Dan Rudolph (University of Maryland)

Relative mixing, isometric extensions and orbit equivalence

Abstract. For measure preserving and ergodic actions of $\mathbb Z$ the general understanding of isometric extensions is rather well understood. In particular one has a well-known list of results that all say if an action T has property A, and $\hat T$ is an isometric extension of T that is weakly mixing, then $\hat T$ must also have property A. Here one can put ``mixing", ``k-fold mixing", ``K", or ``Bernoulli" for property A. It has also been realized for many years that any result for an action should have a ``relativized" generalization, a version stated relative to an invariant factor algebra $\mathcal H$. Our goal is to consider the following question: Is it that case that an isometric extension of an action which is $\mathcal H$-relatively mixing, if it remains $\mathcal H$-relatively weakly mixing must in fact also be $\mathcal H$-relatively mixing?

A first step here is to understand what might be meant by $\mathcal H$-relatively mixing. Speaking vaguely, $\mathcal H$-relative mixing should mean that

$$E(fg(T^n)\vert\mathcal H)- E(f\vert\mathcal H)E(g(T^n)\vert\mathcal H)\underset{n\to\infty}\rightarrow 0.$$

The question is, in what sense should this function be tending to zero. The natural answers are in L¹ or pointwise. To see the difference in these, consider a [T,T^-1]map where T is mixing and $\mathcal H$ is the base i.i.d. Bernoulli action. It is easy to see from the random walk that this action would be $\mathcal H$-relatively mixing if one asked for L¹ convergence. But as the random walk is recurrent, it would be false for pointwise convergence.

What we will show is that the question above about isometric extensions can be answered in the affirmative for the pointwise definition. For the L¹questions it remains open but the methods we use as we understand them will fail.