Invariant $\sigma$-field of a product with a weakly mixing transformation

Question

It is known that an invertible mpt $S$ is weakly mixing if and only if $S \times T$ is ergodic for any ergodic invertible mpt $T$. Is it more generally true that the invariant $\sigma$-field of $S \times T$ is the product of the trivial $\sigma$-field times the invariant $\sigma$-field of $T$ whenever $S$ is weakly mixing ?

Answer 1

The answer is yes (I am assuming here the spaces are standard and the measures are probability measures).

For this kind of questions I find that it is convenient to think about quotient spaces rather then sub-$\sigma$-algebras. In particular, instead of considering the $\sigma$-algebra of invariants you better consider the space of ergodic components.

Recall that given a probability measure preserving transformation $T$ on $(Y,\nu)$ there exists a space $Z$ and a measurable (a.e defined) map $\pi:Y\to Z$ such that the $\sigma$ algebra of $T$-invariants equals the pull back $\sigma$-algebra under $\pi$. Alternatively, $\pi:Y\to Z$ is characterized by:

• $\pi$ is $T$-invariant.

• presenting the disintegration of $\nu$ wrt $\pi$ as $\nu=\int_Z \nu_z d\pi_*\nu(z)$, for $\pi_*\nu$-a.e $z\in Z$, $\nu_z$ is ergodic.

The space $(Z,\pi_*\nu)$ is uniquely defined by these properties (up to a natural isomorphism) and is called the space of ergodic components of $T$.

Its existence is a simple application of Choquet theory. One can also define it as the spectrum of the algebra of $T$-invariants in $L^\infty(Y)$.

Let us go back to the asked question now. Assume $S$ is a transformation of $(X,\mu)$ and $T$ of $(Y,\nu)$, both are probability measure preserving. Assume the $S$ is weakly mixing. Let $Z$ be the space of ergodic components of $T$ and $\pi:Y\to Z$ the corresponding map.

Then $\mu\times \nu=\int_Z \mu\times\nu_z d\pi_*\nu(z)$. Clearly, the map $\pi\circ p_2:X\times Y\to Z$ is $S\times T$-invariant (where $p_2$ is the projection $X\times Y\to Y$) and by the weak mixing assumption, for $\pi_*\nu$-a.e $z\in Z$, $\mu\times\nu_z$ is $S\times T$-ergodic. Thus $Z$ is the space of ergodic components of $S\times T$. It follows that the $\sigma$-algebra of $S\times T$ invariant sets in $X\times Y$ is the pull back $\sigma$-algebra under $\pi\circ p_2$. This clearly provides a positive answer to the question.

Answer 2

Here I will explain (briefly) the proof of the statement using tensor products of unitary operators.

Recall that a unitary operator $U$ on a Hilbert space $H$ is called ergodic if $H^U=\{0\}$, ie there is no $U$-invariant vector in $H$, and weakly mixing if $U\otimes U$ is ergodic on $H\otimes H$ (here and elsewhere refers to the complete hilbertian tensor product). We know that $U$ is weakly mixing iff for every unitary $U'$ on a Hilbert space $H'$, $U\otimes V$ is ergodic.

A probability measure preserving transformation $S$ of $(X,\mu)$ is ergodic / weakly mixing if the associated unitary operator on $L^2_0(X)$ (the orthogonal to the constants) is.

Assume $S$ is a transformation of $(X,\mu)$ and $T$ of $(Y,\nu)$, both are probability measure preserving. Assume that $S$ is weakly mixing. Let $U$ and $V$ denote the corresponding unitary operators. Consider the orthogonal decompositions $L^2(X)=L^2_0(X)\oplus \mathbb{C}$ and $L^2(Y)=H\oplus L^2(Y)^V$. Here $L^2(X)^U=\mathbb{C}$ and $H$ is the orhogonal complement of $L^2(Y)^V$ in $L^2(Y)$.

The claim is that $L^2(X\times Y)^{S\times T}\simeq (L^2(X)\otimes L^2(Y))^{U\otimes V}\simeq \mathbb{C}\otimes L^2(Y)^V$.

Indeed, $L^2(X)\otimes L^2(Y)$ decomposes to four components acording to our direct decompositions above, $\mathbb{C}\otimes L^2(Y)^V$ being one of them, consisting of $U\otimes V$ invariant vectors. In the other three, $\mathbb{C}\otimes H$, $L^2_0(X)\otimes H$ and $L^2_0(X)\otimes L^2(Y)^V$ there are no $U\otimes V$ invariant vectors by the definition of $H$, the weak mixing of $S$ and the ergodicity of $S$ correspondingly.