1
$\begingroup$

$\DeclareMathOperator\SL{SL}$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.

Improve this question
$\endgroup$
2
  • 4
    $\begingroup$ 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
    Sep 29, 2022 at 17:12
  • $\begingroup$ @Asaf, isn't that an answer? $\endgroup$
    – LSpice
    Sep 30, 2022 at 1:01

1 Answer 1

Reset to default
4
$\begingroup$

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.

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.