Let $G$ be a non-uniform lattice Fuchsian group and let $P$ be a Dirichlet region for $G$. In particular $G$ has parabolic elements, $P$ is not compact and has finite area. We are in the unit disc. Is the following statement true? Is the proof correct? Errors? Counterexamples? Thanks.


STATEMENT: If $G$ is a free group then $P$ is an ideal polygon, that is it has all its vertices on the boundary.


PROOF: let $v$ be a vertex of $P$ and assume that $v$ is an interior point of the disc. Look at the tassellation $g(P)$ with $g \in G$ around the vertex v. Let $P\_0, ... , P\_n$ be the elements of the tassellation sharing the vertex $v$, ordered counterclockwise, where $P\_0 = P$. Such elements are finitely many because the tassellation is locally finite. Moreover $n\geq 2$ because all angles are less than $\pi$. We have elements $g\_0 ,..., g\_n$ of $G$ with $g\_k ( P\_k ) = P\_{k+1}$ for $k = 0 , ... , n$, where $P\_{n+1} = P\_0$. This implies that the interior of $g\_n...g\_1g\_0 (P\_0)$ overlaps with the interior of $P\_0$. The last condition implies that $g\_n...g\_1g\_0 = e$ the identity of $G$, thus we have a non trivial relation (recall $n \geq 2$ ), which contradicts that $G$ is free.



Dirichlet region of a free group
The statement is false. Here is a counter-example. Let $T$ be an ideal triangle (say in the unit disk model). Let $S$ be the surface obtained by *doubling* $T$ across it’s boundary: that is, take two copies and glue by the identity on the boundary. Let $x$ be the centre of $T$. The Dirichlet domain based at $x$ has six vertices with three ideal and three material (these glue up to give the copy of $x$ in the other triangle).


I’ll add that this behaviour is generic; all but finitely many Dirichlet domains will have material vertices.


