This is a refined version of Do binary symmetric channels maximize mutual information?, which was answered negatively.

Let the random variables $(X, Y)$ be a doubly symmetric binary source with parameter $0 \le p \le 1/2$, i.e., $X,Y \sim \text{Bernoulli}(1/2)$ and $P(X \neq Y) = p$.

Define the two regions $\mathcal{A}, \mathcal{B} \subseteq \mathbb{R}^3$:

$\mathcal{A}$ consists of all points $(R_0, R_1, R_2)$ such that there exist

**binary**random variables $U,V$, satisfying the Markov chain $U - X - Y - V$ and \begin{align} R_1 &\ge \mathrm{I}(U;X) , \\ R_2 &\ge \mathrm{I}(Y;V) , \\ R_0 &\le \mathrm{I}(U;V) , \end{align} where $\mathrm{I}(\cdot;\cdot)$ is mutual information.$\mathcal{B}$ consists of all points $(R_0, R_1, R_2)$ such that there exist probabilities $a,b \in [0,1]$ with \begin{align} R_1 &\ge 1 - \mathrm{H}(a * p) , \\ R_2 &\ge 1 - \mathrm{H}(b * p) , \\ R_0 &\le 1 - \mathrm{H}(a * p *b) , \end{align} where $a*b := a(1-b)+(1-a)b$ is binary convolution and $\mathrm{H}()$ is the binary entropy function.

My Question: Is $\mathrm{conv}(\mathcal A) = \mathrm{conv}(\mathcal B)$, where $\mathrm{conv}()$ denotes the convex hull?

Some comments:

The definitions of $\mathcal{A}$ and $\mathcal{B}$ are almost the same, except that the channels $X \to U$ and $Y \to V$ are required to be symmetric for $\mathcal{B}$ and hence $\mathcal B \subseteq \mathcal A$.

The question Do binary symmetric channels maximize mutual information? asked whether $\mathcal A = \mathcal B$ and a counterexample was provided for $p=0$, which does not seem to apply here.

Is it true that there exists some constant $A$ such that

$$\int_{2}^x \frac{1}{t}{\Big[\frac{d}{dt}Li(t)\Big]} \mathrm{d}t=\log\log x + A + O(x^{-1/2+\epsilon})$$ for any $\epsilon>0$, where $Li(z)=\int_{2}^z \frac{dz}{\log z} $ ?

For a Banach space $X$, the map $p:X\rightarrow\mathbb{C}$ is called a 2-homogeneous polynomial if there exists a bilinear form $A$ on $X$ such that $p(x)=A(x,x)$, the norm of this polynomial is defined by $sup_{\Vert x\Vert=1}|p(x)|$.

Take $X=l_\infty^n$, a finite dimensional space with maximum norm. I wanna show any 2-homogenous polynomial $p:l_\infty^n\rightarrow\mathbb{C}$ has a norm-preserving extension to $l_\infty$, a complex sequence space with supremum norm, i.e. there is an extension $\hat{p}:l_\infty\rightarrow\mathbb{C}$ so that $\Vert p\Vert=\Vert\hat{p}\Vert$ and $\hat{p}$ is a polynomial on $l_\infty$.

Any related articles are appreciated!

A theorem of Hungerford says that : Every PIR (principal ideal ring , obviously commutative ) is a homomorphic image of a finite direct product of PID s . My question is , is there a similar criteria for Noetherian rings, i.e. : Is every Noetherian ring a homomorphic image of a finite direct product of Noetherian domains ?

This Is every Noetherian Commutative Ring a quotient of a Noetherian Domain? shows that we cannot expect every Noetherian ring to be a homomorphic image of a Noetherian domain.

Recall that the spectrum of an invertible harmonic oscillator is continuous and
covers the real axis with **multiplicity two**.

Thanks a lot

Let $\{Q_{i}: i\in \mathcal{I}\}$ be a family of indecomposible injective modules over a commutative ring $R$ with identity. Is it true that the injective envelope of the direct sum of $Q_{i}$'s equals the direct product $\prod_{i\in \mathcal I}Q_{i}$? ( $R$ may not be Noetherian and $\mathcal{I}$ could be infinite).

Let $X$ be a real Banach space. Apart from the norm topology, we can consider the following weak topologies on $X$:

- the
**weak toplogy**, defined as the initial topology with respect to $X^*$. In other words, it is the coarsest topology for which all $f\in X^*$ are continuous. - the
**weak sequential topology**, which is essentially the topology induced by weak convergence. More precisely, we call a set closed if it is weakly sequentially closed, and this induces a topology.

It is easy to see that the weak topology is the weaker of the two (since every $f\in X^*$ is weakly sequentially continuous). Moreover, it is well known that a weakly sequentially closed set is not necessarily weakly closed. However, the picture is not so clear when it comes to compactness:

By the Eberlein-Smulian theorem, weak compactness coincides with weak sequential compactness. However, it is important to note that *weak sequential compactness* means sequential compactness, not compactness in the weak sequential topology (!!). In particular, this raises the following question:

What does ordinary compactness (i.e., **covering** compactness) look like in the weak sequential topology? Is it equivalent to weak (sequential) compactness?

(Also: if these are not equivalent in general, what about the case of convex sets?)

[This is a duplicate of this question on Stackexchange]

I am trying to figure out how to prove a very basic statement about convolution of $\ell$-adic/perverse sheaves in Katz's "Rigid local systems" (section 2.5.3, (1) ). The fact is fairly obvious and I guess that the proof is purely formal, but I'm not sure about it since my knowledge of the schemes formalism is quite limited.

**Settings**:
$G/k$ is a (smooth, separated) group scheme over a field $k$, of pure relative dimension $d$. Denote $\mu: G\times_k G\rightarrow G$ the multiplication map and $e: k\rightarrow G$ the identity section.
For $K$, $L\in D^b_c(G,\overline{\mathbb{Q}_l})$ (the "derived category" of $\ell$-adic sheaves over $G$, $l\neq \text{char}(p)$) define the product $K\times L\in D^b_c(G\times_k G,\overline{\mathbb{Q}_l})$ as
$$K\times L:=pr_1^{*}K\otimes^{\mathbf{L}}pr_2^{*}L$$
with $pr_1$, $pr_2$ the canonical projections $G\times_k G\rightarrow G$, and $\otimes^{\mathbf{L}}$ the derived tensor product, which I'll just denote by $\otimes$ in the following.

Now define their $\star_{*}$ convolution as $$K\star_{*}L:=R\mu_{*}(K\times L)\in D^b_c(G,\overline{\mathbb{Q}_l})$$

The claim is that, if $G$ is commutative, then the $\star_{*}$ convolution is commutative.

**What I did:**

We want to show that if $K$, $L\in D^b_c(G,\overline{\mathbb{Q}_l})$, then $R\mu_{*}(K\times L)=R\mu_{*}(L\times K)$, i.e., as $\otimes$ is commutative, that $R\mu_{*}(pr_1^* K\otimes pr_2^*L)=R\mu_{*}(pr_2^* K\otimes pr_1^*L)$

First, I think that $$(pr_2,pr_1)^* (pr_2^*K\otimes pr_1^*L)\simeq pr_1^*K\otimes pr_2^*L$$ i.e., $(pr_2,pr_1)^*(L\times K)=K\times L$. This sounds reasonable that, if we switch both factors in $G\times_k G$, then we should replace $K\times L$ by $L\times K$. But I'm not sure this is obvious.

Moreover, we have the following commutative diagram (as $G$ is commutative) \begin{array}{ccc} G\times G &\xrightarrow{(pr_2, pr_1)}& G\times G\\ |& &| \\ \mu & &\mu \\ \downarrow & &\downarrow\\ G &\xrightarrow{id}&G \end{array}

and I think it's also cartesian. I would like to use some kind of "proper base change" to say that $R\mu_*(pr_2,pr_1)^*=R\mu_*$, and to apply $R\mu_*$ to my (claimed) equality $(pr_2,pr_1)^*(L\times K)=K\times L$ to conclude.

But the problem is that I'm not sure that I can do it directly with my diagram, as $\mu$ is maybe not proper.

- Another option could just be to use the commutativity of my diagram to write $R\mu_*R(pr_2,pr_1)_*=R\mu_*$, and invoke that $R(pr_2,pr_1)_*=(pr_2,pr_1)^*$ by proper base change applied to the following diagram \begin{array}{ccc} G\times G &\xrightarrow{(pr_2, pr_1)}& G\times G\\ |& &| \\ id & &(pr_2,pr_1) \\ \downarrow & &\downarrow\\ G\times G &\xrightarrow{id}&G\times G \end{array}

Is there anything right in what I wrote ? Is there some easier proof ? Thanks for your help !

Let $f(x)$ be an irreducible integral polynomial with degree $\text{deg}[f]$. If $p|f(x)$ for some $x$, then denote by $S(p)$ the smallest positive number $x_0$ such that $p | f(x_0)$.

Do there exist infinitely many primes $p$ such that $$ S(p) < cp^{\frac{2}{\text{deg}[f]}} \, ,$$ for some constant $0<c<0.5$ ?

What happens if we replace $2$ by another constant $d$?

Let $\pi$ be an automorphic representation of $GL_2$ over a number field. What can I say concerning the order of the pole at $1$ of the $L$-function $L(s, \pi)$? Can we say more about $L(s, \mathrm{Sym}^2 \pi)$ or other constructions depending on $\pi$?

In the case where only holomorphic cusp forms appear, there is no pole. However, what about possible Maass forms?

Hedetniemi's conjecture is about the chromatic number of the categorical product of simple, finite, undirected graphs $G, H$ : it claims that $\chi(G\times H) = \min \{\chi(G), \chi(H)\}$.

For any simple, undirected graph $G=(V,E)$ we set $\text{col}(G) = \sup\{\delta(H): H\subseteq G\}+1$, where $\delta(\cdot)$ denotes the minimal degree. We have $\chi(G) \leq \text{col}(G)$ for any graph $G$.

Do we have $\text{col}(G\times H) = \min\{\text{col}(G), \text{col}(H)\}$ for all $G, H$?

Let $k$ be an algebraically closed field, $X, Y$ integral $k$-schemes and $Y$ proper over $k$. Let $U$ be a non-empty open subset $U \subset X$ and $f:U \to Y$ a morphism of finite-type. Suppose that for any closed point $x \in X \backslash U$, any DVR $R$ with fraction field $K$, residue field $k$ and any morphism $\phi:\mathrm{Spec}(R) \to X$ with $\phi(\mathrm{Spec}(K)) \in U$ and $\phi(\mathrm{Spec}(k))=x$ we have that the unique morphim $\phi_f:\mathrm{Spec}(R) \to Y$ lifting the morphism $$\mathrm{Spec}(K) \xrightarrow{\phi} U \xrightarrow{f} Y$$ satisfies the property $y(x):=\phi_f(\mathrm{Spec}(k))$ does not depend on the choice of the DVR $R$ or the morphism $\phi$ (i.e., $y(x)$ depends only on the choice of $x$). Does this imply that the morphism morphism $f$ extend to a morphism $\tilde{f}:X \to Y$ such that $\tilde{f}(x)=y(x)$ for all closed $x \in X\backslash U$?

Given three lines $l_a, l_b, l_c$ in $\mathbb {R}^3$ and three positive numbers $a, b, c>0$ I would like to find points $A, B, C$ on $l_a, l_b, l_c$ respectively, such that the side lengths of triangle $ABC$ are $a, b, c$. I know that this problem can not always have a solution and the solution is not necessarily unique but in my case I know that there is a solution and I know that I can start with a good initial guess. One possibility is to do an iteration, however I wonder if there is an analytical solution.

At the end my goal is to estimate the pose of three lines from three intersections with a sphere. This problem can be reduced to the problem above.

The 2-dimensional case would also be interesting. Furthermore I am also interested in the case where the three lines $l_a, l_b, l_c$ intersect in one point.

Consider $f\in L^2(I)$, where $I$ is the unit interval and $L^2$ is w.r.t. Lebesgue measure, and consider an approximation of $f$ denoted by $\tilde{f}\in L^2$.

The error in approximated the moments of $f$ by those of $\tilde{f}$ can be readily bounded by $\|f-\tilde{f}\|_2$, e.g., denoting the respective expectancies as $\mu$ and $\tilde{\mu}$, one has that

$$\left| \mu - \tilde{\mu} \right| \leq \|f-\tilde{f} \|_2 \, , $$

and if we define the variance in the usual way, as ${\rm Var}(f) = \mathbb{E}[(f-\mu)^2]$, then

$$\left| {\rm Var}(f) - {\rm Var}(\tilde{f}) \right| \leq \left(\mu +\tilde{\mu} + \|f\|_2 + \|\tilde{f}\|_2 \right) \|f-\tilde{f}\|_2 \, .$$

**Questions:**

- Are these bounds tight?
- Is there a known formula for the error in $\mathbb{E}[(f-\mu)^n]$ for all $n\in \mathbb{N}$?

*Remark:* The above bounds are obtained using mainly the triangle and Cauchy-Schwartz inequalities.

Let L1 and L2 be two Lie algebras.If U（L1）is isomorphic to U（L2）as associative algebra，then L1 is isomorphic to L2 ？

I am reading the following paper: http://ieeexplore.ieee.org/document/7798534/

For the equation (5):

$$\dot{a}=(1-a_i^2)a_i-\beta\sum_{k-i}(a_k\cos(\theta_k-\theta_i)-a_i)$$
$$\dot{\theta}_i=\beta\sum_{k-i}\frac{a_k}{a_i}\sin(\theta_i-\theta_k)$$

- Both are dynamics for amplitude $a$ and phase $\theta$.
- $k-i$ means there is an edge linking the node $k$ and $i$.
- $\beta$ is just a coefficient

My question is the following:

- Why the amplitude dynamics are invariant?

and the following:

- Why it is "global phase symmetry"
- Why the equilibria are semi-stable with respect to rotation?

Thanks!

Consider following polynomial sequence.

$$\begin{cases}a_{-1}=0,~a_0=1, \\a_{n+1}=x \cdot a_n \pm a_{n-1}\end{cases}$$

Here $+$ or $-$ is taken in such a way that all coefficients in $a_n$ do not exceed $1$ by absolute value (i.e. $\forall k \hookrightarrow |[x^k]a_n(x)|\leq 1$). The sequence seems to be infinite, but what is the strict proof of it and what are the properties of such sequence? Maybe it has unique name?

Also it seems that if we will write $0$ each time we use $-$ and $1$ each time we use $+$ in sequence $s_i$, there will be $2^{\lfloor\log_2 n\rfloor+1}$ sequences $s_i$ which generate correct $a_i$. Here is all $16$ possible patterns of first $15$ terms of $s_i$:

001011001011010 110100110100101

001011010011010 110100101100101

001101001010110 110010110101001

001101010010110 110010101101001

010010101101001 101101010010110

010010110101001 101101001010110

010100101100101 101011010011010

010100110100101 101011001011010

Can you see any pattern in here? Note that in the same row it is the sequence and same sequence reversed.

UPD: It seems that $s_0$ does not matter (since it produces $a_1=x$ in any way) and if we consider sequence $g_i$ which generates $\{s_i\}_{i=1}^{2^{n}-2}$, its prefixes can be generated as

$$\begin{cases}g_1=\varepsilon,\\g_{n+1}=g_n + (01|10) + g_n^r\end{cases}$$

where $g_n^r$ stands for reversed $g_n$.

Assume that $(M, [\lambda, \mu])$ defines an embeddable 3 dimensional CR structure where $\lambda$ is a real form and $\mu$ is a complex 1-form. Because $M$ is embeddable, $\mu=dz$ for some complex variable $z$.

We also assume that the equation $\partial_z \phi=f$ has a complex solution $\phi$ where $f$ is a non zero complex function. Can this equation also have any real solution?

A unit square is divided up with $n$ random lines. The random lines are chosen as follows, we choose one side of the square and pick a random point on that side. From there we choose a random point on one of the three other sides and connect the points. We then continue this process $n$ times.

After $n$ random lines what is the expected number of quadrilaterals formed?

For a scheme $X$, let $LE(X)$ denote the lisse-etale site on $X$. This is the full subcategory of $\textbf{Sch}/X$ consisting of smooth morphisms to $X$, equipped with the etale topology. Let $\mathcal{O}_X$ denote the presheaf on $LE(X)$ sending $$(U\rightarrow X)\mapsto \Gamma(U,\mathcal{O}_U)$$ This is in fact represented by $\mathbb{A}^1_X$, and hence is a sheaf (of sets)

Now let $k$ be an alg. closed field. Let $X := \mathbb{A}^1_k$ with coordinate $t$. Let $pt := \text{Spec }k$.

Consider the map $i : pt\rightarrow X$ sending $pt\mapsto (t=0)$. By Yoneda, it is possible to show that $i^*\mathcal{O}_X = \mathcal{O}_{pt}$ (ie, the inverse image sheaf of the functor of points of $\mathbb{A}^1_X$ in $LE(X)$ is the functor of points of $\mathbb{A}^1_{pt}$ in $LE(pt)$).

In particular, $i^*\mathcal{O}_X(pt) = k$. **I don't see how this agrees with the definition of $i^*$.**

By definition, $i^*\mathcal{O}_X$ is the sheafification of the presheaf defined as follows:

For every $Z\in LE(pt)$, consider the category $I_Z$ consisting of objects $(U,\rho)$ where $U\in LE(X)$ and $\rho : Z\rightarrow U_{pt} := U\times_X pt$ is a morphism in $LE(pt)$. A morphism $(U,\rho)\rightarrow (U',\rho')$ is a morphism $g : U\rightarrow U'$ in $LE(X)$ such that $g_{pt}\circ \rho = \rho'$.

Then, we consider the presheaf on $LE(pt)$: $$(i^*)^{pre}\mathcal{O}_X : (Z\rightarrow pt)\mapsto \varinjlim_{(U,\rho)\in I_Z^{op}} \mathcal{O}_X(U) = \varinjlim_{(U,\rho)\in I_Z^{op}}\Gamma(U,\mathcal{O}_U)$$ where given $h : Y\rightarrow Z$ in $LE(pt)$, we get a restriction map given by the functor $I_Z\hookrightarrow I_Y$ sending $(U,\rho)\mapsto (U,\rho\circ h)$. Then, we define $i^*\mathcal{O}_X$ to be the sheafification of this presheaf.

(So far this is the setup on p45-47 of Olsson's "Algebraic spaces and stacks")

My issue is: by the definition of $i^*\mathcal{O}_X$, there is a natural map $$k[t] = \mathcal{O}_X(X)\rightarrow i^*\mathcal{O}_X(pt) = k$$ Which can't be injective, and for example presumably sends $t,t^2$ both to 0. However, the global sections of the presheaf: $$(i^*)^{pre}\mathcal{O}_X(pt) = \varinjlim_{(U,\rho)\in I_{pt}^{op}}\mathcal{O}_X(U),$$ contains $\mathcal{O}_X(X)$ as a subset (since every $U\rightarrow X$ is smooth, hence flat, hence induces injections on global sections). Thus, the images of $t,t^2$ are distinct in $(i^*)^{pre}\mathcal{O}_X(pt)$. This means that if $t,t^2$ both map to 0 in $i^*\mathcal{O}_X(pt)$, they must agree in some etale covering of $pt$, but this seems impossible since $k$ is alg. closed, so the etale covers carry no additional information.

EDIT: Here is the argument that $i^*\mathcal{O}_X\cong\mathcal{O}_{pt}$. We have, for any sheaf $F$ on $LE(pt)$: $$Hom_{LE(pt)}(i^*h_{\mathbb{A}^1_X},F) = Hom_{LE(X)}(h_{\mathbb{A}^1_X},i_*F) = (i_*F)(\mathbb{A}^1_X) := F(i^*\mathbb{A}^1_X) = F(\mathbb{A}^1_{pt}) = Hom_{LE(pt)}(h_{\mathbb{A}^1_{pt}},F)$$ where we've used Yoneda twice, in the second and final equalities. But then, since this holds for any sheaf $F$ on $LE(pt)$, comparing the last term with the first, by covariant Yoneda we find that $i^*h_{\mathbb{A}^1_X} = h_{\mathbb{A}^1_{pt}}$ Since $\mathcal{O}_X$ is represented by $\mathbb{A}^1_X$ and $\mathcal{O}_{pt}$ is represented by $\mathbb{A}^1_{pt}$, this shows that $i^*\mathcal{O}_X \cong \mathcal{O}_{pt}$.