Recent MathOverflow Questions

Orbits of arithmetic subgroups intersection a compact set

Math Overflow Recent Questions - Wed, 01/17/2018 - 21:56

Let us suppose we have $G$ a connected reductive group over a number field $F$. Consider $G(\mathbb{A})$ the group over the adeles and $G(\mathbb{Q})$ embedded discretely. For $\gamma \in G(\mathbb{Q})$ we denote by $\mathcal{O}(\gamma)$ the orbit $\gamma$ under conjugation by elements of $G(\mathbb{A})$. My question is the following. If $M$ is a compact subset of $G(\mathbb{A})$, is it true that only finitely many $\mathcal{O}(\gamma)$ have the property that intersection with $M$ is nonempty?

Note: I am interested in the finiteness of the number of orbits with such property, not the $\gamma$'s.

Rosenlicht's theorem and rationality questions

Math Overflow Recent Questions - Wed, 01/17/2018 - 21:29

Let $G$ be a connected algebraic group over an algebraically closed field $\overline{k}$ acting on an irreducible variety $X$. A geometric quotient is a morphism of varieties $\pi: X \rightarrow X/\sim$ which on closed points (that is, as a morphism of classical varieties) satisfy the following:

(i): $\pi$ is a surjective open map, and the fibres are exactly the $G$-orbits of $X$.

(ii): for any open set $U$ of $X/\sim$, the ring homomorphism $\pi^{\ast}: \overline{k}[U] \rightarrow \overline{k}[\pi^{-1}U]$ is an isomorphism onto the $G$-fixed points of codomain.

Rosenlicht's theorem says that there exists a $G$-stable open subset of $X$ for which the geometric quotient exists.

Is there any generalization of Rosenlicht's theorem for when $G$ and $X$ are defined over an arbitrary field? The case I'm interested in is when $G$ and $X$ are geometrically connected subgroups of upper triangular unipotent matrices over a $p$-adic field $F$ (so all orbits are closed), and $X$ is normalized by $G$.

Equivalent descriptions of Coherent Groups

Math Overflow Recent Questions - Wed, 01/17/2018 - 21:00

Attending a series of lectures, I have recently been exposed to the notion of Coherent groups, defined as following:

Def: A group $G$ is called $Coherent$ if every finitely generated subgroup $H$ of $G$ is finitely presented.

Examples of such groups are free, surface groups, the fundamental group of 3-manifolds. In contrast $F_2 \times F_2$ (and any group which contains it) is not coherent.

I am just wondering if this notion has any equivalent description, possibly in terms of the module caregory of the group algebra $kG$, or any other area.

References for further reading are highly appreciated.

Explanation of a closed immersion

Math Overflow Recent Questions - Wed, 01/17/2018 - 16:55

Show that a morphism $\varphi:X\rightarrow Y$ is a closed immersion iff $\varphi(|X|)$ is closed in |Y|, $\varphi^*:O_Y\rightarrow\varphi_*O_X$ is a surjection, $|X|\rightarrow\varphi(|X|)$ is a homeomorphism and $\ker\varphi^*$ is of finite type.

We use a quite uncommon definition of closed immersion: $\varphi:X\rightarrow Y$ is a morphism of analytic spaces and there exists a closed analytic subspace Z of Y s.t. $\varphi=\tau\circ\rho$, where $\tau:Z\rightarrow Y$ is the inclusion morphism and $\rho:X\rightarrow Z$ is an isomorphism of analytic spaces.

I feel like "$\rightarrow$" should be clear, but I don't really know where to start and I apologize for already having posted this on MSE.

uniquely determining a distribution using moments

Math Overflow Recent Questions - Wed, 01/17/2018 - 16:55

Let $A$ be a parametric family of probability distributions that include all distributions in the form of $\phi(X)$ where $X\sim\mathcal{N}(0,\mathbf{I})$ is jointly Gaussian and $\phi:\mathbb{R}^d\to \mathbb{R}^d$ belongs to a parametric set of functions $G$.

To uniquely determine a distribution in $A$, how many moments do we need to know? For example, if $G$ is linear, A includes all Gaussian distributions. Thus, we need first and second moments to uniquely determine a distribution in the set $A$. I wonder if such a result can be extended for other $G$ such as polynomials with degree $k$.

Formality of some innner RHom

Math Overflow Recent Questions - Wed, 01/17/2018 - 15:42

Let $Y$ be a scheme and $X$ be a closed subscheme. Consider $\underline{RHom}_{\mathcal O_Y}(\mathcal O_X,\mathcal O_X)$ (inner RHom). We can view this is as a sheaf of dg-algebras on $X$; when both $X$ and $Y$ are (for simplicity) of finite type over a field and smooth then its cohomology is the exterior algebra of the normal bundle to $X$ inside $Y$.

$\mathbf{Question:}$ What kind of results are known about the formality of this algebra in some nice cases? I am especially interested in the case when $Y$ is symplectic and $X$ is Lagrangian in $Y$.

Controlling solutions of a second order linear differential inequality

Math Overflow Recent Questions - Wed, 01/17/2018 - 15:11

A slightly less general version of this question was asked, in a subsequent comment, by the OP of the question at Controlling subsolutions of a second order linear ODE

Let $f:[0,\infty) \to \mathbb{R}$ obey the differential inequality \begin{equation} f'' - 2a f' + (a^2+b^2) f \le 0,\tag{*} \end{equation} where $a\in\mathbb R$ and $b\in(0,\infty)$. Given initial conditions (i) $f(0) = f_0\in(0,\infty)$ and $f'(0) = g_0\in\mathbb R$ or, alternatively, given (ii) $f(0) = 0$ and $f'(0) = g_0\in(0,\infty)$, can we control the smallest real $y>0$ such that $f(y)\le0$?

Recognition of finite simple groups by number of Sylow p-subgroups (3)

Math Overflow Recent Questions - Wed, 01/17/2018 - 15:10

This is a follow-up to this question.

Let $G$ and $G'$ be two finite simple groups of the following structures:

1- $A_{p}$, for some primes $p$;

2- $PSL_{p}(q)$, for some prime $p$ and some prime power $q$;

3- $PSU_{p}(q)$, for some prime $p$ and some prime power $q$;

Also suppose that $s$ is the $\textbf{greatest}$ prime number dividing $|G|$ and $|G'|$ and every Sylow $s$-subgroup of $G$ and $G'$ is a cyclic subgroup of order $s$. If $G$ and $G'$ have the same number of Sylow $s$-subgroups, then can we say that $G\cong G'$?

Analytic Space with No Regular Points

Math Overflow Recent Questions - Wed, 01/17/2018 - 14:19

Define an analytic space to be a topological space $X$ equipped with a sheaf of rings $\mathcal{O}_X$ such that for every point $x \in X$ there is a neighbourhood $U \subseteq X$ such that $(U, \mathcal{O}_X \vert_U)$ is isomorphic (as a ringed space) to $(Y, \mathcal{O}_Y)$, where $Y \subseteq \mathbb{C}^n$ is some domain and $\mathcal{O}_Y = {}_n \mathcal{O} / \mathscr{I}$, where $_n\mathcal{O}$ denotes the sheaf of germs of analytic functions $f: \mathbb{C}^n \to \mathbb{C}$ and $\mathscr{I}$ denotes the ideal sheaf of ${}_n \mathcal{O}$.

Define a regular point of an analytic space to be a point $x \in X$ which admits a neighbourhood $U$ such that $(U, \mathcal{O}_X \vert_U)$ is a complex manifold (isomorphic as a ringed space to $(Y, \mathcal{O}_Y)$, where $Y \subseteq \mathbb{C}^n$ is some domain and $\mathcal{O}_Y$ denotes the sheaf of analytic functions $f: Y \to \mathbb{C}$.

Can someone provide me an example of an analytic space that has no regular points and is very different to a complex manifold. Intuitively, I think of analytic spaces and complex manifolds in a rather similar boat, but analytic spaces permit sharp points. I am aware that this intuition may be terribly erroneous, but it's what I've picked up over the last month or two of studying analytic spaces.

I was also wondering if someone could provide their example (the example they keep in their head) for the difference between a Stein manifold and Stein space? This is essentially the same question as the one just stated.

Any help/remarks/intuitions is really appreciated.


Isomorphism of irreducible R-modules

Math Overflow Recent Questions - Wed, 01/17/2018 - 14:01

Let $R$ be a $k-$algebra and $M,N$ two irreducible $R-$modules, isomorphic as vector spaces. If we know that for every $r\in R$ we have the same eigenvalues on $M$ and $N$ (with multiplicities) is it true that $M\simeq N$?

I have two thoughts:

  1. Denoting $St(m)=\{ rm=m \mid r\in R\}$, if we have $St(m)=St(n)$ for some $m\in M, n\in N$, we could find an isomorphism that sends $m$ to $n$ and is extended by $f(rm)=rf(m)$ (every element of $M$ is $rm$ for some $r$ by irreducibility). I cannot prove the existence of such a pair though.

  2. If $R$ was commutative, we could simultaneously diagonalize all the actions according to some basis and take the isomorphism of vector spaces that sends the one basis to the other. This will then be an isomorphism of modules. I cannot extend that to the general case.

It is true that $\overline{\text{Im}(A)}\otimes \overline{\text{Im}(B)}\subset \overline{\text{Im}(A\otimes B)}$?

Math Overflow Recent Questions - Wed, 01/17/2018 - 12:50

Let $H$ be a complex Hilbert space and $\mathcal{L}(H)$ be the algebra of all bounded linear operators on $E$.

If $A,B\in \mathcal{L}(H)$, It is true that $\overline{\text{Im}(A)}\otimes \overline{\text{Im}(B)}\subset \overline{\text{Im}(A\otimes B)}$?

I try as follows:

Let $z\in \overline{\text{Im}(A)}\otimes \overline{\text{Im}(B)}$, then $z=\displaystyle\sum_{i=1}^dx_i\otimes y_i$, where $x_i\in\overline{\text{Im}(A)},\;y_i\in\overline{\text{Im}(B)}$ and $d\in\mathbb{N}$. So, there exists $(\alpha_i(n))_{n\in\mathbb{N}}\subset \text{Im}(A)$ and $(\beta_i(n))_{n\in\mathbb{N}}\subset \text{Im}(B)$ such that $x_i=\lim_{n\rightarrow\infty}\alpha_i(n)$ and $y_i=\displaystyle\lim_{n\rightarrow\infty}\beta_i(n)$. Moreover, $\alpha_i(n)=A\tilde{\alpha}_i(n),\;\beta_i(n)=B\tilde{\beta}_i(n)$ and $$\alpha_i(n)\otimes\beta_i(n)=(A\otimes B)(\tilde{\alpha}_i(n)\otimes \tilde{\beta}_i(n))\subset \text{Im}(A\otimes B),$$ where $(\tilde{\alpha}_i(n))_{n\in\mathbb{N}}\subset \mathcal{H}$ and $(\tilde{\beta}_i(n))_{n\in\mathbb{N}}\subset \mathcal{H}$. Let $$z_n=\displaystyle\sum_{i=1}^d\alpha_i(n)\otimes \beta_i(n)\in \text{Im}(A\otimes B).$$ Since $\|x\otimes y\|=\|x\|\|y\|$, then the map $(x,y)\longmapsto x\otimes y$ is continuous. So $z_n\longrightarrow z$ as $n\longrightarrow \infty$, this implies that $z\in \overline{\text{Im}(A\otimes B)}$. As a consequence, $\overline{\text{Im}(A)}\otimes \overline{\text{Im}(B)}\subset \overline{\text{Im}(A\otimes B)}$.

Are smooth specializations of smooth hypersurfaces again hypersurfaces

Math Overflow Recent Questions - Wed, 01/17/2018 - 12:35

Let $X\subset \mathbb{P}^n$ be a smooth projective hypersurface of degree $d$ (over the complex numbers).

Assume $n$ is very large compared to $d$, and that $d$ is a prime number (e.g., $d=3$ and $n > 1000000000$). In particular, $X$ is a Fano variety (of dimension $n-1$ and Picard rank $1$).

Let $X_0$ be a smooth projective specialization of $X$. Then $X_0$ is also a Fano variety. However, as far as I can tell, it could happen that $X_0$ is not a hypersurface in $\mathbb{P}^n$. When does this actually happen?

Just to be clear: here we say that $X$ specializes to $X_0$ if there is a smooth proper morphism of schemes $\mathcal{X}\to \mathrm{Spec} \mathbb{C}[[t]]$ whose generic fibre is $X_{\mathbb{C}((t))}$ and whose special fibre is $X_0$.

abundance of "full" couples

Math Overflow Recent Questions - Tue, 01/16/2018 - 18:25

Let $G$ be a finite set, and let $\mathcal A$ be a subset of $\mathcal P(G)$, we denote by $max(\mathcal A)$ the set of maximal elements of $\mathcal A$ (for inclusion)

For any $A\in \mathcal A$ we denote by $\mathcal A_A=\left\{X\in \mathcal A,\, X\subsetneq A\right\}$

We say that $a\in G$ is full in $Y\in \mathcal A$, (or that $(a,Y)$ is a full couple) and write $F_{\mathcal A}(a,Y)$ if $a\in \bigcap max(\mathcal A_Y)$

and we define $F_{\mathcal A}(a,.):=\left\{Y\in \mathcal A,\,\, F_{\mathcal A}(a,Y)\right\}$


Does there exists a constant $C<1$ such that for any finite set $G$ and any $\mathcal A\subset \mathcal P(G)$ there exists $a\in G$, such that

$|F_{\mathcal A}(a,.)|/|\mathcal A|\leq C$ ?

For example if $G_1=\left\{a,b,c,d\right\}$, and $\mathcal A_1= \left\{\emptyset,\left\{a\right\},\left\{a,b\right\},\left\{a,b,c\right\},G_1\right\}$ then, for any $x\in G_1$ and any non empty $Y\in \mathcal A_1$, $x$ is full in $Y$ iff $Y\neq \left\{x\right\}$

Other example : if $\mathcal A_2$ is a finite down-set with ground set $G_2$, then $F_{\mathcal A_2}(x,.)$ is empty for any $x\in G_2$.

One can show that if $\mathcal A'$ is closed under intersection with ground set $G'$ then for ANY $x\in G'$ we have :

$|\left\{A\in \mathcal A', x\in A\right\}|\leq(|\mathcal A'|+|F_{\mathcal A'}(x,.)|)/2$

This last remark shows the link between the question and the Frankl Conjecture

Minimal cardinality of a field where a polynomial has a root

Math Overflow Recent Questions - Tue, 01/16/2018 - 13:19

Let $P(n)$ be the set of all monic polynomials of degree $n$ with integer coefficients, such that all coefficients have absolute value at most $2^n$.

Given a positive integer $n$ let us define $A(n)$ as the minimal number $m$ such that for any $f \in P(n)$ there exists a field $k$ with at most $m$ elements such that for some $x\in k$ we have $\bar f(x)=0$, where $\bar f$ is the reduction of $f$ modulo the characteristic of $k$.

Q: What is the asymptotic behaviour of $A(n)$? In particular is it true that $\liminf_{n} \frac{A(n)}{n} = 0$ ?

Regarding outer functions

Math Overflow Recent Questions - Mon, 01/15/2018 - 02:28

Please see the definition of Hardy spaces on the unit disc here. Let $0<p\leq\infty$. Let $f\in H^p$ with $\|f-1_e\|_p<1$ (Where $1_e$ Is the constant function one). Then is $f$ an outer function?

On prime numbers 2

Math Overflow Recent Questions - Sun, 01/14/2018 - 22:00

$ \text{Prove that } \forall p \in \text{Prime } \text{can have infinitely many zeros at the end} $ $ \text{Excluding the last digit} $

Why does adding a puncture to the moduli space of genus zero curves increase its dimension by $1$?

Math Overflow Recent Questions - Sun, 01/14/2018 - 22:00

Consider $\mathcal{M}_{0,n}$, the moduli space of genus zero curves with $n$-punctures.

Using a combination of the Kodaira-Spencer map, Riemann-Roch, and Serre Duality, I have calculated the dimension of $\mathcal{M}_{0,n}$ to be $n-3$.

Using such heavy theorems, I have lost intuition as to why adding a puncture increases the dimension of the moduli space.

Very roughly, using the interpretation of $\mathcal{M}_{0,n}$ as the configuration space of projectively equivalent $n$-tupels points in $\mathbb{P}^1$, I imagine that the configuration class becomes "finer" as we increase the number of punctures. However, my reasoning isn't all that satisfying.

Why is the dimension of the moduli space increasing by exactly one for each puncture?

On $\{0,\pm1\}$ and $\{0,1\}$ sequences arising from cyclotomic polynomials

Math Overflow Recent Questions - Sun, 01/14/2018 - 20:56

We know that the coefficients of the $n$th cyclotomic polynomial can be large.

However if $n=2^kp^rq^m$ where $p,q$ are odd primes then the coefficients are in $\{0,1,-1\}$.

Given cyclotomic polynomial $\Phi_{2^kp^rq^m}(x)$ denote $\Phi_{abs,2^kp^rq^m}(x)$ to be polynomial with coefficients being absolute value of coefficients of $\Phi_{2^kp^rq^m}(x)$.

Consider map $\pi_{c}:f(x))\rightarrow\Bbb Z^{deg(f(x))+1}$ where $x^t$ coefficient of $f(x)$ are mapped to $deg(f(x))+1$ length vector with integer coefficients at location $(t+c)\bmod deg(f(x))+1$ (the locations are $0$ through $deg(f(x))$).

Are there infinitely many $k,p,q,r,m$ at which

  1. at every $0\leq c<c'\leq deg(\Phi_{2^kp^rq^m}(x))$ we have $$\pi_c(\Phi_{2^kp^rq^m}(x))\neq \pi_{c'}(\Phi_{2^kp^rq^m}(x))?$$

From MTyson's comment it looks like $k=0$, $rm=0$ and $r+m=1$ is needed for this (that is $2^kp^rq^m$ is either $p$ or $q$). Nevertheless we have infinitely many $k,p,q,r,m$ at which $1$. holds.

  1. at every $0\leq c<c'\leq deg(\Phi_{2^kp^rq^m}(x))$ we have $0\leq c<c'\leq deg(\Phi_{abs,2^kp^rq^m}(x))$ do we have $$\pi_c(\Phi_{abs,2^kp^rq^m}(x))\neq \pi_{c'}(\Phi_{abs,2^kp^rq^m}(x))?$$

  2. $\Phi_{abs,2^kp^rq^m}(x))$ does not have more than $\frac{deg(\Phi_{abs,2^kp^rq^m}(x))}{(\log(deg(\Phi_{abs,2^kp^rq^m}(x))))!}$ contiguous repeating coefficient ($0,1$ varies sufficiently enough)?

  3. What are the distributions of $$\langle\pi_c(\Phi_{2^kp^rq^m}(x)), \pi_{c'}(\Phi_{2^kp^rq^m}(x))\rangle$$ $$\langle\pi_{c}(\Phi_{abs,2^kp^rq^m}(x)), \pi_{c'}(\Phi_{abs,2^kp^rq^m}(x))\rangle?$$

Permutations which avoid consecutive entries of the form (m,m+1)

Math Overflow Recent Questions - Sun, 01/14/2018 - 20:16

I found an interesting question on quora and need help in solving this question. I've just started understanding permutations but could not understand as to how I can come up with a general formula for this problem. Consider all permutations of the numbers 1 to n. A good permutation is one where for any number i at position p in the permutation, i+1 is never at position p+1. For a given n, count the number of good permutations.

For example, for n = 3, the good permutations are:

1, 3, 2

2, 1, 3

3, 2, 1

Come up with a form of an answer which can be easily calculated.

Commutative group algebraic spaces

Math Overflow Recent Questions - Sun, 01/14/2018 - 19:08

Is the category of commutative group algebraic spaces (commutative group objects in algebraic spaces) locally of finite type over a field, an abelian category?

I would benefit from a reference


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