6
$\begingroup$

Consider the alphabet $\mathcal{A} = \{0,1\}$ and consider a finite set of words $W = \{\omega_1, \ldots , \omega_n\}$ over $\mathcal{A}$. Then the renewal system $\Sigma_{W}$ generated by $W$ is formed by bi-infinite concatenations of words of $W$. My first question is, does every renewal system is intrinsically ergodic?

Secondly, is there a one sided version of the definition of renewal subshift?

On the other hand, Adler asked the following question: Is every transitive subshift of finite type topologically conjugated to a renewal system? To the best of my knowledge the conjecture stills open. Is the conjecture open when the alphabet is $\{0,1\}$?

My third question is: Are there any examples of renewal systems $\Sigma_W \subset \mathcal{A}^{\mathbb{Z}}$ that are subshifts of finite type and viceversa?

$\endgroup$
6
$\begingroup$

There are some examples related to your third question in "Renewal Systems, Sharp-Eyed Snakes, and Shifts of Finite Type" by Johnson and Madden, Amer. Math. Monthly 109 (2002), 258-272. A long time ago Goldberger, Smorodinsky, and I showed that for every possible entropy of a shift of finite type (or, what amounts to the same thing, for the logarithm of every Perron number), there is a renewal system with that entropy. However, as far as I know, Adler's question is still open.

$\endgroup$
3
  • $\begingroup$ Thanks Douglas. Actually the paper that you mention states that every uniquely decipherable renewal system is conjugated to a shift of finite type. That definitively will help me. $\endgroup$ – Rafael Alcaraz Barrera Mar 4 '14 at 20:21
  • 1
    $\begingroup$ You're welcome. Renewal systems are trivially sofic, and the essential issue is exactly the ambiguity in decomposing "sentences" into "words". $\endgroup$ – Douglas Lind Mar 4 '14 at 20:35
  • $\begingroup$ Thanks again. Actually with your comment in mind, the answer to my first question is yes. Since all renewal systems are transitive and sofic therefore they are intrinsically ergodic. $\endgroup$ – Rafael Alcaraz Barrera Mar 4 '14 at 21:04

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.