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euclaise

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mathoverflow-accepted

like 3

https://mathoverflow.net/questions/453397

2

A normal modal propositional logic $\Delta$ has the disjunction property if and only if > > For any formulas $A\_1,\dotsc,A\_n$, if $\Box A\_1 \vee \dotsb\vee \Box A\_n \in \Delta$ then $A\_k\in \Delta$ for some $k$ with $1\leq k\leq n$. > > > Let us say that $\Delta$ has the *extended* disjunction property if and only if > > For any formulas $A\_1,\dotsc,A\_n$ and non-modal formula $A\_0$, if $A\_0 \vee \Box A\_1 \vee \dotsb \vee \Box A\_n \in \Delta$ then $A\_k\in \Delta$ for some $k$ with $0\leq k\leq n$. > > > Does there exist a normal modal propositional logic with the disjunction property but not the extended disjunction property? I came across the second property in some unpublished notes of Kit Fine's from the '70s, but I do not know of much discussion of it elsewhere.

https://mathoverflow.net/users/78564

An extension of the disjunction property in modal logic

The extended disjunction property is equivalent to plain disjunction property for all normal modal logics $\Delta$. Assume that $\Delta$ contains $$A\_0\lor\bigvee\_{i=1}^n\Box A\_i,$$ where $A\_0$ is box-free. Let $P$ denote the set of propositional variables occurring in the formula. For each $a\colon P\to\{0,1\}$, let $\sigma\_a$ denote the substitution such that $\sigma\_a(p)=p^{a(p)}$ for all $p\in P$, where $p^1=p$, $p^0=\neg p$. Since $\Delta$ is closed under substitution and classical reasoning, $\Delta$ contains $$\bigwedge\_a\Bigl(\sigma\_a(A\_0)\lor\bigvee\_{i=1}^n\Box\sigma\_a(A\_i)\Bigr).\tag{$\*$}\label{star}$$ If $A\_0$ is a classical tautology, then $A\_0\in\Delta$ and we are done. Otherwise, \eqref{star} implies $$\bigvee\_a\bigvee\_{i=1}^n\Box\sigma\_a(A\_i)\tag{$\*\*$}\label{starstar}$$ by classical propositional reasoning: for any assignment $e$, there is $a$ such that $e(\sigma\_a(A\_0))=0$, thus the rest of the disjuction holds under $\sigma\_a$. Applying the disjuction property to \eqref{starstar}, $\sigma\_a(A\_i)\in\Delta$ for some $i=1,\dotsc,n$ and $a\colon P\to\{0,1\}$. Applying closure under substitution once more, $\Delta$ contains $\sigma\_a(\sigma\_a(A\_i))$, which is equivalent to $A\_i$ over **K**. The argument above applies not just to normal modal logics, but also to quasinormal logics, and even more generally, to all logics $\Delta$ that are closed under modus ponens and box-free substitutions, and that contain the tautologies of **E**. Here, **E** is the smallest logic that includes classical logic, and is closed under modus ponens and under the rule $A\leftrightarrow B\mathrel/\Box(A\leftrightarrow B)$.

2

https://mathoverflow.net/users/12705

453410

https://mathoverflow.net/questions/453393

4

The following question is a direct continuation of [this](https://math.stackexchange.com/questions/4758087/if-mathbbcu-v-xn-mathbbcx-y-for-every-n-geq-1-then-already-m) elaborate question; it is mentioned there at the end: Let $u,v \in \mathbb{C}(x,y)$ or $u,v \in \mathbb{C}[x,y]$, if it is easier to answer in this case. Assume that the following condition, call it $D(f)$, is satisfied for every $f \in \mathbb{C}[x]-\mathbb{C}$: $\mathbb{C}(u,v,f)=\mathbb{C}(x,y)$. > > **Question:** Is it true that > $\mathbb{C}(u,v)=\mathbb{C}(x,y)$? > > > The notation is of fields of fractions. Remarks: 1. It seems that $y \in \mathbb{C}(u,v)$, but actually I am not sure about this, since I only have the following argument: We know that $y \in \mathbb{C}(u,v,f)$, so there exist $F,G \in \mathbb{C}[r,s,t]$ such that $y=\frac{F(u,v,f)}{G(u,v,f)}$. For some $\alpha \in \mathbb{C}$, $f(\alpha)=0$, and then $y=\frac{F(u(\alpha,y),v(\alpha,y),f(\alpha))}{G(u(\alpha,y),v(\alpha,y),f(\alpha))}= \frac{F(u(\alpha,y),v(\alpha,y))}{G(u(\alpha,y),v(\alpha,y))}$, which just shows that $y \in \mathbb{C}(u(\alpha,y),v(\alpha,y))$. 2. I do not know how to show that $x \in \mathbb{C}(u,v)$, if this is true. 3. I guess such $u,v$ must be algebraically independent. Thank you very much!

https://mathoverflow.net/users/72288

If $\mathbb{C}(u(x,y),v(x,y),f(x))=\mathbb{C}(x,y)$, for every $f(x) \in \mathbb{C}[x]-\mathbb{C}$, then already $\mathbb{C}(u,v)=\mathbb{C}(x,y)$?

The answer to the Question is ``no". As an example, let $u=xy$, $v=x+y$ be the elementary symmetric functions in $x$ and $y$. It is well-known that $[\mathbb{C}(x,y): \mathbb{C}(u,v)]=2$, so those two fields are not equal. On the other hand, consider $K\_f:=\mathbb{C}(u,v,f(x))$ for any nonconstant polynomial $f(x)\in \mathbb{C}[x]$. Note that $\mathbb{C}(u,v)$ already contains the functions $f(x)+f(y)$ and $\frac{f(x)-f(y)}{x-y}$, since indeed those are symmetric functions in $x$ and $y$. So successively we find that $K\_f$ contains the elements $(f(x)+f(y))-f(x)=f(y)$, thus also $f(x)-f(y)$, and thus also $\frac{f(x)-f(y)}{(f(x)-f(y))/(x-y)}=x-y$. Finally, it also contains $x+y$ and thus $x$ and $y$, showing $K\_f=\mathbb{C}(x,y)$. (**Edit**: One could have of course argued non-constructively even quicker: Since the field of symmetric functions is of index $2$ in $\mathbb{C}(x,y)$ and doesn't contain any nonconstant function $f(x)$, adjoining any such function must give the whole thing by lack of intermediate fields)

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https://mathoverflow.net/users/127660

453411

https://mathoverflow.net/questions/453453

0

Let $\kappa\_1>0$, $\beta\in [0, 1]$ and $b: \mathbb R\_+ \times \mathbb R^d \to \mathbb R^d$ such that for all $t\ge0$ and $x,y \in \mathbb R^d$ we have $|b(t, 0)| \le \kappa\_1$ and $|b(t, x) - b(t, y)| \le \kappa\_1 ( |x-y| \vee |x-y|^\beta )$ for all $x,y \in \mathbb R^d$. I'm reading section 1.2 in the paper [Density and gradient estimates for non degenerate Brownian SDEs with unbounded measurable drift](https://doi.org/10.1016/j.jde.2020.09.004). --- To state our main result, we need to prepare some deterministic regularized flow associated with the drift $b$. Let $\rho$ be a nonnegative smooth function with support in the unit ball of $\mathbb{R}^d$ and such that $\int\_{\mathbb{R}^d} \rho(x) \mathrm{d} x=1$. For $\varepsilon \in(0,1]$, define $$ \rho\_{\varepsilon}(x):=\varepsilon^{-d} \rho\left(\varepsilon^{-1} x\right), \quad b\_{\varepsilon}(t, x):=b(t, \cdot) \* \rho\_{\varepsilon}(x)=\int\_{\mathbb{R}^d} b(t, y) \rho\_{\varepsilon}(x-y) d y, $$ i.e. $\*$ stands for the usual spatial convolution. Then for each $n=1,2, \dotsc$, it is easy to see that \begin{align} \left|\nabla\_x^n b\_{\varepsilon}(t, x)\right| & =\left|\int\_{\mathbb{R}^d}(b(t, y)-b(t, x)) \nabla\_x^n \rho\_{\varepsilon}(x-y) \mathrm{d} y\right| \tag{1}\label{1} \\ & \leqslant \int\_{\mathbb{R}^d}|b(t, y)-b(t, x)|\left|\nabla\_x^n \rho\_{\varepsilon}\right|(x-y) \mathrm{d} y \\ & \leqslant \kappa\_1 \varepsilon^\beta \int\_{\mathbb{R}^d}\left|\nabla\_x^n \rho\_{\varepsilon}\right|(x-y) \mathrm{d} y \leqslant c \varepsilon^{\beta-n} .\tag{2}\label{2} \end{align} --- **My understanding** It seems from (\ref{1}) that we need $\int\_{\mathbb{R}^d}b(t, x) \nabla\_x^n \rho\_{\varepsilon}(x-y) \mathrm{d} y=0$. A sufficient condition is that $\rho$ is symmetric, i.e., $\rho (y) = \rho(-y)$ for all $y \in \mathbb R^d$. We have \begin{align\*} \int\_{\mathbb{R}^d}\left|\nabla\_x^n \rho\_{\varepsilon}\right|(x-y) \mathrm{d} y &= \int\_{\mathbb{R}^d}\left|\nabla\_{y}^n \rho\_{\varepsilon}\right|(y) \mathrm{d} y \\ &= \int\_{\mathbb{R}^d}\left|\nabla\_{y}^n \rho \right|(y) \mathrm{d} y. \end{align\*} > > Could you explain how to get the inequality > $$ > \int\_{\mathbb{R}^d}\left|\nabla\_x^n \rho\_{\varepsilon}\right|(x-y) \mathrm{d} y \leqslant c \varepsilon^{-n} > $$ > in (\ref{2})? > > > Thank you so much for your help!

https://mathoverflow.net/users/99469

An estimate of the integral of the higher order derivative of a bump function

$\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}\newcommand{\na}{\nabla}$Note that \begin{equation} (\na^n\rho\_\ep)(x)=\ep^{-d-n}(\na^n\rho)(\ep^{-1}x). \end{equation} So, \begin{equation} \int\_{\R^d}|(\na^n\rho\_\ep)(x-y)|\,dy =\ep^{-d-n}\int\_{\R^d}|(\na^n\rho)(\ep^{-1}(x-y))|\,dy \\ =\ep^{-n}\int\_{\R^d}|(\na^n\rho)(\ep^{-1}x-v))|\,dv =c\_{n,\rho} \ep^{-n}, \end{equation} where $c\_{n,\rho}:=\int\_{\R^d}|(\na^n\rho)(z)|\,dz$. $\quad\Box$

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https://mathoverflow.net/users/36721

453455

https://mathoverflow.net/questions/453456

3

Suppose I have a 3-manifold obtained via face identifications of a polyhedron (e.g. the Poincaré sphere presented as a dodecahedron with opposite faces glued). Is there a program that exists for easily working with such manifolds? The only thing I really know how to do is to make a triangulation in Regina, but I'd like to avoid having to input a triangulation with an unholy number of simplices, if possible.

https://mathoverflow.net/users/499323

Computer program for polyhedral manifolds

The two standard answers to this question are [Snappy](https://snappy.computop.org/) and [Regina](https://regina-normal.github.io/). If you have a famous manifold, then it will likely be in one of the many censuses that come with the programs. If you have a less-than-famous manifold, then find a surgery description of it and feed that to Snappy.

1

https://mathoverflow.net/users/1650

453457

https://mathoverflow.net/questions/453469

7

I'm interested in asymptotics for the sum $$\sum\_{n\le x}\frac{n}{\text{rad}(n)}.$$ For my research I only need to know whether or not this is $\mathcal{O}(x)$, but I would appreciate more precise asymptotics. Additionally, a reference would be much appreciated. The sum of $\frac{1}{\text{rad}(n)}$ has been discussed on the site, but I cannot find anything on my sum.

https://mathoverflow.net/users/505902

Asymptotics on sum of n/rad(n)

One has \begin{align\*} \sum\_{n \leq x}\frac{n}{\operatorname{rad}n} & = (1+o(1)) \, x \sqrt{\frac{2}{\log x \log \log x}} \exp\left((1+o(1))\sqrt{\frac{8\log x}{\log \log x}}\right) \ (x \to \infty), \\ & = x\exp\left((1+o(1))\sqrt{\frac{8\log x}{\log \log x}}\right) \ (x \to \infty) \\ & \neq O(x\, (\log x)^A) \ (x\to \infty) \end{align\*} for any $A \in \mathbb{R}$, where the first estimate holds according to user "Ofir Gorodetsky"'s detailed answer to a similar question to yours at [Asymptotic behavior of a "strange" arithmetic function](https://mathoverflow.net/questions/445395/asymptotic-behavior-of-a-strange-arithmetic-function)

6

https://mathoverflow.net/users/17218

453476

https://mathoverflow.net/questions/453477

3

I was thinking of the following problem. Let $f$ be a Taylor expansion and $a\_k$ the associated coefficients, $$\forall x\in\mathbb{R},~f(x)\triangleq\sum\_{k=0}^\infty a\_kx^k.$$ Let suppose that we have: $$\forall x\in\mathbb{R},~f(x)>0.$$ Is it possible to find another expansion such that: $$\forall x\in\mathbb{R},~f(x) = \sum\_{k=0}^\infty b\_k\beta\_k(x)$$ with $b\_k$ positive real numbers and $\beta\_k$ (exponential?) positive functions ? Thank you very much!

https://mathoverflow.net/users/510079

Other expansion for positive Taylor expansion

I believe it is not possible. Here is an argument for this: *Disclaimer: All inequalities hereafter are meant elementwise* Let's consider a discrete version of this problem $x\in\{0,1,\dots,N-1\}$. Then we want to find an invertible $N\times N$ matrix $\beta$ (different from identity), such that for every elementwise non-negative vector $\boldsymbol{f} = (f(0), f(1), \dots f(N-1))^T \geq 0$ we have $$ \boldsymbol{b} = \beta^{-1} \boldsymbol{f} \geq 0\,. $$ The necessary and sufficient condition for this is that $\beta^{-1}\geq 0$. Moreover, you require that $\beta\_k(x)\geq 0$, which in discrete case translates to $\beta\geq 0$. However, citing [Nonnegative matrix - Wikipedia:](https://en.wikipedia.org/wiki/Nonnegative_matrix) > > The inverse of a non-negative matrix is usually not non-negative. The exception is the non-negative monomial matrices: a non-negative matrix has non-negative inverse if and only if it is a (non-negative) monomial matrix. > > > Where the [monomial matrix](https://en.wikipedia.org/wiki/Generalized_permutation_matrix) is defined as: > > In mathematics, a generalized permutation matrix (or monomial matrix) is a matrix with the same nonzero pattern as a permutation matrix, i.e. there is exactly one nonzero entry in each row and each column. > > > Thus the only allowed transformation $\beta$ is reshuffling (and possibly rescaling) the elements of $\boldsymbol{f}$.

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https://mathoverflow.net/users/165560

453485

https://mathoverflow.net/questions/453494

0

For any $A\subseteq \mathbb{N}$ we let the (lower) density of $A$ be defined by $$d(A) = \liminf\_{n\to\infty}\frac{|A\cap\{0,\ldots,n\}|}{n+1}.$$ If $\pi:\mathbb{N}\to\mathbb{N}$ is a permutation (bijection), let $\text{ex}(\pi) = \{n\in\mathbb{N}: \pi(n) > n\}$ be the set of [*exceedances*](https://mathoverflow.net/questions/359684/why-excedances-of-permutations), and let $\text{negex}(\pi)=\{n\in\mathbb{N}: \pi(n) < n\}$ be the set of "negative exceedances". To me it seems inconceivable that there is a permutation $\pi:\mathbb{N}\to\mathbb{N}$ with $d\big(\text{ex}(\pi)\big) \neq d\big(\text{negex}(\pi)\big)$ -- but my intuition has let me down many times. Is my intuition correct this time? ${}$

https://mathoverflow.net/users/8628

Is there a lop-sided permutation $\pi:\mathbb{N}\to\mathbb{N}$?

Let $\pi(n) = n+1$ for all $n$ except that $\pi(2^{k}-1) = 2^{k-1}$. In other words, permute cyclically $(2,3)$, $(4,5,6,7)$, $(8,9,10,\ldots,15)$, and so on. Then $d(\operatorname{pos}(\pi))=1$ whereas $d(\operatorname{neg}(\pi))=0$.

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https://mathoverflow.net/users/17064

453495

https://mathoverflow.net/questions/453482

5

Let $E/F$ be a quadratic extension of nonarchimedean local field, and let $G$ be a reductive group over $E$, and $G(E)$ the $E$-points of $G$. (For my question, one may just assume $G=\operatorname{GL}\_n$.) Now $G(E)$ can be viewed in two different way: the $F$ points of the restriction of scalar $G\_1:=\operatorname{Res}\_{E/F}G$, and the $E$-points of just $G\_2:=G$. Their Langlands $L$-groups are respectively $$ {^LG\_1}=(G^\vee(\mathbb{C})\times G^\vee(\mathbb{C}))\rtimes W\_F\qquad\text{and}\qquad{^LG\_2}=G^\vee(\mathbb{C})\times W\_E, $$ where $G^\vee$ is the dual group of $G$ and for the semidirect product the nontrivial element in $W\_F/ W\_E$ acts by switching the two factors. Now, assuming the local Langlands correspondence, for each irreducible admissible representation of $G(E)$, one can associate local Langlands parameters $$ \varphi\_1:\operatorname{WD}\_F\longrightarrow{^LG\_1}\qquad\text{and}\qquad \varphi\_2:\operatorname{WD}\_E\longrightarrow{^LG\_2}, $$ where $\operatorname{WD}\_F$ and $\operatorname{WD}\_E$ are the Weil-Deligne groups. I believe there must be a way to pass from $\varphi\_1$ to $\varphi\_2$. But how?

https://mathoverflow.net/users/32746

Two different local Langlands parameters for quadratic extension

This came up in a paper of mine not so long ago, and my coauthors and I were surprised that it wasn't made explicit in the standard references, so we wrote it out ourselves: *Dembélé, Lassina; Loeffler, David; Pacetti, Ariel*, [**Non-paritious Hilbert modular forms**](https://doi.org/10.1007/s00209-019-02229-5), Math. Z. 292, No. 1-2, 361-385 (2019). [ZBL1446.11084](https://zbmath.org/?q=an:1446.11084). See Proposition 1.3. (Strictly speaking we are describing the analogue for global fields, not local fields, but the recipe is the same.)

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https://mathoverflow.net/users/2481

453506

https://mathoverflow.net/questions/453502

5

Let $f \in L\_1(\mathbb{R})$ be such that $\operatorname{supp} f \subset [0,1]$, and let $K$ be the gaussian kernel $K(t) := \frac{1}{\sigma \sqrt{2 \pi}} \exp(-t^2/2\sigma^2)$, with some small $\sigma < 1/8$. Is it true that for some universal constant $c > 0$, we have $\|\mathbf{1}\_{[0,1]} \cdot (K \ast f)\|\_1 \geq c \|K \ast f\|\_1$? I.e. is it necessary that at least constant fraction of the $L\_1$ mass of $K\ast f$ stays inside the interval $[0,1]$? This is rather easy to show when $f$ is non-negative (say, if $\|f\|\_1 = 1$, we have $\|K \ast f\|\_1 = \|f\|\_1 = 1$, and $\|(1 - \mathbf{1}\_{[0,1]}) \cdot (K \ast f)\|\_1 \leq (\Phi(0) + (1 - \Phi(1/\sigma)) \leq 0.6$ where $\Phi$ is CDF of Gaussian distribution), but in general case some magical cancellations are *a priori* possible. **Discreete case looks false** Experimentally, the statement seems to be false in the discrete case: let us consider functions on integers $\mathbb{Z}$ and a kernel corresponding to $n$ step random walk $K\_n := L\ast L \ast L \ast \cdots \ast L$, where $L(t) = \frac{1}{3}$ for $t \in \{-1, 0, 1\}$ (and $0$ otherwise). I'm chosing non-zero probability for staying in the same place, to avoid period $2$ in the random walk. Let us fix some $\sigma < 1/8$. Now if we consider a function $f : \mathbb{Z} \to \mathbb{R}$ such that $\mathrm{supp} f \subset \{0, \ldots n\} =: D$, we would like to say that for any such function $\|\mathbf{1}\_D \cdot (K\_{\sigma^2 n^2} \ast f)\| \geq C \|K\_{\sigma^2 n^2} \ast f\|\_1$ for some universal constant $C$ that do not depend on $f$ nor $n$. But the matrix $M\_{i,j} = K\_{\sigma^2 n^2}(i-j)$ for $i,j \in \{0, \ldots n\}$ is invertible (although terribly conditioned), so we can just chose some sparse $\tilde{f} : \{0, \ldots n \} \to \mathbb{R}$, which we intent to be $K\_{\sigma^2 n^2} \ast f$ restricted to $\{0, \ldots, n\}$, say $\tilde{f}(\lfloor n/10\rfloor) = 1$, and $0$ otherwise. We can now solve for $f := M^{-1} \tilde{f}$, and extend $f$ to all integers (by setting $f(t) = 0$ for $t \not \in D$). After solving it numerically it seems that $f$ this provides a counterexample. We have $[K\_{\sigma^2 n^2} \ast f]|\_{D} = \tilde{f}$ by construction, hence $\|\mathbf{1}\_D \cdot (K\_{\sigma^2 n^2} \ast f)\|\_1 = 1$, while outside of the region $D$, the $\|(1 - \mathbf{1}\_D)\cdot (K\_{\sigma^2 n^2} \ast f)\|\_1$ is growing with $n$ quite drastically. This exact construction does not seem to lead to a counterexample for the continuous case --- the functions $f$ we get by solving $M^{-1} \tilde{f}$ do not seem to converge to anything.

https://mathoverflow.net/users/468679

Does convolution of a compactly supported function with Gaussian need to have fraction of the $L_1$ mass in the original interval?

No by the usual duality nonsense. Let's say $\sigma=1$ (it doesn't matter). The convolution at $x$ is $e^{-x^2/2}$ times the integral of $g(t)=f(t)e^{-t^2/2}$ against $e^{tx}$. If the estimate $\left|\int\_2^3(f\*K)e^{x^2/2}\,dx\right|\le C\int\_0^1|f\*K|e^{x^2/2}\,dx$ were possible, then there would exist an $L^\infty$ function $h$ on $[0,1]$ such that $\int\_{0}^1 h(x)e^{tx}\,dx=\int\_2^3 e^{tx}\,dx$ for almost all $t\in[0,1]$, which is clearly impossible (both sides are analytic functions in $t$ and the RHS grows faster than the LHS can afford as $t\to+\infty$).

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https://mathoverflow.net/users/1131

453510

https://mathoverflow.net/questions/453069

10

Is there an example of a smooth projective variety $X$ (say, over $\mathbb{C}$) such that the $\mathbb{C}$-algebra $\bigoplus\_{n\geq 0}H^0(X,K\_X^{-n}) $ is not finitely generated?

https://mathoverflow.net/users/40297

Can $\bigoplus_{n\geq 0}H^0(X,K_X^{-n}) $ fail to be finitely generated?

Chenyang Xu claims that $X = \mathbb{P}\_S\left(\mathcal{O}\_S \oplus \mathcal{O}\_S(H) \right)$, where $S$ the blow-up of $\mathbb{P}^2$ along $9$ general points and $H$ is any ample Cartier divisor on $S$, provides such an example. See example 3.8 in his paper [K-stability for varieties with a big anti-canonical class](https://arxiv.org/pdf/2210.16631.pdf).

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https://mathoverflow.net/users/37214

453524

https://mathoverflow.net/questions/453514

0

The following question is a direct continuation of [this](https://mathoverflow.net/questions/453393/if-mathbbcux-y-vx-y-fx-mathbbcx-y-for-every-fx-in-mathbb/453411?noredirect=1#comment1173100_453411) question: Let $u,v \in \mathbb{C}[x,y]$. Assume that for every $f \in \mathbb{C}[x]$ and every $g \in \mathbb{C}[y]$ (excluding the cases where $f+g \in \mathbb{C}$), the following condition, call it $E(f,g)$, is satisfied: $\mathbb{C}(u,v,f+g)=\mathbb{C}(x,y)$. The notation is of fields of fractions. Then we can ask the following question: > > **Question:** Is it true that $\mathbb{C}(u,v)=\mathbb{C}(x,y)$? > > > Notice that the counterexample for the quoted question, $u=x+y,v=xy$, is not a counterexample here, since condition $E(x,y)$ does not hold, since $\mathbb{C}(x+y,xy,x+y)=\mathbb{C}(x+y,xy) \subsetneq \mathbb{C}(x,y)$ (Also, conditions $E(x^n,y^n)$ do not hold, but it is enough for us that one required condition does not hold). **Remarks:** **(1)** I also thought about the following condition: Assume that for every $f \in \mathbb{C}[x]$ and every $g \in \mathbb{C}[y]$ (excluding the cases where $fg \in \mathbb{C}$), the following condition, call it $M(f,g)$, is satisfied: $\mathbb{C}(u,v,fg)=\mathbb{C}(x,y)$. However, it seems that Joachim König also gave a counterexample for conditions $M(f,g)$: $u=x+y^3,v=y+x^3$, although he referred to adding $(ax+by+c)^n$ , $n \geq 1$, $(ax+by+c)^n \notin \mathbb{C}$. I think I can explain why his counterexample also applies for adding $fg$, $f \in \mathbb{C}[x], g \in \mathbb{C}[y]$, $fg \notin \mathbb{C}$. The argument would be (I do not yet have a proof for the argument, see [this](https://math.stackexchange.com/questions/4723640/generalizing-a-result-in-mathbbcx-to-a-result-in-mathbbcx-y?noredirect=1&lq=1)): Given $u,v,w \in \mathbb{C}[x,y]$, if there exist some $a,b,c \in \mathbb{C}$ such that the ideal generated by $u-a,v-b,w-c$ is maximal, then $\mathbb{C}(u,v,w)=\mathbb{C}(x,y)$. **(2)** Notice that $u=x+y^3, v=y+x^3$ is not a counterexample for the current question, since, for example, conditions $E(x,y^3)$ and $E(x^3,y)$ do not hold. Thank you very much!

https://mathoverflow.net/users/72288

$\mathbb{C}(u(x,y),v(x,y),f(x)+g(y))=\mathbb{C}(x,y)$ implies $\mathbb{C}(u(x,y),v(x,y))=\mathbb{C}(x,y)$?

Actually, there are still counterexamples of essentially the same nature as the one given in the previous (linked) thread: Take $u=x+xy$ and $v=x^2y$, so that $\mathbb{C}(u,v)$ is the field of functions symmetric in $x$ and $xy$ (i.e., the fixed field of the involution $\sigma:x\mapsto xy$ and $y = \frac{xy}{x}\mapsto \frac{x}{xy}=\frac{1}{y}$). This is of index $2$ in $\mathbb{C}(x,xy)=\mathbb{C}(x,y)$ (but of course not identical to the field of symmetric functions in $x$ and $y$). If $f(x),g(y)$ are polynomials, not both constant, then $\sigma(f(x)+g(y))=f(xy)+g(\frac{1}{y})$, and this can never equal $f(x)+g(y)$ (indeed, if $g$ is nonconstant, then $f(xy)+g(\frac{1}{y})$ is not even in $\mathbb{C}[x,y]$, whereas if it is constant, then $f(x)+g\in \mathbb{C}[x]$, but $f(xy)+g\notin \mathbb{C}[x]$. Thus, $f(x)+g(y)\notin Fix(\sigma)=\mathbb{C}(u,v)$, and hence $\mathbb{C}(u,v,f(x)+g(y))=\mathbb{C}(x,y)$.

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https://mathoverflow.net/users/127660

453528

https://mathoverflow.net/questions/453522

17

Is there a characterisation of measurability in a forward way similar to the closure characterisation of continuity? A function $f\colon X \to Y$ is continuous if equivalently: 1. If $G\subset Y$ is open, then is $f^{-1}(G)$ open too. 2. $f(\overline A) \subset \overline{f( A)}$ where $\overline A$ is the closure of $A$. A function $f\colon X \to Y$ is measurable if equivalently: 3. If $A\subset Y$ is a measurable set, then is $f^{-1}(A)$ a measurable subset of $X$ too. 4. ?

https://mathoverflow.net/users/57923

Forward definition of measurability

If $(X,M)$ is an algebra of sets, then define a relation $\delta\_M\subseteq P(X)^2$ by setting $A\mathrel{\delta\_M}B$ if there does not exist some $S\in M$ with $A\subseteq S$, $B\subseteq S^c$. Suppose that $(X,M)$, $(Y,N)$ are algebras of sets. Let $f:X\rightarrow Y$ be a function. Then it is not too hard to prove that $f^{-1}[S]\in M$ whenever $S\in N$ precisely when $A\mathrel{\delta\_M} B\Rightarrow f[A]\mathrel{\delta\_N}f[B]$ for $A,B\subseteq X$. **Ideals** Suppose $A,B\subseteq X$. Let $\mathcal{I}\_{A,B}$ be the collection of all $C\in M$ where there is some $D\in M$ where $A\cap C\subseteq D,B\cap C\subseteq D^c$. Then $\mathcal{I}\_{A,B}$ is an ideal in the Boolean algebra $M$ and $\mathcal{I}\_{A,B}$ is a proper ideal in $M$ if and only if $A\delta\_MB$. **More context: proximity spaces** The reader is assumed to be familiar with proximity spaces when reading this answer. The relation $\delta\_M$ defined above is a proximity relation, and by definition $A\mathrel{\delta\_M}B\Rightarrow f[A]\mathrel{\delta\_N}f[B]$ precisely when $f$ is a proximity map. We shall now establish that the category of zero-dimensional proximity spaces is isomorphic to the category of algebras of sets with measurable mappings as morphisms, and the characterization of measurable transformations as mappings where $A\mathrel{\delta\_M}B\Rightarrow f[A]\mathrel{\delta\_N}f[B]$ will immediately follow as a corollary. A proximity space $X$ is said to be zero-dimensional if whenever $A\ll B$, there is some $S$ with $A\ll S\ll S\ll B$. Equivalently, $X$ is zero-dimensional precisely when there is some $C\subseteq X$ with $A\subseteq C,B\subseteq C,C\not\delta C^c$. If $(X,\ll)$ is a proximity space, then let $M\_\ll\subseteq P(X)$ be the set defined by letting $A\in M\_\ll$ precisely when $A\ll A$. Then $(X,M\_\ll)$ is an algebra of sets. If $(X,M)$ is an algebra of sets, then let $\ll\_M$ be the relation where we set $A\ll\_MB$ precisely when $A\subseteq S\subseteq B$ for some $S\in M$. Then $\ll\_M$ is a proximity relation on the set $X$. Proposition: 1. Suppose that $(X,M)$ is an algebra of sets. Then $M=M\_{\ll\_M}$. 2. If $(X,\ll)$ is a zero-dimensional proximity space, then ${\ll}={\ll\_{M\_\ll}}$. Proof: 1. If $R\in M$, then $R\ll\_M R$, so $R\in M\_{\ll\_M}$. Similarly, if $R\in M\_{\ll\_M}$, then $R\ll\_MR$, hence there is some $S\in M$ with $R\subseteq S\subseteq R$. But this is only possible if $R=S\in M$. 2. Suppose $A\ll B$. Then by zero-dimensionality, there is some $S$ with $A\subseteq S\ll S\subseteq B$. Therefore, $S\in M\_{\ll}$, so $S\ll\_{M\_\ll}S$. We can therefore conclude that $A\ll\_{M\_\ll}B$. Now assume that $A\ll\_{M\_\ll}B$. Then there is some $S\in M\_\ll$ with $A\subseteq S\subseteq B$. But this means that $S\ll S$, so $A\ll B$ as well. $\square$ Observation: If $(X,M)$, $(Y,N)$ are algebras of sets, then a function $f:X\rightarrow Y$ is a measurable mapping (which means that $f^{-1}[R]\in M$ whenever $R\in N$) precisely when $f$ is a proximity map from $(X,\ll\_M)$ to $(Y,\ll\_N)$. Proof: $\rightarrow$ Suppose that $f$ is a measurable mapping. Then whenever $A,B\subseteq Y$ and $A\ll\_N B$, there is some $S\in N$ with $A\subseteq S\subseteq B$. Therefore, $f^{-1}[S]\in M$ and $f^{-1}[A]\subseteq f^{-1}[S]\subseteq f^{-1}[B]$. Therefore, $f^{-1}[A]\ll\_M f^{-1}[B]$, so $f$ is a proximity map. $\leftarrow$ Suppose that $f$ is a proximity map. Then whenever $S\in N$, we have $S\ll\_NS$, so $f^{-1}[S]\ll\_Mf^{-1}[S]$, hence $f^{-1}[S]\in M$ as well, so $f$ is a measurable mapping. $\square$ Proposition: If $(X,M)$ is a separating algebra of sets, then the Smirnov compactification of $(X,\ll\_M)$ is simply the Stone space of $M$. Proof: Let $S(M)$ denote the Stone space of $M$. Let $\iota:X\rightarrow S(M)$ be the mapping where $\iota(x\_0)=\{A\in M:x\_0\in A\}$. Then the space $S(M)$ is a compactification of $\iota[X]$. Suppose now that $A,B\subseteq X$. We now need to show that $A\mathrel{\not\delta\_M} B$ if and only if $\overline{\iota[A]}\cap\overline{\iota[B]}=\emptyset$. If $A\mathrel{\not\delta\_M}B$, then there is some $S\in M$ with $A\subseteq S$, $B\subseteq S^c$. Let $C\subseteq S(M)$ be the clopen set consisting of all ultrafilters $\mathcal{U}\subseteq M$ with $S\in\mathcal{U}$. Then $C^c$ is the set of all ultrafilters $\mathcal{U}\subseteq M$ with $S^c\in\mathcal{U}$. If $a\in A$, then $a\in S$, so $S\in\iota(a)$, hence $\iota(a)\in C$. Therefore, $\iota[A]\subseteq C$. Similarly, if $b\in B$, then $b\in S^c$, so $S^c\in\iota(b)$, hence $\iota(b)\in C^c$. Therefore, $\iota[B]\subseteq C^c$. In particular, $\overline{\iota[A]}\cap\overline{\iota[B]}\subseteq C\cap C^c=\emptyset$. Suppose now that $\overline{\iota[A]}\cap\overline{\iota[B]}=\emptyset$. Then there is some clopen set $C\subseteq S(M)$ with $\iota[A]\subseteq C$, $\iota[B]\subseteq C^c$. But by Stone duality, there is some $S\in M$ where $C=\{\mathcal{U}\in S(M):S\in\mathcal{U}\}$. Therefore, if $a\in A$, then $\iota(a)\in C$, so $S\in\iota(a)$, hence $a\in S$. Thus $A\subseteq S$. Similarly, if $b\in B$, then $\iota(b)\in C^c$, so $S^c\in\iota(b)$, hence $b\in S^c$. Therefore, $B\subseteq S^c$. We conclude that $A\mathrel{\not\delta\_M}B$. We may therefore conclude that the Stone space $S(M)$ is the Smirnov compactification of $X$. $\square$ Proposition: A separated proximity space $(X,\ll)$ is zero-dimensional if and only if the Smirnov compactification $K$ is zero-dimensional. Proof: $\DeclareMathOperator\Cl{Cl}$By the above proposition, we know that if $(X,\ll)$ is zero-dimensional, then $K=S(M\_\ll)$ which is zero-dimensional. Suppose that $K$ is zero-dimensional. Let $A,B\subseteq X$ and suppose that $A\mathrel{\not\delta} B$. Then $\Cl\_K(A)\cap\Cl\_K(B)=\emptyset$. Therefore, there is some clopen subset $C\subseteq K$ where $\Cl\_K(A)\subseteq C$, $\Cl\_K(B)\subseteq K\setminus C$. In this case, $\Cl\_K(C\cap X)\cap\Cl\_K(X\setminus C)=\emptyset$, so $A\subseteq C\cap X$, $B\subseteq X\setminus C$ and $(C\cap X)\mathrel{\not\delta}(X\setminus C)$. Therefore, $\delta$ is zero-dimensional. $\square$

13

https://mathoverflow.net/users/22277

453530

https://mathoverflow.net/questions/453531

0

For any $\kappa>0$, we consider the Gaussian heat kernel $$ p^\kappa\_t (x) := (\kappa \pi t)^{-\frac{d}{2}} e^{-\frac{|x|^2}{\kappa t}}, \quad t>0, x \in {\mathbb R}^d. $$ Let $L^0 := L^0 (\mathbb R^d)$ be the space of real-valued measurable functions on $\mathbb R^d$. Let $L^0\_+ := L^0\_{+} (\mathbb R^d)$ the subspace of $L^0$ consisting of non-negative functions. Let $L^0\_b := L^0\_b (\mathbb R^d)$ the subspace of $L^0$ consisting of bounded functions. For $f \in L^0\_b \cup L^0\_+$, let $$ P\_t^\kappa f (x) := \int\_{\mathbb R^d} p^\kappa\_t (x-y) f (y) \, \mathrm d y, \quad t >0, x \in \mathbb R^d. $$ It is stated in the paper [Singular density dependent stochastic differential equations](https://www.sciencedirect.com/science/article/abs/pii/S0022039623002255) that > > It is well-known that for some constant $c>0$, > $$ > \|P^\kappa\_t \|\_{L^p \to L^{p'}} := \sup\_{\|f\|\_{L^p} \le 1} \|P^\kappa\_t f\|\_{L^{p'}} \le c t^{-\frac{d(p'-p)}{2pp'}}, > \quad t>0, 1 \le p \le p' \le \infty. > $$ > > > I would like to ask if the constant $c$ depends on $\kappa$. Any reference is appreciated. Thank you so much for your elaboration!

https://mathoverflow.net/users/99469

Does $c$ in the embedding inequality $\|P^\kappa_t \|_{L^p \to L^{p'}} \le c t^{-\frac{d(p'-p)}{2pp'}}$ depend on $\kappa$?

$\newcommand\ka\kappa$Yes, the best constant $c$ depends on $\ka$ if $p<p'<\infty$. Indeed, the inequality in question can be rewritten as $$t^Q\sup\_{\|f\|\_{L^p} \le 1} \|P^\ka\_t f\|\_{L^{p'}} \le c \tag{1}\label{1}$$ if $t>0$ and $1 \le p \le p' \le \infty$, where $Q:=\frac{d(p'-p)}{2pp'}\in[0,\infty]$. So, the best constant $c$ in inequality \eqref{1} is $$c^\ka:=\sup\_{t>0}t^Q\sup\_{\|f\|\_{L^p} \le 1} \|P^\ka\_t f\|\_{L^{p'}}.$$ Note that $p^\ka\_t=p^1\_{\ka t}$ and hence $P^\ka\_t=P^1\_{\ka t}$. So, $$c^\ka=\sup\_{t>0}t^Q\sup\_{\|f\|\_{L^p} \le 1} \|P^1\_{\ka t} f\|\_{L^{p'}} =\ka^{-Q}\sup\_{u>0}u^Q\sup\_{\|f\|\_{L^p} \le 1} \|P^1\_u f\|\_{L^{p'}} =\ka^{-Q}c^1,$$ which depends on $\ka>0$ if $Q\ne0$, since $c^1\in(0,\infty)$ (the latter fact follows immediately from [Young's convolution inequality](https://en.wikipedia.org/wiki/Young%27s_convolution_inequality#Euclidean_space), with $r=p'$). $\quad\Box$

2

https://mathoverflow.net/users/36721

453535

https://mathoverflow.net/questions/453222

5

Let $G$ be a group and let $K/k$ be a field extension. Suppose that $V$ is a $kG$-module, and let $V\_K = K \otimes\_k V$ be the $KG$-module given by changing the base field. Is it true that $H^n(G,V\_K) \cong K \otimes\_k H^n(G,V)$? If no, what about with some extra assumptions (for example $G$ is finite and $k$ is finite)? My main motivation is in computing some cohomology/ext groups for $KG$-modules, where $K$ is algebraically closed and $G$ is finite. When $k$ is a finite field, there are ways to compute these with GAP for example.

https://mathoverflow.net/users/510150

Extension of base field for modules of groups and cohomology

There is a commutative diagram $\require{AMScd}$ \begin{CD} k[G]\text{-Mod} @>{-\otimes\_{k[G]}K[G]}>> K[G]\text{-Mod}\\ @V{-^G}VV @V{-^G}VV\\ k\text{-Vect} @>{-\otimes\_kK}>> K\text{-Vect}, \end{CD} i.e., $(K[G]\otimes\_{k[G]}V)^G\simeq K\otimes\_kV^G$, since $K$ is a free $k$-vector space (and hence $K[G]\simeq K\otimes\_kk[G]$ is a free $k[G]$-module). Now the horizontal arrows are exact, so the derived functor of the composition $k[G]\text{-Mod}\to K\text{-Vect}$ is isomorphic to both $H^i(G,V\otimes\_{k[G]}K[G])$ and $H^i(G,V)\otimes\_kK$.

5

https://mathoverflow.net/users/123673

453541

https://mathoverflow.net/questions/453542

0

Let $G$ be a graph on $>1$ vertices. Recall that its maximum average degree is defined to be $$M(G) = \max\left\{\frac{2\left|E\_H\right|}{|V\_H|} \colon H \subseteq G, |V\_H| > 1\right\},$$ and its arboricity is equal to $$A(G) = \max\left\{\left\lceil\frac{|E\_H|}{|V\_H|-1}\right\rceil \colon H \subseteq G, |V\_H| > 1\right\}.$$ In [https://arxiv.org/pdf/1802.15267.pdf, near the end of page 2] it is stated that "it is not difficult to show that $2A(G) - 2 \le \lceil M(G)\rceil$". However, I am not sure what is the reasoning.

https://mathoverflow.net/users/512181

Arboricity and average degree

Denote $A(G)=A$. Let $H$ be the subgraph for which $\lceil \frac{|E\_H|}{|V\_H|-1}\rceil=A$. Write $e$ for $|E\_H|$, $v$ for $|V\_H|$. We have $e/(v-1)>A-1$, thus $e>(v-1)(A-1)$. On the other hand, $e\leqslant v(v-1)/2$. Therefore $v/2>A-1$, $v>2A-2$, $v\geqslant 2A-1$. Then $$M(G)\geqslant \frac{2e}v=\frac{2e}{v-1}\cdot \frac{v-1}v>2(A-1)\cdot \frac{2A-2}{2A-1}>2A-3,$$ thus $\lceil M(G)\rceil \geqslant 2A-2$.

4

https://mathoverflow.net/users/4312

453548

https://mathoverflow.net/questions/453539

2

Let $X$ be a curve defined over a number field $K$, and let $G\_K$ be the absolute Galois group of $K$. Let $\text{Aut}(X)$ be the group of $\overline{K}$-defined automorphisms of $X$, and consider the Galois cohomology set $H^1(G\_K, \text{Aut}(X))$. Denote by $\text{Twist}(X/K)$ the set of twists of $X$ up to $K$-defined automorphisms. According to Silverman, Chapter X.2, Theorem 2.2, there is a bijection $$\text{Twist}(X/K)\longrightarrow H^1(G\_K, \text{Aut}(X)).$$ It is well known that there is an equivalence between a Galois cohomology set, such as $H^1(G\_K, \text{Aut}(X))$ and the set of $\text{Aut}(X)$-torsors under $G\_K$. 1. Consider $X$ as the trivial twist of itself. In what way is $X$ an $\text{Aut}(X)$ torsor? I mean, in order for $X$ to be an $\text{Aut}(X)$-torsor, we should have a well defined, bijective, action map $X\times \text{Aut}(X)\longrightarrow X$, yet $\text{Aut}(X)$ doesn't even act on $X$, but rather on $\overline{X}:=X\otimes \overline{K}$. 2. Is there a version of the bijection between twists and cohomology classes which holds in higher dimensions as well? A reference would be appreciated.

https://mathoverflow.net/users/174655

Equivalence between twists of a curve and torsors of its automorphism group

Actually, this has little to do with varieties and is just some general phenomenon in topos theory. For simplicity, let $\mathscr C$ be a (small) category with finite limits, endowed with a Grothendieck pretopology. For a sheaf of groups $\mathscr G$ on $\mathscr C$, write $H^1(\mathscr C,\mathscr G)$ for the set of isomorphism classes of $\mathscr G$-torsors (this has an easy interpretation in terms of cocycles, but I will work only with the torsor point of view). For a sheaf of sets $\mathscr F$ on $\mathscr C$, write $\operatorname{Twist}(\mathscr F)$ for the set of isomorphism classes of twists of $\mathscr F$, i.e. sheaves $\mathscr G$ on $\mathscr C$ for which there exists a covering $\{U\_i \to \*\}\_{i \in I}$ of the terminal object $\*$ of $\mathscr C$ such that $\mathscr G|\_{U\_i} \cong \mathscr F|\_{U\_i}$ for all $i \in I$. Define a map $\operatorname{Twist}(\mathscr F) \to H^1(\mathscr C,\mathscr Aut(\mathscr F))$ by taking a twist $\mathscr G$ of $\mathscr F$ to the sheaf $\mathscr Isom(\mathscr G,\mathscr F)$, which is naturally an $\mathscr Aut(\mathscr F)$-torsor by the map \begin{align\*} \mu \colon \mathscr Aut(\mathscr F) \times \mathscr Isom(\mathscr G,\mathscr F) &\to \mathscr Isom(\mathscr G,\mathscr F)\\ (f,g) &\mapsto f \circ g. \end{align\*} Indeed, this map is clearly a left action of the sheaf of groups $\mathscr Aut(\mathscr F)$ on the sheaf of sets $\mathscr Isom(\mathscr G,\mathscr F)$, and the map \begin{align\*} (\mu,\text{pr}\_2) \colon \mathscr Aut(\mathscr F) \times \mathscr Isom(\mathscr G,\mathscr F) &\to \mathscr Isom(\mathscr G,\mathscr F) \times \mathscr Isom(\mathscr G,\mathscr F)\label{1}\tag{1}\\ (f,g) & \mapsto (fg,g) \end{align\*} is an isomorphism (over opens $U$ where $\mathscr G|\_U \ncong \mathscr F|\_U$, there is nothing to check!). **Lemma.** *The map $\operatorname{Twist}(\mathscr F) \to H^1(\mathscr C,\mathscr Aut(\mathscr F))$ is a bijection.* *Proof.* Injectivity is a little more conceptual than what Silverman shows: let $\mathscr G$ and $\mathscr H$ be twists of $\mathscr F$ such that $\mathscr Isom(\mathscr G,\mathscr F)$ and $\mathscr Isom(\mathscr H,\mathscr F)$ are isomorphic as $\mathscr Aut(\mathscr F)$-torsors. This means that there exists an isomorphism $\alpha \colon \mathscr Isom(\mathscr G,\mathscr F) \stackrel\sim\to \mathscr Isom(\mathscr H,\mathscr F)$ such that the diagram $$\begin{array}{ccc}\mathscr Aut(\mathscr F) \times \mathscr Isom(\mathscr G,\mathscr F) & \to & \mathscr Isom(\mathscr G,\mathscr F) \\ \downarrow & & \downarrow \\ \mathscr Aut(\mathscr F) \times \mathscr Isom(\mathscr H,\mathscr F) & \to & \mathscr Isom(\mathscr H,\mathscr F)\end{array}\tag{2}\label{2}$$ commutes, where the horizontal maps are the actions and the vertical maps are given by $\alpha$. In particular, we get a map $$\psi \colon \mathscr Isom(\mathscr G,\mathscr F) \stackrel\sim\to \mathscr Isom(\mathscr G,\mathscr H)$$ taking a local section $f \colon \mathscr G|\_U \stackrel\sim\to \mathscr F|\_U$ to the composition $\mathscr G|\_U \stackrel f\to \mathscr F|\_U \stackrel{\alpha(f)^{-1}}\longrightarrow \mathscr H|\_U$. We will show that $\psi$ descends to a map $\* \to \mathscr Isom(\mathscr G,\mathscr H)$, i.e. induces an isomorphism of sheaves $\mathscr G \stackrel\sim\to \mathscr H$. For ease of notation, write $\mathscr T$ for the $\mathscr Aut(\mathscr F)$-torsor $\mathscr Isom(\mathscr G,\mathscr F)$. Then the map $\psi$ gives two maps $\psi\_1, \psi\_2 \colon \mathscr T \times \mathscr T \rightrightarrows \mathscr Isom(\mathscr G,\mathscr H)$ by $\psi\_i = \psi \circ \text{pr}\_i$, and it suffices to show that these agree (since $\mathscr T \times \mathscr T \rightrightarrows \mathscr T \to \*$ is a coequaliser). But \eqref{1} gives an isomorphism $\mathscr Aut(\mathscr F) \times \mathscr T \stackrel\sim\to \mathscr T \times \mathscr T$, and the compositions $\mathscr Aut(\mathscr F) \times \mathscr T \rightrightarrows \mathscr Isom(\mathscr G,\mathscr H)$ are given by $(f,g) \mapsto \alpha(fg)^{-1}fg$ and $(f,g) \mapsto \alpha(g)^{-1}g$ respectively. By \eqref{2}, the first map takes $(f,g)$ to $(f\alpha(g))^{-1}fg = \alpha(g)^{-1}g$, which is what we needed to show. Surjectivity, as in Silverman, relies on descent theory (glueing). Let $\mathscr U = \{U\_i \to \*\}\_{i \in I}$ be a covering over which $\mathscr T$ has sections $g\_i$ for all $i \in I$, and define a glueing datum for a sheaf of sets $\mathscr G$ as follows: over $U\_i$ take the sheaf $\mathscr G\_i = \mathscr F|\_{U\_i}$, and for each $i,j \in I$ take the isomorphism $\phi\_{ij} \colon \mathscr G\_j|\_{U\_i \cap U\_j} \stackrel\sim\to \mathscr G\_i|\_{U\_i \cap U\_j}$ as the unique section $\phi\_{ij}$ of $\mathscr Aut(\mathscr F)(U\_i \cap U\_j)$ such that $\phi\_{ij} g\_j = g\_i$. This satisfies the cocycle condition, turning $(\{\mathscr G\_i\},\{\phi\_{ij}\})$ into a glueing datum. Then there exists a sheaf $\mathscr G$ on $\mathscr C$ such that $\mathscr G|\_{U\_i} = \mathscr G\_i$ for all $i \in I$; see [Stacks, Tag [04TR](https://stacks.math.columbia.edu/tag/04TR)]. It is easy to check that the $\mathscr Aut(\mathscr F)$-torsor $\mathscr Isom(\mathscr G,\mathscr F)$ is canonically isomorphic to $\mathscr T$. $\square$ Here is an even more high-brow argument using stacks of groupoids (= 'sheaves of 1-truncated homotopy types). Let $B(\mathscr Aut(\mathscr F))$ be the stackification of the prestack $U \mapsto B(\mathscr Aut(\mathscr F|\_U))$, and let $\mathbf{Twist}$ be the stack taking $U \in \mathscr C$ to the category of twists of $\mathscr F|\_U$ (with only invertible morphisms). The thing we're trying to understand is $\operatorname{Twist}(\mathscr F) = \pi\_0(\mathbf{Twist}(\mathscr F)(\*))$ (the objects up to isomorphism of the category of global sections of the stack $\mathbf{Twist}(\mathscr F)$). There is a natural map $B(\mathscr Aut(\mathscr F|\_U)) \to \mathbf{Twist}(\mathscr F)(U)$ mapping the unique object of $B(\mathscr Aut(\mathscr F|\_U))$ to the trivial twist $\mathscr F|\_U$ of $\mathscr F|\_U$. The construction of stackification [Stacks, Tag [02ZN](https://stacks.math.columbia.edu/tag/02ZN)] shows that the associated morphism of stacks $B(\mathscr Aut(\mathscr F)) \to \mathbf{Twist}(\mathscr F)$ is an equivalence, giving a bijection $\operatorname{Twist}(\mathscr F) \stackrel\sim\to \pi\_0(B\mathscr Aut(\mathscr F)(\*))$. But for any group object $\mathscr G$ in a topos, the category $(B\mathscr G)(\*)$ of global sections of $B\mathscr G$ is the category of $\mathscr G$-torsors. In particular, we've upgraded our bijection of 'objects up to a suitable notion of isomorphism' to an equivalence of groupoids! --- For the scheme case, there is an extra difficulty that a twist $\mathscr G$ of a representable sheaf $\mathscr F$ (i.e. representable by a scheme $X$) need not be representable (by a scheme $Y$). More or less by definition (of algebraic space), the twist $\mathscr G$ is at least an algebraic space $Y$, as it is an étale sheaf quotient of the scheme $X\_{\bar K}$. If $X$ is projective, it has an ample line bundle, which you can use to produce an ample line bundle on $Y$, so $Y$ is projective as well (hence a scheme). (Proof sketch: an ample line bundle $\mathscr L$ on $X$ gives an ample line bundle on $Y\_L$ after some finite Galois extension $K \to L$, and you can apply the norm map $N\_{L/K} \colon \mathbf{Pic}\_Y(L) \to \mathbf{Pic}\_Y(K)$ to get a line bundle on $Y$. It is ample since it is a sum of Galois conjugates of an ample line bundle. Exercise: find some details to worry about.)

5

https://mathoverflow.net/users/82179

453551

https://mathoverflow.net/questions/453505

1

I'm reading an article regarding the condition for the existence of a solution to a constrained Pick interpolation problem, and the author wrote the following equality - for background, we have the following spaces: $$H^2\_{\alpha,\beta}\subset M\_k(H^2)$$ where $M\_k(H^2)$ is the set of $k\times k$ matrices with entries being functions in $H^2$, and inner product given by: $$\langle f,g\rangle=\frac{1}{2\pi}\int\_{\mathbb{T}}tr[f(z)g(z)^\*]|dz|$$ Now given points $z\_1,\ldots,z\_n$, we let: $$B\_FM\_k(H^2)=\{f\in M\_k(H^2):f(z\_i)=0,\forall i=1,\ldots,n\}$$ To the question: the author defines the subspace: $$M\_{\alpha,\beta}=H^2\_{\alpha,\beta}\cap(H^2\_{\alpha,\beta}\cap B\_FM\_k(H^2))^{\perp}\subset H^2\_{\alpha,\beta}$$ Now assuming that $Q\in M\_k(H^{\infty})$ is such that $M\_Q$ (the multiplier operator by $Q$) is invariant for $H^2\_{\alpha,\beta}$, $g\in H^2\_{\alpha,\beta}$ and $f\in (H^2\_{\alpha,\beta}\cap B\_FM\_k(H^2))^{\perp}$, the author claims that: $$\langle Qg,f\rangle=\langle P\_{H^2\_{\alpha,\beta}}Qg,P\_{(H^2\_{\alpha,\beta}\cap B\_FM\_k(H^2))^{\perp}}f\rangle=\langle P\_{M\_{\alpha,\beta}}Qg,P\_{M\_{\alpha,\beta}}f\rangle$$ I understand the first equality, but cannot understand the second equality. Any ideas or explanations on why this is true would be appreciated. (Clarification: $P\_W$ is the orthogonal projection on the subspace $W$).

https://mathoverflow.net/users/489525

An inner product and projection property in RKHS

I don't fully understand the context here, but it seems to me what you need is the following: Let $A = H^2\_{\alpha, \beta}$, $B = (H^2\_{\alpha, \beta} \cap B\_FM\_k(H^2))^\perp$, then $P\_BP\_A = P\_{A \cap B}$. We can show that this would happen if $A \cap (A \cap B)^\perp \perp B$. (Indeed, we observe that $P\_BP\_A$ and $P\_{A \cap B}$ both reduce to 1 on $A \cap B$, so it suffices to show $P\_BP\_A((A \cap B)^\perp) = 0$. Fix any $h \in (A \cap B)^\perp$, write $h = P\_Ah + P\_{A^\perp}h$. Since $A^\perp \subseteq (A \cap B)^\perp$, we have, $P\_Ah = h - P\_{A^\perp}h \in (A \cap B)^\perp$, so $P\_Ah \in A \cap (A \cap B)^\perp$. As $A \cap (A \cap B)^\perp \perp B$, we have $P\_BP\_Ah = 0$.) Note that in this case, $(A \cap B)^\perp = \overline{A^\perp + B^\perp} = \overline{(H^2\_{\alpha, \beta})^\perp + (H^2\_{\alpha, \beta} \cap B\_FM\_k(H^2))} = (H^2\_{\alpha, \beta})^\perp + (H^2\_{\alpha, \beta} \cap B\_FM\_k(H^2))$, as the two summands are orthogonal to each other, so $A \cap (A \cap B)^\perp = H^2\_{\alpha, \beta} \cap ((H^2\_{\alpha, \beta})^\perp + (H^2\_{\alpha, \beta} \cap B\_FM\_k(H^2))) = H^2\_{\alpha, \beta} \cap B\_FM\_k(H^2) = B^\perp$. This proved the claim.

1

https://mathoverflow.net/users/504602

453559

https://mathoverflow.net/questions/453549

0

If $G =(V,E)$ is a simple, undirected graph (finite or infinite), and $\kappa \neq \emptyset$ is a cardinal, we say that the complete graph $K\_\kappa$ is a **minor** of $G$ if there is a collection ${\frak S}$ of connected and pairwise disjoint subsets of $G$ such that 1. $|{\frak S}| = \kappa$, and 2. whenever $S\neq T\in {\frak S}$, then there is $e\in E$ such that $(S\cap e) \neq \emptyset \neq (T\cap e)$. The [Hadwiger-Nelson graph](https://en.wikipedia.org/wiki/Hadwiger%E2%80%93Nelson_problem) has $\mathbb{R}^2$ as its ground set, and two elements $x,y\in\mathbb{R}^2$ form an edge if and only if their Euclidean distance equals $1$. **Question.** Is the set of cardinals $\kappa \neq \emptyset$ such that $K\_\kappa$ is a minor of the Hadwiger-Nelson graph finite? If yes, what is its greatest member? **Note.** Since the [Hadwiger conjecture](https://en.wikipedia.org/wiki/Hadwiger_conjecture_(graph_theory)) is correct for $k\leq 6$, the set asked for in the question is a super-set of $\{1,2,3,4,5\}$, because it is known that the chromatic number of the Hadwiger-Nelson graph is a member of $\{5,6,7\}$.

https://mathoverflow.net/users/8628

Hadwiger number of the Hadwiger-Nelson graph on $\mathbb{R}^2$

No, the set of such cardinals is infinite. Take $\kappa$ points on the interval $(0, 1) \times \{0\}$. Every two points $(u, 0)$ and $(v, 0)$ have a common neighbour $$\left(\frac{u+v}{2}, \sqrt{1 - \frac{(u-v)^2}{4}}\right)$$ which is not adjacent to any other vertex on the interval. Thus the Hadwiger-Nelson graph has a $K\_\kappa$ minor. Note that we can pick $\kappa$ to be any natural number (as well as $\aleph\_0$ or $2^\aleph\_0$), thus there are infinitely many cardinals $\kappa$ such that $K\_\kappa$ is a minor of the Hadwiger-Nelson graph.

3

https://mathoverflow.net/users/502833

453562

https://mathoverflow.net/questions/453442

2

**Definitions and setting** Let $\mathcal{H}$ be a separable, infinite-dimensional, reproducing kernel Hilbert space on a nonempty set $X$. As usual, denote the reproducing kernel on $\mathcal{H}$ by $K$ and let $K\_x:=K(x,\cdot)$. Let $M(\mathcal{H})$ denote the **multiplier algebra** of $\mathcal{H}$, that is, the functions $f:H\rightarrow \mathbb{R}$ satisfying: for all $h\in \mathcal{H}$ one has $x\mapsto f(x)h(x)$ is a function belonging to $H$ (i.e. pointwise multiplication). A sequence $(k\_i)\_{i\in \mathbb{N}}\subseteq X^{\mathbb{N}}$ is said to be an **interpolating sequence** of $M(\mathcal{H})$ if, for any bounded sequence of reals $(r\_i)\_{i\in \mathbb{N}} \in \ell^{\infty}$ one can always find a multiplier $\mu\in M(\mathcal{H})$ satisfying: $$ \phi(k\_i) = w\_i \mbox{ for all i in }\mathbb{N} . $$ As discussed in the recent article [Tsikalas - Interpolating sequences for pairs of spaces](https://doi.org/10.1016/j.jfa.2023.110059), many RKHSs have the property that the set of interpolating sequences can be characterized as the set of sequences $\boldsymbol{k}:=(k\_i)\in X^{\mathbb{N}}$ for which $T\_{\boldsymbol{k}}(\mathcal{H})=\ell^2$ where $$ \begin{aligned} T\_{\boldsymbol{k}}:\mathcal{H} & \rightarrow \mathbb{R}^{\mathbb{N}}\\ T\_{\boldsymbol{k}}(f) & := \big(\frac{f(k\_i)}{\|K\_{k\_i}\|}\big)\_{i\in \mathbb{N}} \end{aligned} $$ for all $f\in \mathcal{H}$. --- I recently came across this very old [MO post](https://mathoverflow.net/questions/126288/do-kernels-provide-a-basis-for-a-rkhs "Do kernels provide a basis for a RKHS?"), claiming that, $(K\_{k\_i})\_{i\in \mathbb{N}}$ is a Schauder, or even a Riesz, basis for $\mathcal{H}$ if $(k\_i)\_{i\in \mathbb{N}}$ is an interpolating sequence of $M(\mathcal{H})$. If this is correct, why is this this the case? --- **Intuition:** I expect this somehow follows from the characterization of "good interpolating sequences", by which I mean, those satisfying $T\_{\boldsymbol{k}}(\mathcal{H})=\ell^2$; however, I don't really see it….

https://mathoverflow.net/users/36886

Orthonormal bases in RKHSs via interpolating sequences

It is true that if a sequence $(k\_n)$ is interpolating for $M(\mathcal{H})$, then the *normalized* Kernel vectors $g\_n:=K\_{k\_n}/\Vert K\_{k\_n} \Vert\_\mathcal{H} $ form a Riesz system in $\mathcal{H}$. Of course it does not have to be a complete system, so it is not always a Riesz basis. To see this, let $g\_n$ be the normalized kernel vectors, and for $t\in \mathbb{R}, j\in \mathbb{N}$ define $w\_j = e^{2\pi i j t} $. Then by hypothesis, for every $t$ there exists a multiplier $\varphi\_t$ such that $\varphi\_t(z\_j) = w\_j,\, j=1,2,\dots$. Furthermore by the open mapping theorem we can choose $\varphi\_t$ such that $\Vert \varphi\_t \Vert\_{M(\mathcal{H})} \leq M < +\infty$. Then consider the adjoint of the multiplication operator by $\varphi\_t$, $M\_{\varphi\_t}^\*:\mathcal{H} \to \mathcal{H}$. By the bound on the multiplier norms of $\varphi\_t$, \begin{equation}\label{eq1} \Vert M^\*\_{\varphi\_t} f \Vert\_\mathcal{H} ^2 \leq M^2 \Vert f \Vert\_\mathcal{H}^2,\,\,\, \forall t\in \mathbb{R}, \,\,\, \forall f \in \mathcal{H}. \end{equation} Now consider $(a\_n)$ a sequence of complex numbers with only finite non zero terms, and $f=\sum\_{n}a\_n g\_n $. Applying the above inequality for this choice of $f$ and using the fact that $M^\*\_{\varphi\_t} g\_n = \overline{\varphi\_t(k\_n)} g\_n $ we arrive at the inequality, $$ \sum\_{n,m} a\_n \overline{a\_m} e^{2\pi i (n-m) t} \langle g\_n , g\_m \rangle \leq M^2 \Vert \sum\_{n} a\_n g\_n \Vert^2, $$ and integrating with respect to $t$ in $(0,1)$ we find that $$ \sum\_n |a\_n|^2 \leq M^2 \Vert \sum\_{n} a\_n g\_n \Vert^2 $$ By replacing $a\_n$ by $e^{-2\pi i tn} a\_n $ and using the same ineqaulity and integrating with respect to $t$ we find that $$\Vert \sum\_{n} a\_n g\_n \Vert^2 \leq M^2 \sum\_n |a\_n|^2. $$ This proves that $g\_n$ is a Riesz system.

1

https://mathoverflow.net/users/153260

453563

https://mathoverflow.net/questions/453556

0

I am interested in $H¹$ right now and the cocycle condition $φ\_{jk} • φ\_{ij} = φ\_{ik}$ because of how it is said to relate to automorphic forms. I can't quite see the relationship between factors of automorphy and $H¹(G,M)$ that is mentioned [here](https://en.wikipedia.org/wiki/Automorphic_form). Let $G$ be a group acting on a complex manifold $X$ and let $f$ be a holomorphic function from $X$ to $ℂ$. Recall that the factor of automorphy has $$f(gx) = j\_g(x) f(x)$$, which already seems to bear some similarity to a cocycle condition. The cohomology of a group $G$ with coefficients in $M$ can be calculated from the chain complex whose nth component is the $G$-module of functions from $Gⁿ$ to $M$. See [here](https://en.wikipedia.org/wiki/Group_cohomology#Cochain_complexes) for the formulas for the differential in group cohomology. I am interested in $$H¹(G, M) = Z¹(G,M)/B¹(G,M)$$ The wikipedia article on automorphic forms mentions that "The formulation requires the general notion of factor of automorphy j for Γ, which is a type of 1-cocycle in the language of group cohomology." Can someone spell this out? Is it possible to interpret the automorphy condition as a cocycle condition?

https://mathoverflow.net/users/490598

Modular forms and the cocycle condition in group cohomology

We have $j\_{gh}(x)f(x)=f((gh)x)=f(g(hx))=j\_g(hx)f(hx)=j\_g(hx)j\_h(x)f(x)$, and so $j\_{gh}(x)=j\_g(hx)j\_h(x)$, which is the one-cocycle condition.

2

https://mathoverflow.net/users/460592

453564

https://mathoverflow.net/questions/453552

11

How does the cardinality of $$\{(a,b): 1 \leq a,b \leq n, \ 2ab/(a+b) \ \mbox{is an integer}\}$$ grow as a function of $n$? What about $$\{(a,b): 1 \leq a,b \leq n, \ \sqrt{ab} \ \mbox{is an integer}\}?$$ For $n=10$, $100$, $1000$, and $10000$ the two functions of $n$ are close and getting closer (in relative terms); one might even surmise that the ratio tends to 1. For those who care, those ratios are $18/14 = 1.2857$, $310/272 = 1.1397$, $4344/3976 = 1.0926$, and $57296/52756 = 1.0861$. Numerators are for the geometric mean and denominators are for the harmonic mean. On the other hand, $10000$ is not very big, so the data from the previous sentence may not reflect the true asymptotic behavior of the two functions.

https://mathoverflow.net/users/3621

Asymptotics for pairs of positive integers whose harmonic (resp. geometric) mean is an integer

[CORRECTED 8/28] I think the asymptotic ratio is $(4/3)\*\ln(2)\approx 0.924196$. $$G(n) = (6/\pi^2)\*n\*\ln(n) + O(n)$$ $$H(n) = (8\ln(2)/\pi^2)\*n\*\ln(n) + O(n)$$ The proof for the geometric mean goes as Noam Elkies posted. For the harmonic mean, we will first characterize "primitive" solutions: pairs $(a,b)$ with $\gcd(a,b)=1$ and $(a+b)|(2ab)$. Let $r=p/q$ be a rational number in lowest terms. Then $(a,a\*r)$ has integer harmonic mean iff $(p+q)|(2ap)$, which is true iff $a$ is a multiple of $(p+q)/\gcd(p+q,2)$. And we also need $a\*r$ to be an integer, which means $q|a$. So we get a primitive solution iff $a=q(p+q)/\gcd(p+q,2)$. So the primitive solutions are: $(p(p+q),q(p+q))$ when $\gcd(p,q)=1$ and $p+q$ is odd, $(p(p+q)/2,q(p+q)/2)$ when $\gcd(p,q)=1$ and $p+q$ is even. Or, equivalently, $(p\*s, (s-p)\*s)$ when $\gcd(p,s)=1$, $s$ is odd, $p<s$, $(p\*s/2, (s-p)\*s/2)$ when $gcd(p,s)=1$, $s$ is even, $p<s$. We will count all solutions bounded by $n$ by assuming $s/2<p<s$ (except $(p,s)=(1,2)$), and then multiplying by 2 for symmetry: $$n+2 \int\_{1}^{\sqrt{n}} \int\_{p}^{\min(2p,n/p)} \frac{1}{2}\frac{8}{\pi^2}\frac{n}{ps} \,ds\,dp + 2 \int\_{1}^{\sqrt{2n}} \int\_{p}^{\min(2p,2n/p)} \frac{1}{2}\frac{4}{\pi^2}\frac{n}{ps/2} \,ds\,dp $$ $$ = n + \int\_{1}^{\sqrt{n/2}} \int\_{p}^{2p} \frac{8}{\pi^2}\frac{n}{ps} \,ds\,dp + \int\_{\sqrt{n/2}}^{\sqrt{n}} \int\_{p}^{n/p} \frac{8}{\pi^2}\frac{n}{ps} \,ds\,dp \\ $$ $$ + \int\_{1}^{\sqrt{n}} \int\_{p}^{2p} \frac{8}{\pi^2}\frac{n}{ps} \,ds\,dp + \int\_{\sqrt{n}}^{\sqrt{2n}} \int\_{p}^{2n/p} \frac{8}{\pi^2}\frac{n}{ps} \,ds\,dp $$ $$ \asymp n + (4\ln(2)/\pi^2)\*n\*\ln(n) + (2\ln(2)^2/\pi^2)\*n $$ $$ + (4\ln(2)/\pi^2)\*n\*\ln(n) + (2\ln(2)^2/\pi^2)\*n$$ $$ = (8\ln(2)/\pi^2)\*n\*\ln(n) + O(n) $$ (Note: the exact integrals should have $\lfloor n/(ps)\rfloor$ in place of $n/(ps)$, but this is bounded above by $n/(ps)$ and bounded below by $n/(ps)-1$, and we can compute both integrals and show that they are asymptotically equivalent.)

12

https://mathoverflow.net/users/2953

453565

https://mathoverflow.net/questions/453270

3

It is quite well known that for any polyhedral set $K$ and any linear mapping $A$, the set $AK$ is polyhedral and hence closed. I am more curious about the converse problem, namely: Suppose $K$ is a closed convex cone and for any linear mapping $A$, the set $AK$ is closed. Is $K$ a polyhedral cone? For closed convex set, counterexample is easy to construct, e.g., $\{(x,y):y\geq x^2\}$. But for closed convex cone, I didn't find any easy counterexample.

https://mathoverflow.net/users/113353

A converse question about the polyhedrality under linear mapping

I think we can argue as in <https://mathoverflow.net/a/423284/32507> to answer the question in the affirmative. Let $\mathcal R\_K(x)$ be the radial cone of $K$ at $x$ (as defined in the other answer). Further, for fixed $x \in K$, let $A$ be the projection onto the orthogonal complement of $\operatorname{span}(x)$. By assumption, $A K$ is closed. Due to $$ \mathcal R\_K(x) = K + \operatorname{span}(x) = A K + \operatorname{span}(x)$$ (and since $A K$ and $\operatorname{span}(x)$ are orthogonal), the radial cone $\mathcal R\_K(x)$ is closed for each $x \in K$. Hence, by Proposition 2 of [Duality of linear conic problems by Shapiro and Nemirovski](http://www.optimization-online.org/DB_FILE/2003/12/793.pdf) we get that $K$ is a polyhedral cone.

2

https://mathoverflow.net/users/32507

453566

https://mathoverflow.net/questions/453574

5

Let A $\subset [0:2]^n$, where $[0:2]=\{0,1,2\}$, then define $2A= \{ a+b\mid a,b \in A \}$. I wanted to know the best known lower-bound estimates for $|2A|$. I intuitively expect that $|2A| \geq |A|^{\log\_3{5}}$ as this estimate works for $n=1$ and if we take $A= [0:2]^n$, then $|2A| = [0:4]^n$ and the bound fits. For $A \subset [0:d]^n$, I expect $|2A| \geq |A|^{\log\_{d+1}{(2d+1)}}$. Are there any known results for such settings?

https://mathoverflow.net/users/506732

Estimate of Minkowski sum

This is only a partial answer to your question; I believe there is more current work, and have forwarded your question to someone working in this area to see if they have more recent results. In Theorem 5 of *Bourgain, Jean; Dilworth, Stephen; Ford, Kevin; Konyagin, Sergei; Kutzarova, Denka*, [**Explicit constructions of RIP matrices and related problems**](https://arxiv.org/abs/1008.4535), Duke Math. J. 159, No. 1, 145-185 (2011). [ZBL1236.94027](https://zbmath.org/?q=an:1236.94027). it is shown (in your notation) that $|2A| \geq |A|^{2\tau}$ for $A \subset [0:d]^n$, where $\tau$ solves the equation $$ (\frac{1}{d+1})^{2\tau} + (\frac{d}{d+1})^\tau = 1.$$ They do not believe this result to be sharp, and (as you do) conjecture that $|2A| \geq |A|^{\log\_{d+1}(2d+1)}$ instead. This is known for $d=1$, see *Woodall, D. R.*, [**A theorem on cubes**](https://doi.org/10.1112/S0025579300008913), Mathematika, London 24, 60-62 (1977). [ZBL0349.05010](https://zbmath.org/?q=an:0349.05010). From the method of compressions one may assume without loss of generality that $A$ is a downset. If one then restricts $2A$ to the set $[0:d]^n$ then a sharp answer to your question was worked out in *Bollobás, Béla; Leader, Imre*, [**Sums in the grid**](https://doi.org/10.1016/S0012-365X(96)00303-2), Discrete Math. 162, No. 1-3, 31-48 (1996). [ZBL0872.11007](https://zbmath.org/?q=an:0872.11007). however I do not see an easy way to pass from this restricted sumset problem to the full sumset problem. Nevertheless, the method of compressions may be a promising technique to attack the problem. There are related results in *Matolcsi, Dávid; Ruzsa, Imre Z.; Shakan, George; Zhelezov, Dmitrii*, [**An analytic approach to cardinalities of sumsets**](https://doi.org/10.1007/s00493-021-4547-0), Combinatorica 42, No. 2, 203-236 (2022). [ZBL1513.11022](https://zbmath.org/?q=an:1513.11022). which generalize a related inequality $|2A + \{0,1\}^n| \geq 2^n |A|$ that I worked out with Ben Green in *Green, Ben; Tao, Terence*, [**Compressions, convex geometry and the Freiman-Bilu theorem**](https://doi.org/10.1093/qmath/hal009), Q. J. Math. 57, No. 4, 495-504 (2006). [ZBL1160.11003](https://zbmath.org/?q=an:1160.11003). and the analogous problem for various additive energies was studied in [this recent paper of de Dios, Greenfeld, Ivanisvili, and Madrid](https://arxiv.org/abs/2112.09352), which would give some lower bounds on $|2A|$, but probably not the optimal ones.

9

https://mathoverflow.net/users/766

453587

https://mathoverflow.net/questions/453511

1

I am looking for a reference for the following. I believe the set of elliptic curves $E/\mathbb{Q}$ admitting a rational 5-isogeny can be parametrized as $$E: y^2 = x^3 + f(t)x + g(t), t \in \mathbb{Q}$$ for some polynomials $f,g \in \mathbb{Z}[t]$. The analogous parametrization for elliptic curves admitting a 7-isogeny is given by $$\displaystyle y^2 = x^3 + u(t) x + v(t),$$ with $$u(t) = -3(t^2 + t + 7)(t^2 - 231t + 735)$$ and $$v(t) = 2(t^2 + t + 7)(t^4 + 518t^3 - 11025t^2 + 6174t - 64826).$$ Does anyone know what $f,g$ correspond to 5-isogenies?

https://mathoverflow.net/users/10898

Family parametrizing elliptic curves with a rational 5-isogeny

Let $\phi:E\to E'$ be an isogeny of degree $5$, and $j, j'$ be the $j$-invariants of $E$ and $E'$, respectively. Let $r$ be the rational function $$r(z)=\frac{(z^2+10z+5)^3}{z}.$$ Then there is $\tau\in\mathbb Q$ such that $$j=r(\tau)\quad\text{and}\quad j'=r(125/\tau),$$ and the converse holds too. This result is more than 100 years old. I don't know a modern and English reference, but you can find it in Kapitel 4 of the book *Die elliptischen Funktionen und ihre Anwendungen, zweiter Teil* by Robert Fricke. To make the example explicit: The family of these elliptic curves (up to isomorphism over $\bar{\mathbb Q}$) is $$y^2=x^3+Ax+B,$$ where $$\begin{aligned}A&=-108(t^2 + 10t + 5)(t^2 + 22t + 125),\\ B&=-432(t^2 + 4t - 1)(t^2 + 22t + 125)^2.\end{aligned}$$ Its $5$-division polynomial has the quadratic factor $$x^2 + (12t^2 + 264t + 1500)x + (36t^4 + 1584t^3 + 25128t^2 + 169488t + 400500),$$ yielding a subgroup of order $5$ which (as a group, not the individual elements) is defined over $\mathbb Q$.

9

https://mathoverflow.net/users/18739

453596

https://mathoverflow.net/questions/453595

3

If $\sigma\_k(n)=\sum\_{d\vert n} d^k$, denote $$F\_1(q)=\sum\_{n\geq1}\sigma\_1(n)\,q^n \qquad \text{and} \qquad F\_3(q)=\sum\_{n\geq1}n\cdot\sigma\_2(n)\,q^n.$$ > > **QUESTION.** Assume the prime $p$ is either $2, 3$ or $5$. Is it true that > $F\_1(q)-F\_3(q)\equiv F\_1(q^p) \pmod p$? Are there other primes for which such a congruence holds? > > > **Convention.** The congruence should be interpreted coefficient-wise. **Postscript.** Sorry, there was a typo in the definition of $F\_3(q)$. Please read as $$F\_3(q)=\sum\_{n\geq1}n\cdot\sigma\_{\mathbf{{\color{red}3}}}(n)\,q^n.$$

https://mathoverflow.net/users/66131

Congruences for power-sum of divisors

I wrote an answer in the comment section, so I'll rewrite it here. --- The claim of $$F\_1(q)-F\_3(q)\equiv F\_1(q^p) \pmod p$$ is equivalent to the following two cases $$p\nmid n: \sum\_{d|n} d-nd^3 \equiv 0 \pmod p$$ $$p|n: \sum\_{d|n} d-nd^3 \equiv \sum\_{d| \frac{n}{p}} d \pmod p$$ The second case is trivial, because if $p|n$, we have $$\begin{split} \sum\_{d|n} d-nd^3 & \equiv \sum\_{d|n} d \pmod p \\ &\equiv \sum\_{d|\frac{n}{p}} d \pmod p\end{split}$$ --- Thus we only need to check $$p\nmid n: \sum\_{d|n} d-nd^3 \equiv 0 \pmod p$$ For $p=2$, this is immediate as it holds for each term in the sum individually. That is, we necessarily have $n \equiv 1 \pmod 2$, and $d-nd^3 \equiv d-d^3 \equiv d(1-d)(1+d) \equiv 0 \pmod 2$. --- For $p=3$ and $p=5$, it is worthwhile to write $$\sum\_{d|n} d-nd^3 = \sum\_{d|n\text{ and } d<\sqrt{n}} (d-nd^3)+\left(\frac{n}{d}-n\left(\frac{n}{d}\right)^3\right)$$ My claim is that each term in the sum on the right is congruent to $0$ mod $3$ and $5$. This follows from $$(d-nd^3)+\left(\frac{n}{d}-n\left(\frac{n}{d}\right)^3\right) = \frac{1}{d^3}(d^2+n)(d^2-nd^4+n^2d^2-n^3)$$ We can now proceed by cases, remembering that $p \nmid n$ and hence $p \nmid d$. --- For $p=3$, first note that $d^2 \equiv 1 \pmod 3$. Then we enjoy $$\begin{split} \frac{1}{d^3}(d^2+n)(d^2-nd^4+n^2d^2-n^3) &\equiv d (1+n)(1-n+n^2-n^3) \pmod 3 \\ \equiv d(1+n)(1-n)(1+n^2) \pmod 3 \end{split}$$ Note that we have $n \equiv 1 \pmod 3$ or $n \equiv 2 \pmod 3$, and plugging those in gives congruence to $0$. Thus we have confirmed the $p=3$ case. --- For $p=5$, first note that $d^2 \equiv 1 \pmod 5$ or $d^2 \equiv 4 \pmod 5$. In the first case, $$\begin{split} \frac{1}{d^3}(d^2+n)(d^2-nd^4+n^2d^2-n^3) & \equiv d(1+n)(1-n)(1+n^2) \pmod 5 \end{split}$$ In the second case, $$\begin{split} \frac{1}{d^3}(d^2+n)(d^2-nd^4+n^2d^2-n^3) & \equiv d(4+n)(4-n)(1+n^2) \pmod 5 \end{split}$$ It's straightforward to check that $n \equiv 1,2,3,4 \pmod 5$ plugged into the above will vanish. Thus we have confirmed the $p=5$ case. --- We can rule out the claim of $$F\_1(q)-F\_3(q)\equiv F\_1(q^p) \pmod p$$ for all other $p$ by noting that for $n=2$, $$\sum\_{d|n} d-nd^3 \equiv -15 \pmod p$$ This is a problem for prime $p>5$, as then we have $p \nmid n$ but $\sum\_{d|n} d-nd^3$ is not congruent to $0$. (As an aside, note that this isn't a problem for $p=2$, since then $p|n$ and it's straightforward to check that it's congruent to $\sum\_{d|\frac{n}{p}} d$, as needed. This isn't a problem for $p=3,5$, since then $p \nmid n$ and we clearly have congruence to $0$, as needed.) --- EDIT: I realized that in proving the $p=3,5$ cases for $p \nmid n$ that there was an edge case I needed to check. I check this edge case here. To prove the cases of $p=3,5$, I used $$\sum\_{d|n} d-nd^3 = \sum\_{d|n\text{ and } d<\sqrt{n}} (d-nd^3)+\left(\frac{n}{d}-n\left(\frac{n}{d}\right)^3\right)$$ This works whenever $\sqrt{n}$ is not an integer. When $\sqrt{n}$ is an integer, it's a valid divisor, and in that case we'll have one more term $$\sum\_{d|n} d-nd^3 = \sqrt{n}-n(\sqrt{n})^3 + \sum\_{d|n\text{ and } d<\sqrt{n}} (d-nd^3)+\left(\frac{n}{d}-n\left(\frac{n}{d}\right)^3\right)$$ So far, I showed that the rightmost sum is congruent to $0$ mod $3$ and $5$, so I only need to check this extra term. This extra term is $$ \sqrt{n}-n(\sqrt{n})^3 = \sqrt{n}(1-n)(1+n)$$ For the case of $p=3$, $n \equiv 1 \pmod 3$, since $n$ is a square number not divisible by $3$. For the case of $p=5$, $n \equiv 1 \pmod 5$ or $n \equiv 4 \pmod 5$, since $n$ is a square number not divisible by $5$. Plugging these values in confirms congruence to $0$.

5

https://mathoverflow.net/users/153549

453600

https://mathoverflow.net/questions/453608

1

Let us consider a variety $X$ over a field $k$ which is a finite field or an algebraic closure thereof. Let $\ell$ be a prime number different from the characteristic of $k$, and let $\Lambda = \mathbb Z/\ell^k\mathbb Z$ where $k\geq 1$. We have a "derived" category $D\_c^b(X,\Lambda)$ of bounded complexes of constructible $\Lambda$-adic sheaves on $X$. It is equipped with a 6 functors formalism. Let $i:Z\hookrightarrow X$ be a closed immersion, and let $j:U \hookrightarrow X$ be the open complement. For any $K \in D\_c^b(X,\Lambda)$, I expect that we should have two distinguished triangles $$i\_\*i^!(K) \to K \to \mathrm Rj\_\*j^\*(K) \xrightarrow{+1} \ldots$$ and $$\mathrm Rj\_!j^\*(K) \to K \to i\_\*i^\*(K) \xrightarrow{+1} \ldots$$ Moreover, one triangle is the dual of the other via the dualizing functor $D\_X := \mathrm R\mathcal{Hom}(\cdot,f^!\Lambda)$ where $f:X\to \mathrm{Spec}(k)$ is the structure morphism. By this, I mean that one triangle is obtained by applying $D\_X$ to the other triangle for the complex $D\_X(K)$. This is stated in the book *Weil conjectures, perverse sheaves and $\ell$-adic Fourier transform* by Kiehl and Weissauer (2001), section II.11, p.123-124 (see Lemma 11.1). However, in the way the paragraph is written, the authors seem to make the underlying assumption that $f$ is smooth. I feel like smoothness is not used in this specific section of the paragraph which I quoted above, and I'd like to ask confirmation of this fact. Besides, is there maybe another reference without this smoothness assumption?

https://mathoverflow.net/users/125617

Reference for localization distinguished triangles in the derived category of $\ell$-adic sheaves

Indeed $X$ does not have to be smooth. These triangles are part of the yoga of "recollement". See Section 1.4 of Beilinson-Bernstein-Deligne. A quotable reference that works in great generality is (4.10) of Laszlo-Olsson.

4

https://mathoverflow.net/users/1310

453611

https://mathoverflow.net/questions/453435

2

Let $\Omega\_1 \subset \mathbb R^2$ be a bounded simply-connected Lipschitz domain, and $f: \bar \Omega\_1 \rightarrow \bar \Omega\_2$ be a homeomorphism, which is a diffeomorphism on $\Omega\_1$ such that $f$ and its inverse have first derivatives in $L^p$ with $p>2$. The Morrey inequality implies that $f$ is Hölder. What can be said about its inverse provided that $p$ is large enough? The Morrey inequality cannot be directly applied to the inverse since the boundary of $\Omega\_2$ might be not Lipschitz, but could it be salvaged with the additional information we have?

https://mathoverflow.net/users/478527

Hölderness of the inverse to a $W^{1,p}$-homeomorphism (with additional conditions) of a Lipschitz domain

What you want is not possible. The basic idea is that it is possible for a diffeo $\Omega\_1\to \Omega\_2$ that extends to a homeo $\bar{\Omega}\_1 \to \bar{\Omega}\_2$ to "almost fold up" $\partial\Omega\_1$. To give an explicit example: let $\Omega\_1$ be the upper half-unit-disk, in polar coordinates the set $\{0< r < 1, 0 < \theta < \pi\}$. Let $f(r,\theta) = (r,(2 - r^q)\theta)$ for some $q \gg 1$. Then $\Omega\_2 = \{0 < r < 1, 0 < \theta < (2-r^q) \pi \}$. Let $g$ be the inverse mapping, we find $g(s,\phi) = (s, (2 - s^q)^{-1} \phi)$. Both $f$ and $g$ are clearly smooth, and extend continuously to the boundary of the domains $\Omega\_1$ and $\Omega\_2$ respectively. Furthermore, $$ |\nabla f|^2 = 1 + q^2 r^{2q} \theta^2 + (2-r^q)^2 $$ and $$ |\nabla g|^2 = 1 + q^2 s^{2q} \phi^2 (2 - s^q)^{-4} + (2 - s^q)^{-2} $$ are uniformly bounded on their domains. So we have $W^{1,p}$ automatically. Now consider the two points $(r,0)$ and $(r,\pi)\in \bar{\Omega}\_1$, these correspond to $(r,0)$ and $(r, 2\pi - r^q\pi)\in \bar{\Omega}\_2$. In the domain their distance is $2r$. In $\bar{\Omega}\_2$ their distance is on the order of $r^{1+q}$. This shows that $g$ is not in $C^{0,\alpha}$ for any $\alpha > \frac{1}{1+q}$. Replacing $r^q$ by a function vanishing to infinite order at $0$ will get you then an example that is not in $C^{0,\alpha}$ for any $\alpha > 0$. (Note that this is firmly an issue related to the fact that $\Omega\_2$ is not Lipschitz, and then fact that we are measuring Hoelder-ness using the ambient metric on $\mathbb{R}^2$.)

1

https://mathoverflow.net/users/3948

453614

https://mathoverflow.net/questions/453609

1

Lemma 2.4 from Robert Griess' paper "Elementary abelian $p$-subgroups of algebraic groups" states: (i) Let $T$ be a finite $p$-group whose Frattini subgroup is cyclic and central. Then $T'$ has order $1$ or $p$ and there are subgroups $X, Y$ such that $T$ = $X\circ Y$ where $X$ is extraspecial and $Y$ has an abelian maximal subgroup and $\Omega\_1(Y)$ is elementary abelian. (ii) If $T/T'$ is elementary abelian, $Y$ is of the form $p^r$ or $\mathbb{Z}\_{p^2} \times p^r$. The proof for (i): Notice that $T'$ has order 1 or $p$. The abelian case is trivial, so we assume that $T'$ has order $p$. Let $U \ge T'$ satisfy $U/T'=\Omega\_1(T/T')$. Let $E$ be an extraspecial subgroup of $U$ such that $U=Z(U)E$ where $Z(U)$ denotes the center. Then, $[E,T]=T'=Z(E)$ implies that $T=C(E)E$ where $C(E)$ denotes the centralizer. Since $T/U$ is cyclic and $T'$ has order $p$, we see that $|C(E):C(E)\cap C(Z(U))|=1$ or $p$. Since the Frattini subgroup of $T$ is cyclic, the same is true for subgroups and quotients, whence $C(E)\cap C(Z(U))$ is central-by-cyclic, hence abelian. Take $X=E$ and $Y=C(E)$. Question: I don't really understand the following: Since $T/U$ is cyclic and $T'$ has order $p$, we see that $|C(E):C(E)\cap C(Z(U))|=1$ or $p$. Did he use some kind of order formula for centralizers or something? Any help would be appreciated.

https://mathoverflow.net/users/488802

$|C(E):C(E)\cap C(Z(U))|=1$ or $p$

One possible argument is to compare this with $T/C(Z(U))$ as follows. We have $$C(E)/C(E)\cap C(Z(U))\cong C(E)C(Z(U))/C(Z(U)) \leqslant T/C(Z(U))$$ Now $T/U$ is cyclic, so $T/C(Z(U))$ is cyclic. It acts faithfully on $Z(U)$ by conjugation, and if $x$ is an element of $T$ whose image is a generator of $T/C(Z(U))$ then $[x,Z(U)]\leqslant T'$ (since $T/T'$ is abelian), so $x^p$ centralises $Z(U)$ since $T'$ has order $p$. Thus $x^p\in C(Z(U))$, which implies that $T/C(Z(U))$ has order $1$ or $p$.

4

https://mathoverflow.net/users/460592

453627

https://mathoverflow.net/questions/453625

5

For $X\_i$, $i\in[n]$ be a sequence of integrable random variables. Is there a universal constant $c>0$ such that $$\mathbb{E}\max\_{i\in[n]}X\_i \le c\left( \max\_{i\in[n]}\mathbb{E}|X\_i| + \mathbb{E}\max\_{i\in[n]}|X\_i-X\_{i+1}| \right), $$ where $X\_{n+1}:=X\_1$? This would certainly be true, with $c=1$, if $ \mathbb{E}\max\_{i\in[n]}|X\_i-X\_{i+1}| $ were replaced by $ \mathbb{E}\max\_{i,j\in[n]}|X\_i-X\_j| $. What I am looking for is a *sparsified* version, with only $n$ cyclical comparisons (rather than $\approx n^2$).

https://mathoverflow.net/users/12518

Bounding expected maximum via adjacent differences

I don't think so. Define the $X\_i$'s as follows. Choose $j \in [n]$ uniformly at random, set $X\_j = n$, set $X\_{j+k} = n-|k|\sqrt{n}$ for $1 \le |k| \le \sqrt{n}$ (with indices taken mod $n$), and set $X\_i = 0$ for the remaining $i$. Then $\max\_{i \in [n]} X\_i = n$ almost surely, so $\mathbb{E}\max\_{i \in [n]} X\_i = n$. For any fixed $i \in [n]$, we have $\mathbb{E}|X\_i| \asymp \sqrt{n}$. And finally, $\max\_{i \in [n]} |X\_i-X\_{i+1}| = \sqrt{n}$ almost surely. In summary, the LHS is $n$ while the RHS is $c\left(c'\sqrt{n}+\sqrt{n}\right)$.

8

https://mathoverflow.net/users/129185

453630

https://mathoverflow.net/questions/453632

2

Consider the graph on $\mathbb{Q}\times\mathbb{Q}$ where two members of $\mathbb{Q}\times\mathbb{Q}$ form an edge if and only if their distance is $1$. Is that graph bipartite? If not, what is its chromatic number? (It is [known](https://en.wikipedia.org/wiki/Hadwiger%E2%80%93Nelson_problem) to be $\leq 7$.)

https://mathoverflow.net/users/8628

Is the Hadwiger-Nelson graph restricted to $\mathbb{Q}\times\mathbb{Q}$ bipartite?

Yes, the chromatic number is $2$. This result is due to Douglas R. Woodall, “Distances realized by sets covering the plane”, *J. Combinatorial Theory* **14** (1973), 187–200. See also Alexander Soifer, *The Mathematical Coloring Book (Mathematics of Coloring and the Colorful Life of Its Creators)*, Springer (2009), §11.2. See also my unpublished¹ note [“The Hadwiger-Nelson problem over certain fields”](https://arxiv.org/abs/1509.07023), esp. ¶1.6. 1. Because it turned out that (unbeknownst to me) most of the results contained therein were already obtained earlier in an (also unpublished) note by Eric Moorhouse, [“On the Chromatic Numbers of Planes”](https://ericmoorhouse.org/pub/chromatic.pdf), which is also relevant here.

3

https://mathoverflow.net/users/17064

453633

https://mathoverflow.net/questions/453641

1

Let $S$ be a closed real surface having two complex structures $c\_1$ and $c\_2$ which are not biholomorphic (so $S$ is a Riemann surface with genus at least 1). Consider $\omega$ a 1-form on $S$ which is holomorphic for both $c\_1$ and $c\_2$. Can we conclude that $\omega=0$ ? Intuitively I would proceed as follows. Since $\omega$ is holomorphic for $c\_1$, I would write $\omega= f(z) dz$ in conformal coordinates around a given point in $S$. Then, I would compare the Cauchy–Riemann equations for $c\_1$ and $c\_2$ : since $c\_1$ and $c\_2$ are not biholomorphic, the Cauchy–Riemann equations should take the form $\partial\_{\bar{z}} f = 0$ for $c\_1$ and some linear combination $\alpha \partial\_x f + \beta \partial\_y f=0$, with $\beta/\alpha \neq -i$ for $c\_2$. So the derivatives of $f$ vanish. So $\omega$ should be proportional to $dz$, in every choice of conformal coordinates for $c\_1$. But the same also holds for $c\_2$. So the only possibility should be $\omega=0$.

https://mathoverflow.net/users/481540

Common holomorphic forms for two distinct complex structures

Any $C^1$ function $g$ on a connected Riemann surface is holomorphic if and only if $dg\wedge \omega=0$, for $\omega$ holomorphic and not the zero form. Where $\omega\ne 0$ this is clear, expanding out in a holomorphic coordinate, say $\omega=f(z)\,dz$, and then integrating along curves to construct a local holomorphic coordinate $Z=\int f(z)\,dz$ so that $\omega=dZ$. Expand out to find that if $g=u+iv$ and $Z=x+iy$ then $$dg\wedge\omega=du\wedge dx-dv\wedge dy+i(du\wedge dy+dv\wedge dx),\\=((-u\_y-v\_x)+i(u\_x-v\_y))dx\wedge dy,$$ vanishing just when the Cauchy--Riemann equations are satisfied. Elsewhere it follows by continuity. Hence the complex structures are identical.

1

https://mathoverflow.net/users/13268

453642

https://mathoverflow.net/questions/453640

2

I'm reading the celebrated paper written by Congming Li and Wenxiong Chen, [Classification of solutions of some nonlinear elliptic equations](https://www.researchgate.net/profile/Wenxiong-Chen/publication/38333132_Classification_of_solutions_of_some_nonlinear_elliptic_equations/links/09e415126d86fa7bf2000000/Classification-of-solutions-of-some-nonlinear-elliptic-equations.pdf), which considered $$\Delta u = -e^u \ \ in \ \ \mathbb{R}^2.$$ After reading this I began to be interested in the higher dimension case so I found the paper [ON THE CLASSIFICATION OF SOLUTIONS OF $-\Delta u=e^u$ ON $\mathbb{R}^N$ : STABILITY OUTSIDE A COMPACT SET AND APPLICATIONS](https://www.ams.org/journals/proc/2009-137-04/S0002-9939-08-09772-4/S0002-9939-08-09772-4.pdf), it states that let $3 \leq N \leq 9$. Equation $$ -\Delta u=e^u \quad \text { on } \mathbb{R}^N, \quad N \geq 2, $$ does not admit any $C^2$ solution stable outside a compact set of $\mathbb{R}^N$, so **no finite Morse index solution**. I want to ask that: 1.If there are infinite Morse index solutions, what are these infinite Morse index solutions? If we consider the solutions as the critical point of the energy functional, what type of critical point are these infinite Morse index solutions? 2.If I want to learn the method of using Morse index to study elliptic PDE by directly reading papers, can you recommend me some appropriate ones?

https://mathoverflow.net/users/469129

What is the infinite Morse index solution?

NEW ANSWER: Yes there exist infinite Morse index solutions in all dimensions $N\geq 3$. For example you can take the solution in the Li Chen paper $$ \phi(x,y) = \frac{\ln(32)}{(4+|(x,y)|^2)^2} $$ and trivially cross with $\mathbb{R}$ to define $$ \phi(x,y,z) = \frac{\ln(32)}{(4+|(x,y)|^2)^2} $$ Clearly this continues to solve the PDE on $\mathbb{R}^3$. Moreover, by the second link (Dancer--Farina) it cannot have finite Morse index. This works for all $N\geq 3$. SOME COMMENTS: This example might feel cheap, but in many situations infinite Morse index comes from having a periodic or quasi periodic structure. For example, the [helicoid](https://en.wikipedia.org/wiki/Helicoid) has infinite Morse index as a critical point of the area functional. The reason for this is that the helicoid has a discrete translation symmetry. Thus, we see that either the Helicoid is stable or it has infinite Morse index. Proof: if the Helicoid is unstable, this means there is a compact piece that's unstable (stability is always considering compactly supported variations). Now, we can take this unstable piece and translate it sufficiently many times upwards to be disjoint from itself. This gives two $L^2$-orthogonal destabilizations. We can repeat this to get arbitrarily many. Thus the Morse index must be infinity. Of course it *could* a priori happen that a solution with symmetries is *stable* (the helicoid is not, but this is just due to the specifics of the problem). For example, the flat plane *is* a stable minimal surface and it of course has lots of symmetries. This is not to say that all infinite Morse index solutions have a discrete symmetry. In fact, one would expect that in general, infinite Morse index solutions can be very disordered. However, a lot of examples that we can cook up do have some symmetries (maybe this is more due to our limited knowledge than any deep truth). For an example of infinite index with only quasi-symmetries see e.g. the [genus 1 helicoid](https://minimal.sitehost.iu.edu/archive/Tori/Tori/GenusOneHelicoid/web/index.html). --- EDIT: As pointed out by Willie Wong, the original answer considers the wrong question. There are finite Morse index solutions for $N\geq 10$. See remark 1(i) in the second paper you reference. > > > > > > [...] for every N ≥ 10 the equation (1.1) possesses a radial stable solution. The > > existence of such a solution is a consequence of the analysis performed in [12], as > > was remarked in [6]. > > > > > > > > >

4

https://mathoverflow.net/users/1540

453648

https://mathoverflow.net/questions/445457

1

Consider a spacetime $(\zeta^{3,1},g)$ where $$g=\frac{du\,dv}{uv}-\frac{dr^2}{r^2}-\frac{dw^2}{w^2} \quad \forall u,v,r,w \in (0,1)$$ Now this is just Minkowski space in different coordinates (related to Dirac/Light cone coordinates/null coordinates). I'm looking to take a [Cauchy foliation](https://ncatlab.org/nlab/show/Cauchy+surface) of $\zeta^{3,1},$ change the metric back to $g'=dudv-dw^2-dr^2,$ and use the induced measure from $g'$ by means of the volume form to transform the foliated past light cone region onto a new spacetime $(\Psi^{~3,1},d).$ Let's look at the steps involved for the $(1+1)$ dimensional case $(\zeta^{1,1},u)$ where $u=\frac{dxdq}{xq}.$ The Cauchy foliation is simply $\ln(b)\ln(y)=t$ (solving for $y$ gives an explicit representation) which can then be transformed via Mellin-like transform: $$\Phi\_s(t)= \int\_{S=(0,1)} \exp {\frac{t s}{\ln b}}~db = \int\_0^1 tb^{t-1} \exp \frac{s}{\ln b}~db = 2\sqrt{ts}K\_1(2\sqrt{ts})$$ Observe that this is an unnormalized [K-distribution](https://en.wikipedia.org/wiki/K-distribution) and it's inverse transform yields an unnormalized distribution call it the "$\zeta$-distribution" (our Cauchy foliation). Observe that the [Fisher information metric](https://en.wikipedia.org/wiki/Fisher_information_metric) of the $\zeta$-distribution is $\Phi\_s(t)$ up to a factor. Observe that the $\zeta$-distribution yields solutions to the Killing field $X=\langle x \ln x, -y\ln y \rangle.$ It also provides a particular distributional solution to the following diffusion equation with diffusivity depending on both space and time: $$\frac{\partial^2}{\partial t^2}\nu(t,b)=-\frac{b}{t}\frac{\partial}{\partial b}\nu(t,b)$$ In short, $\Phi\_1(t)=v$ is a Lorentzian metric on the smooth $(1+0)$ manifold $(\Psi^{~1,0}, v).$ We have that $\Psi^{~3,1}=\Psi^{~1,1}\times \Psi^{~1,0} \times \Psi^{~1,0}$ so we can define the product metric on $\Psi^{~3,1}.$ There are two different constructions going on here. The first construction starts with a bonafide spacetime and transforms the past light cone region onto a new spacetime. The second construction begins by simplifying the spacetime down to $(1+1)$ dimensions, transforms the past light cone region, and then scales that manifold with the Cartesian product, furnishing with the product metric. > > Does restricting to $g'$ and transforming the foliated past light cone preserve all or most of the information contained in $(\zeta^{3,1},g)?$ Is $(\Psi^{~3,1},d)$ also a spacetime, or a component of one? > > >

https://mathoverflow.net/users/411249

Transporting a Cauchy foliation of Minkowski space

We are given a spacetime: $$(\zeta,g)$$ which we recognize as Minkowski space in different coordinates. We start by observing that the product space of the foliations: $$\Omega\_{t,r,\theta}(x,y,z)= \varphi\_t(x)\varphi\_r(y)\varphi\_{\theta}(z) \space\space\space\space\space{x,y,z\in(0,1)} \space\space\space{t,r,\theta >0}$$ Noticeably yields a product space $\Bbb R^{3,3}$ where $(t,r,\theta)$ are time variables and $(x,y,z)$ are space variables. This $6$-manifold is not entirely helpful, so we reduce to $(3+1)$ dimensions by letting $s=t=r=\theta.$ This collapses the time variables down to just one, which is what we want. Now we are left with a Cauchy foliation, $\Omega$, of a $(3+1)$ spacetime, in fact this spacetime is precisely the $\zeta$ we were initially given: $$ \Omega\_s(x,y,z)=\varphi\_s(x)\varphi\_s(y)\varphi\_s(z)\space\space\space\space{s>0}$$ And now we restrict the metric to $h=dudv-dr^2-dw^2$ and hit each leaf with the transform: $$ \int\_{(0,1)^3} \Omega\_s(x,z,y)~dxdydz=\bigg(\int\_{(0,1)} \Omega\_s(x)~dx\bigg)^3= \psi^3(s)$$ We have an explicit function to describe the leaves of the Cauchy foliation and we now have an explicit function for the hyper-volume. This transform maps from the spatial domain to the time domain. Whether that has any physical relevance, I do not know. We can generalize the integral transform to: $$\int\_{(0,1)^3} x^{a}y^{b}z^{c}\Omega\_s(x,z,y)~\frac{dxdydz}{xyz}=\chi\_s(a,b,c).$$ And so we have a mapping given by the integral transform: $$T: \Omega \to \chi $$ We have that $T$ is linear because it is an integral transform on bounded domain, which we know to be linear. And as $s$ varies continuously on $\Omega$ the corresponding change of $s$ is also continuous on $\chi$. The map is smooth as well, because a smooth change in $s$ on $\Omega$ corresponds to a smooth change of $s$ on $\chi$. Therefore the problem is reduced to finding a complete Lorentzian metric that is preserved by an infinitesimally generated flow on $\chi\_s(a,b,c).$ We are looking for a Killing field. Particularly of interest are the cases $\lim\_{s \to 0}\chi\_s$ and $\lim\_{s \to \infty} \chi\_s$ which serve to define boundaries on the resultant spacetime. The spacetime in question is therfore non-compact with boundary.

0

https://mathoverflow.net/users/411249

453650

https://mathoverflow.net/questions/453580

6

This question deals with a concrete exercise from [Geomerty of Schemes](https://www.google.de/url?sa=t&source=web&rct=j&opi=89978449&url=https://www.maths.ed.ac.uk/%7Ev1ranick/papers/eisenbudharris.pdf&ved=2ahUKEwjzsZi6rf-AAxVKxQIHHah3B6oQFnoECBUQAQ&usg=AOvVaw3usdrlkTi1DgXnq_xqASCS) by Eisenbud and Harris but also moreover the general philosophy attacking typical problems in algebraic geometry of following relative nature: say we have a "nice" enough map (having here sloppy said something "fibration like" in mind) $f:X \to Y$ of schemes over base field $K$, and assume $Y$ (wlog we can assume it to be affine) has some property $\mathcal{P}$ and say we have additional assumpion that there exist a fiber $X\_y$ which has also property $\mathcal{P}$. Natural question: in which situations one should "expect" that this property is carried/ "inherited" by neighbored fibers. Or even stronger, the property $\mathcal{P}$ inherited by $X$ *generically*? Let's consider the concrete problem from the book: **Exercise III-74.** Let $K$ be a field, and let $ B =\mathbb{A}^{12}\_K =\operatorname{Spec}K[a,b,c,d,e,f,g,h,i,j,k,l]$. Consider the two conic curves $\mathcal{C}\_i \subset \mathbb{P}^2\_B$ given by $$\mathcal{C}\_1 :=V(aX^2+bY^2+cZ^2+dXY +eXZ+fYZ) \subset \ \\\\\ \operatorname{Proj} (K[a,b,c,d,e,f,g,h,i,j,k,l][X,Y,Z]) = \mathbb{P}^2\_B $$ and similarly $\mathcal{C}\_2 :=V(gX^2+hY^2+iZ^2+jXY +kXZ+lYZ) \subset \ \mathbb{P}^2\_B $. Consider $\mathcal{C}\_1 \cap \mathcal{C}\_2 $. **(a):** Show that the intersection $\mathcal{C}\_1 \cap \mathcal{C}\_2 $ is *generically reduced* by showing that the canonical induced projection $$ p\_1:\mathcal{C}\_1 \cap \mathcal{C}\_2 \subset \mathbb{P}^2\_B \to B=\mathbb{A}^{12}\_K $$ has a fiber consisting of four distinct (hence reduced and K-rational) points. It's easy to see that there exist such fiber with respect this projection map $p\_1$ which has this property. (consider the maximal ideal $(a-x\_a, b-x\_b,..., l-x\_l) $ with $x\_n \in K$ choosen general enough. Essentially that's Bezout's theorem. So we can easily find such fibers, having obviously this property, expecially beeing reduced. **Question:** But how can we conclude from this that $ \mathcal{C}\_1 \cap \mathcal{C}\_2 $ is generically reduced, ie that there exist an open (wlog affine) subscheme $U =\operatorname{Spec} R \subset \mathcal{C}\_1 \cap \mathcal{C}\_2 $ with reduced $R$. **Metaquestion:** going back to the "motivation" at the beginning: Is there general "principle/ philosophy" behind such kind of argumentation techniques? For example, is there a "interesting" class of algebro geometric properties $\mathcal{P}$ known, for which such argumentation technique as in presented exercise go through, ie $f:X \to Y$ "fibration-like", $Y$ and some fiber $X\_y$ have property $\mathcal{P}$, then $X$ has it *generically* too?

https://mathoverflow.net/users/108274

Existence of a reduced fiber implies generically reduced (Exercise III-74 from Geometry of Schemes)

As noted by Jason Starr, the *Generic Principle* has been worked out in details by Grothendieck in EGA IV. If your French is not on top these days, the following version looks quite appealing (and can be found as Theorem 23.9 and 24.4 in Matsumura's *Commutative Ring Theory*) : Let $f : X \longrightarrow Y$ be a **flat** (this is the expected fibration-like hypothesis), finite type morphism of Noetherian schemes with $X$ irreducible and let $y \in Y$. $\bullet$ if $Y$ and $f^{-1}(y)$ satisfy the condition $R\_k$, then there is a dense open subset $X' \subset X$ which satisfies the condition $R\_k$. $\bullet$ if $Y$ and $f^{-1}(y)$ satisfy the condition $S\_k$, then there is a dense open subset $X' \subset X$ which satisfies the condition $S\_k$. Note that reduced is $R\_0$ + $S\_1$ (see for instane [Lemma 10.157.3 in *The Stack Project*](https://stacks.math.columbia.edu/tag/031R)), so that your expectation in the Metaquestion is correct.

3

https://mathoverflow.net/users/37214

453652

https://mathoverflow.net/questions/453655

3

In Montgomery's "[Ten Lectures on the Interface Between Number Theory and Harmonic Analysis](https://doi.org/10.1090/cbms/084)" a bound for the Fourier coefficients of the Selberg polynomial $S^+\_K$ is obtained by using what he calls Vaaler's lemma. I am interested in obtaining a similar bound directly without using Vaaler's lemma (Montgomery states "We now require an estimate for $|\hat{S}\_K^+(k)|$. Since $S\_K^+(x)$ has been explicitly defined, one might argue directly, but ..." so it seems possible.) The reason being I want to bound the Fourier coefficients of something similar to Selberg polynomial (defined below) where I can not apply Vaaler's lemma as in the referenced book. So I wanted to understand how I can do this for Selberg polynomial first. However, I am not getting anything close to the bound $$ |\hat{S}\_K^+(k)| \leq \frac{1}{K+1} + \min (\beta - \alpha, \frac{1}{\pi |k|}) $$ so far. I apologize this is a technical question, but if someone could get me started on this I would greatly appreciate it. Definitions: Let $K$ be a positive integer. Vaaler's polynomial is defined $$ V\_K(x) = \frac{1}{K+1}\sum\_{k=1}^K\left(\frac{k}{K+1} - \frac12\right) \Delta\_{K+1}\left(x - \frac{k}{K+1}\right) + \frac{1}{2 \pi (K+1)}\sin 2 \pi (K+1) x - \frac{1}{2 \pi} \Delta\_{K+1}(x) \sin 2 \pi x, $$ where $$ \Delta\_K(x) = \sum\_{-K}^K \left(1 - \frac{|k|}{K}\right)e(kx) $$ is the Fejér kernel. Then the Selberg polynomial for the interval $[\alpha, \beta]$ is defined $$ S^+(x) = \beta - \alpha + B\_K(x - \beta) + B\_K( \alpha - x) $$ where $$ B\_K (x) = V\_K(x) + \frac{1}{2 (K+1)} \Delta\_{K+1}(x). $$ PS my previous related question [Showing Vaaler polynomial is a good approximation to saw tooth function](https://mathoverflow.net/questions/453561/showing-vaaler-polynomial-is-a-good-approximation-to-saw-tooth-function?noredirect=1#comment1173453_453561) was an attempt to do this...

https://mathoverflow.net/users/84272

Fourier coefficients of Selberg polynomials

As it is odd, we can write the Vaaler polynomial $V\_K(x)=\sum\_{1 \le k \le K}c\_{k,K} \sin 2\pi kx$ and its fundamental property is that $|c\_{k,K}| \le \frac{1}{\pi k}$. This follows easily from its definition in [Vaaler's paper](https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society-new-series/volume-12/issue-2/Some-extremal-functions-in-Fourier-analysis/bams/1183552525.full) referenced before (there it is called $\psi\*j\_N$, where $j\_N$ is defined on page $207$ and its properties are spelled out in Theorem $18$ on page $210$ (odd follows from $7.13$ in the paper as it has the sign of $\psi$ the sawtooth function and the bound because $|\hat J\_{N+1}(n)|= |\hat J(\frac{n}{N+1})| \le 1$ and $\psi\*j\_N(x)=-\sum\_{n=-N, n \ne 0}^N(-2\pi i n)^{-1}\hat J\_{N+1}(n)e(nx)$, where $J, \hat J$, defined on page $191$ are analyzed in Theorem $6$ page $192$ ) Then $V\_K(x - \beta) + V\_K( \alpha - x)=\sum\_{1 \le k \le K}c\_{k,K} (\sin 2\pi k(x-\beta)+\sin 2\pi k(\alpha-x))$, so $ V\_K(x - \beta) + V\_K( \alpha - x)=\sum\_{1 \le k \le K}(2c\_{k,K}\sin \pi k(\alpha-\beta))\cos (\pi k (2x-\alpha-\beta))$ and clearly the latter written as Fourier series (so the cosine term is expanded as the half sum of two exponentials) has coefficients at most $|c\_{k,K}\sin \pi k(\alpha-\beta))| \le \min (|c\_{k,K}|, |c\_{k,K}(\pi k (\beta-\alpha)| $ in absolute value and we have that being at most $\min (\beta - \alpha, \frac{1}{\pi |k|})$ by the above The other part in the $B\_K$ is the sum of two terms, each being $ \frac{1}{2 (K+1)}$ times a Fourier series with coefficients at most $1$ in absolute value, hence the $\frac{1}{K+1}$ term. If $k \ne 0$ we are done by the above as we indeed get $|\hat{S}\_K^+(k)| \leq \frac{1}{K+1} + \min (\beta - \alpha, \frac{1}{\pi |k|})$, while if $k=0$ the Vaaler polynomial term disappears and the inequality is trivial and is an equality since the $0$ th terms of $\Delta\_{K+1}(x-\beta), \Delta\_{K+1}(\alpha-x)$ are both $1$.

3

https://mathoverflow.net/users/133811

453658

https://mathoverflow.net/questions/453659

2

Let $S$ be a symmetric $(0, 2)$ tensor on a Riemannian manifold $M$. Define $E\_S : M \to \mathbb{Z}$ by $E\_S(x) = \left(\text{the number of distinct eigenvalues of } S\_x\right)$. I've seen the following claims in several papers (also see 16.10 in Besse's Einstein Manifolds): > > * $M\_S \doteq \left\{x \in M \ \vert \ E\_S \text{ is constant in a neighbourhood of } x \right\}$ is an open and dense subset of $M$ > * The eigenvalues of $S$ are distinct and smooth in each connected component $U$ of $M\_S$. > > > I'm a having a hard time proving these facts in a rigorous enough manner (openness is clear, but I'm having a hard time with denseness). It "feels" true since in some sense the eigenvalues should be smooth, but I can't see how to formalize this precisely (smooth from where to where? how to prove smoothness?). I'd appreciate any help.

https://mathoverflow.net/users/119418

Why is this subset associated to a $2$-tensor dense?

I claim that the function $E\_S$ is lower semi-continuous, meaning that for any sequence $x\_k \in M$ converging to some $x \in M$, $\lim\_{k \to \infty} E\_S(x\_k) \geq E\_S(x)$. We want to show that the complement of $M\_S$ has empty interior. Points $x$ in this complement are characterized by the fact that there exists a sequence of points $(x\_k)\_k$ converging to $x$ which is not eventually constant. Assume by contradiction that there exists an open subset $U$ contained in $\overline{M\_S}$. Let $n = \max\_{y \in U} E\_S(y)$ (this max is finite because $E\_S(y) \leq \dim(M)$. As $E\_S$ is lower semi-continuous, the set $V = E\_S^{-1}(n) \cap U = E\_S^{-1}(n-1, \infty) \cap U$ is open and non-empty. But all points in $V$ belong to $M\_S$ as the number of distinct eigenvalues of $S$ is constant on $V$ and $V$ is open. This contradicts the fact that $\overline{M\_S}$ has non-empty interior. Let us now return to the lower semi-continuity of $E\_S$. I guess this is more or less standard stuff. A nice way to see it is via Rouché's theorem. Let me work in a coordinate chart. Assume given a point $x\_0 \in M$ and let $n\_0 = E\_S(x\_0)$. Let $\lambda\_1, \ldots, \lambda\_{n\_0}$ be the distinct eigenvalues of $S$ at $x\_0$. You can consider small disjoint intervals $I\_k$ centered at the eigenvalues and view them as the restriction to $\mathbb{R}$ of balls $B\_1, \ldots B\_k$ in the complex plane. If $x$ is close enough to $x\_0$, the characteristic polynomial $\chi\_{S(x)}$ of $S$ at $x$ will be close (in the sup-norm over some large compact set encompassing all the balls) to $\chi\_{S(x\_0)}$. Hence, the number of eigenvalues in each ball $B\_i$ fox $S(x)$ (counted with multiplicity) will be the same as for $S(x\_0)$. This means that locally, distinct eigenvalues cannot merge, i.e. $E\_S(x) \geq E\_S(x\_0)$ for $x$ close enough to $x\_0$. Rouché's theorem is a very nice tool to prove continuity of the eigenvalues even assuming multiplicity. What you also see is that multiplicity of each of the eigenvalues are locally constant on $M\_S$. To prove differentiability of an eigenvalue $\lambda = \lambda(x)$ with multiplicity $k \geq 1$, the easiest way is to remark that the derivatives of the characteristic polynomial satisfy $\chi\_{S(x)}^{(k-1)}(\lambda) = 0$ while $\chi\_{S(x)}^{(k)}(\lambda) \neq 0$. So you can just apply the implicit function theorem to $(x, \lambda) \mapsto \chi\_{S(x)}^{(k-1)}(\lambda)$. Let me know if you need a more detailed answer.

4

https://mathoverflow.net/users/24271

453663

https://mathoverflow.net/questions/453649

9

Let $x, y$ be finite words over totally ordered alphabet and $<$ denote the lexicographical order, i.e for two not necessarily finite words we say $x < y$ iff one of the following holds 1. There are words $u, x^\prime, y^\prime$ and letters $a < b$ such that $x = uax^\prime, y = uby^\prime$ 2. $y = xu$ for some non-empty word $u$ Then the following are equivalent 1. $xy < yx$ 2. $x^\mathbb{N} < y^\mathbb{N}$, where $x^\mathbb{N}$ denotes infinite word $xxxx\ldots$ --- $x <^\* y$ iff $xy < yx$ is a total order(which we're trying to prove by equivalence) This new total order is another extension of "weak lexicographical order" where only first condition considered, i.e $x$ and $xu$ are incomparable for non-empty $u$(note that $xy$ and $yx$ always have the same length so we're not using "strong" order). It is naturally arising from the following problem: > > Let $x\_1, \ldots, x\_n$ be finite words on totally ordered alphabet, find $\sigma \in S\_n$ minimizing $x\_{\sigma(1)}x\_{\sigma(2)}\ldots x\_{\sigma(n)}$ > > > The solution is to sort $x\_i$ with respect to order $<^\*$. If the sequence is not sorted according to this order, then there are adjacent elements $x, y$ such that $xy > yx$, by swapping them we'll decrease word which we are minimizing. But it might be tedious to prove that $<^\*$ is transitive, so instead I suggest equivalent definition of $<^\*$, that is $x <^\* y$ iff $x^\mathbb{N} < y^\mathbb{N}$. Now transitivity is obvious. But is there elegant and simple proof of equivalence not involving much of casework on $|x| <> |y|$, etc?

https://mathoverflow.net/users/509688

Elegant proof for $xy < yx \Leftrightarrow x^\mathbb{N} < y^\mathbb{N}$

Let $x, y \in \Sigma^+$. Observe that for all $n \in \mathbb{N}$ the inequality $xy<yx$ implies that $$x^ny^n<y^nx^n$$ by repeatedly swapping pairs of $x$ and $y$. It follows that $x^\mathbb{N}\leq y^\mathbb{N}$ (i.e. the two infinite sequences are equal, or $y^\mathbb{N}$ is greater at the first position they differ). Note that equality can only hold if the basic period of the infinite word divides both $|x|$ and $|y|$, in which case we would have $xy=yx$. Conversely, $xy \geq yx$ implies $x^\mathbb{N} \geq y^\mathbb{N}$.

5

https://mathoverflow.net/users/502833

453676

https://mathoverflow.net/questions/428634

28

Is there an abelian group $A$ with $A\not\cong A\oplus A\cong A\oplus A\oplus A\oplus\cdots$ (a direct sum of countably many copies of $A$)? --- Edited to add: As no answers are forthcoming, does anyone know what happens if we allow arbitrary modules in place of abelian groups?

https://mathoverflow.net/users/3199

$A^2$ is isomorphic to $A^{(\omega)}$, but not $A$

This is not a complete answer, but a construction that might give an answer. I'll start by constructing a ring with several objects (a.k.a. preadditive category) $\mathcal{C}$ by generators and relations, using similar ideas as in Leavitt's ring with $R\not\cong R\oplus R\cong R\oplus R\oplus R$. The objects $X\_{1}, X\_{2},\dots$ are indexed by the positive integers. For each $m\leq n$ there are generators $\alpha\_{n,m},\beta\_{n,m}: X\_{n}\to X\_{m}$ and $\gamma\_{n,m},\delta\_{n,m}:X\_{m}\to X\_{n}$. The relations are designed to make $$X\_{n}\oplus X\_{n}\cong X\_{1}\oplus X\_{2}\oplus\cdots\oplus X\_{n}$$ for all $n$ once we close under finite direct sums. For example, for $n=3$ impose the relations that make the matrices $$ \begin{pmatrix} \alpha\_{3,1}&\beta\_{3,1}\\\alpha\_{3,2}&\beta\_{3,2}\\\alpha\_{3,3}&\beta\_{3,3} \end{pmatrix} \text{ and } \begin{pmatrix} \gamma\_{3,1}&\gamma\_{3,2}&\gamma\_{3,3}\\\delta\_{3,1}&\delta\_{3,2}&\delta\_{3,3} \end{pmatrix} $$ mutually inverse. Now form a ring by adjoining a unit: $$R=\mathbb{Z}\oplus\bigoplus\_{m,n}\operatorname{Hom}\_{\mathcal{C}}(X\_{m},X\_{n}).$$ For each $n$ let $e\_{n}\in R$ be the idempotent corresponding to the identity endomorphism of $X\_{n}$, let $P\_{n}=e\_{n}R$ be the corresponding projective right $R$-module, and let $P=\bigoplus\_{n}P\_{n}$. Then $P\_{n}\oplus P\_{n}\cong P\_{1}\oplus P\_{2}\oplus\cdots\oplus P\_{n}$ for each $n$, so \begin{align} P\oplus P&\cong (P\_{1}\oplus P\_{1})\oplus(P\_{2}\oplus P\_{2})\oplus(P\_{3}\oplus P\_{3})\oplus\cdots\\ &\cong P\_{1}\oplus (P\_{1}\oplus P\_{2})\oplus(P\_{1}\oplus P\_{2}\oplus P\_{3})\oplus\cdots\\ &\cong P^{(\omega)} \end{align} **Question 1.** Is $P\cong P\oplus P$? Probably not, as there seems no obvious way to produce such an isomorphism. If the answer to Question 1 is ``no'', then this answers the supplementary question in the OP about modules. As for abelian groups: Clearly $R$ is countable. **Question 2.** Is $R$ torsion free? Probably, as there seems no obvious way to produce a torsion element. **Question 3.** Is $R$ reduced? Probably, as there seems no obvious way to produce a divisible element. If the answers to Questions 2 and 3 are both ``yes'', then Corner's theorem applies, and $R\cong\operatorname{End}(B)$ for some abelian group $B$. Let $A\_{n}=e\_{n}(B)$ and $A=\bigoplus\_{n}A\_{n}$. Then $$A\oplus A\cong A^{(\omega)}.$$ **Question 4.** Is $A\cong A\oplus A$? I don't see any reason that it should have to be (note that $B$ is not uniquely determined, so the answer to Question 4 might conceivably depend on the particular choice of $B$).

4

https://mathoverflow.net/users/22989

453685

https://mathoverflow.net/questions/453692

4

$\DeclareMathOperator\GL{GL}$Consider the unimodular group $\GL\_n(\mathbb{Z})$, consisting of integral matrices $A \in \mathbb{Z}^{n \times n}$ such that that $\det(A) =\pm 1$. It is well known that any $A \in \GL\_n(\mathbb{Z})$ can be written as a product of signed permutation matrices and 'Gauss moves', the latter referring to matrices that have 1 on the diagonal and at most one non-zero entry away from the diagonal. This is proven, for example, in M. Newman's book 'Integral Matrices' (Theorem II.7). My question is: is there an upper bound (depending on $n$, but independent of $A$) on the number of moves required? If not, can you provide a sequence of matrices that require an increasing number of factors in the above decomposition? Thank you in advance for your help.

https://mathoverflow.net/users/106337

Diameter of the unimodular group with Gauss moves

They’re usually called “elementary matrices”, not Gauss moves. Your question is equivalent to asking whether the integer special linear group is boundedly generated by elementary matrices. The answer is no for $n=2$ (an easy exercise using the free product with amalgamation description of the group in that case), but amazingly it is “yes” for larger $n$. This is a theorem of Carter, Keller, and Paige. See [here](https://arxiv.org/abs/math/0503083).

10

https://mathoverflow.net/users/317

453693

https://mathoverflow.net/questions/453678

1

The following question appears in [MSE](https://math.stackexchange.com/questions/4760401/if-langle-f-1-f-2-rangle-langle-g-1-g-2-rangle-then-langle-f-1-lambd) without answers. Let $f\_1,f\_2,g\_1,g\_2 \in \mathbb{C}[x,y]-\mathbb{C}$. Assume that $\langle f\_1,f\_2 \rangle = \langle g\_1,g\_2 \rangle \subsetneq \mathbb{C}[x,y]$, where $\langle u,v \rangle$ denotes the ideal in $\mathbb{C}[x,y]$ generated by $u$ and $v$. **Claim:** Given $\lambda,\mu \in \mathbb{C}$, $(\lambda,\mu) \neq (0,0)$ (but one of $\{\lambda,\mu\}$ can be zero), there exist $\delta,\epsilon \in \mathbb{C}$ such that $\langle f\_1-\lambda,f\_2-\mu \rangle = \langle g\_1-\delta,g\_2-\epsilon \rangle$. Notice that in the claim, necessarily $(\delta,\epsilon) \neq (0,0)$; otherwise, $\delta=\epsilon=0$, so $I:= \langle f\_1-\lambda,f\_2-\mu \rangle = \langle g\_1,g\_2\rangle= \langle f\_1,f\_2 \rangle$ and then $I=\mathbb{C}[x,y]$ since w.l.o.g. $\lambda \neq 0$, hence $\lambda=f\_1-(f\_1-\lambda) \in I$. > > **Question 1:** > Is the claim true when $\langle f\_1,f\_2 \rangle = \langle g\_1,g\_2 \rangle$ is a maximal ideal? > > > > > **Question 2:** > Is the claim true when $\langle f\_1,f\_2 \rangle = \langle g\_1,g\_2 \rangle$ is a finite intersection of maximal ideals? > > > > > **Question 3:** > Are the following two sets equal? > $\{ \langle f\_1-\lambda,f\_2-\mu \rangle \}\_{ \lambda,\mu \in \mathbb{C}}$ and $\{ \langle g\_1-\delta,g\_2-\epsilon \rangle \}\_{ \delta,\epsilon \in \mathbb{C}}$. > > > **Examples:** **(i)** $f\_1=x+y,f\_2=x-y,g\_1=x,g\_2=y$, namely, $\langle x+y,x-y \rangle = \langle x,y \rangle$. If we choose $\lambda=1,\mu=2$, then we can take $\delta=\frac{3}{2}, \epsilon=-\frac{1}{2}$ and get $\langle x+y-1,x-y-2 \rangle = \langle x-\frac{3}{2},y+\frac{1}{2} \rangle$. * This example shows that it is generally *not* true that $\langle f\_1-\lambda,f\_2-\mu \rangle = \langle g\_1-\lambda,g\_2-\mu \rangle$. * At least for linear $f\_1,f\_2,g\_1,g\_2$, there exist $\delta,\epsilon$ such that $\langle f\_1-\lambda,f\_2-\mu \rangle = \langle g\_1-\delta,g\_2-\epsilon \rangle$. Linear means of the form $ax+by+c$, for some $a,b,c \in \mathbb{C}$. **(ii)** $f\_1=x,f\_2=y+x^2+5x+3,g\_1=x,g\_2=y+3$, namely, $\langle x,y+x^2+5x+3 \rangle = \langle x,y+3 \rangle$. If we choose $\lambda=0,\mu=3$, then we can take $\delta=0,\epsilon=3$ and get $\langle x,y+x^2+5x \rangle = \langle x,y \rangle$. * In this particular example $(\lambda,\mu)=(\delta,\epsilon)=(0,3)$. * It seems that at least for triangular/linear $f\_1,f\_2,g\_1,g\_2$, there exist $\delta,\epsilon$ such that $\langle f\_1-\lambda,f\_2-\mu \rangle = \langle g\_1-\delta,g\_2-\epsilon \rangle$. Triangular means of the form $ax+F(y)$ or $by+G(x)$, for some $a,b \in \mathbb{C}-\{0\}$, $F(y) \in \mathbb{C}[y]$, $G(x) \in \mathbb{C}[x]$. **Remarks:** **(i)** Every finite intersection of maximal ideals in $\mathbb{C}[x,y]$ is (a radical ideal) that can be generated by two elements, see [this](https://math.stackexchange.com/questions/4399584/generators-of-a-finite-intersection-of-maximal-ideals?noredirect=1&lq=1). **(ii)** For the general case where $\langle f\_1,f\_2 \rangle = \langle g\_1,g\_2 \rangle \subsetneq \mathbb{C}[x,y]$ is not even radical, perhaps it is interesting to consider $f\_1=x+y,f\_2=xy,g\_1=x+y,g\_2=x^2+y^2$. $\langle x+y,xy \rangle \subsetneq \langle x,y \rangle$ and it is not a radical ideal, since $x^2 \in \langle x+y,xy \rangle$, but $x \notin \langle x+y,xy \rangle$. (Indeed, $x^2=(x+y)x-xy$). Thank you very much!

https://mathoverflow.net/users/72288

Ideals: If $\langle f_1,f_2 \rangle = \langle g_1,g_2 \rangle$, then $\langle f_1-\lambda,f_2-\mu \rangle = \langle g_1-\delta,g_2-\epsilon \rangle$?

The answer to your questions is no. The ideals $\langle x, y(1-xy) \rangle$ and $\langle x, y \rangle$ are equal, and maximal; but $$ \langle x-\lambda, y(1-xy)\rangle \neq \langle x-\delta, y-\epsilon \rangle$$ for any $\lambda,\delta,\epsilon \in \mathbb{C}$, $\lambda \neq 0$. (In the notation of your questions, $\mu=0$.) A related example is $\langle x, 1-xy\rangle = \langle 1 \rangle$, but $\langle x-\lambda, 1-xy \rangle \neq \langle 1 \rangle$ for nonzero $\lambda$.

5

https://mathoverflow.net/users/88133

453703

https://mathoverflow.net/questions/453701

1

I have encountered the polynomial equation $$x^{n+1} = (1 - x)^n ( n + x )$$ where $n \geq 0$, and I am interested in its real roots. The number $n$ can be an integer or, more generally, any positive real number. The main reason I am posting this very specific question here is that this could be one of those classical equations in some subarea of analysis or algebra, so perhaps somebody recognizes it by name. Of course, somebody might happen to have a good hint. It seems that this polynomial has only one single positive root.

https://mathoverflow.net/users/2082

What can be said about the roots of the polynomial $x^{n+1} - (1 - x)^n ( n + x )$?

The positive root is indeed unique and single for any real $n>0$. Indeed, 1. for $x>1$, we have $$|(1-x)^n(n+x)|=(x-1)^n(n+x)<\left(\frac{n\cdot (x-1)+(n+x)}{n+1}\right)^{n+1}=x^{n+1}$$ by the (weighted) Arithemtic-Geometric Means inequality. 2. for $0\leqslant x\leqslant 1$ there exists a root in $(0,1)$ since the function $f(x):=(1-x)^n(n+x)-x^{n+1}$ is positive at $x=0$ and negative at $x=1$. To prove that it is unique and simple, note that $$ (1-x)^{-n}x^{-1}f(x)=\frac n{x}+1-\left(\frac{x}{1-x}\right)^n $$ is a strictly decreasing (and with strictly negative derivative) function on $(0,1)$.

8

https://mathoverflow.net/users/4312

453705

https://mathoverflow.net/questions/453710

2

Let $X$ be an infinite-dimensional Banach space and $C\subseteq X$ be a bounded closed convex subset. Let $\{z\_i\}\_{i\in\mathbb{N}}$ be a sequence of linearly independent points in $C$ and for each $n\in\mathbb{N}$, define $V\_{\leq n} = \operatorname{conv}\{z\_i\}\_{i\leq n}$ and $V\_{\geq n} = \operatorname{conv}\{z\_i\}\_{i\geq n}$. For each $z\in V\_{\geq 1}$, let $N(z)$ denote the least positive integer such that $z\notin V\_{\geq N(z)}$. For each $n\in\mathbb{N}$, define: $$ d\_n: C\rightarrow [0, \infty), \hspace{0.3cm} c\mapsto \inf\big\{ \|c-z\|: z\in V\_{\geq n} \big\} $$ and clearly each $d\_n$ is continuous. **My question**: if, for a fixed $m\in\mathbb{N}$, there exists $\lambda>0$ and $M\in\mathbb{N}$ such that $d\_M(z\_i) > \lambda$ for each $i\leq m$, is it true that for each $z\in V\_{\leq m}$, there exists $n\geq \max(N(z), M)$ such that $d\_n(z) > \lambda$? What inspires this question is that, given $m\in\mathbb{N}$ and $\lambda>0$ such that $d\_M(z\_i)>\lambda$ for each $i\leq m$, I need to find $M'\in\mathbb{N}$ such that: $$ \inf\big\{ \|z-z'\|: z\in V\_{\leq m}, z'\in V\_{\geq M'}\big\} > 0 $$ Notice that each $d\_n$ is continuous. If my guess is true, then by the fact that $V\_{\leq m}$ is compact the inequality above will be true. Any hints, as well as other methods to prove the inequality above without answering my question, will be appreciated.

https://mathoverflow.net/users/151332

Distance between convex hulls in a bounded closed convex set

$\newcommand\la\lambda$The answer is no. E.g., suppose that $X=\ell^\infty$, $z\_1=e\_1$, $z\_2=-e\_1+e\_2$, and $z\_k=e\_2/2+e\_k/k$ for $k\ge3$, where $(e\_1,e\_2,\dots)$ is the standard basis of $\ell^\infty$. Let $C$ be the unit ball in $\ell^\infty$. Then $C$ is a bounded closed convex subset of $X$ and the $z\_k$'s are linearly independent points in $C$. Let $m=2$, $M=3$, $\la=1/2$, and $z:=(z\_1+z\_2)/2=e\_2/2$. Then $d\_M(z\_i)=1>\la$ for each $i\le m$. However, for each $n\ge M$ we have $d\_n(z)=0\not>\la$. $\quad\Box$

5

https://mathoverflow.net/users/36721

453724

https://mathoverflow.net/questions/445100

4

Given a set $A \subset \mathbb{R}^n$, this is called d-rectifiable if it can be covered by a countable union of images of lipshitz functions from $\mathbb{R}^d $ to $ \mathbb{R}^n $ and a $\mathcal{H}^d$ null set. So given $A\_j = f\_j (\mathbb{R}^d)$ with $f\_j$ Lipshitz, one has that A is d-rectifiable if $$ A \subset \cup\_{j=1} ^\infty A\_j \cup A\_0 \text{ where } \mathcal{H}^d (A\_0)=0 .$$ Now, consider $d=1$ and $n=2$ for simplicity. If I consider a countable number of "bad points", I wouldn't really need that $A\_0$ in the definition. The only relevant case is when the cardinality of $A\_0$ is more than countable. Now the question is the following. I know there is the structure theorem by Federer-Fleming which tells you that you can decompose any $\mathcal{H}^d $- measurable set of finite measure in a rectifiable part and in a purely unrectifiable part. Let's call this set $A$, the rectifiable part $A^{rect}$ and the unrectifiable part $A^{unr}$. Given the definition above, the only relevant case is when $\mathcal{H}^d (A^{unr}) >0$, otherwise one could consider it to be the null set in the definition, making $A$ rectifiable. The question is: why did we choose to define rectifiability that way and not without $A\_0$? I mean, one could take $A^{rect}$ to be the union of the $A\_j$s above, while discarding the $A\_0$ in the unrectifiable part. It somehow seems to even make more sense, you could partition a set $A$ in a good part which is ALL contained in unions of images of Lipshitz functions, and in a bad part which contains all the rest. Is this just a convention or there's some good reason to have this definition? I guess it has to do with the fact of wanting to have some "compactness" in the class of rectifiable sets, meaning that a perturbation by a null set don't make you go out of the class, but maybe there's a deeper reason.

https://mathoverflow.net/users/109382

Why is there a $\mathcal{H}^d$-null set in the definition of d-rectifiable set?

If $f:\mathbb{R}^n\to\mathbb{R}^m$, $m<n$ is Lipschitz continuous, then for almost all $x\in\mathbb{R}^m$, the set $f^{-1}(x)$ is ($n-m$)-rectifiable according to the definition that includes the null set $A\_0$. (Such sets are often called countably $(n-m)$-rectifiable). This is a nice generalization of the Sard theorem for Lipschitz mappigns. This is one of the main reasons why the definition is stated the way it is stated. When you remove the set $A\_0$ from the definition you will have much less natural and generic examples of rectifiable sets.

4

https://mathoverflow.net/users/121665

453735

https://mathoverflow.net/questions/453688

7

For an abelian group $A$, put $DA=\text{Hom}(A,\mathbb{Z})$ and $D\_{(p)}A=\text{Hom}(A,\mathbb{Z}\_{(p)})$. It is a theorem of Specker that when $A$ is free abelian of countable rank, the natural map $A\to D^2(A)$ is an isomorphism. The proof uses the ordering of $\mathbb{Z}$ in an essential way: at a key step we have integers $n,m$ with $|n|<m$ and $n=0\pmod{m}$ and we conclude that $n=0$. Is it also true that when $A$ is a free module of countable rank over $\mathbb{Z}\_{(p)}$, the natural map $A\to D\_{(p)}^2(A)$ is an isomorphism? As the order-theoretic properties of $\mathbb{Z}\_{(p)}$ are worse than those of $\mathbb{Z}$, it does not seem possible to adapt Specker's argument in any obvious way.

https://mathoverflow.net/users/10366

Double dual of free $\mathbb{Z}_{(p)}$-modules

There is at least one proof of Specker's theorem that can be adapted in an obvious way. I believe that the first half of this proof is due to Sąsiada, and the second half to Łoś. Let $A$ be a free $\mathbb{Z}\_{(p)}$ module of countable rank, and $B=D\_{(p)}A$ its dual, whose elements I will think of as sequences of elements of $\mathbb{Z}\_{(p)}$. Let $e\_{n}\in B$ be the sequence with $n$th term equal to $1$ and all other terms zero. It is easy to see that to prove that the natural map $A\to D^{2}\_{(p)}A$ is an isomorphism, it is enough to prove the following two facts about elements $\varphi:B\to\mathbb{Z}\_{(p)}$ of $D\_{(p)}B$. **Proposition 1.** $\varphi(e\_{n})=0$ for all but finitely many $n$. *Proof.* Suppose not. By ignoring those $n$ for which $\varphi(e\_{n})=0$, we can assume, for simplicity, that $\varphi(e\_{n})\neq0$ for all $n$. Construct a sequence $(x\_{n})\_{n\in\mathbb{N}}\in B$ by inductively choosing $x\_{n}$ to be nonzero but divisible by a higher power of $p$ than $\varphi(x\_{n-1}e\_{n-1})$. Now consider all sequences whose $n$th term is either $x\_{n}$ or $0$. There are uncountably many of these, but $\mathbb{Z}\_{(p)}$ is countable, so two of these sequences must be sent to the same place by $\varphi$. Taking the difference, we get a nonzero sequence $(y\_{n})\_{n\in\mathbb{N}}\in\ker(\varphi)$ with $y\_{n}$ equal to $0$ or to $\pm x\_{n}$. Choosing $m$ minimal such that $y\_{m}\neq0$ we have $$0=\varphi((y\_{n})\_{n\in\mathbb{N}})= \varphi(y\_{m}e\_{m})+\varphi((y\_{n})\_{n\in\mathbb{N}}-y\_{m}e\_{m}),$$ but $(y\_{n})\_{n\in\mathbb{N}}-y\_{m}e\_{m}$, and hence $\varphi((y\_{n})\_{n\in\mathbb{N}}-y\_{m}e\_{m})$, is divisible by a higher power of $p$ than $\varphi(y\_{m}e\_{m})$. $\;\blacksquare$ **Proposition 2.** $\varphi$ is determined by the values of $\varphi(e\_{n})$. *Proof.* If not, there is some nonzero $\varphi\in D\_{(p)}B$ with $\varphi(e\_{n})=0$ for all $n$. Choose a sequence $(a\_{n})\_{n\in\mathbb{N}}\not\in\ker(\varphi)$, and define a homomorphism $\theta:B\to B$ by $$\theta((b\_{n})\_{n\in\mathbb{N}})=((b\_{0}+b\_{1}+\cdots+b\_{n})a\_{n})\_{n\in\mathbb{N}}.$$ For each $m$, $\theta(e\_{m})$ differs from $(a\_{n})\_{n\in\mathbb{N}}$ in only finitely many terms, so $\varphi\theta(e\_{m})=\varphi((a\_{n})\_{n\in\mathbb{N}})\neq0$ for all $n$, contradicting Proposition 1 for $\varphi\theta$. $\;\blacksquare$

9

https://mathoverflow.net/users/22989

453753

https://mathoverflow.net/questions/453764

4

Let $\mathcal M$ and $\mathcal N$ be two von Neumann algebras. If $x$ is a positive element of $\mathcal M$ and $y$ is a positive element of $\mathcal N$, it is known that $x\otimes y$ is a positive element of $\mathcal M\overline\otimes\mathcal N$. So I was wondering : *is every positive element $z$ of $\mathcal M\overline\otimes\mathcal N$ the supremum of the set of all the elements of $\mathcal M\overline\otimes\mathcal N$ of the form $$\sum\_{n\in\mathbf N}x\_n\otimes y\_n$$ where $(x\_n)\_{n\in\mathbf N}$ and $(y\_n)\_{n\in\mathbf N}$ are finitely supported sequences of positive elements of $\mathcal M$ and $\mathcal N$ respectively such that $\sum\_{n\in\mathbf N}x\_n\otimes y\_n\leq z$ ?* I wouldn't be surprised if this was not true in general and only worked when $z\geq \varepsilon\cdot 1\_{\mathcal M\overline\otimes\mathcal N}$ from some real numer $\varepsilon>0$ or under some condition of a similar fashion.

https://mathoverflow.net/users/508539

Approximation from below of positive elements in tensor product of von Neumann algebras

It fails for $M=N=M\_2(\mathbb C)$. Here $M\overline \otimes N = \mathcal B(\mathbb C^2 \otimes \mathbb C^2)$. Let $e\_1 =(1,0), e\_2 = (0,1) \in \mathbb C^2$. The rank 1 projection $z$ onto the span of $\tfrac{1}{\sqrt{2}} ( e\_1 \otimes e\_1 + e\_2 \otimes e\_2)$ gives a counter-example. In fact, let $x\otimes y \leq z$. Suppose $x\otimes y \neq 0$, then we may assume $x\otimes y = z$ (since $z$ has rank 1). Write $x\_{i,j} = \langle x e\_j, e\_i\rangle$ for the $(i,j)$'th matrix entry (similar for $y$). Then \begin{equation} x\_{i,j} y\_{k,l} = \langle (x\otimes y) (e\_j \otimes e\_l) , e\_i \otimes e\_k \rangle = \langle z (e\_j \otimes e\_l) , e\_i \otimes e\_k \rangle = \delta\_{i,k} \delta\_{j,l} /2. \end{equation} Hence $x\_{1,1}y\_{1,1} = 1/2$, $x\_{1,2}y\_{1,2} = 1/2$ and $x\_{1,1} y\_{1,2} = 0$, a contradiction. So the only positive operator of the form $x\otimes y$ below $z$ is $0$.

5

https://mathoverflow.net/users/126109

453774

https://mathoverflow.net/questions/453623

8

$\DeclareMathOperator\SL{SL}$As explained in [this question](https://mathoverflow.net/questions/109642/lambda-ring-structures-on-mathbb-a2) , there are two $\lambda$-ring structures on ${\mathbb Z}[x]$. In layman's terms, both come from a realization of ${\mathbb Z}[x]$ as the representation ring of an algebraic (semi)-group: it is either $R({\mathbb C},\times)$, or $R(\SL\_2({\mathbb C}))$. **Question 1:** What are $\lambda$-ring homomorphisms between these? As we can choose any of the two structures for the range and the domain, it is four separate questions. In particular, I am interested in the semi-group of endomorphisms of $R(\SL\_2({\mathbb C}))$. I can see 3 of its elements: zero, identity and $$P: R(\SL\_2({\mathbb C}))\cong R(\operatorname{PSL}\_2({\mathbb C}))\hookrightarrow R(\SL\_2({\mathbb C})).$$ If one thinks of $x$ as the standard 2-dimensional representation of $\SL\_2$, they are given by $$0: x\mapsto 2, \ \ 1:x\mapsto x, \ \ P:x\mapsto x^2-2.$$ **Question 2:** Is this semigroup cyclic? Note that $P$ will be the generator.

https://mathoverflow.net/users/5301

$\lambda$-ring endomorphisms of ${\mathbb Z}[x]$

No, it is not cyclic. For each of the rings, the semigroup of endomorphisms consists of one element of degree $n$ for each positive integer $n$ plus one or two elements of degree $0$. For a positive integer, this is given by the unique product of Frobenius lifts of that degree, i.e. in $R(\mathbb C)$ by $x \mapsto x^n$ and in $R(SL\_2)$ by the $n$'th (normalized) Chebyshev polynomial of the first kind. That these are endomorphisms follows from the characterization of torsion-free lambda-rings in terms of commuting lifts of Frobenius. The degree $0$ elements are $x\to 0, x\to 1$ for $R(\mathbb C)$ and $x\to 2$ for $R(SL\_2)$. For maps from one ring to the other, the only choices are degree $0$, i.e. $x\to 2$ for $R(SL\_2) \to R(\mathbb C)$ and $x\to 0$ or $x\to 1$ for $R(\mathbb C) \to R(SL\_2)$. To calculate these, we can use the following trick: Both rings embed into $\mathbb Z[t,t^{-1}]$, with the lambda-ring structure arising from viewing it as the representation ring of $\mathbb G\_m$, by the embeddings $x\to t$ and $x\to t+t^{-1}$. So it suffices to find the lambda-ring homomorphisms from each ring to $\mathbb Z[t,t^{-1}]$ and then throw away the ones whose image does not lie inside the desired target rings. Since $\lambda^2(x) = 0 $ for $x$ the standard generator of $R(\mathbb C)$, any homomorphism $R(\mathbb C) \to \mathbb Z[t,t^{-1}]$ must send $x$ to an element $y$ with $\lambda^2(y)=0$. It's easy to see that such an element must have leading coefficient $1$ and no coefficients below the leading coefficient, i.e. must be a momomial $t^n$ for $n\in \mathbb Z$, or have no leading coefficients at all, since otherwise the leading term of $\lambda^2(y)$ would be nonvanishing. Then one can check that $x\to 0$ and $x \to t^n$ are in fact homomorphisms. The image of these homomorphisms is contained in $R(\mathbb C)$ if and only if $n\geq 0$, and contained in $R(SL\_2)$ only if $n=0$ or $x$ is sent to $0$. Similarly, $R(SL\_2(\mathbb C))$ is generated by an element $x$ with $\lambda^2(x)=1$, so $x$ must be sent to an element $y$ with $\lambda^2(y)=1$. Such an element must have leading coefficient $1$ or $2$, with the second case only in degree $0$, or otherwise the leading term of $\lambda^2(y)$ is something other than $1$. In the first case, the leading term of $\lambda^2(y)$ comes from the next term beyond the leading term of $y$, so we must have $y = t^n + t^{-n} + $ lower order terms. and in the second case we have $y = 2+$ lower order terms. Symmetrically, we must have $y= t^n + t^{-n} + $ higher order terms or $y = 2+$ higher order terms, and this is only possible if $y= t^n+ t^{-n}$ for $n\geq 0$ (with $n=0$ encapsulating the $y=2$ case). One checks that $x \mapsto t^n + t^{-n}$ indeed gives a homomorphism for all $n$, and then observes that its image lies in $R(SL\_2)$ for all $n$ and lies in $R(\mathbb C)$ only if $n=0$.

4

https://mathoverflow.net/users/18060

453789

https://mathoverflow.net/questions/453790

1

Let $\psi\_\alpha(x) = \exp(x^\alpha)-1$ for $\alpha\geq 1$. Define $$ \psi\_\infty(x) = \begin{cases}\infty & x>1\\1& x = 1\\ 0 & x <1 \end{cases} $$ to be such that for any $x>0$ $\psi\_\infty(x) = \lim\_{\alpha\to\infty}\psi\_\alpha(x)$. Let $$\lVert X\rVert\_{\psi\_\alpha} = \inf\{k>0\mid \mathbb{E}[\psi\_\alpha(|X|/k)] \leq 1\}$$ be the Orlicz norm associated with $\psi\_\alpha$. I am curious if the set of all random variables of finite $\lVert X\rVert\_{\psi\_\infty}$ norm is equal to the set of all essentially bounded random variables. More generally, I am interested if there are useful notes on Orlicz norms for functions $\psi :[0,\infty)\to[0,\infty]$ that take values in the extended real numbers. My understanding is that from the point of view of convexity it is often useful to work with such functions (they often appear when taking convex conjugates), but I often see Orlicz functions defined as convex functions $$\psi: [0,\infty)\to[0,\infty)$$ along with some other conditions, e.g. monotonicity, and having prescribed limiting behavior around $0$ and $\infty$. I'm curious how essential omitting the potential value of $\psi(x)=\infty$ is to the entire theory.

https://mathoverflow.net/users/101207

Is the space of bounded $\psi_\infty$ Orlicz norm random variables equal to $L^\infty$?

For each real $k>0$, \begin{equation}E\psi\_\infty(|X|/k)=\infty\,P(|X|>k)+P(|X|=k) \\ =\left\{\begin{aligned}\infty\text{ if } P(|X|>k)>0,\\ P(|X|=k)\le1\text{ if } P(|X|>k)=0. \end{aligned}\right. \end{equation} So, indeed, $\|X\|\_{\psi\_\infty}<\infty$ iff $X$ is essentially bounded. Moreover, $\|X\|\_{\psi\_\infty}=\text{ess}\,\text{sup}\,|X|$. Generally, for any non-constant nondecreasing convex function $F\colon[0,\infty)\to[0,\infty]$ such that $F(0)\le1$, the formula \begin{equation}\|X\|\_F:=\inf\{t>0\colon EF(|X|/t)\le1\} \end{equation} defines a norm on the linear space, say $L\_F$, of random variables (r.v.'s) $X$ on a probability space $\mathcal P$ with $\|X\|\_F<\infty$. The proof of this is the same as the one in the case when $F$ is not allowed to take the value $\infty$. (If, in addition, it is assumed that $F(0+)<1$, then all bounded r.v.'s on $\mathcal P$ will be in $L\_F$.)

2

https://mathoverflow.net/users/36721

453791

https://mathoverflow.net/questions/453787

13

Assume for simplicity that sets $A\_i\subset\mathbb{R}$ are compact. If $A\_1$ and $A\_2$ have positive length, then $A\_1+A\_2$ contains an interval. That is a variant of the classical Steinhaus theorem and it easily follows by looking at neighborhoods of a density point of $A\_1$ and a density point of $A\_2$. > > **Question.** Let $\alpha\in (0,1)$. Is it true that then there is $k$ such that if $\mathcal{H}^\alpha(A\_i)>0$, $i=1,2,\ldots,k$, then $A\_1+A\_2+\ldots+A\_k$ contains an interval? > > > Here $\mathcal{H}^{\alpha}$ stands for the Hausdorff measure. > > I am also interested in a version of this question where $A\_1=A\_2=\ldots=A\_k$ i.e., if we add a set $A$ with $\mathcal{H}^\alpha(A)>0$ to itself $k$ times: $A+A+\ldots+A$. > > > If this is not true, but simiar results are true, I would be interested in references. I am particulary interested in sets in a line.

https://mathoverflow.net/users/121665

Steinhaus theorem and Hausdorff dimension

The answer to the question is negative. Körner in [Hausdorff dimension of sums of sets with themselves](https://www.impan.pl/en/publishing-house/journals-and-series/studia-mathematica/all/188/3/90488/hausdorff-dimension-of-sums-of-sets-with-themselves) and Schmeling-Shmerkin in [On the dimension of iterated sumsets](https://arxiv.org/abs/0906.1537) showed that for any increasing sequence $\alpha\_1 < \alpha\_2 < \ldots < 1$, there is a compact set $A$, s.t. for every $k$, we have $\dim\_{H} k A = \alpha\_k$, where $kA = A + A + \ldots + A$ --- Minkowski sum of the set with itself $k$ times, and $\dim\_H$ is the Hausdorff dimension of the set. In particular, choosing any sequence such that $\dim\_H kA < 1$ for all $k$, none of the sum sets $kA$ could contain an interval. See also the recent paper [Dimension growth for iterated sumsets](https://arxiv.org/abs/1802.03324) for some conditions under which we have dimension growth $\dim\_H (A + A) > \dim\_H(A)$. Unfortunately, it seems that the bounds there still fall short from implying that for some $k$, $\dim\_H k A = 1$, much less providing quantitative bound on $k$. On the positive side, Marstand in "Some Fundamental Geometrical Properties of Plane Sets of Fractional Dimensions" showed that for any given pair of sets $A,B$, and almost every angle $\theta$, if we write $\lambda\_1 = \cos \theta, \lambda\_2 = \sin \theta$, then $$ \dim\_H (A\_{\lambda\_1} + B\_{\lambda\_2}) = \min(\dim\_H(A \times B),1) \geq \min(\dim\_H(A) + \dim\_H(B), 1),$$ where $A\_{\lambda}$ is the dilatation $A\_{\lambda} := \{ \lambda x : x \in A\}$. In fact a slightly stronger statement holds: if $\dim\_H (A) = \alpha$, there is a sequence of parameters $\lambda\_1, \lambda\_2, \ldots, \lambda\_k > 0$ for $k = \lceil 1/\alpha \rceil + 1$, such that for $B := A\_{\lambda\_1} + \ldots + A\_{\lambda\_k}$ not only $\dim\_H(B)=1$, but indeed the 1-dimensional measure of this set is positive, and therefore $B+B$ contains an interval (a generic sequence of $\lambda\_i$ will do the job). Unfortunately, this result does not say anything about any specific sequence of $\lambda\_i$ (of which $\lambda\_1 = \ldots = \lambda\_k$ is the most interesting). A more modern exposition of this theorem can be found in Chapter 9 of "Geometry of sets and measures in Euclidean spaces" by Pertti Mattila.

14

https://mathoverflow.net/users/468679

453793

https://mathoverflow.net/questions/453657

0

$\newcommand{\lorentzian}{\mathrm{lorentzian}}\newcommand{\lorentzian}{\mathrm{lorentzian}}\newcommand{\diff}{\mathrm{diff}}\newcommand{\manifold}{\mathrm{manifold}}$Take a time-oriented Lorentzian manifold $(M, g)$ that is not [strongly causal](https://en.m.wikipedia.org/wiki/Causality_conditions). The Lorentzian metric $g$, however can define a topology, that is strictly coarser than that of the manifold topology. Say: $\tau\_{\lorentzian} \subsetneqq \tau\_{\manifold}$. The topology can be found at this post: [Lorentzian Topology](https://mathoverflow.net/a/266912/503363) Assume I have two Lorentzian time-oriented manifolds $(M\_1,g\_1)$, $(M\_2,g\_2)$ that are not strongly causal, but are homeomorphic **under their Lorentzian topologies** $(M\_1, \tau\_{\lorentzian}) \cong (M\_2, \tau\_{\lorentzian} )$. ***My 1st question:*** Is it possible that $M\_1$ and $M\_2$ are not diffeomorphic as differentiable manifolds (**with the manifold topology and NOT the Lorentzian topology**) $(M\_1, \diff) \not\equiv (M\_2, \diff) $ while $(M\_1, \tau\_{\lorentzian}) \cong (M\_2, \tau\_{\lorentzian} )$ as topological spaces? ***My 2nd question:*** Is it possible for the $\tau\_{\lorentzian}$ not to be antidiscrete? **NOTE**: **the differential structure in $(M, \diff)$ is exactly the one underlying the Lorentzian manifold.** **NOTE**: **to be more physically sound, one can name the $\tau\_{lorenzian}$ topology, also $\tau\_{observers}$ topology.** Even an example of such a scenario would be extremely of use! --- **Physical motivation:** If such distinct differentiable manifolds that induce the same topology via their Lorentzian metric exist, one can define an entropy for the Lorentzian topology by counting the distinct diffeomorphism classes say $w$ and write $S=k\_B \ln(w) $ by Boltzmann's relation. And causality will be emergent rather than fundamental, so to speak. **This can potentially give rise to a cosmological arrow of time or entropic time if not also quantum non-locality in case one interprets the differential and topological degrees of freedom as the Hidden Variables of Quantum Mechanics!**

https://mathoverflow.net/users/503363

Non-diffeomorphic but homeomorphic (under Lorentzian topology) Lorentzian manifolds

I would like to argue that the situation considered [in the](https://mathoverflow.net/questions/453657/non-diffeomorphic-but-homeomorphic-under-lorentzian-topology-lorentzian-manifo?noredirect=1#comment1173719_453657) [comments](https://mathoverflow.net/questions/453657/non-diffeomorphic-but-homeomorphic-under-lorentzian-topology-lorentzian-manifo?noredirect=1#comment1174040_453657) is "close to generic". Let $(M,g)$ be a Lorentzian manifold that is not strongly causal; this implies that $(M,g)$ is also not stably causal, and hence a small perturbation of it will admit a closed time-like curve. We may therefore assume *generically* that our $(M,g)$ admits a closed time-like curve $\gamma$. Choose $p\in \gamma$, we have therefore $\gamma\subset I^+(p)$ and $\gamma\subset I^-(p)$; hence the set $U\_\gamma = I^+(p) \cap I^-(p)$ is non-empty, and must be contained in every "Lorentzian neighborhood" of $p$. In particular, the "Lorentzian topology" must be non-Hausdorff. As $\gamma$ is compact, it has a tubular neighborhood within $U\_\gamma$ that is topologically $\mathbb{S}^1 \times B^n$ where $n$ is the spatial dimension. Let $t\in \mathbb{S}^1$ and $x\in B^n$ provide a set of coordinates for this tubular neighborhood; we can arrange for $\partial\_t$ to be time-like on this neighborhood. Now take an $n$-dimensional manifold $D$ with boundary, whose boundary is $\mathbb{S}^{n-1}$, with non-trivial topology. Equip $D$ with a Riemannian metric. Carve out the tubular neighborhood of $\gamma$ and replace it (smoothly glue in) a copy of $\mathbb{S}^1\times D$, equipped with the product Lorentzian metric. As this modification is entirely within $U\_\gamma$, and one easily sees that for every $q\_1, q\_2\in \mathbb{S}^1\times D$, there exists a time-like curve connecting the two, we see that the Lorentzian topology of the modified manifold is the same as the original. But as long as $M$ doesn't have too crazy a topology the modified manifold should not be diffeomorphic to the original.

5

https://mathoverflow.net/users/3948

453799

https://mathoverflow.net/questions/453808

3

Let $X$ be a complex minimal surface of general type, *id est* $K\_X$ is big and nef. It is well-known that $\displaystyle\int\_X3c\_2(X)-c\_1(X)^2\geq0$, and the equality holds if and only if $X$ is uniformized by $\mathbb{B}\subset\mathbb{C}^2$ (the open ball). Ever in this case: $X$ does not contain neither rational smooth curves nor smooth curves of genus $1$. In the opposite case, *id est* assuming that $\displaystyle\int\_X3c\_2(X)-c\_1(X)^2>0$, there are examples of such $X$'s which do not contain any smooth rational curves. **Questions**: let $X$ be a complex minimal surface of general type such that $\displaystyle\int\_X3c\_2(X)-c\_1(X)^2>0$: does $X$ contain any smooth curves of genus $1$? Does the previous answer change if one assumes $\Omega^1\_X$ nef but not ample and $K\_X$ ample?

https://mathoverflow.net/users/57030

Existence of elliptic curves on surfaces of general type

Take a product $X= C\_1\times C\_2$, where $C\_i$ are smooth curves of genus greater than $1$. $X$ has general type, and is uniformized by a product of two disks. Also $X$ won't contain an elliptic curve because it maps trivially to each $C\_i$.

6

https://mathoverflow.net/users/4144

453812

https://mathoverflow.net/questions/453815

2

Let $H: \mathbf{R}^n \rightarrow \mathbf{R}$ be a bounded continuous function. Set $$\tag{1} \int\_{\mathbf{R}^n}\left\{|\nabla \xi|^2+H(x) \xi^2\right\} \mathrm{d} x \geqslant 0, \quad \forall \xi \in C\_c^{\infty}\left(\mathbf{R}^n\right) $$ Then by Theorem 8.38 in Gilbarg & Trudinger's book, in a ball $B\_R$ we should have $$-\Delta \phi\_R + H(x)\phi\_R=\lambda\_R\phi\_R,$$ where $\lambda\_R\ge0$ is the first nonzero eigenvalue, with $\phi\_R>0$. 1. If there is no $H(x)$, it will be simple to prove that $\lambda\_R$ is a decreasing function when $R$ becomes greater, so there will be a limit for $\lambda\_R$ when $R \rightarrow +\infty$. But here we have $H(x)$, so we can't simply let $\phi(x)=\Phi(xR/a)$ to make the ball bigger (with radius $aR$): do we still have the property that the first nonzero eigenvalue is a decreasing function with respect to $R$? 2. What is the value of $\lim\_{R\rightarrow+\infty}\lambda\_R=\lambda\_0$, is it 0? 3. If we let the $R$ tend to $+\infty$, can we also get a limit function $\Phi$, which satisfies that $$-\Delta \Phi + H(x)\Phi=\lambda\_0\Phi, $$ and what type is this convergence? Do we need local convergence or? All these questions have a vague answer in my mind, but I don't know how to figure out the details, if you have any references I would be very glad!

https://mathoverflow.net/users/469129

Given an eigenvalue equation (elliptic PDE) in a ball $B_R$, prove the convergence of the first nonzero $\lambda_R$ and its eigenfunction $\phi_R$

For simplicity let me take $H$ smooth or at least $C^\alpha$ so that I don't have to worry about elliptic estimates: 1. Limit as $R\to\infty$: Use $$ \lambda\_R = \inf\_{\phi \in C^\infty\_c(B\_R)}\frac{\int |\nabla \phi|^2 + H \phi^2}{\int \phi^2} $$ Since $C^\infty\_c(B\_R) \subset C^\infty\_c(B\_{R+s})$ for all $s>0$ we see that $\lambda\_R$ is non-increasing. 2. The limit $\lambda\_\infty$ (which exists since $\lambda\_R$ is non-increasing and $\geq 0$) depends on $H$. For example if $H=0$ then then since we know the first Dirichlet eigenvalue of a ball we can see that $\lambda\_\infty=0$. On the other hand, if $H=1$ then $$ \int\_{B\_R} |\nabla \phi|^2 + \phi^2 \geq \int \phi^2 $$ so $\lambda\_\infty \geq 1$ (actually we should be able to prove $\lambda\_\infty=1$). 3. Yes. Choose the first eigenfunction $\phi\_R$ and normalize so that $\phi\_R(0) = 1$. Then, the Harnack inequality says that for $K\Subset B\_R$ there's $C=C(K,H)$ so that $C^{-1}\leq \phi\_R \leq C$ on $K$. Bootstrapping to $C^{2,\alpha}$ we can take a subsequence to find $\phi\_\infty$ solving $$ -\Delta \phi\_\infty+H\phi\_\infty = \lambda\_\infty\phi\_\infty. $$ Note that, $\phi\_\infty$ will not necessarily (never?) be in $L^2(\mathbb{R}^n)$. (For example, when $H$ is constant, we get $\phi\_\infty = 1$.)

3

https://mathoverflow.net/users/1540

453820

https://mathoverflow.net/questions/453814

2

> > Assume that $ A $ is self-adjoint operator and $ B $ is a bounded self-adjoint operator. The definite domain of $ A,B $, denoted by $ D(A) $ and $ D(B) $ satisfies $ D(A)\subset D(B) $. Show that > \begin{align\*} > \operatorname{dist}(\sigma(A),\sigma(A+B))\leq \|B\|, > \end{align\*} > where $ \sigma(A) $ and $ \sigma(A+B) $ are the spectrums of $ A $, $ A+B $ and $ \|B\| $ is the operator norm of $ B $. > > > It can be obtained by utilizing the Kato-Rellich theorem that $ A+B $ is also a self-adjoint operator. However I cannot go on since the decomposition of $ A $ and $ A+B $ are not the same. Can you give me some hints or refereces?

https://mathoverflow.net/users/241460

Disturbance of self-adjoint operator

Let $C=A+B$, take a point, say $0$, in the spectrum of $A$ and assume that $[-\|B\|, \|B\|]$, hence $[-\|B\|-\epsilon, \|B\|+\epsilon]$ for some $\epsilon>0$ is contained in the resolvent set of $C$. The spectral theorem, applied to $C$ implies that $\|C^{-1}\| \leq (\|B\|+\epsilon)^{-1}$. Then $A=C-B=(I-BC^{-1})C$ would be invertible because $\|BC^{-1}\| <1$.

2

https://mathoverflow.net/users/150653

453826

https://mathoverflow.net/questions/453865

9

Let $G$ be a finite group, let $k$ be a large enough field of characteristic $p>0$. Let $p\mid |G|$. **Broué's abelian defect group conjecture** states the following: Let $B$ be a block of $kG$ with abelian defect group $D$ and let $b\in \text{Bl}(N\_G(D))$ be its Brauer correspondent. Then the derived categories $D^b (\text{mod}(B))$ and $D^b(\text{mod}(b))$ of bounded complexes of finitely generated modules over $B$ and $b$ are equivalent as triangulated categories. **Alperin's weight conjecture (blockwise version)** states the following: Let $B$ be a block of $kG$. Then the number of irreducible Brauer characters belonging to $B$ equals the number of $G$-conjugacy classes of $p$-weights of $B$. > > Is there a relationship between these two conjectures? For example: does one conjecture imply the other? > > > The motivation for this question is: both Alperin's weight conjecture and Broué's abelian defect group conjecture can be expressed via trivial source modules (\*). Thank you very much for the help. (\*) Small footnote: in the case of Broué's abelian defect group conjecture this means that all modules occurring in the cochain complex realising such a derived equivalence are supposed to be $p$-permutation modules. Hence, this is more restrictive than the general conjecture.

https://mathoverflow.net/users/12826

Is there a relationship between Broué's abelian defect group conjecture and Alperin's weight conjecture?

Yes. Broué's conjecture implies the abelian defect case of Alperin's weight conjecture. This makes one think there might be a structural statement like Broué's conjecture, for non-abelian defect groups, that would imply Alperin's conjecture in general. Many attempts have been made, to little avail; Rouquier has conjectured that if the hyperfocal subgroup of the defect group is abelian then there is a derived equivalence as in Broué's conjecture, but that's about it.

11

https://mathoverflow.net/users/460592

453871

https://mathoverflow.net/questions/453881

4

Given a first-order structure $\mathfrak{A}$ and a first-order theory $T$ one can ask if $$ \varphi(\mathfrak{A}, T) := ``\text{there is a substructure } \mathfrak{B} \text{ of } \mathfrak{A} \text{ such that } \mathfrak{B} \models T \text{''} $$ holds. I am interested in whether $\varphi$ is absolute between inner and outer models. More formally, let $\mathfrak{A}$, $T$, $\mathbf{V}$, $\mathbf{W}$ be such that 1. $\mathbf{V}, \mathbf{W} \models \mathsf{ZFC}$, 2. $\mathfrak{A}$ is a first-order structure and $T$ a first-order theory in $\mathbf{V}$, 3. $\mathbf{V}$ is an inner model of $\mathbf{W}$, and 4. $\mathbf{V} \models \neg \varphi(\mathfrak{A}, T)$. Must it be the case that $\mathbf{W} \models \neg\varphi(\mathfrak{A}, T)$?

https://mathoverflow.net/users/29231

Is the existence of substructures satisfying a theory absolute?

**Assuming $T$ is countable** (in $V$), the answer is yes. By downward Lowenheim-Skolem applied to $\mathfrak{B}$, $\varphi(\mathfrak{A},T)$ is equivalent to "$\mathfrak{A}$ has a *countable* substructure satisfying $T$." This lets us interpret $\varphi(\mathfrak{A},T)$ as asking about the ill-foundedness of a certain tree $S$; intuitively, a height-$n$ node on $S$ consists of $(i)$ a choice of the first $n$ elements of a substructure of $\mathfrak{A}$ which will satisfy $T$ and $(ii)$ some data telling us how that structure will satisfy $T$. By Mostowski absoluteness, $S$ is ill-founded in $W$ iff $S$ is ill-founded in $V$. Countability of $T$ got used twice in the above argument; most obviously when we applied downward Lowenheim-Skolem at the start, but also in the breezed-through bulletpoint $(ii)$. If $T$ is uncountable, both these steps break down. --- **What if $T$ is uncountable in $V$?** *EDIT: Below I construct a counterexample where $W$ and $V$ have different cardinals. [Joel's answer](https://mathoverflow.net/a/453887/8133) shows that with more care we can avoid this, and additionally that it's the uncountability of the language as opposed to any of the structures involved that is crucial.* Consider the language $\{c\_i:i\in\omega\}\cup\{d\_i: i\in\omega\_1^V\}\cup\{E\}$ where the $c$s and $d$s are constants and $E$ is a ternary relation symbol, and let $\mathfrak{A}$ be the two-sorted structure in $V$ consisting of $(i)$ a "point" for each $c\_i$ and $d\_j$ and $(ii)$ a "line" $l\_{i,j}$ for each pair $i\in\omega,j\in\omega\_1^V$ with $E^\mathfrak{A}=\{(c\_i,d\_j,l\_{i,j}): i\in\omega,j\in\omega\_1^V\}.$ Basically, we're looking at $K\_{\omega,\omega\_1}$ with edges being given by objects, not just relations. Note that any substructure contains all the $c\_i$s and $d\_j$s. Now let $T$ be the theory saying that the "lines" provide a bijection between the $c$s and $d$s. This is realized in a substructure of $\mathfrak{A}$ after forcing with $Col(\omega,\omega\_1^V)$ over $V$, but isn't realized in $V$ itself.

8

https://mathoverflow.net/users/8133

453882

https://mathoverflow.net/questions/453837

6

It is standard from work of Joyal and Lurie that there is a Quillen equivalence between the model category of simplicially enriched categories $Cat\_\Delta$ and $\mathcal{S}\text{et}\_\Delta$ with the Joyal model structure. This implies that for any fibrant simplicially enriched category, equivalently a topological category (in the sense of Segal), one gets a corresponding $\infty$-category by the coherent nerve functor $N$. My question (coming out of some practical purposes) is whether one can establish an equivalence between $N$ and the standard coend construction using the *usual* nerve of a topological category. This sounds somewhat problematic, since the former preserves limits while the latter preserves colimits, but in the following we view $N$ as an equivalence $(Cat\_\Delta)^{cf}\overset{\sim}{\to} (\mathcal{S}\text{et}\_\Delta)^{J,cf}$ as $\infty$-categories, where $cf$=cofibrant and fibrant objects and $J$ stands for the Joyal model structure. So this seems not unreasonable. More explicitly, (1) for a topological category $\mathsf{T}$, the usual nerve is a simplicial space $\mathcal{P}\_\bullet(\mathsf{T})$, with $\mathcal{P}\_0(\mathsf{T})$ equal to the discrete set of objects in $\mathsf{T}$, and $$\mathcal{P}\_n(\mathsf{T})=\coprod\_{t\_0,\cdots,t\_n\in \mathcal{P}\_0(\mathsf{T})} Maps\_{\mathsf{T}}(t\_0,t\_1)\times\cdots\times Maps\_{\mathsf{T}}(t\_{n-1},t\_n), n\geq 1$$ Here $$\mathcal{P}\_n(\mathsf{T})\to \mathcal{P}\_1(\mathsf{T})\underset{\mathcal{P}\_0(\mathsf{T})}{\times}\cdots\underset{\mathcal{P}\_0(\mathsf{T})}{\times} \mathcal{P}\_1(\mathsf{T}), n\geq 1$$ is a strict isomorphism of topological spaces. But one can add the flexibility demanding it to be a weak homotopy equivalence, for which we call a "weak" topological category. From here one can take the coend $\int^{[n]\in \Delta}(\text{Sing}\_\bullet\mathcal{P}\_n(\mathsf{T}))\times N(\Delta^n)$ in $\text{Cat}\_\infty$ (the $\infty$-category of $\infty$-categories); (2) conversely, for any $\infty$-category $\mathcal{C}$, by the work of Rezk, we can view it as a complete Segal space $\widetilde{\mathcal{P}}\_{\bullet}$. One can cook up a simplicial space $\mathcal{P}\_\bullet$ with a discrete 0-space (up to homotopy) as follows. Take a base point $x\_i$ in each connected component of $\widetilde{\mathcal{P}}\_0$, and let $\mathcal{P}\_0\to \widetilde{\mathcal{P}}\_0$ be the standard fibration from the (disjoint union of) path spaces based at $x\_i$. Set $$\mathcal{P}\_n=\widetilde{\mathcal{P}}\_n\underset{\widetilde{\mathcal{P}}\_0^{\times (n+1)}}{\times}\mathcal{P}\_0^{\times (n+1)}.$$ Then $\mathcal{P}\_\bullet$ gives a "weak" topological category. My question is: (i) Do the functors (1) and (2) give inverse equivalences between the $\infty$-category of "weak" topological categories and $Cat\_\infty$, both modeled as full subcategories of the $\infty$-category of simplicial spaces? If I'm not mistaken, they give an adjunction pair. (ii) If so, does the above coincide with the standard equivalence between $(Cat\_\Delta)^{cf}$ and $(\mathcal{S}\text{et}\_\Delta)^{J,cf}$? Intuitively, one would expect this to be true.

https://mathoverflow.net/users/37400

From the *usual* nerve of topological categories to $\infty$-categories

The answer to Question (ii) is positive. That is to say, there is a weak equivalences between the following functors from Segal topological categories to quasicategories: the composition of the singular complex functor with the homotopy coherent nerve functor, and the realization of the singular complex. Indeed, the first step in both functors is the same: we take the singular complex of a Segal topological category, which yields a Segal category. Therefore, the problem reduces to establishing a weak equivalence between the homotopy coherent nerve functor N and the realization functor R, both considered as functors from Segal categories to quasicategories. Both functors are homotopy cocontinuous: the homotopy coherent nerve is a Quillen equivalence, and the realization functor by construction. Since the quasicategory of quasicategories is a reflective localization of ∞-presheaves on Δ, it suffices to construct a weak equivalence between the restructions of N and R along the Yoneda embedding of Δ into Segal categories. Indeed, both restrictions are weakly equivalent to the Yoneda embedding of Δ into quasicategories, by construction. Question (i) is not formulated rigorously, but there are rigorously defined functors from quasicategories to Segal categories. For example, one can use the left adjoint of the homotopy coherent nerve functor, which tautologically provides a positive answer. Other constructions can be obtained by passing from quasicategories to Segal spaces, and then to Segal categories using the constructions of Joyal–Tierney and Bergner. Bergner's book has details of these construction. To see that the resulting functor is indeed the inverse of the functors considered above, consider the two compositions, and show they are weakly equivalent to the corresponding identity functor by restricting along the Yoneda embedding and observing that the resulting restrictions are weakly equivalent by construction.

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https://mathoverflow.net/users/402

453891

https://mathoverflow.net/questions/453768

5

Heinrich Martin Weber and David Hilbert created their own class fields in [1891](https://www.sciencedirect.com/science/article/abs/pii/B9780125996617500197) and [1898](https://en.wikipedia.org/wiki/Hilbert_class_field) respectively. In the past, Weber continued to name $K={Q}(\sqrt{-m}, j(\omega))$, the Kronecker class field of $k={Q}(\sqrt{-m})$, called "species associated with a field k" created by Leopold Kronecker, as 'class field', but later in a series of papers in 1896, Weber defined the concept of his class field. He extended it to the field K related to the congruence class group. Therefore, he used the class field K as a general class field, which is related to the law of decomposition of prime numbers at k. And Hilbert predicted his own famous class field, the 'Hilbert class field', which was later realized by Philipp Furtwangler. As a result, My question is what Hilbert thought was lacking in Weber's class field, so there was a need to create a new class field concept.

https://mathoverflow.net/users/167507

Compare with Weber and Hilbert class field

Keith Conrad discusses the history of class fields in these [lecture notes](https://kconrad.math.uconn.edu/blurbs/gradnumthy/cfthistory.pdf). Weber's and Hilbert's definitions are equivalent, but Weber only considered class fields for ideal groups over $\mathbf{Q}$ or imaginary quadratics.

1

https://mathoverflow.net/users/11260

453897