# Does there exist a finite-index subgroup of SL2Z with all cusps irregular?

Recall that if $\Gamma$ is a finite-index subgroup of $\operatorname{SL}_2(\mathbf{Z})$, then a cusp of $\Gamma$ is an orbit of $\Gamma$ on the set $\mathbf{P}^1_{\mathbf{Q}}$. If $-1\notin \Gamma$, then for any cusp $c$, the stabilizer of $c$ in $\Gamma$ (well-defined up to conjugacy) is an infinite cyclic group generated by an element conjugate in $\operatorname{SL}_2(\mathbf{Z})$ to either $\begin{pmatrix} 1 & h \\ 0 & 1\end{pmatrix}$ or $\begin{pmatrix} -1 & h \\ 0 & -1 \end{pmatrix}$ for some $h \in \mathbf{N}$ (the width of the cusp $c$), and we say that $c$ is regular in the first case and irregular in the second.

My question, as in the title, is:

Does there exist a finite-index subgroup $\Gamma \le \operatorname{SL}_2(\mathbf{Z})$ for which every cusp is irregular?

(It would suffice to find a finite-index normal subgroup with at least one irregular cusp, since the cusps of a normal subgroup are either all regular or all irregular; but I don't know if that helps!)

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How about the group generated by $\begin{pmatrix} -1 & 2 \\ 0 & -1 \end{pmatrix}$ and $\begin{pmatrix} -1 & 0 \\ 2 & -1 \end{pmatrix}$? – S. Carnahan Nov 24 '11 at 11:58
@Algori: The problem is that the action of $\Gamma$ on $H$ factors through the projective image of $\Gamma$ and thus it doesn't see the sign. – David Loeffler Nov 24 '11 at 15:02
@Scott: Wow, now I feel stupid! You should post that as an answer and I'll accept it. – David Loeffler Nov 24 '11 at 15:19
Regarding Scott's example, it is a more general fact that for any fuchsian group uniformizing a once-punctured torus, the commutator has trace $-2$, and not $+2$. – Richard Kent Nov 24 '11 at 15:37

The thrice-punctured sphere can be represented as a quotient of the upper half-plane by $\Gamma(2)$. One may take as a fundamental domain the region contained in the geodesics between $i\infty, 0, 1, -1$, where $1$ and $-1$ are identified by the "translation by 2" operator. The fundamental group of the punctured sphere is free on 2 generators, and we may choose these to be $\begin{pmatrix} -1 & 2 \\ 0 & -1 \end{pmatrix}$ (which induces translation by -2) and $\begin{pmatrix} -1 & 0 \\ 2 & -1 \end{pmatrix}$ (which rotates around the cusp at zero - you can check this by noting where the endpoints of the geodesics go). One usually sees the negatives of these matrices in descriptions of generators of $\Gamma(2)$, so they may be unfamiliar.

At any rate, this group has index 12 in $SL_2(\mathbb{Z})$, since $\Gamma(2)$ is a direct product of $\pm 1$ with this group (see e.g., Andy Putman's answer here). To check that it's normal, I conjugated the above matrices against the standard S and T generators by hand and looked at the results mod 4, where this group has only four representatives out of 48 invertible matrices. I'd be interested to hear a computation-free method.

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Just to record what I mentioned in the comments, you may argue as follows:

Let $X$ be the orbifold $\mathrm{SL}_2(\mathbb{Z})\backslash \mathbb{H}^2$. The curve $X$ has two singular points, one a cone point $x$ of order $2$ (the image of the fixed point of an elliptic of order two), and a cone point $y$ of order $3$ (the image of the fixed point of an elliptic of order three).

There's a $6$--fold branched cover $T$ of $X$ which is smooth and diffeomorphic to a punctured torus. (First take a $3$--fold cover $X_3$ of $X$ branched over $x$. Then take the $2$--fold cover of $X_3$ branched over the three preimages of $y$ to obtain $T$.)

If $\Gamma$ is the subgroup of $\mathrm{SL}_2(\mathbb{Z})$ corresponding to $T$ (of index twelve to avoid the center), then we may write $\Gamma = \langle a , b \rangle$ so that the parabolic conjugacy class is represented by the commutator $[a,b]$, and a simple computation reveals that $\mathrm{trace}([a,b]) = - 2$. (As mentioned in my comment, you don't need to compute $a$ and $b$ to compute their trace.)

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+1: I like how this example has only one cusp to worry about. – S. Carnahan Nov 25 '11 at 3:09