Recent MathOverflow Questions

computing a determinant involving roots of unity

Math Overflow Recent Questions - Fri, 06/23/2017 - 09:32

Let $d \geq 2$ be an integer and $\xi=\exp(\frac{2\pi i}{d})$. I am trying to compute the determinant of the matrix $$ (\xi^{ij}-1)_{1 \leq i, j \leq d-1}. $$ Let me call it $\Delta(d)$. For small values of $d$ I get:

$\Delta(2)=-2$

$\Delta(3)=-3\sqrt{3}i$

$\Delta(4)=-16i$

but I don't manage to prove a general formula. Could anyone help me?

Second Chern class of Stable Higgs sheaf

Math Overflow Recent Questions - Fri, 06/23/2017 - 09:06

Let $X$ be a smooth projective surface, $\mathcal{E}$ a stable torsion free Higgs sheaf of degree 0. Consider the following short exact sequence: $$0\rightarrow \mathcal{E}\rightarrow \mathcal{E}^{**}\rightarrow \mathcal{S}\rightarrow 0$$ where $\mathcal{E}^{**}$ is the double dual of $\mathcal{E}$. Then $\mathcal{S}$ is a skyscraper sheaf.

Q1: Why is the second Chern class $c_2(\mathcal{S})\leq 0$?

Q2: We know that since $E$ stable, so $E$ admits a Hermitian-Yang-Mills metric, so is $E^{**}$. Is it easy to explain that $c_2(E^{**})\geq0$, given the existence of Hermitian-Yang-Mills metric?

On the series $\sum_{\rho}x^{\rho}\Gamma(\rho)/\Gamma(\rho+k),\,0<k<1$

Math Overflow Recent Questions - Fri, 06/23/2017 - 08:22

Let $x>1$ be a real number. For a work I need to find an uniform estimation of the series the series $$\sum_{\rho}x^{\rho}\frac{\Gamma\left(\rho\right)}{\Gamma\left(\rho+k\right)}\tag{1}$$ where $\rho$ runs over the non-trivial zeros of the Riemann Zeta function and $0<k<1$ is a real number.

My attempt: I tried to use the classical estimation for the ratio of Gamma function $$\left|\frac{\Gamma\left(\rho\right)}{\Gamma\left(\rho+k\right)}\right|\leq\frac{1}{\left|\rho\right|^{k}}$$ but since $0<k<1$ it does not work. So I tried to to use the residue theorem. Since, if $c>1,$ we have $$\frac{1}{\Gamma\left(k\right)}\sum_{n\leq x}\Lambda\left(n\right)\left(1-\frac{n}{x}\right)^{k-1}=-\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}x^{w}\frac{\Gamma\left(w\right)}{\Gamma\left(w+k\right)}\frac{\zeta'}{\zeta}\left(w\right)dw$$ from the residue theorem we get $$\frac{1}{\Gamma\left(k\right)}\sum_{n\leq x}\Lambda\left(n\right)\left(1-\frac{n}{x}\right)^{k-1}=\frac{x}{\Gamma\left(1+k\right)}-\sum_{\rho}x^{\rho}\frac{\Gamma\left(\rho\right)}{\Gamma\left(\rho+k\right)}-\frac{\zeta'}{\zeta}\left(0\right)\frac{1}{\Gamma\left(k\right)}$$ $$-\frac{1}{2\pi i}\int_{-1/2-i\infty}^{-1/2+i\infty}x^{w}\frac{\Gamma\left(w\right)}{\Gamma\left(w+k\right)}\frac{\zeta'}{\zeta}\left(w\right)dw$$ but now I don't see how to evaluate the integral. I tried to use the Stirling's approximation but it does not work and I'm not able to see that if the real part of $w$ goes to $-\infty$ the integral vanish or not.

Remark: For $k=1$ we have the classical explicit formula for $\psi(x).$

Question: How can I evaluate $(1)$?

Conics, string art, and Bezier-like curves

Math Overflow Recent Questions - Fri, 06/23/2017 - 08:09

It is well documented that certain string-art patterns generate quadratic Bezier curves: let $x, y_1, y_2$ be three points in $\mathbb{E}^2$, consider the family of line segments joining $x + (1-t) (y_1 - x)$ to $x + t (y_2 - x)$ for $t\in [0,1]$, its envelope forms a quadratic Bezier curve (and hence a parabola). Furthermore, at the parameter $t$, the intersection of the line segment with the parabolic curve is at $$ (1-t)[x + (1-t)(y_1 - x)] + t[x + t(y_2 - x)] $$ the linear interpolant with parameter $t$ on the line segment.

It is also not hard to verify that given the same three points, considering the family of line segments joining $x + e^t(y_1 - x)$ to $x + e^{-t}(y_2 - x)$, its envelope forms a hyperbola. The intersection of the line segments with the hyperbola always occur at the midpoint.

In both cases the construction changes appropriately under affine transformations of $\mathbb{E}^2$, and in the case $x, y_1, y_2$ being colinear, recovers the degenerate conics of the appropriate types.

Given that the parabolas can be thought of as a limit of hyperbolas.

Question 1: Is there a way to take a "limit" of the construction above for the hyperbolas to get the one for the parabola?

Questions 2: Can this be taken further to include elliptical arcs as part of the family of constructions?

Equivalences of representation-finite selfinjective algebras

Math Overflow Recent Questions - Fri, 06/23/2017 - 08:01

Over an algebraically closed field two selfinjective representation-finite algebras are derived equivalent iff they are stably equivalent. Is there a good reason for that without using the known classification of such algebras up to isomorphism? Is this also true for arbitrary fields?

rational cohomology of classifying spaces of complex reductive Lie groups

Math Overflow Recent Questions - Fri, 06/23/2017 - 07:35

I am looking for a reference or an ad-hoc proof of the following fact, which seems to be known to experts: Let $\mathbf{G}$ be a complex algebraic group with maximal (algebraic) torus $\mathbf{T}$ and Weyl group $W$. Let $G=\mathbf{G}(\mathbb{C})$ and let $\mathrm{B}G$ denote the classifying space of $G$; define $T$ and $\mathrm{B}T$ similarly. Then the canonical map between the rational cohomology rings $\mathrm{H}^*(\mathrm{B}G,\mathbb{Q})\to\mathrm{H}^*(\mathrm{B}T,\mathbb{Q})$ is injective and its image is the $W$-fixed elements in $\mathrm{H}^*(\mathrm{B}T)$.

There are numerous proofs of the analogous result for compact (real) Lie groups, i.e. when $G$ is a compact Lie group and $T$ is maximal compact torus in $G$.

It should be possible to deduce the complex case from the real one because every complex reductive Lie group $G$ contains a maximal compact real Lie group $G_0$ which is a retract, but this fact alone is not sufficient as one has to take into consideration the corresponding maximal tori, and make sure that the corresponding Weyl groups of $G$ and $G_0$ relative to these tori are canonically isomorphic.

As $N$ shrinks to $\{1\}$, the functions $F_N$ tend to $H$ in $L^2$

Math Overflow Recent Questions - Fri, 06/23/2017 - 07:24

Let $G$ be a compact group together with normalized Haar measure. Let $H\in L^2(G)$. For any open neighbourhood $N$ of the identity element $1$ of $G$, we define $$F_N(x)=\frac{1}{|N|}\int_G I_N(y)H(h^{-1}x)dy$$ where $I_N$ is the indicator function of $N$ and $|N|$ is the Haar measure of $N$. I try to show that as $N$ shrinks to $\{1\}$, the functions $F_N$ tend to $H$ in $L^2$.

$Attempt$: For any open neighbourhood $N$ of $1$, we have \begin{align*} \|F_N-H\|_2^2 &=\int_G|(F_N-H)(x)|^2dx\leq \int_G(|F_N(x)|+|H(x)|)^2dx \qquad \text{by triangle inequality} \\& =\int_G|F_N(x)|^2dx+2\int_G|F_N(x)||H(x)|dx+\int_G |H(x)|^2dx\qquad (1) \end{align*} Note that using Schwartz inequality we have $$\int_G|F_N(x)|^2dx=\frac{1}{|N|^2}\int_G|(I_N*H)(x)|^2dx\leq \frac{1}{|N|^2}\int_G\|I_N\|_2^2 \|H\|_2^2dx=\frac{1}{|N|^2}\|I_N\|_2^2 \|H\|_2^2$$ and $$\int_G|F_N(x)||H(x)|dx\leq \frac{1}{|N|}\|I_N\|_2\|H\|_2\|H\|_1$$ Therefore (1) becomes \begin{align*} \|F_N-H\|_2^2 &\leq \frac{1}{|N|^2}\|I_N\|^2\|H\|_2^2+\frac{2}{|N|}\|I_N\|_2\|H\|_2\|H\|_1+\|H\|_2^2 \\&=\underbrace{\frac{1}{|N|^2}|N|}_{(*)}\|H\|_2^2+\underbrace{\frac{2}{|N|}|N|^{1/2}}_{(**)}\|H\|_2\|H\|_1+\|H\|_2^2 \end{align*} $Question$: As $N$ shrinks to $\{1\}$, the term $(**)\to 0$ but $(*)\to\infty$. Could anyone help me to find my mistake? Thanks!

Spiral lattice random walk

Math Overflow Recent Questions - Fri, 06/23/2017 - 07:07

Execute a random walk from the origin on the integer lattice, but bias the four compass-direction probabilities from $\frac{1}{4}$ each to prefer to step in a spiraling direction. Calling the four step vectors $c_0,c_1,c_2,c_3$, with $c_i=(\cos (i \, \pi/2), \sin (i \, \pi/2))$, adjust the probabilities as follows. Let $v$ be the vector from the origin to the last point on the path, and $n$ the unit normal to $v$, counterclockwise $90^\circ$ to $v$. Then select step $c_i$ with probability $\frac{1}{4} (1+ c_i \cdot n)$.          
          $\theta=60^\circ$. $\frac{1}{4}(1+c_2 \cdot n) = (1+\sqrt{3}/2)/4 \approx 0.47$. Unsurprisingly, the random walks spiral around the origin:    
    $2000$-step random walks. Origin: green. Last point: red.

Q. Does Pólya's recurrence theorem hold for these walks? Do the walks return to the origin with probability $1$?

All three of the above examples returned to the origin, but rather quickly: within $12$ steps.

Are triangulated equivalence detected at compact level?

Math Overflow Recent Questions - Fri, 06/23/2017 - 06:15

Suppose that $D$ and $E$ are compactly generated triangulated categories, even algebraic (i.e. equivalent to derived categories of small dg categories) if we want, and asume that their subcategories $D^c$ and $E^c$ of compact objects are triangle equivalent. Are $D$ and $E$ triangle equivalent?. By Theorem 9.2 of Keller's 'Deriving dg categories', we know that the answer is 'yes' when either $D$ or $E$ is the derived category of (the category of modules over) a small $k$-linear category, but I do not know the general answer. Any help would be highly appreciated.

Relaxed Condition for Rayleigh Quotient Theorem?

Math Overflow Recent Questions - Fri, 06/23/2017 - 06:08

According to Matrix Analysis by Roger A. Horn and Charles R. Johnson, a theorem related to Rayleigh Quotient is as follows (my compact version):

[Theorem 4.2.2 (Rayleigh Quotient) ] Let $A \in M_n(\mathbb{C})$ be Hermitian, let the eigenvalues of $A$ be ordered as $\lambda_1(A) \le \lambda_2(A) \le \ldots \le \lambda_n(A)$, let $i_1,...,i_k$ be given integers with $1 \le i_1 < ... < i_k \le n$, let $ \boldsymbol{x_{i_1},...,x_{i_k}}$ be orthonormal and such that $A x_{i_p} = \lambda_{i_p} x_{i_p}$ for each $p=1,...,k$, and let $S=span\{x_{i_1},...,x_{i_k}\}$. Then \begin{equation}\label{eq34} \lambda_{i_1}=\min_{\{x:0 \ne x \in S\}} \frac{x^{*}Ax}{x^{*}x} \end{equation} where $x^{*}$ denotes the conjugate transpose of $x$. The minimum value is achieved if and only if $Ax=\lambda_{i_1}x$.

Given the special spectrum properties of a Hermitian matrix, I just wonder is the theorem still valid if I remove the parts in bold, i.e., I don't require the eigenvectors of $A$ to be orthonormal?

Let me put in another way. Suppose $x_{i_1},...,x_{i_k}$ are orthonormal eigenvectos while $\hat{x}_{i_1},...,\hat{x}_{i_k}$ are non-orthonormal$^*$ eigenvectors of the Hermitian matrix $A$. Then is the following equation correct? \begin{equation} span(x_{i_1},...,x_{i_k})=span(\hat{x}_{i_1},...,\hat{x}_{i_k})? \end{equation}

$^*$ I know that eigenvectos of the Hermitian matrix $A$ associated with different eigenvalues are automatically orthogonal. So non-orthonormal applies only to the case when $A$ has multiple eigenvectors associated with the same eigenvalue; these eigenvectors are not automatically orthogonal.

Unirationality of moduli spaces of marked elliptic curves

Math Overflow Recent Questions - Fri, 06/23/2017 - 05:58

Let us consider the moduli space of genus one curves with an effective divisor of degree d; this space is birational to the quotient of the usual moduli space of genus one curves with $d$ marked points $M_{1,d}$ by the symmetric group $S_d$ action which permutes the marked points.

I'd like to figure out whether $M_{1,d}/S_d$ is unirational for large $d$, and I am pretty sure that this must be known.

By googling the keywords I found out that $M_{1,d}$ is unirational for $d \le 10$, so that the quotient $M_{1,d}/S_d$ is unirational is well, and $M_{1,d}$ not unirational for $d \ge 11$.

My motivation to ask this is the following: in small degree, genus one curves with an effective degree $d$ divisor are obtained as complete intersections in the projective space $P^d$. For instance genus one curves with a degree $3$ divisor are parametrized by a choice of a cubic and a line in $P^2$, genus one curves with a degree $4$ divisor are intersections of two quadrics in $P^3$, and in degree $5$ one takes plane sections of the Grassmannian $Gr(2,5) \subset P^9$. This goes on for a little while (actually I am not sure for how long, perhaps for $d \le 10$), and shows that in small degree the moduli space $M_{1,d}/S_d$ is unirational. Non-unirationality in large degree will morally mean that there is no universal geometric construction of genus one curves with a degree $d$ divisor.

Eigenvalues of a sample covariance matrix [on hold]

Math Overflow Recent Questions - Fri, 06/23/2017 - 04:19

Do you know any good pedagogical course about the eigenvalues and eigenvectors of sample covariance matrices.

Thank you for your attention,

A simple but curious determinantal inequality

Math Overflow Recent Questions - Fri, 06/23/2017 - 04:16

Let $A$ and $B$ be $n\times n$ Hermitian positive definite matrices and $k>0$ real. Then $A^k$ is well-defined and experimentally, we have $$\det(A^k+BABA^{-1})\geqslant \det(A^k+BA^{-1}BA),$$or equivalently $$\det(A^k+A^{-1}BAB)\geqslant \det(A^k+ABA^{-1}B).$$ Note that this would imply the inequality $\color{green}{\det (A^2+BAAB)\geqslant \det(A^2+ABBA)}$ found (but presumably unproved) by M. Lin, which is mentioned in a comment here.

One striking fact is that $B$ (or $A$, same outcome) can be multiplied by a positive factor, which means that the inequality remains valid for any "proportionality" between the two terms of each sum. Indeed, for any $\lambda>0$, we have equivalently $$\det(A^k+\color{red}\lambda BABA^{-1})\geqslant \det(A^k+\color{red}\lambda BA^{-1}BA).$$

Any ideas how to prove this conjecture?

need help on the proof of the Christoffel–Darboux formula of Laguerre Polynomial

Math Overflow Recent Questions - Fri, 06/23/2017 - 02:08

I found the Christoffel–Darboux formula of the Laguerre polynomials on wikipedia, https://en.wikipedia.org/wiki/Laguerre_polynomials

\begin{align} K_n^{(\alpha)}(x,y) &:= \frac{1}{\Gamma(\alpha+1)} \sum_{i=0}^n \frac{L_i^{(\alpha)}(x) L_i^{(\alpha)}(y)}{{\alpha+i \choose i}}\\ &{=}\frac{1}{\Gamma(\alpha+1)} \frac{L_n^{(\alpha)}(x) L_{n+1}^{(\alpha)}(y) - L_{n+1}^{(\alpha)}(x) L_n^{(\alpha)}(y)}{\frac{x-y}{n+1} {n+\alpha \choose n}} \\ &{=}\frac{1}{\Gamma(\alpha+1)}\sum_{i=0}^n \frac{x^i}{i!} \frac{L_{n-i}^{(\alpha+i)}(x) L_{n-i}^{(\alpha+i+1)}(y)}{{\alpha+n \choose n}{n \choose i}}; \end{align}

but there is no reference for the proof of the last equation. I have tried for several days, but failed.

I used the definition of Laguerre polynomials

\begin{equation} L_n^{(\alpha)} (x) = \sum_{i=0}^n (-1)^i {n+\alpha \choose n-i} \frac{x^i}{i!} \end{equation}

and the expansions

\begin{equation}\frac{x^n}{n!}= \sum_{i=0}^n (-1)^i {n+ \alpha \choose n-i} L_i^{(\alpha)}(x),\end{equation} \begin{equation}L_n^{(\alpha)}(x)= \sum_{i=0}^n L_{n-i}^{(\alpha+i)}(y)\frac{(y-x)^i}{i!},\end{equation} \begin{equation}L_n^{(\alpha+1)}(x)= \sum_{i=0}^n L_i^{(\alpha)}(x)\end{equation}

The generating function for Laguerre polynomials is

\begin{equation}\sum_n^\infty t^n L^{(\alpha)}_n(x)= \frac{1}{(1-t)^{\alpha+1}} e^{-\frac{tx}{1-t}}.\end{equation}

Some other references are listed as follows. I am not sure if they would help.

[1] https://academic.oup.com/qjmath/article-abstract/os-11/1/18/1574925/EXPANSIONS-AND-INTEGRAL-TRANSFORMS-FOR-PRODUCTS-OF?redirectedFrom=fulltext

[2] doi.org/10.1016/j.jat.2009.07.006

Can anyone help me on the proof of it?

Generalization of topological functor which need not be fibre-complete

Math Overflow Recent Questions - Fri, 06/23/2017 - 01:24

Is there a useful property for a concrete category $(\mathcal A, U : \mathcal A \to \mathcal X)$ such that in this case $U$ is fibre-complete iff $U$ is topological? (kind of like what "flat" is too "left exact).

The motivation is the following: given a finitely complete category $\mathcal X$ we can define the concrete category of $\mathcal X$-objects equipped with relations / reflexive relations / preorders / congruences / etc. . These seem to be topological iff the functor $U : (X,R) \mapsto X$ is fibre-complete (unless I'm mistaken).

Almost All Hyperplanes are Not Tangent [on hold]

Math Overflow Recent Questions - Thu, 06/22/2017 - 18:12

I am teaching a summer qualifying exam class, and a student was looking up practice questions online. I cannot figure one of them out.

I am trying to solve this problem: Let $M$ be an $n$-dimensional manifold embedded in $\mathbb{R}^{n + 1}$. Then almost every hyperplane in $\mathbb{R}^{n + 1}$ is not tangent to $M$ at any point.

The hint given is to consider the map $f: M \to S^n$ that takes $x \in M$ to the unit normal at $x$.

I first thought to use Sard's Theorem and analyze the critical values of this map. Then I found examples where there are points with tangent hyperplanes, but $f$ has no critical values. I've tried to define other maps and analyze them, but can't produce a map that has critical values precisely where I want them.

I'd like if the solution was via the hint, but any solution is welcome.

Which manifolds are sensitive to the cocycle in the Dijkgraaf-Witten model?

Math Overflow Recent Questions - Thu, 06/22/2017 - 04:45

Often, TQFTs are defined in families, parametrised by some algebraic data. For example, the Turaev-Viro-Barrett-Westbury TQFTs are parametrised by spherical fusion categories, the Crane-Yetter TQFTs are parametrised by ribbon fusion categories, and the $n$-dimensional Dijkgraaf-Witten theory is parametrised by a finite group $G$ and an $n$-cocycle $\omega$.

Instead of regarding TQFTs with a fixed datum as an invariant of manifolds (and cobordisms), one can also fix a manifold and regard a family of TQFTs as invariant of the parametrising data. For example, the Crane-Yetter invariant of $\mathbb{CP}^2$ is the "Gauss sum" $\sum_X d(X)^2 \theta_X$ of the ribbon fusion category, where $X$ ranges over simple objects and $\theta$ is the twist eigenvalue. (I'm thanking Ehud Meir for making me appreciate this viewpoint.)

From this viewpoint, my question is: In the Dijkgraaf-Witten TQFT, which manifolds give invariants that are sensitive to the cocycle?

In details, let us define the following invariant: $$ DW_{G,\omega}(M) = \sum_{\phi\colon \pi_1(M) \to G} \int_M \phi^* (\omega)$$ Here, $M$ is an $n$-manifold, $\omega \in H^n(G,U(1))$ is an element of the $n$-th group cohomology, and $\phi^*\colon H^n(G,U(1)) \to H^n(M,U(1))$ is induced by the flat $G$-connection $\phi$.

I'm looking for a manifold $M$ such that $DW_{G,\omega}(M) \neq DW_{G,\omega'}(M)$ for some $\omega \neq \omega'$. Ideally, the example would be in 3 or 4 dimensions.

Is every transitive ZF-model of inaccessible height a truncation of an inner model?

Math Overflow Recent Questions - Wed, 06/21/2017 - 08:44

Let $\kappa$ be an inaccessible cardinal and let $M \subseteq V_{\kappa}$ be an inner model of $V_{\kappa}$, i.e., a transitive model of $\mathsf{ZF}$ containing all the ordinals up to $\kappa$.

My question is whether such a model is always a rank-truncation of an inner model of $\mathbf V$ (defined using $M$ as parameter). Equivalently this question can be phrased as follows:

If $M \subseteq V_{\kappa}$ is an inner model of $V_\kappa$ is then $\mathbf L(M) \cap V_\kappa = M$, where $\mathbf L(M)$ is the minimal inner model of $\mathbf V$ containing $M$ ?

If the answer to this question is negative, I am curious whether there can be a counterexample $M$ which is definable in $V_\kappa$ without parameters and/or which is a model of $\mathsf{ZFC}$. If the answer to this question is positive, I am curious whether it is still positive if we weaken the assumption that $\kappa$ is inaccessible to $\kappa$ being worldly.

The minimum of a sum of absolute values of inner products in $\mathbb{R}^d$

Math Overflow Recent Questions - Wed, 06/21/2017 - 05:02

Consider a collection of unit vectors $v_1, \ldots, v_n$ in $\mathbb{R}^d$ (we think of $n$ being much larger than $d$). I would like to minimize the sum:

$$\sum_{i\neq j}|\langle v_i,v_j\rangle|.$$

Clearly, if $n=d$, the minimum is attained by taking $v_i=e_i$. Could it be that for $n>d$ in order to minimize the latter expression it is still best to take the vectors $v_i=e_{i\, \text{mod}\, d}$?

Characters of a quadratic extension and convergence

Math Overflow Recent Questions - Tue, 06/20/2017 - 09:31

Let $F$ be a non-archimedean local field, $\chi$ a quasi-character of $F^\star$ and $\psi$ a positive character of $E^\star$. I would like to understand why the usual Rankin-Selberg zeta integrals converge, and this question reduces to the convergence of

$$\int_{E^\star} \chi(a) \psi(a) |a|^s d^\times a$$

this convergence seems to be true in a half-plane $Re(s)>s_0$ for some $s_0 \in \mathbb{R}$, depending only on $\chi$ and $\psi$.

Why is that true? What do we know about characters of $E^\star$ to be able to conclude that straigthforwardly? And why, when converges, this integral gives a polynomial in $q^{—s}$ where $q$ is the cardinality of the residue field?

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