3
$\begingroup$

$\DeclareMathOperator\GL{GL}$Let $ \overline{\mathbb{Q}}_{p} $ be an algebraic closure of $ p $-adic numbers $ \mathbb{Q}_{p} $. A closed subgroup $ H $ of a general linear group $ \GL_{n}(\overline{\mathbb{Q}}_{p}) $ is called completely reducible if the inclusion map $ i:H\to \GL_{n}(\overline{\mathbb{Q}}_{p}) $ is a completely reducible representation of $ H $.

Now let $ G $ be a topologically finitely generated closed subgroup of $ \GL_{n}(\overline{\mathbb{Q}}_{p}) $ and assume that $ G $ is completely reducible. Suppose that $ H $ is a closed subgroup of $ G $ and the union of the conjugacy classes of $ H $ in $ G $ is closed. My question is the following:

Is $ H $ also completely reducible?

For example, if $ H $ is a normal subgroup of $ G $, then it's true. Moreover, if we replace the above $ p $-adic topology with Zariski topology, then the question has a affirmative answer, cf. Theorem 4.1 in Algebraic groups and $G$-complete reducibility: A geometric approach by Benjamin Martin.

Improve this question
$\endgroup$
5
  • $\begingroup$ Just to check, since complete reducibility makes sense for arbitrary subgroups of reductive groups: when asking about complete reducibility of $H$, you mean, in your sense, of the composite $H \to G \to \operatorname{GL}_n(\overline{\mathbb Q_p})$? And, as you seem implicitly to confirm, when you speak of $H$ being closed, you mean in the $p$-adic, not the Zariski, topology, right? Since complete reducibility is really a property of algebraic subgroups—and, in particular, is preserved under Zariski closure—you almost (but maybe not quite) might assume that $H$ and $G$ are Zariski closed. $\endgroup$
    – LSpice
    Oct 14, 2022 at 23:55
  • $\begingroup$ One more question: when you ask about "the union of the conjugacy classes of $H$ in $G$", you mean "the union $\bigcup_{g \in G} g H g^{-1}$ of the conjugates of $H$ in $G$", right? $\endgroup$
    – LSpice
    Oct 14, 2022 at 23:56
  • 1
    $\begingroup$ @LSpice All my answers to your questions are yes. $\endgroup$
    – stupid boy
    Oct 14, 2022 at 23:58
  • 1
    $\begingroup$ The statement "the union of the conjugacy classes of $H$ in $G$" seems like an inaccurate summary of the actual statement of Theorem 4.1, which is that "for some topological generating set $g_1,\dots, g_n$ of $H$, the conjugacy class of $(g_1,\dots, g_n)$ in $G^n$ is closed". $\endgroup$
    – Will Sawin
    Oct 14, 2022 at 23:59
  • $\begingroup$ I agree with @WillSawin. Also, do you have an interesting example of subgroups $H$ and $G$ satisfying your condition where $H$ is not of finite index in a normal subgroup of $G$? $\endgroup$
    – LSpice
    Oct 14, 2022 at 23:59

1 Answer 1

Reset to default
9
$\begingroup$

Based on your answer to LSpice's comment, the answer is negative in both the Zariski and p-adic topology.

Take $G$ to be $GL_n$ and $H$ to be the group of upper-triangular unipotent matrices. Then $\bigcup_{g\in G} g Hg^{-1}$ is just the set of unipotent matrices, since every unipotent matrix in $GL_n$ is conjugate to an upper-triangular one. This is closed, since it is the locus where the characteristic polynomial is equal to $(T-1)^n$, a Zariski-closed condition. However, it is not completeley reducible.

Improve this answer
$\endgroup$

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.