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}.


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

Our main technical result is

Proposition 5 Let {M<Y=X\times(B_1^{\Omega_1}\times\cdots\times B_t^{\Omega_t})} 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<B} 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.


About metric2011

metric2011 is a program of Centre Emile Borel, an activity of Institut Henri Poincaré, 11 rue Pierre et Marie Curie, 75005 Paris, France. See
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