Stable points and the fundamental domain of the modular group

Question

Let $\mathbb{\Gamma} = \mathrm{SL_2}(\mathbb{Z})$ be the modular group, $\mathcal{F} = \{z \in \mathbb{C} ;\; \lvert z \rvert \geq 1,\; \lvert \Re (z) \rvert \leq 1/2\}$ its fundamental domain.

How do I prove that if $z$ is in $\mathcal{F}$ and $\mathbb{\alpha}$ is in the modular group and $\mathbb{\alpha} z=z$, that is $z$ is a fixed point, then either $\mathbb{\alpha}$ is the identity or one of $S, ST, (ST)^2, TS, (TS)^2$ where $S$ and $T$ are the generators of $\Gamma$?

Answer 1

Let $z\in\cal F$, $\alpha=\bigl(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\bigr)\in\Gamma$ be such that $\alpha(z)\in\cal F$. Wlog suppose ${\rm Im}(\alpha(z))\le{\rm Im}(z)$. Since

$${\rm Im}\,\alpha(z)=\frac{{\rm Im}\,z}{|cz+d|^2},$$

we must have $|cz+d|\le1$. Geometrically, $z$ must be in the disk of radius $1/|c|$ around $-d/c$. It is easy to check that $-d/c$ is on the real axis, everything in $\cal F$ is at least $\sqrt{3}/2>1/2$ above the real axis, so that necessarily $|c|<2$. We must check three cases: $c=0,\pm1$.

If $c=0$, then $1=\det\alpha=ad$ so $a=d=\pm1$, which ($\alpha$ being a translation by $\pm b$) yields $|b|<2$, and if $b=\pm1$ then $|{\rm Re}(z)|=1/2$ and $\alpha(z)\ne z$ (think geometrically again), yielding $b=0$ so $\alpha={\rm Id}$.

If $c=\pm1$, then $|cz+d|\le1$ so the disk of radius $1$ around $\pm d$ intersects $\cal F$, so $|d|<2$. It is fairly straightforward to check that the only matrices in ${\rm SL}_2({\bf Z})$ that meet the conditions $c=\pm1,|d|\le1$ up to sign are given for some $k\in\bf Z$ by

$$\begin{pmatrix}k & -1 \\ 1 & 0\end{pmatrix},\quad \begin{pmatrix}k+1&k\\1&1\end{pmatrix},\quad \begin{pmatrix}1-k&k\\-1&1\end{pmatrix}.$$

Hopefully I didn't miss any. Given these, you need to check (using geometric arguments; these are all translations) that the only $k$ for which these things can possibly fix something in $\cal F$ are those that correspond to matrices in $\langle S\rangle$, $\langle ST\rangle$ or $\langle TS\rangle$, which requires writing out all these matrices - this involves two multiplications and two $2\times2$ inverses, which is pretty speedy. Do keep in mind I suppressed some signs when I wrote the above three types of matrices (e.g. those with $d=-1$).

In fact, the only points in $\cal F$ with nontrivial stabilizers are

• ${\rm Stab}(i)=\langle S\rangle$,
• ${\rm Stab}(e^{2\pi i/3})=\langle ST\rangle$,
• ${\rm Stab}(w^{i\pi /3})=\langle TS\rangle$.