Let $(X, \sigma)\subset (\{0, 1, 2, 3\}^\mathbb{N},\sigma)$ be a subshift of finite type. Let $P_n$ be the set of $n$-periodic points. If $|P_n|=2^n$ for all $n\ge 1$, then it is true that $(X, \sigma)$ is conjugated to a full shift of two letters?
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$\begingroup$ Do you mean if $|P_n|=2^n$? $\endgroup$– Ale De LucaCommented Oct 27 at 15:01
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$\begingroup$ Ornstein extended his result about Bernoulli shifts to weak Bernoulli shifts, which include mixing Markov shifts... $\endgroup$– AsafCommented Oct 27 at 23:58
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3$\begingroup$ @Asaf Could you please give more details? How is it related to this question? Thanks. $\endgroup$– user119197Commented Oct 28 at 13:08
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1$\begingroup$ I believe @Asaf is talking about measurable isomorphism, while the question is, I think, about topological conjugacy $\endgroup$– Anthony QuasCommented Oct 28 at 15:53
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1$\begingroup$ You can do something cheap and have symbols 2 and 3 only followed by 0; 0 and 1 can be followed by 0 and 1.. This is still a (reducible) shift of finite type, and it satisfies your conditions. This issue might be improved by asking for a 2-sided irreducible SFT $\endgroup$– Anthony QuasCommented Oct 28 at 15:56
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1 Answer
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If $A$ is the transition matrix, the number of periodic points is $\text{Tr}(A^n)$. If you conjugate the matrix to Jordan normal form, you can see $ \text{Tr}(A^n)$ is the sum of the $n$th powers of the eigenvalues, counted with algebraic multiplicity (including generalized eigenvalues). The only choice of eigenvalues with the correct sum are 2,0,0,0. So you require $A$ to be a 0,1-valued matrix that has a single eigenvalue of 2 and three eigenvalues of 0 (possibly in a non-trivial Jordan block).
If you take $A$ to be, for example $$ \begin{pmatrix} 1&1&0&0 \cr 1&1&0&0 \cr 1&0&0&0 \cr 1&0&0&0 \end{pmatrix}, $$ you get a matrix with the correct number of periodic points, but for which the SFT is not topologically conjugate to the full shift on 2 symbols (because of the additional points starting with 2 and 3).
If you require the SFT to be two-sided, there are no points with 2's and 3's; and the two SFTs are conjugate. And any example is essentially of this type (possibly with some relabelling).
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$\begingroup$ Thanks a lot. Could you explain a little more for the last paragraph? Do you mean if the SFT is supposed to be two-sided, then it is topological conjugate to full shift? How to show that " any example is essentially of this type "? $\endgroup$ Commented Oct 29 at 23:50
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$\begingroup$ What I mean is that in the two-sided version, even though the symbols 2 and 3 are in the alphabet, they do not appear in the actual 2-sided points because 2 and 3 do not follow any other symbol in the alphabet. In this case, the 2-sided shift with the $4\times 4$ matrix is equal to the original 2-sided shift. The same will be true for any SFT with these properties: there will be exactly two symbols that appear in the 2-sided shift; and they can occur in arbitrary concatenations. $\endgroup$ Commented Oct 30 at 2:49
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$\begingroup$ How to see all of this? The matrix is supposed to have an eigenvalue 2; and an eigenvalue 0 with algebraic multiplicity 3. This means the Jordan form of the matrix has a one-dimensional block with eigenvalue 2; and one of {three one-dimensional blocks with eigenvalue 0; a two-dimensional block with eigenvalue 0 and a one-dimensional block with eigenvalue 0; a three-dimensional block with eigenvalue 0}. Going through the cases is a bit painful, and I have to admit I have not written out full details (although I am quite confident in the outcome) $\endgroup$ Commented Oct 30 at 2:56
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$\begingroup$ Quas are you saying there are just a few matrices with the correct number of periodic points? The 2-shift has a pretty large orbit under the automorphism group of the 4-shift, so I don't get how such a naive approach can work. $\endgroup$ Commented Oct 30 at 6:47
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1$\begingroup$ Oh, you seem to be restricting to vertex shifts. Then it might work but it's not the same question. $\endgroup$ Commented Oct 30 at 7:01