# 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 $G$ equipped with a total order $\leq$ which is compatible with (left) multiplication in the following sense:

\(g_1 \leq g_2 \quad \implies \quad hg_1 \leq hg_2\)

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 $\text{Homeo}_+(\mathbb{R})$, the group of orientation preserving homeomorphisms of $\mathbb{R}$.

The order on $\text{Homeo}_+(\mathbb{R})$ is as follows:

Enumerate \(\mathbb{Q} = \{q_n\}\). Then we say $f \lt g$ exactly when $f q_i \lt g q_i$, where $i$ is least with $f q_i \neq g q_i$ (this is more or less the lex order on $\prod_{\omega} \mathbb{R}$).

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 $\lt \mathfrak{c}$? Do they all have to embed in $\text{Homeo}_+(\mathbb{R})$ as well?

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 $G$! This is a summary of a proof from Clay and Rolfsen’s
“Ordered Groups and Topology” (see here), Theorem 2.23.

$\ulcorner$ Let $G$ be countable and LO. Then by looking at $G \times \mathbb{Q}$ with the lex order, we can assume $G$’s ordering is dense. Moreover, it is easy to see that $G$ is torsion free, so for any element $g$, there is always some $g_L \lt g$ and some $g_R \gt g$ ($g^{-1}$ works for one, and $g^2$ for the other).

So $G$ is a countable dense linear order without endpoints! If you’re a
logician your heart should be leaping now. Cantor’s famed
back and forth argument shows that *any* such ordering is isomorphic
(as an order) to $(\mathbb{Q}, \lt)$. It was really exciting to see this
familiar face pop up in this proof! But since $(G, \lt) \cong (\mathbb{Q}, \lt)$
embeds densely in $(\mathbb{R}, \lt)$, we can extend the left action of $G$ on
itself to a homeomorphism of $\mathbb{R}$.
$\lrcorner$

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 $\kappa$ there are $2^\kappa$ nonisomorphic
dense linear orders without endpoints of cardinality $\kappa$
(see here, for instance). This made me start wondering if the theorem is
actually *false* for groups of size, say, $\aleph_1$.

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 $\mathbb{R}$ doesn’t contain any chains of length $\omega_1$. That is, there’s no monotone function $f : \omega_1 \to \mathbb{R}$.

I thought if we could find a LO group $G$ with some $\omega_1$ chain, then we should be done. My thought process was baiscally:

- If \(G \hookrightarrow \text{Homeo}_+(\mathbb{R})\), then $G \curvearrowright \mathbb{R}$.
- If \(\{ g_\alpha \}_{\omega_1}\) is a chain in $G$, then \(\{ g_\alpha x \}_{\omega_1}\) should be a chain in $\mathbb{R}$ once we pick some initial value $x$.

Of course, this turned out to be wrong too. It’s not hard to find homeomorphisms $f \lt g$ where $fx \not \lt gx$. It was a good start, though, on the way to the right answer.

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 \(\omega_1\) many ~ bonus constants ~ \(x_\alpha\).
- Look at the theory which includes the sentences
- “I am an ordered group”
- ”\(x_\alpha \lt x_\beta\)” for each $\alpha \lt \beta$.

- Now each finite subtheory only refers to finitely many constants, so $\mathbb{Z}$ is a model.
- Then compactness buys us a model of the whole theory – an ordered group with a chain of length $\omega_1$.
- Now by Lowenheim-Skolem, we look at the elementary submodel containing this chain. This is also an ordered group with a chain of length $\omega_1$, but it’s guaranteed to have cardinality $\aleph_1$.

So we’ve successfully found a LO group of size $\aleph_1$ which contains an increasing chain of length $\omega_1$… But didn’t we say this doesn’t actually solve our problem?

This is where I remembered a fact from Descriptive Set Theory: For a
compact metric space $X$, we actually know that \(\text{Homeo}(X)\) is
polish (see Kechris’s “Classical Descriptive Set Theory”, I.9B, example 8).
There’s a classic argument that $\mathbb{R}$ doesn’t contain any chains of
length $\omega_1$ which seems to only use separability^{1}, and the
dream would be to show this continues to hold in a general polish space.

Of course, we *also* need to check that \(\text{Homeo}_+(\mathbb{R})\) is
actually polish. The above theorem only guarantees polishness for *compact*
spaces $X$, and the reals are (among other things) not compact.

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 want^{2}, but it’s phrased
in terms of subsets of $\mathbb{R}$… But now we know what to search for,
and we quickly find a mse question which cites the paper and makes me
feel confident that I’m not misinterpreting it.

All that’s left is to show $\text{Homeo}_+(\mathbb{R})$ is really polish, but our journey ends like it began, on mse.

As a nice exercise, can you show that the order on $\text{Homeo}_+(\mathbb{R})$ really is borel? That is, can you show

\[\{ (f,g) ~|~ f \leq g \} \subseteq \text{Homeo}_+(\mathbb{R}) \times \text{Homeo}_+(\mathbb{R})\]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 ($\lnot \mathsf{CH}$):

There exists a LO group of size $\aleph_1$ which does not embed in $\text{Homeo}_+(\mathbb{R})$

$\ulcorner$ Since $(\mathbb{Z}, \lt)$ is an infinite LO group, a standard logical compactness argument furnishes an LO group of size $\aleph_1$ which contains an increasing sequence \(\{g_\alpha\}\) of length $\omega_1$. Call such a group $G$.

Then since \(\text{Homeo}_+(\mathbb{R})\) is a polish space (cf. here) and its ordering is borel, a theorem of Shelah and Harrington (cf. Corollary 3.2 here) shows that no chain of length $\omega_1$ can exist in $\text{Homeo}_+(\mathbb{R})$.

Since $G$ contains such a chain, it cannot embed into $\text{Homeo}_+(\mathbb{R})$. $\lrcorner$

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.