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Is center of a fundamental group of finite volume-hyperbolic n-orbifold trivial? Is there a good reference that the proof is wriiten?

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The center is abelian. Any infinite abelian dicrete group of isometries of the hyperbolic space either fixes a point at infinity, or stabilizes a pair of points at infinity. Since the center is a normal subgroup in a lattice, the whole lattice must stabilize a point or a pair of points at infinity, which in turn would imply that the lattice is virtually abelian, which is nonsense. Thus the center cannot be infinite. Much more generally, the center of an (non-elementary and not parabolic) relatively hyperbolic group is finite. –  Igor Belegradek Jan 12 '12 at 17:35
@Igor, why don't you write it as an answer? –  Anton Petrunin Jan 12 '12 at 18:54

3 Answers 3

Any lattice in a hyperbolic space acts as a convergence group on the sphere at infinity. Thus it suffices to prove the following:

Lemma If $G$ has a minimal convergence group action on a set $S$ of cardinality $>2$ (e.g. a sphere), then any normal abelian subgroup $A$ of $G$ is finite.

Proof. If $A$ is infinite, then $A$ fixes one or two points of $S$, and this is the whole fixed point set of $A$. Now any group that contains $A$ as a normal subgroup must stabilize the fixed point set of $A$, but $G$ has a dense orbit in $S$, a contradiction.

Basic facts about convergence groups that were used above can be found e.g. in "Convergence groups and configuration spaces" by B.H. Bowditch available here.

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Let $G$ be the fundamental group of your orbifold, and recall that $G$ is a lattice in the group of isometries of hyperbolic space. An element in the center of $G$ has to fix every fixed point of any loxodromic and any parabolic element of $G$. The fixed points of such elements are dense in the boundary at infinity of hyperbolic space (this is a consequence of the fact that the orbifold has finite volume, I think that a proof of this fact can be found in Ratcliffe's book, for example).

This implies that any element in the center of $G$ acts as the identity on the boundary of hyperbolic space, so the center of $G$ is trivial.

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The center of your group $G$ has to be finite (since it is finitely generated, abelian, and has no element of infinite order -- if it did, the group would fail to be word-hyperbolic, since that element together with some other element of $G$ of infinite order would generate a $\mathbb{Z} \oplus \mathbb{Z}$). But now, Lemma 3.3 of states that no discrete group of motions of an Hadamard manifold ($\mathbb{H}^n$ certainly qualifies) has a finite normal subgroup.

EDIT If the orbifold has cusps, and the center has an element of infinite order $\gamma,$ then all the elements of $G$ fix the fixed point (which has to be on the sphere at infinity) of $\gamma$ (since they all commute with $\gamma$) which contradicts the finite volume assumption. In fact, if $\gamma$ has finite order, it has a fixed point in $\mathbb{H}^n,$ that fixed point has to be fixed by every element of $G,$ and so $G$ is a subgroup of $SO(n)$ (and the quotient is, again, not finite volume) -- the argument before the edit is unnecessary, but is left there for historical reason.

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@Igor, the orbifold is only assumed to have finite volume but it need not be compact. so its fundamental group is not word hyperbolic. –  Vitali Kapovitch Jan 12 '12 at 17:11
Oops. Careless reading :( –  Igor Rivin Jan 12 '12 at 17:33

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