Let $M/L/K$ be a tower of local fields such that $M/L$ is abelian with Galois group $G$. The Artin map $\psi_{M/L}$ restricted to $K^\times$ is a continuous map to $G$ and thus corresponds to some abelian extension $T/K$ with an embedding $\operatorname{Gal}(T/K) \hookrightarrow G$. Challenge: Find $T$. (I am primarily interested in the mixed characteristic case, i.e. $K$ extends $\mathbb{Z}_p.$)

I conjecture the following Galois-theoretic description. Let $\Gamma = \operatorname{Gal}(\bar K/K)$ be the absolute Galois group; then $L$ and $M$, respectively, give us maps $\Gamma \to S_n$ and $\Gamma \to S_n \wr G$; and then, postcomposition with the summation map $S_n \wr G \to G$ gives the map $\Gamma \to G$ corresponding to the $T$ we seek.

Using Kummer theory, I was able to prove this when $\mu_m \subseteq K$, where $m$ is the exponent of $G$, or when $m$ is squarefree. Also the global analogue, where $M/L/K$ is a tower of number fields and we restrict $\psi_{M/L}$ to ideals of $K$, yields readily to manipulation of Frobenius elements. But there we are aided by having to consider only ideals composed of unramified primes. Indeed, the local *unramified* case is not hard.

Given $n,m=2^{poly(n)}$ what is $d\in\Bbb N$ needed so that there is $M_1,\dots,M_d\in\{-1,+1\}^{n\times n}$ with $$\sum_{\substack{q\in\{\mbox{prime }\leq n\}\\i\in\{1,\dots,\lfloor\frac nq\rfloor\}}}Tr\big(T_{_{(i,q)}}^{^{iq}}\big)=m$$ where $\Bbb 1_{n\times 1}$ is all $1$ vector and each $T_{_{(i,q)}}\in\{M_1,\dots,M_d\}$?

Is $d=O(polylog(n))$?

The answer to the question at Does almost every pair of elements in a compact Lie group generates the connected component? says there must be countably many conjugacy classes of closed subgroups of compact connected Lie groups. One of the comments (Does almost every pair of elements in a compact Lie group generates the connected component?) says that this is a direct result of inclusions of closed subgroups being (real) algebraic. However, I don't see how the result follows without using something fancy. Is there an obvious or nice proof I'm missing?

Consider 2-dimensional solution of

$$dX_{\varepsilon,t}=\mu(X_{\varepsilon,t})dt+\sigma(X_{\varepsilon,t})dB_{\varepsilon,t},$$ where $dB_{\varepsilon,t}$ is one-dimensional but the $X_{\varepsilon,t},\mu,\sigma$ could be complex.

For positive $\varepsilon$ and in one dimension instead, this is just an ODE, so I am wondering if there could be any Ito lemma that converges to the usual one as $\varepsilon\to 0$ since the only reason of the second derivative term in Ito's lemma was the quadratic variation of $dB_{t}$. For example, is it true that

$$df(X_{\varepsilon,t})=(f_{x}\mu(X_{\varepsilon,t})+\frac{1}{2}f_{xx}\sigma^{2}(X_{\varepsilon,t}))dt+f_{x}\mu(X_{\varepsilon,t})dB_{\varepsilon,t}.$$

I read in a paper this sentence, and I did not understand: the sullivan algebras do not distinguish classifying spaces of rational groups from contractibles spaces. Please if someone tells me where this comes from?

I am writing a piece on curve shortening flow and lots of my sources have referenced Alexander Polden's honours thesis 'Evolving Curves' from the Australian National University. I have tried to find this paper to no avail, even through ANU's thesis catalogue here which returns nothing. I was wondering whether anyone could provide me access to this paper (I don't know if there are any copyright issues with such theses) - I would very much like to read it and most likely cite it. Thanks.

I have a polytope formed by intersection of 'n' half spaces. How to quickly find if a given point lies inside the polytope?

My question is an amalgamation of two previous questions. The first question I'd like to draw attention to is here. It asks whether there can exist a non trivial semigroup defined on $\mathbb{C}$

$$\phi(s,z):\mathbb{C}_{\Re(s) > 0}\times\mathbb{C} \to \mathbb{C}$$ $$\phi(s_0, \phi(s_1,z)) = \phi(s_0+s_1,z)$$

The answer is no, because such a semigroup would necessarily have a repelling fixed point for some $s_0$, and each function commutes with each other, and an entire function with a repelling fixed point can only have a countable number of functions that commute with it; which forms a contradiction. A big thanks goes to Alexandre Eremenko for referencing I.N. Baker and solving this.

The second question is found here and asks whether an entire function $g$ of finite order can have a composite square-root. The answer is yes. Again, another big thanks goes to Alexandre Eremenko for referencing I.N. Baker again, and solving this.

A great question which hybridizes the above two questions, but laying insoluble in both instances:

For an entire function $f:\mathbb{C} \to \mathbb{C}$, can there exist a sequence of entire functions $\{f_n\}_{n=1}^\infty$ such that: $$f_n(f_n(...(n\,times)...f_n(z))) = f_n^{\circ n}(z) = (f_n \circ f_n \circ ...(n\,times)...\circ f_n)(z) = f(z)$$

This first answer fails to prove this can't happen, this is only a countable list of functions that commute with $f$, so the proof provided by Baker does not work. The second answer suggests this might be possible, but only refers to a square root of $f$, not an arbitrary $n$'th root for all $n$.

Essentially I'm asking this because I've learnt that if $f:\mathbb{C} \to \mathbb{C}$ there is no valid way of constructing

$$(f \circ f \circ...(s\,times)...\circ f)(z) : \mathbb{C}_{\Re(s) > 0}\times\mathbb{C} \to \mathbb{C}$$

so I'm wondering if we can weaken this to, sure there is no semigroup, but there can be

$$(f\circ f\circ... (q\,times)...\circ f)(z):\mathbb{Q} \times \mathbb{C} \to \mathbb{C}$$

My money is on the fact this is impossible and I'm crossing my fingers that Alexandre Eremenko knows why because he's read so much Baker.

Given a universe set $U$ and $n$ sets $A_i$ ($i=1, \cdots, n$). Each set $A_i$ contains $k_i$ subsets of $U$, i.e., $A_i=\{B_j\}$ ($j=1, \cdots, k_i$) where $B_j$ is a subset of $U$. I want to choose one set $B_m$ from each set $A_m$ to cover $U$ and I want to minimize the number of chosen set. I can only choose at most one set from each $A_i$.

Let $f=\sum_{n\geq 1} a_nq^n$ and $g=\sum_{n\geq 1} b_nq^n$ be two distinct cusp eigenforms of same weight $k\geq 2$ with real Fourier coefficients. We normalize the coefficients by defining: $\widetilde {a_n}:=\frac{a_n}{n^{(k-1)/2}}$ and similarly $\widetilde{b_n}:=\frac{b_n}{n^{(k-1)/2}}$. I was wondering if there is a quick way to see that $$\sum_{p-prime}\frac {\widetilde{a_p}-\widetilde{b_p}}{p^s}$$ is bounded as $s\to 1^{+}$.

Suppose $M$ is a 2-dimensional smooth Riemannian manifold and $P\subset M$ is an open and connected subset with compact closure and a piecewise geodesic boundary.

My question is: What further conditions must $P$ (and $M$) satisfy such that the Gauss-Bonnet theorem is fulfilled for $P$?

I have found a lot of different versions of the Gauss-Bonnet theorem. All of them demand for instance that $M$ is orientable. Sometimes the boundary of $\partial P$ is supposed to be simple, closed and without cusps. But can't we do better than that? Is there a good survey which treats also more general cases?

Best wishes

Maybe, I am being stupid, but when I consider ramified extension of a perfectoid field with the characteristic $0$, I cannot find the correspondent field with characteristic $p$. Let me put it more precisely.

Let $K$ be the completed field of $\mathbb{Q}_p(p^{-\infty})$. Consider $\zeta_p$ be a $p$th root of unity, and let $L=K(\zeta_p)$. According to Scholze's theorem (or Fontaine-Winterberger maybe enough), there exists a correspondent field $L^\flat$ which is of degree $p-1$ over $K^{\flat}$. However, I cannot find such $L^\flat$.

I tried to follow the definition to construct $L^\flat$ which is $$ L^\flat=\varprojlim L^{\circ}/p $$ where the transition map is Forbenius, and then I meet the problem that I cannot find the $p$th root of $\zeta_p$ in $L$. Similar problem happens if I consider $K$ to be the completion of $\mathbb{Q}_p(\zeta_{p^{\infty}})$ and adjoint $p$th root of $p$. However, the extension of a perfectoid field is a perfectoid field according to Scholze, so I do not think this should happen.

I would like to thanks all of you for your help.

Let $G$ be a *simply connected* simple algebraic group over the field of complex numbers $\mathbb C$. Let $H$ be a *symmetric subgroup* of $G$. This means that there exists an automorphism of order 2 $\sigma\colon G\to G$ such that $H$ is the group of fixed points of $\sigma$ in $G$. It is known that $H$ is connected.

Let us assume that $H$ is *not semisimple*. It is known that then the center of $H$ is one-dimensional. Consider the derived subgroup $H^{\rm der}:=[H,H]$.

**Question.** Is it true that the semisimple group $H^{\rm der}$ is always simply connected?

It seems that this is true when $G$ is a classical group (say, when $G={\rm Sp}_{2n}$, $H=U_n$). What about $E_6$ and $E_7$? If the answer is always "yes", I would prefer a classification-free proof.

Consider $X=\{2,3,4\}$. This set has some interesting properties:

- The number of even numbers in $X$ is 2, an even number
- The number of odd numbers in $X$ is 1, an odd number
- The number of primes in $X$ is 2, a prime
- The number of squares in $X$ is 1, a square

We might wonder if this example can be extended to any number of such similar "properties." Hence:

**Question**:

Suppose $\mathcal{A} = \{A_1,A_2,\ldots,A_k\} \subseteq 2^{\mathbb{N}}$ is a finite family of infinite subsets of nonnegative integers (that is, each $A_i$ is infinite). Does there exist $X \subseteq \mathbb{N}$ such that $|A_i \cap X| \in A_i$ for all $i=1,\ldots,k$?

Note that it is not necessarily the case that such an $X$ exists if the $A_i$ are allowed to be finite, even if $\mathcal{A}$ satisfies the following obvious necessary condition:

- $\mathrm{min}(A) \leq |A|$ for all $A \in \mathcal{A}$ (where $\mathrm{min}(\emptyset):=\infty$);

consider the example $\mathcal{A}=\{\{1\},\{2,4\},\{1,2,4\}\}$.

Also note that there is not necessarily such an $X$ if $\mathcal{A}$ itself is allowed to be infinite, as in the example $\mathcal{A}=\{A_1,A_2,\ldots\}$ where $A_i =\{i,i+1,i+2,\ldots\}$.

Let $u\in \mathcal{D}'(\mathbb{R}^d)$, $f\in C_c^\infty(\mathbb{R}^d)$ and $f(x)=1$ if $|x|\leq 1$; $f(x)=0$ if $|x|>2$. Can we get the following conclusion: there exists $v\in \mathcal{D'}(\mathbb{R}^d)$, such that, for any $\phi\in \mathcal{D}(\mathbb{R}^d)$,

$$ \lim_{n\to\infty} \langle u \ f(n\cdot),\phi\rangle = \langle v,\phi\rangle, $$ i.e. $u\ f(n\cdot) \stackrel{\mathcal{D'}(\mathbb{R}^d)}{\longrightarrow} v$.

It is easy to see that there is a surjective lattice homomorphism $s:({\omega+1})^\omega \to 2^\omega$ (construction see after the horizontal line below). Is there a surjective lattice homomorphism $s:2^\omega \to ({\omega+1})^\omega$?

Assign to every $f\in ({\omega+1})^\omega$ the map $s(f)\in 2^\omega$ defined by setting for every $n\in\omega$: $\big(s(f)\big)(n) = 0$ if $f(n) = 0$ and $\big(s(f)\big)(n) = 1$ if $f(n)>0 $.

Consider the inhomogeneous boundary value problem on the infinite strip $(x,y)\in \mathbb{R}\times [0,1]$ defined by

$$\begin{cases}\partial_{x}u + \partial_{y}v=f & {(x,y)\in \mathbb{R}\times (0,1)}\\ \partial_{y}u-\partial_{x}v=g & {(x,y)\in\mathbb{R}\times(0,1)} \\ v|_{y=0}=u|_{y=1}=0 & {}\end{cases}\tag{BVP} $$

Define a Banach space of analytic functions $Y_{\sigma,s}(\mathbb{R})$ by the norm

$$\|U\|_{Y_{\sigma,s}}^{2} := \int_{\mathbb{R}}e^{2\sigma|k|}(1+k^{2})^{s}|\widehat{U}(k)|^{2}dk$$

Let $H^{r}$ denote the usual $L^{2}$ based Sobolev space with regularity parameter $r$. Now define the Banach space $$\mathbb{K}_{\sigma,s}^{r} := H^{r}((0,1); Y_{\sigma,s}(\mathbb{R}))$$

In the boundary value problem (BVP), we impose the condition $f,g\in \mathbb{K}_{\sigma,s}^{0}\cap\mathbb{K}_{\sigma,s-1}^{1}$, which we equip with the max norm

$$\|f\|_{\mathbb{K}_{\sigma,s}^{0}\cap \mathbb{K}_{\sigma,s-1}^{1}} := \max\{\|f\|_{\mathbb{K}_{\sigma,s}^{0}}, \|f\|_{\mathbb{K}_{\sigma,s-1}^{1}}\}$$

One can show using the Fourier transform/Fourier series the following result:

**Claim.** If $f,g\in \mathbb{K}_{\sigma,s}^{0}\cap \mathbb{K}_{\sigma,s-1}^{1}$, then the solutions $u,v$ to (BVP) belong to $\mathbb{K}_{\sigma,s+1}^{0}\cap\mathbb{K}_{\sigma,s}^{1}$.

My interest lies in the image of the trace operator $v\mapsto v|_{y=1}$. In Appendix A of the paper which I'm reading, the authors make reference to "the trace theorem" at several points to assert that $v|_{y=1}\in Y_{\sigma,s}$. Actually, they make the general claim (see p. 48) that the trace operator is a bounded linear operator

$$\mathbb{K}_{\sigma, s}^{0}\cap \mathbb{K}_{\sigma,s-1}^{1} \rightarrow Y_{\sigma,s-\frac{1}{2}}$$

However, they neither include the statement of a specific trace theorem in the paper, nor include a reference to a specific trace theorem.

Reading between the lines, I'm guessing they have the classical Sobolev trace theorem in mind and an argument something like as follows. If $u\in \mathbb{K}_{\sigma,s+1}^{0}\cap\mathbb{K}_{\sigma,s}^{1}$, then by complex interpolation

$$u\in \mathbb{K}_{\sigma,s+\frac{1}{2}}^{\frac{1}{2}} = H^{\frac{1}{2}}((0,1); Y_{\sigma,s+\frac{1}{2}})$$

This is too crude, though, as $H^{\frac{1}{2}}\not\subset C_{b}^{0}$, where $C_{b}^{0}$ is the space of bounded continuous functions. The only argument that I can think of to make their claim is the following real interpolation equality

$$(L^{2}, H^{1})_{\theta,q} = B_{2,q}^{\theta}, \quad 0<\theta<1, 1\leq q\leq \infty$$

Taking $\theta=\frac{1}{2}$ and $q=1$, we have $(L^{2}, H^{1})_{1,\frac{1}{2}} = B_{2,1}^{\frac{1}{2}}$. So "morally",

$$(\mathbb{K}_{\sigma,s+1}^{0}, \mathbb{K}_{\sigma,s}^{1})_{\frac{1}{2},1} = B_{2,1}^{\frac{1}{2}}((0,1); Y_{\sigma,s+\frac{1}{2}})$$

Since $B_{2,1}^{\frac{1}{2}} \subset C_{b}^{0}$, it follows that $v|_{y=1} \in Y_{\sigma,s+\frac{1}{2}}$.

I tried proving $(\mathbb{K}_{\sigma,s+1}^{0}, \mathbb{K}_{\sigma,s}^{1})_{\frac{1}{2},1} \subset B_{2,1}^{\frac{1}{2}}Y_{\sigma,s+\frac{1}{2}}$ using the definition of the real interpolation K-functional, but it's not clear to me how to arrive at the desired result.

**Question 1.** Is it true that the $(\mathbb{K}_{\sigma,s+1}^{0}, \mathbb{K}_{\sigma,s}^{1})_{\frac{1}{2},1}\subset B_{2,1}^{\frac{1}{2}}Y_{\sigma,s+\frac{1}{2}}$?

I consider real interpolation to be a nontrivial tool, and I would prefer to avoid and thus be very grateful for a simpler argument

**Question 2.** Is there a non-interpolation argument for arriving at the conclusion that $v|_{y=1}\in Y_{\sigma,s+\frac{1}{2}}$?

**Terminology:**

For a bounded closed convex (bcc for short) set $A$, define $w(A)$ to be the infimum of the distance between pairs of parallel hyperplanes supporting $A$.

We say that a bcc set $A$ is

*reduced*, if for every bcc set $A'$ strictly contained in $A$, we have $w(A') < w(A)$.We say that a bcc set $A$ is

*diametrically minimal*, if every bcc set $A'$ strictly contained in $A$ has a smaller diameter.A set is

*of constant width*means the distance between two supporting hyperplanes is the same, in any direction.

**Question:** Is it true that every reduced diametrically minimal bcc set is of constant width?

Thurston's 1982 article on three-dimensional manifolds1 ends with $24$ "open questions":

$\cdots$

Two naive questions from an outsider: (1) Have all $24$ now been resolved? (2) If so, were they all resolved in his lifetime?

1Thurston, William P. "Three dimensional manifolds, Kleinian groups and hyperbolic geometry." *Bull. Amer. Math. Soc*, 6.3 (1982).
Also: In *Proc. Sympos. Pure Math*, vol. 39, pp. 87-111. 1983.

It is known that if $D:H_1 \to H_2$ is a bounded operator between Hilbert spaces, then there exists an adjoint operator $D^* : H_2 \to H_1$ (the field is just $\mathbb R$ rather than $\mathbb C$, so we can identify $H^*$ with $H$) such that $$ \langle Dx,y \rangle_{H_2}=\langle x,D^*y\rangle_{H_1} $$

Nevertheless, let us think a more complicated case as follows.

Let $p>2$ and then $X=W^{1,p}(S^2, \mathbb R^n)$ is a Banach space. We define $$ \langle f,g\rangle:= \int_{S^2} \langle f(x),g(x) \rangle_{\mathbb R^n} d\mathrm{vol}_{S^2}(x) $$for $f,g\in X$. By Sobolev embedding we know functions in $X$ are continuous. Thus, this bilinear pairing is positive (i.e.$\langle f,f\rangle =0$ implies $f =0$ ) and becomes an inner product.

**Question 1**: Is the topology on $X$ induced by this inner product "compatible" with the original $W^{1,p}$-topology on $X$ in some sense?

**Question 2**: If there is a bounded linear operator $D: X\to X$, then can we define an adjoint operator $D^*: X\to X$ using this inner product? For example, a possibility is that for $f\in X$ we can "define" $D^*f$ pointwisely by $\langle Dg(x), f(x)\rangle_{\mathbb R^n}= \langle g(x), D^*f(x)\rangle_{\mathbb R^n}$

**Quesition 3**: By the way, the purpose of this question is to understand the adjoint operator $D_u^*$ defined in McDuff-Salamon's book J-holomorphic curves and symplectic topology, 2nd edition (cf. page 48 or page 582). Briefly, let $E$ and $F$ be vector bundles over $S^2$, and let $D$ be an operator
$$
D: W^{l,p}(S^2, E) \to W^{l-1,p}(S^2, F)
$$
then how to see that the formal adjoint will be an operator
$$
D^*: W^{l,p}(S^2, F) \to W^{l-1,p}(S^2,E)
$$? Why not an operator from $W^{l-1,p}(S^2, F)$ to $W^{l,p}(S^2,E)$?