Recent MathOverflow Questions

What does the torsion-free condition for a connection mean in terms of its horizontal bundle?

Math Overflow Recent Questions - Sat, 04/13/2019 - 19:01

I must have read and re-read introductory differential geometry texts ten times over the past few years, but the "torsion free" condition remains completely unintuitive to me.

The aim of this question is to try to finally put this uncomfortable condition to rest.

Ehresmann Connections

Ehresmann connections are a very intuitive way to define a connection on any fiber bundle. Namely, an Ehressmann connection on a fiber bundle $E\rightarrow M$ is just a choice of a complementary subbundle to $ker(TE \rightarrow TM)$ inside of $TE$. This choice is also called a horizontal bundle.

If we are dealing with a linear connection, then $E=TM$, and the Ehresmann connection is a subbundle of $TTM$. This makes intuitive sense -- basically it's saying that for each point in $TM$ it tells you how to move it to different vectors at the tangent spaces of different points. ($ker(TTM \rightarrow TM)$ will mean moving to different vectors at the same tangent space; so that is precluded.)

I like this definition -- it makes more intuitive sense to me than the definition of a Koszul connectionan $\mathbb{R}$-linear map $\Gamma(E)\rightarrow\Gamma(E\otimes T^*M)$ satisfies some condition. Unlike that definition it puts parallel transport front and center.

Torsion-Freeness

A Levi-Civita connection is a connection that: 1. It preserves with the Riemannian metric. (Basically, parallel transporting preserves inner products.) 2. It is torsion-free. Torsion free means $\nabla_XY - \nabla_YX = [X,Y]$.

This definition very heavily uses the less intuitive notion of connection.

So:

Questions
  1. How can you rephrase the torsion-free condition in terms of the horizontal bundle of the connection? (Phrased differently: how can it be phrased in terms of parallel transports?)
  2. I realized that I don't actually have handy an example of a connection on $\mathbb{R}^2$ that preserves the canonical Riemannian metric on $\mathbb{R}^2$ but that does have torsion. I bet that would help elucidate the answer to my first question.

Moduli spaces of arithmetic varieties with isomorphic $l$-adic cohomology

Math Overflow Recent Questions - Sat, 04/13/2019 - 09:58

Given a positive integer $d$, a rational prime $l$ and a number field $K$, is it sensible to consider the moduli stack of $d$-dimensional varieties over $K$ whose $l$-adic cohomology rings are isomorphic as Galois modules?

Categories with every indecomposable object being uniserial

Math Overflow Recent Questions - Sat, 04/13/2019 - 08:04

Let $A$ be an abelian category. A Jordan-Hölder series for an object $X$ is a filtration $0<X_0<X_1<\cdots<X_n=X$ such that $X_i/X_{i-1}$ are simple. Call $X$ uniserial in case it has a unique Jordan-Hölder series up to isomorphism.

Questions:

  1. Is the endomorphism ring of a uniserial object local?

  2. Assume $A$ has the property that every indecomposable object is uniserial. Does there exist a projective-injective object $M$ such that for every projective object $P$ there is a monomorphism $P \rightarrow M$?

  3. Is there a name for such $A$ with the property that every indecomposable object is uniserial? Have they been studied in this generality? For Artin algebras those are exactly the Nakayama algebras.

  4. Is there a classification of such abelian categories with every indecomposable object being uniserial and such that the abelian category has global dimension equal to one? For finite dimensional algebras those are exactly direct products of matrix rings over division fields and upper triangular matrix rings over division rings. Are there easy examples that are not of this form?

  5. Is there even a general classification for abelian categories such that any indecomposable object is uniserial?

A singular holomorphic foliation of $\mathbb{C}^2$ with a bounded leaf

Math Overflow Recent Questions - Fri, 04/12/2019 - 13:26

Is there a polynomial vector field on $\mathbb{C}^2$ which possesses a bounded regular leaf? By bounded, I mean a bounded subset of $\mathbb{C}^2$.

Substitute Concrete Value in Conditional Expectation

Math Overflow Recent Questions - Fri, 04/12/2019 - 11:00
  • Let $(\Omega, \mathcal{G}, \mathbb{P})$ be a probability space.
  • Let $$ X, Y : \Omega \rightarrow \mathbb{R} $$ be random variables.
  • Furthermore, let
    $$ f: \mathbb{R}^2 \rightarrow \mathbb{R} $$ be a $\mathcal{B}(\mathbb{R}^2)/\mathcal{B}(\mathbb{R})$-measurable function such that, for all $y \in \mathbb{R}$, the random variables $f(X,y)$ and $f(X,Y)$ have finite expectation.

Now let $y \in \mathbb{R}$ be arbitrary. Under the above assumptions, the expected value $\mathbb{E}[f(X,y)]$ and a $\mathbb{P}$-unique conditional expectation $\mathbb{E}[f(X,Y) \mid Y]$ do exist.

Furthermore, since $\mathbb{E}[f(X,Y) \mid Y]$ is $\sigma(Y)/\mathcal{B}(\mathbb{R})$-measurable, there exists a $\mathbb{P}_Y$-unique $\mathcal{B}(\mathbb{R})/\mathcal{B}(\mathbb{R})$-measurable function $$ \varphi : \mathbb{R} \rightarrow \mathbb{R} $$ such that $\varphi(Y) = \mathbb{E}[f(X,Y) \mid Y]$.

Under which circumstances does it hold, that $\varphi$ can be chosen such that $$ \varphi (y) = \mathbb{E}[f(X,y)] $$ and why?

Thanks in advance for any advice!

Maximal dimension of a vector subspace of real matrices with a special spectral property

Math Overflow Recent Questions - Fri, 04/12/2019 - 08:52

What is the maximal $n$ for which there exists a linear map $L:\mathbb{R}^n\to\text{End}\left(\mathbb{R}^k\right)$ such that 1 is always an eigenvalue of $L_v$ for every $v\in\mathbb{S}^{n-1}$?

Guess: would $n=\max\left\{m\in\mathbb{Z}:mk^2-{\left(\kern-.3em\left(\genfrac{}{}{0pt}{}{m}{2k}\right)\kern-.3em\right)}\geq0\right\}$?

Motivation for the guess: the assumption implies that $v\cdot v$ is an eigenvalue of $L_v^2$ for every $v\in\mathbb{R}^n$. In other words, $\det\left(L_v^2-v\cdot v I\right)=0$, which is a homogeneous polynomial equation of degree $2k$ in the components of $v$. Since the number of monomials in such a polynomial is ${\left(\kern-.3em\left(\genfrac{}{}{0pt}{}{n}{2k}\right)\kern-.3em\right)}$ and their coefficients are expressions in the $nk^2$ entries of $L_{e_i},\ 1\leq i\leq n$, the condition $nk^2-{\left(\kern-.3em\left(\genfrac{}{}{0pt}{}{n}{2k}\right)\kern-.3em\right)}\geq0$ says the system is not overdetermined.

Explicit formula for Neumann heat kernel

Math Overflow Recent Questions - Fri, 04/12/2019 - 08:26

It is well-known that $u(x,y,t)=(4\pi t)^{-n/2}(e^{-|x-y|/4t}+e^{-|x-y'|/4t})$, $x,y\in \mathbb{R}^n_+=\{x\in \mathbb{R}^n|x_n\geq 0\}$, $y'=(y_1,\dots,y_{n-1},-y_n)$, is Neumann heat kernel of $\partial_tu-\Delta_yu=0$ in $\mathbb{R}^n_+$, $\partial_{y_n}u=0$ on $\partial\mathbb{R}^n_+$.

Is there any other explicit example of Neumann heat kernels?

Computation of limit [on hold]

Math Overflow Recent Questions - Fri, 04/12/2019 - 08:02

What is the following limit equal to: $$\lim_{n\to \infty}\frac{1}{n}(\sum_{k=1}^\frac{n}{2}a_{1}a_{1}'+\sum_{k=\frac{n}{2}+1}^na_{2}a_{2}')$$ where $a_{1}=(a_{11},a_{12},…,a_{1p})′$ with $a_{1i}$ $i.i.d. Uniform(c_{1},c_{2})$ and $a_{2}=(a_{21},a_{22},…,a_{2p})′$ with $a_{2i} \; i.i.d. Uniform(d_{1},d_{2}), \; i=1,2,...,p$, and $c_{1},c_{2},d_{1},d_{2}$ are given numbers.

On cyclicity of fixed point algebra of flip automorphism

Math Overflow Recent Questions - Fri, 04/12/2019 - 01:48

Let $M$ be a von Neumann algebra having a cyclic vector in $\mathcal{H}$, is the fixed point subalgebra under the flip automorphism on $M\otimes M$ has a cyclic vector in $\mathcal{H}\otimes \mathcal{H}$. Furthermore, what is the commutant of the fixed point algebra under flip?

open book decompositions of $\Sigma\times S^1$

Math Overflow Recent Questions - Thu, 04/11/2019 - 13:50

Let $\Sigma$ be a closed orientable surface. Is there a standard open book decomposition on the $3$-manifold $M=\Sigma\times S^1$?

If the binding is allowed to be empty in the definition of an open book decomposition, then this is obvious, since $M$ is the mapping torus of the identity of $\Sigma$. The literature is not clear on this, however.

If the binding must be non-empty, then since it is contained in every page, it seems that the obvious fibering $M\to S^1$ is never the fibering of an open book.

Prove that the flow of a divergence-free vector field is measure preserving

Math Overflow Recent Questions - Thu, 04/11/2019 - 09:01

On page 3 of this preprint, after recalling the definition of flow generated by a vector field, the authors remark that "a necessary condition for a flow $\varphi_t(\cdot)$ generated by $a(t, \cdot)$ to be measure preserving is: $\mathrm{div}\, a = 0$ in a suitable sense".

Assuming $a(t,\cdot) \in BV(\mathbb{R}^N; \mathbb{R}^N)$, is the "suitable sense" the sense of distributions? If yes, how can one prove that $\mathrm{div}\, a = 0$ in the sense of distributions implies $\phi_t$ measure preserving?

Regularity of the pdf of partial Birkhoff sums

Math Overflow Recent Questions - Wed, 04/10/2019 - 17:21

Suppose that $T: X \to X$ is some measurable map on a Riemannian manifold $X$ (possibly with boundary). Let $\mu$ denote the Riemannian measure on $X$. For measurable, real-valued $g$ we may consider the partial Birkhoff sums $S_n := \sum_{k=0}^{n-1} g \circ T^k$. I am specifically interested in the case where $T$ possesses some hyperbolicity e.g. $T$ could be a piecewise $\mathcal{C}^2$ expanding map on $[0,1]$, or a $\mathcal{C}^2$ Anosov map on $\mathbb{T}^2$. Also, $g$ may be as regular as needed.

If $g$ and $T$ have enough regularity, then $S_n$ has a probability density function $f_n$. My question is whether anybody has studied the regularity of $f_n$: under what conditions does $f_n'$ exist, and what can we say about it as $n \to \infty$. If we define the Fischer information of $S_n$ to be $$ I(S_n) = \int_{\mathbb{R}} \left[\frac{\mathrm{d}}{\mathrm{d}x} \log f_n\right]^2 f_n(x) \, \mathrm{d}x, $$ can we say whether $I(S_n)$ is finite? If so, can we say anything about the growth of $I(S_n)$ as $n \to \infty$?

Uniqueness of the scheme structure for a given Hilbert polynomial

Math Overflow Recent Questions - Wed, 04/10/2019 - 17:02

If we have two lines in $P^3$ which are skewed, then we can take the union of those lines as a subscheme of $P^3$ in order to obtain a subscheme of $P^3$ with a Hilbert Polynomial given by $2m+2$.

Suppose you want to define a map from the product of the grasmannians $G(1,3)\times G(1,3)$ away from the diagonal to the component of the Hilbert scheme where the generic point is precisely two skew lines with the reduced structure. Then one possible flat limit is two different lines intersecting in a point and in order for the map to be defined there must be exactly one scheme structure that can be obtained as a flat limit for this configuration.

More precisely, take $P^3$ with coordinate $x, y, z$ and $w$, and a pair of skew lines defined one by $<x,y>$ and the other one $L_t$ defined by $<z,y-tw>$ for $t\neq 0$, then the flat limit of the union of these two skewed lines as $t\to 0$ is a pair of reduced lines intersecting at a point, plus an embedded point at $[0:0:0:1]$ whose defining ideal is $<y,xz>\cap<x,y^2,z> $, and that once again gives us a subscheme of $P^3$ with Hilbert Polynomial $2m+2$.

My question is this: Is the given structure for this singular conic $<y,xz>$ in $P^3$ the only one we can obtain as a flat limit of skew lines? And if so why, and how general is this result?

Groups With Arbitrarily Large Torsion

Math Overflow Recent Questions - Wed, 04/10/2019 - 16:34

Thompson's Group has two well known presentations:

$\langle x_0,x_1, ... $ | $ x_k^{-1} x_n x_k = x_{n+1}\forall k < n \rangle$

$\langle A,B $ | $ [AB^{-1}, A^{-1}BA], [AB^{-1}, A^{-2}BA^2] \rangle$

where $x_0=A$ and $x_n = A^{1-n}BA^{n-1}$

It is also known that every finite group exists as a subgroup of Thompson's Group. In particular, the Thompson Group has arbitrarily large torsion, as one can find any finite subgroup within it.

A simple construction like $G = \prod_{n \in \mathbb{N}} G_n$, where each $G_n$ is a finite group yields another group with arbitrarily large torsion.

What are good examples of other groups with arbitrarily large torsion? Can they be finitely generated or presented?

Thank you =)

Does the Kummer sequence of an elliptic curve could be represented by a scheme under etale topology?

Math Overflow Recent Questions - Wed, 04/10/2019 - 16:31

Given an elliptic curve $E$ over a field $K$ and an integer $n$, we have the Kummer sequence (1): $$0\longrightarrow E(K)[n]\longrightarrow E(K)\stackrel{[n]}{\longrightarrow}E(K)$$

On the other hand, over a small etale site $S_{etale}$, we also have Kummer sequence (2): $$ 0\longrightarrow \mu_{n,S}\longrightarrow \mathbb{G}_{m,S}\stackrel{[n]}{\longrightarrow}\mathbb{G}_{m,S}\longrightarrow 0$$

The question: are there schemes $X$ over $S$, such that the evaluation of (2) on $X$ exactly gives the sequence (1), which satisfies some functoriality?

A strange sequence of numbers that reversed are prime [on hold]

Math Overflow Recent Questions - Wed, 04/10/2019 - 16:07

$1, 17, 751, 967, 32953,...$ is a sequence that I invented.

The first term of the sequence divides $10^1+12345679$. The second term divides $10^{17}+12345679$.

The n-th term of the sequence a(n) divides $10^{a(n)}+12345679$.

Apart the first term, if you reverse the other four terms of the sequence, you have a prime. I don't know if other terms of the sequence could be calculated. $12345679=37\cdot 333667$. Is this sequence present in Oeis and has some importance? Do you believe that if other terms exist, reversed are primes?

A complex limit cycle not intersecting the real plane(2)

Math Overflow Recent Questions - Wed, 04/10/2019 - 15:53

Inspired by this question and the counter example provided in its answer we ask:

Is there a polynomial vector field on $\mathbb{R}^2$ such that after complexification of the equation, the corresponding singular holomorphic foliation of $\mathbb{C}^2$ possess a regular complex leaf $L$ whose holonomy is nontrivial and $L$ does not intersect the real part of $\mathbb{C}^2$? That is $L$ does not intersect $\{(z,w)\in \mathbb{C}^2 \mid im(z)=im (w)=0\}$.

Note: The counter example in the above linked post shows that this situation can occur if the coefficient of polynomial vector field are complex. But what about if the coefficients are real

Optimal solution to a swapping problem

Math Overflow Recent Questions - Wed, 04/10/2019 - 14:57

If I had two red boxes and two blue boxes stacked so that the first column is Red then Blue, the second column is Blue then Red and the third column is empty (room to move) The minimum number of moves so that it ends up with all reds in one stack, and all blues in the other, whilst only allowing for a maximum of 3 to be stacked at any one time is 3. What is the method of solving this generically? Is there an algorithm? Thanks

The (co)tangent sheaf of a topological space

Math Overflow Recent Questions - Wed, 04/10/2019 - 14:24

Let $X$ be a topological space (assume additional assumptions if needed) and denote by $\mathcal O _X$ its sheaf of $\Bbbk$-valued continuous functions where $\Bbbk$ is $\mathbb{R}$ or $\mathbb{C}$ with standard topology.

Then, as it is done in the differentiable setting or in algebraic geometry, one can define the following objects $$T_X:=\mathscr{Der}_\Bbbk (\mathcal O_X,\mathcal O_X)$$ the tangent sheaf, i.e. the sheaf of $\Bbbk$-linear derivations of $\mathcal O_X$ with values in $\mathcal O_X$ (on local sections, $\Bbbk$-linear maps $D:\mathcal O_X(U)\to\mathcal O_X(U)$ satisfying Leibniz: $D(f\cdot g)=f\cdot Dg + g\cdot Df$), and $$\Omega_X^1:=\mathcal I/\mathcal I^2$$

the sheaf of differentials, where $\mathcal I$ is the ideal sheaf of $X$ embedded diagonally $\Delta:X\hookrightarrow X\times X$ into $X\times X$ (i.e. $\mathcal I(U)=$ functions in $\mathcal O_{X\times X}(U)$ that are zero on every point of $\Delta(X)\subset X\times X$).

Well, what can be said about these two sheaves? Anything interesting at all?

Also, is there any relationship between $T_X$ and the "tangent microbundle" $\tau_X$ in case $X$ is a topological manifold?

Trotter-Kato approximation theorem for uniformly continuous approximants

Math Overflow Recent Questions - Wed, 04/10/2019 - 14:20

Let

  • $E$ be a $\mathbb R$-Banach space
  • $(T_n(t))_{t\ge0}$ and $(T(t))_{t\ge0}$ be strongly continuous contraction semigroups on $E$ with generators $(\mathcal D(A_n),A_n)$ and $(\mathcal D(A),A)$, respectively
  • $D$ be a core of $(\mathcal D(A),A)$

Consider the following assertions:

  1. $D\subseteq\mathcal D(A_n)$ for all $n\in\mathbb N$ and $$\left\|A_nx-Ax\right\|_E\xrightarrow{n\to\infty}0\;\;\;\text{for all }x\in E\tag1$$
  2. For each $x\in D$ there is a sequence $x_n\in\mathcal D(A_n)$, n$\in\mathbb N$, with $$\left\|x_n-x\right\|_E+\left\|A_nx_n-Ax\right\|_E\xrightarrow{n\to\infty}0\tag2$$
  3. For each bounded interval $I\subseteq[0,\infty)$ and $x\in E$, $$\sup_{t\in I}\left\|T_n(t)x-T(t)x\right\|_E\xrightarrow{n\to\infty}0\tag3$$

By the Trotter-Kato Approximation Theorem, the following implications hold: $$\text{1.}\Rightarrow\text{2.}\Leftrightarrow\text{3.}\tag4$$

Now assume $(T_n(t))_{t\ge0}$ is not a semigroup on $E$, but on another $\mathbb R$-Banach space $E_n$. Moreover, assume there is a bounded linear operator $\iota_n:E\to E_n$ such that $$\sup_{n\in\mathbb N}\left\|\iota_n\right\|<\infty\tag5$$ (if necessary, assume that $\left\|\iota_n\right\|\le1$). Now consider the following assertions:

  1. For each $x\in D$ there is a sequence $x_n\in\mathcal D(A_n)$, n$\in\mathbb N$, with $$\left\|x_n-\iota_nx\right\|_{E_n}+\left\|A_nx_n-\iota_nAx\right\|_{E_n}\xrightarrow{n\to\infty}0\tag5$$
  2. For each bounded interval $I\subseteq[0,\infty)$ and $x\in E$, $$\sup_{t\in I}\left\|T_n(t)\iota_nx-\iota_nT(t)x\right\|_{E_n}\xrightarrow{n\to\infty}0\tag6$$

Question 1: Are we able to infer the equivalence $$\text{4.}\Leftrightarrow{5.}\tag7$$ by the known result $(4)$ or do we need to mimic its proof from scratch?

Question 2: Is there an easier proof of $\text{2.}\Leftrightarrow{3.}$ (in the former setting) available, if we assume that each $(T_n(t))_{t\ge0}$ is even uniformly continuous (and hence each $A_n$ is bounded)?

In the context of question 2 I've got the Hille-Yosida approximation theorem in mind where something similar is shown. The crucial fact therein seems that $A_n$ and $T(t)$ commute. While this should be wrong in our general setting (is it?), there might still be an easier proof available.

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