2
$\begingroup$

I've read the following question:

Finite subgroups of ${\rm SL}_2(\mathbb{Z})$ (reference request)

and it made me wonder. It's easy to see that $\operatorname{SL}_2(\mathbb{Z})=\operatorname{Sp}_2(\mathbb{Z})$. So does it remain true that $\operatorname{Sp}_{2n}(\mathbb{Z})$ has only finitely many finite subgroups, if $n$ is general? If not, can we still say something about the possible orders of its finite subgroups? (For example, must they be divisible by only finitely many primes?)

$\endgroup$
7
  • 1
    $\begingroup$ Yes, there's a bound on the order that depends on $n$. The idea is to show that most congruence subgroups are torsion-free. $\endgroup$ Mar 20, 2016 at 7:43
  • 4
    $\begingroup$ For any Chevalley group $G$ (e.g., ${\rm{Sp}}_{2n}$) there is a systematic approach to getting a uniform upper bound (in terms of the root system) on the size of finite subgroups, even working inside $G(k)$ for a fixed number field $k$. See Theorem 5 in section 5.4 of part II of college-de-france.fr/media/jean-pierre-serre/… (where $t=\ell-1$ for $\ell$ unramified in $k$). The conjugacy aspects seem to be much more subtle (especially if you only work with $\mathbf{Z}$-points); see Theorem 8 in section 6.6 of part II for a sample. $\endgroup$
    – nfdc23
    Mar 20, 2016 at 7:51
  • 2
    $\begingroup$ It has infinitely many finite subgroups. Probably you have another question in mind, namely: does it have finitely isomorphism type of subgroups (yes, because it's virtually torsion-free, and you have bounds on the possible orders using the embedding into $\mathrm{SL}_{2n}(\mathbb{Z})$), or, more interesting, does it have finitely many conjugacy classes of finite subgroups. The latter is still true, but less obviously and I'm not sure of a reference (and unlike the previous question, it does not boil down to $\mathrm{SL}_{2n}(\mathbb{Z})$). $\endgroup$
    – YCor
    Mar 20, 2016 at 9:15
  • 3
    $\begingroup$ As mentioned in MO106338, Zassenhaus proved the finiteness result ( following results of Blichfeldt,Schur, Jordan etc), and there is a proof in the 1962 edition of Curtis and Reiner. $\endgroup$ Mar 20, 2016 at 10:00
  • 3
    $\begingroup$ @GeoffRobinson: just to make your statement more precise: the link is mathoverflow.net/questions/106338, and it refers to the finiteness of the number of conjugacy classes of finite subgroups in $\mathrm{GL}_m(\mathbb{Z})$. But this does not imply that all its subgroups also have finitely many conjugacy classes of finite subgroups; it's probably false in general, while it's certainly true for $\mathrm{Sp}_{2n}(\mathbb{Z})$. $\endgroup$
    – YCor
    Mar 20, 2016 at 10:23

1 Answer 1

2
$\begingroup$

This seems to be done by Markus Kirschner (see these 2011 talk notes).

$\endgroup$
1
  • $\begingroup$ The problem (of the classification, not just the finiteness result) is indeed tackled these slides. It provides some kind of procedure; he obtains classification of maximal finite subgroups up to dimension $2n=22$. $\endgroup$
    – YCor
    Mar 20, 2016 at 18:24

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.