According to ERT8 and ERT10, absolute notions of compact and weakly mixing can be also characterized in terms of eigenfunctions. Contrary to ERT15, in which we defined weakly mixing extensions via the behavior of ergodic averages, this post is devoted to the definition of compact extensions from the spectral viewpoint, whatever this means. To this matter, we investigate what happens to functions when we disintegrate then. By this we mean: what is the relation between and the various , ? Here, is an extension, is the disintegration of with respect to and , .

Obviously, by Rohklin’s theorem (see ERT14), if then for -a.e. . If is equal to regarded as a function in , the above implication simply means that there is an injective map

which is not onto. Actually, observe that satisfies

and not every element of has this property. Restricted to this condition, we’ll see that is bijective, by considering as a Borel Hilbert bundle over and proving that is an isomorphism between and the measurable square integrable sections of this bundle. By this bijection we mean that is a direct integral of the Borel Hilbert bundle .

Firstly, we develop general properties of Borel Hilbert bundles, for which is our toy model. Secondly, we investigate by means of its -modules and describe their respective “linear algebra”. Once this is done, three points will become clear:

- The relation between and .
- The concept of generalized eigenfunction.
- The qualitative concepts of compactness and weak mixing.

**1. Hilbert bundles **

Let be a measure space. A *Borel bundle* over is a pair where is a measure space and is a measurable onto map. is called the *base space* and the *bundle projection*. For each , is the *fiber* over , and it will be denoted by .

A map is a *section* if is the identity map on .

Let be a Borel bundle over such that each fiber is a complex Hilbert space, with , and denoting addition, scalar product and multiplication by a scalar respectively (). Then we say that is a *Borel Hilbert bundle* if

- The map is a measurable map from
into .

- is measurable on to .
- For each , the map is measurable on .

Now let us put the definition in our context. Let be a homomorphism. As usual, we let be the decomposition of into measure on the fibres over . We then define

and by the relation

Here as in the sequel an assertion made about is to be understood as valid for almost every with respect to the measure . Throughout our discussion we will assume that is ergodic. This will become necessary for each to have the same dimension, as we explain in the next paragraph.

The uniqueness of the decomposition of implies that and thus

so that is an isometry of to . By the ergodicity assumption on it follows that the spaces are isomorphic as Hilbert spaces, all having the same dimension .

**2. -modules **

In this section we investigate the structure of the Hilbert space , seen as a module over . This space is separable, by the usual assumption that is a Lebesgue space. Remember that each element defines a section defined almost everywhere simply by regarding as a function . From now on, we implicitly consider as a subset of and denote its elements simply as instead of .

Definition 1A closed subspace is called a-moduleif for each and any measurable function for which the product we have .

Alternatively, the above definition is equivalent to the requirement that is a –module: whenever and . The reason we take products with elements of is that this is the only way of making sure the result is again in .

If is a -module we denote by the image of in . (Formally, if is the canonical projection, then is the image of under . The maps and are both homomorphisms and so is a linear subspace of .) may be regarded too as a “bundle” of linear subspaces over . We say is of *finite rank* if a.e. and of rank if a.e..

The next results of this section develop the linear algebra of these modules.

*Proof:* Given , let . For a fixed , define

Clearly, is measurable and , that is, does the job in the set . If ranges over a dense set in , then . Moreover, if are such that and , then and . Thus, consecutive combinations in of the functions will give integral 1 on a set arbitrarily close to , and since is closed we can achieve 1 almost everywhere.

This lemma can be iterated to construct basis for modules of finite rank.

Definition 3Let be a -module. A set such that a.e. for every is called-orthonormal. If also spans as a -module, then we call it a-orthonormal basisof .

To say that spans as a -module means that every function in can be expressed as a convergent linear combination with coefficient function on .

Lemma 4If is a -module with a.e. then has a -orthonormal basis .

*Proof:* Choose using Lemma 2. Let . is also a -module and a.e. Now use induction. Once have been found, they form an orthonormal basis at each and for arbitrary we have

which proves the lemma.

In the case the “local dimension” is infinite, for example when , one has to exercise more care in the construction of a -orthonormal basis, since in passing to infinity one may exhaust certain subspaces before others. To avoid this we use the following

Lemma 5Let be a -module with a.e., and let be functions in . Then there exists a -submodule of rank such that .

*Proof:* We use induction on . If , let be the function obtained in Lemma 2 and consider the rank one -module generated by the function . Observe that a.e. and belongs to .

Now suppose the lemma has been proved for , and is a rank submodule containing . Let . Observe that if is a -orthonormal basis for , the function

belongs to . Consider a rank one submodule of containing (this follows from the case ). Then is a submodule of whose rank is . It obviously contains , and also .

Lemma 6Every -module with a.e. has a -orthonormal basis. In particular, has a -orthonormal basis.

*Proof:* The finite dimensional case is Lemma 4. Assume and let be a dense subset of (remember has a countable dense subset and so does any of its closed subspaces). Define inductively submodules such that a.e. and . Set

is a rank one submodule and thus we can find such that a.e. The are -orthonormal and spans . Hence the -module spanned by contains all and so coincides with .

**3. as a direct integral of **

We now proceed to obtain the first part of our goals: there is a bijective correspondence between and the measurable square integrable sections of . Fix a -orthonormal basis of . For each section , form the coefficient functions by

Definition 7ismeasurableif each is measurable. It ismeasurable square integrableif it is measurable and

We can equip the linear space of measurable square integrable sections of with the inner product

If we put , then and so is well-defined. It actually turns into an isometry from to its image.

Proposition 8There is a bijective correspondence between and the space of measurable square integrable sections of . Such correspondence is given by .

*Proof:* Given , each function is measurable, because it is the composition of measurable functions. Since

it follows that every is indeed measurable square integrable. Conversely, let be measurable square integrable. Then satisfies

which is finite, proving that . Clearly, and so the proof is complete.

Each defines a rank one -module and each section of defines a bundle of one dimensional subspaces of . By the previous proposition, rank one -module are in correspondence with bundles of one dimensional subspaces of . We can go further on this correspondence and obtain representatives for -modules of any finite rank of the same nature.

Let be the set of endomorphisms of . The union is again a Borel Hilbert bundle over . A section of this bundle will be called an *operator valued section*. We say an operator valued section is *measurable* if, for any measurable square integrable section , the section is measurable.

Definition 9Let be a map such that each is an -dimensional subspace of . We say is ameasurable -plane sectionif the operator valued section , with the orthogonal projection, is measurable.

Compare this definition with that of a metric in a Grassmanian: each subspace is identified to its orthogonal projection. Observe that one can not ask more than measurability for operator valued sections, since they are defined in . On the other hand, observers must be in the -space we are working with, that is, they are measurable square integrable sections.

Theorem 10For any fixed , there is a bijective correspondence between the collection of rank -modules in and the space of measurable -plane sections.

*Proof:* Let be a rank -module. Choose a -orthonormal basis for and observe that the operator valued section is measurable, since it is given by

Conversely, let be a measurable -plane section. Let be a -orthonormal basis for and define sections by . By assumption, each is square integrable and so, according to Proposition 8, is the image under of some . Let be the -module generated by . It is clear that .

**4. Generalized eigenfunctions **

We keep the setup of the previous section and assume is a -module invariant under , that is, a.e. By the ergodicity assumption on , the dimension of is constant. In particular, a finite rank -module is of rank .

Definition 11A function is ageneralized eigenfunctionwith respect to or a -eigenfunction if the module spanned by is of finite rank. Equivalently, is a generalized eigenfunction if it belongs to a -invariant -module of rank .

Let be a -orthonormal basis for . belongs to and so we can write

with . If is the vector valued function with components and , then

that is, is a vector valued eigenfunction, the “eigenvalue” being a matrix valued function on . In addition, is again a -orthonormal basis for and so is a unitary matrix. The conclusion is that all generalized eigenfunctions belong to -modules spanned by *unitary generalized eigenfunctions*. This fact implies that the bounded generalized eigenfunctions form an algebra containing . Indeed:

- Each function in is a generalized eigenfunction: obvious.
- is a generalized eigenfunction whenever are generalized eigenfunctions: assume with and with , and being both unitary. Then
belongs to the -module generated by . If is the vector valued function with components , then , where

is unitary.

- is a generalized eigenfunction whenever are generalized eigenfunctions: let and be finite rank -modules with bounded -orthonormal basis and , respectively. Then the -module spanned by the set has finite rank and contains the product of any two bounded eigenfunctions with and .

Definition 12If is a homomorphism, denote the subspace of spanned by the generalized eigenfunctions.

is clearly a -module. It contains as well as the subspace of -invariant functions on . In addition, we have the following

Lemma 13contains the dense algebra of bounded generalized eigenfunctions.

*Proof:* Let be a -module of finite rank and a vector valued function such that , with unitary. Consider the function and observe that

since the vectors and are related by a unitary matrix. This proves that is -invariant and so, for each , the set is -invariant. In these sets are bounded and then each is bounded. Furthermore, letting ,

and so generates a finite rank -module, proving that they are generalized eigenfunctions. Obviously, .

Algebras with the property of Lemma 13 can always be realized by intermediate factors.

Lemma 14Let be a Lebesgue space and an algebra, invariant under conjugation. If is the smallest -algebra of with respect to which all functions in are measurable, then

where the closure is taken over .

This result dates back to R. Zimmer in the paper

Extensions of ergodic group actions, *Illinois J. Math* **20** (1976), no. 3, 373–409,

where, independently, he has developed similar ideas to those of Furstenberg. We invite the reader to see the original work for a proof of this lemma.

**5. Compact extensions **

Definition 15We say that is acompact extensionif .

The last two result directly give the following.

Theorem 16Given , there exists an intermediate factor such that is a compact extension, with .

By definition, is the largest intermediate compact extension of in . In order to keep accordance with the theory already developed, a compact extension should represent the complementary structure of a weakly mixing extension. This is indeed the case.

Proposition 17The extension is weakly mixing if and only if it has no non-trivial generalized eigenfunctions.

In other words, weakly mixing extensions are those for which . This being true, compact and weakly mixing extensions are actually complementary when looked through the spectral lenses. It is a matter of fact that this dichotomy turns Furstenberg-Zimmer structural theorem direct, as we’ll prove in ERT17.

We postpone the proof of Proposition 17 to a future post. Basically, it follows the ideas of Theorem 8 from ERT8: once there is a generalized eigenfunction, we construct a non-trivial -invariant function, but the technicalities are much harder and require the machinery of Hibert-Schmidt operators.

This post of Terence Tao has much in common with our discussion.

**Previous posts:** ERT0, ERT1, ERT2, ERT3, ERT4, ERT5, ERT6, ERT7, ERT8, ERT9, ERT10, ERT11, ERT12, ERT13, ERT14, ERT15.

Dear Matheus, I’m confused about the definition of $M_y$ in this post. It’s defined as the image of (a Y-module) $M$ under $p_y\circ \Gamma$, where $\Gamma$ is the bijection between $L^2(X)$ and the set of cross-sections from $Y$ to the Hilbert bundle (mod zero measure sets). What I can’t see is the well-definition of $p_y$ (even on a full measure set of y’s) since it must be defined on equivalence classes of cross-sections.

By:

Jaramilloon July 28, 2015at 9:34 pm

Dear Jaramillo,

I’m not sure that I understood your question: the canonical projection is well-defined independently of measure-theoretical considerations.

Indeed, if we think of each as an abstract space , then the canonical projection is well-defined because the same is true for the projection from the abstract product space to consisting into taking the -th coordinate of a sequence .

Best, Matheus

By:

matheuscmsson July 29, 2015at 11:45 am

Dear Matheus,

I meant that the image of is not , but rather where is the -a.s. equal equivalence relationship. On this set the projection is not defined.

Put differently, if and we decompose (i.e., ) then the “projection” of at () is not defined (of course, for each fixed it is defined for a.e. , but these full measures ‘s depend on , and there are uncountably many of them -and I don’t see how taking a dense, countable subset of $M$ will do the job-).

By:

Jaramilloon July 29, 2015at 9:34 pm

Dear Jaramillo,

I see your point now. I think you are right that the map is not well-defined as it is presented in the post. For this reason, I wrote to the author (Yuri Lima) of this guest post asking for clarifications and he promise to reply to your comment in the near future.

Best, Matheus

By:

matheuscmsson July 31, 2015at 3:07 pm