Let $H$ be a simply-connected, complete space of constant negative curvature, that is, a hyperbolic space, $\Gamma$ a discrete group of isometries, and and $M=H/\Gamma$ its quotient space; we assume that $M$ has finite volume.

It is known, I think, that $M$ is "of finite geometric type", that is, there exists a fundamental domain $D\subset H$ which is a finite hyperbolic polyhedron.

There is a universal way of building such polyhedra: we take a point $x\in H$, and consider its $\Gamma$-orbit $\Gamma\cdot x$. Define the Dirichlet-Voronoi polyhedron $P$ of $\Gamma\cdot x$ as the set of all $y\in H$ such that the distance from $y$ to $x$ is less than the distance to any other point of $\Gamma\cdot x$.

It is clear that $P$ is a fundamental domain.

Is it true that $P$ is a finite polyhedron for all $x\in H$? I think I can find this statement in the literature for dimension 3. Somebody told me this can be false in dimension more than 3.

A proof (or disproof) of this statement or a reference to it would be highly appreciated.

  • $\begingroup$ Note that the Dirichlet-Voronoi polyhedron you describe (more commonly called the Dirichlet domain) is only a fundamental domain when $\mathrm{Stab}_{\Gamma}(x)=\{1\}$. This is not guaranteed for an arbitrary discrete group of isometries $\Gamma$ of $H$. If $M$ is a manifold though, as suggested by your title, $\Gamma$ will be torsion-free in which case the condition is guaranteed. $\endgroup$
    – j0equ1nn
    Sep 14, 2015 at 22:49

1 Answer 1


B.Bowditch, Geometrical finiteness for hyperbolic groups. J. Funct. Anal. 113 (1993), no. 2, 245–317. The statement that you need is a corollary of his Proposition 5.6 in conjunction with finiteness of the holonomy group of a compact flat manifold (one of the Bieberbach's theorems). The statement which is false for $H^n$, $n\ge 4$, is "geometric finiteness implies that every Dirichlet domain is finitely-sided". (It is a nice example to work out, starting with an infinite cyclic groups of skew motions in $R^3$; this observation is originally due to Margulis.)


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