What do linearly ordered abelian groups look like?

Recently a post on MathOverflow connected a theorem about ordered abelian groups to the axiom of choice, and it made me want to try and play with these objects a bit.

But it appears to me that I don't know enough about linearly ordered abelian groups. In particular I am interested in the structure of their Archimedean classes. Let's establish some ad-hoc terminology first. Let $(G,+,\leq)$ be a linearly ordered abelian group.

1. $g\in G$ is positive if $g>0$.
2. The absolute value of $g$ is $|g|=\max\{g,-g\}$.
3. If $g,h\in G$ then $g\sim h$ if and only if $|h|\leq|g|$ and for some natural number $n$, $|g|\leq n|h|$; or $|g|\leq|h|$ and for some natural number $n$, $|h|\leq n|g|$.
4. An Archimedean class of $G$ is an equivalence class of $\sim$. $\Gamma(G)$ is the set of equivalence classes, and $\preceq$ is the natural ordered induced from $\leq$. If $[g],[h]\in\Gamma(G)$ then $[g]\preceq[h]$ if and only if $|g|\leq|h|$ or $g\sim h$.
5. If $g\in G$ then the positive part of the Archimedean class of $g$ is the set $\{h\in G\mid h\sim g\land h>0\}$.
6. If $G$ is an ordered group, an Archimedean extension is a group $K$ such that $G\leq K$, and $\Gamma(G)=\Gamma(K)$. If $G$ has no [nontrivial] Archimedean extensions we say that $G$ is Archimedean complete.

In "Ordered Abelian Groups", Gravett points out that given an ordered group $G$, we can replace it with a "rational group" $G^*$ such that every $x\in G^*$ has some $n$ such that $nx\in G$ (I believe this is "divisible group" in modern terms, and $G^*$ is the injective hull of $G$). It is quite clear that this rational group is a vector space over $\Bbb Q$, and that it is an Archimedean extension of $G$.

So we can assume that an ordered group is a linearly ordered vector space over the rationals.

My questions are:

1. What do Archimedean classes of an ordered group look like? Are they isomorphic to subgroups of $\Bbb R$ (or the positive parts, to subgroups of $\Bbb R^+$)?

2. If not, what sort of counterexamples can we find?

3. Gravett proves the Hahn completeness theorem which states that a group $G$ is Archimedean complete if and only if it is isomorphic to its Hahn group (which is $\Bbb R^{\Gamma(G)}$ where the exponentiation is similar to ordinal exponentiation). Can we conclude in that case that $G$ is actually an ordered vector space over $\Bbb R$?

4. Can we prove that $G$ is Archimedean complete if and only if it is a vector space over $\Bbb R$? Does this mean that $G$ is Archimedean complete if and only if each Archimedean class is isomorphic to $\Bbb R$ (or the positive part to the positive reals), or is this a separate fact that we can prove?

Solution 1:

These are my answers (edited after @Shane O Rourke's answer):

1. No (as is indicated in @Shane O Rourke's answer).
2. counterexample in @Shane O Rourke's answer.
3. Yes. On one side, the Hahn's Embedding theorem states that every abelian ordered group $G$ can be embedded in its Archimedean completion defined by $$\mathbb{R}((\Gamma(G))):=\{g\in\mathbb{R}^{\Gamma(G)}:\{q\in\Gamma(G):g(q)\neq0\}\mbox{ is well-ordered}\}.$$ On the other side, $\mathbb{R}((\Gamma(G)))$ is a $\mathbb{R}$-vector space (actually it is a field extension of $\mathbb{R}$ when $\Gamma(G)$ is an ordered group). Thus, whenever $G$ is Archimedean complete, it is isomorphic to $\mathbb{R}((\Gamma(G)))$, and thus the scalar multiplication can be defined through isomorphism.
4. No and No. A counterexample is $$G:=\{g\in \mathbb{R}^\mathbb{Q}:\{q\in\mathbb{Q}:g(q)\neq0\}\cap(-\infty,n]\mbox{ is finite for all }n\in\mathbb{Z}\}.$$ Here $G$ is a $\mathbb{R}$-vector space, each Archimedean class is isomorphic to $\mathbb{R}$ and $\Gamma(G)=\mathbb{Q}$, but $\mathbb{R}((\mathbb{Q}))$ is a proper Archimedean immediate extension of $G$, actually, it is the Archimedean completion of $G$.

Solution 2:

An earlier answer given by Chilote used a different definition of archimedean class to the one given in the question, and this impacted on the answers to the questions. (It's been corrected now.)

In the sense of the question, an archimedean class consists of those $g$ that span a common convex subgroup of $G$. Here a subgroup $H$ of $G$ is convex if $$\left(a,c\in H \mbox{ and }a\leq b\leq c\right) \Rightarrow\ b\in H.$$ So for example if $G$ is the non-archimedean ordered abelian group $\mathbb{R}\times\mathbb{R}$ with the (left to right) lexicographic order (so $(-4,9)<(2,0)<(3,-8)<(3,0)$), then any pair of elements of $G$ with non-zero first entry are equivalent and span $G$ as a convex subgroup. However the classes $[g]$ do not form a subgroup, since for example $(3,2)\sim(-3,1)$ and their sum is $(0,3)\prec(3,2)$. And indeed, typically $0\notin[g]$.

For what it's worth, one does get naturally defined archimedean groups from this set-up by taking the convex subgroup spanned by $g$ and taking the quotient by the largest convex subgroup not containing $g$. Different $h$ that are equivalent to $g$ will give rise to the same archimedean group.

So the answers to (1) and (2) are no. I agree that the answer to (3) is yes and to (4) is no (with the sense of archimedean class as in the question). Chilote has given an example for this; another is $K=\mathbb{R}^{\mathbb{N}}$, and the proper subgroup $G$ consisting of those elements with finite support. Here $G$ and $K$ have `the same' archimedean classes, so $G$ is not archimedean complete, even though it is an $\mathbb{R}$-vector space.

By the way I've used inverted commas here since this is slight abuse of notation, as there is in the equation $\Gamma(G)=\Gamma(K)$. For $G\leq K$ this should be taken to mean that every archimedean class in $K$ is represented by an element of $G$.