2
$\begingroup$

Consider $(X,\sigma_X)$ and $(Y, \sigma_{Y})$ be subshifts of the one sided shift in two symbols. Assume that $(X,\sigma_X)$ is a transitive subshift of finite type and $(Y, \sigma_{Y})$ is a transitive sofic subshift such that $h_{top}(\sigma_X) < h_{top}(\Sigma_{Y})$.

My question is: What are the conditions (if any) to be sure that there is a proper embedding from X to Y?

It is known that if $(X,\sigma_X)$ and $(Y, \sigma_{Y})$ are both transitive subshifts of finite type then such embedding exists if and only if $h_{top}(\sigma_X) < h_{top}(\sigma_{Y})$ and $Per_n(\sigma_X) \leq Per_n(\sigma_Y)$ for every $n \in \mathbb{N}$ where $Per_n$ denotes the cardinality of the set of points having least period $n$. I believe that $h_{top}(\sigma_X) < h_{top}(\Sigma_{Y})$ is a necessary condition. Also, the examples that I am working with satisfy that $Per(\sigma_X) \cap Per(\sigma_Y) = \emptyset$.

$\endgroup$
2
$\begingroup$

Take a look at Klaus Thomsen's paper: On the structure of a sofic shift space. He discusses generalizations of Krieger's embedding theorem and gives conditions when it still holds respectively when it fails.

Not sure what your last condition (the intersection of periodic point sets is empty? for some $n$ or for all $n$?) means. You could have $X$ containing the fixed point of all 1s and $Y$ the fixed point of all 0s and the two shifts could even be conjugate. E.g. think of $X$ as the Fibonacci and $Y$ the "flipped Fibonacci", where you interchanged the role of 0s and 1s etc.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Hi, About the last condition, basically it was a bad attempt (sorry about that) to explain my setup. Actually, in my setup $X$ is not a subset of $Y$ and $X \cap Y = \emptyset$. I'll check the paper, thanks a lot! $\endgroup$ – Rafael Alcaraz Barrera Mar 31 '15 at 20:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.