Timeline for free subgroups of $SL_2(\mathbb{Z[i]})$
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Sep 12, 2016 at 18:57 | comment | added | Ofir | @AlainValette By the growth I meant the number of points in a ball of radius R as R goes to infinity, for some given matrix norm. In the $SL_2(\mathbb{Z})$ case it grows as $\Theta(R^2)$ (for any fixed finite index subgroup). | |
| Sep 12, 2016 at 15:13 | comment | added | Alain Valette | As was mentioned earlier, there is no free subgroup of finite index in $SL_2(\mathbb{Z}[i])$: my favorite proof uses virtual cohomological dimension, which is 2 here (as opposed to 1 for virtually free groups). See the wonderful paper by J.-P. Serre, "Cohomologie des groupes discrets", especially Proposition 21. Question for Prometheus: what do you mean by "free subgroups which are large in the sense that they have large growth"? | |
| Sep 12, 2016 at 12:50 | comment | added | Venkataramana | As Misha and Uri Bader have observed, you cannot get free subgroups of finite index. If you view $SL_2(\mathbb{Z}[i])$ as a subgroup of $SL_2(\mathbb{C})$ the latter sen as a REAL group, then free subgroups of infinite index which are Zariski dense in the real group do exist (again, by Margulis Soifer, as referenced by Uri Bader, if you like). Then a result of Nori and Weisfeiler says that these free subgroups surject onto $SL_2(O_K/p)$ for any non-zero prime ideal in $O_K$ where $O_K$ is the ring of integers in the imaginary quadratic $\mathbb{Q}(i)$. | |
| Sep 12, 2016 at 12:13 | comment | added | Uri Bader | Prometheus, just to emphasize @Misha 's answer, you should know that for every number field $k$ other then $\mathbb{Q}$, $\Gamma=\text{SL}_2(\mathcal{O})$ will not be virtually free. Nevertheless, $\Gamma$ will contain "large" free subgroups, for various notions of "largeness". However, I suppose that these large free subgroups will give you nothing but obvious lower bounds for lattice points counting problems. | |
| Sep 12, 2016 at 11:40 | comment | added | Misha | @Prometheus: One way to see this is to observe that $SL(2, Z[i])$ contains free abelian subgroups of rank 2. | |
| Sep 12, 2016 at 10:46 | history | edited | Ofir | CC BY-SA 3.0 |
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| Sep 12, 2016 at 10:39 | comment | added | Ofir | @Misha I was hoping that the principal congruence subgroups in $SL_2(\mathbb{Z}[i])$ would be free, or at least virtually free. Why aren't there any finite index free subgroups? | |
| Sep 12, 2016 at 10:29 | history | edited | Ofir | CC BY-SA 3.0 |
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| Sep 12, 2016 at 10:19 | comment | added | Uri Bader | A possible interpretation of the question is whether every lattice contains a profinitly dense free group. Then the answer should be yes, but I am not sure what is the right refernce (Margulis-Soifer?). | |
| Sep 12, 2016 at 10:14 | comment | added | Misha | You should be more specific, the answer depends on the exact meaning of "phenomena". Are you asking for existence of free subgroups? Subgroups generated by powers of two given unipotent elements not generating a solvable subgroup? Then yes. Are you asking for free subgroups of finite index? Then no (ever). | |
| Sep 12, 2016 at 9:28 | history | asked | Ofir | CC BY-SA 3.0 |