.net/questions/453007

5

Consider the Fermat cubic $$ X = \{x\_0^3+\dots +x\_n^3 = 0\}\subset\mathbb{P}^n\_{\mathbb{F}\_{q}} $$ over a finite field $\mathbb{F}\_{q}$ with $q$ elements. If $q \equiv 2 \mod 3$ then the projection $\pi:X\rightarrow \mathbb{P}^{n-1}\_{\mathbb{F}\_{q}}$ defined by $\pi(x\_0,\dots,x\_n) = (x\_0,\dots,x\_{n-1})$ yields a bijection $X(\mathbb{F}\_{q})\rightarrow \mathbb{P}^{n-1}\_{\mathbb{F}\_{q}}(\mathbb{F}\_{q})$ so that the number of $\mathbb{F}\_{q}$-points of $X$ is $\frac{q^n-1}{q-1}$. Does there exist a formula for the number of $\mathbb{F}\_{q}$-points of $X$ when $q \not\equiv 2 \mod 3$?

https://mathoverflow.net/users/510696

Fermat cubic hypersurfaces over finite fields

A. Weil, in "Numbers of solutions of equations in finite fields" (Bull. Am. Math. Soc. 55, 497-508 (1949)), proved that the number of $\mathbb{F}\_q$-points when $q\equiv 1 \bmod 3$ is $$\frac{q^n-1}{q-1} +\frac{1}{q-1} \sum\_{\chi\_0,\ldots,\chi\_n} J\_0(\chi\_0,\ldots,\chi\_n)$$ where the sum is over $(n+1)$-tuples of non-trivial characters $\chi\_i \colon \mathbb{F}\_q^{\times} \to \mathbb{C}^{\times}$ of order $3$ such that $\prod\_i \chi\_i$ is trivial. Here $J\_0$ is the Jacobi sum $$J\_0(\chi\_0,\ldots,\chi\_n):=\sum\_{t\_0+\ldots+t\_n=0} \prod\_i \chi\_i(t\_i).$$ ($\chi\_i(0)$ is defined to be $0$.) The Jacobi sum is expressible as $$\frac{q-1}{q}\prod\_i g(\chi\_i)$$ where $g$ is the Gauss sum $g(\chi\_i)=\sum\_{t \in \mathbb{F}\_q^{\times}}\chi\_i(t)e^{\frac{2\pi i}{p}\mathrm{Tr}(t)}$. In particular, $$|J\_0(\chi\_0,\ldots,\chi\_n)|=(q-1)q^{\frac{n+1}{2}-1}.$$ In fact Weil studied more general diagonal hypersurfaces and his results inspired him to formulate his famous conjectures. See also Chapters 8 and 10 in Ireland and Rosen's book "A Classical Introduction to Modern Number Theory" (particularly Theorem 2 in Chapter 10). Finally, if $q \equiv 0 \bmod 3$ then the number of $\mathbb{F}\_q$-points is like in the $q \equiv 2 \bmod 3$ case, since the equation reads $(x\_0+\ldots+x\_n)^3=0$. *Addendum:* When $q \equiv 1 \bmod 3$, the total contribution of the Jacobi sums is $O(q^{\frac{n-1}{2}}q^n)$ which is exponentially smaller then the main term $(q^n-1)/(q-1)$ as long as $q \neq 4$. But for $q=4$ this is quite useless. However, for $q=4$, $x\_i^3$ is either $1$ (if $x\_i \neq 0$) or $0$ (if $x\_i=0$) so one can parametrize the points as tuples $(x\_0,\ldots,x\_n)$ (up to a multiplicative scalar) where the number of non-zero entries is even. This leads by the binomial theorem to the formula $$\frac{4^{n+1}+(-2)^{n+1}-2}{6}.$$ It can also be derived by computing the Gauss sums explicitly.

12

https://mathoverflow.net/users/31469

453010

https://mathoverflow.net/questions/452903

3

*Based upon discussion at [Math.SE](https://math.stackexchange.com/questions/4751580/what-separation-is-required-to-ensure-extremally-disconnected-spaces-are-sequent)* Consider the property [extremally disconnected](https://topology.pi-base.org/properties/P000049), for which the closure of any open set remains open. Frequently, this property is paired with the assumption of Hausdorff. This allows nice results like [all extremally disconnected spaces are totally separated](https://topology.pi-base.org/theorems/T000045) and disallows silliness like [all hyperconnected spaces are extremally disconnected](https://topology.pi-base.org/theorems/T000096). But I like silliness, so how much of the theory can we recover without Hausdorff? In [this Math.SE post](https://math.stackexchange.com/questions/4751580/what-separation-is-required-to-ensure-extremally-disconnected-spaces-are-sequent) it was pointed out that while all Hausdorff extremally disconnected spaces are sequentially discrete (only trivial sequences converge), the cofinite topology on an infinite set is a $T\_1$ extremally disconnected space that is not sequentially discrete. On the other hand, the only example I know of a sequentially discrete, $T\_1$, extremally disconnected space is US (limits of sequences are unique): the cocountable topology on an uncountable set. In fact, this is pretty immediate: all spaces where countable sets are closed are sequentially discrete, and all sequentially discrete spaces are US. Might it be the case that the theorem that all Hausdorff extremally disconnected spaces are sequentially discrete can be improved to only assume US? If not, what counterexample can be constructed?

https://mathoverflow.net/users/73785

Must US extremally disconnected spaces be sequentially discrete?

Here's an example, I think. Let $X=(\omega\_1+1)\times(\omega\_0+1)$ and let $Y$ denote the subset $\omega\_1\times(\omega\_0+1)$. Define a topology on $X$ by declaring every set of the form $Y\setminus C$ with $C$ countable open. So $Y$ will be an open subspace. We ensure that $Y$ has the co-countable topology by specifying local bases at the other points. At $\langle\omega\_1,n\rangle$ (with $n\in\omega\_0$) the basic neighbourhoods are of the form $\{\langle\omega\_1,n\rangle\}\cup(Y\setminus C)$ with $C$ countable. At $\langle\omega\_1,\omega\_0\rangle$ the basic neighbourhoods are of the form $\{\{\omega\_1\}\times(n,\omega\_0]\cup(Y\setminus C)$ with $n\in\omega\_0$ and $C$ countable. If $A$ is a countable subset of $X$ then $A\cap Y$ is closed in $X$ and so if $A\cap Y$ is infinite then $A$ cannot form a convergent sequence under any enumeration. Hence if $A$ is an infinite convergent sequence then $A\cap Y$ must be finite and upon deleting that finite set $A\subseteq \{\omega\_1\}\times(\omega\_0+1)$; the latter has the natural topology of $\omega\_0+1$ so $A$ converges to $\omega\_0$ (and no other point). Finally: every nonempty open set is co-countable, and so every nonempty open set is dense and that makes $X$ extremally disconnected.

4

https://mathoverflow.net/users/5903

453013

https://mathoverflow.net/questions/453016

3

The following lemma is due to Campana, [The class $\mathcal C$ is not stable by small deformations](https://link.springer.com/article/10.1007/BF01459236) > > Let $\mathcal X\rightarrow \Delta$ be a smooth family, if $K\_{X\_0}$ is nef and big, then so is every $K\_{X\_t}$ for $t$ sufficiently small. > > > Here is the step. Invoking Kawamata-Viehweg vanishing theorem, one has $$H^i(X\_0,NK\_{X\_0})=0$$ for $N\geq 2$. Then the similarly higher cohomology vanishing holds for $K\_{X\_t}$ by Grauert direct image theorem. Therefore, $$h^0(X\_t,NK\_{X\_t})=h^0(X\_0,NK\_{X\_0})\,\,\,\,\,\,(\*)$$ for $N\geq 2$ thanks to deformation invariance of Euler characteristic. Also, the base point free theorem tells us that $|NK\_{X\_0}|$ is globally generated for sufficiently big $N$. Then the author claims that this fact combines (\*) can infer that $|NK\_{X\_s}|$ is also globally generated for sufficiently big $N$ (this will imply the nefness of $X\_t$ then). Why this holds? How the size of global sections determine the base locus with respect to $X\_t$ in our setting? Any suggestions will be appreciated. Thanks in advance.

https://mathoverflow.net/users/141609

semiample of canonical bundle in a smooth family (Campana's proof)

Let $f:X\rightarrow \Delta $ be your family. $\ (\*)$ implies that $f\_\*K\_{X/\Delta }^{N}$ is a vector bundle on $\Delta $, with fiber $H^0(X\_t, K\_{X\_t}^N)$ at $t\in\Delta$. The canonical homomorphism $\ f^\*f\_\*K\_{X/\Delta }^{N}\rightarrow K\_{X/\Delta }^{N}$ is surjective on $X\_0$ by hypothesis, hence also on $X\_t$ for small $t$.

5

https://mathoverflow.net/users/40297

453017

https://mathoverflow.net/questions/451528

0

We have proposed a new approach to solve the maximum edge bi-clique problem, however, we couldn't succeed to find real-world datasets (graph or bipartite graph datasets) to test our approach. Does anyone have an idea where I could find some datasets for this problem?

https://mathoverflow.net/users/509350

Real-world datasets for testing the maximum edge bi-clique problem

Since this problem has many applications in data mining, you probably should take a look at datasets used as benchmarks in papers on the topic. As an example, [this recent paper](https://ieeexplore.ieee.org/abstract/document/10121476) mostly uses graphs from the [KONECT Project](http://konect.cc/).

0

https://mathoverflow.net/users/8193

453023

https://mathoverflow.net/questions/453004

10

Do there exist positive integers $x,y,z$ such that $$ (x+y)(xy-1)=z^2+1 $$ In my previous question [Can you solve the listed smallest open Diophantine equations?](https://mathoverflow.net/questions/400714), I discuss the smallest equations for which we do not know if they have any integer solutions. Some equations (think about Fermat Last Theorem) have some trivial integer solutions, and the correct question to ask is whether they have a solution in **positive** integers. The equation in the title is the smallest one for which I do not know the answer. It is known that all positive divisors of $z^2+1$ must be $1$ or $2$ modulo $4$, but if we take $x$ and $y$ to be $2$ and $3$ modulo $4$, this condition is satisfied, so no contradiction. On the other hand, search returns no results.

https://mathoverflow.net/users/89064

Positive integers such that $(x+y)(xy-1)=z^2+1$

As you have already noticed, we may assume that $x \equiv 3 \pmod{4}$, $y \equiv 2 \pmod{4}$. Let $p \equiv 3 \pmod{4}$ be a prime divisor of $y + 1 \equiv 3 \pmod{4}$ such that $\nu\_p(y+1) \equiv 1 \pmod{2}$. Consider the equation $$(x + y)(xy - 1) = z^2 + 1$$ modulo $p$. $$(x + y)(xy - 1) \equiv (x - 1)(-x - 1) \equiv -x^2 +1 \pmod{p}$$ $$x^2 + z^2 \equiv 0 \pmod{p}$$ Therefore $p \mid x, z$ since $-1$ is a quadratic nonresidue modulo $p$. Let $Y = y + 1$. $$x^2Y - x^2 + xY^2 - 2xY - Y = z^2$$ $$Y(x^2 + xY - 2x - 1) = x^2 + z^2$$ There is a contradiction, because $\nu\_p(\cdot) $ is odd for left side of the equation and is even for right side of the equation.

12

https://mathoverflow.net/users/507773

453024

https://mathoverflow.net/questions/453006

4

Is there a general algorithm that can compute in finite time all rational points on any curve of genus $g\geq 2$ whose Jacobian has rank $0$? If not, for what special cases such algorithm is known? For genus $g=2$, we have Chabauty0 command in Magma, is it always guaranteed to work? For genus $g=3$, natural families to consider are hyperelliptic curves $y^2=P(x)$ for $P$ of degree $7$ or $8$, or Picard curves $y^3=P(x)$ for $P$ quartic. Is there an algorithm for computing rational points on them if rank if the Jacobian is $0$? Is there a corresponding Magma code? If yes, can you provide a reference? In Cohen's book, I found the following deep theorem Demyanenko-Manin. Let C be a curve defined over a number field K. Assume that A is a K-simple Abelian variety such that $A^m$ occurs in the decomposition of the Jacobian J of C up to isogeny over K and that $$ m > \frac{rk(A(K))}{rk(End\_K(A))}, $$ where as usual rk denotes the rank. Then C(K) is finite and can be determined explicitly. Can we derive the result from here with, say, $K={\mathbb Q}$, $m=1$, and $A=J$? Even if yes, is there a better reference, ideally book/paper where the existence of an algorithm for the rank $0$ case is stated explicitly?

https://mathoverflow.net/users/89064

Algorithm for computing rational points if the rank of Jacobian is 0

There is an algorithm due to Bjorn Poonen (Computing torsion points on curves, Experiment. Math. 10 (2001), no.3, 449–465) that, given a (not necessarily rational) base-point $P\_0$ on the curve, finds all (geometric) points $P$ on the curve such that $[P-P\_0]$ is torsion in the Jacobian. This solves an even harder problem! (Although, as far as I know, this has been implemented only for curves of genus 2 with a Weierstraß point as base-point.)

5

https://mathoverflow.net/users/21146

453031

https://mathoverflow.net/questions/453043

5

How many solutions are there to $n\_1^2-n\_2^2-n\_3^2+n\_4^2=k$ where $n\_i\in [1,N]$ and $n\_i,k\in \mathbb Z$? For $k=0$ by paucity we know it should be $\ll N^2 \log N$ but what about different $k$?

https://mathoverflow.net/users/479223

How many solutions are there to $n_1^2-n_2^2-n_3^2+n_4^2=k$?

Let us allow $n\_i\in[-N,N]$ for simplicity; this will not change the order of magnitude of the number of solutions. So let us consider $$R(k,N):=\#\left\{(n\_1,n\_2,n\_3,n\_4)\in\{-N,\dotsc,N\}^4:n\_1^2-n\_2^2-n\_3^2+n\_4^2=k\right\}.$$ Let us also assume that $k$ is odd, and $N\gg\sqrt{k}$ with a sufficiently large implied constant. If $r(n)$ denotes the number of ways $n$ can be written as a sum of two squares, then clearly $$\sum\_{m\leq N^2-k}r(m)r(m+k)\leq R(k,N)\leq\sum\_{m\leq 2N^2}r(m)r(m+k).$$ Hence it follows from Theorem 12.5 in Iwaniec: Spectral methods of automorphic forms (2nd ed.) that $$R(k,N)\asymp \Bigl(\sum\_{d\mid k}d^{-1}\Bigr)N^2.$$ The same result would also follow from Theorem 4 in Heath-Brown: A new form of the circle method, and its application to quadratic forms, J. reine angew. Math. 481 (1996), 149-206, upon calculating the singular series $\sigma(x\_1^2-x\_2^2-x\_3^2+x\_4^2,k)$ there. Surely, with more work, an asymptotic formula can also be derived, and the result should be extendable to even $k$ as well.

6

https://mathoverflow.net/users/11919

453048

https://mathoverflow.net/questions/452924

5

Let $p$ be a prime number and let $k$ be a field with $char(k)\neq p$ such that all finite extensions have degree coprime to $p$. (For example, we can take $k=\mathbb{R}$ and $p\neq 2$ or let $k$ the union of $\mathbb{F}\_{l^{(p^n)}}$ for all $n\geq 1$.) Let $C$ be a regular affine curve over $k$ with (regular) projective compactification $\widetilde{C}$. Then my question is as follows: > > Is $\operatorname{Pic}(C)$ a $p$-divisible group? In other words, is every divisor on $C$ linearly equivalent to $pD$ for a divisor $D$ on $C$? > > > In a result I have proven related to strong approximation over function fields, the $p$-divisibilty is one of the necessary and sufficient conditions, so a positive answer here would allow me to simplify the statement. One of the reasons why I suspect the above statement could be true is that the statement is true if $k$ is algebraically closed and the assumption on $k$ implies that $l^\times$ is a $p$-divisible group for all finite extensions $l/k$. If the statement is false, then I would be very interested in a counterexample and under what additional assumptions the statement is true. **My attempt thus far**: If $\widetilde{C}\setminus C$ contains a rational point, or more generally, there exists a divisor of degree $1$ supported in $\widetilde{C}\setminus C$, then the restriction map $\operatorname{Pic}^0(\widetilde{C})\rightarrow \operatorname{Pic}(C)$ is surjective so under this assumption we can reduce the question to > > Is $\operatorname{Pic}^0(\widetilde{C})$ a $p$-divisible group? In other words, is every divisor on $C$ of degree $0$ linearly equivalent to $pD$ for a divisor $D$ on $\widetilde{C}$? > > > If $\widetilde{C}$ is smooth (automatic if $k$ is perfect) and has a rational point, then $\operatorname{Pic}^0(\widetilde{C})$ are the $k$-points of the Jacobian of $\widetilde{C}$. This may allow us to use tools from the theory of abelian varieties. Maybe $A(k)$ is $p$-divisible for any (principally polarized) abelian variety $A$?

https://mathoverflow.net/users/127489

$p$-divisibility of Picard groups

$\newcommand{\wt}{\widetilde}$ $\newcommand{\mr}{\mathrm}$ The question has a positive answer, in fact, regularity of $C$ is not needed. The proof as written below works under the assumption that $C$ is (geometrically) irreducible and reduced, but both these conditions could be relaxed to some extent. (Note that $\mr{Pic}(C)$ depends only on the scheme $C$, not the base field.) Let $\wt{C}$ be a projective compactification of $C$ such that $\wt{C}$ is regular at all points of $\wt{C} \setminus C$; this is possible since we assume $C$ is reduced. Let $J = \mr{Jac}(\wt{C})$, i.e., the identity component of the Picard scheme of $\wt{C}$. Since $p \neq \mr{char}(k)$, the multiplication by $p$ map $[p]:J \to J$ is finite and etale (consider the induced map on the Lie algebra of $J$). The other assumption on $k$ implies that $J(k)$ is $p$-divisible: This follows from the fact that for any finite $G$-module $M$ of order a power of $p$, where $G$ is the absolute Galois group of $k$, $H^1(G, M) = 0$. (This is well-known and follows easily from the inflation-restriction sequence since any finite quotient of $G$ has order prime to $p$.) Applying this to the $G$-module $J(k^{\mr{sep}})[p]$ one sees that $H^1(G, J(k^{\mr{sep}})[p]) = 0$. The short exact sequence of $G$-modules $$ 0 \to J(k^{\mr{sep}})[p] \to J(k^{\mr{sep}}) \stackrel{[p]}{\to} J(k^{\mr{sep}}) \to 0 $$ (for surjectivity we use that $[p]$ is finite etale) gives rise to a long exact sequence of Galois cohomology from which the $p$-divisibility of $J(k)$ follows by the vanishing of $H^1(G, J(k^{\mr{sep}})[p])$. Now I claim that $\mr{Pic}^0(\wt{C})$ is also $p$-divisible: If $L$ is a line bundle of degree $0$ on $\wt{C}$ then it follows from the $p$-divisibility of $J(k)$ that the obstruction to existence of a $p$-th root of $L$ is an element of $\mr{Br}(k)$ of order $p$, but $\mr{Br}(k)[p] = 0$ from the Kummer sequence and the assumption on $k$. Since $\wt{C}$ is regular at all points of $\wt{C} \setminus C$, the restriction map $\mr{Pic}(\wt{C}) \to \mr{Pic}(C)$ is a surjection. Since all the points in $\wt{C} \setminus C$ are of degree prime to $p$ (by the assumption on $k$), it follows that there is an exact sequence $$ \mr{Pic}^0(\wt{C}) \to \mr{Pic}(C) \to Q \to 0 $$ where $Q$ is a cyclic group of order prime to $p$. Then $p$-divisibility for $\mr{Pic}^0(\wt{C})$ and $Q$ implies it for $\mr{Pic}(C)$. Note: If we drop the assumption that $p \neq \mr{char}(k)$, then the above proof still works if $\wt{C}$ is smooth. (In fact, the assumption on $k$ would imply that it is perfect, so regularity implies smoothness.)

3

https://mathoverflow.net/users/519

453055

https://mathoverflow.net/questions/453036

9

We work in ZFC. Let $C(X)$ be the ring of continuous functions $f:X\to\mathbb{R}$, and $M$ a maximal ideal. We call $C(X)/M$ a hyperreal field if it's not isomorphic to $\mathbb{R}$. A field $E$ is called real-closed if it has unique ordering $x\geq 0$ iff $\exists\_{y\in E}\ x = y^2$ and all polynomials of odd degree have a root. An ordered field $E$ is called an $\eta\_1$-field if for all countable $A, B\subseteq E$ such that $A < B$ there is $x\in E$ such that $A < v < B$. All hyperreal fields are real-closed $\eta\_1$-fields. We have the following result: **Theorem.** Assume CH. Then all real-closed $\eta\_1$-fields of size continuum are isomorphic. **Corollary.** Assume CH. Then all hyperreal fields of size continuum are isomorphic. Let P be "All hyperreal fields $C(\mathbb{R})/M$ are isomorphic". I'm wondering if one can get some insight into what kind of statement is P in relation to known set-theoretic principles. Is P independent of ZFC+$\lnot$CH? Above theorem shows that ZFC+CH implies P. Note: The choice of $X = \mathbb{R}$ was arbitrary, to not consider *every* possible topological space $X$ such that $C(X)/M$ is a hyperreal field of size continuum. If this is helpful, please consider a different space $X$ instead.

https://mathoverflow.net/users/150060

Is "All hyperreal fields $C(\mathbb{R})/M$ are isomorphic" independent of ZFC+$\lnot$CH?

In [On ultra powers of Boolean algebras](http://topology.nipissingu.ca/tp/reprints/v09/tp09205.pdf) (Topology Proceedings 9 (1984) 269-291) Alan Dow proved (Corollary 2.3) that $\neg\mathsf{CH}$ implies there are two fields of the form $C(\mathbb{N})/M$ that are not isomorphic (as ordered sets, and hence as ordered fields). To be a bit more precise: Alan constructed two ultrafilters $u$ and $v$ on $\mathbb{N}$ with non-isomorphic ultrapowers. These yield maximal ideals $M\_u$ and $M\_v$ in $C(\mathbb{N})$ whose quotients are (isomorphic to) the respective ultrapowers. And because $\mathbb{N}$ is a closed discrete subspace of the metric space $\mathbb{R}$ these determine maximal ideals $N\_u$ and $N\_v$ in $C(\mathbb{R})$ whose quotients are isomorphic to those by $M\_u$ and $M\_v$ respectively.

13

https://mathoverflow.net/users/5903

453063

https://mathoverflow.net/questions/453059

3

The integral I want to calculate is defined as $$ P(s)=\int\_{-\infty}^{\infty}{\rm d}x\int\_{-\infty}^{\infty}{\rm d}y\ \delta\left(\frac{(x+y)^2+4x^2y^2}{(x+y)^2+(x+y)^4}-s\right)e^{-\left(x^2+y^2\right)/2a^2} $$ This integral seems no explicit solution (?) since the quartic terms in the $\delta$ function. So I just want to find the asymptotic behavior for small and positive $s$, i.e $P(s)$ for $s\to0^+$. I am trying to find the answer by the changing of variables as $x=p+q,y=p-q$, then I have $$ P(s)=2\int\_{-\infty}^{\infty}{\rm d}p\int\_{-\infty}^{\infty}{\rm d}q\ \delta\left(\frac{p^2+(p^2-q^2)^2}{p^2+4p^4}-s\right)e^{-\left(p^2+q^2\right)/a^2}. $$ To get small $s$, the variable $q$ should closed to $\pm p$, and $p$ should be as large as possible. So I directly let $q=p$, neglect the integral over $q$, and I have $$ P(s)\propto\int\_{-\infty}^{\infty}{\rm d}p\ \delta\left(\frac{1}{4p^2}-s\right)e^{-p^2/a^2}, $$ where I used the assumption $p\gg1$ so that $1+4p^2\approx 4p^2$. Using the properties of $\delta$ function, I finally obtain $$ P(s)\propto s^{-2}e^{-1/(2a^2s)}. $$ I am probably wrong, especially for the step that I neglect the integration over $q$. But this is all I can get. Any suggestion will be helpful.

https://mathoverflow.net/users/482984

Asymptotic behavior of the integral that contains $\delta$ function

$$P(s)=2\int\_{-\infty}^{\infty}{\rm d}p\int\_{-\infty}^{\infty}{\rm d}q\ \delta\left(\frac{p^2+(p^2-q^2)^2}{p^2+4p^4}-s\right)e^{-\left(p^2+q^2\right)/a^2}$$ $$\qquad=2\int\_{0}^{\infty}\frac{{\rm d}x}{\sqrt{x}}(x+4x^2)\int\_{0}^{\infty}\frac{{\rm d}y}{\sqrt{y}}\ \delta\left(x+(x-y)^2-s(x+4x^2)\right)e^{-\left(x+y\right)/a^2}$$ $$\qquad=2\int\_{0}^{\infty}\frac{{\rm d}x}{\sqrt{x}}e^{-x/a^2}\frac{x+4x^2}{z\_+-z\_-}\left[\frac{\theta(z\_+)}{\sqrt{z\_+}}e^{-z\_+/a^2}+\frac{\theta(z\_-)}{\sqrt{z\_-}}e^{-z\_-/a^2}\right]\theta\bigl(x-(1-s)/4s\bigr)$$ $$\qquad=\int\_{(1-s)/4s}^{\infty}dx\,e^{-x/a^2}\frac{1+4x}{\sqrt{4 s x+s-1}}\left[\frac{1}{\sqrt{z\_+}}e^{-z\_+/a^2}+\frac{1}{\sqrt{z\_-}}e^{-z\_-/a^2}\right],$$ with $z\_\pm=x\pm\sqrt{x (4 s x+s-1)}$ and $\theta(z)$ the unit step function. In the final equality I assumed $0<s<1/4$. For $s\ll 1$ this approximates to $$P(s)\approx\frac{4}{\sqrt{s}}\int\_{1/4s}^{\infty}dx\,e^{-2x/a^2}\frac{1}{\sqrt{4 s x-1}}=\sqrt{2\pi}(a/s)\exp\left(-\frac{1}{2a^2s}\right).$$

5

https://mathoverflow.net/users/11260

453064

https://mathoverflow.net/questions/453071

3

Let $X$ be a smooth, complex projective variety of dimension $n$. Let $F$ be a vector bundle over $X$ with rank$F=r \leq n$. Let $s$ be a regular global section of $F$ and $Z$ be the scheme of zeroes of $s$ [Here by regular I mean codim$Z$=r]. Then we have the corresponding exact Koszul complex given by : $0 \to \wedge^rF^\vee \to \wedge^{r-1}F^\vee \to \dots \to F^\vee \to \mathcal{O}\_X \to \mathcal{O}\_Z\to 0$ Do we have an analogous sequence if rank$F=r > n$? There is a description in Fulton's Intersection theory book Page $431$, but I'm not sure if that is irrespective of any relation between $r$ and $n$. More precisely, what I want to ask is that for $r>n$ if one has a regular global section $s$ [then $Z = \emptyset$] of $F$, then can one use the first complex mentioned in Fulton's book to conclude that there is a sequence $0 \to \wedge^rF^\vee \to \wedge^{r-1}F^\vee \to \dots \to F^\vee \to \mathcal{O}\_X \to 0$ which is exact on $X-Z=X$? Any suggestion is appreciated.

https://mathoverflow.net/users/187857

Koszul complex for bundles of rank higher than the dimension

Yes, this complex is exact. To prove exactness at point $x \in X$, note that the question is local, so one may assume that $F \cong \mathcal{O}^{\oplus r}$ is a trivial bundle and $s = (s\_1,\dots,s\_r)$ is a sequence of functions. Note further that the Koszul complex can be written as a tensor product $$ \bigotimes\_{i=1}^r(\mathcal{O} \stackrel{s\_i}\to \mathcal{O}). $$ Since $s(x) \ne 0$, we have $s\_i(x) \ne 0$ for some $i$, hence one of the factors above is acyclic at $x$, and therefore so is the tensor product.

7

https://mathoverflow.net/users/4428

453072

https://mathoverflow.net/questions/452598

5

I'm interested in connected matroids $M$ on the ground set $[n]$ for which there is no connected matroid on $[n]$ of the same rank but with a strictly smaller set of bases (by inclusion). Equivalently, the matroid polytope of $M$ has dimension $n-1$ and does not contain any other matroid polytope of dimension $n-1$. Obvious examples are uniform matroids of ranks $1$ and $n-1$. A less trivial example for $n=4$ is a matroid of rank 2 with 5 bases. What is known about such matroids? Is there some classification or interesting alternative characterization?

https://mathoverflow.net/users/19864

"Minimal" connected matroids

In general there does not seem to be a nice classification. You're asking for connected matroids of a given rank which are minimal in the "weak map order" (a weak map between matroids of the same rank is an inclusion of bases). In [Weak maps of combinatorial geometries, Theorem 6.10](https://www.ams.org/journals/tran/1975-206-00/S0002-9947-1975-0371693-2/S0002-9947-1975-0371693-2.pdf), Dean Lucas shows that every connected binary matroid is minimal. In particular, every connected graphic matroid is minimal. I believe there are examples which are not binary.

3

https://mathoverflow.net/users/94696

453074

https://mathoverflow.net/questions/314946

4

Let $k$ be a $p$-adic field, $G$ a connected reductive group over $k$ with minimal parabolic $P\_0$ containing a maximal split torus $A\_0$. Let $W = N\_G(A\_0)(k)/Z\_G(A\_0)(k)$ be the Weyl group, and $S \subset W$ the simple reflections from $P\_0$. For $\theta, \Omega \subset S$, we have the standard parabolic subgroups $P\_\theta, P\_\Omega$. Each double coset $P\_{\theta}w P\_{\Omega}$, for $w \in W$ with a $k$-rational representative, is a locally closed subvariety of $G$, with $P\_\theta wP\_\Omega(k) = P\_\theta(k)wP\_\Omega(k)$. How do we know that the quotient $P\_\theta \backslash P\_\theta wP\_\Omega$ is an algebraic variety over $k$? This variety and its dimension are considered in [Casselman's notes](https://www.math.ubc.ca/~cass/research/pdf/p-adic-book.pdf) on representation theory, Chapter 6. I do not know the general theory of quotients of algebraic group actions which would make sense out of something like this. Once this variety is made sense of, can we say that its $k$-rational points coincides with $P\_\theta(k) \backslash P\_\theta(k) wP\_\Omega(k)$?

https://mathoverflow.net/users/38145

The quotients of double cosets $P_\theta \backslash P_\theta w P_\Omega$ are algebraic varieties over $k$

I originally gave an [answer](https://mathoverflow.net/a/315163) that relied on a claim in [Casselman - Introduction to the theory of admissible representations of $p$-adic reductive groups](https://personal.math.ubc.ca/%7Ecass/research/pdf/p-adic-book.pdf) that, as you pointed out by linking to [Errata for Casselman's unpublished notes](https://mathoverflow.net/questions/306675/errata-for-casselmans-unpublished-notes), is wrong; so I deleted the answer. Here's another try. I hope that I may use $\Theta$ instead of $\theta$ for a subset of the set of simple roots. The natural map $(w^{-1}P\_\Theta w \cap P\_\Omega)\backslash P\_\Omega \to P\_\Theta\backslash P\_\Theta w P\_\Omega$ is an isomorphism of varieties, so $P\_\Theta\backslash P\_\Theta w P\_\Omega$ is a quasi-projective variety. That seems like it must not be what you were asking for; if not, please let me know, so that I can try to fix this part of the answer. Since $w^{-1}P\_\Theta w \cap P\_\Omega$ is smooth, we have that $P\_\Omega(k^\text{sep}) \to ((w^{-1}P\_\Theta w \cap P\_\Omega)\backslash P\_\Omega)(k^\text{sep})$, and hence $P\_\Omega(k^\text{sep}) \to (P\_\Theta\backslash P\_\Theta w P\_\Omega)(k^\text{sep})$, is surjective. Since $(w^{-1}P\_\Theta w \cap P\_\Omega)U\_\Omega$ is a pseudo-parabolic subgroup of $P\_\Omega$, in the sense of [Conrad, Gabber, and Prasad - Pseudo-reductive groups, 2nd edition](https://doi.org/10.1017/CBO9781316092439), Definition 2.2.1, we have that $P\_\Omega(k) \to ((w^{-1}P\_\Theta w \cap P\_\Omega)U\_\Omega\backslash P\_\Omega)(k)$ is surjective by Lemma C.2.1 there. Let $g$ be a point of $(P\_\Theta w P\_\Omega)(k^\text{sep})$ whose image in $(P\_\Theta\backslash P\_\Theta w P\_\Omega)(k^\text{sep})$ is ($k$-)rational. Then there is some $h \in P\_\Omega(k)$ such that $g$ belongs to $(P\_\Theta w U\_\Omega h)(k^\text{sep}) = (P\_\Theta w U\_\Omega)(k^\text{sep})h$. (If you're uncomfortable with pseudo-parabolic subgroups, then it's just a bit of fiddling to handle the genuine parabolic subgroup $w^{-1}P\_\Theta w \cap M\_\Omega$ of $M\_\Omega$ instead.) It remains to show that $U\_\Omega(k) \to (P\_\Theta\backslash P\_\Theta w U\_\Omega)(k)$ is surjective. This is where we come to the [claim](https://personal.math.ubc.ca/%7Ecass/research/pdf/p-adic-book.pdf#page=11) of Casselman that isn't true for the full quotient $P\_\Theta\backslash P\_\Theta w P\_\Omega$—but we've got down to, as it were, ‘the unipotent part’, where it's fine! We have that $P\_\Theta\backslash P\_\Theta w U\_\Omega$ is isomorphic, as a variety, to $(w^{-1}P\_\Theta w \cap U\_\Omega)\backslash U\_\Omega$. Since both $w^{-1}P\_\Theta w \cap U\_\Omega$ and $U\_\Omega$ are smooth, connected, unipotent subgroups of $G$ that are normalized by $M\_\emptyset = C\_G(A\_\emptyset)$, and since the weights of $A\_\emptyset$ on $w^{-1}P\_\Theta w \cap U\_\Omega$, and on $U\_\Omega$, form closed subsets of the relative root system of $G$ with respect to $A\_\emptyset$, each is directly spanned by relative root groups ([Borel - Linear algebraic groups](https://doi.org/10.1007/978-1-4612-0941-6), Proposition 21.9(ii)). In particular, since both sets of roots are divisible (in the sense that, if $a$ is a relative root such that $2a$ is a weight of $A\_\emptyset$ on either group, then $a$ is also a weight of $A\_\emptyset$ on that group), we have that there is a section, *as varieties*, of $U\_\Omega \to (w^{-1}P\_\Theta w \cap U\_\Omega)\backslash U\_\Omega$, hence of $U\_\Omega \to P\_\Theta\backslash P\_\Theta w U\_\Omega$. All together, we have shown that $P\_\Omega(k) \to (P\_\Theta\backslash P\_\Theta w P\_\Omega)(k)$ is surjective, so that we may identity $(P\_\Theta\backslash P\_\Theta w P\_\Omega)(k)$ with $P\_\Theta(k)\backslash P\_\Theta(k)w P\_\Omega(k)$, as you desired.

2

https://mathoverflow.net/users/2383

453078

https://mathoverflow.net/questions/453077

0

Let $\mathbf{G}$ be a connected semisimple algebraic group defined over a field $k$. Let $T$ be a maximal torus of $\mathbf{G}$ defined over $k$, and let $S \subset T$ be a maximal $k$-split torus. Let $\Phi = \Phi(T,\mathbf{G})$ be the absolute root system of $\mathbf{G}$, and let $\Phi\_{\text{rel}} = \Phi(S,\mathbf{G})$ be the relative root system. Let $T^{\ast}$ and $S^{\ast}$ be the groups of characters of $S$ and $T$. The restriction map $T^{\ast} \rightarrow S^{\ast}$ restricts to a surjection $\Phi \sqcup \{0\} \rightarrow \Phi\_{\text{rel}} \sqcup \{0\}$. We can view $\Phi$ and $\Phi\_{\text{rel}}$ as subsets of $T^{\ast} \otimes \mathbb{R}$ and $S^{\ast} \otimes \mathbb{R}$. The root system structures on $\Phi$ and $\Phi\_{\text{rel}}$ come from Euclidean inner products on these Euclidean spaces that are invariant under the actions of the Weyl group and the relative Weyl group. Let's say that I am working with $k$-forms of a semisimple algebraic group that I understand well. I can therefore compute the Euclidean inner product on $T^{\ast} \otimes \mathbb{R}$ as well as the kernel $K$ of the restriction map $T^{\ast} \rightarrow S^{\ast}$. This is enough to determine $\Phi\_{\text{rel}}$ as a subset of $S^{\ast} \otimes \mathbb{R}$. But what I don't know how to do is compute the Euclidean inner product on $S^{\ast} \otimes \mathbb{R}$, which is the last thing I need to actually determine the root system $\Phi\_{\text{rel}}$. Another way of saying what I want to be able to compute is the inner product between elements of $\Phi\_{\text{rel}}$. My guess is that what you do is as follows. Consider elements $r,r' \in \Phi\_{\text{rel}}$. There are then unique lifts $\tilde{r},\tilde{r}' \in T^{\ast} \otimes \mathbb{R}$ that are orthogonal to the kernel $K$, and I suspect that the inner product of $r$ and $r'$ equals the inner product of $\tilde{r}$ and $\tilde{r}'$. **Question**: Is my guess correct, and is there a source that spells it out like this? I apologize for two things: 1. I have never really worked with algebraic groups over fields that are not algebraically closed before, so since I am self-taught in that area I might be writing things in non-standard ways. 2. I know that instead of $\Phi\_{\text{rel}}$ I should write $\Phi$ with the field $k$ as a left subscript. I know several ways to do this in LaTeX, but they all seem to break the in-line LaTeX processor used by MathOverflow.

https://mathoverflow.net/users/510773

Calculating relative root systems

Since $W(G, S)(k) = (N\_G(S)/C\_G(S))(k) = N\_G(S)(k)/C\_G(S)(k)$ is finite, there is what Borel, in [Linear algebraic groups](https://doi.org/10.1007/978-1-4612-0941-6), §14.7, calls an admissible scalar product on $S^\* \otimes\_{\mathbb Z} \mathbb R$, obtained simply by averaging any random scalar (i.e., inner) product over its $W(G, S)(k)$-orbit. Then his Theorem 21.6 says that any such choice of admissible scalar product makes $\Phi\_\text{rel}$ a root system, with Weyl group $W(G, S)(k)$. Per your [comment](https://mathoverflow.net/questions/453077/calculating-relative-root-systems/453080#comment1171972_453080) and my [response](https://mathoverflow.net/questions/453077/calculating-relative-root-systems/453080#comment1171974_453080), if you wish to do this without knowing $W(G, S)(k)$ directly, then Borel’s Corollary 21.4 says that you can just work with the subgroup $W(G, T, S)(k^\text{sep})$ of $W(G, T)(k^\text{sep})$ that stabilises $S$. You can typeset a left subscript in a fairly naïve way, using $\_k\Phi$ `_k\Phi`. This does not work well if it's not at the beginning of a formula, as in $\Phi(G, S) = \_k\Phi$ `\Phi(G, S) = _k\Phi`. I prefer to shut this off with surrounding braces, like $\Phi(G, S) = {\_k\Phi}$ `\Phi(G, S) = {_k\Phi}`, but many people just use an extra atom, like $\Phi(G, S) = {}\_k\Phi$ `\Phi(G, S) = {}_k\Phi`. (A full solution might be something like `\newcommand\lsub[2]{\vphantom{#2}_{#1}#2}`, and then $\newcommand\lsub[2]{\vphantom{#2}\_{#1}#2}\lsub k\Phi$ `\lsub k\Phi`. I suspect that packages like [`leftidx`](https://ctan.org/pkg/leftidx) do something like that, but cleverer.)

3

https://mathoverflow.net/users/2383

453080

https://mathoverflow.net/questions/453091

5

I've read in a couple of different places (a paper and a blog) the following fact: if $F$ is a surface, properly embedded in a three-dimensional handlebody of genus at least two, then $F$ is either compressible or boundary compressible. Neither of these sources included a reference or a proof and I am a bit of a three-manifold newbie. Could somebody please direct me to reference for this fact? Or provide a proof?

https://mathoverflow.net/users/156387

Properly embedded surfaces in handlebodies are compressible or boundary compressible?

Suppose that the surface $F$ is properly embedded in the handlebody $V$. We induct on the genus of $V$. In the base case $V$ has genus zero and so is a three-ball. In this case the desired result follows from a theorem of Alexander. (Briefly, suppose that $F$ is not a sphere or a disk. Sweepout the ball $V$ by planes $(P\_t)$ and track the pattern of intersections with the surface $F$. Then there is at least one time $t'$ where $F \cap P\_{t'}$ contains a compressing curve or a boundary compressing arc.) Suppose that $V$ has genus greater than zero. Suppose that $(D, \partial D) \subset (V, \partial V)$ is a non-separating compressing disk for $V$. Properly isotope $F$ to minimise the intersection between $F$ and $D$. If $F$ is disjoint from $D$ we are done by induction. Suppose instead that $F \cap D$ contains a component $\alpha$ which is either an innermost curve or an outermost arc. If $\alpha$ is not a (boundary-)compression for $F$ in $V$ then there is an isotopy that further reduces $|F \cap D|$, a contradiction.

8

https://mathoverflow.net/users/1650

453100

https://mathoverflow.net/questions/453103

10

I could name on the spot a bunch of abelian categories satisfying AB5 but I cannot think of any that satisfies AB5\*. That is, it should have all limits and the cofiltered limits are exact. Is there any example that I should have in mind?

https://mathoverflow.net/users/496941

Abelian categories satisfying AB5*

The snarky response would be "the opposite category of any of the categories you could name on the spot". The less-snarky response is to observe that some of these are quite natural. For example, $\mathrm{Vect}^{\mathrm{op}}\_{\mathbb{F}\_p}$ agrees with the category of profinite $\mathbb{F}\_p$ vector spaces, since $\mathrm{Vect}\_{\mathbb{F}\_p}$ is the $\mathrm{Ind}$-category of finite $\mathbb{F}\_p$ vector spaces, finite vector spaces are self-dual, and profinite $\mathbb{F}\_p$ vector spaces are the $\mathrm{Pro}$-category of finite vector spaces.

14

https://mathoverflow.net/users/39747

453104

https://mathoverflow.net/questions/453081

8

Let $k=\mathbb{C}$ be the field of complex numbers. I consider the (DG) algebra $A:=k[x]/(x^2)$ such that $\deg(x)=-1$. My question is how to compute the periodic cyclic homology, Hochschild homology and Hochschild cohomology of this (DG) algebra? There are some references on computing those (co)homology of the ring of dual numbers $k[x]/(x^2)$. But in my case, $\deg(x)=-1$. Is there any reference on such computations?

https://mathoverflow.net/users/41650

How to compute the periodic cyclic homology of this algebra

You can use a derived version of the HKR theorem, i.e. $HH(A) = \text{Sym}\_A^\bullet(\mathbb{L}\_A[1])$ where $\mathbb{L}\_A$ is the cotangent complex (over $k$). I'm not sure about a reference though. Since $A$ is quasi-smooth, its cotangent complex will be in degrees $[-1, 0]$, and one can check using the fact that $\text{Spec}(A) = \{0\} \times\_{\mathbb{A}^1} \{0\}$ that $\mathbb{L}\_A \simeq \mathcal{N}^\vee\_{\{0\}/\mathbb{A}^1}|\_A \simeq A[1]$. Then, $HH(A) \simeq \text{Sym}\_A^\bullet(A[2])$ (note that this is in even degree, so as a chain complex it is $\bigoplus\_{n \geq 0} A[2n]$). You can do something similar for Hochschild cohomology using the tangent complex. Now, the Connes $B$-operator is given by the de Rham differential. The module $\mathbb{L}\_A$ is generated by $dx$, and the de Rham differential takes $x$ to $dx$. You can now calculate the various cyclic homologies, and in particular $HP(A) \simeq k(u)$ where $u \in H^2(BS^1; k)$ is the Chern class (note that since $dx$ is in degree $-2$, it behaves like a symmetric variable not an exterior one).

8

https://mathoverflow.net/users/6059

453106

https://mathoverflow.net/questions/453099

1

I'm considering a problem about time varying domain in Chen Wenxiong and Li Congming 's study on $-\Delta u=\exp u$ in $\mathbb R^2$ and $\int\_{\mathbb R^2} \exp u(x) \, d x< +\infty$. LEMMA 1.1 (Ding). If $u$ is a solution of $-\Delta u=\exp u$ in $\mathbb R^2$ and $\int\_{\mathbb R^2} \exp u(x) \, d x<$ $+\infty$, then $\int\_{\mathbb R^2} \exp u(x) \, d x \geqslant 8 \pi$. Proof. For $-\infty<t<\infty$ let $\Omega\_t=\{x \mid u(x)>t\}$; one can obtain $$ \begin{aligned} \int\_{\Omega\_t} \exp u(x) \, d x & =-\int\_{\Omega\_t} \Delta u=\int\_{\partial \Omega\_t}|\nabla u| \,d s \\ -\frac{d}{d t}\left|\Omega\_t\right| & =\int\_{\partial \Omega\_t} \frac{d s}{|\nabla u|} . \end{aligned} $$ By the Schwarz inequality and the isoperimetric inequality, $$ \int\_{\partial \Omega\_t} \frac{d s}{|\nabla u|} \cdot \int\_{\partial \Omega\_t}|\nabla u| \geqslant\left|\partial \Omega\_t\right|^2 \geqslant 4 \pi\left|\Omega\_t\right| . $$ Hence $$ -\left(\frac{d}{d t}\left|\Omega\_t\right|\right) \cdot \int\_{\Omega\_t} \exp u(x) \, d x \geqslant 4 \pi\left|\Omega\_t\right| $$ and so $$ \frac{d}{d t}\left(\int\_{\Omega\_t} \exp u(x)\, d x\right)^2=2 \exp t \cdot\left(\frac{d}{d t}\left|\Omega\_t\right|\right) \cdot \int\_{\Omega\_t} \exp u(x)\, d x \leqslant-8 \pi \exp t \cdot\left|\Omega\_t\right| . $$ Integrating from $-\infty$ to $\infty$ gives $$\tag{1} -\left(\int\_{\mathbb R^2} \exp u(x) \,d x\right)^2 \leqslant-8 \pi \int\_{\mathbb R^2} \exp u(x) \,d x $$ which implies $\int\_{\mathbb R^2} \exp u(x) \,d x \geqslant 8 \pi$ as desired. **The proof details before (1) is presented by the discussions between me and other users in [Differentiation in time varying domain](https://math.stackexchange.com/questions/4528507/differentiation-in-time-varying-domain/4677803#4677803), I'm confused about the last step, how they integrate from $-\infty$ to $\infty$?** In detail, for the LHS, how to compute the limit of $\lim\_{t \rightarrow +\infty} \int\_{\Omega\_t} \exp u(x) \, d x$ and $\lim\_{t \rightarrow -\infty} \int\_{\Omega\_t} \exp u(x)\, d x$, I think $\lim\_{t \rightarrow -\infty} \int\_{\Omega\_t} \exp u(x)\, d x$ must be $\int\_{\mathbb R^2} \exp u(x) \,d x$ but I don't know how to deal with $\lim\_{t \rightarrow +\infty} \int\_{\Omega\_t} \exp u(x) \,d x$, from the fact that $\lim\_{t \rightarrow -\infty} \int\_{\Omega\_t} \exp u(x) \,d x = \int\_{\mathbb R^2} \exp u(x) \,d x$ which is bounded and $\frac{d}{d t}\left(\int\_{\Omega\_t} \exp u(x) \,d x\right)^2=2 \exp t \cdot\left(\frac{d}{d t}\left|\Omega\_t\right|\right) \cdot \int\_{\Omega\_t} \exp u(x) \,d x \leqslant 0 $ we know that when $t \rightarrow +\infty$, $\lim\_{t \rightarrow +\infty} \int\_{\Omega\_t} \exp u(x) \, d x$ is also bounded, but based on the proof, it seems that it is 0. For the RHS, how to integrate $\exp t \cdot\left|\Omega\_t\right|$ from $-\infty$ to $\infty$? I try to use Reynolds transport theorem \begin{align} & \frac{\mathrm{d}}{\mathrm{d} t} \int\_{\Omega(t)} \mathbb{A}(\boldsymbol{x}, t)\, \mathrm{d}^n \boldsymbol{x} \\[8pt] = {} & \int\_{\partial \Omega(t)}(\boldsymbol{V}(\boldsymbol{x}, t) \cdot \boldsymbol{n}(\boldsymbol{x}, t)) \mathbb{A}(\boldsymbol{x}, t) \,\mathrm{d}^{n-1} \boldsymbol{x}+\int\_{\Omega(t)} \partial\_t \mathbb{A}(\boldsymbol{x}, t) \,\mathrm{d}^n \boldsymbol{x} \end{align} I set $A\_1=e^u$ and $A\_2=e^t$, because on $\partial \Omega(t)$ we have $u=t$ so we have $$\exp t \cdot\left|\Omega\_t\right|= \int\_{\Omega t} e^t \,d x =\frac{d}{d t} \int\_{\Omega t} e^t \,d x - \frac{d}{d t}\int\_{\Omega t} e^u \,d x $$ Now we have to face another question which is $\lim\_{t \rightarrow +\infty} \exp t \cdot\left|\Omega\_t\right|$ and $\lim\_{t \rightarrow -\infty} \exp t \cdot\left|\Omega\_t\right|$

https://mathoverflow.net/users/469129

Time varying domain in Chen Wenxiong and Li Congming 's study on $-\Delta u=\exp u$ in $\mathbb R^2$ and $\int_{\mathbb R^2}\exp u(x) \, dx< +\infty$

I think your question is just basic real analysis, not at the research level, you should ask this question on stack exchange instead of overflow. Since $\int\_{\mathbb{R}^2} e^u \, dx <\infty$, for every $\epsilon>0$, there exists $\delta>0$ such that if $|A|<\delta$, then $\int\_A e^u \, dx <\epsilon$. By Chebyshev's Inequality, it is easy to see that $|\Omega\_t|\to 0$ as $t\to +\infty$. So $\lim\_{t\to 0} \int\_{\Omega\_t} e^u \, dx=0$. On the other hand, by the description of the $L^1$ norm in terms of the distribution function, we have $$\int\_{\mathbb{R}^2} e^u \, dx =\int\_0^\infty |\Omega\_{\ln s}| \, ds =\int\_{-\infty}^{+\infty}e^t|\Omega\_{t}| \, dt. $$ In the last step, we use the formula of change of variables by setting $s=e^t.$

2

https://mathoverflow.net/users/166368

453107

https://mathoverflow.net/questions/453133

3

### Problem I am trying to invert an equation of the form: $R(l\_0)=(\int\_{0}^{l\_0} \rho(x) \, dx)(\int\_{l\_0}^{l} \rho(x) \, dx)$ where $0\leq l\_0 \leq l$ I.e. I want to find $\rho(x)$ given $R(l\_0)$ via some kind of integral transform, series solution, or discretized approximation. What I do not want to do is just blind curve fitting. Everything is real and positive. Is this possible, and if so, what should technique should I use for a solution? I suspect it may be helpful to convert this into some kind of product of two integral transforms where the kernels are Heaviside step functions. That might look like this: $R(l\_0)=(\int\_{0}^{l} K\_1(x,l\_0)\rho(x) \, dx)(\int\_{0}^{l} K\_2(x,l\_0)\rho(x) \, dx)$ where $K\_1(x,l\_0) = H(l\_0-x)$ and $K\_2(x,l\_0) = H(x-l\_0)$. ### What I have done so far: * I have looked in the [*Handbook of Integral Equations*](https://hupaa.com/book/HANDBOOK_OF_INTEGRAL_EQUATIONS.pdf) and think it has to do with difference kernels on finite intervals, but none of their integrals take the form of the product of two integral equations, even in the nonlinear sections as far as I can tell. * I have tried to find double integrals with two kernels like in [this paper](https://www.jstor.org/stable/2037988?seq=1), but it seems like this is just proving the existence of something and not actually solving anything. Is this better to solve? $R(l\_0)=\int\_{0}^{l}\int\_{0}^{l}K\_1(x,l\_0)\rho(x)K\_2(y,l\_0)\rho(y) dx dy$ * Circular convolution looked promising, but I cannot see how I might apply it here. * I have seem some sources that convert integral equations into differential equations, but the product rule seems to introduce additional problems here ### Motivation The problem motivation is simple can be understood as follows: You have a loop of wire of length $l$ with resistivity along it $\rho(x)$. Without cutting the wire you want to figure out $\rho(x)$ using only a multimeter. Assume that $\int\_{0}^{l}\rho(x) = 1$ (i.e, the total resistance if you did cut the wire is 1). The only way you can take measurements is by probing it with an ohmmeter at $0$ and $l\_0$, which gives $R(l\_0)$. I am quite new to functional analysis, so any other tips on even approximation techniques would be appreciated. I suspect but cannot prove a solution to this exists at all, so steps toward proof that there is no unique solution for a given $R(l\_0)$ would also be helpful.

https://mathoverflow.net/users/510808

How to find the inverse of a product of two integral equations

Let $F(y) = \int\_0^y \rho(x) dx$. Then your equation reduces to: $$R(y) = F(y)(F(l) - F(y))$$ Note that for any value of $F(y)$, we get that $R(y) \leq \frac{F(l)^2}{4}$; this is easily proven by completing the square. As $F$ is an antiderivative, it is continuous, so by the intermediate value theorem (using $F(0) \leq \frac{F(l)}{2} \leq F(l)$) there is some $c$ such that $F(c) = \frac{F(l)}{2}$. Then $R(c) = \frac{F(l)^2}{4}$. Therefore $\max\_{y \in [0, l]} R(y) = \frac{F(l)^2}{4}$, so $F(l) = 2 \sqrt{\max\_{x \in [0, l]} R(x)}$ With $F(l)$ known, we can now solve for $F(y)$ easily: $F(y) = \frac{F(l) \pm \sqrt{F(l)^2 - 4 R(y)}}{2}$; we can easily check whether to use the positive or negative square root by whether $x$ is greater than or less than $c$. Finally, $\rho(x) = F'(x) = \frac{|R'(x)|}{2\sqrt{\left(\max\_{x \in [0, l]} R(x)\right) - R(x)}}$, as we have assumed that $\rho(x)$ is positive for all $x$.

4

https://mathoverflow.net/users/44191

453139

https://mathoverflow.net/questions/453122

2

I would like to prove the following but I couldn't manage to do it. Let $a>b>0$ be two real numbers. Let $f$ be the function defined as: $$\forall \sigma>0, ~\forall x\in\mathbb{R},~f\_\sigma(x):=\sigma e^{-x^2/\sigma^2}+\sqrt{\pi}x\text{erf}\left(\frac{x}{\sigma}\right).$$ **I would like to show that $f\_a-f\_b$ is log-concave.** One can see $f\_\sigma$ as the function that, differentiated two times, gives the gaussian function (with a scaling factor) of mean $0$ and deviation $\sigma$. More formally, one can compute the second derivative and show that: $$f\_\sigma''(x) = \frac{e^{-x^2/\sigma^2}}{2\sigma}.$$ Numerically, I am pretty sure that this statement is true. I tried to compute the second derivative of the $\log(f\_a-f\_b)$ but without sucess (the derivatives are too diffucult to analyse). Any hints or solutions will be highly appreciated! Thank you.

https://mathoverflow.net/users/510079

Log-concavity of the difference of the second anti-derivative of Gaussians

This conjecture is true. Indeed, for $h:=\ln(f\_a-f\_b)$ and real $x\ne0$ we have \begin{equation} h'(x)=R(x):=\frac{F(x)}{G(x)}, \end{equation} where \begin{equation} F(x):=\sqrt{\pi } \left(\text{erf}\left(\frac{x}{a}\right)-\text{erf}\left(\frac{x}{b}\right)\right), \end{equation} \begin{equation} G(x):= a e^{-x^2/a^2}+\sqrt{\pi } x\, \text{erf}\left(\frac{x}{a}\right)-b e^{-x^2/b^2}-\sqrt{\pi } x \, \text{erf}\left(\frac{x}{b}\right); \end{equation} note that $G(0)=a-b>0$ and $G'(x)=\sqrt{\pi } \left(\text{erf}\left(\frac{x}{a}\right)-\text{erf}\left(\frac{x}{b}\right)\right)>0$ for real $x>0$, so that $G>0$ on $[0,\infty)$. Since the function $h$ is smooth and even, it is enough to show that $R$ is decreasing on $(0,\infty)$. In what follows, $a>b>0$ and $x>0$, unless otherwise indicated. Let \begin{equation} R\_1(x):=\frac{F'(x)}{G'(x)}, \end{equation} \begin{equation} R\_2(x):=\frac{F''(x)}{G''(x)} = -\frac{2 x \left(a^3 e^{x^2/a^2}-b^3 e^{x^2/b^2}\right)}{a^2 b^2 \left(a e^{x^2/a^2}-b e^{x^2/b^2}\right)}; \end{equation} note that $R\_2(x)$ is undefined at $x=x\_{a,b}$, where \begin{equation} x\_{a,b}:=a b \sqrt{\frac{\ln a-\ln b}{a^2-b^2}}. \end{equation} Then \begin{equation} \begin{aligned} H(t)&:=R'\_2(x)\frac{1}{2} a^3 b^3 \left(a e^{x^2/a^2}-b e^{x^2/b^2}\right)^2 \\ & =a^4 \left(b^2-2 t\right)-a b^5 e^{\left(\frac{1}{b^2}-\frac{1}{a^2}\right)t}+a^2 \left(b^4+4 b^2 t\right)-a^5 b e^{\left(\frac{1}{a^2}-\frac{1}{b^2}\right)t}-2 b^4 t, \end{aligned} \end{equation} where $t:=x^2>0$, and \begin{equation} H'(t)\frac{ab}{a^2-b^2}=H\_1(u):=\frac{a^4}{u}-2 a^3 b+2 a b^3-b^4 u, \end{equation} where $u:=e^{\left(\frac{1}{a^2}-\frac{1}{b^2}\right)t}>1$. The only root $u$ of the equation $H\_1(u)=0$ that may be positive is \begin{equation} u\_{a,b}:=\frac{a b^3-a^3 b+\sqrt{a^6 b^2-a^4 b^4+a^2 b^6}}{b^4}. \end{equation} Replacing now $e^{\left(\frac{1}{a^2}-\frac{1}{b^2}\right)t}$ and $e^{\left(\frac{1}{b^2}-\frac{1}{a^2}\right)t}$ in the expression for $H(t)$ by $u\_{a,b}$ and $1/u\_{a,b}$, respectively, we see that the condition $H'(t)=0$ for some real $t>0$ will imply \begin{equation} \begin{aligned} H(t)&=a^4 \left(-\frac{b^4}{-a^2+\sqrt{a^4-a^2 b^2+b^4}+b^2}+2 b^2-2 t\right) \\ &-a^2 b^2 \left(\sqrt{a^4-a^2 b^2+b^4}-4 t\right)-2 b^4 t, \end{aligned} \end{equation} which is a rather simple algebraic expression, which is actually $<0$ (still assuming $a>b>0$ and $t>0$). So, $H(t)<0$ at any critical point $t>0$ of $H$. Also, $H(0)=-a (a - b)^2 b (a^2 + a b + b^2)<0$ and $H(\infty-):=\lim\_{t\to\infty}H(t)=-\infty$. So, $H(t)<0$ for all real $t\ge0$. So, $R\_2$ is decreasing on $(0,x\_{a,b})$ and on $(x\_{a,b},\infty)$. Also, $R\_1>0$ on $(0,x\_{a,b})$, $R\_1(x\_{a,b})=0$, $R\_1<0$ on $(x\_{a,b},\infty)$, $R\_1(0+)=\infty$, and $R\_1(\infty-)=-\infty$. So, by [Table 1.1](https://eudml.org/doc/128522), $R\_1$ is decreasing on $(0,\infty)$. Also, $R'(0)=-\frac2{ab}<0$ and $R(\infty-)=-\infty$. So, again by [Table 1.1](https://eudml.org/doc/128522), $R$ is decreasing on $(0,\infty)$. $\quad\Box$

2

https://mathoverflow.net/users/36721

453145

https://mathoverflow.net/questions/453123

39

One of the cornerstones of the mathematical formulation of General Relativity (GR) is the result (due to Choquet-Bruhat and others) that the initial value problem for the Einstein field equations is solvable and it admits a essentially unique maximal (globally hyperbolic) development (GHD). The proofs I have seen of this all seem to make use of the Axiom of Choice (AC) in a nontrivial way (usually, Zorn's lemma is applied twice: first to prove the existence of a setwise maximal solution, then to prove the existence of a common GHD given two solutions, from which one then infers the result). Hence the question as in the title: how much AC is really needed for all of this? The question can be interpreted mainly in two ways: 1. Literally. Since a lot of analysis is dependent from some form of AC, and a few results from PDE theory are needed for the theorem, I'm skeptical that this interpretation has an easy answer: backtracking all the applications of AC seems unfeasible. 2. Restricting the question to "natural scenarios": maybe one needs the full force of AC to prove the theorem for completely general initial data, but restricting to a natural class of data it turns out that we can make explicit choices in the arguments necessary. Any reference, observation or comment is welcome. EDIT: It seems that Zorn's lemma is not really needed after all. I'm fairly convinced by now that the complete theorem can be proved in ZF+DC (i.e. that all the analysis needed for the theorem can be done in that context) but also that actually checking everything to see that this is the case would be a painstakingly long process. I am still very interested in the absoluteness approach hinted at by Gro-Tsen, Dorais, Chow and Hanson, but that is material for a new question.

https://mathoverflow.net/users/493244

How much of mathematical General Relativity depends on the Axiom of Choice?

The dependence on AC through the use of Zorn's lemma in the proof of the Choquet-Bruhat–Geroch theorem on the existence of a maximal globally hyperbolic development for solutions of the Einstein equations has been eliminated around the same time by Jan Sbierski, in [On the Existence of a Maximal Cauchy Development for the Einstein Equations — a Dezornification](https://arxiv.org/abs/1309.7591), and Willie Wong, in [A comment on the construction of the maximal globally hyperbolic Cauchy development](https://arxiv.org/abs/1310.1318) (the latter having been mentioned by Asaf Karagila in the [comments](https://mathoverflow.net/questions/453123/how-much-of-mathematical-general-relativity-depends-on-the-axiom-of-choice#comment1172084_453123)). More generally, an analogous, but non-exhaustive, discussion has appeared earlier over at [M.SE10102](https://math.stackexchange.com/q/10102 "Where do we need the axiom of choice in Riemannian geometry?") and [MO45928](https://mathoverflow.net/q/45928 "Does Arzelà-Ascoli require choice?") in the context of Riemannian geometry. The conclusion seems to be that several fundamental results in functional analysis anyway use the axiom of choice or some version of it. Mathematical GR certainly uses the theory of linear (and nonlinear) elliptic and hyperbolic PDEs, which in particular relies on the theory of Sobolev spaces (and possibly also Schwartz distributions). The basic results where (to my knowledge) AC creeps in include * the Hahn–Banach theorem, * the countable additivity of Lebesgue measure (may be used in Sobolev theory), * the Arzelà–Ascoli and Fréchet–Kolmogorov theorems (used in Sobolev embeddings), * the Banach–Alaoglu theorem (as a step in applying Schauder-like fixed-point theorems), * the uniform boundedness principle, * and maybe others. Most likely, the list is non-exhaustive. From the notes [Zornian Functional Analysis or: How I Learned to Stop Worrying and Love the Axiom of Choice](https://arxiv.org/abs/2010.15632) by Asaf Karagila and [the answer by Cloudscape](https://mathoverflow.net/a/218133/2622) to [MO45928](https://mathoverflow.net/q/45928 "Does Arzelà-Ascoli require choice?"), it seems that there are versions of these theorems that do not require AC under some hypothesis (like restricting to separable Banach spaces) or only a weakend version of it (like countable choice). I guess that it might be an open question, whether all of this dependence on AC could be eliminated under sufficiently reasonable hypothesis (say sufficient for the mathematical study of gravitational waves from astrophysical sources). If one could write a reasonable textbook on PDE theory without AC, then probably one could adapt the same methods to Mathematical GR. For reference, below is the currently functional link to the online database of the consequence of AC. Not all the above theorems appear there by name, but maybe only in some equivalent version (cf. the previously mentioned [M.SE](https://math.stackexchange.com/q/10102 "Where do we need the axiom of choice in Riemannian geometry?") and [MO](https://mathoverflow.net/q/45928 "Does Arzelà-Ascoli require choice?") discussions). * <https://cgraph.inters.co/>

32

https://mathoverflow.net/users/2622

453152

https://mathoverflow.net/questions/275409

8

I asked this question on [math.stackexchange](https://math.stackexchange.com/q/2348425/275190), but without much success. Assume that $R$ is a ring (commutative, with unit) and that $M$, $N$ are two $R$-modules. Let $(e\_i)\_{i\in I}$ and $(f\_j)\_{j\in J}$ be families of elements in $M$ and $N$, respectively. In $M\otimes\_RN$, consider the family $$(e\_i\otimes f\_j)\_{(i,j)\in I\times J}\;.$$ It is not difficult to prove that if $(e\_i)$ and $(f\_j)$ are generating families, then $(e\_i\otimes f\_j)$ generates $M\otimes\_RN$. And if both are bases, then so is $(e\_i\otimes f\_j)$. However, it does not follow immediately that if $(e\_i)$ and $(f\_j)$ are linearly independent families (not necessarily generating), then so is $(e\_i\otimes f\_j)$. My question is therefore, does anybody know a counterexample? Or is it true and there is a proof? Note: If $R$ was a field, then one can extend $(e\_i)$ and $(f\_j)$ to bases and see that $(e\_i\otimes f\_j)$ becomes part of a basis. If $R$ is an integral domain mit quotient field $\Bbbk=Q(R)$, then $(e\_i)$ and $(f\_j)$ remain linear independent in $M\otimes\_R\Bbbk$, $N\otimes\_R\Bbbk$, and we are in the situation above. So $R$ has to be "weird" enough to start with.

https://mathoverflow.net/users/70808

linear independent families in a tensor product

[This answer](https://mathoverflow.net/a/232343/22989) to a related question gives a way of constructing counterexamples. For a similar but more concrete example, let $k$ be a field and $R=k[x,y]/(x^2,xy,y^2)$, so $R$ is a $3$-dimensional algebra over $k$, spanned by $1$, $x$ and $y$. Let $M$ be the $5$-dimensional module with basis $\{a,b,c,d,e\}$, where $x$ acts by $$ax=0, bx=d, cx=e, dx=0, ex=0$$ and $y$ acts by $$ay=d, by=e, cy=0, dy=0, ey=0.$$ Then $\operatorname{Ann}\_R(b)=\{0\}$, so $\{b\}\subset M$ is linearly independent. However, in $M\otimes\_RM$, $$(b\otimes b)x=(bx)\otimes b=d\otimes b=(ay)\otimes b=a\otimes (by)=a\otimes e=a\otimes(cx)=(ax)\otimes c=0\otimes c=0,$$ so $\{b\otimes b\}\subset M\otimes\_RM$ is not linearly independent.

6

https://mathoverflow.net/users/22989

453155

https://mathoverflow.net/questions/453167

2

Let $\mathcal{C}$ be a small category and let $\mathcal{M} = \operatorname{sPre}(C)$ be the model category of simplicial presheaves on $\mathcal{C}$ with the injective model structure. Let $S$ be a set of morphisms in $\mathcal{M}$ that is stable under pullback and let $L\_S\mathcal{M}$ be the left Bousfield localization of $\mathcal{M}$ with respect to $S$. By [this nlab page](https://ncatlab.org/nlab/show/locally+cartesian+closed+model+category), $\mathcal{M}$ is a locally cartesian closed model category, and any right proper left Bousfield localization of $\mathcal{M}$ is as well. Is this true for any other left Bousfield localization $L\_S\mathcal{M}$? If not, is it true when $S$ is (the saturation of) the set of covering sieves $R \hookrightarrow j(X)$ for some Grothendieck topology $t$ on $\mathcal{C}$, where $j(X)$ is the simplicial Yoneda embedding applied to $X \in \mathcal{C}$ (note that the local objects of $L\_S\mathcal{M}$ are simplicial presheaves satisfying $t$-descent but not necessarily $t$-hyperdescent)? This is really the case I care about. If not, is there a counterexample? It should be noted that $L\_S\mathcal{M}$ is at least Quillen equivalent to a locally cartesian closed model category because the $\infty$-category it presents is locally cartesian closed by a result of Hoyois.

https://mathoverflow.net/users/133676

Is a left Bousfield localization of simplicial presheaves a locally cartesian closed model category?

Section 2 of [this paper of Rezk](https://arxiv.org/abs/0901.3602) addresses exactly the question of when the localization by S yields a Cartesian model category. For that the relevant property is that that if you take the product of a map in S and a representable object, the result must be in the saturation of S. There are examples of S which fail this condition where the resulting localization is not a Cartesian model category. (An example is the n-fold Segal space model category constructed by Barwick - it is a simplicial model category but not Cartesian). For *local* Cartesian-ness you are essentially talking about Cartesian-ness in the over categories $\mathcal{M}\_{/X}$. There the same argument works where stability under products with representables is replaced with stability under fiber products of representables over X. So if your set of morphisms S is stable under all pullbacks, as mentioned in the OP, that is sufficient to give the local cartesian model category structure on $L\_S\mathcal{M}$.

4

https://mathoverflow.net/users/184

453176

https://mathoverflow.net/questions/453175

1

Fix $p \in [1, \infty)$. Let $(L^p (\mathbb R^d), \|\cdot\|\_{L^p})$ be the Lesbesgue space of $p$-integrable real-valued functions on $\mathbb R^d$. Let $\tilde L^p (\mathbb R^d)$ be the space of Lebesgue measurable functions $f:\mathbb R^d \to \mathbb R$ such that $$ \|f\|\_{\tilde L^p} := \sup\_{x \in \mathbb R^d} \|1\_{B(x, 1)} f\|\_{L^p} < \infty, $$ where $B(x, 1)$ is the open unit ball centered at $x$. I have [verified](https://math.stackexchange.com/questions/4750669/the-space-lp-textloc-mathbb-rd-is-complete-w-r-t-the-norm-f) that $(\tilde L^p (\mathbb R^d), \|\cdot\|\_{\tilde L^p})$ is complete. Let $(x\_m)$ be a countable dense subset of $\mathbb R^d$. The sphere has Lebesgue measure $0$. Then by dominated convergence theorem (DCT), $$ \|f\|\_{\tilde L^p} = \sup\_{m \in \mathbb N} \|1\_{B(x\_m, 1)} f\|\_{L^p} \quad \forall f \in \tilde L^p (\mathbb R^d). $$ I would like to ask if a version of DCT holds for $L^p (\mathbb R^d)$, i.e., > > Let $f\_n, g \in \tilde L^p (\mathbb R^d)$ such that $|f\_n| \le g$ a.e. for all $n$. If there is $f:\mathbb R^d \to \mathbb R$ such that $f\_n \to f$ a.e., then $f \in \tilde L^p (\mathbb R^d)$ and $\|f\_n-f\|\_{\tilde L^p} \to 0$. > > > Thank you so much for your elaboration!

https://mathoverflow.net/users/99469

Is there a version of dominated convergence theorem for local $L^p$ spaces?

A counterexample is given by $f\_n=1\_{(n,\infty)}$, $g=1$, and $f=0$. Then all the conditions on $f\_n,g,f$ hold, but $\|f\_n-f\|\_{\tilde L^p} \not\to0$.

5

https://mathoverflow.net/users/36721

453180

https://mathoverflow.net/questions/453185

6

Given $f \in L^1 (\mathbb R^d)$, and $\varepsilon > 0$, define the *minimal function* $m\_\varepsilon f$ by $$m\_\varepsilon f(x) := \inf\_B \frac1{|B|} \int\_B |f| ,$$ where the infimum is taken over all balls $B$ containing $x$ of radius less than or equal to $\varepsilon$, and the integral is with respect to Lebesgue measure. **Question:** Is it true that for every $\varepsilon > 0$, there exists a constant $C\_\varepsilon > 0$ depending only on $\varepsilon$ and the dimension $d$ such that for all $f \in L^1(\mathbb R^d)$, $$\|m\_\varepsilon f\|\_{L^1} \geq C\_\varepsilon \|f\|\_{L^1}\text{?}$$ Further, is it true that the optimal constants $C\_\varepsilon$ converge to $1$ as $\varepsilon \to 0$? *Remark: By the Lebesgue differentiation theorem we have $|m\_\varepsilon f| \leq |f|$ a.e., and hence $\|m\_\varepsilon f\|\_{L^1} \leq \|f\|\_{L^1}$. Thus if the conjectured result is true then we would have $m\_\varepsilon f \sim\_\varepsilon f$, with scale tending to $1$ as $\varepsilon \to 0$.*

https://mathoverflow.net/users/173490

Is the Hardy Littlewood “minimal function” comparable to the original function in $L^1$ norm?

This is not true. Take $\epsilon=1$ and, on the real line, $f\_\delta=\delta^{-1} \chi\_{(0, \delta)}$, so that $\|f\_\delta\|\_1=1$. Then $0 \leq m\_1 f\_\delta \leq \chi\_{(0, \delta)}$, by choosing small balls outside $(0, \delta)$ and balls of radius 1 otherwise. Thus $\|m\_1 f\_\delta\|\_1 \leq \delta$.

4

https://mathoverflow.net/users/150653

453189

https://mathoverflow.net/questions/452544

3

**Problem:** Let $f\colon \mathopen[0,1\mathclose] \to \mathbb{R}$ be defined as $$ f(x) = \frac{e^{\rho x}-1}{e^{\rho x}-1+e^{\rho (1-\gamma x)}-e^{\rho (1-\gamma) x}} $$ where $\rho$ and $\gamma$ are strictly positive real parameters. I am interested in showing that, for any fixed $\rho$, there exists a value $\gamma\_{\rho}$ such that for any $\gamma\geq \gamma\_{\rho}$, the function $f$ is first convex on an interval $[0,\bar{x}[$ and then concave on $[\bar{x},1]$, where $\bar{x}$ is the inflection point of $f$. This essentially boils down to showing that $f''$ is positive before $\bar{x}$ and negative afterwards. **What I have tried:** I observed that for all $x \in \mathopen]0,1\mathclose[$, the function $f$ satisfies the differential equation $$ f'(x) = \alpha(x) f(x) (1-f(x)) $$ where $$ \alpha(x) = \rho \left( \frac{1}{e^{\rho x}-1} + \frac{1}{e^{\rho (1-x)}-1} + (1+\gamma) \right). $$ From this, I deduced that $$ f(x) = \frac{f(x\_{0})}{ f(x\_{0}) + \big( 1-f(x\_{0}) \big) \, e^{ -\int\_{x\_{0}}^{x} \alpha(s) \mathrm{d}s } } $$ for some $x\_{0} \in \mathopen]0,1\mathclose[$. Consequently, $f''(x)$ has the same sign as $$ \frac{\sigma''(A(x))}{\sigma'(A(x))} + \frac{\alpha'(x)}{\alpha(x)^2} $$ where $$ \sigma(x) = \frac{f(x\_{0})}{ f(x\_{0}) + ( 1-f(x\_{0}) ) e^{ -x} } $$ is the sigmoid function (which is a convex-concave function) and $$ A(x) = \int\_{x\_{0}}^{x} \alpha(s) \mathrm{d}s. $$ The challenge is that the signs of $\frac{\sigma''(A(x))}{\sigma'(A(x))}$ and $\frac{\alpha'(x)}{\alpha(x)^2}$ go in opposite directions, making it difficult to determine the sign of $f''$. A similar problem arises when you directly take the second derivative of $f$. **Question:** How can we determine the sign of $f''(x)$ and thus establish the regions of convexity and concavity of $f$? Are there any other techniques or insights that can help navigate this problem?

https://mathoverflow.net/users/510325

How to establish regions of convexity/concavity of a ratio of exponential polynomials?

This conjecture is true. Indeed, letting $r:=\rho$, $a:=\gamma$, $t:=rx$, and $u:=e^r-1$, we see \begin{equation\*} f(x)=R(t):=\frac{e^t-1}{(1+u)e^{-a t}-e^{t-a t}+e^t-1}. \tag{10}\label{10} \end{equation\*} So, the problem can be restated as follows: > > Show that for each real $u>0$ there is some real $a\_u>0$ such that for each $a\ge a\_u$ there is some $t\_{u,a}\in(0,\ln(1+u))$ such that $R''\ge0$ on $[0,t\_{u,a}]$ and $R''\le0$ on $[t\_{u,a},\ln(1+u)]$. > > > In what follows, it is assumed that $u>0$, $a$ is a large enough (depending on $u$) positive real number, and $t\in(0,\ln(1+u)]$ -- unless otherwise specified. Note that $(1+u)e^{-a t}-e^{t-a t}+e^t-1>e^{-a t}-e^{t-a t}+e^t-1=(e^t-1)(1-e^{-a t})>0$, so that \eqref{10} makes sense. We have \begin{equation\*} \begin{aligned} g(t)&:=R''(t)e^{-a t} \left(-e^{a t}+e^{a t+t}-e^t+u+1\right)^3 \\ &=-a^2 (u+1) e^{a t}+e^{a t+2 t} \left(-\left(a^2 (u+3)\right)-2 a u-u\right) \\ &+e^t (u+1) \left(a^2 (u+3)+2 a u+u\right)+e^{2 t} \left(-\left(a^2 (2 u+3)\right)-2 a u+u\right) \\ &+e^{a t+t} \left(a^2 (2 u+3)+2 a u-u\right)+a^2 e^{3 t}+a^2 e^{a t+3 t}-a^2 (u+1)^2. \end{aligned} \end{equation\*} Note that $-e^{a t}+e^{a t+t}-e^t+u+1>-e^{a t}+e^{a t+t}-e^t+1=(e^t-1)(e^{a t}-1)>0$, so that $g(t)$ equals $R''(t)$ in sign. So, the problem can be further restated as follows: > > Show that for each real $u>0$ there is some real $a\_u>0$ such that for each $a\ge a\_u$ there is some $t\_{u,a}\in(0,\ln(1+u))$ such that $g\ge0$ on $[0,t\_{u,a}]$ and $g\le0$ on $[t\_{u,a},\ln(1+u)]$. > > > To prove this, kill the exponentials of the form $e^{ct}$ for constant $c$'s one by one, by letting \begin{equation\*} D(t):=\frac{g'(t)}{e^t},\quad D\_2(t):=\frac{D'(t)}{e^t},\quad D\_3(t):=\frac{D\_2'(t)}{e^t}, \end{equation\*} \begin{equation\*} D\_4(t):=\frac{D\_3'(t)}{e^{(a-3)t}},\quad D\_5(t):=\frac{D\_4'(t)}{e^t},\quad D\_6(t):=\frac{D\_5'(t)}{e^t}. \end{equation\*} Then \begin{equation\*} D\_6(t)=6 a^3 (a+1) (a+2) (a+3) e^t \\ -2 a (a+1) \left(a^2+a-2\right) \left(a^2 (u+3)+2 a u+u\right), \end{equation\*} which is clearly increasing in $t$, from $D\_6(0)=-2 a (a+1) (a+2) \left(a^3 u+a^2 (u-12)-a u-u\right)<0$ eventually (that is, for all large enough $a$, depending on $u>0$) and $D\_6(\ln(1+u))=2 a (a+1) (a+2) \left(2 a^3 u+4 a^2 (2 u+3)+a u+u\right)>0$ eventually (and, in fact, for all real $a>0$ and $u>0$). So, eventually $D\_6$ is $-+\,$: that is, for some $t\_{6;a,u}\in(0,\ln(1+u))$ we have $D\_6<0$ on $[0,t\_{6;u,a})$ and $D\_6>0$ on $(t\_{6;u,a},\ln(1+u)]$. So, eventually $D\_5$ is down-up -- that is, for some $t\_{6;a,u}\in(0,\ln(1+u))$ we have $D\_5$ is decreasing on $[0,t\_{6;u,a}]$ and increasing on $[t\_{6;u,a},\ln(1+u)]$. Also, eventually $D\_5(0)=-a (a+1) \left(10 a^3 u+a^2 (5 u-36)-13 a u-2 u\right)<0$ and $D\_5(\ln(1+u))=a (a+1) \left(a^4 u^2+a^3 u (9 u+14)+a^2 \left(16 u^2+43 u+36\right)+a u (6 u+13)+2 u (2 u+1)\right)>0$. So, eventually $D\_5$ is $-+$. So, eventually $D\_4$ is down-up. Also, eventually $D\_4(0)=2 a^2 \left(-11 a^2 u+6 a (u+2)+5 u\right)<0$ and $D\_4(\ln(1+u))=a (u+1) \left(2 a^4 u^2+7 a^3 u (u+2)+8 a^2 \left(u^2+3 u+3\right)+5 a u (u+2)+2 u^2\right)>0$. So, eventually $D\_4$ is $-+$. So, eventually $D\_3$ is down-up. Also, eventually $D\_3(0)=-a \left(14 a^2 u+3 a (3 u-4)+u\right)<0$ and $D\_3(\ln(1+u))\frac{(1+u)^2}{a}=(u+1)^a\left(a^3 u^2+2 a^2 u (u+2)+a \left(-u^2+3 u+6\right)-u (2 u+1)\right) +6 a (u+1)^2>0$. So, eventually $D\_3$ is $-+$. So, eventually $D\_2$ is down-up. Also, eventually $D\_2(0)=-12 a (a+1) u<0$ and $D\_2(\ln(1+u))(1+u)^2 =(u+1)^a\left(-a^2 u (u+1) (3 u+2)-a u (u+1) (7 u+8)-2 u (u+1)^2\right) +2 a^2 u (u+1)^2-4 a u (u+1)^2+2 u (u+1)^2<0$. So, eventually $D\_2<0$. So, eventually $D$ is decreasing. Also, eventually $D(0)=u \left(a^2 u+2 a (u-3)+u\right)>0$ and $D(\ln(1+u))(1+u)^2 =(u+1)^a\left(-a^2 u^2 (u+1)^2-a u (5 u+4) (u+1)^2-u (2 u+3) (u+1)^2\right) -2 a u (u+1)^3+3 u (u+1)^3<0$. So, eventually $D$ is $+-\,$: that is, for some $t\_{1;a,u}\in(0,\ln(1+u))$ we have $D>0$ on $[0,t\_{1;u,a})$ and $D<0$ on $(t\_{1;u,a},\ln(1+u)]$. So, eventually $g$ is up-down -- that is, for some $t\_{1;a,u}\in(0,\ln(1+u))$ the function $g$ is increasing on $[0,t\_{1;u,a})$ and decreasing on $(t\_{1;u,a},\ln(1+u)]$. Also, eventually $g(0)=(2 a+1) u^2>0$ and $g(\ln(1+u))(1+u)^2 =(u+1)^a\left(-2 a u^2 (u+1)^3-u (u+2) (u+1)^3\right) +2 u (u+1)^4<0$. So, eventually $g$ is $+-$. $\quad\Box$

1

https://mathoverflow.net/users/36721

453192

https://mathoverflow.net/questions/453157

3

I am a bit rusty in my differential geometry and I would like to confirm that my reasoning below holds, and I have some related questions (and all references to related concepts are of interest to me). Let $G$ be a compact, connected Lie group endowed with the Riemannian metric induced by the Killing form, I would like to show that there is a global basis of the cotangent bundle composed by $1$-eigenforms of the Hodge Laplacian. My argument goes as follows. 1. We can pick a basis $(\xi\_i)\_i$ of the Lie algebra of $G$ and obtain $1$-forms $\theta\_i\in\Omega^1(G)$ by moving the dual elements via the group action. 2. Since both the exterior derivative and the Hodge star commute with the group action, we have that $\Delta\theta\_i$ is in the $\mathbb{C}$-span of $(\xi\_i)$, meaning that the Hodge Laplacian is a linear automorphism of $\mathrm{span}\_\mathbb{C}(\xi\_i)\_i$. 3. We can now diagonalize this linear map and its eigenvectors $\tilde{\theta}\_i$ are the desired basis. In particular, this tells us that a basis of $1$-eigenforms of the Hodge Laplacian on $G$ is given by the $1$-forms $f\_j\tilde\theta\_i$ for $(f\_j)\_j$ a basis of eigenfunctions (by a Stone-Weierstrass type of argument). Similar conclusions can be made about higher degree forms. Assuming the above is correct, I have a few questions: * I expect the result above to be well known. Is there any good reference? * What can be said about the eigenvalues? I would expect them to be related to the curvature of the metric. * Is all of the diagonalization procedure really necessary? It feels a bit clunky and dependent on the initial choice of basis. E.g. would the original $\theta\_i$ already be eigenforms is we take the basis $\xi\_i$ to be orthonormal with respect to the Killing metric?

https://mathoverflow.net/users/44134

Eigenforms of the Laplacian on Lie groups

Here are a few brief comments, but, as you suspect, an enormous amount is known about the Laplacian on functions and forms on compact Lie groups. • Presumably, you know that the Killing form is non-degenerate only in the semi-simple case, i.e., when the center of $G$ is discrete, so I assume that you are only interested in the compact, semi-simple case. Moreover, in that case, you might as well consider the simply-connected cover of $G$, in which case $G$ is the product of its simple factors and the Laplacian 'preserves' the factors, so you might as well assume $G$ is compact, simple, and simply-connected. So I'll assume that henceforth. • Because the Laplacian preserves the left-invariant 1-forms on $G$, which are an irreducible representation of the right action of $G$, it must be a multiple of the identity, so, in fact, all of the left-invariant 1-forms on $G$ are eigenforms with the same eigenvalue. (The eigenvalue depends on the specific simple Lie group, and this is calculable from the root diagram.) You'll want to look at some of Weyl's original papers on this or at some of the many expositions of this material. A good place to start is S. Helgason's "Differential Geometry, Lie groups, and Symmetric Spaces". • The eigenfunctions are well-known and can be calculated explicitly, though the combinatorics can be quite involved. See any book that discusses the Peter-Weyl Theorem. The important thing is that the precise eigenvalues and their multiplicities are determined by the Casimir element and combinatorial formulae involving roots and weights. In the simplest case, $\mathrm{SU}(2)$, you can make everything very explicit. (It's a standard exercise in the geometry of Lie groups.) • Looking at the products $f\theta$ where $f$ comes from an eigenspace of the Laplacian and $\theta$ is a left-invariant $1$-form gets you a start, but these are not (usually) eigenforms themselves. Instead the set of such products spans a (finite-dimensional) tensor product of representations of $G$ and you usually have to decompose this tensor product into its irreducible pieces (or at least determine the possible highest weights and their multiplicities) before you find the actual eigenspaces of the Laplacian on this tensor product. (There's nearly always more than one eigenvalue of the Laplacian in this tensor product.) • Once you know what you are looking for, you can find a lot of this information in Tony Knapp's "Lie groups: Beyond an Introduction", or you can calculate most of what one wants to know from the material in Knapp's book if he doesn't give it explicitly.

2

https://mathoverflow.net/users/13972

453195

https://mathoverflow.net/questions/452916

5

I am interested in the following graph invariant: for a given graph $G=(V,E)$, $c(G)$ is defined to be the smallest number of vertices such that I can recreate the connectivity of $G$ by disconnecting and relabelling vertices for reuse once they have been connected to all vertices in their neighborhood in $G$. That is, once a vertex, $v$, has been connected to all vertices $u \in \mathcal{N}(v)$, all edges incident to $v$ can be removed so that the vertex $v$ can be relabeled and used to recreate connectivity elsewhere. Note that we allow vertices $u\in\mathcal{N}(v)$ to have already been disconnected and relabelled prior to disconnecting and relabelling $v$, so long as they have been connected previously. Some simple observations include: * For a clique $K\_n$, $c(K\_n)=n$. * For any path graph $P\_\ell$, $c(P\_\ell)=2$ * For any cycle $C\_m$ with $m>2$ $c(C\_m)=3$. * For a star graph $K\_{1,n}$ $c(K\_{1,n})=2$ * For a complete bipartite graph $K\_{m,n}$ $c(K\_{m,n})=\min(m,n)+1$ * The number is lower bounded by the order of the largest clique $K\subset G$ I would be interested to know if this graph invariant has been studied before. This problem seems to be related to graph bubbling in the sense that we wish to reduce the number of vertices present by cutting across the minimal number of edges. Worked Example -------------- Consider the graph $G=(V,E)$ with $V=\{v\_i\}\_{i=0}^5$ and edge set $E=\{(v\_0,v\_4), (v\_1,v\_2), (v\_1,v\_3), (v\_1,v\_4), (v\_2,v\_3), (v\_3,v\_5),(v\_4,v\_5)\}$ [Graph $G$](https://i.stack.imgur.com/lesVr.png) The connectivity of this graph can be recreated using a minimum of three vertices. To do so, we create the following induced subgraphs in turn: * create [$G[\{v\_0,v\_4,v\_5\}]$](https://i.stack.imgur.com/ch87i.png) * remove $v\_0$, reassign to $v\_1$ * reconnect $v\_1$ as [$G[\{v\_1,v\_4,v\_5\}]$](https://i.stack.imgur.com/ZMkhl.png) * remove $v\_5$ and reassign to $v\_3$ * reconnect $v\_3$ as [$G[\{v\_1,v\_3,v\_5\}]$](https://i.stack.imgur.com/a7YGC.png) * remove $v\_5$ and reassign to $v\_2$ * reconnect $v\_2$ as [$G[\{v\_1,v\_2,v\_3\}]$](https://i.stack.imgur.com/vAngB.png). Alternative definition ====================== for a given graph $G=(V,E)$, $c(G)$ is defined to be the smallest number of counters that we can use to cover all vertices in $V$, with following prescription 1. Place the counters on a subset of vertices $S\subseteq V$ 2. A counter may be lifted from a vertex $v$ if and only if each vertex $u\in \mathcal{N}(v)$ also has a counter 3. When a counter is lifted we update the graph by removing the vertex from which the counter was lifted: $G\gets G[V\setminus v]$ 4. Repeat until all vertices have been removed or have a counter Worked Example -------------- Consider the graph $G=(V,E)$ with $V=\{v\_i\}\_{i=0}^5$ and edge set $E=\{(v\_0,v\_4), (v\_1,v\_2), (v\_1,v\_3), (v\_1,v\_4), (v\_2,v\_3), (v\_3,v\_5),(v\_4,v\_5)\}$ [Graph $G$](https://i.stack.imgur.com/lesVr.png) We have $c(G)=3$ using the following method * place counters on vertices $\{v\_0,v\_4,v\_5\}]$ * lift counter from $v\_0$ * Set $G\gets G[V\setminus v\_0]$ * Place counter on $v\_1$ * lift counter from $v\_5$ * Set $G\gets G[V\setminus v\_5]$ * place counter on $v\_3$ * lift counter from $v\_3$ * Set $G\gets G[V\setminus v\_3]$ * place counter on $v\_2$ Now all vertices have either been removed or have a counter

https://mathoverflow.net/users/510619

Connectivity graph invariant

I think what you describe is exactly the "node-search number" as described by Kirousis and Papadimitriou (<https://www.sciencedirect.com/science/article/pii/0012365X85900469>). As shown in that paper, it is indeed equivalent to the interval thickness, or (which is the same) the pathwidth plus one.

2

https://mathoverflow.net/users/45545

453203

https://mathoverflow.net/questions/453210

8

Let $f(T) = \sum a\_n T^n \in \mathbf{F}\_p [[ T ]]$ be a power series. We'll say that the coefficients of $f(T)$ are equidistributed modulo $p$ if for every residue class $a$ modulo $p$, we have $$ \lim\_{X \to \infty} \dfrac{1}{X} \cdot \# \{n<X : a\_n \equiv a \mod p \} = \dfrac{1}{p}.$$ Suppose that $f(T), g(T) \in \mathbf{F}\_p [[ T ]]$ are power series whose coefficients are equidistributed modulo $p$ in the above sense. Is it true that the coefficients of the product $f(T) \cdot g(T)$ are also equidistributed modulo $p$?

https://mathoverflow.net/users/394740

Is the product of two equidistributed power series equidistributed?

No. Let $f(T)$ be chosen uniformly at random from all power series with constant term nonzero and let $g(T) = 1/ f(T) $. Then $g(T)$ is also chosen uniformly at random from all power series with constant term nonzero. (Here uniformly means each possible sequence of first $n$ coefficients has equal probability, and the proof is just that the first $n$ coefficients of $f$ determine the first $n$ coefficients of $g$ and vice versa.) So $f(T)$ and $g(T)$ are each equidistributed with probability $1$ but $f(T)g(T)=1$ is not equidistributed. The same argument shows that any condition on $f$ and $g$ that is "generic" in the sense of holding for a random power series is not sufficient.

14

https://mathoverflow.net/users/18060

453211

https://mathoverflow.net/questions/453117

3

Is there a reasonable/canonical way to mollify a Borel probability measure without changing its marginals. Let $\pi \in \mathcal{P}(\mathbb{R}^2)$ with marginals $\mu,\nu$. I want to smooth out $\pi$ up to scale $\varepsilon$, which is equivalent to cutting it off at frequency $|k|\approx \varepsilon^{-1}$ in Fourier space, but I do not want to change is marginals. The resulting measure should have a smooth density. It goes without saying that the usual procedure (convolution with a smooth, positive, bump function) does not work. Of course, you can assume $\mu,\nu$ are absolutely continuous with smooth densities.

https://mathoverflow.net/users/106281

Mollifying a measure without changing its marginals

$\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}\newcommand{\tX}{\tilde X}\newcommand{\tY}{\tilde Y}\newcommand{\tpi}{\tilde\pi} $Here is how this can be done in an explicit way, at least when the densities (say $p$ and $q$) of $\mu$ and $\nu$ are everywhere $>0$. Let $(X,Y)$ be a random point in $\R^2$ with distribution $\pi$ and thus marginals $\mu$ and $\nu$. Let $F$ and $G$ be the c.d.f.'s of $X$ and $Y$ (respectively). Let $\pi\_\ep$ be a distribution over $\R^2$ with a smooth density such that $\pi\_\ep$ is close to $\pi$. Let $(X\_\ep,Y\_\ep)$ be a random point in $\R^2$ with distribution $\pi\_\ep$. Let $F\_\ep$ and $G\_\ep$ be the c.d.f.'s of $X\_\ep$ and $Y\_\ep$ (respectively), so that $F\_\ep$ and $G\_\ep$ are close to $F$ and $G$, and hence $F^{-1}\circ F\_\ep$ and $G^{-1}\circ G\_\ep$ are smooth maps each close to the identity map. Let $\tX:=F^{-1}(F\_\ep(X\_\ep))$ and $\tY:=G^{-1}(G\_\ep(Y\_\ep))$. Then the distribution (say $\tpi$) of the random point $(\tX,\tY)$ in $\R^2$ will have a smooth density and marginals $\mu$ and $\nu$, and $\tpi$ will be close to $\pi$.

3

https://mathoverflow.net/users/36721

453237

https://mathoverflow.net/questions/453242

2

I've been trying to understand various questions to do with sigma algebras on uncountable product spaces. Let $T$ be an uncountable set and for each $t \in T$, let $\Omega\_t$ be a topological space. Let $\Omega := \prod\_{t \in T}\Omega\_t$ be the product space equipped with the product topology in the usual way. The Borel sigma-algebra $\mathcal{Borel}(\Omega)$ is the sigma-algebra generated by the open sets in this topology. The Baire sigma-algebra is the sigma-algebra generated by sets of the form {$\omega \in \Omega : f(\omega) > 0$} for continuous functions $f : \Omega \to \mathbf{R}$. (Equivalently use zero sets if you prefer but this makes it look a little more on par with open sets) EDT: The Baire sigma algebra is contained in the Borel sigma algebra but presumably(??) it isn't the case that they are equal in general? i.e. Is it the case that the Baire is strictly smaller and if so how can we show it?

https://mathoverflow.net/users/122587

Relationship between Baire sigma algebra and Borel sigma algebra of an uncountable product

Let all factor spaces be nontrivial compact Hausdorff spaces. Then every continuous function is [determined by countably many coordinates](https://mathoverflow.net/questions/364062/can-a-continuous-real-valued-function-on-a-large-product-space-depend-on-uncount), and so is, consequently, every Baire measurable set. It follows that singletons are not Baire sets. However, they are closed and, consequently, Borel. The same argument works if all factor spaces are nontrivial [separable metrizable spaces](https://mathoverflow.net/a/364126/35357).

3

https://mathoverflow.net/users/35357

453246

https://mathoverflow.net/questions/452382

13

We endow ${\cal P}(\omega)$ with an equivalence relation by saying that $A\simeq\_{\text{fin}} B$ iff the symmetric difference $A\Delta B$ is finite. The resulting set of equivalence classes is denoted ${\cal P}(\omega)/(\text{fin})$, and by $$[A]\_{\simeq\_{\text{fin}}} \leq [B]\_{\simeq\_{\text{fin}}} \text{ if and only if } A\setminus B \text{ is finite } $$ for $A,B\in {\cal P}(\omega)$, we get a partial order on ${\cal P}(\omega)/(\text{fin})$, resulting in an atomless Boolean algebra. Can any poset $(P,\leq)$ with $|P|\leq 2^{\aleph\_0}$ be order-embedded into ${\cal P}(\omega)/(\text{fin})$?

https://mathoverflow.net/users/8628

Can any poset of cardinality $\leq 2^{\aleph_0}$ be embedded in ${\cal P}(\omega)/(\text{fin})$?

Here is an attempt at a 'definitive summary'. To begin with positive results: $\mathsf{CH}$ implies a “yes” answer to this question. The fastest way to see this is to first embed a given partial order $(P,\le)$ (of cardinality at most $\mathfrak{c}$) into its power set via $p\mapsto\{x:x\le p\}$ and then take the Boolean subalgebra generated by the image under this embedding. An application of Parovichenko's theorem that every Boolean algebra of cardinality at most $\aleph\_1$ embeds into $\mathcal{P}(\omega)/\mathit{fin}$ finishes the argument. To elaborate a bit on Will Brian's [answer to this question](https://mathoverflow.net/questions/441234): Laver's proof of the consistency with $\neg\mathsf{CH}$ with the statement that was published in [*Linear orders in $(\omega)^\omega$ under eventual dominance*](https://zbmath.org/0464.03046), in *Logic colloquium ’78*. The proof and the proof of the generalization to Boolean algebras that Will mentioned both proceed by embedding a $\mathfrak{c}$-saturated linear order or Boolean algebra of cardinality $\mathfrak{c}$ respectively into $\mathcal{P}(\omega)/\mathit{fin}$. The saturated order/algebra contains all orders/algebras of cardinality at most $\mathfrak{c}$. This establishes the consistency of a ``yes'' answer with $\neg\mathsf{CH}$. It is easy to see that every separable compact space is a continuous image of $\beta\omega\setminus\omega$ so that, dually, every $\sigma$-centered Boolean algebra embeds into $\mathcal{P}(\omega)/\mathit{fin}$. There are many consistent counterexamples: Kunen's result that establishes that in the Cohen model (with any allowable value of $2^{\aleph\_0}$) the ordinal $\omega\_2$ is not embeddable into $\mathcal{P}(\omega)/\mathit{fin}$ shows that Parovichenko's result is sharp. It also shows that the result on $\sigma$-centered Boolean algebras does not generalize to partial orders: a linear order is ($\sigma$-)centered. Other negative results: 1. The Open Colouring Axiom implies that the [*Measure Algebra does not embed into $\mathcal{P}(\omega)/\mathit{fin}$*](http://matwbn.icm.edu.pl/ksiazki/fm/fm163/fm16325.pdf) and, in fact, that many 'natural candidates' are [*not continuous images of $\beta\omega\setminus\omega$*](http://dx.doi.org/10.1007/BF02775024) and so, dually, many 'natural candidates' cannot be embedded into $\mathcal{P}(\omega)/\mathit{fin}$. 2. It is even consistent that [*there is no universal Boolean algebra of cardinality $\mathfrak{c}$*](http://dx.doi.org/10.1090/S0002-9947-00-02601-5). The proof is topological and yields, given a compact space of weight $\mathfrak{c}$, a linearly ordered compact space that is not a continuous image of the given space; it dualizes to yield given a Boolean algebra of cardinality $\mathfrak{c}$ a linearly ordered set that cannot be embedded into the given Boolean algebra.

7

https://mathoverflow.net/users/5903

453258

https://mathoverflow.net/questions/453231

8

Let $E/\mathbb{Q} = E\_{a,b}$ $$\displaystyle y^2 = x^3 + ax + b, a,b \in \mathbb{Z}$$ be an elliptic curve defined over the field of rational numbers, and let $n \geq 3$ be an integer. Let $K\_n$ be the $n$-torsion field of $E$, namely the field obtained by adjoining all of the coordinates of $n$-torsion points of $E$ to $\mathbb{Q}$. Let $p$ be a prime divisor of the discriminant $\Delta(E)$ of $E$. Does it follow that $K\_n$ is ramified at $p$? If so, what can be said about the splitting behaviour of $p$ in $K\_n$?

https://mathoverflow.net/users/10898

$n$-torsion fields of an elliptic curve defined over $\mathbb{Q}$

No. The modular curve $X\_E(n)$, which paramaterizes pairs $(E',\iota)$, where $E'$ is another elliptic curve and $\iota \colon E'[n] \to E[n]$ is an isomorphism (of group schemes, or equivalently, of Galois modules) is a twist of $X(n)$, and thus has genus 0 for $n = 2,3,4,5$ and genus 1 for $n = 6$. In particular, for $n = 3,4,5$, $X\_E(n)$ is isomorphic to $\mathbb{P}^1$ and has infinitely many rational points; i.e., there are infinitely many $E'/\mathbb{Q}$ such that $\mathbb{Q}(E'[n]) = \mathbb{Q}(E[n])$. But $\mathbb{Q}(E[n])$ is unramified outside of the primes dividing the conductor of $E$ or $n$, and most such $E'$ have bad reduction at new primes. For primes greater than 6, the genus of $X(p)$ is at least 3. Anyway, this is related to the Frey--Mazur conjecture; but even if that conjecture is true, I don't think it implies what you're asking for $n > 6$. I do think what you are asking is true for large $n$ (independent of $E$), but as far as I know this is unknown.

10

https://mathoverflow.net/users/2

453261

https://mathoverflow.net/questions/453255

9

Let $K$ be an imaginary quadratic number field (so there are no real embeddings) with ring of integers $\mathcal{O}\_K$ . Let $w$ be the number of units in $K$ and $h$ be the class number of $K$. Let $\mathfrak{f}$ be a non-zero integral ideal in $K$, and let $\mathcal{I}\_{\mathfrak{f}}$ be the ray class group mod $\mathfrak{f}$, i.e. $$ \mathcal{I}\_{\mathfrak{f}} = \frac{J\_{\mathfrak{f}}}{P\_{\mathfrak{f}}}, $$ where $J\_{\mathfrak{f}}$ is the group of fractional ideals coprime to $\mathfrak{f}$ and, for imaginary quadratic fields, $P\_{\mathfrak{f}}$ is the group of principal fractional ideals $(\alpha)$ with $\alpha \equiv 1 \mod{\mathfrak{f}}$. Let $h(\mathfrak{f})$ denote the order of $\mathcal{I}\_{\mathfrak{f}}$, $\varphi(\mathfrak{f})$ the order of the multiplicative group $(\mathcal{O}\_K/\mathfrak{f})^\*$, and $g$ the number of units $\varepsilon$ with $\varepsilon \equiv 1 \mod{\mathfrak{f}}$. I have two questions pertaining to some literature I have read recently (namely [this paper of Coleman](https://doi.org/10.1112/plms/s3-61.3.433) and [this paper by Khale et. al.](https://doi.org/10.1016/j.jnt.2021.04.004)) 1. In both papers linked above, there appears the term "ideal class group mod $\mathfrak{f}$." Is this meant to be the *ray* class group mod $\mathfrak{f}$, or is there some other object this term is referring to? From both papers, it seems the answer is yes, it's the ray class group, but I want to make sure I'm not missing something. 2. In the first paper linked above, there appears implicit in the arguments the identity $$ h(\mathfrak{f}) = \frac{h\varphi(\mathfrak{f})g}{w}, $$ This appears specifically between displays (4.4) and (4.6). I would like a reference and/or an explanation of why this identity holds. My algebraic number theory is not the strongest, and my familiarity with the literature and common jargon is low, so forgive me if these are "dumb" questions.

https://mathoverflow.net/users/307675

Questions about ray class groups

**1.** Surely, by "class group modulo $\mathfrak{f}$", the authors mean "ray class group modulo $\mathfrak{f}$". I guess the authors omit "ray", because $K$ has no real embedding, so there are no "rays" here. **2.** Let $K^\times\_\mathfrak{f}$ be the multiplicative group of elements $\alpha\in K^\times$ congruent to $1$ modulo $\mathfrak{f}$, so that $P\_\mathfrak{f}$ is the multiplicative group of fractional ideals generated by such elements. Analogously, let $K^\times\_{\ast,\mathfrak{f}}$ be the multiplicative group of elements $\alpha\in K^\times$ coprime to $\mathfrak{f}$, and let $P\_{\ast,\mathfrak{f}}$ be the multiplicative group of fractional ideals generated by such elements. Then we have the following exact sequence of finite abelian groups, with the obvious homomorphisms: $$1\to U/U\_\mathfrak{f}\to K^\times\_{\ast,\mathfrak{f}}/K^\times\_\mathfrak{f}\to J\_\mathfrak{f}/P\_\mathfrak{f}\to J\_\mathfrak{f}/P\_{\ast,\mathfrak{f}}\to 1.$$ Indeed, the image of the second map and the kernel of the third map are equal to $UK^\times\_\mathfrak{f}/K^\times\_\mathfrak{f}$, while the image of the third map and the kernel of the fourth map are equal to $P\_{\ast,\mathfrak{f}}/P\_\mathfrak{f}$. Comparing the sizes of the groups in the exact sequence, we get that $$|U/U\_\mathfrak{f}|\cdot|J\_\mathfrak{f}/P\_\mathfrak{f}|=|K^\times\_{\ast,\mathfrak{f}}/K^\times\_\mathfrak{f}|\cdot|J\_\mathfrak{f}/P\_{\ast,\mathfrak{f}}|.$$ On the other hand, it follows from weak approximation that $J\_\mathfrak{f}/P\_{\ast,\mathfrak{f}}$ is isomorphic to the class group of $K$, hence our equation becomes $$(w/g)\cdot h(\mathfrak{f})=\varphi(\mathfrak{f})\cdot h.$$

8

https://mathoverflow.net/users/11919

453269

https://mathoverflow.net/questions/453274

2

Let $M$ be the projective cover (e.g, [Gleason1958](https://doi.org/10.1215/ijm/1255454110)) of the Cantor set $\{-1,1\}^{\mathbb{N}}$. Let $\textrm{homeo}(M)$ denote the group of all homeomorphisms of $M$. Some of the $\gamma\in\textrm{homeo}(M)$ are extensions of the homeomorphisms of the Cantor set. Some others are the extensions of the homeomorphisms of Polish spaces $K$ for which $M$ is the projective cover of $K$. > > **Q:** Is there a complete explicit description of the homeomorphisms of $M$? > > > I'd be grateful if you point me to a reference about the properties of $\textrm{homeo}(M)$. Thanks in advance.

https://mathoverflow.net/users/164350

Homeomorphisms of the projective cover of the Cantor set

I claim that every autohomeomorphism of $M$ is conjugate to an extension of an autohomeomorphism of the Cantor space. Let $B$ be the Stone space of the Cantor space. Let $\overline{B}$ denote its Boolean completion. We shall show that every automorphism of $\overline{B}$ is conjugate to an automorphism of $B$. Recall that $\overline{B}$ is up-to-isomorphism the only complete atomless Boolean algebra with a countable basis (by a basis, I mean a subset $A\subseteq\overline{B}$ where $\overline{B}=\{\bigvee R:R\subseteq A\}$; for example, $B$ is a countable basis for $\overline{B}$). Suppose that $h:\overline{B}\rightarrow \overline{B}$ is an automorphism. Then let $D\subseteq \overline{B}$ be the smallest Boolean subalgebra closed under the operations $h,h^{-1}$ and which contains the countable basis $A$. Then $h$ restricts to an automorphism $g:D\rightarrow D$, and $h$ is the unique automorphism of $\overline{B}$ that extends $g$. Recall that every countable atomless Boolean algebra is isomorphic (the theory of countable atomless Boolean algebras is $\omega$-categorical), so $B$ and $D$ are isomorphic. Let $\phi:B\rightarrow D$ be an isomorphism. Then $\phi$ extends to an automorphism $\Phi:\overline{B}\rightarrow\overline{B}$. In this case, $\Phi^{-1}h\Phi:\overline{B}\rightarrow\overline{B}$ is an automorphism that uniquely extends an automorphism of $B$. One can extend this argument to conclude that if $(h\_n)\_{n=0}^\infty$ is a collection of automorphisms of $\overline{B}$, then there is some automorphism $\Phi:\overline{B}\rightarrow\overline{B}$ where for each $j$, the mapping $\Phi^{-1}h\_j\Phi$ uniquely extends an automorphism of $B$.

2

https://mathoverflow.net/users/22277

453286

https://mathoverflow.net/questions/453067

5

Suppose $C\leq A,B$ are finitely generated groups of finite exponent. Can $A$ and $B$ be [amalgamated](https://en.wikipedia.org/wiki/Amalgamation_property) over $C$ in a group of finite exponent? What about if $A,B$ are periodic, can we find a periodic amalgam? Note that if we assume that $A$ and $B$ are finite, then there is a finite amalgam by a classical construction of Neumann. If we do not assume that the groups be finitely generated, then taking $C$ to be an infinite dimensional vector space over the field with $2$ elements, and $A,B$ to be appropriate expansions of $C$ by involutive automorphisms, we get an example where $A,B$ are of exponent 4 and every amalgam has an element of infinite order. Thus, if this is true, the proof must use local finiteness, and if not, counterexamples must be (necessarily countably) infinite. We could try to emulate the counterexample if we could find a f.g. group $C$ of finite exponent (or periodic) with two automorphisms of finite order whose product has infinite order (or generate jointly automorphisms of arbitrarily high order). But even then we would need to ensure that the group generated by $C$ together with each of these automorphisms is still finite exponent/periodic. (I did not check carefully, but possibly, the conclusion is also true if we assume that $C$ is one of finite, finite index in one of $A,B$, or central in one of $A,B$, but I do not want to assume anything like that.)

https://mathoverflow.net/users/54415

Amalgamation of finitely generated finite exponent groups

It's not hard to construct counter-examples for large exponents. Consider the following two automorphisms $\xi, \eta$ of the free group $F=F(x,y)$ of rank $2$, defined by $$\xi(x)=y,~\xi(y)=x \text{ and } \eta(x)=x,~\eta(y)=yx.$$ The composition $\xi \circ \eta \in Aut(F)$ is the so-called Fibonacci automorphism $\varphi$, given by $$\varphi(x)=y,~\varphi(y)=xy.$$ Now, for every $n \in \mathbb{N}$ the free Burnside group $B(2,n)$, of exponent $n$, is the quotient of $F$ by the verbal subgroup generated by all $n$-th powers. It follows that every automorphism of $F$ induces an automorphism of $B(2,n)$ (there is a natural homomorphism $Aut(F) \to Aut(B(2,n))$, $\alpha \mapsto \bar\alpha$). Now, take $n$ be to sufficiently large and odd. Then it is known that the Fibonacci automorphism $\bar\varphi$ has infinite order in $Aut(B(2,n))$. This result was proved in [*Cherepanov, E. A.*, [**Free semigroup in the group of automorphisms of the free Burnside group.**](https://doi.org/10.1081/AGB-200047435), Commun. Algebra 33, No. 2, 539-547 (2005). [ZBL1121.20028](https://zbmath.org/?q=an:1121.20028).[, and was improved (for smaller exponents) in [*Pajlevanyan, Ashot S.*, [**The Fibonacci automorphism of free Burnside groups.**](https://doi.org/10.1051/ita/2011118), RAIRO, Theor. Inform. Appl. 45, No. 3, 301-309 (2011). [ZBL1227.20038](https://zbmath.org/?q=an:1227.20038).] Evidently $\bar\xi$ has order at most $2$ and $\bar\eta$ has order dividing $n$ in $Aut(B(2,n))$, because the order of the image of $x$ divides $n$ (in fact, the orders of $\bar\xi$ and $\bar\eta$ will be exactly $2$ and $n$). Therefore, the semidirect products $G=B(2,n) \rtimes\_{\bar\xi} C\_2$ and $H=B(2,n)\rtimes\_{\bar\eta} C\_n$, where $C\_n$ is the cyclic group of order $n$, have exponents dividing $2n$ and $n^2$ respectively. It's easy to see that both $G$ and $H$ are generated by $2$ elements. However, as the OP suggested in the post, since the composition $\bar \varphi$, of the automorphisms $\bar\xi$ and $\bar \eta$, has infinite order in $Aut(B(2,n))$, the groups $G$ and $H$ cannot be amalgamated over $B(2,n)$ in a periodic group.

3

https://mathoverflow.net/users/7644

453288

https://mathoverflow.net/questions/453268

5

Suppose one has a filter (a collection of subsets closed under increasing the size of the set and under finite intersection) $F$ on a ring $R$. Say that $F$ is (ring) ideal-like if for every set $U \subset R$ in $F$, there exists a $U'\in F$ such that $U' + U' \subset U$, and there exists a $U'' \in F$ such that $U'' R \subset U$. (This corresponds to the filtered intersection of elements of $F$ giving rise to a pro-set which satisfies the ideal conditions.) Does this imply that $F$ is generated as a filter by actual ring ideals of $R$? In other words, for any $U \in F$, does there exist an ideal $I$ such that $I \in F$ and $I \subset U$? If such a situation is possible, is it still possible if $R$ is Noetherian? If $R = \mathbb{Z}$?

https://mathoverflow.net/users/484180

Ideal-like filter on a ring not generated by ring ideals

This does not hold for all commutative rings, but it does hold for Noetherian rings and for valuation rings (assuming the convention that filters don't contain $\emptyset,$ or else $\mathcal{P}(R)$ is a trivial counterexample for any ring). Suppose $R$ is a commutative ring, $F$ is an ideal-like filter on $R,$ and $U\_0 \in F$ is such that there is no ideal $I \in F \cap \mathcal{P}(U\_0).$ Recursively choose $U\_n \in F$ such that $U\_{n+1}+U\_{n+1} \subset U\_n$ and $U\_{n+1}=U\_{n+1}R.$ Since no $U\_n$ is empty or an ideal, we immediately have $U\_{n+1} \subsetneq U\_n.$ For $n \ge 1,$ we recursively choose $u\_n \in U\_n$ for $n \ge 1$ as follows: first, let $u\_1$ be any element of $U\_1 \setminus U\_2.$ Suppose $u\_1, \ldots, u\_n$ have been chosen. An inductive argument shows that $I\_n:=\langle u\_1, \ldots, u\_n \rangle \subset U\_0,$ so $I\_n \not \in F.$ Let $u\_{n+1}$ be any element of $U\_{n+1} \setminus I\_n.$ Then $\{I\_n\}$ is a strictly increasing sequence of ideals, so $R$ is non-Noetherian. Furthermore, $\langle u\_1 \rangle$ and $\langle u\_2 \rangle$ are incomparable, so $R$ is not a valuation ring. Now we'll construct a commutative ring $R$ with an ideal-like filter $F$ not generated by ring ideals. Let $R = C^{\infty}(S^1)$ be the ring of smooth real-valued functions on the circle. Every maximal ideal in $R$ is of the form $\mathfrak{m}\_x=\{f \in R: f(x)=0\}$ for some $x \in S^1$ (see <https://math.stackexchange.com/a/183213/210610>). Let $A\_n=\{f \in R: \lambda(f^{-1}(0))> 1- \frac{1}{n}\},$ and let $F$ be the filter on $R$ generated by the $A\_n$'s, i.e. $$F=\bigcup\_{n=1}^{\infty} \{X \subset R: A\_n \subset X\}.$$ It's clear that every $X \in F$ is infinite (in fact, $|X|=|\mathbb{R}|$) and that for every finite $S \subset R \setminus \{0\},$ $R \setminus S \in F.$ We will show that there is no proper ideal in $F.$ Suppose $I \in F$ is a proper ideal on $R.$ There is $x$ such that $I \subset \mathfrak{m}\_x \in F.$ But for every $n,$ there is a bump function in $A\_n \setminus \mathfrak{m}\_x,$ contradiction. Thus, $U=R \setminus \{1\} \in F$ does not contain any ideal in $F.$ Finally, we verify that $F$ is ideal-like. Fix $U \in F.$ Let $n$ be such that $A\_n \subset U.$ Then $V=A\_{2n}$ satisfies $V=VR$ and $V+V \subset A\_n \subset U.$ --- Original answer: This does not hold for all commutative rings, but if $R$ is a Dedekind domain, and we include the filter convention that $\emptyset \not \in F,$ then an ideal-like filter $F$ is generated by ring ideals. Suppose $F$ is a counterexample. Let $U \in F$ be such that there is no ideal $I \in F \cap \mathcal{P}(U).$ Let $V, W \in F$ be such that $V+V+V \subset U$ and $WR \subset V.$ We may assume $W=WR.$ Since $W$ is nonempty and not an ideal, we can fix a nonzero non-unit $w \in W,$ and prime ideals $P\_i$ such that $\langle w \rangle = \prod\_{i=1}^n P\_i.$ Let $A \subset \{1,\ldots, n\}$ be maximal such that $\prod\_{i \in A} P\_i \in F.$ Let $X \in F$ be such that $X=XR$ and $\sum\_{i=1}^n X \subset W.$ For $j \in A,$ set $x\_j=0,$ and otherwise let $x\_j$ be an arbitrary element of $X \cap \prod\_{i \in A} P\_i \setminus \prod\_{i \in A \cup \{j\}} P\_i,$ which is nonempty since $X \cap \prod\_{i \in A} P\_i \in F.$ Let $x, y$ be such that $\langle x, y \rangle = \langle x\_i: i \le n \rangle \subset W.$ Consider $$I:=\langle w, x, y \rangle \subset (W + W + W) \cap \prod\_{i \in A} P\_i \subset U \cap \prod\_{i \in A} P\_i.$$ Since $\prod\_{i=1}^n P\_i \subset I \subsetneq \prod\_{i \in A} P\_i \in F,$ there is $B \supsetneq A$ such that $I = \prod\_{i \in B} P\_i.$ Fix $j \in B \setminus A.$ Then $A \cup \{j\} \subset B$ and $x\_j \in \langle x, y \rangle,$ so $x\_j \in \prod\_{i \in B} P\_i \setminus \prod\_{i \in A \cup \{j\}} P\_i = \emptyset,$ contradiction.

6

https://mathoverflow.net/users/109573

453291

https://mathoverflow.net/questions/453282

9

Given a cardinal $\kappa\geq \aleph\_0$, is there a poset $(P,\leq)$ with $|P| = \kappa$ such that every poset of cardinality $\kappa$ can be order-embedded into $(P,\leq)$?

https://mathoverflow.net/users/8628

Universal poset for cardinals $\kappa \geq \aleph_0$

The easy case occurs when $\kappa^{<\kappa}=\kappa$. Under GCH, this includes every regular cardinal. As Emil mentions in the comments, the result for this case follows from general model-theoretic considerations of saturated models. But for definiteness, let me describe a concrete construction. We construct a universal partial order in a process of length $\kappa$. Start with nothing $P\_0=\emptyset$. At each stage $\alpha$, consider all possible subsets $A,B,C\subseteq P\_\alpha$ each of size less than $\kappa$, such that every element of $A$ is strictly below every element of $B$, and no element of $C$ is below an element of $A$ or above an element of $B$. Add a new point to $P\_{\alpha+1}$ that is above every element of $A$, below every element of $B$, and incomparable with every element of $C$. That is, for each possible way of relating to a given fewer than $\kappa$ many elements we have so far, we add such a point realizing that possibility. Since $\kappa^{<\kappa}=\kappa$, there are at most $\kappa$ many $A,B,C$, and so we have added at most $\kappa$ many points at each stage, and so inductively every $P\_\alpha$ has size at most $\kappa$. At limit stages, we take the union of everything we've added up to that stage. Consider the ultimate order $P\_\kappa$ that we've built by stage $\kappa$. This order has size $\kappa$, and I claim it is universal for all orders of size $\kappa$. To see this, suppose that you have an order $\langle Q,\leq\rangle$ of size $\kappa$. Enumerate it in a $\kappa$ sequence $q\_\alpha$ for $\alpha<\kappa$, and build an embedding into $P\_\kappa$ in stages. We will ensure that each $q\_\alpha$ finds its image in $P\_{\alpha+1}$. At each stage, $q\_\alpha$ relates to the previous elements $q\_\beta$ for $\beta<\alpha$ in a certain way — above some, below others, and incomparable with the rest. But at stage $\alpha$ in the universal construction, we added a point of exactly that type relative to the images of those points. So there is a point in $P\_{\alpha+1}$ to which to map $q\_\alpha$ so as to extend the embedding one more step. Thus, $Q$ embeds into $P\_\kappa$, and so our partial order is universal for all orders of size continuum. The model we've constructed is not only universal, but also homogeneous, since any given embedding of a less-than-size-$\kappa$ order can be extended. But if CH fails, then it is easy to see that there can be no such order of size $\aleph\_1$, since for a given copy of $\mathbb{Q}$ it would have to realize all the cuts, but there are too many. More generally, whenever $2^\gamma>\kappa$ for some $\gamma<\kappa$, then there would have to be an antichain of size $\gamma$, and every subset of it would have to have a point above it and not the other points in the antichain, but there are too many subsets. It is much more subtle, however, to rule out merely universal orders of these sizes. To get an idea of the difficulties, consider the following paper of Kojman and Shelah concerning the corresponding case of universal linear orders. * *Kojman, Menachem; Shelah, Saharon*, [**Nonexistence of universal orders in many cardinals**](https://doi.org/10.2307/2275437), J. Symb. Log. 57, No. 3, 875-891 (1992). [ZBL0790.03036](https://zbmath.org/?q=an:0790.03036). They prove, for example (theorem 3.6), that if $\aleph\_1<\lambda=\text{cof}(\lambda)<2^{\aleph\_0}$, then there is no $\lambda$-universal linear order of size $\lambda$. I am unsure of the extent to which those results apply to partial orders.

6

https://mathoverflow.net/users/1946

453305

https://mathoverflow.net/questions/453034

4

Do there exist 3+1 dimensional spacetimes (i.e. Lorentzian manifolds with signaure (1,3)), for which the Chern–Pontryagin scalar \begin{equation} K\_2= \epsilon^{\mu\nu\rho\sigma}R^{\alpha}{}\_{\beta\mu\nu}R^{\beta}{}\_{\alpha\rho\sigma}\,, \end{equation} is a non-zero constant? $R^{\alpha}{}\_{\beta\mu\nu}$ is Riemann's curvature tensor and $\epsilon^{\mu\nu\rho\sigma}$ is the contravariant Levi-Civita tensor. To be specific, I would be interested in finding a metric of such a spacetime. In 3+1 dimensional theories of chiral (Weyl) fermions in a curved space-time, $K\_2$ is at source of gravitational anomaly -- a violation of charge conservation, where particles are produced at a rate proportional to $K\_2$ ([Jensen, K., Loganayagam, R. & Yarom, A. Thermodynamics, gravitational anomalies and cones](https://link.springer.com/article/10.1007/JHEP02(2013)088)). In chiral anomaly, a closely related phenomenon due to gauge fields, particle production rate is proportional to $\epsilon^{\mu\nu\rho\sigma}F\_{\mu\nu}F\_{\rho\sigma}=\mathbf{E}\cdot\mathbf{B}$, for which it's very easy to find a case where it's constant and non-zero. My best effort so far was to find a power-series perturbation around a Minkowski spacetime in Cartesian coordinates $(t,x,y,z)$ \begin{equation} g\_{\mu\nu}=\left( \begin{array}{cccc} y^2-y z-z^2+1 & y^2-z^2 & 0 & 0 \\ y^2-z^2 & y^2-y z-z^2-1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{array} \right)\,, \end{equation} for which \begin{equation} K\_2=\frac{32 \left(y z \left(2 y^2-y z-2 z^2\right)-1\right)}{\left(y z \left(2 y^2-y z-2 z^2\right)+1\right)^{5/2}} = -32 + O\left((y^2+z^2)^2\right)\,. \end{equation} I can push the calculation to higher orders, but I'm not too hopeful in finding a closed-form expression this way.

https://mathoverflow.net/users/165560

Metric with a constant Chern–Pontryagin scalar

I found a solution: \begin{equation} g\_{\mu\nu} = \left( \begin{array}{cccc} \frac{A}{(1-t z)^2 (x y-1)^2} & 0 & 0 & 0 \\ 0 & -\frac{A}{(1-t z)^2 (x y-1)^2} & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{array} \right)\,. \end{equation} The Chern-Pontryagin scalar is $K\_2 = -16/|A|$ for $xy>1,tz>1$. **Update:** Under a conformal transformation $g'=\Omega g$, the Chern-Pontryagin scalar transforms as $K\_2' = K\_2/\Omega^2$, so for any metric with a non vanishing $K\_2$, one can take $\Omega = \sqrt{|K\_2|}$ and produce a metric, for which it's constant.

3

https://mathoverflow.net/users/165560

453306

https://mathoverflow.net/questions/453281

2

According to the [Wikipedia](https://en.wikipedia.org/wiki/Vertex_enumeration_problem) page on the issue, the vertex enumeration problem is NP-hard. However, double description and reverse linear search are algorithms listed to solve the problem. Moreover, Fukuda and Avis have published a library for it that works well for really large dimensional problems. There's even a [question](https://mathoverflow.net/questions/203966/computionally-efficient-vertex-enumeration-for-convex-polytopes) on this site claiming it's pseudo-polynomial. So which one is it?

https://mathoverflow.net/users/159040

Is the problem of vertex enumeration from an H-representation of a polytope NP-hard?

As noted in the comments this is not a decision problem. But more than that typically we discuss P and NP in terms of polynomial time with respect to input size. In this problem, the input is the collection of inequalities/half-spaces. However, the output is the collection of vertices which can be exponential in the number of inequalities defining the polytope. So, no algorithm could output the vertices using only polynomial time in the input. As a result, people often then talk about polynomial time in terms of the input and output size. I recommend the paper [Generating All Vertices of a Polyhedron Is Hard](https://doi.org/10.1007/s00454-008-9050-5) for information on the problem. Lastly, note the algorithms discussed on Wikipedia talk about things like $O(ndv)$ where $n$ is the number of facets, $d$ is the dimension, and $v$ is the number of vertices. So, $n+v$ gives the combined size of the input and output, but there is also the $d$. This means you need to fix a dimension, talk about parameterized complexity, etc. If you try to apply the algorithm to any dimension you have an issue for large $d$ (theoretically, even if it seems to work on your examples).

2

https://mathoverflow.net/users/51668

453324

https://mathoverflow.net/questions/453198

13

There seems to be some combinatorial fact that every subset $A$ of $G=(\mathbb{Z}/p)^{\times n}$ of cardinality $\frac{p^n-1}{p-1}+1$ containing $\vec{0}$ satisfies $(p-1)A=G$. ($p$ is a prime number.) Is that true and known in the literature? I would appreciate a reference or a proof. Thanks!

https://mathoverflow.net/users/510890

Subsets of $(\mathbb{Z}/p)^{\times n}$

There is an alternative elementary way to prove this and see where the number $\frac{p^n-1}{p-1}+1$ comes from. **Lemma:** If $|A|\geq \frac{p^n-1}{p-1}+1$ then every point in $\mathbb F\_p^n$ lies in a line that passes from two distinct points in $A$. Proof: If this was false then there would be a point $P\notin A$ such that every line through $P$ contains at most one point of $A$. However the number of distinct lines passing through $P$ is $\frac{p^n-1}{p-1}$, the size of the corresponding projective space. This gives a contradiction, since it implies $|A|\le \frac{p^n-1}{p-1}$. Returning to the statement in the OP, starting with an arbitrary point $P\in \mathbb F\_p^n$ the lemma tells us that there are points $M,N\in A$ such that $-P$ lies in the line passing through $M$ and $N$. Now consider the line consisting of the points $kM+(p-1-k)N$, for $0\le k\le p-1$. This line contains $(p-1)M$ and $(p-1)N$, therefore it also contains $(p-1)(-P)=P$. This means that $P=kM+(p-1-k)N$ for some $k$, which is in $(p-1)A$, as desired.

14

https://mathoverflow.net/users/2384

453328

https://mathoverflow.net/questions/453323

5

I think my question applies to most games, but for the sake of concreteness, I shall consider one specific game in this question. We consider the game posed by Ilias Kastanas in his paper *On the Ramsey property for sets of reals*: We first fix some $X \subseteq [\omega]^\omega$. 1. Player I start by choosing an infinite subset $A\_0 \subseteq \omega$. 2. Player II responds by choosing $x\_0 \in A\_0$, and then some $B\_0 \subseteq A\_0$ with $x\_0 < \min(B\_0)$. 3. Player I responds by choosing some $A\_1 \subseteq B\_0$. 4. Player II responds by choosing $x\_1 \in A\_1$, and then some $B\_1 \subseteq A\_1$ with $x\_1 < \min(B\_1)$. 5. Etc. Player I wins iff $\{x\_0,x\_1,\dots\} \in X$. Assume that Player I has a winning strategy $\sigma$. If $(A\_0,x\_0,B\_0,\dots,A\_n,x\_n,B\_n)$ is a partial play, then clearly the only things that determine what Player I should respond with are the finite sequence $(x\_0,\dots,x\_n)$ and $B\_n$, Player II's last played set. However, it appears to me that we cannot assume that $\sigma$ satisfies this property. Let's say that $\sigma$ is *uniform* if $\sigma(A\_0,x\_0,B\_0,\dots,A\_n,x\_n,B\_n)$ only depends on $x\_0,x\_1,\dots,x\_n,B\_n$. > > If Player I has a winning strategy, does Player I necessarily have a uniform strategy? > > > I'm also interested in how much the axiom of choice plays a part in the above statement.

https://mathoverflow.net/users/146831

Uniform strategy on Kastanas' game

This is a great question — definitely enjoyed. Assuming the axiom of choice, then the answer is yes. **Theorem.** Assume there is a well ordering of the real numbers. If player I has a winning strategy, then there is a uniform winning strategy. **Proof.** Suppose that $\sigma$ is a winning strategy for player I. Fix a well ordering of the collection of sets $B\subseteq\omega$. Now, suppose that we are playing the game, faced with the actual position $(A\_0, x\_0, B\_0, \ldots,A\_n,x\_n,B\_n)$. But let us look now only at the allowed information for a uniform strategy $(x\_0,\ldots,x\_n,B\_n)$. Consider the least with respect to the well ordering subset $B\_n'\subseteq B\_n$, such that there is an imaginary play of $B\_i'$ that accords with $\sigma$ and the data $(x\_0,\ldots,x\_n,B\_n')$, such that furthermore each $B\_i'\subseteq A\_i'$ is chosen as least at that stage, subject to being compatible with the data up to $n$. Let $\sigma$ respond to this imaginary play to give us a set $A\_{n+1}$, which we now play as the result for this new strategy. Since the play depended only on $(x\_0,\ldots,x\_n,B\_n)$, I have described a uniform strategy. Let me prove that it is winning. What I claim is that every sequence $x\_0,x\_1,x\_2,\ldots$ arising from a play according to this uniform strategy can be realized as arising in a play according to $\sigma$. The key point is that as $n$ increases, the imaginary sets $B\_i'$ will eventually stabilize for each $i$. To see this, consider what happens at stage $n+1$, after we've played $A\_{n+1}$. The actual play will respond with some $x\_{n+1}\in A\_{n+1}$ and $B\_{n+1}\subseteq A\_{n+1}$, and we will forget about the actual $B\_n$ and find some minimal $B\_{n+1}''\subset B\_{n+1}$ and new imaginary sets $B\_i''$ that produce a play according with $\sigma$ and the data $(x\_0,\ldots,x\_{n+1},B\_{n+1}'')$, and is minimal at each stage $i$. But the point now is that the requirement on $B\_i''$ is a bit lighter than that on the imaginary set $B\_i'$ chosen at stage $n$, provided that $x\_{n+1}$ is in $B\_n$, which it is. Earlier, we needed $B\_i'$ to contain $B\_n'$, but now we only need it to contain $B\_{n+1}''$, which is strictly smaller. So the set $B\_i''$ will be either equal to $B\_i'$ or preceed it in the well order. But it can only go down in the well order finitely often, and so eventually, for any given $i$, the imaginary versions of $B\_i'''$ will stabilize. In other words, if we have an infinite play according to the uniform strategy, we can find choices of $B\_i'''$ that give rise to the whole play. And since this imaginary play accords with $\sigma$, the set $\{x\_0,x\_1,\ldots\}$ will be in $X$, as desired. $\Box$

6

https://mathoverflow.net/users/1946

453330

https://mathoverflow.net/questions/453312

19

Last year, Andrej Bauer [gave a talk](https://www.youtube.com/watch?v=4CBFUojXoq4&list=FLuLR2oAJOXHh0nBVDDuj3Lg&index=24&t=71s) showing that there is a topos in which the set of Dedekind reals is (sub)countable, and thus, you cannot prove that $\mathbb{R}$ is uncountable without LEM. He claims in [this abstract](https://researchseminars.org/talk/ToposInstituteColloquium/57/) that this was an open problem. However, [it was already known](https://mathoverflow.net/questions/410703/subcountability/410755?_gl=1*8q3akn*_ga*NDIwNzk0NDg2LjE2NDA5NzU5Njg.*_ga_S812YQPLT2*MTY5MjgwMzczMC4yOTguMS4xNjkyODA0MzQzLjAuMC4w#410755) that it is consistent with CZF that every set (and in particular, the Dedekind reals) is subcountable. So is Bauer's and Hanson's result actually new? Also note that Bauer's and Hanson's result is not yet published, but it should supposedly be published soon according to [Bauer's blog](https://math.andrej.com/). Edit: Thanks to the answers by Andrej Bauer and James Hanson, I have realized my biggest misunderstanding. I for some reason implicitly thought subcountability and countability are equivalent, but this is not the case in the intuitionist setting. Thus, the consistency of the Dedekind reals being subcountable does not yield that it is consistent that the Dedekind reals are countable. Something else several others pointed out is that CZF is a generally weaker system than "ZFC without choice or LEM". So, even if the question were to be solved in CZF, this wouldn't mean the question "Can the real numbers be shown uncountable without excluded middle and without the axiom of choice?" was really settled. Bauer's and Hanson's topos construction gives a negative answer in the setting of higher-order intuitionist logic (and in IZF as per Hanson's comment). Thanks again to everyone for sharing your input!

https://mathoverflow.net/users/502850

Is Bauer–Hanson’s result “there is a topos where the Dedekind reals are countable” novel?

Please allow me to list some basic observations that might clear up things. I work constructively (without excluded middle) and without the axiom of choice, and assuming powersets are available. (One can contemplate working in a predicative setting such as CZF, where the Dedekind reals form a class, but I fail to see how it is useful to discuss countability of reals in a setting that does not even allow the formation of reals as a bona fide set.) #### Which reals? There are three standard constructions of reals, which differ constructively: * **Cauchy reals** $\mathbb{R}\_c$ are constructed as a quotient of rational Cauchy sequences. * **Dedekind reals** $\mathbb{R}\_d$ are constrcuted as (double-sided) Dedekind cuts of rationals. * **MacNeille reals** $\mathbb{R}\_m$ are constructed as a certain weaker version of Dedekind cuts of rationals. We have $\mathbb{R}\_c \subseteq \mathbb{R}\_d \subseteq \mathbb{R}\_m$, where all three inclusions might be proper. The MacNeille reals fail to satisfy $0 < x \lor x < 1$, which makes them less useful. Without countable choice, the Cauchy reals are not nice, either. One cannot even show that they are Cauchy-complete. So the canonical construction for reals is by Dedekind cuts, so most constructive mathematics is done with $\mathbb{R}\_d$. (Note also that $\mathbb{R}\_c = \mathbb{R}\_d$ in the presence of countable choice.) ### Countability and subcountability The definition of countability that works well is: $A$ is **countable** if there is a surjection $\mathbb{N} \to 1 + A$. When $A$ is inhabited this is equivalent to having a surjection $\mathbb{N} \to A$. A set $A$ is **subcountable** if there is $S \subseteq \mathbb{N}$ and a surjection $S \to A$. In particular, every subset of $\mathbb{N}$ is subcountable. A set $A$ is **uncountable** if it is not countable. A stronger property is **sequence-avoiding**: for every sequence $\mathbb{N} \to A$ there is an element of $A$ that is not a term of the sequence. **Theorem:** *The MacNeille reals are sequence-avoiding, thus uncountable.* *Proof.* See [A constructive Knaster–Tarski proof of the uncountability of the reals](https://arxiv.org/abs/1902.07366) by Ingo Blechschmidt and Matthias Hutzler. $\Box$ **Theorem:** *$\lbrace 0, 1\rbrace^\mathbb{N}$ and $\mathcal{P}(\mathbb{N})$ are sequence-avoiding, thus uncountable.* *Proof.* Cantor's diagonal method is constructive. Given $f : \mathbb{N} \to \lbrace 0, 1\rbrace^\mathbb{N}$, the sequence $n \mapsto 1 - f(n)(n)$ differs from $f(n)$ in the $n$-th place. Similarly, given $g : \mathbb{N} \to \mathcal{P}(\mathbb{N})$, the set $\lbrace n \in \mathbb{N} \mid n \notin f(n) \rbrace$ differs from $f(n)$ at $n$. $\Box$ Caveats: 1. Constructively the set of binary sequences $\lbrace 0, 1\rbrace^\mathbb{N}$, the powerset $\mathcal{P}(\mathbb{N})$, and the reals (of any kind) *cannot* be shown to be in bijective correspondence. 2. It cannot be shown constructively that every real has a digit expansion, so we cannot carry out the diagonal method on $\mathbb{R}$ that way. (This is also a good reason for *not* teaching uncountability of the reals using decimal expansions. The method of nested intervals is to be preferred, as it works with either excluded middle or countable choice.) **Theorem:** *If excluded middle holds then $\mathbb{R}\_c = \mathbb{R}\_d = \mathbb{R}\_m$, and they are all sequence-avoiding, thus uncountable.* *Proof.* See notes from your freshman year in analysis. $\Box$ **Theorem:** *If countable choice holds then $\mathbb{R}\_c = \mathbb{R}\_d$ and they are both sequence-avoiding, thus uncountable.* *Proof.* See for example Bishop's book “Foundations of constructive analysis" (1967, Section 2.2), where the method of nested intervals is employed, using countable choice. $\Box$ Contrary to classical mathematics, subcountability has very little to do with countability, apart from the obvious observation that every countable set is subcountable. **Theorem:** *There is a topos in which there is an injection $\mathbb{R}\_d \to \mathbb{N}$ and $\mathbb{R}\_d$ is sequence-avoiding, so there is no surjection $\mathbb{N} \to \mathbb{R}\_d$.* *Proof.* This happens in the realizability topos on Joel Hamkins's infinite-time Turing machines. I have not actually written this down, but the embedding is done much the same way as the embedding $\mathbb{N}^\mathbb{N} \to \mathbb{N}$ in [An injection from the Baire space to natural numbers](https://doi.org/10.1017/S0960129513000406). The idea is that infinite-time Turing machines can compute from any code of a rational Cauchy sequence converging to $x$ a canonical such code. $\Box$ It has also been known since at least the 1980's that in the effective topos $\mathbb{R}\_d$ is subcountable and sequence-avoiding. ### Further remarks about “sizes” of sets When you enter the constructive world, you should leave classical ideas about size behind. **Theorem:** *Suppose the following principle holds: if there are injections $A \to B$ and $B \to A$ then there is a bijection $A \to B$. Then excluded middle holds.* *Proof.* See [Cantor–Bernstein implies Excluded Middle](https://arxiv.org/abs/1904.09193) by Chad Brown and Cécilia Pradic. $\Box$ **Theorem:** *Suppose the following principle holds: every subset of a finite set is finite. Then excluded middle holds.* *Proof.* See the Anger stage of [Five stages of accepting constructive mathematics](http://dx.doi.org/10.1090/bull/1556). $\Box$ **Theorem:** *If every subcountable set is countable and Markov principle holds, then excluded middle holds.* *Proof.* See Proposition 2.6 of [Every metric space is separable in function realizability](https://doi.org/10.23638/LMCS-15%282%3A14%292019) by Andrej Bauer and Andrew Swan. $\Box$

24

https://mathoverflow.net/users/1176

453340

https://mathoverflow.net/questions/453097

4

Let $I$ be the (nerve of the) walking isomorphism (a simplicial set). Consider the inclusion $\Delta^1 \to I$ and the inclusion $\partial \Delta^1 \to \Delta^1$. Take their [pushout-product](https://ncatlab.org/nlab/show/pushout-product) to obtain a map $$\Delta^1 \times \Delta^1 \cup\_{\partial \Delta^1 \times \Delta^1} \partial \Delta^1 \times I \to \Delta^1 \times I$$ **Question 1:** Is this map a Joyal equivalence? **Question 2:** Is this map (cellularly) inner anodyne? If we reflect to $Cat$, then the map becomes an isomorphism. The content of this assertion is that if I have a commtutative square in a category $C$ and two opposite ends are isomorphisms, then we have an isomorphism in the arrow category $C^{\Delta^1}$. I am fairly certain this is also true in quasicategories, so the answer to (1) is presumably *yes*. I'd like a direct proof of this fact, and the most straightforward proof would be to explicitly fill some horns and see that it's cellularly inner anodyne -- i.e. obtained just by gluing on some inner horns -- no retracts required, and no fancier Joyal trivial cofibrations required. But perhaps this is not possible.

https://mathoverflow.net/users/2362

Is $\Delta^1 \times \Delta^1 \cup_{\partial \Delta^1 \times \Delta^1} \partial \Delta^1 \times I \to \Delta^1 \times I$ inner anodyne?

As requested, here is a proof that the map in question is inner anodyne that does not proceed by explicitly filling inner horns. By definition (more or less), a map is inner anodyne iff it has the LLP wrt all inner fibrations; but if moreover the codomain of that map is a quasi-category, as in our example, it suffices by pull-back to show that it has the LLP wrt all inner fibrations between quasi-categories. So let $p \colon X\to Y$ be an inner fibration between quasi-categories. By adjointness, the pushout-product of the boundary inclusion $i \colon \partial\Delta^1 \to \Delta^1$ with the inclusion $j \colon \Delta^1 \to \mathbb{I}$ has the LLP wrt $p$ iff the inclusion $j$ has the LLP wrt to the "pullback-internal hom" of $i$ and $p$, $\{i.p\} \colon \operatorname{Fun}(\Delta^1,X) \to \operatorname{Fun}(\partial\Delta^1,X) \times\_{\operatorname{Fun}(\partial\Delta^1,Y)}\operatorname{Fun}(\Delta^1,Y)$. Now, since $i$ is a bijective-on-0-simplices monomorphism and $p$ is an inner fibration between quasi-categories, this map $\{i,p\}$ is a conservative inner fibration between quasi-categories. So the result follows from the fact that any conservative inner fibration between quasi-categories has the RLP wrt to the inclusion $j \colon \Delta^1 \to \mathbb{I}$. This is a consequence of Joyal's special outer horn lifiting theorem, for this inclusion $j \colon \Delta^1 \to \mathbb{I}$ can be built up by filling outer horns $\Lambda^n\_n \to \Delta^n$ for $n > 1$ all of whose final edges are the edge of the original domain $\Delta^1$.

1

https://mathoverflow.net/users/57405

453341

https://mathoverflow.net/questions/453349

2

I'm considering Gaussian process on open domain $T$ in $\mathbb{R}^n$ and I tried to follow the abstract Wiener space construction of Gross. Since my sample paths are meant to be continuous, I thought that $sup$ norm would be nice measurable norm to complete the RKHS of mine (just as classical Wiener measure on $C[0,1]$), but since $T$ is open, this is not a good choice. Instead I considered countable family of semi norms $sup\_{K\_j}, j=1,2,...$ where compact sets $K\_j$ exhaust $T$. However with this family of semi norms $C(T)$ is merely Frechet space, not Banach space, hence I cannot use Gross's construction. Hence I like to know is there a generalized version of Gorss's construction, for example where measurable norms are replaced by family of semi norms, Banach space $W$ is replaced by Frechet space etc. One thing I noticed is that on "Encyclopedia of Mathematics" page about Wiener measure, the comment mentions about Wiener measure defined on $C[0,\infty)$, which seems to be closely related to my question, but I couldn't find the reference about it.

https://mathoverflow.net/users/508638

Reference for Wiener type measure on $C(T)$ when $T$ is open

I think you can adapt the proof in [these notes of mine](https://arxiv.org/abs/1607.03591v2), Theorem 4.44. The first step in the proof of my notes is to pick, for each $n$, a finite-rank projection $P\_n$ such that for all finite-rank projections $P \perp P\_n$, we have $$\tilde{\mu}\left(\left\{ h \in H : \|Ph\|\_W > 2^{-n}\right\}\right) < 2^{-n}.$$ Suppose instead we have a countable sequence of measurable seminorms $|\cdot|\_k$ $k=1,2,3,\dots$, which define the topology of $W$. Then for each $n$, we can apply this argument $n$ times (enlarging the projection $P\_n$ as we go) to instead find $P\_n$ such that for $P \perp P\_n$ we have $$\tilde{\mu}\left(\left\{ h \in H : |Ph|\_k > 2^{-n}\right\}\right) < 2^{-n} , \qquad k = 1, 2, \dots, n.$$ A few steps later, we then obtain for each $k$ that $\mathbb{P}(|S\_n - S\_{n-1}|\_k > 2^{-n}) < 2^{-n}$ for all $n \ge k$. This again implies that $\sum\_n \mathbb{P}(|S\_n - S\_{n-1}|\_k > 2^{-n}) < \infty$. Borel-Cantelli gives us that for each $k$, $\mathbb{P}$-a.s., $S\_n$ is Cauchy in $|\cdot|\_k$. Taking a countable intersection, we in fact have that, $\mathbb{P}$-a.s., $S\_n$ is Cauchy in $|\cdot|\_k$ for all $k$. That is, on some event $E$ with $\mathbb{P}(E)=1$, we have for all $\omega \in E$ that $S\_n(\omega)$ is Cauchy in $|\cdot|\_k$ for all $k$. Since our space is complete, this implies that $S\_n(\omega)$ converges in $W$ to some $S(\omega)$. By standard arguments, $S$ is measurable, and its law $\mu = \mathbb{P} \circ S^{-1}$ is the measure we want.

2

https://mathoverflow.net/users/4832

453353

https://mathoverflow.net/questions/453361

3

Suppose that graphs $A$ and $B$ with $V(A)=V(B)$ have Hadwiger numbers $a$ and $b$. That is, $K\_a$ and $K\_b$ are the largest clique minors of $A$ and $B$, respectively. Are there upper bounds on the Hadwiger number of the union graph $A\cup B$ (with edge set $E(A\cup B)=E(A)\cup E(B)$) using $a$ and $b$ like $a+b$ or $a\cdot b$?

https://mathoverflow.net/users/511540

Bounding the size of clique minor of the union of two graphs

Take a complete graph $G\_n$, replace each edge $uv$ to a path $uxv$. Imagine that $ux\in A$, $vx\in B$, and that's for every edge. It looks that both $A,B$ are forests, so have small Hadwiger number.

7

https://mathoverflow.net/users/4312

453362

https://mathoverflow.net/questions/453343

20

Cross post from [Maths stack exchange](https://math.stackexchange.com/questions/4495174/a-difficult-integral-for-the-chern-number?noredirect=1#comment10099463_4495174) --- The integral $$ I(m)=\frac{1}{4\pi}\int\_{-\pi}^{\pi}\mathrm{d}x\int\_{-\pi}^\pi\mathrm{d}y\phantom{,} \frac{m\cos(x)\cos(y)-\cos x-\cos y}{\left( \sin^2x+\sin^2y +(m-\cos x-\cos y)^2\right)^{3/2}} $$ gives the Chern number of a certain vector bundle [1](https://math.stackexchange.com/questions/4495174/a-difficult-integral-for-the-chern-number?noredirect=1#comment10099463_4495174) over a torus. It can be shown using the theory of characteristic classes that $$ I(m) = \frac{\mathrm{sign}(m-2)+\mathrm{sign}(m+2)}{2}-\mathrm{sign}(m) = \begin{cases}1 & -2< m < 0 \\ -1 & 0 < m < 2 \\0 & \text{otherwise}\end{cases}. $$ Is there any way to evaluate this integral directly (i.e. without making use of methods from differential geometry) to obtain the above result? I should mention that the above integral can be written as ($1/4\pi$ times) the solid angle subtended from the origin of the unit vector $\hat{\mathbf{n}}$, $$ I(m)=\frac{1}{4\pi}\int\_{-\pi}^{\pi}\mathrm{d}x\int\_{-\pi}^\pi\mathrm{d}y\phantom{,} \hat{\mathbf{n}}\cdot\left(\partial\_x \hat{\mathbf{n}}\times \partial\_y \hat{\mathbf{n}}\right), $$ where $\mathbf{n}(m)=(\sin x, \sin y, m- \cos x-\cos y)$. While this form makes it very straightforward to evaluate $I(m)$, I am interested in whether there is a way to compute this integral using more standard techniques. --- [1](https://math.stackexchange.com/questions/4495174/a-difficult-integral-for-the-chern-number?noredirect=1#comment10099463_4495174) B. Bernevig *Topological Insulators and Topological Superconductors* Chapter 8

https://mathoverflow.net/users/490549

A difficult integral for the Chern number

If Stokes' theorem counts as a standard technique, then here's an answer: Introduce a "vector potential" \begin{equation} A\_i = \frac{1-\hat n\_z}{\hat n\_x^2 + \hat n\_y^2}(\hat n\_x \partial\_i \hat n\_y - \hat n\_y \partial\_i \hat n\_x)\,,\qquad i=x,y\,, \end{equation} which is defined everywhere, except where $\hat n\_x^2 + \hat n\_y^2=0$, which is $(x,y)\in\{(0,0),(0,\pi),(\pi,0),(\pi,\pi)\}$ (assuming $-\pi<x,y\leq \pi$). One can check that \begin{equation} \Omega(x,y) \equiv \hat{\mathbf{n}}\cdot(\partial\_x \hat{\mathbf{n}} \times \partial\_y \hat{\mathbf{n}}) = \partial\_xA\_y -\partial\_y A\_x = \nabla\times \mathbf{A}\,. \end{equation} Now, going back to the integral, since the integrand is periodic, we can move the integration limit, such that the singular points of $\mathbf{A}$ are not at the boundary of the integration region, for instance \begin{equation} I(m) = \frac{1}{4\pi}\int\_{-\pi+1}^{\pi+1}\mathrm{d} x \int\_{-\pi+1}^{\pi+1}\mathrm{d} y\, \Omega(x,y)\,. \end{equation} We can now exclude small disks centered at the singular points from the intragral and treat them separately: $$R = [-\pi+1,\pi+1]\times[-\pi+1,\pi+1] \\ = R' \setminus D(0,0,\epsilon)\setminus D(0,\pi,\epsilon)\setminus D(\pi,0,\epsilon)\setminus D(\pi,\pi,\epsilon)$$ where $D(x,y,r)$ is a disk centered at $(x,y)$ with radius $r$, and $0<\epsilon<1$. $R'$ is now a square with 4 holes, and the integral is written as \begin{align} I(m) &= \frac{1}{4\pi}\int\_{R'} \mathrm{d} x \mathrm{d} y\, \Omega(x,y) + \frac{1}{4\pi}\int\_{D(0,0,\epsilon)} \mathrm{d} x \mathrm{d} y\, \Omega(x,y) + \dots\\ &= \frac{1}{4\pi}\int\_{R'} \mathrm{d} x \mathrm{d} y\, \nabla\times \mathbf{A} + \frac{1}{4\pi}\int\_{D(0,0,\epsilon)} \mathrm{d} x \mathrm{d} y\, \Omega(x,y) + \dots\\ &= \frac{1}{4\pi}\oint\_{\partial R'} \mathrm{d} \mathbf{l}\cdot \mathbf{A} + \frac{1}{4\pi}\int\_{D(0,0,\epsilon)} \mathrm{d} x \mathrm{d} y\, \Omega(x,y) + \dots\\ &= \left(\frac{1}{4\pi}\oint\_{\partial R} \mathrm{d} \mathbf{l}\cdot \mathbf{A} -\frac{1}{4\pi}\oint\_{\partial D(0,0,\epsilon)} \mathrm{d} \mathbf{l}\cdot \mathbf{A} -\dots\right) + \frac{1}{4\pi}\int\_{D(0,0,\epsilon)} \mathrm{d} x \mathrm{d} y\, \Omega(x,y) + \dots\\ \end{align} In the last two steps we used Stokes' theorem and that $\partial R' = \partial R \cup \partial D(0,0,\epsilon) \cup \dots$, and $\pm$ signs account for the orientation of the boundary. Since $\Omega(x,y)$ is finite everywhere, when $\epsilon\rightarrow 0$ \begin{equation} \left|\int\_{D(x\_0,y\_0,\epsilon)} \mathrm{d} x \mathrm{d} y\, \Omega(x,y)\right| \leq \pi\epsilon^2 \max\_{x,y} |\Omega(x,y)| \xrightarrow{\epsilon\rightarrow 0} 0 \end{equation} Furthermore, $\oint\_{\partial R} \mathrm{d} \mathbf{l}\cdot \mathbf{A}=0$, because $\mathbf{A}$ is periodic, so the integrals over opposite sides of the square cancel out. We're left with \begin{equation} I(m) = -\frac{1}{4\pi}\lim\_{\epsilon\rightarrow 0}\oint\_{\partial D(0,0,\epsilon)} \mathrm{d} \mathbf{l}\cdot \mathbf{A} -\dots \end{equation} We have \begin{align} \frac{1}{4\pi}\lim\_{\epsilon\rightarrow 0}\oint\_{\partial D(x\_0,y\_0,\epsilon)} \mathrm{d}\mathbf{l}\cdot\mathbf{A} = \frac{1}{4\pi}\int\_0^{2\pi} \mathrm{d} \phi \lim\_{\epsilon\rightarrow 0} \epsilon (-\sin\phi,\cos\phi)\cdot\mathbf{A}(x\_0 + \epsilon\cos\phi, y\_0 + \epsilon\sin\phi)\,. \end{align} Now one just has to evaluate this for all 4 disks. The results are \begin{align} (0,0)&: \frac{-m+\sqrt{(m-2)^2}+2}{2\sqrt{(m-2)^2}} = \theta(-m+2)\\ (0,\pi)&: \frac{m-\sqrt{m^2}}{2\sqrt{m^2}} = -\theta(-m)\\ (\pi,0)&: \frac{m-\sqrt{m^2}}{2\sqrt{m^2}} = -\theta(-m)\\ (\pi,\pi)&: \frac{-m+\sqrt{(m+2)^2}-2}{2\sqrt{(m+2)^2}} = \theta(-m-2)\,, \end{align} where $\theta$ is Heaviside step function. The total integral then is \begin{equation} I(m) = 2\theta(-m) - \theta(-m+2) - \theta(-m-2) = \begin{cases} -1 & 0<m<2 \\ 1 & -2<m<0 \\ 0 & |m|>2 \\ \end{cases}\,. \end{equation}

24

https://mathoverflow.net/users/165560

453369

https://mathoverflow.net/questions/453236

3

Let $G$ and $H$ two connected linear Lie groups. Is $G\ltimes H$ also linear?

https://mathoverflow.net/users/509535

Semidirect product of two linear groups

No. There is a semidirect product $\mathbf{R}\ltimes (\mathbf{R}\times\mathbf{R}/\mathbf{Z})$ that is not linear. Namely, the quotient of the Heisenberg group by a lattice of its center has this form. Replacing $\mathbf{R}$ with $\mathbf{C}$ also yields an example in the realm of complex Lie groups. These are classical examples of non-linear Lie groups. Indeed, if $G$ is a solvable connected Lie group then (the closure of) its derived subgroup is simply connected (indeed, taking a complex representation, one can triangulate it, and this directly follows since every closed connected subgroup of the upper unipotent group is simply connected).

6

https://mathoverflow.net/users/14094

453378

https://mathoverflow.net/questions/453370

10

We say a measurable function $f: \mathbb R^n \to \mathbb R$ is *essentially continuous* if the inverse image of any open set $O$ differs from an open set by a set of null measure, in the sense that there exists an open subset $U$ of $\mathbb R^n$ such that $\mu(f^{-1} (O) \, \Delta \, U) = 0$ where $\Delta$ denotes the symmetric difference and $\mu$ denotes the Lebesgue measure. **Question:** Is it true that $f$ is essentially continuous if and only if $f$ agrees a.e. with a function that is continuous a.e.? *Remark: The chosen open set $U$ is allowed to be empty.*

https://mathoverflow.net/users/173490

A topological characterisation of a.e. continuity

I adapt the proof of a more general result: Theorem 4.12 in Chapter XI of K. Kuratowski and A. Mostowski’s book “*Set Theory: with an introduction to descriptive set theory*” (see page 408). **Claim.** *$f$ is essentially continuous iff there exists a null set $N$ such that $f|({\bf R}^n-N)$ is continuous.* **Proof.** ($\Rightarrow$) Let $U\_1$, $U\_2$, … be an open base of $\bf R$. By assumption, $$f^{-1}(U\_n)=(O\_n-A\_n)\cup B\_n,$$ where $O\_n$ is open and $A\_n$, $B\_n$ are null sets. Set $N=\bigcup\_nA\_n\cup\bigcup\_nB\_n$, so that $N$ is a null set. We show that $g=f|({\bf R}^n-N)$ is continuous. Let $H\subset\bf R$ be open; we prove that $g^{-1}(H)=f^{-1}(H)-N$ is open in ${\bf R}^n-N$. Now $H=U\_{k\_1}\cup U\_{k\_2}\cup\cdots$, and so $$g^{-1}(H)=\bigcup\_nf^{-1}(U\_{k\_n})-N=\bigcup\_n\bigl((O\_{k\_n}-A\_{k\_n})\cup B\_{k\_n}\bigr)-N.$$ Since $A\_{k\_n}\cup B\_{k\_n}\subset N$, it follows that $$g^{-1}(H)=\bigcup\_nO\_{k\_n}-N,$$ which completes the proof since $\bigcup\_nO\_{k\_n}$ is open. ($\Leftarrow$) Let $N$ be a null set and suppose $g=f|({\bf R}^n-N)$ is continuous. Then, if $H\subset\bf R$ is open, the set $g^{-1}(H)=f^{-1}(H)-N$ is open in ${\bf R}^n-N$; that is, there exists open $O\subset{\bf R}^d$ such that $f^{-1}(H)-N=O-N$. Therefore, $$\begin{align\*} f^{-1}(H) &= (f^{-1}(H)-N)\cup(f^{-1}(H)\cap N)\\ &=(O-N)\cup(f^{-1}(H)\cap N). \end{align\*}$$ Since $N$ is a null set, $f^{-1}(H)\cap N$ is also a null set, and it follows that $f^{-1}(H)\triangle O$ is a null set. $\square$

2

https://mathoverflow.net/users/118745

453383

https://mathoverflow.net/questions/452795

3

I have a question about following argument I found in [these notes](https://www.google.de/url?sa=t&source=web&rct=j&opi=89978449&url=https://people.math.rochester.edu/faculty/doug/otherpapers/WebbMF.pdf&ved=2ahUKEwiiu9Lu7d6AAxVg_rsIHU-KCSEQFnoECBAQAQ&usg=AOvVaw1OZ96p6a5V9Y-wW04DRo45) on Mackey functors: > > **(2.1) LEMMA.** (page 6) Let $G$ be a finite group and $J$ any subgroup. Whenever $H$ and $K$ are subgroups of $J$, there is a pullback diagram of $G$-sets > > > $$ \require{AMScd} \begin{CD} \Omega @>{} >> G/K \\ @VVV @VVV \\ G/H @>{}>> G/J \end{CD} $$ where we have Mackey double coset decomposition $\Omega= \coprod\_{x \in [H\backslash J/K]}G/(H \cap x^{-1}Kx)$ The proof shows by hand for the special case $G=J$ that the pullback in $$ \require{AMScd} \begin{CD} J/K \times J/H @>{} >> J/K \\ @VVV @VVV \\ J/H @>{}>> J/J = \{\*\} \end{CD} $$ is given by decomposition $J/K \times J/H= \coprod\_{x \in [H\backslash J/K]}J/(H \cap x^{-1}Kx)$ and then claims then claims that the result follows from application of the usual induction functor $\text{Ind}^G \_J: (J-\text{Set}) \to (G-\text{Set})$ sending a $J$-set $S$ to $G$-set $G \times\_J S$. **Question:** Why this functor preserves *limits* (eg a pullback as in this case). It is well known that the induction functor $\text{Ind}^G \_J$ is *left-adjoint* to usual restriction functor $\text{Res}^J \_G$, and therefore definitely preserves colimits. But the pullback in the question is a limit, so I not understand why the induction functor applied to $J/K \times J/H$ from the special case preserves the limit/ pullback property. A remark: I found googling this problem several alternative proofs of the claimed decomposition formula, but the objection of this question is really to understand why *this* argument given by Webb in the linked notes works, ie why in this situation the induction functor commutes with the pullback.

https://mathoverflow.net/users/108274

Mackey coset decomposition formula

The induction functor preserves all *(weakly) connected* limits (note that this covers pullbacks, but not products - and of course, because it doesn't preserve the terminal object, it couldn't do both !). There is a more general way to argue, but in this case you can say as follows (there is also probably a more concrete one) : one can check on underlying objects whether this preserves a given limit diagram. But forgetting the $G$-structure on $Ind\_J^G$ is equivalent to forgetting the $J$-structure and taking a $G/J$-indexed constant coproduct. Thus it suffices to check that coproducts on the category of sets commute with weakly connected limits. To prove this, note that for a given set $I$ (the indexing set), the equivalence $Set^I \simeq Set\_{/I}$ intertwines the "coproduct" functor and the forgetful functor. But now it is a general fact that if $C$ has $K$-shaped limits, and $K$ is weakly connected, then the forgetful functor $C\_{/x} \to C$ preserves $K$-shaped limits.

3

https://mathoverflow.net/users/102343

453392

https://mathoverflow.net/questions/417448

23

Cardinalities in non-standard models of $\mathsf{ZF}$ are not generally reflected externally, but certain internal facts about cardinalities are always externally reflected (e.g., any internally infinite set is externally infinite; if $|A| \leq |B|$ internally, then the same holds externally; the internal powerset of a given set is no larger than the external powerset; etc.). (By an abuse of notation, I'll write $|A|$ for $|\{a \in M : M \models a \in A\}|$.) In $\mathsf{ZFC}$, $\omega$ embeds into any infinite set, so for any $M \models \mathsf{ZFC}$, we always have that $|\omega^M| \leq |A|$ for any $A \in M$ such that $M \models \text{“}A\text{ is infinite"}$ (i.e., $A$ is internally infinite). In $\mathsf{ZF}$ on the other hand, it can be the case that there are infinite sets admitting no injection from $\omega$. At most we know that each initial segment of $\omega$ injects into a given infinite set. Externally, this means that if $M \models \mathsf{ZF}$ and $M \models \text{“}A\text{ is infinite"}$, then $|A| \geq |n|$ for each $n \in \omega^M$. This leaves open the possibility that $A$ is actually strictly smaller than $\omega^M$, but not that much smaller. In particular, we must have that $|\omega^M| \leq |A|^+$. My question is whether that bound is ever exact. > > **Question:** Is there a model $M$ of $\mathsf{ZF}$ with an $A \in M$ such that $M\models\text{“}A\text{ is infinite"}$ and $|\omega^M| = |A|^+$? > > > Certain basic model-theoretic facts about two-cardinal models and Vaughtian pairs tell us that this is equivalent to the existence of a Vaughtian pair of models of $\mathsf{ZF}$ in which $\omega^M$ grows but some internally infinite $A$ does not grow. ​Stated more formally: **Fact:** The following are equivalent: * The above question has a positive answer. * There is a model $M\models \mathsf{ZF}$ with an internally infinite element $A$ such that $|A| = \aleph\_0$ and $|\omega^M| = \aleph\_1$. * There is a model $N \models \mathsf{ZF}$ with an internally infinite element $B$ and an elementary extension $N' \succ N$ satisfying that there is an $n \in N' \setminus N$ such that $N' \models n \in \omega$ but for all $b \in N'$, if $N' \models b \in B$, then $b \in N$. It's possible to get a more detailed equivalent with the omitting types theorem, but it's a bit awkward to state.

https://mathoverflow.net/users/83901

Can a non-standard model $M$ of $\mathsf{ZF}$ contain an internally infinite set that is externally smaller than $\aleph_0^M$?

The answer is yes. Let $M$ be a countable transitive model of $\mathsf{ZF}$ containing an amorphous set $A$ (i.e., every subset of $A$ is either finite or co-finite). Let $T$ be the theory consisting of the elementary diagram of $M$ together with a fresh constant $c$ and the sentences $c \in \omega$ and $n < c$ for each standard natural $n$. Let $(a\_i)\_{i<\omega}$ be an (external) enumeration of $A$. By the omitting types theorem and the argument in the question, it is sufficient to show that the type $\Sigma(x) = \{x \in A\}\cup\{x \neq a\_i : i < \omega\}$ is not principal (i.e., for each consistent formula $\varphi(x)$, it is not the case that $\varphi(x) \vdash \Sigma(x)$). (If we can show this then we can build a model of $T$ in which $\Sigma(x)$ is omitted, which is precisely the Vaughtian pair condition we need.) Suppressing constants from $M$, we'll write these formulas as $\varphi(x,c)$ to emphasize the role of $c$. By comprehension, we have that for any $M$-formula $\varphi(x,y)$, there is a set $Q \subseteq A \times \omega$ such that for any $a \in A$ and $n < \omega$, $\varphi(a,n)$ holds if and only if $\langle a,n \rangle \in Q$. This means that it is sufficient to consider formulas of the form $\langle x,c \rangle \in Q$ for $Q \subseteq A\times \omega$. Given $Q \subseteq A \times \omega$ and $n < \omega$, we'll write $Q(n)$ for $\{a \in A : \langle a,n \rangle \in Q\}$. > > **Lemma.** For any $Q \subseteq A \times \omega$, there is a finite sequence $B\_0,\dots,B\_{k-1}$ of subsets of $A$ such that for all $n < \omega$, $Q(n) = B\_i$ for some $i<k$. > > > *Proof.* (This is probably standard somewhere.) Let $Q'$ be the set satisfying that for each $n < \omega$, $Q'(n) = Q(n)$ if $Q(n)$ is finite and $Q'(n)= A \setminus Q(n)$ if $Q(n)$ is infinite. Let $A\_n = Q'(n) \setminus \bigcup\_{m<n}Q'(m)$. It must be the case that $A\_n = \varnothing$ for all sufficiently large $n$ (otherwise we would have a surjection from a subset of $A$ onto $\omega$, which cannot happen). This implies that there is a finite $C \subset A$ such that $Q'(n) \subseteq C$ for all $n$. Let $B\_0,\dots,B\_{k-1}$ be an enumeration of all subsets of $C$ and all complements of subsets of $C$ (which is necessarily finite). $\square$ Now fix a set $Q \subseteq A \times \omega$. We just need to show that either $T \vdash \neg (\exists a \in A) \langle a,c \rangle \in Q$ or there is an $i<\omega$ such that $T \cup \{\langle a\_i,c \rangle \in Q\}$ is consistent. An immediate compactness argument shows that $T \vdash \neg (\exists a \in A) \langle a,c \rangle \in Q$ if and only if $Q(n) = \varnothing$ for all sufficiently large $n<\omega$, so assume that $Q(n)$ is not eventually empty. By the lemma, there must be some non-empty $B \subseteq A$ such that $Q(n) = B$ for infinitely many $n<\omega$. Fix some $a\_i \in B$. We now have by compactness that $T\cup \{\langle a\_i,c \rangle \in Q\}$ is consistent, whence the formula $\langle x,c \rangle \in Q$ does not isolate $\Sigma(x)$. Since we can do this for any $Q \subseteq A \times \omega$, we have by the omitting types theorem that there is an elementary extension $N \succ M$ with the property that $A^N = A^M$ and $\omega^N \supset \omega^M$. To finish we just apply the standard Vaughtian pair argument. Using the pair $(N,M)$, we can build a pair $(D\_1,D\_0) \equiv (N,M)$ with $D\_0$ and $D\_1$ isomorphic and $\omega$-homogeneous. Then we can build an elementary chain $(D\_\alpha)\_{\alpha < \omega\_1}$ starting with $D\_0$ and $D\_1$ satisfying that $(D\_{\alpha+1},D\_\alpha)\cong (D\_1,D\_0)$ for each $\alpha$. The union $D = \bigcup\_{\alpha < \omega\_1}$ then has the property that $|A^{D}|=\aleph\_0$ yet $|\omega^{D}| = \aleph\_1$, which is what we were after.

8

https://mathoverflow.net/users/83901

453394

https://mathoverflow.net/questions/453267

4

I have a second-order ODE with an unknown parameter $p$, $$\frac{y''}{(1+y'^{2})^{\frac{3}{2}}} -p - A(x-B)^2 =0,$$ where $x$ is the independent variable, $y$ is the unknown function, $p$ is unknown parameter to be determined, and $A$ and $B$ are known constants. I am given $y$ values at three $x$ points, $y\_0(x=x\_0)$, $y\_1(x=x\_1)$ and $y\_2(x=x\_2)$. I have seen some boundary value problem examples from MATLAB (bvp4c) or Scipy (solve\_bvp), where the boundary conditions must be on the boundary exactly such as $y(a)$, $y(b)$, $y'(a)$. [Solve BVP with Unknown Parameter](https://www.mathworks.com/help/matlab/math/solve-bvp-with-unknown-parameter.html#responsive_offcanvas) [scipy.integrate.solve\_bvp](https://docs.scipy.org/doc/scipy/reference/generated/scipy.integrate.solve_bvp.html#scipy.integrate.solve_bvp) Is there a way to solve my above ODE equation given $y$ values not on two boundaries? Currently I am using an optimizing method to search best solution of parameter p and $y'(x\_0)$ by solving the IVP problem, but it is time consuming. Could you kindly give me a more efficient way to solve it? Thank you.

https://mathoverflow.net/users/511129

How to solve an ODE with an unknown parameter but given y values at three points

According to the suggestions from @AVK, I built the problem in MATLAB(but I have no MATLAB to test) I have read that example and try to adapt it to my case . $$ \frac{y''}{(1+y'^{2})^{\frac{3}{2}}} -p - A(x-B)^2 =0 $$ given $$ \begin{equation} \begin{cases} y(x\_a)=YA \\ y(x\_c)=YC \\ y(x\_b)=YB \end{cases}\, \end{equation} $$ where $ x\_a \lt x\_c \lt x\_b$ Let $y\_1=y,\quad y\_2=y'$ $$ \begin{equation} \begin{cases} y\_1'=y\_2, \\ y\_2'=(1+y\_2^{2})^{\frac{3}{2}}\cdot (p + A(x-B)^2) =0 \end{cases}\, \end{equation} $$ ``` function dydx = fcn(x,y,regin,p) % equation to solve A = 0.001 B = 1.0e-3 dydx = zeros(2,1); dydx = [y(2) (1+y(2)^2)^(3/2)*(p+A*(x-B)^2)]; end ``` I apply the following BCs. ``` function res = bcfcn(ya,yb,p) % boundary conditions YA = 0.0 YB = 5.0 YC = 3.0 res = [ya(1,1)-YA % y1(xa)=ya ya(1,2)-YC % y1(xc)=yc Continuity yb(1,1)-YC % y1(xc)=yc Continuity ya(2,2)-yb(2,1) % y2(xc)=y2(xc) Continuity yb(1,2)-YB]; % y1(xb)=yb end ``` initialization ``` xc = 1.0; xa = 0.0; xb = 2.0 p = 0.0 xmesh = [0 0.25 0.5 0.75 xc xc 1.25 1.5 1.75 2]; yinit = [0.0; 0.0]; solinit = bvpinit(xmesh,yinit,p); ``` solve ``` sol = bvp5c(@fcn, @bcfcn, solinit); fprintf('sol.parameters %7.3f.\n',... sol.parameters); %check solution ypred = deval(sol,[xa,xc,xb]) fprintf("%7.3f",ypred(1,:)) ``` I don't know if it is correct. Because I have no MATLAB on hand, I just check it in mathworks online site. I want to try a Fortran solver [COLNEW](https://people.sc.fsu.edu/%7Ejburkardt/f77_src/colnew/colnew.html) to do similar work. If someone has experience on this, please help me. [ADD] I also found a newer solver [`BVP_SOLVER`](https://cs.smu.ca/%7Emuir/BVP_SOLVER_Webpage.shtml). Though it is easier to use, it only supports two-point BVPs and separated BCs, it is not convenient to use on my problem. [Update] Finally, I have solved the problem(MP-BVP with unknown parameters) using COLNEW, which is really a great solver.

2

https://mathoverflow.net/users/511129

453405

https://mathoverflow.net/questions/453093

2

Let $R$ be a commutative Noetherian ring, and $\phi: R \to R$ be a ring homomorphism. For an $R$-module $M$, let $^{\phi}M$ be the $R$-module defined via restriction of scalars via $\phi$, i.e., as abelian groups, $M$ and $^{\phi}M$ are the same, but the $R$-action on $^{\phi}M$ is given by $r\*x:=\phi(r)x, \forall r \in R, x \in M$. For any $R$-modules $M,N$, it is clear that $\text{Hom}\_R(M,N)\subseteq \text{Hom}\_R(^{\phi}M, ^{\phi}N)$. In particular, any exact sequence of $R$-modules $M\xrightarrow{f} N \xrightarrow{g} L$ gives rise to an exact sequence of $R$-modules $^{\phi}M\xrightarrow{f} \space^{\phi}N \xrightarrow{g} \space ^{\phi}L$. Now I want to apply the above scenario to the case where $(R,\mathfrak m,k)$ is a prime characteristic, $F$-finite, local Cohen--Macaulay ring of dimension $1$, $\phi$ is some $e$-th iteration of the Frobenius map, $M=R$, and $f$ is the multiplication by some non-zero-divisor $x\in \mathfrak m$. If I am not mistaken, then $^{\phi}R$ is exactly what is usually written as $F^e\_\*R$. The exact sequence $0\to R \xrightarrow{x} R\to R/xR\to 0$ then gives rise to exact sequence $0\to F^e\_\* R \xrightarrow{x} F^e\_\* R \to F^e\_\*(R/xR)\to 0$ implying $F^e\_\*R/xF^e\_\*R \cong F^e\_\*(R/xR)$. Now, $R/xR$ is an Artinian local ring, so $\mathfrak m^{p^e}(R/xR)=0$ for all large enough $e$, hence $\mathfrak m F^e\_\*(R/xR)=0$ for all large enough $e$, hence $F^e\_\* (R/xR)$ is a finite dimensional $k$-vector space. So, $F^e\_\*R/xF^e\_\*R \cong F^e\_\*(R/xR)$ is a finite dimensional $k$-vector space for large enough $e$. Now, take $R$ to be $F$-split, for instance take $R=\mathbb F\_p[[u,v]]/(uv)$, and $x:=u+v$ a non-zero-divisor on $R$. Then, $R$ is a direct summand of $F^e\_\*R$ for all $e>0$, hence $R/xR$ is a direct summand of $F^e\_\*R/xF^e\_\*R \cong F^e\_\*(R/xR)$, but this is a finite dimensional $k$-vector space for large enough $e$, hence $R/xR$ is annihilated by $\mathfrak m$, i.e., $\mathfrak m \subseteq xR$, hence $\mathfrak m=xR$, so $R$ must be regular! But clearly, $R$ is not regular!! My question is: Where am I going wrong in the above argument?

https://mathoverflow.net/users/135389

Finding the mistake in an argument concerning $F$-finite $F$-split local Cohen--Macaulay ring of dimension $1$

@uno, you are correct in the comments above, $F^e\_\*(-)$ is not $R$-linear. In particular, the mistake is the line right here: > > The exact sequence $0\to R \xrightarrow{x} R\to R/xR\to 0$ then gives rise to exact sequence $0\to F^e\_\* R \xrightarrow{x} F^e\_\* R \to F^e\_\*(R/xR)\to 0$ implying $F^e\_\*R/xF^e\_\*R \cong F^e\_\*(R/xR)$. > > > You have to apply the functor $F^e\_\*$ to the map $\cdot x$ as well. In particular, what you get is an exact sequence: $$0\to F^e\_\* R \xrightarrow{F^e\_\* (\cdot x)} F^e\_\* R \to F^e\_\*(R/xR)\to 0.$$ The $F^e\_\* (\cdot x)$-map is not the same as multiplication by $x$. Multiplication by $x$ is the same as the $F^e\_\* (\cdot x^{p^e})$-map though.

1

https://mathoverflow.net/users/3521

453408

https://mathoverflow.net/questions/453398

1

*(This question is a repost of a deleted question I asked, because the previous version had several elements missing)* **Setting** For fixed $N \in \mathbb{N}$, **I wish to compute the entries of a matrix $P\_N$** that is $2^N \times 3^N$-dimensional, real-valued and satisfies: 1. $P\_N^T \textbf{1} = \textbf{1}$ 2. $\textbf{1} P\_N^T = \frac{3^N}{2^N} \textbf{1}^T$ where $\textbf{1}$ is the all 1's vector in the appropriate dimension. Additionally, $P\_N$ is such that: (i) its first $2^N$ columns are the first $2^N$ standard basis vectors (ii) the remaining columns **contain 0's in all positions except two, which are unknown entries, $w\_{i,j}$, that I need to solve for.** After some analysis, we can conclude that there are (i) $2(3^N - 2^N)$ total variables to solve for, and (ii) $3^N$ equations that they satisfy, which implies that for $N > 3$, the system is "undetermined", so has either no solutions or infinitely many solutions for the $(w\_{i,j})\_{i,j}$ (In actuality, I know everything about where the non-zero entries of $P$ are, as I know the underlying system of equations it is representing. However, I have not included this, as it may not be necessary to my questions, and more importantly it's really complicated and, as I'm in the middle of formalizing it, the structure is not entirely clear yet either to me). **Question** Are there any computational methods or matrix tricks to show that such a system has a solution, for arbitrary $N$? The reason I ask is because the *form* of the matrix makes it really tricky to do so explicitly in this case, and so I wonder if there exist other methods to do this (either computational or theoretical), as admittedly I don't have much background in matrix analysis. I apologize in this question is vague, I wasn't sure and so I can certainly expand if needed, but really just providing me with the most general suggestions is alright too! Thank you all!

https://mathoverflow.net/users/511514

Methods to solve for a matrix whose entries satisfy certain properties

There are zero or infinitely many solutions depending on where the non-zero entries have to be. So there is no general-purpose answer. I don't think your equations are properly stated, as $\boldsymbol 1$ seems to have at least two different meanings. I'll interpret your intention as that every row sums to $3^N/2^N$ and every column sums to $1. If that is incorrect, please let us know. For a subset $R$ of the rows, let $N(R)$ be the set (not multiset!) of columns $c$ such some entry in column $c$ is allowed to be nonzero for some row in $R$. That is, for some $r\in R$, position $(r,c)$ in the matrix is one of those positions not required to be 0. Here is a necessary and sufficient condition for there to be a non-negative solution: For every subset $R$ of the rows, $|N(R)| \ge (3/2)^N |R|$. There is a similar dual nsc with the rows and columns reversed. Whether this is useful will depend on where your non-zero entries have to be, and without that information little more can be said.

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https://mathoverflow.net/users/9025

453409