## Notes of Shahar Mozes’s Cambridge lecture 06-04-2017

Finite topological generation

Joint with Marc Burger.

We are interested in cocompact lattices in products of trees. They come with two projections in ${Aut(T_i)}$, and their closures ${H_i}$. The vertex stabilizers ${H_i(x)}$ are automatically topologically finitely generated.

Since ${K=H_1(x)}$ is a compact open subgroup in ${Aut(T_1)}$, ${\Gamma\cap(K\times Aut(T_2))}$ is a cocompact lattice in ${K\times Aut(T_2)}$, hence finite topological generation follows.

It follows that projections of ${\Gamma}$ are never dense. Indeed, the automorphism group of a regular rooted tree is not topologically finitely generated.

1. Groups of coloured tree automorphisms

Let us color edges of a ${d}$-reguler tree ${T}$ and fix a subgroup ${F<\mathfrak{S}_d}$. Define

$\displaystyle \begin{array}{rcl} U(F)=\{g\in Aut(T)\,;\, \textrm{the local action at each vertex nelongs to }F\}. \end{array}$

Question. When is ${U(F)}$ topologically finitely generated?

Assume that ${F}$ is transitive. Let ${L}$ be the stabilizer of colour 1 in ${F}$. Let

$\displaystyle \begin{array}{rcl} L_n =L\wr\cdots\wr L\quad n\textrm{ times}, \quad L_\infty=\lim L_n. \end{array}$

${L_n}$ acts on the ${n}$-sphere, and ${L_\infty}$ on the set of ends of ${T}$.

Observe the following necessary conditions for top fin generation.

1. ${L}$ is perfect (${L=[L,L]}$),
2. ${L}$ has no fixed points at infinity.

Theorem 1 ${L_\infty}$ is topologically finitely generated iff

1. ${L}$ is perfect,
2. ${L}$ has no fixed points at infinity.

Definition 2 Let ${K}$ be a compact group. Say ${K}$ is PFG (for positively finitely generated) if for some ${k}$, the probability that a random ${k}$-tuple of elements topologically generate ${K}$ is positive.

This is stronger than top fin generation. For instance the profinite completion of ${F_2}$ does not have it (Kantor-Lubotzky). We shall prove that ${L_\infty}$ is PFG.

2. Proof

Relies on an idea of Meenaxi Bhattachargee. Let ${p_k(X)}$ denote the probability that ${k}$ elements generate finite group ${X}$. We want to estimate this for ${L_n}$.

Theorem 3 (Bhattachargee 1994) Let ${Y\rightarrow X}$ be an epimorphism of finite groups. Then

$\displaystyle \begin{array}{rcl} p_k(Y)\geq (1-\sum_{[M]\in \mathcal{G}(Y|X)}\frac{1}{[Y:M]^{k-1}})p_k(X). \end{array}$

where we sum over conjugacy classes of maximal subgroups ${[M]}$ of ${Y}$ which map onto ${X}$.

Indeed, being confined in a maximal subgroup is the obstacle to generate ${Y}$.

Set

$\displaystyle \begin{array}{rcl} \zeta_{Y|X}=\sum_{[M]\in \mathcal{G}(Y|X)}\frac{1}{[Y:M]^{s}}. \end{array}$

Corollary 4 Assume that for some ${k}$,

$\displaystyle \begin{array}{rcl} \sum_n \zeta_{L_{n+1}|L_n}(k-1)<\infty, \end{array}$

then ${L_\infty}$ is PFG.

2.1. Set up

${X}$ is a finite group (it will be ${L_n}$). Let ${X}$ act on some finite set ${\Omega}$, split into orbits ${\Omega_i}$. Let ${B_1,\ldots,B_t}$ be perfect groups, and

$\displaystyle \begin{array}{rcl} Y=X\times(B_1^{\Omega_1}\times\cdots\times B_t^{\Omega_t}) \end{array}$

Say a normal subgroup ${N is standard if it is a product of normal subgroups of ${B_i}$‘s. A subgroup ${M is clean if it does not contain any nontrivial standard subgroup.

Our main technical result is

Proposition 5 Let ${M be a clean maximal subgroup. Then one of the following holds.

1. ${t=2}$, ${B_1}$ and ${B_2}$ are nonabelian simple groups,
1. ${M\cap B_1^{\Omega_1}\times B_2^{\Omega_2}}$ is the graph of an isomorphism ${B_1^{\Omega_1}\rightarrow B_2^{\Omega_2}}$.
2. ${M}$ is the normalizer in ${Y}$ of ${M\cap B_1^{\Omega_1}\times B_2^{\Omega_2}}$ ; there are at most ${|\Omega_1||Out(B_1)|}$ conjugacy classes of such subgroups and

$\displaystyle \begin{array}{rcl} [Y:M]\geq |B_1|^{|\Omega_1|}. \end{array}$

2. ${t=1}$, ${M\cap B^\Omega=\{1\}}$, ${pr_\omega(M\cap B^\Omega)=B}$ for all ${\omega\in\Omega}$, there is a normal subgroup ${U of the form ${U=T^r}$ where ${T}$ is an abelian simple group, and

$\displaystyle \begin{array}{rcl} [Y:M]\geq |T|^{r|\Omega|/2}. \end{array}$

3. ${t=1}$, ${M\cap B^\Omega\not=\{1\}}$, ${pr_\omega(M\cap B^\Omega)}$ is a proper subgroup…

2.2. Questions

Caprace: from general principles, it follows that the set of ${k}$-tuples generating ${L_\infty}$ always has empty interior.