This post is about an equivariant integration formula in a famous paper https://arxiv.org/pdf/alg-geom/9701016.pdf by Alexander Givental, where the author presented the formula without proof or reference. It turns equivariant localization into calculation of residues I am trying to combine it with global residue theorem to derive some vanishing result of some invariant.

$$ \int_X f(p,\lambda)=\sum_{\alpha}Res_{\alpha}\frac{f(p,\lambda)dp_1\wedge\cdots\wedge dp_k}{u_1(p,\lambda)\cdots u_n(p,\lambda)} $$ where $X$ is a toric symplectic variety, $p_k$ forms base of $H^2(X)$ (the Picard group) and $u_i(p,\lambda)=\sum_ip_im_{ij}-\lambda_j$, where $m_{ij}$ is a integer matrix describing how $p_i$ span all the invariant divisors and $\lambda_i$ is the ordinary equivariant index for torus action. The RHS sums over fixed point or pole specified by some of the $u_i=0$.

This paper is hard to read because I am not an expert on symplectic geometry but I really need to re-derive this explicit formula. Maybe some experts could kindly provide me some hints or references and any other discussion is welcomed. Thanks in advance.

With a friend I made a program in the GAP-package QPA to check whether a given finite dimensional quiver algebra is quasi-hereditary. It is very slow since it has to go through all permutations of points in the quiver but in principle it works by using just linear algebra. Now a similar class as quasi-hereditary algebras are cellular algebras: https://en.wikipedia.org/wiki/Cellular_algebra. A cellular algebra is quasi-hereditary iff it has finite global dimension.

Question: Is it possible to have a programm in QPA that checks whether a given finite dimensional quiver algebra is cellular (by using given commands that preferably only use linear algebra)?

I have no real experience with cellular algebras and the definitions make it look like the answer is no. But maybe there is an equivalent definition of cellular algebras that makes such a QPA-program easy?

We know that primes of the form $p=4k+1$ can be written as sum of two squares, i.e., $p=x^2+y^2$ (uniquely iff $0<x<y$). However, this expression holds for composite numbers that all of the prime factors are of the form $4k+1$, or if they have prime factors of the form $4k+3$, these factors are of even power.

I am looking for some expression to represent prime numbers of the form $p=4k+1$ such that it does not hold for composite numbers? Is it possible to have such expression?

P.S.: If there are some conditions (not necessarily expressing as polynomials) that determine such prime numbers would be desirable.

Thank you.

It is basic that the norm map $N:\mathbf{F}_{q^n}^* \to \mathbf{F}_q^*$ is surjective for finite fields. In fact $N(x) = x^{(q^n-1)/(q-1)}$. How well does this simple fact extend to subspaces?

A basic example is an intermediate extension $\mathbf{F}_{q^d}$. On $\mathbf{F}_{q^d}^*$ we have $$N(x) = \left(x^{(q^d-1)/(q-1)}\right)^{(q^n-1)/(q^d-1)} = \left(x^{(q^d-1)/(q-1)}\right)^{n/d}$$ since the term in the brackets is in $\mathbf{F}_q^*$ and $(q^n-1)/(q^d-1) \equiv n/d \pmod {q-1}$. So $N$ is surjective on $\mathbf{F}_{q^d}^*$ if and only if $(n/d, q-1) = 1$. In particular $N$ fails to be surjective on a subspace of dimension $n/2$ whenever $n$ is even and $(n/2, q-1) > 1$.

As a sort of converse note that if $(n,q-1)=1$ then $N$ is surjective on every one-dimensional subspace.

Is it true that if $V \leq \mathbf{F}_{q^n}$ is a $\mathbf{F}_q$-rational subspace of dimension $>n/2$ then $N$ is surjective on $V$?

Equivalently, if $\dim_{\mathbf{F}_q} V > n/2$, can we always find $x^{q-1} \in V$?

Vakil gives two equivalent definitions of associated points in his "Rising Sea":

- a prime ideal $p$ of a ring $A$ is called an associated prime for module $M$ if it is the annihilator of an element $ m \in M$, i.e. $p = \mathrm{Ann}(m) $
- a point p in $\mathrm{Spec} A$ is called an associated point for module $M$ if there is an element $ m \in M$ such that $p$ is a generic point of $\mathrm{Supp } \text{ }m$

Where $A$ is a Noetherian, $M$ is finite generated over A.

Vakil asks readers to check this equivalence, and he gives a hint:

if $p$ is an associated point, then there is an element $m$ with Support $\bar{p}$

If I prove this then I finish the exercise, but I can't.

• A special die is designed with n faces, enumerated 1 through n with all faces being equally likely.
If X is the observed number when this die is thrown, what is the expected value of X? [C]

(a) $\frac n2+1$

(b) $\frac n2$

(c) $\frac{n+1}2$

(d) $\frac{n−1}2$

Why is the answer c correct? I know that it is a discrete uniform distribution.

Does anyone know whether bipartite symmetric graphs are hamiltonian? I'm not sure whether anyone have proved it before, but a nonhamiltonian symmetric bipartite graph would lead to a counterexample to the Lovasz conjecture. I would appreciate any references or ideas.

Let $\mathcal{B}_{\mathbb{R}}$ be the Borel $\sigma$-algebra on $\mathbb{R}$ and $\mu_L$ be the Lebesgue measure on $\mathbb{R}$.

Define a new $\sigma$-algebra $\mathcal{B}_0$ as follows: $$\mathcal{B}_0=\{A\in \mathcal{B}_{\mathbb{R}}:\mu_L(E)=0\ \text{or}\ +\infty\}.$$ I want to prove that the family of all locally measurable sets of the measure space $(\mathbb{R},\mathcal{B}_0,\mu_L|_{\mathcal{B}_0})$, that is, $$\{E\subset \mathbb{R}:E\cap A\in \mathcal{B}_0\ \text{for all $A\in \mathcal{B}_0$ such that $\mu_L(A)<\infty$}\}$$ is not the family of all subsets of $\mathbb{R}$.

So I want to ask whether there exists a Lebesgue nonmeasurable set $E$ in $\mathbb{R}$ satisfies that $E\cap A$ is a Borel null set for every Borel null set $A$.

I have the following first-order difference equation

$$y_{t} = \frac{x_{t}}{1-\rho L} + \epsilon_{t}$$

where $L$ denotes the backshift operator, i.e., $L(x_{t}) = x_{t-1}$. I can obtain a solution to this difference equation heuristically, but I am wondering if there is a general procedure. A solution (the solution?) is

$$y_{t} = \sum_{i = 0}^{\infty} \rho^{i}X_{t-i} + \epsilon_{t}$$

I looked in Goldberger's text and couldn't find it there. Any reference is appreciated also. Thanks.

As the question title asks for, how do others "visualize" Witt vectors? I just think of them as algebraic creatures. Bonus points for pictures.

Take a category $C$, and take all endofunctors of $C$, so the set $E= \{ M| M: C \rightarrow C \}$. $E$ forms the objects of a category with morphisms given by all natural transformations $\mu : M \rightarrow N$ for $M,N \in E$. Let $\mathcal{C}$ be the endofunctor category as defined. What are the internal categories in $\mathcal{C}$?

Further suppose $C$ is a symmetric monoidal dagger category, in this case, what are the internal categories in $\mathcal{C}$?

Related to Relations between coefficients of expansions of a rational function at 0 and infinity

I commented at the linked question that the question seemed less about what happened "at infinity", and more about what happened "away from zero". And to some extent, the answer confirmed it by discussing the rank of (the biinfinite matrix corresponding to) the biinfinite sequence - which seems to be the degree of the rational function "away from zero and infinity". This is related to the functions $t$ and $t^{-1}$ on $\mathbb{P}^1$; they are the unique functions with a single zero and single pole at $0$ and $\infty$, or vice versa.

So that brings up a corresponding question:

Are there similar algebraic relations that could be made between the expressions of a rational function $f(t)$ when it is expressed as $A(t), B(t) \in F((t))$ such that $f(t) = A(t), f(t - c) = B(t)$ for some constant $c$?

This isn't quite a generalization of the original question, but gets into another interesting situation - when the zeroes don't match, but the poles do. More generally, we can separate the zeroes as well, leading to:

Are there similar algebraic relations that could be made between $A(t), B(t) \in F((t))$ such that $f(t) = A(t), B(t) = f(\frac{at + b}{ct + d})$?

As in the above question, there is no expectation of a finite algebraic relation. Instead, I'm hoping for a series of "loosening" conditions (similar to the conditions that the biinfinite matrix be rank $n$ for some $n \in \mathbb{Z})$, though I would expect that more than $1$ parameter would be necessary.

I am somewhat a beginner in the field of operator algebras and was wondering about the following:

Let $T$ be a linear map between the space of bounded operators $B(H)$ on some Hilbert space and $S$ a map between the space of trace-class operators that we denote by $N(H)$ in the sequel.

Then, one defines maps $T_n: M_n(B(H)) \rightarrow M_n(B(H))$ by $$T_n((a_{ij})):=(T(a_{ij}))$$

Then, $T$ is completely bounded if and only if $\sup_n \left\lVert T_n \right\rVert< \infty.$

And analogously for $S.$

I would like to know whether a map $T$ is completely bounded if and only if $$\left\lVert T \otimes \operatorname{id}_{X} \right\rVert< \infty$$ or in case of $S$ whether complete boundedness is equivalent to $$\left\lVert S \otimes \operatorname{id}_{X} \right\rVert< \infty$$

for some $X$?

For $T$ I think I found a reference on math.stackexchange saying that $X=B(K)$ works for any infinite-dimensional Hilbert space $K$ (nothing on whether $K$ needs to be separable discussed in this thread.) This should imply that $X=N(H)$ works for $S$ as well.

Assuming this to be true, I am wondering about two things now:

1.)Assume I can take $X=B(H)$ for $T$ and $X=N(H)$ for $S$, then $T \otimes \operatorname{id}_{B(H)}$ is a map from $B(H) \otimes B(H)$ into itself. Can the space $B(H) \otimes B(H)$ be identified with $B(H \otimes H)$?- I assume that by a duality argument this would be equivalent to asking whether $N(H) \otimes N(H)$ is isomorphic to $N(H\otimes H)$.

2.) Are there any other choices of $X$ allowed in the above two examples?

Some computational problems have variants that appear to be harder. For instance, Graph Automorphism (GA) problem has quasi-polynomial time algorithm ( by Babai's Graph Isomorphism result) while the fixed-point free GA problem is NP-complete.

Partition problem is weakly NP-complete problem since it has pseudo-polynomial time algorithm. I am interested in variants that are strongly NP-complete.

Here is a variant of partition problem:

Restricted partition problem

**Input**: Set $S$ of $2N$ integers, and a collection of pairs $P$ from $S$

**Query**: Is there a partition of $S$ into two equal cardinality parts $A$ and $S-A$ such that both parts have the same sum and no pair in $P$ has both elements in one side of the partition?

Is this variant of partition problem NP-complete in the strong sense?

Is the following claim correct (Chapter 13 before Theorem 87 of Todorcevic's book: Notes on forcing axioms): Let $\alpha$ be an infinite countable indecomposable ordinal and $U$ be an uniform ultrafilter on $\alpha$ (namely elements in $U$ have order type $\alpha$). Then for any collection $\{B_i\in U: i<\mathfrak{m}\}$ it is true that there exists $B\subset \alpha$ of order type $\alpha$ such that $B-B_i$ is finite for all $i<\mathfrak{m}$. Here $\mathfrak{m}$ is the least cardinal such that Martin's Axiom holds below $\mathfrak{m}$ (i.e. meeting any $\beta$ many dense sets for any $\beta<\mathfrak{m}$).

In fact, it will be interesting to know if this is true at all for countable collection $\{B_i: i<\omega\}$.

It is proved there that the statement is true if we relax the conclusion asking only $B-B_i$ to be bounded in $B$. Any thoughts?

Factoring x^15−1 into irreducible polynomials over GF(3) How can i find this? Thanks in advance.

Are there two symplectic structures $\omega_1, \omega_2$ on $M_{2n}(\mathbb{R})$ such that the function $Det:M_{2n}(\mathbb{R})\to \mathbb{R}$ is completely integrable with respect to $\omega_{1}$ but is not completely integrable with respect to $\omega_2$

Note that we do not limit the symplectic structures to structures with constant coefficients.

I had already asked a weaker version of this question in the following link but I confess that I did not understand the details(How does the action of orthonormal group guarantee's the integrability?)

http://mathforum.org/kb/thread.jspa?forumID=253&threadID=1653483&messageID=5990478#5990478

Consider the following system of coupled differential equations $$ \dot{x}_1(t) = -x_1(t) - \cos(\omega t)x_1(t) + \cos(\omega t)x_2(t), \ x_1(0)\in\mathbb{R},\\ \dot{x}_2(t) = -\gamma x_2(t) - \cos(\omega t)x_2(t) + \cos(\omega t)x_1(t), \ x_2(0)\in\mathbb{R}, $$ where $\omega$ and $\gamma$ are positive real constants. Observe that $\bar{x}=(\bar{x}_1,\bar{x}_2)=(0,0)$ is an equilibrium of the above system.

It is almost trivial to see that if $\gamma=1$ then $\bar x$ is attractive. Indeed, in this case, we have that $x(t)=[x_1(t), x_2(t)]^\top$ can be explicitly computed as $$ x(t) = \exp\left(\begin{bmatrix}-t &0\\ 0 & -t\end{bmatrix} + \frac{1}{\omega}\sin(\omega t)\begin{bmatrix}-1 &1\\ 1 & -1\end{bmatrix}\right)x(0), $$ so that $x(t)\to 0$ for $t\to \infty$.

However in case $\gamma\ne 1$ proving the attractiveness of the origin is not obvious (and perhaps not even true!).

In particular, numerical simulations seem to suggest that for $\gamma$ and $\omega$ sufficiently small (e.g. $\gamma=0.001$ and $\omega=10$) the equilibrium $\bar{x}$ is **not** attractive.

I've struggled a lot to find a way of formally proving this, with no luck. So I decided to post the problem here hoping that some of you will provide some useful suggestions or tips. Thank you!

I post here the Mathematica code that I've used in my simulations:

(* nominal values for simulation *) values = {gamma -> 0.001, w -> 10}; equations = { {x1'[t], x2'[t]} == {-x1[t] - Cos[w*t]*x1[t] + Cos[w*t]*x2[t], -gamma*x2[t] - Cos[w*t]*x2[t] + Cos[w*t]*x1[t]}, {x1[0], x2[0]} == {0.1, 0.1}}; {x1t, x2t} = NDSolveValue[equations /. values, {x1[t], x2[t]}, {t, 0, 1000}]; Plot[x1t, {t, 0, 1000}, PlotRange -> {-0.2, 0.2}] Plot[x2t, {t, 0, 1000}, PlotRange -> {-0.2, 0.2}]Let $M^n$ be compact, connected, oriented $n$-dimensional smooth manifold without boundary, the Hopf degree theorem states that the homotopy class of continuous maps from $M^n$ to $S^n$ is classified by its degree. What about continuous map from $M^n$ to $S^{n-1}$ (assume $n>0$) ? I am mostly interested in the case $n>2$.

More specially, when does there exist a non null-homotopic map in the case $n=3$? If $H^2(M,\Bbb Z) \not=0$ then it's true as $[M,\Bbb{C}P^{\infty}]=H^2(M,\Bbb Z)$ and $\Bbb CP^1 \cong S^2$. This is not necessary as $M=S^3$ shows.

We say that a sequence $(\mathcal X_n)$ of families of subsets of a topological space $X$ is a $\sigma$-disjoint cover of $X$ if every family $\mathcal X_n$ consists of mutually disjoint sets and $\bigcup\limits_n\bigcup\mathcal X_n=X$.

Let us say that a space $X$ is weakly Lindelof, if every open cover of $X$ admits a $\sigma$-disjoint subcover. Clearly, every Lindelof space is weakly Lindelof.

**Question 1.** Is there any well-known in the literature name for the class of "weakly Lindelof" spaces?

The following question concerns weaker property than "weak Lindeloffness".

**Question 2.** Does there exist a $\sigma$-disjoint cover of a Banach space $X$ by open balls of diameters $\le 1$?