Timeline for Finite subgroups of ${\rm SL}_2(\mathbb{Z})$ (reference request)
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| S Sep 25, 2015 at 6:20 | history | suggested | Philip | CC BY-SA 3.0 |
The notation for group insertes in dollar signs.
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| Sep 25, 2015 at 5:51 | review | Suggested edits | |||
| S Sep 25, 2015 at 6:20 | |||||
| Jun 8, 2010 at 14:06 | comment | added | Peter Samuelson | Thanks for the reference, it looks like a nice read. I mainly asked for a reference because I thought that the result was probably well-known enough to be written down nicely somewhere, and it seems easier to reference an answer than write it down. But reading people's answers here is often nicer. In particular, thanks for all the good answers. | |
| Jun 8, 2010 at 13:55 | vote | accept | Peter Samuelson | ||
| Jun 7, 2010 at 15:04 | answer | added | user2490 | timeline score: 3 | |
| Jun 7, 2010 at 14:05 | comment | added | Skip | You asked for a reference and got real and interesting answers instead. I am sure there are a million places to find this information, but one gentle entree to the question (with pointers to other references) is J. Kuzmanovich and A. Pavlichenkov "Finite groups of matrices whose entries are integers" from American Math. Monthly, vol. 109, #2 (2002) 173-186. jstor.org/stable/2695329 | |
| Jun 7, 2010 at 7:43 | answer | added | Roland Bacher | timeline score: 9 | |
| Jun 6, 2010 at 23:59 | comment | added | Qiaochu Yuan | Regarding (iii), note more generally that any finite subgroup of the multiplicative group of a field is cyclic. | |
| Jun 6, 2010 at 23:40 | answer | added | Allen Knutson | timeline score: 23 | |
| Jun 6, 2010 at 19:17 | answer | added | Torsten Ekedahl | timeline score: 4 | |
| Jun 6, 2010 at 18:19 | comment | added | Pete L. Clark | Here is a sketch of a proof: (i) every compact group of $\operatorname{GL}_n(\mathbb{R})$ is conjugate to a subgroup of the orthogonal group $O_n(\mathbb{R})$ (e.g. Thm. 3 of math.uga.edu/~pete/8410Chapter9.pdf). (ii) So any finite subgroup of $SL_2(\mathbb{R})$ is conjugate to a subgroup of $SO_2(\mathbb{R}) \cong S^1$. (iii) The finite subgroups of $S^1$ are $\langle \zeta_n \rangle$, where $\zeta_n$ is an nth root of 1. (iv) The charpoly of $\zeta_n$ is $T^2 - (\zeta_n + \zeta_n^{-1})T + 1$. The values of $n$ for which the coeff. of $T$ is rational are exactly those you give. | |
| Jun 6, 2010 at 18:19 | answer | added | Robin Chapman | timeline score: 10 | |
| Jun 6, 2010 at 18:05 | history | asked | Peter Samuelson | CC BY-SA 2.5 |