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## Centraliser of an absolute Galois groupLet $K$ be a finite extension of $Q_p$. Is the centraliser of $Gal(\overline{K}/K)$ in $Gal(\overline{Q_p} / Q_p)$ trivial ? If yes, how can I show it ? | |

## Invariant subspace in infinite dimensionsLet $A(t)$ be a family of skew self-adjoint operator defined on some Hilbert space $H$ with common domain $D(A).$ The dependence on $t$ is in the strongly continuous sense, i.e. for all $x \in D(A)$ the map $t \mapsto A(t)x$ is continuous. Consider the initial value problem $$\varphi'(t)=A(t)\varphi(t)$$ with $\varphi(0)=\varphi_0.$ Assume that there is a dense subspace $X$ in $H$ that is contained in the domain of $A(t)$ for all $t$ and $A(t)X \subset X.$ I ask: Let $\varphi_0 \in X$. Does this imply that $\varphi(t) \in X$ for all $t>0?$ This sounds very natural but I do not have any tools/ideas to show this at the moment. | |

## Schur multiplier for a semidirect productIf $G=N\rtimes H$ is the semi-direct product of two finite groups, and $|N|,|H|$ are coprime, then if we know the Schur multiplier of $N$ and $H$ then we can find the Schur multiplier for $G$. Let $N=C_n^n/C_n$, where $C_n$ is the cyclic group on $n$ elements and we are modding out by the diagonal copy of $C_n$, and let $H=S_n$ act on $N$ via permutations (if it is easier, assume that $n$ is prime). In this case $|N|$ and $|H|$ are never coprime ($n$ divides both), but is it still possible to compute the Schur and Bogomolov multipliers, or at least to understand if they are trivial/non-trivial? If I did not divide by the central $C_n$, the answer would be well known, see the chapter on wreath products of Karpilovsky's book on group representations. | |

## intersection of free/affine submodules, comparison with vector spacesIf $W_1,W_2 \subset V$ are finite-dimensional $k$-vector spaces of dimensions $d_1, d_2 \leq d$, respectively, then $d_1 + d_2 > d$ suffices to guarantee $W_1 \cap W_2 \neq \{0\}$. There are similar results for affine subspaces of a spaces of a $k$-vector space, $E_1,E_2 \subset V$. I'm looking for analogous results for submodules of a free $R$-module $N_1,N_2 \subset M$ of finite ranks $r_1,r_2 \leq r$, and for "affine submodules" (i.e. torsors/cosets/translates of submodules), $A_1,A_2 \subset R$. The main question I'm interested in is I would be interested in E.g. Consider the translates of rank-2 $\mathbb{Z}$-modules $$A_1 = (u_1,v_2,w_1)+\langle (a_1,b_1,c_1),(a_2,b_2,c_2) \rangle$$ $$A_2=(u_2,v_2,w_2)+\langle (a_3,b_3,c_3),(a_4,b_4,c_4) \rangle \subset \mathbb{Z}^3$$ What is $A_1 \cap A_2$? Lastly I have a Thank you. | |

## If $A,B$ are upper triangular matrices such that $AX=XA\implies BX=XB$ for upper triangular $X$, is $B$ a polynomial in $A$?A professor of mine told me that this is true, but he doesn't remember what the proof was or where to find it, and I haven't been able to find a source for it yet. As such I am looking for one here. In the theorem as stated, $\mathbb{F}$ is any field and $T_n(\mathbb{F})$ denotes the algebra of upper triangular $n\times n$ matrices over $\mathbb{F}$.
Does anyone know of a source for this result? I have searched Google, MSE, MO, and the like to no avail. If we replace $T_n(\mathbb{F})$ by $M_n(\mathbb{F})$, the question is answered in this paper. Unfortunately, the argument doesn't seem to translate directly, as I can't find a way to force the $M_i$ maps to be upper-triangular. Also, I have already asked this question here on MSE. As the question is for an undergraduate research project, it felt appropriate to ask it here as well. Thanks for any help! | |

## Bernstein-Zelevinsky classification: viewing a representation as a subrepresentation or a quotient$\DeclareMathOperator{\GL}{GL}$ $\DeclareMathOperator{\Ind}{Ind}$I have a question on the details of the Bernstein-Zelevinsky classification. This classification allows us to obtain irreducible representations of $\operatorname{GL}_n(F)$ over a $p$-adic field $F$ as unique quotients of certain induced representations coming from supercuspidals, in an essentially unique way. It also allows us to obtain these representations as unique subrepresentations in an essentially unique way. Suppose $\pi$ is a smooth irreducible representation of $\GL_n(F)$, and we know how to obtain $\pi$ as a quotient $Q(\Delta_1, ... , \Delta_r)$ in the B.Z. classification. Then do we also know how to obtain $\pi$ as a subrepresentation $Z(\Delta_1', ... , \Delta_{r'}')$ in B.Z. classification? Example: suppose $G = \operatorname{GL}_2(F)$, and $\chi = \chi_1$ is a character of $F^{\ast}$, $\chi_2(x) = \chi_1(x)|x|_F$. Then $\Ind_{TU}^G \chi_1 \otimes \chi_2$ has a unique irreducible quotient $\pi$. If we want to get $\pi$ as subrepresentation instead of a quotient, then we swap $\chi_1$ and $\chi_2$ and use $\Ind_{TU}^G \chi_2 \otimes \chi_1$. | |

## Inertial decomposition of graphsThe problem is this: given a graph $G$, to find a decomposition of $G$, i.e. a set F of disjoint proper subgraphs of $G$ such that: $$\text{inertia}(G) = \sum_{H \in F} \operatorname{inertia}(H)$$ where $\operatorname{inertia}(G)=(a,d,c)$, with $a$ equals the number of negative eigenvalues of $A(G)$, the adjacency matrix of $G$, $b$ equals the dimension of null space of $A(G)$, and $c$ equals the number of positive eigenvalues of $A(G)$. I can not find any paper about this problem. Does anyone know something about this problem? | |

## Lie brackets of automorphismsLet $F$ be the vector fields of a differential manifold $M$, let $[X,Y]$ be the Lie brackets of $F$, now let $a$ be an automorphism of $F$ for the structure of real vector space of $F$. I consider now the bracket: $$[X,Y]_a= a^{-1}[a(X),a(Y)]$$ A simple calculus shows that it satisfies the Jacobi identities, so $(F,[,]_a)$ is a new Lie algebra. My question is: when does this new Lie bracket come from the vector fields of a manifold $M_a$? For example, if I take $a=g_*$, with $g$ a diffeomorphism of $M$, I could say that $M_{g_*}=M$. Perhaps that a condition of a certain smoothness could be added for the automorphism $a$? | |

## How to prove positivity of determinant for these matrices?Let $g(x) = e^x + e^{-x}$. For $x_1 < x_2 < \dots < x_n$ and $b_1 < b_2 < \dots < b_n$, I'd like to show that the determinant of the following matrix is positive, regardless of $n$: $\det \left (\begin{bmatrix} \frac{1}{g(x_1-b_1)} & \frac{1}{g(x_1-b_2)} & \cdots & \frac{1}{g(x_1-b_n)}\\ \frac{1}{g(x_2-b_1)} & \frac{1}{g(x_2-b_2)} & \cdots & \frac{1}{g(x_2-b_n)}\\ \vdots & \vdots & \ddots & \vdots \\ \frac{1}{g(x_n-b_1)} & \frac{1}{g(x_n-b_2)} & \cdots & \frac{1}{g(x_n-b_n)} \end{bmatrix} \right ) > 0$. Case $n = 2$ was proven by observing that $g(x)g(y) = g(x+y)+g(x-y)$, and $g(x_2 - b_1)g(x_1-b_2) = g(x_1+x_2 - b_1-b_2)+g(x_2-x_1+b_2-b_1) > g(x_1+x_2 - b_1-b_2)+g(x_2-x_1-b_2+b_1) = g(x_1-b_1)g(x_2-b_2)$ However, things get difficult for $n \geq 3$. Any ideas or tips? Thanks! | |

## Amenable Thompson-like groups
I realise that my question is vague, but defining and studying groups which look like Thompson's groups $F$, $T$ and $V$ seems to be an independent field of research in the litterature, so my question should make sense. Among the examples of such groups I know, typically either they contain a non-abelian free group or they contain Thompson's group $F$. In the former case, of course the group is not amenable; and in the latter case, the amenability of the entire group would imply the amenability of $F$, so a proof of the amenability should not be available. | |

## Symplectic structures in rigid geometryLet $K$ be a non-archimedean valued field (with any further adjectives attached as necessary). I'm looking for references or information about symplectic structures on rigid $K$-spaces. For example, if $X$ is an affinoid $K$-variety, then there is a tangent sheaf $\mathcal{T}_X$ on $X_{\text{rig}}$, with sections $$\mathcal{T}_X(U) = \text{Der}_K(U)$$ over affinoid subdomains $U$. It gives rise to a sheaf of $\mathcal{O}_X$-algebras $\mathcal{A} = \text{Sym}_{\mathcal{O}_X} \mathcal{T}_X$. There is then a "relative analytification'' space $Y = \text{Spec}^{\text{an}} \mathcal{A}$, as described by Conrad in Section 2.2 here. As defined, this $Y$ should be the correct candidate for the cotangent space $T^*X$, and so one might try to make sense of a symplectic structure on $Y$. However, I haven't read anything about tangent spaces or symplectic forms in the rigid setting. Is that because these notions are problematic or because they are obvious? Should tangent spaces to points on $X$ be defined in the same way as if $X$ were a scheme? Thanks for any insight or pointers to the literature. | |

## Thom class in motivic cohomologyLet $E$ be vector bundle over smooth scheme $X$. Thom space of $E$ is $Th(E)=E/E-i(X)$ where $i\colon X \longrightarrow E$ is zero section. This space is $\mathbb{A}^{1}$ isomorphic to $\mathbb{P}(E \oplus \mathscr{O})/\mathbb{P}(E)$ As same as in algebraic topology, Thom class of vector bundle $E$ (it is denoted $t_{E}$) was defined in motivic homotopy theory and satisfied the property as follow (Th) Thom class give the isomorphism $\tilde{H}^{*,*}(F_{･} \wedge X_{+}) \simeq \tilde{H}^{*+2\dim E,*+ \dim E}(F_{･} \wedge Th(E)) $ In Reduced power operations in motivic cohomology, V.Voevodsky was defined Thom class. But I don't understand it. So I think definition of Thom class as follow. Natural monomorphism between vector bundles $f \colon E \longrightarrow E\oplus \mathscr{O}$ give morphism $\mathbb{P}(f)\colon \mathbb{P}(E) \longrightarrow \mathbb{P}(E \oplus \mathscr{O})$ and $\mathbb{P}(f)^{*}(\mathscr{\sigma}_{E\oplus\mathscr{O}})=\sigma_{E}$ where $\sigma_{E}(resp. \sigma_{E\oplus \mathscr{O}}) =\mathscr{O}(-1) \in H^{2,1}(\mathbb{P}(E),\mathbb{Z})(resp.H^{2,1}(\mathbb{P}(E\oplus \mathscr{O}),\mathbb{Z}))$ if $\dim E=d $, then since $\{1,\sigma_{E},\cdots \sigma_{E}^{d-1}\}$ and $\{1,\sigma_{E\oplus \mathscr{O}},\cdots,\sigma_{E\oplus \mathscr{O}}^{d}\}$ are basis of $H^{*,*}(\mathbb{P}(E),\mathbb{Z})$ and $H^{*,*}(\mathbb{P}(E\oplus \mathscr{O}),\mathbb{Z})$, $\mathbb{P}(f)^{*}$ is epimorphism and dimension of its Kernel 1. so Thom class $t_{E}$ of $E$ defined as genereted element in its Kernel. As $Th(E) \simeq \mathbb{P}(E\oplus \mathscr{O})/\mathbb{P}(E)$ it belong to $H^{*,*}(Th(E),\mathbb{Z})$. Question. Is this definition right? if isn't please tell me true def of thom class. | |

## Classification of a system of two second order PDEs with two dependent and two independent variablesIf we have a second order quasilinear PDE of the form $A\frac{\partial^2 u}{\partial x^2}+B\frac{\partial^2 u}{\partial y^2}+2C\frac{\partial^2 u}{\partial x\partial y}+ lower\,\, order \,\, terms=0$ where $A,B,C$ are functions of $x,y,u$, then the equation is called elliptic if $det=\begin{vmatrix}A &C \\C & B\end{vmatrix}>0$, parabolic if $det=0$ and hyperbolic if $det<0$. Now what happens if we have a system of two coupled PDEs of the form $A\frac{\partial^2 u}{\partial x^2}+B\frac{\partial^2 u}{\partial y^2}+2C\frac{\partial^2 u}{\partial x\partial y}+D\frac{\partial^2 v}{\partial x^2}+E\frac{\partial^2 v}{\partial y^2}+2F\frac{\partial^2 v}{\partial x\partial y}+ lower\,\, order \,\, terms=0 \\ G\frac{\partial^2 u}{\partial x^2}+H\frac{\partial^2 u}{\partial y^2}+2K\frac{\partial^2 u}{\partial x\partial y}+L\frac{\partial^2 v}{\partial x^2}+M\frac{\partial^2 v}{\partial y^2}+2N\frac{\partial^2 v}{\partial x\partial y}+ lower\,\, order \,\, terms=0$ with $A,B,C,...$ being functions of $x,y,u,v$. Does it make sense to construct the determinant $det=\begin{vmatrix}A &C & D&F\\C & B&F&E\\G &K & L&N\\K &H & N&M\end{vmatrix}$ and investigate its sign, or something like that? --------- Update ---------- In the chapter 10 of the book [Seiler, Werner M., Involution. The formal theory of differential equations and its applications in computer algebra] the author states that since ellipticity is a property defined at points and depends only on the principal symbol it then suffices to study it for linear equations. I understand that the same procedure holds for quasilinear ones, the only difference is that the coefficients are allowed to be functions of the dependent variables. Then regarding the specific example above if one implement the theory for linear equations the principal symbol should be $\tau[\xi,\eta]=\begin{bmatrix}B \eta^2 + 2 C \eta\xi + A \xi^2& E\eta^2 + 2 F \eta\xi + D \xi^2\\ H\eta^2 + 2 K \eta\xi + G \xi^2 & M\eta^2 + 2 N \eta\xi + L \xi^2\end{bmatrix}$ Ellipticity requires $det(\tau)\neq 0$ that is $(B M-E H) \eta^4 + 2 (C M + B N-F H - E K ) \eta^3 \xi + (B L + A M + 4 C N-E G - D H - 4 F K ) \eta^2 \xi^2 + 2 ( C L + A N-F G - D K ) \eta \xi^3 + (A L-D G ) \xi^4\neq 0$ well, when this condition is to be satisfied? Is it true that it is hyperbolic in direction $x$ or $y$ direction, if $(B M-E H) \eta^4 + 2 (C M + B N-F H - E K ) \eta^3 \xi + (B L + A M + 4 C N-E G - D H - 4 F K ) \eta^2 \xi^2 + 2 ( C L + A N-F G - D K ) \eta \xi^3 + (A L-D G ) \xi^4 = 0$ possesses only simple real roots? | |

## Which primes $\mathfrak{p} \in \mathbb{Z}[i]$ can be represented as $\mathfrak{p} = x^2 + 2y^2$?Consider the quadratic form $x^2 + 2y^2$ over the ring $\mathbb{Z}[i]$. It is irreducible, so I wanted to know which primes could be represented by this quadratic form. $$ \mathfrak{p} = x^2 + 2y^2 $$ There's two slightly different questions here. One about the integer $\mathfrak{p} \in \mathbb{Z}[i]$ and one about the ideals $(\mathfrak{p}) \subseteq \mathbb{Z}[i]$. This is still a PID so I can write just one element. This hopefully in analogy to Fermat's theorem on primes as the sum of two squares, where it is solved by a congruence condition. The analogous question for $x^2 + y^2 = (x+iy)(x-iy)$ is degenerate. | |

## A conjecture on simplexLet $A_0A_1...A_n$ be a simplex in $\Bbb E^n.$ Let $B_{ij}$ be the points on the edges $A_iA_j,\ 0\le i\not=j\le n.$ Denote by $(\beta_i)$ the hyperplane passing through points $B_{i0},$ $B_{i1},$ $B_{ii-1},$ $B_{ii+1},$ $...,$ $B_{in}.$ Assume that the reflections of $A_i$ in hyperplane $(\beta_i)$ lies on the hyperplane $(A_0A_1...A_{i-i}A_{i+1}...A_n).$
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## About the product of finite subsets of a torsion free groupLet $G$ be a torsion free group with identity $e$. For a subset $X$ of $G$, denote by $X^\#$ the set $X\setminus\{e\}$. Let $A$ be a finite subset of $G$ containing $e$. Is there a finite subset $B$ containing $e$ such that $$A\subset B^\#A\quad\text{and}\quad B\subset BA^\#$$ It is obvious there is no such $B$ when $A=\{e,x\}$; this follows from the right inclusion and that $G$ is torsion free. | |

## distribution of supremum of quadratic formFor any given vector $w$, we have score test statistic $\frac{w^{T}Aw}{\sigma^{2}}$, A is some symmetric matrix, from asymptotic theory, we know this test statistic asymptotically follows $\chi^{2}$ distribution with some degree of freedom,e.g. d, my question is: what is the distribution of $sup_{w}\frac{w^{T}Aw}{\sigma^{2}}$? | |

## Cyclic quadrilateral in metric spaceConsider a metric space $(\Bbb M,d).$ If $X,Y,Z\in \Bbb M.$ We define cosin of angle by $$\cos(\angle YXZ)=\frac{d(X,Y)^2+d(X,Z)^2-d(Y,Z)^2}{2d(X,Y)\cdot d(X,Z)}.$$ If we have four points $A,$ $B,$ $C$ and $D$ in $\Bbb M$ satify the indentity $$d(A,B)\cdot d(C,D)+d(A,D)\cdot d(B,C)=d(A,C)\cdot d(B,D).$$ Then we say that $A,$ $B,$ $C,$ $D$ are concyclic.
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## Higher degree of Hilbert's irreducibility theoremA basic form of Hilbert's irreducibility theorem can be formulated as follows: Let $f(t,x)\in\mathbb{Q}[t,x]\setminus\mathbb{Q}[t]$ be an irreducible polynomial. There exist infinitely many linear polynomials $p(t)\in\mathbb{Q}[t]$, $p(t)=t-t_0$ such that $(p(t),f(t,x))$ is a maximal ideal of $\mathbb{Q}[t,x]$. My question is: can we drop the "linear" assumption above and have polynomials $p(t)$ of arbitrarily high degree? More precisely, Let $f(t,x)\in\mathbb{Q}[t,x]\setminus\mathbb{Q}[t]$ be an irreducible polynomial. Let $n$ be a positive integer. There exists a polynomial $p(t)\in\mathbb{Q}[t]$ such that $\deg p(t)>n$ and $(p(t),f(t,x))$ is a maximal ideal of $\mathbb{Q}[t,x]$. Of course $p(t)$ must be irreducible itself, but that's far from enough. | |

## Decomposition of $\widehat{k^{\times}}$ occuring in local class field theoryLet $k$ be a finite extension of $\mathbb{Q}_p$ very often we use the isomorphism that $Gal(\overline{k}/k)^{ab} \simeq \hat{(k^{\times})}$ given by local class field theory. My question would be do we have (and if yes how can I prove it) $ \hat{O_k^{\times}} \times \hat{\mathbb{Z}} \simeq \hat{(k^{\times})}$ ? and is the image of the inertia subgroup of $Gal(\overline{k}/k)$ in $Gal(\overline{k}/k)^{ab}$ isomorphic to $ \hat{O_k^{\times}}$ ? I imagine that the proof of the first point can come from the fact that $\hat{k^{\times}} \simeq \widehat{O_k^{\times} \times \mathbb{Z}}$ using the decomposition with an uniformizer. But then can I split the product ? Do we have $\widehat{O_k^{\times}} = O^{\times}_k$ ? |