The $T,T^{1}$ transformation is an example of a $K$ automorphism which is not Bernoulli (not isomorphic to a shift of an I.I.D. sequence). Hoffman in http://www.math.washington.edu/~hoffman/publications/scenery.pdf constructed a factor of this transformation which he calls the scenery factor and showed that the $T,T^{1}$ is isomorphic to this factor times a Bernoulli shift. The question I have is what is the entropy of the scenery factor? The only thing I seem to be able to say is that it is a number in $(0,\log2)$. The reasoning for this trivial bounds is as follows: Since the $T,T^{1}$ is a $K$ automorphism then it has no factors of $0$ entropy hence the entropy of the scenery factor (and the Bernoulli shift in Hoffman's theorem) is positive. The upper bound is given by $\log2$ which is the entropy of the $T,T^{1}$ transformation.
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1$\begingroup$ I think it's hopeless to come up with a number for this transformation. The gist of the theory for this transformation is that you can look at the past of the scenery transformation and recover the entire scenery. The entropy is then trying to recover the steps you're making based on what you see. Clearly if you have a long block of 1's it's essentially impossible to see what steps you're making. In this way, you can in principle obtain weak, but explicit upper bounds on the entropy below $\log 2$. $\endgroup$– Anthony QuasSep 11 '15 at 18:01

$\begingroup$ Thanks for the reply. I am wandering if there is a simple argument I am missing. $\endgroup$– user78465Sep 13 '15 at 20:35
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