6
$\begingroup$

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!

$\endgroup$
11
  • 2
    $\begingroup$ 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
  • 3
    $\begingroup$ 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
  • 3
    $\begingroup$ 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
  • 2
    $\begingroup$ For maximal compacts, you must catch some Bruhat-Tits theory. $\endgroup$
    – Uri Bader
    Commented Dec 15, 2016 at 7:45
  • 2
    $\begingroup$ @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

1 Answer 1

1
$\begingroup$

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!

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.