I'm trying to prove the following result.

If $G$ is a non compact semisimple Lie group with no compact factors (lying in some $SL(l,\mathbb{R})$), and $\Gamma$ is a lattice in $G$, then $\Gamma$ is not solvable, and $[\Gamma, \Gamma]$ is infinite.

So far, I've managed to prove that $\Gamma$ can't be abelian, since $\Gamma$ projects densely onto the maximal compact factor $\{e\}$, and a version of the Borel density theorem tells us that $\mathcal{C}_G(\Gamma) = Z(G)$, and $Z(G)$ is finite, but $\Gamma$ isn't, which rules out the possibility that $\Gamma$ is abelian. My question is whether I can refine this argument to prove that $\Gamma$ isn't solvable as well, or do I need some additional machinery to prove that? Thanks!

  • 2
    $\begingroup$ Yes you can refine the argument to get solvability, and also use the Borel density for the derived subgroup being infinite. However, this seems to better fit MathSE. $\endgroup$ – YCor May 18 at 6:13
  • 1
    $\begingroup$ Already crossposted at MSE here. $\endgroup$ – Dietrich Burde May 18 at 16:59
  • $\begingroup$ One can prove this as a consequence of the Tits alternative, or maybe of its proof. en.wikipedia.org/wiki/Tits_alternative $\endgroup$ – Ian Agol May 22 at 3:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.