Timeline for Zariski dense subgroups and conjugates
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 18, 2015 at 14:46 | comment | added | YCor | @Venkataramana: yes, indeed I want no rational eigenvalue at all. | |
| Jan 18, 2015 at 13:39 | comment | added | Venkataramana | @YCor: a small correction to your construction; you must assume that $g$ has no rational eigenvalues, and is diagonalisable over over $\mathbb R$; otherwise, you might end up in the integer points of a proper parabolic subgroup | |
| Jan 18, 2015 at 13:35 | comment | added | Venkataramana | It is a result of Jacques Tits (an old paper in Comptes Rendus). Hee Oh's result is about any Zariski dense discrete subgroup which intersects the upper triangular unipotents in a finite index subgroup. | |
| Jan 18, 2015 at 8:43 | comment | added | YCor | @Pablo I think so, but it's harder. Isn't it due to Hee Oh? I'm not sure of the reference but I learnt it from Misha Kapovich on MO. | |
| Jan 18, 2015 at 8:11 | comment | added | Pablo | @YCor Is Venkataramana right? Does the index become finite for every $a \in \mathrm{SL}_3(\mathbb{Z})$ ? | |
| Jan 18, 2015 at 7:57 | comment | added | Venkataramana | I am assuming that $a\in SL_3({\mathbb Z})$. | |
| Jan 18, 2015 at 7:34 | comment | added | Venkataramana | The group $<H,aga^{-1}> $ of YCor actually has finite index in $SL_3({\mathbb Z})$. | |
| Jan 17, 2015 at 16:47 | comment | added | YCor | In $SL_3(\mathbf{Z})$, Zariski dense is equivalent to having profinite closure of finite index, as I told you in a comment yesterday. | |
| Jan 17, 2015 at 15:49 | comment | added | Pablo | @YCor What if we replace "Zariski dense" by "profinite closure of infinite index"? Is the profinite closure of your example of finite index? | |
| Jan 17, 2015 at 15:44 | comment | added | YCor | No. If $H$ is the Heisenberg group over $\mathbf{Z}$ (the upper unipotent matrices with integral entries) and $g\in\mathrm{SL}_3(\mathbf{Z})$ is diagonalizable over $\mathbf{R}$ and not $\mathbf{Q}$, it is easy to show that for every $a\in\mathrm{SL}_3(\mathbf{Q})$, the group $\langle H,aga^{-1}\rangle$ is Zariski-dense. Hint: classify $\mathbf{Q}$-defined Zariski closed subgroups containing the upper unipotent subgroup. | |
| Jan 17, 2015 at 14:57 | history | asked | Pablo | CC BY-SA 3.0 |