# Antichains and measure-preserving actions on Boolean algebras

Let $G$ be a group of automorphisms of the countable atomless Boolean algebra $B$. Suppose that every orbit of $G$ on $B$ is an antichain. Does it follow that $G$ preserves a non-zero (probability) measure on $B$?

Does the answer change if we extend $B$ to some complete or $\sigma$-complete algebra, and the action of $G$ extends to one in which orbits are still antichains?

I'm also interested in group actions that satisfy a very different condition: $G$ is a group such that for some (all) $a \in B \setminus \{0,1\}$ and for all $b \in B \setminus \{0,1\}$ there is some $g \in G$ such that $ga < b$. Do such actions have a name and has anything been proved about them (or about groups that have such actions)?

Edit: To clarify, by 'antichain' I just mean a set of pairwise incomparable elements. I didn't know about the stronger meaning used by set theorists. For what it's worth I am mainly interested in using actions to understand algebraic properties of the group $G$, so I probably don't need to consider any exotic algebras of the kind set theorists would find interesting; the most obvious examples, such as the countable atomless Boolean algebra or the standard Borel $\sigma$-algebra, are probably good enough. I definitely do not want to assume that $G$ is the whole automorphism group, however.

• Could you clarify your sense of "antichain"? In set theory, an antichain has $a\wedge b=0$ for all its elements, but elsewhere, one need only that $a$ and $b$ are incomparable with respect to the order. – Joel David Hamkins Sep 26 '11 at 18:17

Regarding the question is your last paragraph, there is the following often-studied but not-quite-equivalent-to-your property:

• A Boolean algebra $\mathbb{B}$ is almost homogeneous if for every nonzero $a,b\in\mathbb{B}$ there is an automorphism $\pi$ with $0\lt \pi(a)\wedge b$. Similarly, a group $G$ acts almost homogeneously if such $\pi$ can be found in $G$. This concept is also applied to partial orders in the natural way.

This is a weakening of your property, but it turns out to be critically important in the forcing context, since the theory forced by an almost homogeneous Boolean algebra does not depend on the generic filter.

• What does 'homogeneous' mean in this context? – Colin Reid Sep 26 '11 at 15:44
• Homogeneous means that you can attain $\pi(a)=b$. This is very rarely the case, for the Boolean algebras that arise in forcing, and so almost homogeneity is the principal notion. – Joel David Hamkins Sep 26 '11 at 18:11

I think the answer to first part of your question is affirmative. Begin by noting that the requirement on the automorphism $\Phi$ that all orbits are antichains implies that $\Phi^2$ is the identity. To see this suppose $\Phi$ is an automorphism acting on $B$ and let $a\in B$ be such that $\Phi(a) \neq a$. Since every orbit is an antichain, it must be that $a\cap \Phi(a) = \emptyset$. But now $\Phi(a\cup \Phi(a)) = \Phi(a)\cup \Phi(\Phi(a))$ and hence the orbit of $a\cup \Phi(a)$ is not an antichain unless $\Phi^2(a) = a$.

Now construct the probability measure $\mu$ by choosing $a_0\in B$ such that $\Phi(a_0)\neq a_0$ (and hence $\Phi(a_0)\cap a_0 = \emptyset$). It follows that if $b_0$ is the complement of $a_0\cup\Phi(a_0)$ then $\Phi(b_0) = b_0$ because $$\emptyset = \Phi(b_0\cap (a_0\cup\Phi(a_0))) = \Phi(b_0)\cap (\Phi(a_0)\cup\Phi^2(a_0)) = \Phi(b_0)\cap (\Phi(a_0)\cup a_0)$$ and hence $\Phi(b_0)\cap b_0\neq \emptyset$ --- so the antichain property yields that $\Phi(b_0) = b_0$. Let $\mu(a_0) = \mu(\Phi(a_0)) = 1/2$. So $\mu(b_0) = 0$ and this much of $\mu$ is preserved by $\Phi$.

Next choose a partition of $a_0$ into disjoint sets $a_{0,0}$ and $a_{0,1}$ and let $\mu(a_{0,i} )= \mu(\Phi(a_{0,i})) = 1/4$ Continuing in this spirit will define a probability measure on $B$ that is $\Phi$ invariant.

This is a follow-up to Juris’ answer, but it is a bit too long for a comment.

There is in fact no nontrivial group $G$ of automorphisms of $B$ such that all orbits are antichains, where $B$ is any Boolean algebra with more than $4$ elements.

Assume for contradiction that $f\in G$ is not the identity. If $a$ is any element of $B$ such that $f(a)\ne a$, then $a\land f(a)=0$ as the orbit of $a$ is an antichain; the orbit of $-a$ is also an antichain, hence $-a\land-f(a)=0$, which together imply $f(a)=-a$. Assume there is $b\in B$ such that $f(b)=b$ and $b\ne0,1$. Then $f(a\lor b)=-a\lor b$, but on the other hand we already know $f(a\lor b)\in\{a\lor b,-(a\lor b)\}$, which is easily seen to imply $b=1$ or $b=0$, a contradiction. Thus, $f(b)=-b$ for every $b\ne0,1$. However, this means that $f$ is order-reversing on $B\smallsetminus\{0,1\}$, whereas being an isomorphism, it is also order-preversing. This is a contradiction as long as we can find $0< a< b< 1$ in $B$, which we always can if $|B|>4$.

• The argument by Emil does not seem to use the term "antichain" in the usual way, i.e., a set of pairwise incomparable elements. For a finite boolean algebra (or any finite poset), every orbit of every automorphism is an antichain. – Richard Stanley Sep 26 '11 at 17:38
• I used the term in the same sense as in Juris Steprans’ answer, i.e., a set of pairwise incompatible elements. I was under the impression that this is the standard definition in forcing. – Emil Jeřábek Sep 26 '11 at 17:53
• Emil, this certainly does the trick! And yes, I had assumed that"antichain" was used in the forcing sense. But the other version of the question also makes sense. – Juris Steprans Sep 26 '11 at 18:15
• Emil's is the standard meaning of antichain within set theory, but it seems that the OP may not intend this meaning, and the other meaning of antichain is very common outside set theory. The Boolean algebra under consideration, however, has a dense subset that is a tree, where the two concepts of "antichain" coincide. – Joel David Hamkins Sep 26 '11 at 18:16
• @Joel David Hamkins: Indeed I mean the weaker sense of 'antichain'. It is interesting that you mention trees - I would certainly like to have some non-trivial criteria for when such a tree is preserved setwise by $G$ (whether or not the orbits on it are antichains), as the question of whether a given group possesses well-behaved actions on trees (especially locally finite trees) is of considerable group-theoretic interest. – Colin Reid Sep 28 '11 at 11:29

I think the answer to your question is yes when $G$ is finite, but I guess this is way too restrictive (since the condition on the orbits would be automatically true in that case).

This doesn´t answer your question either, but since you are "mainly interested in using actions to understand algebraic properties of the group $G$", you might want to take a look at the paper "Free continuous actions on zero-dimensional spaces" which you can find at one of the authors´ webpage. Among other things it is shown that any countable group acts freely on a Cantor space (and hence on the countable atomless Boolean algebra $B$) with an action that admits an invariant probability measure.