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 $X_i$?


1 Answer 1


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.

  • $\begingroup$ 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 $\endgroup$ Feb 9, 2012 at 12:11
  • $\begingroup$ Anthony, I just realized about why does the sequence converges. Sorry! $\endgroup$ Feb 9, 2012 at 15:35

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