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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.

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  • $\begingroup$ 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. $\endgroup$
    – HJRW
    Commented Apr 3, 2021 at 12:37
  • $\begingroup$ 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? $\endgroup$
    – user178149
    Commented Apr 3, 2021 at 12:53
  • $\begingroup$ 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. $\endgroup$
    – HJRW
    Commented Apr 3, 2021 at 13:02
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    $\begingroup$ 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? $\endgroup$
    – HJRW
    Commented Apr 3, 2021 at 13:03
  • $\begingroup$ Yes, I don't know how to move the question to another forum, but I have no problems if moderators do. $\endgroup$
    – user178149
    Commented Apr 3, 2021 at 13:07

1 Answer 1

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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.

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  • $\begingroup$ Thanks! But what's the Fuchsian group underlying your example? Isn't it the fundamental group of the 3-punctures sphere? In this case it is not free. Wrong? $\endgroup$
    – user178149
    Commented Apr 3, 2021 at 15:17
  • $\begingroup$ Yes, that’s wrong. The fundamental group of the three-punctured sphere is free of rank two. $\endgroup$
    – HJRW
    Commented Apr 3, 2021 at 15:42

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