Posted by: yglima | May 1, 2011

## ERT16: Compact extensions

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 ${L^2(\mathbb X)}$ and the various ${L^2(\mathbb X_y)}$, ${y\in Y}$? Here, ${\mathbb X\rightarrow\mathbb Y}$ is an extension, ${\mu=\int_Y \mu_yd\nu(y)}$ is the disintegration of ${\mu}$ with respect to ${\nu}$ and ${\mathbb X_y=(X,\mathcal A,\mu_y,T)}$, ${y\in Y}$.

Obviously, by Rohklin’s theorem (see ERT14), if ${f\in L^2(\mathbb X)}$ then ${f\in L^2(\mathbb X_y)}$ for ${\nu}$-a.e. ${y\in Y}$. If ${f_y}$ is equal to ${f}$ regarded as a function in ${L^2(\mathbb X_y)}$, the above implication simply means that there is an injective map

$\displaystyle \begin{array}{ccrcl} \Gamma&:&L^2(\mathbb X)&\longrightarrow&\displaystyle\prod_{y\in Y}L^2(\mathbb X_y)\\ & & & &\\ & & f&\longmapsto &\displaystyle\prod_{y\in Y}\{f_y\} \end{array}$

which is not onto. Actually, observe that ${\Gamma(f)}$ satisfies

$\displaystyle \int_Y\left(\int_X |f_y|^2d\mu_y\right)d\nu(y)=\int_X |f|^2d\mu<\infty$

and not every element of ${\prod_{y\in Y}L^2(\mathbb X_y)}$ has this property. Restricted to this condition, we’ll see that ${\Gamma}$ is bijective, by considering ${\bigcup_{y\in Y}L^2(\mathbb X_y)}$ as a Borel Hilbert bundle over ${Y}$ and proving that ${\Gamma}$ is an isomorphism between ${L^2(\mathbb X)}$ and the measurable square integrable sections of this bundle. By this bijection we mean that ${L^2(\mathbb X)}$ is a direct integral of the Borel Hilbert bundle ${\bigcup_{y\in Y}L^2(\mathbb X_y)}$.

Firstly, we develop general properties of Borel Hilbert bundles, for which ${\bigcup_{y\in Y}L^2(\mathbb X_y)}$ is our toy model. Secondly, we investigate ${L^2(\mathbb X)}$ by means of its ${\mathbb Y}$-modules and describe their respective “linear algebra”. Once this is done, three points will become clear:

1. The relation between ${L^2(\mathbb X)}$ and ${\bigcup_{y\in Y}L^2(\mathbb X_y)}$.
2. The concept of generalized eigenfunction.
3. The qualitative concepts of compactness and weak mixing.

1. Hilbert bundles

Let ${(Y,\mathcal B)}$ be a measure space. A Borel bundle over ${Y}$ is a pair ${(B,p)}$ where ${B}$ is a measure space and ${p:B\rightarrow Y}$ is a measurable onto map. ${Y}$ is called the base space and ${p}$ the bundle projection. For each ${y\in Y}$, ${p^{-1}(y)}$ is the fiber over ${y}$, and it will be denoted by ${B_y}$.

A map ${\gamma:Y\rightarrow B}$ is a section if ${p\circ\gamma}$ is the identity map on ${Y}$.

Let ${(\dot{\mathcal H},p)}$ be a Borel bundle over ${Y}$ such that each fiber ${\mathcal H_y}$ is a complex Hilbert space, with ${+}$, ${\langle\ \,,\ \rangle=\langle\ \,,\ \rangle_y}$ and ${\lambda\cdot h}$ denoting addition, scalar product and multiplication by a scalar respectively (${\lambda\in\mathbb C,h\in\mathcal H_y}$). Then we say that ${(\dot{\mathcal H},p)}$ is a Borel Hilbert bundle if

1. The map ${(g,h)\mapsto\langle g,h\rangle_y}$ is a measurable map from

$\displaystyle \dot{\mathcal H}\times_Y\dot{\mathcal H}=\bigcup_{y\in Y}\mathcal H_y\times\mathcal H_y$

into ${\mathbb C}$.

2. ${(g,h)\mapsto g+h}$ is measurable on ${\dot{\mathcal H}\times_Y\dot{\mathcal H}}$ to ${\dot{\mathcal H}}$.
3. For each ${\lambda\in\mathbb C}$, the map ${h\mapsto \lambda\cdot h}$ is measurable on ${\dot{\mathcal H}}$.

Now let us put the definition in our context. Let ${\mathbb X\stackrel{\pi}\longrightarrow\mathbb Y}$ be a homomorphism. As usual, we let ${\mu=\int_Y \mu_yd\nu(y)}$ be the decomposition of ${\mu}$ into measure on the fibres over ${Y}$. We then define

$\displaystyle \mathcal H_y=L^2(\mathbb X_y)\ \ ,\ \ \dot{\mathcal H}=\bigcup_{y\in Y}L^2(\mathbb X_y)$

and ${p:\dot{\mathcal H}\rightarrow Y}$ by the relation

$\displaystyle p(f)=y\ \iff\ f\in \mathcal H_y.$

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

The uniqueness of the decomposition of ${\mu}$ implies that ${\mu_{Sy}=T_*\mu_y}$ and thus

$\displaystyle \int_X fd\mu_{Sy}=\int_X Tfd\mu_y,$

so that ${f\mapsto Tf}$ is an isometry of ${\mathcal H_{Sy}}$ to ${\mathcal H_y}$. By the ergodicity assumption on ${\mathbb Y}$ it follows that the spaces ${\mathcal H_y}$ are isomorphic as Hilbert spaces, all having the same dimension ${\le\infty}$.

2. ${\mathbb Y}$-modules

In this section we investigate the structure of the Hilbert space ${L^2(\mathbb X)}$, seen as a module over ${L^\infty(\mathbb Y)}$. This space is separable, by the usual assumption that ${\mathbb X}$ is a Lebesgue space. Remember that each element ${f\in L^2(\mathbb X)}$ defines a section ${\Gamma(f):y\rightarrow\mathcal H_y}$ defined almost everywhere simply by regarding ${f}$ as a function ${f_y\in\mathcal H_y}$. From now on, we implicitly consider ${L^\infty(\mathbb Y)}$ as a subset of ${L^2(\mathbb X)}$ and denote its elements simply as ${h}$ instead of ${h^\pi}$.

Definition 1 A closed subspace ${M\subset L^2(\mathbb X)}$ is called a ${\mathbb Y}$-module if for each ${f\in M}$ and any measurable function ${h\in L^2(\mathbb Y)}$ for which the product ${hf\in\mathcal H}$ we have ${hf\in M}$.

Alternatively, the above definition is equivalent to the requirement that ${M}$ is a ${L^\infty(\mathbb Y)}$module: ${hf\in M}$ whenever ${f\in M}$ and ${h\in L^\infty(\mathbb Y)}$. The reason we take products with elements of ${L^\infty(\mathbb Y)}$ is that this is the only way of making sure the result is again in ${L^2(\mathbb X)}$.

If ${M}$ is a ${\mathbb Y}$-module we denote by ${M_y\subset\mathcal H_y}$ the image of ${M}$ in ${\mathcal H_y}$.  (Formally, if ${p_y:\prod_{z\in Y}L^2(\mathbb X_z)\rightarrow L^2(\mathbb X_y)}$ is the canonical projection, then ${M_y}$ is the image of ${M}$ under ${p_y\circ\Gamma}$. The maps ${p_y}$ and ${\Gamma}$ are both homomorphisms and so ${M_y}$ is a linear subspace of ${\mathcal H_y}$.) ${M}$ may be regarded too as a “bundle” of linear subspaces over ${Y}$. We say ${M}$ is of finite rank if ${{\rm dim }M_y\le r<\infty}$ a.e. and of rank ${r}$ if ${{\rm dim }M_y=r<\infty}$ a.e..

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

Lemma 2 If ${{\rm dim }M_y\ge 1}$ a.e., then there exists ${f\in M}$ with ${\|f\|_y=1}$ a.e..

Proof: Given ${g\in M}$, let ${A_g=\left\{y\in Y\,;\, \|g\|_y>0\right\}}$. For a fixed ${g}$, define

$\displaystyle \begin{array}{ccll} h(y)&=&1/\|g\|_y&\text{, if }y\in A_g,\\ & & &\\ &=&0 &\text{, if }y\not\in A_g. \end{array}$

Clearly, ${h}$ is measurable and ${\|gh\|_y=\chi_{A_g}(y)}$, that is, ${f=gh}$ does the job in the set ${A_g}$. If ${\{g_n\}}$ ranges over a dense set in ${M}$, then ${\nu\left(\bigcup A_{g_n}\right)=1}$. Moreover, if ${f_1,f_2\in M}$ are such that ${\|f_1\|_y=\chi_{A_{g_1}}(y)}$ and ${\|f_2\|_y=\chi_{A_{g_2}}(y)}$, then ${f_1+\chi_{A_{g_2}\backslash A_{g_1}}f_2\in M}$ and ${\|f_1+\chi_{A_{g_2}\backslash A_{g_1}}f_2\|_y=\chi_{A_{g_1}\cup A_{g_2}}(y)}$. Thus, consecutive combinations in ${M}$ of the functions ${\{f_n\}}$ will give integral 1 on a set arbitrarily close to ${Y}$, and since ${M}$ is closed we can achieve 1 almost everywhere. $\Box$

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

Definition 3 Let ${M}$ be a ${\mathbb Y}$-module. A set ${L\subset M}$ such that ${\langle f,g\rangle_y=\delta_{fg}}$ a.e. for every ${f,g\in L}$ is called ${\mathbb Y}$-orthonormal. If ${L}$ also spans ${M}$ as a ${\mathbb Y}$-module, then we call it a ${\mathbb Y}$-orthonormal basis of ${M}$.

To say that ${L}$ spans ${M}$ as a ${\mathbb Y}$-module means that every function in ${M}$ can be expressed as a convergent linear combination with coefficient function on ${Y}$.

Lemma 4 If ${M}$ is a ${\mathbb Y}$-module with ${{\rm dim }M_y=r<\infty}$ a.e. then ${M}$ has a ${\mathbb Y}$-orthonormal basis ${\{f_1,\ldots,f_r\}}$.

Proof: Choose ${f_1}$ using Lemma 2. Let ${M'=\{f\in M\,;\, \langle f,f_1\rangle_y=0\text{ a.e.}\}}$. ${M'}$ is also a ${\mathbb Y}$-module and ${{\rm dim }M_y=r-1}$ a.e. Now use induction. Once ${f_1,\ldots,f_r}$ have been found, they form an orthonormal basis at each ${M_y}$ and for arbitrary ${f\in M}$ we have

$\displaystyle f=\sum_{i=1}^r\langle f,f_i\rangle_y f_i=\sum_{i=1}^r c_i(y)f_i\ ,$

which proves the lemma. $\Box$

In the case the “local dimension” ${{\rm dim }M_y}$ is infinite, for example when ${M=L^2(\mathbb X)}$, one has to exercise more care in the construction of a ${\mathbb Y}$-orthonormal basis, since in passing to infinity one may exhaust certain subspaces before others. To avoid this we use the following

Lemma 5 Let ${M}$ be a ${\mathbb Y}$-module with ${{\rm dim }M_y\ge r}$ a.e., and let ${g_1,\ldots,g_r}$ be functions in ${M}$. Then there exists a ${\mathbb Y}$-submodule ${N\subset M}$ of rank ${r}$ such that ${g_1,\ldots,g_r\in N}$.

Proof: We use induction on ${r}$. If ${r=1}$, let ${f\in M}$ be the function obtained in Lemma 2 and consider ${N\subset M}$ the rank one ${\mathbb Y}$-module generated by the function ${f_1=g_1+\chi_{Y\backslash A_{g_1}}f}$. Observe that ${{\rm dim}N_y=1}$ a.e. and ${g_1=\chi_{A_{g_1}}f_1}$ belongs to ${N}$.

Now suppose the lemma has been proved for ${r-1}$, and ${N'\subset M}$ is a rank ${r-1}$ submodule containing ${g_1,\ldots,g_{r-1}}$. Let ${M'=\{f\in M\,;\,\langle f,g\rangle_y=0\text{ a.e. for all }g\in N'\}}$. Observe that if ${f_1,\ldots,f_{r-1}}$ is a ${\mathbb Y}$-orthonormal basis for ${N'}$, the function

$\displaystyle \tilde g_r=g_r-\sum_{i=1}^{r-1}\langle g_r,f_i\rangle_y f_i=g_r-\sum_{i=1}^{r-1}c_i(y)f_i$

belongs to ${M'}$. Consider ${M''}$ a rank one submodule of ${M'}$ containing ${\tilde g_r}$ (this follows from the case ${r=1}$). Then ${N=N'+M''}$ is a submodule of ${M}$ whose rank is ${r}$. It obviously contains ${g_1,\ldots,g_{r-1}}$, and also ${g_r=\sum_{i=1}^{r-1}c_i(y)f_i+\tilde g_r}$. $\Box$

Lemma 6 Every ${\mathbb Y}$-module ${M\subset L^2(\mathbb X)}$ with ${{\rm dim}M_y=\text{const.}\le\infty}$ a.e. has a ${\mathbb Y}$-orthonormal basis. In particular, ${L^2(\mathbb X)}$ has a ${\mathbb Y}$-orthonormal basis.

Proof: The finite dimensional case is Lemma 4. Assume ${{\rm dim}M_y=\infty}$ and let ${\{g_n\}}$ be a dense subset of ${M}$ (remember ${L^2(\mathbb X)}$ has a countable dense subset and so does any of its closed subspaces). Define inductively submodules ${N_1\subset N_2\subset\cdots}$ such that ${{\rm dim}N_{n,y}=n}$ a.e. and ${g_1,\ldots,g_n\in N_n}$. Set

$\displaystyle M_n'=\{f\in N_n\,;\,\langle f,g\rangle_y=0\text{ a.e. for all }g\in N_{n-1}\}.$

${M_n'}$ is a rank one submodule and thus we can find ${f_n\in M_n}$ such that ${\|f_n\|_y=1}$ a.e. The ${\{f_n\}}$ are ${\mathbb Y}$-orthonormal and ${\{f_1,\ldots,f_n\}}$ spans ${N_n}$. Hence the ${\mathbb Y}$-module spanned by ${\{f_n\}}$ contains all ${g_n}$ and so coincides with ${M}$. $\Box$

3. ${L^2(\mathbb X)}$ as a direct integral of ${\dot{\mathcal H}}$

We now proceed to obtain the first part of our goals: there is a bijective correspondence between ${L^2(\mathbb X)}$ and the measurable square integrable sections of ${\dot{\mathcal H}}$. Fix a ${\mathbb Y}$-orthonormal basis ${\{e_n\}}$ of ${L^2(\mathbb X)}$. For each section ${\gamma:Y\rightarrow\dot{\mathcal H}}$, form the coefficient functions ${\{c_n\}\in L^\infty(\mathbb Y)}$ by

$\displaystyle c_n(y)=\langle\gamma(y),e_{n,y}\rangle_y,\ \ y\in Y.$

Definition 7 ${\gamma}$ is measurable if each ${c_n}$ is measurable. It is measurable square integrableif it is measurable and

$\displaystyle \int_Y\|\gamma(y)\|_y^2d\nu=\sum\int_Y|c_n(y)|^2d\nu(y)<\infty.$

We can equip the linear space of measurable square integrable sections of ${\dot{\mathcal H}}$ with the inner product

$\displaystyle \langle\gamma,\delta\rangle_{\rm sec}=\int_Y\langle\gamma(y),\delta(y)\rangle_yd\nu(y).$

If we put ${\|\gamma\|_{\rm sec}=\sqrt{\langle\gamma,\gamma\rangle_{\rm sec}}}$, then ${\langle\gamma,\delta\rangle_{\rm sec}\le \|\gamma\|_{\rm sec}\cdot\|\delta\|_{\rm sec}}$ and so ${\langle\ ,\ \rangle_{\rm sec}}$ is well-defined. It actually turns ${\Gamma}$ into an isometry from ${L^2(\mathbb X)}$ to its image.

Proposition 8 There is a bijective correspondence between ${L^2(\mathbb X)}$ and the space of measurable square integrable sections of ${\dot{\mathcal H}}$. Such correspondence is given by ${\Gamma}$.

Proof: Given ${f\in L^2(\mathbb X)}$, each function ${\langle f_y,e_{n,y}\rangle_y}$ is measurable, because it is the composition of measurable functions. Since

$\displaystyle \|f\|^2=\int_Y\left(\int_X|f|^2d\mu_y\right)d\nu(y)=\int_Y\|f_y\|_y^2d\nu(y)=\|\Gamma(f)\|_{\rm sec}^2\ ,$

it follows that every ${\Gamma(f)}$ is indeed measurable square integrable. Conversely, let ${\gamma:Y\rightarrow\dot{\mathcal H}}$ be measurable square integrable. Then ${f=\sum c_n(y)e_n}$ satisfies

$\displaystyle \begin{array}{rcl} \|f\|^2&=&\displaystyle\int_Y\left(\int_X |f|^2d\mu_y\right)d\nu(y)\\ &&\\ &=&\displaystyle\int_Y\left(\int_X\left|\sum c_n(y)e_n\right|^2d\mu_y\right)d\nu(y)\\ &&\\ &=&\displaystyle\int_Y\left\|\sum c_n(y)e_{n,y}\right\|_y^2d\nu(y)\\ &&\\ &=&\displaystyle\int_Y\left\|\gamma(y)\right\|_y^2d\nu(y)\\ &&\\ &=&\|\gamma\|_{\rm sec}^2\ , \end{array}$

which is finite, proving that ${f\in L^2(\mathbb X)}$. Clearly, ${\gamma=\Gamma(f)}$ and so the proof is complete. $\Box$

Each ${f\in L^2(\mathbb X)}$ defines a rank one ${\mathbb Y}$-module and each section of ${\dot{\mathcal H}}$ defines a bundle of one dimensional subspaces of ${\mathcal H_y}$. By the previous proposition, rank one ${\mathbb Y}$-module are in correspondence with bundles of one dimensional subspaces of ${\mathcal H_y}$. We can go further on this correspondence and obtain representatives for ${\mathbb Y}$-modules of any finite rank of the same nature.

Let ${{\rm End}(\mathcal H_y)}$ be the set of endomorphisms of ${\mathcal H_y}$. The union ${\bigcup_{y\in Y}{\rm End}(\mathcal H_y)}$ is again a Borel Hilbert bundle over ${Y}$. A section of this bundle will be called an operator valued section. We say an operator valued section ${y\mapsto A(y)}$ is measurable if, for any measurable square integrable section ${\gamma:Y\rightarrow\dot{\mathcal H}}$, the section ${y\mapsto A(y)\gamma(y)}$ is measurable.

Definition 9 Let ${V:y\mapsto V(y)}$ be a map such that each ${V(y)}$ is an ${r}$-dimensional subspace of ${\mathcal H_y}$. We say ${V}$ is a measurable ${r}$-plane section if the operator valued section ${y\mapsto P_{V(y)}}$, with ${P_{V(y)}:\mathcal H_y\rightarrow V(y)}$ 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 ${Y}$. On the other hand, observers must be in the ${L^2}$-space we are working with, that is, they are measurable square integrable sections.

Theorem 10 For any fixed ${r<\infty}$, there is a bijective correspondence between the collection of rank ${r}$ ${\mathbb Y}$-modules in ${L^2(\mathbb X)}$ and the space of measurable ${r}$-plane sections.

Proof: Let ${M}$ be a rank ${r}$ ${\mathbb Y}$-module. Choose a ${\mathbb Y}$-orthonormal basis ${\{f_1,\ldots,f_r\}}$ for ${M}$ and observe that the operator valued section ${y\mapsto P_{M_y}}$ is measurable, since it is given by

$\displaystyle P_{M_y}f_y=\sum_{i=1}^r\langle f_y,f_{i,y}\rangle_y f_{i,y}\ ,\ \ f\in L^2(\mathbb X).$

Conversely, let ${V:y\mapsto V(y)}$ be a measurable ${r}$-plane section. Let ${\{e_n\}}$ be a ${\mathbb Y}$-orthonormal basis for ${L^2(\mathbb X)}$ and define sections ${\{s_n\}}$ by ${s_n(y)=P_{V(y)}e_{n,y}}$. By assumption, each ${s_n}$ is square integrable and so, according to Proposition 8, is the image under ${\Gamma}$ of some ${f_n\in L^2(\mathbb X)}$. Let ${M}$ be the ${\mathbb Y}$-module generated by ${\{f_n\}}$. It is clear that ${M_y=V(y)}$. $\Box$

4. Generalized eigenfunctions

We keep the setup of the previous section and assume ${M}$ is a ${\mathbb Y}$-module invariant under ${T}$, that is, ${TM_{Sy}=M_y}$ a.e. By the ergodicity assumption on ${\mathbb Y}$, the dimension of ${M_y}$ is constant. In particular, a finite rank ${\mathbb Y}$-module is of rank ${r<\infty}$.

Definition 11 A function ${f\in L^2(\mathbb X)}$ is a generalized eigenfunction with respect to ${\mathbb Y}$ or a ${\mathbb Y}$-eigenfunction if the module spanned by ${\{T^nf\}_n}$ is of finite rank. Equivalently, ${f}$ is a generalized eigenfunction if it belongs to a ${T}$-invariant ${\mathbb Y}$-module of rank ${r<\infty}$.

Let ${\{f_1,\ldots,f_r\}}$ be a ${\mathbb Y}$-orthonormal basis for ${M}$. ${Tf_i}$ belongs to ${M}$ and so we can write

$\displaystyle Tf_i=\sum_{j=1}^r \lambda_{ij}(y)f_j,$

with ${\lambda_{ij}\in L^\infty(\mathbb Y)}$. If ${F(x)}$ is the vector valued function with components ${f_i(x)}$ and ${\Lambda(y)=(\lambda_{ij}(y))}$, then

$\displaystyle TF=\Lambda(y)F,$

that is, ${F}$ is a vector valued eigenfunction, the “eigenvalue” being a matrix valued function on ${\mathbb Y}$. In addition, ${\{Tf_1,\ldots,Tf_r\}}$ is again a ${\mathbb Y}$-orthonormal basis for ${M}$ and so ${\Lambda(y)}$ is a unitary matrix. The conclusion is that all generalized eigenfunctions belong to ${\mathbb Y}$-modules spanned by unitary generalized eigenfunctions. This fact implies that the bounded generalized eigenfunctions form an algebra containing ${L^\infty(\mathbb Y)}$. Indeed:

1. Each function in ${L^\infty(\mathbb Y)}$ is a generalized eigenfunction: obvious.
2. ${f+g}$ is a generalized eigenfunction whenever ${f,g\in L^2(\mathbb X)}$ are generalized eigenfunctions: assume ${f=\sum_{i=1}^r c_i(y) f_i}$ with ${TF=\Lambda_1(y)F}$ and ${g=\sum_{j=1}^s d_j(y)g_j}$ with ${TG=\Lambda_2(y)G}$, ${\Lambda_1(y)}$ and ${\Lambda_2(y)}$ being both unitary. Then

$\displaystyle f+g=\sum_{i=1}^r c_i(y)f_i+\sum_{j=1}^s d_j(y)g_j$

belongs to the ${\mathbb Y}$-module generated by ${\{f_1,\ldots,f_r,g_1,\ldots,g_s\}}$. If ${H}$ is the vector valued function with components ${f_1,\ldots,f_r,g_1,\ldots,g_s}$, then ${TH=\Lambda(y)H}$, where

$\displaystyle \Lambda(y)= \left(\begin{array}{cc} \Lambda_1(y)& 0\\ 0&\Lambda_2(y) \end{array}\right)$

is unitary.

3. ${fg}$ is a generalized eigenfunction whenever ${f,g\in L^\infty(\mathbb X)}$ are generalized eigenfunctions: let ${M}$ and ${N}$ be finite rank ${\mathbb Y}$-modules with bounded ${\mathbb Y}$-orthonormal basis ${\{f_1,\ldots,f_r\}}$ and ${\{g_1,\ldots,g_s\}}$, respectively. Then the ${\mathbb Y}$-module spanned by the set ${\{f_ig_j\}_{i,j}}$ has finite rank and contains the product of any two bounded eigenfunctions ${fg}$ with ${f\in M}$ and ${g\in N}$.

Definition 12 If ${\mathbb X\rightarrow\mathbb Y}$ is a homomorphism, denote ${\mathcal E(\mathbb X/\mathbb Y)}$ the subspace of ${L^2(\mathbb X)}$ spanned by the generalized eigenfunctions.

${\mathcal E(\mathbb X/\mathbb Y)}$ is clearly a ${\mathbb Y}$-module. It contains ${L^2(\mathbb Y)}$ as well as the subspace of ${T}$-invariant functions on ${L^2(\mathbb X)}$. In addition, we have the following

Lemma 13 ${\mathcal E(\mathbb X/\mathbb Y)}$ contains the dense algebra of bounded generalized eigenfunctions.

Proof: Let ${M}$ be a ${\mathbb Y}$-module of finite rank and ${F=(f_1,\ldots,f_r)}$ a vector valued function such that ${TF=\Lambda(y)F}$, with ${\Lambda(y)}$ unitary. Consider the function ${p(x)=\sum_{i=1}^r |f_i(x)|^2}$ and observe that

$\displaystyle Tp(x)=\sum_{i=1}^r|Tf_i(x)|^2=\sum_{i=1}^r|f_i(x)|^2=p(x),$

since the vectors ${F}$ and ${TF}$ are related by a unitary matrix. This proves that ${p}$ is ${T}$-invariant and so, for each ${n\ge 1}$, the set ${A_n=\{x\in X\,;\,p(x)\le n\}}$ is ${T}$-invariant. In these sets ${f_1,\ldots,f_r}$ are bounded and then each ${f_{i,n}=\chi_{A_n}f_i}$ is bounded. Furthermore, letting ${F_n=(f_{1,n},\ldots,f_{r,n})}$,

$\displaystyle TF_n=T\chi_{A_n}TF=\chi_{A_n}\Lambda(y)F=\Lambda(y) F_n$

and so ${\{f_{1,n},\ldots,f_{r,n}\}}$ generates a finite rank ${\mathbb Y}$-module, proving that they are generalized eigenfunctions. Obviously, ${f_i=\lim_{n}f_{i,n}}$. $\Box$

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

Lemma 14 Let ${(X,\mathcal A,\mu)}$ be a Lebesgue space and ${B\subset L^\infty(X,\mathcal A,\mu)}$ an algebra, invariant under conjugation. If ${\mathcal B}$ is the smallest ${\sigma}$-algebra of ${X}$ with respect to which all functions in ${B}$ are measurable, then

$\displaystyle \overline{B}=L^2(X,\mathcal B,\mu),$

where the closure is taken over ${L^2(\mathbb X)}$.

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 15 We say that ${\mathbb X\rightarrow\mathbb Y}$ is a compact extension if ${\mathcal E(\mathbb X/\mathbb Y)=L^2(\mathbb X)}$.

The last two result directly give the following.

Theorem 16 Given ${\mathbb X\rightarrow\mathbb Y}$, there exists an intermediate factor ${\mathbb X\rightarrow\mathbb Z\rightarrow\mathbb Y}$ such that ${\mathbb Z\rightarrow\mathbb Y}$ is a compact extension, with ${\mathcal E(\mathbb X/\mathbb Y)=L^2(\mathbb Z)}$.

By definition, ${\mathbb Z}$ is the largest intermediate compact extension of ${\mathbb Y}$ in ${\mathbb X}$. 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 17 The extension ${\mathbb X\rightarrow\mathbb Y}$ is weakly mixing if and only if it has no non-trivial generalized eigenfunctions.

In other words, weakly mixing extensions are those for which ${\mathcal E(\mathbb X/\mathbb Y)=L^2(\mathbb Y)}$. 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 ${\mathbb X\times_{\mathbb Y}\mathbb X}$-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.

## Responses

1. 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.

• Dear Jaramillo,

I’m not sure that I understood your question: the canonical projection $p_y:\prod\limits_{z\in Y}L^2(\mathbb{X}_z)\to L^2(\mathbb{X}_y)$ is well-defined independently of measure-theoretical considerations.

Indeed, if we think of each $L^2(\mathbb{X}_z)$ as an abstract space $E_z$, then the canonical projection $p_y$ is well-defined because the same is true for the projection from the abstract product space $\prod\limits_{z\in Y} E_z$ to $E_y$ consisting into taking the $y$-th coordinate $e_y$ of a sequence $(e_z)_{z\in Y}\in\prod\limits_{z\in Y} E_z$.

Best, Matheus

2. Dear Matheus,
I meant that the image of $\Gamma$ is not $\prod_{z\in Y}L^2(X_z)$, but rather $\prod_{z\in Y}L^2(X_z)/R$ where $R$ is the $\nu$-a.s. equal equivalence relationship. On this set the projection is not defined.

Put differently, if $f\in L^2(Y\times K)$ and we decompose $X:=Y\times K=\coprod_{z\in Y}X_z=\coprod_{z\in Y}\{z\}\times K$ (i.e., $X_z=\pi^{-1}(z)$) then the “projection” of $f$ at $z$ ($f(\cdot,\cdot)\mapsto f(z,\cdot)$) is not defined (of course, for each fixed $f$ it is defined for a.e. $z$, but these full measures $z$‘s depend on $f$, and there are uncountably many of them -and I don’t see how taking a dense, countable subset of $M$ will do the job-).

• Dear Jaramillo,

I see your point now. I think you are right that the map $\Gamma$ 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