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Are there any congruence subgroups other than $\operatorname{SL}_2(\mathbb Z)$ which have exactly 1 cusp? By congruence subgroup, I mean a subgroup of $\operatorname{SL}_2(\mathbb Z)$ containing $\Gamma(N)$ for some $N$.

This question was motivated by the comments on the answer to Generators of the graded ring of modular forms. In particular, if there are no such subgroups, then all congruence subgroups other than $\operatorname{SL}_2(\mathbb Z)$ are generated in degree at most 5, with relations in degree at most 10, as follows from the comments on the answer to Generators of the graded ring of modular forms.

I have a feeling the answer to this question may be well known, but I could not find it in Diamond and Shurman or by a Google search.

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  • $\begingroup$ My previous answer had an error. I'll write a replacement when I get time. $\endgroup$ – David E Speyer Jan 9 at 16:24
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Yes, there are plenty of others with the best known being the one corresponding to the hexagonal torus. For a list of low-genus congruence subgroups see here. (Sebastian Pauli at UNC Greensboro).

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$\def\SL{\mathrm{SL}}\def\GL{\mathrm{GL}}\def\PSL{\mathrm{PSL}}\def\ZZ{\mathbb{Z}}\def\FF{\mathbb{F}}\def\PP{\mathbb{P}}$I previously gave a wrong answer to this question. In fact, such groups exist for $N$ any prime which is $3 \bmod 4$ and in many other cases, as I will now spell out.

If we have $G_1 \subset \SL_2(\ZZ/n_1 \ZZ)$ and $G_2 \subset \SL_2(\ZZ/n_2 \ZZ)$ with $n_1$ and $n_2$ relatively prime, then $G_1 \times G_2 \subset \SL_2(\ZZ/n_1 \ZZ) \times \SL_2(\ZZ/n_2 \ZZ) \cong \SL_2(\ZZ/(n_1 n_2) \ZZ)$ also has this property. So I'll concentrate on prime powers.


I'll concentrate first on the case of $N$ a prime $p$. As in my wrong answer, if $\Gamma \subset \SL_2(\FF_p)$ has the required property, so does any larger group containing $\Gamma$, so it is enough to restrict our attention to the maximal proper subgroups of $\SL_2(\FF_p)$. These are classified; see Corollary 2.2 in this paper by King. Note that types (d) and (e) only occur for $p$ a prime power, not a prime, so they are irrelevant to us. Group (a) has a two element orbit and group (c) is $\Gamma_0(p)$, with two cusps. This leaves (b) and (f), (g), (h).

Type (b) This is the one I got wrong; I misread King's paper. The correct description of Group (b) is as follows: Let $q$ be an isotropic quadratic form on $\FF_p^2$. For example, if $p \equiv 3 \bmod 4$, then we can take $q(x,y) = x^2+y^2$. Then Group (b) is the group of matrices which either preserve the quadratic form or multiply it by $-1$. (I misread this as the index $2$ subgroup that preserve the form.) For example, when $q(x,y) = x^2+y^2$, this is matrices of the form $\left[ \begin{smallmatrix} a & b \\ -b & a \end{smallmatrix} \right]$ with $a^2+b^2 = 1$ or $\left[ \begin{smallmatrix} c & d \\ d & -c \end{smallmatrix} \right]$ with $c^2+d^2 = -1$.

Now, $q$ descends to a map $\mathbb{P}^1(\FF_p) \to \FF_p^{\times}/(\FF_p^{\times})^2$. The two classes in $\FF_p^{\times}/(\FF_p^{\times})^2$ thus divide $\PP^1(\FF_p)$ into two orbits for the action of the index two subgroup that preserves $q$. What about the elements that switch the sign of $q$? If $-1$ is a square, they preserve the split of $\PP^1(\FF_p)$ into two orbits; but, if $-1$ is not square, then they switch the orbits, and we get just one cusp. Of course, $-1$ is not square if and only if $p \equiv 3 \bmod 4$. Conveniently, in this case, we can take $q(x,y) = x^2+y^2$, so all of my examples are relevant.

Lifting to $\ZZ/p^k \ZZ$ Take an isotropic quadratic form on $\FF_p^2$ and lift it to an quadratic form $Q: \ZZ_p^2 \to \ZZ_p$ where $\ZZ_p$ is the $p$-adics. Let $\Sigma \subset \SL_2(\ZZ_p)$ be the matrices which rescale $Q$ by (respectively) $\pm 1$ and any scalar. Then $Q$ descends to a surjection $\PP^1(\ZZ_p) \to \ZZ_p^{\times} / (\ZZ_p^{\times})^2$ (here $\ZZ_p^{\times}$ is the unit group of $\ZZ_p$ and $\PP^1(\ZZ_p)$ is pairs $[u:v]$ in $\ZZ_p^2$, not both in $p \ZZ_p$, modulo simultaneous rescaling by $\ZZ_p^{\times}$). The group $\Sigma$ acts transitively on $\PP^1(\ZZ_p)$; if $-1$ generates the quotient $\ZZ_p^{\times} / (\ZZ_p^{\times})^2$, which happens for $p \equiv 3 \bmod 4$. Descending modulo $p^k$ gives us examples in $\ZZ/p^k \ZZ$.

Types (f), (g), (h) These types arise from the exceptional subgroups $A_4$, $S_4$ and $A_5$ of $\PSL_2(\FF_p)$ (for $p \geq 5$). They can only act transitively on $\PP^1(\FF_p)$ if $p+1$ divides the order of these groups, which happens for $p \in \{ 5,11 \}$, $p \in \{ 5,7,11,23 \}$ and $p \in \{ 5,11,19,29 \}$ rspectively. I believe all of those cases occur; many of them occur in the list of near fields. I am not sure which of these cases can be lifted modulo powers of primes.

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