Show that the Ford domain may be seen as the Dirichlet domain around $z_0 \to \infty$ (Series 2010)

Question

I am working through Hyperbolic geometry by C. Series, ch 5.2. I came across the following which I do not understand.

Consider the upper half plane model of the hyprbolic plane, $\mathbb{H}$.

We consider a discrete group $G \subset \operatorname{PSL}(2, \mathbb{C})$ acting on $\mathbb{H}$, where we assume that $G$ has a subgroup $G_0 = <T>$ where $T$ is a parabolic transformation fixing $\infty$, i.e., $T = z + t$.

Then for $g = \begin{bmatrix} a & b \\ c &d\end{bmatrix} \in G - G_0$, we have that $c \neq 0$ we construct the isometric circle $$I_g = \{ z : |g'(z)| = 0 \} = \{z : |cz + d| = 1\} $$ and define $$ E_g = \{z : |cz + d| > 1\} $$ Then let $\Sigma$ be a strip of width $t$, i.e., $c < \Re(z) < c +t $, and we define the Ford domain as $$ F = \Sigma \cap \bigcap\limits_{g \in G - \operatorname{id}} E_g$$ This domain is a fundamental domain. We can also define the Dirichlet domain around a center $z_0 \in \mathbb{H}$, defined as $$ D(z_0) = \{ z : d(z,z_0) < d(z,gz_0) \ \forall g \in G \}.$$

Now on page 52, Ex 5.7 it is claimed that we may view the Ford domain as the Dirichlet domain under $z_0 \to \infty$.

She gives some hints: First, for some $g \neq \operatorname{id}$ let $C_R(z_0)$ be the unique circle such that $C_R(z_0)$ and $g(C_R(z_0))=C_R(gz_0)$ are tangent. Then, $L_g$, the perpendicular to the line from $z_0$ to $gz_0$ touches both circles in the point of tangency and is tangent to both circles. Now, as we let $z_0 \to \infty \in \mathbb{R}$ she claims that $L_g $ converges to a horocycle $H(d) = \{ z : \Im (z) = d \}$ for some $d > 0$.

I do not see why this is true, for instance in the case that $g = [z \to z + 1]$ we have that $L_g = \{ z : \Re(z) = \Re(z_0) + \frac12 \}$. I may have missed some implicit assumption about the absence of parabolic elements in this method. Even then, I cannot really see why the claim is true. Also, I do not see how to then continue the proof, even if we ignore for the moment parabolic translations.

Answer 1

I think you might have misunderstood some of the objects discussed in the passage that you are reading.

I think the object $L_g$ is a line, i.e. a bi-infinite geodesic in $\mathbb{H}$. More specifically, it is the unique line that is a perpendicular bisector of the segment from $z_0$ to $g z_0$. As $z_0$ varies (but keeping $g$ fixed), the line $L_g$ varies as a function of $z_0$. Let me write this function as $L_g(z_0)$.

Now, it makes no sense to say that $L_g(z_0)$ converges to a horocycle. If it converges to anything, it can only converge to another line (for example, one can ask whether the pair of ideal endpoints $\partial L_g(z_0) = \{\zeta(z_0),\zeta'(z_0)\} \subset \partial\mathbb{H}$ converges to a pair of distinct points $\{\zeta,\zeta'\} \subset \partial\mathbb{H}$, and if so then unique line with ideal endpoints $\{\zeta,\zeta'\}$ is the limit of the lines $L_g(z_0)$).

What does make sense, and is in fact true, is that the circle $C_R(z_0)$ approaches a horocycle as $z_0$ approaches $\infty$, as long as the approach happens along a vertical line in the upper half plane; to be precise, if we fix the real part $x_0$, and write $z_0 = x_0+ti$, and let $t \to +\infty$, then $C_R(z_0)$ approaches a horocycle as $t \to +\infty$. Furthermore, the pair of circles $C_R(g z_0)$ and $C_R(z_0)$ remain tangent as $t \to +\infty$ (by definition), they approach a pair of tangent horocycles, one based at $+\infty$ and the other based at $g \cdot +\infty$, and the unique line tangent to that pair of horocycles is the limit of the lines $L_g(z_0)$.