I am interested in realizing linear orders as orderings of abelian groups. In particular, can Suslin lines (and other classes of line) be realised in this way?


Let $\mathcal{C}$ be a class of (torsion-free) abelian groups, and $L$ be the first-order language of linear orderings. Let $LO(\mathcal{C})$ be the theory of ordered abelian groups belonging to the class $\mathcal{C}$. Recall an abelian group $A$ is ordered by the binary relation $<$ if $<$ is a linear order on $A$ such that whenever $g < h$ then $g+k < h+k$ for all $g, h, k \in A$. 


Let $\mathcal{K}\_{LO(\mathcal{C})} = \lbrace A \mid L : A \models LO(\mathcal{C}) \rbrace$ be the class of all linear orderings which are orderings of members in $\mathcal{C}$. 


Suppose $\mathcal{C}$ is the class of all abelian groups. So $\mathcal{K}\_{LO(\mathcal{C})}$ is a pseudo-elementary class. From A.C. Morel's work (\textit{Structure and order structure in Abelian groups}, Colloq. Math. 19 (1968), 199-209), the $\aleph\_{1}$ countable members of $\mathcal{K}\_{LO(\mathcal{C})}$ are known up to isomorphism.


What uncountable lines can turn up in $\mathcal{K}\_{LO(\mathcal{C})}$?


Is it consistent that $\mathcal{K}\_{LO(\mathcal{C})}$ contains a Suslin line (or some but not all Suslin lines, or all Suslin lines if these exist)? If one adds a Cohen real, does $\mathcal{K}\_{LO(\mathcal{C})}$ contain a Suslin line?


Does $\mathcal{K}\_{LO(\mathcal{C})}$ contain an Aronszajn line, i.e. a line obtained by squashing an Aronszajn tree?


Thank you for any help.



Can Suslin (or Aronszajn) lines ever be orderings of abelian groups?
Let me give a simple direct proof that there is no ordered abelian group $\langle S,+,<\rangle$ whose order type is a Suslin line (Dedekind-complete, densely ordered, and without a countable dense subset; the argument doesn’t need ccc).


Since $S$ is nontrivial and densely ordered, we can find a strictly decreasing sequence of positive elements $\{a\_n:n\in\omega\}$. By subtracting $\inf\_na\_n$ if necessary, we can ensure $\inf\_na\_n=0$.
Let $G$ be the countable subgroup of $S$ generated by $\{a\_n:n\in\omega\}$. 


I claim that $G$ is dense in $S$, hence $S$ is not a Suslin line after all. If $(u,v)$ were an interval disjoint from $G$, there are elements of $G$ below $u$ or above $v$. W.l.o.g. assume the former, and put $u'=\sup\{a\in G:a\le u\}$. By construction, there exists $a\_n\in G$ such that $0<a\_n<v-u$. If $a\in G$ is such that $a\le u$, then $a+a\_n<v$, hence $a+a\_n\le u$ by assumption, hence $a+a\_n\le u'$. It follows that $u'+a\_n\le u'$, a contradiction.




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On the other hand,



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> **Proposition:** There exists a divisible ordered abelian group whose order type is an Aronszajn line (i.e., it has cardinality $\aleph\_1$, but contains no copy of $\omega\_1$, $\omega\_1^\*$, or of an uncountable real type).
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**Proof:** Let $L$ be any Aronszajn line, and let $G$ be the lexicographically ordered direct sum $\bigoplus\_{a\in L}\mathbb Q$. We need to check that the order of $G$ is still Aronszajn. Let $P\subseteq G$ be an uncountable subset isomorphic to one of the offending order types. Since $\aleph\_1$ is regular and every element of $G$ has finite support, we may assume that all elements of $P$ have support of the same size $n\in\omega$.


Let $P'=\{\max(\mathrm{supp}(x)):x\in P\}\subseteq L$, and for any $q\in\mathbb Q$, let $P'\_q$ be its subset consisting of those $a=\max(\mathrm{supp}(x))$ where the $a$th coefficient of $x$ is $q$. If $P'$ is uncountable, then also $P'\_q$ is uncountable for some $q\in\mathbb Q$; but then $P$ contains an isomorphic copy of $P'\_q$ or $P'^\*\_q$ (depending on the sign of $q$), hence $L$ is not Aronszajn, a contradiction.


On the other hand, if $P'$ is countable, there is $a\in P'$ and $q\in\mathbb Q$ such that the set of $x\in P$ with $\max(\mathrm{supp}(x))=a$ and the corresponding coefficient $q$, is uncountable. We may replace $P$ with this subset, and kill the leading monomials (which is an order-preserving shift), hence we reduce $n$ to $n-1$. Continuing in this fashion, we can reduce $n$ to $0$, which is a contradiction. QED


Note that being a divisible torsion-free group, $G$ is a $\mathbb Q$-linear space. We can in fact get even more structure:



> 
> **Corollary:** There exists an ordered integral domain and $\mathbb Q$-algebra whose order type is an Aronszajn line.
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**Proof:** Repeat the construction $G=\bigoplus\_{a\in L}\mathbb Q$ with $L$ being an Aronszajn abelian group, which we have just shown to exist. Then $G$ is not just an ordered $\mathbb Q$-linear space, but it also carries a natural structure of an ordered ring, namely the group ring $\mathbb Q[L]$. (Notice that since every finitely generated subgroup of $L$ is free abelian, $\mathbb Q[L]$ as a ring is isomorphic to a direct limit of multivariate Laurent polynomial rings over $\mathbb Q$.)




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As Joel mentioned in the comments, it is unfortunate to base the argument for Suslin lines solely on the rather accidental Dedekind completeness property. This can be circumvented by a different argument:



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> **Proposition:** Every ccc ordered abelian group has a countable dense subset.
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**Proof:** Let $G$ be such a group. The main property we need is the following:



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> **Lemma.** If $H$ is a nontrivial convex subgroup of $G$, then $G/H$ is countable.
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**Proof:** Fix $0<h\in H$, and let $\{g\_\alpha:\alpha<\omega\_1\}\subseteq G$ be representatives for uncountably many cosets of $H$. Then $\{(g\_\alpha-h,g\_\alpha+h):\alpha<\omega\_1\}$ are uncountably many disjoint nonempty intervals. QED


The ccc also implies that $G$ contains no strictly decreasing $\omega\_1$-sequence, hence $G^{>0}$ has (a least element or) countable downward cofinality. Let us fix a sequence $h\_0\ge h\_1\ge h\_2\ge\cdots$ of positive elements cofinal in $G^{>0}$, and let $H\_n$ be the smallest convex subgroup of $G$ containing $h\_n$, so that $H\_0\supseteq H\_1\supseteq H\_2\supseteq\cdots$.


**Case 1:** The $H\_n$ sequence is eventually constant. Let $H$ be its limit (which is the least archimedean class of $G$). Then $H$ is archimedean, hence embeddable in $\mathbb R$, hence it has a countable dense subset $C$. (In fact, one can show directly that the subgroup generated by $\{h\_n:n\in\omega\}\cap H$ is dense in $H$.) Also, $G/H$ is countable by the Lemma, hence we can find a countable set $D$ of coset representatives for $H$. Then $C+D$ is a countable dense subset of $G$: if $(g,g')$ is a nonempty interval, there is $d\in D$ such that $g-d\in H$; either $g-d<g'-d\in H$ as well, or $g'-d>H$. Either way, there is $c\in C$ such that $g-d<c<g'-d$, hence $c+d\in(C+D)\cap(g,g')$.


**Case 2:** The $H\_n$ sequence has no least element, which also implies that $G$ is densely ordered. Using the Lemma, let $D$ be a countable set which contains coset representatives for each $H\_n$. Then $D$ is dense in $G$: if $g<g'$, find $g<g''<g'$, and $n\in\omega$ such that $g''-g,g'-g''>H\_n$. There is $d\in D$ such that $g''-d\in H\_n$, which implies $g<d<g'$.


