MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $(X_i)$ be a sequence of compact metric spaces and $(f_i)$ a sequence of transitive transformations $f_i:X_i \to X_i$ with $0 < h_{top}(f_i) < \infty$.

The sequence of dynamical systems satifies:

  • $X_i \subset X_{i+1}$, $h_{top}(f_i) < h_{top}(f_{i+1}) $;
  • $X_i$ converges to a compact metric space $X$;
  • $f_{i+1}\mid_{X_i} = f_i$ for every $i$;
  • Besides, there is a transformation $f:X \to X$ such that f is transitive, $0 < h_{top}(f) < \infty$ and $f\mid_{X_i} = f_i$.
  • $h_{top}(f_i)$ converges to $h_{top}(f)$

Assume now that the system $(X_i, f_i)$ is intrinsically ergodic for all $i\ge0$, i.e., it has a unique measure of maximal entropy.

QUESTION. Is $(X,f)$ intrinsically ergodic?

(If it helps, each $(X_i,f_i)$ in my set-up is a transitive subshift of finite type (SFT), but $(X,f)$ is not an SFT.)

If the answer is yes, does there exist a natural way to project the (unique) measure of maximal entropy $\mu$ on $X$ onto $X_i$ so that the projection of $\mu$ is the measure of maximal entropy $\mu_i$ on $X_i$?

share|cite|improve this question
up vote 6 down vote accepted

The answer is no. It's based on a (un?)published example of Crannell, Rudolph and Weiss.

The example is the following shift: $X$ is the subset of $\lbrace 0,\pm 1\rbrace ^{\mathbb Z}$ with the property that $x_k\cdot x_{k+2^n}$ is not allowed to be $-1$ for any values of $k$ and $n$.

What they prove is that there are 2 measures of maximal entropy for $X$: one the Bernoulli (1/2,1/2) measure living on sequences of 0's and 1's; the other the Bernoulli (1/2,1/2) measure living on sequences of 0's and $-1$'s. In fact I showed with Ayse Şahin that these are the unique measures of maximal entropy.

Now if you let $X_i$ be the subset of $X$ where you can't have $i$ consecutive $-1$'s, then $X_i$ is intrinsically ergodic, but $X$ is not.

share|cite|improve this answer
Thank you very much for your answer Anthony. I'd like to know why the sequence of subshifts that you are constructing converges to the non-intrinsically ergodic system that you are mentioning. Best, Rafa – Rafael Alcaraz Feb 9 '12 at 12:11
Anthony, I just realized about why does the sequence converges. Sorry! – Rafael Alcaraz Feb 9 '12 at 15:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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