Timeline for Has dynamics on $G/\Gamma$ ever been used to prove interesting things about $\Gamma$?
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| Sep 28, 2018 at 0:32 | history | edited | Venkataramana | CC BY-SA 4.0 |
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| Sep 21, 2018 at 15:06 | history | made wiki | Post Made Community Wiki by S. Carnahan♦ | ||
| Sep 21, 2018 at 14:51 | comment | added | Ian Agol | @Kim for consequences of arithmeticity I would recommend Dave Witte-Morris’ book “introduction to arithmetic groups”. Also note that Margulis showed (using similar techniques) that rank one lattices are arithmetic iff they are infinite index in their commensurator. | |
| Sep 21, 2018 at 10:01 | vote | accept | Kim | ||
| Sep 21, 2018 at 4:13 | history | edited | Venkataramana | CC BY-SA 4.0 |
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| Sep 20, 2018 at 20:31 | comment | added | Kim | @Venkataramana Thanks for the reference, by the way. I'll take a look.. | |
| Sep 20, 2018 at 20:16 | comment | added | Venkataramana | it does not matter. An arithmetic group somewhere satisfies $comm (\Gamma)/\Gamma$ is infinite. | |
| Sep 20, 2018 at 20:15 | comment | added | Kim | in the assertion about the commensurator. | |
| Sep 20, 2018 at 20:11 | history | edited | Venkataramana | CC BY-SA 4.0 |
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| Sep 20, 2018 at 19:58 | comment | added | Kim | @Venkataramana Are you using the assumption that $\Gamma = G(\mathbf{Z})$ (as opposed to a general arithmetic group) somewhere? | |
| Sep 20, 2018 at 19:58 | comment | added | Venkataramana | @Kim: Yes. Moreover, the original proof of arithmeticity assumed that lattices were finitely generated (Kazhdan, Garland-Raghunathan). This can be avoided, and you can deduce that superrigidity implies arithmeticity directly, thereby proving that higher rank lattices are finitely generated (I have a note in Comptes Rendus where this is proved). | |
| Sep 20, 2018 at 19:55 | comment | added | Kim | @Venkataramana Okay, so you are saying that this is something we can prove about $\Gamma$ if we know it can be realized as a higher rank lattice, by applying the arithmeticity theorem. | |
| Sep 20, 2018 at 19:52 | comment | added | Venkataramana | @Kim: I am saying that if $\Gamma $ is a higher rank lattice, then $Comm (\Gamma)/\Gamma$ is infinite. | |
| Sep 20, 2018 at 19:50 | comment | added | Kim | @Venkataramana Are you saying that, when some group $\Gamma$ satisfies this condition about abstract commensurator, then it may be realized as an arithmetic subgroup in some algebraic group? | |
| Sep 20, 2018 at 19:44 | history | edited | Venkataramana | CC BY-SA 4.0 |
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| Sep 20, 2018 at 19:41 | comment | added | Venkataramana | @Kim: Well, arithmeticity of $\Gamma =G(\mathbb Z)$ can be interpreted (plus Mostow rigidity, which can be proved by using superrigidity (dynamical considerations again) ) as saying that the abstract commensurator of $\Gamma $ contains $\Gamma$ as an infinite index subgroup. | |
| Sep 20, 2018 at 19:29 | comment | added | YCor | @Kim I don't know any systematic survey. For distortion of cyclic subgroups, the main article is by Lubotzky-Mozes-Raghunathan. | |
| Sep 20, 2018 at 19:27 | comment | added | Kim | @YCor this is very interesting. Would you happen to know a reference that describes the kind of intrinsic consequences we can get for $\Gamma$ from arithmeticity? | |
| Sep 20, 2018 at 19:05 | comment | added | YCor | @Kim technically speaking you're right about arithmeticity, but indeed it allows to prove plenty of things about the underlying group structure (distorsion of elements, description of solvable subgroups, etc) | |
| Sep 20, 2018 at 18:27 | comment | added | Kim | I'm happy with the result about normal subgroups having finite index, but the other examples are not what I have in mind. Oppenheim is related to the quotient space, but it is not really a statement about the group $\text{SL}(3,\mathbf{Z})$. Similarly, Borel density and arithmeticity are statements about the position of $\Gamma$ as a subgroup inside a larger group $G$. But these do not seem to be intrinsic properties of $\Gamma$ itself. Although the following is unclear to me: is arithmeticity an intrinsic property of a discrete group $\Gamma$? (Zariski density obviously is not) | |
| Sep 20, 2018 at 17:56 | history | answered | Venkataramana | CC BY-SA 4.0 |