# Is a “uniformly minimal” dynamical system ergodic?

Let $$X$$ be a compact metric space, and $$\mu$$ a probability measure on $$X$$ with $$\text{supp} \ \mu = X$$. Suppose $$T: X \to X$$ is continuous, measure preserving and uniformly transitive in the sense that it satisfies the following two conditions:

i) The orbit of every point is dense in $$X$$.

ii) For every $$\varepsilon > 0$$, there exists some $$\delta > 0$$ such that for every $$x \in X$$, and $$r > 0$$ with $$\mu(B_r (x)) > \varepsilon$$, we have $$T(B_s (T^{-1} (x)) \subset B_r (x)$$ for all $$s > 0$$ such that $$\mu(B_s (T^{-1} (x)) < \delta$$.

Question: Does it follow that $$T$$ is ergodic?

• Isn't an invertible minimal system typically of type ii? Even always? Jun 10, 2021 at 9:56
• If the answer is yes, I don't understand your choice of terminology but the answer to your question is "no". Jun 10, 2021 at 9:58

The answer is no, I think any of the usual examples works. Some argument below.

Lemma. Suppose $$X$$ is a compact metric space, $$\mu$$ a nonatomic probability measure on $$X$$, and $$T : X \to X$$ is a minimal measure-preserving homeomorphism. Then $$(X, \mu, T)$$ is "uniformly transitive" (did you mean to write "uniformly minimal" as in the title?)

Proof. i) The orbit of every point is dense, that's the definition of minimality.

ii) Let $$S_r$$ be the set of measures of all balls $$B_r(x)$$ as $$x$$ ranges over $$X$$. Clearly $$S_r$$ does not contain $$0$$ for any $$r > 0$$: apply compactness to the cover by $$B_{r/2}(x)$$'s and you get that for any $$x$$, every point in $$X$$ enters $$B_r(x)$$ after some uniform time $$N$$ meaning the measure of $$B_r(x)$$ is at least $$1/N$$. In fact this means even the closure of $$S_r$$ does not contain $$0$$. On the other hand, $$\sup S_r$$ tends to zero as $$r \rightarrow 0$$ by nonatomicity.

The fact $$0$$ is not in the closure means $$\mu(B_r(x)) > \epsilon$$ puts some lower bound on $$r$$. On the other hand by uniform continuity of $$T^{-1}$$, $$\mu(B_s(T^{-1}(x)) < \delta$$ puts a uniform upper bound on $$s$$. By uniform continuity of $$T$$, there is then a uniform upper bound on the diameter of $$T(B_s(T^{−1}(x))$$. This set contains $$x$$, so we have $$T(B_s(T^{−1}(x)) \subset B_r(x)$$ for small enough $$s$$, meaning any small enough $$\delta$$ works. Square.

Then we just recall the classical result.

Theorem. There is a compact metric space $$X$$, a nonatomic probability measure $$\mu$$ on $$X$$, and a minimal measure-preserving homeomorphism $$T : X \to X$$, such that $$(X, \mu, T)$$ is not ergodic.

Proof. It's enough to find a system $$(X, T)$$ with two different invariant measures. Take words $$w_0^0 = 0$$ and $$w_1^0 = 1$$, and construct $$w^{n+1}_0$$ and $$w^{n+1}_1$$ inductively from $$w^n_0$$ and $$w^n_1$$ by concatenating them around. Do this inductively so that the frequency of $$a$$-symbols in $$w^{n+1}_a$$ is strictly more than $$2/3$$, and take at least $$3$$ copies of both $$w^n_0$$ and $$w^n_1$$ when constructing $$w^n_a$$ for $$a \in \{0,1\}$$. Let $$X \subset \{0,1\}^{\mathbb{Z}}$$ be the subshift (topologically closed set invariant under the left shift action $$T(x)_i = x_{i+1}$$) consisting of all bi-infinite words whose finite subwords all appear in the words $$w^n_a$$. By a compactness argument $$X$$ is nonempty, and $$X$$ is minimal by the following standard argument: if a word $$w$$ appears in some point of $$X$$, it appears in some $$w^n_a$$, thus it appears in $$w^{n+1}_0$$ and $$w^{n+1}_1$$, and every point in $$X$$ is a concatenation of these, so $$w$$ appears with bounded gaps in every point.

Now, for a fixed $$n$$ construct two measures $$\mu^n_0, \mu^n_1$$ on $$X$$ as follows: for $$\mu^n_a$$ take any configuration $$x$$ in the cylinder $$[w^n_a]$$ (any word containing $$w^n_a$$ at the origin), and if $$w^n_a$$ has length $$k$$ define $$\mu^n_a = \sum_{i = 0}^{k-1} T^i(\delta_x)/k$$ where $$\delta_x$$ is the atomic measure on $$\{x\}$$. Clearly $$\mu^n_a([a]) > 2/3$$ and $$\mu^n_a([1-a]) < 1/3$$. Any weak-* limit point of $$\mu^n_a$$ is $$T$$-invariant by a standard computation, and by the definition of the topology we get a measure $$\mu_a$$ with $$\mu_a([a]) \geq 2/3$$ and $$\mu_a([1-a]) \leq 1/3$$. So we have two different invariant measures on $$X$$. Square.

• Ah, very nice - I did miss that any invertible minimal system is already “uniformly transitive” or uniformly minimal might be the better terminology like you say. Jun 10, 2021 at 12:12