# DEGREES OF ORDERS ON TORSION-FREE ABELIAN GROUPS

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1 DEGREES OF ORDERS ON TORSION-FREE ABELIAN GROUPS ASHER M. KACH, KAREN LANGE, AND REED SOLOMON Abstract. We construct two computable presentations of computable torsion-free abelian groups, one of isomorphism type Q ω and one of isomorphism type Z ω, having computable orders but not having orders of every Turing degree. 1. Introduction In this paper, we present two results concerning computability theoretic properties of the spaces of orderings on abelian groups. To motivate these properties, we compare the nown results on computational properties of orderings on abelian groups with those for fields. We refer the reader to [6] and [8] for a more complete introduction to ordered abelian groups and to [9] for bacground on ordered fields. Definition 1.1. An ordered abelian group consists of an abelian group G = (G; +, 0) and a linear order G on G such that a G b implies a + c G b + c for all c G. An abelian group G which admits such an order is orderable. Definition 1.2. The positive cone P (G; G ) of an ordered abelian group (G; G ) is the set of positive elements P (G; G ) := {g G 0 G G g}. Because a G b if and only if b a P (G; G ), there is an effective one-to-one correspondence between positive cones and orderings. Furthermore, an arbitrary subset X G is the positive cone of an ordering on G if and only if X is a semigroup such that X X 1 = G and X X 1 = {0 G }, where X 1 := { g g X}. We let X(G) denote the space of all positive cones on G. Notice that the conditions for being a positive cone are Π 0 1. The definitions for ordered fields are much the same, and we let X(F) denote the space of all positive cones on the field F. (We will not give the definitions here as the results for fields are only used as motivation.) As in the case of abelian groups, the conditions for a subset of F to be a positive cone are Π 0 1. Classically, a field F is orderable if and only if it is formally real, i.e., if 1 F is not a sum of squares in F; and an abelian group G is orderable if and only if it is torsion-free, i.e., if g G and g 0 G implies ng 0 G for all n N with n > 0. In both cases, the effective version of the classical result is false: Rabin [12] constructed a computable formally real field that does not admit a computable order, and Downey and Kurtz [4] constructed a computable torsion-free abelian group (in fact, isomorphic to Z ω ) that does not admit a computable order. Date: January 12, Mathematics Subject Classification. Primary: 03D45; Secondary: 06F20. Key words and phrases. ordered abelian group, degree spectra of orders. 1

2 2 KACH, LANGE, AND SOLOMON Despite the failure of these classifications in the effective context, we have a good measure of control over the orders on formally real fields and torsion-free abelian groups. Because the conditions specifying the positive cones in both contexts are Π 0 1, the sets X(F ) and X(G) are closed subsets of 2 F and 2 G respectively, and hence under the subspace topology they form Boolean topological spaces. If F and G are computable, then the respective spaces of orders form Π 0 1 classes, and therefore computable formally real fields and computable torsion-free abelian groups admit orders of low Turing degree. For fields, one can say considerably more. Craven [1] proved that for any Boolean topological space T, there is a formally real field F such that X(F) is homeomorphic to T. Translating this result into the effective setting, Metaides and Nerode [11] proved that for any nonempty Π 0 1 class C, there is a computable formally real field F such that X(F) is homeomorphic to C via a Turing degree preserving map. Friedman, Simpson, and Smith [5] proved the corresponding result in reverse mathematics that WKL 0 is equivalent to the statement that every formally real field is orderable. Most of the corresponding results for abelian groups fail because a countable torsion-free abelian group either has two orderings (if the group has ran one) or has continuum many orderings. Even if one only considers Π 0 1 classes of separating sets (which have size continuum except in trivial cases) and only requires that the map from X(F) into the Π 0 1 class be degree preserving, one cannot represent all such classes by spaces of orders on computable torsion-free abelian groups. (See Solomon [15] for a precise statement and proof of this result.) However, the connection to Π 0 1 classes is preserved in the context of reverse mathematics as Hatziiriaou and Simpson [7] proved that WKL 0 is equivalent to the statement that every torsion-free abelian group is orderable. Because torsion-free abelian groups are a generalization of vector spaces, it is not surprising that notions such as linear independence play a large role in studying these objects. Definition 1.3. Let G be a torsion-free abelian group. Elements g 0,..., g n are linearly independent (or just independent) if for all c 0,..., c n Z c 0 g 0 + c 1 g c n g n = 0 G implies each c i = 0. An infinite set of elements is independent if every finite subset is independent. A maximal independent set is a basis and the cardinality of any basis is the ran of G. It is nown that if G is a computable torsion-free abelian group of ran at least two and B is a basis for G, then G has orders of every Turing degree greater than or equal to the degree of B. (See Solomon [15] and Dabowsa, Dabowsi, Harizanov, and Tonga [2].) Therefore, the set deg(x(g)) := {d d = deg(p ) for some P X(G)} contains all the Turing degrees when the ran of G is finite (but not one) and contains cones of degrees when the ran is infinite. Furthermore, Dobritsa [3] proved that for every computable torsion-free abelian group G, there is a computable H = G such that H has a computable basis. Therefore, every computable torsionfree abelian group has a computable copy that has orders of every Turing degree.

3 DEGREES OF ORDERS ON TORSION-FREE ABELIAN GROUPS 3 These facts bring us to the motivating question for this paper if G is a computable torsion-free abelian group, is deg(x(g)) necessarily closed upwards in the Turing degrees? We show that the answer to this question is no by constructing computable copies of Q ω and Z ω that have computable orders but not orders of every Turing degree. Theorem 1.4. There is a computable presentation G of the group Q ω and a noncomputable, computably enumerable set C such that: The group G has exactly two computable orders. Every C-computable order on G is computable. Thus, the set of degrees of orders on G is not closed upwards. Theorem 1.5. There is a computable presentation G of the group Z ω and a noncomputable, computably enumerable set C such that: The group G has exactly two computable orders. Every C-computable order on G is computable. Thus, the set of degrees of orders on G is not closed upwards. We demonstrate Theorem 1.4 and Theorem 1.5 in Section 2 and Section 3, respectively. We finish with additional remars and open questions in Section 4. Our notation is mostly standard. In particular we use the following convention from the study of linear orders: If G is a linear order on G, then G denotes the linear order defined by x G y if and only if y G x. Note that if (G; G ) is an ordered abelian group, then (G; G ) is also an ordered group. 2. Proof of Theorem 1.4 We divide the proof into several steps. First, we describe our general method of building the computable presentation G = (G; + G ). Second, we describe how the computable ordering G on G is constructed. (The second computable order on G is G.) Third, we give the construction of C and the diagonalization process to ensure the only C-computable orders on G are G and G. Part 1. General Construction of G. The group G is constructed in stages, with G s denoting the finite set of elements in G at the end of stage s. We maintain G s G s+1 and let G := s G s. We define a partial function + s on G s giving the addition facts declared by the end of stage s. For G to be a computable group, we cannot change any addition fact once it is declared, so we need to maintain x + s y = z = ( t s) [x + t y = z] for all x, y, z G s. Furthermore, for any pair of elements x, y G s, there must eventually be a stage t and an element z G t such that we declare x + t y = z. Of course, our declared addition facts must also satisfy the axioms for a torsion-free abelian group. To define the addition function, we use an approximation {b s 0, b s 1,..., b s s} G s to an initial segment of our eventual basis for G. During the construction, each approximate basis element b s i will be redefined at most finitely often (in fact, at most once), so each will eventually reach a limit. We let b i := lim s b s i denote this limit. The first approximate basis element b s 0 will never be redefined, so although we often use the notation b s 0 (for uniformity), we have b 0 = b s 0 for all s.

4 4 KACH, LANGE, AND SOLOMON At stage 0, we begin with G 0 := {0, 1}. We let 0 denote the identity element 0 G and we assign 1 the label b 0 0. We declare 0 G G = 0 G, 0 G + 0 b 0 0 = b 0 0, and b G = b 0 0. More generally, at stage s, each element g G s is assigned a Q-linear sum over the stage s approximate basis of the form q s 0b s q s nb s n where n s, qi s Q for i n, and qs n 0. This assignment is required to be oneto-one, and the identity element 0 G is always assigned the empty sum. It will often be convenient to extend such a sum by adding more approximate basis elements on the end of the sum with coefficients of zero. For example, the element 0 G can be represented by any sum in which all the coefficients are zero. We trust that using such extensions will not cause confusion. We define the partial function + s on G s by letting x + s y = z (for x, y, z G s ) if the sums for x and y add together to form the sum for z. Thus, the function + s is commutative and associative (on the elements for which it is defined) and satisfies x + s 0 G = x for all x G s. During stage s + 1, we do one of two things either we leave our approximate basis unchanged or we add a dependency relation for a single b s l (for some l s). The diagonalization process dictates which happens. (1) If we leave our approximate basis unchanged, then we define b s+1 i := b s i for all i s. For each g G s (viewed as an element of G s+1 ), we define q s+1 i := qi s and assign g the same sum with bs+1 i and q s+1 i in place of b s i and qi s, respectively. It follows that x + s+1 y = z (for x, y, z G s ) if x + s y = z. We add two new elements to G s+1, labeling the first by b s+1 s+1 and labeling the second by q0 s+1 b s qn s+1 b s+1 n, where q0 s+1,..., qn s+1 is the least tuple of rationals (under some fixed computable enumeration of all tuples of rationals) such that n s, qn s+1 0, and this sum is not already assigned to any element of G s+1. This completes the description of G s+1 in this case. (2) If we redefine an approximate basis element b s l (for the sae of diagonalizing) by adding a new dependency relation, then we proceed as follows. We define b s+1 i := b s i for all i s with i l. The diagonalization process will tell us either to set b s l = qbs+1 0 for some rational q or to set b s l = b s+1 j + qb s+1 0 for some rational q and some index j < l. (That is, we mae the element b s l dependent either on bs+1 0 or on b s+1 0 and b s+1 j.) For each g G s (viewed as an element of G s+1 ), we assign g the same sum except we replace each b s i by bs+1 i (for i s and i l) and we replace b s l by either qb s+1 0 or b s+1 j +qb s+1 0 (depending on how we made b s l dependent). For example, if the diagonalization process tells us to mae b s l = bs+1 j + qb s+1 0, then the sum for g G s changes from g = q s 0b s 0 + q s j b s j + + q s l b s l + + q s sb s s at stage s (where we have added zero coefficients if necessary) to g = q0b s s qj s b s+1 j + + ql s (b s+1 j + qb s+1 0 ) + + qsb s s+1 s = (q0 s + ql s q)b s (qj s + ql s )b s+1 j + + qsb s s+1 s

5 DEGREES OF ORDERS ON TORSION-FREE ABELIAN GROUPS 5 at stage s + 1. Therefore, we set q0 s+1 := q0 s + ql s q s+1 l := 0, while leaving q s+1 i := qi s for all i {0, j, l}. q, qs+1 j := q s j + qs l, and There are four points to notice about these transformations. First, the approximate basis element b s+1 l (which we have yet to define) does not appear in the new sum for any element of G s viewed as an element of G s+1. Second, the approximate basis elements b s+1 0 and b s+1 j may have nonzero coefficients in the new sum for g even if b s 0 and b s j had zero coefficients in the old sum. However, these are the only approximate basis elements that can have coefficients that change from zero to nonzero. Third, we need to mae sure that the assignment of sums to elements of G s (viewed as elements of G s+1 ) is one-to-one when we choose q. The diagonalization process will place some restrictions on the value of q, but as long as there are infinitely many possible choices for q, we can assume q is chosen to maintain the oneto-one assignment of sums to elements of G s+1. Fourth, by the linearity of this substitution, if x + s y = z, then x + s+1 y = z. To complete stage s + 1, we add three new elements to G s+1. The first is labelled b s+1 l, the second is labelled b s+1 s+1, and the third is labelled q0 s+1 b s qn s+1 b s+1 n, where q0 s+1,..., qn s+1 is the least tuple of rationals such that n s, qn s+1 0, and this sum is not assigned to any element of G s+1. This completes the description of G s+1 in our second case. It remains to chec various properties of this construction. We wor under the assumption that each approximate basis element is only redefined finitely often and hence b i := lim s b s i exists for all i. Lemma 2.1. For each g G there is a stage t such that g is assigned a sum q t 0b t q t nb t n that is not later changed in the sense that, for all stages u t, the element g is assigned the sum q u 0 b u q u nb u n with b u i = b t i and qu i = q t i for all i n. Proof. When g enters G, it is assigned a sum. The coefficients in this sum only change when a diagonalization occurs. In this case, some approximate basis element b s l with nonzero coefficient in the sum for g is made dependent via a relation of the form b s l = qbs+1 0 or b s l = bs+1 j + qb s+1 0 with j < l. Therefore, each time the sum for g changes, some approximate basis element with nonzero coefficient is replaced by rational multiples of approximate basis elements with lower indices. This process can only occur finitely often before terminating. We refer to the sum in Lemma 2.1 as the limiting sum for g and denote it by q 0 b q n b n. Lemma 2.2. For each rational tuple q 0,..., q n with q n 0, there is an element g G such that the limiting sum for g is q 0 b q n b n. Proof. For a contradiction, suppose there is a rational tuple violating this lemma. Fix the least such tuple q 0,..., q n in our fixed computable enumeration of rational tuples. Let s n be a stage such that b s 0,..., b s n have reached their limits and each tuple before q 0,..., q n has appeared as the limiting sum of an element in G s. By our construction, at stage s + 1, either there is an element that is assigned the sum q 0 b s q n b s+1 n or else we add a new element to G s+1 and assign it this sum. In either case, this element has the appropriate limiting tuple since b s+1 0,..., b s+1 n have reached their limits (and thus we obtain our contradiction).

6 6 KACH, LANGE, AND SOLOMON Lemma 2.3. If x+ s y = z, then x+ t y = z for all t s. In particular, if x+ s y = z, then the limiting sums for x and y add to form the limiting sum for z. Proof. In both cases of stage s + 1 of our construction, we checed that x + s y = z implies x + s+1 y = z. Thus, the result follows by induction. Lemma 2.4. For each pair x, y G s, there is a stage t s and an element z G t such that x+ t y = z. For each x G s, there is a stage t s and an element z G t such that x + t z = 0 G. Proof. For the first statement, fixing x, y G s, let s s be a stage at which x and y have been assigned their limiting sums and let q 0 b s 0 + q n b s n be the sum of these limiting sums. By Lemma 2.2, let t s be a stage such that there is a z G t assigned to the sum q 0 b t q n b t n. Then x + t y = z. The proof of the second statement is similar. Using Lemma 2.3 and Lemma 2.4, we define the additive function + G on G by putting x + y = z if and only if there is a stage s such that x + s y = z. Lemma 2.5. The group G is a computable copy of Q ω. Proof. The domain and addition function on G are computable. By Lemma 2.4, every element of G has an inverse, and it is clear from the construction that the addition operation satisfies the axioms for a torsion-free abelian group. Notice that x + y = z in G if and only if the limiting sums for x and y add to form the limiting sum for z. Therefore, the limiting sums for each element give us a 0 2 map from G into the standard computable presentation of the divisible torsion-free abelian group of countably infinite ran. Part 2. Defining the Computable Orders on G. We define the computable ordering of G in stages by specifying a partial binary relation s on G s at each stage s. To mae the ordering relation computable, it suffices to satisfy x s y = ( t s) [x t y] (1) for all x, y G s. Typically, the relation s will not describe the ordering between every pair of elements of G s, but it will have the property that for every pair of elements x, y G s, there is a stage t s at which we declare x t y or y t x, and not both unless x = y. Since we will be considering several orderings on G, for an ordering on G, we let (g 1, g 2 ) denote the set {g G g 1 g g 2 }. Moreover, given a 1, a 2 R, we let (a 1, a 2 ) R denote the interval {a R a 1 < R a < R a 2 }. To specify the computable order on G, we build a 0 2-map from G into R. (Thus our order will be archimedean.) To describe this order, let {p i } i 1 enumerate the prime numbers in increasing order. Our map will send the basis element b 0 to the real 1. For i 1, we will assign (in the limit of our construction) a real number r i to the basis element b i such that r i is a rational multiple of p i. We choose the r i in this manner simply so that they are algebraically independent over Q. In the construction of G, each element g G is assigned a limiting sum g = q 0 b q n b n. We define our 0 2-map into R to send g to the real q 0 r 0 + +q n r n (and to send 0 G to 0). Since r 0 = 1, the rationals will be in the image of this map. We need to approximate this 0 2-map during the construction. At each stage s, we eep a real number ri s as an approximation to r i, viewing ri s as our current target

7 DEGREES OF ORDERS ON TORSION-FREE ABELIAN GROUPS 7 for the image of b i. The real r0 s is always 1 and the real ri s is always a rational multiple of p i. Exactly which rational multiple may change during the course of the diagonalization process. We could generate a computable order on G s by mapping G s into R using a linear extension of the map sending each b s i to ri s. However, this would restrict our ability to diagonalize. Therefore, at stage s, we assign each b s i (for i 1) an interval (a s i, âs i ) R where a s i, âs i Q, rs i (as i, âs i ) R, and â s i as i 1/2s. Because b s 0 is always assigned 1 and b 0 = b s 0 for all s, these intervals give us a range of ordering relations between b s i and b 0 in G. That is, we thin of them as declaring that in G, the element b s i is contained in the interval (as i b 0, â s i b 0) s. Because each x G s is assigned a sum describing its relationship to the current approximate basis, we can generate an interval approximating the image of x in R under the 0 2-map. That is, suppose x is assigned the sum x = q s 0b s q s nb s n at stage s. The interval constraints on the image of each b s i in R translate into a rational interval constraint on the image of x in R. (The endpoints of this constraint can be calculated using the coefficients of the sum for x and the rationals a s i and âs i. The exact form depends on the signs of these numbers.) If n = 0, then the image of x in R is q 0 since b s 0 = b 0 = 1. To define s on G s at stage s, we loo at the interval constraints for each pair of distinct elements x, y G s. If the interval constraint for x is disjoint from the interval constraint for y, then we declare x s y or y s x (depending on which inequality is forced by the constraints). If the interval constraints are not disjoint, then we do not declare any ordering relation between x and y at stage s. Of course, we also declare x s x for each x G s. It remains to see how these constraints interact with the construction of G and to verify that we obtain the appropriate properties during the construction. In particular, to maintain the implication in (1), we need to chec that x s y implies x s+1 y. It suffices to ensure that for each x G s, the interval constraint for x at stage s + 1 is contained within the interval constraint for x at stage s. Recall that at stage s+1, there are two possible cases in the construction of G s+1. (1) Suppose that we leave the approximate basis unchanged. We set r0 s+1 := 1. For each 1 i s, we define r s+1 i := ri s (so we maintain our guess at the target rational multiple of p i for b i ) and define a s+1 i and â s+1 i so that (a s+1 i, â s+1 i ) R (a s i, â s i ) R, r s+1 i (a s+1 i, â s+1 i ) R, and â s+1 i a s+1 i < 1/2 s+1. Because we nest the constraint intervals for these approximate basis elements, the interval constraints for each x G s at stage s + 1 are nested within the constraints at stage s. For the approximate basis element b s+1 s+1 introduced at this stage, we set r s+1 s+1 to be a rational multiple of p s+1 (as directed by the diagonalization process) and define a s+1 s+1, âs+1 Q so that rs+1 s+1 (as+1, âs+1 ) R and â s+1 s+1 as+1 < 1/2s+1. (2) Suppose that we add a dependency for an approximate basis element b s l at stage s + 1 by maing b s l = qbs+1 0 or b s l = bs+1 j + qb s+1 0. In this case, our ordering conditions require us to place some restrictions on the possible choices of q. Recall that we can carry out our general group construction as long as there are infinitely many possible choices for q. At this point, we

8 8 KACH, LANGE, AND SOLOMON place some restrictions on q for the sae of the ordering and later we will place additional restraints on the choice of q for the sae of the diagonalization action. In the end, we only need to verify that there remain infinitely many possible choices for q. For our declared ordering relations from stage s to continue to hold at stage s + 1, we need that the image of b s l at stage s + 1 remains constrained to lie in (a s l, âs l ) R. Therefore, in the case when b s l = qbs+1 0, we need to choose q (a s l, âs l ) R since the image of b s+1 0 in R is 1. There are obviously infinitely many choices of q satisfying this condition. In the case when b s l = bs+1 j + qb s+1 0, we need to choose q, a s+1 j, and â s+1 j so that (q + a s+1 j, q + â s+1 j ) R (a s l, â s l) R. By choosing a s+1 j and â s+1 j so that the diameter of (a s+1 j, â s+1 j ) R is suitably small, we can guarantee that â s+1 j a s+1 j < 1/2 s+1 and that there are infinitely many choices of q for which the relation above holds. We handle the ordering approximations for all other approximate basis elements b s+1 i for 1 i s with i l (and i j in the case when b s l = bs+1 j + qb s+1 0 ) as in the case when our approximate basis is unchanged. For the new approximate basis elements b s+1 l and b s+1 s+1, we can assign rs+1 l and rs+1 s+1 to be any rational multiple of p l and p s+1 (respectively) and pic a l, â s+1 l, a s+1 s+1, and âs+1 accordingly. (The diagonalization strategy will constrain our choices of these values.) We verify several properties of this general ordering construction based on the assumptions that each approximate basis element b s i eventually reaches a limit and that we choose our intervals and associated rationals in the manner described above. Lemma 2.6. For every pair of elements x, y G s, if x s y, then x s+1 y. Hence, we have the implication given in (1). Proof. This follows from the definition of s and the fact that for each x G s the constraining interval for x at stage s + 1 is contained in the constraining interval for x at stage s. By construction, we have r0 s = 1 for all s, so r 0 := lim s r0 s = 1 exists. We verify that the other ri s also reach limits. Lemma 2.7. For each i 1, the limit r i := lim s ri s exists and is a rational multiple of p i. Furthermore, once ri s reaches its limit, the rational intervals (at i, ât i ) R for t s form a nested sequence converging to r i. Proof. We have r s+1 i ri s only when b s+1 i b s i. Since the latter happens only finitely often, each ri s reaches a limit. The remainder of the statement is immediate from the construction. Lemma 2.8. For each pair x, y G s, there is a stage t s for which either x t y or y t x. Proof. Since x s x for all x G s, we consider distinct elements x, y G s. Let t s be a stage such that x and y have reached their limiting sums and such that for each b t i occurring in these sums, the real rt i has reached its limit r i. Because the reals r i are algebraically independent over Q and the nested approximations

9 DEGREES OF ORDERS ON TORSION-FREE ABELIAN GROUPS 9 (a u i, âu i ) R (for u t) converge to r i, there is a stage at which the interval constraints for x and y are disjoint. At the first such stage, we declare an ordering relation between x and y. We define the order G on G by x G y if and only if x s y for some s. By Lemma 2.6 and Lemma 2.8, this relation is computable and every pair of elements is ordered. By construction, the 0 2-map from G to R that sends q 0 b q n b n q 0 + q 1 r q n r n is order preserving. Therefore, our ordered group G is classically isomorphic to the divisible subgroup of (R; +) with basis {1} { p i : i 1} under the standard ordering. Part 3. Building C and Diagonalizing. It remains to show how to use this general construction method to build the ordered group (G; G ) together with a noncomputable c.e. set C such that the only C-computable orders on G are G and G. The requirements S e : C Φ e to mae C noncomputable are met in the standard finitary manner. The strategy for S e chooses a large witness x, eeps x out of C, and waits for Φ e (x) to converge to 0. If this convergence never occurs, the requirement is met because x C. If the convergence does occur, then S e is met by enumerating x into C and restraining C. The remaining requirements are R e : If Φ C e (x, y) is an ordering on G, then Φ C e is either G or G. We explain how to meet a single R e in a finitary manner, leaving it to the reader to assemble the complete finite injury construction in the usual manner. In particular, note that each approximate basis element b s i eventually reaches a limit. To simplify the notation, we let C e be the binary relation on G computed by Φ C e. We will assume throughout that C e never directly violates any of the Π 0 1 conditions in the definition of a group order. For example, if we see at some stage s that C e has violated transitivity, then we can place a finite restraint on C to preserve these computations and win R e trivially. The strategy to meet R e is as follows. We wait for a stage s at which C e declares either b 0 C e 0 G or b 0 C e 0 G. While we wait, the construction of G proceeds as in the general description, leaving the approximate basis unchanged at each stage (that is, no dependencies are added) and defining rs+1 s+1 = p s+1. If C e never gives either computation, then R e is won trivially. Therefore, we assume that C e eventually orders 0 G and b 0. We restrain C to preserve this ordering relationship. To simplify the exposition, we assume that b 0 C e 0 G. By restraining C, we now C e is not G, so it only remains to ensure that if C e is an order, then C e is G. (If we have b 0 C e 0 G, then we can wor with C e and show that if C e is an order, then C e is G.) If G and C e are distinct, then we claim there must eventually be a stage s, an approximate basis element b s j (for some j > 0), and rationals q 0 < q 1 such that we have declared q 0 b s 0 < s b s j < s q 1 b s 0 in G s, and the order C e has declared b s j <C e q 0 b s 0 or q 1 b s 0 < C e b s j.

10 10 KACH, LANGE, AND SOLOMON (If g G s is assigned the sum q 0 b s 0 or q 1 b s 0, then it has reached its limiting sum since b s 0 = b 0.) Why must such a situation occur? If not, then in C e, each b i := lim s b s i sits in the same rational cut (relative to b 0 ) as in G. Since G is generated by the basis {b i : i ω}, the orderings G and C e are generated by their respective orderings of these basis elements. It then follows that the order C e is archimedean because G is archimedean, and furthermore, that C e is the same as G. Note that if C e states that b s j equals some endpoint in the above inequalities, for example b s j = q 0b s 0, then C e fails to be an ordering if we continue building G as stated. Thus, the requirement R e is met with no further action beyond restraining C. Therefore, we assume that we eventually see such an element b s j at some stage s + 1. At this stage, we begin our diagonalization action. We restrain C to preserve the ordering relations that C e has given so far. In the general group construction, we leave our approximate basis unchanged (i.e., b s+1 i := b s i for all i s) and we leave their target images unchanged (i.e., r s+1 i := ri s for all i s). For the new approximate basis element b s+1 s+1, we want to have bs+1 (q 0b s 0, q 1 b s 0) s+1. That is, we want to choose rs+1 s+1 to be a rational multiple of p s+1 that sits in the interval (q 0, q 1 ) R and we want to choose a s+1 s+1 and âs+1 so that (as+1, âs+1 ) R (q 0, q 1 ) R. Notice that both b s+1 j and b s+1 s+1 are then in the interval (q 0b s 0, q 1 b s 0) s+1. Furthermore, we want if b s j C e q 0 b s 0, then b s+1 s+1 < s+1 b s+1 j, and if q 1 b s 0 C e b s j, then bs+1 j < s+1 b s+1 s+1. We can meet these additional constraints by choosing rs+1 s+1 to be a rational multiple of p s+1 for which there are appropriate choices of a s+1 s+1 and âs+1 satisfying either â s+1 s+1 < as+1 j (in the first case) or â s+1 j < a s+1 s+1 (in the second case). Now, we wait for a stage t > s+1 at which C e declares whether or not b t s+1 is in the interval (q 0 b t 0, q 1 b t 0) C e. (We assume b u j = bs+1 j while we are waiting. If not, then this strategy would be injured and would start over again. Note that our current requirement is the only one that could cause b u s+1 b s+1 s+1, so we automatically maintain b u s+1 = b s+1 s+1 while we wait.) If C e never specifies whether or not b t s+1 is in this interval, then C e is not an order on G and R e is won. Thus, we assume that C e eventually gives us this information. There are two cases to consider. (1) Suppose at stage t, the order C e says that b t s+1 (q 0 b t 0, q 1 b t 0) C e. At stage t + 1, we redefine b t s+1 by adding the dependency b t s+1 = qb t+1 0 for a rational q. As described in the construction of the ordering, the rational q must be consistent with our current declared interval for b t s+1, that is, we must have q (a t s+1, â t s+1) R. As there are infinitely many such rationals, we can choose one that wors in the general group construction (i.e., eeps the assignment of sums one-to-one). Furthermore, by our nesting of interval constraints, we have (a t s+1, â t s+1) R (a s+1 s+1, âs+1 s+1 ) R (q 0, q 1 ) R. Therefore, we must have q 0 < q < q 1 and any ordering of G in which b 0 is positive must satisfy q 0 b t 0 < qb t 0 < q 1 b t 0 and hence must satisfy q 0 b t 0 < b t s+1 < q 1 b t 0. As C e maes b 0 positive but does not satisfy this ordering relation, we can win R e by restraining C one last time to preserve the ordering computation that specifies that b t s+1 (q 0 b t 0, q 1 b t 0) C e.

11 DEGREES OF ORDERS ON TORSION-FREE ABELIAN GROUPS 11 (2) Suppose at stage t, the order C e says that b t s+1 (q 0 b t 0, q 1 b t 0) C e. In this case, there are two subcases. (a) Suppose that when we started the diagonalization process at stage s + 1, we had b s j C e q 0 b s 0. In this case, we made b s+1 s+1 < s+1 b s+1 j and hence have b t s+1 < t b t j. Because we restrained C at stage s + 1 and because b t j = bs j, we have bt j C e q 0 b t 0. At stage t + 1, we restrain C to has declared b t s+1 (q 0 b t 0, q 1 b t 0) C e. There- preserve the fact that C e fore, the order C e is committed to maing b t j <C e b t s+1 unless it violates transitivity (and gives us a trivial win). We pic a positive rational q and add the dependency relation 0 at stage t + 1. (For each i t with i s + 1, b t s+1 = b t+1 j qb t+1 we leave b t+1 i = b t i. In particular, bt+1 j = b t j.) In the description of the ordering, we explained how to shrin the constraining interval around r t+1 j = rj t so that there are infinitely many choices of q for which this added dependency will not disrupt our current ordering facts..) We have successfully diagonalized because the relation b t s+1 = b t+1 j qb t+1 0 forces b t s+1 to be less than b t+1 j in any order in which b 0 is positive. But, (The rational q has to be positive because b t s+1 < t b t j = bt+1 j the order C e made b 0 positive and is committed to b t+1 j = b t j <C e b t s+1. These facts are inconsistent with being an order, so we have won R e in this case. (b) Suppose that when we started the diagonalization process at stage s + 1, we had q 1 b s 0 C e b s j. In this case, we made bs+1 j < s+1 b s+1 s+1 and hence have b t j < t b t s+1. Restraining C to preserve the fact that b t s+1 (q 0 b t 0, q 1 b t 0) C e and reasoning as above, the order C e is committed to b t s+1 < C e b t j. At stage t + 1, we pic a positive rational q and add the dependency relation b t s+1 = b t+1 j + qb t+1 0. (As above, there are infinitely many choices for an appropriate q.) Because of this added dependency, any order that maes b 0 positive must mae b t+1 j than b t s+1. However, the order C e is committed to b t s+1 < C e b t j = bt+1 j and hence we have successfully diagonalized to meet R e. Combining the requisite parts, we see that the construction of G, the construction of the computable orders, and the construction of the noncomputable, computably enumerable set C yield the theorem. 3. Proof of Theorem 1.5 The construction of G of isomorphism type Z ω is similar to when the desired isomorphism type is Q ω. Again, the proof is divided into several steps. First, we describe our general method of building the computable presentation G = (G; +). Second, we describe how the computable order G on G is constructed. (As before, the order G is the second computable order on G.) Third, we give the construction of C and the diagonalization process that ensures that the only C-computable orders on G are G and G. Part 1. General Construction of G. This construction proceeds in a manner similar to Part 1 of Theorem 1.4. As done there, we build G as the union of less

12 12 KACH, LANGE, AND SOLOMON an increasing sequence of sets {G s } s ω, we approximate + G as the union of an increasing sequence of partial relations {+ s } s ω, we maintain an approximation {b s 0, b s 1,..., b s s} G s to an initial segment of our eventual basis for G and every element g G s is assigned a Q-linear sum over this approximate basis of the form g = q s 0b s q s nb s n where n s, qi s Q for 0 i n, and qs n 0. The major difference in the construction of the group is that, at each stage s, we will maintain a positive integer Ni s for each i s. This positive integer will restrain the nonzero coefficients qi s allowed in the Q-linear sum for an element g G s by requiring that d s i N i s, where ds i is the denominator of qs i when expressed in lowest terms. (Throughout this construction, we will use d q, d s i and d i to denote the denominators of rational numbers q, qi s and q i, respectively, when written in lowest terms.) It will be the case that N i := lim s Ni s exists and is finite for all i. These properties guarantee that G is a presentation of the group Z ω since each b i is not divisible by any m > N i. (Later we will introduce a basis restraint K ω that will prevent us from changing Ni s too often.) At a stage s + 1, we do one of two things: we leave our approximate basis unchanged or we add a dependency relation for an approximate basis element b s l (for some l s). The diagonalization process dictates which happens. (1) If we leave the basis unchanged, then we proceed as in (1) of Part 1 of Theorem 1.4, adding two new elements b s+1 s+1 and qs+1 0 b s qn s+1 b s+1 n to G s+1 as done there. Before doing so, we set N s+1 i := Ni s for all i s and Ns+1 s+1 s+1 := 1. By setting Ns+1 = 1, we are currently building G so that b s+1 s+1 has no divisors. We also now require that the tuple qs+1 0,..., qn s+1 satisfy d s+1 i N s+1 i for all i n with q s+1 i 0. (2) If we redefine the approximate basis element b s l (for the sae of diagonalizing) by adding a new dependency relation, then we proceed as in (2) of Part 1 of Theorem 1.4. The difference is that the diagonalization process will tell us either to set b s l = qbs+1 for some rational q, or to set b s l = m 1b s+1 j + m 2 b s+1 for some integers m 1 and m 2. In either case, the index will be even and greater than the basis restraint K and j, < l. We assign g G s the same sum except we replace each b s i by bs+1 i (for i s and i l) and we replace b s l by either qbs+1 or m 1 b s+1 j +m 2 b s+1 (as dictated by the diagonalization process). We add three new elements b s+1 l, b s+1 s+1, and q0 s+1 b s qn s+1 b s+1 n to G s+1. Before adding these new elements, we set N s+1 i := Ni s for i s and i {l,, j}. We set N s+1 l := 1 and Ns+1 s+1 s+1 := 1. For N, we set N s+1 := N s N l s d q if the dependency was b s l = qbs+1 and N s+1 := N s N l s if the dependency was b s l = m 1b s+1 j +m 2 b s+1. For N s+1 j, we set N s+1 j := Nj s if the dependency was b s l = qbs+1 and N s+1 j := Nj s N l s if the dependency was b s l = m 1b s+1 j + m 2 b s+1. We require that the triple q0 s+1,..., qn s+1 satisfy d s+1 i N s+1 i for all i n with q s+1 i 0. It remains to chec various properties of this construction. We wor under the assumption that the limits b i := lim s b s i and N i := lim s Ni s exist for all i. These assumptions follow from the diagonalization process.

13 DEGREES OF ORDERS ON TORSION-FREE ABELIAN GROUPS 13 Lemma 3.1. Each g G is eventually assigned a sum q t 0b t 0 + +q t nb t n such that at every stage u t, the element g is assigned the sum q u 0 b u 0 + +q u nb u n with the same coefficients and basis elements. Moreover, the nonzero coefficients in the limiting sum q 0 b q n b n satisfy d i N i for all 0 i n. Proof. When g enters G, it is assigned a sum. The coefficients and basis elements in this sum only change when the diagonalization process declares a dependency relation for some approximate basis element b s l having a nonzero coefficient in g. Since the dependencies are either b s l = qbs+1 or b s l = m 1b s+1 j +m 2 b s+1 with j, < l, this reassignment of sums can only occur finitely often before terminating. To see that a nonzero coefficient q i satisfies d i N i, it suffices to show by induction on s that d s i N i s whenever qs i 0. When g first enters G and is assigned a sum, we have d s i N i s by construction. The coefficient of bs i can only change if bs i is involved in an added dependency relation. If b s l = qbs+1 i, the coefficient of b s+1 i in g changes from qi s to qs+1 i = qi s +qqs l. The denominator d s+1 i divides the product of the denominators d s i, ds l, and d q. Since N s+1 i = Ni s N l s d q, we have that d s+1 i N s+1 i by the induction hypothesis. Notice that this analysis is valid even when qi s = 0 or qs l = 0. If b s l = m 1 b s+1 i + m 2 b s+1, the coefficient of b s+1 i in g changes from qi s to q s+1 i = qi s + m 1ql s. Since m 1 Z, the denominator d s+1 i divides the product of the denominators d s i and ds l. The fact that ds+1 i N s+1 i follows from N s+1 i = Ni s N l s and the inductive hypothesis. Again, note that this analysis is valid even when qi s = 0 or qs l = 0. This leaves only the cases when b s i = qbs+1 and b s i = m 1b s+1 j + m 2 b s+1. In these cases, q s+1 i = 0 and we have the result at stage s + 1 trivially. Lemma 3.2. For each rational tuple q 0,..., q n with q n 0 and d i N i for all 0 i n with q i 0, there is an element g G such that the limiting sum for g is q 0 b q n b n. Proof. The proof is identical to the proof of Lemma 2.2. As in Part 1 of Theorem 1.4, we define the additive function + G on G by setting x + G y = z if there is a stage s such that x + s y = z. Since Lemma 2.3 and Lemma 2.4 remain true (with identical proofs) in this setting, this is a well-defined and total function. As a consequence of Lemma 3.2, every element has an additive inverse. Lemma 3.3. The group G is a computable presentation of Z ω. Proof. By construction, together with the preceding comments, the domain G and addition function + G are computable, with the addition function satisfying the axioms for a torsion-free abelian group. Since the limit N i := lim s Ni s exists and is finite for all i, viewing Z ω as having basis {e i } i ω with elements g = γ 0 e γ n e n (where γ 0,..., γ n Z), we have that the map e i b i /N i induces an isomorphism. Thus, the isomorphism type of G is indeed Z ω. Part 2. Defining the Computable Orders on G. This proceeds in a manner similar to Part 2 of Theorem 1.4. As done there, we approximate the computable ordering G as the union of an increasing sequence of binary relations { s } s ω.

14 14 KACH, LANGE, AND SOLOMON We again specify the order on G by building a 0 2-map from G into R. Again, we always map b 0 to the real 1 and assign each b s i (for i 1) an interval of the form (a s i, âs i ) R with a s i, âs i Q and a real rs i (as i, âs i ) R that is a rational multiple of pi. The difference is that we will require 0 a s i âs i < 1/2i for all even i and s. The reason we do so is that the element b 0 no longer provides a means of measuring the size of arbitrary elements as only integer multiples of b 0 exist. If r i were large (say, greater than one) for all i, the only elements in G that would be small (close to zero in size) would be complicated linear combinations of basis elements. By requiring r < 1/2 for all even, we can measure the size of an arbitrary basis element b s i in terms of b 0 through an intermediary basis element b s with even index. We describe the interaction between the order and changes to the basis approximation. (1) If we leave the approximate basis unchanged at stage t + 1, then we proceed as in Case 1 of Part 2 of Theorem 1.4. We let rt+1 t+1 be a rational multiple of p t+1 such that if t + 1 is even, then rt+1 t+1 < 1/2t+1. We pic positive rationals a t+1 t+1 and ât+1 such that at+1 < rt+1 < ât+1 and â t+1 t+1 at+1 < 1/2t+1. If t + 1 is even, we also require â t+1 t+1 < 1/2t+1. (2) If we wish to add the dependency b t l = qbt+1 at stage t + 1, then is even and l is odd. This dependency relation is consistent with t provided we can choose q, a t+1, and â t+1 so that (qa t+1, qâ t+1 ) R (a t l, ât l ) R. It is immediate that there infinitely many such choices for q as we need only ensure a t at+1 â t+1 â t and ât+1 a t+1 1/2 t+1. (3) If we wish to add the dependency b t l = m 1b t j m 2b t at stage t + 1, then we will be in one of two contexts. 3(a). If we are in a context in which 0 < na t < nâ t < a t l < â t l < a t j < â t j < (n + 1)a t < (n + 1)â t, (2) then we will choose m 1, m 2 N such that m 1 m 2 /n and (m 1 a t+1 j m 2 â t+1, m 1 â t+1 j m 2 a t+1 ) R (a t l, â l t ) R. (3) 3(b). If we are in a context in which 0 < na t < nâ t < a t j < â t j < a t l < â t l < (n + 1)a t < (n + 1)â t, (4) then we will choose m 1, m 2 N such that m 1 m 2 (n + 1) and (m 1 a t+1 m 2 â t+1 j, m 1 â t+1 m 2 a t+1 j ) R (a t l, â t l) R. (5) By Lemma 3.5, in each of these contexts, there are infinitely many such choices for m 1 and m 2 satisfying the given conditions. To explain why appropriate m 1, m 2 N exist for the two contexts above, we rely on the following fact from ordered abelian group theory. Lemma 3.4. Let r 1 and r 2 be reals that are linearly independent over Q. For any rational numbers q 1 < q 2, there are infinitely many m 1, m 2 N such that m 1 r 1 m 2 r 2 (q 1, q 2 ) R. Moreover, one can require m 1 and m 2 to be divisible by any given n N. Lemma 3.5. If we are in the context of (2) (respectively (4)), then there are infinitely many choices for m 1 and m 2 that satisfy (3) (respectively (5)).

15 DEGREES OF ORDERS ON TORSION-FREE ABELIAN GROUPS 15 Proof. First, suppose we are in the context of (2). We have that b t j and bt are currently identified with the rational multiples rj t and rt of p j and p respectively, so rj t and rt are linearly independent over Q. Hence, by Lemma 3.4 (requiring m 2 to be divisible by n), there are infinitely many choices of m 1, m 2 N such that m 1 rj t m 2nr t (at l, ât l ) R. (We let m 2 := m 2 n.) We can choose a t+1 j, â t+1 j, a t+1, â t+1 Q with a t+1 j < rj t < ât+1 j and a t+1 < r t < ât+1 satisfying (3) by shrining the intervals (a t j, ât j ) R and (a t, ât ) R appropriately. It remains to see why we must have m 1 m2 n = m 2. Suppose m 1 > m2 n = m 2, so m 1 1 m 2. Then m 1 r t j m 2 nr t = r t j + (m 1 1)r t j m 2 nr t rj t + m 2 rj t m 2 nr t = rj t + m 2 (rj t nr) t > r t j because rj t nrt > 0 by (2). We have reached a contradiction since m 1 rj t m 2nr t (at l, ât l ) R and rj t (at j, ât j ) R but â t l < at j. So, m 1 m2 n = m 2 as desired. Now suppose we are in the context of (4). Since rj t and rt are linearly independent over Q, by Lemma 3.4 (requiring m 1 to be divisible by n + 1) there are infinitely many choices of m 1, m 2 N such that m 1 (n + 1)r t m 2rj t (at l, ât l ) R. (We let m 1 := m 1 (n+1).) As before, we can choose a t+1 j, â t+1 j, a t+1, â t+1 Q satisfying (5). It remains to see why m 1 = m 1 (n + 1) m 2 (n + 1). Suppose m 1 = m 1 (n + 1) > m 2 (n + 1), so m 1 1 m 2. Then m 1 r t m 2 r t j = m 1 (n + 1)r t m 2 r t j m 1 (n + 1)r t ( m 1 1)r t j > m 1 (n + 1)r t ( m 1 1)(n + 1)r t = (n + 1)r t. The first inequality follows because m 1 1 m 2 and rj t is positive, and the second inequality follows because rj t < (n + 1)rt by (4). We have reached a contradiction since m 1 r t m 2rj t (at l, ât l ) R but â t l < (n + 1)at. Since none of the differences from Part 2 of Theorem 1.4 impact the verification, again we can define an order G on G by x G y if x s y for some s. As in Part 2 of Theorem 1.4, this relation is computable and a group order on G. Part 3. Building C and Diagonalizing. This proceeds in a manner slightly different than Part 3 of Theorem 1.4. As done there, we meet the requirements S e and R e. Though we satisfy S e in the same manner, we satisfy R e in a different manner. Again, we use C e to denote the binary relation on G computed by Φ C e, and we assume it never violates the axioms of an ordered group. For R e, we set the basis restraint K := e. The strategy to satisfy R e is as follows. If G C e and G C e, then there must eventually be a stage s, an approximate basis element b s j, a nonnegative integer n, and an even index > K such that: we have declared 0 < s nb s < s b s j < s (n + 1)b s in G s, and

16 16 KACH, LANGE, AND SOLOMON the order C e has declared either (a) b s >C e 0 G and either b s j <C e nb s or b s j >C e (n + 1)b s, or (b) bs <C e 0 G and either b s j >C e nb s or bs j <C e (n + 1)b s. In the latter case, we wor with the ordering C e, transforming the latter case into the former case. We therefore assume that we are in the former case. When such s, b s j, n, and are found, we say R e is activated, and we restrain C to preserve the computations ordering b s j and nbs and (n + 1)bs. In Lemma 3.7, we verify that such s, b s j, n, and must exist if G C e and G C e. As R e is trivially satisfied otherwise, we assume such parameters are found at some stage s + 1. At stage s+1 (without loss of generality, we assume s+1 is odd), we order the new approximate basis element b s+1 s+1 depending on whether bs j <C e nb s or bs j >C e (n+1)b s. If b s j <C e nb s, we order bs+1 s+1 so that nbs < s+1 b s+1 s+1 < s+1 b s j choose rs+1 s+1 to be a rational multiple of p s+1 and a s+1 s+1 nâ s < as+1 < rs+1 < âs+1 < as j and âs+1 as+1 < 1/2s+1. if b s j >C e (n + 1)b s, we order bs+1 so that bs j < s+1 b s+1, that is, we so that and âs+1 s+1 s+1 < s+1 (n + 1)b s, that is, we choose r s+1 s+1 to be a rational multiple of p s+1 and a s+1 s+1 that â s j < as+1 s+1 < rs+1 s+1 < âs+1 s+1 < (n + 1)as and âs+1 s+1 as+1 and âs+1 so s+1 < 1/2s+1. We then wait for a stage t + 1 so that C e declares b t s+1 < C e nb s or nbs <C e b t s+1 < C e (n + 1)b s or bt s+1 > C e (n + 1)b s. Again, we assume that such a stage t + 1 is found, else R e is trivially satisfied. If C e declares b t s+1 < C e nb t or bt s+1 > C e (n+1)b t, then we declare bt s+1 = qb t for some q Q with q (n, n + 1) R (a t s+1, â t s+1) R. This ensures C e violates the order axioms as n was a nonnegative integer and b t >C e 0 G. The fact that there are infinitely many q (n, n + 1) R (a t s+1, â t s+1) R allows us to eep the assignments of sums one-to-one in the general group construction. If C e declares nb t <C e b t s+1 < C e (n+1)b t, then we act depending on whether b t s+1 < t b t j or bt s+1 > t b t j. If b t s+1 < t b t j, then we declare bt s+1 = m 1 b t j m 2b t for some positive integers m 1, m 2 with m 1 m 2 /n that satisfy our ordering constraints. This ensures C e violates the order axioms as this necessitates b t s+1 = m 1 b t j m 2 b t C e (m 2 /n)b t j m 2 b t < C e (m 2 /n)(nb t ) m 2 b t = 0 G < C e nb t < C e b t s+1. (The first inequality in the line above holds because we can conclude b t j >C e 0 G from the first equality since b t s+1 > C e 0 G and b t >C e 0 G.) By Lemma 3.5 there are infinitely many choices for m 1, m 2, so we can choose them to eep the assignments of sums one-to-one in the general group construction. If b t s+1 > t b t j, then we declare bt s+1 = m 1 b t m 2b t j for some positive integers m 1, m 2 with m 1 m 2 (n + 1) that satisfy our ordering constraints. This ensures C e violates the order axioms as this necessitates b t s+1 = m 1 b t m 2 b t j C e (m 2 (n+1))b t m 2 b t j < C e m 2 b t j m 2 b t j = 0 G < C e nb t < C e b t s+1. Again, by Lemma 3.5 there are infinitely many choices for m 1, m 2, so we can choose them to eep the assignments of sums one-to-one in the general group construction.

17 DEGREES OF ORDERS ON TORSION-FREE ABELIAN GROUPS 17 In all cases, we restrain C to preserve the computations ordering b t s+1 and nb t and (n+1)b t under C e. This completes our description of building C and diagonalizing. Having explained how to meet a single R e in a finitary manner, we again leave it to the reader to assemble the complete finite injury construction in the usual manner. As before, the finitary nature of R e and the action of the construction guarantees that the limits b i := lim s b s i and r i := lim s ri s exist for all i. We demonstrate N i := lim s Ni s exists for all i. Lemma 3.6. The limit N i := lim s N s i exists for all i. Proof. For i = 0, this is immediate as Ni s = 1 for all s by construction. For i > 0, the important observation is that if N t+1 i Ni t, then there was a basis dependency introduced of the form b t l = qbt, bt l = m 1b t j m 2b t, or bt l = m 1b t m 2b t j with i being either j, or l. By construction, in these dependency relations, is even, l is odd, and j is either even or odd. First, we claim that if i is odd, then Ni s = 1 for all s after it is defined. For a contradiction, let t be the least stage at which N t+1 i 1 for some odd i, and let i be the least odd number such that N t+1 i 1. (When Ni s is initially defined, its value is set to 1. Therefore, we now Ni t t+1 = 1 and hence Ni Ni t.) Since i is odd, the index i must be either j or l in the introduced dependency relation. However, since the construction sets N t+1 l = 1, we cannot have i = l. Therefore, the new dependency relation is of the form b t l = m 1b t i m 2b t, or bt l = m 1b t m 2b t i. In either case, N t+1 i = Ni t N l t. Since i and l are odd, we have N i t = N l t = 1, so N t+1 i = 1 for the desired contradiction. Second, we consider the case when i is even. In this case, the index i is either or j in the added dependency relation. If i =, then the dependency relation can only be added by a requirement R e with e i (because of the basis restraint). These requirements can only cause Ni s exists. If i = j in the dependency relation, then N t+1 i index l is odd, so Nl t t+1 = 1, and hence Ni = Ni t to change finitely often, so N i = lim s Ni s = Ni t N l t. However, the does not actually change. We now show that R e is activated at some stage s if G C e and G C e. Lemma 3.7. If we fail to find a stage s where R e is activated for an approximate basis element b s j, a nonnegative integer n, and an even index > K, then either C e is not an order or G = C e or G = C e. Proof. Assume that C e is an order on G. First, we claim that if we fail to find a stage s at which R e is activated, then either 0 G < C e b for all even or b < C e 0 G for all even. (In both cases, we include = 0.) To prove this claim, suppose that R e is never activated and that j 0, j 1 are even with b j1 < C e 0 G < C e b j0. Fix a stage s such that R e is the highest priority requirement left to act after stage s, b s j 1 = b j1, b s j 0 = b j0 and b j1 < C e 0 G < C e b j0 is permanently fixed by stage s. Consider a stage t s and an even index greater than the basis restraint for R e such that b t = b has reached its limit and there are n 0, n 1 ω for which 0 G < t n 0 b t < t b t j 0 < t (n 0 + 1)b t and 0 G < t n 1 b t < t b j1 < t (n 1 + 1)b t. Since C e is an order, there must be a stage u t at which it declares either 0 G < C e b u or bu < 0 G permanently.

18 18 KACH, LANGE, AND SOLOMON If 0 G < C e b u, then we must eventually see bv j 1 < C e 0 G < C e n 1 b v for some v u. Therefore, R e is activated at stage v (with j = j 1, =, and n = n 1 ) for the desired contradiction. Alternately, if b u <C e 0 G, then we must eventually see n 0 b v <C e 0 G < C e b v j 0 for some v u. Again, R e is activated at stage v (with j = j 0, =, and n = n 0 ) for the desired contradiction. This completes the proof of the claim. To complete the proof of this lemma, assume that R e is never activated and 0 G < C e b for all even. We show that C e = G. (It follows by a similar argument that if b < C e 0 G for all even, then C e = G.) By construction, (G; + G, G ) can be embedded (as an ordered group) into (R; + R, R ) by sending each basis element b i G to r i R. To show that G = C e, it suffices to show that the same map is an ordered group embedding of (G; + G, C e ) into (R; + R, R ). For each even index, we fix n 0, ω such that n 0, b G b 0 G (n 0, + 1)b. By the construction, this condition is equivalent to n 0, r R r 0 R (n 0, + 1)r. Since is even, we have (n 0, + 1)r n 0, r = r 1/2 and hence lim n 0,r = lim (n 0, + 1)r = r 0 = 1 where the limits (and all limits throughout this lemma) are taen over even indices. More generally, for each index i ω and each even index, we fix n i, ω such that n i, b G b i G (n i, + 1)b. As above, this condition is equivalent to n i, r R r i R (n i, + 1)r and we have Combining these limits, we have and lim lim n i,r = lim (n i, + 1)r = r i. n i, n 0, + 1 = lim n i, r (n 0, + 1)r = r i 1 = r i n i, + 1 (n i, + 1)r lim = lim n 0, n 0, r = r i 1 = r i. We now translate these results to (G, C e ). Because R e is never activated and 0 G < C e b for all even, the inequalities n i, b C e b i C e (n i, + 1)b hold for all i and all even such that is greater than the basis restraint for R e. In particular, combining the inequalities n 0, b C e b 0 C e (n 0, + 1)b and n i, b C e b i C e (n i, + 1)b, we have n i, n 0, + 1 b 0 C e b i C e n i, + 1 n 0, b 0 where this inequality is interpreted as representing the corresponding inequality after multiplying through by the denominators so all the coefficients are integers. (Alternately, this inequality can be viewed in the divisible closure of G using the fact that an order on an abelian group has a unique extension to an order on its divisible closure.) The limits above show that the map sending b i to r i defines an embedding of (G; + G, C e ) into (R; + R, R ) as required. We have completed the proof of Theorem 1.5.

19 DEGREES OF ORDERS ON TORSION-FREE ABELIAN GROUPS Remars and Open Questions Since both Theorem 1.4 and Theorem 1.5 are typical finite injury constructions, modifications to the constructions are straightforward. Remar 4.1. Rather than building G so that there are exactly two computable orders, it is an easy modification to build exactly any even number or an infinite number of computable orders (with no other C-computable orders). For example, to build Q ω with four computable orders, we double the number of R e requirements. We build a computable order 0 G in which 0 0 G b 0 0 G b 1 and a computable order 1 G in which 0 1 G b 1 1 G b 0. For each of these orders, we meet a slightly modified requirement: R i e : If b i C e b 1 i and Φ C e is an ordering of G, then Φ C e is either i G or i G. Since these requirements are still finitary (both restraint and injury) in nature, these combine easily to yield the desired result. Remar 4.2. We note that the computably enumerable set C cannot be complete. The reason is that 0 can compute a basis for any computable torsion-free abelian group G, and hence G has orders of degree 0. We also note that, as long as the construction remains finitary (both restraint and injury), additional requirements on C can be added. Thus lowness requirements can be added, for example, though this would be counter-productive (the weaer C is computationally, the weaer the result). Remar 4.3. The computable orders in our constructions were always archimedean. If non-archimedean computable orders are desired, it suffices to tae G Z (for a copy of Z ω ) or G Q (for a copy of Q ω ), where G is as constructed and every element x can be computably decomposed into x G and x Z (for Z ω ) or x Q (for Q ω ). The reason, of course, is that any (C-)computable order on G Z or G Q restricts to a (C-)computable order on G. Certain natural questions remain open. Question 4.4. Do Theorems 1.4 and 1.5 remain true when the isomorphism types Q ω and Z ω are replaced by an arbitrary computable isomorphism type of a torsion-free abelian group? Finally, we wonder what degree spectra for orders on a computable torsion-free abelian group are possible. As noted in Section 1, if the ran of G is 1, then G has only two orders both of which are computable. If the ran is finite but greater than 1, then G has orders of every degree. However, if G has infinite ran, then the only constraint we now of (beyond Π 0 1 class basis theorems and including the cone above 0, or more generally the cone above the degree of each basis of G) is the following. Proposition 4.5 (With Daniel Turetsy). If G is a computable presentation of a torsion-free abelian group with infinite ran, then deg(x(g)) contains infinitely many low degrees. Proof. We inductively show deg(x(g)) must contain at least n-many low degrees for all n. Fix two linearly independent elements g, h G and let T 0 be a computable

20 20 KACH, LANGE, AND SOLOMON tree such that [T 0 ] (the set of infinite paths through T 0 ) contains exactly the orders G on G satisfying 0 G < G g < G h < G 4g. Note that the set of orders on G satisfying this constraint is a Π 0 1 class and hence can be represented in this manner. The Low Basis Theorem applied to T 0 yields a low order of some degree d 0. To get a second order of low degree d 1 d 0, it suffices (as low over low is low) to build a nonempty d 0 -computable subtree T 1 of T 0 having no d 0 -computable paths. From this, we obtain a low (low over d 0 ) order of some degree d 1 not computable from d 0. The subset T 1 of T 0 is constructed (using an oracle of degree d 0 ) by illing paths that agree with the e th (candidate) d 0 -computable order e on the relative ordering of g and h for a sufficiently large amount of precision. In particular, to diagonalize against e, we attempt to find positive rationals q 0 < Q q 1 such that q 1 q 0 < 2 e and q 0 g < e h < e q 1 g. If and when such rationals are found, we ill initial segments of T 0 that specify q 0 g < G h < G q 1 g (if any exist). Notice that [T 1 ] as e=0 2 e = 2 < 4. To get a third order of low degree d 2 {d 0, d 1 }, we repeat this process to construct a (d 0 d 1 )-computable subtree T 2 of T 1 such that T 2 has no d 1 -computable paths. We note that T 2 cannot have any d 0 -computable paths as it is a subtree of T 1. The only change we need to mae is to require the rationals q 0 and q 1 (being used to diagonalize against the e th (candidate) d 1 -computable order e ) to satisfy q 1 q 0 < 2 (e+1). Since e=0 2 (e+1) = 1 < 2, we guarantee that [T 2 ]. Continuing to repeat this process in the obvious way yields the proposition. We close by asing the following general question. Question 4.6. Describe the possible degree spectra of orders X(G) on a computable presentation G of a computable torsion-free abelian group. Acnowledgements The first author was partially supported by a grant from the Pacard Foundation through a Post-Doctoral Fellowship. The second author was partially supported by NSF DMS and NSF DMS The authors thans Daniel Turetsy for allowing them to include Proposition 4.5. References [1] T.C. Craven, The Boolean space of orderings of a field, Transactions of the American Mathematical Society, vol. 209, 1975, [2] M.A. Dabowsa, M.K. Dabowsi, V.S. Harizanov and A.A. Togha, Spaces of orders and their Turing degree spectra, Annals of Pure and Applied Logic, vol. 161, 2010, [3] V. Dobritsa, Some constructivizations of abelian groups, Siberian Journal of Mathematics, vol. 24, 1983, [4] R. Downey and S.A. Kurtz, Recursion theory and ordered groups, Annals of Pure and Applied Logic, vol. 32, 1986, [5] H.M. Friedman, S.G. Simpson and R.L. Smith, Countable algebra and set existence axioms, Annals of Pure and Applied Logic, vol. 25, 1983, [6] L. Fuchs, Partially ordered algebraic systems, Pergamon Press, Oxford, [7] K. Hatziiriaou and S.G. Simpson, WKL 0 and orderings of countable abelian groups, in Logic and computation (Pittsburgh, PA, 1987), vol. 106 of Contemporary Mathematics, , American Mathematical Society, Providence RI, [8] A. Koorin and V. Kopytov, Fully ordered groups, Halsted Press, New Yor, 1974.

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