## Notes of Pierre-Emmanuel Caprace’s fifth Cambridge lecture 04-05-2017

Exotic lattices and simple locally compact groups, V

Today, I conclude the discussion and explain how to get simple cocompact lattices in products of trees.

Before, I need to complete the issue of irreducibility.

1. Recap

We consider leafless trees with at least three ends. Let ${\Gamma}$ be a cocompact lattice in the product. What we have proven.

The normal subgroup theorem. A sufficient condition in order that ${\Gamma}$ be hereditarily just infinite: the closure of the projections to both factors are simple by compact.

A sufficient condition for ${\Gamma}$ to be nonresidually finiteness: the projection of ${\Gamma}$ to one factor is noninjective and ${\Gamma}$ is irreducible. It also follows that ${\Gamma}$ is not just infinite (because of the kernel of the projection), so we will have to deal with incompatible criteria.

Sufficient conditions for ${\Gamma}$ to be irreducible: the projection to one factor is non-discrete or the stabilizer of some vertex of one factor is inseparable.

The second criterion is due to Wise, the first to Mozes.

2. Irreducibility

2.1. Discreteness criteria

Definition 1 Let ${X}$ be a graph and ${G}$ a group of automorphisms of ${X}$. The local action of ${G}$ at a vertex ${v}$ is the permutation group ${G_v/G_v[1]}$ acting on edges emanating from ${v}$.

Theorem 2 (Trofimov-Weiss 1995) Let ${X}$ be a connected, locally finite graph. Let ${G be vertex-transitive, whose local action at every vertex is 2-transitive. If ${G}$ is discrete, then the fixator of every ball of radius 5 is trivial.

The figure 5 arises from the classification of 2-transitive permutation groups, which in turn follows from the classification of finite simple groups. The fact that 5-ball fixators are trivial can be checked by examination of one 6-ball.

A longstanding conjecture is wether 2-transitive can be weakened to primitive.

With a stronger assumption, one gets a simpler result:

Theorem 3 (Burger-Mozes 2000) Let ${T}$ be a ${d}$-regular tree, ${d\geq 6}$, and ${G be vertex transitive, and the local action is alternating group ${\mathfrak{A}_d}$. Then

1. Either

$\displaystyle \begin{array}{rcl} |G_v/G_v^{[2]}|\leq \frac{d!}{2}\frac{(d-1)!}{2}, \end{array}$

and then ${G}$ is discrete.

2. Or ${G_v/G_v^{[2]}}$ is a wreath product ${\mathfrak{A}_{d-1}\wr\mathfrak{A_d}}$, and then ${G}$ is non-discrete, its closure has a simple subgroup of index 2.

Up to conjugation in ${Aut(T)}$, there is a unique closed non-discrete vertex transitive subgroup whose local action at every vertex is alternating, denoted by ${U(\mathcal{A}_d)}$.

The list of 2-transitive groups is ${\mathfrak{S}_d}$, ${\mathfrak{A}_d}$ and a few simple groups of Lie type and of exceptional type. So the alternating case is the most frequent. The second most frequent is ${\mathfrak{S}_d}$.

Theorem 4 (Radu 2016) A similar statement holds for local action the full symmetric group ${\mathfrak{S}_d}$. The differences are

2. in the non-discrete case, ${\bar G}$ has a simple subgroup of index 8.
3. infinitely many conjugacy classes.

2.2. Computing the local actions on factors

This is easy, given a BMW action. For ${\Gamma_2}$ generated by ${a,b,x,y}$, in the local action on the first factor (edges labelled ${x,y}$), ${a}$ is mapped to the product of transpositions ${(x,y^{-1})(x^{-1},y)}$ and ${b}$ to the 4-cycle ${(x,y,x^{-1},y^{-1})}$. The group they generate is ${D_4}$. Similarly, the local action on the second factor is ${\mathfrak{A}_4}$. For large enough ${n}$, ${(xy)^n}$ fixes the 5-ball but not the 6-ball.

From Trofimov-Weiss’ theorem, we conclude that the projection to the second factor is non-discrete, hence ${\Gamma_2}$ is irreducible.

3. Piling up intermediate results

3.1. Step 1

Pick a BMW complex ${Y_0}$ with non-residually finite fundamental group. E.g. of order ${(2,2)}$ with fundamental group ${\Gamma_2}$.

3.2. Step 2

Build a BMW complex ${Y}$ which contains isometrically ${Y_0}$, such that the local actions of ${\pi_1(Y)}$ on both factors contain alternating groups.

For this, merely add generators horizontally and vertically. Generically, local actions become full.

3.3. Step 3

Since ${\pi_1(Y)}$ contains ${\pi_1(Y_0)}$, it is non-residually finite. For the same reason, it is irreducible. By Burger-Mozes or Radu, plus the normal subgroup theorem, ${\pi_1(Y)}$ is hereditarily just infinite. Thus it is virtually simple. Indeed, the intersection of all finite index subgroups of ${\pi_1(Y)}$ is normal, therefore it has finite index, and no normal subgroups.

An explicit example, of order ${(4,3)}$, with local actions ${\mathfrak{A}_8}$ and ${\mathfrak{S}_6}$, has been found by Radu, assisted by computer. The methods produces tons of examples.

The resulting group is virtually simple. In fact, we know that the simple finite index subgroup is the normal closure of an explicit element. However, we do not know what the index is.