Recent MathOverflow Questions

By an unorientable Riemann surface C, I mean a compact unorientable two-manifold without boundary that is endowed with a conformal structure.

Such objects have a moduli space that is somewhat like the more familiar moduli space of oriented Riemann surfaces, but this moduli space is itself unorientable. See corollary 2.3 of https://arxiv.org/abs/1309.0383, where this is proved with a simple explicit example.

My question is this: suppose that C is endowed with a pin^+ structure. (This is one of the two possible analogs of a spin structure in the unorientable case, the other being a pin^- structure.) I've come to suspect that the moduli space of unorientable Riemann surfaces with a pin^+ structure is itself orientable. I wonder if this is a known result and where it might be found.

Question: Let $\Gamma$ be a countable group. Is the intersection of two thickly syndetic sets still thickly syndetic?

I've only seen the proof for the group $\mathbb{Z}$ (and I believe this method generalises to amenable groups). (there is an "extra" question below)

Background:

• a set $S \subset \Gamma$ is syndetic if there is a finite $A \subset \Gamma$ to that $AS = \Gamma$.
• $T$ is thick if for all finite $F \subset \Gamma$, $\cap_{f \in F} fT \neq \emptyset$.
• $U$ is thickly syndetic, if for any finite $F \subset \Gamma$, $U' = \cap_{f \in F} fL$ is syndetic.

Partial answer: It's an exercise to see that if $S$ is syndetic and $T$ is thick, then their intersection is non-empty. Indeed, if $A$ is the set for which $S$ is syndetic, then $$ \begin{array}{rl} \displaystyle \bigcap_{a' \in A} a'T &= \displaystyle \Gamma \cap \big( \bigcap_{a' \in A} a'T \big) \\ &= \displaystyle \big( \cup_{a \in A} aS \big) \cap \big( \bigcap_{a' \in A} a'T \big)\\ &= \displaystyle \bigcup_{a \in A} \bigg( aS \cap \big( \bigcap_{a' \in A} a'T \big) \bigg)\\ &= \displaystyle \bigcup_{a \in A} a \bigg( S \cap \big( \bigcap_{a' \in A} a^{-1}a'T \big) \bigg)\\ & \subseteq \displaystyle \bigcup_{a \in A} a \big( S \cap T \big) \end{array} $$ Since $T$ is thick the $S \cap T$ is non-empty. Now if $T$ is thickly syndetic, then $S \cap T$ is even syndetic. However this argument does not yield thickly syndetic.

extra question: it's another exercise to show that a set $S$ is syndetic iff its intersection with any thick is non-empty. Likewise a set is syndetic iff its intersection with any thick is non-empty. Is a set thickly syndetic iff its intersection with any thickly syndetic set is thickly syndetic?

Let $f_k$ be the number of $k$-sided faces in a polyhedron and $v$,$e$ numbers of vertices and edges. Assume that all vertices are $3$-valent. Then by Euler characteristic and elementary combinatorics $$\sum_k f_k -e+v=2\quad \text{and}\quad 2e=3v=\sum_k kf_k.$$ What else can be said about $e,v,f_3,f_4,...$? I am pretty sure the above equalities do not characterize $e,v,f_3,f_4,...$ completely. For example, the above implies that $$\sum_k (k-6)f_k=12,$$ but I suspect that there is an upper bound on $f_6$ in terms of $f_3,f_4,f_5.$

More generally, I am interested in these questions for $3$-valent topological graphs $\Gamma$ in $S^2$ ("faces" being the connected components of $S^2-\Gamma$). All of the above applies here, but $f_1$ and $f_2$ maybe non-zero and an edge maybe a side of the same face twice.

I am trying to determine the eigenvalues and eigenvectors of the following matrix:

$$M_{ij} = 4^{-j}\binom{2j}{i}$$

where it is understood that the binomial coefficient $\binom{m}{k}$ is zero if $k<0$ or $k>m$. The indices $i,j$ traverse a discrete finite range, $i,j \in \{a, a+1, \dots, b\}$, from $a$ to $b$, where $a,b$ are non-negative integers with $0\le a\le b$. Therefore the matrix $M_{ij}$ has dimensions $(b-a+1) \times (b-a+1)$.

I would content myself with an approximate expression (if there is no exact analytical result), valid for large $a,b$. I am mostly interested in the largest eigenvalue (not in absolute value, but the largest positive eigenvalue) and the corresponding eigenvector.

Also a recurrence relation would be useful. Anything that helps...

To be explicit, we want to solve the following eigenquation:

$$\sum_{j=a}^b 4^{-j} \binom{2j}{i} x_j = \lambda x_i,\quad i=a, a+1, ..., b$$

for the eigenvalues $\lambda$ and eigenvectors $x_i$.

given graph

Description:

Coelan would like to cross river isar,

there are 13 bridges connencting left river bank l and right river bank r, between two banks there are 6 insels a, b, c, d, e and f.

however, after flood, every bridges is independently destroyed with probability 1/2.

Determine the probability which there is an intact path for Coelan to cross the river.

i found there are too many cases to calculate the probability if i list all possibilities, (which should be Markov chains), at the moment i tried to solve it by using indicator variable(is it in the right way?)

and i was wondering whether there is a general way to solve this kind of "random" graph problem?

I study PDEs that arise in fluid dynamics in an infinite dimensional Riemannian geometric perspective. For example, Ebin-Marsden(1970) showed that the group of volume preserving diffeomorphisms has an infinite dimensional Banach manifold structure, and the solution of Euler equations for perfect fluid can be described as a geodesic equation. In this perspective, the Euler equations can be realized as a stationary point of the energy functional, i.e. an Euler-Lagrangian equation.

Now, there is another infinite dimensional space that I'm interested in studying, which is Met(M), the space of Riemannian metrics on M. Many people already worked on this space and it has an infinite dimensional manifold structure as well.

From the Exterior Differential System(EDS), we know that a given PDE is an Euler-Lagrange equation if and only if it satisfy the Helmholtz condition. This is a criterion on the finite dimenisonal setting. So my question is, is there is an infinite dimensional analogue of the Helmholtz condition to test whether a given PDE is an Euler-Lagrange equation or not? Thanks in advance!

Let $A$ and $B$ two Hermitian matrices and $D$ a diagonal matrix. Does exist any inequality involving the trace for this diagonal term:

$$ (D A D A D A B)_{i,i} $$

I am looking for something like that: $$ |(D A D A D A B)_{i,i}| \leq | \text{Tr}\left(D^3\right)^{p} \text{Tr}\left(A^3\right)^{q} \text{Tr}\left(B\right)^{r} | $$

Thank for your help.

The $C_{\le k}$-free 2-factor problem asks whether an input graph $G$ admits a 2-factor which forbids cycles with length $\le k$. It is a relaxation of the Hamiltonian cycle problem. Hell et al shown that the problem is NP-hard unless the set of forbidden cycle lengths $F$ is a subset of $\{3,4\}$.

It is known that the problem is polynomial time decidable for $F=\{\phi\}$ and $F=\{3\}$ . However, the complexity is still unknown for the cases where $F=\{4\}$ (square-free 2-factor problem) and $F=\{3,4\}$. I wonder if restricting input graphs to be cubic would resolve these two open problems.

What is the complexity of the problem for cases $F=\{4 \}$ and $F=\{3,4\}$ if input graph is cubic?

Let $K$ be a field of characteristic zero, $\bar{K}$ its algebraic closure and $X$ a smooth, projective $K$-scheme. We know the Galois descent theory for quasi-coherent sheaves defined on $X_L$ for a finite extension $L$ of $K$. This gives a cocycle condition on a quasi-coherent sheaf on $X_L$ to descend to $X$ (see for example page 139, section 6.2 of "Neron Models" by S. Bosch and others).

I am looking for an analogous result for the absolute Galois group i.e., if we take a coherent sheaf $E$ on $X_{\bar{K}}$ which satisfies the analogous cocycle condition for any pair of elements $\sigma, \tau \in \mathrm{Gal}(\bar{K}/K)$, does there exist a (quasi)-coherent sheaf $F$ on $X$ such that $F \otimes_K \bar{K} \cong E$?

Any hint/reference on this topic will be most helpful.

Is it enough to say that both sets are infinite? or is there any other way to prove it?

Within my research I found an important doubt and that prevents me from advancing, the context of my doubt is as follows:

We considerer the following optimization problem $$ \left\{\begin{array}{cl} \max\limits_{x\in\mathcal{C}} & f(x) \\[2pt] \text{s.t.} & \mathcal{A}x-b \in K \end{array} \right. \tag{1} $$

where $f:\mathbb{R}^{n}\rightarrow \mathbb{R}$ and $\mathcal{A}:\mathbb{R}^{n}\rightarrow \mathbb{R}^{m}$ are linear functions non-zero, $b\in\mathbb{R}^{m}$, $\mathcal{C}$ is a convex cone in $\mathbb{R}^{n}$ and $K$ is a closed convex cone in $\mathbb{R}^{m}$.

Question: If $\mathcal{C}$ is the convex cone generated by a set $U\subset\mathbb{R}^{n}$, can we say that problem $(1)$ is equivalent to the following problem $$ \left\{\begin{array}{cl} \max\limits_{x\in U} & f(x) \\[2pt] \text{s.t.} & \mathcal{A}x-b \in K \end{array} \right. \tag{2} ?$$

My Idea: My idea is to show that if there is an optimal point $x^{*}\in\mathcal{C}$ for the problem $(1)$ then that point must belong to $U$. The problem is the condition $\mathcal{A}x-b \in K$, I feel that I must put conditions in $ K $ because I see no guarantee that points in $A$ will satisfy the condition, I do not know if I am wrong.

Let $\xi_1,\xi_2,\ldots$ be i.i.d. positive random variables with infinite mean $\mathbb E[\xi_i]=\infty$, and consider the random walk $$T_n=\sum_{i=1}^n\xi_i,\qquad n\in\mathbb N.$$ Here's an example that illustrates why I'm interested in such questions.

Example. If $(S_n)_{n\in\mathbb N}$ is a simple symmetric random walk on $\mathbb Z$, then $\xi_1,\xi_2,\ldots$ could represent the time intervals between consecutive returns to zero of $S_n$.

Q. Does there exist general local limit theorems for the random walk $T_n$?

In the case of the example I have above, everything can be computed explicitly quite nicely, and it can be argued that $$n\cdot \mathbb P\{T_{\sqrt{n}x}=n\}\asymp xe^{-x^2/2}.$$ I'm interested in more general statements of this form.

This question already has an answer here:

• bounding the trace of a matrix product by the operator norms; generalized Hölder inequality? 2 answers

Is there an equality for Hermitian matrices $H_1,\dots,H_n$ such as the following

$$ \left| \text{Tr} \left( \prod_{i=1}^n H_i \right) \right| = \prod_{i=1}^n |\text{Tr}(H_i^{p_i})|^{\frac{1}{p_i}} $$

for $\sum\limits_{i=1}^n \frac{1}{p_i} = 1$?

Thank you for your help.

Let $W$ be a Coxeter group (finite or infinite) with (finite) set $S$ of Coxeter generators, and let $I \subseteq S$ be some subset. If $w\in W$ then I call $m_I(w)$ the minimum total number of occurrences of elements of $I$ in some expression for $w$ as a product of elements of $S$, or formally: $$ m_I(w) := \min\{m : \exists x_1,\ldots,x_l\in S\; (w = x_1\cdots x_l \,\land\, m=\#\{i : x_i\in I\})\} $$ Evidently, $m_S$ is the usual length function on $W$, and also, $m_I(w)$ is the minimum total number of occurrences of elements of $I$ in some reduced expression for $w$ (since any expression for $w$ has a reduced subexpression).

I don't know anything else about $m_I$, including:

Question 0: Does this function have a standard name?

Now consider the would-be-Poincaré series associated to $m_I$, namely:

$$ P_I(q) := \sum_{w\in W} q^{m_I(w)} $$

This doesn't make sense in general, but if the ("parabolic") subgroup generated by $S\setminus I$ is finite, which I now assume, then there are only finitely many $w \in W$ having a given value of $m_I$, and $P_I \in \mathbb{Z}[[q]]$.

Question 1: Is this enumerating function rational? (I.e., does it belong to $\mathbb{Q}(q)$?)

Question 2: Assuming yes, how can I compute (a rational form for) it algorithmically?

I'm thinking maybe there's a standard reduced form for elements of $W$ that guarantees that they have the minimum number of elements of $I$, and maybe this standard form is recognizable by a finite automaton, which would be very much in line with other results about Coxeter groups, but I didn't find anything (presumably for lack of knowledge of the correct name for $m_I$). Of course, it would be even better to be able to compute $P_I$ without going through an automaton (along the lines of proposition 7.1.7 in Björner & Brenti's book Combinatorics of Coxeter Groups).

Let $X$ and $Y$ are Calabi-Yau varieties and mirror to each other. Then from HMS the Fukaya Floer category of Lagrangian intersections in $X$, is equivalent to bounded derived category of coherent sheaves on $Y$. Candelas showed that the number of rational curves in Fermat CY threefold can be calculated by periods of its mirror.

Now instead Calabi-Yau varieties, take $X$ and $Y$ are Fano or of general type and mirror to each other, then what information mirror part give us? How can we formulate HMS for Fano variety? Is there any SYZ conjecture when we are in deal with Fano variety or varieties of general type?.

What is the geometric interpretation of Large Complex Structure Limit of varieties of Fano or general type family ?

Kelly's theorem states:

Every finite point set of complex space such that the line joining any two points from this set contains at least one more point from this set (every Sylvester-Gallai configaration) is confined to the plane.

The proof is rather short, however it used rather deep result (Bogomolov–Miyaoka–Yau inequality, more accurately, its corollary of F. Hirzebruch).

Are there more simpler proof of Kelly's theorem?

For which $n$ we may mark $n$ red and $n$ blue points on the Euclidean plane, not all on a line, so that any line which passes through two points of different colour contains another point?

For $n=1991$ this was proposed in a not-up-to-date edition of Prasolov's problem book on planimetry, but the suggested solution actually solves a different problem (in the newest edition this is fixed.)

I am having some difficulties with the following limit

$$ L(\alpha) = \lim_{N\to \infty} \frac{(2N)!!}{(2N-1)!!}N^2 \left( \frac{2N-1}{2N} \int_{\frac{1}{2}}^{\infty} \frac{1}{x^\alpha} \cos(\pi x)^{2N-2}dx - \int_{\frac{1}{2}}^{\infty} \frac{1}{x^\alpha} \cos(\pi x)^{2N}dx\right)$$ with $\alpha >1$ Can someone help me out? Is it even convergent? Are there any particular values (i.e positive integers) of $\alpha$ for which it is possible to express $L(\alpha)$? Or a nicer form for $L$ ? Thank you!

I obtained this limit upon writing

$$ \zeta(\alpha) = \lim_{N \to \infty} \frac{(2N)!!}{ (2N-1)!!}\int_{\frac{1}{2}}^{\infty}\frac{1}{x^\alpha} \cos(\pi x)^{2N} dx$$ and integrated two times by parts in my attempt to find a relation between $\zeta(\alpha+2)$ and $\zeta(\alpha)$

Let $$z_{2N}(\alpha) = \frac{(2N)!!}{ (2N-1)!!}\int_{\frac{1}{2}}^{\infty}\frac{1}{x^\alpha} \cos(\pi x)^{2N} dx \to \zeta(\alpha)$$ for $N \to \infty$, hence $$L(\alpha) = \lim_{N \to \infty} \frac{z_{2N-2}(\alpha) - z_{2N}(\alpha)}{\frac{1}{N^2}} = \frac{0}{0}$$ ...

If I got it right, there is a theorem due to M. Artin that on a projective complex manifold any point has a Zariski open neighbourhood which is a $K(\pi, 1)$ space. I have two questions about it.

1. Is there a proof of this result written in English? (I am really bad at French.)

2. Is it true that a point has an arbitrarily small neighbourhood with this property (i.e. smaller than a given one)?

What is a complete description for the configuration of zero locus of the algebraic curve $C$ defined by $$yP(x,y)-xQ(x,y)=0$$

where $P,Q \in \mathbb{R}[x,y]$ are arbitrary polynomials of degree $2$.

What is the (sharp) maximum number of connected components of $\mathbb{R}^2 \setminus C$?

This question is motivated by "EDITED" part of the following answer: Finding a 1-form adapted to a smooth flow