Timeline for Examples of discrete subgroups of $PSL_2(\mathbf{R})$ with finite covolume and which are not co-compact
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Jul 3, 2016 at 14:30 | answer | added | Lee Mosher | timeline score: 1 | |
| Jul 2, 2016 at 19:59 | history | edited | Hugo Chapdelaine | CC BY-SA 3.0 |
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| Jul 2, 2016 at 16:36 | answer | added | Igor Rivin | timeline score: 2 | |
| Jul 2, 2016 at 15:24 | review | Close votes | |||
| Jul 2, 2016 at 17:43 | |||||
| Jul 2, 2016 at 12:35 | vote | accept | Hugo Chapdelaine | ||
| Jul 2, 2016 at 12:30 | comment | added | Hugo Chapdelaine | Thanks @YCor for the clarification. So what I thought was indeed wrong. | |
| Jul 2, 2016 at 12:30 | answer | added | ThiKu | timeline score: 1 | |
| Jul 2, 2016 at 12:07 | comment | added | YCor | You mean a angle that is zero (i.e., a vertex at infinity). But I think Anton is right. Indeed, if a subgroup is commensurate to $\mathrm{SL}_2(\mathbf{Z})$, then it is contained in its commensurator $\mathrm{SL}_2(\mathbf{Q})$, whose elements of finite order only have order dividing 12. Hence a triangle group containing an element of another order, say 5, cannot be commensurate to a conjugate of $\mathrm{SL}_2(\mathbf{Z})$. [Note: I distinguish "commensurate" subgroups, meaning: the interesection has finite index in both, with "commensurable"= have isomorphic finite index subgroups] | |
| Jul 2, 2016 at 11:34 | answer | added | coudy | timeline score: 10 | |
| Jul 2, 2016 at 10:07 | comment | added | Hugo Chapdelaine | Hi @Anton, if your triangle group has three finite angles, then it seems to me that it will be co-compact. Therefore, this means that you need to choose one angle which has to be infinite. But I thought that all such triangle groups would be commensurable to a conjugate of $PSL_2(Z)$, so may be I was wrong. | |
| Jul 2, 2016 at 9:48 | comment | added | YCor | Yes I know that my examples are not really explicit at the group level (although it's probably doable at the cost of a lengthy exercise), that's why I posted as a comment rather than an answer. | |
| Jul 2, 2016 at 9:46 | comment | added | Hugo Chapdelaine | Hi @YCor. I see, so you are appealing here to some kind of uniformization result for non-compact Riemann surfaces. In practice is it possible to give an explicit example of such a group $\Gamma$, by explicit I mean: explicit generators where the entries of each matrix can be viewed as the zeros of some "reasonable function" (e.g. hyper-geometric functions, or some function which satisfies a simple differential equation) ? | |
| Jul 2, 2016 at 9:45 | comment | added | user1688 | Hecke triangle groups provide many examples. | |
| Jul 2, 2016 at 9:39 | history | edited | Hugo Chapdelaine | CC BY-SA 3.0 |
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| Jul 2, 2016 at 9:39 | comment | added | Hugo Chapdelaine | Hi @Andreas, sure, I mean commensurable. I'll reedit it. | |
| Jul 2, 2016 at 9:38 | history | edited | YCor |
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| Jul 2, 2016 at 9:37 | comment | added | YCor | A way to produce hyperbolic surfaces consists in gluing pants along the boundaries. You can allow pants with cusps. In this way you can get hyperbolic surfaces of finite volume with cusps, and with closed geodesics of arbitrary length. While only countably many values for the length of closed geodesics are allowed in the arithmetic case. | |
| Jul 2, 2016 at 9:30 | history | asked | Hugo Chapdelaine | CC BY-SA 3.0 |