Let $A \mapsto S$ and $B \mapsto S$ be two schemes over the scheme $S$. Is there a connection between the automorphism group of the scheme $A \otimes_{S} B$ and the automorphism groups of $A$ and $B$ ? What about special cases such as:

the case where $A$ and $B$ are affine schemes over $\mathrm{Spec}(k)$ with $k$ a field, or

specific examples which behave good/bad ?

(Pierre Gillibert asked me this question and I post it with his permission.)

Let $X$ be an infinite set, and $f\colon[X]^\omega\to[X]^\omega$. We say that $\{x_n\mid n<\omega\}\subseteq X$ is a *free decreasing sequence* (for $f$) if for all $n$, $x_n\notin f(\{x_k\mid k>n\})$.

Is there an infinite set $X$ such that for every isotone $f\colon[X]^\omega\to[X]^\omega$ there exists a free decreasing sequence? Is it at least consistent from assumptions such as $V=L$, large cardinals or strong forcing axioms?

Some observations:

It is clear that $X$ is uncountable, because otherwise $f(A)=X$ for all $A\in[X]^\omega$ would pose a counterexample.

If there is no such set of size $\kappa$, then there is no such example of size $\kappa^+$.

If $\kappa$ has uncountable cofinality, and there is no such set of size $<\kappa$, then there is no example of size $\kappa$, since we can "glue" counterexamples and use the fact that every countable set is bounded (this is in effect the same proof for the previous observation).

In “Simple group of Lie type” by R. W. Carter there is a remark (after Theorem 13.7.4):

It is not known whether $H^1$ coincides with the set of $\sigma$-invariant elements of $H$ if $\mathfrak{L}$ has type $G_2$ and $K$ is infinite.

Here $H$ is the maximal split torus of $G_2$, $\sigma$ is an automorphism of $K$ such that $\sigma^2\varphi=1$, $\varphi$ is the Frobenius, $U^1$ and $V^1$ are the subgroups of upper and lower unitriangular matrices stable under the exceptional automorphism of $G_2$ induced by $\sigma$ and the root length changing symmetry of the Dynkin diagram, the subgroup $G^1=\langle U^1, V^1\rangle$ and $H^1=H\cap G^1$.

My question: is this still the case?

In the celebrating paper "Completely continuous endomorphisms of p-adic Banach spaces", Serre established a Fredholm-Riesz theory for compact endomorphisms of Banach spaces over (spherically complete) non-Archimedean field.

Later, people have two different generalizations. In one direction, people fix the ring and generalize the vector spaces , i.e people establish the Fredholm-Riesz theory for nuclear endomorphisms of locally convex vector spaces. On the other direction, people establish the Fredholm-Riesz theory for compact operators of (ON-able) Banach modules over general (Noetherian) Banach rings.

Did there exist a combination of these two generalization?Say, a Fredholm-Riesz theory for "Nuclear endomorphisms" of certain "locally convex" modules over general Banach ring? If this kind of generalization does not exist in general, what's the essential obstruction?

What is the motivation of an open set in a metric space? I understand how open balls are motivated, but I do not understand why one would want to define an interior point or an open set.

I am only talking about metric spaces, not topological spaces.

I'm working on a special kind of graphs. To prove some uniqueness, I need to prove that the polynomial \begin{equation} x^{8}-7x^{6}+14x^{4}-8x^{2}+1 \end{equation} does not have any root of the form \begin{equation} 2\cos\frac{(2k+1)\pi}{2n} \quad k\in \lbrace 0,1,\cdots , n-1 \rbrace , n \in \mathbb{E}. \end{equation} Can anybody help me?

Bests.

The definition of a "connected Shimura datum" (as in Milne's notes) is a pair $(G, X)$, where $G$ is a reductive algebraic group and $X$ is a $G(\mathbb{R})$-conjugacy class of morphisms $$ x: \mathbb{S}^1 \to G_\mathbb{R}, $$ where $\mathbb{S}^1$ is the norm one subtorus of $\text{Res}_{\mathbb{C}/\mathbb{R}}$ satisfying a short list of axioms. Given such a morphism $x$, one gets a family of morphisms $$ x_n: \text{Res}_{\mathbb{C}/\mathbb{R}} \mu_n \to G_\mathbb{R} $$ compatible with the natural inclusions $\mu_n \hookrightarrow \mu_{mn}$, and if I'm not mistaken, by Zariski density, $x$ is determined uniquely by the $x_n$, and moreover, two maps $x$, $x'$ are $G(\mathbb{R})$ conjugate if and only if their associated families $x_n$, $x_n'$ are. It also makes sense to demand that Deligne's axioms hold for the $x_n$, and one sees that if they hold for $x$, they hold for $x_n$.

From the perspective of special points and canonical models, it is not clear to me where one uses the morphism $x$; on the level of real points, it seems that only the action of roots of unity are used, and so much of the theory should be recoverable from only the $x_n$; however, there is a lot of it that I haven't understood yet.

My questions are:

if we define a generalized Shimura datum to be a family of $x_n$ compatible with the natural inclusions (equivalently, a conjugacy class of maps from the direct limit of the $\mu_n$), do we get generalized Shimura varieties?

If so, are there generalized Shimura varieties which do not come from Shimura varieties?

We know that $ \lim_{n \to \infty} \int\limits^{+\infty}_{-\infty}\frac{\sin (nx)}{x \pi} \cos (nx) dx$ is equal to $\cos (0)$ since $\frac{\sin (nx)}{x \pi}$ is delta function when n-->infinity.

But if we write $\sin (nx) ×\cos( nx)=\frac{\sin (2nx)}{2}$, then the integral becomes $\int\limits^{+\infty}_{-\infty}\frac{\sin(2nx)}{2x \pi} dx$. And if we pull out constant 1/2π out of integral, the remaining integrand is just a sinc function. Which has its integral value π. So now value of $\int\limits^{+\infty}_{-\infty}\frac{\sin(2nx)}{2x \pi} dx$= 1/2. Which is wrong. So my question is where i am going wrong?

Is there a compact Riemann surface other than sphere with an atlas consisting of open sets bioholomorphic to $\mathbb{C}$? Is there a compact Riemann surface other than sphere which possess an open subset bioholomorphic to $\mathbb{C}$?

Let $(E_b)_{b\in B}$ be a family of vector bundles on a smooth projective variety $X$, parameterized by a smooth curve $B$. Let $\mathrm{o}\in B$. Assume that $E_b$ is decomposable (= direct sum of two lower rank bundles) for $b\neq \mathrm{o}$. Can we conclude that $E_{\mathrm{o}}$ is decomposable? If not, what would be a counter-example?

*Background* : For $X=\mathbb{P}^n$, $n\geq 2$, I believe this is an open problem (see Mohan's answer to this MO question). I am primarily interested in the case of rank 2 vector bundles on a curve.

Let stable matrix (i.e., its eigenvalues have negative real parts) $B \in \mathbb R^{n \times n}$ and anti-symmetric matrix $T \in \mathbb R^{n \times n}$ satisfy

$$B^\top - T B^\top = B + B T$$

Prove that $\mbox{tr}(TB) \leq 0$.

What are necessary and sufficient conditions on $B$ such that $\mbox{tr}(TB) = 0$. (e.g. it is sufficient for $B$ to be symmetric. Is that also a necessary condition?)

Note that for each $B$, there is at most one $T$ satisfying this condition.

**Motivation**:
There is a claim in this paper that nonorthogonality of the eigenvectors of linear stability operator of a stochastic dynamical system amplifies the effect of noise, which is proven for the $2\times2$ case, and stated for the general case. This claim can be reduced to the above statement. Here is how it goes:

Consider a linear stochastic dynamics described by $$d x = Ax\, dt+\sigma dW,$$ where $\sigma>0$, $t\in\mathbb R^+$, $x(t)\in\mathbb R^n$, $A\in\mathbb R^{n\times n}$ with eigenvalues in left half plane, and $W$ is the $n$-dimensional Wiener process. This is an $n$-dimensional Ornstein-Uhlenbeck process.

If $A$ is symmetric, the distribution of $x$ at long time approaches a multivariate normal distribution with its covariance given by $A^{-1}$, and

$$\left\langle ||x||^2 \right\rangle = -\frac12\sigma^2\mbox{tr}(A^{-1}).$$

When $A$ is not symmetric, the covariance can be written as the inverse of a symmetric matrix $GA$, where $$\frac12(G^{-1}+(G^{-1})^\top) = I_{n\times n}.$$ This relationship along with the symmetry of $GA$ uniquely defines $G$. In this case

$$\left\langle ||x||^2 \right\rangle = -\frac12\sigma^2\mbox{tr}(G^{-1}A^{-1}).$$

The ratio of mean squared norm of $x$ to its value for a symmetric matrix with the same eigenvalues is what is called the amplification $$\mathcal H=\frac{\mbox{tr}(G^{-1}A^{-1})}{\mbox{tr}(A^{-1})}.$$ The claim is that $\mathcal H\geq 1$.

Let $B = A^{-1}$, and $T$ be the anti-symmetric part of $G^{-1}$. Now, $GA$ is symmetric iff $B^\top-TB^\top=B+BT$, and $\mathcal H\geq 1$ iff $\mbox{tr}(TB)\leq 0$: $$ \begin{cases} T = \frac12 (G^{-1}-(G^{-1})^\top)\\ I_{n\times n} = \frac12 (G^{-1}+(G^{-1})^\top) \end{cases}\implies G^{-1} = I_{n\times n}+T\\ (GA)^\top=GA\Leftrightarrow (G^{-1})^\top(A^{-1})^\top=A^{-1}G^{-1}\Leftrightarrow B^\top-TB^\top=B+BT.$$ $$\mathcal H=\frac{\mbox{tr}(G^{-1}A^{-1})}{\mbox{tr}(A^{-1})} = \frac{\mbox{tr}(B+TB)}{\mbox{tr}(B)}= 1+\frac{\mbox{tr}(TB)}{\mbox{tr}(B)}\geq 1\Leftrightarrow \mbox{tr}(TB)\leq 0.$$

I am having difficulty understanding how the equation $B^\top - T B^\top = B + B T$, puts a restriction on the $\mbox{tr}(TB)$, since taking the trace of both side of this equation gives no new information.

Let us consider the following conjecture:

**Conjecture**: There are no integer solutions of the equation $$x^{y-z}z^{x-y}=y^{x-z}$$ with $x,y,z$ distinct positive integers greater than or equal to $2$.

I came across this result when studying some diophantine equations. Several attempts were made to find a solution, but without any success. By this question I want to see if someone can give me a conterexample to this conjecture.

I am looking for a simple illustration of generating functions in graph theory.

So far, the matching polynomial seems to be the best. But I want something bit richer; at least a derivative should show up.

*Please help.*

For any square matrix $Y$ let $\chi_x(Y) = det(xI -Y)$ denote its characteristic polynomial.

Say $A$ and $B$ are two $n-$dimensional symmetric matrices with constant row sums $a$ and $b$. Lets define the polynomials $p_1(x)$ and $p_2(x)$ as follows : $\chi_x(A)=(x-a)p_1(x)$ and $\chi_x(B)=(x-b)p_2(x)$. Then over uniform sampling from the permutation group $S_n$ one can show (quite a non-trivial proof) that ``finite free convolution" (denoted as $\boxplus$) satisfies the following identity,

$$\mathbb{E}_{P \sim S_n} [\chi_x(A + PBP^T)] = (x-(a+b))[p_1(x) \boxplus p_2(x)]$$

Given a $n-$dimensional symmetric matrix $M$ such that, $\chi_x(M) = (x-1)^{\frac {n}{2}}(x+1)^{\frac {n}{2}}$ define a polynomial $p$ such that $\chi_x(M)=(x-1)p(x)$. Now apparently the following identity holds for any positive integer $d$,

$$\underset{P_1,P_2,..,P_d \sim S_n}{\mathbb{E}}[\chi_x(P_1MP_1^T+ P_2MP_2^T+..+P_dMP_d^T)]\\ = (x-d)[p(x)\boxplus p(x) ..(d \text{ times})..\boxplus p(x)]$$

Can someone kindly help derive the second equality from the first?

I believe this is some kind of an induction but I am unable to get it work. As in even at $d=3$ I cant get this explicitly.

Suppose we have a machine which takes the input $x_{in}$. In this machine the variable $x_{in}$ is converted to $y_{in}$ with the function $f(x)$, $f(x_{in})=y_{in}$. $f(x)$ is a known function, but not very easy to evaluate.

Secondly the machine is externally measured. This gives a measurement $x_{out}$. Assuming the measurement device has no errors, then there is phenomenon that converts $y_{in}$ to $x_{out}$ by a function, which we call $g(y)$. Since we don't know this phenomenon, $g(y)$ is unknown.

The machine works correct when for every $x_{in} > 0$, $x_{in}$ and $x_{out}$ are close together. To accomplish this, it is possible to set two parameters into the machine. Let's call those parameters $a$ and $b$. These paremeters are used in the following way: we take $y_{in}$ and set $y_{new} = a\times y_{in} + b$.

Since we do not know the function $g(y)$, we do not know what effect those parameters have on the measured output $x_{out}$. So in fact the problem here is about minimizing the following:

$||x_{in} - x_{out}|| = ||x_{in} - g(a\times y_{in} + b)||$

over $a$ and $b$, with the unknown function $g(y)$.

At the moment those parameters are set by using trial and error, but it can take up to two days to get the best set of parameters.

Now I've read some things about

- Simulated Annealing
- Black box optimization
- Surrogate modelling

But I'm not sure if I am looking in the right direction. Or if this problem is even solvable without trial and error. If it is solvable is there someopne who can give me some good referecences to this type of problems?

A famous theorem of Gromov says that the set of compact Riemannian manifolds with $Ric \geq c$ and $\text{diam} \leq D$ is relatively compact in the Gromov-Hausdorff metric. Chapter 10 of the book by Burago-Burago-Ivanov lists a few other such compactness theorems, but they are all dependent on some variant of global curvature bounds.

My question is, are there examples of such compactness theorems which do not use assumptions on the curvature? In other words, is it reasonable to expect some kind of compactness theorem under uniform control of a combination of other geometric quantities like volumes of balls, curve lengths, diameter, injectivity radius etc., but not curvature?

This is mainly a reference request, I am trying to get a feel for whether there is a metamathematical principle which tells us that one cannot expect relative compactness in the Gromov-Hausdorff metric without curvature restrictions.

Let $M$ be a compact kahler manifold with kahler form $\omega$. For any line bundle $\mathcal{L}$ on $M$,we define the '$\omega - degree$' of $\mathcal{L}$ to be $deg(\mathcal{L}) :=\int_{M}c_{1}(\mathcal{L})\wedge \omega ^{n-1}$.

How does one show that if $D$ is an effective cartier divisor on $M$ and $\mathcal{O}(D)$ be the corresponding line bundle, then $deg \mathcal{O}(D)\geq0$ ? The book I'm reading (Kobayashi's differential geometry on complex vector bundles) states that the integral above actually becomes $\int_{D} \omega^{n-1}$ and thus concludes from there, but I can't figure out why it's true. Thanks in advance!

Suppose $X$ is a smooth variety defined over $\mathbb{Q}$. There are (at least) two automorphisms of cohomology groups of $X$ that are called "Frobenius", and I would like to understand how they are related.

If $p$ is a prime of good reduction for $X$, then $H^n_{dR}(X)\otimes\mathbb{Q}_p$ depends functorially on the special fiber of an integral model of $X$. The Frobenius endomorphism of the special fiber induces an automorphism $F_p\in GL\big(H^n_{dR}(X)\otimes\mathbb{Q}_p\big)$.

Fix a prime $\ell$ and an algebraic closure $\bar{\mathbb{Q}}$ of $\mathbb{Q}$. Then $H^n_{et}(X_{\bar{\mathbb{Q}}},\mathbb{Q}_{\ell})$ comes with an action of $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ that is unramified for all but finitely many primes $p$. For unramified $p$, there is a well-defined conjugacy class $\Phi_p\subset GL\big(H^n_{et}(X_{\bar{\mathbb{Q}}},\mathbb{Q}_{\ell})\big)$ coming from the conjugacy class of the Frobenius at $p$ in $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$.

What is the relationship between $F_p$ and $\Phi_p$?

Let $\kappa$ be an infinite cardinal. We say $E\subseteq {\cal P}(\kappa)$ is an *infinite projective plane* on $\kappa$ if

- $e_1\neq e_2\in E$ implies $|e_1\cap e_2| = 1$, and
- whenever $n\neq m\in \kappa$, there is $e\in E$ with $\{n,m\}\subseteq e$.

Is it possible to find an infinite projective plane $E$ on an infinite cardinal $\kappa$ such that for all members of $E$ we have $|e|<\kappa$?

Let $F \in \mathbb{Q}[x_1, \ldots, x_n]$ be a homogeneous polynomial of degree $d > 1$ and $n \geq 2$. I am interested in finding an example (or maybe that such $F$ doesn't exist?) with the following properties.

1) $W = \{ \mathbf{x} \in \mathbb{R}^n : F(\mathbf{x}) = 0, \nabla F \not = \mathbf{0} \} \cap (\mathbb{R}_{>0})^n \not = \emptyset$.

2) For all $\mathbf{z} \in W$, we have $\frac{\partial F}{\partial x_i} \cdot \frac{\partial^2 F}{\partial x_i^2} \leq 0$ for every $i \in \{1, ..., n\}$.

I would greatly appreciate if someone could provide me an example of such polynomial, or an argument on why such polynomials do not exist. My guess is that ``most'' random choice of polynomial with non-empty $W$ will not satisfy 2) but I wasn't really sure... Any helpful comments are appreciated as well. Thank you very much.