# How to find the Fuschian group associated with a region of the complex plane

Any connected open set in the complex plane (except for the plane and punctured plane) is covered conformally by the unit disk. This in turn means that it is conformally covered by the hyperbolic plane (since the disk and hyperbolic plane are conformally equivalent).

For a given region $U$ in the complex plane that satisfies this criteria, let $q : H^2 \to U$ be the that covering map. He can then form a quotient of the hyperbolic plane as following

$$H_U = H^2/\{(a,b):p(a) = p(b)\}$$

That is, we identify points that are sent to same place by $p$. This will be a hyperbolic surface, and therefore will have a Fuchsian model. This Fuschian model will have an associated Fuchsian group $\Gamma$.

So my question is starting with a region $U$, how do you find the group $\Gamma$?

EDIT: If I understand the idea of Fuschsian groups correctly, my question boils down to finding a group of conformal maps that "doesn't affect" the hyperbolic plane/unit disk. Namely, $g \in \Gamma$ iff $q = g \circ q$.

• What kind of answer do you want? In what form do you expect $\Gamma$ to be given to you? – Lee Mosher Jan 18 '18 at 19:46
• @LeeMosher uhm, maybe via a generating set. – PyRulez Jan 18 '18 at 19:49
• Even in the case of $U$ equal to the complement to a finite subset $A$, an "explicit" computation of $\Gamma$ in terms of $A$ is a notorious open problem, going back to Poincare (the problem of "accessory parameters"). – Moishe Kohan Jan 18 '18 at 22:16
• @MoisheCohen That would be a valid answer. – PyRulez Jan 18 '18 at 22:20

Even in the case of $U$ equal to the complement to a finite subset $A$, an "explicit" computation of $\Gamma$ in terms of $A$ is a notorious open problem, going back to Poincare (the problem of "accessory parameters"). See for instance this paper for some background and references: