Timeline for Property of lattices in Lie groups
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 4, 2013 at 17:54 | comment | added | YCor | @Aakumadula: great; this means it works (i.e. Anton's question has a negative answer) for an arbitrary f.g. group having a linear representation (in finite dimension) over any field with non-virtually-solvable image. | |
| Feb 25, 2013 at 19:02 | comment | added | Ian Agol | @Aakumadula: Yes, your answer settles the question. | |
| Feb 25, 2013 at 14:10 | comment | added | Venkataramana | to add to the previous comment, strong approximation for these Zariski dense f.g. groups is known thanks to Nori and Weisfeiler. | |
| Feb 25, 2013 at 13:59 | comment | added | Venkataramana | @Cornulier I think the only thing that needs to be used is that $\Gamma$ is finitely generated and Zariski dense in a semisimple group $G$ s.t. $G({\mathbb C})$ is simply connected. You don't need arithmeticity. | |
| Feb 25, 2013 at 13:23 | comment | added | YCor | @Aakumadula: higher rank is used because of arithmeticity. So your argument gives an alternative proof in the case of arithmetic lattices of rank 1 (in addition including arithmetic non-cocompact ones). | |
| Feb 25, 2013 at 7:41 | comment | added | Venkataramana | @Agol. The above proof that the image of the $n$-th power map on an arithmetic lattice $\Gamma$ does not contain a finite index subgroup $\Sigma$ , really does not use higher rank. All it uses is strong approximation. If such a $\Sigma$ existed, then for almost all congruence quotients $G_p=G({\mathbb F}_p)$ of $\Gamma$ the $n$-th power map would be surjective and hence injective. You can choose infinitely many primes for which on $G_p$ the $n$-th power map is not injective. | |
| Feb 24, 2013 at 21:45 | comment | added | YCor | @anon: Agol's answer concerns all cocompact torsion-free lattices in rank 1, including arithmetic ones, not just a single one. | |
| Feb 24, 2013 at 21:40 | comment | added | Ian Agol | @ Martel: I realize that, I was just pointing out that the same argument wouldn't work in higher rank. | |
| Feb 24, 2013 at 21:39 | history | edited | Ian Agol | CC BY-SA 3.0 |
added 102 characters in body
|
| Feb 24, 2013 at 19:53 | comment | added | JHM | I think the answer to the OPs question is no because finitely generated (nonabelian) free groups fail to have the property, and many (i don't know exactly which) lattices admit finite index f.g. free subgroups. | |
| Feb 24, 2013 at 19:49 | comment | added | JHM | The normal subgroup theorem does not quite answer the question for higher rank, because the OP is looking to know that every element in $\Sigma$ is exactly say, a $100^{th}$ power, and not just generated by $100^{th}$ powers. The silly point here being that $a^{100}b^{100}$ is not typically a $100^{th}$-power. | |
| Feb 24, 2013 at 18:55 | comment | added | Ian Agol | @anon: you mean arithmetic groups in rank 1? In higher rank (semisimple), Margulis' superrigidity theorem implies that all lattices are arithmetic. Also, Margulis' normal subgroup theorem implies that all normal subgroups are finite index, so for example the subgroup generated by nth powers is finite-index, as opposed to the rank 1 case. | |
| Feb 24, 2013 at 18:17 | comment | added | anon | The rank one case is exceptional in that there are lots of nonarithmetic irreducible lattices. No one was saying that it isn't important. | |
| Feb 24, 2013 at 13:44 | comment | added | YCor | @anon: I wouldn't say that rank one is exceptional, at least in terms of importance in mathematics. | |
| Feb 24, 2013 at 8:12 | comment | added | user1688 | @Algol: can you give a reference for this result? I am particularly interested in the p-adic case of higher rank, anything known there? | |
| Feb 24, 2013 at 7:26 | comment | added | anon | The rank one case tends to be a bit exceptional. Is anything known about arithmetic subgroups? | |
| Feb 24, 2013 at 7:16 | history | answered | Ian Agol | CC BY-SA 3.0 |