Is there any example of a complete atomless Boolean algebra with a non-trivial abelian automorphism group?

This is equivalent, by Stone duality, to asking for an extremally disconnected compact Hausdorff space with no isolated points, having abelian homeomorphism group. Note that no such example can be metrizable.

I only know of examples of rigid complete Boolean algebras, i.e., those having trivial automorphism groups. For examples, see

  • Jónsson, B., A Boolean Algebra Without Proper Automorphisms, Proc. AMS, 1951.

Another rigid example coming from forcing appears in

  • McAloon, K., Consistency Results About Ordinal Definability, Ann. Math. Log, 1971.
  • 2
    $\begingroup$ What about $A\times A$ when $A$ has a trivial automorphism group (and maybe some additional assumption)? Here $(A,+)$ acts by automorphisms, namely $a\cdot (b,c)=((1-a)b+ac,(1-a)c+ab)$, and we could expect there's nothing more. $\endgroup$ – YCor Aug 25 '18 at 21:08
  • $\begingroup$ I should mention, de Groot (in the same paper I mentioned in my last comment), gives an example of a 0-dim, I believe non-compact, subset of $\mathbb{R}$, with non-trivial homeomorphism group where every element is order two (and hence, is abelian). Maybe this can be adapted to what I want by taking the Cech-Stone compactification? $\endgroup$ – Iian Smythe Aug 25 '18 at 22:15
  • $\begingroup$ YCor, Oops, I should read comments with a bit more care. For some reason I read your comment as suggesting the automorphism group was cyclic, which of course you are not. $\endgroup$ – Iian Smythe Aug 25 '18 at 23:45
  • $\begingroup$ Indeed for a few seconds I first expected that the automorphism group could be cyclic, but the topological intuition makes it clear that it's much larger. $\endgroup$ – YCor Aug 26 '18 at 12:33

If A is a rigid complete BA, then the automorphism group of AxA is isomorphic to the direct sum of |A| copies of the two-element group.This is proved in the remark following Lemma 1.9 in a paper of McKenzie and Monk (Colloq. Math. Soc. Janos Bolyai, 1973, 951-988); the result is attributed to de Groot; see the reference above.

  • $\begingroup$ Can we say anything like this for the sum $A\oplus A$ (i.e., the completion of the corresponding product partial order) where $A$ is complete and rigid? $\endgroup$ – Iian Smythe Aug 27 '18 at 13:39
  • $\begingroup$ @IianSmythe No idea, except that I would rather have expected that this looks like a tensor product $A\otimes A$. $\endgroup$ – YCor Aug 27 '18 at 13:50

Don Monk's answer confirms my expectation; let me now also provide a more precise statement and a proof:

If $A$ is a Boolean algebra with a trivial automorphism group, then the automorphism group of $A\times A$ is canonically isomorphic to $(A,+)$, acting by $a\cdot(b,c)=((1-a)b+ac,(1-a)c+ab)$.

By Stone duality, this amounts to proving: if $X$ is a Stone space (totally disconnected, Hausdorff compact topological space) with trivial self-homeomorphism group, then any self-homeomorphism of $X\times\{-1,1\}$ is given by $(x,t)\mapsto (x,u(x)t)$ where $u$ is a continuous function $X\to\{-1,1\}$.

To prove the latter, let us first check that for any two clopen subsets $Y,Z$ of $X$ and homeomorphism $h:Y\to Z$, we have $Y=Z$ and $h$ is the identity. Indeed, otherwise there exists $y\in Y$ such that $h(y)\neq y$. By passing to smaller clopen subsets, we can then assume that $Y$ and $Z$ are disjoint. Then we can extend $h$ to a self-homeomorphism of $X$, as equal to $h^{-1}$ on $Z$ and identity elsewhere. (Note that in this part, we use that $X$ is Hausdorff with a basis of clopen subsets, but compactness does not play any role).

Then it's easy to conclude. Let $h$ be a self-homeomorphism of $X\times\{-1,1\}$. Call "component" the two subsets $X\times\{t\}$. We find a finite clopen partition of $X$ such that for each part $Y$ of the partition, each of $Y\times\{t\}$ is mapped into a single component. Then by the previous fact on partial homeomorphism, we have $h(y,t)=(y,s(y,t))$ for some $s(y,t)\in\{-1,1\}$ (which is constant when $y$ varies in $Y$). By injectivity, we have $s(y,1)=-s(y,-1)$. So we can write $s(y,t)=u(y)t$. $\Box$

Similarly, for a finite set $F$, the automorphism group of $X\times F$ consists of the $(x,t)\mapsto (x,\sigma(x)(t))$, where $\sigma$ ranges over the continuous functions $X\to\mathfrak{S}(F)$. The corresponding Boolean algebra action can be described accordingly.

  • $\begingroup$ I commented on Don Monk's answer with a follow up question: Can we say anything like this for the sum $A\oplus A$ (i.e., the completion of the corresponding product partial order, or Stone dual of the product of the corresponding Stone space) where $A$ is complete and rigid? $\endgroup$ – Iian Smythe Aug 27 '18 at 13:48
  • $\begingroup$ (I don't think it's a good idea to copy questions twice in comments.) $\endgroup$ – YCor Aug 27 '18 at 13:49

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