Ramsey and Ultrafilters 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.

Special thanks to Ivan Khatchatourian for some elaborations.

Title: Ramsey and Ultrafilters 2 (of 2)

Lecturer: Slawomir Solecki

Date: Friday October 21, 2016.

Main Topics: Abstraction of Gowers/Lupini/Furstenberg-Katznelson, worked example with Furstenberg-Katznelson, slides about tensors

Definitions: Forestation of a Poset, Semigroup generated by a poset, [Specific to these constructions], $\hat{Y^{\prime\prime}(M)}$

Lecture 1 – Lecture 2

You might also be interested in Solecki’s independent lecture about projective Fraisse limits [link soon].


In the previous lecture we looked at three Ramsey theorems all of the same flavour: Gowers’ theorem, Lupini’s theorem and the Furstenberg-Katznelson theorem. Our goal is to express these three theorems in a common framework. We will do this with the language of semigroups and monoids, introduced in the first lecture.

In this lecture we focus on building a finite structure ($\hat{Y^{\prime\prime} (M)}$) that captures a lot of the information of the action of a monoid on a semigroup. We will construct this object in general, and then see what it looks like for the Furstenberg-Katznelson monoid.

After this we provide slides that explain how to get the full Ramsey behaviour from this object.

Important definitions

Observation. Given an action of $M$ on $\gamma S$, there is a finite structure defined from $M$ that captures a lot of the Ramsey information.

This object will be defined in four steps, by repeatedly taking posets related to the collection of ideals $X(M) = \{m M : m \in M\}$, which itself is a poset (with $\subseteq$).

Definition (Kurepa, 1935) . Let $\mathbb{P} = (P, \leq_P)$ be a (finite) poset. Let $\text{Forest}(\mathbb{P}) = \text{Fr}(\mathbb{P})$ be the poset with point set $\{x \subseteq P : x \neq \emptyset, \text{ and } x \text{ is linearly ordered by } \leq_P\}$ and the order is $x \leq_{\text{Fr}(\mathbb{P})} y$ iff $x \subseteq y$ and $\forall j \in y \setminus x, \forall i \in x, i \leq_P j$.

This is called the forestation of $\mathbb{P}$.

For ease of notation, if $M$ is a monoid, then for what remains we will call $Y(M) = \text{Fr}(X(M))$.

This is end extension. For example, take $\mathbb{P} = (\{0,1,2\}, \leq)$ with its usual order. We’ll use $F=\text{Fr}(\mathbb{P})$ for notational simplicity. Then $\{0\} \leq_F \{0,1\} \leq_F \{0,1,2\}$ and $\{0\} \leq_F \{0,2\} \not\leq_F \{0,1,2\}$.

$\mathbb{P}$ and its forestation.

$\mathbb{P}$ and its forestation.

Exercise. Show that if $M$ acts on $\mathbb{P}$ in an ordering preserving way, then $M$ acts on $\text{Fr}(\mathbb{P})$ in an ordering preserving way.

Show that $\text{Fr}(\mathbb{P})$ does not contain any cycles (so if it is connected it will be a tree). Show that if $\mathbb{P}$ contains a unique minimal element $m$, then the collection $\{x \in \text{Fr}(\mathbb{P}) : m \in x\}$ is a tree.

In general $Y(M)$ is more manageable than $X(M)$.

We now get into a slight annoyance. We’re going to want to refer to the semigroup $\gamma S$, but for notational convenience we only want to use a single letter. So we’re going to fix $T$ a compact left topological semigroup, which will be a $\gamma S$ in practice.

Fact. There is an $I(T)$ a smallest (under $T$) compact two-sided ideal.
Proof. If $I,J \in \mathcal{I}$ the collection of all compact two-sided ideals then $\emptyset \neq I \cdot J \subseteq I \cap J$. (The $\neq$ is by Ellis’ Lemma.) So $I$ has the finite intersection property.
Definition. Consider $E(T)$ the collection of all idempotents of $T$ (which is non-empty by Ellis’ Lemma). Say $x \leq_T y$ iff $xy=yx=x$. This is a partial ordering (idempotent is needed for the relfexive condition).

For in depth analysis of this object, see the 1998 book “Algebra in the Stone-Čech Compactification: Theory and Applications” by Hindman and Strauss or “???” by Todorcevic.

Homomorphism lemma

Theorem (Solecki). Let $M$ be an almost-R-trivial monoid acting by continuous endomorphisms on a left topological compact semigroup $T$. There is an $f: Y(M) \longrightarrow T$ such that

  1. $f$ is $M$-equivariant (i.e. $f(my) = mf(y)$).
  2. $f$ is order-reversing.
  3. $f$ maps maximal elements of $Y(M)$ to $I(T)$, the smallest compact two-sided ideal.

Note that $I(T) \not \subseteq E(T)$, but Ellis’ lemma says $\emptyset \neq I(T) \cap E(T)$. Historically, $f$ is order reversing because of the definition of the idempotent ordering. This order was meant to mimic the behaviour of projections.

Definition. Define $Y^\prime(M)$ to be the collection of all $y \in Y(M)$ such that

  • $\min y$ is a minimal element of $X(M)$.
  • $y = \{i_0 <_{X(M)} i_1 <_{X(M)} \ldots <_{X(M)} i_k \}$, where $i_j$ is an ($X(M)$) immediate predecessor of $i_{j+1}$.

That is, the $Y^\prime(M)$ is initial segments of $X(M)$.

Interestingly, in all the cases studied so far $Y^\prime(M)$ is $M$-invariant. The guess is that this is not a general phenomenon because if it was, a simple proof should exist.

The Furstenberg-Katznelson monoid

We’ve already seen that $X(M)$ is the partial order below. The complete forest of $Y(M)$ is unruly (Exercise; see example computation above), but we will skip right to $Y^\prime (M)$.

This gives the following poset $Y^\prime(M)$ (assuming the minimal element of $X(M)$ is in each element of $Y^\prime (M)$).

We pick one of these copies of the identity to keep (and call $1_\text{top}$ and discard the rest. This gives us $Y^{\prime\prime}(M)$.

(Note that this definition of $Y^{\prime\prime}(M)$ is specific to this case. In other settings there will be other non-canonical choices for this object. It seems like in other settings we merely choose one of the copies of $1_M$ to keep and discard the rest. Different choices will yield different objects and hence different Ramsey results.)

$X(M)$, $Y^\prime (M)$ and $Y^{\prime\prime}(M)$.

Note that $M$ acts on $W(M)$ by multiplication on the left. Multiplying by $a \in A$ is like replacing all other elements of $A$ with this $a$ and fixing all elements of $B$.

Let $I$ be all words in $w \in W(M)$ such that $1_M$ occurs in $w$. Note $I \subseteq S$. Also note that $I$ is a two-sided ideal (exercise). Let $\gamma I = \{\mathcal{U} \in \gamma S : I \in \mathcal{U}\}$. This is a compact two-sided ideal of $\gamma S$, so $\gamma I \supseteq I(\gamma S)$.

So there is an $f: Y^{\prime\prime}(M) \longrightarrow E(\gamma S)$ such that $f(1_\text{top}) \in \gamma I$, that is $I \in f(1_\text{top})$. Also $f$ is $M$-equivariant and order reversing.

Using $Y^{\prime\prime}(M)$ to find an invariant

We’re almost at our final finite object that will be used to identify a Ramsey invariant. We describe a way to go from a poset to a semigroup.

Definition. Let $\mathbb{P} = (P, \leq_P)$ be a poset. The semigroup $\hat{\mathbb{P}}$ generated by $\mathbb{P}$ is the free semigroup on $P$ with the relations $pq=qp=q$ if $p \leq_P q$.

Let’s say we’re given an $M$-equivariant semigroup homomorphism $f: \hat{Y^{\prime\prime}(M)} \longrightarrow E(\gamma S)$. What happens to words from $\hat{Y^{\prime\prime}(M)}$?

Runs of a single $a \in A$ disappear ($aa=aa=a$), runs of a single $b \in B$ disappear. If a $b$ is next to an $a$, the $b$ disappears. So $f(w) = \overline{w}$ as in the statement of the Furstenberg-Katznelson theorem. You can extract the invariant from this homomorphism.

This gives us a way to find invariants, but the Ramsey part is still missing. Where does the sequence of words come from in the statement of these theorems?

Looking at $E(\gamma S)$ will not be enough to give us all invariant types. In the language of the Furstenberg-Katznelson theorem: just looking at the one homomorphism will only give us some (not all) finite sets $F$ (as in the theorem).

To fix this we go “up a category” and introduce tensor products.

The slides

From here we move to slides. (I’ve tried to keep some helpful notes here. Slide X will be replaced by the appropriate slide number when I get the slides. -Mike)

The slides are available here [PDF – 48 Slides].

Slide 4. A $\Lambda$-algebra is just a list of functions from $X$ to $S$.

Slide 5. Such a homomorphism is “exactly what it should be”. If $X = \{\bullet\}$ is a single point, then this is just a rooting of $Y$.

Slides 7-8. $\{x \in X : s \lambda(x) \text{ is defined }\} \in \mathcal{V}$. This reflects our earlier definition replacing $t$ with $\lambda(x)$.

Slide 9. $X \longrightarrow^{\Lambda} S$ goes to $\gamma X \longrightarrow^{\Lambda} \gamma S$.

Slide 11. Tameness is like the $\overline{w}$ type. (At first this isn’t general enogh to capture all $F$.)

Slide 12. $f(\bullet)$ is an ultrafilter. So there will be lots of choices for a basic sequence.

Slide 14. $\Lambda(\bullet)$ is not enough generality, so we take tensor products to get more.

Slide 20. $\Lambda_{<r}(\bullet)$ is what we want. (We skipped over a lot in between.)


Repeat the construction of $\hat{Y^{\prime\prime}(M)}$ for $I_4$ in Lupini’s example. Explicitly construct all relevant objects (except for with $Y^\prime(M)$ just do it until you give up). Get to the point where you need a homomorphism from $\hat{Y^{\prime\prime}(M)}$ to $\gamma S$.

At the point where you take $\hat{\bullet}$ you will need to make a choice of linear suborder of $Y^{\prime\prime}(M)$. Different choices will yield different objects and different Ramsey results. You will also need to define the second $\prime$ in $Y^{\prime\prime}(M)$ using your brain; we did not tell you how to do this in general.

This entry was posted in Course Notes, Ramsey DocCourse Prague 2016 and tagged , , , , , , , , , . Bookmark the permalink. Both comments and trackbacks are currently closed.

One Trackback