7
$\begingroup$

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?

$\endgroup$

1 Answer 1

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. $\endgroup$ Commented Mar 4, 2014 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". $\endgroup$ Commented Mar 4, 2014 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. $\endgroup$ Commented Mar 4, 2014 at 21:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .