There is one sentence I don't understand in some paper.

"A simply connected and conformally flat three mainifold can be conformally immersed into $S^3$" by the means of a developing map.

Is any reference about this short argument? Maybe it is a direct consequence from definition. Could anyone explain a little bit to me?

**Question:** Is there an approach to $\partial G \cong S^1$ implies virtually Fuchsian using bounded cohomology of $\mathrm{Homeo^+} (S^1)$? If not is there a reason to believe it wouldn't work, or maybe just a lot harder than known proofs?

I recently decided I would like to attempt to learn a proof of the theorem that for a hyperbolic group $G$ with $\partial G \cong S^1$ implies that the group is virtually Fuchsian. From what I understand the proof goes through showing that convergence group acting on $S^1$ conjugate to Fuchsian groups, and then show that hyperbolic group act on their boundary as convergence groups.

Something else I would like to learn more about is applications of bounded cohomology. It is my understanding that it is useful in determining things like conjugacy of representations into $\mathrm{Homeo}^+(S^1)$. So, I would guess (perhaps naively) that it would be useful or an alternative approach to the above question. If there was it would be a cool application...

I am having a hard time finding information on this (and can't find a copy of *Groups acting on the circle* by Ghys for some reason which is where I would think it would be discussed if it was a plausible approach) so I am guessing there isn't much connections, but maybe someone has thought about this before or is an expert who can see why bounded cohomology would not be useful.

I'm wondering if the following argument is correct:

Consider optimizing a complex functional S[x(t)]. Since S is complex, it only has an optimum with respect to the lexicographic order of the complex numbers. (This is different from Cauchy-Riemann stability, where δS = 0, but there is no notion of optimality).

To optimize with respect to the lexicographic order of the complex numbers, we first optimize Real(S)

0 = δ(Real S)

Next, we optimize imaginary(S) while keeping δ(Real S) = 0.

This is the part I'm unsure of: The Lagrange multiplier method is not suitable here, and in general, we cannot say that optimizing Imaginary S will also stabilize it. However, we can say that the difference between stabilizing Imaginary S and optimizing it is unpredictable noise, which is 0 on average.

δ(Imaginary S) = noise

Adding these together, gives δS = i noise.

Is my argument that δ(Imaginary S) = noise correct? Furthermore, what can we say about this noise?

If $a_n$ satisfies the linear recurrence relation $a_n = \sum_{i=1}^k c_i a_{n-i}$ for some constants $c_i$, then is there an easy way to find a linear recurrence relation for $b_n = a_n^2$ ?

For example, if $a_n = a_{n-1} + a_{n-3}$, then $b_n=a_n^2$ seems to satisfy $b_n=b_{n-1}+b_{n-2}+3b_{n-3}+b_{n-4}-b_{n-5}-b_{n-6}$.

I remember reading somewhere that the complement of a meagre set in a Baire space is also a Baire space and this is in fact easy to prove. Looking for this result in the standard collection of Topology books I could not find it. Can anyone help me locate a reference?

I have a doubt, Roth's discrepancy theorem [1] says that there is a subset of arithmetic progressions $A \in [n]$ where any function $f:N\rightarrow \left \{ -1,1 \right \} $ implies

$ \left | \sum _{a\in A}f(a) \right |\geq cn^{1/4} $

(Does not specify if the arithmetic progression is homogeneous or not homogeneous)

In 2015 Terence Tao [2] proved for any homogeneous arithmetic progression we have

$Sup_{n,d\in N}\left | \sum^{n} _{j=1}f(jd) \right |=\infty$

It occurred to me to ask if the discrepancy is infinite for non-homogeneous arithmetic progressions or is it an open question

[1]Roth, K. F. Remark concerning integer sequences. Acta Arith. 9 1964 257–260.

[2] Tao, Terence (2016). "The Erdős discrepancy problem". Discrete Analysis: 1–29. arXiv:1509.05363

Let $X$ be a proper scheme over $k$ where the characteristic of $k$ is $p>0$.

Consider the etale sheaves $\mathbb{G}_m$ over $X$ and consider the $p$-th power map from $\mathbb{G}_m \to \mathbb{G}_m$ ( $x \in \mathbb{G}_m(U)$ goes to $x^p \in \mathbb{G}_m(U)$. When is this map of map of etale sheaves? It is true $X$ is smooth over $k$, is it if $X$ is just proper?

I am trying to identify a solution for a multiplicative problem that involves taking two large numbers and multiplying them together.

Has the 3-tag system investigated by Emil Post $(0->00, 1->1101)$ been solved? Is there a decision algorithm to determine which starting strings terminate, which end up in a cycle, and which (if any) grow without bound?

Also, what cycle structures are there? Setting a='00' and b='1101', the only cycles I know of begin with $ab, b^2 a^2$, combinations of these, and $a^2 b^3 (a^3 b^3)^n$. Are there any more?

Define: $$H^*_{\alpha}=\{x | \ \forall y \in TC(\{x\}) (|y| \leq |\alpha|)\}$$

Where: $TC(x)= \{y| \ \forall t (transitive(t) \wedge x \subseteq t \to y \in t)\},$

$transitive(t) \iff \forall r,s (r \in s \in t \to r \in t)$, $``| x|"$ signify the cardinality of $x$ which is the smallest Von Neumann ordinal bijective to $x$.

In English: $H^*_{\alpha}$ is the set of all sets that are hereditarily subnumuerous $(\leq)$ to $\alpha$.

Define recursively: $$\daleth_0=\omega_0$$

$$\daleth _{i+1}= |H^*_{\daleth_i}|$$

$$\ \ \daleth_j = \bigcup_{i<j} (\daleth_i) , \text { if }\not \exists k(k+1=j)$$ .

Questions:

1) is $\daleth_1 = \aleph_1$ satisfied in $L$?

2) is $\daleth_i = \aleph_i$ satisfied in $L$?

where $L$ is Gödel's constructible universe.

In paper: **Three manifolds with positive ricci curvature**

In theorme 9.4, p.281, What does ``Suppose the null eigenvalue of $M_{ij}$ occurs in the top position" mean? Does it mean the null eignvector of $M_{ij}$ is $e_1$ (first vector in the standard ordered basis)? Also diagonalizing with respect to some basis gives us $R_{ij}=\begin{bmatrix} \lambda&&\\ &\mu&\\ &&\nu\\ \end{bmatrix}$, why does diagonalizing with respect to the same basis gives us $g_{ij}=I_3$?

I was wondering... Is every symplectic connection $\nabla$ on some symplectic manifold $(M,ω)$ the Levi-Civita connection of some metric $g$ on $M$? What about the local statement?

**Definition.** A finite group $G$ is called *multifactorizable* if for any positive integer numbers $a_1,\dots,a_n$ with $a_1\cdots a_n=|G|$ there are subsets $A_1,\dots,A_n\subset G$ such that $A_1\cdots A_n=G$ and $|A_i|=a_i$ for all $i\le n$.

In this case we shall write that the group $G$ is *$a_1{\times}\cdots{\times}a_n$-factorizable.*

It can be shown that each finite Abelian group is multifactorizable.

**Problem 1.** Is each finite (simple) group multifactorizable?

As was observed by Geoff Robinson in his answer to this question, each finite nilpotent group is multifactorizable.

**Problem 2.** Is each finite solvable group multifactorizable?

**Added in Edit.** It turns out that the alternating (solvable) group $A_4$ is not multifactorizable, more precisely, $A_4$ is not $2{\times}3{\times}2$-factorizable.

Now it remains to find an example of a finite simple group which is not multifactorizable.

**Problem 3.** Is the alternating group $A_5$ multifactorizable? In particular, is $A_5$ $2{\times}15{\times}2$-factorizable?

Suppose $B$ is the ball of center $a$ and radius $R>0$ in $ \mathbb{R}^{n} $ $n>1$. Suppose also that $u$ is subharmonic and real analytic on a neighborhood of the closure of $B$. We know that according to Riesz decomposition theorem there is a harmonic function $h$ on $B$ and a potential $p$ such that \begin{equation} u(x)=h(x)-p(x) \end{equation} Here, the potential is given by $p(x)=\int_{B}G(x,\zeta) \Delta u(\zeta)d\zeta$, where $G(x,\zeta)=\log|x-\zeta|$ for $n=2$, and $G(x,\zeta)=|x-\zeta|^{2-n}$ for $n>2$.

My problem is the following: since $u$ is real analytic, by the Riesz decomposition theorem, so is $p$. But if I try to expand $p(x)$, about, say $x=a$, it is impossible that the expansion of $G(.,\zeta)$ about $a$ be convergent (because $\zeta$ varies in the whole $B$ and among other things, it can even be $\infty$ at $a$) and the laplacian $\Delta u(\zeta)$ under the integral sign has no effect on the expansion of $p(x)$ (which is in terms of $x$). So: where does the power series expansion of $p$ come form? From that of $h$ only??? Is there someone who can explain? Thanks.

Let $C_{jk}^l$ be the number of times class $l$ is generated from the classes $j$, $k$ and $c_j$ be order of class $j$. See for example finite groups by Jansen and Boon for the notation and some description if not familiar and/or many other texts. For example of one relation $c_j c_k=\sum_l C_{jk}^l$. All $c_j , C_{jk}^l$ are non negative integers. Anyway I derive if $C_{j,k}^l>0$ then $C_{j,k}^l \ge\frac {\max(c_j,c_k)}{c_l}$ and $c_l$ divides $\max(c_j,c_k)$? I derive it as follows: if $C_{j,k}^l>0$ then by conjugation can express it so that every element of either one but not generally both of the two classes contributes, possibly more than once, to every element of class $l$ because if $g(j) g(k)=g(l)$ then every other element of class $j$ can be written as $g^{-1}g(j)g$ and $g^{-1}g(j)g g^{-1}g(k)g=g^{-1}g(l)g$ for some $g$ of group $G$.

Anyway I haven't seen it in any texts but would presume if it is true something similar is likely stated somewhere. First wanted to make sure everyone agrees with this and if not give an example of contradiction. Otherwise has anyone seen this or similar statement in any texts?

Let P be a distribution on a finite set of size $k$ and let $\hat{P}_N$ be the empirical distribution (frequencies) from a samples of size $N$. Consider the Hellinger distance between $P$ and $\hat{P}_N$, namely

$$ D_{\text{Hell}}(P\|\hat{P}_N) := \left(\sum_{i=1}k\left(\sqrt{p_i}-\sqrt{N_i/N}\right)^2 \right)^{1/2}=2\left(1-\sum_{i=1}^k\sqrt{Ni/N}p_i\right)^{1/2}. $$

Using the Markov inequality, (Matusita 1995) showed **non-asymptotic** bound

**Theorem I.** For all $t>0$, it holds that
$$
P(D_{\text{Hellinger}}(P\|\hat{P}_N)^2 \ge (k-1)t/N) \le 1/t.
$$

The author also proved the convergence convergence in law

**Theorem II.**
$$4ND_{\text{Hellinger}}(P\|\hat{P}_N)^2 \overset{\mathcal L}{\longrightarrow}\chi_{(k-1)}^2.
$$

Combined with an sub-exponential tail bound for the chi-squared distribution and doing a bit of algebra, gives the **asymptotic** tail bound

**Corollary.** For confidence level $\beta$ with every $0 < \beta \le exp(1-k) \le 1$, it holds that
$$
P(D_{\text{Hellinger}}(P\|\hat{P}_N)^2 \ge 5\log(\beta^{-1})/N) \le \beta.
$$

Can the **non-asymptotic** tail bound in Theorem I above be improved (ideally, to something close the the exponential bound in above Corollary) ?

**Definition.** A zero-dimensional topological space $X$ is called *base zero-dimensional* if for any base $\mathcal B$ of the topology that consists of closed-and-open sets in $X$, any open cover $\mathcal U$ of $X$ has a disjoint refinement $\mathcal V\subset\mathcal B$.

It can be shown that each countable regular space is base zero-dimensional.

**Problem.** Is the Cantor cube $\{0,1\}^\omega$ base zero-dimensional?

For a field $F$ let $p(F)=p$ if the characteristic of $F$ is a prime $p$, and $p(F)=+\infty$ if $F$ is of characteristic zero.

In 2007 I considered the linear extension of the Erdos-Heilbronn conjecture, and conjectured that (cf. http://arXiv.org/abs/0810.0467) for any nonzero elements $\lambda_1,\ldots,\lambda_n$ of a field $F$ with $p(F)\not=n+1$ and a finite subset $A$ of $F$ we have \begin{align}&|\{\lambda_1a_1+\ldots+\lambda_n a_n:\ a_1,\ldots,a_n\ \text{are distinct elements of}\ A\}| \\&\qquad \qquad \ge\min\{p(F)-\delta,\, n(|A|-n)+1\},\end{align} where $\delta$ is $1$ if $n=2$ and $\lambda_1+\lambda_2=0$, and $\delta=0$ otherwise.

Motivated by the above as well as Question 316142 of mine, here I ask the following question.

QUESTION: Is my following conjecture true?

**Conjecture**. Let $\lambda_1,\ldots,\lambda_n\ (n\ge3)$ be positive integers with $\lambda_1\le\ldots\le\lambda_n\le n$ and $\gcd(\lambda_1,\ldots,\lambda_n)=1$. Let $F$ be a field with $p(F)>n+1$. Then, for any finite subset $A$ of $F$ with $|A|\ge n+\delta_{n,3}$ we have
\begin{align}&\bigg|\bigg\{\sum_{k=1}^n\lambda_ka_k:\ a_1,\ldots,a_n\ \text{are distinct elements of}\ A\bigg\}\bigg|
\\\ge&\min\bigg\{p(F),\ (\lambda_1+\ldots+\lambda_n)(|A|-n)+\sum_{k=1}^{\lfloor n/2\rfloor}(n+1-2k)(\lambda_{n+1-k}-\lambda_k)+1\bigg\}.\end{align}

Now let me explain where the lower bound comes from. Suppose that $A$ is just the subset $\{1,\ldots,m\}$ of the rational field $\mathbb Q$. For the set $$S=\{\lambda_1a_1+\ldots+\lambda_na_n:\ a_1,\ldots,a_n\ \text{are distinct elements of}\ A\},$$ its minimal element should be $\sum_{k=1}^n\lambda_k(n+1-k)$ while its maximal element should be $\sum_{k=1}^n\lambda_k(m-n+k)$. Note that \begin{align}&\bigg|\bigg\{\sum_{k=1}^n\lambda_k(n+1-k),\ \ldots,\ \sum_{k=1}^n\lambda_k(m-n+k)\bigg\}\bigg| \\=&(\lambda_1+\ldots+\lambda_n)(m-n)+\sum_{k=1}^{\lfloor n/2\rfloor}(n+1-2k)(\lambda_{n+1-k}-\lambda_k)+1. \end{align} If $\lambda_k=k$ for all $k=1,\ldots,n$, then $$\sum_{k=1}^{\lfloor n/2\rfloor}(n+1-2k)(\lambda_{n+1-k}-\lambda_k)=\sum_{k=1}^{\lfloor n/2\rfloor}(n+1-2k)^2=\frac{n(n^2-1)}6.$$

Any comments are welcome!

Motivated by Question 315568 of mine, I'm interested in the set $$S(n):=\bigg\{\sum_{k=1}^n k\pi(k):\ \pi\in S_n\bigg\}.$$ It is easy to see that $$S(1)=\{1\},\ S(2)=\{4,5\}\ \text{and}\ S(3)=\{10,11,13,14\}.$$ By the Cauchy-Schwarz inequality, for any $\pi\in S_n$ we have $$\bigg(\sum_{k=1}^nk\pi(k)\bigg)^2\le\bigg(\sum_{k=1}^nk^2\bigg)\bigg(\sum_{k=1}^n\pi(k)^2\bigg)$$ and hence $$\sum_{k=1}^nk\pi(k)\le\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6.$$ If we let $\sigma(k)=n+1-\pi(k)$ for all $k=1,\ldots,n$, then $\sigma\in S_n$ and \begin{align}\sum_{k=1}^n k\pi(k)=&\sum_{k=1}^nk(n+1-\sigma(k))=(n+1)\sum_{k=1}^nk-\sum_{k=1}^nk\sigma(k) \\\ge&\frac{n(n+1)^2}2-\frac{n(n+1)(2n+1)}6=\frac{n(n+1)(n+2)}6. \end{align} Thus $$S(n)\subseteq T(n):=\left\{\frac{n(n+1)(n+2)}6,\ldots,\frac{n(n+1)(2n+1)}6\right\}.$$ My computation indicates that $S(n)=T(n)$ whenever $n\not=3$. Note that $|T(n)|=n(n^2-1)/6+1$.

Inspired by the above analysis, here I pose the following new conjecture in additive combinatorics.

**Conjecture**. Let $n$ be a positive integer and let $F$ be a field with $p(F)>n+1$, where $p(F)=p$ if the characteristic of $F$ is a prime $p$, and $p(F)=+\infty$ if the characteristic of $F$ is zero. Let $A$ be any finite subset of $F$ with $|A|\ge n+\delta_{n,3}$, where $\delta_{n,3}$ is $1$ or $0$ acording as $n=3$ or not. Then, for the set
$$S(A):=\bigg\{\sum_{k=1}^n ka_k:\ a_1,\ldots,a_n\ \text{are distinct elements of}\ A\bigg\},$$
we have $$|S(A)|\ge\min\left\{p(F),\ (|A|-n)\frac{n(n+1)}2+\frac{n(n^2-1)}6+1\right\}.$$

QUESTION: Is my above conjecture true?

PS: I'll soon pose another question which extends the current one to the general case.

Let $a,b$ be two positive integer numbers. A group $G$ is called *$a{\times}b$-decomposable* if there are subsets $A,B\subset G$ of cardinality $|A|=a$ and $|B|=b$ such that $AB=G$ where $AB=\{xy:x\in A,\;y\in B\}$.

I am looking for an example of a finite group $G$ which is not $a{\times}b$-decomposable for some numbers $a,b$ with $a\cdot b=|G|$.

**Remark.** It is easy to see that a group $G$ is $a{\times}b$-decomposable if $a\cdot b=|G|$ and $G$ contains a subgroup of order or index equal to $a$ or $b$. Consequently, any Abelian group $G$ is $a{\times}b$-decomposable for any numbers $a,b$ with $a\cdot b=|G|$.

According to the answer of Geoff Robinson to this MO-problem the alternating group $A_9$ contains no subgroups of order or index equal to 35.

**Question 1.** Is the group $A_9$ $35{\times}5184$-decomposable?

By the comments of @YCor to the same MO-problem,

$\bullet$ the group $PSL_2(11)$ has cardinality $|PSL_2(11)|=15\times 44$ but contains no subgroups of order or index 15;

$\bullet$ the group $PSL_2(13)$ has cardinality $|PSL_2(13)|=21\times 52$ but contains no subgroups of order or index 21.

**Question 2.** Is the group $PSL_2(11)$ $15{\times}44$-decomposable?

**Question 3.** Is the group $PSL_2(13)$ $21{\times}52$-decomposable?