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Let $X=\{0,1\}^\mathbb{Z}$ with measure $\mu=(p,1-p)^{\mathbb{Z}}$.

Let $(\phi(x))_i=(x_i+x_{i+1})$mod$2$.

If $p=1/2$, then $\phi(X)=X$. If $p \not = 1/2$, then $\phi(X)$ is not a Bernoulli scheme (i.i.d.).

For $x \in X$, define $x^*$ so that $x^*_i=(x_i+1)$mod$2$. Then $\phi(x)=\phi(x^*)$.

A factor map $\psi$ is finitary if for almost every $x \in X$ there exists integers $m \leq n$ such that the zero coordinates of $\psi(x)$ and $\psi(x')$ agree for almost all $x' \in X$ with $x[m,n]=x'[m,n]$.

In the case that $p \not = 1/2$, is it possible to construct a finitary map $\psi:(X, \mu) \to (X, \mu)$, such that for almost all $x \in X$, $\psi(x)=\psi(x^*)$?

Thank you.

share|cite|improve this question
You didn't say this, but I presume you also want that $\psi$ should preserve the measure $\mu$? (Otherwise your previous map $\phi$ will do the job). – Anthony Quas Jan 25 '11 at 22:32
Yes, I'd like $\psi$ to preserve $\mu$. Thank you Anthony. – Stephen Shea Jan 25 '11 at 23:35
Question has been edited to clarify that $\psi$ should preserve $\mu$. – Stephen Shea Jan 30 '11 at 12:05

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