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Let $k$ be a finite extension of $\mathbb{Q}_p$ and let $(A,A^0)$ be a $k$-affinoid algebra, where $A^0$ is the subring of power bounded elements. Suppose given a compact subgroup, L, of $GL_n(A)$, is it true that there exists an element $g \in GL_n(A)$ such that $gLg^{-1} \subset GL_n(A^0)$?

This is true in the (well known) case of $A$ a finite extension of $\mathbb{Q}_p$. I tried to reproduce the same argument in the general case but I ran into problems at some point. Basically, we want to consider the submodule $T = \sum_{g \in L/G} g \cdot (A^0)^n$, where $G$ is a suitable open subgroup of $L$. In the case of a local field this is easily seen to be an $A^0$-lattice but I am not sure that it also holds for a general affinoid.

Any suggestions will be of great interest to me.

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  • $\begingroup$ I need to consider a general $k$-affinoid algebra $A = \mathbb{Q}_p \langle T_1, \dots, T_n \rangle / I$. $\endgroup$
    – user105552
    Commented Mar 1, 2017 at 16:11
  • $\begingroup$ It is harmless to pass to reduced $A$, but without requiring the reduced $A$ to be normal, probably it is false. (In the normal case $A^0$ is a noetherian normal ring, so the gulf between $A$ and $A^0$ is governed by finitely many dvr's, for generic points of $A^0/m_kA^0$.) I don't see a counterexample offhand (as it seems hard to control $A^0$ beyond the normal case), but it feels unlikely that for $A$ with arbitrary singularities there is just a single conjugacy class of maximal compact subgroups of ${\rm{GL}}_n(A)$. Hopefully someone else will have something more useful to say. $\endgroup$
    – nfdc23
    Commented Mar 1, 2017 at 21:07
  • $\begingroup$ Have you tried the case when $A$ is a non-smooth curve over $k$? Even for that I guess that one should find counterexamples. $\endgroup$
    – nfdc23
    Commented Mar 1, 2017 at 21:10
  • $\begingroup$ I think that I was asked this question a few years ago and that I came up with a simple counterexample. Let's see if I can manage to remember what it was... $\endgroup$
    – Laurent Berger
    Commented Mar 3, 2017 at 15:59
  • $\begingroup$ Ah, not quite. The question I was actually asked was the following. Let $D$ be the closed unit disk, and let $A^0 = Z_p\langle T \rangle$. Let $A$ be the ring of functions on $D \setminus \{ 0 \}$. Then is every compact subgroup of $GL_n(A)$ conjugate to a subgroup of $GL_n(A^0)$? The answer to this is no, and there is a reasonably simple counterexample. $\endgroup$
    – Laurent Berger
    Commented Mar 3, 2017 at 16:25

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