3
$\begingroup$

This is Theorem 10.1.1 of Lind & Marcus's book, An Introduction to Symbolic Dynamics and Coding. They say that is "straightfordward" to go from

Let $X$ a shift of finite type and $Y$ a mixing shift of finite type such that $\text{Per}(X)\hookrightarrow\text{Per}(Y)$ and $h(X)<h(Y)$. Then, $X\hookrightarrow Y$.

to

Let $X$ and $Y$ irreducible shift of finite type such that $\text{Per}(X)\hookrightarrow\text{Per}(Y)$ and $h(X)<h(Y)$. Then, $X\hookrightarrow Y$.

How we can drop the mixing hypothesis on $Y$? I have thought in this all new year!

$\endgroup$
3
$\begingroup$

the difference between the two statements is rather subtle. Of course proving the result with $X$ any SFT is more general than assuming $X$ to be irreducible, so there is nothing to do there. For $Y$, going from irreducible to mixing seems to be a stronger condition, however the structure threoy of (one-dimensional) SFTs tells us that a non-mixing irreducible SFT is merely a finite union of mixing ones, all conjugate to each other. If you have not seen this, take a look at Chapter 4.5 of Lind-Marcus. Hence the conditions of the embedding theorem still hold and we might look at one of the cyclic components (i.e. a mixing SFT) of $Y$ instead.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks! I didnt know about that decomposition $\endgroup$ – Veridian Dynamics Jan 4 '19 at 19:08

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.