2019/10/21 by Emily Stark, Stark, Emily, Daniel J. Woodhouse +1
Mathematics · #20E06 #20F65 #20F67 #57M07 #57M10 #FOS: Mathematics #Geometric Topology (math.GT) #Geometric and Algebraic Topology #Group Theory (math.GR)
paper · pdf · doi:10.48550/arxiv.1910.09609
openalex publication_date 2019/10/21 · openalex created_date 2025/10/10 · openalex updated_date 2026/07/28
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.