# Recent MathOverflow Questions

### A question related to bousfield localization and nilpotent completion

Math Overflow Recent Questions - Tue, 06/04/2019 - 11:29

I am reading Bousfield's paper entitled "The localization of spectra with respect to homology" (MSN). In that paper, Corollary 6.13 states that, if a ring spectrum $E$ with countable homotopy and satisfies some vanishing conditions in the associated Adams spectral sequence, then the localization is equivalent to nilpotent completion.

So, my question is the following:

What are the ring spectra $E$ (I need an updated list) satisfying the above conditions?

### Is the strong topology of a locally convex space always barrelled?

Math Overflow Recent Questions - Tue, 06/04/2019 - 11:07

For a locally convex space $E$ let $E_\beta$ be the space $E$ endowed with the locally convex topology $\beta(E,E')$ whose neighborhood base at zero consists of barrels, i.e., closed absolutely convex absorbing sets.

Observe that a space $E$ is barrelled if and only if $E=E_\beta$.

Question 1. Is $(E_\beta)_\beta=E_\beta$ for any locally convex space? Equivalently, is the space $E_\beta$ always barrelled?

If not, then we can ask a more restricted version of Question 1.

Question 2. Let $E$ be a locally convex space such that the space $E_\beta$ is normable (and separable). Is $E_\beta$ barrelled?

### Practical example of Hamiltonian reduction

Math Overflow Recent Questions - Tue, 06/04/2019 - 09:11

I know what is the Liouville integrability: given a Hamiltonian with $n$ degrees of freedom, with $n$ independent constants of motion in involution, the Hamiltonian can be brought to the form $H(p_1, \dots, p_n)$ (i.e. independent on the $q$s) by a canonical transformation.

Many persons told me that something similar can be done in the case of one constant of motion, i.e., given a Hamiltonian, if we know that it has one constant of motion (independent on $H$ itself), then we bring the Hamiltonian to the form $H(p_1, q_1, \dots, p_{n-1}, q_{n-1}, p_n)$ (i.e. we remove the dependence on one of the $q$s, $q_n$) by means of a canonical transformation.

It should be called "Hamiltonian reduction", however, no one was able to provide me a reference.

Someone says that this is called "Marsden-Weinstein-Meyer reduction". I tried to read the corresponding theorem but it looks something different, moreover, it is written in a very technical way, too hard for a simple physicist like me. I found this, saying "reduction by stages", but it refers to a completely different idea of "stages", not the reduction of the degrees one by one.

So, I'm looking for one of the following: a reference where I can find the theorem (clearly stated in terms of Hamiltonians and constants of motion), or an example of how to carry out the reduction (also in this case, giving an explicit Hamiltonian and an explicity constant of motion).

### Describing fiber products in stable $\infty$-categories

Math Overflow Recent Questions - Tue, 06/04/2019 - 08:25

Let $f\colon X \rightarrow Z$ and $g\colon Y \rightarrow Z$ be two morphisms in a stable infinity category $\mathcal{C}$. How does one show that the $\infty$-categorical fiber product $X \times_Z Y$ can be canonically identified with the fiber of the map $f - g \colon X \oplus Y \rightarrow Z$?

I am sure that the question is quite basic (for instance, something quite similar is used implicitly in the proof of Proposition 1.1.3.4 of Lurie's "Higher algebra"), but I would nevertheless appreciate an explanation. The type of answer I am not looking for is "this is trivial consequence of the definitions" because I believe that there is something to check: $X \times_Z Y$ makes sense in any $\infty$-category that has fiber products, whereas the other description uses the stability assumption.

While we are at it, let me ask a related bonus question: with $f$ and $g$ as above, suppose that we also have maps $a\colon F \rightarrow X$ and $b\colon F \rightarrow Y$ that make the evident diagram commute and such that $\mathrm{fib}(a) \simeq \mathrm{fib}(g)$ via the induced map (a five lemma type of setting where one of the vertical maps is an isomorphism). Do these conditions imply that the map $F \rightarrow X \times_Z Y$ is an isomorphism?

### Shortest possible good codes?

Math Overflow Recent Questions - Tue, 06/04/2019 - 06:46

Good codes (those with positive rate $r=k/n$ and positive relative distance $\delta=d/n$) will achieve capacity on $BSC$ (binary symmetric channel) if the codes have lower rates than capacity where positive relative distance is seen. However this requires very long codes to drive the error to reasonably low value.

To achieve an error rate of $e$ if capacity is $C$ then what is the shortest good code that is possible over $BSC$ as a function of $e$? I am just looking for an upper bound and a lower bound.

### Holomorphic versus algebraic $\mathbb C^*$-actions

Math Overflow Recent Questions - Tue, 06/04/2019 - 05:06

I believe that the following is true:

Statement. A holomorphic $\mathbb C^*$-action on a complex projective manifold is algebraic if and only if it has a fixed point.

Where can I find a proof of this statement? Is it really true that no one ever wrote down this statement in such a form?

### Karamata's proof of Hardy-Littlewood Tauberian theorem

Math Overflow Recent Questions - Tue, 06/04/2019 - 03:37

I understand Karamata's proof of the Hardy-Littlewood Tauberian theorem as in http://individual.utoronto.ca/jordanbell/notes/karamata.pdf, but what on earth is the motivation behind Lemma 4 - i.e, what would be the motivation to look at the space of all functions $g(x)$ for which $\displaystyle \lim_{x \rightarrow 1^-} (1-x)\sum_{0 \leq i} a_i x^i g(x^i) = \int_{0}^{1} g(x) dx$ ?

Trying to prove the theorem by myself first then failing and seeing the proof, I find that to be coming out of nowhere and miraculously solving the problem.

### Leafwise de Rham cohomology (A true definition of differential forms along leaves)

Math Overflow Recent Questions - Tue, 06/04/2019 - 03:07

For a foliated space $(M, \mathcal{F})$, one associate a leafwise de Rham cohomology. This cohomology and trace-class operators on this cohomology and trace interpretations for closed orbits of certain flow on $M$ is the main object of this paper "Number theory and dynamical system of foliated manifolds.

But in the later paper, I did not find a very precise definition of "Differential forms along a leaf".

So I try to find other papers or talks to find a precise definition for this concept. Then I found a definition at page 8 of this talk "Lefschetz trace formula for flow on foliated manifolds" which gives a local representation for such forms. But my problem is the following:

I think that such representation of a differential form along leaves of a $k$-dimensional foliation of a $n$-manifold, which is quoted below, is NOT invariant under foliation charts $(x,y)\mapsto (f(x,y),g(y)),\quad x\in \mathbb{R}^k, y\in \mathbb{R}^{n-k}$:

$$\omega=\sum_{\alpha_1<\alpha_2<\ldots<\alpha_k} a_{\alpha}(x,y) dx_{\alpha_1}\wedge dx_{\alpha_2}\wedge \ldots\wedge dx_{\alpha_k}.$$

Am I mistaken?

What is a precise definition and precise local representations of "Differential forms along leaves"?

### Upper bound on the length of chordless cycles in d-regular graphs

Math Overflow Recent Questions - Tue, 06/04/2019 - 00:52

Given a $d$-regular graph with $n$ vertices is there a known (non-trivial) upper bound on the length of chordless cycles in it (presumably as a function of $d$ and $n$)? I wasn't able to find anything after some online searches. Thank you.

### Up and down (sequence defined by recurrence relation)

Math Overflow Recent Questions - Mon, 06/03/2019 - 10:36

This question is based on previous one by me, which has been easily solved by Will Sawin. First, lets redefine some functions in compliance with their purpose ($\ell$ for logarithm, $p$ for product and $s$ for sum) $$\ell_k(n)=\left\lfloor\log_{k}n\right\rfloor=\begin{cases} 0,&\text{if n<k}\\ \ell_k\left(\left\lfloor{n \over k}\right\rfloor\right)+1,&\text{otherwise} \end{cases}$$ also $$p_k(n)=\prod_{i=0}^{\ell_k(n)} (1+\frac{1}{q}a_k\left(\left\lfloor{n \over k^i}\right\rfloor\right))$$ and finally $$s_k(m) = \sum_{n=0}^{k^m-1}p_k(n)$$ So we can see new parameter $q\in\mathbb{N}$ and we also need to redefine $$a_k(n)=a_k\left(\left\lfloor{n \over k}\right\rfloor\right)+(n \operatorname{mod} k), \qquad a_k(0)=0.$$ In fact, this sequence can be interpreted as sum of digits of $n$ in base $k$. What is nice here, it is the fact, that for $k=2$ $$s_2(m)=\frac{1}{q^m}\sum_{j=1}^{m+1} (j-1)! {m+1 \brace j}T_{q-1}(j)$$ with $$T_{k}(n)=T_{k-1}(n)+2\sum_{j=1}^{n-1} T_{k-1}(j), \qquad T_0(n)=n.$$ There exist A064984, where this polynomials interpreted as "counting up to $n$ and down again" (exactly for $T_1(n)$).

How can one prove it? How it can be extended for $k>2$? Is there a way to nicely define $T_k(n)$ in terms of Bernoulli or Eulerian numbers?

### Finiteness of $H_1 \backslash G / H_2$ and the geometry of the orbits

Math Overflow Recent Questions - Sun, 06/02/2019 - 13:40

Let $G$ be a connected reductive group over an algebraically closed field $k$. By the Bruhat decomposition, $P \backslash G/P \cong W_P \backslash W / W_P$ is a finite set for any parabolic subgroup $P$ of $G$. So there is some connection between the combinatorics of the Weyl group and the geometry of the orbits.

Now consider two closed subgroups $H_i$ of $G$.

When is $H_1 \backslash G / H_2$ finite? In such cases, can we describe the $H_1$-orbits in $G/H_2$ using some group theoretic data or combinatoric data in general?

If $H_1$ is a Borel subgroup, the finiteness is equivalent to $G/H_2$ being a spherical variety. And things become trivial if $H_1=G$.

If $H_1=H_2$, is the finiteness equivalent to $H_1=H_2$ being a parabolic subgroup? For low rank $G$, can we classify all such pair $(H_1,H_2)$?

### Are Turaev-Viro invariants holonomic?

Math Overflow Recent Questions - Sat, 06/01/2019 - 17:19

Consider a 3-manifold $M$ with a boundary, which is a genus $g\geq 1$ surface $\Sigma$. Fix a triangulation $T$ of $\Sigma.$ Then Turaev-Viro invariants $TV_q(M)$ are functions, assigning to integer labelings of the edges of $T$ certain $q-$polynomials.

For a closed $3-$manifolds $q$ needs to be a root of unity for the invariant to be defined, but I think that for a manifold with boundary it can be viewed as a formal variable.

Question: is function $TV_n(M)$ known (expected) to be $q-$holonomic?

### Distribution of inner product involving tensor product of Gaussian distributions?

Math Overflow Recent Questions - Sat, 06/01/2019 - 16:36

Take $\alpha\in[-1,+1]^k$ and $\beta\in\{-1,+1\}^k$ and take $d(\sigma)\in\mathbb R^k$ where each $d(\sigma)_i\in\mathcal N(0,\sigma^2)$.

How is $\langle\alpha\otimes\beta,d(\sigma_1)\otimes d(\sigma_2)\rangle$ distributed and what is its mean and variance?

### Is bounded graph isomorphism $NP$ complete?

Math Overflow Recent Questions - Sat, 06/01/2019 - 15:58

Fix a matrix $M\in(\mathbb Z\backslash\{0\})^{n\times n}$ where $\|M\|_\infty\leq 2^{poly(n)}$.

Given symmetric $A,B\in\{0,1\}^{n\times n}$ and $U,V>0$ is there a permutation $P$ with $$PAP'=B$$ $$U<\|P\odot M\|_F<V$$ where $\odot$ is Hadamard product and $F$ is Frobenius norm?

### Weight monodromy conjecture for surfaces?

Math Overflow Recent Questions - Sat, 06/01/2019 - 15:41

Related question: weight monodromy conjecture for curves?

In order to compute the local zeta functions of two classes of Shimura surfaces at primes of bad reduction, Rapoport and Zink proved the weight monodromy conjecture in the case of a smooth projective surface over a mixed characteristic local field (and other things). The reference is U ̈ber die lokale Zetafunktion von Shimuravarieta ̈ten. Monodromiefiltration und verschwindende Zyklen in ungleicher Charakteristik (German), Invent. Math. 68 (1982), no. 1, 21–101.

I don't know any German language hence have difficulty reading the orginal reference. What is the idea of the original proof (or is there any other good reference)? Is there a simple proof of the weight monodromy conjecture in the case of a smooth projective surface over a mixed characteristic local field now?

The excellent proof by Scholze works in the case of complete intersection inside a smooth projective toric variety. But I don't know whether every surface is of such kind, for example how about Enriques surfaces?

### Projectile motion function [on hold]

Math Overflow Recent Questions - Sat, 06/01/2019 - 15:20

An artillery crew is trying to hit multiple different targets over the course of a battle. Where and when these targets show up is initially unknown but they all appear at different ranges (x) and different heights (h). The crew needs to calculate the angle ( Θ ) to aim the gun at to hit a target. Assume the velocity (v) is 100m/s and gravity (g) is 7.5.

Find the equation for Θ where x and h can remain variables so that the gun crew can use them to deduce the angle they need to aim their gun at.

I really need this to be cookie cutter, plug in the x and h and it rolls out theta if you crunch the numbers.

### How can i determine the elements of this matrix? [on hold]

Math Overflow Recent Questions - Sat, 06/01/2019 - 13:04

i am new to matrices, i am learning with a book but i am stuck at this task, i need to determine the elements of this matrix:

and this is the solution :

i dont understand how to get there, any help is appreciated

### Linear algebra proof involving the Birkhoff polytope

Math Overflow Recent Questions - Sat, 06/01/2019 - 13:03

Setup:

Fix $n$. Let $\mathcal{B}$ denote the $n$-dimensional Birkhoff polytope, i.e. the set of all $n \times n$ doubly stochastic matrices. There is a fixed natural number $N(n)$. We drop the argument $n$ of $N$ for the rest of the question.

Suppose $\Gamma=\mathcal{B}^N$, i.e. the set of all ordered sets of $N$ points, each chosen independently from the Birkhoff polytope. Let a typical element of $\Gamma$ be denoted by $x \equiv (x^1,\cdots,x^N)$, where $x^1,\cdots,x^N$ are bistochastic matrices. Let $x^k_{ij}$ denote the $(i,j)$-th entry of the bistochastic matrix $x^k$.

Now we define $S=\{x \in \Gamma: Ax=0, Bx \geq 0\}$, where $A$ and $B$ are given linear transformations ($A$ and $B$ depend on $n$). Each row of each of $A$ and $B$ is of the form $[0, \cdots, 0, 1, -1, 0, \cdots 0]$, i.e. each linear restriction imposed by $A$ and $B$ only involve difference between two terms. It may also be worth mentioning that each of the difference-based restrictions contained in $A$ ($B$) is of the form $x^k_{ij}-x^l_{ij}=0\;(\geq 0)$, i.e. difference between corresponding entries of two bistochastic matrices from $\Gamma$. Other than these restrictions, $A$ and $B$ can be completely general.

What I'm trying to prove:

$S$ is obviously a polytope. I'm trying to prove it has integral extreme points, i.e. its extreme points consist only of entries in $\{0,1\}$. It is my conjecture that this holds for any $A$ and $B$ of the form described above.

My approach:

Consider $I=\{x \in \Gamma: Ax=0\}$. Clearly $S \subseteq I$. Then I have used the exact same logic as given in the proof of the Birkhoff-von Neumann theorem here to argue that $I$ cannot have fractional extreme points. Essentially the logic is as follows: If there exists any entry of any element of $\Gamma$ which is in $(0,1)$, then there has to exist a cycle - of even length - of entries in $(0,1)$, such that each of the entries in this cycle can be increased and decreased by the same $\epsilon$ amount, while still staying inside $\Gamma$. In my case I know the cycle has to be of even length because the part of the cycle inside each bistochastic matrix has to be of even length, by the Birkhoff-von Neumann proof, and each point $x \in \Gamma$ is just an ordered set of bistochastic matrices. Therefore $I$ has integral extreme points.

By the above logic, any subset of $\Gamma$, defined by equalities of the form $x^k_{ij}-x^l_{ij}=0$ has extreme points consisting only of entries in $\{0,1\}$.

Questions:

First question - is this argument correct?

Secondly, how do we bring the inequality constraints, $Bx \geq 0$, into this framework and make any conclusions about integral or fractional extreme points?

Thank you for your help.

### What is an example of a quasicategory with an outer 4-horn which has no filler?

Math Overflow Recent Questions - Sat, 06/01/2019 - 13:02

A quasicategory has fillers for all inner horns $\Lambda^i[n]$ where $n\geq 2$ and $0<i<n$, but it need not have fillers for $i=0$ (or $i=n)$. In particular, for $n=2$ and $n=3$ there are easy counterexamples.

For n=2, let $X=\Delta$ (which is the nerve of a category): there is a map $\Lambda^0\rightarrow\Delta$ that maps vertices $0\mapsto 1$, $1\mapsto 0$, and $2\mapsto 2$, which can't have a filler because there is no arrow from $1$ to $0$ in $\Delta$.

For $n=3$, any small category $X$ which has a morphism $f$ which is not an epimorphism will give us an unfillable $\Lambda^0$. For example, let $X=N(\Delta)$, and consider the maps $f=d^0\colon  \rightarrow $ and $g=d^0\circ s^0\colon \rightarrow $. Then we can construct the horn $\Lambda^0$ where the $d_1$-face witnesses the composition $g\circ f = d^0 \circ s^0 \circ d^0 = d^0 = f$, and the $d_2$ and $d_3$ faces witness $\text{id}_{}\circ f = f$. If there were a filler for this horn, then we'd have $\text{id}_{}=g\circ\text{id}_{}=g$, but $g=d^0\circ s^0$ is not the identity.

For $n\geq 4$, a counterexample can't come from the nerve of a category, because nerves of categories are $2$-coskeletal. I've tried to look at other examples of quasicategories—for example, those given in section 8 of Rezk's notes on quasicategories—but I believe I can show the existence of outer horns for $n\geq 4$ in every example given there.

So, are there any known examples of quasicategories with an unfillable $\Lambda^0$?

### Finishing a phD in USA? [migrated]

Math Overflow Recent Questions - Sat, 06/01/2019 - 12:06

I studied in Europe some times ago but didn't finish my phD dissertation. Recently I finished it and I published in a math journal. I never studied in USA, but I have been living and working here. Is it possible to finish this phD with the same results that I have found, or I have to start all over again? Any idea would be welcome. Thanks for your help.

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