Notes of Nicolas Matte Bon’s Cambridge lecture 23-05-2017

Uniformly recurrent subgroups and rigidity of non-free minimal actions

Joint work with A. Le Boudec and T. Tsankov.

1. The Chabauty space of a group

Let {G} be a locally compact group. The Chabauty space of {G} is the set of subgroups of {G}, where two subgroups are nearby if their intersections with every compact set are. If {G} is countable, the topology coincides with that induced from {\{ 0,1 \}^G}.

The {G} action by conjugation on {Sub(G)} is usually interesting. Glasner and Weiss suggested to study uniformly recurrent subgroups, aka URS, i.e. minimal invariant subsets of this action.


  1. Normal subgroups.
  2. Conjigacy classes of cocompact subgroups.
  3. Stabilizers of actions on compact spaces.

Indeed, the closure of the set of stabilizers of a {G}-action on {X} contains a unique minimal subset, called the stabilizer URS of the action, {S_G(X)} (Glasner-Weiss).

Theorem 1 (Matte Bon-Tsankov, Elek) Conversely, every URS of {G} is the stabilizer URS of some action of {G} on a compact space.

Our initial motivation was to study {C^*} simplicity of countable groups, following this theorem.

Theorem 2 (Kalantar-Kennedy, Kennedy) If {G}is countable, the followng are equivalent.

  1. {G} is {C^*}-simple.
  2. {G} acts freely on its universal Furstenberg boundary.
  3. {G} has no non-trivial URS consisting of amenable subgroups.

Our interest has shifted to examples.

2. Examples: Thomson’s groups

These are three groups {F<T<V} acting respectvely on {[0,1]}, on the circle and on the Cantor set {\{ 0,1 \}^{\mathbb N}}. Each of them consists of all homeomorphisms acting locally like {x\mapsto 2^n x+q}. {T} and {V} act minimally (but {F} does not), with large stabilizers.

Theorem 3 (Le Boudec-Matte Bon) Let {T} act on a compact space {X}. Assume action is minimal and not topologically free. Then there is a continuous surjective equivariant map {\phi:X\rightarrow S^1}. Moreover, for all {y\in S^1} but countably many, the action of the stabilizer of {y} on the fiber {\phi^{-1}(y)} is trivial.

Topologically free means that the set of points with trivial stabilizer is a dense {G_\delta}.

Free actions on the circle can be blown up: certain points are replaced with intervals. This gives mny examples.

Bounded cohomology method allow to show that certain group actions on the circle cannot be topologically free.

A similar statement holds fro {V}, and for {F}, it says that there are no such actions at all.

2.1. Proof

  1. Classify all URS of {T}. There are only three: trivial, {T} and the stabilizer URS of the action on the circle.
  2. Construct a map to the circle in the third case.

The first step has a more general character. Let {G} be a countable group acting faithfully on a compact Hausdorff space {Z}. We shall focus on subgroups fixing pointwise the complement of open sets {U}, {G_U}, called rigid stabilizers.

Proposition 4 Let {H<G} be a subgroup. Either a sequence of conjugates of {H} converges to the trivial subgroup, or there exists an open subset {U\subset Z} and a subgroup {K<H} stabilizing {U} and whose action on {U} coincides with that of a finite index subgroup of {G_U}. Moreover, if the the {G} action on {Z} is extremely proximal (every proper closed set can be shrinked to a point), then there exists a finite index subgroup of {G_U} whose commutator subgroup is contained in {H}.

3. Non-discrete locally compact groups without URS nor IRS

This is work in progress.

3.1. Invariant random subgroups

An IRS is an invariant probability measure on Chabauty space. IRS are stabilizers of probability measure preserving actions.

Can a group have no IRS, or no URS, at all ? For instance, Neretin’s group is simple and has no lattices, this rules out the most obvious examples of IRS. Does it have others?

3.2. Answers in the discrete case

Tarski monsters have only countably many subgroups, so no URS nor IRS.

Finitary alternating group {\mathfrak{A}_f({\mathbb N})} has no URS but plenty of IRS (Vershik).

Thompson’s groups {T} and {V} have 1 URS and no IRS (Dudko-Medynets).

The commutator subgroup of Thompson’s group {F} has no URS, no IRS.

3.3. The non-discrete case

Compactly generated examples can be obtained, using dense subgroups in them obtained by slightly enlarging a known group. Neretin’s group is of this sort, starting from Thompson’s group {V}.

We describe non compactly generated examples. Let {\Gamma} be a countable group action faithfully on a set {\Omega}. Let {D_i} be finite subsets of {\Gamma} with pairwise dsjoint supports, and such that for every {g\in\Gamma}, the support of {g} has only finitely many points in common with any support{(D_i)}. Consider the closed subgroup generated by {\Gamma} and the product of {D_i}, in the topology where {\prod D_i} is open.

Such constructions arise in Willis, Akim-Glasner-Weiss, Caprace-Cornulier to produce various counterexamples.

We take {\Omega={\mathbb N}}, {\Gamma=\mathfrak{A}_f({\mathbb N})} and {D_i=\mathfrak{A}([n_i,n_{i+1}])} for {i} even, and trivial if {n} s odd. Then the resulting group has no URS, but many IRS.

Let us take {\Omega={\mathbb Q}_2}, {\Gamma=} the group of homeos locally modelled on {x\mapsto 2^nx+q}. It is a non-discrete variant of Thompson’s Then the resulting group has no IRS, but many URS.


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