Posted by: yglima | April 12, 2011

## ERT15: Weakly mixing extensions

Two common examples of ergodic measure-preserving systems, an irrational rotation of the circle with Lebesgue measure and the Bernoulli shift on ${\{0,1\}^{\mathbb Z}}$ with the ${(1/2,1/2)}$ product measure are typical members of the classes of compact and weak mixing systems, respectively. Due to the theory developed in ERT8, ERT9 and ERT10, we consider these two classes as well understood. Can one hope to describe general mps in terms of such building blocks? The surprising answer to this question is that if one allows for the relative notions of compact and weak mixing to play the part of building blocks, then this is indeed the case. This vague statement is made precise in the Furstenberg-Zimmer structure theorem, which we intend to prove soon.

Let ${\pi:\mathbb X\rightarrow\mathbb Y}$ be a homomorphism, with ${\mathbb X=(X,\mathcal A,\mu,T)}$ and ${\mathbb Y=(Y,\mathcal B,\nu,S)}$. The present and next two posts are devoted to the definitions and study of the notions of compact and weakly mixing extensions, by adapting the usual notions of compact and weak mixing mps to a “relative” or “conditional” setting. The ideas are essentially the same as in the absolute cases, the main new trick being not to work in the Hilbert space ${L^2(\mathbb X)}$ over the complex numbers, but rather viewing the Hilbert space ${L^2(\mathbb X)}$ as a direct integral of a Hilbert bundle over the von Neumann algebra ${L^\infty(\mathbb Y)}$. Specifically, we consider the fiber Hilbert spaces ${L^2(\mathbb X,\mu_y)}$, where ${\mu=\int \mu_yd\nu(y)}$ is the disintegration associated to ${\pi}$, form the Hilbert bundle ${\bigsqcup_{y\in Y}L^2(\mathbb X,\mu_y)}$ over ${L^\infty(\mathbb Y)}$ and identify ${L^2(\mathbb X)}$ with integrable sections of this Hilbert bundle. This will be properly explained in ERT16 (for weakly mixing extensions do not make use of such machinery).

As in the absolute case, there are two ways of defining relative notions of compactness and weak mixing. The first one, which I see as a qualitative one, is intrinsically more natural. As a matter of fact, once the two definitions are complementary, Furstenberg-Zimmer structure theorem basically follows from a transfinite induction. This definition relies on the analysis of generalized eigenfunctions.

On the other hand, to obtain Furstenberg multiple recurrence theorem, it is necessary a quantitative characterization of such objects. This is the other way of defining the concepts, in terms of ergodic averages and its expectations. See this post of Terry Tao for further details on this viewpoint.

In the present post, just like we did in the absolute case, we will define ergodicity and weak mixing from the quantitative viewpoint. This has two advantages: it is more concrete and demonstrates its usefulness for the purpose of multiple recurrence. ERT16 will describe the qualitative viewpoint. This will require the machinery of Hilbert bundles and generalized eigenfunctions. Then Furstenberg-Zimmer structure theorem will be established and, finally, for the purpose of multiple recurrence, we will show the two definitions are equivalent.

1. Ergodic extensions

Let ${\mathbb X=(X,\mathcal A,\mu,T)}$ and ${\mathbb Y=(Y,\mathcal B,\nu,S)}$ a factor of ${\mathbb X}$. We write ${\mathbb X\stackrel{\pi}{\rightarrow}\mathbb Y}$ for the corresponding homomorphism and, when ${\pi}$ is implicit, just ${\mathbb X\rightarrow\mathbb Y}$.

Definition 1 ${\mathbb X}$ is an ergodic extension of ${\mathbb Y}$ if the only ${T}$-invariant sets of ${\mathcal A}$ are images (modulo nulls sets) under ${\pi^{-1}}$ of ${S}$-invariant sets of ${\mathcal B}$.

If we denote the trivial one-point mps by ${\mathbb X_0}$, then ${\mathbb X}$ is an extension of ${\mathbb X_0}$. Clearly, saying that ${\mathbb X}$ is ergodic now becomes the assertion that ${\mathbb X\rightarrow\mathbb X_0}$ is an ergodic extension.

Ergodic extensions behave well under composition and restriction. Specifically, if ${\mathbb X\stackrel{\pi_1}\longrightarrow\mathbb Y\stackrel{\pi_2}\longrightarrow\mathbb Z}$, then ${\mathbb X\rightarrow\mathbb Z}$ is ergodic if and only if ${\mathbb X\rightarrow\mathbb Y}$ and ${\mathbb Y\rightarrow\mathbb Z}$ are both ergodic. This is straightforward, but for the purpose of completeness, I will give the proof.

1. ${(\Longrightarrow)}$ Assume ${\mathbb X\rightarrow\mathbb Z}$ is ergodic and let ${A}$ be ${\mathbb X}$-invariant (this means ${\mathbb X=(X,\mathcal A,\mu,T)}$ and ${A\in\mathcal A}$ satisfies ${T^{-1}A=A}$). By hypothesis, ${A=(\pi_2\pi_1)^{-1}C={\pi_1}^{-1}{\pi_2}^{-1}C}$, where ${C}$ is ${\mathbb Z}$-invariant. Observe that ${{\pi_2}^{-1}C}$ is invariant under ${\mathbb Y}$ and so ${\mathbb X\rightarrow\mathbb Y}$ is ergodic. Now, let ${B}$ be ${\mathbb Y}$-invariant. Then ${{\pi_1}^{-1}B}$ is ${\mathbb X}$-invariant. By assumption, there is ${C}$ invariant under ${\mathbb Z}$ such that
$\displaystyle {\pi_1}^{-1}B={\pi_1}^{-1}{\pi_2}^{-1}C\ \ \Longrightarrow\ \ \ B={\pi_2}^{-1}C.$
2. ${(\Longleftarrow)}$ Assume ${\mathbb X\rightarrow\mathbb Y}$ and ${\mathbb Y\rightarrow\mathbb Z}$ are ergodic and let ${A}$ be ${\mathbb X}$-invariant. Then ${A={\pi_1}^{-1}B}$ for some ${\mathbb Y}$-invariant ${B}$. Then ${B={\pi_2}^{-1}C}$, where ${C}$ is ${\mathbb Z}$-invariant and so ${A=(\pi_2\pi_1)^{-1}C}$.

Remember that ${\pi}$ induces a natural imbedding of ${L^p(\mathbb Y)\cong L^p(\mathbb Y)^\pi}$ into ${L^p(\mathbb X)}$, ${1\le p\le\infty}$. It is evident ${\mathbb X\stackrel{\pi}{\rightarrow}\mathbb Y}$ is ergodic if and only if every ${T}$-invariant function of ${L^p(\mathbb X)}$ belongs to ${L^p(\mathbb Y)^\pi}$, that is, iff ${f=\mathbb E(f|\mathbb Y)^{\pi}}$ for any ${T}$-invariant ${f\in L^p(\mathbb X)}$. By von Neummann’s theorem (see ERT1), the set of ${T}$-invariant functions in ${L^p(\mathbb X)}$ is equal to ${\{\tilde f\,;\,f\in L^p(\mathbb X)\}}$, where

$\displaystyle \tilde f=\lim_{N\rightarrow\infty}\dfrac{1}{N}\sum_{n=1}^NT^nf.$

We thus have an alternative definition for ergodic extensions.

Definition 2 ${\mathbb X}$ is an ergodic extension of ${\mathbb Y}$ if and only if ${\tilde f=\mathbb E(\tilde f|\mathbb Y)^\pi}$, for every ${f\in L^p(\mathbb X)}$.

Observe that, as ${\mathbb E:L^2(\mathbb X)\rightarrow L^2(\mathbb Y)}$ is continuous in the strong topology (see ERT14), it is enough to check the above equality in the weak topology.

Proposition 3 Let ${\mathbb X\rightarrow\mathbb Y}$ be an ergodic extension and let ${f,g\in L^2(\mathbb X)}$ with ${\mathbb E(g|\mathbb Y)=0}$. Then

$\displaystyle \lim_{N\rightarrow\infty}\dfrac{1}{N}\sum_{n=1}^N\int_XfT^ngd\mu=0 \ \ \ \ \ (1)$

or, equivalently,

$\displaystyle \lim_{N\rightarrow\infty}\dfrac{1}{N}\sum_{n=1}^N\int_Y\mathbb E(fT^ng|\mathbb Y)d\nu=0. \ \ \ \ \ (2)$

Proof: Note that ${fT^ng\in L^1(\mathbb X)}$ and so the above expression makes sense. Firstly,

$\displaystyle \dfrac{1}{N}\sum_{n=1}^N\int_Y\mathbb E(fT^ng|\mathbb Y)d\nu= \dfrac{1}{N}\sum_{n=1}^N\int_X fT^ngd\mu=\left\langle f,\dfrac{1}{N}\sum_{n=1}^N T^ng\right\rangle$

and so the lemma reduces to saying that ${g_N=N^{-1}\sum_{n=1}^N T^ng}$ converges in the weak topology to zero. Let ${h\in L^2(\mathbb X)}$ be a weak limit. Since ${Tg_N-g_N}$ converges to zero in the strong topology, ${Th=h}$ and so, by ergodicity, ${h=\mathbb E(h|\mathbb Y)^\pi}$. Finally, as each ${g_N}$ has null ${\mathbb Y}$-expectation,

$\displaystyle \mathbb E(g_N|\mathbb Y)=\dfrac{1}{N}\sum_{n=1}^N\mathbb E(T^ng|\mathbb Y)= \dfrac{1}{N}\sum_{n=1}^N S^n\mathbb E(g|\mathbb Y)=0,$

a passage of the weak limit (see the exercise below) gives that ${\mathbb E(h|\mathbb Y)=0}$ and so ${h=0}$. $\Box$

Exercise 1 Prove that ${\mathbb E:L^2(\mathbb X)\rightarrow L^2(\mathbb Y)}$ is continuous in the weak topology.

Although equations (1) and (2) are equivalent, the natural place to work with then is downstairs, in ${\mathbb Y}$. This is corroborated when we write the equations in the inhomogeneous case, which is the content of the next result. Actually, it characterizes ergodicity.

Corollary 4 The extension ${\mathbb X\stackrel{\pi}\longrightarrow\mathbb Y}$ is ergodic if and only if, for every ${f,g\in L^2(\mathbb X)}$, one has

$\displaystyle \lim_{N\rightarrow\infty}\dfrac{1}{N}\sum_{n=1}^N\int_Y\{\mathbb E(fT^ng|\mathbb Y)- \mathbb E(f|\mathbb Y)S^n\mathbb E(g|\mathbb Y)\}d\nu=0. \ \ \ \ \ (3)$

Proof:
(${\Longrightarrow}$) Just apply the previous proposition to the functions ${f}$ and ${g-\mathbb E(g|\mathbb Y)^\pi}$, noting that

$\displaystyle \mathbb E(g-\mathbb E(g|\mathbb Y)^\pi|\mathbb Y)=\mathbb E(g|\mathbb Y)- \mathbb E(\mathbb E(g|\mathbb Y)^{\pi}|\mathbb Y)=0$

and

$\displaystyle \begin{array}{rcl} \mathbb E(fT^n(g-\mathbb E(g|\mathbb Y)^\pi)|\mathbb Y)&=&\mathbb E(fT^ng|\mathbb Y)- \mathbb E(f(S^n\mathbb E(g|\mathbb Y))^\pi|\mathbb Y)\\ & & \\&=&\mathbb E(fT^ng|\mathbb Y)- \mathbb E(f|\mathbb Y)S^n\mathbb E(g|\mathbb Y). \end{array}$

(${\Longleftarrow}$) Let ${g\in L^2(\mathbb X)}$ be ${T}$-invariant. First assume that ${\mathbb E(g|\mathbb Y)=0}$. Equation (3) gives that

$\displaystyle \lim_{N\rightarrow\infty}\dfrac{1}{N}\sum_{n=1}^N\int_Y\mathbb E(fg|\mathbb Y)=0\ \ \ \Longrightarrow\ \ \ \int_X fgd\mu=0\ ,\forall\,f\in L^2(\mathbb X),$

and so ${g=0}$. If ${g}$ is an arbitrary, the above argument implies that ${g-\mathbb E(g|\mathbb Y)^\pi=0}$. $\Box$

2. Weakly mixing extensions

Given ${\mathbb X\stackrel{\pi}{\longrightarrow}\mathbb Y}$, remember from ERT14 that the relative product mps ${\mathbb X\times_{\mathbb Y}\mathbb X}$ is also an extension of ${\mathbb Y}$, via any of the compositions  ${\pi\circ\theta_1}$ or ${\pi\circ\theta_2}$ (${\theta_1}$ and ${\theta_2}$ are the natural projections).

Definition 5 ${\mathbb X}$ is a weakly mixing extension of ${\mathbb Y}$ if ${\mathbb X\times_{\mathbb Y}\mathbb X\rightarrow\mathbb X}$ is ergodic.

Clearly, this extends Theorem 6 of ERT8: ${\mathbb X}$ is a weakly mixing system iff ${\mathbb X}$ is a weakly mixing extension of ${\mathbb X_0}$. Also, if ${\mathbb X\times_{\mathbb Y}\mathbb X\rightarrow\mathbb X\rightarrow\mathbb Y}$ is ergodic, then ${\mathbb X\rightarrow\mathbb Y}$ is ergodic, that is: every weakly mixing extension is ergodic. What we’ll do in the remaining of this post is to relativize the results of ERT8.

Lemma 6 Let ${\mathbb X\stackrel{\pi}{\longrightarrow}\mathbb Y}$ be a weakly mixing extension and let ${f,g\in L^\infty(\mathbb X)}$. Then

$\displaystyle \lim_{N\rightarrow\infty}\dfrac{1}{N}\sum_{n=1}^N\int_Y \left\{\mathbb E(fT^ng|\mathbb Y)-\mathbb E(f|\mathbb Y)S^n\mathbb E(g|\mathbb Y)\right\}^2d\nu=0.$

Proof: The idea is to apply Lemma 3 to the extension ${\mathbb X\times_{\mathbb Y}\mathbb X\rightarrow\mathbb Y}$. We divide the proof in two parts.

Part 1. ${\mathbb E(g|\mathbb Y)=0}$: as ${f,g\in L^\infty(\mathbb X)}$, we have ${f\otimes f,g\otimes g\in L^2(\mathbb X\times_{\mathbb Y}\mathbb X)}$. Observing that

$\displaystyle \mathbb E((f\otimes f)T^n(g\otimes g)|\mathbb Y)=\mathbb E(fT^ng\otimes fT^ng|\mathbb Y)= \mathbb E(fT^ng|\mathbb Y)^2,$

where in the last equality we used Proposition 14 of ERT14, Lemma 3 guarantees that

$\displaystyle \lim_{N\rightarrow\infty}\dfrac{1}{N}\sum_{n=1}^N\int_Y\mathbb E(fT^ng|\mathbb Y)^2d\nu=0.$

Part 2. ${g}$ is arbitrary: apply Part 1 to the functions ${f}$ and ${g-\mathbb E(g|\mathbb Y)^{\pi}}$. $\Box$

We now prove the relative version of Theorem 6 of ERT8.

Proposition 7 If ${\mathbb X_1\rightarrow\mathbb Y}$ is weakly mixing and ${\mathbb X_2\rightarrow\mathbb Y}$ ergodic, then ${\mathbb X_1\times_{\mathbb Y}\mathbb X_2\rightarrow\mathbb Y}$ is ergodic.

Proof: Let ${\mathbb X_i=(X_i,\mathcal A_i,\mu_i,T_i)}$, ${\mathbb Y=(Y,\mathcal B,\nu,S)}$, ${\mathbb X_i\stackrel{\pi_i}\longrightarrow\mathbb Y}$ and ${\mathbb X_1\times_{\mathbb Y}\mathbb X_2=(X,\mathcal A,\mu,T)\stackrel{\pi}\longrightarrow\mathbb Y}$. We’ll check (3). As ${L^\infty(\mathbb X_1)\otimes L^\infty(\mathbb X_2)}$ generates a dense subset of ${L^2(\mathbb X)}$, assume without loss of generality that ${f=f_1\otimes f_2,g=g_1\otimes g_2\in L^\infty(\mathbb X_1)\otimes L^\infty(\mathbb X_2)}$. Due to the bilinearity in the functions ${f_1,f_2}$, we can divide the proof into two parts.

Part 1. ${\mathbb E(f|\mathbb Y)=0}$: we want to check equation (3). Observe that

$\displaystyle \begin{array}{rcl} \displaystyle\int_Y \mathbb E(fT^ng|\mathbb Y)d\nu &=&\displaystyle\int_X (f_1\otimes f_2)\cdot({T_1}^ng_1\otimes {T_2}^ng_2)d\mu_1\times_{\mathbb Y}\mu_2\\ & &\\&=&\displaystyle\int_X (f_1{T_1}^ng_1)\otimes(f_2{T_2}^ng_2)d\mu_1\times_{\mathbb Y}\mu_2\\ & &\\&=&\displaystyle\int_Y \mathbb E((f_1{T_1}^ng_1)\otimes(f_2{T_2}^ng_2)|\mathbb Y)d\nu\\ & &\\ &=&\displaystyle\int_Y \mathbb E(f_1{T_1}^ng_1|\mathbb Y)\cdot\mathbb E(f_2{T_2}^ng_2|\mathbb Y)d\nu\\ & &\\ &=&\displaystyle\int_Y\{\mathbb E(f_1{T_1}^ng_1|\mathbb Y)- \mathbb E(f_1|\mathbb Y) S^n\mathbb E(g_1|\mathbb Y)\}\mathbb E(f_2{T_2}^ng_2|\mathbb Y)d\nu\\ & &\\ & &+\displaystyle\int_Y\mathbb E(f_1|\mathbb Y) S^n\mathbb E(g_1|\mathbb Y)\mathbb E(f_2{T_2}^ng_2|\mathbb Y)d\nu. \end{array}$

By Lemma 6, the sequence ${(\|\mathbb E(f_1{T_1}^ng_1|\mathbb Y)- \mathbb E(f_1|\mathbb Y)S^n\mathbb E(g_1|\mathbb Y)\|)_{n\ge 1}}$ converges a Cesàro to zero, and so, as ${(\|\mathbb E(f_2{T_2}^ng_2|\mathbb Y)\|)_{n\ge 1}}$ is uniformly bounded, (2) is equivalent to

$\displaystyle \lim_{N\rightarrow\infty}\dfrac{1}{N}\sum_{n=1}^N\int_Y\mathbb E(f_1|\mathbb Y) S^n\mathbb E(g_1|\mathbb Y)\mathbb E(f_2{T_2}^ng_2|\mathbb Y)d\nu=0.$

Now observe that

$\displaystyle \begin{array}{rcl} \mathbb E(f_1|\mathbb Y) S^n\mathbb E(g_1|\mathbb Y)\mathbb E(f_2{T_2}^ng_2|\mathbb Y) &=&\mathbb E\left(\mathbb E(f_1|\mathbb Y)^{\pi_2} \left(S^n\mathbb E(g_1|\mathbb Y)\right)^{\pi_2} (f_2{T_2}^ng_2)|\mathbb Y\right)\\ & &\\ &=&\mathbb E\left(\mathbb E(f_1|\mathbb Y)^{\pi_2}{T_2}^n\mathbb E(g_1|\mathbb Y)^{\pi_2}(f_2{T_2}^ng_2)|\mathbb Y\right)\\ & &\\ &=&\mathbb E((\mathbb E(f_1|\mathbb Y)^{\pi_2}f_2 {T_2}^n(\mathbb E(g_1|\mathbb Y)^{\pi_2}g_2)|\mathbb Y). \end{array}$

Then we want to prove that

$\displaystyle \lim_{N\rightarrow\infty}\dfrac{1}{N}\sum_{n=1}^N\int_X(\mathbb E(f_1|\mathbb Y)^{\pi_2}f_2){T_2}^n (\mathbb E(g_1|\mathbb Y)^{\pi_2}g_2)d\mu=0$

and this follows from the ergodicity assumption on ${\mathbb X_2}$ and Proposition 3, because

$\displaystyle \mathbb E(\mathbb E(f_1|\mathbb Y)^{\pi_2}f_2|\mathbb Y)=\mathbb E(f_1|\mathbb Y)\mathbb E(f_2|\mathbb Y) =\mathbb E(f|\mathbb Y)=0.$

Part 2. ${f_1, f_2}$ are arbitrary. Let ${h_1=f_1-\mathbb E(f_1|\mathbb Y)^{\pi_1}}$ and ${h_2=f_2-\mathbb E(f_2|\mathbb Y)^{\pi_2}}$. Then

$\displaystyle \begin{array}{rcl} f&=&h_1\otimes h_2+h_1\otimes\mathbb E(f_2|\mathbb Y)^{\pi_2}+\mathbb E(f_1|\mathbb Y)^{\pi_1}\otimes h_2+ \mathbb E(f_1|\mathbb Y)^{\pi_1}\otimes\mathbb E(f_2|\mathbb Y)^{\pi_2}\\ &=:&f_{11}+f_{12}+f_{21}+f_{22} \end{array}$

By Part 1, (2) holds for ${f_{11},f_{12}}$ and ${f_{21}}$. Finally, as ${f_{22}=\mathbb E(f_1\otimes f_2|\mathbb Y)^{\pi}}$, we have ${\tilde f_{22}=\mathbb E(\tilde f_{22}|\mathbb Y)^\pi}$ and in this case (2) is clearly valid. This concludes the proof. $\Box$

Corollary 8 If ${\mathbb X\rightarrow\mathbb Y}$ is weakly mixing, then ${\mathbb X\times_{\mathbb Y}\mathbb X\rightarrow\mathbb Y}$ is weakly mixing.

Proof: The assumption gives in particular that ${\mathbb X\rightarrow\mathbb Y}$ is ergodic. Three successive applications of Proposition 7 give that

$\displaystyle (\mathbb X\times_{\mathbb Y}\mathbb X)\times_{\mathbb Y}(\mathbb X\times_{\mathbb Y}\mathbb X)= (((\mathbb X\times_{\mathbb Y}\mathbb X)\times_{\mathbb Y}\mathbb X)\times_{\mathbb Y}\mathbb X)$

is ergodic and so ${\mathbb X\times_{\mathbb Y}\mathbb X\rightarrow\mathbb Y}$ is weakly mixing. $\Box$

Previous posts: ERT0, ERT1, ERT2, ERT3, ERT4, ERT5, ERT6, ERT7, ERT8, ERT9, ERT10, ERT11, ERT12, ERT13, ERT14.