Timeline for Subgroup of $SL_2(O)$ with nice fundamental domain in complex upper half-plane
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 9, 2016 at 15:37 | vote | accept | Ciro | ||
| Dec 9, 2016 at 10:39 | answer | added | David Loeffler | timeline score: 4 | |
| Dec 9, 2016 at 10:27 | comment | added | Matthias Wendt | Is it really possible for an arithmetic subgroup of $SL_2(O)$ to act properly discontinuously on $\mathbb{H}$? Usually one would let such a group act on two copies of $\mathbb{H}$ for the two real embeddings... | |
| Dec 9, 2016 at 10:05 | comment | added | Ciro | @DavidLoeffler My apologies for the confusion. I was first not going to involve "$S$" and only added that at the end (but then forgot to correct the rest of the question). | |
| Dec 9, 2016 at 10:04 | comment | added | Ciro | @YCor Yes, that's a good point. Can we say something if we add the condition that $G$ acts properly discontinuously on $\mathbb H$? What if we also assume $G$ torsion-free? | |
| Dec 9, 2016 at 10:03 | comment | added | Ciro | @znt That's a good question. I don't have a specific example unfortunately. | |
| Dec 9, 2016 at 10:03 | history | edited | Ciro | CC BY-SA 3.0 |
Hopefully clarified question a bit
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| Dec 9, 2016 at 8:04 | comment | added | David Loeffler | This question seems confused. If $G$ is $S$-arithmetic for a nonempty set $S$ then by definition it will be commensurable with $SL_2(O_S)$ which is far larger than $SL_2(O)$, so it cannot be a subgroup of $SL_2(O)$. If we take $S = \varnothing$ then $G$ is automatically commensurable with SL_2(O) and it follows immediately that $G \cap SL_2(Z)$ is commensurable with $SL_2(O) \cap SL_2(Z) = SL_2(Z)$. | |
| Dec 8, 2016 at 19:16 | comment | added | YCor | The first question is possibly to understand when the action of $G$ on the hyperbolic plane is proper? if you're interested in the non-proper case (e.g. $G=\mathrm{SL}_2(\mathbf{Z}[\sqrt{2}])$), possibly it would be useful to specify what is meant by "fundamental domain". | |
| Dec 8, 2016 at 18:44 | comment | added | znt | Out of interest, can you give an example of an $S$-arithmetic subgroup whose intersection with $SL_2(Z)$ does not have finite index in $SL_2(Z)$? | |
| Dec 8, 2016 at 18:41 | history | asked | Ciro | CC BY-SA 3.0 |