On the complexity of the quasi-isometry and virtual isomorphism problems for finitely generated groups

Abstract

Abstract. We study the Borel complexity of the quasi-isometry and virtual isomorphism problems for the class of finitely generated groups.

On the concept of “largeness” in group theory

“…Combining Theorem 1.4 with the earlier results of Thomas [18,19], it follows that ∼ =<B ≈ VI < B ≈ p . In particular, the Borel complexity of ≈ VI is strictly less than that of Pride's equivalence relation ≈ p .…”

“…For example, the isomorphism relation, the virtual isomorphism relation and the quasi-isometry relation are all K σ equivalence relations on G. (See Thomas [19].) By Kechris [8] and Louveau and Rosendal [10], there also exists a universal K σ equivalence relation.…”

“…If X is a Polish space, then a Borel equivalence relation on X is an equivalence relation E ⊆ X × X which is a Borel subset of X × X . For example, the isomorphism relation, the virtual isomorphism relation and the quasi-isometry relation are all Borel equivalence relations on G. (See Thomas [19].) If E, F 1 If C is any countably infinite set, then the Cantor space 2 C = { f | f : C → {0, 1}} with the natural product topology is a compact space.…”

“…By Kechris [8] and Louveau and Rosendal [10], there also exists a universal K σ equivalence relation. In fact, Rosendal [17] has recently shown that the relation of Lipschitz equivalence between compact metric spaces is a universal K σ equivalence relation; and Thomas [19] has conjectured that the quasi-isometry relation on G is also universal K σ . In Section 5, we will prove the following result, which provides the first purely grouptheoretic example of a complete K σ equivalence relation.…”

The bi-embeddability relation for finitely generated groups

“…In Thomas [12], it was conjectured that the quasi-isometry relation ≈ QI on the space of finitely generated groups is a universal K σ equivalence relation. Of course, if this is true, then there does not exist a Borel reduction from ≈ QI to ∼ = and so Conjecture 10.3 holds.…”