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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.

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  • $\begingroup$ Fantastic question! $\endgroup$ – Joel David Hamkins Aug 28 '14 at 23:00
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    $\begingroup$ Suppose you start with a large saturated ordered abelian group, whose order type was universal, and then take the subgroup generated by a Suslin suborder. Is it still Suslin? $\endgroup$ – Joel David Hamkins Aug 29 '14 at 0:38
  • $\begingroup$ If $G$ is the large saturated ordered abelian group, could you pin down why its ordering will be universal for linear orders, in particular, why will it embed a Suslin line? I had tried with Ehrenfeucht-Mostowski models but worried that the subgroup generated by the Suslin line would add new points to the Suslin line and destroy the Suslin property. Maybe your proposal gets around the problem. $\endgroup$ – Avshalom Aug 29 '14 at 1:53
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    $\begingroup$ If the structure is countably saturated, then the order will be universal for linear orders of size $\omega_1$, as a consequence (think back-and-forth style, and write down the type of realizing the next element, which can surely be done in an ordered abelian group). So the Suslin line will order-embed. My question is whether the subgroup generated by this suborder remains Suslin. $\endgroup$ – Joel David Hamkins Aug 29 '14 at 1:55
  • $\begingroup$ Dear Joel, thank you very much for the new comment. Let $S$ be the Suslin line. To be clear, your claims are: (A): the order $<_{G}$ of $G$ is universal for orders of size $\aleph_{1}$. (B): the subgroup generated by $S$ gives rise to a Suslin line, maybe not $S$. By (A), the order-type $\zeta(1+\eta)$ embeds in $<_{G}$, where $\zeta$ and $\eta$ are the integer and rational order-types. But $\zeta(1+\eta)$ is not the ordering of any abelian group (by Morel's classification). So (B) is risky if $S$ contains $\zeta(1+\eta)$ or similar, unless maybe $S$ is decomposed very smoothly (under CH). $\endgroup$ – Avshalom Aug 29 '14 at 11:37
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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.


On the other hand,

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).

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.

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 $\mathbb Q$.)


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:

Proposition: Every ccc ordered abelian group has a countable dense subset.

Proof: Let $G$ be such a group. The main property we need is the following:

Lemma. If $H$ is a nontrivial convex subgroup of $G$, then $G/H$ is countable.

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 $g<d<g'$.

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    $\begingroup$ This is very clear. I notice that this and the other arguments all use completeness in a strong way. Can you say what happens if we drop completeness, but assume c.c.c.? That is, the order would have a Suslin line as its completion, but it would not be complete itself. In other words, can the Dedekind-completion of an ordered abelian group be a Suslin line? $\endgroup$ – Joel David Hamkins Sep 7 '14 at 17:23
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    $\begingroup$ @Joel: I believe this is indeed a more natural question to ask, and I’m pondering it at the moment. I think the answer is still no, but I have yet to finish the argument. $\endgroup$ – Emil Jeřábek supports Monica Sep 7 '14 at 17:25
  • $\begingroup$ Thank you, Emil; these are genially transparent arguments. I would like also to extend to modules over commutative rings; it seemed better to start with the special tighter questions. $\endgroup$ – Avshalom Sep 7 '14 at 20:09
  • $\begingroup$ You’re welcome. The Aronszajn group construction was already a module (vector space), but I’ve expanded it to make it a ring as well. $\endgroup$ – Emil Jeřábek supports Monica Sep 7 '14 at 22:36
  • $\begingroup$ Let me see. The proof shows that $\mathbb Q^{(L)}$ is Aronszajn iff $L$ is, and I believe a simple EF argument shows that if $L\simeq L'$, then $\mathbb Q^{(L)}\simeq\mathbb Q^{(L')}$. So, this means that ordered abelian groups with Aronszajn order are $L_{\infty\omega}$-axiomatizable iff Aronszajn orders as such are $L_{\infty\omega}$-axiomatizable, of which I know nothing about, but I take your question to imply that it is false. $\endgroup$ – Emil Jeřábek supports Monica Sep 8 '14 at 14:40
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Edit: As Emil Jeřábek has pointed out in the comments, there is a stronger version of the proposition available that gives a more direct proof.

Theorem. If $(S, +, <)$ is an ordered abelian group whose order type is dense and complete, then $S$ is isomorphic to $(\mathbb{R}, +, <)$.

It follows immediately that no ordered group can be Suslin. Emil credits this to Hölder, and gives a proof of the theorem in his answer here: mathoverflow.net/a/179520/12705

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I worked through this problem with Geoff Galgon. For Suslin lines at least the answer is negative: no such line can be the order type of an ordered abelian group. A key point is the following:

Proposition. If $(S, +, <)$ is an ordered abelian group whose order type is dense and complete, then there is an embedding of $(\mathbb{R}, +, <)$ into $S$.

From this it follows that no such $S$ can be Suslin. For now we'll delay the proof of the proposition and give the remainder of the argument.

Suppose toward a contradiction that $(S, +, <)$ is a linearly ordered abelian group whose order type is Suslin. For every $c \in S$, the map $x \mapsto x+c$ is an order-automorphism of $S$. It follows that no interval $(a, b)$ of $S$ can be separable (where we allow $a=-\infty$ or $b=\infty$, though in those cases we need to modify the following slightly). If there were such an interval, then by translating to the left and right in increments of $b-a$ and using the fact that $S$ is ccc, we would be able to cover $S$ (minus some endpoints) with countably many copies of $(a, b)$ (though we may need more than $\omega$-many in each direction: if, say, $\omega$ copies of $(a,b)$ does not cover $S$ to the right, translate $(a,b)$ so that $a$ sits at the limit point of this sequence of intervals, which exists since $S$ is complete, and continue covering incrementally). This is impossible, as then $S$ would be separable.

Now, by our proposition there exists a subgroup of $S$ that is isomorphic to $\mathbb{R}$. This subgroup is cofinal and coinitial in some open interval around $0$. Since this interval is not separable, our copy of $\mathbb{R}$ cannot cover it completely. That is, there must be some cut in $\mathbb{R}$ that contains a point in $S \setminus \mathbb{R}$. This cut either has the form $(-\infty, r] \cup (r, \infty)$ or $(-\infty, r) \cup [r, \infty)$. We may assume it's of the former kind, and for simplicity let's suppose $r=0$. So there is at least one point in $S \setminus \mathbb{R}$ that sits above $0$ but below every real number greater than $0$. In fact, since the order type of $S$ is dense, there must be an entire interval $I_0 \subset S \setminus \mathbb{R}$ sitting in that region. We take $I_0$ to denote the maximal such interval, that is, all points in $S \setminus \mathbb{R}$ between $0$ and the positive reals. This interval has the form $(0, s]$, where $s$ is the infimum in $S$ of the positive reals.

Now, for any real $r \in \mathbb{R}$, consider the translation of $S$ by $r$. This is an order-automorphism of $S$ that restricts to an order-automorphism of $\mathbb{R}$, and it sends $0$ to $r$. Necessarily then, it sends the interval $I_0$ onto some interval $I_r$ above $r$ but below ever real greater than $r$ (indeed it sends it onto the maximal such interval). For $r < r'$, the intervals $I_r$ and $I_{r'}$ are disjoint, since if they were not we would have $r' \in I_r$, impossible as $I_r$ contains no reals.

But now we have a contradiction, since we have found an uncountable collection of disjoint intervals in $S$, namely $\{I_r: \, r \in \mathbb{R} \}$.

Proof of the proposition: It seems likely that similar arguments to the following have been done before. If anyone knows of standard results to cite that would streamline things, I would be glad to hear them.

Let $(S, +, <)$ be an ordered abelian group whose order type is dense and complete. Completeness is a strong hypothesis: we'll show that such an $S$ is actually a vector space over $\mathbb{Q}$. This goes by first showing, for every $s \in S$ and $n \in \mathbb{N}$, there is an element $y$ (necessarily unique), which we denote as $\frac{1}{n}s$, such that $ny=s$. We may then define $\frac{m}{n} s = m (\frac{1}{n}s)$, and check that these fractional coefficients satisfy the vector space axioms for scalars.

Once we've done this, we can show that $S$ contains the desired copy of the ordered group $\mathbb{R}$. First, fix any element greater than $0$ in $S$ and label it $1$. Let $Q$ denote the one-dimensional subspace of $S$ generated by $1$. Then $Q$ is isomorphic (as a linear order, group, subspace) to $\mathbb{Q}$. Since $S$ is complete, every unfilled cut in $Q$ contains at least one point, and possibly many. As such there may be many ways to take a Dedekind completion of $Q$ within $S$ to get a suborder isomorphic to $\mathbb{R}$. To ensure that what we end up with is actually a subgroup, we build it as a limit of subspaces within $S$ as one would do within $\mathbb{R}$ itself when searching for a basis for $\mathbb{R}$ over $\mathbb{Q}$. At every stage we choose some linearly independent element to fill a new cut in $Q$, never choosing more than one point from a single cut, and then close to form a new subspace.

Explicitly, we do an induction of length continuum. At stage $\alpha$, choose some $s_{\alpha} \in S$ to be any element in some as yet unfilled cut in $Q$, such that $s_{\alpha} \not \in Q_{\alpha}$, where $Q_{\alpha}$ denotes the subspace of $S$ generated by $Q \cup \{s_{\beta}: \, \beta <\alpha\}$. There is some element $r_{\alpha}$ in $\mathbb{R}$ that sits in the corresponding cut in $\mathbb{Q}$, and this element is also linearly independent over the corresponding subspace $\mathbb{Q}_{\alpha} \subset \mathbb{R}$. After continuum many steps we will have filled all gaps in $Q$, and constructed the subspace $R \subset S$ generated by $Q \cup \{s_{\alpha}: \alpha < 2^{\aleph_0} \}$. This space is necessarily isomorphic to $\mathbb{R}$ as an additive group. To see this, note every point in $R$ has a unique representation as $q_01 + q_1s_{\beta_1} + \ldots + q_ks_{\beta_k}$, where the $q_i \in \mathbb{Q}$. Then the isomorphism is the obvious one: just extend the map that takes each $s_{\alpha}$ to $r_{\alpha}$.

It remains to verify that $S$ is a vector space over $\mathbb{Q}$. We'll sketch how to show that $\frac{1}{2}x$ exists for ever $x \in S$. Assume for simplicity that $x$ is positive. Proceed as one would guess: pick some $x_0$ between $0$ and $x$. Then $x_0 + x_0$ is either equal to, less than, or greater than $x$. In the first case we are done; in the second, pick an $x_1$ between $x_0$ and $x$; in the third, pick $x_1$ to be between $0$ and $x_0$. And so on, in this way building two sequences, one increasing and one decreasing (and one possibly empty), that seek to converge to $\frac{1}{2}x$. At successor stages go as before, and at limit stages take limit points, which always exist as $S$ is complete. This process will terminate, in the sense that either at some stage we find $\frac{1}{2}x$, or we keep going until our upper sequence converges to some $y \leq \frac{1}{2}x$, or the lower sequence converges to some $y \geq \frac{1}{2}x$. There are some details to check, but one can show that in each case the point to which the relevant sequence converges must be $\frac{1}{2}x$.

Analogously, one may find $\frac{1}{n}x$, and so $\frac{m}{n}x$. From there it is not too difficult to show that these rational scalars satisfy the vector space axioms. We are done.

It's worth noting that we use in an essential way the ccc-ness and completeness of Suslin lines to get a contradiction. Nothing like this argument can be used in the case of Aronszajn lines, and it seems plausible to me that such a line could be the order type of an abelian group.

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  • $\begingroup$ +1. It's very nice! $\endgroup$ – Joel David Hamkins Sep 6 '14 at 1:30
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    $\begingroup$ A densely ordered Dedekind-complete group is isomorphic to the reals, see e.g. mathoverflow.net/a/179520/12705 . This is a classical result of Hölder, which states more generally that an archimedean ordered group can be embedded in $\mathbb R$. $\endgroup$ – Emil Jeřábek supports Monica Sep 6 '14 at 10:45
  • $\begingroup$ @Emil: THAT's where this sort of result seemed familiar! Thanks! $\endgroup$ – Asaf Karagila Sep 6 '14 at 18:46
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This answer might be late, but I found the following papers related:

(A) Koppelberg, Sabine, ``Groups cannot be Souslin ordered'', Arch. Math. (Basel) 29 (1977), no. 3, 315–317.

In this paper it is shown that there are no Suslin ordered groups.

I may also mention that the problem of whether a Suslin order may be the underlying order of an ordered field or of some more general algebraic structure is also discussed in the following paper:

(B) Felgner, Ulrich, ``Das Problem von Souslin für geordnete algebraische Strukturen''. Set theory and hierarchy theory (Proc. Second Conf., Bierutowice, 1975), pp. 83–107. Lecture Notes in Math., Vol. 537, Springer, Berlin, 1976.

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  • $\begingroup$ Thanks Mohammad; I am grateful for the extra references. $\endgroup$ – Avshalom Mar 12 '15 at 18:48
  • $\begingroup$ @Avshalom Your welcome, I am happy if they were useful. $\endgroup$ – Mohammad Golshani Mar 14 '15 at 4:20

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