Welcome to MathOverflow

Question

1

Let $H \subset \mathrm{Sp}(\mathbb{Z})$ be a subgroup of the simplectic group of (square of even dimension) matrices with integer entries, and let $H^{(\ell)}$ denote its pro-$\ell$ completion for some prime $\ell$. Then does $H$ being Zariski dense in $\mathrm{Sp}(\mathbb{Z})$ imply that the abelianization $(H^{(\ell)})^{ab}$ is finite?

Or does the converse hold (i.e. $(H^{\ell})^{ab}$ is finite for some $\ell$ implies $H < \mathrm{Sp}(\mathbb{Z})$ is Zariski dense)?

I have no idea how to go about solving this, and half expect that there are easy counterexamples that I'm not thinking of.

flag	asked Nov 10 2011 at 20:45 Jeff 51●2

1

The converse seems to be disproved by considering e.g. any finite subgroup of Sp$_n(\mathbb{Z})$. – Tom Church Nov 10 2011 at 21:29

2

The Tits alternative gives lots of free subgroups which are Zariski dense, and their abelianizations are long way from being finite (see Theorem 3 of Tits' paper). For the converse, Tom Church gave examples of finite groups, but there are infinite examples as well, for example, $\mathrm{Sp}_{2n}(\mathbf{Z})$ naturally contains a copy of $\mathrm{SL}_n(\mathbf{Z})$, and the latter has finite abelianization for all $n \ge 3$. – Lavender Honey Nov 10 2011 at 23:44

@Jeff: Could you please give some motivation for the question? – Alain Valette Nov 11 2011 at 11:09

Wadim Zudilin · Answer 1 · 2011-11-23 02:24:47Z UTC

Your question is related to an open problem about finiteness of the index of an "arithmetic" group in its Zariski closure (see, for example, Peter Sarnak's lecture Thin Groups and the Affine Sieve as well as arXiv:math/0605675 and arXiv:0803.3322).

Question related to the abelianization of simplectic groups

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.