Fundamental domain for congruence subgroup.

Question

Let $\Gamma$ be a congruence subgroup of the modular group $\mathrm{SL}_{2}(\mathbb{Z})$. Let $R$ be coset representatives of the quotient $\Gamma\setminus\mathrm{SL}_{2}(\mathbb{Z})$ and let $\mathcal{D}$ be the standard fundamental domain for $\mathrm{SL}_{2}(\mathbb{Z})$ (actually we could consider any fundamental domain for $\mathrm{SL}_{2}(\mathbb{Z})$). I want to prove that

$$\mathcal{D}_{\Gamma}=\bigcup_{\gamma\in R}\gamma\mathcal{D}$$

is a fundamental domain for the congruence subgroup $\Gamma$, i.e. that for every $z$ in the upper complex half-plane $\mathbb{H}$ there exists $\gamma\in\Gamma$ such that $\gamma z\in\mathcal{D}_{\Gamma}$ and also that if $z,\gamma z\in\mathcal{D}_{\Gamma}^{\circ}$ then this $\gamma$ is unique up to multiplication by an element of $\Gamma\cap\{\pm I\}$, where $I$ is the identity matrix.

It is simple to prove the existence of $\gamma\in\Gamma$. I have a problem showing the "uniqueness". Here is my proof:

Consider $z\in\mathbb{H}$ and $\gamma_{1},\gamma_{2}\in\Gamma$ such that $\gamma_{1}z,\gamma_{2}z\in\mathcal{D}_{\Gamma}^{\circ}$. This implies that there exist $\gamma_{1}',\gamma_{2}'\in R$ and $z_{1},z_{2}\in\mathcal{D}$ such that $\gamma_{1}z=\gamma_{1}'z_{1}$ and $\gamma_{2}z=\gamma_{2}'z_{2}$. Therefore, \begin{equation*} \gamma_{1}^{-1}\gamma_{1}'z_{1}=z=\gamma_{2}^{-1}\gamma_{2}'z_{2}. \end{equation*} Set $w_{1}=\gamma_{1}'z_{1}\in\mathcal{D}_{\Gamma}^{\circ}$, $w_{2}=\gamma_{2}'z_{2}\in\mathcal{D}_{\Gamma}^{\circ}$, and $\gamma=\gamma_{1}\gamma_{2}^{-1}$. Then, we have that $w_{1}=\gamma w_{2}$. We claim that $\gamma=I$. Indeed, since $w_{1}\in\mathcal{D}_{\Gamma}^{\circ}$, there exists $\epsilon>0$ such that $B_{\epsilon}(z)\subseteq\mathcal{D}_{\Gamma}^{\circ}$. $B_{\epsilon}(z)$ intersects some translations of $\mathcal{D}$ in $\mathcal{D}_{\Gamma}$. Let $R'$ be the set of representatives of such translations, i.e. $S\subseteq R$ and \begin{equation*} \delta\in S\iff B_{\epsilon}(z)\cap \delta \mathcal{D}\neq\emptyset. \end{equation*} Now we have that $\gamma z\in\mathcal{D}_{\Gamma}^{\circ}$ and so there exists $\delta\in R$ such that $\gamma B_{\epsilon}(z)\cap \delta \mathcal{D}\neq \emptyset$. This implies that \begin{equation*} B_{\epsilon}(z)\cap\gamma^{-1}\delta\mathcal{D}\neq\emptyset \end{equation*} and so $\gamma^{-1}\delta=\delta'$ for some $\delta'\in R'$. Hence, \begin{equation*} \Gamma \delta'=\Gamma\gamma^{-1}\delta=\Gamma\delta \end{equation*} and so $\delta=\delta'$, which implies that $\gamma=I$.

My problems is the following:

1. First, the above proof works if $\gamma B_{\epsilon}(z)=B_{\epsilon}(\gamma z)$, which I suspect holds but I cannot prove, or at least $B_{\epsilon}(\gamma z)\subseteq \gamma B_{\epsilon}(z)$.
2. The second is that I miss the multiplication by an element in $\Gamma\cap\{\pm 1\}$. Where is the error?

Answer 1

I know this is an old question, but it is sad to see a decent question go unanswered. In your proof, you at some point swap between $w_1$ and $z$ for the same variable, and some confusion with $R'$ and $S$, but other than that it seems fine. I'll answer your two specific questions.

1. $\gamma B_\epsilon(z) = B_\epsilon(\gamma z)$ holds as the fractional linear transformations with which $\Gamma$ acts on $\Bbb{H}$ are isometries; they preserve the hyperbolic distance. This can be checked directly, and the consequence above follows from writing out the definitions.
2. Note that $I$ and $-I$ both act as the identity on $\Bbb{H}$. Thus, the actions of $g$ and $-g$ are the same, for $g\in SL_2(\Bbb{R})$. Concluding that $\gamma\delta=\delta'$ for some $\delta'\in S$ is not correct, since you may have $\gamma\delta \mathcal{D}=\delta'\mathcal{D}$ without $\gamma\delta=\delta'$; it could be $-\delta'$.