Fundamental domain of the modular group

Example

Let the modular group $\Gamma = \mathrm{SL}_2(\Z)$ act on the upper half-plane $\mathbb H$ by Möbius transformations, generated by $T \tau = \tau + 1$ and $S \tau = -1/\tau$. Let $$\mathcal F = \set{\tau \in \mathbb H : \abs{\operatorname{Re} \tau} \le \tfrac12, \ \abs{\tau} \ge 1}.$$ Then $\mathcal F$ is a fundamental domain for $\Gamma$:

1. every orbit of $\Gamma$ in $\mathbb H$ meets $\mathcal F$;
2. two distinct points of $\mathcal F$ lie in the same orbit only if they lie on the boundary of $\mathcal F$.

Check of 1: for $\gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, one has $\operatorname{Im}(\gamma \tau) = \operatorname{Im} \tau / \abs{c\tau + d}^2$. Only finitely many pairs $(c, d)$ of integers have $\abs{c\tau + d} \le 1$, so some $\gamma$ makes $\operatorname{Im}(\gamma \tau)$ maximal. Apply a power of $T$ to move $\gamma \tau$ into the strip $\abs{\operatorname{Re} \tau} \le \tfrac12$; this does not change the imaginary part. If the result $\tau'$ had $\abs{\tau'} < 1$, then $\operatorname{Im}(S \tau') = \operatorname{Im} \tau' / \abs{\tau'}^2 > \operatorname{Im} \tau'$, against maximality, so $\tau' \in \mathcal F$.

Check of 2: there are two boundary identifications. The map $T$ sends the line $\operatorname{Re} \tau = -\tfrac12$ to the line $\operatorname{Re} \tau = \tfrac12$. The map $S$ sends the arc $\abs{\tau} = 1$ to itself by $\tau \mapsto -\bar\tau$. A short computation with $\abs{c\tau + d} \ge 1$ on $\mathcal F$ shows there are no other identifications.