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 and , we want to investigate when they are compatible in some sense. There are three main notions in this respect:
- Conjugacy between boolean algebras.
- Spectral equivalence.
The main reference of this lecture is Lectures on Ergodic Theory, from Paul Halmos.
Definition 1 Two mps and are similar if there exists an invertible bimeasurable transformation such that
In this case, is called an isomorphism.
Notice that only needs to be defined in almost every point of . The main invariant of similarity is metric entropy.
Theorem 2 If and are similar, then the entropy of with respect to is equal to the entropy of with respect to .
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
of all subsets of measure zero. This is an ideal of in the following sense:
- 1. If and , then .
- 2. If , then .
These properties allow the definition of the quocients
is a set of equivalence classes defined by the equivalence relation
and, if is the equivalence class of ,
defines a function on . Obviously, these do not depend on the representant of the class.
that is, is the only element of zero measure and, by the same reason, is the only element of total measure.
Definition 4 Two probability spaces and are conjugate if there exists an invertible bimeasurable map , called boolean isomorphism, such that
- is an isomorphism in the boolean algebra category: it preserves union, intersection and complement.
A measure-preserving transformation on induces in a natural way a mapping of into itself: the image of an equivalence class under this mapping is defined by selecting a representative and forming the equivalence class of ,
The measure-preserving character of 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 preserves , this is well-defined. Observe that is not a map sending points of to . Instead, it sends classes of subsets of to classes of subsets of .
Definition 5 Two mps and are conjugate if there exists an invertible bimeasurable map such that
is also called a boolean isomorphism, being implicit and .
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
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 and are spectrally equivalent if there is a unitary invertible operator such that
Proposition 8 Conjugacy implies spectral equivalence.
Proof: Let and be two conjugate mps and the conjugacy. The map defines an isometry of the characteristic functions of into . Because the characteristic functions generate a dense set in , the linear extension of the above association defines an an isometry , which is surjective because is so.
If is an isomorphism between the mps and , then the induced unitary map from to preserves more than the norm and the linear structure of the spaces. The more comes from the fact that the elements of are not merely abstract vectors: being functions, they also have multiplicative properties. If are bounded, then also belongs to , as well as , are bounded. In this case, sends bounded function to bounded functions and preserves their product. These properties turn out to be sufficient when going from spectral equivalence to similarity.
- (a) and send bounded functions to bounded functions;
- (b) , for every bounded functions .
Then and are conjugate.
Proof: Let be the characteristic function of . Because , and so is the characteristic function of a set . Also, if , then , such that . That is clear, because is unitary. is surjective because (as is multiplicative), is a characteristic function, for every .
preserves intersection because it is multiplicative. Also, if are the characteristic functions of , then is the characteristic function of , which shows that preserves union.
Exercise 2 In the notation of the above proposition, prove that is multiplicative.
Ok. If we have such in hands, the systems are conjugate. But when does exist?
Definition 10 The spectrum of is the set
If conjugates and and , say , then
that is, . This means spectral equivalence implies the equality . In the case of ergodic compact systems (discrete spectrum), this is a sufficient condition.
Exercise 3 Suppose and are unitary operators with discrete spectrum, where , are Hilbert spaces. Prove that and are equivalent iff .
- and are similar;
- and are conjugate;
- and are spectrally equivalent;
Proof: (a) (b) (c) (d) are clear. (b) (a) is Theorem 6. (d) (c) is the previous exercise. Let us prove that (c) implies (b) by checking the conditions of Proposition 9. For each , let , be the unitary eigenfunctions of , associated to , respectively. Because and are ergodic, they are unique up to scalar multiple. We make the following
Observe that, in general, this is not the case. Actually, what we have is that and are both eigenfunctions of so that . The assumption says we can assume . The reader can check this on page 46 of Lectures on Ergodic Theory.
Let linear such that
By definition, . Also,
and so, by linearity and a denseness argument is multiplicative.
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 be the mps and . Also, let be the character group of . Consider the element such that
It is clear that . It defines a rotation , , that preserves the Haar measure of . is discrete because of the properties of the characters of . In fact, they form an orthonormal basis of and, if is one of them, then
that is, is an eigenfunction whose eigenvalue is . By Pontryagin duality, the dual group is canonically isomorphic to . Such correspondence shows that the spectrum of is exactly . Moreover, by the same reason, every eigenvalue is simple, so that is ergodic. This concludes the proof.