Timeline for Direct product of free groups in $\mathrm{SL}_3(\mathbb{Z})$
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 19, 2022 at 13:01 | comment | added | Roland Bacher | @AndyPutman Andy, Thanks for this precision. | |
| Oct 19, 2022 at 12:54 | comment | added | Andy Putman | @RolandBacher: The Tits alternative does not say that all subgroups of fg linear groups are either virtually solvable or free. | |
| Oct 19, 2022 at 12:48 | comment | added | Roland Bacher | Is the negative answer not a consequence of the Tits alternative: Otherwise ${\mathrm{SL}}_3(\mathbb Z)$ (which is linear and finitely generated) contains $F_2\times \mathbb Z$ which is neither virtually solvable nor free? | |
| Oct 19, 2022 at 11:45 | answer | added | Giles Gardam | timeline score: 4 | |
| Jan 8, 2019 at 18:13 | vote | accept | burtonpeterj | ||
| Jan 8, 2019 at 18:13 | vote | accept | burtonpeterj | ||
| Jan 8, 2019 at 18:13 | |||||
| Jan 8, 2019 at 14:33 | answer | added | YCor | timeline score: 13 | |
| Jan 8, 2019 at 9:26 | comment | added | YCor | No; more generally, let $G$ be a non-virtually-solvable subgroup of $\mathrm{GL}_3(\mathbf{C})$. Then the centralizer of $G$ is abelian. Indeed its Zariski closure contains a Zariski-closed copy of $\mathrm{(P)SL}_2(\mathbf{C})$. There are two such subgroups up to conjugation: the irreducible $\mathrm{PSL}_2(\mathbf{C})(=\mathrm{SO}_3)$ and the upper-left block. The first has a trivial centralizer, and the second has centralizer equal to the diagonal matrices $(a,a,b)$. [I'm pretty sure this argument already exists somewhere on this site.] | |
| Jan 8, 2019 at 5:03 | answer | added | Ian Agol | timeline score: 11 | |
| Jan 8, 2019 at 0:58 | history | asked | burtonpeterj | CC BY-SA 4.0 |