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})}.


— 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


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

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2 Responses to Factors and joinings of measure preserving systems

  1. Pingback: Szemerédi’s Theorem Part II – Overview of the proof | I Can't Believe It's Not Random!

  2. Pingback: Entropy of measure preserving systems | I Can't Believe It's Not Random!

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