Exotic lattices and simple locally compact groups, IV
1. Subgroup separability and residual finiteness
Theorem 1 (Kropholler-Reid-Wesolek-Caprace) Let be a finitely generated group, let be a commensurated subgroup (every conjugate commensurates ). Let be the profinite closure of , i.e. the intersection of all finite index subgroups of containing . Then the intersection of all conjugates of has finite index in .
is commensurated as well. Therefore one can assume that is closed. For , let be the intersection of conjugates of by elements of . We must show that has finite index in .
Since is commensurated, for every finite set containing 1, has finite index in . Since is closed, each is separable, hence where is a finite index in . Let be a finite index normal subgroup of contained in . Then , hence .
We apply this to a fixed generating set containing 1, and get a normal subgroup . Let be the set of finite subsets containing 1 such that . Then is stable under finite unions. Similarly, if and , then . These two properties imply that all balls belong to . Therefore , hence , contains a finite index subgroup of .
Our inital proof with Monod used more structure of locally compact groups.
1.2. Application to lattices in products of trees
Proposition 2 Let be two leafless trees, let act discretely cocompactly on their product. Then the following are equivalent.
- is reducible.
- The projection of on a factor is discrete.
- For every vertex of one of the trees, the stabilizer is separable in .
- There exists a vertex of one of the trees whose stabilizer is separable in .
The implication (2)(1) uses Burger-Mozes’ result that for all normal subgroups of , is a lattice in iff is a lattice in .
The implication (4)(2) uses Theorem 1.
Corollary 3 Let be a finitely generated group such that every infinite normal subgroup of has trivial centralizer. If is residually finite, then every commensurated subgroup of has trivial quasi-centralizer (i.e. no element commutes with a finite index subgroup of ).
This gives a fast proof that Baumslag-Solitar group is residually finite iff or or (Meskin). Indeed, this group acts faithfully on its Bass-Serre tree with no fixed points at infinity. Therefore normal subgroups have trivial centralizers. The cyclic group generated by is commensurated and quasicentralizes itself, thus the ambient group cannot be residually finite.
Corollary 4 Let be a finitely generated group such that every infinite normal subgroup of has trivial centralizer. Let be infinite and commensurated. If is residually finite, then every normal subgroup of cuts an infinite subgroup of .
This gives a proof of the following result of Burger and Mozes.
Proposition 5 (Burger-Mozes) Let be discrete and cocompact on product of two trees. If is irreducible and residually finite, then the projection to each factor is injective.
Recall the group I introduced in the first lecture. Then and both commute with an index 2 subgroup of . Therefore they act trivially on the vertical tree.
Next time, I will explain the proof that is irreducible. We will then be able to conclude that is not residually finite.