Let $G$ be a finitely generated Fuchsian group, and let $\mathcal{F}$ denote the Dirichlet fundamental domain of $G$ with respect to $0$ in the Poincaré disc model. Assume throughout that $\mathcal{F}$ is non-compact.

I am interested in properties of $G$, and how these properties are connected with Poincaré's theorem. Here are my questions:

  1. Is $G$ the free product of elementary hyperbolic, parabolic and elliptic subgroups? It is true for the modular group PSL(2,Z). Moreover, it holds under the stronger assumptions in 2).

  2. Assume that $G$ has no elliptic elements. Then $G$ is free (because $G$ is fundamental group of a non-compact surface) and thus, $\mathcal{F}$ has vertices only at the boundary of hyperbolic space. Can we derive this property of the vertices from Poincaré's theorem?

  3. Assume that $G$ is of the second kind (that is, the limit set of $G$ is not equal to the boundary of hyperbolic space). Then in particular $\mathcal{F}$ is non-compact. Is it true that the sides of $\mathcal{F}$ are pairwise disjoint, except (possibly) the sides paired by elliptic elements?

UPDATE: Sam Nead answered 3) in the negative. Is the answer in 3) also negative if we allow an arbitrary reference point for the Dirichlet fundamental domain? Or does there always exist a suitable reference point for a Dirichlet fundamental domain such that the sides are pairwise disjoint, except (possibly) the sides paired by elliptic elements?

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    $\begingroup$ SL(2,Z) is not a free product, just PSL(2,Z) is the free product of two elliptic subgroups. $\endgroup$
    – ThiKu
    Commented Aug 14, 2021 at 7:22

1 Answer 1


For (1) the answer is "yes". Since the surface has finitely generated fundamental group, there is a finite collection of disjoint embedded bi-infinite geodesics that cut the surface into a collection of ideal (or hyperideal) polygons, each with at most one cone point in its interior. (There is a special case when the cone point has angle $\pi$, which I leave as an exercise.) The statement now follows from the Seifert-van Kampen theorem.

For (2) I read your question as "can a Dirichlet domain for a surface (as above) have material vertices?" and the answer is "yes, it may". To see this, consider the surface made by doubling an ideal triangle across its boundary. If we take the origin to be the centre of one of the triangles, then the Dirichlet domain has three ideal vertices and three material vertices.

For (3) the answer is "not in general". To see this, consider the surface of (2) where we "open" the cusps, replacing them by (identical) funnels. Again the surface has a three-fold symmetry about the centres of the (now hyperideal) triangles. The Dirichlet domain has three material vertices, six ideal vertices, six material edges, and three ideal edges.

The property you want - disjoint material edges glued in pairs - feels very similar to "Schottky, with all circles perpendicular to a single circle". Even this only gets you some Dirichlet domain with the desired form, not all. I am not sure if this property has a name, but I would look at Marden's book, or Maskit's, as a first reference.

For (3) "updated" the answer is "not in general". That is, there is a hyperbolic surface $S$ with infinite volume, with finitely generated (and free) fundamental group, so that all Dirichlet domains have material vertices.

Consider the surface $T$ given by (3) (first version) above. Let $x$ in $T$ be one of the points with three-fold symmetry. Let $D \subset T$ be a very small round disk about $x$. Let $S$ be the surface obtained by doubling $T - D$ across $\partial D$. Uniformise $S$. Let $\gamma$ be the image of $\partial D$ in $S$. Since $\gamma$ is fixed by a reflection in $R$, it is also a hyperbolic geodesic. Since the radius of $D$ was very small, the geodesic $\gamma$ is very short.

I claim that $S$ has the desired property - that is, all Dirichlet domains have material vertices. The proof is harder than that for (3).

  • $\begingroup$ Could you please clarify "surface made by doubling an ideal triangle across its boundary"? $\endgroup$
    – JackTodd
    Commented Aug 14, 2021 at 14:41
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    $\begingroup$ Let $T$ be an ideal triangle. Define $T_i = T \times \{i\}$ for $i = 0, 1$. So each of $T_0$ and $T_1$ are also ideal triangles. We define a new topological space $S$ by gluing $(x, 0) \in T_0$ to $(x, 1) \in T_1$ if and only if $x$ is a boundary point of $T$. It is now an exercise to define the natural hyperbolic structure on $S$. (In fact, there is a subtle point here about the induced orientation on $S$. But I think it is fair to simplify the exposition for the first presentation of the idea.) $\endgroup$
    – Sam Nead
    Commented Aug 14, 2021 at 16:24
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    $\begingroup$ By the way, it is polite to upvote an answer if you find it helpful. And, if it answers your question, then you should "accept" it (by ticking the tick mark). $\endgroup$
    – Sam Nead
    Commented Aug 14, 2021 at 16:28
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    $\begingroup$ I assume you are asking about the double of an ideal triangle. Let $p_0$ and $p_1$ be the centres of the two triangles $T_0$ and $T_1$. Suppose that we place the basepoint of our Dirichlet domain at $p_0$. Let $L_j$ be the three geodesic rays in $T_1$ that emanate from $p_1$ and "end" at the three ideal points of $T_1$. We cut the surface along $\cup L_j$ to obtain the Dirichlet domain. It helps to draw a picture! $\endgroup$
    – Sam Nead
    Commented Aug 15, 2021 at 2:10
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    $\begingroup$ There are no relations in that group. And yes, a fundamental domain is given by a pair of triangles. But that fundamental domain is not the Dirichlet domain based at the center of one of the triangles... $\endgroup$
    – Sam Nead
    Commented Aug 15, 2021 at 8:49

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