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$\DeclareMathOperator\GL{GL}$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.

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  • $\begingroup$ 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
    Commented 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
    Commented Oct 14, 2022 at 23:56
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    $\begingroup$ @LSpice All my answers to your questions are yes. $\endgroup$
    – stupid boy
    Commented Oct 14, 2022 at 23:58
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    $\begingroup$ 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
    Commented 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
    Commented Oct 14, 2022 at 23:59

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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.

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