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First, I'd like to understand what the compact open subgroups of $H(\mathbb{Q}_p)$ are, where $H$ is an inner form of $GL_n$ over $\mathbb{Q}_p$.

Second, I'd like to know where I can read about this for other reductive groups.

Any pointers would be greatly appreciated. Thanks!

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    $\begingroup$ What do you mean by "inner form of $GL_n(Q_p)$"? "inner form of $GL_n$ over $Q_p$" makes sense (in case it's what you mean), not what is written. $\endgroup$
    – YCor
    Commented Dec 15, 2016 at 1:25
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    $\begingroup$ I assume you don't really mean to ask for all the compact open subgroups, but only the interesting ones; in which case what you want is the Moy–Prasad theory, as first expounded in ams.org/mathscinet-getitem?mr=1253198 and ams.org/mathscinet-getitem?mr=1371680 . A much more user-friendly introduction appears in Joe Rabinoff's lovely senior thesis. $\endgroup$
    – LSpice
    Commented Dec 15, 2016 at 1:51
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    $\begingroup$ For the question in the title (that is all compact subgroups), see Richard Pink, Compact subgroups of linear algebraic groups, J. Algebra 206 (1998), no. 2, 438-504. $\endgroup$
    – Uri Bader
    Commented Dec 15, 2016 at 7:43
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    $\begingroup$ For maximal compacts, you must catch some Bruhat-Tits theory. $\endgroup$
    – Uri Bader
    Commented Dec 15, 2016 at 7:45
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    $\begingroup$ @KConrad pointed out that my link is no longer working, and supplied an updated link to Rabinoff's thesis. Thanks! $\endgroup$
    – LSpice
    Commented Jul 18, 2019 at 10:42

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You get compact subgroups by taking compact-open subgroups of algebraic subgroups. My understanding is that they are more-or-less all compact subs, e.g. any compact subgroup $H$ should have a finite index subgroup of this form.

Uri Bader's reference to Pink's 1998 paper is a good start. Pink proves this sort of rigidity theorem over a local field of positive characteristic. I believe he should explain why it is "eazzzy" over a local field of zero characteristic but I have no intention of checking it: no pink till Xmas!

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  • $\begingroup$ Your "should" in the first paragraph is true: for a smooth affine group $G$ over the fraction field $k$ of a discrete valuation ring $O$, any bounded subgroup $K$ of $G(k)$ lies in a bounded open subgroup (maybe also for rank-1 valuation rings?). Indeed, realizing $G$ as a closed $k$-subgroup scheme of some GL$_n$, $G(k)$ is topologically a closed subgroup of GL$_n(k)$ and bounded subsets of $G(k)$ are bounded in GL$_n(k)$. That reduces us to the case of GL$_n$ in place of $G$. Clearly $L := K.O^n \subset c O^n$ for some $c\in k^{\times}$, so $L$ is finite free and $K\subset {\rm{GL}}(L)$. $\endgroup$
    – nfdc23
    Commented Dec 15, 2016 at 14:58
  • $\begingroup$ What does "compact-open subgroup of algebraic subgroups" mean (if not just a compact, open subgroup of the group of rational points, which is the kind of object that we are trying to describe)? $\endgroup$
    – LSpice
    Commented Dec 15, 2016 at 19:07

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