Vtyjkyuk

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. $gin G$ is positive if $g>0$.


2. The absolute value of $g$ is $|g|=maxg,-g$.


3. If $g,hin G$ then $gsim h$ 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 $[g],[h]inGamma(G)$ then $[g]preceq[h]$ if and only if $|g|leq|h|$ or $gsim h$.


5. If $gin G$ then the positive part of the Archimedean class of $g$ is the set $hin Gmid hsim gland h>0$.


6. If $G$ is an ordered group, an Archimedean extension is a group $K$ such that $Gleq K$, 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 $xin G^*$ has some $n$ such that $nxin G$ (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?

These are my answers:

1. An Archimedean class is by definition an Archimedean ordered subgroup of $G$, and every Archimedean ordered group is isomorphic to a subgroup of $(mathbbR,+)$.


2. no apply.


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 $$mathbbR((Gamma(G))):=ginmathbbR^Gamma(G):qinGamma(G):g(q)neq0mbox is well-ordered.$$ On the other side, $mathbbR((Gamma(G)))$ is a $mathbbR$-vector space (actually it is a field extension of $mathbbR$ when $Gamma(G)$ is an ordered group). Thus, whenever $G$ is Archimedean complete, it is isomorphic to $mathbbR((Gamma(G)))$, and thus the scalar multiplication can be defined through isomorphism.


4. No and No. A counterexample is
   $$G:=gin mathbbR^mathbbQ:qinmathbbQ:g(q)neq0cap(-infty,n]mbox is finite for all ninmathbbZ.$$
   Here $G$ is a $mathbbR$-vector space, each Archimedean class is isomorphic to $mathbbR$ and $Gamma(G)=mathbbQ$, but $mathbbR((mathbbQ))$ is a proper Archimedean immediate extension of $G$, actually it is the Archimedean completion of $G$.