Timeline for What are the finite subgroups of $\operatorname{Sp}_{2n}(\mathbb{Z})$?
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11 events
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| Mar 20, 2016 at 15:16 | answer | added | Igor Rivin | timeline score: 2 | |
| Mar 20, 2016 at 13:47 | comment | added | YCor | Anyway, the question does not seem to be about the conjugacy classification. | |
| Mar 20, 2016 at 11:46 | review | Close votes | |||
| Mar 21, 2016 at 21:38 | |||||
| Mar 20, 2016 at 10:23 | comment | added | YCor | @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 $\mathrm{Sp}_{2n}(\mathbb{Z})$. | |
| Mar 20, 2016 at 10:00 | comment | added | Geoff Robinson | 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. | |
| Mar 20, 2016 at 9:15 | comment | added | YCor | 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 $\mathrm{SL}_{2n}(\mathbb{Z})$). | |
| Mar 20, 2016 at 7:53 | comment | added | nfdc23 | The bound through congruence subgroups being torsion-free amounts to using a bound from ${\rm{GL}}_{2n}$. The methods in Serre's paper give better bounds because they uses more refined information about the structure of the chosen algebraic group (and the bounds are even shown to be optimal in many cases when restricted to $\ell$-groups for prime $\ell$). | |
| Mar 20, 2016 at 7:51 | comment | added | nfdc23 | 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. | |
| Mar 20, 2016 at 7:43 | comment | added | Qiaochu Yuan | Yes, there's a bound on the order that depends on $n$. The idea is to show that most congruence subgroups are torsion-free. | |
| Mar 20, 2016 at 5:10 | history | edited | Quinlan Aktaş | CC BY-SA 3.0 |
edited title
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| Mar 20, 2016 at 5:02 | history | asked | Quinlan Aktaş | CC BY-SA 3.0 |