Let $\mathcal{B}$ be the (unique up to isomorphism) countable atomless boolean algebra, and $\mathrm{Aut}(\mathcal{B})$ its automorphism group, with pointwise convergence topology.

My question: Does $\mathrm{Aut}(\mathcal{B})$ contain a closed (non-abelian) free subgroup?

This question arises from reading about the following two results (whose proofs I am not overly familiar with):

$\mathrm{Aut}(\mathcal{B})$ contains many free subgroups; by a result of Machpherson (Groups of Automorphisms of $\aleph_0$-Categorical Structures, QJM, 1986), it has a dense subgroup freely generated by countably many generators.

$\mathrm{Aut}(\mathbb{Q},<)$ contains a closed non-abelian free group; this is a result of Pestov (On free actions, minimal flows, and a problem by Ellis, TAMS, 1998). In particular, closed subgroups of extremely amenable topological groups need not be amenable.


3 Answers 3



This group is widely documented as "(self-)homeomorphism group of the Cantor set" (by Stone duality).

It admits, for every prime $p$ and $p$-adic field $K$, and $d\ge 2$, the group $\mathrm{PGL}_d(K)$ as a closed subgroup (viewed as acting on the projective space $\mathbb{P}^{d-1}(K)$ which is homeomorphic to a Cantor space), and this group contains a non-abelian free discrete subgroup.

Added: here's a direct "Schottky" construction. Suppose that we have a partition of the Cantor set $X=A_+\sqcup A_-\sqcup B_+\sqcup B_-\sqcup Y$ into nonempty clopen subsets, and self-homeomorphisms $a,b$ of $X$ such that $a^{\pm 1}A_{\mp}^c\subset A_\pm$ and $b^{\pm 1}B_{\mp}^c\subset B_\pm$ (ping-pong condition). Then $(a,b)$ freely generates a discrete free subgroup (discrete because the stabilizer of the clopen subset $Y$ is reduced to $\{1\}$). There are plenty of such pairs, for instance coming from suitable loxodromic isometries of trees acting on the boundary.

  • $\begingroup$ Can you say a word about why these copies of $PGL_d(K)$ are closed in the group? $\endgroup$ May 29, 2018 at 20:01
  • 1
    $\begingroup$ For $d=2$ it's the stabilizer of the cross-ratio, hence it's closed. For $d\ge 3$ the stabilizer of the alignment relation is closed, and equal (fundamental theorem of projective geometry) to $PGL_d(K)\rtimes Aut(K)$ where Aut means topological field automorphisms, so its open subgroup of finite index $PGL_d(K)$ is closed as well. $\endgroup$
    – YCor
    May 29, 2018 at 22:32
  • $\begingroup$ (PS if $K$ is a local (locally compact) field of finite characteristic then it still works: $Aut(K)$ is (infinite) compact rather than finite, but this still proves closedness; Uri's answer shows that this could be deduced automatically by the way.) $\endgroup$
    – YCor
    May 31, 2018 at 10:34

The answer is "yes", as YCor had already answered, but I wish to add another perspective.

I think the easiest example to have in mind is the the action of a free group on its Gromov boundary (the set of infinite reduced words) endowed with the standard Markov measure.

Moreover, every Gromov hyperbolic group $\Gamma$ has a closed image in $\text{Aut}(\partial\Gamma,\nu)$ where $\nu$ is any $\Gamma$-quasi-invariant measure on the Gromov boundary $\partial \Gamma$, eg a Patterson-Sullivan or a Furstenberg-Poisson measure. The fact that the image of $\Gamma$ is closed follows at once from the "convergence group action property".

Let me also add another remark concerning Yves' answer: in fact, the groups $\text{PGL}_d(K)$ have the property that their image under any continuous homomorphism into any other (Hausdorff) topological group is closed, see Proposition 5.4 here (self promotion alert).


One more answer: The atomless Boolean algebra is just the algebra of clopen subsets of $\{0,1\}^{\mathbb N}$, now pick any discrete embedding $F_2 \subset {\rm Sym}(\mathbb N)$ - for example letting $F_2$ act on itself by translation, and consider the induced action on $\{0,1\}^{\mathbb N}$.

It is an open problem if the full group of the hyperfinite equivalence relation (or the unitary group of the hyperfinite $II_1$-factor) contains a discrete free subgroup. (Those groups are SIN, which makes the problem more complicated.) Some people (including myself) believed that maybe all closed subgroups of amenable polish SIN groups are again amenable, but this was disproved by example in https://arxiv.org/abs/1603.00691.


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