What do linearly ordered abelian groups look like?

2.7k Views Asked by At

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?