Action rigidity for free products of hyperbolic manifold groups

Abstract

Two groups have a common model geometry if they act properly and cocompactly by isometries on the same proper geodesic metric space. The Milnor-Schwarz lemma implies that groups with a common model geometry are quasi-isometric; however, the converse is false in general. We consider free products of uniform lattices in isometry groups of rank-1 symmetric spaces and prove, within each quasi-isometry class, residually finite groups that have a common model geometry are abstractly commensurable. Our result gives the first examples of hyperbolic groups that are quasi-isometric but do not virtually have a common model geometry. Indeed, each quasi-isometry class contains infinitely many abstract commensurability classes. We prove that two free products of closed hyperbolic surface groups have a common model geometry if and only if the groups are isomorphic. This result combined with a commensurability classification of Whyte yields the first examples of torsion-free abstractly commensurable hyperbolic groups that do not have a common model geometry. An important component of the proof is a generalization of Leighton's graph covering theorem. The main theorem depends on residual finiteness, and we show that finite extensions of uniform lattices in rank-1 symmetric spaces that are not residually finite would give counterexamples.

arXiv commentv2: 30 pages, 3 figures. Explained how the main theorem depends on residual finiteness and added Section 7 to show if there exists a nonresidually finite finite extension of a uniform lattice in the isometry group of a rank-1 symmetric space, then there are free products of hyperbolic manifold groups that are not action rigid. Other minor corrections. To appear in Annales de l'Institut Fourier