Timeline for Uniform lattice in semidirect product
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Jan 23, 2017 at 16:52 | history | edited | YCor |
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| Jan 23, 2017 at 16:16 | answer | added | HackVlix | timeline score: 2 | |
| Feb 18, 2016 at 12:30 | comment | added | YCor | (After talking with Yves Benoist): there are much simpler examples. $V\rtimes S$ has cocompact lattices as soon as $V$ is definable over some $\mathbb{Q}$-isotropic form of $S$. In particular, $\mathbb{R}^3\rtimes\mathrm{SL}_2(\mathbb{R})$ and $(\mathbb{R}^2\oplus\mathbb{R}^2)\rtimes\mathrm{SL}_2(\mathbb{R})$ both have cocompact lattices. | |
| Feb 18, 2016 at 10:56 | comment | added | YCor | By the way, as $SL_2(\mathbb{R})$-module, $M_2(\mathbb{R})$ is just twice the 2-dimensional module. So we eventually get $[(\mathbb{R}^2\oplus\mathbb{R}^2)\rtimes\mathrm{SL}_2(\mathbb{R})]^2$. | |
| Feb 18, 2016 at 10:48 | comment | added | YCor | You both wrote typos :) I guess you mean $\mathrm{Mat}_2(\mathbb{R})^2\rtimes \mathrm{SL}_2(\mathbb{R})^2$ (which is the square $[\mathrm{Mat}_2(\mathbb{R})\rtimes \mathrm{SL}_2(\mathbb{R})]^2$ ). | |
| Feb 18, 2016 at 9:11 | comment | added | Venkataramana | @ADe: you are right: $G=Mat _2({\mathbb R}^2\rtimes SL_2({\mathbb R})^2$ admits a co-compact lattice as written. | |
| Feb 18, 2016 at 9:07 | comment | added | Venkataramana | @YCor: thanks you are right. I was hasty in writing this. What happens is that this ${\mathbb R}^8$ splits into two copies of ${\mathbb R}^4$ into each of which the image of $O$ projects (non-injectively) as a lattice. (the representation of $M_2({\mathbb R})$ as a left module is a direct sum of 2 copies of ${\mathbb R}^2$). | |
| Feb 17, 2016 at 16:18 | comment | added | user1688 | So it should be $G=Mat_2({\mathbb R})^2\rtimes SL_2({\mathbb R})$ then? | |
| Feb 17, 2016 at 15:49 | comment | added | YCor | @Venkataramana I understand $O$ to be additively isomorphic to $\mathbb{Z}[\sqrt{2}]^4$, which would be a lattice in $\mathbb{R}^8$ and not $\mathbb{R}^4$. Do I miss something? | |
| Feb 16, 2016 at 7:41 | comment | added | user1688 | @Venkataramana: Thanks! That puts things in perspective. | |
| Feb 15, 2016 at 23:29 | comment | added | Venkataramana | Let $D$ be a quaternionic central division algebra over $K={\mathbb Q }[{\sqrt 2}]$ which splits over all the real places of $K$. Let $O$ be an order in $D$. Then the semi-direct product of $O$ wth $SL_1(O)$ is a co-compact lattice in the group I have written. | |
| Feb 15, 2016 at 15:41 | comment | added | YCor | @Venkataramana how do you construct it? | |
| Feb 15, 2016 at 15:02 | comment | added | Venkataramana | It is a little intriguing. There do exist uniform lattices in $({\mathbb R }^2 \times {\mathbb R}^2 ) \rtimes SL_2({\mathbb R})^2$ but not in the example you have asked (as YCor's answer tells you) | |
| Feb 15, 2016 at 13:24 | comment | added | YCor | Btw, it is unknown whether $G$ is quasi-isometric to any finitely generated group (probably it isn't). | |
| Feb 15, 2016 at 10:40 | vote | accept | CommunityBot | ||
| Feb 15, 2016 at 9:56 | answer | added | YCor | timeline score: 6 | |
| Feb 15, 2016 at 8:24 | history | asked | user1688 | CC BY-SA 3.0 |