Finite index subgroup of Fuchsian group of the first type is also Fuchsian of the first type

Question

Just trying to see why the statement in the title is true. For $\Gamma$ a Fuchsian group of the first type on the complex upper half plane $\mathbb{H}$, a subgroup $\Gamma'$ of finite index is also Fuschian of the first type. A Fuchsian group on $\mathbb{H}$ is a discrete subgroup of PSL$_2(\mathbb{R})$. To say that $\Gamma$ is of the first type means that every $x\in\mathbb{R}\cup\{\infty\}$ is the limit of an orbit $\Gamma z$ for some $z\in\mathbb{H}$ (using the topology of the Riemann sphere $\mathbb{C}\cup\{\infty\}$). So for every $x\in\mathbb{R}\cup\{\infty\}$ we can take a sequence $\gamma_i\in\Gamma,i\in\mathbb{N}$ and some $z\in\mathbb{H}$ so that $\lim_{i\rightarrow\infty}\gamma_iz=x.$ In Topics In Classical Automorphic Forms it is said that $\Gamma'$ is also Fuschian of the first type. I can't see an obvious subsequence of $\gamma_i$ contained in $\Gamma'$ that would work, but perhaps there is one, or I'm looking at it from the wrong angle.

Answer 1

Here's a proof. I'm going to represent the hyperbolic plane $\mathbb H$ using the Poincaré disc model $\mathbb D$ instead of the upper half plane model. So with $\mathbb H = \mathbb D$ we also have $\partial\mathbb H = S^1$. Convergence in $\mathbb H \cup \partial \mathbb H = \mathbb D \cup S^1 = \overline{\mathbb D}$ is therefore governed by the Euclidean metric $d_{\mathbb E}$, so it is easier to talk about convergence to boundary points using $\mathbb D$ than it is using the upper half plane model.

Now I'll state a geometric fact that one can use to resolve this problem.

Lemma: For any $z \in \mathbb H$ there exists $r > 0$ such that for any $\gamma \in \Gamma$ there exists $\delta \in \Gamma'$ such that $d_{\mathbb H}(\gamma \cdot z,\delta \cdot z) < r$.

In this statement, the distance function $d_{\mathbb H}$ refers to distance in $\mathbb H$. In words, this lemma says that the entire $\Gamma$-orbit of $z$ is contained in the $r$ neighborhood of the $\Gamma'$ orbit of $z$. This lemma is true for any group acting by isometries on any metric space; I'll leave the proof as an exercise for you to ponder (hint: first pick right coset representatives).

To see how this proves what you want, consider a point $x \in \partial\mathbb H = S^1$. Using that $\Gamma$ is of the first type, choose $z \in \mathbb H$ and a sequence $\gamma_i \in \Gamma$ such that $\lim_{i \to \infty} \gamma_i \cdot x = z$. Applying the lemma, choose $r > 0$ and a sequence $\delta_i \in \Gamma'$ such that $d(\gamma_i \cdot z, \delta_i \cdot z) < r$.

We want to prove that $\lim_{i \to \infty} \delta_i \cdot z = x$. To do this, we use a fact about the Poincaré disc model $\mathbb D$: as the center of a hyperbolic ball of radius $r$ approaches $S^1$, the Euclidean diameter of that ball approaches zero. So, knowing that $\lim_{i \to \infty} \gamma_i \cdot z = x \in S^1$, and that $d_{\mathbb H}(\gamma_i \cdot z,\delta_i \cdot z) < r$, it follows that $\lim_{i \to \infty} d_{\mathbb E}(\gamma_i \cdot z,\delta_i \cdot z ) = 0$. Therefore, $$\lim_{i \to \infty} \delta_i \cdot z = \lim_{i \to \infty} \gamma_i \cdot z = x $$