Recent MathOverflow Questions

Which cluster algebras where the existence of maximal green sequences is still unknown?

Math Overflow Recent Questions - Mon, 05/08/2017 - 03:16

Maximal green sequences are studied in many papers. For example, Maximal Green Sequences for Cluster Algebras Associated to the n-Torus by Eric Bucher, On Maximal Green Sequences by Thomas Brüstle, Grégoire Dupont, Matthieu Pérotin, Minimal length maximal green sequences by Alexander Garver, Thomas McConville, Khrystyna Serhiyenko, Maximal green sequences for quivers of finite mutation type by Matthew R. Mills, Maximal Green Sequences of Exceptional Finite Mutation Type Quivers by Ahmet I. Seven.

For which cluster algebras is the existence of maximal green sequences still unknown?

Squareful values of polynomials

Math Overflow Recent Questions - Sun, 05/07/2017 - 13:04

Recall that an integer $n$ is called squareful if for every prime $p$ with $p \mid n$, we also have $p^2 \mid n$.

Any squareful number can be written uniquely as $n= x^2 y^3$ where $y$ is squarefree. From this, it is easy to see that $$\#\{ n \in \mathbb{Z}: |n| \leq X, \, n \text{ is squareful} \} \ll X^{1/2}.$$

I would like a version of this for polynomials.

Let $f \in \mathbb{Z}[x]$ be non-constant and separable. Then does there exist $\delta > 0$ such that $$\#\{ n \in \mathbb{Z}: |n| \leq X, \, f(n) \text{ is squareful} \} \ll X^{1 - \delta} \quad ?$$

Hopefully there are not some necessary local conditions here that I overlooked. In my application I am happy to change $f$ as required so that one can assume that $f$ is sufficiently "general". Moreover, I can even assume that $f$ is of very large degree if necessary to simplify things.

I would normally try to prove something like this using the large sieve, however the large sieve gives the poor upper bound $X/(\log X)$, whereas I would like a power saving.

If necessary, I'm happy to assume some standard conjectures (e.g. the abc conjecture).

A $K/k$-image of an abelian variety and finiteness of the Galois invariants

Math Overflow Recent Questions - Sat, 05/06/2017 - 01:49

Let $k$ be an algebraically closed field.

Let $A$ be an abelian variety over a primary extension $K$ of $k$ and consider a $K/k$-image, $\text{Im}_{K/k}(A)$.

Suppose that $\text{Im}_{K/k}(A)=0$.

Then, by the Mordell-Weil theorem $A(K)_{\text{tors}}$ is finite.

Now let $p=\text{char}(k)\nmid n$.

Why is it true that the points of order $n$ on $A(\overline{K})$ are separable over $K$ and why this implies that the invariants of the Galois group $\text{Gal}(\overline{K}/K)$ in $A(\overline{K})_{p-\text{tors}}$ are also finite?

Here $p-\text{tors}$ means torsion or order prime to $p$.

Does each compact topological group admit a discontinuous homomorphism to a Polish group?

Math Overflow Recent Questions - Sat, 05/06/2017 - 00:12

A compact topological group $G$ is called Van der Waerden if each homomorphism $h:G\to K$ to a compact topological group is continuous. By a classical result of Van der Waerden (1933) the groups $SO(2n+1,\mathbb R)$ are Van der Waerden, i.e., admit no discontinuous homomorphisms to compact topological groups.

On the other hand, Kallman proved that for every $n\in\mathbb N$ the Lie group $GL(n,\mathbb R)$ admits an injective homomorphism to the symmetric group $Sym(\mathbb N)$.

Problem. Does each infinite compact topological group $K$ admit a discontinuous homomorphism $h:K\to P$ to a Polish group $P$?

The answer to this problem is affirmative in three cases:

1) $K$ is solvable. In this case $K$ admits a homomorphism into an infinite abelian-by-finite group in which it is easy to construct a normal subgroup of countable index and the quotient homomorphism by this subgroup will be a discontinuous homomorphism to a countable discrete (and hence Polish) group;

2) $K$ is non-metrizable. In this case $K$ is not Wan der Waerden (see page 412 in the paper of Comfort) and hence $K$ admits a discontinuous homomorphism to some compact Lie group;

3) $K$ is not zero-dimensional. In this case it is possible to construct a continuous homomorphism $h:K\to O(n)\subset SL(n,\mathbb R)$ for some $n$ whose image $h(K)$ is not finite and being a Lie group is not zero-dimensional. Then the Kallman injective homomorphism $\varphi:k(K)\to Sym(\mathbb N)$ is necessarily discontinuous and so is the homomorphism $\varphi\circ h:K\to Sym(\mathbb N)$.

Those partial answers reduce the problem to the following its partial case.

Problem 0. Let $K$ be an infinite compact metrizable zero-dimensional topological group. Does $K$ admit a dicontinuous homomorphism to a Polish group?

Remark. If a compact topological group $K$ has no discontinuous homomorphism to a finite group, then for every $n\in\mathbb N$ it has only finitely many subgroups of index $n$. This can be shown using the limits by ultrafilters (see p.9 of An example of a compact group with finitely many subgroups of a given finite index is the product $\prod_{n\ge 5}A_n$.

Question. Let $(k_n)_{n=5}^\infty$ be a sequence of natural numbers. Does the compact group $\prod_{n\ge 5}A_n^{k_n}$ admit a discontinuous homomorphism to a Polish group?

How to minimize a homogeneous polynomial with unknown parameters?

Math Overflow Recent Questions - Fri, 05/05/2017 - 20:24

We are given a homogeneous polynomial $$f(x_1,x_2)=2x_1^4-6x_1^3x_2+5x_1^2x_2^2-2bx_1x_2^3-ax_2^4$$ where $a$ and $b$ are parameters. Find the largest scalar $a$ for which there exists a scalar $b$ such that the optimal value of the problem of minimizing $f$ over $\mathbb R^2$ is finite (i.e., not $-\infty$).

Please see the link above for the problem. I know the optimal value is 0 but how shall I go about finding the parameter values?

Subcategories of the Verdier quotient?

Math Overflow Recent Questions - Fri, 05/05/2017 - 19:41

Let $\mathcal T$ be a triangulated category and $\mathcal C$ a thick triangulated subcategory. We consider the Verdier quotient $\mathcal T/\mathcal C$.

Is there a bijective correspondence between thick triangulated subcategories of the quotient $\mathcal T/\mathcal C$ and thick triangulated subcategories of $\mathcal T$ containing $\mathcal C$?

It "feels" true, but I have learned never to take chances with technicalities of triangulated categories.

Random iteration of a set of monotone maps until fixed point

Math Overflow Recent Questions - Fri, 05/05/2017 - 18:35

Let $P$ be a poset with a least element $\bot$ ($\forall x \in P.\ \bot \le x$). Let $M$ be a set of monotone maps $P \to P$. Call $x \in P$ reachable if $x = f_1(f_2(...f_n(\bot)...))$ for some sequence $f_i \in M$. Call $x$ fixed if $x = f(x)$ for all $f \in M$.

If $x$ reachable, $y$ fixed, then $x \le y$. For then $x = F(\bot)$ for $F$ some composition of functions in $M$. Since $F$ is monotone and $\bot \le y$, we have $x = F(\bot) \le F(y) = y$. Thus there is at most one reachable fixed point.

Question #1: If there is a reachable fixed point $x$, will applying functions in $M$ uniformly at random (starting from $\bot$) eventually find $x$ with probability 1?

This requires a uniform probability distribution over $M$, so is ill-defined if $M$ is infinite. So:

Question #2: If there is a reachable fixed point $x$, will any schedule which eventually includes every sequence of $M$s infinitely often eventually find $x$ (starting from $\bot$)?

For our purposes, $M$ is at most countable. I'd even be interested in a solution to the case where $M$ is finite.

Representability of the sum of homology classes

Math Overflow Recent Questions - Fri, 05/05/2017 - 18:08

This is probably a very simple question, but I have not found it addressed in the references that I know. Let $M$ be a closed orientable $d$-dimensional manifold and let $[\alpha_{i}] \in H_{k}(M,\mathbb{Z}), i=1,2,$ be $k$-homology classes of $M$ that admit representations $\iota_{i}\colon Z_{i}\hookrightarrow M$ in terms of closed oriented submanifolds $Z_{i}$, that is:

$\iota_{i \ast}[Z_{i}] = [\alpha_{i}]$

Let us assume also that $[\alpha] := [\alpha_{1}] + [\alpha_{2}]\in H_{k}(M,\mathbb{Z})$ is also representable by a closed oriented submanifold $\iota\colon Z\hookrightarrow M$. Is there any relation between $Z$ and $Z_{i}$? Can $Z$ be "constructed" in terms of $Z_{1}$ and $Z_{2}$, for example by a connected sum operation $Z = Z_{1}\sharp Z_{2}$?


How to prove infinity inequality? [on hold]

Math Overflow Recent Questions - Fri, 05/05/2017 - 17:59

How can I prove or mathematically represent the idea that there infinitely more irrational numbers than there are rational numbers?

I've been sharing this idea with people as a conversation starter. It'd be nice to actually be able to express the idea in more concrete mathematical terms.

Optimal constant for Diophantine irrational number $n\alpha$

Math Overflow Recent Questions - Fri, 05/05/2017 - 17:38

Let $\alpha \in DC(C,\epsilon)$ be a Diophantine irrational; that is, $|q\alpha - p| > \frac{C}{q^{1+\epsilon}}$ for any $p,q \in \mathbb{Z}$. Then, for $n \in \mathbb{N}$, $|qn\alpha - p|> \frac{C}{(nq)^{1+\epsilon}} = \frac{C}{n^{1+\epsilon}}\cdot \frac{1}{q^{1+\epsilon}}$. This shows that $n\alpha \in DC(\frac{C}{n^{1+\epsilon}},\epsilon)$. I am wondering if $\frac{C}{n^{1+\epsilon}}$ is the optimal constant for $n\alpha$? That is, can we replace $\frac{C}{n^{1+\epsilon}}$ by a larger constant? I am particularly interested when $n = q_m$ for some $m$ where $q_m$ is the denominator for the $m$-th convergent of $\alpha$ (here, $\alpha = [a_0,a_1,a_2,\ldots]$, then $\frac{p_m}{q_m} = [a_0,a_1,\ldots,a_m]$).

Ideally, I would like to know if I can replace $\frac{C}{n^{1+\epsilon}}$ by a larger constant via replacing $\epsilon$ by a smaller constant. But, this might be too much of an ask. Still, I would like to know if the constant $\frac{C}{n^{1+\epsilon}}$ can be improved for both cases (1) $n =q_m$ and (2) arbitrary positive integer.

Thank you so much,

Ito lemma for manifold semimartingales

Math Overflow Recent Questions - Fri, 05/05/2017 - 17:23

I'm looking for a generalization of the usual Ito lemma to manifolds $M$, preferably not under the assumption that $M$ is embedded in $\mathbb{R}^d$. Unfortunately any reference I've found either assumes $M$ is a submanifold of Euclidean space.

Divisibility properties of Lucas sequences

Math Overflow Recent Questions - Fri, 05/05/2017 - 17:01

I'm trying to prove this theorem: If p is an odd prime not dividing PQ, then

  1. $u_{p-(D/p)} \equiv 0$ mod $p$,
  2. $u_p \equiv (D/p)$ mod $p$,
  3. $v_p \equiv v_1 \equiv P$ mod $p$ and if also gcd$(p,D)=1$, then
  4. $v_{p-(D/p)} \equiv 2Q^{(1-(D/p))/2}$ mod $p$.

I am able to prove the first, second and third congruence easily, and also half of the fourth one. If $(D/p)=1$, I can prove that $v_{p-1} \equiv 2$ mod $p$. But if $(D/p)=-1$, how can I prove that $v_{p+1} \equiv 2Q$ mod $p$? I can get to a point where I have this congruence: $2v_{p+1} \equiv P^2+D$ mod $p$, but the wanted congruence is $2v_{p+1} \equiv P^2-D$ mod $p$. I don't know if I'm making a mistake somewhere or if I'm missing some theorem that would solve this.

I use this information: The Lucas sequences with parameters P and Q are the two sequences {u$_n$} and {$v_n$} defined by $u_0=0$, $u_1=1$, $v_0=2$, $v_1=P$, $u_n=Pu_{n-1}-Qu_{n-2}$, $v_n=Pv_{n-1}-Qv_{n-2}$, for $n\ge2$. The recurrence polynomial associated to the Lucas sequences is $x^2-Px+Q$ and $D=P^2-4Q$ is the discriminant of the polynomial.

Brauer groups and field extensions

Math Overflow Recent Questions - Fri, 05/05/2017 - 16:26

Let $k$ be a field and $\mathrm{Br}(k)$ the Brauer group of $k$. Let $k \subset L$ be a field extension. Let $b \in \mathrm{Br}(k)$ and denote by $b \otimes L \in \mathrm{Br}(L)$ the base-change of $b$ to $L$.

If $b \otimes L = 0$, then does this exist a subextension $k \subset K \subset L$ such that $K/k$ has finite degree and such that $b \otimes K = 0$?

i.e. if $b$ is killed by some field extension $L$, then must $b$ be killed by some finite field extension of $k$ which is contained in $L$?

Lyapunov Functions [on hold]

Math Overflow Recent Questions - Fri, 05/05/2017 - 16:26



Hint: Find a,b,m such that $V(x,y)=ax^{2m}+by^{2n}$ is a Lyapunov function

So I have:


But I'm not sure how to proceed in picking my constants such that $\dot{V}<0$

Mountain Pass Theorem on a non-vector space

Math Overflow Recent Questions - Fri, 05/05/2017 - 15:46

(Reposted from Math StackExchange because this may be more suited for professional mathematicians.)

The Mountain Pass Theorem roughly says the following. Consider a differentiable functional $I: H \rightarrow \mathbb{R}$ from a Hilbert space $H$ to the reals which satisfies an appropriate compactness/properness condition (the Palais-Smale condition). Assume the following are true:

  • $I[0] = 0$;
  • There exist positive constants $r$ and $a$ such that $I[q] \geq a$ for all $\|q\| = r$;
  • There exists a $p$ (with $\|p\| > r$) at which $I[p] \leq 0$.

Then $I$ has a critical point $c$ at which $I[c] > 0$ (in fact $I[c] \geq a$).

My question is the following: can an analogous statement be made for functionals whose domain is not a vector space? For a concrete example (which is the one I care about), let $\mathcal{M}$ be the space of all differentiable Riemannian metrics on the (two-dimensional) sphere, and let $I: \mathcal{M} \rightarrow \mathbb{R}$ be a functional from $\mathcal{M}$ to the reals. Obviously $\mathcal{M}$ is not a vector space and so the Mountain Pass Theorem can't be applied. But say the following properties hold:

  • $I$ obeys some appropriate properness condition;
  • $I[g^\mathrm{round}] = 0$, where $g^\mathrm{round}$ is the metric of the round sphere;
  • There exists a neighborhood $U$ of $g^\mathrm{round}$ in $\mathcal{M}$ such that $I[g] > 0$ for any $g \in U \setminus \{g^\mathrm{round}\}$;
  • There exists a metric $\bar{g} \in \mathcal{M}$ at which $I[\bar{g}] < 0$.

Then it seems like one should still be able to apply the spirit of the proof of the Mountain Pass Theorem to conclude that $I$ must have some critical point $g^\mathrm{crit}$ at which $I[g^\mathrm{crit}] > 0$. Obviously making this statement precise would require understanding what the "appropriate properness condition" above is, but in principle it seems to me like the spirit of the theorem should still apply.

Does anyone know if this is indeed the case?

Concavity of product and ratio of sums

Math Overflow Recent Questions - Fri, 05/05/2017 - 15:30

Apologies if this question is not appropriate for MathOverflow. I have asked at Math.StackExchange without success.

Consider the function $f:\mathbb{R}^n\rightarrow \mathbb{R}$ defined as $$ f(x)=\left(\sum_{j=1}^{n}x_{j}\right)\left(1-\sum_{j=1}^{n}\frac{x_{j}}{1+\sum_{i=1}^{n}A_{ij}x_{i}}\right) $$ where $0\leq x_i\leq 1$ and $0\leq A_{ij}\leq 1$ are constants. Assume also the the second parenthesis is always positive.

I am looking for necessary conditions on the matrix $A$ such that $f$ exhibits some form of concavity (pseudo, log, quasi, standard). For instance, if $A_{ij}$ equals some constant then $f$ is log concave. I am hoping to find that the concavity result extends to more general $A$'s.

Bialgebra pairing on ring polynomial $K[x]$

Math Overflow Recent Questions - Fri, 05/05/2017 - 12:56

I need your help on how to show the existence of a bialgebra pairing: for the polynomial ring $k[x]$ over a field $k$, there is a bialgebra pairing $t:k[x]\otimes k[x]→k$ such that $t(x,x)=1$. What is the unique bialgebra pairing satisfying $t(x,x)=1$?

Exponentially weighted spaces: Effect on spectrum

Math Overflow Recent Questions - Fri, 05/05/2017 - 10:13

My question is somewhat broad, but I do not know how to precisely state the issue. I am investigating stability of certain class of scalar PDE on $\mathbb{R}$.

Previous work in this topic has introduced exponentially weighted space to look into this, e.g.


I am trying to understand the effect of introducing such spaces on the discrete and continuous spectrum of a Linear operator (coming from linearizing the said PDE).

1). For example, does the point spectra remain same ?

Are there any references (papers/books/monographs) which discuss this issue ?

2). A related question is this. If we work in $H_a$, a natural way to tackle is introduction of Hermite basis, which is countable.

If we work in $H_b=L^2[\mathbb{R},dx]$, one would use Fourier transform and have an uncountable number of basis functions. Does this difference in number of basis relate to 1).

To give a specific example, one can consider the following operator $L$, where


If one is trying to find the essential spectrum of this operator acting on $H_b$, then the usual technique of looking at dispersion relation of asymptotic operator (i.e. $x\rightarrow \pm\infty$) fails since the first term in the operator blows up.

So there is hope (?) that if we work in exponentially decaying weight, i.e work in $H_a$ instead, we can again use this technique of looking at asymptotic operator, however I am not sure how to justify this or actually prove anything here.

Estimate of heat kernel via blowing up of a singular variety

Math Overflow Recent Questions - Fri, 05/05/2017 - 10:12

We know from the book of Aubin:Let $(X, \omega)$ be an $n$-dimensional compact Kahler manifold and $\pi : Y \to X$ a blowing up with non-singular center. Fix an arbitrary Kahler metric $\theta$ on $Y$ and set $\omega_\epsilon = \pi^*\omega+ \epsilon\theta$ for $0 < \epsilon \leq 1$. Let $H_\epsilon$ be the heat kernel with respect to the metric $\omega_\epsilon$, then we have a Cheng-Yau uniform estimate $0\leq H_\epsilon \leq C(t^{-n} + 1)$ with a positive constant $C$ outside of orthogonal set.

My question is

  • Which type of bound we can get for the heat kernel with respect of $\omega_\epsilon$ on orthogonal part which is non-smooth

Bi-Lipschitz version of Kirszbraun's extension theorem

Math Overflow Recent Questions - Fri, 05/05/2017 - 07:30

Kirszbraun's theorem for $\mathbb{R}^2$ states the following:

Given any set $S\subset \mathbb{R}^2$ and any Lipschitz function $f:S\rightarrow \mathbb{R}^2$ with Lipschitz constant $k$, $0< k< \infty$, for any set $F$ which contains $S$ there exists a function $\tilde f:F\rightarrow C$ such that $Lip(\tilde{f})=k$, $\tilde{f}|_{S}=f$ and $C$ is any closed convex set containing $f(S)$.

I'm wondering if something similar could hold true for bi-Lipschitz functions, for example is the following claim true?

Given any two discrete sets of $n\ge 3$ linearly independent points $S_1,S_2\subset \mathbb{R}^2$, and any bi-Lipschitz function $f:S_1\rightarrow S_2$ with bi-Lipschitz constant $k$, $0<k<\infty$, there exists a function $\tilde{f}:C_1\rightarrow C_2$ which extends $f$ and has bi-Lipschitz constant $k$, where $C_i$ is the convex envelope of $S_i$.

This problem seems more complicated than what I expected since the only article I've found on this subject is this which computes the bi-Lipschitz extension only in the case $S$ is the border of the unit square and the extended function has the bi-Lipschitz constant multiplied by a "big" constant factor.

Do you know any other reference on this problem? Maybe one should start with the simpler case: is my claim true when $S$ is composed of 3 linearly independent points? (even this doesn't seem trivial to me)


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