## Notes of Pierre-Emmanuel Caprace’s fourth Cambridge lecture 27-04-2017

Exotic lattices and simple locally compact groups, IV

1. Subgroup separability and residual finiteness

Theorem 1 (Kropholler-Reid-Wesolek-Caprace) Let ${G}$ be a finitely generated group, let ${H be a commensurated subgroup (every conjugate commensurates ${H}$). Let ${\tilde H}$ be the profinite closure of ${H}$, i.e. the intersection of all finite index subgroups of ${G}$ containing ${H}$. Then the intersection of all conjugates of ${\tilde H}$ has finite index in ${\tilde H}$.

1.1. Proof

${\tilde H}$ is commensurated as well. Therefore one can assume that ${H}$ is closed. For ${X\subset G}$, let ${H_X}$ be the intersection of conjugates of ${H}$ by elements of ${X}$. We must show that ${H_G}$ has finite index in ${H}$.

Since ${H}$ is commensurated, for every finite set ${X}$ containing 1, ${H_X}$ has finite index in ${H}$. Since ${H}$ is closed, each ${H_X}$ is separable, hence ${H_X=W\cap H}$ where ${W}$ is a finite index in ${G}$. Let ${V}$ be a finite index normal subgroup of ${G}$ contained in ${W}$. Then ${H_X\subset VH_X\cap H\subset W\cap H=H_X}$, hence ${VH_X\cap H=H_X}$.

We apply this to a fixed generating set containing 1, and get a normal subgroup ${V}$. Let ${\mathcal{Y}}$ be the set of finite subsets containing 1 such that ${VH_Y\cap H=H_Y}$. Then ${\mathcal{Y}}$ is stable under finite unions. Similarly, if ${Y\in\mathcal{Y}}$ and ${\{1,t\}\in\mathcal{Y}}$, then ${Y\cup tY\in \mathcal{Y}}$. These two properties imply that all balls ${X^n}$ belong to ${\mathcal{Y}}$. Therefore ${H\cap V\subset VH_{X^n}\cap H=H_{X^n}}$, hence ${H\cap V\subset H_G\subset H}$, ${H_G}$ contains a finite index subgroup of ${H}$.

Our inital proof with Monod used more structure of locally compact groups.

1.2. Application to lattices in products of trees

Proposition 2 Let ${T_i}$ be two leafless trees, let ${\Gamma}$ act discretely cocompactly on their product. Then the following are equivalent.

1. ${\Gamma}$ is reducible.
2. The projection of ${\Gamma}$ on a factor is discrete.
3. For every vertex ${v}$ of one of the trees, the stabilizer ${\Gamma_v}$ is separable in ${\Gamma}$.
4. There exists a vertex ${v}$ of one of the trees whose stabilizer ${\Gamma_v}$ is separable in ${\Gamma}$.

The implication (2)${\Rightarrow}$(1) uses Burger-Mozes’ result that for all normal subgroups ${N}$ of ${G}$, ${\Gamma N/N}$ is a lattice in ${G/N}$ iff ${\Gamma\cap N}$ is a lattice in ${N}$.

The implication (4)${\Rightarrow}$(2) uses Theorem 1.

Corollary 3 Let ${G}$ be a finitely generated group such that every infinite normal subgroup of ${G}$ has trivial centralizer. If ${G}$ is residually finite, then every commensurated subgroup ${H}$ of ${G}$ has trivial quasi-centralizer (i.e. no element commutes with a finite index subgroup of ${H}$).

This gives a fast proof that Baumslag-Solitar group ${BS(m,n)}$ is residually finite iff ${|m|=1}$ or ${|n|=1}$ or ${|m|=|n|}$ (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 ${a}$ is commensurated and quasicentralizes itself, thus the ambient group cannot be residually finite.

Corollary 4 Let ${G}$ be a finitely generated group such that every infinite normal subgroup of ${G}$ has trivial centralizer. Let ${H be infinite and commensurated. If ${G}$ is residually finite, then every normal subgroup of ${G}$ cuts an infinite subgroup of ${H}$.

This gives a proof of the following result of Burger and Mozes.

Proposition 5 (Burger-Mozes) Let ${\Gamma}$ be discrete and cocompact on product of two trees. If ${\Gamma}$ is irreducible and residually finite, then the projection to each factor is injective.

Recall the group ${\Gamma_2=\langle a,b,x,y\;|\;\ldots\rangle}$ I introduced in the first lecture. Then ${x^3}$ and ${y^3}$ both commute with an index 2 subgroup of ${\langle a,b \rangle}$. Therefore they act trivially on the vertical tree.

Next time, I will explain the proof that ${\Gamma_2}$ is irreducible. We will then be able to conclude that ${\Gamma_2}$ is not residually finite.