Linearly Ordered Groups and CH
26 Feb 2021 - Tags: featured
Earlier today Jonathan Alcaraz gave a GSS talk about Linearly Ordered (LO) Groups, which are a fun topic with connections to dynamics, topology, geometric group theory, etc. This reminded me of a problem I told myself to think about a while ago, and so I decided to finally do that. After a bit of thought, a friend from CMU (Pedro Marun) and I were able to figure it out. This post is going to be somewhat more meandering than usual (if you can imagine such a thing), because I want to showcase what the flow of thoughts was in solving the problem. At the end I’ll clean things up and write them linearly.
I guess we should start with what a LO Group is, but it’s pretty much what it says on the tin:
A (Left) Linearly Ordered Group is a
group
I first heard of LO Groups from an exercise in Marker’s “Model Theory: An Introduction”, where an exercise has you use compactness to show every torsion free abelian group admits a compatible linear order. I heard about them again on mse, to the surprise of nobody. Somebody asked for examples of finitely generated left orderable groups. I knew about the abelian example because of Marker, but I was curious about nonabelian examples.
This led me down a rabbit hole of papers to skim, including Katheryn Mann’s “Left Orderable Groups that Don’t Act on the Line” (see here). This paper mentions a classical result:
A countable group is LO if and only if it embeds in
The order on
Enumerate
As soon as I saw this, I wondered if anything was special about “countable” here.
If we assume the Continuum Hypothesis (CH) fails, what can we say about other
LO groups of size
I keep a list of “problems to think about”, so I added this and left some brief thoughts before going back to answering mse questions.
When Jonathan brought up this theorem in his talk, it reminded me to think
about that problem. I found a proof of the result to see if it immediately
worked for larger cardinalities, and much to my surprise it relies heavily
on the countability of
So
This theorem relies on a back and forth argument for most of the heavy lifting,
and that argument fails spectacularly for uncountable cardinalities.
In fact, for any uncountable
I texted Pedro, a close friend and set theorist, with some ideas that he
pretty quickly found flaws in. He had a good idea, though, and reminded me
that
I thought if we could find a LO group
- If
, then . - If
is a chain in , then should be a chain in once we pick some initial value .
Of course, this turned out to be wrong too. It’s not hard to find homeomorphisms
If nothing else, we should just build such a group to show we know how, right? This is a simple compactness argument:
- Look at the language of ordered groups, but add
many ~ bonus constants ~ . - Look at the theory which includes the sentences
- “I am an ordered group”
- ”
” for each .
- Now each finite subtheory only refers to finitely many constants, so
is a model. - Then compactness buys us a model of the whole theory – an ordered group with a chain of length
. - Now by Lowenheim-Skolem, we look at the elementary submodel containing this chain.
This is also an ordered group with a chain of length
, but it’s guaranteed to have cardinality .
So we’ve successfully found a LO group of size
This is where I remembered a fact from Descriptive Set Theory: For a
compact metric space
Of course, we also need to check that
First, I searched for “borel ordering” in Kechris’s book, and found a
reference to Harrington, Marker, and Shelah’s “Borel Orderings”
(see here). Corollary 3.2 gives exactly what we want2, but it’s phrased
in terms of subsets of
All that’s left is to show
As a nice exercise, can you show that the order on
is a borel subset?
Ok. Now that the exposition is out of the way, we’re holding a draft of a proof in our heads. It was a wandering path, but look how deceptively simple it looks once we organize our thoughts and write it down:
Theorem (
There exists a LO group of size
Then since
Since
Can you believe that teeny little proof took hours of reading and thinking (times two people, no less!) to figure out? It really makes you appreciate how much work goes into some of the longer and tricker theorems you come across.