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Let $ G $ be a simple Lie group and let $ \Gamma $ be a lattice in $ G $. If $ \Gamma $ is not contained in any connected subgroup of $ G $ does that imply that $ \Gamma $ is not contained in any positive dimensional subgroup of $ G $?

As pointed out in the comments, an equivalent way of saying this is: If $ \Gamma $ is not contained in any connected subgroup of $ G $ does that imply that the only Lie subalgebras $ \mathfrak{h} \subset \mathfrak{g} $ normalized by $ \Gamma $ are $ \mathfrak{h}=0 $ and $ \mathfrak{h}=\mathfrak{g} $?

Context: I was thinking about the lattice $ SL(2,\mathbb{Z}) $ in $ SL(2,\mathbb{C}) $. $ SL(2,\mathbb{Z}) $ is not contained in any proper connected subgroup of $ SL(2,\mathbb{C}) $. And it is also true that $ SL(2,\mathbb{Z}) $ is not contained in any positive dimensional proper subgroup.

Cross-posted on Math.se: it has been up on MSE for over a week with 1 upvote and no comments or answers so I thought MO might be better.

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  • $\begingroup$ If $G$ is connected noncompact, Borel density says that $\Gamma$ is Zariski-dense. In particular, if $\Gamma\subset H$ with $H$ closed, then $\Gamma$ normalizes $\mathfrak{h}$ and hence $G$ normalizes $\mathfrak{h}$, so $\mathfrak{h}$ is either $\{0\}$ or $\mathfrak{g}$, which means that $H$ is either discrete or equal to $G$. $\endgroup$
    – YCor
    Commented Dec 24, 2023 at 22:46
  • $\begingroup$ @YCor I see that takes care of the noncompact case. What if $ G $ is compact? Then the lattice $ \Gamma $ is just any finite subgroup. Does the ame conclusion still hold? $\endgroup$ Commented Dec 25, 2023 at 14:40
  • $\begingroup$ No, take some large enough dihedral group in $\mathrm{SO}(3)$. $\endgroup$
    – YCor
    Commented Dec 25, 2023 at 18:49
  • $\begingroup$ @YCor ah good point. What if $ \Gamma $ is perfect like $ SL(2, \mathbb{Z}) $? Then does $ \Gamma $ not contained in any connected subgroup imply that $ \Gamma $ is not contained in any positive dimensional subgroup? $\endgroup$ Commented Dec 25, 2023 at 23:23
  • $\begingroup$ $\mathrm{SL}(2,\mathbf{Z})$ is not perfect. $\endgroup$
    – YCor
    Commented Dec 26, 2023 at 0:04

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