Introduction to the KPT Correspondence 2 – Ramsey DocCourse Prague 2016

The following notes are from the Ramsey DocCourse in Prague 2016. The notes are taken by me and I have edited them. In the process I may have introduced some errors; email me or comment below and I will happily fix them.

Title: Introduction to the KPT correspondence 2 (of 3).

Lecturer: Lionel Ngyuen Van Thé.

Date: November 16, 2016.

Main Topics: Computing universal minimal flows, $M(S_\infty)$, why precompactness is important.

Definitions: Minimal flow, universal flow, Logic action, $G$-equivariant.

Lecture 1 – Lecture 2 – Lecture 3


Last time we looked at how the Ramsey property of a structure $\mathbb{K}$ ensures that $\text{Aut}(\mathbb{K})$ is extremely amenable.

Today we will look at what can be said about the dynamics of $\text{Aut}(\mathbb{K})$ when $\text{Age}(\mathbb{K})$ is not Ramsey?

Examples of extremely amenable groups

Last lecture we did not provide many examples of extremely amenable groups, so let us fix that now.

Theorem (Pestov, 1998). $\text{Aut}(\mathbb{Q}, <)$ is extremely amenable.

The underlying Ramsey principle here is the classical Ramsey theorem. This was the first known example of an extremely amenable group. Note that it comes seven years before the 2005 KPT paper.

The following examples were shown to be extremely amenable using the 2005 KPT correspondence, although the underlying Ramsey principles were already known.

Theorem (KPT, 2005). The folowing groups are extremely amenable. The needed Ramsey principle is in brackets.

  1. $\text{Aut}(\mathcal{R}, <)$, where $\mathcal{R}$ is the Rado-graph. (Nešetřil-Rödl 1977, Abramson-Harrington 1978)
  2. $\text{Aut}(H_n, <)$, where $H_n$ is the random $K_n$ free graph. (Nešetřil-Rödl 1977)
  3. $\text{Iso}(\mathbb{U}_\mathbb{Q}, <)$, where $\mathbb{U}_\mathbb{Q}$ is the rational Urysohn space. (Nešetřil)
  4. $\text{Aut}(F^{< \omega}, <)$, where $F$ is a finite field. (Graham-Rothschild)
  5. $\text{Aut}( \mathcal{B}, <)$, where $\mathcal{B}$ is the generic countable atomless Boolean algebra. (Graham-Rothschild)


In order to analyze what happens to $\text{Aut}(\mathbb{K})$ when $\mathbb{K}$ is not Ramsey, we will introduce the notion of a universal minimal flow, which at its heart is a canonical compact object we can associate to a group. The size (both topologically and in terms of cardinality) of a group’s universal minimal flow will be determined by the “amount of Ramsey” that the group has.

Definition. A $G$-flow $G \curvearrowright X$ in minimal when it has no proper invariant subflow.
Fact. For all $x \in X$, $\overline{G \cdot X}$ is an invariant subflow of $G \curvearrowright X$. So $G \curvearrowright X$ is minimal when $\forall x \in X$, $\overline{G \cdot x} = X$, i.e. when every orbit is dense.

Here are two exercises to play around with these concepts.

Exercise. Show that every $G$-flow admits a minimal subflow. (Use Zorn’s Lemma.)
Definition. $S_\infty$ acts on the set of binary relations on $\mathbb{N}$, i.e. $X = \{0,1\}^{\mathbb{N}^2}$ which is compact. It acts on $X$ by the logic action, where if $R \in X$ then $$\forall x,y \in \mathbb{N}, \forall g \in S_\infty, g \cdot R(x,y) \Leftrightarrow R(g{-1}(x), g{-1}(y)).$$
Exercise. Show that the collection of linear orders $\text{LO}(\mathbb{N})$ is a closed $S_\infty$-invariant subset of $X$ that is minimal. (Where this is with the logic action, and $X$ is defined as above.)

For a fixed $G$, the object that is universal in the class of minimal $G$-flows will be a canonical object we can associate to $G$, called the universal minimal flow of $G$. To make sense of this, we introduce the concept of universality and flow homomorphism.

Definition. Given $G$-flows $G \curvearrowright X$ and $G \curvearrowright Y$, a flow homomorphism is a map $\pi: X \rightarrow Y$ that is continuous and $G$-invariant.

A map $\pi: X \rightarrow Y$ is $G$-invariant if $\forall g \in G, \forall x \in X$ we have $$\pi(g \cdot x) = g \cdot \pi(x).$$

These universal objects always exist, although the proof is non-constructive.

Theorem. Let $G$ be a topological gorup. There is a minimal $G$-flow $G \curvearrowright M(G)$ that is universal in the sense that for all $G \curvearrowright Y$ minimal there is an onto flow homomorphism $\pi: M(G) \rightarrow Y$.

In addition, $M(G)$ is unique (up to flow isomorphism). So $M(G)$ is called the universal minimal flow of $G$.

Typically $M(G)$ will be hard to describe. The following facts show cases where they are easily understood.


  1. $G$ is extremely amenable iff $\vert M(G) \vert = 1$.
  2. If $G$ is compact, then $M(G) = G$. (Here the action is by left translation.)

Two other examples where $M(G)$ is known.

Fact. If $G$ is discrete, then $M(G)$ is any minimal flow of $\beta(G)$.

The first known example of a non-trivial metrizable universal minimal flow is the following.

Theorem (Pestov 1998). If $G= \text{Homeo}(S^1)$, then $M(G) = S^1$, where the action is the natural one.


We will compute the universal minimal flow of $S_\infty$. The original proof is due to Glasner-Weiss in 2002, but we will present proof that is easier to generalize. You should compare this with their original proof.

Theorem (Glasner-Weiss, 2002). $M(S_\infty) = \text{LO}(\mathbb{N})$.

Proof. By an earlier exercise, $\text{LO}(\mathbb{N})$ is a minimal flow, so we need “only” show that it is universal. So let $G = S_\infty$ and let $G \curvearrowright X$.

Step 1: Use extreme amenability of a smaller group.

Fix a linear ordering $<^\mathbb{Q} \in \text{LO}(\mathbb{N})$ such that $(\mathbb{N}, <^\mathbb{Q}) \cong (\mathbb{Q}, <)$.

In this way we have that $G^\star = \text{Aut}(\mathbb{N}, <^\mathbb{Q}) \cong \text{Aut}(\mathbb{Q}, <)$ which is extremely amenable by Pestov’s theorem. Note that $G^\star \leq G$. So $G \curvearrowright X$ induces an action $G^\star \curvearrowright X$. By extreme amenability of $G^\star$, there is a $G^\star$-fixed point $x \in X$.

Step 2: Use uniform spaces to extend the group action.

Now consider the map $\pi: G \rightarrow X$ that sends $g \mapsto g \cdot x$. Since $G^\star = \text{stab}(<^\mathbb{Q})$ we have that $\pi(g)$ only depends on $[g] \in G / G^\star$. Thus $$G / G^\star \cong G \cdot <^\mathbb{Q}.$$

We also see that $$G \cdot <^\mathbb{Q} = \{\preceq \in \text{LO}(\mathbb{N}) : (\mathbb{N}, \preceq) \cong (\mathbb{N}, <^\mathbb{Q}) \cong (\mathbb{Q}, <)\}.$$

So, in this way we can think of, $\pi: G \cdot <^\mathbb{Q} \rightarrow X$.

Exercise. Show that this map $\pi$ is $G$-equivariant.

Assume for the moment that $\pi$ can be continuously extended to a map $\tilde{\pi}$ on all of $\text{LO}(\mathbb{N})$. In this case $\tilde{\pi}[\text{LO}(\mathbb{N})]$ is a compact subspace of $X$ containing $x$ (the $G^\star$ fixed point), hence $G \cdot x$. Since $X$ is minimal, $X = \overline{G \cdot x} \subseteq \tilde{\pi}[\text{LO}(\mathbb{N})] \subseteq X$. So we are done.

Claim. $\pi$ can be continuously extended to a map $\tilde{\pi}$ on all of $\text{LO}(\mathbb{N})$.

Proof of claim. We would like to show first that $\pi$ is uniformly continuous. What does that even mean in the non-metric setting? How do we capture the interplay between the topology of $\text{LO}(\mathbb{N})$ and the group $G$?

We can’t assume that $X$ has a metric, but it will always have a unique uniformization, which will act like a metric for the purposes of defining uniform continuity.

To extend $\pi$ continuously, if you are familiar with uniform spaces:

  1. Show that $\pi: G \rightarrow X$ is uniformly continuous when $G$ is equipped with $d_R$ (defined in lecture 1).
  2. Show that $\tilde{\pi}: G / G^\star \rightarrow X$ is uniformly continuous when $G/G^\star$ is equipped with the natural projection of $d_R$.
  3. Show that the identification $G/G^\star \cong G \cdot <^\mathbb{Q}$ also holds when $G / G^\star$ is equipped with the natural projection of $d_R$ and $G \cdot <^\mathbb{Q}$ is equipped with the uniform structure from $\text{LO}(\mathbb{N})$.

If you aren’t familiar with uniform spaces, then just pretend that $X$ has a metric and do the same as above.

This part shows why this type of argument doesn’t always work.

Precompact expansions

This proof works directly when you replace $S_\infty$ by $\text{Aut}(\mathbb{K})$ and $(\mathbb{N}, <^\mathbb{Q})$ is replaced by a closed subgroup $G^\star \leq G$ such that

  • $G^\star$ is extremely amenable, and
  • $G / G^\star$ is precompact (i.e. the completion is compact) when equipped with the projection of $d_R$.

Question: What does “$G/G^\star$ is precompact” mean combinatorially? Put another way, what do such $G^\star$ look like?

Since $G^\star \leq G = \text{Aut}(\mathbb{K})$ we can think of $G^\star = \text{Aut}(\mathbb{K}^\star)$ as an expansion of $\mathbb{K}$ where $\mathbb{K}^\star = (\mathbb{K}, (R_i^\star)_{i \in I}) = (\mathbb{K}, \vec{R^\star})$, where $I$ is possibly infinite.

If the parity of $R_i^\star$ is denoted by $a(i)$, then $$\vec{R^\star} \in \prod_{i \in I} \{0,1\}^{\mathbb{N}^{a(i)}} =: P^\star$$ is compact.

Here are two exercises to help you understand the interplay of these objects.

Exercise. Show that $G / G^\star$, equipped with the projection of $d_R$, is isometric to $G \cdot \vec{R^\star} \subset P^\star$, where $P^\star$ is equipped with the following metric $d^\star$, and action is componentwise. $$d^\star = \text{sup}_{i \in I} d_i$$ where $d_i(S,T) = 2^{-n}$ and $n = \min\{k \in \mathbb{N} : S \upharpoonright [k+1]^{a(i)} \neq T \upharpoonright [k+1]^{a(i)}\}.$

A priori, $d^\star$ gives the box topology which could be different than the product topology. However, precompactness guarantees that these are the same.

Exercise. Show that $(G / G^\star, \text{proj}_R)$ is precompact iff $d^\star$ generates the product topology on $P^\star$, and every element of $\text{Age}(\mathbb{K})$ has only finitely many expansions in $\text{Age}(\mathbb{K}^\star)$.

That is, $\mathbb{K}^\star$ is a precompact expansion of $\mathbb{K}$, hence the name.

In this case, we write $$X^\star:= \overline{G \cdot \vec{R^\star}} \subset P^\star.$$

Theorem (KPT 05, NVT 13). Assume that $\mathbb{K}^\star$ is a precompact Ramsey expansion of $\mathbb{K}$. Then $\text{Aut}(\mathbb{K}) \curvearrowright X^\star$ is universal for minimal $\text{Aut}(\mathbb{K})$-flows.

Recall that $G \curvearrowright X$ is minimal iff there is a flow homomorphism $\pi: X^\star \rightarrow X$. Now for $Y \subseteq X^\star$ any minimal flow we take $y \in Y$ and see that $\pi(y) \supseteq \overline{G \cdot \pi(y)} = X$.

Corollary. Under the same assumptions, any minimal subflow of $\text{Aut}(\mathbb{K}) \curvearrowright X^\star$ is the universal minimal flow.

In particular, $M(\text{Aut}(\mathbb{K}))$ is metrizable.

In practice, computing this requires understanding what the minimal subflows of $\text{Aut}(\mathbb{K}) \curvearrowright X^\star$ look like. This amounts to understanding when $\text{Aut}(\mathbb{K}) \curvearrowright \overline{G \cdot \vec{R^\star}}$ is minimal.


These are our overarching references

Here are the references to specific theorems we mentioned. (Mike: I’m missing a couple.)

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