5
$\begingroup$

Margulis' normal subgroup theorem states that any normal subgroup of a higher rank lattice is either finite or of finite index. The obvious question is: can one classify finite normal subgroups of such lattices? (even $SL(n, \mathbb{Z})$ and $Sp(2n, \mathbb{Z})$ would be a good start).

$\endgroup$
  • $\begingroup$ Central subgroups only (if you take lattices in semi-simple connected Lie groups)? $\endgroup$ – Mark Sapir Apr 18 '11 at 2:46
  • $\begingroup$ @Mark: That's what I thought, but was always puzzled why no one actually stated it this way, so thought that maybe I was missing something... $\endgroup$ – Igor Rivin Apr 18 '11 at 2:49
13
$\begingroup$

These are the central subgroups, see http://www.mathematik.uni-regensburg.de/loeh/seminars/normal_subgroup_thm.pdf . It is proved that every non-central normal subgroup has finite index (page 7).

$\endgroup$
  • 4
    $\begingroup$ The fact that finite normal subgroups $N$ in lattices are central in the ambient Lie group, follows by invoking Zariski density of lattices (to the effect that $N$ is actually normal in the ambient Lie group), and then the classical fact that a discrete normal subgroup in a connected group must be central. $\endgroup$ – Alain Valette May 2 '11 at 8:46

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.