Roughly speaking, this question asks whether there is a measure-conjugacy between two transformations if there are measure-conjugacies between their ergodic components.

Suppose $(X,\mu)$ is a standard probability space and $T$ and $S$ are measure-preserving transformations of $(X,\mu)$. By the ergodic decomposition theorem, there are standard probability spaces $(Y,\nu), (Z,\zeta)$ and Borel maps $\phi:Y \to M_1(X), \psi:Z\to M_1(X)$ (where $M_1(X)$ is the space of Borel probability measures on $X$) such that for a.e. $y \in Y$, $\phi(y)$ is ergodic, $T$-invariant and $\int \phi(y)~d\nu(y)=\mu$. Similarly, for a.e. $z\in Z$, $\psi(z)$ is ergodic and $S$-invariant and $\int \psi(z)~d\zeta(z)=\mu$.

Now suppose there is a measure-space isomorphism $\Omega:(Y,\nu) \to (Z,\zeta)$ such that for a.e. $y \in Y$, $(T,X,\phi(y))$ is measurably conjugate to $(S,X,\psi(\Omega(y)))$. Then is $T$ measurably conjugate to $S$?


This is probably related to the following question (and, most likely, can be obtained from it or from a modification of the argument): let M and N be two finite von Neumann algebras with centers $Z(M)$ and $Z(N)$ and faithful traces $\tau_M$, $\tau_N$. Assume that there is a (trace-preserving) isomorphism $(Z(M),\tau_M)\cong L^\infty(X,\mu)\stackrel{\alpha}{\to} L^\infty(Y,\nu)\cong (Z(N),\tau_N)$ so that the central components $M_x$ and $N_{\alpha(x)}$ are a.e. isomorphic. Does it follow that $M\cong N$?

This was proved in this form by Effros [Trans. Amer. Math. Soc. 121 (1966), 434--454; MR0192360 (33 #585)] with a later (shorter) proof by Elliott [MR0310659 (46 #9757) An extension of some results of Takesaki in the reduction theory of von Neumann algebras. Pacific J. Math. 39 (1971), 145–148.]

The proofs rely on existence of Borel structure on von Neumann algebras and Borel selection theorems (so, I would guess, they should go through in your context as well?).

  • $\begingroup$ Thanks! It seems reasonable that the same ideas should work in my context; I'll look into it. $\endgroup$ – Lewis Bowen Jul 30 '11 at 18:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.