A direct solution to the Generic Point Problem

TL;DR

This paper offers a new proof demonstrating that for certain Polish groups and minimal flows with meager orbits, the universal minimal flow is inherently non-metrizable, highlighting a fundamental property of these mathematical structures.

Contribution

It provides a direct proof of a recent theorem relating to the non-metrizability of universal minimal flows for specific Polish groups and flows.

Findings

Universal minimal flow $M(G)$ is non-metrizable under given conditions.

Universal highly proximal extension of certain flows is non-metrizable.

The proof offers a new perspective on the structure of minimal flows.

Abstract

We provide a new proof of a recent theorem of Ben-Yaacov, Melleray, and Tsankov. If $G$ is a Polish group and $X$ is a minimal, metrizable $G$-flow with all orbits meager, then the universal minimal flow $M(G)$ is non-metrizable. In particular, we show that given $X$ as above, the universal highly proximal extension of $X$ is non-metrizable.

Full text

1 Introduction

In this paper, we are concerned with actions of a topological group $G$ on a compact space $X$. All groups and spaces are assumed Hausdorff. A compact space $X$ equipped with a continuous $G$-action $a:G\times X\rightarrow X$ is called a $G$-flow. The action $a$ is often suppressed in the notation, i.e. $gx$ is written for $a(g,x)$. A $G$-flow $X$ is called minimal if every orbit is dense. It is a fact that every topological group $G$ admits a universal minimal flow $M(G)$, a minimal flow which admits a $G$-map onto any other minimal flow. A $G$-map is a continuous map respecting the $G$-action. The flow $M(G)$ is unique up to $G$-flow isomorphism.

We can now recall the following theorem of Ben-Yaacov, Melleray, and Tsankov [4].

Theorem 1.1**.**

Let $G$ be a Polish group, and let $M(G)$ be the universal minimal flow of $G$. If $M(G)$ is metrizable, then $M(G)$ has a comeager orbit.

The question of whether or not metrizability of $M(G)$ was enough to guarantee a comeager orbit was first asked by Angel, Kechris, and Lyons [5]. In [6], the current author proved Theorem 1.1 in the case when $G$ is the automorphism group of a first-order structure. The proof given there used topological properties of the largest $G$-ambit $S(G)$ along with combinatorial reasoning about the structures. In [4], the authors also use topological properties of $S(G)$, but the combinatorics is replaced by the following theorem due to Rosendal; see [4] for a proof.

Theorem 1.2**.**

Let $G$ be a Polish group acting continuously on a compact metric space $X$. Assume the action is topologically transitive. Then the following are equivalent.

$G$* has a comeager orbit.* 2. 2.

For any open $1\in V\subseteq G$ and any open $B\subseteq X$, there is open $C\subseteq B$ so that for any $D\subseteq C$, the set $C\setminus VD$ is nowhere dense.

It is proven in [5] that comeager orbits push forward; namely, if $X$ is a minimal $G$-flow, $x\in X$ is a point whose orbit is generic, and if $\pi:X\rightarrow Y$ is a surjective $G$-map, then $\pi(x)$ has generic orbit in $Y$. Theorem 1.1 then becomes equivalent to the following: whenever $G$ is a Polish group and $X$ is a minimal metrizable flow with all orbits meager, then $G$ must admit some minimal, non-metrizable flow. Remarkably, neither [4] nor [6] prove Theorem 1.1 in this direct fashion.

We provide a direct proof of Theorem 1.1. For any topological group $G$ and any $G$-flow $X$, we construct a new $G$-flow denoted $S_{G}(X)$. We then show that if $X$ is minimal, then so is $S_{G}(X)$. Lastly, if $G$ is Polish and $X$ is metrizable and has all orbits meager, we use Theorem 1.2 to show that $S_{G}(X)$ is non-metrizable.

After providing our new proof of 1.1, we investigate the flow $S_{G}(X)$ in more detail. For any $G$-flow $X$, there is a natural map $\pi_{X}:S_{G}(X)\rightarrow X$. When $X$ is minimal, we show that $\pi_{X}$ is the universal highly proximal extension of $X$. The notion of a highly proximal extension was introduced by Auslander and Glasner in [2]. If $X$ and $Y$ are minimal $G$-flows, a $G$-map $\varphi:Y\rightarrow X$ is highly proximal if for any $x\in X$ and non-empty open $U\subseteq Y$, there is $g\in G$ with $g\pi^{-1}(\{x\})\subseteq U$. Auslander and Glasner prove in [2] that for every minimal $G$-flow $X$, there is a universal highly proximal extension $\pi:\widehat{X}\rightarrow X$. This means that $\pi$ is highly proximal, and for every other highly proximal $\varphi:Y\rightarrow X$, there is a $G$-map $\psi:\widehat{X}\rightarrow Y$ so that $\pi=\varphi\circ\psi$. The map $\pi$ is unique up to $G$-flow isomorphism over $X$. Our construction of the flow $S_{G}(X)$ provides a new construction of the universal highly proximal extension of $X$ and hints at a generalization of this notion even when $X$ is not minimal.

1.1 Acknowledgments

I would like to thank Eli Glasner for many helpful discussions, including the initial suggestion that $\pi_{X}$ was the universal highly proximal extension of $X$. I would also like to thank Todor Tsankov for helpful discussions, and I would like to thank the Casa Matemática Oaxaca for their hospitality while some of this work was being completed.

2 The flow $S_{G}(X)$ and proof of Theorem 1.1

All groups and spaces will be assumed Hausdorff. In this section, fix a topological group $G$ and a $G$-flow $X$. Write $\mathcal{N}_{G}$ for the collection of symmetric open neighborhoods of the identity in $G$, and write $\mathrm{op}(X)$ for the collection of nonempty open subsets of $X$.

Definition 2.1**.**

A near filter is any $\mathcal{F}\subseteq\mathrm{op}(X)$ so that for any $A_{1},...,A_{k}\in\mathcal{F}$ and any $U\in\mathcal{N}_{G}$, we have $UA_{1}\cap\cdots\cap UA_{k}\neq\emptyset$. A near ultrafilter is a maximal near filter.

Near ultrafilters exist by an application of Zorn’s lemma. Near ultrafilters on a uniform space have been considered in [1] and [3]. Two aspects of our approach are slightly different. First, the notion of nearness is not given by the natural uniform structure on the compact Hausdorff space $X$. Second, instead of working with a notion of nearness on $\mathcal{P}(X)$, we are more or less working with the regular open algebra on $X$ (see item (2) in Lemma 2.2).

Let $S_{G}(X)$ denote the space of near ultrafilters on $\mathrm{op}(X)$.

Lemma 2.2**.**

**

Let $p\in S_{G}(X)$, and let $A\subseteq X$ be open. If $A\not\in p$, then there is some $V\in\mathcal{N}_{G}$ with $VA\not\in p$. 2. 2.

Let $A\subseteq X$ be open, and let $B_{1},...,B_{k}\subseteq A$ be open with $B_{1}\cup\cdots\cup B_{k}$ dense in $A$. If $p\in S_{G}(X)$ and $A\in p$, then $B_{i}\in p$ for some $i\leq k$.

Proof.

As $A\not\in p$, find $B_{1},...,B_{n}\in p$ and $U\in\mathcal{N}_{G}$ with $UA\cap UB_{1}\cap\cdots\cap UB_{n}=\emptyset$. Let $V\in\mathcal{N}_{G}$ with $VV\subseteq U$. Then $V(VA)\cap VB_{1}\cap\cdots\cap VB_{n}=\emptyset$. 2. 2.

Towards a contradiction, assume $B_{i}\not\in p$ for each $i\leq k$. For each $i\leq k$, find $B^{i}_{1},...,B^{i}_{n_{i}}\in p$ and a $U\in\mathcal{N}_{G}$ so that $UB_{i}\cap UB^{i}_{1}\cap\cdots\cap UB^{i}_{n_{i}}=\emptyset$. We can take the same $U\in\mathcal{N}_{G}$ for each $i\leq k$ by intersecting. Let $C=\bigcap_{i\leq k}\bigcap_{j\leq n_{i}}UB^{i}_{j}$. Then since $A\in p$, we have $UA\cap C\neq\emptyset$. Let $ga\in UA\cap C$, where $g\in U$ and $a\in A$. Since $UA\cap C$ is open, there is open $A^{\prime}\subseteq A$ with $gA^{\prime}\subseteq UA\cap C$. As $B_{1}\cup\cdots\cup B_{k}$ is dense in $A$, there is some $i\leq k$ and some $b\in B_{i}$ with $gb\in UA\cap C$. Since $gb\in UB_{i}$, this is a contradiction. ∎

Definition 2.3**.**

If $A\in\mathrm{op}(X)$, set $N_{A}:=\{p\in S_{G}(X):A\not\in p\}$. We endow $S_{G}(X)$ with the topology whose typical basic open neighborhood is $N_{A}$ for $A\in\mathrm{op}(X)$.

Proposition 2.4**.**

The topology from Definition 2.3 is compact Hausdorff.

Proof.

To show that $S_{G}(X)$ is Hausdorff, let $p\neq q\in S_{G}(X)$. Find some $A\in p\setminus q$. As $A\not\in q$, find some $V\in\mathcal{N}_{G}$ so that $VA\not\in q$. Set $B=\mathrm{int}(X\setminus VA)$. Then $B\not\in p$. So $p\in N_{B}$, $q\in N_{VA}$, and $N_{VA}\cap N_{B}=\emptyset$.

To show that $S_{G}(X)$ is compact, suppose $\mathcal{C}:=\{N_{A_{i}}:i\in I\}$ is a collection of basic open sets without a finite subcover. Then for any $i_{1},...,i_{k}\in I$, we can find $p\in\bigcap_{j\leq k}S_{G}(X)\setminus N_{A_{i_{j}}}$, equivalently, with $A_{i_{1}},...,A_{i_{k}}\in p$. But this implies that $\{A_{i}:i\in I\}$ is a near filter, and can be extended to a near ultrafilter $q$. Therefore $\mathcal{C}$ is not an open cover. ∎

Definition 2.5**.**

If $p\in S_{G}(X)$ and $g\in G$, we let $gp\in S_{G}(X)$ be defined by declaring $A\in gp$ iff $g^{-1}A\in p$ for each $A\in\mathrm{op}(X)$.

Proposition 2.6**.**

The action in Definition 2.5 is continuous.

Proof.

First note that for a fixed $g\in G$, the map $p\rightarrow gp$ is continuous. So let $p_{i},p\in S_{G}(X)$ and $g_{i}\in G$ with $p_{i}\rightarrow p$ and $g_{i}\rightarrow 1$. Suppose $A\not\in p$. Find $V\in\mathcal{N}_{G}$ with $VA\not\in p$. So eventually $VA\not\in p_{i}$. Also, as $g_{i}\rightarrow 1$, eventually we have $g_{i}^{-1}\in V$. Whenever $g_{i}^{-1}A\subseteq VA$, we must have $g_{i}^{-1}A\not\in p_{i}$. So eventually $A\not\in g_{i}p_{i}$. ∎

Up until now, no assumptions on $G$ and $X$ have been needed. In fact, we did not even need $X$ to be compact to construct $S_{G}(X)$. We now begin adding extra assumptions to $G$ and $X$ to obtain stronger conclusions about $S_{G}(X)$.

Proposition 2.7**.**

Suppose $X$ is a minimal $G$-flow. Then so is $S_{G}(X)$.

Proof.

Let $p\in S_{G}(X)$, and let $A\in op(X)$ with $N_{A}\neq\emptyset$. Find some $V\in\mathcal{N}_{G}$ with $N_{VA}\neq\emptyset$. Then $B:=\mathrm{int}(X\setminus VA)\neq\emptyset$. As $X$ is minimal, find $g_{1},...,g_{k}$ with $X=\bigcup_{i\leq k}g_{i}B$. For some $i\leq k$, we must have $g_{i}B\in p$. Then $B\in g_{i}^{-1}p$, so we must have $A\not\in g_{i}^{-1}p$, and the orbit of $p$ is dense as desired. ∎

Before proving Theorem 1.1, we need a sufficient criterion for when $S_{G}(X)$ is non-metrizable.

Proposition 2.8**.**

Suppose there are $\{A_{n}:n<\omega\}\subseteq\mathrm{op}(X)$ and $V\in\mathcal{N}_{G}$ so that the collection $\{VA_{n}:n<\omega\}$ is pairwise disjoint. Then $S_{G}(X)$ is non-metrizable.

Proof.

If $S\subseteq\omega$, let $A_{S}=\bigcup_{n\in S}A_{n}$, and let $Y=\{p\in S_{G}(X):A_{\omega}\in p\}$. Then $Y\subseteq S_{G}(X)$ is a closed subspace. To show that $S_{G}(X)$ is non-metrizable, we will exhibit a continuous surjection $\pi:Y\rightarrow\beta\omega$. First note that if $S\subseteq\omega$, then $VA_{S}\cap VA_{\omega\setminus S}=\emptyset$. Therefore, if $p\in Y$, $p$ contains exactly one of $A_{S}$ or $A_{\omega\setminus S}$ for each $S\subseteq\omega$. We let $\pi:Y\rightarrow\beta\omega$ be defined so that for $S\subseteq\omega$, $S\in\pi(p)$ iff $A_{S}\in p$. It is immediate that $\pi$ is continuous. To see that $\pi$ is surjective, let $q\in\beta\omega$. Then $\{A_{S}:S\in q\}$ is a near filter; any near ultrafilter $p$ extending it is a member of $Y$ with $\pi(p)=q$. ∎

Proof of Theorem 1.1.

We now fix a Polish group $G$ and a minimal $G$-flow $X$ whose orbits are all meager. Then by Theorem 1.2, there is $U\in\mathcal{N}_{G}$ and open $B\subseteq X$ so that for any open $C\subseteq B$, there is open $D\subseteq C$ with $C\setminus UD$ somewhere dense (since $C$ and $UD$ are open, this is the same as $C\setminus UD$ having nonempty interior).

Let $V\in\mathcal{N}_{G}$ with $VV\subseteq U$. We now produce $\{A_{n}:n<\omega\}\subseteq\mathrm{op}(X)$ with $\{VA_{n}:n<\omega\}$ pairwise disjoint. First set $B_{0}=B$. As $B_{0}\subseteq B$, there is $A_{0}\subseteq B_{0}$ so that $B_{0}\setminus UA_{0}$ has nonempty interior. Suppose open sets $B_{0},...,B_{n-1}$ and $A_{0},...,A_{n-1}$ have been produced so that $A_{i}\subseteq B_{i}$ and $\mathrm{int}(B_{i}\setminus UA_{i})\neq\emptyset$. We continue by setting $B_{n}=\mathrm{int}(B_{n-1}\setminus UA_{n-1})$. As $B_{n}\subseteq B$, there is $A_{n}\subseteq B_{n}$ so that $B_{n}\setminus UA_{n}$ has nonempty interior. Notice that for any $m\leq n$, we also have $A_{n}\subseteq B_{m}$. It follows that if $m<n$, we have $UA_{m}\cap A_{n}=\emptyset$. This implies that $VA_{m}\cap VA_{n}=\emptyset$ as desired. We can now apply Proposition 2.8 to conclude that $S_{G}(X)$ is not metrizable. ∎

3 Universal highly proximal extensions

Let $\varphi:Y\rightarrow X$ be a $G$-map between minimal flows. There are several equivalent definitions which all say that $\varphi$ is highly proximal. The definition we will use here is that $\varphi$ is highly proximal iff every non-empty open $B\subseteq Y$ contains a fiber $\varphi^{-1}(\{x\})$ for some $x\in X$. Define the fiber image of $B$ to be the set $\varphi_{fib}(B):=\{x\in X:\varphi^{-1}(\{x\})\subseteq B\}$. Notice that $\varphi_{fib}(B)$ is open, and $\varphi$ is highly proximal iff $\varphi_{fib}(B)\neq\emptyset$ for every non-empty open $B\subseteq Y$. It follows that this definition is the same as the one given in the introduction.

Now let $X$ be a $G$-flow, and form $S_{G}(X)$. We define the map $\pi_{X}:S_{G}(X)\rightarrow X$ as follows. For each $p\in S_{G}(X)$, there is a unique $x_{p}\in X$ so that every neighborhood of $x_{p}$ is in $p$. The existence of such a point is an easy consequence of the compactness of $X$ and the second item of 2.2. For uniqueness, notice that if $x\neq y\in X$, we can find open $A\ni x$, $B\ni y$ and $U\in\mathcal{N}_{G}$ with $UA\cap UB=\emptyset$. We set $\pi_{X}(p)=x_{p}$. This map clearly respects the $G$-action. To check continuity, one can check that if $K\subseteq X$ is closed, then $\pi_{X}^{-1}(K)=\{p\in S_{G}(X):A\in p\text{ for every open }A\supseteq K\}$, and this is a closed condition.

Proposition 3.1**.**

Let $X$ be minimal. Then the map $\pi_{X}:S_{G}(X)\rightarrow X$ is highly proximal.

Proof.

By 2.7, $S_{G}(X)$ is a minimal flow. So let $N_{A}\subseteq S_{G}(X)$ be a nonempty basic open neighborhood. This implies that $\mathrm{int}(X\setminus A)\neq\emptyset$. Let $x\in\mathrm{int}(X\setminus A)$. Then there are open $B\ni x$ and $U\in\mathcal{N}_{G}$ with $UB\cap A=\emptyset$. It follows that any $p\in S_{G}(X)$ containing $B$ cannot contain $A$. In particular, we have $\pi_{X}^{-1}(\{x\})\subseteq N_{A}$. ∎

Theorem 3.2**.**

Let $X$ be minimal. Then the map $\pi_{X}:S_{G}(X)\rightarrow X$ is the universal highly proximal extension of $X$.

Proof.

Fix a highly proximal extension $\varphi:Y\rightarrow X$. For each $y\in Y$, let $\mathcal{F}_{y}:=\{\varphi_{fib}(B):B\ni y\text{ open}\}$. Then $\mathcal{F}_{y}\subseteq\mathrm{op}(X)$ is a filter of open sets, so in particular it is a near filter. We will show that for each $p\in S_{G}(X)$, there is a unique $y\in Y$ with $\mathcal{F}_{y}\subseteq p$. This will define the map $\psi:S_{G}(X)\rightarrow Y$.

We first show that for each $p\in S_{G}(X)$, there is at least one such $y\in Y$. To the contrary, suppose for each $y\in Y$, there were $B_{y}\ni y$ open so that $\varphi_{fib}(B_{y})\not\in p$. Find $y_{1},...,y_{k}$ so that $\{B_{y_{1}},...,B_{y_{k}}\}$ is a finite subcover. Let $A_{i}=\varphi_{fib}(B_{y_{i}})$. Each $A_{i}$ is open, so we will reach a contradiction once we show that $\bigcup_{i\leq k}A_{i}$ is dense. Let $A\subseteq X$ be open. Then $C:=B_{y_{i}}\cap\varphi^{-1}(A)\neq\emptyset$ for some $i\leq k$. As $C$ is open, $\varphi_{fib}(C)\neq\emptyset$, and $\varphi_{fib}(C)\subseteq A\cap A_{i}$.

Now we consider uniqueness. Let $p\in S_{G}(X)$, and consider $y\neq z\in Y$. Find open $B\ni y$ and $C\ni z$ and some $V\in\mathcal{N}_{G}$ so that $VB\cap VC=\emptyset$. It follows that $\varphi_{fib}(VB)\cap\varphi_{fib}(VC)=\emptyset$. Now notice that $V\varphi_{fib}(B)\subseteq\varphi_{fib}(VB)$, and likewise for $C$. Hence $p$ cannot contain both $\mathcal{F}_{y}$ and $\mathcal{F}_{z}$.

The map $\psi$ clearly respects the $G$-action and satisfies $\pi_{X}=\varphi\circ\psi$. To show continuity, let $K\subseteq Y$ be closed. Let $\mathcal{F}_{K}:=\{\varphi_{fib}(B):B\supseteq K\text{ open}\}$. We will show that $\psi(p)\in K$ iff $\mathcal{F}_{K}\subseteq p$. From this it follows that $\psi^{-1}(K)$ is closed. One direction is clear. For the other, suppose $\psi(p)=y\not\in K$. Find open sets $B\ni y$, $C\supseteq K$, and $V\in\mathcal{N}_{G}$ with $VB\cap VC=\emptyset$. As in the proof of uniqueness, $p$ cannot contain both $\mathcal{F}_{y}$ and $\mathcal{F}_{K}$. ∎

By combining the main results of the previous two sections, we obtain the following.

Corollary 3.3**.**

Let $G$ be a Polish group, and let $X$ be a minimal, metrizable $G$-flow with all orbits meager. Then the universal highly proximal extension of $X$ is non-metrizable.