# Recent MathOverflow Questions

### JSJ decomposition and classification of 3-manifolds

Math Overflow Recent Questions - Wed, 09/19/2018 - 03:25

I need some philosophical explanation for JSJ decomposition theorem. It says that closed orientable irreducible 3-manifold can be cut along set of incompressible tori onto pieces which are:

• atoroidal or Seifert-fibered
• hyperbolic or Seifert-fibered
• hyperbolic or spherical or Seifert-fibered with infinite fundamental group.

The three bullets are just different wording of the the same theorem.

My question is what is the method for producing all irreducible closed 3-manifolds ? We start with the ones having no incompressible torus in JSJ theorem. Let's call them family zero manifolds which are closed hyperbolic, spherical or Seifert-fibered with infinite $\pi_1$. In Friedl, Introduction to 3 manifolds, I read that Seifert-fibered manifolds are finitely covered by $S^1$-bundle over surface. The hyperbolic closed manifolds are kind of mystery for me yet. In the same place I read that closed hyperbolic manifold is finitely covered by surface bundle over circle (famous virtually fibered conjecture). Still it does not help me to understand hyperbolic manifolds but this would be the topic for another question.

Next we go family one i.e. irreducible closed manifold which has one uncompressible torus. When we cut along this torus we obtain either one or two pieces. How can we describe these pieces ? How can we prepare hyperbolic or Seifert-fibered manifold having one or two tori as boundary ? The good candidate seems to be knot complement in sphere $S^3$. How about circle complement in family zero member ? Say, we have such pieces in hand, what are possible ways of gluing the boundary tori ? The automorphism group of torus up to homotopy is $SL_2(\mathbb Z)$ - it is named mapping class group - MCG in Friedl introduction. Thus we have a lot of freedom in gluing these pieces. How can we control this ? Our goal is to enlist all manifolds in family one.

Next we continue with family two i.e. irreducible closed manifolds having two incompressible tori in JSJ decomposition. In this family we again define all possible pieces and all possible gluings. And so on...

Can someone explain me whether this procedure works and we can name all irreducible closed 3-manifolds by such method ?

### If $A$ is a (shifted) Poisson algebra, what does $A[\varepsilon]$ represent?

Math Overflow Recent Questions - Wed, 09/19/2018 - 02:44

I have a question which is not really precise, unfortunately.

Let $A$ be a Poisson $n$-algebra, i.e. a graded commutative algebra with a Lie bracket of degree $n-1$ s.t. the bracket is a biderivation w.r.t. the commutative product. Add a new square-zero variable $\varepsilon$ to $A$, possibly of positive degree, to get $A[\varepsilon]$. Clearly this is a graded commutative algebra, and there is also a shifted Lie bracket $[a+\varepsilon b, a'+\varepsilon b'] := [a,a'] + \varepsilon ([a,b'] \pm [a',b])$.

If $A$ has some geometric meaning (and this is the imprecise part), does $A[\varepsilon]$ also has some geometric meaning? I'm open to various notions of "geometric meaning". I found this appearing in an algebro-topological setting, and I'm wondering if there is a bigger picture I don't know about. This may be something very simple. My best guess is that $A[\varepsilon]$ is twice the dimension of $A$, so perhaps this is related to some (co)tangent bundle...?

### When is the category of models of a limit theory a topos?

Math Overflow Recent Questions - Tue, 09/18/2018 - 19:24

If $\mathcal{E}$ is a Grothendieck topos on a small base, then it is locally presentable, and hence is equivalent to the category of models of some limit theory.

Is there a characterization of limit theories $\mathcal{T}$ such that the category of models of $\mathcal{T}$ is a topos? The category of models of a $\mathcal{T}$ is locally presentable and hence a reflective subcategory of presheaves on $\mathcal{T}$. If there is a characterization of these theories, is it automatic that the reflector is finitely continuous, and hence gives the theories that underly Grothendieck toposes?

I know that Johnstone characterizes the product theories whose models are toposes. Here $\mathcal{T}$ must have no pseudoconstants and be sufficiently unary. I guess I'm curious if anyone has generalized this result.

### monochromatic convex hull

Math Overflow Recent Questions - Tue, 09/18/2018 - 16:02

Suppose we have $n^2$ red points and $n^2$ blue points in the plane. Is it possible to find a subset $S$ of red points such that the convex hull of $S$ does not contain any blue points, where $|S|>n^{\epsilon}$?

### How big can the index inside the root lattice of the lattice generated by a subset of roots be?

Math Overflow Recent Questions - Tue, 09/18/2018 - 15:34

Let $\Phi$ be an irreducible crystallographic root system in a Euclidean vector space $V$. Let $S\subseteq \Phi$ be some subset of roots for which $\mathrm{Span}_{\mathbb{R}}(S)=V$.

Question: How big can $[\mathrm{Span}_{\mathbb{Z}}(\Phi):\mathrm{Span}_{\mathbb{Z}}(S)]$ be?

If $\Phi$ is of Type A, then I believe we always have $[\mathrm{Span}_{\mathbb{Z}}(\Phi):\mathrm{Span}_{\mathbb{Z}}(S)]=1$ because of total unimodularity.

But for instance, if $\phi=D_4$ and $S=\{\alpha_1,\alpha_1+2\alpha_2+\alpha_3+\alpha_4,\alpha_3,\alpha_4\}$, where $\alpha_2$ corresponds to the trivalent node, then I get $[\mathrm{Span}_{\mathbb{Z}}(\Phi):\mathrm{Span}_{\mathbb{Z}}(S)]=2$. (With the standard realization of $D_4$ this is $S=\{(1,-1,0,0),(1,1,0,0),(0,0,1,-1),(0,0,1,1)\}$.)

In particular I'd like to know if $[\mathrm{Span}_{\mathbb{Z}}(\Phi):\mathrm{Span}_{\mathbb{Z}}(S)]$ is absolutely bounded or not.

### On the Fourier-Laplace transform of compactly supported distributions

Math Overflow Recent Questions - Mon, 09/17/2018 - 02:56

Let $\mathcal{E}'(\mathbb{R})$ be the space of all compactly supported distributions on $\mathbb{R}$.

For $f\in \mathcal{E}'(\mathbb{R})$, let $\widehat{f}$ denote the entire extension of the Fourier transform of $f$.

Question: If $f_n\stackrel{n\rightarrow\infty}{\longrightarrow}f$ in $\mathcal{E}'(\mathbb{R})$, then does $\widehat{f}_n$ converge to $\widehat{f}$ uniformly on compact subsets of $\mathbb{C}$ as $n\rightarrow\infty$?

### Updated background on the Hilbert 16th problem?

Math Overflow Recent Questions - Sun, 09/16/2018 - 14:56

What is the current situation of the second part of the Hilbert 16th problem? What are the most updated news on this problem?

### Seifert fiberings of zero euler number which are semi-bundles

Math Overflow Recent Questions - Sun, 09/16/2018 - 14:42

Let M be a closed oriented manifold which has the structure of a "semi-bundle" (See Section 1.2. of Hatcher's notes on three-manifolds) over an interval I. Assume that M is Seifert fibered over a base B and that the Euler number of the Seifert fibering is zero. Suppose finally that M has a unique Seifert fibering, so B is uniquely determined. Is it always the case that B is non-orientable?

### Nonlocal integral

Math Overflow Recent Questions - Sun, 09/16/2018 - 14:12

I have a little problem with the next integral, $$\int{d^3{\bf r^{\prime}}\left[\frac{exp(-ar^{\prime})}{r^\prime}\right]u({\bf r}-{\bf r^\prime}})=\int{4\pi r^\prime dr^{\prime}exp(-ar^{\prime}) u({\bf r}-{\bf r^\prime}})$$

where u(r) is an unknown wave function. I found that the integral can be writen as $$4\pi X u({\bf r})$$

where I assume X is given by $$X=\int{ r^\prime{ dr^{\prime}}exp(-ar^{\prime}) }$$ It is not clear to me why u(r-r')=u(r), perhaps a Taylor expansion? Is my assumption wrong? This integral is a second term in a temperature modified Schrodinger equation (D1,D2,D3 are not important for the discussion)

$$\left(E+\frac{\nabla}{2 \mu}\right) u({\bf r})=\frac{e^2D_1}{2r}u({\bf r})+\frac{e^2D_2D_3}{8\pi r}\int{d^3{\bf r^{\prime}}}\left[\frac{exp(-\sqrt{D_2}r^{\prime})}{r^\prime}\right]u({\bf r}-{\bf r^\prime})$$ Thanks in advance for the help

### On non-representability of certain hom schemes

Math Overflow Recent Questions - Sun, 09/16/2018 - 14:00

Let $k$ be an algebraically closed field. It is well-known that the isom-sheaf Isom$(\mathbb{A}^1_k,\mathbb{A}^1_k)$ is not representable by an algebraic space. (To be clear, the functor Isom$(\mathbb{A}^1,\mathbb{A}^1)$ associates to a $k$-scheme $S$, the set of isomorphisms $\mathbb{A}^1_S\to \mathbb{A}^1_S$ of $S$-schemes.)

Now, I expect that the non-representability of this isom-functor implies the non-representability of many hom-functors. My question is about how to make this precise.

Let X be a (positive-dimensional) variety and let $f:\mathbb{A}^1_k\to X$ be a finite morphism. How does one show that the hom-functor $\mathrm{Hom}_k(\mathbb{A}^1_k,X)$ is not representable by an algebraic space?

What have I tried? Well, there is a natural morphism Isom$(\mathbb{A}^1,\mathbb{A}^1) \to \mathrm{Hom}(\mathbb{A}^1_k,X)$ which sends $g$ to $f\circ g$. My expectation is that this is an open or closed immersion of functors. But how to make this precise?

### Is nash equilibrium the same a perpetual game where a cooperative action always takes place?

Math Overflow Recent Questions - Sun, 09/16/2018 - 13:45

First question in this forum so apologies if something I'm asking is not very clear.

In a game where many players interact is it possible for one player to establish a non finite cooperative outcome even if the game is based on non cooperative actions? To sum it up = permanent cooperative outcome based or not on a cooperative game.

### An analytical zero divisor

Math Overflow Recent Questions - Sun, 09/16/2018 - 12:49

Let $G$, $\mathbb C[G]$ and $\text{vN}(G)$ be a torsion free group, it's group ring and group von Neumann algebra, resp.. Let $0\neq\alpha\in\mathbb C[G]$ and $0\neq p\neq1$ is a projection in the group von Neumann algebra $\text{vN}(G)$, is it true that $\alpha p\neq0$?

### Generating correlated random variables: why the covariance result depends on the length of the variables?

Math Overflow Recent Questions - Sun, 09/16/2018 - 12:25

Suppose I have a vector $X$ of random independent time series (uniform distribution) and I'd like to transform them into dependent time series having a covariance $Q$.

So, first I standardize $X$ to have $cov(X)=I$ and I decompose $Q$ such that $Q=L.L^T$ using Cholesky decomposition. $Q$ is positive definite.

Hence, the new vector $Y$ having the covariance $Q$ is simply the product $X.L$ because: $cov(X.L)=L.cov(X).L^T=L.I.L^T=Q$

The question: using simulated data (Octave GNU), I noticed that $cov(X.L)=Q$ is true when the size of $X$ is very high $(10^8)$, but the smallest the size of $X$ the most different $cov(X.L)$ is from $Q$.

Might you have an explanation for this ?

### Strong estimates for the zeta function on natural numbers

Math Overflow Recent Questions - Sun, 09/16/2018 - 11:24

Let $$\zeta(s) = \sum_{n = 1}^\infty \frac{1}{n^s}$$ be the Riemann zeta function (here we just consider real $s$). We do have a description given by $$\zeta(s) = \frac{s}{s-1}-s\int_{1}^\infty \frac{u-[u]}{u^{s+1}} du$$ Can we deduce by this formula or another one some fairly "precise" estimates for giving upper and lower bounds for $\zeta(s)$ for all natural numbers $s$ (or just an infinite subset of the natural numbers).

### Mathematical Justification for reduced probability of an event [on hold]

Math Overflow Recent Questions - Sun, 09/16/2018 - 11:04

Beginner in the field so apologies if the question is not totally clear.

What is the mathematical explanation for something that is highly unlikely to occur? Is the 1% on a 5% likely to occur event equal to the 1% on a 95% likely to occur event?

Hope I can get some feedback on this.

### Are there partially algebraic Hecke characters?

Math Overflow Recent Questions - Sun, 09/16/2018 - 10:46


Let $\chi\colon \mathbb{A_F}^\times/F^\times \to \C^\times$ be a Hecke character with local components $\chi_v$ for each place $v$ of $F$. When $v$ is an archimedean place, then there are $m_v\in\Z$, $\sigma_v,\varphi_v\in\R$ such that for all $x\in F_v^\times$, $$\chi_v(x) = \left(\frac{x}{|x|}\right)^{m_v}|x|^{\sigma_v+i\varphi_v}.$$ Recall that $\chi$ is algebraic if for every archimedean place $v$ the local component $\chi_v$ is a rational function, i.e. if there exists integers $p_v,q_v\in\Z$ such that $$\chi_v(x) = x^{p_v}\bar{x}^{q_v}.$$ We understand well when algebraic characters exists: up to a finite order character, they factor through the norm to the maximal CM subfield of $F$ (or to $\Q$ if there is no subfield).

For an algebraic character, all $\varphi_v = 0$, and up to finite index and multiplication by a power of the norm this characterises algebraic characters.

Now let $\Sigma$ be a subset of the set of archimedean places of $F$.

Does there exists Hecke characters as above such that $\varphi_v=0$ if and only if $v\in\Sigma$?

If $|\Sigma|\le 1$ there you can obtain such characters by multiplying any Hecke character by a suitable power of the norm.

If $\Sigma$ is the set of all archimedean places, we know the answer by the usual theory of algebraic characters.

Otherwise, since I do not see an obvious reason why such characters should exist, the "obvious guess" is that they do not. Is it true?

I was able to rule out a couple more cases in an ad-hoc way, and it seems to quickly involve transcendence problems.

### Checking Planar Convexity of 4 Points with Stewart's Formula

Math Overflow Recent Questions - Sun, 09/16/2018 - 09:44

Is the following conjecture correct?

Conjecture:

If $A,B,C,D$ are four points in general position in the euclidean plane, with

$a:=\|C-B\|,\ \ b:=\|C-A\|,\ \ c:=\|B-A\|$
$a':=\|D-A\|,\ b':=\|D-B\|,\ c':=\|D-C\|$ ,

\begin{align} a+b+c\ &>\ a+b'+c',\ a'+b+c',\ a'+b'+c\\ c+c'\ &>\ a+a',\ b+b'\\ \end{align},

then $A,B,C,D$ are in strict convex configuration iff $$(a'+b')((a+c')(a-c')-a'b')+b'(b+a)(b-a) \ \lt 0$$

If so, then checking planar convexity of four points is possible with 6 length measurements and 5 multiplications. This checks convexity via an inequality, so it is stable under small displacements of the points into 3-space.

By contrast, using the equality

$$area(ABC) + area(ACD) = area(ABD) + area(BCD)$$

in a check of convexity, which also can be done with only six length measurements, makes the check unstable under those small displacements. The first approach is based on Stewart's Theorem, where the lengths $a',b',c'$ are known as $d,m,n$; the second approach is based on Heron's Formula instead.

### On a topological group on $\Bbb N\,?$ [on hold]

Math Overflow Recent Questions - Sun, 09/16/2018 - 08:09

Suppose $r:\Bbb N\to(0,1)$ is a function given by $r(n)$ is obtained by putting a point at the beginning of $n$ instance $r(34880)=0.34880$.

now let $\lt_1$ be a total order relation (not well ordering) on $\Bbb N$ as: $\forall m,n\in\Bbb N,\,m\lt_1n$ iff

$\begin{cases} r(m)\lt r(n),\,m=m_1\times10^r,\,n=n_1\times10^r,\,10\nmid m_1,\,10\nmid n_1,\\m_1,n_1\in\Bbb N,\,r\in\Bbb N\cup\{0\}, & \text{ or}\\\\\\m=m_1\times10^s,\,n=n_1\times10^t,\,s\lt t,\,m_1,n_1,t\in\Bbb N,\,s\in\Bbb N\cup\{0\}\end{cases}$

then assume $\mathfrak T$ is a topology on $\Bbb N$ induced by $\lt_1$,

Question $1$: Is $\Bbb P$ dense in $(1,10)$? $$\\$$ on the other hand $\Bbb N$ is a cyclic group by:

$\begin{cases} \forall m,n\in\Bbb N\\ e=1\\ (2n)\star (2n+1)=1\\ (2n)\star (2m)=2n+2m\\ (2n+1)\star (2m+1)=2n+2m+1\\ (2n)\star (2m+1)=\begin{cases} 2m-2n+1 & 2m+1\gt 2n\\ 2n-2m & 2n\gt 2m+1\end{cases}\\\langle 2\rangle =\langle 3\rangle =(\Bbb N,\star)\simeq (\Bbb Z,+)\end{cases}$ $$\\$$ Question $2$: Is $\Bbb N$ a topological group?

Thanks a lots!

(Goldbach's conjecture: $\forall n\in\Bbb N,\,\exists p,q\in\Bbb P$ such that $2n+3=p\star q$.)

### Basis theorem for finite abelian groups question (Pinter's "A book of Abstract Algebra") [migrated]

Math Overflow Recent Questions - Sun, 09/16/2018 - 07:27

I am stuck at the exercise Q5 of Chapter 16. Here is a summary of the question:

Let $G$ be a finite Abelian $p$-group where $p$ is a prime. Let $a \in G$ be an element of highest possible order in $G$. Let $H$ be the subgroup generated by $a$. Now $G/H$ = [$Hb_1$, $Hb_2$, ... $Hb_n$] and we can choose $b_i$ such that $ord(b_i) = ord(Hb_i)$ (this is proven in another exercise O).

The question is to prove that for any intergers $l_0, l_1, l_2,..., l_n$, if $a^{l_0}b_1^{l_1}b_2^{l_2}...b_n^{l_n} = e$, then $a^{l_0}=b_1^{l_1}=b_2^{l_2}=...=b_n^{l_n} = e$.

What I have tried: it looks like I need to use induction. So I started with simple cases.

If $n=0$ this is true.

If $n=1$, $a^{l_0}b_1^{l_1}=e$ means $(Ha^{l_0})(Hb_1^{l_1})=H$, which in turn gives $(Hb_1)^{l_1}=H$. So $l_1$ is a multiple of $ord(Hb_1)$, which is also a multiple of $ord(b_1)$. This indicates $b_1^{l_1}=1$.

If $n=2$, assume $a^{l_0}b_1^{l_1}b_2^{l_2}=e$. There are two cases, if $b_2^{l_2}=1$, by the $n=1$ case it's done. If, however, if $b_2^{l_2} \neq 1$, I am stuck. Trying the logic $(Hb_1^{l_1})(Hb_2^{l_2})=H$ and I cannot use the information that $ord(b_i) = ord(Hb_i)$.

I also notice that I haven't used the information that $G$ is a $p$-group, but I have struggled a few days without any progress. I've tried searching for proof and found a question on Q2, which unfortunately does not seem to help in my case. Could anyone shed some light on how to prove this? Appreciate your help!

### Entropy and total variation distance

Math Overflow Recent Questions - Sun, 09/16/2018 - 06:41

Let $X$, $Y$ be discrete random variables taking values within the same set of $N$ elements. Let the total variation distance $|P-Q|$ (which is half the $L_1$ distance between the distributions of $P$ and $Q$) be at most $\epsilon$. What is a standard, easily provable bound on the difference between the entropies of $X$ and $Y$?

(It is easy to see that such a bound must depend not only on $\epsilon$ but also on $N$.)