# Dirichlet region of a free group

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.

• There are certainly some problems with your proof. For instance, the existence of an equation in a free group isn’t a contradiction; it just means that the expression you obtained wasn’t reduced.
– HJRW
Commented Apr 3, 2021 at 12:37
• Thanks a lot! I still don't understand why the expression $g_n...g_1g_0$ isn't reduced. For me, since the angles of $P$ are all less than $\pi$, I cannot have $g_kg_{k+1}=e$, the identity of $G$. Where am I wrong? Commented Apr 3, 2021 at 12:53
• A priori your $g_i$ aren't part of a basis for your free group, but just words in the generators, so although they may not cancel independently, they may still satisfy a non-trivial relation. (E.g. you could have $g_1=a$, $g_2=b$, $g_3=b^{-1}a^{-1}$.) I think your question is very close to showing that the side-pairing isometries form a free basis for the fundamental group.
– HJRW
Commented Apr 3, 2021 at 13:02
• In any case, my feeling is that this question is not quite at the right level for Mathoverflow. Can I suggest that you try math.stackexchange first, and then come back here if you don't get a good answer?
– HJRW
Commented Apr 3, 2021 at 13:03
• Yes, I don't know how to move the question to another forum, but I have no problems if moderators do. Commented Apr 3, 2021 at 13:07

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).