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This has nothing to do with Witt vectors, smoothness over $\mathbf{Z}_p$ etc: formation of the $\mathfrak{g}_n$ commutes with base-extension to $\mathbf{Q}_p$ or even $\overline{\mathbf{Q}}_p$, so we …

Let $S$ be some base scheme, $H$ a finite flat group scheme over $S$, and $\alpha: \mu_p \to H$ a homomorphism of group schemes ($p$ a prime). Is the kernel of $\alpha$ necessarily flat over $S$? (I …

(This is a follow-up to this question of mine.) Is there an example of a connected reductive algebraic group $G$ over $\mathbb{R}$ such that: $G$ is not isomorphic to a product $G_1 \times G_2$ of …

Let $F$ be a $p$-adic field, $E / F$ a quadratic extension, and $n \ge 1$. Let $V = E^n$ with the obvious diagonal Hermitian form, $$ \langle (u_1, \dots, u_n), (v_1, \dots, v_n) \rangle = \sum_{i = …

Yes, this also follows from the same approximation theorems. One very concrete way of stating the strong approximation theorem for $Sp_{2g}$ is that $Sp_{2g}(\mathbf{Z})$ surjects onto $Sp_{2g}(\math …

Theorem: Let $G$ be a reductive algebraic group over a local field $F$, let $K$ be any maximal compact subgroup of $G(F)$, and let $Z = Z(G)$. Then $K \cap Z(F)$ is the unique maximal compact subgroup …

I find this question somewhat strange; you ask "what is the meaning of the L-group?", but the survey article of Casselman which you link to is largely devoted to explaining the historical and conceptu …

Regarding the second part of question 1, which is what the fibres of the integral model given by a choice of L will look like: you might find the paper of Gan, Hanke and Yu, On an exact mass formula o …

No, there is not a well-defined subgroup "$G(\mathcal{O}_K)$" for a semisimple algebraic group over $K$; if you define it using embeddings into $GL_n$ then the subgroup you get will depend on the embe …

First question (do non-strictly-arithmetic subgroups exist?): Any "strictly arithmetic" subgroup in your sense will, in particular, be a congruence subgroup, i.e. the intersection of $G(\mathbb{Q})$ w …

This is much easier than it looks. The point is that any reductive group $G$ is isogenous to the product of its radical, which is its centre $Z(G)$, and its commutator subgroup, which is a semisimple …

(Expanded version of my earlier comment, reposted as an answer): If $F^2(a) = a$, $F^2(b) = b$, and $a F(a) + \pi b F(b) = 1$, then it is not too difficult to show that $a$ and $b$ have to be in $\ma …

Let $G$ be a split semisimple algebraic group scheme over $\mathbf{Z}$ (I'm mostly interested in the case $G = Sp_4$). Let $V$ be an irreducible representation of the generic fibre $G_{\mathbf{Q}}$, …

$\DeclareMathOperator\SL{SL}\DeclareMathOperator\PGL{PGL}\DeclareMathOperator\Br{Br}\DeclareMathOperator\U{U}\DeclareMathOperator\disc{disc}\DeclareMathOperator\Nm{Nm}\DeclareMathOperator\diag{diag}\D …

Symplectic case: Here are two reasons (not necessarily the only ones) why $\operatorname{GSp}_{2n}$ is more convenient to work with than $\operatorname{Sp}_{2n}$. Firstly: there is no Shimura datum w …