Extensibility and denseness of periodic semigroup actions

Raimundo Briceño , Álvaro Bustos-Gajardo and Miguel Donoso-Echenique Facultad de Matemáticas, Pontificia Universidad Católica de Chile. Santiago, Chile {raimundo.briceno, abustog, miguel.donosoe}@uc.cl

Abstract.

We study periodic points and finitely supported invariant measures for continuous semigroup actions. Introducing suitable notions of periodicity in both topological and measure-theoretical contexts, we analyze the space of invariant Borel probability measures associated with these actions. For embeddable semigroups, we establish a direct relationship between the extensibility of invariant measures to the free group on the semigroup and the denseness of finitely supported invariant measures. Applying this framework to shift actions on the full shift, we prove that finitely supported invariant measures are dense for every left amenable semigroup that is residually a finite group and for every finite-rank free semigroup.

Key words and phrases:

Countable semigroup; embeddable semigroup; natural extension; free semigroup; periodic measure; Markov tree chain

2010 Mathematics Subject Classification:

Primary 22D40, 37A15, 20M05; Secondary 20M30, 37B10, 60J10.

R.B. was partially supported by ANID/FONDECYT Regular 1240508. A.B.-G. was supported by ANID/FONDECYT Postdoctorado 3230159 (year 2023). M.D. was partially supported by Beca de Magíster Nacional ANID 22230917 (year 2023).

Introduction

The study of periodic orbits is a fundamental aspect of the qualitative theory of dynamical systems. In particular, the question of whether periodic points are dense in a given system is relevant to topological and smooth dynamics [3, 11]. The measure-theoretical analogue of this problem is whether periodic measures are dense in the set of invariant measures, a topic that has been extensively addressed in classical ergodic theory [27, 29].

In the context of group actions, a concrete formulation of the first question is whether, for a countable group $G$ acting continuously on a compact metric space $X$ , the set of periodic points $\operatorname{Per}(X,G)$ is dense in $X$ . In this setting, the concept of periodicity must be adapted to the group framework, and the potential denseness of periodic points imposes algebraic constraints on $G$ , specifically requiring it to be residually finite. Moreover, denseness of periodic points characterizes residual finiteness in the following sense: a countable group $G$ is residually finite if and only if $\operatorname{Per}(X,G)$ is dense in $\mathcal{A}^{G}$ for the shift action [8, Theorem 2.7.1].

More recently, interest has grown [7, 22, 26, 28] in the measure-theoretical analogue of this characterization: determining for which countable groups $G$ periodic measures are weak-* dense in the space of $G$ -invariant Borel probability measures $\mathcal{M}_{G}(\mathcal{A}^{G})$ . This leads to a natural dichotomy among countable groups, distinguishing those for which periodic measures are weak-* dense from those for which they are not. The groups for which periodic measures are dense are said to have the pa property, and they are necessarily residually finite. However, not every residually finite group has the pa property.

We may further ask if this still holds true for the subset of ergodic periodic measures; the groups with this property will be said to have the epa property. Clearly, the epa property implies the pa property, but it is not known if the converse implication holds in general. It is known that all amenable, residually finite groups have the epa property (see [26], where denseness is established for more general systems with specification). On the other hand, Bowen showed that free groups on finitely many generators have the pa property [5]. However, the construction outlined in this work does not suffice to conclude that these groups have the epa property.

Notice that these properties are intrinsic to the group, as the action is prescribed once the group is set. This is not the only case where the structure of the space of invariant measures tell us something about the algebraic properties of the acting group (e.g., see [18]).

In this work, we aim to extend these questions and results to the more general setting of semigroup actions. To do so, we first need to examine the notion of periodicity, which is more subtle in the semigroup setting than in the group case, as pre-periodic behavior that is not periodic appears. With the appropriate definitions in place, our first result is as follows.

{restatable}

thmfirstproposition Let $S$ be a left reductive semigroup. Then:

(i)

$S$ is residually a finite semigroup if and only if the set of pre-periodic points of $\mathcal{A}^{S}$ is dense in $\mathcal{A}^{S}$ for every finite alphabet $\mathcal{A}$ .
(ii)

$S$ is residually a finite group if and only if the set of periodic points of $\mathcal{A}^{S}$ is dense in $\mathcal{A}^{S}$ for every finite alphabet $\mathcal{A}$ .

Next, we turn our attention to the measure-theoretical case and introduce the (e)pa property for semigroups, extending the notion developed for groups. To study this notion, we make use of the tool of natural extensions. Given a continuous action $S\curvearrowright X$ of an embeddable semigroup $S$ and a receiving $S$ -group $\mathbf{G}$ , we consider its topological $\mathbf{G}$ -extension $G\curvearrowright X_{\mathbf{G}}$ , as studied in [6]. Then, for the projection map $\pi\colon X_{\mathbf{G}}\rightarrow X$ , we let $\pi_{*}\colon\mathcal{M}_{G}(X_{\mathbf{G}})\rightarrow\mathcal{M}_{S}(X)$ be the push-forward map and define the set of $\mathbf{G}$ -extensible measures as $\operatorname{Ext}_{\bm{\Gamma}}(X,S)=\operatorname{im}(\pi_{*})$ . With this in place, we establish the following connection between the set of $\mathbf{G}$ -extensible measures and the weak-* closure of the set of $S$ -periodic measures (resp. ergodic $S$ -periodic measures).

{restatable}

thmsecondtheorem Let $S$ be an embeddable semigroup, and let $\bm{\Gamma}=(\Gamma,\gamma)$ be a realization of the free $S$ -group. If $S\curvearrowright X$ is a continuous action and $\mathcal{P}(X_{\bm{\Gamma}},\Gamma)$ (resp. $\mathcal{P}^{\mathrm{erg}}(X_{\bm{\Gamma}},\Gamma)$ ) is weak-* dense in $\mathcal{M}_{\Gamma}(X_{\bm{\Gamma}})$ , then

\operatorname{Ext}_{\bm{\Gamma}}(X,S)=\overline{\mathcal{P}(X,S)}\quad\text{(resp. $\operatorname{Ext}_{\bm{\Gamma}}(X,S)=\overline{\mathcal{P}^{\mathrm{erg}}(X,S)}$)}.

In particular, if

(i)

$\Gamma$ has the (e)pa property, and
(ii)

$\textup{Ext}_{\bm{\Gamma}}(\mathcal{A}^{S},S)=\mathcal{M}_{S}(\mathcal{A}^{S})$ for every finite alphabet $\mathcal{A}$ ,

then $S$ has the (e)pa property.

We then investigate the $\bm{\Gamma}$ -extensibility of measures for the free $S$ -group $\bm{\Gamma}$ , and use Theorem Introduction to establish the (e)pa property for some classes of semigroups. First, we consider the class of left amenable semigroups and prove the following result.

{restatable}

thmthirdtheorem Let $S$ be a left amenable semigroup that is residually a finite group. Then, $S$ has the epa property.

Next, we consider the free case, and obtain the following.

{restatable}

thmfourththeorem Let $\mathbb{F}_{d}$ be the free group on generators $\{a_{1},\dots,a_{d}\}$ . For $\Sigma\subseteq\{a_{1}^{\pm 1},\dots,a_{d}^{\pm 1}\}$ , consider the subsemigroup $S=\left<\Sigma\right>^{+}$ , so that $\mathbf{F}_{d}=(\mathbb{F}_{d},\iota)$ is the free $S$ -group for the inclusion $\iota\colon S\to\mathbb{F}_{d}$ . Then, $\textup{Ext}_{\mathbf{F}_{d}}(\mathcal{A}^{S},S)=\mathcal{M}_{S}(\mathcal{A}^{S})$ and $S$ has the pa property.

In particular, Theorem Introduction allows us to conclude that the free semigroup $\mathbb{F}_{d}^{+}$ on $d$ generators has the pa property. Finally, we provide a characterization of the free $S$ -group for actions of certain subsemigroups $S$ of free groups of finite rank $\mathbb{F}_{d}$ , which is reminiscent of the main result in [6], but in a measure-theoretical context.

{restatable}

thmfifththeorem Let $d$ be a positive integer and consider the free group $\mathbb{F}_{d}$ on the generators $\{a_{1},\dots,a_{d}\}$ . Given $\Sigma\subseteq\{a_{1}^{\pm 1},\dots,a_{d}^{\pm 1}\}$ such that $\Sigma\supseteq\{a_{1},\dots,a_{d}\}$ , consider the subsemigroup $S=\left<\Sigma\right>^{+}$ , and let $\mathbf{G}$ be a receiving $S$ -group. The following are equivalent:

(i)

$\mathbf{G}$ is a realization of the free $S$ -group, i.e., $\mathbf{G}\simeq\mathbf{F}_{d}$ .
(ii)

Every $S$ -invariant measure in $\mathcal{M}(\mathcal{A}^{S})$ is $\mathbf{G}$ -extensible.
(iii)

Every fully supported $S$ -invariant Markov measure in $\mathcal{M}(\mathcal{A}^{S})$ is $\mathbf{G}$ -extensible.

The paper is organized as follows. In §1, we introduce the basic concepts of semigroup actions. In §2, we discuss the notion of periodicity for semigroup actions—in the topological and measure-theoretical settings—and its connection to residual finiteness, leading to the proof of Theorem Introduction. In §3, we consider natural extensions and explore the relationship between the (e)pa property for a semigroup $S$ and its free $S$ -group, establishing Theorem Introduction. In §4, we prove the (e)pa property for certain semigroups, as given in Theorem Introduction and Theorem Introduction, and conclude with a measure-theoretical characterization of the free $S$ -group for certain subsemigroups $S$ of free groups of finite rank $\mathbb{F}_{d}$ , as given in Theorem Introduction.

1. Preliminaries

1.1. Semigroups and embeddings into groups

A semigroup is a set $S$ together with an associative binary operation $(s,t)\mapsto st$ . A monoid is a semigroup which has a—necessarily unique—identity element $1_{S}$ . Throughout this work, we shall only deal with countable, discrete semigroups, with a special emphasis on monoids.

A subsemigroup of $S$ is a subset $T\subseteq S$ such that $tt^{\prime}\in T$ for all $t,t^{\prime}\in T$ . In this scenario, we write $T\leq S$ . A common way to identify a subsemigroup $T$ is to take this set to be comprised of all possible products of elements of a subset $K\subseteq S$ . More precisely, we define the subsemigroup generated by $K$ as follows:

\langle K\rangle^{+}=\{k_{0}\cdots k_{n-1}:n\in\mathbb{N},k_{0},\dots,k_{n-1}\in K\}\subseteq S,

where the empty product is the identity $1_{S}$ when $S$ is a monoid. We say that $K$ generates a subsemigroup $T\leq S$ if $T=\langle K\rangle^{+}$ .

A semigroup homomorphism is a function $\theta\colon S\rightarrow T$ satisfying $\theta(st)=\theta(s)\theta(t)$ for every $s,t\in S$ . A semigroup homomorphism between monoids is called a monoid homomorphism if, additionally, $\theta(1_{S})=1_{T}$ . An embedding will be an injective homomorphism, and an isomorphism will be a bijective homomorphism. If there is an isomorphism $S\to T$ , we say $S$ and $T$ are isomorphic and denote this by $S\simeq T$ . In this case, $S$ and $T$ will be, in most cases, essentially the same for our purposes.

1.1.1. Free semigroups and presentations

We want to give a semigroup structure to the quotient of a semigroup by an equivalence relation. An equivalence relation $\mathcal{R}$ on a semigroup $S$ is said to be a congruence if for every $a,b,c\in S$ , $a\mathbin{\mathcal{R}}b$ implies both $ca\mathbin{\mathcal{R}}cb$ and $ac\mathbin{\mathcal{R}}bc$ . Given such a relation, we denote by $\pi_{\mathcal{R}}\colon S\rightarrow S/\mathcal{R}$ the quotient map $\pi_{\mathcal{R}}(s)=[s]_{\mathcal{R}}$ .

If $\mathcal{R}$ is a congruence on $S$ and $a\mathbin{\mathcal{R}}a^{\prime}$ , $b\mathbin{\mathcal{R}}b^{\prime}$ , transitivity implies that $ab\mathbin{\mathcal{R}}a^{\prime}b^{\prime}$ , and thus the binary operation $S/\mathcal{R}\times S/\mathcal{R}\longrightarrow S/\mathcal{R}$ given by $[a]_{\mathcal{R}}\cdot[b]_{\mathcal{R}}=[ab]_{\mathcal{R}}$ is well-defined and gives $S/\mathcal{R}$ a semigroup structure, for which $\pi_{\mathcal{R}}$ is a semigroup homomorphism. For groups, the notions of quotient by a congruence and quotient by a normal subgroup are equivalent; this is easily seen by noting that there is a bijective correspondence associating to a congruence $\mathcal{R}$ the normal subgroup $N_{\mathcal{R}}=\{g\in G:g\mathrel{\mathcal{R}}1_{G}\}\trianglelefteq G$ and vice versa.

Given any set $B$ , the free semigroup generated by $B$ is the set $\mathbb{F}(B)^{+}$ of all finite words of elements from $B$ , with concatenation as its binary operation. There is a canonical inclusion $\iota\colon X\hookrightarrow\mathbb{F}(B)^{+}$ , which comes with a universal property, analogous to that of free groups. If $\lvert B_{1}\rvert=\lvert B_{2}\rvert$ , then $\mathbb{F}(B_{1})^{+}\simeq\mathbb{F}(B_{2})^{+}$ . Thus, we may talk of the free semigroup generated by $d$ elements, $\mathbb{F}_{d}^{+}:=\mathbb{F}(\{a_{1},\dotsc,a_{d}\})^{+}$ for an arbitrary, fixed set $\{a_{1},\dotsc,a_{d}\}$ , and then $\mathbb{F}(B)^{+}\cong\mathbb{F}_{d}^{+}$ for any set $B$ of $d$ elements. The empty word $\varepsilon$ will be the identity element of $\mathbb{F}(B)^{+}$ .

If $S=\langle B\rangle^{+}$ for some $B\subseteq S$ , then $S$ is isomorphic to a quotient of $\mathbb{F}(B)^{+}$ . This allows us to give a combinatorial description of a semigroup: given a set $B$ of generators and a collection $R\subseteq\mathbb{F}(B)^{+}\times\mathbb{F}(B)^{+}$ of relations, the semigroup presentation $\langle B\mid R\rangle^{+}$ corresponds to the “largest” semigroup generated by $B$ for which all equations $u=v$ , for $(u,v)\in R$ , hold true. This is formally the quotient $\mathbb{F}(B)^{+}/\mathcal{R}$ , where $\mathcal{R}$ is the smallest congruence in $\mathbb{F}(B)^{+}$ containing $R$ as a subset. Note that many different sets of relations may describe the same semigroup.

1.1.2. The free group on a semigroup

A pair $\mathbf{G}=(G,\eta)$ is an $S$ -group if $G$ is a group and $\eta\colon S\rightarrow G$ is a semigroup morphism with $\langle\eta(S)\rangle=G$ . If $\eta$ is an embedding, $\mathbf{G}$ is called a receiving $S$ -group, and whenever a semigroup admits an embedding into a group, we will say $S$ is embeddable. A morphism between two receiving $S$ -groups $(G,\eta)$ and $(G^{\prime},\eta^{\prime})$ is a group morphism $\theta\colon G\rightarrow G^{\prime}$ such that $\theta\circ\eta=\eta^{\prime}$ .

Definition 1.1.

Let $S$ be a semigroup. A free group on the semigroup $S$ , or a free $S$ -group, is an $S$ -group $\bm{\Gamma}=(\Gamma,\gamma)$ such that for every $S$ -group $(G,\eta)$ there is a unique morphism $\theta\colon\Gamma\rightarrow G$ with $\theta\circ\gamma=\eta$ .

Note that, if $(\Gamma,\gamma)$ is the free $S$ -group and $(G,\eta)$ is any $S$ -group, the morphism $\theta\colon\Gamma\to G$ granted by the universal property of $(\Gamma,\gamma)$ must be surjective, as its image contains a generating set for $G$ .

The free $S$ -group always exists and it is unique up to isomorphism of $S$ -groups (see [10, §12] for more details). Accordingly, we will speak of the free $S$ -group whenever we talk about properties that are stable under isomorphisms of $S$ -groups, and of a realization of the free $S$ -group when we want to refer to a specific pair $(\Gamma,\gamma)$ .

If we know a presentation $\langle B\mid R\rangle^{+}$ for $S$ , the group $\langle B\mid R\rangle$ provides a concrete way of viewing the free $S$ -group. More precisely, if $\mathcal{R}^{+}$ is the congruence generated by $R$ in $\mathbb{F}(B)^{+}$ and $\mathcal{R}$ is the congruence generated by $R$ in $\mathbb{F}(B)$ , let

S=\mathbb{F}(B)^{+}/\mathcal{R}^{+}\quad\text{and}\quad\Gamma=\mathbb{F}(B)/\mathcal{R}.

Then, if $\iota\colon S\rightarrow\Gamma$ is the semigroup morphism given by $\iota([b]_{\mathcal{R}^{+}})=[b]_{\mathcal{R}}$ for $b\in B$ , extended homomorphically to all of $S$ , the pair $(\langle B\mid R\rangle,\iota)$ will be a realization of the free $S$ -group. In this case, the pair $(\langle B\mid R\rangle,\iota)$ will be referred to as the canonical realization of the free $S$ -group (see [6, Proposition 3.2]). For example, the following are the associated groups to canonical realizations of semigroups:

(1)

$\mathbb{Z}^{d}=\langle a_{1},\dots a_{d}\mid a_{i}a_{j}=a_{j}a_{i}\rangle$ for $\mathbb{N}^{d}=\langle a_{1},\dots a_{d}\mid a_{i}a_{j}=a_{j}a_{i}\rangle^{+}$ ;
(2)

$\text{BS}(m,n)=\langle a,b\mid ab^{m}=b^{n}a\rangle$ for $\text{BS}(m,n)^{+}=\langle a,b\mid ab^{m}=b^{n}a\rangle^{+}$ ;
(3)

$\mathbb{F}_{d}=\langle a_{1},\dotsc,a_{d}\mid\varnothing\rangle$ for $\mathbb{F}_{d}^{+}=\langle a_{1},\dotsc,a_{d}\mid\varnothing\rangle^{+}$ .

In these three instances, the semigroup morphism $\iota$ associated to each canonical realization is the one that sends generators to generators in the natural way. Moreover, in every case, $\iota$ is an embedding. In particular, all these cases correspond to embeddable semigroups. If $\langle B\mid R\rangle^{+}$ turns out to be embeddable into $\langle B\mid R\rangle$ via $\iota$ , we can always assume that $\langle B\mid R\rangle^{+}$ is a monoid by artifically adding an identity $1_{S}$ if necessary. In this case, defining $\iota(1_{S})=1_{\Gamma}$ yields an embedding as well. Thus, in any setting where embeddability plays a role, the semigroup $\langle B\mid R\rangle^{+}$ will be understood as a monoid. For instance, $\mathbb{F}_{d}^{+}$ will be understood as the free monoid on $d$ elements by adjoining the empty word $\varepsilon$ to the free semigroup generated by $d$ elements, etc.

Embeddable semigroups are necessarily bicancellative, meaning that $a=b$ whenever $ac=bc$ or $ca=cb$ hold. Nonetheless, not every bicancellative semigroup can be embedded into a group [23]. In fact, given a semigroup presentation $S=\langle B\mid R\rangle^{+}$ , determining whether $S$ can be embedded into its free $S$ -group, or more generally into an arbitrary group, is not an easy matter. The relevant thing about the free $S$ -group is that it characterizes embeddability, meaning that a semigroup $S$ can be embedded into a group if and only if it can be embedded into its free $S$ -group [10, Theorem 12.4].

A useful result due to Adian [1] states that bicancellative one-relator finitely presented semigroups can be embedded into a group.

1.1.3. The left reversible case

A semigroup $S$ is said to be left reversible if for every $s,t\in S$ there are $x,y\in S$ such that $sx=ty$ , or equivalently if for all $s,t\in S$ , $sS\cap tS\neq\varnothing$ . Groups and Abelian semigroups are left reversible semigroups. In contrast, the free monoid $\mathbb{F}_{d}^{+}$ is not left reversible for $d\geq 2$ , since $a_{1}\mathbb{F}_{d}^{+}\cap a_{d}\mathbb{F}_{d}^{+}=\varnothing$ .

In [25, Theorem 1], Ore showed that reversibility is a sufficient condition for a bicancellative semigroup to be embeddable into a group. In this scenario, the possible receiving $S$ -groups have a nice characterization, for which we need the following definition:

Definition 1.2.

Let $S$ be a semigroup. A receiving $S$ -group $\mathbf{G}=(G,\eta)$ is a group of right fractions of $S$ if for every $g\in G$ there exist $s,t\in S$ such that $g=\eta(s)\eta(t)^{-1}$ .

If $\mathbf{G}$ and $\mathbf{G}^{\prime}$ are groups of right fractions of $S$ , then they are isomorphic as $S$ -groups [9, Theorem 1.25], and the group of right fractions may thus be denoted by $\bm{\Gamma}_{S}=(\Gamma_{S},\gamma)$ .

Remark 1.3.

In [13], Dubreil proved that if $S$ is a bicancellative semigroup, then $S$ is left reversible if and only if $\bm{\Gamma}_{S}$ exists. Whenever the group of right fractions of $S$ exists, it is the only receiving $S$ -group up to isomorphism of $S$ -groups, and it is hence isomorphic to the free $S$ -group [6, Corollary 2.20 and Remark 3.4].

Remark 1.4.

If $S$ is embeddable but not left reversible, then receiving $S$ -groups are not necessarily unique modulo isomorphism of $S$ -groups. This is shown in [6, Example 2.27], where both $\text{BS}(1,2)$ and $\mathbb{F}_{2}$ —paired with adequate embeddings—play the role of receiving $\mathbb{F}_{2}^{+}$ -groups, but are non isomorphic even as groups.

1.2. Semigroup actions and topological natural extensions

We will understand an action $\alpha$ of $S$ over a set $X$ , denoted as $S\overset{\alpha}{\curvearrowright}X$ , by a function $\alpha\colon S\times X\rightarrow X$ such that $\alpha(s,\alpha(t,x))=\alpha(st,x)$ for all $x\in X$ and $s,t\in S$ . If $S$ is a monoid, we additionally request that $\alpha(1_{S},x)=x$ for all $x\in X$ . Most of the time $\alpha(s,x)$ will be written simply as $s\cdot x$ , and the function $\alpha(s,\cdot)\colon X\rightarrow X$ as simply $\alpha_{s}$ or $s$ . An action $S\curvearrowright X$ is said to be surjective if $\alpha_{s}$ is surjective for each $s\in S$ . Given two monoids $S$ and $T$ , a monoid morphism $\theta\colon S\to T$ and two actions $S\curvearrowright X$ and $T\curvearrowright Y$ , a function $\varphi\colon Y\rightarrow X$ will be called $\theta$ -equivariant if $\varphi(\theta(s)\cdot y)=s\cdot\varphi(x)$ for all $y\in Y$ and $s\in S$ . If $S=T$ and $\theta=\text{id}_{S}$ , we say that $\varphi$ is equivariant. A subset $A\subseteq X$ is $S$ -invariant for an action $S\overset{\alpha}{\curvearrowright}X$ if $sA:=\{s\cdot x:x\in A\}\subseteq A$ for all $s\in S$ , and completely $S$ -invariant if $sA=A$ for every $s\in S$ . In this case it makes sense to consider the action $S\curvearrowright A$ given by restricting $\alpha|_{S\times A}\colon S\times A\rightarrow A$ . Given $x\in X$ , we define its $S$ -orbit as the $S$ -invariant set $Sx=\{s\cdot x:s\in S\}$ .

1.2.1. Structure preserving actions

Suppose that $X$ is a compact metric space. We will denote by $\mathcal{B}(X)$ the associated Borel $\sigma$ -algebra. Given a Borel probability measure $\mu$ on $(X,\mathcal{B}(X))$ , we will denote by $(X,\mu)$ the corresponding Borel probability space.

A continuous $S$ -action (or simply continuous action, if $S$ is implicit) will be an action $S\overset{\alpha}{\curvearrowright}X$ such that $\alpha_{s}$ is continuous for each $s\in S$ . Given two continuous actions $S\overset{\alpha}{\curvearrowright}X$ and $S\overset{\beta}{\curvearrowright}Y$ , a continuous equivariant function $\varphi\colon Y\to X$ is called a topological factor map if it is surjective, and a topological conjugacy if it is a homeomorphism. In the first case, we say that $\alpha$ is a topological factor of $\beta$ (and that $\beta$ is a topological extension of $\alpha$ ), while in the second we say that both actions are topologically conjugate.

Given two Borel probability spaces $(X,\mu)$ and $(Y,\nu)$ , a measurable map $\varphi\colon X\rightarrow Y$ is measure preserving if $\mu(\varphi^{-1}A)=\nu(A)$ for all $A\in\mathcal{B}(Y)$ . Given a continuous action $S\overset{\alpha}{\curvearrowright}X$ , we say that $\mu$ is $S$ -invariant if each $\alpha_{s}$ is measure-preserving, i.e., if $\mu(s^{-1}A)=\mu(A)$ for all $A\in\mathcal{B}(X)$ and $s\in S$ . In this case, we also say that $\alpha$ is a probability measure-preserving (p.m.p.) action, and denote it by $S\curvearrowright(X,\mu)$ . Given two p.m.p. actions $S\overset{\alpha}{\curvearrowright}(X,\mu)$ and $S\overset{\beta}{\curvearrowright}(Y,\nu)$ , we say that $\alpha$ is factor of $\beta$ (and that $\beta$ is an extension of $\alpha$ ) if there exist a full measure $S$ -invariant sets $X^{\prime}\subseteq X$ and $Y^{\prime}\subseteq Y$ and a measure preserving equivariant map $\varphi\colon Y^{\prime}\rightarrow X^{\prime}$ . In this case, $\varphi$ is called a factor map. If $\varphi$ is moreover a bi-measurable map, we say that $\alpha$ and $\beta$ are (measure-theoretically) conjugate and that $\varphi$ is a (measure-theoretical) conjugacy.

The space of Borel probability measures on $(X,\mathcal{B}(X))$ will be denoted by $\mathcal{M}(X)$ . We can endow $\mathcal{M}(X)$ with the weak-* topology, which is metrizable and compact. Convergence is characterized by:

\mu_{n}\to\mu\iff\int_{X}f\,d\mu_{n}\to\int_{X}f\,d\mu\quad\text{for all }f\in\mathcal{C}(X),

and a basis for the topology consists of the sets

V(\mu,\mathcal{F},\varepsilon)=\left\{\mu^{\prime}\in\mathcal{M}(X):\left|\int f\,d\mu^{\prime}-\int f\,d\mu\right|<\varepsilon,\forall f\in\mathcal{F}\right\},

where $\varepsilon>0$ , $\mathcal{F}\subseteq\mathcal{C}(X)$ is finite, and $\mu\in\mathcal{M}(X)$ .

Given a continuous action $S\curvearrowright X$ , the space of $S$ -invariant measures on $(X,\mathcal{B}(X))$ , denoted by $\mathcal{M}_{S}(X)$ , is convex and weak-* closed in $\mathcal{M}(X)$ . Depending on $S\curvearrowright X$ , the space $\mathcal{M}_{S}(X)$ may be empty or not. A measure $\mu\in\mathcal{M}_{S}(X)$ is ergodic if every set $A\in\mathcal{B}(X)$ such that $\mu(s^{-1}A\triangle A)=0$ for all $s\in S$ satisfies $\mu(A)\in\{0,1\}$ .

Given $\mu\in\mathcal{M}(X)$ , we define the support of $\mu$ as

\operatorname{supp}(\mu)=\{x\in X:\mu(U)>0\text{ for every open neighborhood }U\text{ of }x\}.

The set $\operatorname{supp}(\mu)$ is closed, and $\mu(A)=0$ whenever $A\cap\operatorname{supp}(\mu)=\varnothing$ .

Remark 1.5.

If $\mu\in\mathcal{M}_{S}(X)$ , then $\operatorname{supp}(\mu)$ is an $S$ -invariant subset. Indeed, take $x\in\operatorname{supp}(\mu)$ and $s\in S$ . Given an open neighborhood $U$ of $s\cdot x$ , we have that $s^{-1}U$ is an open neighborhood of $x$ , so $\mu(U)=\mu(s^{-1}U)>0$ . This proves that $s\cdot x\in\operatorname{supp}(\mu)$ , so $\operatorname{supp}(\mu)$ is $S$ -invariant.

1.2.2. Shift actions

Given a compact metric space $\mathcal{A}$ , we consider the product space $\mathcal{A}^{S}$ of all functions $x\colon S\to\mathcal{A}$ endowed with the product topology. There is a natural continuous and surjective action $S\curvearrowright\mathcal{A}^{S}$ given by

(s\cdot x)(t)=x(ts)\quad\text{ for all }s,t\in S,

which will be called the shift action and denoted by $\sigma$ . An important special case is when $\mathcal{A}$ is a finite set—an alphabet—endowed with the discrete topology. If $|\mathcal{A}|\geq 2$ , the space of configurations $\mathcal{A}^{S}$ is a Cantor space, and together with the action $S\curvearrowright\mathcal{A}^{S}$ is called the full $S$ -shift. A closed $S$ -invariant subset of $\mathcal{A}^{S}$ will be called an $S$ -subshift.

Given subsets $Q\subseteq P\subseteq S$ and $p\in\mathcal{A}^{P}$ , we define the cylinder set induced by $p$ at $Q$ as

[p;Q]=\{x\in\mathcal{A}^{S}:x(t)=p(t)\text{ for all }t\in Q\}.

The collection of all cylinder sets $[x;Q]$ with $x\in\mathcal{A}^{S}$ and $Q$ finite is a basis of clopen sets for the topology of $\mathcal{A}^{S}$ .

1.3. Periodic configurations in groups

Let $G$ be a group and $\mathcal{A}^{G}$ be the full $G$ -shift for some alphabet $\mathcal{A}$ . An element $x\in\mathcal{A}^{G}$ is periodic if it has a finite $G$ -orbit, that is, if $|Gx|<\infty$ . Equivalently, $x\in\mathcal{A}^{G}$ is periodic if and only if its stabilizer

\text{Stab}_{G}(x)=\{g\in G:g\cdot x=x\}

is a finite-index subgroup of $G$ , i.e., $[G:\text{Stab}_{G}(x)]<\infty$ . These notions extend to any group action $G\curvearrowright X$ , and we denote by $\operatorname{Per}(X,G)$ the set of periodic points in $X$ .

A group $G$ is residually finite if for every pair of distinct elements $g,h\in G$ there is a finite group $F$ and a group morphism $\theta\colon G\rightarrow F$ such that $\theta(g)\neq\theta(h)$ . More generally, if $(\textsc{p})$ is a class of semigroups, we say that a semigroup $S$ is residually $(\textsc{p})$ if for any distinct $s,t\in S$ there exists a semigroup $T$ in the class $(\textsc{p})$ and a semigroup morphism $\theta\colon S\to T$ such that $\theta(s)\neq\theta(t)$ . Residually finite groups have several equivalent characterizations, some of which are listed below:

Proposition 1.6.

Let $G$ be a group. The following are equivalent:

(i)

$G$ is residually finite.
(ii)

For any $g\in G$ there exists a finite group $F$ and a group morphism $\theta\colon G\to F$ such that $\theta(g)\neq 1_{F}$ ,
(iii)

$G$ is isomorphic to a subgroup of a Cartesian product of (possibly infinitely many) finite groups.
(iv)

The set $\operatorname{Per}(\mathcal{A}^{G},G)$ is dense in $\mathcal{A}^{G}$ for every finite alphabet $\mathcal{A}$ .

Evidently, every finite group is automatically residually finite. Some less trivial examples of residually finite groups include $\mathbb{Z}^{d}$ and all free groups, including $\mathbb{F}_{d}$ for all $d\in\mathbb{N}$ ; similarly, the linear groups $\mathrm{GL}_{d}(\mathbb{Z})$ for any $d\geq 1$ have this property. Furthermore, the class of residually finite groups is closed under taking subgroups, Cartesian products and inverse limits (but not under quotients). In contrast, divisible groups, that is, those for which the equation $x^{n}=g$ has a solution for any $n\geq 1$ and $g\in G$ , are never residually finite; these include $\mathbb{Q},\mathbb{R}$ and $\mathbb{C}$ . Similarly, non-Hopfian groups, that is, those for which a surjective, non-injective morphism $G\to G$ exists, cannot be residually finite either; an example of such a group is the Baumslag–Solitar group $\mathrm{BS}(2,3)$ [21].

2. Periodicity for semigroup actions

In general, periodicity is a concept that tries to capture both finiteness and circularity. In the case of group actions, due to invertibility, the finite orbit condition is sufficient for precluding the existence of any transient portion in the orbit, thus assuring a circular behavior. However, even in the case of a single non-invertible surjective endomorphism, that is, an action of the semigroup $\mathbb{N}$ , there could exist finite orbits with a transient initial section, yielding the distinction between periodic and pre-periodic points.

2.1. Periodic points and finitely supported invariant measures

The definition of periodicity in the context of semigroup actions is subtle and there may be various notions that capture it. However, one of our main purposes is to study finitely supported invariant measures. With this goal in mind, we introduce the following definition of periodicity.

Definition 2.1.

Let $S\curvearrowright X$ be a semigroup action. An element $x\in X$ will be called pre-periodic if $|Sx|<\infty$ , and periodic if $|Sx|<\infty$ and the set $Sx$ is completely $S$ -invariant, that is, $tSx=Sx$ for all $t\in S$ . The set of $S$ -periodic points of $X$ will be denoted as $\operatorname{Per}(X,S)$ .

Example 2.2.

Complete $S$ -invariance does not necessarily imply $|Sx|<\infty$ , although in the $S=\mathbb{N}$ case it does. For instance, the element $x\in\{0,1\}^{\mathbb{N}^{2}}$ given by $x(n,m)=1$ if and only if $n=m$ , has infinite translates, and every element of $\mathbb{N}^{2}$ acts bijectively upon its orbit.

A natural question is whether, as in the case of $\mathbb{N}$ actions, the orbit of every pre-periodic point contains a periodic point or, equivalently, the support of an $S$ -invariant measure. However, this is not necessarily the case for general semigroup actions.

Example 2.3.

Consider $\mathbb{F}_{2}^{+}=\langle a,b\rangle^{+}$ acting on the full $\mathbb{F}_{2}^{+}$ -shift $\{0,1\}^{\mathbb{F}_{2}^{+}}$ . Let $x\in\{0,1\}^{\mathbb{F}_{2}^{+}}$ be the configuration defined by $x(\varepsilon)=0$ and, for all $s\in\mathbb{F}_{2}^{+}$ , $x(as)=1$ and $x(bs)=0$ . The orbit of this configuration does not contain a periodic point, as can be seen in Figure 1, so there is no subset of this orbit which can support an invariant measure. Although these kind of orbits exhibit a valid form of periodicity, they will not be taken in consideration here.

Figure 1. A diagram of the

\mathbb{F}_{2}^{+}

-orbit

\{x,y\}

x

and the representation of

x

as a labeling of

\mathbb{F}_{2}^{+}

Remark 2.4.

In the context of group actions, we saw that every periodic element induces a finite-index subgroup (namely, $\text{Stab}(x)$ ). For semigroup actions, if $x\in X$ has finite $S$ -orbit, it induces a congruence on $S$ by

s\mathrel{\mathcal{R}_{x}}t\iff s\cdot y=t\cdot y\text{ for all }y\in Sx\cup\{x\}.

Moreover, $\mathcal{R}_{x}$ has finite index, i.e., finitely many equivalence classes, so in order to admit periodic points, a semigroup must contain finite-index congruences. Note that, in the group case, the analogous to the congruence $\mathcal{R}_{x}$ is the largest normal subgroup contained in the stabilizer, i.e., the intersection of the stabilizers of all elements in $Gx$ .

Note that every element $t\in S$ defines a function $t\colon Sx\to Sx$ by $y\mapsto t\cdot y$ . The full transformation monoid of $Sx$ will be the set of all functions $Sx\to Sx$ together with function composition, denoted by $\mathcal{T}(Sx)$ . We have that $s\mathrel{\mathcal{R}_{x}}t$ if and only if $s$ and $t$ define the same elements of $\mathcal{T}(Sx)$ , obtaining an embedding $S/\mathcal{R}_{x}\to\mathcal{T}(Sx)$ . By a subgroup of $\mathcal{T}(Sx)$ , we will understand a submonoid of $\mathcal{T}(Sx)$ that is a group.

Proposition 2.5.

Let $S\curvearrowright X$ be a semigroup action and $x\in X$ . Then, $x$ is pre-periodic if and only if $S/\mathcal{R}_{x}$ is a finite semigroup, and $x$ is periodic if and only if $S/\mathcal{R}_{x}$ is a finite subgroup of $\mathcal{T}(Sx)$ .

Proof.

If $x$ is pre-periodic, there are finitely many functions $Sx\to Sx$ , hence finitely many equivalence classes for $\mathcal{R}_{x}$ . Conversely, if $x$ is not pre-periodic, there exists an infinite subset $Q\subseteq S$ such that $s\cdot x\neq s^{\prime}\cdot x$ for all $s,s^{\prime}\in Q$ with $s\neq s^{\prime}$ . This immediately implies that $|S/\mathcal{R}_{x}|\geq|Q|=\infty$ .

If $x$ is periodic, every $t\in S$ defines a bijection $Sx\to Sx$ , so there is an injective semigroup morphism $S/\mathcal{R}_{x}\to\mathrm{Sym}(Sx)$ . Since $\mathrm{Sym}(Sx)$ is a finite group, $S/\mathcal{R}_{x}$ must be a finite group as well. Conversely, suppose that $S/\mathcal{R}_{x}$ is a subgroup of $\mathcal{T}(Sx)$ and let $t\in S$ . There exists some $t^{\prime}\in S$ such that $[tt^{\prime}]_{\mathcal{R}_{x}}$ is the identity class in $S/\mathcal{R}_{x}$ , that is, $t\circ t^{\prime}$ is the identity as a function $Sx\to Sx$ . Thus, for any $s\in S$ we have $t\cdot(t^{\prime}s\cdot x)=(tt^{\prime})\cdot(s\cdot x)=s\cdot x$ , i.e., $tSx=Sx$ . ∎

Notice that the action of $\mathbb{F}_{2}^{+}$ upon the orbit of the point $x$ defined in Example 2.3 is transitive, i.e., for all $s,t\in S$ , $s\cdot x\in S(t\cdot x)$ . This holds true for every periodic point.

Proposition 2.6.

If $x\in X$ is periodic, then the action $S\curvearrowright Sx$ is transitive. In particular, two $S$ -periodic orbits are either equal or disjoint.

Proof.

Suppose that $S\curvearrowright Sx$ is not transitive. Then, there exists $t\in S$ such that $x\not\in S(t\cdot x)$ . Let

k=\min\{i\geq 1:\text{ there is a }n>i\text{ with }t^{n}\cdot x=t^{i}\cdot x\},

which exists because $Sx$ is finite. By definition, there is an $n>k$ with $t^{n}\cdot x=t^{k}\cdot x$ , and we must have $t^{n-1}\cdot x\neq t^{k-1}\cdot x$ . Note that

t(t^{n-1}\cdot x)=t^{n}\cdot x=t^{k}\cdot x=t(t^{k-1}\cdot x),

contradicting the injectivity of $t$ upon $Sx$ . Therefore, $S\curvearrowright Sx$ is transitive.

For the final statement, let $x_{1},x_{2}\in X$ be two $S$ -periodic elements. If $y\in Sx_{1}\cap Sx_{2}$ , there are elements $s_{1},s_{2}\in S$ such that $y=s_{1}\cdot x_{1}=s_{2}\cdot x_{2}$ . Since the action $S\curvearrowright Sx$ is transitive, we may choose an element $t\in S$ with $ts_{1}\cdot x_{1}=x_{1}$ , so $ts_{2}\cdot x_{2}=x_{1}$ , which implies $Sx_{1}\subseteq Sx_{2}$ . By symmetry, we get $Sx_{2}\subseteq Sx_{1}$ , and so both orbits coincide. ∎

Remark 2.7.

Notice that if $x$ is pre-periodic for an $\mathbb{N}$ -action, then the transitivity of $\mathbb{N}\curvearrowright Sx$ is equivalent to $x$ being periodic. However, Proposition 2.6 combined with Example 2.3 show that periodicity is a strictly stronger notion for general semigroup actions.

A periodic measure in $\mathcal{M}_{S}(X)$ will be a measure $\mu\in\mathcal{M}_{S}(X)$ such that $\operatorname{supp}(\mu)$ is finite. The set of periodic measures will be denoted by $\mathcal{P}(X,S)$ , and the set of periodic ergodic measures in $\mathcal{M}_{S}(X)$ will be denoted as $\mathcal{P}^{\mathrm{erg}}(X,S)$ . The following result justifies our chosen definition of periodicity.

Lemma 2.8.

Let $S\curvearrowright X$ be a continuous action, and $x\in X$ a pre-periodic point. Then, there is an $S$ -invariant measure $\mu\in\mathcal{M}_{S}(X)$ with $\textup{supp}(\mu)=Sx$ if and only if $x$ is periodic.

Proof.

Suppose $x$ is not periodic and choose $t\in S$ such that $t\colon Sx\rightarrow Sx$ is not bijective. Then, since $Sx$ is finite, there exists $y\in Sx$ such that $t^{-1}(y)\cap(Sx)=\varnothing$ . Take any $S$ -invariant measure $\mu\in\mathcal{M}_{S}(X)$ with $\textup{supp}(\mu)\subseteq Sx$ . Then, $\mu(X-Sx)=0$ , so

\mu(\{y\})=\mu(t^{-1}(y))=\mu(t^{-1}(y)\cap Sx)=0.

Thus, $\text{supp}(\mu)\subsetneq Sx$ , so no invariant measure can be supported upon the whole of $Sx$ . Conversely, assume each $t\in S$ acts as a bijection of $Sx$ and consider the measure given by

\mu(A)=\frac{|A\cap Sx|}{|Sx|}\quad\text{ for every }A\in\mathcal{B}(X).

Define $\varphi_{t}\colon t^{-1}A\cap Sx\rightarrow A\cap Sx$ by $y\mapsto t\cdot y$ . For each $t\in S$ , the function $\varphi_{t}$ is injective as a consequence of $t$ being injective upon $Sx$ . Choose any $y\in A\cap Sx$ . Since $tSx=Sx$ , there exists $y^{\prime}\in Sx$ with $t\cdot y^{\prime}=y\in A$ , so $y^{\prime}\in t^{-1}A$ as well, showing surjectivity of $\varphi_{t}$ . Thus, $\mu$ satisfies

\mu(t^{-1}A)=\frac{|t^{-1}A\cap Sx|}{|Sx|}=\frac{|A\cap Sx|}{|Sx|}=\mu(A),

as we wanted. ∎

We have the following characterizations of periodic measures.

Proposition 2.9.

Let $S\curvearrowright X$ be a continuous action and $\mu\in\mathcal{M}_{S}(X)$ . Then,

(i)

$\mu\in\mathcal{P}(X,S)$ if and only if $\operatorname{supp}(\mu)$ is a finite disjoint union of periodic $S$ -orbits,
(ii)

$\mu\in\mathcal{P}^{\mathrm{erg}}(X,S)$ if and only if $\operatorname{supp}(\mu)$ corresponds to a single periodic $S$ -orbit.

Proof.

To prove (i), first assume that $\mu\in\mathcal{P}(X,S)$ . Then, by Remark 1.5 we have that $\operatorname{supp}(\mu)$ is $S$ -invariant, which means $Sx\subseteq\operatorname{supp}(\mu)$ for all $x\in\operatorname{supp}(\mu)$ . If it were the case that an element $x\in\operatorname{supp}(\mu)$ does not belong to the orbit of an element of $\operatorname{supp}(\mu)$ , then $s^{-1}(x)\subseteq X-\operatorname{supp}(\mu)$ for every $s\in S$ , yielding $\mu(\{x\})=0$ , a contradiction. This means that $\operatorname{supp}(\mu)=\bigcup_{x\in\operatorname{supp}(\mu)}Sx$ , which implies the statement, since periodic orbits are either equal or disjoint by Proposition 2.6. The converse is direct.

To prove (ii), first assume $\mu\in\mathcal{P}^{\mathrm{erg}}(X,S)$ . Then, by part (i) there exist $S$ -periodic elements $x_{1},\dots,x_{m}\in X$ such that

\operatorname{supp}(\mu)=\bigsqcup_{i=1}^{m}Sx_{i}.

Given $t\in S$ and $i\neq j$ , if $x\in t^{-1}(Sx_{i})\cap Sx_{j}$ , then $t\cdot x\in Sx_{i}\cap tSx_{j}=Sx_{i}\cap Sx_{j}=\varnothing$ , so $t^{-1}(Sx_{i})\cap Sx_{j}=\varnothing$ . Also note that $Sx_{i}\subseteq t^{-1}(Sx_{i})$ for every $t\in S$ and $i$ , as $x_{i}$ is $S$ -periodic. Therefore,

\mu(t^{-1}(Sx_{i})\triangle Sx_{i})=\mu(t^{-1}(Sx_{i})-Sx_{i})\leq\mu(X-\operatorname{supp}(\mu))=0.

Hence, since $\mu$ is ergodic, $\mu(Sx_{i})\in\{0,1\}$ , so $\operatorname{supp}(\mu)$ consists of a single orbit.

Conversely, if $\operatorname{supp}(\mu)=Sx$ for an $S$ -periodic point $x$ and $A\in\mathcal{B}(X)$ is such that $\mu(s^{-1}A\triangle A)=0$ for all $s\in S$ , then $Sx\cap(A-s^{-1}A)=\varnothing$ for all $s\in S$ , and we have two cases. First, if $Sx\cap A=\varnothing$ , then $\mu(A)=0$ . Second, if $t\cdot x\in Sx\cap A$ for some $t\in S$ , necessarily we must have $t\cdot x\in s^{-1}A$ for every $s\in S$ as well, obtaining $st\cdot x\in A$ for every $s\in S$ . Since the action of $S$ upon $Sx$ is transitive, this implies $Sx\subseteq A$ , and so $\mu(A)=1$ . Thus, $\mu$ is ergodic. ∎

As a consequence of last proposition, a periodic measure can always be written in the form

\mu=\sum_{i=1}^{m}\frac{\mu(Sx_{i})}{|Sx_{i}|}\sum_{x\in Sx_{i}}\delta_{x},

where $\sum_{i=1}^{m}\mu(Sx_{i})=1$ , with $m=1$ if and only if the measure is moreover ergodic.

2.2. Residual finiteness and periodic orbits

We aim to relate algebraic properties of semigroups and both periodic and pre-periodic orbits. In order to do this, let us start by introducing two properties that are semigroup analogues to residual finiteness in groups.

Definition 2.10.

A semigroup $S$ is residually a finite group (resp. residually a finite semigroup) if for every pair of distinct elements $t,t^{\prime}\in S$ there is a finite group (resp. semigroup) $F$ and a semigroup morphism $\theta\colon S\rightarrow F$ such that $\theta(t)\neq\theta(t^{\prime})$ .

Remark 2.11.

If $S$ is residually a finite group (resp. residually a finite semigroup) and $Q\subseteq S$ , then for every $t,t^{\prime}\in Q$ there exist a group (resp. semigroup) $F_{t,t^{\prime}}$ and a semigroup morphism $\theta_{t,t^{\prime}}\colon S\to F_{t,t^{\prime}}$ such that $\theta_{t,t^{\prime}}(t)\neq\theta_{t,t^{\prime}}(t^{\prime})$ if $t\neq t^{\prime}$ , and $F_{t,t^{\prime}}$ is the trivial group if $t=t^{\prime}$ . Thus, we may define a semigroup morphism $\theta_{Q}\colon S\to\prod_{t,t^{\prime}\in Q}F_{t,t^{\prime}}$ such that $\theta(t)\neq\theta(t^{\prime})$ for any $t,t^{\prime}\in Q$ , $t\neq t^{\prime}$ , by sending any $s\in S$ to the tuple $(\theta_{t,t^{\prime}}(s))_{t,t^{\prime}\in Q}$ .

As a consequence of this, $S$ is residually a finite group (resp. residually a finite semigroup) if and only if for every finite subset $Q\subseteq S$ , there is a finite group (resp. semigroup) $F$ , namely $\prod_{t,t^{\prime}\in Q}F_{t,t^{\prime}}$ , and a semigroup morphism $\theta\colon S\to F$ , such that $\theta\rvert_{Q}$ is injective.

Remark 2.12.

It is clear that being residually a finite group always implies being residually a finite semigroup. Furthermore, we have the converse in the case where $S$ is a group, since the image of a group via a semigroup morphism is always a group. In this situation, both notions coincide with the classic notion of residual finiteness for groups. However, for general semigroups these notions differ: every finite non-bicancellative semigroup is residually a finite semigroup but not residually a finite group. Indeed, if $S$ a non-bicancellative semigroup, it cannot be residually a finite group. Indeed, if $\theta\colon S\rightarrow F$ is a morphism to a group $F$ , and $a,b,c\in S$ are such that $ab=ac$ and $b\neq c$ , then $\theta(a)\theta(b)=\theta(a)\theta(c)$ so $\theta(b)=\theta(c)$ , so $b$ and $c$ cannot be distinguished by such morphism.

The following result is analogous to the characterization of residually finite groups provided in Proposition 1.6 (iii).

Proposition 2.13.

A semigroup $S$ is residually a finite group (resp. residually a finite semigroup) if and only if it can be embedded into a Cartesian product of finite groups (resp. semigroups). In particular, a semigroup is residually a finite group if and only if it can be embedded into a residually finite group.

Proof.

Assume $S$ is residually a finite group (resp. residually a finite semigroup). The map $\theta_{Q}$ for $Q=S$ , following the notation from Remark 2.11, is an embedding into a product of finite groups (resp. semigroups). ∎

Remark 2.14.

Notice that Proposition 2.13 implies that any semigroup that is residually a finite group is necessarily bicancellative, and moreover embeddable into some residually finite group.

A semigroup $S$ such that for every pair of distinct elements $s,s^{\prime}\in S$ , there is some $t\in S$ with $ts\neq ts^{\prime}$ , is called a left reductive semigroup. In particular, every monoid and every left cancellative semigroup is left reductive. On the other hand, a semigroup action $S\curvearrowright X$ is said to be faithful if for every $s,s^{\prime}\in S$ with $s\neq s^{\prime}$ there exists $x\in X$ such that $s\cdot x\neq s^{\prime}\cdot x$ . The following result relates these two notions.

Lemma 2.15.

A semigroup $S$ is left reductive if and only if the shift action $S\curvearrowright\mathcal{A}^{S}$ is faithful for every (resp. any) alphabet $\mathcal{A}$ with $|\mathcal{A}|\geq 2$ .

Proof.

If $S$ is left reductive and $s,s^{\prime}\in S$ are distinct elements, there is a $t\in S$ with $ts\neq ts^{\prime}$ , so the configuration $x\in\mathcal{A}^{S}$ given by $x(t)=1$ and $x(t^{\prime})=0$ if $t^{\prime}\neq t$ satisfies $s\cdot x\neq s^{\prime}\cdot x$ . Conversely, if $S\curvearrowright\mathcal{A}^{S}$ is faithful for some alphabet $\mathcal{A}$ with $|\mathcal{A}|\geq 2$ , given $s\neq s^{\prime}$ in $S$ there exist $x\in\mathcal{A}^{S}$ and $t\in S$ with $(s\cdot x)(t)\neq(s^{\prime}\cdot x)(t)$ , so $ts$ and $ts^{\prime}$ cannot be equal. ∎

Our first main result is the following.

\firstproposition

Proof.

Let us prove (ii), as the proof of (i) is essentially the same.

Suppose that $S$ is residually a finite group, $x\in\mathcal{A}^{S}$ is an arbitrary point, and let $Q\subseteq S$ be an arbitrary finite set. By Remark 2.11, there exist a finite group $F$ and a semigroup morphism $\theta\colon S\to F$ with $\theta\rvert_{Q}$ injective.

Note that every $w\in\mathcal{A}^{F}$ defines a point $\bar{w}\in\mathcal{A}^{S}$ by $\bar{w}(s)=w(\theta(s))$ for all $s\in S$ . It holds true that $s\cdot\bar{w}=\overline{\theta(s)\cdot w}$ for any $s\in S$ , since for every $t\in S$ we have

(s\cdot\bar{w})(t)=\bar{w}(ts)=w(\theta(ts))=w(\theta(t)\theta(s))=(\theta(s)\cdot w)(\theta(t))=\big{(}\overline{\theta(s)\cdot w}\big{)}(t).

Thus, for any $w\in\mathcal{A}^{F}$ , the element $\bar{w}\in\mathcal{A}^{S}$ is $S$ -periodic. Indeed, the above proven equality shows that $S\bar{w}\subseteq\{\bar{w^{\prime}}:w^{\prime}\in Fw\}$ , which together with the fact that $F$ is finite prove that $|S\bar{w}|<\infty$ . Moreover, given $t\in S$ , since $\theta(S)$ is a group, for every $s\in S$ , there is an $s^{\prime}\in S$ such that $\theta(t)\theta(s^{\prime})=\theta(s)$ , yielding

ts^{\prime}\cdot\bar{w}=\overline{\theta(ts^{\prime})\cdot w}=\overline{\theta(s)\cdot w}=s\cdot\bar{w}.

Thus, $tS\bar{w}=S\bar{w}$ for every $t\in S$ . Define a configuration $w_{x}\colon F\to\mathcal{A}$ by $w_{x}(\theta(s))=x(s)$ for all $s\in Q$ , extending it arbitrarily to the rest of $F$ if needed. As the map $\theta\rvert_{Q}\colon Q\to F$ is injective, this is well-defined. The corresponding point $\overline{w_{x}}$ in $\mathcal{A}^{S}$ is $S$ -periodic and satisfies the equality $\overline{w_{x}}\rvert_{Q}=x\rvert_{Q}$ . This shows that, for any $x\in\mathcal{A}^{S}$ and any finite $Q\subseteq S$ , the cylinder $[x;Q]$ must contain a periodic point, and thus the set of periodic points is dense.

Conversely, if $\mathcal{A}^{S}$ has dense set of periodic points and $s,s^{\prime}\in S$ are distinct elements, by left reductiveness, there exists $t\in S$ such that $ts\neq ts^{\prime}$ . Choose two distinct elements from $\mathcal{A}$ , which, for simplicity, will be denoted by $0$ and $1$ , and consider any $x\in\mathcal{A}^{S}$ such that $x(ts)=0$ and $x(ts^{\prime})=1$ . By our hypothesis of denseness of periodic points, we may assume that $x$ is $S$ -periodic and consider $F=\text{Sym}(Sx)$ , which is a finite group. The definition of $S$ -periodicity yields a morphism $\theta\colon S\rightarrow F$ by sending $a\in S$ to the function $a\rvert_{Sx}\colon Sx\rightarrow Sx$ given by $a\rvert_{Sx}(b\cdot x)=ab\cdot x$ for all $b\in S$ . By our choices of $x$ and $t$ , we must have $(s\cdot x)(t)=x(ts)=0$ and $(s^{\prime}\cdot x)(t)=x(ts^{\prime})=1$ . Hence, $s\cdot x\neq s^{\prime}\cdot x$ , and thus $s\rvert_{Sx}$ and $s^{\prime}\rvert_{Sx}$ correspond to different permutations of $Sx$ , that is, $\theta(s)\neq\theta(s^{\prime})$ .

The proof of (i) is identical, but replacing the finite group $F$ by a finite semigroup satisfying that $\theta\rvert_{Q}$ is injective, and omitting the proof of complete $S$ -invariance; and for the opposite direction, by replacing the symmetric group $\textup{Sym}(Sx)$ by the monoid of all transformations $Sx\to Sx$ . ∎

Since a left cancellative semigroup is automatically left reductive, any semigroup that is residually a finite group is left reductive. In the non-left reductive case, however, the associated full $S$ -shift may have dense periodic points, but this does not imply that the semigroup is residually a finite group.

Example 2.16 (Non-reductive case).

Let $L=\{a,b\}$ be the left zero semigroup, where $st=s$ for all $s,t\in L$ . The semigroup $L$ is not left reductive, but satisfies that $s\cdot x=x$ for every $s\in L$ and $x\in\mathcal{A}^{L}$ , so every element of $\mathcal{A}^{L}$ is periodic. Thus, it has a dense set of periodic points for the shift action $L\curvearrowright\mathcal{A}^{L}$ , but $L$ is not residually a finite group since it is not bicancellative. Similarly, the semigroup $S=\mathbb{Z}\times L$ provides an example of an infinite semigroup with dense set of periodic points in $\mathcal{A}^{S}$ , which is not residually a finite group.

2.3. The periodic approximation property

In view of Theorem Introduction, it is natural to ask under what conditions upon the acting semigroup $S$ the periodic measures on $\mathcal{A}^{S}$ are weak-* dense. Considering this, we now introduce a definition which will be fundamental throughout this work.

Definition 2.17.

A semigroup $S$ has the periodic approximation property, or simply the pa property, if $\mathcal{P}(X,S)$ is weak-* dense in $\mathcal{M}_{S}(\mathcal{A}^{S})$ for every finite alphabet $\mathcal{A}$ . A semigroup $S$ has the ergodic periodic approximation property, or simply the epa property, if $\mathcal{P}^{\mathrm{erg}}(X,S)$ is weak-* dense in $\mathcal{M}_{S}(\mathcal{A}^{S})$ for every finite alphabet $\mathcal{A}$ .

This definition extends the one discussed for groups in the Introduction. In the case of groups, the pa property is equivalent to the property md described by Kechris [7, 22] in the context of weak containment of group actions. It is known that all amenable, residually finite groups have the epa property [26]. Similarly, Bowen showed that free groups on finitely many generators have the pa property [5]. Examples of residually finite groups without the pa property include the special linear group $\operatorname{SL}_{n}(\mathbb{Z})$ for $n\geq 3$ [22] (in contrast to $\operatorname{SL}_{2}(\mathbb{Z})$ , which has the pa property) and $\mathbb{F}_{2}\times\mathbb{F}_{2}$ as a consequence of the negative answer to the Connes’ Embedding Problem (see [7, p. 27] and [20]).

We stress that, although the pa property has been attentively studied for groups, this has not been the case with general semigroups. As a first preliminary result, we show that residual finiteness turns out to be a key necessary condition for having the pa property, as seen below.

Proposition 2.18.

Let $S$ be a left reductive semigroup. If $S$ has the pa property, then it is residually a finite group.

Proof.

Assume $S$ is not residually a finite group. By Theorem Introduction, there is a configuration $x\in\mathcal{A}^{S}$ and a finite subset $F\subseteq S$ such that $[x;F]\cap\operatorname{Per}(\mathcal{A}^{S},S)=\varnothing$ . Choose a measure $\mu\in\mathcal{M}_{S}(\mathcal{A}^{S})$ such that $\mu([x;F])>0$ (take, for instance, the Bernoulli measure associated to a positive probability vector), and let $(\mu_{n})_{n}$ be a sequence in $\mathcal{P}(X,S)$ that weak-* converges to $\mu$ . By Proposition 2.9, we can assume that for every $n\geq 1$ ,

\mu_{n}=\sum_{i=1}^{m_{n}}\frac{\mu_{n}(Sx^{n}_{i})}{|Sx^{n}_{i}|}\sum_{y\in Sx^{n}_{i}}\delta_{y},

where each $x^{n}_{i}$ is $S$ -periodic for every $1\leq i\leq m_{n}$ . By weak-* convergence, we have that $\mu_{n}([x;F])\to\mu([x;F])$ . However, for all $n\geq 1$ and $1\leq i\leq m_{n}$ , the set $[x;F]\cap Sx^{n}_{i}$ is empty, meaning that $\mu_{n}([x;F])=0$ , a contradiction with our choice of $\mu$ . ∎

Remark 2.19.

Based on Example 2.16, it is not difficult to construct an example of a semigroup with the pa property that is not left reductive, and thus cannot be residually a finite group.

3. A sufficient condition for the pa property

We want to understand the pa property in the more general landscape of semigroups, taking advantage of what is already known for groups. With this goal in mind, we restrict ourselves to the class of embeddable monoids. The purpose of this section is to establish a direct connection between the pa property of an embeddable monoid $S$ and the pa property of the corresponding free $S$ -group. In order to do this, the tool of choice will be the natural extension construction, which associates to an $S$ -action a corresponding invertible action of the free $S$ -group which extends the original action in a natural way.

3.1. Topological natural extensions

In [6], topological natural extensions were profusely discussed. Based on this work, we consider the following definition.

Definition 3.1.

Let $S$ be an embeddable monoid and $S\overset{\alpha}{\curvearrowright}X$ a continuous $S$ -action. Given a receiving $S$ -group $\mathbf{G}=(G,\eta)$ , the topological natural $\mathbf{G}$ -extension is the tuple $(X_{\mathbf{G}},\sigma,\pi)$ , where

X_{\mathbf{G}}=\left\{(x_{h})_{h\in G}\in X^{G}:s\cdot x_{h}=x_{\eta(s)h}\textup{ for all }s\in S\textup{ and }h\in G\right\}

is endowed with the subspace topology of the product topology, $\sigma$ denotes the restriction of the shift action $G\overset{\sigma}{\curvearrowright}X^{G}$ to $X_{\mathbf{G}}$ , and $\pi\colon X_{\mathbf{G}}\rightarrow X$ is the projection $\pi_{1_{G}}\colon X^{G}\to X$ restricted to $X_{\mathbf{G}}$ . If $\pi\colon X_{\mathbf{G}}\rightarrow X$ is surjective, $\alpha$ is said to be $\textbf{topologically }\mathbf{G}$ -extensible, and if $X_{\mathbf{G}}\neq\varnothing$ , $\alpha$ is said to be topologically partially $\mathbf{G}$ -extensible.

Since $X_{\mathbf{G}}$ is closed in $X^{G}$ , and $X^{G}$ is compact—as we are assuming $X$ is a compact metric space—, we have that $X_{\mathbf{G}}$ compact as well. In addition, the map $\pi$ is continuous and $\eta$ -equivariant in the sense that

s\cdot\pi(\overline{x})=\pi(\eta(s)\cdot\overline{x})\quad\text{ for all }\overline{x}\in X_{\mathbf{G}}\text{ and }s\in S.

In [6] it was proven that the natural $\mathbf{G}$ -extension of $S\curvearrowright X$ comes with a universal property: if $(\hat{X},\beta,\tau)$ is any tuple such that $\hat{X}$ is a topological space, $\beta$ is a continuous action $G\curvearrowright\hat{X}$ , and $\tau\colon\hat{X}\rightarrow X$ is surjective, continuous, and $\eta$ -equivariant, then there is a unique equivariant continuous function $\varphi\colon\hat{X}\rightarrow X_{\mathbf{G}}$ satisfying $\pi\circ\varphi=\tau$ , so that the following diagram commutes.

A key consequence of the main result in [6] is the following:

Theorem 3.2 ([6]).

If $\mathbf{G}$ is the free $S$ -group, then every surjective continuous $S$ -action is topologically $\mathbf{G}$ -extensible.

3.2. Measure-theoretical extensions

We will make use of topological natural extensions to define measure-theoretical natural extensions. Observe that, given a receiving $S$ -group $\mathbf{G}=(G,\eta)$ and a topologically $\mathbf{G}$ -extensible continuous action $S\curvearrowright X$ , the projection map $\pi\colon X_{\mathbf{G}}\rightarrow X$ induces a push-forward map $\pi_{*}\colon\mathcal{M}(X_{\mathbf{G}})\rightarrow\mathcal{M}(X)$ given by $\pi_{*}\hat{\mu}(A)=\mu(\pi^{-1}(A)),$ for all $A\in\mathcal{B}(X)$ . Moreover, since $\pi$ is $\eta$ -equivariant, the image of a $G$ -invariant measure on $X_{\mathbf{G}}$ via $\pi_{*}$ is an $S$ -invariant measure on $X$ , so the operator $\pi_{*}\colon\mathcal{M}_{G}(X_{\mathbf{G}})\rightarrow\mathcal{M}_{S}(X)$ is well-defined. From now on, $\pi_{*}$ will denote this restricted version of the push-forward. It is standard that $\pi_{*}$ is weak-* continuous (see, e.g., [14, Appendix B]).

Definition 3.3.

Let $S\curvearrowright X$ be a continuous action and $\mu\in\mathcal{M}_{S}(X)$ . If there exists $\bar{\mu}\in\mathcal{M}_{G}(X_{\mathbf{G}})$ such that $\pi_{*}\bar{\mu}=\mu$ , then $\mu$ will be called $\mathbf{G}$ -extensible and $\bar{\mu}$ a $\mathbf{G}$ -extension of $\mu$ . The set $\operatorname{im}(\pi_{*})$ will be denoted by $\operatorname{Ext}_{\mathbf{G}}(X,S)$ .

Remark 3.4.

Observe that if $\mathbf{G}=(G,\eta)$ is a receiving $S$ -group, $S\curvearrowright(X,\mu)$ is a p.m.p. action, and $\mu\in\mathcal{M}_{S}(X)$ is $\mathbf{G}$ -extensible, there is a $G$ -invariant measure $\bar{\mu}$ on $(X_{\mathbf{G}},\mathcal{B}(X_{\mathbf{G}}))$ such that $\pi\colon(X_{\mathbf{G}},\bar{\mu})\to(X,\mu)$ is measure-preserving. Thus, the shift action $G\curvearrowright(X_{\mathbf{G}},\bar{\mu})$ is a p.m.p. action, and the tuple $(X_{\mathbf{G}},\sigma,\bar{\mu},\pi)$ may be regarded as a measure-theoretical natural $\mathbf{G}$ -extension of $S\curvearrowright(X,\mu)$ .

Remark 3.5.

Let $\mathbf{G}=(G,\eta)$ be a receiving $S$ -group. We may identify the topological natural $\mathbf{G}$ -extension of $\mathcal{A}^{S}$ with $(\mathcal{A}^{G},\pi)$ , where $\pi\colon\mathcal{A}^{G}\rightarrow\mathcal{A}^{S}$ is given by $\overline{x}\mapsto\overline{x}\circ\eta$ (see [6]). With this identification in mind, a measure $\mu\in\mathcal{M}_{S}(\mathcal{A}^{S})$ is $\mathbf{G}$ -extensible if and only if there is a measure $\bar{\mu}\in\mathcal{M}_{G}(\mathcal{A}^{G})$ such that $\pi_{*}\bar{\mu}=\mu$ . In the forthcoming, whenever we deal with natural extensions in the symbolic case, we will proceed with this identification, so $\pi$ will denote the map $\overline{x}\mapsto\overline{x}\circ\eta$ .

Proposition 3.6.

The subset $\operatorname{Ext}_{\mathbf{G}}(X,S)\subseteq\mathcal{M}_{S}(X)$ is weak-* closed and convex.

Proof.

Let $\mu\in\mathcal{M}_{S}(X)$ , and let $(\mu_{n})_{n}$ be a sequence in $\operatorname{Ext}_{\mathbf{G}}(X,S)$ converging to $\mu$ . Consider, for every $n\in\mathbb{N}$ , a measure $\bar{\mu}_{n}\in\mathcal{M}_{G}(X_{\mathbf{G}})$ such that $\pi_{*}\bar{\mu}_{n}=\mu_{n}$ . Since $X_{\mathbf{G}}$ is compact, $\mathcal{M}_{G}(X_{\mathbf{G}})$ is weak-* compact. Thus, we can find a subsequence $(\bar{\mu}_{n_{k}})_{k}$ of $(\bar{\mu}_{n})_{n}$ weak-* converging to some $\bar{\mu}\in\mathcal{M}_{G}(X_{\mathbf{G}})$ . By continuity of $\pi_{*}$ , we get

\mu=\lim_{k\to\infty}\mu_{n_{k}}=\lim_{k\to\infty}\pi_{*}\bar{\mu}_{n_{k}}=\pi_{*}\left(\lim_{k\to\infty}\bar{\mu}_{n_{k}}\right)=\pi_{*}\bar{\mu}.

Therefore, $\mu\in\operatorname{Ext}_{\mathbf{G}}(X,S)$ . Finally, $\operatorname{Ext}_{\mathbf{G}}(X,S)$ is convex, as it is the image of the convex set $\mathcal{M}_{G}(X_{\mathbf{G}})$ under the linear function $\pi_{*}$ . ∎

Proposition 3.7.

Let $\mathbf{G}=(G,\eta)$ be a receiving $S$ -group. Let $S\overset{\alpha}{\curvearrowright}(X,\mu),S\overset{\beta}{\curvearrowright}(Y,\nu)$ be two p.m.p. actions such that $\alpha$ is a factor of $\beta$ . Then, if $\nu$ is $\mathbf{G}$ -extensible, so is $\mu$ .

Proof.

Let $X^{\prime}\subseteq X$ and $Y^{\prime}\subseteq Y$ be full measure $S$ -invariant sets and let $\varphi\colon Y^{\prime}\rightarrow X^{\prime}$ be a measure preserving equivariant map. Let $(X_{\mathbf{G}},\pi_{\alpha})$ and $(Y_{\mathbf{G}},\pi_{\beta})$ be the topological $\mathbf{G}$ -extensions of $\alpha$ and $\beta$ , respectively, and $\bar{\nu}\in\mathcal{M}_{G}(X_{\mathbf{G}})$ such that $(\pi_{\beta})_{*}\bar{\nu}=\nu$ . The first thing we need to check is that $X_{\mathbf{G}}$ is non-empty (and hence $\alpha$ is topologically partially $\mathbf{G}$ -extensible). Indeed, since $\bar{\nu}$ is $G$ -invariant, $\bar{\nu}(g\pi_{\beta}^{-1}(Y^{\prime}))=\mu(Y^{\prime})=1$ for all $g\in G$ , so the fact that $G$ is countable implies

\bar{\nu}\left(\bigcap_{g\in G}g\pi_{\beta}^{-1}(Y^{\prime})\right)=1.

Thus, there is an element $(y_{h})_{g\in G}\in Y_{\mathbf{G}}$ with $y_{h}\in Y^{\prime}$ for all $h\in G$ . The element $(\varphi(y_{h}))_{h\in G}$ is an element of $X_{\mathbf{G}}$ .

We want to construct a measure $\bar{\mu}\in\mathcal{M}_{G}(X_{\mathbf{G}})$ satisfying $(\pi_{\alpha})_{*}\bar{\mu}=\mu$ . Consider the function

	$\displaystyle\bar{\varphi}\colon\bigcap_{g\in G}g\pi_{\beta}^{-1}(Y^{\prime})$	$\displaystyle\longrightarrow\bigcap_{g\in G}g\pi_{\alpha}^{-1}(X^{\prime})$
	$\displaystyle(y_{h})_{h\in G}$	$\displaystyle\longmapsto(\varphi(y_{h}))_{h\in G},$

which is well-defined, as $\varphi(Y^{\prime})\subseteq X^{\prime}$ and, given $(y_{h})_{h\in G}$ in $\bigcap_{g\in G}g\pi_{\beta}^{-1}(Y^{\prime})$ , $y_{h}\in Y^{\prime}$ for every $h\in G$ . As we already mentioned, $\bar{\varphi}$ is defined upon a full-measure subset of $Y_{\mathbf{G}}$ . Also, $\bar{\varphi}$ is clearly $G$ -equivariant. Define, for $\bar{A}\in\mathcal{B}(X_{\mathbf{G}})$ ,

\bar{\mu}\left(\bar{A}\right)=\bar{\nu}\left(\bar{\varphi}^{-1}\left(\bar{A}\cap\bigcap_{g\in G}g\pi_{\alpha}^{-1}(X^{\prime})\right)\right).

The set function $\bar{\mu}$ is a probability measure on $X_{\mathbf{G}}$ . The $G$ -invariance of $\bar{\mu}$ comes as a consequence of the $G$ -equivariance of $\bar{\varphi}$ and the $G$ -invariance of $\bigcap_{g\in G}g\pi_{\alpha}^{-1}(X^{\prime})$ . To see that $\pi_{\alpha}$ is a measure-preserving map, note that $\pi_{\alpha}\circ\bar{\varphi}=\varphi\circ\pi_{\beta}$ , so for every $A\in\mathcal{B}(X)$ ,

	$\displaystyle\bar{\mu}(\pi_{\alpha}^{-1}(A))$	$\displaystyle=\bar{\nu}\left(\bar{\varphi}^{-1}\left(\pi_{\alpha}^{-1}(A)\cap\bigcap_{g\in G}g\pi_{\alpha}^{-1}(X^{\prime})\right)\right)$
		$\displaystyle=\bar{\nu}\left(\bar{\varphi}^{-1}(\pi_{\alpha}^{-1}(A))\cap\bigcap_{g\in G}g\bar{\varphi}^{-1}(\pi_{\alpha}^{-1}(X^{\prime}))\right)$
		$\displaystyle=\bar{\nu}\left(\pi_{\beta}^{-1}\left(\varphi^{-1}(A)\right)\cap\bigcap_{g\in G}g\pi_{\beta}^{-1}(Y^{\prime})\right)$
		$\displaystyle=\bar{\nu}\left(\pi_{\beta}^{-1}\left(\varphi^{-1}(A)\right)\right)=\nu(\varphi^{-1}(A))=\mu(A).$

∎

In view of Proposition 3.7, if $S\curvearrowright(X,\mu)$ and $S\curvearrowright(Y,\nu)$ are measure-theoretically conjugate, then $\mu$ is $\mathbf{G}$ -extensible if and only if $\nu$ is.

3.3. Periodicity and extensibility

A first key observation is that periodic measures are always $\bm{\Gamma}$ -extensible when $\bm{\Gamma}$ is the free $S$ -group.

Proposition 3.8.

Assume that $S$ is an embeddable monoid, that $\bm{\Gamma}=(\Gamma,\gamma)$ is a realization of the free $S$ -group, and let $S\curvearrowright X$ be a continuous action. Then, for every $\mu\in\mathcal{P}^{\mathrm{erg}}(X,S)$ , there exists $\bar{\mu}\in\mathcal{P}^{\mathrm{erg}}(X_{\bm{\Gamma}},\Gamma)$ such that $\pi_{*}\bar{\mu}=\mu$ . In particular, $\mathcal{P}(X,S)\subseteq\operatorname{Ext}_{\bm{\Gamma}}(X,S)$ .

Proof.

Let $\mu\in\mathcal{P}^{\mathrm{erg}}(X,S)$ be given by

\mu=\frac{1}{|Sx|}\sum_{y\in Sx}\delta_{y}\quad\text{ for some }x\in\operatorname{Per}(X,S).

Consider a disjoint copy $\hat{S}=\{\hat{s}:s\in S\}$ of $S$ , with the operation $\hat{t}\hat{s}=\hat{st}$ for $\hat{s},\hat{t}\in\hat{S}$ . We define an action of $\hat{S}$ on $Sx$ as follows: for every $\hat{s}\in\hat{S}$ and $y\in Sx$ , we let $\hat{s}\cdot y$ to be the only element of $s^{-1}(\{y\})\cap Sx$ . Notice that this is indeed an action, because $\hat{st}\cdot y$ is the only element of

(st)^{-1}(\{y\})\cap Sx=t^{-1}(s^{-1}(\{y\}))\cap Sx=t^{-1}(s^{-1}(\{y\})\cap Sx)\cap Sx,

and the only element of the latter is exactly $\hat{t}\cdot(\hat{s}\cdot y)$ .

Since both $S$ and $\hat{S}$ act upon $Sx$ , we immediately get an action $S\ast\hat{S}\curvearrowright Sx$ by concatenation, where $S\ast\hat{S}$ denotes the free product semigroup, where the empty word $\varepsilon$ acts as the identity permutation.

We want to see the action of $S\ast\hat{S}$ descends to an action of $\Gamma$ . To do this, we define, for every $\hat{s}\in\hat{S}$ , the element $\hat{\gamma}(\hat{s})=\gamma(s)^{-1}\in\Gamma$ . Note that this defines a morphism $\hat{\gamma}\colon\hat{S}\to\Gamma$ , since $\hat{\gamma}(\hat{t}\hat{s})=\hat{\gamma}(\hat{st})=\gamma(st)^{-1}=\gamma(t)^{-1}\gamma(s)^{-1}=\hat{\gamma}(\hat{t})\hat{\gamma}(\hat{s})$ for all $\hat{s},\hat{t}\in\hat{S}$ . This induces a morphism $\gamma_{*}\colon S\ast\hat{S}\to\Gamma$ by

\gamma_{*}(s_{1}\hat{t}_{1}\cdots s_{n}\hat{t}_{n})=\gamma(s_{1})\hat{\gamma}(\hat{t}_{1})\cdots\gamma(s_{n})\hat{\gamma}(\hat{t}_{n}).

Given $w_{1},w_{2}\in S\ast\hat{S}$ , declare $w_{1}\sim w_{2}$ if and only if $w_{1}$ and $w_{2}$ induce the same element in $\mathrm{Sym}(Sx)$ , i.e., if $w_{1}\cdot y=w_{2}\cdot y$ for every $y\in Sx$ . We want to check that, if $\gamma_{*}(w_{1})=\gamma_{*}(w_{2})$ , then $w_{1}\sim w_{2}$ .

First, observe that if $w_{1}\sim w_{2}$ , then for any $u,v\in S\ast\hat{S}$ ,

uw_{1}v\cdot y=u\cdot(w_{1}\cdot(v\cdot y))=u\cdot(w_{2}\cdot(v\cdot y))=uw_{2}v\cdot y\quad\text{ for every }y\in Sx,

thus making $\sim$ a congruence on $S\ast\hat{S}$ . Next, consider the quotient semigroup $S\ast\hat{S}/\sim$ . Take an arbitrary element $w=s_{1}\hat{t}_{1}\cdots s_{n}\hat{t}_{n}$ in $S\ast\hat{S}$ and define $\hat{w}=t_{n}\hat{s}_{n}\cdots t_{1}\hat{s}_{1}$ . Since $s\hat{s}\sim\varepsilon\sim\hat{s}s$ for every $s\in S$ , inductively we have $w\hat{w}\sim\varepsilon\sim\hat{w}w$ . This means

[w]_{\sim}[\hat{w}]_{\sim}=[w\hat{w}]_{\sim}=[\varepsilon]_{\sim}=[\hat{w}w]_{\sim}=[\hat{w}]_{\sim}[w]_{\sim}.

Hence, $G:=S\ast\hat{S}/\sim$ is actually a group, where the inverse of $[w]_{\sim}$ is just $[\hat{w}]_{\sim}$ .

Let $q\colon S\ast\hat{S}\to G$ be the quotient map associated to $\sim$ . The map $q\rvert_{S}\colon S\rightarrow G$ is a semigroup morphism, as equal elements of $S$ define equal elements in $\textup{Sym}(Sx)$ . Analogously, $q\rvert_{\hat{S}}\colon\hat{S}\rightarrow G$ is a semigroup morphism, too. Since $[s_{1}\hat{t}_{1}\cdots s_{n}\hat{t}_{n}]_{\sim}=[s_{1}]_{\sim}[t_{1}]_{\sim}^{-1}\cdots[s_{n}]_{\sim}[t_{n}]_{\sim}^{-1}$ for every $s_{1}\hat{t}_{1}\cdots s_{n}\hat{t}_{n}\in S\ast\hat{S}$ , we find that $(G,q\rvert_{S})$ is an $S$ -group (note that, in general, the morphism $q\rvert_{S}$ may not be injective). By the universal property of the free $S$ -group, we get a group morphism $\theta\colon\Gamma\rightarrow G$ such that $\theta\circ\gamma=q\rvert_{S}$ , and it is verified that $\theta\circ\hat{\gamma}=q\rvert_{\hat{S}}$ as well. Therefore, $\theta\circ\gamma_{*}=q$ , which implies $w_{1}\sim w_{2}$ whenever $\gamma_{*}(w_{1})=\gamma_{*}(w_{2})$ , as desired.

We can now define an action $\Gamma\curvearrowright Sx$ by setting $\gamma_{*}(w)\cdot y=w\cdot y$ for every $w\in S\ast\hat{S}$ and $y\in Sx$ , as a consequence of the fact that $\gamma_{*}(S\ast\hat{S})=\Gamma$ . For a given $y\in Sx$ , let $\bar{y}=(y_{h})_{h\in G}\in X_{\bm{\Gamma}}$ be given by $y_{h}=h\cdot y$ for every $h\in\Gamma$ . Note that if $g\in\Gamma$ , then

g\cdot\bar{x}=g\cdot(h\cdot x)_{h\in G}=(hg\cdot x)_{h\in G}=(h\cdot(g\cdot x))_{h\in G}=\overline{g\cdot x}.

This shows $\Gamma\bar{x}\subseteq\{\bar{y}:y\in Sx\}$ . Also, if $y\in Sx$ , then $y=s\cdot x$ for some $s\in S$ , which implies $\bar{y}=\overline{s\cdot x}=\overline{\gamma(s)\cdot x}=\gamma(s)\cdot\bar{x}$ . Notice that in this last equation there are three different actions involved. Thus, $\{\bar{y}:y\in Sx\}\subseteq\Gamma\bar{x}$ . Since $\bar{y}\neq\bar{y}^{\prime}$ whenever $y,y^{\prime}\in Sx$ differ, we have that $|\Gamma\bar{x}|=|\{\bar{y}:y\in Sx\}|=|Sx|$ . Thus, $\bar{x}$ is $\Gamma$ -periodic, and the measure

\bar{\mu}=\frac{1}{|\Gamma\bar{x}|}\sum_{\bar{y}\in\Gamma\,\bar{x}}\delta_{\bar{y}}=\frac{1}{|Sx|}\sum_{\bar{y}\in\Gamma\,\bar{x}}\delta_{\bar{y}}

is $\Gamma$ -periodic and ergodic, as it is supported on a single orbit. Finally, note that

\bar{\mu}(\pi^{-1}A)=\frac{1}{|Sx|}\sum_{\bar{y}\in\Gamma\,\bar{x}}\delta_{\bar{y}}(\pi^{-1}A)=\frac{1}{|Sx|}\sum_{\bar{y}\in\Gamma\,\bar{x}}\delta_{\pi(\bar{y})}(A)=\frac{1}{|Sx|}\sum_{y\in Sx}\delta_{y}(A)=\mu(A)

for every $A\in\mathcal{B}(X)$ , so $\pi_{*}\overline{\mu}=\mu$ . Hence $\mu\in\textup{Ext}_{\bm{\Gamma}}(X,S)$ .

Finally, as every $\mu\in\mathcal{P}(X,S)$ is a convex combination of elements in $\mathcal{P}^{\mathrm{erg}}(X,S)$ , $\pi_{*}$ is linear, and, by Proposition 3.6, $\operatorname{Ext}_{\bm{\Gamma}}(X,S)$ is a convex subset of $\mathcal{M}_{S}(X)$ , we conclude. ∎

Example 3.9.

Note that if we remove the assumption that $\bm{\Gamma}$ is the free $S$ -group then Proposition 3.8 does not necessarily hold, as the set $X_{\mathbf{G}}$ might be empty for an arbitrary receiving $S$ -group $\mathbf{G}$ .

It is important to know the behaviour of images of $G$ -orbits via $S$ -equivariant maps.

Lemma 3.10.

Let $\mathbf{G}=(G,\eta)$ be a receiving $S$ -group, and let $G\curvearrowright\hat{X}$ and $S\curvearrowright X$ be two actions. If $\phi:\hat{X}\rightarrow X$ is an $\eta$ -equivariant map and $\hat{x}\in\hat{X}$ is $G$ -periodic (i.e., $|G\hat{x}|<\infty$ ), then $\phi(\hat{x})$ is $S$ -periodic and $\phi(G\hat{x})=S\phi(\hat{x})$ .

Proof.

If $s\in S$ , it is clear that $\phi(\eta(s)\cdot\hat{x})=s\cdot\phi(\hat{x})\in S\phi(\hat{x})$ . Also, $\eta(s)$ has finite order as an element of $\textup{Sym}(G\hat{x})$ , so there exists $n>1$ such that $\eta(s^{n})$ acts as the identity of $G\hat{x}$ . Thus,

\phi(\eta(s)^{-1}\cdot\hat{x})=\phi(\eta(s^{n-1})\cdot\hat{x})=s^{n-1}\cdot\phi(\hat{x})\in S\phi(\hat{x}).

Since $G=\langle\eta(S)\rangle$ , we conclude that $\phi(G\hat{x})\subseteq S\phi(\hat{x})$ , which directly implies that $\phi(G\hat{x})=S\phi(\hat{x})$ .

It is now clear that $\phi(\hat{x})$ is pre-periodic. To see that $S\phi(\hat{x})$ is completely $S$ -invariant, let $s,t\in S$ . Since the set $\{\eta(t)^{n}\eta(s)\cdot\hat{x}:n\geq 1\}$ is finite, there are $k>j$ such that $\eta(t)^{k}\eta(s)\cdot\hat{x}=\eta(t)^{j}\eta(s)\cdot\hat{x}$ . The action on $\hat{X}$ is given by a group, so $\eta(s)\cdot\hat{x}=\eta(t)^{k-j}\eta(s)\cdot\hat{x}$ and $t^{k-j}\in S$ . Then, due to the $\eta$ -equivariance of $\phi$ ,

s\cdot\phi(\hat{x})=t\cdot(t^{k-j-1}s\cdot\phi(\hat{x}))\in tS\phi(\hat{x}),

so $t\colon S\phi(\hat{x})\rightarrow S\phi(\hat{x})$ is surjective, and hence bijective by finiteness of $S\phi(\hat{x})$ .

∎

Proposition 3.11.

Let $S\curvearrowright X$ be a continuous action, and let $\mathbf{G}$ be a receiving $S$ -group. If $\bar{\mu}\in\mathcal{M}_{G}(X_{\mathbf{G}})$ , then $\operatorname{supp}(\pi_{*}\bar{\mu})=\pi(\operatorname{supp}(\bar{\mu}))$ . In particular, we have the following:

(i)

$\pi_{*}(\mathcal{P}^{\mathrm{erg}}(X_{\mathbf{G}},G))\subseteq\mathcal{P}^{\mathrm{erg}}(X,S)$ and $\pi_{*}(\mathcal{P}(X_{\mathbf{G}},G))\subseteq\mathcal{P}(X,S)$ .
(ii)

If $\overline{\mathcal{P}(X_{\mathbf{G}},G)}=\mathcal{M}_{G}(X_{\mathbf{G}})$ , then $\operatorname{Ext}_{\mathbf{G}}(X,S)\subseteq\overline{\mathcal{P}(X,S)}$ .
(iii)

If $\overline{\mathcal{P}^{\mathrm{erg}}(X_{\mathbf{G}},G)}=\mathcal{M}_{G}(X_{\mathbf{G}})$ , then $\operatorname{Ext}_{\mathbf{G}}(X,S)\subseteq\overline{\mathcal{P}^{\mathrm{erg}}(X,S)}$ .

Proof.

First, we prove that $\operatorname{supp}(\pi_{*}\bar{\mu})=\pi(\operatorname{supp}(\bar{\mu}))$ . Let $x\in\operatorname{supp}(\pi_{*}\bar{\mu})$ and fix a decreasing sequence $(U_{n})_{n}$ of open neighborhoods of $x$ with $\bigcap_{n}U_{n}=\{x\}$ . Since $x\in\operatorname{supp}(\pi_{*}\bar{\mu})$ , $\pi_{*}\bar{\mu}(U_{n})>0$ for every $n$ . In particular, for every $n$ , $\pi^{-1}(U_{n})\cap\operatorname{supp}(\bar{\mu})$ is non-empty, so there exists $\bar{x}_{n}\in\operatorname{supp}(\bar{\mu})$ with $\pi(\bar{x}_{n})\in U_{n}$ . By compactness of $X_{\mathbf{G}}$ , we can take a subsequence $(\bar{x}_{n_{k}})_{k}$ converging to $\bar{x}\in\operatorname{supp}(\bar{\mu})$ as $k\to\infty$ . By continuity of $\pi$ , we get $\pi(\bar{x}_{n_{k}})\to\pi(\bar{x})$ , but since $\pi(\bar{x}_{n_{k}})\in U_{n_{k}}$ , we also have that $\pi(\bar{x}_{n_{k}})\to x$ . Thus, we conclude that $\pi(\bar{x})=x$ , so $x\in\pi(\operatorname{supp}(\bar{\mu}))$ . For the opposite inclusion, let $\bar{x}\in\operatorname{supp}(\bar{\mu})$ . If $U$ is an open neighborhood of $\pi(\bar{x})$ , then $\bar{x}\in\pi^{-1}U$ and $\pi_{*}\bar{\mu}(U)=\bar{\mu}(\pi^{-1}U)>0$ , so $\pi(x)\in\operatorname{supp}(\pi_{*}\bar{\mu})$ , following the desired inclusion.

To prove (i), let $\bar{\mu}\in\mathcal{P}(X_{\mathbf{G}},G)$ . There exist elements $x_{1},\dots,x_{m}\in\operatorname{Per}(X_{\mathbf{G}},G)$ such that

\operatorname{supp}(\bar{\mu})=\bigcup_{i=1}^{m}Gx_{i},

so by the equality proven above and Lemma 3.10, we have $\operatorname{supp}(\pi_{*}\bar{\mu})=\bigcup_{i=1}^{m}\pi(Gx_{i})=\bigcup_{i=1}^{m}S\pi(x_{i})$ . Taking $m=1$ proves the ergodic case.

To prove (ii) and (iii), by (i) and weak-* continuity of $\pi_{*}$ we have $\operatorname{Ext}_{\mathbf{G}}(X,S)=\operatorname{im}(\pi_{*})=\pi_{*}\left(\overline{\mathcal{P}(X_{\mathbf{G}},G)}\right)\subseteq\overline{\mathcal{P}(X,S)}$ , and the same argument shows that $\operatorname{Ext}_{\mathbf{G}}(X,S)\subseteq\overline{\mathcal{P}^{\mathrm{erg}}(X,S)}$ . ∎

Combining the previous results together yields the following theorem.

\secondtheorem

Proof.

We prove the periodic case; the ergodic periodic case is identical. By Proposition 3.8, $\mathcal{P}(X,S)\subseteq\operatorname{Ext}_{\bm{\Gamma}}(X,S)$ . By Proposition 3.11(ii), $\operatorname{Ext}_{\bm{\Gamma}}(X,S)\subseteq\overline{\mathcal{P}(X,S)}$ . By Proposition 3.6, $\operatorname{Ext}_{\bm{\Gamma}}(X,S)$ is weak-* closed, so $\operatorname{Ext}_{\bm{\Gamma}}(X,S)=\overline{\mathcal{P}(X,S)}$ .

Finally, since $(\mathcal{A}^{S})_{\bm{\Gamma}}$ is topologically conjugate to $\mathcal{A}^{\Gamma}$ for every finite alphabet $\mathcal{A}$ , if (i) and (ii) are satisfied, then $\overline{\mathcal{P}(\mathcal{A}^{S},S)}=\operatorname{Ext}_{\bm{\Gamma}}(\mathcal{A}^{S},S)=\mathcal{M}_{S}(\mathcal{A}^{S})$ , so $S$ has the pa property. ∎

4. Semigroups with the pa property

In this section we aim to prove that some particular families of semigroups have the (e)pa property. The strategy will be to take an embeddable semigroup $S$ and its free $S$ -group $\bm{\Gamma}=(\Gamma,\gamma)$ , check that $\Gamma$ has the (e)pa property and that every measure in $\mathcal{M}_{S}(\mathcal{A}^{S})$ is $\bm{\Gamma}$ -extensible, and then appeal to Theorem Introduction to conclude.

First, we observe that, due to Proposition 2.18, a necessary condition for a semigroup to have the pa property is to be residually a finite group. In particular, every group that has the pa property must be residually finite. Thus, we will focus on embeddable semigroups $S$ that are residually a finite group and such that the underlying group $\Gamma$ of the free $S$ -group is residually finite. To check these two conditions, it suffices to check that $\Gamma$ is residually finite, as this directly implies that $S$ is residually a finite group. However, the converse might not be the case. An example of this would be the Baumslag–Solitar semigroup

\textup{BS}(2,3)^{+}=\langle a,b\mid ab^{2}=b^{3}a\rangle,

which is residually a finite group [19, Theorem 4.5], while the corresponding Baumslag-Solitar group $\textup{BS}(2,3)$ is known to be non-Hopfian, and hence non-residually finite [4].

4.1. The left amenable case

The following result was proven in [26].

Theorem 4.1 ([26, Theorem 1.1]).

Let $G$ be a discrete countable residually finite amenable group acting on a compact metric space $X$ with specification property. Then $\mathcal{P}^{\mathrm{erg}}(X,G)$ is dense in $\mathcal{M}_{G}(X)$ in the weak-* topology.

Since the shift action on the full $G$ -shift $\mathcal{A}^{G}$ trivially satisfies the specification property, this result immediately implies that residually finite amenable groups have the epa property. We want to establish an analogous result for semigroups.

A bicancellative semigroup $S$ is left amenable if there exists a left Følner sequence, namely a sequence $(F_{n})_{n}$ of finite subsets of $S$ such that

\lim_{n\to\infty}\frac{|sF_{n}\triangle F_{n}|}{|F_{n}|}=0\quad\text{ for every }s\in S.

Check [2, 16, 24] for further details.

We need to check that when $S$ is left amenable, residually a finite group, and $\bm{\Gamma}=(\Gamma,\gamma)$ is the free $S$ -group, $\Gamma$ has the epa property. In order to do this, we prove that $\Gamma$ is amenable and residually finite.

Proposition 4.2 ([12, Lemma 1], [6, Proposition 4.2]).

If $S$ is a bicancellative and left amenable semigroup, then $S$ is left reversible. In addition, if $\bm{\Gamma}_{S}=(\Gamma_{S},\gamma)$ is the group of right fractions of $S$ , then $\Gamma_{S}$ is amenable.

Similarly, residual finiteness translates in the reversible case as well.

Proposition 4.3.

Let $S$ be a left reversible semigroup, and $(\Gamma_{S},\gamma)$ be the group of right fractions of $S$ . If $S$ is residually a finite group, then $\Gamma_{S}$ is residually finite.

Proof.

Assume $S$ is residually a finite group. By Proposition 2.13 we may assume $S\leq H$ , where $H$ is a residually finite group. Hence, the subgroup of $H$ generated by $S$ , together with the natural embedding $\iota\colon S\to H$ , is a receiving $S$ -group and thus isomorphic to the group of right fractions of $S$ ; see Remark 1.3. As a consequence, $\Gamma_{S}$ is isomorphic to a subgroup of the residually finite group $H$ and is thus residually finite as well. ∎

Corollary 4.4.

If $S$ is left amenable, residually a finite group and $\bm{\Gamma}=(\Gamma,\gamma)$ is the free $S$ -group, then $\Gamma$ has the epa property.

Proof.

By Remark 1.3, $\bm{\Gamma}$ is isomorphic to $\bm{\Gamma}_{S}=(\Gamma_{S},\gamma)$ . In particular, $\Gamma\simeq\Gamma_{S}$ . By Proposition 4.2 and Proposition 4.3, we know $\Gamma_{S}$ is amenable and residually finite, hence so is $\Gamma$ . By Theorem 4.1, we conclude $\Gamma$ has the epa property. ∎

Regarding $\bm{\Gamma}$ -extensibility of $S$ -invariant measures in the reversible case, we have the following.

Theorem 4.5.

Let $S\curvearrowright X$ be a continuous action. If $S$ is left reversible and bicancellative, and $\bm{\Gamma}$ is the free $S$ -group, then $\textup{Ext}_{\bm{\Gamma}}(X_{\bm{\Gamma}},S)=\mathcal{M}_{S}(X)$ .

Remark 4.6.

While this work was in preparation, a version of Theorem 4.5 was announced in [15, Theorem 2.9]. For completeness, we include our own proof in the Appendix.

A consequence of the previous discussion is the following.

\thirdtheorem

Proof.

By Corollary 4.4, the free $S$ -group $\bm{\Gamma}=(\Gamma,\gamma)$ is such that $\Gamma$ has the epa property. Since $S$ is residually a finite group, it is embeddable and, in particular, bicancellative. By Proposition 4.2, $S$ is left reversible. Therefore, by Theorem 4.5, $\textup{Ext}_{\bm{\Gamma}}(X_{\bm{\Gamma}},S)=\mathcal{M}_{S}(X)$ . We conclude by appealing to Theorem Introduction. ∎

4.2. The free case

Let $d$ be a positive integer and consider the free group $\mathbb{F}_{d}$ on the generators $\{a_{1},\dots,a_{d}\}$ . The following result was proven in [5].

Theorem 4.7 ([5, Theorem 3.4]).

For every $d\geq 1$ , the group $\mathbb{F}_{d}$ has the pa property.

Given $\Sigma\subseteq\{a_{1}^{\pm 1},\dots,a_{d}^{\pm 1}\}$ , consider the subsemigroup $S=\left<\Sigma\right>^{+}$ . Notice that if $\Sigma=\{a_{1},\dots,a_{d}\}$ , $S$ is the free semigroup $\mathbb{F}_{d}^{+}$ . If $|\Sigma|=1$ , then $S$ is $\mathbb{N}$ ; if $\Sigma=\Sigma^{-1}$ , then $S$ is $\mathbb{F}_{k}$ with $2k=|\Sigma|$ ; otherwise, $S$ is a non-reversible semigroup. Without loss of generality, we will assume that $\Sigma\supseteq\{a_{1},\dots,a_{d}\}$ , and therefore $k=d$ . Denote the set $\{a_{1}^{\pm 1},\dots,a_{d}^{\pm 1}\}$ by $\Sigma^{\pm}$ . The free $S$ -group is $\mathbf{F}_{d}=(\mathbb{F}_{d},\iota)$ with $\iota\colon S\to\mathbb{F}_{d}$ the inclusion. When dealing with elements of $S$ as elements of $\mathbb{F}_{d}$ , we will omit any mention of $\iota$ .

We want to use Theorem 4.7 and appeal Theorem Introduction to establish the pa property for the semigroups $S=\left<\Sigma\right>^{+}$ . It remains to show that every $S$ -invariant measure on the full $S$ -shift is $\mathbb{F}_{d}$ -extensible.

Let $\text{Cay}(S,\Sigma)$ be the right Cayley graph of $S$ with respect to $\Sigma$ . The identity $1_{S}$ will be denoted by $\varepsilon$ , and the ball of radius $r$ with center on $\varepsilon$ will be denoted by $B_{r}$ . A subset $T\subseteq S$ will be called a tree if it induces a connected subgraph $(T,E(T))$ in $\text{Cay}(S,\Sigma)$ . The root of a tree $T$ will be the unique element $r(T)\in T$ with minimal distance to $\varepsilon$ .

Fix a finite alphabet $\mathcal{A}$ . A Markov $\Sigma$ -tree chain is a pair $(\mathbf{p},\mathbf{P})$ consisting of a positive probability vector $\mathbf{p}$ (i.e., $\mathbf{p}>\mathbf{0}$ and $\sum_{\ell\in\mathcal{A}}\mathbf{p}_{\ell}=1$ ) and a family $\mathbf{P}=\{\mathbf{P}^{a}:a\in\Sigma\}$ of real $|\mathcal{A}|\times|\mathcal{A}|$ stochastic matrices $\mathbf{P}^{a}$ (i.e., $\mathbf{P}^{a}\geq\mathbf{0}$ and $\sum_{\ell\in\mathcal{A}}\mathbf{P}^{a}_{k,\ell}=1$ for all $k\in\mathcal{A}$ ).

Definition 4.8.

Given a Markov $\Sigma$ -tree chain $(\mathbf{p},\mathbf{P})$ , a measure $\mu\in\mathcal{M}(\mathcal{A}^{S})$ is said to be $(\mathbf{p},\mathbf{P})$ -Markov if, for every $x\in\mathcal{A}^{S}$ and every finite tree $T\subseteq S$ with root $\varepsilon$ , we have

\mu([x\,;T])=\mathbf{p}_{x(\varepsilon)}\cdot\prod_{(t,at)\in E(T)}\mathbf{P}^{a}_{x(t),x(at)}.

A measure $\mu\in\mathcal{M}(\mathcal{A}^{S})$ is said to be Markov if it is $(\mathbf{p},\mathbf{P})$ -Markov for some Markov $\Sigma$ -tree chain $(\mathbf{p},\mathbf{P})$ .

Proposition 4.9.

For every Markov $\Sigma$ -tree chain $(\mathbf{p},\mathbf{P})$ , there exists a unique $(\mathbf{p},\mathbf{P})$ -Markov measure on $\mathcal{A}^{S}$ . Moreover, this measure is $S$ -invariant if and only if

(1)

$\mathbf{p}$ is a left eigenvector of each $\mathbf{P}^{a}$ , i.e., $\mathbf{p}\mathbf{P}^{a}=\mathbf{p}$ for every $a\in\Sigma$ , and
(2)

if $a,a^{-1}\in\Sigma$ , then $\mathbf{p}_{k}\mathbf{P}^{a^{-1}}_{k,\ell}=\mathbf{p}_{\ell}\mathbf{P}^{a}_{\ell,k}$ for every $k,\ell\in\mathcal{A}$ .

Proof.

See [17, p. 240]. ∎

Proposition 4.10.

For every Markov $\Sigma$ -tree chain $(\mathbf{p},\mathbf{P})$ that induces an $S$ -invariant measure $\mu$ , there exists a family $\hat{\mathbf{P}}=\{\hat{\mathbf{P}}^{a}:a\in\Sigma^{\pm}\}$ such that the Markov $\Sigma^{\pm}$ -tree chain $(\mathbf{p},\hat{\mathbf{P}})$ induces an $\mathbb{F}_{d}$ -invariant measure $\bar{\mu}_{\mathbb{F}_{d}}$ that is an $\mathbf{F}_{d}$ -extension of $\mu$ .

Proof.

For $a\in\Sigma$ , let $\hat{\mathbf{P}}^{a}=\mathbf{P}^{a}$ , and for $a\notin\Sigma$ , let $\hat{\mathbf{P}}^{a}$ be defined coordinate-wise as

\hat{\mathbf{P}}^{a}_{k,\ell}=\frac{\mathbf{p}_{\ell}}{\mathbf{p}_{k}}\mathbf{P}^{a^{-1}}_{\ell,k}\quad\text{ for }k,\ell\in\mathcal{A}.

First, let’s check that $(\mathbf{p},\hat{\mathbf{P}})$ is a Markov $\Sigma^{\pm}$ -tree chain. For every $a\not\in\Sigma$ , we have $a^{-1}\in\Sigma$ and $\mathbf{p}\mathbf{P}^{a^{-1}}=\mathbf{p}$ . Thus, for all $k\in\mathcal{A}$ ,

\sum_{\ell\in\mathcal{A}}\hat{\mathbf{P}}^{a}_{k,\ell}=\sum_{\ell\in\mathcal{A}}\frac{\mathbf{p}_{\ell}}{\mathbf{p}_{k}}\mathbf{P}^{a^{-}1}_{\ell,k}=\frac{1}{\mathbf{p}_{k}}\sum_{\ell\in\mathcal{A}}\mathbf{p}_{\ell}\mathbf{P}^{a^{-}1}_{\ell,k}=\frac{1}{\mathbf{p}_{k}}\mathbf{p}_{k}=1,

showing that $\hat{\mathbf{P}}^{a}$ is stochastic. Let’s check that $(\mathbf{p},\hat{\mathbf{P}})$ induces an $\mathbb{F}_{d}$ -invariant Markov measure. Indeed,

(1)

if $a\in\Sigma$ , then $\mathbf{p}\hat{\mathbf{P}}^{a}=\mathbf{p}\mathbf{P}^{a}=\mathbf{p}$ and, if $a\notin\Sigma$ , then $a^{-1}\in\Sigma$ and $\mathbf{p}\hat{\mathbf{P}}^{a}=\mathbf{p}\mathbf{P}^{a^{-1}}=\mathbf{p}$ ;
(2)

for every $a\in\mathbb{F}_{d}$ , $\mathbf{p}_{k}\hat{\mathbf{P}}^{a^{-1}}_{k,\ell}=\mathbf{p}_{\ell}\hat{\mathbf{P}}^{a}_{\ell,k}$ for every $k,\ell\in\mathcal{A}$ , by construction of $\hat{\mathbf{P}}$ .

Let $\bar{\mu}_{\mathbb{F}_{d}}$ be the $\mathbb{F}_{d}$ -invariant $(\mathbf{p},\hat{\mathbf{P}})$ -Markov measure on $\mathcal{A}^{\mathbb{F}_{d}}$ . It remains to show that $\bar{\mu}_{\mathbb{F}_{d}}$ is an $\mathbf{F}_{d}$ -extension of $\mu$ . To see this, it suffices to check that $\pi_{*}\bar{\mu}_{\mathbb{F}_{d}}=\mu$ on a cylinder set supported upon a finite tree $T\subseteq S$ , where $\pi\colon\mathcal{A}^{\mathbb{F}_{d}}\rightarrow\mathcal{A}^{S}$ is the restriction map $x\mapsto x|_{S}$ (see Remark 3.5; precomposing with $\iota$ is the same as restricting to $S$ ). Notice that, due to $S$ -invariance of $\pi_{*}\bar{\mu}_{\mathbb{F}_{d}}$ and $\mu$ , we can assume that $T$ has root $\varepsilon$ . Then, it is direct that $\pi_{*}\bar{\mu}_{\mathbb{F}_{d}}=\mu$ , since both $\mu$ and $\bar{\mu}_{\mathbb{F}_{d}}$ assign the same measure to every cylinder $[x;T]$ —as presented in Definition 4.8—, because $\mathbf{P}^{a}$ and $\hat{\mathbf{P}}^{a}$ coincide for $a\in\Sigma$ . ∎

Figure 2. The Cayley graph of

\mathbb{F}_{2}

for the definition of a Markov measure on

\mathcal{A}^{\mathbb{F}_{2}}

. Dashed lines represent inverses of the generators of

\mathbb{F}_{2}

Remark 4.11.

A feature worth highlighting from the last proof is that, if $a\in\Sigma$ , and $\mathbf{P}^{a}$ is a real $|\mathcal{A}|\times|\mathcal{A}|$ stochastic matrix with left eigenvector $\mathbf{p}$ , then the matrix $\mathbf{P}^{a^{-1}}$ defined by

\mathbf{P}^{a^{-1}}_{k,\ell}=\frac{\mathbf{p}_{\ell}}{\mathbf{p}_{k}}\mathbf{P}^{a}_{\ell,k}\quad\text{ for }k,\ell\in\mathcal{A}.

is also a real $|\mathcal{A}|\times|\mathcal{A}|$ stochastic matrix with left eigenvector $\mathbf{p}$ .

We want to use the fact that all $S$ -invariant Markov measures are $\mathbf{F}_{d}$ -extensible to show that every $S$ -invariant measure is $\mathbf{F}_{d}$ -extensible. To do this, we consider appropriate Markovizations and recodings of these measures.

Given alphabets $\mathcal{A},\mathcal{B}$ , and $S$ -subshifts $X\subseteq\mathcal{A}^{S}$ and $Y\subseteq\mathcal{B}^{S}$ , we say that $\varphi:X\to Y$ is a sliding block code if there exists a finite subset $F\subseteq S$ and a map $\Phi:\mathcal{A}^{F}\to\mathcal{B}$ such that $\varphi(x)(s)=\Phi((s\cdot x)|_{F})$ for every $x\in X$ and $s\in S$ . A sliding block code is always continuous and $S$ -equivariant and, due to a straightforward generalization of Curtis-Hedlund-Lyndon theorem, these two properties characterize them. Given a finite subset $F\subseteq S$ , the higher $F$ -block code will be the particular sliding block code $\phi_{F}:\mathcal{A}^{F}\to(\mathcal{A}^{F})^{S}$ given by $\Phi=\mathrm{Id}_{\mathcal{A}^{F}}$ . It is direct to check that $\phi_{F}$ is injective and a conjugacy between $\mathcal{A}^{S}$ and $\phi_{F}(\mathcal{A}^{S})$ . In particular, $\phi_{F}(\mathcal{A}^{S})$ is an $S$ -subshift.

Proposition 4.12.

Every measure $\mu\in\mathcal{M}_{S}(\mathcal{A}^{S})$ is $\mathbf{F}_{d}$ -extensible.

Proof.

Fix $\mu\in\mathcal{M}_{S}(\mathcal{A}^{S})$ and let $X=\text{supp}(\mu)$ , which is an $S$ -subshift of $\mathcal{A}^{S}$ . For each $m\geq 0$ , we abbreviate by $\phi_{m}$ the higher $B_{m}$ -block code $\phi_{B_{m}}:\mathcal{A}^{S}\to(\mathcal{A}^{B_{m}})^{S}$ . Fix $m$ , define $\left.X\right|_{B_{m}}=\{x\rvert_{B_{m}}:x\in X\}$ , which is a finite set and will play the role of an alphabet. Set for each $\alpha,\beta\in\left.X\right|_{B_{m}}$ and $a\in\Sigma$ ,

\mathbf{p}_{\alpha}=\mu([\alpha\,;B_{m}])\quad\text{ and }\quad\mathbf{P}^{a}_{\alpha,\beta}=\dfrac{\mu\big{(}[\alpha\,;B_{m}]\cap a^{-1}[\beta\,;B_{m}]\big{)}}{\mathbf{p}_{\alpha}}.

Notice that $\mathbf{p}_{\alpha}>0$ , since $[\alpha\,;B_{m}]\cap\text{supp}(\mu)\neq\emptyset$ . It is clear that $\mathbf{p}$ is a probability vector, and for each $\alpha\in\left.X\right|_{B_{m}}$ ,

\sum_{\beta\in\left.X\right|_{B_{m}}}\mathbf{P}^{a}_{\alpha,\beta}=\frac{1}{\mathbf{p}_{\alpha}}\sum_{\beta\in\left.X\right|_{B_{m}}}\mu\big{(}[\alpha\,;B_{m}]\cap a^{-1}[\beta\,;B_{m}]\big{)}=1,

since the sets $\big{\{}a^{-1}[\beta;B_{m}]:\beta\in\left.X\right|_{B_{m}}\big{\}}$ form a partition of $(\left.X\right|_{B_{m}})^{S}$ . Thus, $\mathbf{P}^{a}$ is a stochastic matrix for all $a\in\Sigma$ . For each $\beta\in\left.X\right|_{B_{m}}$ , we get

\sum_{\alpha\in\left.X\right|_{B_{m}}}\mathbf{p}_{\alpha}\mathbf{P}^{a}_{\alpha,\beta}=\sum_{\alpha\in\left.X\right|_{B_{m}}}\mu\big{(}[\alpha;B_{m}]\cap a^{-1}[\beta;B_{m}]\big{)}=\mu\big{(}a^{-1}[\beta\,;B_{m}]\big{)}=\mathbf{p}_{\beta},

so $\mathbf{p}\mathbf{P}^{a}=\mathbf{p}$ . Finally, if $a,a^{-1}\in\Sigma$ , then

\mathbf{p}_{\alpha}\mathbf{P}^{a}_{\alpha,\beta}=\mu\big{(}[\alpha\,;B_{m}]\cap a^{-1}[\beta\,;B_{m}]\big{)}=\mu\big{(}a[\alpha\,;B_{m}]\cap[\beta\,;B_{m}]\big{)}=\mathbf{p}_{\beta}\mathbf{P}^{a^{-1}}_{\beta,\alpha},

and, by Proposition 4.9, we conclude $\mathbf{p}$ and $\{\mathbf{P}^{a}:a\in\Sigma\}$ define an $S$ -invariant Markov measure $\nu_{m}$ on $(\left.X\right|_{B_{m}})^{S}$ .

We want to see that $\operatorname{supp}(\nu_{m})\subseteq\phi_{m}(\mathcal{A}^{S})$ . Notice that, if $y\in(\left.X\right|_{B_{m}})^{S}-\phi_{m}(\mathcal{A}^{S})$ , then there exist $s\in S$ and $a\in\Sigma$ such that $[y(s)\,;B_{m}]\cap a^{-1}[y(as)\,;B_{m}]=\varnothing$ . Indeed, if we assume otherwise, for every $s\in S$ and $a\in\Sigma$ , there is an $x\in[y(s);B_{m}]\cap a^{-1}[y(as);B_{m}]$ , so for all $\ell<m$ and $t\in B_{\ell}$ ,

y(s)(ta)=x(ta)=(a\cdot x)(t)=y(as)(t).

Define $z\in\mathcal{A}^{S}$ by $z(s)=y(s)(\varepsilon)$ . Applying iteratively the identity just proven implies that, for $s\in S$ and $t\in B_{m}$ ,

(\varphi_{m}(z)(s))(t)=(s\cdot z)(t)=z(ts)=y(ts)(\varepsilon)=y(s)(t),

so $\varphi_{m}(z)=y$ , contradicting that $y\not\in\phi_{m}(\mathcal{A}^{S})$ . Hence, $[y(s)\,;B_{m}]\cap a^{-1}[y(as)\,;B_{m}]=\varnothing$ for some $s\in S$ and $a\in\Sigma$ , which directly implies that $\nu_{m}([y\,;\{s,as\}])=\mathbf{p}_{y(s)}\mathbf{P}^{a}_{y(s),y(as)}=0$ , so there is an open set containing $y$ with null measure, i.e., $y\not\in\operatorname{supp}(\nu_{m})$ . Thus $\operatorname{supp}(\nu_{m})\subseteq\phi_{m}(\mathcal{A}^{S})$ .

Notice that the Markov measure $\nu_{m}\in\mathcal{M}_{S}((\left.X\right|_{B_{m}})^{S})$ is $\mathbf{F}_{d}$ -extensible (by Proposition 4.10) and trivially conjugate to a measure $\nu_{m}^{\prime}\in\mathcal{M}_{S}(\phi_{m}(\mathcal{A}^{S}))$ . Then, since $\mathbf{F}_{d}$ -extensibility is preserved under factors (by Proposition 3.7), the measure $\nu^{\prime}_{m}$ is $\mathbf{F}_{d}$ -extensible. Let $\mu_{m}=(\phi_{m}^{-1})_{*}\nu^{\prime}_{m}$ , i.e., $\mu_{m}(A)=\nu^{\prime}_{m}(\phi_{m}(A))$ for all $A\in\mathcal{B}(\mathcal{A}^{S})$ . Since $\phi^{-1}_{m}:\phi_{m}(\mathcal{A}^{S})\to\mathcal{A}^{S}$ is a conjugacy and the measure $\nu^{\prime}_{m}$ is $\mathbf{F}_{d}$ -extensible, the measure $\mu_{m}$ is $\mathbf{F}_{d}$ -extensible (again by Proposition 3.7).

We just need to prove that $\mu_{m}\to\mu$ in the weak-* topology. If $F\subseteq S$ is finite and $x\in\mathcal{A}^{F}$ , there is some $m_{0}\geq 0$ such that $F\subseteq B_{m}$ for every $m\geq m_{0}$ , so

\displaystyle\phi_{m}([x\,;F])=\phi_{m}\left(\bigsqcup_{y\in\mathcal{A}^{B_{m}-F}}[y\wedge x\,;B_{m}]\right)=\phi_{m}(\mathcal{A}^{S})\cap\ \bigsqcup_{y\in\mathcal{A}^{B_{m}-F}}[y\wedge x\,;\varepsilon],

whence

	$\displaystyle\mu_{m}([x\,;F])$	$\displaystyle=\overline{\nu}_{m}\left(\bigsqcup_{y\in\mathcal{A}^{B_{m}-F}}[y\wedge x\,;\varepsilon]\right)=\nu_{m}\left(\bigsqcup_{y\in\mathcal{A}^{B_{m}-F}:y\wedge x\in\left.X\right\|_{B_{m}}}[y\wedge x\,;\varepsilon]\right)$
		$\displaystyle=\sum_{y\in\mathcal{A}^{B_{m}-F}:y\wedge x\in\left.X\right\|_{B_{m}}}\mathbf{p}_{y\wedge x}=\sum_{y\in\mathcal{A}^{B_{m}-F}}\mu([y\wedge x\,;B_{m}])=\mu([x\,;F]).$

As a result, $\mu_{m}([x;F])$ is eventually $\mu([x;F])$ as $m\to\infty$ for every cylinder, implying that $\mu_{m}\to\mu$ in the weak-* topology.

∎

We have the following result.

\fourththeorem

Proof.

By Theorem 4.7, we know $\mathbb{F}_{d}$ has the pa property. By Proposition 4.12, we have that $\textup{Ext}_{\mathbf{F}_{d}}(\mathcal{A}^{S},S)=\mathcal{M}_{S}(\mathcal{A}^{S})$ for every finite alphabet $\mathcal{A}$ . We conclude that $S$ has the pa property by appealing to Theorem Introduction. ∎

4.3. Non-extensibility

Another consequence of the main result in [6] is the following:

Theorem 4.13 ([6]).

Let $\mathbf{G}$ be a receiving $S$ -group, $\bm{\Gamma}=(\Gamma,\gamma)$ be the free $S$ -group, and assume that $\Gamma$ is residually finite. Then, $\mathbf{G}$ is the free $S$ -group if and only if every surjective continuous $S$ -action is topologically $\mathbf{G}$ -extensible.

We want to establish a measure-theoretical analog of this theorem. Observe that in the case where $S$ is left amenable this is trivial, as there is only one receiving $S$ -group up to isomorphism of $S$ -groups, namely, the group of right fractions $\bm{\Gamma}_{S}$ , and we have already shown that every $S$ -invariant measure is $\bm{\Gamma}_{S}$ -extensible. We focus on the free case. Let us consider the following proposition.

Proposition 4.14.

Let $S\curvearrowright(X,\mu)$ be a p.m.p. action, and let $\mathbf{G}$ be a receiving $S$ -group. Then, for every $\bar{\mu}\in\mathcal{M}_{G}(X_{\mathbf{G}})$ such that $\pi_{*}\bar{\mu}=\mu$ , we have that $\operatorname{supp}(\bar{\mu})\subseteq(\operatorname{supp}(\mu))_{\mathbf{G}}$ . In particular, if $S\curvearrowright\operatorname{supp}(\mu)$ is not partially $\mathbf{G}$ -extensible, then $\mu$ cannot be $\mathbf{G}$ -extensible.

Proof.

Note that if $\bar{x}=(x_{h})_{h\in G}\in\operatorname{supp}(\bar{\mu})\subseteq X_{\mathbf{G}}$ and $U_{h}\subseteq X$ is an open neighborhood of $x_{h}$ for $h\in G$ , then $h^{-1}\pi^{-1}U_{h}$ is an open neighborhood of $\bar{x}$ , which means

\mu(U_{h})=\bar{\mu}(\pi^{-1}U_{h})=\bar{\mu}(h^{-1}\pi^{-1}U_{h})>0,

so $x_{h}\in\operatorname{supp}(\mu)$ for all $h\in G$ . Hence $\operatorname{supp}(\bar{\mu})\subseteq X_{\mathbf{G}}\cap(\operatorname{supp}(\mu))^{G}=(\operatorname{supp}(\mu))_{\mathbf{G}}$ . Finally, if $S\curvearrowright\operatorname{supp}(\mu)$ is not partially $\mathbf{G}$ -extensible and $\bar{\mu}\in\mathcal{M}_{G}(X_{\mathbf{G}})$ is such that $\pi_{*}\bar{\mu}=\mu$ , then $(\operatorname{supp}(\mu))_{\mathbf{G}}=\varnothing$ , so $\operatorname{supp}(\bar{\mu})=\varnothing$ , which is absurd. ∎

So far, we have not provided an example of a receiving $S$ -group $\mathbf{G}$ and an $S$ -invariant measure which is not $\mathbf{G}$ -extensible. In view of Proposition 4.14, taking a continuous action $S\curvearrowright X$ that is not partially $\mathbf{G}$ -extensible and admits an $S$ -invariant measure would provide an example of this. This can be done, as for every $S$ such that the free $S$ -group $\bm{\Gamma}=(\Gamma,\gamma)$ satisfies that $\Gamma$ is residually finite, if $\mathbf{G}\not\simeq\bm{\Gamma}$ , there exist a finite alphabet $\mathcal{A}$ and an $S$ -periodic point $x\in\mathcal{A}^{S}$ such that $S\curvearrowright Sx$ is not partially $\mathbf{G}$ -extensible, so we may take the periodic measure associated with $x$ (see [6]).

Our last main result characterizes $\mathbf{G}$ -extensibility of $S$ -invariant measures in the symbolic context. In particular, it tells us that if $\mathbf{G}\not\simeq\mathbf{F}_{d}$ , there is a fully-supported Markov measure on $\mathcal{A}^{S}$ —which is trivially topologically $\mathbf{G}$ -extensible—that is not $\mathbf{G}$ -extensible.

\fifththeorem

Before proving Theorem Introduction, we need the following general result, which in particular will allow us to translate $\mathbf{F}_{d}$ -extensibility to arbitrary realizations of the free $S$ -group.

Proposition 4.15.

Let $S\curvearrowright X$ be a surjective continuous action, and let $\theta\colon\mathbf{G}\to\mathbf{G}^{\prime}$ be an isomorphism between two receiving $S$ -groups $\mathbf{G}=(G,\eta)$ and $\mathbf{G}^{\prime}=(G^{\prime},\eta^{\prime})$ . Then, there is a $\theta$ -equivariant homeomorphism $\varphi\colon X_{\mathbf{G}^{\prime}}\to X_{\mathbf{G}}$ such that $\pi\circ\varphi=\pi^{\prime}$ . Moreover, the push-forward $\varphi_{*}\colon\mathcal{M}_{G^{\prime}}(X_{\mathbf{G}^{\prime}})\to\mathcal{M}_{G}(X_{\mathbf{G}})$ is a bijection and, in particular, $\textup{Ext}_{\mathbf{G}}(X,S)=\textup{Ext}_{\mathbf{G}^{\prime}}(X,S)$ .

Proof.

Define $\varphi\colon X_{\mathbf{G}^{\prime}}\to X_{\mathbf{G}}$ by

(x_{h^{\prime}})_{h^{\prime}\in G^{\prime}}\mapsto(x_{\theta(h)})_{h\in G},

which is well-defined, since for all $s\in S$ and $(x_{h^{\prime}})_{h^{\prime}\in G^{\prime}}\in X_{\mathbf{G}^{\prime}}$ we have

s\cdot x_{\theta(h)}=x_{\eta^{\prime}(s)\theta(h)}=x_{\theta(\eta(s))\theta(h)}=x_{\theta(\eta(s)h)},

so $(x_{\theta(h)})_{h\in G}\in X_{\mathbf{G}}$ . Continuity of $\varphi$ is easily verified, since $X_{\mathbf{G}}$ and $X_{\mathbf{G}^{\prime}}$ have the topology of pointwise convergence: for every $g\in G$ there is a unique $h\in G^{\prime}$ with $\theta(g)=h$ and vice versa, and thus for any sequence $(x^{(n)})_{n}$ in $X_{\mathbf{G}}$ , convergence of $x^{(n)}$ at the coordinate $h$ is equivalent to the convergence of $\varphi(x^{(n)})$ at coordinate $g$ , so $\varphi$ maps convergent sequences to convergent sequences. Similarly, the map $X_{\mathbf{G}}\to X_{\mathbf{G}^{\prime}}$ given by $(x_{h})_{h\in G}\mapsto(x_{\theta^{-1}(h^{\prime})})_{h^{\prime}\in G^{\prime}}$ is well-defined, so it defines an inverse for $\varphi$ , which is thus bijective. The map $\varphi^{-1}$ is, as well, continuous by the same argument as above; thus, $\varphi$ is a homeomorphism. Since $\theta(1_{G^{\prime}})=1_{G}$ , we have $\pi^{\prime}=\pi\circ\varphi$ and $\pi=\pi^{\prime}\circ\varphi^{-1}$ . For any $g\in G$ and $(x_{h^{\prime}})_{h^{\prime}\in G}\in X_{\mathbf{G}^{\prime}}$ , we have

\varphi(\theta(g)\cdot(x_{h^{\prime}})_{h^{\prime}\in G^{\prime}})=\varphi((x_{h^{\prime}\theta(g)})_{h^{\prime}\in G^{\prime}})=(x_{\theta(h)\theta(g)})_{h\in G}=g\cdot(x_{\theta(h)})_{h\in G}=g\cdot\varphi((x_{h})_{h\in G}),

so $\varphi$ is $\theta$ -equivariant.

It is readily checked that the push-forward $\varphi_{*}\colon\mathcal{M}_{G^{\prime}}(X_{\mathbf{G}^{\prime}})\to\mathcal{M}_{G}(X_{\mathbf{G}})$ is well-defined and satisfies $\pi_{*}\circ\varphi_{*}=\pi_{*}^{\prime}$ . The inverse of $\varphi_{*}$ is $(\varphi^{-1})_{*}$ . ∎

Figure 3. In this cycle in

G

, transitions corresponding to solid edges all have high probability. As

A_{i_{1}}^{\varepsilon_{1}}\cdots A_{i_{n}}^{\varepsilon_{n}}\mathbf{v^{*}}\neq\mathbf{v^{*}}

, the final transition is thus forced to be of very low probability.

Proof of Theorem Introduction.

It follows from Proposition 4.12 together with Proposition 4.15 that (i) implies (ii). That (ii) implies (iii) is direct. To see that (iii) implies (i), assume that (i) does not hold and let us show that (iii) does not hold either. Indeed, since $\mathbf{G}\not\simeq\mathbf{F}_{d}$ , there is a surjective morphism $\theta\colon\mathbb{F}_{d}\to G$ with non-trivial kernel and such that $\theta\circ\iota=\eta$ , so we may choose an element $w\in\text{ker}(\theta)-\{\varepsilon\}$ . Consider a $2\times 2$ matrix representation of $\mathbb{F}_{d}$ generated by $d$ matrices $A_{1},\dots,A_{d}$ . Write $w=A_{i_{1}}^{\varepsilon_{1}}\cdots A_{i_{n}}^{\varepsilon_{n}}$ for $n\geq 1$ and $\varepsilon_{j}\in\{-1,1\}$ , $j=1,\dots,n$ , and choose a sufficiently large positive prime $p$ such that $A_{i_{1}}^{\varepsilon_{1}}\cdots A_{i_{n}}^{\varepsilon_{n}}\neq I_{d}\bmod p$ . Let $\mathcal{A}=\mathbb{Z}_{p}^{2}$ and $0<\delta<1/(p^{2}-1)$ . Define, for $A\in\Sigma$ and $\mathbf{u},\mathbf{v}\in\mathcal{A}$ ,

\mathbf{P}^{A}_{\mathbf{u},\mathbf{v}}=\begin{cases}1-(p^{2}-1)\delta&\text{if }\mathbf{v}=A\mathbf{u}\bmod p,\\ \delta&\text{otherwise}.\end{cases}

and let $\mathbf{p}\in[0,1]^{p^{2}}$ be the uniform distribution upon $\mathcal{A}$ , i.e., $\mathbf{p}_{\mathbf{u}}=1/p^{2}$ for all $\mathbf{u}\in\mathcal{A}$ . Since $A$ is invertible and $p$ is prime, $A$ is invertible modulo $p$ as well. Then, for a fixed $\mathbf{v}\in\mathcal{A}$ , there is a unique $\mathbf{u}\in\mathcal{A}$ such that $\mathbf{v}=A\mathbf{u}\bmod p$ , so

(\mathbf{p}\mathbf{P}^{A})_{\mathbf{v}}=\sum_{\mathbf{u}\in\mathcal{A}}\mathbf{p}_{\mathbf{u}}\mathbf{P}^{A}_{\mathbf{u},\mathbf{v}}=\frac{(p^{2}-1)\delta}{p^{2}}+\frac{1-(p^{2}-1)\delta}{p^{2}}=\frac{1}{p^{2}},

yielding $\mathbf{p}\mathbf{P}^{A}=\mathbf{p}$ . It is also readily checked that $\mathbf{P}^{A^{-1}}$ is the transpose of $\mathbf{P}^{A}$ , which implies that $\mathbf{p}_{\mathbf{u}}\mathbf{P}^{A^{-1}}_{\mathbf{u},\mathbf{v}}=\mathbf{p}_{\mathbf{v}}\mathbf{P}^{A}_{\mathbf{v},\mathbf{u}}$ for all $\mathbf{u},\mathbf{v}\in\mathcal{A}$ , since $\mathbf{p}$ is uniform. Thus, letting $\mathbf{P}=\{\mathbf{P}^{A}:A\in\Sigma\}$ , the $(\mathbf{p},\mathbf{P})$ -Markov measure $\mu$ on $\mathcal{A}^{S}$ is $S$ -invariant by Proposition 4.9. Since $\mathbf{p}>0$ and $\mathbf{P}^{A}>0$ for every $A\in\Sigma$ , $\mu$ is fully supported.

Assume that there is a $\mathbf{G}$ -extension $\bar{\mu}\in\mathcal{M}_{G}(\mathcal{A}^{G})$ of $\mu$ . If $\overline{x}\in\mathcal{A}^{G}$ , $g\in G$ , and $A\in\Sigma$ , then

	$\displaystyle\bar{\mu}\big{(}[\overline{x}\,;\{g,\eta(A)g\}]\big{)}$	$\displaystyle=\bar{\mu}\big{(}g^{-1}[g\cdot\overline{x}\,;\{1_{G},\eta(A)\}]\big{)}$
		$\displaystyle=\bar{\mu}\big{(}[g\cdot\overline{x}\,;\{1_{G},\eta(A)\}]\big{)}$
		$\displaystyle=\mu\left(\left[\pi(g\cdot\overline{x})\,;\left\{\varepsilon,A\right\}\right]\right)$
		$\displaystyle=\mathbf{p}_{\hskip 1.0pt\overline{x}(g)}\mathbf{P}^{A}_{\overline{x}(g),\overline{x}(\eta(A)g)},$

as $\pi(g\cdot\overline{x})(t)=\overline{x}(tg)$ for all $t\in S$ . Analogously, we have that

\bar{\mu}\big{(}[\overline{x}\,;\{g,\eta(A)^{-1}g\}]\big{)}=\mathbf{p}_{\hskip 1.0pt\overline{x}(\eta(A)^{-1}g)}\mathbf{P}^{A}_{\overline{x}(\eta(A)^{-1}g),\overline{x}(g)}.

Since $A_{i_{1}}^{\varepsilon_{1}}\cdots A_{i_{n}}^{\varepsilon_{n}}\neq I_{d}\bmod p$ , there exists $\mathbf{v^{*}}\in\mathcal{A}$ such that $A_{i_{1}}^{\varepsilon_{i_{1}}}\cdots A_{i_{n}}^{\varepsilon_{i_{n}}}\mathbf{v^{*}}\neq\mathbf{v^{*}}\bmod p$ . Consider the finite subset $C=\{\eta(A_{i_{k}})^{\varepsilon_{k}}\cdots\eta(A_{i_{n}})^{\varepsilon_{n}}:1\leq k\leq n\}$ of $G$ . Since $\theta\circ\iota=\eta$ , we have that $\eta(A_{i_{1}})^{\varepsilon_{1}}\cdots\eta(A_{i_{n}})^{\varepsilon_{n}}=\theta(w)=1_{G}$ , so $C$ can be viewed as a cycle in $G$ . Let $\tau\in\mathcal{A}^{C}$ be a configuration such that $\tau(1_{G})=\mathbf{v}^{*}$ , and assume that $\tau(\eta(A_{i_{k}})^{\varepsilon_{k}}\cdots\eta(A_{i_{n}})^{\varepsilon_{n}})=A_{i_{k}}^{\varepsilon_{k}}\tau(\eta(A_{i_{k+1}})^{\varepsilon_{k+1}}\cdots\eta(A_{i_{n}})^{\varepsilon_{n}})$ for every $0\leq k<n$ , where we consider $A_{i_{0}}=A_{i_{n}}$ and $\varepsilon_{0}=\varepsilon_{n}$ . Iteratively, we see that

\mathbf{v^{*}}=\tau(1_{G})=\tau(\eta(A_{i_{1}})^{\varepsilon_{1}}\cdots\eta(A_{i_{n}})^{\varepsilon_{n}})=A_{i_{1}}^{\varepsilon_{1}}\cdots A_{i_{n}}^{\varepsilon_{n}}\mathbf{v}^{*}\neq\mathbf{v}^{*},

a contradiction, so it must be the case that there exist $g\in C$ and $0\leq k<n$ such that $\eta(A_{i_{k}})^{\varepsilon_{k}}g\in C$ and $\tau(\eta(A_{i_{k}})^{\varepsilon_{k}}g)\neq A_{i_{k}}^{\varepsilon_{k}}\tau(g)$ . Hence,

\bar{\mu}([\tau;C])\leq\bar{\mu}\big{(}[\tau\,;\{g,\eta(A_{i_{k}})^{\varepsilon_{k}}g\}]\big{)}=\mathbf{p}_{\tau(g)}\mathbf{P}^{A_{i_{k}}}_{\tau(g),\tau(\eta(A_{i_{k}})g)}=\frac{\delta}{p^{2}}

if $\varepsilon_{k}=1$ , and similarly $\bar{\mu}([\tau;C])\leq\delta/p^{2}$ if $\varepsilon_{k}=-1$ as well. Therefore,

\frac{1}{p^{2}}=\mu([\mathbf{v^{*}}\,;\varepsilon])=\bar{\mu}([\mathbf{v^{*}}\,;1_{G}])=\sum_{\tau\in\mathcal{A}^{C}:\tau(1_{G})=\mathbf{v^{*}}}\bar{\mu}([\tau;C])\leq(p^{2})^{n-1}\frac{\delta}{p^{2}}=p^{2n-4}\delta,

which leads to a contradiction if we take $\delta<\frac{1}{p^{2n-2}}$ .

∎

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Appendix A Proof of Theorem 4.5

Assume that $S$ is left reversible and bicancellative, and $\bm{\Gamma}$ is the free $S$ -group. Recall that $\bm{\Gamma}$ is isomorphic to $\bm{\Gamma}_{S}$ , the group of right fractions of $S$ . Define the preorder $\leq_{S}$ on $\Gamma$ by

g\leq_{S}h\iff hg^{-1}\in\gamma(S).

Denote by $\mathcal{F}(\Gamma)$ the collection of all finite subsets of $\Gamma$ . We say $\leq_{S}$ is downward directed if for every $F\in\mathcal{F}(\Gamma)$ there is an $m\in\Gamma$ such that $m\leq_{S}t$ for all $g\in F$ .

Lemma A.1 ([6, Lemma 2.22]).

The pre-order $\leq_{S}$ is downward directed. Equivalently, the subset $\gamma(S)$ is thick, that is, for every $F\in\mathcal{F}(\Gamma)$ there exists $g\in\Gamma$ such that $Fg\subseteq\gamma(S)$ .

Let $S\overset{\alpha}{\curvearrowright}X$ be a continuous action, and $\mu\in\mathcal{M}_{S}(X)$ . Given $F\in\mathcal{F}(\Gamma)$ , fix any lower bound $m_{F}$ for $F$ (i.e., such that $m_{F}\leq_{S}t$ for all $t\in F$ ). Note that, by definition, this means $s_{t,F}\in\gamma(S)$ for each $t\in F$ , so that there is a unique element $s_{t,F}\in S$ with $\gamma(s_{t,F})=s_{t,F}$ . Define the following collection of subsets of $\mathcal{B}(X^{F})$ :

\mathcal{C}_{F}=\left\{\prod_{t\in F}A_{t}:A_{t}\in\mathcal{B}(X)\text{ for every }t\in F\right\},

and the set function $\mu_{F}\colon\mathcal{C}_{F}\rightarrow[0,1]$ by

\mu_{F}\left(\prod_{t\in F}A_{t}\right)=\mu\left(\bigcap_{t\in F}(s_{t,F})^{-1}(A_{t})\right).

Observe that, due to the $S$ -invariance of $\mu$ , the value of the function $\mu_{F}$ does not depend on the choice of $m_{F}$ .

Lemma A.2.

The function $\mu_{F}$ extends to a finitely additive probability measure on the algebra of sets $\mathcal{A}_{F}$ generated by $\mathcal{C}_{F}$ .

Proof.

Since $\mathcal{A}_{F}$ consists of finite disjoint unions of elements of $\mathcal{C}_{F}$ , we just need to check that, for any $C\in\mathcal{A}_{F}$ and finite partitions $\mathcal{P},\mathcal{Q}$ of $C$ by elements of $\mathcal{C}_{F}$ ,

\sum_{A\in\mathcal{P}}\mu_{F}(A)=\sum_{B\in\mathcal{Q}}\mu_{F}(B).

First, if $\mathcal{Q}$ is a refinement of $\mathcal{P}$ , take any $A\in\mathcal{P}$ and write it as a union of a collection $\{B^{(i)}\}_{i=1}^{n}\subseteq\mathcal{Q}$ :

A=\prod_{t\in F}A_{t}=\bigsqcup_{i=1}^{n}B^{(i)},\quad B^{(i)}=\prod_{t\in F}B^{(i)}_{t}.

Since this union is disjoint, for $1\leq i<j\leq n$ there must be a $t_{ij}\in F$ such that $B_{t_{ij}}^{(i)}\cap B_{t_{ij}}^{(j)}=\varnothing$ , which implies

(s_{t_{ij},F})^{-1}\left(B_{t_{ij}}^{(i)}\right)\cap(s_{t_{ij},F})^{-1}\left(B_{t_{ij}}^{(j)}\right)=\varnothing.

In particular, this means

\left[\bigcap_{t\in F}(s_{t,F})^{-1}\left(B_{t}^{(i)}\right)\right]\cap\left[\bigcap_{t\in F}(s_{t,F})^{-1}\left(B_{t}^{(j)}\right)\right]=\varnothing,

and therefore the collection $\left\{\bigcap_{t\in F}(s_{t,F})^{-1}\left(B_{t}^{(i)}\right):1\leq i\leq n\right\}$ is pairwise disjoint. Hence,

\mu\left(\bigsqcup_{i=1}^{n}\bigcap_{t\in F}(s_{t,F})^{-1}\left(B_{t}^{(i)}\right)\right)=\sum_{i=1}^{n}\mu_{F}\left(B^{(i)}\right).

Now, for every $1\leq i\leq n$ and $t\in F$ , $B^{(i)}_{t}\subseteq A_{t}$ , from where we obtain

\bigsqcup_{i=1}^{n}\bigcap_{t\in F}(s_{t,F})^{-1}\left(B_{t}^{(i)}\right)\subseteq\bigcap_{t\in F}(s_{t,F})^{-1}(A_{t}).

We want to see that this last inclusion is an equality. Take any $x\in\bigcap_{t\in F}(s_{t,F})^{-1}(A_{t})$ . Then, the tuple $(s_{t,F}\cdot x)_{t\in F}$ belongs to $A$ , which means it belongs to some $B^{(i)}$ . Thus,

x\in\bigcap_{t\in F}(s_{t,F})^{-1}\left(B^{(i)}_{t}\right),

and we get the desired opposite inclusion. Putting all together yields

\mu_{F}(A)=\mu\left(\bigcap_{t\in F}(s_{t,F})^{-1}(A_{t})\right)=\mu\left(\bigsqcup_{i=1}^{n}\bigcap_{t\in F}(s_{t,F})^{-1}\left(B_{t}^{(i)}\right)\right)=\sum_{i=1}^{n}\mu_{F}\big{(}B^{(i)}\big{)}.

Finally, summing over $\mathcal{P}$ :

\sum_{A\in\mathcal{P}}\mu_{F}(A)=\sum_{A\in\mathcal{P}}\sum_{\begin{subarray}{c}B\in\mathcal{Q}\\ B\subseteq A\end{subarray}}\mu_{F}(B)=\sum_{B\in\mathcal{Q}}\mu_{F}(B).

The remaining case, where $\mathcal{Q}$ need not be a refinement of $\mathcal{P}$ , follows by considering a refinement $\mathcal{P}\lor\mathcal{Q}$ of both $\mathcal{P}$ and $\mathcal{Q}$ . ∎

We want to prove that $\mu_{F}$ is $\sigma$ -additive on $\mathcal{A}_{F}$ . We recall a result which will help us.

Lemma A.3.

Let $\nu$ be a finite measure on an algebra $\mathcal{A}$ which is finitely additive and continuous at $\varnothing$ . Then, $\nu$ is $\sigma$ -additive in $\mathcal{A}$ .

Lemma A.4.

The measure $\mu_{F}\colon\mathcal{A}_{F}\rightarrow[0,1]$ is continuous at $\varnothing$ , hence $\sigma$ -additive.

Proof.

Let $A_{n}\downarrow\varnothing$ in $\mathcal{A}_{F}$ , that is, $A_{n}\supseteq A_{n+1}$ and $\bigcap_{n}A_{n}=\varnothing$ . For each $n\in\mathbb{N}$ write

A_{n}=\bigsqcup_{i=1}^{k_{n}}\prod_{t\in F}A_{t}^{(i,n)}.

Observe that

	$\displaystyle\mu_{F}(A_{n})$	$\displaystyle=\sum_{i=1}^{k_{n}}\mu_{F}\left(\prod_{t\in F}A_{t}^{(i,n)}\right)=\sum_{i=1}^{k_{n}}\mu\left(\bigcap_{t\in F}(s_{t,F})^{-1}\left(A_{t}^{(i,n)}\right)\right)$
		$\displaystyle=\mu\left(\bigsqcup_{i=1}^{k_{n}}\bigcap_{t\in F}(s_{t,F})^{-1}\left(A_{t}^{(i,n)}\right)\right).$

These last sets are decreasing in $n$ . Suppose there exists an element

x\in\bigcap_{n}\left[\bigsqcup_{i=1}^{k_{n}}\bigcap_{t\in F}(s_{t,F})^{-1}\left(A_{t}^{(i,n)}\right)\right].

Then, for each $t\in F$ define $x_{t}:=s_{t,F}\cdot x$ . We would have that for all $n\geq 1$ there is a $1\leq i_{n}\leq k_{n}$ such that $x_{t}\in A_{t}^{(i_{n},n)}$ for all $t\in F$ , meaning

(x_{t})_{t\in F}\in\prod_{t\in F}A_{t}^{(i_{n},n)}\subseteq\bigsqcup_{i=1}^{k_{n}}\prod_{t\in F}A_{t}^{(i,n)}

for all $n\geq 1$ , which contradicts the fact that $A_{n}\downarrow\varnothing$ . Thus, the intersection was empty, and by continuity of $\mu$ we conclude

\lim_{n\to\infty}\mu_{F}(A_{n})=\lim_{n\to\infty}\mu\left(\bigsqcup_{i=1}^{k_{n}}\bigcap_{t\in F}(s_{t,F})^{-1}\left(A_{t}^{(i,n)}\right)\right)=0.

∎

By applying Carathéodory’s Extension Theorem, we obtain a unique extension of $\mu_{F}$ to the $\sigma$ -algebra $\sigma(\mathcal{C}_{F})=\mathcal{B}(X^{F})$ generated by $\mathcal{C}_{F}$ .

Corollary A.5.

Let $F\in\mathcal{F}(\Gamma)$ and $m_{F}\leq_{S}t$ for all $t\in F$ . Then, there is a unique probability measure $\mu_{F}\colon\mathcal{B}(X^{F})\rightarrow[0,1]$ such that

\mu_{F}\left(\prod_{t\in F}A_{t}\right)=\mu\left(\bigcap_{t\in F}(s_{t,F})^{-1}(A_{t})\right)

for every $\prod_{t\in F}A_{t}\in\mathcal{B}(X^{F})$ .

We want to extend this collection of measures to a measure on $X^{\Gamma}$ , via Kolmogorov’s Extension Theorem. If $F\subseteq K\subseteq\Gamma$ , we define $\pi_{F}^{K}\colon X^{K}\rightarrow X^{F}$ as the canonical projection, and we omit the super-index if $K=\Gamma$ . Recall that a family of measures $\{\nu_{F}\colon\mathcal{B}(X^{F})\rightarrow[0,1]\mid F\in\mathcal{F}(\Gamma)\}$ is called consistent if whenever $F\subseteq K\in\mathcal{F}(\Gamma)$ , we have $(\pi^{K}_{F})_{*}\mu_{K}=\mu_{F}$ .

Lemma A.6.

The family of probability measures $\{\mu_{F}:F\in\mathcal{F}(\Gamma)\}$ is consistent.

Proof.

Let $F\subseteq K\in\mathcal{F}(\Gamma)$ . Then, a lower bound $m_{K}$ for $K$ is a lower bound for $F$ as well. Let $\prod_{t\in F}A_{t}$ be an arbitrary element of $\mathcal{C}_{F}$ , and define, for $g\in K-F$ , $A_{g}:=X$ . Then,

	$\displaystyle\left(\pi_{F}^{K}\right)_{*}\mu_{K}\left(\prod_{t\in F}A_{t}\right)$	$\displaystyle=\mu_{K}\left(\prod_{g\in K}A_{g}\right)=\mu\left(\bigcap_{g\in K}(s_{g,K})^{-1}(A_{g})\right)$
		$\displaystyle=\mu\left(\bigcap_{t\in F}(s_{t,K})^{-1}(A_{t})\right)=\mu_{F}\left(\prod_{t\in F}A_{t}\right).$

Now, the sets in $\mathcal{B}(X^{F})$ which satisfy the formula $\left(\pi_{F}^{K}\right)_{*}\mu_{K}=\mu_{F}$ form a $\sigma$ -algebra, which implies the result for all sets in $\mathcal{B}(X^{F})$ . ∎

By Kolmogorov’s Extension Theorem, we obtain a unique probability measure $\bar{\mu}_{\Gamma}$ on $\mathcal{B}(X^{\Gamma})$ satisfying the condition $\mu_{F}=\left(\pi_{F}^{\Gamma}\right)_{*}\bar{\mu}_{\Gamma}$ for every finite subset $F\subseteq\Gamma$ . This allows to establish the following result.

Proposition A.7.

The measure $\bar{\mu}_{\Gamma}$ is $\Gamma$ -invariant, and we have $\bar{\mu}_{\Gamma}(X_{\bm{\Gamma}})=1$ . Therefore, $\mu_{\Gamma}:=\bar{\mu}_{\Gamma}|_{\mathcal{B}(X_{\bm{\Gamma}})}$ is a $\Gamma$ -invariant probability measure satisfying $\pi_{*}\mu_{\Gamma}=\mu$ .

Proof.

To show $\Gamma$ -invariance of $\bar{\mu}_{\Gamma}$ , let $g\in\Gamma$ and define $\nu=g_{*}\bar{\mu}_{\Gamma}$ . It suffices to check that, for a finite subset $F\subseteq\Gamma$ , $\mu_{F}=\left(\pi_{F}^{\Gamma}\right)_{*}\nu$ on cylinders, so that by the uniqueness granted by Kolmogorov’s Extension Theorem we get $\nu=\overline{\mu}_{\Gamma}$ .

It can be easily verified that, if $m_{F}$ is a lower bound for $F$ , then so is $m_{F}g$ for $Fg$ . Let $\prod_{t\in F}A_{t}$ be any element of $\mathcal{C}_{F}$ , and define $A_{g}=X$ for $g\in\Gamma-F$ . We obtain the following.

	$\displaystyle\left(\pi_{F}^{G}\right)_{*}\nu\left(\prod_{t\in F}A_{t}\right)$	$\displaystyle=\bar{\mu}_{\Gamma}\left(g^{-1}\prod_{t\in\Gamma}A_{t}\right)=\bar{\mu}_{\Gamma}\left(\prod_{t\in Fg}A_{tg^{-1}}\right)=\mu_{Fg}\left(\prod_{t\in Fg}A_{tg^{-1}}\right)$
		$\displaystyle=\mu\left(\bigcap_{t\in Fg}(s_{tg^{-1},F})^{-1}A_{tg^{-1}}\right)=\mu\left(\bigcap_{t\in F}(s_{t,F})^{-1}A_{t}\right)=\mu_{F}\left(\prod_{t\in F}A_{t}\right).$

Now we prove $X_{G}$ has full measure. We already know $X_{\bm{\Gamma}}$ is closed, thus measurable. As it can be written as

\displaystyle X_{\bm{\Gamma}}

\displaystyle=\bigcap_{t\in G}\bigcap_{s\in S}\left\{(x_{t})_{t\in\Gamma}\in X^{\Gamma}:s\cdot x_{t}=x_{\gamma(s)t}\right\},

defining, for each $s\in S$ and $t\in\Gamma$ , $A_{s,t}=\{(x_{t})_{t\in\Gamma}\in X^{\Gamma}:s\cdot x_{t}=x_{\gamma(s)t}\}$ (which is also closed, by continuity of $S\curvearrowright X$ and of the coordinate projections), it suffices to check that each $A_{s,t}$ has full measure. To see this, fix $s\in S$ , $t\in\Gamma$ . The set $X^{\Gamma}-A_{s,t}$ is open, and can hence be written as a countable union of cylinders of $X^{\Gamma}$ :

X^{\Gamma}-A_{s,t}=\bigcup_{n}\prod_{g\in\Gamma}C_{g}^{(n)},

where for each $n$ , $C^{(n)}_{g}=X$ if $g\in\Gamma-F_{n}$ . We must have that $C_{t}^{(n)}\cap s^{-1}\big{(}C_{\gamma(s)t}^{(n)}\big{)}=\varnothing$ (in particular $t,\gamma(s)t\in F_{n}$ for every $n\geq 1$ ), and therefore, for all $n\geq 1$ ,

\big{(}s_{t,F_{n}}\big{)}^{-1}\left[C_{t}^{(n)}\cap s^{-1}\big{(}C_{\gamma(s)t}^{(n)}\big{)}\right]=\big{(}s_{t,F_{n}}\big{)}^{-1}\left(C_{t}^{(n)}\right)\cap\big{(}s_{\gamma(s)t,F_{n}}\big{)}^{-1}\big{(}C_{\gamma(s)t}^{(n)}\big{)}=\varnothing.

This last fact directly implies that

\displaystyle\bar{\mu}_{\Gamma}\left(X_{\bm{\Gamma}}-A_{s,t}\right)\leq\sum_{n}\mu_{F_{n}}\left(\prod_{h\in F_{n}}C_{h}^{(n)}\right)=\sum_{n}\mu\left(\bigcap_{h\in F_{n}}(s_{h,F_{n}})^{-1}\left(C_{h}^{(n)}\right)\right)=0,

obtaining $\bar{\mu}_{\Gamma}(A_{s,t})=1$ , as desired.

Finally, let $\mu_{\Gamma}:=\bar{\mu}_{\Gamma}|_{\mathcal{B}(X_{\bm{\Gamma}})}$ . This measure is $\Gamma$ -invariant, as $\mu_{\Gamma}(A)=\bar{\mu}_{\Gamma}(A)$ for every $A\in\mathcal{B}(X_{\bm{\Gamma}})$ . Since $(\pi_{F}^{\Gamma})_{*}\bar{\mu}_{\Gamma}=\mu_{F}$ and $\pi^{\Gamma}_{1_{\Gamma}}(\overline{x})=\pi(\overline{x})$ for every $\overline{x}\in X_{\bm{\Gamma}}$ , we have

\mu_{\Gamma}(\pi^{-1}(A))=\mu_{\Gamma}\big{(}[\pi^{\Gamma}_{1_{\Gamma}}]^{-1}(A)\cap X_{\bm{\Gamma}}\big{)}=\overline{\mu}_{\Gamma}\big{(}[\pi^{\Gamma}_{1_{\Gamma}}]^{-1}(A)\big{)}=\mu(A)

for all $A\in\mathcal{B}(X_{\bm{\Gamma}})$ , i.e., $\pi_{*}\mu_{\Gamma}=\mu$ . ∎

Theorem 4.5 follows directly from Proposition A.7.