# Fricke groups and Fricke curves

The congruence subgroup $\Gamma_{0}(n) \subset PSL_{2}(\mathbb{Z})$ is normalized by the Fricke involution $F_n: z \mapsto -1/nz$ and so we may form the Fricke modular group $\Gamma_{0}^{+}(n)\langle \Gamma_{0}(n), F_n \rangle \subset PSL_{2}(\mathbb{R})$. (Sometimes people focus on the case $p=n$, since then I think that $\Gamma_{0}^{+}(p)$ is the full normalizer of $\Gamma_{0}(p) \subset PSL_{2}(\mathbb{R})$.)

Questions:

1. Is it known that and how $\Gamma_{0}^{+}(n)$ splits as a free product of cyclic groups? (This is known for $\Gamma_{0}(n)$ and I can check it by hand for $n=2$.)

2. Are convenient fundamental domains for $\Gamma_{0}^{+}(n)$ acting on the upper half-plane $\mathbb{H}$ known?

3. Are the orbifold points of $\mathbb{H}/\Gamma_{0}^{+}(n)$ completely understood?

For questions of the form `Is it known or understood... ?', I'm looking for answers of the form 'Yes, this is well-known and works as follows... or is written down here...' or 'Yes, it is well-known, but scattered over the literature...' or 'No, this is not obvious.' Also, any good reference is welcome, since I've only found bits and pieces addressing these questions.

• With question 1. I assume you are asking whether $\Gamma_0^+(n)$ splits as a free product of cyclic groups. – Arend Bayer May 25 '11 at 19:05
• I don't understand Q1 but it seems to me that Q2 and Q3 are just exercises, unless I'm missing something. The answers for $\Gamma_0(N)$ are well-understood in both cases and then it's just a case of explicitly working out what the quotient is doing... – Kevin Buzzard May 25 '11 at 19:08
• I added 'free product of cyclic groups' as Arend points out. I have in mind something like $PSL_{2}(\mathbb{Z}) \simeq \mathbb{Z}/2 *\mathbb{Z}/3\mathbb{Z}$. For $\Gamma_{0}^{+}(2)$, I think that you can show it splits as a free product $\mathbb{Z}/2 *\mathbb{Z}/4\mathbb{Z}$. As for Kevin's comment, I'm glad to hear that this is well known. Where do I learn it? – Chris Brav May 25 '11 at 20:07
• I know that you can get some fundamental domain by writing down cosets for $\Gamma_{0}(n)$, but it would be nice to have a fundamental domain described by (say) inequalities. Also, is it clear what to do once you add in the Fricke involution? I apologize if this obvious, but I am new at this business. – Chris Brav May 25 '11 at 20:18
• @Chris: I think we're at cross purposes with (2) -- probably I don't understand what your use of "convenient" means. As for (3) you're asking for points for which the stabilizer is finite but non-trivial, so you want to find the finite subgroups of $\Gamma_0(n)^+$. The intersection of such a finite subgroup with $\Gamma_0(n)$ will still be finite and the size will have gone down by at most a factor of 2. So you'll get the elliptic points for $\Gamma_0(n)$, which are well-understood, and you'll get points stabilised by order 2 elements of $\Gamma_0(n)^+$ that aren't in $\Gamma_0(n). – Kevin Buzzard May 25 '11 at 20:48 ## 2 Answers The answer to question 1. is yes. The quotient$\mathbb{H}^2/\Gamma_0^+(n)$is an orbifold with cusps. Take a separating non-trivial arc connecting a cusp to itself. One may show that such an arc exists using the fact that the orbifold has negative Euler characteristic, and therefore has at least three cone points or cusps, or has positive genus (and at least one cone point). One may take an arc from a cusp to one cone point, and take the boundary of a regular neighborhood to get a separating arc which is non-trivial. By Van-Kampen's theorem, the fundamental group is a free product. For question 2., check out Cummin's tables. Cummins has a list of fundamental domains for$\Gamma_0(n)$for many such$n$which he sent me once, and I think were used or are related to the computations of matrix generators. As Kevin Buzzard says, once you have the fundamental domain for$\Gamma_0(N)$, you can take half of it to get one for$\Gamma_0^+(N)$. For question 3., I think this is explicitly known in terms of arithmetic data. See Johansson. • After submitting my answer, I saw that you had modified the question to a free product of cyclic groups. Such a group would have to be genus 0. The arithmetic groups of this sort have been classified by Cummins, and you could go through them case-by-case to see which ones split as a free product of cyclic groups (see the above tables). – Ian Agol May 25 '11 at 20:55 • All the comments and answers are helpful, but I accepted this one since it seemed to have the most useful information. – Chris Brav May 26 '11 at 8:39 Following Agol's answer, you can find a list of genus zero groups of$n|h$-type together with a discussion of fundamental domain computations in Ferenbaugh's paper: The Genus-zero problem for n|h-type groups, Duke Math. J. 72, no. 1 (1993), 31-63 (sorry, subscription required). From that, you can find all integers$n$for which the group$\Gamma_0(n)+n$is genus zero. Here's the list of possible$n$: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 29, 31, 32, 35, 36, 39, 41, 47, 49, 50, 59, 71. • A free product of cyclic groups gives an orbifold of area <2π since it is a turnover, so this should cut down the search even more. – Ian Agol May 26 '11 at 7:03 • Okay, Table 2 column D in Conway and Norton's Monstrous Moonshine is a list of areas. Your new condition drops the possibilities to: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 17, 19. – S. Carnahan May 26 '11 at 13:45 • Perhaps someone could expand on Agol's above comment. Since$PSL_{2}(\mathbb{Z}) \simeq \mathbb{Z}/2\mathbb{Z}*\mathbb{Z}/3\mathbb{Z}$, it follows from Kurosh's theorem that every subgroup also splits as a free product of cyclic groups. In particular,$\Gamma_{0}(n)$splits as a free product of cyclic groups. Is this consistent with the comment about area? – Chris Brav May 26 '11 at 19:08 • Now that I think about it, I don't see how any of the groups$\Gamma_0(n)+n\$ could fail to be free products of cyclic groups. – S. Carnahan May 27 '11 at 2:39
• I have a similar feeling but not yet a proof, though I haven't tried very hard yet. Will think about it more. – Chris Brav May 27 '11 at 15:07