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$\DeclareMathOperator\SL{SL}$Let $ G $ be a noncompact simple Lie group. For example $ \SL_n $. Let $ \Gamma $ be a lattice in $ G $. Consider the action of $ \Gamma $ on the Lie algebra of $ G $ by conjugation. Is this representation of $ \Gamma $ always irreducible?

For example, I think it is true that all lattices in $ \SL_2(\mathbb{R}) $ and $ \SL_2(\mathbb{C}) $ are Ad-irreducible.

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    $\begingroup$ Well we do know that $Ad(G)$ acts irreducibly, and $\Gamma$ is Zariski dense by Borel's density theorem, so yes. Probably one can get a more ``geometric'' direct proof in several cases, for example for $SL_{2}(Z)$, one has the raising/lowering operators given by the unipotents... $\endgroup$
    – Asaf
    Commented Sep 29, 2022 at 17:12
  • $\begingroup$ @Asaf, isn't that an answer? $\endgroup$
    – LSpice
    Commented Sep 30, 2022 at 1:01

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Per the request to post it as an answer.

Notice that the Ad representation is a polynomial representation into $\operatorname{GL}(\operatorname{Lie}(G))$. We do know that $\operatorname{Ad}(G)$ acts irreducibly, and $\Gamma$ is Zariski dense by Borel's density theorem. Hence $\operatorname{Ad}\rvert_{\Gamma}$ is also irreducible.

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