# Subshifts with the same entropy

It is known that two Markov subshifts with the same entropy are "almost isomorphic" (up to a subset of measure 0) if the entropy is a logarithm of an integer (see R. L. Adler, L. W. Goodwyn, and B.Weiss. Equivalence of topological markov shifts. Israel J. Math, 27(1):49--63, 1977). Is it true (known) if the entropy is not a logarithm of an integer?

Update Basically I want to know if the result of AGW is true without the integer assumption. I would appreciate an answer of the form "yes plus reference" or "no plus reference" or "it is still unknown".

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This is a bit vague... What "measure 0" do you mean if we have two topological subshifts? Ornstein's theorem for any two Bernoulli shifts is of course still valid if you have two Markov subshifts endowed with Markov measures. –  Nikita Sidorov May 1 '11 at 3:01
@Nikita: I mean the statement in AGW's paper. They have a precise definition there. The paper seems to be well known to specialists (I am not one of them, though), so I decided not to repeat it here. My question is whether the statement (as formulated by AGW) is still valid with the assumption that $n$ is an integer. –  Mark Sapir May 1 '11 at 3:18

Look at Section 9 of http://www-users.math.umd.edu/users/mmb/papers/openfinalsub3nov2007.pdf

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There are several versions of equivalence in this section. These are not the same as in Adler,Goodwyn and Weiss. Right? –  Mark Sapir May 1 '11 at 0:56
Yes. You are right. The two notions are different. I thought the section also talked about almost topological conjugacies. There is one extension that I know of books.google.ca/… –  Nishant Chandgotia May 1 '11 at 1:46
This is just saying that the period and the entropy of the topological markov chain are the complete invariant for almost topological conjugacy. –  Nishant Chandgotia May 1 '11 at 2:01
Thanks! Do you know who can answer this question? I can ask Weiss, of course, but I thought the answer should be well known and did not want to bother him. –  Mark Sapir May 1 '11 at 2:29
The notion of isomorphism discussed in "Equivalence of topological markov shifts" is almost topological conjugacy.This fully classifies topological markov shifts(shifts of finite type) upto almost topological conjugacy. The result that you stated above was proved for aperiodic or mixing topological markov chains. So the answer to your question is almost a yes except for trivialities where the periods are different.e.g. Let X be the orbit of a point of period 3 and Y to consist only of a single fixed point. Then X and Y have the same entropy but they are not almost topologically conjugate. –  Nishant Chandgotia May 1 '11 at 6:49

The main theorem in Adler, Roy L.; Marcus, Brian

"Topological entropy and equivalence of dynamical systems"

doesn't use the fact that the entropy is log of a natural number. However it states that two ergodically supported topological Markov shifts are almost topologically conjugated if and only if them have the same topological entropy and the same ergodic period.

There is a work of Wenxiang Sun which extends this result for general expansive, ergodically supprted maps that have the shadowing property:

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Although you have answered your own question completely, I'd like to add a reference to a closely related open question from Mike Hochman. I believe readers interested in almost isomorphisms will also be interested in this open question. One can consider two equal entropy mixing shifts of finite type, $X$ and $Y$, that are not topologically conjugate. Let $Per(X)$ and $Per(Y)$ be the periodic points of $X$ and $Y$ respectively. Are $X$ \$Per(X)$ and $Y$ \$Per(Y)$ topologically conjugate?

Here is a link with a formal description of the problem and some remarks.

From Open Problems section of the 3rd Pingree Park Workshop (2010):

http://www.math.princeton.edu/%7Ehochman/open-problems/pingree-open-problems.pdf

I'd also like to add that there is a closely related theory of finitary isomorphisms of Markov processes (Keane and Smorodinsky for finite state, Rudolph for countable state) that the reader may be interested in.

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Here is an answer of Benjy Weiss. This answers my question completely:

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The general case is treated in: