A note on minimal models for pmp actions

TL;DR

This paper presents a new streamlined construction of a minimal model-universal flow for countable groups, demonstrating that such flows are not unique, and advancing the understanding of invariant measures in dynamical systems.

Contribution

It introduces a simplified method for constructing minimal model-universal flows, showing their non-uniqueness and expanding the theoretical framework for measure-preserving group actions.

Findings

A new streamlined construction of minimal model-universal flows.

Proof that minimal model-universal flows are not unique.

Enhanced understanding of invariant measures in dynamical systems.

Abstract

Given a countable group $G$, we say that a metrizable flow $Y$ is model-universal if by considering the various invariant measures on $Y$, we can recover every free measure-preserving $G$-system up to isomorphism. Weiss has constructed a minimal model-universal flow. In this note, we provide a new, streamlined construction, allowing us to show that a minimal model-universal flow is far from unique.

Equations

$$Y_Q=\{s\in 2^G:s^{-1}(\{1\})\text{ is a maximal Q-spaced set}\}.$$

$$z{|}_F={\mathrm{Conf}}_Y({\mathrm{Trans}}_Y(F,E),(h\cdot y){|}_{E^5\setminus E^3},E^5){|}_F$$

$$\overline{(A\cap B)}\setminus (A\cap B)$$

$$\subseteq (\overline{A}\setminus A)\cup (\overline{B}\setminus B).$$

$$\{y\in Y:|{\pi }^{-1}(\{y\})|=1\}=\bigcap _{A\in \mathcal{B}}A\cup (Y\setminus \overline{A}).$$

Full text

In this paper, we consider actions of an infinite countable group $G$ on a standard Borel probability space $(X,\mu)$ by Borel, measure-preserving bijections. When an action $a\colon G\times X\to X$ is understood, we will suppress the action notation, and given $g\in G$ and $x\in X$ just write $gx$ or $g\cdot x$ for $a(g,x)$. We will refer to $(X,\mu)$ as a $G$-system. A $G$-system is free if for $\mu$-almost every $x\in X$, we have $G_{x}=\{1_{G}\}$, where $G_{x}:=\{g\in G:gx=x\}$ is the stabilizer of $x\in X$. By passing to a subset of measure $1$, we will often implicitly assume that every point in a free $G$-system has trivial stabilizer. If $(X,\mu)$ and $(Y,\nu)$ are two $G$-systems, we say that $(Y,\nu)$ is a factor of $(X,\mu)$ if there is a Borel $X^{\prime}\subseteq X$ with $\mu(X^{\prime})=1$ and a Borel $G$-equivariant map $f\colon X^{\prime}\to Y$ with $\nu=f^{*}\mu$. If we can find $f$ as above that is also injective, then we call $(X,\mu)$ and $(Y,\nu)$ isomorphic $G$-systems.

A $G$-flow is an action of $G$ by homeomorphisms on a compact Hausdorff space. We similarly suppress the action notation. Given a $G$-system $(X,\mu)$, a model for $(X,\mu)$ is a compact metric $G$-flow $Y$ and an invariant Borel probability measure $\nu$ so that $(X,\mu)$ and $(Y,\nu)$ are isomorphic $G$-systems. We will be most interested in minimal $G$-flows, those $G$-flows in which every orbit is dense. Notice that any minimal model of a free $G$-system must be essentially free, where a $G$-flow $Y$ is essentially free if for each $g\in G\setminus\{1_{G}\}$, the set $\{y\in Y:gy=y\}$ is nowhere dense.

We say that a metrizable $G$-flow $Y$ is model-universal if by considering the various invariant measures $\nu$ on $Y$, the $G$-systems $(Y,\nu)$ recover every (standard) free $G$-system up to isomorphism. In [5], Weiss constructs for every countable group $G$ a minimal model-universal flow. It is natural to ask in what sense a minimal model-universal flow must be unique. Here, we prove a strong negative result. Given a family $\{Y_{i}:i\in I\}$ of minimal $G$-flows, we say that $\{Y_{i}:i\in I\}$ is mutually disjoint if the product $\prod_{i\in I}Y_{i}$ is minimal. In particular, this implies that the $Y_{i}$ are pairwise non-isomorphic $G$-flows.

Let us call a $G$-flow $Y$ weakly model-universal if for every free $G$-system $(X,\mu)$, there is an invariant measure $\nu$ on $Y$ so that $(Y,\nu)$ is a factor of $(X,\mu)$. In [5], Weiss first constructs a minimal, essentially free, weakly model-universal flow, then proves that any flow with these properties admits an almost 1-1 extension which is model-universal. We instead build our model-universal flows in one step.

A recent result of Elek in [1] shows the existence of a free minimal model-universal flow. Recall that a $G$-flow $Y$ is free when for any $y\in Y$ and any $g\in G\setminus\{1_{G}\}$, we have $gy\neq y$. In the last section of this paper, we show how one can deduce this result using rather soft arguments.

As almost 1-1 extensions always preserve minimality and disjointness, we can strengthen Theorem 1 as follows.

I would like to thank Benjamin Weiss for many helpful comments on an earlier draft, as well as the anonymous referee for suggesting many improvements.

1 Basic examples of model-universal flows

We briefly collect a few simple examples which will be important in what follows. Let $K$ be a compact space. Then $K^{G}$ is a $G$-flow with the right shift action, where given $g,h\in G$ and $s\in K^{G}$, we have $g\cdot s(h)=s(hg)$. Mostly we take $K=2^{n}$ or $2^{\omega}$.

Proposition 4**.**

The flow $(2^{\omega})^{G}$ is model-universal.

Proof.

Let $(X,\mu)$ be a free $G$-system, and fix $\varphi\colon X\to 2^{\omega}$ a Borel bijection. Now define $\psi\colon X\to(2^{\omega})^{G}$ via $\psi(x)(g)=\varphi(g\cdot x)$. Then $\psi$ is injective, and $(X,\mu)\cong((2^{\omega})^{G},\psi^{*}\mu)$. ∎

A subshift of $K^{G}$ is a closed, $G$-invariant subspace. The following family of subshifts of $2^{G}$ will be an important source of weakly model-universal flows. Let $Q\subseteq G$ be a finite symmetric set. We say that $S\subseteq G$ is $Q$-spaced if whenever $g,h\in S$ with $g\neq h$, then $Qg\cap Qh=\emptyset$. We say that $S$ is $Q$-syndetic if we have $\bigcup_{g\in Q}gS=\bigcup_{g\in S}Qg=G$. Notice that maximal $Q$-spaced sets exist and are $Q^{2}$-syndetic. Conversely, any $Q^{2}$-syndetic $Q$-spaced set is a maximal $Q$-spaced set. We define

[TABLE]

Proposition 5**.**

The flow $Y_{Q}$ is weakly model-universal.

Remark*.*

This proposition is also one of the key ingredients used by Weiss (see [5], Lemma 2.2).

Proof.

Let $(X,\mu)$ be a free $G$-system. By freeness, we can find for every Borel $B\subseteq X$ with $\mu(B)>0$ a Borel subset $A\subseteq B$ with $\mu(A)>0$ and with $gA\cap A=\emptyset$ for any $g\in Q^{2}$. Let us call a Borel set $A$ with this property a $Q^{2}$-disjoint set. Now if $\bigcup_{g\in Q^{2}}gA$ doesn’t have full measure, we can find a $Q^{2}$-disjoint Borel set $A^{\prime}\subseteq X$ with $\mu(A^{\prime})>0$ and $gA\cap A^{\prime}=\emptyset$ for every $g\in Q^{2}$. As $Q$ is assumed symmetric, it follows that $A\cup A^{\prime}$ is also $Q^{2}$-disjoint.

Thus using a measure exhaustion argument, we can find $A\subseteq X$ a $Q^{2}$-disjoint Borel set so that $\mu\left(\bigcup_{g\in Q^{2}}gA\right)=1$. We now let $\varphi\colon X\to 2^{G}$ be the map given by $\varphi(x)(g)=1$ iff $gx\in A$. Then for almost every $x\in X$, $\varphi(x)^{-1}(\{1\})$ is both $Q^{2}$-syndetic and $Q$-spaced, so a maximal $Q$-spaced set. It follows that $Y_{Q}$ contains the closed support of $\varphi^{*}\mu$, so $(Y_{Q},\varphi^{*}\mu)$ is a factor of $(X,\mu)$. ∎

We end the section by noting a simple closure property of (weakly) model-universal flows.

Proposition 6**.**

Let $Y_{n}$ be weakly model-universal $G$-flows. Then $Y:=\prod_{n}Y_{n}$ is weakly model-universal. If at least one of the $Y_{n}$ is model-universal, then so is $Y$.

Proof.

Let $(X,\mu)$ be a free $G$-system, and for each $n<\omega$, let $\varphi_{n}\colon X_{n}\to Y_{n}$ be a Borel, $G$-equivariant map, where $X_{n}\subseteq X$ satisfies $\mu(X_{n})=1$. Set $X^{\prime}=\bigcap_{n}X_{n}$. Then $\mu(X^{\prime})=1$, and the map $\varphi\colon X^{\prime}\to\prod_{n}Y_{n}$ given by $\varphi(x)=(\varphi_{n}(x))_{n<\omega}$ is Borel and $G$-equivariant. If for some $n<\omega$, the map $\varphi_{n}$ is injective, then $\varphi$ will also be injective. ∎

2 Strongly irreducible subshifts

The key technical tool we use here is the notion of a strongly irreducible subshift. First, we introduce some general terminology. Write $\mathrm{Fin}(G)$ for the collection of finite subsets of $G$. Given $S_{1},S_{2}\subseteq G$ and symmetric $D\in\mathrm{Fin}(G)$ with $1_{G}\in D$, we say that $S_{1}$ and $S_{2}$ are $D$-apart if $DS_{1}\cap DS_{2}=\emptyset$. Let $A$ be a finite set. If $Y\subseteq A^{G}$ is a subshift and $F\in\mathrm{Fin}(G)$, we define the $F$-patterns of $Y$ to be the set $S_{F}(Y):=\{s|_{F}:s\in Y\}\subseteq A^{F}$. Given $\alpha\in S_{F}(Y)$, we define the basic clopen neighborhood $N_{Y}(\alpha):=\{y\in Y:y|_{F}=\alpha\}$. If $F\in\mathrm{Fin}(G)$, $S\subseteq G$, $\alpha\in A^{F}$, and $\beta\in A^{S}$, we say that $\alpha$ appears in $\beta$ if there is $g\in G$ with $Fg\subseteq S$ and $\beta(fg)=\alpha(f)$ for each $f\in F$. We say in this case that $\alpha$ appears at $g\in G$.

We say that $Y$ is strongly irreducible if there is $D\in\mathrm{Fin}(G)$ so that for any $F_{0},F_{1}\in\mathrm{Fin}(G)$ which are $D$-apart and any $\alpha_{i}\in S_{F_{i}}(Y)$, there is $y\in Y$ with $y|_{F_{i}}=\alpha_{i}$. We sometimes say that $Y$ is $D$-irreducible. We will frequently use the following facts about strongly irreducible subshifts. Here $A$ and $B$ are finite sets.

If $Y\subseteq A^{G}$ is $D_{Y}$-irreducible and $Z\subseteq B^{G}$ is $D_{Z}$-irreducible, then $Y\times Z\subseteq(A\times B)^{G}$ is $(D_{Y}\cup D_{Z})$-irreducible. 2. 2.

Suppose $Y\subseteq A^{G}$ is $D$-irreducible and $\varphi\colon Y\to B^{G}$ is continuous and $G$-equivariant. By continuity, there is $F\in\mathrm{Fin}(G)$ so that $\varphi(y)(1_{G})$ depends only on $y|_{F}$. Then $Z:=\varphi[Y]$ is $DF$-irreducible.

We will also need a method of making explicit choices of patterns in $S_{F}(Y)$. To that end, suppose that $A$ is linearly ordered, and enumerate the group $G$ in some fashion. This allows us to order $S_{F}(Y)$ lexicographically. We will use this ordering in the following two ways. Fix $Y\subseteq A^{G}$ a $D$-irreducible subshift.

If $F_{0},...,F_{n-1}\in\mathrm{Fin}(G)$ are pairwise $D$-apart, $\alpha_{i}\in S_{F_{i}}(Y)$, and $E\in\mathrm{Fin}(G)$ contains each $F_{i}$, then we let $\mathrm{Conf}_{Y}(\alpha_{0},...,\alpha_{n-1},E)\in S_{E}(Y)$ be the lexicographically least $E$-pattern $\beta$ satisfying $\beta|_{F_{i}}=\alpha_{i}$. 2. 2.

Every strongly irreducible subshift is topologically transitive. In particular, fix $F\in\mathrm{Fin}(G)$. Then for any $E\in\mathrm{Fin}(G)$ containing at least $|S_{F}(Y)|$ many disjoint right translates of $DF$, there is $\beta\in S_{E}(Y)$ so that every $\alpha\in S_{F}(Y)$ appears in $\beta$. We let $\mathrm{Trans}_{Y}(F,E)$ be the lexicographically least $E$-pattern with this property.

Most of the time, we take $A=2^{n}$ for some $n<\omega$, and we take the lexicographic ordering on $2^{n}$ as the ordering on $A$.

3 The operator $\Phi$

A subset $S\subseteq G$ is called syndetic if $S$ is $Q$-syndetic for some $Q\in\mathrm{Fin}(G)$. Given $F\in\mathrm{Fin}(G)$ with $1_{G}\in F$ and $Y\subseteq A^{G}$ a subshift, we say that $Y$ is $F$-minimal if for every $y\in Y$, every $\alpha\in S_{F}(Y)$ appears in $y$. Equivalently, for every $y\in Y$, every $\alpha\in S_{F}(Y)$ appears syndetically often. The following observation will be useful; suppose $Y\subseteq A^{G}$ is $F$-minimal and that every $\alpha\in S_{F}(Y)$ appears $E$-syndetically for some $E\in\mathrm{Fin}(G)$. Then every such $\alpha$ appears in every $\beta\in S_{FE}(Y)$.

The following is our main method of producing strongly irreducible, $F$-minimal flows. First, recalling the flow $Y_{Q}$ from section 1, we note that $Y_{Q}$ is $Q^{3}$-irreducible. Now let $Y\subseteq A^{G}$ be $D$-irreducible. Let $E\in\mathrm{Fin}(G)$ be symmetric, contain $D$, and be large enough to contain at least $|S_{F}(Y)|\leq|A|^{|F|}$ many disjoint right translates of $DF$. Let $C\in\mathrm{Fin}(G)$ be symmetric with $E^{5}\subseteq C$. We define a continuous, $G$-equivariant map $\varphi\langle Y,F,E,C\rangle=\varphi\colon Y\times Y_{C}\to A^{G}$ as follows. Suppose $(y,s)\in Y\times Y_{C}$, and write $z=\varphi(y,s)$. Let $g\in G$.

• •

If $g=kh$, where $s(h)=1$ and $k\in E$, set $z(g)=\mathrm{Trans}_{Y}(F,E)(k)$.

• •

If there are not $k\in E^{3}$ and $h\in G$ with $s(h)=1$ and $g=kh$, set $z(g)=y(g)$

• •

If $g=kh$, where $s(h)=1$ and $k\in E^{3}\setminus E$, set

$z(g)=\mathrm{Conf}_{Y}(\mathrm{Trans}_{Y}(F,E),\,(h\cdot y)|_{E^{5}\setminus E^{3}},\,E^{5})(k)$.

The idea behind this definition is to reprint $y$ most of the time, using $s$ to tell us where to overwrite with the pattern $\mathrm{Trans}_{Y}(F,E)$, and using strong irreducibility to blend everything together. This construction is a slight modification of a construction in [2]; see their Figure 3 for a good illustration.

It is routine to verify that $\varphi$ as defined is continuous and $G$-equivariant. Denote by $\Phi(Y,F,E,C)$ the image of $\varphi=\varphi\langle Y,F,E,C\rangle$. Then $\Phi(Y,F,E,C)$ is $C^{5}$-irreducible.

Lemma 7**.**

We have $S_{F}(Y)=S_{F}(\Phi(Y,F,E,C))$.

Proof.

The $\subseteq$ direction is clear. For the $\supseteq$ direction, suppose $z\in\Phi(Y,F,E,C)$ with $z=\varphi(y,s)$. It is enough to show that $z|_{F}\in S_{F}(Y)$. If there is $h\in G$ with $s(h)=1$ and $F\cap E^{3}h\neq\emptyset$, then $F\subseteq E^{5}h$, so we have

[TABLE]

If there is no such $h\in G$, then we have $z|_{F}=y|_{F}$. ∎

For any $z\in\Phi(Y,F,E,C)$, the $E$-pattern $\mathrm{Trans}_{Y}(F,E)$ appears in $z$, so in particular every pattern in $S_{F}(Y)$ appears in $z$. Hence $\Phi(Y,F,E,C)$ is $F$-minimal. Indeed, every $F$-pattern appears $C^{3}$-syndetically, since maximal $C$-spaced sets are $C^{2}$-syndetic. So every pattern in $S_{F}(Y)$ appears in every pattern in $S_{C^{4}}(\Phi(Y,F,E,C))$.

4 A tree of subshifts

We now use the operator $\Phi$ to produce a tree of strongly irreducible flows. We will construct for each $s\in 2^{<\omega}$ a strongly irreducible flow $X_{s}\subseteq(2^{|s|})^{G}$ by induction. This tree will be controlled by rapidly increasing sequences $\{D_{k}:k<\omega\}$, $\{E_{k}:k<\omega\}$, and $\{F_{k}:k<\omega\}$ of finite symmetric subsets of $G$. We will continue to add assumptions about how rapid this needs to be, but for now, we assume that

• •

$\bigcup_{n}D_{n}=\bigcup_{n}E_{n}=\bigcup_{n}F_{n}=G$.

• •

$E_{n}$ contains at least $2^{|D_{n}|(n+1)}$-many pairwise disjoint translates of $D_{n}^{2}$

• •

$F_{n}\supseteq E_{n}^{5}$

• •

$D_{n+1}\supseteq F_{n}^{5}$

Let $X_{\emptyset}$ be the trivial flow. If $s\in 2^{<\omega}$ and $X_{s}$ is defined, and $t=s^{\frown}0$, then we set $X_{t}=X_{s}\times 2^{G}$. Suppose we are given $k<\omega$, $s\in 2^{k}$, and $t=s^{\frown}1\in 2^{k+1}$. Then we set $X_{t}=\Phi(X_{s}\times 2^{G},D_{k},E_{k},F_{k})$.

In order to discuss the key properties of this construction, we think of $(2^{n})^{G}$ as embedded into $(2^{\omega})^{G}$ by adding zeros to the end. In this way, we can refer to the $(n\times F)$-patterns of a subflow $Y\subseteq(2^{N})^{G}\cong 2^{N\times G}$, the set $S_{n\times F}(Y):=\{y|_{n\times F}:y\in Y\}$, whenever $N\geq n$.

Each $X_{s}$ is $D_{|s|}$-irreducible. 2. 2.

For any $s\sqsubseteq t\in 2^{<\omega}$ with $|s|=n$, we have $S_{n\times D_{n}}(X_{s})=S_{n\times D_{n}}(X_{t})$. 3. 3.

Suppose $s\in 2^{<\omega}$ is such that $|s|>n$ and $s(n)=1$. Then every pattern in $S_{(n+1)\times D_{n}}(X_{s})$ appears in every pattern in $S_{(n+1)\times D_{n+1}}(X_{s})$. 4. 4.

Suppose $s\in 2^{n}$. Then $S_{(n+1)\times D_{n+1}}(X_{s^{\frown}0})\neq S_{(n+1)\times D_{n+1}}(X_{s^{\frown}1})$. This is because the conclusion of item 3 is true for $X_{s^{\frown}1}$ and false for $X_{s^{\frown}0}=X_{s}\times 2^{G}$.

We can now consider taking limits along the branches. It follows from item 2 above that for any $\alpha\in 2^{\omega}$, the flow $X_{\alpha}\subseteq(2^{\omega})^{G}$ is well defined. We can think of $X_{\alpha}$ as a point in the space $K((2^{\omega})^{G})$ of compact subsets of $(2^{\omega})^{G}$. The subshifts form a closed subspace, and given subshifts $\{Z_{n}:n<\omega\}\subseteq K((2^{\omega})^{G})$ and $Z\in K((2^{\omega})^{G})$, we have $Z_{n}\to Z$ iff for each finite $F\subseteq G$ and $k<\omega$, we eventually have $S_{k\times F}(Z_{n})=S_{k\times F}(Z)$. With this topology, the map $\Theta\colon 2^{\omega}\to K((2^{\omega})^{G})$ given by $\Theta(\alpha)=X_{\alpha}$ is continuous. Item 4 shows that $\Theta$ is injective. Whenever $\alpha\in 2^{\omega}$ has $\alpha^{-1}(\{1\})$ infinite, then item 3 implies that $X_{\alpha}$ is a minimal flow.

Proposition 8**.**

For any $\alpha\in 2^{\omega}$ with $\alpha^{-1}(\{0\})$ and $\alpha^{-1}(\{1\})$ infinite, the flow $X_{\alpha}$ is a minimal, model-universal flow.

Proof.

Having already discussed minimality, we focus on model-universality. Write $T=\alpha^{-1}(\{1\})$, and form the flow $Y_{\alpha}:=(2^{G})^{\omega}\times\prod_{n\in T}Y_{F_{n}}$. Then $Y_{\alpha}$ is model-universal. We have a continuous $G$-map $\psi_{\alpha}\colon Y_{\alpha}\to\prod_{n}X_{\alpha|_{n}}$ given inductively as follows. First let $f_{\omega}\colon\omega\to(\omega\setminus T)$ and $f_{T}\colon T\to T$ be infinite-to-one surjections. Let $y\in Y_{\alpha}$, and write $y=\{(y_{n})_{n<\omega},(s_{n})_{n\in T}\}$ with $y_{n}\in 2^{G}$ and $s_{n}\in Y_{F_{n}}$. Then we write $\psi_{\alpha}(y)=(\psi_{\alpha}(y)_{n})_{n<\omega}$ with each $\psi_{\alpha}(y)_{n}\in X_{\alpha|_{n}}$. We let $\psi_{\alpha}(y)_{0}$ be the unique member of the trivial flow $X_{\emptyset}$. If $\psi_{\alpha}(y)_{n}$ has been defined and $n\not\in T$, then $\psi_{\alpha}(y)_{n+1}=(\psi_{\alpha}(y)_{n},y_{f_{\omega}(n)})$. If $n\in T$, then $\psi_{\alpha}(y)_{n+1}=\varphi_{n}((\psi_{\alpha}(y)_{n},s_{f_{T}(n)}),s_{n})$, where $\varphi_{n}=\varphi\langle X|_{\alpha|_{n}}\times 2^{G},D_{n},E_{n},F_{n}\rangle$.

Notice that if the sequence $(\psi_{\alpha}(y)_{n})_{n<\omega}$ converges to some $x\in(2^{\omega})^{G}$, then $x\in X_{\alpha}$. Let $Y_{\alpha}^{\prime}\subseteq Y_{\alpha}$ be the subset of those $y$ for which $\psi_{\alpha}(y)_{n}$ is convergent. Then the map $\eta\colon Y^{\prime}_{\alpha}\to X_{\alpha}$ with $\eta(y)=\lim_{n}\psi_{\alpha}(y)_{n}$ is Borel. It suffices to show that if the $D_{n}$ grow rapidly enough, then $Y_{\alpha}^{\prime}$ has measure $1$ for any $G$-invariant measure on $Y_{\alpha}$. To that end, fix $y=((y_{n})_{n<\omega},(s_{n})_{n\in T})$, and consider some $g\in G$. A sufficient condition for the sequence $\psi_{\alpha}(y)_{n}(g)$ to be convergent is that for a tail of $n\in T$, we have $s_{n}(h)=0$ whenever $h\in E_{n}^{3}g$. This condition ensures that for suitably large $n\in T$, we have $\psi_{\alpha}(y)_{n+1}(g)=(\psi_{\alpha}(y)_{n}(g),s_{f_{T}(n)}(g))$. Define $Y^{\prime\prime}_{\alpha}\subseteq Y^{\prime}_{\alpha}$ to be those $y$ for which on a tail of $n\in T$, we have $s_{n}(g)=0$ for any $g\in E_{n}^{4}$. Notice that $Y^{\prime\prime}_{\alpha}$ is also Borel and $G$-invariant.

Fix $\nu$ an invariant measure on $Y_{F_{n}}$. Then letting $U=\{s\in Y_{F_{n}}:s(1_{G})=1\}$, we have $\nu(U)\leq 1/|F_{n}|$. This is because $g\cdot U=\{s\in Y_{F_{n}}:s(g^{-1})=1\}$, so by definition of the subshift $Y_{F_{n}}$, we have that the collection $\{g\cdot U:g\in F_{n}\}$ is pairwise disjoint. Then by invariance and a union bound, we have $\nu(\{s\in Y_{F_{n}}:s(g)=1\text{ for some }g\in E_{n}^{4}\})\leq|E_{n}^{4}|/|F_{n}|$. We now add our last assumption to the growth of the $D_{n}$.

• •

$|E_{n}^{4}|/|F_{n}|<1/2^{n}$.

From this assumption, it follows from the Borel-Cantelli lemma that for any invariant measure $\mu$ on $Y_{\alpha}$ that $\mu(Y_{\alpha}^{\prime\prime})=1$.

Furthermore, we claim that $\eta$ is injective on $Y_{\alpha}^{\prime\prime}$. To see this, suppose that $y\neq y^{\prime}\in Y_{\alpha}^{\prime\prime}$, with $y=\{(y_{n})_{n<\omega},(s_{n})_{n\in T}\}$ and $y^{\prime}=\{(y^{\prime}_{n})_{n<\omega},(s^{\prime}_{n})_{n\in T}\}$. First suppose that $y_{n}(g)\neq y_{n}^{\prime}(g)$ for some $n<\omega$ and $g\in G$. Then for some large enough $N<\omega$ and any $k,\ell\geq N$, we have $\psi_{\alpha}(y)_{k}(g)=\psi_{\alpha}(y)_{\ell}(g)$, and same for $y^{\prime}$. Now pick some suitably large $k\in\omega\setminus T$ with $f_{\omega}(k)=n$. Then $\psi_{\alpha}(y)_{k+1}(g)=\psi_{\alpha}(y)_{k}(g)\times y_{n}(g)$, and similarly for $y^{\prime}$. It follows that $\eta(y)\neq\eta(y^{\prime})$. In the case that $s_{n}(g)\neq s_{n}^{\prime}(g)$ for some $n\in T$, the argument is almost the same. For a suitably large $k\in T$ with $f_{T}(k)=n$, we use the assumption that $y$ and $y^{\prime}$ are in $Y_{\alpha}^{\prime\prime}$ to see that $\psi_{\alpha}(y)_{k+1}(g)=\psi_{\alpha}(y)_{k}(g)\times s_{n}(g)$, and similarly for $y^{\prime}$. Once more, we have $\eta(y)\neq\eta(y^{\prime})$. ∎

To prove Theorem 1, we need to recall some results from [3] (in particular, see Corollary 6.8). There, it is shown that every minimal flow is disjoint from every strongly irreducible subshift. From this, it follows that every minimal flow is disjoint from any $X_{\alpha}$ where $\alpha$ has a tail of zeros. Since disjointness is a $G_{\delta}$ condition ([3], Proposition 6.4), it follows that every minimal flow is disjoint from $X_{\alpha}$ for comeagerly many $\alpha\in 2^{\omega}$. We are now in a position to apply Mycielski’s theorem (see [4], 19.1) to find our mutually disjoint family $\{X_{\alpha_{i}}:i<\mathfrak{c}\}$ of minimal, model-universal shifts.

5 From essentially free to free

Recall that if $Y$ is a minimal metrizable flow, then an extension $\pi\colon Z\to Y$ is called almost 1-1 if the set $\{z\in Z:|\pi^{-1}(\{\pi(z)\})|=1\}$ is comeager. Notice that $Z$ must also be minimal. To see this, let $z\in Z$ and $V\subseteq Z$ be non-empty open. Then find $z^{\prime}\in V$ with $|\pi^{-1}(\{\pi(z^{\prime})\})|=1$. We can find a net $g_{i}\in G$ with $g_{i}\cdot\pi(z)\to\pi(z^{\prime})$. It follows that $g_{i}\cdot z\to z^{\prime}$. In particular, the orbit of $z$ meets $V$.

One method of producing almost 1-1 extensions of a given minimal $G$-flow is to consider $\mathrm{Reg}(Y)$, the Boolean algebra of regular open subsets of $Y$. Recall that $A\subseteq Y$ is regular open if $\mathrm{Int}(\overline{A})=A$. We remind the reader that in this Boolean algebra, we have $A^{c}=Y\setminus\overline{A}$, $A\vee B=\mathrm{Int}(\overline{A\cup B})$, and $A\wedge B=A\cap B$. If $\mathcal{B}\subseteq\mathrm{Reg}(Y)$ is a subalgebra, then $\mathrm{St}(\mathcal{B})$, the space of ultrafilters on $\mathcal{B}$, is a compact, zero-dimensional space whose basic clopen neighborhood has the form $\{p\in\mathrm{St}(\mathcal{B}):A\in p\}$, where $A\in\mathcal{B}$. If $\mathcal{B}$ is also $G$-invariant, then $\mathrm{St}(\mathcal{B})$ is a $G$-flow. If $\mathcal{B}$ is countable, then $\mathrm{St}(\mathcal{B})$ is homeomorphic to Cantor space. Now suppose that $\mathcal{B}$ contains a basis for the topology on $Y$. Then we have a $G$-map $\pi\colon\mathrm{St}(\mathcal{B})\to Y$ given by $\pi(p)=y$ iff every $A\in\mathcal{B}$ with $A\ni y$ satisfies $A\in p$. Furthermore, the map $\pi$ is pseudo-open, meaning that images of open sets have non-empty interior. For $y\in Y$, we have $|\pi^{-1}(\{y\})|=1$ iff for every $A\in\mathcal{B}$, we have $y\in A$ or $y\in Y\setminus\overline{A}$. So when $\mathcal{B}$ is countable, the set $\{y\in Y:|\pi^{-1}(y)|=1\}$ is comeager. Since $\pi$ is pseudo-open, it follows that $\{z\in Z:|\pi^{-1}(\pi(z))|=1\}$ is also comeager.

In general, an almost 1-1 extension can have very different measure-theoretic behavior than the base flow. Indeed, this fact is heavily exploited in [5]. For us however, we will seek to build almost 1-1 extensions which preserve the measure-theoretic properties of the base flow. For the remainder of the section, fix $Y$ a minimal, model-universal flow whose underlying space is a Cantor set. Recall that this implies that $Y$ is essentially free. We will call an invariant measure $\mu$ on $Y$ free if for every $g\in G$, we have $\mu(Y_{g})=0$, where $Y_{g}=\{y\in Y:gy=y\}$.

Definition 9**.**

Given $A\subseteq Y$, we call $A$ strongly regular open if $A$ is regular open and for every free invariant measure $\mu$, we have $\mu(A)+\mu(Y\setminus\overline{A})=1$. Denote by $\mathrm{SReg}(Y)$ the collection of strongly regular open sets.

Proposition 10**.**

$\mathrm{SReg}(Y)$* is a $G$-invariant subalgebra of $\mathrm{Reg}(Y)$.*

Proof.

Clearly $\mathrm{SReg}(Y)$ is $G$-invariant and closed under complements, so it is enough to check closure under intersection. Given $A,B\in\mathrm{SReg}(Y)$, we have

[TABLE]

Since $A$ and $B$ are both strongly regular open, the last entry must have measure zero for any free invariant measure $\mu$. ∎

Of course, we have yet to prove the existence of any interesting strongly regular open sets. We do this in the next lemma.

Lemma 11**.**

For every $g\in G\setminus\{1_{G}\}$, there is a partition of $Y\setminus Y_{g}$ into three relatively clopen pieces $A_{g}$, $B_{g}$, and $C_{g}$ with the property that $gA_{g}\cap A_{g}=\emptyset$, and likewise for $B_{g}$ and $C_{g}$. In particular, $A_{g}$, $B_{g}$, and $C_{g}$ are all strongly regular open sets.

Proof.

Write $Y\setminus Y_{g}=\bigcup_{n}U_{n}$ with each $U_{n}$ compact open. We may assume that the $U_{n}$ are pairwise disjoint, and by further partitioning each $U_{n}$ into finitely many clopen pieces if needed, we may assume that $gU_{n}\cap U_{n}=\emptyset$ for each $n<\omega$. We will inductively partition $V_{n}:=\bigcup_{k<n}U_{n}$ into pieces $A_{n}$, $B_{n}$, and $C_{n}$ with the property that $A_{N}\cap V_{n}=A_{n}$ for $N\geq n$, likewise for $B_{N}$ and $C_{N}$. We then set $A_{g}=\bigcup_{n}A_{n}$, and likewise for $B_{g}$ and $C_{g}$.

We set $A_{0}=B_{0}=C_{0}=\emptyset$. Assume $A_{k}$, $B_{k}$, and $C_{k}$ have been defined for some $k<\omega$. We will form clopen sets $A_{k}^{\prime}$, $B_{k}^{\prime}$, and $C_{k}^{\prime}$ so that $U_{k}=A_{k}^{\prime}\cup B_{k}^{\prime}\cup C_{k}^{\prime}$. Partition $U_{n}$ into finitely many clopen sets $\{W_{j}:j<m\}$ with the property that for each $j<m$ and for each $h\in\{g^{-1},g\}$, we either have $hW_{j}\subseteq A_{k}$, $hW_{j}\subseteq B_{k}$, $hW_{j}\subseteq C_{k}$, or $hW_{j}\cap(A_{k}\cup B_{k}\cup C_{k})=\emptyset$. Add each $W_{j}$ to the set $A_{k}^{\prime}$, $B_{k}^{\prime}$, or $C_{k}^{\prime}$ in such a way so that if $hW_{j}\subseteq A_{k}$ for some $h$ as above, then $W_{j}$ is not added to $A_{k}^{\prime}$, and likewise for $B_{k}^{\prime}$ and $C_{k}^{\prime}$. We then set $A_{k+1}=A_{k}\cup A_{k}^{\prime}$, and likewise for $B_{k+1}$ and $C_{k+1}$.

Notice that for each $n<\omega$, we have $gA_{n}\cap A_{n}=\emptyset$, and likewise for $B_{n}$ and $C_{n}$. Hence $A_{g}$ will also satisfy $gA_{g}\cap A_{g}=\emptyset$ as desired, and likewise for $B_{g}$ and $C_{g}$. ∎

The last lemma we will need shows that metrizable, almost 1-1 extensions of $Y$ using strongly regular open sets preserve the measure-theoretic properties of $Y$.

Lemma 12**.**

Let $\mathcal{B}$ be a countable $G$-invariant subalgebra of $\mathrm{SReg}(Y)$ extending the clopen algebra of $Y$. Let $Z=\mathrm{St}(\mathcal{B})$, and let $\pi\colon Z\to Y$ be the associated almost 1-1 extension. Then for any free invariant measure $\mu$ on $Y$, we have $\mu(\{y:|\pi^{-1}(\{y\})|=1\})=1$.

Proof.

By the discussion at the beginning of the section, we have

[TABLE]

Since $\mathcal{B}$ is a countable collection of strongly regular open sets, this set must have measure $1$ for any free $\mu$. ∎

Proof of Theorem 2.

Let $\mathcal{B}\subseteq\mathrm{SReg}(Y)$ be a countable, $G$-invariant subalgebra containing all of the sets $A_{g}$, $B_{g}$, $C_{g}$ from Lemma 11. Then $\mathrm{St}(\mathcal{B})$ will be the desired flow. To see that $\mathrm{St}(\mathcal{B})$ is free, let $p\in\mathrm{St}(\mathcal{B})$ and $g\in G\setminus\{1_{G}\}$. Then $p$ contains one of $A_{g}$, $B_{g}$, or $C_{g}$, WLOG say $A_{g}\in p$. Then since $gA_{g}\cap A_{g}=\emptyset$, we must have $gp\neq p$. To see that $\mathrm{St}(\mathcal{B})$ is model-universal, we note that on the set $Y_{0}:=\{y\in Y:|\pi^{-1}(\{y\})|=1\}$, the map $\pi^{-1}\colon Y_{0}\to Z$ is well defined. By Lemma 12, this set has measure $1$ for all free invariant measures on $Y$. ∎