Recent MathOverflow Questions

How to find the Eulerian circuit with the minimum accumulative angular distance within a Eulerian graph?

Math Overflow Recent Questions - Tue, 04/16/2019 - 21:33

Note: I originally posed this question to Mathematics, but it was recommended that I try here as well.

Context

For context, this problem is part of my attempt to determine the path of least inertia for a free and open-source laser scanner DAC API I am developing. The following problem arises during the vector image optimisation pass. I convert the 2D vector image into a graph of 2D positions and add blank edges (i.e. transparent lines) to represent the image as a strongly connected, undirected Eulerian graph from which I should be able to determine the optimal Eulerian circuit.

Problem

Given a strongly connected, undirected Eulerian graph (i.e. each vertex has an even degree), I'm trying to determine the Eulerian circuit that results in the minimum possible accumulative angular distance, where each vertex is a position in 2D space and each edge describes a straight line between the vertices.

My Solution Attempt

Edit: Since posting, I've since realised that my solution attempt is incorrect as the resulting traversal is not guaranteed to form a Eulerian circuit. Feel free to skip this section and the My Solution Attempt Issues section.

My attempt at solving this was to first simplify the problem by looking at each vertex individually. We know that each vertex must have an even degree, and thus for each vertex there must be an optimal set of incoming/outgoing edge pairs (where each edge is used once) that results in a minimum accumulative angular distance. By minimum accumulative angular distance, I'm referring to the sum of the difference between the result of the difference between the angle of each incoming/outgoing edge pair and a straight line. For example, given the following vertex A and its neighbours B, C, D and E:

an example of optimal pairs would be (DA, AB) and (EA, AC) as they are cumulatively the least sharp angles through which A may be traversed (and in turn would induce the least inertia), whereas the pairs (EA, AD) and (BA, AC) would be the least optimal as cumulatively they contain the sharpest angles to be traversed (resulting in the highest inertia).

Once the set of optimal pairs is determined for each vertex, I suspect the Eulerian Circuit can be created by starting at one of the vertices, picking a direction to begin and following the optimal pairs until the beginning is reached again.

My Solution Attempt Issues

Currently however I'm running into two issues.

  1. I don't know for sure whether or not my assumption holds true for all Euler graphs (where all nodes have an even degree).
  2. I'm unsure of the best approach for determining the set of optimal edge pairs for each vertex. I suspect it may be possible to represent each vertex and its edges as a sub-graph and treat the problem as finding the shortest path (where the "shorter" distances are the paths through the vertex that result in the straightest angles), but I'm struggling to come up with a sub-graph representation that would allow me to do this.
Related Research

In section 3.4 of Accurate and Efficient Drawing Method for Laser Projection the paper describes using Hierholzer’s algorithm for finding an optimal Eulerian circuit with the amendment that during traversal of each vertex you select the unvisited edge along the angle closest to a straight line. One issue that occurs to me with this approach is that it is not clear to me that this always results in the absolute optimal circuit, only one that is probably more optimal than a naive construction without this added amendment.

Questions
  1. Is there an existing solution to the original Problem stated above? If so, is there somewhere I might read further on this?
  2. If not, does my attempted solution sound like a reasonable approach? If so, do you have an idea of how I might represent the sub-graph for determining the set of edge pairs resulting in the minimum accumulative angular distance for each vertex? Approach determined to be invalid.
  3. If not, can you recommend an approach I might be able to take to make progress on solving the previously mentioned Problem?

Any advice appreciated!

Formal sums on groups with symmetric finite presentations [on hold]

Math Overflow Recent Questions - Tue, 04/16/2019 - 07:59

Suppose we have a group $G$ with a "symmetric" finite presentation: \begin{equation} G = \langle g_i, i=1\ldots m | R_\alpha (\{g_i\}) ,\, \alpha=1\ldots n \rangle \end{equation} \begin{equation} R_\alpha (\{g_i\}) \iff R_\alpha(\{g_{\sigma(i)}\}) ,\,\,\,\forall \sigma \in S_m \,. \end{equation} That is, the set of relations is symmetric under every permutation of the generators.

Now I want to make a formal sum over elements of $G$. Can I do so by taking a formal power series in $\sum_i (g_i + g_i^{-1})$? And is there a well-defined way of determining said power series from the set of relations $R_\alpha$?

I would like to create the formal sum $\sum_{g\in G} g$ in the group algebra $\mathbb{R}[G]$, as a formal power series $\sum_{p=-\infty}^{\infty} c_p (\sum_i g_i+g_i^{-1})^p$. Is this possible for all such group presentations, and is there a way to determine the power series from the relations $R$?

How to calculate the dual of this SDP?

Math Overflow Recent Questions - Tue, 04/16/2019 - 07:17

Consider the problem: \begin{eqnarray} &\min\langle Q, X \rangle \\ s.t. \quad & X_{ij} - u_j x_i \leq 0, \quad i \leq j \\ & X_{ij} - u_i x_j \leq 0, \quad i \leq j \\ & -X_{i,j} + u_jx_i + u_ix_j \leq u_i u_j, \quad x \leq j\\ & -X_{ij} \leq 0, \quad i \leq j\\ &\begin{bmatrix} 1 & x^T \\ x & X \end{bmatrix} \succeq 0, \\ & x \in \mathbb{R}^n, X \in \mathcal{S}^n. \end{eqnarray} Here $u \in \mathbb{R}^n$ is given. It turns out the dual problem is the following: \begin{eqnarray} & \max -\langle \Phi_3 ,uu^T\rangle \\ s.t. \quad & Q + \Phi \succeq 0 \\ & -(\Phi_1 + \Phi_2 - 2 \Phi_3)^T u \geq 0 \\ & \Phi = \Phi_1 + \Phi_2 - \Phi_3 -\Phi_4 \\ & \Phi \in \mathcal{S}_n, \; \Phi_1, \Phi_2, \Phi_3, \Phi_4 \in \mathcal{S}_n^+ \end{eqnarray}

I don't understand how they calculate the dual problem, the dual variables for the affine constraints are supposed to be nonnegative, how did they end up being positive semidefinite? It is said $\Phi_1,...,\Phi_4$ are built from dual variables associated with the constraints.

This is a paper from mathematical programming, but especially I don't understand how they get the second constraint in the dual, if replacing the semidefinite constraint with $X-xx^T\succeq 0$, then in the Lagrangian, there is a xx^t tem, you would expect something quadratic in the dual.

Overapproximating $\binom{n/t + m}{m}^t$ [on hold]

Math Overflow Recent Questions - Mon, 04/15/2019 - 20:27

It is well-known that

$$ \left(1+\frac{n}{t}\right)^t\ \le\ e^n $$

for positive integers $n$ and $t$.

You can also write the basis on the LHS in a more cumbersome way as

$$ 1+\frac{n}{t}\quad =\quad \binom{\frac{n}{t}+m}{m}\quad\text{for}\ m=1 $$

Here, you have to somehow generalize the binomial coefficients to accept rational numbers in the upper position, e.g.:

$$ \binom{x}{m}\ ≝\ \prod_{0\le i < m} \frac{x-i}{i+1}\quad\text{for}\ x\in ℚ\ \text{and}\ m\in ℕ_{\ge 0} $$ with the convention that the empty product evaluates to 1.

Now, is it possible to generalize the inequality

$$ \binom{\frac{n}{t}+m}{m}^t\ \le\ e^n\quad\text{for}\ m=1 $$

from $m=1$ to arbitrary positive integers $m$ such that the new upper bound (i.e., the new RHS), whatever it might be, would be elementary, would not involve the binomial coefficients (and possibly not involve the factorials), and would simplify to $e^n$ (i.e., the old RHS) for $m=1$? Of course, we'd like to have the upper bound on the generalization to be reasonably low. (I admit that the term "reasonably" is diffuse; my apologies.)

Literature references are also welcome.

Endomorphism rings of ordinary elliptic curves

Math Overflow Recent Questions - Sun, 04/14/2019 - 16:43

Let's say $p$ is a prime and $t\neq 0$ is a trace of Frobenius that occurs over $\mathbb{F}_p$. The discriminant of the Frobenius polynomial is $\Delta:=t^2-4p.$ So we obtain $4p=t^2-\Delta.$ If $E$ is an elliptic curve over $\mathbb{F}_p$ with trace of Frobenius $t$, then the Frobenius, call it $\sigma$, generates an order in the endomorphism ring of $E$, $End(E)$. Symbollically, $\mathbb{Z}[\sigma]\subseteq End(E).$ Put another way, the discriminant of $End(E)$ divides $\Delta$.

Now $\Delta$ may not be the discriminant of a maximal order. We write $\Delta=B\Delta_{\mathcal{O}_K}$ where $\Delta_{\mathcal{O}_K}$ is the discriminant of the ring of integers of some imaginary quadratic field $K$ and $B\geq 1$. Suppose there is an integer $b>1$ with $b^2\mid B$ and $b$ prime to $p$. Then $b$, the conductor, is the index of the order of discriminant $b\Delta_{\mathcal{O}_K}$ in $\mathcal{O}_K$. There could be an elliptic curve over $\mathbb{F}_p$ whose endomorphism ring is the order of discriminant $b\Delta_{\mathcal{O}_K}$.

My question is, in general, does every possible order occur as an endomorphism ring of an elliptic curve over $\mathbb{F}_p$? Kohel seems to indicate this is the case in his thesis, but I can't find a proof. By work of Deuring and because the $j$-invariants in question are algebraic, we know that over some finite extension of $\mathbb{F}_p$ there is an elliptic curve with an endomorphism ring that is isomorphic to each possible order. For my applications though, I'd like to have this elliptic curve be defined over $\mathbb{F}_p.$

Quadratic covariation of two not independent Brownian motions

Math Overflow Recent Questions - Sun, 04/14/2019 - 07:02

Given two not independent Brownian motions, $X$ and $Y$. I was wondering if we can say anything about the quadratic covariation of $X$ and $Y$, $\langle X,Y \rangle_t$. I know that for two independent Brownian motions, that this quadratic covariation is zero, but does this also hold when we cannot say whether or not the Brownian motions are independent? I'm familiar with the definition of the quadratic (co)variation, but it didn't help me in finding the answer. I'm starting to suspect that we cannot do much with $\langle X,Y \rangle_t$ when $X$ and $Y$ are dependent.

Any help is appreciated!

Inequality in L2

Math Overflow Recent Questions - Sun, 04/14/2019 - 06:59

Is following inequality true:

$\int\limits_{0}^{l} (u^{'}(x))^{2} dx \leq C \cdot\int\limits_{0}^{l} (u(x))^{2} dx $

for $u \in L^2(0,l)$ with $u(0)=u(l)=0$

On solution to the equation $x_{1}^{2}x_{2}^{2}x_{3}^{2}x_{4}^{2}x_{5}^{2}\left(\sum_{i=1}^{6}a_{i}x_{i}\right)^{2}=1$

Math Overflow Recent Questions - Sun, 04/14/2019 - 06:41

For any $a_{1},a_{2},\cdots a_{6}$ in $\mathbb{R}$ with $\sum_{i=1}^{6}a_{i}^{2}=1$, is it true that there always exist $x_{1},x_{2},\cdots x_{6}$ in $\mathbb{R}$ with $\sum_{i=1}^{6}x_{i}^{2}=6$, such that $$ x_{1}^{2}x_{2}^{2}x_{3}^{2}x_{4}^{2}x_{5}^{2}\left(\sum_{i=1}^{6}a_{i}x_{i}\right)^{2}=1? $$

Any helpful answer that could lead to a correct answer to this question would be highly appreciated!

Classification of bundles, Postnikov towers, obstruction theory, local coefficients

Math Overflow Recent Questions - Sun, 04/14/2019 - 06:23

RECAP on classification of bundles

We want to classify $G$-principal bundles over $X$ (smooth manifold, G compact Lie). These are in 1-1 correspondence with homotopy classes of maps $[X,BG]$ (where $EG\to BG$ is the universal bundle as usual). If $BG \simeq K(\pi,n)$ then it's easy: $$[X, BG]\leftrightarrow H^n(X,\pi),$$ therefore there is a cohomology class that gives us the classification (e.g. the 1st Chern class for the frame bundles of complex line bundles). In general, $BG$ can be more complicated, in any case $BG$ has a Postnikov tower which induces a factorization of the classifying map $f\in [X, BG]$ in $(f_i)_i$ $\require{AMScd}$ \begin{CD} \vdots@. \vdots\\ @| @VVV \\ X@>>f_2> P_2(BG)\\ @| @VVp_2V \\ X @>>f_1> P_1(BG)@.\simeq K(\pi_1(BG),1) \end{CD}

The homotopy type of $f$ is given by the homotopy type of the $f_i$s. Since $P_1(BG)\simeq K(\pi_1(BG),1)$ $f_1$ is given by a cohomology class in $H^1(X, \pi_1(BG))$. However, not any choice of $f_2$ works, because it must lift $f_1$. From this answer* I learned that there is a Cartesian diagram $\require{AMScd}$ \begin{CD} X@>>f_2> P_{2}(BG) @>>> K(\pi_0G,1)\\ @| @VVp_2V @VVV\\ X @>>f_1> P_1(BG) @>>> K(\pi_1 G,1)_{h\pi_0G} \end{CD}

1)Explanation/references for this? I was expecting the second column to be something like $K(\pi_2(BG),2)\to K(\pi_1(BG),1)$, it reminds me of principal fibrations.

2)How to see that these lifts are parametrized by $H^{2}(X,\pi_{2}(BG))$ cohomology with local coefficients twisted by $f_1\in H^1(X,\pi_1(BG))$? Obstruction theory tells the necessary conditions to lift $f_1$ but not how many lifts there are.

In the end we get that the principal bundle is classified by $f_1\sim \alpha_1 \in H^1(X,\pi_1(BG))$ and a sequence of cohomology classes $\alpha_k \in H^{k}(X, \pi_k(BG)) $ in the cohomology with local coefficients twisted by $\alpha_1$.

3) How to compute them? Is there any example for say $G=O(2)$? Any link with invariant polynomials in $\mathfrak{g}$ or the Weyl algebra of $\mathfrak{g}$?

This is essentially Denis Nardin's answer. In his comment Nardin, says another interesting thing if $G=O(n)$, then $\alpha_1 = 0 $ iff the bundle is orientable, $\alpha_1, \alpha_2 = 0$ iff the bundle is spin and so on climbing the Whitehead tower of $O(n)$ $$O(n)\leftarrow SO(n)\leftarrow Spin(n)\leftarrow String(n)\leftarrow ...$$

4)Is this true for any Whitehead tower of groups? Does this implicitly say that the Postnikov tower of $SO(n)$ $(Spin(n))$ is the one of $O(n)$ without the first (second) term?

BG as a twisted product

If $\pi_1(BG)$acts on $\pi_{n+1}(BG)$ trivially then

  1. the Postnikov tower gives us an expression for $BG$ in terms of a twisted product of Eilenberg-MacLane spaces $BG \simeq K(\pi_1(BG),1)\times_{k_1} K(\pi_2(BG),2)\times_{k_2} \dots$
  2. There is no need of local coefficients for the $\alpha_k$ above.

In the same answer*, Mark Grant says that in the case of $G=O(2)$:

there is a fibration $$ K(\mathbb{Z},2)\to E\mathbb{Z}/2\times_{\mathbb{Z}/2} K(\mathbb{Z},2)\to B\mathbb{Z}/2 $$ given by the twisting of $w_1$ on the universal $SO(2)$ bundle, and this fibration agrees up to homotopy with the fibration $$BSO(2)\to BO(2)\to BO(1).$$

Question:

5) Can you explain this in the more general setting of a fibration $F\to E\to B$? Also I do not understand $E\mathbb{Z}/2\times_{\mathbb{Z}/2} K(\mathbb{Z},2)$, what is the $\mathbb{Z}/2$ action on $K(\mathbb{Z},2)$? Grant says $w_1$ is involved but I cannot imagine how. (I know that in general $w_1 \in H^1(X, \pi_1(BO(2))$ gives me an action of $\pi_1(X)$ on $\pi_n(BO(2))$). How does this relate to the Whitehead tower above?

*Classification of $O(2)$-bundles in terms of characteristic classes

Finding coefficient of multivariate polynomial

Math Overflow Recent Questions - Sun, 04/14/2019 - 06:13

$f(x_1,x_2,\ldots x_n)$ is polynomial with integer coefficients. $f$ is rather large to be computed explicitly, but an algorithm can compute it efficiently at integers and complex number and "lazy" form using parenthesis. The degree of $f$ is $n$. Let $C$ be the coefficient of the monomial $\prod_{i=1}^n x_i$.

Q1: What is the complexity of deciding if $C$ is zero or not?

Q2: What is the complexity of computing $C$?

Can we do better than $\exp(o(n())$ (small oh)?

Possible approaches:

  • Find some domain $K$ with many nilpotent elements and work over $K[x_1,x_2,\ldots x_n]/(x_1^2,x_2^2 \ldots x_n^2)$
  • Using Automatic differentiation compute the partial derivative.

Any holomorphic vector bundle over a compact Riemann surface can be defined by only one transition function?

Math Overflow Recent Questions - Sun, 04/14/2019 - 06:01

It is known that any holomorphic bundle of any rank over a noncompact Riemann surface is trivial. A proof can be found in Forster's "Lectures on Riemann surfaces", section 30.

Let $E$ be a holomorphic vector bundle over a compact Riemann surface $X$ with gauge group $G$. A consequence of the above theorem is the restriction $E|_{X-\{p\}}$ for any point $p\in X$ is a trivial bundle. Thus $E$ can be recovered by specifying the transition function $g: D\cap (X-\{p\}) \rightarrow G$ where $D$ is a small disk containing $p$.

Is this correct? If not, could you give a counter-example? I am mainly interested in learning about the moduli space of holomorphic bundles over $X$ in a concrete way, e.g. using transition functions.

Fourier coeffients of Cantor measure

Math Overflow Recent Questions - Sun, 04/14/2019 - 05:41

For $0<\theta<\frac{1}{2}$, denote by $\mu_\theta$ the uniform Cantor measure with dissection ratio $\theta$. It is not hard to show that the Fourier–Stieltjes transform of $\mu_\theta$ is $$ \widehat{\mu}_\theta(\xi)=\prod_{k=1}^{\infty} \cos(\pi\theta^k\xi) $$ (up to scaling and constant multiple). It is well known that $\widehat{\mu}_\theta(\xi)\to 0$ as $\xi\to\infty$ if and only if $\theta$ is a Pisot number.

Now we are concentrated on the case where $1/\theta$ is an integer. We know that $\limsup_{n\in \mathbb{Z}, n\to\infty} \widehat{\mu}_\theta(n)>0$. It is not hard to see that $\liminf_{n\in \mathbb{Z}, n\to\infty} \widehat{\mu}_\theta(n)=0$. My question is whether we can characterize (or approximate) the set $$ A_{\epsilon}:=\{n\in \mathbb{Z}: |\widehat{\mu}_\theta(n)|>\epsilon\}, $$ for $0<\epsilon<1$. My second question is what the "maximal" set $B$ is such that $$ \liminf_{n\in \mathbb{Z}\setminus B, n\to\infty} \widehat{\mu}_\theta(n)>0. $$ Thanks.

Relation between groups An, Bn, Dn and Sn or inversions of random elements in coxeter groups

Math Overflow Recent Questions - Sun, 04/14/2019 - 04:40

First of of all I'm trying to find a general interperetation to the following facts below.

  1. Let's look at the property of Kendall-Mann numbers $M(n)$ which are row maxima of Triangle of Mahonian numbers $T(n,k)$ (the number of permutations of {1..n} with k inversions). According to Richard Stanley $$ \left| P\left( \frac{\mathrm{inv}(\pi)-\frac 12{n\choose 2}}{\sqrt{n(n-1)(2n+5)/72}}\leq x\right)-\Phi(x)\right| \leq \frac{C}{\sqrt{n}}, $$ where $\Phi(x)$ denotes the standard normal distribution. From this it is immediate that $M(n+1)/M(n)=n-\frac 12+o(1)$

  2. Looking at combinatorial proof for the property of Kendall-Mann numbers numbers at MO $M(n) \approx c n!/n^{3/2}$ and $$\frac{M(n+1)}{M(n)} \approx \frac{(n+1)(n+1)^{-3/2}}{n^{-3/2}} = n (1+1/n)^{-1/2} \approx n-1/2.$$ This is pretty the same result as #1.

  3. Reading through Counting inversions and descents of random elements in finite Coxeter groups I noticed Corollary 3.2 (page 6 of the article, pls have a look at it) that the mean and variance of W-Mahonian distribution depend on the types of groups, i.e. $A_n$, $B_n$, $D_n$. By and large it's about $n^{3/2}$ like for #1 and #2.

This results in the similar 'structure': $\approx n-1/2$ .

So I wonder why? I am looking for a general explanation to the facts. I guess that it is needed to study relations between groups: $A_n$, $B_n$, $D_n$ and $S_n$. Any help in explanation of the facts are higly welcomed.

Airy stress, Beltrami stress and gauge fields

Math Overflow Recent Questions - Sun, 04/14/2019 - 03:58

The following problem comes from the theory of elasticity, but reduces to a pure geometric problem. Consider a $d$-dimensional Riemannian manifold $(M,g)$ with boundary representing the intrinsic geometry of an elastic body. A configuration of that body is an immersion $f:M\to R^d$. For simplicity assume that $M$ is compact, orientable and simply-connected.

Equilibrium configurations are critical points of an energy functional; the Euler-Lagrange equation is of the form $$ \delta^\nabla \tau = 0, $$
where $\tau\in\Omega^1(M;f^*TR^d)$ is the tension field and $\delta^\nabla$ is the covariant codifferential associated with the induced connection on $f^*TR^d$. The tension field $\tau$ is determined by the precise form of the energy functional. A typical choice of the energy functional yields $$ \tau = df((df)^Tdf - Id_{TM}). $$

Regardless of the form of the energy functional a tension field $\tau$ satisfying the Euler-Lagrange equation can be represented as $$ \tau = df\circ \delta^\nabla(\delta^\nabla\Psi)^T\circ(df^Tdf)^{-1} \, \det df, $$ where $\Psi\in\Omega^2(M;\wedge^2 TM)$ and the connection here is the pullback of the Euclidean connection on $TM$. The ``stress function" $\Psi$ is of the type of a curvature operator, i.e., has $d^2(d^2-1)/12$ degrees of freedom. In turn, $\tau$ determines $df^Tdf$ which determines the pullback metric on $TM$; since the latter must be flat, one obtains a system of $d^2(d^2-1)/12$ compatibility equations for the vanishing of the Riemann curvature tensor of the pullback metric. These equations are highly nonlinear, but this doesn't matter for the sake of the ensuing discussion.

For $d=2$, $\Psi$ can be represented by a scalar field (known as the Airy stress function) and one obtains a single scalar compatibility condition for the vanishing of the Gaussian curvature of the pullback metric. For $d\ge 3$, one has a gauge freedom, as the operator $\Psi \mapsto \delta^\nabla(\delta^\nabla \Psi)^T$ has a non-trivial kernel. For $d=3$, this has been known for over a century, yielding the so-called Maxwell and Morera representations of the stress function.

I am trying to figure out the reduced representation in general dimension. The first two questions that come in mind are:

  1. What is the symmetry at the heart of this gauge freedom?
  2. Is there a natural coordinate-free way to obtain a reduced set of compatibility equations for a reduced representation of $\Psi$.

Covariant derivative of determinant of the metric tensor

Math Overflow Recent Questions - Sun, 04/14/2019 - 03:51

Let $(M,g)$ be a Riemannian manifold and $g$ the Riemannian metric in coordinates $g=g_{\alpha \beta}dx^{\alpha} \otimes dx^{\beta}$, where $x^{i}$ are local coordinates on $M$. Denote by $g^{\alpha \beta}$ the inverse components of the inverse metric $g^{-1}$. Let $\nabla$ be the Levi-Civita connection of the metric $g$. Consider, locally, the function $\det((g_{\alpha \beta})_{\alpha \beta})$. It is known that $\nabla \det((g_{\alpha \beta})_{\alpha \beta}) = 0$ by using normal coordinates etc...

I would like to show this fact without using normal coordinates. Just by computation. Here is what I have so far:

$\nabla \det((g_{\alpha \beta})_{\alpha \beta}) = \left [ g^{\gamma \delta} \partial_{\delta} \det((g_{\alpha \beta})_{\alpha \beta}) \right ] \partial_{\gamma} = \left [ \det((g_{\alpha \beta})_{\alpha \beta}) g^{\gamma \delta} g^{\beta \alpha} \partial_{\delta} g_{\alpha \beta}\right ] \partial_{\gamma}.$`

Here: the first equality sign follows from the definition of the gradient of a function and the second equality sign is the derivative of the determinant.

Question: How do I continue from here without using normal coordinates? Or are there any mistakes? If yes, where and which?

Greetings, Phil

limit of n root question

Math Overflow Recent Questions - Sun, 04/14/2019 - 03:33

I would like to know if the following statement is true or false:

Let $(a_n),(\tilde{a}_n),(b_n),(\tilde{b}_n),(c_n),(\tilde{c}_n),(d_n),(\tilde{d}_n)\subset \mathbb{C}.$ If $$\limsup_{n \rightarrow \infty}|a_n b_n-c_n d_n|^{1/n}<\frac{1}{4},$$ $$\limsup_{n \rightarrow \infty}|a_n-\tilde{a}_n|^{1/n}<1/2,$$ $$\limsup_{n \rightarrow \infty}|b_n-\tilde{b}_n|^{1/n}<1/2,$$ $$\limsup_{n \rightarrow \infty}|c_n-\tilde{c}_n|^{1/n}<1/2,$$ and $$\limsup_{n \rightarrow \infty}|d_n-\tilde{d}_n|^{1/n}<1/2,$$ then $$\limsup_{n \rightarrow \infty}|\tilde{a}_n \tilde{b}_n-\tilde{c}_n \tilde{d}_n|^{1/n}<\frac{1}{4}.$$

Thank you very much.

Masih

Positivity of q-analogs of central binomial coefficients?

Math Overflow Recent Questions - Sun, 04/14/2019 - 02:52

With the usual $q-$notations $[n]_q=1+q+\cdots+q^{n-1}=\frac{\,\,1-q^n}{1-q},$ $[n]_q!=[1]_q[2]_q\cdots[n]_q$ and $\binom{n}k_q=\frac{[n]_q!}{[k]_q!\cdot[n-k]_q!}$ let $$b(n,k,r,q)=\det\left(q^{r\binom{i-j}2}\frac{[2i+k+1]_q}{[i+j+k]_q}\binom{i+j+k}{i-j+1}_q\right)_{i,j=0}^{n-1}.$$

It can be shown that $b(n,k,1,q)=\binom{2n+k-1}{n}_q.$ Therefore $b(n,k,1,q)$ has positive coefficients as a polynomial in $q$ for each positive integer $k.$

Computations suggest that also $b(n,k,0,q)$ and $b(n,k,2,q)=q^{n(n+k-1)}b(n,k,0,1/q)$ have positive coefficients.

Any idea how to prove this?

I need help with the following [on hold]

Math Overflow Recent Questions - Sun, 04/14/2019 - 01:07

enter image description here

need help with derivatives and general function questions

Properties of coefficients of cohomology theories

Math Overflow Recent Questions - Sun, 04/14/2019 - 00:08

Coefficients of cohomology theories can come from a variety of categories: fields, rings, sheaves ...

I wonder: what are the properties an object must satisfy in order to be a legitimate candidate for coefficients of a cohomology theory? What is the most general such object or structure in some suitable sense?

Modulus of continuity of flow for non-Lipschitz vector fields satisfies Osgood condition

Math Overflow Recent Questions - Sat, 04/13/2019 - 21:51

An Osgood modulus of continuity is an increasing function $\omega:(0,1]\to(0,1]$ such that $\int_0^1\frac{dt}{\omega(t)}=\infty$.

We say a vector field $X$ satisfies Osgood condition with modulus $\omega$ if locally there is a and $|X(x)-X(y)|\le\omega(|x-y|)$ (here I don't want a constant).

When $X$ satisfies Osgood condition, we know the ODE has uniqueness hence the flow $\mathfrak F_X(t,x)$ of $X$, given by $\begin{cases}\frac d{dt}\mathfrak F_X(t;x)=X(\mathfrak F_X(t;x))\\\mathfrak F_X(t;x)=x\end{cases}$ is well-defined locally in a small time interval.

My question is, given a Osgood modulus $\omega$, is there any estimate under modulus of continuity, that for any Osgood vector fields $X$ of modulus $\omega$, what is the modulus of continuity for $\mathfrak F_X$?

In DiPerna-Lions people mostly treat the cases where the vector field has bounded divergence, and the flow is not a regular Lagrange flow. But this is not true for general Osgood vector fields. I found a paper here of Clop the talk about the general Osgood vector fields but the regularity result seems still rough if we focus on some "smoother" modulus.

Consider a log-Lipschitz ODE $x'=x\log|x|$, $\begin{cases}\frac d{dt}\phi(t;s)=\phi\log|\phi|\\\phi(0;s)=s\end{cases}$, then $\phi(t;s)=|s|^{e^t}\operatorname{sgn}s$. This flow is not regular Lagrange.

Here globally $\phi$ is $\bigcup_{\alpha>0}C^\alpha$. And locally near 0 we have for any $\epsilon>0$ there is a neighborhood $0\in U_\epsilon\subset\mathbb R\times\mathbb R$, such that $\phi$ is $C^{1-\epsilon}(U_\epsilon)$.

For general log-Lipschitz vector field, is the flow (global or local) regularity in this case sharp?

I think in Saric's paper Here, for Zygmund vector field in 1-dim, the global result is $\bigcup_{\alpha>0}C^\alpha$ holds (so-called a quasisymmetric flow). What about the higher dimension?

If we consider log-log-Lipschitz case: Consider $x'=x\log|x|\log|\log|x||$, $\begin{cases}\frac d{dt}\phi(t;s)=\phi\log|\phi|\log|\log|\phi||\\\phi(0;s)=s\end{cases}$, then near 0 we have $\phi(t;s)=|s|^{|\log s|^{e^t-1}}\operatorname{sgn}s$. It's not even locally Holder near 0.

It's likely that the modulus of the flow depends on the asymtoptic behavior of $\int_\delta^1\frac{dt}{\omega(t)}$ as $\delta\to0$.

Pages

Subscribe to curious little things aggregator