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?

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?

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 $(1)$?

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?

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?

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.

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!

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.

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.

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*.

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.

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

Thank you for your attention,

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?

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

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

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

Can anyone help me on the proof of it?

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).

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.

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.

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.

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 $v_i=e_{i\, \text{mod}\, d}$?

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?