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.



Dirichlet polyhedra for hyperbolic manifolds
B.Bowditch, [Geometrical finiteness for hyperbolic groups](http://www.sciencedirect.com/science/article/pii/S0022123683710529#). 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.) 


