## Factors and joinings of measure preserving systems

The joining of two measure preserving systems is a third measure preserving system that has the two original systems as factors. In analogy with classical arithmetic, using joinings it is possible to have a notion of a common multiple of two systems, and two (nonequivalent) notions of relatively prime systems. In this post I will explore some basic properties of joinings and touch the notion of disjoint systems.

Joinings were introduced by Furstenberg and have become a fundamental tool in ergodic theory. For a complete treatment of the topic see the book of Glasner.

— 1. Factors —

For the sake of completeness I will briefly state the notions I will use.

A measure preserving system (or m.p.s.) is a quadruple ${{\bf X}=(X,{\cal B},\mu, T)}$, where ${(X,{\cal B},\mu)}$ is a probability space and ${T:X\rightarrow X}$ is a measurable map such that ${\mu(T^{-1}A)=\mu(A)}$ for all ${A\in{\cal B}}$.

Definition 1 We say that two m.p.s. ${{\bf X}=(X,{\cal B},\mu, T)}$ and ${{\bf Y}=(Y,{\cal C},\nu, S)}$ are isomorphic if there exists a map ${\phi:X\rightarrow Y}$ such that

• We have ${\phi\circ T=S\circ\phi}$ almost everywhere.
• For every ${f\in L^1(X,{\cal B},\mu)}$ there exists ${g\in L^1(Y,{\cal C},\nu)}$ such that ${f=g\circ\phi}$.
• For every ${g\in L^1(Y,{\cal C},\nu)}$ we have ${g\circ\phi\in L^1(X,{\cal B},\mu)}$ and

$\displaystyle \int_Xg\circ\phi d\mu=\int_Ygd\nu$

Example 1 For any invertible m.p.s. ${{\bf X}=(X,{\cal B},\mu, T)}$, the map ${T:X\rightarrow X}$ is an isomorphism between ${{\bf X}}$ and itself. More generally, for any ${n\in{\mathbb Z}}$ the map ${T^n:{\bf X}\rightarrow{\bf X}}$ is an isomorphism.

Definition 2 (Factor map) Let ${{\bf X}=(X,{\cal B},\mu, T)}$ and ${{\bf Y}=(Y,{\cal C},\nu, S)}$, be two m.p.s. A map ${\pi:X\rightarrow Y}$ is a factor map if:

• It is surjective (up to measure ${0}$),
• ${\mu\big(\pi^{-1}(C)\big)=\nu(C)}$ for every ${C\in{\cal C}}$,
• ${\pi\circ S=T\circ\pi}$.

In this situation we call ${{\bf Y}}$ a factor of ${{\bf X}}$ and we call ${{\bf X}}$ an extension of ${{\bf Y}}$.

Example 2 Any isomorphism of measure preserving systems is a factor map.

Example 3 For any m.p.s. ${{\bf X}=(X,{\cal B},\mu, T)}$, the map ${T:X\rightarrow X}$ is a factor map between ${{\bf X}}$ and itself. More generally, the map ${T^n}$ (with ${n\in{\mathbb N}}$ or ${n\in{\mathbb Z}}$ for invertible ${T}$) is a factor map.

Example 4 If ${{\bf X}=(X,{\cal B},\mu,T)}$ and ${{\bf Y}=(Y,{\cal C},\nu, S)}$ are measure preserving systems we denote by ${{\bf X}\times{\bf Y}}$ the product system ${(X\times Y,{\cal B}\otimes{\cal C},\mu\otimes\nu,T\times S)}$ (the ${\sigma}$-algebra ${{\cal B}\otimes{\cal C}}$ is the smallest ${\sigma}$-algebra over ${X\times Y}$ which contains all rectangles ${B\times C}$ with ${B\in{\cal B}}$ and ${C\in{\cal C}}$, and the measure ${\mu\otimes\nu}$ is the unique measure on ${{\cal B}\otimes{\cal C}}$ satisfying ${(\mu\otimes\nu)(B\times C)=\mu(B)\nu(C)}$ for every ${B\in{\cal B}}$ and ${C\in{\cal C}}$).

In this situation, both projections ${\pi_X:{\bf X}\times{\bf Y}\rightarrow{\bf X}}$ and ${\pi_Y:{\bf X}\times{\bf Y}\rightarrow{\bf Y}}$ are factor maps.

Example 5 For a more concrete example, take ${X={\mathbb Z}/6{\mathbb Z}}$ and ${Y={\mathbb Z}/3{\mathbb Z}}$, both with the discrete ${\sigma}$-algebra and normalized counting measure and let ${Tx=x+1\mod6}$ and ${Sy=y+1\mod3}$. Then the reduction${\mod3}$ ${\pi:X\rightarrow Y}$ is a factor map.

Example 6 Let ${X=\mathbb{T}^2}$ with the usual topology, the Borel ${\sigma}$-algebra and the Lebesgue (or Haar) measure. Let ${\alpha\in\mathbb{T}}$ and define ${T:X\rightarrow X}$ by ${T(x_1,x_2)=(x_1+\alpha,x_2+x_1)}$. Let ${Y=\mathbb{T}}$, also with the Borel ${\sigma}$-algebra and the Haar measure, and define ${S:Y\rightarrow Y}$ by ${Sy=y+\alpha}$. Then the map ${\pi:X\rightarrow Y}$ defined by ${\pi(x_1,x_2)=x_1}$ is a factor map. Note that this example does not fit in Example 4: even though the set ${X}$ is the product of two copies of ${Y}$, ${T}$ is not the product of ${S}$ with another map on ${Y}$.

If ${{\bf X}=(X,{\cal B},\mu, T)}$ is a m.p.s. and ${{\cal C}\subset{\cal B}}$ is a ${\sigma}$-algebra such that ${T^{-1}C\in{\cal C}}$ for every ${C\in{\cal C}}$ (in other words if ${T:X\rightarrow X}$ is measurable with respect to ${{\cal C}}$; we call such ${{\cal C}}$ an invariant ${\sigma}$-algebra under ${T}$), then the system ${{\bf Y}=(X,{\cal C},\mu, T)}$ is a factor of ${{\bf X}}$ with the identity of ${X}$ as factor map. Conversely, given a factor map ${\pi:{\bf X}\rightarrow{\bf Y}}$, the ${\sigma}$-algebra ${\pi^{-1}({\cal C})}$ of ${{\cal B}}$ in invariant under ${T}$. Therefore we can (and will) identify factors of a system ${{\bf X}}$ with the invariant ${\sigma}$-subalgebras of ${{\cal B}}$.

An intuitive (but not completely trivial) property of factors is that if ${\pi_1:{\bf X}\rightarrow{\bf X}_1}$ and ${\pi_2:{\bf X}\rightarrow{\bf X}_2}$ are factor maps such that ${\pi_1^{-1}({\cal B}_1)\supset\pi_2^{-1}({\cal B}_2)}$ (up to sets of measure ${0}$) then ${{\bf X}_2}$ is also a factor of ${{\bf X}_1}$, in the sense that there exists a factor map ${\pi_3:{\bf X}_1\rightarrow{\bf X}_2}$.

Given two ${\sigma}$-algebras ${{\cal B}}$ and ${{\cal C}}$ on the same set ${X}$, we denote by ${{\cal B}\vee{\cal C}}$ the smallest ${\sigma}$-algebra on ${X}$ containing both ${{\cal B}}$ and ${{\cal C}}$. Note that if ${{\cal B}}$ and ${{\cal C}}$ are factors of a m.p.s. ${{\bf X}}$, then also ${{\cal B}\vee{\cal C}}$ is a factor of ${{\bf X}}$; we call this the factor generated by ${{\cal B}}$ and ${{\cal C}}$.

If ${{\cal C}}$ is a factor of a m.p.s. ${(X,{\cal B},\mu,T)}$, the disintegration of ${\mu}$ over ${{\cal C}}$ is the family ${(\mu_x)_{x\in X}}$ of probability measures defined by

$\displaystyle \int_Xfd\mu_x=\mathop{\mathbb E}[f\mid{\cal C}](x)$

where ${\mathop{\mathbb E}}$ denotes the conditional expectation. Observe that the measure ${\mu_x}$ is supported in the atom of ${{\cal C}}$ which contains ${x}$. I posted about disintegration of measures before.

— 2. Joinings —

Given two m.p.s. ${{\bf X}=(X,{\cal B},\mu, T)}$ and ${{\bf Y}=(Y,{\cal C},\nu, S)}$, a joining (of ${{\bf X}}$ and ${{\bf Y}}$) is a probability measure ${\lambda}$ on the product measurable space ${(X\times Y,{\cal B}\otimes{\cal C})}$ preserved under ${T\times S}$ and with marginals ${\mu}$ and ${\nu}$. To be clear, we say that ${\lambda}$ is preserved under ${T\times S}$ if ${\lambda\big((T\times S)^{-1}(D)\big)=\lambda(D)}$ for all ${D\in{\cal B}\otimes{\cal C}}$, and we say that ${\lambda}$ has marginals ${\mu}$ and ${\nu}$ if for ${B\in{\cal B}}$ we have ${\lambda(B\times Y)=\mu(B)}$ and for ${C\in{\cal C}}$ we have ${\lambda(X\times C)=\nu(C)}$.

In other words, ${\lambda}$ is a measure on ${{\cal B}\otimes{\cal C}}$ such that ${{\bf Z}:=(X\times Y,{\cal B}\otimes{\cal C},\lambda,T\times S)}$ is a m.p.s. and the projections ${\pi_X:X\times Y\rightarrow X}$ and ${\pi_Y:X\times Y\rightarrow Y}$ are factor maps. This interpretation gives the trivial example ${\mu\otimes\nu}$ of a joining between two arbitrary systems. We will use the abuse of language to denote both the measure ${\lambda}$ and the m.p.s. ${{\bf Z}}$ as the joining of ${{\bf X}}$ and ${{\bf Y}}$.

A more abstract (yet equivalent) definition of a joining of two m.p.s. ${{\bf X}}$ and ${{\bf Y}}$ is an arbitrary m.p.s. ${{\bf Z}}$ together with two factor maps ${\pi_X:{\bf Z}\rightarrow{\bf X}}$ and ${\pi_Y:{\bf Z}\rightarrow{\bf Y}}$ which generate the whole ${\sigma}$-algebra of ${{\bf Z}}$.

Proposition 3 The two definitions of joining above coincide.

Proof: From the rewording of the first definition it is clear that such a joining satisfies the second definition. Now let ${{\bf X}=(X,{\cal B},\mu,T)}$ and ${{\bf Y}=(Y,{\cal C},\nu,S)}$ be two factors of the m.p.s. ${{\bf Z}=(Z,{\cal D},\tilde\lambda,R)}$ which generate the whole ${\sigma}$-algebra ${{\cal D}}$. This means that ${{\cal D}=\pi_X^{-1}({\cal B})\vee\pi_Y^{-1}({\cal C})}$, where ${\pi_X:Z\rightarrow X}$ and ${\pi_Y:Z\rightarrow Y}$ are the respective factor maps.

Let ${\phi:Z\rightarrow X\times Y}$ be defined by ${\phi(z)=\big(\pi_X(z),\pi_Y(z)\big)}$. Let ${\lambda}$ be the pushforward of ${\tilde\lambda}$ under ${\phi}$ (this means that ${\lambda(A)=\tilde\lambda\big(\phi^{-1}(A)\big)}$ for every set ${A\in{\cal B}\otimes{\cal C}}$). Note that, for all ${B\in{\cal B}}$ and ${C\in{\cal C}}$,

$\displaystyle \begin{array}{rcl} \displaystyle\lambda\big((T\times S)^{-1}(B\times C)\big)&=&\displaystyle\lambda\big((T^{-1}B)\times(S^{-1}C)\big)=\tilde\lambda\big(\pi_X^{-1}(T^{-1}B)\cap\pi_Y^{-1}(S^{-1}C)\big)\\ &=&\displaystyle\tilde\lambda\big(R^{-1}\pi_X^{-1}(B)\cap R^{-1}\pi_Y^{-1}(C)\big)=\tilde\lambda\big(\pi_X^{-1}(B)\cap\pi_Y^{-1}(C)\big)\\&=&\displaystyle\tilde\lambda\big(\phi^{-1}(B\times C)\big) =\lambda(B\times C) \end{array}$

Thus ${\lambda}$ is invariant under ${T\times S}$. Since ${\pi_X}$ is a factor map we have, for any ${B\in{\cal B}}$,

$\displaystyle \lambda(B\times Y)=\tilde\lambda\big(f^{-1}(B\times Y)\big)=\tilde\lambda(\pi_X^{-1}B)=\mu(B)$

and analogously we have ${\lambda(X\times C)=\nu(C)}$ for every ${C\in{\cal C}}$. This shows that ${\lambda}$ is a joining of ${{\bf X}}$ and ${{\bf Y}}$ in the sense of the first definition.

Finally we need to show that the systems ${\phi:{\bf Z}\rightarrow(X\times Y,{\cal B}\otimes{\cal C},\lambda,T\times S)}$ is an isomorphism. The only property of Definition 1 which does not follow directly from the construction is the second, but that property holds for ${\phi}$ because ${{\cal D}=\pi_X^{-1}({\cal B})\vee\pi_Y^{-1}({\cal C})}$.

$\Box$

— 3. Analogies with arithmetic —

The use of the word factor and Example 4 suggest an arithmetic structure on measure preserving system. However one needs to be careful not to push the analogy with the arithmetic on ${{\mathbb N}}$ too far. For instance, in ${{\mathbb N}}$, if ${d}$ is a factor if ${n}$, then there exists another factor ${q}$ of ${n}$ such that ${n}$ is the product of ${d}$ and ${q}$. It is not clear at all how to do this with m.p.s.; in other words, with m.p.s. we can not divide by a factor.

With the analogy in mind, a joining of two systems ${{\bf X}}$ and ${{\bf Y}}$ is a common extension (“multiple”) of both ${{\bf X}}$ and ${{\bf Y}}$. Thus, in analogy with the notion of relatively prime numbers, we say that the systems ${{\bf X}}$ and ${{\bf Y}}$ are disjoint if their only joining is (isomorphic to) the product ${{\bf X}\times{\bf Y}}$.

The first question that pops to mind is whether being disjoint systems is equivalent to having no nontrivial factor in common. The answer is, unfortunately, no, (the first counter-example was given by Rudolph) but one implication is still true.

Theorem 4 If ${{\bf X}_1}$ and ${{\bf X}_2}$ are disjoint measure preserving systems, then they have no non-trivial factor in common.

I will give a proof of this theorem in the end of this section.

An interesting example of a non-trivial joining is the relatively independent joining over a common factor:

Definition 5 Let ${{\bf X}_1}$, ${{\bf X}_2}$ and ${{\bf Y}}$ be m.p.s. and let ${\pi_1:{\bf X}_1\rightarrow{\bf Y}}$ and ${\pi_2:{\bf X}_2\rightarrow{\bf Y}}$ be factor maps. Then the relatively independent joining of ${{\bf X}_1}$ and ${{\bf X}_2}$ over ${{\bf Y}}$ is the measure ${\lambda}$ on ${X_1\times X_2}$ defined by

$\displaystyle \lambda(D)=\int_Y(\mu_y^{(1)}\otimes\mu_y^{(2)})(D)d\nu(y)$

where ${(\mu_y^{(1)})_{y\in Y}}$ and ${(\mu_y^{(2)})_{y\in Y}}$ are disintegrations of ${{\bf X}_1}$ and ${{\bf X}_2}$, respectively, over ${{\bf Y}}$.

Equivalently, we can define ${\lambda}$ for functions of the form ${f_1(x_1)f_2(x_2)}$ the following way:

$\displaystyle \begin{array}{rcl} \displaystyle\int_{X_1\times X_2}f_1(x_1)f_2(x_2)d\lambda(x_1,x_2)&=&\displaystyle\int_Y\mathop{\mathbb E}[f_1\mid{\bf Y}](y)\mathop{\mathbb E}[f_2\mid{\bf Y}](y)d\nu(y)\\&=&\displaystyle\int_Y\left(\int_{X_1}f_1d\mu_y^{(1)}\right)\left(\int_{X_2}f_2d\mu_y^{(2)}\right)d\nu(y) \end{array}$

Since linear combinations of such functions forms a dense set in ${L^2(X_1\times X_2)}$, this definition extends to all ${L^2}$.

An important observation is that the set

$\displaystyle Z=\big\{(x_1,x_2)\in X_1\times X_2:\pi_1(x_1)=\pi_2(x_2)\big\}\ \ \ \ \ (1)$

supports the measure ${\lambda}$:

Lemma 6 The set ${Z}$ in (1) satisfies ${\lambda(Z)=1}$, where ${\lambda}$ is the relatively independent joining defined in Definition 5.

Proof: For almost every ${y\in Y}$ and ${i\in\{1,2\}}$, the measure ${\mu_y^{(1)}\otimes\mu_y^{(2)}}$ on ${X_1\times X_2}$ is supported on ${\big\{(x_1,x_2)\in X_1\times X_2:\pi_1(x_1)=\pi_2(x_2)=y\big\}\subset Z}$. Thus ${(\mu_y^{(1)}\otimes\mu_y^{(2)})(Z)=1}$ almost everywhere, and therefore

$\displaystyle \lambda(Z)=\int_{Y}(\mu_y^{(1)}\otimes\mu_y^{(2)})(Z)d\nu(y)=1$

$\Box$

We can now give a proof of Theorem 4

Proof: Assume that the systems ${{\bf X}_1}$ and ${{\bf X}_2}$ shared a common factor ${{\bf Y}}$. Let ${{\bf Z}}$ denote the relatively independent joining defined in Definition 5. Since ${{\bf X}_1}$ and ${{\bf X}_2}$ are disjoint, the joining ${\lambda}$ must be isomorphic to the product ${\mu_1\otimes\mu_2}$. On the other hand, we know that the subset ${Z}$ constructed in (1) has full measure. If ${{\bf Y}}$ is not a trivial system, there exists some ${C\in{\cal C}}$ with ${\nu(C)>0}$ and ${\nu(Y\setminus C)>0}$. Then ${\big(\pi_1^{-1}(C)\big)\times\big(\pi_2^{-1}(Y\setminus C)\big)}$ would have positive measure and be disjoint from ${Z}$, which contradicts Lemma 6. Thus ${{\bf Y}}$ is trivial as desired. $\Box$