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],
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 () 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.
This object will be defined in four steps, by repeatedly taking posets related to the collection of ideals , which itself is a poset (with ).
Definition (Kurepa, 1935) . Let be a (finite) poset. Let be the poset with point set and the order is iff and .
This is called the forestation of .
For ease of notation, if is a monoid, then for what remains we will call .
This is end extension. For example, take with its usual order. We’ll use for notational simplicity. Then and .
Exercise. Show that if acts on in an ordering preserving way, then acts on in an ordering preserving way.
Show that does not contain any cycles (so if it is connected it will be a tree). Show that if contains a unique minimal element , then the collection is a tree.
In general is more manageable than .
We now get into a slight annoyance. We’re going to want to refer to the semigroup , but for notational convenience we only want to use a single letter. So we’re going to fix a compact left topological semigroup, which will be a in practice.
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.
Theorem (Solecki). Let be an almost-R-trivial monoid acting by continuous endomorphisms on a left topological compact semigroup . There is an such that
- is -equivariant (i.e. ).
- is order-reversing.
- maps maximal elements of to , the smallest compact two-sided ideal.
Note that , but Ellis’ lemma says . Historically, is order reversing because of the definition of the idempotent ordering. This order was meant to mimic the behaviour of projections.
Definition. Define to be the collection of all such that
- is a minimal element of .
- , where is an () immediate predecessor of .
That is, the is initial segments of .
Interestingly, in all the cases studied so far is -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 is the partial order below. The complete forest of is unruly (Exercise; see example computation above), but we will skip right to .
This gives the following poset (assuming the minimal element of is in each element of ).
We pick one of these copies of the identity to keep (and call and discard the rest. This gives us .
(Note that this definition of 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 to keep and discard the rest. Different choices will yield different objects and hence different Ramsey results.)
Note that acts on by multiplication on the left. Multiplying by is like replacing all other elements of with this and fixing all elements of .
Let be all words in such that occurs in . Note . Also note that is a two-sided ideal (exercise). Let . This is a compact two-sided ideal of , so .
So there is an such that , that is . Also is -equivariant and order reversing.
Using 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.
Let’s say we’re given an -equivariant semigroup homomorphism . What happens to words from ?
Runs of a single disappear (), runs of a single disappear. If a is next to an , the disappears. So as in the statement of the Furstenberg-Katznelson theorem. You can extract the invariant from this homomorphism.
Looking at 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 (as in the theorem).
To fix this we go “up a category” and introduce tensor products.
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)
Slide 4. A -algebra is just a list of functions from to .
Slide 5. Such a homomorphism is “exactly what it should be”. If is a single point, then this is just a rooting of .
Slides 7-8. . This reflects our earlier definition replacing with .
Slide 9. goes to .
Slide 11. Tameness is like the type. (At first this isn’t general enogh to capture all .)
Slide 12. is an ultrafilter. So there will be lots of choices for a basic sequence.
Slide 14. is not enough generality, so we take tensor products to get more.
Slide 20. is what we want. (We skipped over a lot in between.)
Repeat the construction of for in Lupini’s example. Explicitly construct all relevant objects (except for with just do it until you give up). Get to the point where you need a homomorphism from to .
At the point where you take you will need to make a choice of linear suborder of . Different choices will yield different objects and different Ramsey results. You will also need to define the second in using your brain; we did not tell you how to do this in general.