Expansive actions with specification on uniform spaces, topological entropy, and the Myhill property

TL;DR

This paper proves that expansive actions with the weak specification property of an amenable group on a compact space have the Myhill property, meaning pre-injective maps commuting with the action are surjective, extending previous metrizable results.

Contribution

It extends the Myhill property to expansive actions with weak specification on non-metrizable compact spaces for amenable groups.

Findings

Pre-injective maps are surjective under the given conditions.

The result generalizes previous metrizable space cases.

It links expansive actions, weak specification, and the Myhill property.

Abstract

We prove that every expansive continuous action with the weak specification property of an amenable group $G$ on a compact Hausdorff space $X$ has the Myhill property, i.e., every pre-injective continuous self-mapping of $X$ commuting with the action of $G$ on $X$ is surjective. This extends a result previously obtained by Hanfeng Li in the case when $X$ is metrizable.

Equations

$${\Delta }_{\epsilon }(X,d):=\{(x,y)\in X\times X:d(x,y)<\epsilon \},$$

$$Y^{(E)}:=\bigcap _{g\in E}{g}^{-1}Y.$$

$$gx:=(g{x}_k{)}_{k\in K}$$

$$gx(h):=x(g^{-1}h)\text{ for all g,h∈G and x∈AG.}$$

$$W(\Omega ,U):=\{(x,y)\in A^G\times A^G:(x(g),y(g))\in U\text{ for all }g\in \Omega \},$$

$$W(\Omega ):=\{(x,y)\in A^G\times A^G:x{|}_{\Omega }=y{|}_{\Omega }\},$$

$$\underset{j\in J}{\lim }\frac{|g{F}_j\setminus F_j|}{|F_j|}=0\text{for all g∈G}.$$

$$(X\times X)\setminus V\subset \bigcup _{g\in E}((X\times X)\setminus g^{-1}{U}_0).$$

$$N(\alpha \vee \beta )\le N(\alpha )\cdot N(\beta ).$$

$$N(g\alpha )=N(\alpha ).$$

$${\alpha }^{(F)}:=\underset{g\in F}{\bigvee }{g}^{-1}\alpha .$$

$$h(F\cup F^{\prime })=\log N({\alpha }^{(F\cup F^{\prime })})=\log N({\alpha }^{(F)}\vee {\alpha }^{(F^{\prime })})\le \log N({\alpha }^{(F)})+\log N({\alpha }^{(F^{\prime })})=h(F)+h(F^{\prime })$$

$$h(Fg)=\log N({\alpha }^{(Fg)})=\log N\left(\underset{k\in Fg}{\bigvee }{k}^{-1}\alpha \right)=\log N\left(g^{-1}{\alpha }^{(F)}\right)=\log N\left({\alpha }^{(F)}\right)$$

$${\mathrm{h}}_{\text{top}}(X,G,\alpha ):=\underset{j\in J}{\lim }\frac{\log N({\alpha }^{(F_j)})}{|F_j|}$$

$${\mathrm{h}}_{\text{top}}(X,G):=\underset{\alpha }{\sup }{\mathrm{h}}_{\text{top}}(X,G,\alpha ),$$

$${\mathrm{h}}_{\text{top}}(X,G)=\underset{\alpha \in \mathcal{C}}{\lim }{\mathrm{h}}_{\text{top}}(X,G,\alpha ),$$

$${\mathrm{h}}_{\text{top}}(Y,G,\alpha )=\underset{j\in J}{\lim }\frac{\log N_Y({\alpha }^{(F_j)})}{|F_j|}\le \underset{j}{\lim }\frac{\log N_X(({\alpha }^{\prime }{)}^{(F_j)})}{|F_j|}={\mathrm{h}}_{\text{top}}(X,G,{\alpha }^{\prime })\le {\mathrm{h}}_{\text{top}}(X,G).$$

$${\mathrm{h}}_{\text{top}}(Y,G,\alpha )=\underset{j\in J}{\lim }\frac{\log N_Y({\alpha }^{(F_j)})}{|F_j|}=\underset{j\in J}{\lim }\frac{\log N_X(({\alpha }_X{)}^{(F_j)})}{|F_j|}={\mathrm{h}}_{\text{top}}(X,G,{\alpha }_X)\le {\mathrm{h}}_{\text{top}}(X,G).$$

$$U^{(F)}:=\bigcap _{g\in F}{g}^{-1}U.$$

$$\operatorname{sep}(X,G,F,U)\coloneqq\max\{|Z|:\mbox{$Z\subset X$ is $(F,U)$-separated}\}.$$

$$\operatorname{span}(X,G,F,U)\coloneqq\min\{|Z|:\mbox{$Z\subset X$ is $(F,U)$-spanning}\}.$$

$$\begin{array}{rl}\mathrm{sep}(X,G,Fg,U)& =\mathrm{sep}(X,G,F,U)\\ \mathrm{span}(X,G,Fg,U)& =\mathrm{span}(X,G,F,U)\\ \mathrm{cov}(X,G,Fg,U)& =\mathrm{cov}(X,G,F,U).\end{array}$$

$$\begin{gathered} \mathrm{cov}(X,G,F,V\circ V)\le \mathrm{span}(X,G,F,V)\le \mathrm{sep}(X,G,F,V) \\ \le \mathrm{span}(X,G,F,U)\le \mathrm{cov}(X,G,F,U). \end{gathered}$$

$$\mathrm{cov}(X,G,E\cup F,U)\le \mathrm{cov}(X,G,E,U)\cdot \mathrm{cov}(X,G,F,U).$$

$${\mathrm{h}}_{\text{sep}}(X,G,\mathcal{F},U):=\underset{j\in J}{\mathrm{limsup}}\frac{\log \mathrm{sep}(X,G,F_j,U)}{|F_j|}$$

$${\mathrm{h}}_{\text{spa}}(X,G,\mathcal{F},U):=\underset{j\in J}{\mathrm{limsup}}\frac{\log \mathrm{span}(X,G,F_j,U)}{|F_j|}.$$

$${\mathrm{h}}_{\text{cov}}(X,G,U):=\underset{j\in J}{\lim }\frac{\log \mathrm{cov}(X,G,F_j,U)}{|F_j|}$$

$$\begin{gathered} {\mathrm{h}}_{\text{cov}}(X,G,V\circ V)\le {\mathrm{h}}_{\text{spa}}(X,G,\mathcal{F},V)\le {\mathrm{h}}_{\text{sep}}(X,G,\mathcal{F},V) \\ \le \mathrm{span}(X,G,\mathcal{F},U)\le {\mathrm{h}}_{\text{cov}}(X,G,U). \end{gathered}$$

$${\mathrm{h}}_{\text{sep}}(X,G,\mathcal{F})$$

$${\mathrm{h}}_{\text{spa}}(X,G,\mathcal{F})$$

$${\mathrm{h}}_{\text{cov}}(X,G)$$

Full text

1. 1 Introduction

2. 2 Background material

3. 2.1 General notation

4. 2.2 Subsets of $X\times X$

5. 2.3 Uniform spaces

6. 2.4 Ultrauniform spaces

7. 2.5 Actions

8. 2.6 Covers

9. 2.7 Shift spaces

10. 2.8 Amenable groups

11. 3 Expansiveness

12. 4 Topological entropy

13. 4.1 Topological entropy

14. 4.2 $(F,U)$-separated subsets, $(F,U)$-spanning subsets, and $(F,U)$-covers

15. 4.3 Uniform approaches to topological entropy

16. 4.4 Topological entropy and expansiveness

17. 5 Homoclinicity

18. 6 Weak specification

19. 7 Proof of the main result

1. Introduction

A topological dynamical system is a pair $(X,G)$, where $X$ is a topological space and $G$ is a group acting continuously on $X$. Analogously, a uniform dynamical system is a pair $(X,G)$, where $X$ is a uniform space and $G$ is a group acting uniformly continuously on $X$. Every uniform dynamical system may be regarded as a topological dynamical system. Indeed, every uniform space has an underlying topology and every uniformly continuous self-mapping of a uniform space is continuous with respect to this topology. In the other direction, every topological dynamical system $(X,G)$ with $X$ compact Hausdorff can be regarded as a uniform dynamical system. Indeed, a compact Hausdorff space admits a unique uniform structure compatible with its topology and every continuous self-mapping of the space is uniformly continuous with respect to this uniform structure.

An endomorphism of a topological dynamical system $(X,G)$ is a continuous $G$-equivariant self-mapping of $X$, that is, a continuous map $f\colon X\to X$ such that $f(gx)=gf(x)$ for all $g\in G$ and $x\in X$. One says that the topological dynamical system $(X,G)$ is surjunctive if every injective endomorphism of $(X,G)$ is surjective. The term “surjunctive” was created by Gottschalk [20] in the early 1970s and the search for conditions guaranteeing surjunctivity of certain classes of dynamical systems has attracted much interest, especially in the last two decades since the seminal work of Gromov [23].

Let $(X,G)$ be a uniform dynamical system. Two points $x,y\in X$ are called homoclinic if their orbits are asymptotically close, i.e., for every entourage $U$ of $X$, there exists a finite subset $\Omega\subset G$ such that $(gx,gy)\in U$ for all $g\in G\setminus\Omega$. Homoclinicity is an equivalence relation on $X$. An endomorphism of $(X,G)$ is called pre-injective if its restriction to each homoclinicity class is injective.

One says that a uniform dynamical system $(X,G)$ satisfies the Myhill property if every pre-injective endomorphism of $(X,G)$ is surjective. As injectivity clearly implies pre-injectivity, a uniform dynamical system is surjunctive whenever it has the Myhill property.

One says that a uniform dynamical system $(X,G)$ is expansive if there is an entourage $U_{0}$ of $X$ such that there is no pair $(x,y)\in X\times X$ with $x\not=y$ satisfying $(gx,gy)\in U_{0}$ for all $g\in G$. Given any set $A$, the full $G$-shift with alphabet $A$, i.e., the system $(A^{G},G)$, where $A^{G}$ is equipped with its uniform prodiscrete structure and $G$ acts on $A^{G}$ by the $G$-shift (see Formula (2.3)), yields an example of an expansive uniform dynamical system.

One says that the uniform dynamical system $(X,G)$ has the weak specification property if for every entourage $U$ of $X$, there is a finite subset $\Lambda=\Lambda(U)\subset G$ satisfying the following condition: for every finite sequence $\Omega_{1},\dots,\Omega_{n}$ of finite subsets of $G$ such that $\Omega_{j}\cap\Lambda\Omega_{k}=\varnothing$ for all distinct $j,k\in\{1,\dots,n\}$ and every sequence $x_{1},\dots,x_{n}$ of points of $X$, there exists a point $x\in X$ such that $(gx,gx_{i})\in U$ for all $g\in\Omega_{i}$ and $1\leq i\leq n$.

The main goal of the present paper is to establish the following result.

Theorem 1.1**.**

Let $X$ be a compact Hausdorff space equipped with a continuous action of an amenable group $G$. Suppose that $(X,G)$ is expansive and has the weak specification property. Then $(X,G)$ has the Myhill property. In particular, $(X,G)$ is surjunctive.

Theorem 1.1 has been previously established by Li [30, Theorem 1.1] under the additional hypotheses that $X$ is metrizable and $G$ is countable. By virtue of the Bryant-Eisenberg theorem, if a compact Hausdorff space $X$ admits an expansive continuous action of a countable group $G$, then $X$ is necessarily metrizable (see Theorem 3.4 below). Therefore Theorem 1.1 reduces to Theorem 1.1 in [30] when the group $G$ is countable. However, for any uncountable group $G$, there exist expansive topological dynamical systems with the weak specification property $(X,G)$, with $X$ compact Hausdorff but non-metrizable (e.g. the full shift $(A^{G},G)$ with $A$ a finite set with more than one element).

We cannot drop the weak specification hypothesis in Theorem 1.1. Indeed, consider a discrete space with two distinct points $X=\{x_{1},x_{2}\}$ and a group $G$ fixing each point of $X$. Observe that the space $X$ is compact metrizable and $(X,G)$ is expansive. Moreover, each homoclinicity class is reduced to a single point so that the endomorphism $f\colon X\to X$ given by $f(x_{1})=f(x_{2})=x_{1}$ is pre-injective. As $f$ is not surjective, we deduce that $(X,G)$ fails to have the Myhill property. Note however that this dynamical system is surjunctive since $X$ is finite. A non-surjunctive example is provided by the subshift $(X,\mathbb{Z})$, where $X\subset\{0,1\}^{\mathbb{Z}}$ consists of all bi-infinite sequences of [math]s and $1$s containing at most one chain of $1$s. Here also $X$ is compact metrizable and $(X,\mathbb{Z})$ is expansive. However, the map $f\colon X\to X$ which replaces the word $10$ if it appears in a configuration $x\in X$ by the word $11$ is an injective endomorphism of $(X,\mathbb{Z})$. This endomorphism is not surjective since any configuration with exactly one occurrence of the symbol $1$ cannot be in the image of $f$. Therefore $(X,\mathbb{Z})$ is not surjunctive.

The expansiveness assumption cannot be either dropped in Theorem 1.1. Indeed, take any topological dynamical system $(X,G)$ with the weak specification property and $X$ compact Hausdorff with more than one point (e.g. the $G$-shift on $\{0,1\}^{G}$). Consider the topological dynamical system $(X^{\mathbb{N}},G)$, where $X^{\mathbb{N}}=\prod_{n\in\mathbb{N}}X$ is equipped with the product topology and $G$ acts diagonally on $X^{\mathbb{N}}$. Then $X^{\mathbb{N}}$ is compact Hausdorff and the topological dynamical system $(X^{\mathbb{N}},G)$ has the weak specification property (cf. Proposition 6.3). However, $(X^{\mathbb{N}},G)$ is not surjunctive (and hence does not have the Myhill property) since the map $f\colon X^{\mathbb{N}}\to X^{\mathbb{N}}$ defined by $f(x)(0)=x(0)$ and $f(x)(n)=x(n-1)$ for all $n\geq 1$, is an injective but not surjective endomorphism of $(X^{\mathbb{N}},G)$.

It follows from a result of Bartholdi [2] that, for any non-amenable group $G$, there exists a finite set $A$ such that the full $G$-shift $(A^{G},G)$ does not have the Myhill property. This shows in particular that the amenability assumption cannot be removed from Theorem 1.1.

A subshift $X\subset A^{G}$ has the weak specification property for the $G$-shift if and only if $X$ is strongly irreducible (see Proposition 6.7 below). Therefore, Theorem 1.1 yields the following.

Corollary 1.2**.**

Let $A$ be a finite set and let $G$ be an amenable group. Let $X\subset A^{G}$ be a strongly irreducible subshift. Equip $X$ with the shift action of $G$ and the topology induced by the prodiscrete topology on $A^{G}$. Then $(X,G)$ has the Myhill property. In particular, $(X,G)$ is surjunctive. ∎

In the case when $X$ is the full shift $A^{G}$ and $G=\mathbb{Z}^{d}$ (the free abelian group of finite rank $d$), Corollary 1.2 is due to Myhill in [32]. Myhill’s result was subsequently extended to full shifts over finitely generated amenable groups by Machì, Scarabotti,and the first author in [12] and then to strongly irreducible subshifts of finite type over finitely generated amenable groups by Fiorenzi in [18]. Finally, Corollary 1.2 was established in its full generality by the authors in [9].

The paper is organized as follows. In Section 2, we introduce notation and review some background material on uniform spaces, ultrauniform spaces, group actions, shift spaces, and amenable groups. Section 3 investigates expansiveness for uniform dynamical systems. We prove the Bryant-Eisenberg theorem stating that any compact Hausdorff space admitting an expansive continuous action of a countable group is necessarily metrizable (see Theorem 3.4). Section 4 is devoted to the definition and basic properties of the topological entropy $\operatorname{h_{\scriptstyle\text{\rm top}}}(X,G)$ of a topological dynamical system $(X,G)$ with $G$ amenable. In Section 5, we introduce the notion of homoclinicity for uniform dynamical systems. We prove that two configurations in a shift space are homoclinic if and only if they coincide outside of a finite subset of the underlying group (Proposition 5.2). In Section 6, we study weak specification for uniform dynamical systems. Our definition of weak specification coincides with the one in [13] and [30] in the compact metrizable case. Weak specification for subshifts with discrete alphabet is equivalent to strong irreducibility (see Proposition 6.7). We also show that if $X$ is a compact Hausdorff space with more than one point equipped with a continuous action with the weak specification property of an amenable group $G$, then $\operatorname{h_{\scriptstyle\text{\rm top}}}(X,G)>0$ (see Theorem 6.8). The proof of Theorem 1.1 is given in Section 7 and is divided into two parts, which are of independent interest. We first show (Theorem 7.1) that if $X$ is a compact Hausdorff space equipped with a continuous action with the weak specification property of an amenable group $G$ and $Y\subsetneqq X$ is a proper closed $G$-invariant subset such that the action of $G$ restricted to $Y$ is expansive, then $\operatorname{h_{\scriptstyle\text{\rm top}}}(Y,G)<\operatorname{h_{\scriptstyle\text{\rm top}}}(X,G)$. Next we establish Theorem 7.2 which says that if $X$ and $Y$ are compact Hausdorff spaces equipped with expansive continuous actions of an amenable group $G$ such that the action of $G$ on $X$ has the weak specification property and $\operatorname{h_{\scriptstyle\text{\rm top}}}(Y,G)<\operatorname{h_{\scriptstyle\text{\rm top}}}(X,G)$, then any continuous $G$-equivariant map $f\colon X\to Y$ fails to be pre-injective. These two results combined together immediately yield Theorem 1.1.

Acknowledgments. We express our deepest gratitude to the anonymous referee for the most careful reading of our manuscript and the numerous suggestions and remarks.

2. Background material

2.1. General notation

Given sets $A$ and $B$, we write $A\subset B$ if every element in $A$ is also in $B$. We denote by $A^{B}$ the set consisting of all maps $x\colon B\to A$. When $A$ is a finite set, we write $|A|$ its cardinality.

2.2. Subsets of $X\times X$

Let $X$ be a set.

We denote by $\Delta_{X}$ the diagonal of $X$, that is, the subset $\Delta_{X}\subset X\times X$ consisting of all pairs $(x,x)$ with $x\in X$.

Let $U\subset X\times X$. Given a point $x\in X$, we denote by $U[x]$ the subset of $X$ consisting of all points $y\in X$ such that $(x,y)\in U$.

One says that $U$ is reflexive if $\Delta_{X}\subset U$.

The inverse of $U$ is the subset $U^{*}\subset X\times X$ consisting of all pairs $(x,y)$ such that $(y,x)\in U$. One says that $U$ is symmetric if $U^{*}=U$.

The composite of $U$ with another subset $V\subset X\times X$ is the subset $U\circ V\subset X\times X$ consisting of all pairs $(x,y)\in X\times X$ such that there exists $z\in X$ with $(x,z)\in V$ and $(z,y)\in U$. One says that $U$ is transitive if $U\circ U\subset U$.

Observe that the set consisting of all subsets of $X\times X$ is a monoid for $\circ$ with identity element $\Delta_{X}$ and that the map $U\mapsto U^{*}$ is an anti-involution of this monoid. This monoid is ordered for inclusion in the sense that $U\circ V\subset U^{\prime}\circ V^{\prime}$ whenever $U,V,U^{\prime},V^{\prime}\subset X\times X$ satisfy $U\subset U^{\prime}$ and $V\subset V^{\prime}$. This implies in particular that $U\subset U\circ V$ if $\Delta_{X}\subset V$ and that $V\subset U\circ V$ if $\Delta_{X}\subset U$.

Note also that a relation $\sim$ on $X$ is reflexive (resp. symmetric, resp. transitive) if and only if its graph $\Gamma(\sim)\coloneqq\{(x,y):x\sim y\}$ is a reflexive (resp. symmetric, resp. transitive) subset of $X\times X$.

2.3. Uniform spaces

The theory of uniform spaces we briefly review here was introduced by André Weil in [37]. The reader is referred to the monographs [3, Ch. 2], [25], and [26, Ch. 6] for a more detailed exposition.

Let $X$ be a set. A uniform structure on $X$ is a nonempty set $\mathcal{U}$ of subsets of $X\times X$, whose elements are called the entourages of $X$, satisfying the following conditions:

1. (UNI-1)

every $U\in\mathcal{U}$ is reflexive; 2. (UNI-2)

if $U\in\mathcal{U}$ and $U\subset V\subset X\times X$, then $V\in\mathcal{U}$; 3. (UNI-3)

if $U\in\mathcal{U}$ and $V\in\mathcal{U}$, then $U\cap V\in\mathcal{U}$; 4. (UNI-4)

if $U\in\mathcal{U}$, then $U^{*}\in\mathcal{U}$; 5. (UNI-5)

if $U\in\mathcal{U}$, then there exists $V\in\mathcal{U}$ such that $V\circ V\subset U$.

Remark 2.1**.**

It immediately follows from conditions (UNI-3), (UNI-4), and (UNI-5) that, given any entourage $U\in\mathcal{U}$, there exists a symmetric entourage $V\in\mathcal{U}$ such that $V\circ V\subset U$.

Let $X$ be a set and let $U\subset X\times X$.

A set equipped with an uniform structure is called a uniform space.

If $X$ is a uniform space and $Y\subset X$, then the uniform structure on $X$ naturally induces a uniform structure on $Y$. The entourages of the uniform structure induced by $X$ on $Y$ are the sets of the form $U\cap(Y\times Y)$, where $U$ runs over all entourages of $X$.

A subset $\mathcal{B}$ of a uniform structure $\mathcal{U}$ on a set $X$ is called a base of entourages if for every entourage $U\in\mathcal{U}$, there exists an entourage $V\in\mathcal{B}$ such that $V\subset U$.

The discrete uniform structure on a set $X$ is the uniform structure on $X$ whose entourages are all the reflexive subsets of $X\times X$.

If $X$ is a uniform space, there is an induced topology on $X$ characterized by the fact that the neighborhoods of an arbitrary point $x\in X$ consist of the sets $U[x]$, where $U$ runs over all entourages of $X$. This topology is Hausdorff if and only if the intersection of all the entourages of $X$ is reduced to the diagonal $\Delta_{X}$.

One says that a topological space $X$ is uniformizable if there is a uniform structure on $X$ inducing its topology.

Every compact Hausdorff space $X$ is uniquely uniformizable, that is, there is a unique uniform structure on $X$ inducing its topology. The entourages of this uniform structure are the neighborhoods of the diagonal in $X\times X$ (see [3, Théorème 1, TG II.27]).

When $X$ is a compact Hausdorff space, it immediately follows from the normality of $X\times X$ that the closed entourages (i.e., the entourages that are closed in $X\times X$) form a base of entourages of the uniform structure on $X$.

If $d$ is a metric on a set $X$, then $d$ defines a uniform structure on $X$. A base of entourages for this uniform structure consists of all the symmetric sets

[TABLE]

with $\varepsilon>0$. The topology associated with this uniform structure coincides with the topology defined by the metric $d$.

A uniform space $X$ is called metrizable if there exists a metric $d$ on $X$ which defines the uniform structure on $X$. The metrization theorem for uniform spaces (see, e.g., [26, Chapter 6, Theorem 13]), states that a uniform space is metrizable if and only if it is Hausdorff and admits a countable base of entourages.

A map $f\colon X\to Y$ between uniform spaces is said to be uniformly continuous if $(f\times f)^{-1}(W)$ is an entourage of $X$ for every entourage $W$ of $Y$. One says that $f\colon X\to Y$ is a uniform isomorphism if $f$ is bijective with $f$ and $f^{-1}$ both uniformly continuous.

If $(X_{k})_{k\in K}$ is a family of uniform spaces, indexed by a set $K$, the product uniform structure on the product set $X\coloneqq\prod_{k\in K}X_{k}$ is the smallest (with respect to inclusion) uniform structure on $X$ such that all projection maps $X\to X_{k}$, $k\in K$, are uniformly continuous. In the case when the uniform structure on each $X_{k}$ is the discrete uniform structure, the product uniform structure on $X$ is called the prodiscrete uniform structure.

2.4. Ultrauniform spaces

A uniform structure $\mathcal{U}$ on a set $X$ is called a ultrauniform structure if $\mathcal{U}$ admits a base of entourages consisting of graphs of equivalence relations. In other words, $\mathcal{U}$ is a ultrauniform structure if for every entourage $U\in\mathcal{U}$, there exists a symmetric and transitive entourage $V\in\mathcal{U}$ such that $V\subset U$. A set equipped with a ultrauniform structure is called a ultrauniform space.

Examples 2.2**.**

1. A set equipped with its discrete uniform structure is a ultrauniform space.

2. The product of a family of ultrauniform spaces, equipped with its product uniform structure, is a ultrauniform space. In particular, the prodiscrete uniform structure on a product of sets is ultrauniform.

3. If $X$ is a ultrauniform space and $Y\subset X$, then the uniform structure induced on $Y$ by the uniform structure on $X$ is also ultrauniform.

Let $(X,d)$ be a metric space and suppose that $d$ satisfies the ultrametric inequality, i.e., $d(x,y)\leq\max(d(x,z),d(y,z))$ for all $x,y,z\in X$. Then the uniform structure defined by $d$ on $X$ is ultrauniform. Indeed, if $d$ satisfies the ultrametric inequality, then the sets $\Delta_{\varepsilon}(X,d)$ defined in (2.1) are clearly transitive.

2.5. Actions

An action of a group $G$ on a set $X$ is a map $(g,x)\mapsto gx$ from $G\times X$ into $X$ such that $g_{1}(g_{2}x)=(g_{1}g_{2})x$ and $1_{G}x=x$ for all $g_{1},g_{2}\in G$ and $x\in X$.

Let $X$ be a set equipped with an action of a group $G$. Let $Y\subset X$. For $g\in G$, we write $gY\coloneqq\{gy:y\in Y\}$. Given a subset $E\subset G$, we write

[TABLE]

Thus $x\in Y^{(E)}$ if and only if $gx\in Y$ for all $g\in E$. One says that $Y$ is $G$-invariant if $gY=Y$ for all $g\in G$.

Suppose that $(X_{k})_{k\in K}$ is a family of sets and that a group $G$ acts on each of the sets $X_{k}$, $k\in K$. Then $G$ naturally acts on the product set $\prod_{k\in K}X_{k}$ via the diagonal action defined by

[TABLE]

for all $x=(x_{k})_{k\in K}\in\prod_{k\in K}X_{k}$.

In particular, if a group $G$ acts on a set $X$, then $G$ naturally acts on $X\times X$ via the diagonal action.

If a group $G$ acts on two sets $X$ and $X^{\prime}$, one says that a map $f\colon X\to X^{\prime}$ is $G$-equivariant if $f(gx)=gf(x)$ for all $g\in G$ and $x\in X$.

An action of a group $G$ on a topological space $X$ is said to be continuous if the map $x\mapsto gx$ is continuous on $X$ for every $g\in G$. This amounts to saying that $g^{-1}U$ is an open subset of $X$ for every open subset $U\subset X$ and any $g\in G$.

An action of a group $G$ on a uniform space $X$ is called uniformly continuous if the map $x\mapsto gx$ is uniformly continuous on $X$ for each $g\in G$. This amounts to saying that $g^{-1}U$ is an entourage of $X$ for every entourage $U\subset X\times X$ of $X$ and any $g\in G$.

Let $X$ and $Y$ be topological spaces equipped with a continuous action of a group $G$. One says that the topological dynamical system $(Y,G)$ is a topological factor of the topological dynamical system $(X,G)$ if there exists a continuous $G$-equivariant surjective map from $X$ onto $Y$. One says that the topological dynamical systems $(X,G)$ and $(Y,G)$ are topologically conjugate if there exists a $G$-equivariant homeomorphism between $X$ and $Y$. Similarly, if $X$ and $Y$ are uniform spaces equipped with a uniformly continuous action of a group $G$, one says that the uniform dynamical system $(Y,G)$ is a uniform factor of the uniform dynamical system $(X,G)$ if there exists a uniformly continuous $G$-equivariant surjective map from $X$ onto $Y$. One says that the uniform dynamical systems $(X,G)$ and $(Y,G)$ are uniformly conjugate if there exists a $G$-equivariant uniform isomorphism between $X$ and $Y$.

2.6. Covers

Let $X$ be a set. A cover of $X$ is a set of subsets of $X$ whose union is $X$. Let $\alpha$ and $\beta$ be covers of $X$. The join of $\alpha$ and $\beta$ is the cover $\alpha\vee\beta$ of $X$ consisting of all the subsets of the form $A\cap B$ with $A\in\alpha$ and $B\in\beta$. One says that $\beta$ is a subcover of $\alpha$ if $\beta\subset\alpha$. One says that $\beta$ is a refinement of $\alpha$ if for every $B\in\beta$, there exists $A\in\alpha$ such that $B\subset A$. Note that $\alpha\vee\beta$ is a refinement of both $\alpha$ and $\beta$.

When the set $X$ is equipped with an action of a group $G$, given a cover $\alpha$ of $X$ and an element $g\in G$, we denote by $g\alpha$ the cover of $X$ consisting of all the subsets of the form $gA$ with $A\in\alpha$.

When $X$ is a topological space, an open cover of $X$ is a cover of $X$ whose elements are all open subsets of $X$.

2.7. Shift spaces

Let $G$ be a group and let $A$ be a set called the alphabet. The set $A^{G}$ consisting of all maps $x\colon G\to A$ is called the set of configurations over the group $G$ and the alphabet $A$.

Given a subset $\Omega\subset G$ and a configuration $x\in A^{G}$, we shall denote by $x|_{\Omega}$ the restriction of $x$ to $\Omega$, i.e., the map $x|_{\Omega}\in A^{\Omega}$ given by $x|_{\Omega}(g)=x(g)$ for all $g\in\Omega$.

The shift on $A^{G}$ is the action of $G$ on $A^{G}$ defined by

[TABLE]

Suppose now that $A$ is a uniform space. We equip $A^{G}=\prod_{g\in G}A$ with the product uniform structure. A base of entourages for this uniform structure consists of all the sets

[TABLE]

where $\Omega$ runs over all finite subsets of $G$ and $U$ runs over all entourages of $A$. The shift action on $A^{G}$ is clearly uniformly continuous.

A $G$-invariant closed subset $X\subset A^{G}$ is called a subshift of $A^{G}$.

In the case when $A$ is equipped with the discrete uniform structure, the corresponding product uniform structure on $A^{G}$ is the prodiscrete uniform structure and a base of entourages of $A^{G}$ is formed by the sets

[TABLE]

where $\Omega$ runs over all finite subsets of $G$ (see [7]).

Note that if $G$ is uncountable and $A$ is a Hausdorff (e.g. discrete) uniform space with more than one element, then $A^{G}$ is not metrizable (not even first countable).

2.8. Amenable groups

There are many equivalent definitions of amenability for groups in the literature (see for example [22], [34], [7]). In the present paper, we shall only use the following one.

Definition 2.3**.**

A group $G$ is called amenable if there exists a net $\mathcal{F}=(F_{j})_{j\in J}$ of nonempty finite subsets $F_{j}\subset G$ such that

[TABLE]

Such a net $\mathcal{F}$ is called a left Følner net for $G$.

The class of amenable groups includes all finite groups, all abelian groups, all solvable groups, and all finitely generated groups with subexponential growth. Moreover, it is closed under taking subgroups, quotients, extensions, and directed limits.

Given a group $G$, we denote by $\mathcal{P}_{\text{fin}}(G)$ the set of all finite subsets of $G$. A map $h\colon\mathcal{P}_{\text{fin}}(G)\to\mathbb{R}$ is said to be subadditive if $h(A\cup B)\leq h(A)+h(B)$ for all $A,B\in\mathcal{P}_{\text{fin}}(G)$. It is said to be right-invariant if $h(Ag)=h(A)$ for all $A\in\mathcal{P}_{\text{fin}}(G)$ and $g\in G$. The following convergence result is due to Ornstein and Weiss [33] (see also [24, Section 1.3.1], [27], [11]).

Theorem 2.4** (Ornstein-Weiss lemma).**

Let $G$ be an amenable group and let $h\colon\mathcal{P}_{\text{fin}}(G)\to\mathbb{R}$ be a subadditive right-invariant map. Then there exists a real number $\lambda\geq 0$, depending only on the map $h$, with the following property: if $(F_{j})_{j\in J}$ is a left Følner net for $G$ then the net of real numbers $\left(\dfrac{h(F_{j})}{|F_{j}|}\right)_{j\in J}$ converges to $\lambda$.

3. Expansiveness

An action of a group $G$ on a uniform space $X$ is called expansive if there exists an entourage $U_{0}$ of $X$ satisfying the following property: for all points $x,y\in X$ with $x\not=y$, there exists an element $g\in G$ such that $(gx,gy)\notin U_{0}$. Such an entourage $U_{0}$ is then called an expansiveness entourage for the dynamical system $(X,G)$.

Example 3.1**.**

Let $G$ be a group and let $A$ be a set. Equip $A^{G}$ with its prodiscrete uniform structure and the shift action of $G$. Then $(A^{G},G)$ is expansive. Indeed, the set $W(\{1_{G}\})$ defined by (2.5) is clearly an expansiveness entourage for the system.

Example 3.2**.**

Let $(X,G)$ be an expansive uniform dynamical system and let $Y\subset G$ be a $G$-invariant subset. Then the uniform dynamical system $(Y,G)$ is also expansive. Indeed, if $U_{0}$ is an expansiveness entourage for $(X,G)$, then $U_{0}\cap(Y\times Y)$ is an expansiveness entourage for $(Y,G)$.

Example 3.3**.**

Combining the two previous examples, we deduce that if $G$ is a group, $A$ a set, and $X\subset A^{G}$ a subshift, then $(X,G)$ is expansive.

Expansiveness for uniformly continuous actions on uniform spaces has been investigated in [6], [19] for iterates of uniformly continuous maps and in [17], [29], [8] for actions of general groups.

Observe that if a uniform space $X$ admits a uniformly continuous and expansive action of a group $G$ then the topology on $X$ is necessarily Hausdorff. Indeed, if $U_{0}$ is an expansiveness entourage then the diagonal in $X\times X$ is the intersection of the entourages $g^{-1}U_{0}$, $g\in G$.

The following result was first established by Bryant [6, Theorem 1] for $G=\mathbb{Z}$ and then extended to general countable groups by Eisenberg [17, Theorem 1] (see also [29, Corollary 2.8]).

Theorem 3.4**.**

Let $X$ be a compact Hausdorff space equipped with an expansive continuous action of a countable group $G$. Then $X$ is metrizable.

For the proof, we shall use the following result.

Lemma 3.5**.**

Let $X$ be a compact Hausdorff space equipped with an expansive continuous action of a group $G$ and let $U_{0}$ be a closed expansiveness entourage for $(X,G)$. Then, for every entourage $U$ of $X$, there exists a finite subset $E=E(U)\subset G$ such that $\bigcap_{g\in E}g^{-1}U_{0}\subset U$.

Proof.

Let $U$ be an entourage of $X$ and let $V$ be an open neighborhood of the diagonal in $X\times X$ such that $V\subset U$. Since $U_{0}$ is a closed expansiveness entourage, the open sets $(X\times X)\setminus g^{-1}U_{0}$, $g\in G$, cover $(X\times X)\setminus V$. By compactness of $(X\times X)\setminus V$, there exists a finite subset $E\subset G$ such that

[TABLE]

This implies $\bigcap_{g\in E}g^{-1}U_{0}\subset V\subset U$. ∎

Proof of Theorem 3.4.

By the metrization theorem for uniform spaces mentioned above, it suffices to show that $X$ admits a countable base of entourages. Let $U_{0}$ be a closed expansiveness entourage. By Lemma 3.5, the set $\mathcal{B}$ consisting of all the entourages of the form $\bigcap_{g\in E}g^{-1}U_{0}$, where $E$ runs over all finite subsets of $G$, is a base of entourages of $X$. As the group $G$ is countable, the set of its finite subsets is also countable. Thus $\mathcal{B}$ is countable as well. This shows that $X$ is metrizable. ∎

4. Topological entropy

4.1. Topological entropy

Topological entropy for continuous self-mappings of topological spaces was first introduced by Adler, Konheim and McAndrew [1]. Their definition was directly inspired by the one given by Kolmogorov for measure-theoretic entropy.

Let $X$ be a topological space.

If $\alpha$ is a finite open cover of $X$, we denote by $N_{X}(\alpha)$, or simply $N(\alpha)$ if there is no risk of confusion on the ambient space, the minimal cardinality of a subcover of $\alpha$. Observe that if $\alpha$ and $\beta$ are finite open covers of $X$, then $\alpha\vee\beta$ is also a finite open cover of $X$. Moreover, it satisfies

[TABLE]

Note also that if $\beta$ is a refinement of $\alpha$, then $N(\beta)\geq N(\alpha)$.

Suppose now that $X$ is equipped with a continuous action of a group $G$. If $\alpha$ is a finite open cover of $X$ and $g\in G$, then $g\alpha$ is also a finite open cover of $X$ and

[TABLE]

If $\alpha$ is an open cover of $X$ and $F$ is a finite subset of $G$, we define the open cover $\alpha^{(F)}$ by

[TABLE]

Lemma 4.1**.**

Let $\alpha$ be a finite open cover of $X$. Then the map $h\colon\mathcal{P}_{\small fin}(G)\to\mathbb{R}$ defined by $h(F)\coloneqq\log N(\alpha^{(F)})$ is subadditive and right $G$-invariant.

Proof.

If $F$ and $F^{\prime}$ are finite subsets of $G$, then the covers $\alpha^{(F\cup F^{\prime})}$ and $\alpha^{(F)}\vee\alpha^{(F^{\prime})}$ refine each other so that

[TABLE]

by applying (4.1). This shows that $h$ is subadditive.

On the other hand, for every $g\in G$, we have

[TABLE]

by applying (4.2). This shows that $h$ is right $G$-invariant. ∎

Let $\alpha$ be a finite open cover of $X$. Let $(F_{j})_{j\in J}$ be a left Følner net for $G$. By Theorem 2.4 and Lemma 4.1, the limit

[TABLE]

exists, is finite, and does not depend on the choice of the Følner net for $G$.

We define the topological entropy $\operatorname{h_{\scriptstyle\text{\rm top}}}(X,G)\leq\infty$ of the topological dynamical system $(X,G)$ by

[TABLE]

where $\alpha$ runs over all finite open covers of $X$. Observe that

[TABLE]

where $\mathcal{C}$ is the set consisting of all finite open covers of $X$ partially ordered by the relation defined by $\alpha\leq\beta$ if $\beta$ is a refinement of $\alpha$.

Theorem 4.2**.**

Let $G$ be an amenable group. Then the following hold.

1. (i)

Suppose that $X$ is a topological space equipped with a continuous action of $G$ and let $Y\subset X$ be a closed invariant subset. Then one has $\operatorname{h_{\scriptstyle\text{\rm top}}}(Y,G)\leq\operatorname{h_{\scriptstyle\text{\rm top}}}(X,G)$. 2. (ii)

Let $X$ and $Y$ be topological spaces equipped with a continuous action of $G$. Suppose that the system $(Y,G)$ is a topological factor of $(X,G)$. Then one has $\operatorname{h_{\scriptstyle\text{\rm top}}}(Y,G)\leq\operatorname{h_{\scriptstyle\text{\rm top}}}(X,G)$.

Proof.

Let $\mathcal{F}=(F_{j})_{j\in J}$ be a left Følner net for $G$.

(i) Consider a finite open cover $\alpha$ of $Y$. For each $A\in\alpha$, we can find an open subset $A^{\prime}$ of $X$ such that $A=A^{\prime}\cap Y$. Then $\alpha^{\prime}\coloneqq\{A^{\prime}:A\in\alpha\}\cup\{X\setminus Y\}$ is a finite open cover of $X$ and $N_{Y}(\alpha)\leq N_{X}(\alpha^{\prime})$. Also, since $Y$ is $G$-invariant, so is $X\setminus Y$, and, for every finite subset $F\subset G$, we have $\alpha^{(F)}=\left(\alpha^{\prime}\right)^{(F)}\cap Y$.

It follows that

[TABLE]

Taking the supremum over all finite open covers $\alpha$ of $Y$, this gives us $\operatorname{h_{\scriptstyle\text{\rm top}}}(Y,G)\leq\operatorname{h_{\scriptstyle\text{\rm top}}}(X,G)$.

(ii) As $(Y,G)$ is a topological factor of $(X,G)$, there is a continuous $G$-equivariant surjective map $f\colon X\to Y$. Let $\alpha$ be a finite open cover of $Y$. Then $\alpha_{X}\coloneqq f^{-1}(\alpha)=\{f^{-1}(A):A\in\alpha\}$ is a finite open cover of $X$. Moreover, if $\beta$ is a subcover of $\alpha$, then $\beta_{X}$ is a subcover of $\alpha_{X}$ and, since $f$ is surjective, $N_{Y}(\alpha)=N_{X}(\alpha_{X})$. Also, for every finite subset $F\subset G$ one has $\left(\alpha^{(F)}\right)_{X}=\left(\alpha_{X}\right)^{(F)}$.

We then have

[TABLE]

Taking the supremum over all finite open covers $\alpha$ of $Y$, we get $\operatorname{h_{\scriptstyle\text{\rm top}}}(Y,G)\leq\operatorname{h_{\scriptstyle\text{\rm top}}}(X,G)$. ∎

4.2. $(F,U)$-separated subsets, $(F,U)$-spanning subsets, and $(F,U)$-covers

Let $X$ be a compact uniform space equipped with a uniformly continuous action of a group $G$.

Denote by $\mathcal{U}$ the directed set consisting of all entourages of $X$ partially ordered by reverse inclusion.

Let $U\in\mathcal{U}$ and let $F$ be a finite subset of $G$. We define the entourage $U^{(F)}$ by

[TABLE]

A subset $Z\subset X$ is said to be $(F,U)$-separated if $(x,y)\notin U^{(F)}$ for all distinct $x,y\in Z$. This amounts to saying that if $x,y\in Z$ satisfy $(gx,gy)\in U$ for all $g\in F$, then $x=y$.

Lemma 4.3**.**

Every $(F,U)$-separated subset $Z\subset X$ is finite. More precisely, there is an integer $N=N(X,F,U)$ such that every $(F,U)$-separated subset $Z\subset X$ has cardinality at most $N$.

Proof.

Let $V$ be a symmetric entourage of $X$ such that $V\circ V\subset U^{(F)}$. For each $x\in X$, the set $V[x]$ is a neighborhood of $x$. By compactness of $X$, there is a finite subset $K\subset X$ such that $X=\bigcup_{x\in K}V[x]$. Now let $Z\subset X$ be an $(F,U)$-separated subset. As each $V[x]$ can contain at most one point of $Z$, the set $Z$ is finite with cardinality at most $N\coloneqq|K|$. ∎

We define the integer $\operatorname{sep}(X,G,F,U)$ as being the maximal cardinality of an $(F,U)$-separated subset contained in $X$:

[TABLE]

A subset $Z\subset X$ is said to be $(F,U)$-spanning if for every $x\in X$ there exists $z=z(x)\in Z$ such that $(z,x)\in U^{(F)}$. This amounts to saying that for every $x\in X$, there exists $z\in Z$ such that $(gz,gx)\in U$ for all $g\in F$.

Lemma 4.4**.**

There exists a finite $(F,U)$-spanning subset $Z\subset X$.

Proof.

The set $U^{(F)}[x]$ is a neighborhood of $x$ in $X$ for each $x\in X$. By compactness of $X$, there exists a finite subset $Z\subset X$ such that $X=\bigcup_{z\in Z}U^{(F)}[z]$. Then $Z$ is a finite $(F,U)$-spanning subset for $X$. ∎

We define the integer $\operatorname{span}(X,G,F,U)$ as being the minimal cardinality of an $(F,U)$-spanning subset for $X$:

[TABLE]

A cover $\alpha$ of $X$ is called an $(F,U)$-cover if for each $A\in\alpha$ and all $x,y\in A$ one has $(x,y)\in U^{(F)}$, that is, $(gx,gy)\in U$ for all $g\in F$.

Lemma 4.5**.**

There exists a finite $(F,U)$-cover $\alpha$ of $X$.

Proof.

Let $V$ be a symmetric entourage of $X$ such that $V\circ V\subset U$. The set $A_{x}\coloneqq V^{(F)}[x]$ is a neighborhood of $x$ in $X$ for each $x\in X$. By compactness, we can find a finite subset $Z\subset X$ such that the set $\alpha\coloneqq\{A_{z}:z\in Z\}$ is a cover of $X$. If $z\in Z$ and $x,y\in A_{z}$ we have $(x,z),(z,y)\in V^{(F)}$ so that $(x,y)\in V^{(F)}\circ V^{(F)}\subset(V\circ V)^{(F)}\subset U^{(F)}$. Therefore $\alpha$ is a finite $(F,U)$-cover of $X$. ∎

We define the integer $\operatorname{cov}(X,G,F,U)$ as being the minimal cardinality of an $(F,U)$-cover of $X$.

Lemma 4.6**.**

Let $g\in G$. Then one has

[TABLE]

Proof.

A subset $Z\subset X$ is $(F,U)$-separated if and only if $g^{-1}Z$ is $(Fg,U)$-separated. As $|Z|=|g^{-1}Z|$, this gives us the first equality. Similarly, the second one follows from the fact that $Z\subset X$ is $(F,U)$-spanning if and only if $g^{-1}Y$ is $(Fg,U)$-spanning. The last equality follows from the fact that a cover $\alpha$ of $X$ is an $(F,U)$-cover if and only if $g^{-1}\alpha$ is an $(Fg,U)$-cover. ∎

Lemma 4.7**.**

The maps $U\mapsto\operatorname{sep}(X,G,F,U)$, $U\mapsto\operatorname{span}(X,G,F,U)$, and $U\mapsto\operatorname{cov}(X,G,F,U)$ are non-decreasing on $\mathcal{U}$.

Proof.

It suffices to observe that if $U_{1},U_{2}\in\mathcal{U}$ satisfy $U_{1}\subset U_{2}$, then every $(F,U_{2})$-separated subset of $X$ is also $(F,U_{1})$-separated, every $(F,U_{1})$-spanning subset of $X$ is also $(F,U_{2})$-spanning, and every $(F,U_{1})$-cover of $X$ is also an $(F,U_{2})$-cover. ∎

Lemma 4.8**.**

Let $U$ and $V$ be entourages of $X$ such that $V$ is symmetric and $U\circ U^{*}\subset V$. Then one has

[TABLE]

Proof.

Let $Y\subset X$ be an $(F,V)$-spanning subset with minimal cardinality. Then $\alpha\coloneqq\{V^{(F)}[y]:y\in Y\}$ is a finite cover of $X$. Moreover, if $x,x^{\prime}\in V^{(F)}[y]$, then $(y,x),(y,x^{\prime})\in V^{(F)}$ and hence $(x,x^{\prime})\in V^{(F)}\circ V^{(F)}\subset(V\circ V)^{(F)}$. This shows that $\alpha$ is an $(F,V\circ V)$-cover of $X$. We deduce that $\operatorname{cov}(X,G,F,V\circ V)\leq|\alpha|\leq|Y|=\operatorname{span}(X,G,F,V)$.

Let now $Z\subset X$ be an $(F,V)$-separated subset with maximal cardinality. If $y\in X\setminus Z$, then $y\in V^{(F)}[z]$ for some $z\in Z$ by maximality. Thus $Z$ is an $(F,V)$-spanning subset of $X$ and hence $\operatorname{span}(X,G,F,V)\leq|Z|=\operatorname{sep}(X,G,F,V)$.

Suppose now that $S\subset X$ is an $(F,U)$-spanning subset for $X$ with minimal cardinality. Thus for each $x\in X$ there exists $s(x)\in S$ such that $(gs(x),gx)\in U$ for all $g\in F$. If $z_{1},z_{2}\in Z$ are distinct, then $s(z_{1})\neq s(z_{2})$, otherwise, since $U\circ U^{*}\subset V$, we would have $(gz_{1},gz_{2})\in V$ for all $g\in F$, contradicting the fact that $Z$ is $(F,V)$-separated. It follows that $\operatorname{sep}(X,G,F,V)=|Z|\leq|S|=\operatorname{span}(X,G,F,U)$.

Finally, let $\alpha$ be an $(F,U)$-cover of $X$ with minimal cardinality. For each $A\in\alpha$ pick $t_{A}\in A$ and set $T\coloneqq\{t_{A}:A\in\alpha\}$. Given $x\in X$ we can find $A\in\alpha$ such that $x\in A$. Since $\alpha$ is an $(F,U)$-cover, we have $(t_{A},x)\in U^{(F)}$. This shows that $T$ is an $(F,U)$-spanning subset. We deduce that $\operatorname{span}(X,G,F,U)\leq|T|\leq|\alpha|=\operatorname{cov}(X,G,F,U)$. ∎

Lemma 4.9**.**

Let $U\in\mathcal{U}$ and let $E,F$ be two finite subsets of $G$. Then

[TABLE]

Proof.

Let $\alpha$ (resp. $\beta$) be an $(E,U)$-cover (resp. $(F,U)$-cover) of $X$ with minimal cardinality. Let $A\in\alpha$ and $B\in\beta$. If $x,y\in A\cap B$, then $(x,y)\in U^{(E)}\cap U^{(F)}=U^{(E\cup F)}$. Therefore $\alpha\vee\beta$ is an $(E\cup F,U)$-cover of $X$. It follows that $\operatorname{cov}(X,G,E\cup F,U)\leq|\alpha\vee\beta|\leq|\alpha|\cdot|\beta|=\operatorname{cov}(X,G,E,U)\cdot\operatorname{cov}(X,G,F,U)$. ∎

4.3. Uniform approaches to topological entropy

The computation of topological entropy for uniformly continuous actions of amenable groups on compact uniform spaces we discuss now is a uniform analogue of the Bowen-Dinaburg approach in the metrizable case [4], [16].

Let $X$ be a compact uniform space equipped with a uniformly continuous action of an amenable group $G$. Let $\mathcal{F}=(F_{j})_{j\in J}$ be a left Følner net for $G$.

Let $U$ be an entourage of $X$. We define the quantities $\operatorname{h_{\scriptstyle\text{\rm sep}}}(X,G,\mathcal{F},U)$ and $\operatorname{h_{\scriptstyle\text{\rm spa}}}(X,G,\mathcal{F},U)$ by

[TABLE]

and

[TABLE]

It follows from Lemma 4.6 and Lemma 4.9 that the map $F\mapsto\log\operatorname{cov}(X,G,F,U)$ is right-invariant and subadditive on $\mathcal{P}_{\text{fin}}(G)$. Thus, we deduce from Theorem 2.4 that the limit

[TABLE]

exists, is finite, and does not depend on the choice of the left Følner net $\mathcal{F}$.

Lemma 4.10**.**

Let $U$ and $V$ be entourages of $X$ such that $V$ is symmetric and $U\circ U^{*}\subset V$. Then one has

[TABLE]

Proof.

This is immediately deduced from Lemma 4.8 after taking limits. ∎

Lemma 4.11**.**

The maps $U\mapsto\operatorname{h_{\scriptstyle\text{\rm sep}}}(X,G,\mathcal{F},U)$, $U\mapsto\operatorname{h_{\scriptstyle\text{\rm spa}}}(X,G,\mathcal{F},U)$, and $U\mapsto\operatorname{h_{\scriptstyle\text{\rm cov}}}(X,G,U)$ are non-decreasing on $\mathcal{U}$.

Proof.

This immediately follows from Lemma 4.7 by taking limits. ∎

By Lemma 4.11, the maps $U\mapsto\operatorname{h_{\scriptstyle\text{\rm sep}}}(X,G,\mathcal{F},U)$, $U\mapsto\operatorname{h_{\scriptstyle\text{\rm spa}}}(X,G,\mathcal{F},U)$, and $U\mapsto\operatorname{h_{\scriptstyle\text{\rm cov}}}(X,G,U)$ admit (finite or infinite) limits on $\mathcal{U}$ that are equal to their upperbounds, namely

[TABLE]

Lemma 4.12**.**

One has

[TABLE]

Proof.

Equalities (4.16) follow from (4.12) after taking limits in $\mathcal{U}$. ∎

We deduce from Lemma 4.12 that $\operatorname{h_{\scriptstyle\text{\rm sep}}}(X,G,\mathcal{F})$ and $\operatorname{h_{\scriptstyle\text{\rm spa}}}(X,G,\mathcal{F})$ do not depend on the choice of the left Følner net $\mathcal{F}$. In the sequel, we shall simply write $\operatorname{h_{\scriptstyle\text{\rm sep}}}(X,G)$ instead of $\operatorname{h_{\scriptstyle\text{\rm sep}}}(X,G,\mathcal{F})$ and $\operatorname{h_{\scriptstyle\text{\rm spa}}}(X,G)$ instead of $\operatorname{h_{\scriptstyle\text{\rm spa}}}(X,G)$.

Theorem 4.13**.**

Let $X$ be a compact uniform space equipped with a uniformly continuous action of an amenable group $G$. Then one has

[TABLE]

For the proof, we shall need some auxiliary results. The first one is a uniform version of Lebesgue’s covering lemma (see e.g. [36, Theorem 0.20]).

Lemma 4.14**.**

Let $X$ be a compact uniform space. Let $\alpha$ be an open cover of $X$. Then there exists an entourage $U=U(\alpha)$ of $X$ such that for every $x\in X$ there exists $A\in\alpha$ such that $U[x]\subset A$.

Such an entourage $U$ is then called a Lebesgue entourage for the open cover $\alpha$.

Proof.

Since $\alpha$ is an open cover of $X$, there exists, for every point $x\in X$, an open subset $A_{x}\in\alpha$ and an entourage $V_{x}$ of $X$ such that $V_{x}[x]\subset A_{x}$. Choose, for each $x\in X$, an entourage $W_{x}$ such that $W_{x}\circ W_{x}\subset V_{x}$. By compactness of $X$, there exists a finite subset $Y\subset X$ such that the sets $W_{y}[y]$, $y\in Y$, cover $X$. Consider the entourage $U\coloneqq\bigcap_{y\in Y}W_{y}$. Then, for every $x\in X$, there is a point $y\in Y$ such that $x\in W_{y}[y]$. This implies

[TABLE]

Thus the entourage $U$ has the required property. ∎

Lemma 4.15**.**

Let $X$ be a compact uniform space equipped with a uniformly continuous action of a group $G$. Let $\alpha$ be a finite open cover of $X$ and let $F$ be a finite subset of $G$. Suppose that $U$ is a Lebesgue entourage for $\alpha$. Then one has

[TABLE]

Proof.

Let $Z\subset X$ be an $(F,U)$-spanning subset with minimal cardinality. Since $U$ is a Lebesgue entourage for $\alpha$, given $z\in Z$ and $g\in F$, we can find an open set $A_{z,g}\in\alpha$ such that $U[gz]\subset A_{z,g}$. As $Z$ is $(F,U)$-spanning, given any point $x\in X$ we can find $z=z(x)\in Z$ such that $(gz,gx)\in U$ for all $g\in F$. This implies $gx\in U[gz]\subset A_{z,g}$ for all $g\in F$, that is, $x\in\bigcap_{g\in F}g^{-1}A_{z,g}$. This shows that $\{\bigcap_{g\in F}g^{-1}A_{z,g}:z\in Z\}$ is a subcover of $\alpha^{(F)}$. Therefore $N(\alpha^{(F)})\leq|Z|=\operatorname{span}(X,G,F,U)$. ∎

Lemma 4.16**.**

Let $X$ be a compact uniform space equipped with a uniformly continuous action of a group $G$. Let $U$ and $V$ be entourages of $X$ with $V$ symmetric and $V\circ V\subset U$. Let $\alpha$ be a finite open cover of $X$ and let $F$ be a finite subset of $G$. Suppose that for every $A\in\alpha$, there exists $x\in X$ such that $A\subset V[x]$. Then one has

[TABLE]

Proof.

Let $Z\subset X$ be an $(F,U)$-separated subset with maximal cardinality. It follows from the hypotheses that each open set in $\alpha^{(F)}$ can contain at most one point of $Z$. Therefore $\operatorname{sep}(X,G,F,U)=|Z|\leq N(\alpha^{(F)})$. ∎

Proof of Theorem 4.13.

Let $\mathcal{F}=(F_{j})_{j\in J}$ be a left Følner net for $G$.

Let $\alpha$ be a finite open cover of $X$. Choose a Lebesgue entourage $U$ for $G$. By Lemma 4.15, we have

[TABLE]

for all $j\in J$. It follows that

[TABLE]

and hence

[TABLE]

Let now $U$ be an entourage of $X$. Choose a symmetric entourage $V$ of $X$ such that $V\circ V\subset U$. For each $x\in X$, there is an open neighborhood $A_{x}$ of $x$ such that $A_{x}\subset V[x]$. By compactness of $X$, we can find a finite subset $K\subset X$ such that $\alpha\coloneqq\{A_{x}:x\in K\}$ covers $X$. Lemma 4.16 gives us

[TABLE]

for all $j\in J$. Therefore

[TABLE]

and hence

[TABLE]

Combining Lemma 4.12 with inequalities (4.18) and (4.19), we finally get (4.17). ∎

4.4. Topological entropy and expansiveness

Theorem 4.17**.**

Let $X$ be a compact Hausdorff space equipped with an expansive continuous action of an amenable group $G$. Let $U_{0}$ be a closed expansiveness entourage for $(X,G)$. Then one has

[TABLE]

In the proof of Theorem 4.17, we shall use the following result.

Lemma 4.18**.**

Let $G$ be an amenable group and let $(F_{j})_{j\in J}$ be a left Følner net for $G$. Suppose that $E\subset G$ is a nonempty finite subset. Then the following hold:

1. (i)

$\lim_{j\in J}\frac{|EF_{j}\setminus F_{j}|}{|F_{j}|}=0$. 2. (ii)

$\lim_{j\in J}\frac{|EF_{j}|}{|F_{j}|}=1$; 3. (iii)

the net $(EF_{j})_{j\in J}$ is a left Følner net for $G$.

Proof.

As $EF_{j}\setminus F_{j}=\bigcup_{g\in E}(gF_{j}\setminus F_{j})$, we have

[TABLE]

so that Assertion (i) follows from (2.6).

As the set $E$ is nonempty, we have $|EF_{j}|\geq|F_{j}|$ and hence $|EF_{j}\setminus F_{j}|\geq|F_{j}\setminus EF_{j}|$. Therefore, we deduce from (i) that

[TABLE]

Now, using $|EF_{j}|=|F_{j}|+|EF_{j}\setminus F_{j}|-|F_{j}\setminus EF_{j}|$, we get

[TABLE]

which gives (ii).

Let $g\in G$ and fix some arbitrary element $e\in E$. By applying (i) with $E$ replaced by $e^{-1}gE$, we obtain

[TABLE]

Now observe that

[TABLE]

where the last inequality follows from the fact that $F_{j}\subset e^{-1}EF_{j}$. Thus, we deduce from (4.21) that

[TABLE]

Using (ii), this gives us

[TABLE]

which shows (iii). ∎

Proof of Theorem 4.17.

Let $U$ be an entourage of $X$. By Lemma 3.5, there exists a nonempty finite subset $E\subset G$ such that $U_{0}^{(E)}\coloneqq\bigcap_{g\in E}g^{-1}U_{0}\subset U$.

Observe that if $F\subset G$ is a finite subset, then a cover $\alpha$ of $X$ is an $(F,U_{0}^{(E)})$-cover if and only if $\alpha$ is an $(EF,U_{0})$-cover. Therefore

[TABLE]

Let now $(F_{j})_{j\in J}$ be a left-Folner net for $G$. Then we have

[TABLE]

where the last equality follows from the fact that the net $(EF_{j})_{j\in J}$ is a left Følner net by Lemma 4.18.(iii). As

[TABLE]

by Theorem 4.13, this shows that $\operatorname{h_{\scriptstyle\text{\rm top}}}(X,G)=\operatorname{h_{\scriptstyle\text{\rm cov}}}(X,G,U_{0})$. ∎

Since $\operatorname{h_{\scriptstyle\text{\rm cov}}}(X,G,U_{0})<\infty$ by the Ornstein-Weiss lemma, an immediate consequence of Theorem 4.17 is the following.

Corollary 4.19**.**

Let $X$ be a compact Hausdorff space equipped with an expansive continuous action of an amenable group $G$. Then $\operatorname{h_{\scriptstyle\text{\rm top}}}(X,G)<\infty$. ∎

Example 4.20**.**

Let $G$ be an amenable group, $A$ a finite set, and $X\subset A^{G}$ a subshift. Equip $X$ with the action of $G$ induced by the $G$-shift on $A^{G}$. Then $X$ is compact Hausdorff and $(X,G)$ is expansive. A closed expansiveness entourage for $(X,G)$ is $U_{0}\coloneqq\{(x,y)\in X\times X:x(1_{G})=y(1_{G})\}$. Observe that if $F\subset G$ is a finite subset, then $(x,y)\in U_{0}^{(F)}$ if and only if $x(g)=y(g)$ for all $g\in F^{-1}$. It follows that

[TABLE]

where $\pi_{F^{-1}}\colon A^{G}\to A^{F^{-1}}$ is the projection map. We deduce from Theorem 4.17 that if $(F_{j})_{j\in J}$ is a left Følner net for $G$, then

[TABLE]

(cf. [7], [10, Proposition 2.7]). This gives us $\operatorname{h_{\scriptstyle\text{\rm top}}}(X,G)=\log|A|$ if $X=A^{G}$ is the full shift.

Corollary 4.21**.**

Let $X$ be a compact Hausdorff space equipped with an expansive continuous action of an amenable group $G$. Let $\mathcal{F}$ be a left Følner net for $G$ and let $V_{0}$ be a closed symmetric entourage of $X$ such that $V_{0}\circ V_{0}$ is an expansiveness entourage for $(X,G)$. Then one has

[TABLE]

In particular, the quantities $\operatorname{h_{\scriptstyle\text{\rm sep}}}(X,G,\mathcal{F},V_{0})$ and $\operatorname{h_{\scriptstyle\text{\rm spa}}}(X,G,\mathcal{F},V_{0})$ do not depend on the left Følner net $\mathcal{F}$.

Proof.

Let $U_{0}$ be a closed expansiveness entourage such that $U_{0}\circ U_{0}^{*}\subset V_{0}$. By applying Lemma 4.10 with $V=V_{0}$ and $U=U_{0}$, we get

[TABLE]

where the first and last equalities follow from Theorem 4.17. ∎

5. Homoclinicity

Let $X$ be a uniform space equipped with a uniformly continuous action of a group $G$. One says that the points $x,y\in X$ are homoclinic if they satisfy the following condition: for every entourage $U$ of $X$, there exists a finite subset $\Omega=\Omega(U)\subset G$ such that $(gx,gy)\in U$ for all $g\in G\setminus\Omega$.

Proposition 5.1**.**

Let $X$ be a uniform space equipped with a uniformly continuous action of a group $G$. Then homoclinicity is an equivalence relation on $X$.

Proof.

Homoclinicity is a reflexive relation since every entourage of $X$ contains the diagonal by (UNI-1). It is symmetric since every entourage contains a symmetric one by (UNI-3) and (UNI-4). Transitivity follows from the fact that if $U$ is an entourage of $X$, then there is an entourage $V$ of $X$ such that $V\circ V\subset U$ by (UNI-5). ∎

Proposition 5.2**.**

Let $G$ be a group and let $A$ be a set. Equip $A^{G}$ with its prodiscrete uniform structure and the shift action of $G$. Then two configurations in $A^{G}$ are homoclinic if and only if they are almost equal.

Proof.

Let $x,y\in A^{G}$ and suppose first that $x$ and $y$ are almost equal. This means that there exists a finite subset $\Omega\subset G$ such that $x(g)=y(g)$ for all $g\in G\setminus\Omega$. Now let $W\subset A^{G}\times A^{G}$ be an entourage of $A^{G}$. By definition of the prodiscrete uniform structure, there is a finite subset $\Lambda\subset G$ such that $W(\Lambda)\subset W$, where $W(\Lambda)\subset A^{G}\times A^{G}$ is as in (2.5). Observe now that $(gx,gy)\in W(\Lambda)\subset W$ for all $g\in G\setminus\Lambda\Omega^{-1}$. As the set $\Lambda\Omega^{-1}$ is finite, we deduce that $x$ and $y$ are homoclinic.

Conversely, suppose that $x$ and $y$ are homoclinic. Then there exists a finite subset $\Omega\subset G$ such that $(gx,gy)\in W(\{1_{G}\})$ for all $g\in G\setminus\Omega$. This implies that $x(g)=y(g)$ for all $g\in G\setminus\Omega^{-1}$. Therefore $x$ and $y$ are almost equal. ∎

The following result is a uniform version of [13, Lemma 6.2] (see also [6, Lemma 1] and [21, Theorem 10.36]).

Proposition 5.3**.**

Let $X$ be a compact Hausdorff space equipped with a continuous action of a group $G$. Suppose that the action of $G$ on $X$ is expansive and let $U_{0}$ be a closed expansiveness entourage for $(X,G)$. Let $x,y\in X$. Then the following conditions are equivalent:

1. (a)

$x$* and $y$ are homoclinic;* 2. (b)

there exists a finite subset $\Omega_{0}\subset G$ such that $(gx,gy)\in U_{0}$ for all $g\in G\setminus\Omega_{0}$.

Proof.

The implication (a) $\implies$ (b) follows immediately from the definition of homoclinicity.

Conversely, suppose (b) and let $U$ be an entourage of $X$. By Lemma 3.5, there exists a finite subset $F=F(U)\subset G$ such that $\bigcap_{h\in F}h^{-1}U_{0}\subset U$. Consider the finite subset $\Omega\subset G$ defined by $\Omega\coloneqq F^{-1}\Omega_{0}$. Then for $g\in G\setminus\Omega$ and $h\in F$, we have $hg\in G\setminus\Omega_{0}$ so that $(gx,gy)\in h^{-1}U_{0}$. It follows that $(gx,gy)\in\bigcap_{h\in F}h^{-1}U_{0}\subset U$ for all $g\in G\setminus\Omega$. This shows that $x$ and $y$ are homoclinic. ∎

6. Weak specification

The concept of specification for dynamical systems was first introduced by Bowen in [5, Section 2.9]. Several variants and extensions of Bowen’s original definition of specification appear in the literature (see [13, Definition 6.1], [15, Chapter 21], [28], [31, Definition 5.1], [35]). Roughly speaking, specification is a property allowing to approximate sufficiently separated pieces of orbits by a single (sometimes required to be periodic) orbit. Connections of specification with chaos for iterates of uniformly continuous maps on uniform spaces was investigated in [14]. The definition of weak specification below is equivalent to the one given in [13] and [30] when restricted to continuous group actions on compact metrizable spaces.

Definition 6.1**.**

Let $X$ be a uniform space equipped with a uniformly continuous action of a group $G$. We say that the action of $G$ on $X$ has the weak specification property if it satisfies the following condition:

1. (WSP)

for every entourage $U$ of $X$, there exists a finite subset $\Lambda=\Lambda(U)\subset G$ such that the following holds: for any finite family $(\Omega_{i})_{i\in I}$ of finite subsets of $G$ such that $\Omega_{j}\cap\Lambda\Omega_{k}=\varnothing$ for all distinct $j,k\in I$ and for any family of points $(x_{i})_{i\in I}$ in $X$, there exists a point $x\in X$ such that $(x,x_{i})\in U^{(\Omega_{i})}$ for all $i\in I$.

Such a subset $\Lambda\subset G$ is then called a specification subset for $(X,G,U)$.

Proposition 6.2**.**

Let $X$ be a uniform space equipped with a uniformly continuous action of a group $G$. Suppose that the action of $G$ on $X$ has the weak specification property. Then every uniform factor of $(X,G)$ also has the weak specification property.

Proof.

Let $Y$ be a uniform space equipped with a uniformly continuous action of $G$ and suppose that $(Y,G)$ is a uniform factor of $(X,G)$. This means that there exists a $G$-equivariant uniformly continuous surjective map $f\colon X\to Y$.

Let $V$ be an entourage of $Y$. Since $f$ is uniformly continuous, we can find an entourage $U$ of $X$ such that

[TABLE]

Let $\Lambda\subset G$ be a specification subset for $(X,G,U)$. Let $(\Omega_{i})_{i\in I}$ be a finite family of finite subsets of $G$ such that $\Omega_{j}\cap\Lambda\Omega_{k}=\varnothing$ for all distinct $j,k\in I$, and let $(y_{i})_{i\in I}$ be a family of points in $Y$. Since $f$ is surjective, we can find, for each $i\in I$, a point $x_{i}\in X$ such that $y_{i}=f(x_{i})$. On the other hand, as $\Lambda$ is a specification subset for $(X,G,U)$, there exists a point $x\in X$ such that

[TABLE]

for all $i\in I$. Setting $y\coloneqq f(x)\in Y$, we then have

[TABLE]

and hence $(y,y_{i})\in V^{(\Omega_{i})}$ for all $i\in I$. This shows that $\Lambda$ is a specification subset for $(Y,G,V)$ and hence that the action of $G$ on $Y$ has the weak specification property. ∎

Proposition 6.3**.**

Let $(X_{k})_{k\in K}$ be a (possibly infinite) family of uniform spaces and let $G$ be a group. Suppose that each $X_{k}$, $k\in K$, is equipped with an action of $G$ having the weak specification property. Then the diagonal action of $G$ on the uniform product $X\coloneqq\prod_{k\in K}X_{k}$ also has the weak specification property.

Proof.

For each $k\in K$, denote by $p_{k}\colon X\to X_{k}$ the projection map onto $X_{k}$. Let $U$ be an entourage of $X$. Then one can find a finite set $L\subset K$ and, for every $k\in L$, and entourage $U_{k}$ of $X_{k}$ such that the entourage $V$ of $X$ defined by

[TABLE]

satisfies $V\subset U$. As the action of $G$ on $X_{k}$ has the weak specification property, one can find, for every $k\in L$, a finite subset $\Lambda_{k}\subset G$ such that $\Lambda_{k}$ is a specification subset for $(X_{k},G,U_{k})$. Clearly $\Lambda\coloneqq\bigcup_{k\in L}\Lambda_{k}$ is a specification subset for $(X,G,V)$ and hence for $(X,G,U)$. ∎

The following result will be used in the proof of Theorem 7.2. It says that, when $X$ is compact Hausdorff, we get an equivalent definition for weak specification if we remove the finiteness hypotheses for the index set $I$ and the subsets $\Omega_{i}$ in Definition 6.1.

Proposition 6.4**.**

Let $X$ be a compact Hausdorff space equipped with a continuous action of a group $G$ satisfying the weak specification property. Let $U$ be a closed entourage of $X$ and let $\Lambda\subset G$ be a specification subset for $(X,G,U)$. Let $(\Omega_{i})_{i\in I}$ be a (possibly infinite) family of (possibly infinite) subsets of $G$ such that

[TABLE]

and let $(x_{i})_{i\in I}$ be a family of points in $X$. Then there exists a point $x\in X$ such that $(x,x_{i})\in U^{(\Omega_{i})}$ for all $i\in I$.

Proof.

Denote by $\mathcal{P}_{\text{fin}}(G)$ (resp. $\mathcal{P}_{\text{fin}}(I)$) the set of all finite subsets of $G$ (resp. $I$). Consider, for $A\in\mathcal{P}_{\text{fin}}(G)$ and $J\in\mathcal{P}_{\text{fin}}(I)$, the subset $X(A,J)\subset X$ consisting of all $x\in X$ such that $(x,x_{i})\in U^{(A\cap\Omega_{i})}$ for all $i\in J$. First observe that since $U$ is closed then $X(A,J)$ is closed in $X$ for all $A\in\mathcal{P}_{\text{fin}}(G)$ and $J\in\mathcal{P}_{\text{fin}}(I)$. Indeed,

[TABLE]

On the other hand, if we fix $A\in\mathcal{P}_{\text{fin}}(G)$ and $J\in\mathcal{P}_{\text{fin}}(I)$, then the subsets $A\cap\Omega_{i}$, $i\in J$, are finite and satisfy

[TABLE]

by (6.3). Since $\Lambda$ is a specification subset for $(X,G,U)$, we have $X(A,J)\neq\varnothing$. As

[TABLE]

we deduce that $X(A_{1},J_{1})\cap X(A_{2},J_{2})\cap\dots\cap X(A_{n},J_{n})\neq\varnothing$ for all $A_{1},A_{2},\ldots,A_{n}\in\mathcal{P}_{\text{fin}}(G)$, $J_{1},J_{2},\ldots,J_{n}\in\mathcal{P}_{\text{fin}}(I)$, and $n\geq 0$.

Thus $(X(A,J))_{A\in\mathcal{P}_{\text{fin}}(G),J\in\mathcal{P}_{\text{fin}}(I)}$ is a family of closed subsets of $X$ with the finite intersection property. By compactness of $X$, the intersection of this family is not empty. This means that there exists a point $x\in X$ such that $x\in X(A,J)$ for all finite subsets $A\subset G$ and $J\subset I$. Clearly, such an $x$ satisfies $(x,x_{i})\in U^{(\Omega_{i})}$ for all $i\in I$. ∎

Proposition 6.5**.**

Let $X$ be a ultrauniform space equipped with a uniformly continuous action of a group $G$. Then the action of $G$ has the weak specification property, that is, satisfies condition (WSP), if and only if the following condition is satisfied:

1. (WSP’)

for every entourage $U$ of $X$, there exists a finite subset $\Lambda\subset G$ such that the following holds: if $\Omega_{1}$ and $\Omega_{2}$ are two finite subsets of $G$ such that $\Omega_{1}\cap\Lambda\Omega_{2}=\varnothing$, then, given any two points $x_{1},x_{2}\in X$, there exists a point $x\in X$ such that $(x,x_{k})\in U^{(\Omega_{k})}$ for all $k\in\{1,2\}$.

Proof.

As (WSP) trivially implies (WSP’), we only need to prove the converse implication. Suppose (WSP’). In order to show (WSP), we proceed by induction on $|I|$. The base case corresponds to $|I|=2$ and this is indeed our assumption. Suppose that (WSP) holds whenever $|I|\leq n$ and let $\Lambda=\Lambda(W,n)\subset G$ denote a finite subset guaranteeing (WSP) for any equivalence entourage $W$ of $X$ and any index set $I$ with $|I|\leq n$. Fix an entourage $U$ of $X$ and let $I^{\prime}$ be a finite index set with $|I^{\prime}|=n+1$. Since $X$ is ultrauniform, we can find an equivalence entourage $W\subset U$. Let us show that $\Lambda\coloneqq\Lambda(W,n)$ also satisfies (WSP) for $I^{\prime}$. Let $(\Omega_{i})_{i\in I^{\prime}}$ be a family of finite subsets of $G$ such that $\Omega_{j}\cap\Lambda\Omega_{k}=\varnothing$ for all distinct $j,k\in I^{\prime}$, and let $(x_{i})_{i\in I^{\prime}}$ be a family of points in $X$. Fix $i^{\prime}\in I^{\prime}$ and set $I\coloneqq I^{\prime}\setminus\{i^{\prime}\}$ so that $|I|=n$. Then, by the inductive hypothesis, we can find a point $x^{\prime}\in X$ such that

[TABLE]

Set $\Omega_{1}\coloneqq\cup_{i\in I}\Omega_{i}$ and $\Omega_{2}\coloneqq\Omega_{i^{\prime}}$ as well as $x_{1}\coloneqq x^{\prime}$ and $x_{2}\coloneqq x_{i^{\prime}}$. Then $\Omega_{1}\cap\Lambda\Omega_{2}=\varnothing$ so that we can find $x\in X$ such that $(x,x_{k})\in W^{(\Omega_{k})}$ for all $k\in\{1,2\}$, that is,

[TABLE]

and

[TABLE]

Since $W^{(\Omega_{i})}$ is an equivalence entourage, we deduce from (6.4) and (6.5) that $(x,x_{i})\in W^{(\Omega_{i})}$ for all $i\in I$. This, together with (6.6), yields $(x,x_{i})\in W^{(\Omega_{i})}\subset U^{(\Omega_{i})}$ for all $i\in I^{\prime}$. This completes the inductive argument and shows the implication (WSP’) $\implies$ (WSP). ∎

Let $G$ be a group and let $A$ be a set. Recall the following definitions [9, Section 3]. Given a finite subset $\Delta\subset G$, one says that a subset $X\subset A^{G}$ is $\Delta$-irreducible if it satisfies the following condition:

1. (SI)

if $\Omega_{1}$ and $\Omega_{2}$ are two finite subsets of $G$ such that $\Omega_{1}\cap\Omega_{2}\Delta=\varnothing$, then, given any two configurations $x_{1},x_{2}\in X$, there exists a configuration $x\in X$ such that $x|_{\Omega_{k}}=x_{k}|_{\Omega_{k}}$ for all $k\in\{1,2\}$.

One says that a subset $X\subset A^{G}$ is strongly irreducible if there exists a finite subset $\Delta\subset G$ such that $X$ is $\Delta$-irreducible.

Proposition 6.6**.**

Let $A$ be a uniform space and let $G$ be a group. Equip $A^{G}$ with the product uniform structure and the shift action of $G$. Let $X\subset A^{G}$ be a strongly irreducible $G$-invariant subset. Then the uniform dynamical system $(X,G)$ has the weak specification property. In particular, the full shift $(A^{G},G)$ has the weak specification property.

Proof.

Let $\Delta\subset G$ be a finite subset such that $X$ is $\Delta$-irreducible. Fix an entourage $U$ of $X$. Then we can find an entourage $V$ of $A$ and a finite subset $\Omega\subset G$ such that the entourage

[TABLE]

is contained in $U$. We claim that the finite subset $\Lambda\coloneqq\Omega\Delta^{-1}\Omega^{-1}\subset G$ is a specification subset for $(X,G,U)$. To see this, let $(\Omega_{i})_{i\in I}$ be a finite family of finite subsets of $G$ such that $\Omega_{j}\cap\Lambda\Omega_{k}=\varnothing$ for all distinct $j,k\in I$ and let $(x_{i})_{i\in I}$ be a family of points in $X$. Setting $\Omega_{i}^{\prime}\coloneqq\Omega_{i}^{-1}\Omega\subset G$ for all $i\in I$, we have that $\Omega_{j}^{\prime}\cap\Omega_{k}^{\prime}\Delta=\varnothing$ for all distinct $j,k\in I$. Since $X$ is $\Delta$-irreducible, using an immediate inductive argument on $|I|$, we can find a configuration $x\in X$ such that $x|_{\Omega_{i}^{\prime}}=x_{i}|_{\Omega_{i}^{\prime}}$ for all $i\in I$. This implies that $(gx)|_{\Omega}=(gx_{i})|_{\Omega}$ for all $g\in\Omega_{i}$ and $i\in I$. It follows that $(x,x_{i})\in W^{(\Omega_{i})}\subset U^{(\Omega_{i})}$ for all $i\in I$. This proves our claim and shows that $(X,G)$ has the weak specification property. ∎

The following result extends Proposition A.1 in [30].

Proposition 6.7**.**

Let $G$ be a group and let $A$ be a set. Equip $A^{G}$ with the prodiscrete uniform structure and the shift action of $G$. Let $X\subset A^{G}$ be a $G$-invariant subset. Then the following conditions are equivalent:

1. (a)

$X$* is strongly irreducible;* 2. (b)

the $G$-shift action on $X$ has the weak specification property.

Proof.

The implication (a) $\implies$ (b) follows from Proposition 6.6 after equipping $A$ with the discrete uniform structure.

Conversely, suppose (b). Let $U=W(1_{G})\coloneqq\{(x,y)\in X\times X:x(1_{G})=y(1_{G})\}$ and let $\Lambda=\Lambda(U)\subset G$ be a specification subset for $(X,G,U)$. Let us show that $X$ is $\Delta$-irreducible with $\Delta\coloneqq\Lambda^{-1}\subset G$. Let $\Omega_{1},\Omega_{2}\subset G$ such that $\Omega_{1}\cap\Omega_{2}\Delta=\varnothing$ and let $x_{1}$ and $x_{2}$ be two configurations in $X$. As $\Omega_{1}^{-1}\cap\Lambda\Omega_{2}^{-1}=\varnothing$ and $\Lambda$ is a specification subset for $(X,G,U)$, we can find a configuration $x\in X$ such that $(x,x_{k})\in U^{(\Omega_{k}^{-1})}$ for all $k\in\{1,2\}$. This implies $x(g_{k})=(g_{k}^{-1}x)(1_{G})=(g_{k}^{-1}x_{k})(1_{G})=x_{k}(g_{k})$ for all $g_{k}\in\Omega_{k}$ and $k\in\{1,2\}$. It follows that $x|_{\Omega_{k}}=x_{k}|_{\Omega_{k}}$ for all $k\in\{1,2\}$. This shows that $X$ is $\Delta$-irreducible. The implication (b) $\implies$ (a) follows. ∎

The following result extends Corollary 5.2 in  [30].

Theorem 6.8**.**

Let $X$ be a compact Hausdorff space equipped with a continuous action of an amenable group $G$. Suppose that $X$ has more than one point and that the action of $G$ on $X$ has the weak specification property. Then one has $\operatorname{h_{\scriptstyle\text{\rm top}}}(X,G)>0$.

Proof.

Let $x_{1}$ and $x_{2}$ be two distinct points in $X$. Then we can find a symmetric entourage $U$ of $X$ such that

[TABLE]

Let $\Lambda\subset G$ be a specification subset for $(X,G,U)$. Up to enlarging $\Lambda$ if necessary, we can assume that $\Lambda=\Lambda^{-1}$.

Let $\mathcal{F}=(F_{j})_{j\in J}$ be a left Følner net for $G$. Given $j\in J$, let $\Delta_{j}\subset F_{j}$ be a maximal subset subject to the condition that

[TABLE]

By maximality of $\Delta_{j}$ and the fact that $\Lambda=\Lambda^{-1}$, we have $F_{j}\subset\Lambda\Delta_{j}$. This implies $|F_{j}|\leq|\Lambda\Delta_{j}|\leq|\Lambda|\cdot|\Delta_{j}|$ and hence

[TABLE]

As $\Lambda$ is a specification subset for $(X,G,U)$ and $\Delta_{j}$ satisfies (6.8), for every $x\in\{x_{1},x_{2}\}^{\Delta_{j}}$, we can find $z=z(x)\in X$ such that

[TABLE]

Comparing (6.10) and (6.7), we deduce that the set $Z_{j}\coloneqq\{z(x):x\in\{x_{1},x_{2}\}^{\Delta_{j}}\}\subset X$ is $(\Delta_{j},U)$-separated and has cardinality

[TABLE]

Since $\Delta_{j}\subset F_{j}$, the set $Z_{j}$ is also $(F_{j},U)$-separated, so that

[TABLE]

We conclude that

[TABLE]

This implies $\operatorname{h_{\scriptstyle\text{\rm top}}}(X,G)>0$. ∎

7. Proof of the main result

The following result extends Proposition 3.1 in [30].

Theorem 7.1**.**

Let $X$ be a compact Hausdorff space equipped with a continuous action of an amenable group $G$ and let $Y\subsetneqq X$ be a proper closed $G$-invariant subset of $X$. Suppose that the action of $G$ on $X$ has the weak specification property and that the action of $G$ on $Y$ is expansive. Then one has $\operatorname{h_{\scriptstyle\text{\rm top}}}(Y,G)<\operatorname{h_{\scriptstyle\text{\rm top}}}(X,G)$.

Proof.

To simplify, if $U$ is an entourage of $X$, we shall also write $U$ to denote the entourage of $Y$ obtained by intersecting $U$ with $Y\times Y$.

Fix a point $x_{0}\in X\setminus Y$. Since $X$ is compact Hausdorff and therefore regular, we can find an entourage $W$ of $X$ such that $W[x_{0}]$ does not meet $Y$, that is, $(x_{0},y)\notin W$ for all $y\in Y$.

Let $U_{0}\subset X\times X$ be an expansiveness entourage for the action of $G$ on $Y$ and take a closed symmetric entourage $V$ of $X$ such that $V\circ V\circ V\circ V\subset W\cap U_{0}$. Choose also a symmetric entourage $U$ of $X$ such that $U\circ U\circ U\subset V$.

Let $\Lambda\subset G$ be a specification subset for $(X,G,U)$. Up to enlarging $\Lambda$ if necessary, we can assume that $\Lambda=\Lambda^{-1}$ and $1_{G}\in\Lambda$.

Let $\mathcal{F}=(F_{j})_{j\in J}$ be a left Følner net for $G$. Given $j\in J$, let $\Delta_{j}\subset F_{j}$ be a maximal subset subject to the condition that $g\notin\Lambda h$ for all distinct $g,h\in\Delta_{j}$. By maximality of $\Delta_{j}$ and the fact that $\Lambda=\Lambda^{-1}$ and $1_{G}\in\Lambda$, we have $F_{j}\subset\Lambda\Delta_{j}$. This implies $|F_{j}|\leq|\Lambda\Delta_{j}|\leq|\Lambda|\cdot|\Delta_{j}|$ and hence

[TABLE]

Fix now $j\in J$ and $A\subset\Delta_{j}$. Take a minimal $(F_{j}\setminus(\Lambda A),V)$-spanning subset $Y_{A}\subset Y$ for $Y$, so that

[TABLE]

Choose also, for each $a\in A$, a minimal $(\Lambda a,V)$-spanning subset $Z_{a}\subset Y$ for $Y$, so that

[TABLE]

where the last equality follows from (4.6).

Consider now a point $y\in Y$. Since $Y_{A}$ is $(F_{j}\setminus(\Lambda A),V)$-spanning for $Y$, we can find a point $y_{A}=y_{A}(y)\in Y_{A}$ such that $(gy,gy_{A})\in V$ for all $g\in F_{j}\setminus(\Lambda A)$. On the other hand, since $Z_{a}$ is $(\Lambda a,V)$-spanning for $Y$, we can find, for each $a\in A$, a point $z_{a}=z_{a}(y)\in Z_{a}$ such that $(gy,gz_{a})\in V$ for all $g\in\Lambda a$. Now, if $y,y^{\prime}\in Y$ satisfy $y_{A}(y)=y_{A}(y^{\prime})$ and $z_{a}(y)=z_{a}(y^{\prime})$ for all $a\in A$, then $(gy,gy^{\prime})\in V\circ V$ for all $g\in F_{j}$. Therefore the map

[TABLE]

is injective on each $(F_{j},V\circ V)$-separated subset of $Y$. By taking cardinalities: we deduce that

[TABLE]

This gives us

[TABLE]

Let now $S_{A}\subset Y$ be an $(F_{j}\setminus(\Lambda A),V)$-separated subset of $Y$ with maximal cardinality, so that

[TABLE]

Let $y\in S_{A}$. Take as an index set $I\coloneqq A\cup\{i_{0}\}$, where $i_{0}$ is an index element not in $A$, and consider the family $(\Omega_{i})_{i\in I}$ of finite subsets of $F_{j}$ defined by $\Omega_{a}\coloneqq\{a\}$ for all $a\in A$ and $\Omega_{i_{0}}\coloneqq F_{j}\setminus\Lambda A$. Observe that $\Omega_{i}\cap\Lambda\Omega_{k}=\varnothing$ for all distinct $i,k\in I$. Indeed, $\{a\}\cap\Lambda\{a^{\prime}\}=\varnothing$ for all distinct $a,a^{\prime}\in A\subset\Delta_{j}$, while, obviously, $F_{j}\setminus(\Lambda A)\cap\Lambda\{a\}=\varnothing$ for all $a\in A$. Thus, if we consider the family of points $(x_{i})_{i\in I}$ in $X$ defined by $x_{a}\coloneqq a^{-1}x_{0}$ for all $a\in A$ and $x_{i_{0}}\coloneqq y$ (the given point in $S_{A}$), condition (WSP) ensures the existence of a point $x=x(y,A)\in X$ such that $(g_{i}x,g_{i}x_{i})\in U$ for all $g_{i}\in\Omega_{i}$ and $i\in I$, that is, $(ax,x_{0})=(ax,a(a^{-1}x_{0}))\in U$ for all $a\in A$ and $(gx,gy)\in U$ for all $g\in F_{j}\setminus(\Lambda A)$.

Now, if $y,y^{\prime}\in S_{A}$ are distinct, we can find a group element $g\in F_{j}\setminus(\Lambda A)$ such that $(gy,gy^{\prime})\notin V$. Setting $x=x(y,A)$ and $x^{\prime}=x(y^{\prime},A)$, this implies $(gx,gx^{\prime})\notin U$. Indeed, otherwise, from $(gy,gx),(gx,gx^{\prime}),(gx^{\prime},gy^{\prime})\in U$ we would deduce

[TABLE]

a contradiction.

Moreover, for $A,B\subset\Delta_{j}$ distinct, $y\in S_{A}$ and $y^{\prime}\in S_{B}$, and for $c\in A\triangle B$, where $A\triangle B\coloneqq(A\setminus B)\cup(B\setminus A)$ denotes the symmetric difference set of $A$ and $B$, we have $(cx(y,A),cx(y^{\prime},B))\notin U$. Otherwise, if $c=:a\in A\setminus B$ (so that $a\in F_{j}\setminus\Lambda B$) from $(x_{0},ax),(ax,ax^{\prime}),(ax^{\prime},ay^{\prime})\in U$, we would deduce

[TABLE]

a contradiction since $ay^{\prime}\in Y$. Similarly, assuming $c=:b\in B\setminus A$ we would get $(x_{0},by)\in W$, again a contradiction since $by\in Y$.

It follows that the set $\{x(y,A):\;y\in S_{A},A\subset\Delta_{j}\}\subset X$ is $(F_{j},U)$-separated with cardinality $\sum_{A\subset\Delta_{j}}|S_{A}|=\sum_{A\subset\Delta_{j}}\operatorname{sep}(Y,G,F_{j}\setminus(\Lambda A),V)$. We deduce that

[TABLE]

This gives us

[TABLE]

where the second inequality follows from (7.1).

Finally, we obtain

[TABLE]

where the last equality follows from Theorem 4.17, since $V\circ V\circ V\circ V\subset U_{0}$ is a closed expansiveness entourage for $(Y,G)$. This shows that $\operatorname{h_{\scriptstyle\text{\rm top}}}(X,G)>\operatorname{h_{\scriptstyle\text{\rm top}}}(Y,G)$. ∎

The following result extends Proposition 3.2 in [30].

Theorem 7.2**.**

Let $X$ and $Y$ be compact Hausdorff spaces equipped with expansive continuous actions of an amenable group $G$. Suppose that the action of $G$ on $X$ has the weak specification property and that $\operatorname{h_{\scriptstyle\text{\rm top}}}(Y,G)<\operatorname{h_{\scriptstyle\text{\rm top}}}(X,G)$. Let $f\colon X\to Y$ be a continuous $G$-equivariant map. Then $f$ is not pre-injective. In fact, the restriction of $f$ to any homoclinicity class of $X$ fails to be injective.

Proof.

Let $\mathcal{F}=(F_{j})_{j\in J}$ be a left Følner net for $G$. Let $U_{0}$ (resp. $V_{0}$) be an expansiveness entourage for $(X,G)$ (resp. $(Y,G)$). Choose a closed symmetric entourage $V$ of $Y$ such that $V\circ V\circ V\circ V\subset V_{0}$. Since $f$ is uniformly continuous, we can find a closed symmetric entourage $U$ of $X$ such that $U\circ U\circ U\circ U\subset U_{0}$ and

[TABLE]

Fix a point $x\in X$. Let $\Lambda\subset G$ be a specification subset for $(X,G,U)$. It is not restrictive to suppose that $\Lambda=\Lambda^{-1}$.

Since $U\circ U$ is a closed symmetric entourage of $X$ and $U\circ U\circ U\circ U$ is an expansiveness entourage for $(X,G)$, we have

[TABLE]

On the other hand, We have

[TABLE]

Fixing some constant $\eta>0$ such that $3\eta<\operatorname{h_{\scriptstyle\text{\rm top}}}(X,G)-\operatorname{h_{\scriptstyle\text{\rm top}}}(Y,G)$, we deduce that for all $j_{0}\in J$, there exists $j\in J$ with $j\geq j_{0}$ such that

[TABLE]

so that

[TABLE]

By virtue of Assertion (i) in Lemma 4.18, we can furthermore assume that

[TABLE]

so that

[TABLE]

Let now $Z_{j}\subset X$ be an $(F_{j},U\circ U)$-separated subset with maximal cardinality, so that

[TABLE]

and let $S_{j}\subset Y$ be an $(F_{j},V)$-spanning subset with minimal cardinality, so that

[TABLE]

The fact that $S_{j}$ is $(F_{j},V)$-spanning, implies that, for each $z\in Z_{j}$, we can find $s\in S_{j}$ such that $(f(z),s)\in V^{(F_{j})}$. By (7.7), (7.9), and (7.10), we have

[TABLE]

Consequently, it follows from the pigeon-hole principle that there exists a subset $\widetilde{Z}_{j}\subset Z_{j}$ such that

[TABLE]

and an element $s_{0}\in S_{j}$ such that all $z\in\widetilde{Z}_{j}$ satisfy

[TABLE]

Since $\Lambda$ is a specification subset for $(X,G,U)$ such that $\Lambda=\Lambda^{-1}$, so that $F_{j}\cap\Lambda(G\setminus\Lambda F_{j})=\varnothing$, it follows from Proposition 6.4 that for each $z\in\widetilde{Z}_{j}$, we can find $z^{\prime}\in X$ satisfying

[TABLE]

and

[TABLE]

As $U$ is a closed expansiveness entourage for $(X,G)$, we deduce from Proposition 5.3 and (7.14) that $z^{\prime}$ is homoclinic to $x$.

On the other hand, it follows from (7.13), (7.6), and the $G$-equivariance of $f$ that

[TABLE]

Similarly, (7.14), (7.6), and the $G$-equivariance of $f$ imply that

[TABLE]

Now, as

[TABLE]

by (7.11) and (7.8), it follows again from the pigeon-hole principle that we can find two distinct points $z_{1},z_{2}\in\widetilde{Z}_{j}$ such that

[TABLE]

By (7.15) and (7.12), we have

[TABLE]

On the other hand, by using (7.16), we get

[TABLE]

Combining (7.17), (7.18), and (7.19), we obtain

[TABLE]

which implies $f(z_{1}^{\prime})=f(z_{2}^{\prime})$ since $V\circ V\circ V\circ V$ is an expansiveness entourage for $(Y,G)$. As $z_{1}^{\prime}$ and $z_{2}^{\prime}$ are in the homoclinicity class of $x$, it remains only to show that the points $z_{1}^{\prime}$ and $z_{2}^{\prime}$ are distinct. But this is clear since otherwise (7.13) would then imply $(z_{1},z_{2})\in(U\circ U)^{(F_{j})}$, a contradiction since $z_{1}$ and $z_{2}$ are distinct points in $\widetilde{Z}_{j}\subset Z_{j}$ and $Z_{j}$ is $(F_{j},U\circ U)$-separated. ∎

Combining the two previous theorems, we get the following.

Theorem 7.3**.**

Let $X$ and $Y$ be compact Hausdorff spaces equipped with expansive continuous actions of an amenable group $G$. Suppose that the actions of $G$ on $X$ and $Y$ have the weak specification property and that $\operatorname{h_{\scriptstyle\text{\rm top}}}(X,G)=\operatorname{h_{\scriptstyle\text{\rm top}}}(Y,G)$. Then every pre-injective continuous $G$-equivariant map $f\colon X\to Y$ is surjective. ∎

Proof of Theorem 1.1.

This is Theorem 7.3 with $X=Y$. ∎

When $G$ is a group, $A$ and $B$ are finite sets, $X\subset A^{G}$ and $Y\subset B^{G}$ are subshifts, a continuous $G$-equivariant map $f\colon X\to Y$ is also called a cellular automaton (this terminology is widely used among computer scientists, see e.g. [7]).

Corollary 7.4**.**

Let $G$ be an amenable group, $A$ and $B$ be finite sets, and $X\subset A^{G}$ and $Y\subset B^{G}$ be subshifts. Suppose that $X$ and $Y$ are strongly irreducible and $\operatorname{h_{\scriptstyle\text{\rm top}}}(X,G)=\operatorname{h_{\scriptstyle\text{\rm top}}}(Y,G)$. Then every pre-injective cellular automaton $f\colon X\to Y$ is surjective.

Proof.

This is an immediate consequence of Theorem 7.3 since $(X,G)$ and $(Y,G)$ are expansive (cf. Example 3.3) and have the weak specification property by Proposition 6.7. ∎