Timeline for Reference request for statement concerning free subgroups of $ \mathrm{SL}_2(\mathbb{Z}). $
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 15, 2022 at 13:16 | comment | added | Sam Nead | @IanAgol - I agree that the math question is not research level, but what about the reference request? I’ve had a few times when I’ve tried (and failed) to find a perfect reference for something a bit easy… which I still did not want to prove myself… | |
| Jun 15, 2022 at 12:17 | review | Close votes | |||
| Jun 20, 2022 at 3:06 | |||||
| Jun 15, 2022 at 11:59 | comment | added | Ian Agol | It’s a fine question, but not appropriate for mathoverflow which is aimed at research-level questions. | |
| Jun 15, 2022 at 8:44 | comment | added | YCor | It seems to be the easiest instance of the ping-pong lemma, which implies that every Zariski-dense subgroup of $\mathrm{PSL}(2,\mathbf{R})$ contains a subgroup isomorphic to $F_2$ and QI-embedded in $\mathrm{PSL}(2,\mathbf{R})$. That it is QI-embedded prevents the existence of unipotents. | |
| Jun 15, 2022 at 7:55 | answer | added | Sam Nead | timeline score: 2 | |
| Jun 15, 2022 at 5:13 | comment | added | Moishe Kohan | Much more is true: Every Zariski dense subgroup of $SL(n,R)$ contains a free subgroup without nontrivial unipotents. This is due to J.Tits | |
| Jun 15, 2022 at 1:25 | comment | added | Venkataramana | one can show that the commutator subgroup of the congruence subgroup $\Gamma (2)$ of level 2 in $SL_2(\mathbb Z)$ does not contain unipotent elements, but is free on infinitely many generators. You can then take any two elements in the generating set. | |
| S Jun 14, 2022 at 21:34 | review | First questions | |||
| Jun 14, 2022 at 21:51 | |||||
| S Jun 14, 2022 at 21:34 | history | asked | Georgi Kocharyan | CC BY-SA 4.0 |