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Let $\Sigma$ be a two-sided subshift on a finite alphabet $A$. Let $\Sigma_n$ denote all words $x_{-n}\dots x_n\in A^{2n+1}$ such that $(x_k)_{-\infty}^\infty \in \Sigma$ for some $x_k, |k|>n$.

Its topological entropy is defined as follows: $$ h(\Sigma) = \lim_{n\to\infty}\frac{\log \#\Sigma_n}{2n+1}. $$ Now let $Per_n$ denote all periodic sequences in $\Sigma$ of period $n$. Assume $$ \widetilde h(\Sigma) = \lim_{n\to\infty}\frac{\log \#Per_n}{n}=0. $$

Question. I need sufficient conditions for $h(\Sigma)=0$. It is known to hold for the sofic subshifts (Lind & Marcus, Exercise 4.4.3) but in my examples they are not sofic. Nor are they transitive.

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    $\begingroup$ If I'm correct, there are Toeplitz subshifts that are minimal (so $\tilde{h}=0$) and have positive topological entropy. $\endgroup$
    – YCor
    Mar 27, 2016 at 13:15
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    $\begingroup$ Oh I'm sure there are examples like that. What I need some property which ensures this does not happen. Specification works, for instance, but what if my subshift is not even transitive, for instance? $\endgroup$
    – Nikita Sidorov
    Mar 27, 2016 at 13:17
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    $\begingroup$ For sofic shifts (not necessarily irreducible), there is the equality $h(\Sigma)=\tilde{h}(\Sigma)$ as long as the limit in $\tilde{h}(\Sigma)$ exists. This is exercise 4.4.3 of the book of Lind and Marcus. $\endgroup$
    – Algernon
    Mar 27, 2016 at 16:34
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    $\begingroup$ Vaughn, here's a toy example. Let $t$ be the Thue-Morse constant, i.e., its binary expansion is $0110 1001 1001 0110\dots$. Consider the open map which is the doubling map with the hole $(t,1-t)$. Here any sequence is of the form $t_1^{k_1}t_2^{k_2}\dots$, where $t_k$ is the prefix of the Thue-Morse sequence of length $2^k$ and $k_j\ge0$ and may equal $\infty$. The only admissible cycles are $t_k^\infty$, which easily implies $\tilde h=0$. We also have $h=0$, which follows from the description of the subshift. Clearly, $t_k$ cannot be followed by $t_j$ with $j<k$, so it's not transitive. $\endgroup$
    – Nikita Sidorov
    Feb 19, 2017 at 16:34
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    $\begingroup$ Vaughn, thanks. I guess the bottom line is that there is no general result which I could use straight away. Personally I think this is a very natural question in a general setting, since it is usually easier to describe the cycles than all admissible sequences. Hopefully, such general results will be proved in the near future! $\endgroup$
    – Nikita Sidorov
    Feb 24, 2017 at 21:18

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