Posted by: yglima | April 16, 2010

## ERT11: Conjugation, equivalence and similarity of measure-preserving systems

In order to characterize compact systems, we discuss the notions of conjugacy between measure-preserving systems. We will do much more than needed for this matter, as I think these notions are important for any ergodic theorist.

Given two mps ${\mathbb X=(X,\mathcal A,\mu,T)}$ and ${\mathbb Y=(Y,\mathcal B,\nu,S)}$, we want to investigate when they are compatible in some sense. There are three main notions in this respect:

1. Similarity.
2. Conjugacy between boolean algebras.
3. Spectral equivalence.

The main reference of this lecture is Lectures on Ergodic Theory, from Paul Halmos.

1. Similarity

Definition 1 Two mps ${\mathbb X}$ and ${\mathbb Y}$ are similar if there exists an invertible bimeasurable transformation ${\Phi:(X,\mathcal A,\mu)\rightarrow(Y,\mathcal B,\nu)}$ such that$\displaystyle \nu=\Phi_{*}\mu\ \ \text{ and }\ \ \Phi\circ T=S\circ\Phi.$

In this case, ${\Phi}$ is called an isomorphism.

Notice that ${\Phi}$ only needs to be defined in almost every point of ${X}$. The main invariant of similarity is metric entropy.

Theorem 2 If ${\mathbb X}$ and ${\mathbb Y}$ are similar, then the entropy of ${T}$ with respect to ${\mu}$ is equal to the entropy of ${S}$ with respect to ${\nu}$.

The above result gave the first proof that the two-symbol and three-symbol shifts, both endowed with the natural metrics, are not similar. Similarity is the most usual notion we will discuss. Because of this, many can be found in any book of ergodic theory.

2. Conjugacy between boolean algebras

In many situations, the difficulty in constructing isomorphic bijections from isomorphisms is to deal with sets of measure zero. Because of this, we discard them by considering the set

$\displaystyle \mathcal I=\{A\in\mathcal A\,;\;\mu(A)=0\}$

of all subsets of measure zero. This is an ideal of ${\mathcal A}$ in the following sense:

1. 1. If ${A\subset B}$ and ${B\in\mathcal I}$, then ${A\in\mathcal I}$.
2. 2. If ${(A_n)_n\subset\mathcal I}$, then ${\bigcup_n A_n\in\mathcal I}$.

These properties allow the definition of the quocients

$\displaystyle \hat{\mathcal A}=\mathcal A/\mathcal I\text{\ \ \ and\ \ \ }\hat\mu=\mu/\mathcal I.$

${\hat{\mathcal A}}$ is a set of equivalence classes defined by the equivalence relation

$\displaystyle A\sim B\ \iff\ \mu(A\Delta B)=0$

and, if ${[A]}$ is the equivalence class of ${A}$,

$\displaystyle \hat\mu([A])=\mu(A).$

defines a function on ${B}$. Obviously, these do not depend on the representant of the class.

Definition 3 ${\hat{\mathcal A}}$ is the boolean algebra and ${(\hat{\mathcal A},\hat\mu)}$ the measure algebra of the probability space ${(X,\mathcal A,\mu)}$.

${\hat{\mathcal A}}$ is a boolean algebra under the natural boolean operations for sets: union, intersection and complement. It is clear that

$\displaystyle \hat\mu([A])=0\ \iff\ [A]=[\emptyset], \ \ \ \ \ (1)$

that is, ${[\emptyset]}$ is the only element of zero measure and, by the same reason, ${[X]}$ is the only element of total measure.

Definition 4 Two probability spaces ${(X,\mathcal A,\mu)}$ and ${(Y,\mathcal B,\nu)}$ are conjugate if there exists an invertible bimeasurable map ${\Phi:(\hat{\mathcal A},\hat\mu)\rightarrow(\hat{\mathcal B},\hat\nu)}$, called boolean isomorphism, such that

1. ${\Phi}$ is an isomorphism in the boolean algebra category: it preserves union, intersection and complement.
2. ${\Phi_*\hat\mu=\hat\nu}$.

A measure-preserving transformation ${T}$ on ${X}$ induces in a natural way a mapping of ${\hat{\mathcal A}}$ into itself: the image of an equivalence class under this mapping is defined by selecting a representative ${A}$ and forming the equivalence class of ${T^{-1}A}$,

$\displaystyle \hat T[A]=[T^{-1}A].$

The measure-preserving character of ${T}$ implies that the image class is unambiguously determined by this process and that the measure of the image class is the same as the measure of the original one. Because ${T}$ preserves ${\mu}$, this is well-defined. Observe that ${\hat T}$ is not a map sending points of ${X}$ to ${X}$. Instead, it sends classes of subsets of ${X}$ to classes of subsets of ${X}$.

Definition 5 Two mps ${\mathbb X=(X,\mathcal A,\mu,T)}$ and ${\mathbb Y=(Y,\mathcal B,\nu,S)}$ are conjugate if there exists an invertible bimeasurable map ${\Phi:(\hat{\mathcal A},\hat\mu)\rightarrow(\hat{\mathcal B},\hat\nu)}$ such that$\displaystyle \Phi_*\hat\mu=\hat\nu\ \ \ \text{ and }\ \ \ \Phi\circ\hat T=\hat S\circ\Phi.$

${\Phi}$ is also called a boolean isomorphism, being implicit ${T}$ and ${S}$.

Exercise 1 Prove that similarity implies conjugacy.

Conjugacy does not imply similarity in general. There exist some highly pathological measure spaces that are in a certain vague sense absolutely non-measurable. One way that this pathology shows up is that these measure spaces do not possess sufficiently many measure-preserving transformations to induce all the desired boolean isomorphisms, that is, if the boolean algebra is richer than the original space. On the other hand, conjugagy does imply similarity in all decent measure spaces. By decent I mean Lebesgue spaces, which is a usual assumption in ergodic theory. From now on, we only consider Lebesgue spaces, so that we have the

Theorem 6 (von Neumann) Conjugacy implies similarity.

The proof of the above theorem may be found in the book Ergodic Theory and Differentiable Dynamics by Ricardo Mañé.

3. Spectral equivalence

Definition 7 Two mps ${\mathbb X=(X,\mathcal A,\mu,T)}$ and ${\mathbb Y=(Y,\mathcal B,\nu,S)}$ are spectrally equivalent if there is a unitary invertible operator ${U:L^2(\mu)\rightarrow L^2(\nu)}$ such that$\displaystyle U_S\circ U=U\circ U_T.$

Proposition 8 Conjugacy implies spectral equivalence.

Proof: Let ${\mathbb X=(X,\mathcal A,\mu,T)}$ and ${\mathbb Y=(Y,\mathcal B,\nu,S)}$ be two conjugate mps and ${\Phi:(\hat{\mathcal A},\hat\mu)\rightarrow(\hat{\mathcal B},\hat\nu)}$ the conjugacy. The map ${\chi_A\mapsto \chi_{\Phi[A]}}$ defines an isometry of the characteristic functions of ${L^2(\mu)}$ into ${L^2(\nu)}$. Because the characteristic functions generate a dense set in ${L^2(\mu)}$, the linear extension of the above association defines an an isometry ${U:L^2(\mu)\rightarrow L^2(\nu)}$, which is surjective because ${\Phi}$ is so. $\Box$

If ${\Phi}$ is an isomorphism between the mps ${\mathbb X}$ and ${\mathbb Y}$, then the induced unitary map from ${L^2(\mu)}$ to ${L^2(\nu)}$ preserves more than the norm and the linear structure of the spaces. The more comes from the fact that the elements of ${L^2(\mu)}$ are not merely abstract vectors: being functions, they also have multiplicative properties. If ${f,g\in L^2(\mu)}$ are bounded, then ${fg}$ also belongs to ${L^2(\mu)}$, as well as ${Uf}$, ${Ug}$ are bounded. In this case, ${U}$ sends bounded function to bounded functions and preserves their product. These properties turn out to be sufficient when going from spectral equivalence to similarity.

Proposition 9 (Multiplicative theorem) Let ${\mathbb X=(X,\mathcal A,\mu,T)}$ and ${\mathbb Y=(Y,\mathcal B,\nu,S)}$ be mps and ${U:L^2(\mu)\rightarrow L^2(\nu)}$ a surjective unitary linear map such that

1. (a) ${U}$ and ${U^{-1}}$ send bounded functions to bounded functions;
2. (b) ${U(fg)=Uf\cdot Ug}$, for every bounded functions ${f,g}$.

Then ${\mathbb X}$ and ${\mathbb Y}$ are conjugate.

Proof: Let ${f}$ be the characteristic function of ${A\in\mathcal A}$. Because ${f^2=f}$, ${(Uf)^2=Uf}$ and so ${Uf}$ is the characteristic function of a set ${\Phi A}$. Also, if ${[A]=[A']}$, then ${[\Phi A]=[\Phi A']}$, such that ${\Phi:(\hat{\mathcal A},\hat\mu)\rightarrow(\hat{\mathcal B},\hat\nu)}$. That ${\Phi_*\hat\mu=\hat\nu}$ is clear, because ${U}$ is unitary. ${\Phi}$ is surjective because (as ${U^{-1}}$ is multiplicative), ${U^{-1}B}$ is a characteristic function, for every ${B\in\mathcal B}$.

${\Phi}$ preserves intersection because it is multiplicative. Also, if ${f,g}$ are the characteristic functions of ${A,B\in\mathcal A}$, then ${f+g-fg}$ is the characteristic function of ${A\cup B}$, which shows that ${U}$ preserves union. $\Box$

Exercise 2 In the notation of the above proposition, prove that ${U^{-1}}$ is multiplicative.

Ok. If we have such ${U}$ in hands, the systems are conjugate. But when does ${U}$ exist?

Definition 10 The spectrum of ${T}$ is the set$\displaystyle \sigma(T)=\{\lambda\in\mathbb C\,;\,\exists\,f\in L^2(\mu)\text{ such that }U_Tf=\lambda f\}.$

If ${U}$ conjugates ${U_T}$ and ${U_S}$ and ${\lambda\in\sigma(T)}$, say ${U_T=\lambda f}$, then

$\displaystyle U_SUf=UU_Tf=U(\lambda f)=\lambda Uf,$

that is, ${\lambda\in\sigma(S)}$. This means spectral equivalence implies the equality ${\sigma(T)=\sigma(S)}$. In the case of ergodic compact systems (discrete spectrum), this is a sufficient condition.

Exercise 3 Suppose ${U:\mathcal H\rightarrow\mathcal H}$ and ${U':\mathcal H'\rightarrow\mathcal H'}$ are unitary operators with discrete spectrum, where ${\mathcal H}$, ${\mathcal H'}$ are Hilbert spaces. Prove that ${U}$ and ${U'}$ are equivalent iff ${\sigma(U)=\sigma(U')}$.

Theorem 11 (Discrete spectrum theorem) If ${\mathbb X=(X,\mathcal A,\mu,T)}$ and ${\mathbb Y=(Y,\mathcal B,\nu,S)}$ are ergodic transformations with discrete spectrum, the following assertions are equivalent:

1. ${T}$ and ${S}$ are similar;
2. ${T}$ and ${S}$ are conjugate;
3. ${T}$ and ${S}$ are spectrally equivalent;
4. ${\sigma(T)=\sigma(S)}$.

Proof: (a) ${\Longrightarrow}$ (b) ${\Longrightarrow}$ (c) ${\Longrightarrow}$ (d) are clear. (b) ${\Longrightarrow}$ (a) is Theorem 6. (d) ${\Longrightarrow}$ (c) is the previous exercise. Let us prove that (c) implies (b) by checking the conditions of Proposition 9. For each ${\lambda\in\sigma(T)}$, let ${f_\lambda}$, ${g_\lambda}$ be the unitary eigenfunctions of ${U_T}$, ${U_S}$ associated to ${\lambda}$, respectively. Because ${T}$ and ${S}$ are ergodic, they are unique up to scalar multiple. We make the following

Assumption. ${f_{\lambda\eta}=f_\lambda f_\eta}$.

Observe that, in general, this is not the case. Actually, what we have is that ${f_{\lambda\eta}}$ and ${f_\lambda f_\eta}$ are both eigenfunctions of ${\lambda\eta}$ so that ${f_{\lambda\eta}=r(\lambda,\eta)f_\lambda f_\eta}$. The assumption says we can assume ${r(\lambda,\nu)=1}$. The reader can check this on page 46 of Lectures on Ergodic Theory.

Let ${U:L^2(\mu)\rightarrow L^2(\nu)}$ linear such that

$\displaystyle Uf_\lambda=g_\lambda,\ \lambda\in\sigma(T).$

By definition, ${U_T\circ U=U\circ U_S}$. Also,

$\displaystyle U(f_\lambda f_\eta)=Uf_{\lambda\eta}=g_{\lambda\eta}=g_\lambda g_\eta$

and so, by linearity and a denseness argument ${U}$ is multiplicative. $\Box$

4. The representation theorem

We finally arrive at the expected result.

Theorem 12 (Representation theorem) An ergodic and compact mps is similar to a rotation on a compact abelian group.

Proof: The idea is to construct an ergodic and compact rotation with the same spectrum as the original one. The discrete spectrum theorem will guarantee the conclusion. Let ${\mathbb X=(X,\mathcal A,\mu,T)}$ be the mps and ${\Lambda=\sigma(T)}$. Also, let ${Y}$ be the character group of ${\Lambda}$. Consider the element ${z\in Y}$ such that

$\displaystyle z(\lambda)=\lambda,\ \lambda\in\Lambda.$

It is clear that ${z\in Y}$. It defines a rotation ${S:Y\rightarrow Y}$, ${Sx=zx}$, that preserves the Haar measure ${\nu}$ of ${Y}$. ${\sigma(S)}$ is discrete because of the properties of the characters of ${Y}$. In fact, they form an orthonormal basis of ${L^2(\nu)}$ and, if ${f}$ is one of them, then

$\displaystyle f(Sx)=f(zx)=f(z)\cdot f(x),$

that is, ${f}$ is an eigenfunction whose eigenvalue is ${f(z)}$. By Pontryagin duality, the dual group ${\hat{Y}}$ is canonically isomorphic to ${\Lambda}$. Such correspondence shows that the spectrum of ${S}$ is exactly ${\Lambda}$. Moreover, by the same reason, every eigenvalue is simple, so that ${S}$ is ergodic. This concludes the proof. $\Box$

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