Timeline for Are there noncongruence subgroups (of finite index) of the modular group generated only by 2 or 3 elements?
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| S Apr 17, 2021 at 7:03 | history | suggested | Maarten Derickx | CC BY-SA 4.0 |
Fixed formatting of math symbols
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| Apr 16, 2021 at 20:12 | review | Suggested edits | |||
| S Apr 17, 2021 at 7:03 | |||||
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jan 10, 2013 at 17:19 | comment | added | Ian Agol | Nice Lee. It's not hard to check that an orbifold with 2 cusps and cone points of orders 2 and 3, and euler characteristic $-7/6$ must have fundamental group $Z\ast Z/2\ast Z/3$, which is what I referred to in my answer. | |
| Jan 10, 2013 at 15:54 | comment | added | Lee Mosher | Draw a circle of two edges. To one of its vertices, attach one endpoint of an arc of two edges. From the opposite endpoint of that arc, attach an arc of one edge with opposite vertex labelled $\mathbb{Z}/2$, and attach another arc of two edge with opposite vertex labelled $\mathbb{Z}/3$. | |
| Jan 10, 2013 at 15:54 | comment | added | Lee Mosher | Wohlfart also mentions that in his example, one cusp has amplitude 1 and the other has amplitude 6. If you trace a closed curve in the graph of groups corresponding to the cusp, the amplitude equals half the edge length. Exactly one of the two $\mathbb{Z} * \mathbb{Z}/3 * \mathbb{Z}/2$ examples exhibits this possibility. I'll describe the graph. | |
| Jan 10, 2013 at 15:48 | comment | added | Lee Mosher | Wohlfart's example has index 7 and two cusps. He gives a formula for index which amounts to $index = 6 \cdot (\mathbb{Z}-rank) + 4 e_3 + 3 e_2 - 6$ where the $\mathbb{Z}$-rank is the number of $\mathbb{Z}$-free factors and equals $2 \cdot (genus) + (number of cusps) - 1$, $e_3$ is the number of $\mathbb{Z}/3$ free factors, and $e_2$ is the number of $\mathbb{Z}/2$ free factors. From this I believe it follows that Wohlfart's example is in one of the two $\mathbb{Z} * \mathbb{Z}/3 * \mathbb{Z}/2$ conjugacy classes mentioned above. | |
| Jan 10, 2013 at 14:40 | comment | added | Lee Mosher | three of index $9$ isomorphic to $\mathbb{Z} * \mathbb{Z} * \mathbb{Z}/2$; three of index $10$ isomorphic to $\mathbb{Z} * \mathbb{Z} * \mathbb{Z}/3$; and five of index $12$ isomorphic to $\mathbb{Z}*\mathbb{Z}*\mathbb{Z}$. The only one of this list which is itself normal is the index $12$ group isomorphic to $\mathbb{Z}*\mathbb{Z}*\mathbb{Z}$ (arising from the barycentric subdivision of the 1-skeleton of the tetrahedron) whose quotient is the alternating group of order 12. | |
| Jan 10, 2013 at 14:39 | comment | added | Lee Mosher | Using the graphical method described in my earlier comments, I enumerated conjugacy classes of finite index subgroups of $PSL(2,Z) = \mathbb{Z}/2 * \mathbb{Z}/3$ having rank $\le 3$ and index $\ge 7$. There are fifteen of them, here is a summary. Each of the six classes of rank 2 subgroups has index $\le 6$. Of rank 3 classes there are twenty one altogether, fifteen of which have index $\ge 7$: two of index $7$ isomorphic to $\mathbb{Z}*\mathbb{Z}/2*\mathbb{Z}/3$; two of index $8$ isomorphic to $\mathbb{Z} * \mathbb{Z}/3 * \mathbb{Z}/3$; .... | |
| Jan 9, 2013 at 17:36 | comment | added | Ian Agol | By Theorem 5 of this paper, the smallest index of a non-congruence subgroup is 7 (an example is given right before the theorem). By Misha's estimate, there are no rank 2 non-congruence subgroups. One could then check Wohlfahrt's example to see if it is rank 3. projecteuclid.org/… | |
| Jan 9, 2013 at 3:24 | comment | added | Lee Mosher | Oops, OK. But anyway, that's the quickest way to enumerate the relevant finite index subgroups. On top of which, the number of generators is easy to compute: it's the sum of the rank of the underlying graph plus the number of points labelled with a nontrivial group. | |
| Jan 9, 2013 at 2:49 | comment | added | Ian Agol | @Lee: principal congruence subgroups are normal, but not a general congruence subgroup, which need only contain a principal one as a subgroup. | |
| Jan 9, 2013 at 2:10 | comment | added | Lee Mosher | Using graphs of groups, it is easy to enumerate the finitely many conjugacy classes of subgroups of rank $\le 3$. Each one corresponds to a finite connected bi-partite graph; each vertex in the first set is either of valence 2 or of valence 1 labelled with $\mathbb{Z}/2\mathbb{Z}$; and each vertex in the second is either of valence 3 or of valence 1 labelled with $\mathbb{Z}/3\mathbb{Z}$. Congruence subgroups being normal, each such graph whose self-isomorphism group is NOT transitive on edges is not a congruence subgroup, and the vast majority satisfy that. | |
| Jan 8, 2013 at 19:20 | comment | added | Misha | Noam: Perhaps. Then the question becomes more interesting and I do not know examples, but: On general grounds, if $\Gamma$ is an $N$-generated Fuchsian group of finite covolume then $Area(H^2/\Gamma)\le 2\pi(N-1)$, which gives an upper bound on index of $\Gamma$ in $PSL(2,Z)$. Tim Hsu in "Identifying congruence subgroups of the modular group", Proc. Amer. Math. Soc. 124, 1351-1359, gives an algorithm to determine if given $\Gamma$ is a congruence subgroup or not. Putting two things together, somebody can then search for non-congruence subgroups of rank $\le 3$. | |
| Jan 8, 2013 at 18:27 | comment | added | Noam D. Elkies | Perhaps oxeimon is interested in noncongruence subgroups of finite index, which would make this a harder problem (but probably still feasible). | |
| Jan 8, 2013 at 18:19 | history | answered | Misha | CC BY-SA 3.0 |