Finitary factors of Bernoulli schemes that pair duals - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T18:44:24Z http://mathoverflow.net/feeds/question/77392 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/77392/finitary-factors-of-bernoulli-schemes-that-pair-duals Finitary factors of Bernoulli schemes that pair duals Stephen Shea 2011-10-06T19:55:31Z 2011-10-06T19:55:31Z <p>This question is related to my question: </p> <p><a href="http://mathoverflow.net/questions/53276/entropy-preserving-finitary-factor-maps-of-bernoulli-schemes" rel="nofollow">http://mathoverflow.net/questions/53276/entropy-preserving-finitary-factor-maps-of-bernoulli-schemes</a>.</p> <p>Hopefully, this one is a bit easier. </p> <p>Let <code>$X=\{0,1\}^\mathbb{Z}$</code> with measure $\mu=(p,1-p)^{\mathbb{Z}}$, where $p \not = 1/2$.</p> <p>For $x \in X$, define <code>$x^*$</code> so that $x^*_i=(x_i+1)$mod$2$. </p> <p>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 other words, a factor map is finitary if it is continuous after removing a subset of measure zero.</p> <p>Let $\psi:X \to Y$, such that for almost all $x \in X$, $\psi(x)=\psi(x^*)$. </p> <p>Can we find an example of such a $\psi$ and $Y$ where $h(Y)=h(X)$ ($\psi$ does not decrease entropy), and $Y$ is Markov (or even variable length Markov)?</p> <p>Thank you.</p>