Devaney chaos and topological entropy

Question

I am searching for dynamical systems on compact spaces which are Devaney chaotic but have topological entropy zero. On the interval such systems do not exist. I think on the Cantor space and on the torus it should be possible to construct such systems, but I do not find a reference to such results.

Answer 1

An example of a dynamical system exhibiting Devaney chaos with zero topological entropy is constructed in the paper "Entropy and Exact Devaney Chaos on Totally Regular Continua". The key ideas behind the construction are outlined below.

Let $(X,d)$ be a compact metric space. The goal is to define a continuous map $f: X \rightarrow X$ that is Devaney chaotic yet has topological entropy $h(f) = 0$.

For $f$ to be Devaney chaotic means satisfying:

1. $f$ has a dense set of periodic points.
2. $f$ is topologically transitive.
3. $f$ displays sensitive dependence on initial conditions.

It's shown such an $f$ cannot exist when $X$ is the unit interval. Instead, $X$ is taken as a totally regular continuum. Conditions are derived under which P-Lipschitz maps on such spaces can achieve zero entropy.

In particular, $A$ be a finite splitting of $X$. For a P-Lipschitz map $f$, the topological entropy is bounded by:

$h(f) \leq \log^+ L_B + 2\theta_B \log^+ L_A$

Here, $L_A = \max_{A \in A} L_A$ is the maximal Lipschitz constant over $A \in A$, $L_B = \max_{A \in B} L_A$ for $B \subseteq A$, and $\theta_B$ quantifies asymptotic transition frequencies in the graph $G_f$ of $f$.

By carefully selecting the Lipschitz constants $\{L_A\}$ and frequencies $\theta_B$, the author shows that it may be possible to construct $f$ satisfying conditions 1), 2), and 3) while still having $h(f) = 0$. I am not sure of an actual explicit construction of such a function but there are other examples of such functions where the Devaney chaos condition is slightly relaxed (e.g. see "Exact Devaney chaos and entropy").

Answer 2

It's fairly easy to make such a system symbolically (i.e. on the Cantor set), but I'm not sure of examples in print. Here is the simplest one I could think of. Our system $(X, T)$ will be a subshift, meaning that $X \subset A^\mathbb{Z}$ for some finite alphabet $A$ (here $A = \{0,1\}$), the dynamics $T$ are given by the left shift map, and $X$ is closed and $T$-invariant.

We recursively define sets $A_k$ of $0$-$1$ words, beginning with $A_0 = \{0,1\}$. We will see from the definition that $|A_k| = k+2$ for all $k$. Our recursive rule is: for all $k$, if $A_k = \{w_1, \ldots, w_{k+2}\}$, then $A_{k+1} = \{w_1^{k+2}, w_2^{k+2}, \ldots, w_m^{k+2}, w_1 w_2 w_3 \ldots w_{k+2}\}$, where $w^i$ is shorthand for the concatenation $ww\ldots w$ of $i$ copies of $w$. For example, $A_1 = \{00, 11, 01\}$ and $A_2 = \{000000, 111111, 010101, 001101\}$. Define $X$ to be the subshift consisting of all $x \in \{0,1\}^{\mathbb{Z}}$ in which every subword of $x$ is a subword of some word in $A_k$ for some $k$.

We first claim that $X$ has dense periodic points. To see this, assume that $v$ is a word in the language of $X$, i.e. some $x \in X$ contains $v$. Then by definition, there exists $k$ and $w \in A_k$ so that $w$ contains $v$ as a subword. By definition, $A_{k+1}$ contains $w^{k+2}$, $A_{k+2}$ contains $(w^{k+2})^{k+3} = w^{(k+2)(k+3)}$, etc. This means that the infinite periodic sequence $\ldots www \ldots \in X$, since every subword of it is a subword of $w^m$ for some $m$, which is itself in $A_{n}$ for large enough $n$. Since $w$ was arbitrary, periodic points of $X$ are dense.

To see that $X$ is transitive, let's denote, for each $k$, $v_k$ by the word in $A_k$ which is the concatenation of all words from $A_{k-1}$ (rather than just a power of one). Clearly $v_k$ is a subword of $v_{k+1}$ for all $k$, so we can take a limit of the $v_k$ to obtain $z \in X$ which contains all $v_k$. Since each $v_{k+1}$ contains all words in $A_k$ as subwords, $z$ contains every word in every $A_k$, and so by definition of $X$ it has dense orbit.

Sensitive dependence is known for all subshifts without isolated points. The key is that the usual metric on $X$ is $d(x,y) = 2^{-k}$, where $k$ is the minimum absolute value of a coordinate $i$ where $x(i) \neq y(i)$. In particular, for any $x \neq y$, there is $i$ s.t. $x(i) \neq y(i)$, then $(T^i x)(0) \neq (T^i y)(0)$, so $d(T^i x, T^i y) = 1$. Now, for any $x \in X$, you can find arbitrarily close $y$ not equal to $x$ (if $x$ is periodic, take an appropriate shift of $z$ above, and if $x$ is not periodic, take a very close periodic points). By the above argument, some shifts of $x,y$ have distance $1$, verifying sensitive dependence.

The entropy of any subshift $X$ is given by the exponential growth rate of the number of words in the language of $X$ as a function of length. In other words, $h(X) = \lim_n \frac{1}{n} \log |L_n(X)|$, where $L_n(X)$ is the set of $n$-letter subwords of points of $X$.

A simple induction shows that all words in $A_k$ have length $(k+1)!$, and we recall that $|A_k| = k + 2$ for all $k$. Now, for any $n$, take the minimal $k$ so that $n \leq (k+1)!$; then $n > k!$. Since every word in every $L_m(X)$ for $m \geq k$ is a concatenation of words in $A_k$, and since $n < (k+1)!$, every $w \in L_n(X)$ is a subword of a concatenation of two words in $A_k$. It is therefore determined by those two words, and the location within the first word at which $w$ begins. Therefore, $|L_n(X)| \leq (k+2)^2 (k+1)! = (k+1)(k^2)^2 k!$. It's clear that for large $k$, $(k+1)(k+2)^2$ is much smaller than $k!$, which in turn is smaller than $n$. So, for large $n$, $|L_n(X)| \leq n^2$, which means that $h(X) \leq \lim_n \frac{\log(n^2)}{n} = 0$.