In this post I discuss a calculation involving topological entropy and expansive maps.

Set Up

Recall that there are a few definitions for the topological entropy of a dynamical system on a compact metric space \(f: X \rightarrow X\). We’ll go through them now. First, it is useful to have the so-called Bowen metric. This is a metric defined by

\[d_n^f(x,y) := \max_{0 \leq i \leq n} d(f^i(x), f^i(y)).\]

In some sense, this is the metric which measures the distance between finite orbit segments. Open balls with respect to this metric will be defined by

\[B_f(x, \epsilon, n) := \{y \in X \ | \ d_n^f(x,y) < \epsilon\}.\]

Next, there are a few kinds of sets we care about. The first of which is an \((n,\epsilon)\)-spanning set. We say that \(E\) is an \((n,\epsilon)\)-spanning set if we have

\[X \subseteq \bigcup_{x \in E} B_f(x,\epsilon,n).\]

On the other end of the spectrum is an \((n,\epsilon)\)-separated set. We say that a set \(E\) is \((n,\epsilon)\)-separated if for every \(x,y \in E\) with \(x \neq y\) we have \(d_n^f(x,y) \geq \epsilon\). Let \(S_d(f,\epsilon,n)\) be the minimal cardinality of an \((n,\epsilon)\)-set. Then we can define the topological entropy as

\[h(f) := \lim_{\epsilon \rightarrow 0^+} \limsup_{n \rightarrow \infty} \frac{1}{n} \log S_d(f,\epsilon,n).\]

As an alternative, if we let \(D_d(f,\epsilon,n)\) be the minimum number of sets whose diameter in the \(d_n^f\) metric is at most \(\epsilon\) and whose union covers \(X\), then we also have

\[h(f) = \lim_{\epsilon \rightarrow 0^+} \limsup_{n \rightarrow \infty} \frac{1}{n}\log D_d(f,\epsilon,n).\]

Finally, if we let \(N_d(f,\epsilon,n)\) be the cardinality of the largest \((n,\epsilon)\)-separated set, then we also have

\[h(f) = \lim_{\epsilon \rightarrow 0^+} \limsup_{n \rightarrow \infty} \frac{1}{n} \log N_d(f,\epsilon,n).\]

We would like to establish a bound involving the topological entropy of an expansive map. Recall that a continuous map \(f : X \rightarrow X\) is expansive if there is a constant \(\delta > 0\) such that if \(d(f^n(x), f^n(y)) < \delta\) for all \(n \geq 0\), then \(x = y\).

Result

Define

\[p(f) := \limsup_{n \rightarrow \infty} \frac{1}{n} \log \# P(n),\]

where

\[P(n) := \{x \in X \ | \ f^n(x) = x\}.\]

Then we’d like to show that \(p(f) \leq h(f)\) for expansive maps. This turns out to be an easy argument. Let \(f : X \rightarrow X\) be an expansive map with expansivity constant \(\delta\). let \(x,y \in P(n)\) with \(x \neq y\). Notice that

\[d_n^f(x,y) = \max_{0 \leq i \leq n} d(f^i(x), f^i(y)) \geq \delta.\]

If this was less than \(\delta\), then we see that \(d(f^k(x), f^k(y)) < \delta\) for all \(k \geq 0\) by periodicity, so we would have \(x = y\), a contradiction. Thus we see that \(P(n)\) is an \((n,\delta)\)-separated set. In particular, it is \((n,\epsilon)\) separated for all \(0 < \epsilon \leq \delta.\) Taking \(\epsilon\) sufficiently small, we see that

\[\# P(n) \leq N_d(f,\epsilon,n)\]

by maximality, hence taking log and limits we get the desired inequality.