Antichains and measure-preserving actions on Boolean algebras - MathOverflow

Antichains and measure-preserving actions on Boolean algebras

2011-09-26T13:27:41Z

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.

Answer by Joel David Hamkins for Antichains and measure-preserving actions on Boolean algebras

2011-09-26T15:03:09Z

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.

Answer by Juris Steprans for Antichains and measure-preserving actions on Boolean algebras

2011-09-26T16:56:03Z

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.

Answer by Emil Jeřábek for Antichains and measure-preserving actions on Boolean algebras

2011-09-26T17:30:23Z

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 $|B|>4$.

Answer by Ramiro de la Vega for Antichains and measure-preserving actions on Boolean algebras

2011-10-19T21:29:39Z

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.