Is there any continuous transformation on a compact metric space with infinite topological entropy?

Question

It is well-known that any expansive homeomorphism on a compact metric space has finite topological entropy (Walters' Ergodic theory book, page 177). And any homeomorphism on easy spaces like unit circle or unit interval has even zero entropy.

I am trying to find some not-hard-to-prove examples for a continuous transformation (or better, a homeomorphism) on a compact metric space having infinite topological entropy (I can possible make some really weird examples, on say infinite product of compact spaces, but hard to convince myself that the entropy is really infinite).

It would be great if anyone could show me an example of homeomorphism satisfying such but I am also happy with any continuous transformation (but must be on compact metric spaces). Please feel free to use the covering set, spanning set or separating set definitions for the topological entropy.

Answer 1

Let $X$ be some infinite compact metric space and let $$\varphi\colon X^\Bbb N\to X^\Bbb N$$ be the shift map $(\varphi(x))_n=x_{n+1}$. Then $h(\varphi)=\infty$.

This example (with computation in a finite case, see the remark following it) is example 2 in the Adler, Konheim and McAndrew paper in which topological entropy was first defined.

There is also a 2020 paper by Peter Hazard constructing one-dimensional examples, the introduction of his paper gives a historical overview with references for constructions of more maps with infonite entropy.

Answer 2

Here is a more detailed answer; the simplest example is the one Alessandro Codenotti gave; I'll give a few instantiations of it below as well as make some broad comments.

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First of all, the most general obstruction to infinite entropy is by Kushnirenko (see my answer https://math.stackexchange.com/a/4341581/169085, where the point was to guarantee finiteness of entropy). This means that for infinite entropy we can't have differentiable self-maps of compact manifolds or Lipschitz self-maps of compact metric spaces of finite lower box dimension.

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The simplest example that comes to mind is the shift acting on bi-infinite sequences taking values in an infinite compact space. Note that from the ergodic theory point of view these are not weird spaces (in fact, any space traditional ergodic theory is interested in is measurably isomorphic to a sequence space) Let's set up some notation. For any compact metric space $X$ let's put $\Omega_{_X}= F(\mathbb{Z}; X)$, so that $\Omega_{_X}$ is the set of all bi-infinite sequences taking values in $X$. Endowing $\Omega_{_X}$ with product topology gives us a compact metrizable space; as a distance we may use

$$d(\omega_\bullet,\eta_\bullet)=\sum_{n\in\mathbb{Z}}\dfrac{1}{2^{|n|}}d_{_X}(\omega_n,\eta_n).$$

The translation action of $\mathbb{Z}$ induces the shift $\sigma_{_X}$ on $\Omega_{_X}$; it's straightforward that $\sigma_{_X}\in\operatorname{Homeo}(\Omega_{_X})$.

We have that if $\#(X)<\infty$, then $\operatorname{topent}(\sigma_{_X})=\log(\#(X))$ (This is Thm.7.12 on p.178 in Walters' book; as Alessandro pointed out there is also a proof in the Adler-Konheim-McAndrews paper.). Heuristically, $\operatorname{topent}(\sigma_{_X})$ should approach to infinity as $\#(X)$ approaches infinity. To show this one can use the definitions directly. Alternatively, considering specific infinite compact metric spaces one can give arguments for infinite entropy.

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Claim: Let $X$ be compact metric. If $\#(X)=\infty$, then $\operatorname{topent}(\sigma_{_X})=\infty$.

Proof: Let's use $(n,\epsilon)$-separated sets for $\sigma_{_X}$. First note that for any $k\in\mathbb{Z}_{\geq1}$, there is a subset $S_k\subseteq X$ and a number $\delta_k\in\mathbb{R}_{>0}$ such that $\#(S_k)=k$ and $S_k$ is $\delta_k$-separated (that is, $\forall s_1,s_2\in S_k: s_1\neq s_2\implies$ $d_{_X}(s_1,s_2)\geq \delta_k$). Indeed, since $X$ is infinite it has a subset $S_k$ with $k$ elements, and we may choose $\delta_k=\min\{d_{_X}(s_1,s_2)\,|\, s_1,s_2\in S_k\}$.

Fix $n\in\mathbb{Z}_{>0}$, $p\in\mathbb{Z}_{>0}$ and $k\in\mathbb{Z}_{\geq1}$. Consider the open sets

$\cdots X\times X\times \boxed{B\left(s_{i_0},\dfrac{\delta_k}{2}\right)} \times B\left(s_{i_1},\dfrac{\delta_k}{2}\right) \times \cdots \times B\left(s_{i_{n+p-1}},\dfrac{\delta_k}{2}\right) \times X\times X\times \cdots \subseteq \Omega_{_X}$,

where $s_{i_0},s_{i_1},...,s_{i_{n+p-1}}\in S_k$ and $B(x,\rho)$ is the open ball in $X$ w/r/t $d_{_X}$ centered at $x$ with radius $\rho$, and the boxed coordinate is the zeroth entry. Pick one sequence from each one of these open sets and with these form the set $\widehat{S_k}\subseteq X$. Note that $\#\left(\widehat{S_k}\right)= k^{n+p}$.

We claim that $\widehat{S_k}$ is an $\left(n,\dfrac{\delta_k}{2^p} \right)$-separated set for $\sigma_{_X}$. Indeed, let $\alpha_\bullet,\beta_\bullet\in \widehat{S_k}$ be distinct elements. Then for some $k\in\{0,1,...,n+p-1\}$, $d_{_X}(\alpha_k,\beta_k)\geq \dfrac{\delta_k}{2}$. Either $k\in\{0,1,...,n\}$, in which case $d_n^{\sigma_{_X}}(\alpha_\bullet,\beta_\bullet)\geq \dfrac{\delta_k}{2}\geq \dfrac{\delta_k}{2^p}$ (here $d_n^{\sigma_{_X}}$ is the $n$th Bowen metric associated to $\sigma_{_X}$), or else $k\in\{n+1,n+2,...,n+p-1\}$, in which case $d_n^{\sigma_{_X}}(\alpha_\bullet,\beta_\bullet)\geq \dfrac{\delta_k}{2^p}$. In any event $\widehat{S}$ is thus an $\left(n,\dfrac{\delta_k}{2^p} \right)$-separated set for $\sigma_{_X}$, whence we have

$$\operatorname{maxsep}\left(\sigma_{_X};n,\dfrac{\delta_k}{2^p}\right)\geq \#\left(\widehat{S_k}\right)=k^{n+p}.$$

This implies that $\limsup_{n\to \infty}\dfrac{1}{n} \log\left(\operatorname{maxsep}\left(\sigma_{_X};n,\dfrac{\delta_k}{2^p}\right)\right)\geq \log(k)$, and taking $p$ to infinity gives

$$\operatorname{topent}(\sigma_{_X})\geq \log(k).$$

$k$ was arbitrary; so $\operatorname{topent}(\sigma_{_X})=\infty$.

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If we consider more specific examples of infinite compact spaces $X$ it becomes much easier to verify that $\operatorname{topent}(\sigma_{_X})=\infty$. The key to these is to keep in mind that under topological factors topological entropy can't increase and under measure theoretical factors metric entropy can't increase.

Example 1: Consider $X=[0,1]$ and fix $k\in\mathbb{Z}_{\geq1}$. We have a measurable partition $P_k=\left\{\left[0,\dfrac{1}{k}\right[,\left[\dfrac{1}{k},\dfrac{2}{k}\right[,\cdots,\left[\dfrac{k-2}{k},\dfrac{k-1}{k}\right[,\left[\dfrac{k-1}{k},1\right]\right\}$. This defines a measurable quotient map $X\to \{0,1,...,k-1\}$; let $\pi_k: \Omega_{_X}\to \Omega_{k}$ be the associated surjection (here we abbreviated $\Omega_{\{0,1,...,k-1\}}$ by $\Omega_k$). It's clear that via $\pi_k$ becomes $\sigma_k$ is a measurable factor of $\sigma_{_X}$. Further, in we endow $\Omega_{_X}$ with the infinite product probability measure $\mu=\operatorname{leb}^{\otimes \mathbb{Z}}$, $\sigma_{_X}$ becomes measure preserving and $\nu_k=(\pi_k)_\ast(\mu)$ is the infinite product uniform probability measure. Under measurable factors metric entropy can't increase, so

$$\operatorname{ent}_\mu(\sigma_{_X})\geq \operatorname{ent}_{\nu_k}(\sigma_k)= \log(k).$$

Thus $\operatorname{ent}_\mu(\sigma_{_X})=\infty$, since $k$ was arbitrary. By the Goodwyn Theorem (or the variational principle; see my answer https://math.stackexchange.com/a/4341581/169085 ) topological entropy is not less than any metric entropy, so we have that $\operatorname{topent}(\sigma_{_X})=\infty$.

(Note that this is essentially Ex.(8) on p.103 of Walters' book.)

The only reason to evoke metric entropy in the previous example is to be able to make a comparison with a shift system with finitely many states. In the measurable category such a factor is significantly easier to write. With a more algebraic example we can do everything in the topological category. We'll use the inverse limit construction (see e.g. Uniqueness of a homomorphism in a projective limit dynamical system; showing that a projective limit system is compact - Is my proof correct?).

Example 2: Take $X=\lim_{k\to \infty} \mathbb{Z}/k\mathbb{Z}$ as the compact abelian group of profinite integers (we'll only make use of the topology). For any $k\in\mathbb{Z}_{\geq1}$, there is a continuous surjection $X\to \mathbb{Z}/k\mathbb{Z}$; which induces a continuous surjection of the associated shift spaces. Again we have $\operatorname{topent}(\sigma_{_X})\geq \log(k)$ with $k$ arbitrary.

Example 3: As a variant of the previous example, take $X=\lim_{m\to \infty} \mathbb{Z}/p^m\mathbb{Z}$, with $p\in\mathbb{Z}_{\geq2}$ a fixed prime number, so that now $X$ is the compact abelian group of $p$-adic integers. Same comparison argument gives $\operatorname{topent}(\sigma_{_X})\geq m\log(p)$ with $m$ arbitrary.

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Underlying all these examples is the so-called inverse limit theorem for topological entropy (see Goodwyn's paper "The product theorem for topological entropy"):

Theorem (Goodwyn): Let $\{(X_\alpha,\phi_\alpha)\}_{\alpha\in A}$ be a directed system of pairs of compact Hausdorff topological spaces and continuous self-maps with bonding maps $\{\pi_{\alpha\beta}\}_{\alpha\beta}$. Then

$$\operatorname{topent}\left(\lim_\alpha \phi_\alpha\right)\leq \sup_\alpha \operatorname{topent}(\phi_\alpha),$$

and if the bonding maps are surjective, then

$$\operatorname{topent}\left(\lim_\alpha \phi_\alpha\right)= \lim_\alpha \operatorname{topent}(\phi_\alpha).$$

Theorem (Goodwyn): Let $\{(X_k,\phi_k)\}_{k\in\mathbb{Z}_{\geq1}}$ be a countable collection of pairs of compact Hausdorff topological spaces and continuous self-maps. Then

$$\operatorname{topent}\left(\phi_\infty\right)= \sum_{k\in\mathbb{Z}_{\geq1}}\operatorname{topent}(\phi_k),$$

where $\phi_\infty=\prod_{k\in\mathbb{Z}_{\geq1}} \phi_k : \prod_{k\in\mathbb{Z}_{\geq1}} X_k \to \prod_{k\in\mathbb{Z}_{\geq1}} X_k$ is defined by $x_\bullet\mapsto [k\mapsto \phi_k(x_k)]$.

The proofs of these use some abstract machinery, as expected.

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Finally I'd like to return to the matter of obstructions on infinite topological entropy. As mentioned above, Kushnirenko Theorem gives the obstruction on regularity, and dimension one of the underlying manifold is also an obstruction (as we discussed in my answer to another question of yours at https://math.stackexchange.com/a/4363597/169085 no homeomorphism of the closed interval has positive entropy (let alone infinite entropy); and a similar argument gives that no homeomorphism of the circle has positive entropy). Apart from these two obstructions, topological entropy being infinite is a common phenomenon:

Theorem (Yano): Let $M$ be a compact $C^0$ manifold. If $\dim(M)>1$, then

$$\{f\in\operatorname{Homeo}(M)\,|\, \operatorname{topent}(f)=\infty\}$$

is a dense $G_\delta$ subset.

In other words, as long as the regularity and dimension obstructions are bypassed infinite topological entropy is a generic property. The argument is a Baire category argument; the group $\operatorname{Homeo}(M)$ is completely metrizable space with distance

$$d_{\operatorname{Homeo}(M)}(f,g)=d_{C^0}(f,g)+d_{C^0}(f^{-1},g^{-1}) $$

See Yano's paper "A remark on the topological entropy of homeomorphisms" for details.