DEGREES OF ORDERS ON TORSIONFREE ABELIAN GROUPS


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1 DEGREES OF ORDERS ON TORSIONFREE ABELIAN GROUPS ASHER M. KACH, KAREN LANGE, AND REED SOLOMON Abstract. We show that if H is an effectivey competey decomposabe computabe torsionfree abeian group, then there is a computabe copy G of H such that G has computabe orders but not orders of every (Turing) degree. 1. Introduction A recurring theme in computabe agebra is the study of the compexity of reations on computabe structures. For exampe, fix a natura mathematica reation R on some cass of computabe agebraic structures such as the successor reation in the cass of inear orders or the atom reation in the cass of Booean agebras. One can consider whether each computabe structure in the cass has a computabe copy in which the reation is particuary simpe (say computabe or ow or incompete) or whether there are structures for which the reation is as compicated as possibe in every computabe presentation. For the successor reation, Downey and Moses [9] show there is a computabe inear order L such that the successor reation in every computabe copy of L is as compicated as possibe, namey compete. On the other hand, Downey [5] shows every computabe Booean agebra has a computabe copy in which the set of atoms is incompete. Aternatey, one can expore the connection between definabiity and the computationa properties of the reation R. More abstracty, one can start with a set S of Turing (or other) degrees and as whether there is a reation R on a computabe structure A such that the set of degrees of the images of R in the computabe copies of A is exacty S. For exampe, Hirschfedt [13] proved that this is possibe if S is the set of degrees of a uniformy c.e. coection of sets. One can aso consider reations such as being a cooring for a computabe graph or being a basis for a torsionfree abeian group. In these exampes, for each fixed computabe structure, there are many subsets of the domain (or functions on the domain) satisfying the property. It is natura to as whether there are computabe structures for which a of these instantiations are compicated and whether this compexity depends on the computabe presentation. In the case of coors of a panar graph, Remme [22] proves that one can code arbitrary Π 0 1 casses (up to permuting the coors) by the coection of coorings. For torsionfree abeian groups, there is a computabe group G such that every basis computes 0. However, for any computabe H, one can find a computabe copy of the given group in which there is a computabe basis (see Dobritsa [4]). Therefore, whie every basis can be compicated in one computabe presentation, there is aways a computabe presentation having a computabe basis. Date: Juy 27, Mathematics Subject Cassification. Primary: 03D45; Secondary: 06F20. Key words and phrases. ordered abeian group, degree spectra of orders. 1
2 2 KACH, LANGE, AND SOLOMON In this paper, we present a resut concerning computabiitytheoretic properties of the spaces of orderings on abeian groups. To motivate these properties, we compare the nown resuts on computationa properties of orderings on abeian groups with those for fieds. We refer the reader to [11] and [14] for a more compete introduction to ordered abeian groups and to [16] for bacground on ordered fieds. Definition 1.1. An ordered abeian group consists of an abeian group G = (G; +, 0) and a inear order G on G such that a G b impies a + c G b + c for a c G. An abeian group G that admits such an order is orderabe. Definition 1.2. The positive cone P (G; G ) of an ordered abeian group (G; G ) is the set of nonnegative eements P (G; G ) := {g G 0 G G g}. Because a G b if and ony if b a P (G; G ), there is an effective onetoone correspondence between positive cones and orderings. Furthermore, an arbitrary subset X G is the positive cone of an ordering on G if and ony if X is a semigroup such that X X 1 = G and X X 1 = {0 G }, where X 1 := { g g X}. We et X(G) denote the space of a positive cones on G. Notice that the conditions for being a positive cone are Π 0 1. The definitions for ordered fieds are much the same, and we et X(F) denote the space of a positive cones on the fied F. We suppress the definitions here as the resuts for fieds are ony used as motivation. As in the case of abeian groups, the conditions for a subset of F to be a positive cone are Π 0 1. Cassicay, a fied F is orderabe if and ony if it is formay rea, i.e., if 1 F is not a sum of squares in F; and an abeian group G is orderabe if and ony if it is torsionfree, i.e., if g G and g 0 G impies ng 0 G for a n N with n > 0. In both cases, the effective version of the cassica resut is fase: Rabin [21] constructed a computabe formay rea fied that does not admit a computabe order, and Downey and Kurtz [6] constructed a computabe torsionfree abeian group (in fact, isomorphic to Z ω ) that does not admit a computabe order. Despite the faiure of these cassifications in the effective context, we have a good measure of contro over the orders on formay rea fieds and torsionfree abeian groups. Because the conditions specifying the positive cones in both contexts are Π 0 1, the sets X(F) and X(G) are cosed subsets of 2 F and 2 G respectivey, and hence under the subspace topoogy they form Booean topoogica spaces. If F and G are computabe, then the respective spaces of orders form Π 0 1 casses, and therefore computabe formay rea fieds and computabe torsionfree abeian groups admit orders of ow Turing degree. For fieds, one can say consideraby more. Craven [2] proved that for any Booean topoogica space T, there is a formay rea fied F such that X(F) is homeomorphic to T. Transating this resut into the effective setting, Metaides and Nerode [20] proved that for any nonempty Π 0 1 cass C, there is a computabe formay rea fied F such that X(F) is homeomorphic to C via a Turing degree preserving map. Friedman, Simpson, and Smith [10] proved the corresponding resut in reverse mathematics that WKL 0 is equivaent to the statement that every formay rea fied is orderabe. Most of the corresponding resuts for abeian groups fai. For exampe, a countabe torsionfree abeian group G satisfies either X(G) = 2 (if the group has ran one) or X(G) = 2 ℵ0 and X(G) is homeomorphic to 2 ω. For a computabe
3 DEGREES OF ORDERS ON TORSIONFREE ABELIAN GROUPS 3 torsionfree abeian group G, even if one ony considers infinite Π 0 1 casses of separating sets (which are cassicay homeomorphic to 2 ω ) and ony requires that the map from X(G) into the Π 0 1 cass be degree preserving, one cannot represent a such casses by spaces of orders on computabe torsionfree abeian groups. (See Soomon [25] for a precise statement and proof of this resut.) However, the connection to Π 0 1 casses is preserved in the context of reverse mathematics as Hatziiriaou and Simpson [12] proved that WKL 0 is equivaent to the statement that every torsionfree abeian group is orderabe. Because torsionfree abeian groups are a generaization of vector spaces, notions such as inear independence pay a arge roe in studying these groups. Definition 1.3. Let G be a torsionfree abeian group. Eements g 0,..., g n are ineary independent (or just independent) if for a c 0,..., c n Z, c 0 g 0 + c 1 g c n g n = 0 G impies c i = 0 for 0 i n. An infinite set of eements is independent if every finite subset is independent. A maxima independent set is a basis and the cardinaity of any basis is the ran of G. Soomon [25] and Dabowsa, Dabowsi, Harizanov, and Tonga [3] estabished that if G is a computabe torsionfree abeian group of ran at east two and B is a basis for G, then G has orders of every Turing degree greater than or equa to the degree of B. Therefore, the set deg(x(g)) := {d d = deg(p ) for some P X(G)} contains a the Turing degrees when the ran of G is finite (but not one) and contains cones of degrees when the ran is infinite. As mentioned earier, Dobritsa [4] proved that every computabe torsionfree abeian group has a computabe copy with a computabe basis. Therefore, every computabe torsionfree abeian group has a computabe copy that has orders of every Turing degree, and hence has a copy in which deg(x(g)) is cosed upwards. Our broad goa, which we address one aspect of in this paper, is to better understand which Π 0 1 casses can be reaized as X(G) for a computabe torsionfree abeian group G and how the properties of the space of orders changes as the computabe presentation of G varies. Specificay, is deg(x(g)) aways upwards cosed? If not, does every group H have a computabe copy in which it fais to be upwards cosed? We show that if H is effectivey competey decomposabe, then there is a computabe G = H such that deg(x(g)) contains 0 but is not cosed upwards. We conjecture that this statement is true for a computabe infinite ran torsionfree abeian groups. Definition 1.4. A computabe infinite ran torsionfree abeian group H is effectivey competey decomposabe if there is a uniformy computabe sequence of ran one groups H i, for i ω, such that H is equa to i ω H i (with the standard computabe presentation). There are a number of recent resuts concerning computabiity theoretic properties of cassicay competey decomposabe groups in, for exampe, [7], [8], and [19]. Our main resut is the foowing theorem.
4 4 KACH, LANGE, AND SOLOMON Theorem 1.5. Let H be an effectivey competey decomposabe computabe infinite ran torsionfree abeian group. There is a computabe presentation G of H and a noncomputabe, computaby enumerabe set C such that: The group G has exacty two computabe orders. Every Ccomputabe order on G is computabe. Thus, the set of degrees of orders on G is not cosed upwards. If H is effectivey competey decomposabe, then deg(x(h)) contains a Turing degrees because H has a computabe basis formed by choosing a nonidentity eement h i from each H i. Therefore, athough the group G in Theorem 1.5 is competey decomposabe in the cassica sense, it cannot be effectivey competey decomposabe. In genera, one does not expect the coection of degrees reaizing a reation on a fixed computabe copy of an agebraic structure to be upwards cosed and hence this resut is not surprising from that perspective. However, the corresponding resut for the basis of a computabe torsionfree abeian group fais. Proposition 1.6. Let H be an infinite ran torsionfree abeian group with a computabe basis B. For every set D, there is a basis B D of H such that deg(b D ) = deg(d). Proof. Let B = {b 0, b 1,...} be effectivey isted such that b i < N b i+1. Fix a set D. Let B D = {n 0 b 0, n 1 b 1,...} where the n i N are chosen so that n i b i < N n i+1 b i+1 and n i is even if and ony if i D. It is cear that B D is a basis for H and that B D T D. To compute D from B D, et B D = {c 0, c 1,...} be isted in increasing order. For each i, we can find c i effectivey in B D, and then we can effectivey (with no orace) find b i and n i such that c i = n i b i. By testing whether n i is even or odd, we can determine whether i D. In Section 2, we present bacground agebraic information. In Section 3, we give the proof of Theorem 1.5. In Section 4, we state some generaizations of our resuts, present some reated open questions, and finish with remars concerning the foowing genera question. Question 1.7. Describe the possibe degree spectra of orders X(G) on a computabe presentation G of a computabe torsionfree abeian group. Our notation is mosty standard. In particuar we use the foowing convention from the study of inear orders: If G is a inear order on G, then G denotes the inear order defined by x G y if and ony if y G x. Note that if (G; G ) is an ordered abeian group, then (G; G ) is aso an ordered group. 2. Agebraic bacground In our proof of Theorem 1.5, we wi need two facts from abeian group theory. The first fact is that every computabe ran one group can be effectivey embedded into the rationas. To define this embedding for a ran one H, fix any nonidentity eement h H. Every nonidentity eement g H satisfies a unique equation of the form nh = mg where n N, m Z, n, m 0, and gcd(n, m) = 1. Map H into Q by sending 0 H to 0 Q, sending h to 1 Q, and sending g satisfying nh = mg (with n constraints as above) to the rationa m. Because this map is effective, the image of H in Q is computaby enumerabe and hence we can view H as a computaby
5 DEGREES OF ORDERS ON TORSIONFREE ABELIAN GROUPS 5 enumerabe subgroup of Q. Athough the image need not be computabe, it does contain Z and, more generay, is cosed under mutipication by any integer. If H = i ω H i is effectivey competey decomposabe, we can effectivey map H into Q ω = i ω Q (with its standard computabe presentation) by fixing a nonidentity eement h i H i for each i and mapping H i into Q as above. Therefore, we wi often treat H as a computaby enumerabe subgroup of Q ω, and, in particuar, treat eements in each H i subgroup as rationas. The second fact we need is Baer s Theorem (see [1]) giving cassica agebraic invariants for ran one groups. The Baer sequence of a ran one group is a function of the form f : ω ω { } moduo the equivaence reation defined on such functions by f g if and ony if f(n) g(n) for at most finitey many n and ony when neither f(n) nor g(n) is equa to. To define the Baer sequence of a ran one group H, fix a nonidentity eement h H and et {p i } i ω denote the prime numbers in increasing order (ater, for notationa convenience, we ater the indexing to start with one). For a prime p, we say p divides h (in H) if p g = h for some g H. We define the pheight of an eement h by { if is greatest such that p divides h, ht p (h) := otherwise, i.e., if p divides h for a. The Baer sequence of h is the function B H,h (n) = ht pn (h). If h, ĥ H are nonidentity eements, then B H,h B H, ĥ. The Baer sequence B H of the group H is (any representative of) this equivaent cass. Baer s Theorem states that for ran one groups, H 0 = H1 if and ony if B H0 B H1. 3. Proof of Theorem 1.5 Fix an effectivey competey decomposabe group H = i ω H i as in the statement of Theorem 1.5. We divide the proof into three steps. First, we describe our genera method of buiding the computabe copy G = (G; + G, 0 G ) which is 0 2 isomorphic to H. Second, we describe how the computabe ordering G on G is constructed. (The second computabe order on G is G.) Third, we give the construction of C and the diagonaization process to ensure the ony Ccomputabe orders on G are G and G. Part 1. Genera Construction of G. The group G is constructed in stages, with G s denoting the finite set of eements in G at the end of stage s. We maintain G s G s+1 and et G := s G s. We define a partia binary function + s on G s giving the addition facts decared by the end of stage s. To mae G a computabe group, we do not change any addition fact once it is decared, so we maintain x + s y = z = ( t s) [x + t y = z] for a x, y, z G s. Furthermore, for any pair of eements x, y G s, we ensure the existence of a stage t and an eement z G t such that we decare x + t y = z. To define the addition function, we use an approximation {b s 0, b s 1,..., b s s} G s to an initia segment of our eventua basis for G. During the construction, each approximate basis eement b s i wi be redefined at most finitey often, so each wi eventuay reach a imit. We et b i := im s b s i denote this imit. If is an even
6 6 KACH, LANGE, AND SOLOMON index then the approximate basis eement b s wi never be redefined, so athough we often use the notation b s (for uniformity), we have b = b s for a s. Athough G wi not be effectivey decomposabe, the group G wi decompose cassicay into a countabe direct sum using the basis B = {b 0, b 1, b 2,...}. At stage 0, we begin with G 0 := {0, 1}. We et 0 denote the identity eement 0 G and we assign 1 the abe b 0 0. We decare 0 G G = 0 G, 0 G + 0 b 0 0 = b 0 0, and b G = b 0 0. More generay, at stage s, each eement g G s is assigned a Qinear sum over the stage s approximate basis of the form q0b s s qnb s s n where n s, qi s Q for i n, and qs n 0. (Later there wi be further restrictions on the vaues of qi s to ensure that G is isomorphic to H.) This assignment is required to be onetoone, and the identity eement 0 G is aways assigned the empty sum. It wi often be convenient to extend such a sum by adding more approximate basis eements on the end of the sum with coefficients of zero. For exampe, the eement 0 G can be represented by any sum in which a the coefficients are zero. We trust that using such extensions wi not cause confusion. We define the partia function + s on G s by etting x + s y = z (for x, y, z G s ) if the assigned sums for x and y add together to form the assigned sum for z. Thus, the function + s is commutative and associative (on the eements for which it is defined) and satisfies x + s 0 G = x for a x G s. For each i ω, we fix a nonidentity eement h i H i and embed H i into Q by sending h i to 1 Q as described in Section 2. We equate H i with its image in Q in the sense of treating eements of H i as rationas. For exampe, if a H i and q Q, we et qa Q denote the product of q with the image of a under this (fixed) embedding of H i. (Reca that whie we cannot determine effectivey whether the rationa qa is in H i, if qa is in H i, then we wi eventuay see this fact.) In particuar, since h i is mapped to 1 Q, if a H i and a = qh i, we view a as being the rationa q. At each stage s, we maintain positive integers Ni s for i s. These integers restrain the (nonzero) coefficients qi s of b s i aowed in the Qinear sum for each eement g G s by requiring that qi sn i s H i and that we have seen this fact by stage s. Using the fact that N i := im s Ni s exists and is finite for a i, we wi show (using Baer s Theorem) that in the imit, the ith component of G is isomorphic to H i, and hence that G is a computabe copy of H. (Later we wi introduce a basis restraint K ω that wi prevent us from changing Ni s too often.) During stage s + 1, we do one of two things either we eave our approximate basis unchanged or we add a dependency reation for a singe b s for some odd index s. The diagonaization process dictates which happens. Case 1. If we eave the basis unchanged, then we define b s+1 i := b s i for a i s. For each g G s (viewed as an eement of G s+1 ), we define q s+1 i := qi s and assign g the same sum with b s+1 i and q s+1 i in pace of b s i and qs i, respectivey. It foows that x + s+1 y = z (for x, y, z G s ) if x + s y = z. We set N s+1 i := Ni s for a i s and Ns+1 s+1 := 1. We add two new eements to G s+1, abeing the first by b s+1 s+1 and abeing the second by q0 s+1 b s qn s+1 b s+1 n, where q0 s+1,..., qn s+1 is the first tupe of rationas (under some fixed computabe enumeration of a tupes of rationas) we find such that n s, qn s+1 0, q s+1 i N s+1 i H i at stage s for a i n, and this
7 DEGREES OF ORDERS ON TORSIONFREE ABELIAN GROUPS 7 sum is not aready assigned to any eement of G s+1. (We can effectivey search for such a tupe.) This competes the description of G s+1 in this case. Case 2. If we redefine the approximate basis eement b s (for the sae of diagonaizing) by adding a new dependency reation, then we proceed as foows. We define b s+1 i := b s i for a i s with i. The diagonaization process wi te us either to set b s = qbs+1 for some rationa q, or to set b s = m 1b s+1 j + m 2 b s+1 for some integers m 1 and m 2. (We wi specify properties of these integers beow.) In either case, the index wi be even and greater than the basis restraint K and j, <. We assign g G s the same sum except we repace each b s i by bs+1 i (for i s and i ) and we repace b s by either qbs+1 or m 1 b s+1 j + m 2 b s+1 (as dictated by the diagonaization process). For exampe, if the diagonaization process tes us to mae b s = m 1b s+1 j + m 2 b s+1, then the sum for g G s changes from q s 0b s 0 + q s j b s j + + q s b s + + q s b s + + q s sb s s at stage s (where we have added zero coefficients if necessary) to q0b s s qj s b s+1 j + + qb s s q s (m 1 b s+1 j + m 2 b s+1 ) + + qsb s s+1 s = q0b s s (qj s + q s m 1 )b s+1 j + + (q s + q s m 2 )b s qsb s s+1 s at stage s+1. Therefore, we set q s+1 j := qj s+qs m 1, q s+1 := q s+qs m 2, and q s+1 := 0, whie eaving q s+1 i := qi s for a i {j,, }. Simiary, if the diagonaization process tes us to mae b s = qbs+1, then we set q s+1 := q s + qqs and qs+1 := 0 whie eaving q s+1 i = qi s for a i {, }. We define N s+1 i, for i s, as foows. If b s = m 1b s+1 j +m 2 b s+1, then N s+1 i := Ni s i := Ni s for a i s with i and for a i s. If b s = qbs+1, then N s+1 N s+1 := d q dn s where d q is the denominator of q (when written in owest terms) and d is the product of a the (finitey many) denominators of coefficients q s for g G s. In either case, set Ns+1 s+1 := 1. We add three new eements to G s+1, abeing the first by b s+1, abeing the second by b s+1 s+1, and abeing the third by qs+1 0 b s qn s+1 b s+1 n where q0 s+1,..., qn s+1 is the first tupe of rationas we find such that n s, qn s+1 0, q s+1 i N s+1 i H i at stage s for a i n, and this sum is not aready assigned to any eement of G s+1. This competes the description of G s+1 in this case. We note severa trivia properties of the transformations of sums in Case 2. First, the approximate basis eement b s+1 does not appear in the new sum for any eement of G s viewed as an eement of G s+1. Second, for any eement g G s, if q s = 0, then the coefficients q s+1 j and q s+1 satisfy q s+1 j = qj s and qs+1 = q s. Third, by the inearity of the substitutions, if x + s y = z, then x + s+1 y = z. We aso require two additiona properties which pace some restrictions on the rationa q or the integers m 1 and m 2. The first property is that the assignment of sums to eements of G s (viewed as eements of G s+1 ) remains onetoone. The diagonaization process wi pace some restrictions on the vaue of either q or m 1 and m 2, but as ong as there are infinitey many possibe choices for these vaues (which we wi verify when we describe the diagonaization process), we can assume they are chosen to maintain the onetoone assignment of sums to eements of G s+1.
8 8 KACH, LANGE, AND SOLOMON The second property is that for each g G s+1, we need each coefficient q s+1 i to satisfy q s+1 i N s+1 i H i. We wi verify this property beow under the assumption that when we set b s = m 1b s+1 j +m 2 b s+1, the integers m 1 and m 2 are chosen so that they are divisibe by the denominator of each q s coefficient of each g G s. (Again, we wi verify this property of m 1 and m 2 in the description of the diagonaization process.) We now chec various properties of this construction under these assumptions and the assumption that the imits b i := im s b s i and N i := im s Ni s exist for a i (which wi be verified in the diagonaization description). Lemma 3.1. For g G s, the coefficients in the assigned sum q s 0b s q s nb s n satisfy q s i N s i H i. Proof. The proof proceeds by induction on s. If g is added at stage s, then the resut for g foows triviay. Therefore, fix g G s and assume the condition hods at stage s. We show that the condition continues to hod at stage s + 1. Note that if we do not add a dependency reation (i.e., we are in Case 1), then the condition at stage s + 1 foows immediatey since q s+1 i = qi s and N s+1 i = Ni s. Assume we add a new dependency reation; we spit into cases depending on the form of this dependency. If b s = qbs+1, then for a i {, }, the condition hods since q s+1 i = qi s and = Ni s. For the index, we have qs+1 triviay. For the index, we have q s+1 = q s + qqs N s+1 i = 0 and hence the condition hods s+1 and N = d q dn s. Therefore, q s+1 N s+1 = (q s + qq s )d q dn s = q s d q dn s i + qq s d q dn s. Since q sn s H and d q d Z, we have q sd qdn s H. By definition, qd q Z and q sd Z, and hence qqs d qdn s Z H. Therefore, we have the desired property when b s = qbs+1. If b s = m 1b s+1 j + m 2 b s+1, then for a i {j, } the condition hods as above. For the index j, we have q s+1 j = qj s + qs m 1 and N s+1 j = Nj s. By assumption, the integer m 1 is divisibe by the denominator of q s and hence qs m 1 Z. Therefore, q s+1 j N s+1 j = (qj s + q s m 1 )Nj s = qj s Nj s + q s m 1 Nj s H j since qj sn j s H j by the induction hypothesis and q sm 1Nj s the index is identica. Z. The anaysis for Lemma 3.2. 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 ater changed in the sense that, for a stages u t, the eement 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 a i n. Proof. When g enters G, it is assigned a sum. The coefficients in this sum ony change when a diagonaization occurs. In this case, some approximate basis eement b s with nonzero coefficient in the sum for g is made dependent via a reation of the form b s = qbs+1 or b s = m 1b s+1 j + m 2 b s+1 with j, <. Therefore, each time the sum for g changes, some approximate basis eement with nonzero coefficient is repaced by rationa mutipes of approximate basis eements with ower indices. This process can ony occur finitey often before terminating.
9 DEGREES OF ORDERS ON TORSIONFREE ABELIAN GROUPS 9 We refer to the sum in Lemma 3.2 as the imiting sum for g and denote it by q 0 b 0 + +q n b n. It foows from Lemma 3.1 and Lemma 3.2 that each coefficient q i in a imiting sum satisfies q i N i H i. Lemma 3.3. For each rationa tupe q 0,..., q n such that q n 0 and q i N i H i for a i n, there is an eement g G such that the imiting sum for g is q 0 b q n b n. Proof. For a contradiction, suppose there is a rationa tupe vioating this emma. Fix the east such tupe q 0,..., q n in our fixed computabe enumeration of rationa tupes. Let s n be a stage such that b s 0,..., b s n and N0 s,..., Nn s have reached their imits, each tupe before q 0,..., q n which satisfies the conditions in the emma has appeared as the imiting sum of an eement in G s, and we have seen by stage s that q i N i H i for each i n. By our construction, at stage s + 1, either there is an eement that is assigned the sum q 0 b s q n b s+1 n or ese we add a new eement to G s+1 and assign it this sum. In either case, this eement has the appropriate imiting tupe since b s+1 0,..., b s+1 n have reached their imits (and thus we obtain our contradiction). By Lemma 3.3 and the remars foowing Lemma 3.2, the imiting sums of eements of G are exacty the sums q 0 b q n b n with q n 0 and q i N i H i for a i n. Lemma 3.4. If x+ s y = z, then x+ t y = z for a t s. In particuar, if x+ s y = z, then the imiting sums for x and y add to form the imiting sum for z. Proof. In both cases of stage s + 1 of our construction, we checed that x + s y = z impies x + s+1 y = z. Thus, the resut foows by induction. Lemma 3.5. For each pair x, y G s, there is a stage t s and an eement z G t such that x+ t y = z. For each x G s, there is a stage t s and an eement z G t such that x + t z = 0 G. Proof. For the first statement, fixing x, y G s, et u s be a stage at which x and y have been assigned their imiting sums x = q u 0 b u q u nb u n and y = ˆq u 0 b u ˆq u nb u n, adding zero coefficients if necessary to mae the engths equa. By Lemma 3.1, for a t u and i n, we have that q t i N t i H i and ˆq t i N t i H i. Therefore, (q t i + ˆqt i )N t i H i. As in the proof of Lemma 3.3, there must be a stage t u and an eement z G t assigned to the sum z = (q t 0 + ˆq t 0)b t (q t n + ˆq t n)b t n. Then x + t y = z. The proof of the second statement is simiar. Using Lemma 3.4 and Lemma 3.5, we define the addition function + G on G by putting x + y = z if and ony if there is a stage s such that x + s y = z. Lemma 3.6. The group G is a computabe copy of H. Proof. The domain and addition function on G are computabe. By Lemma 3.5, every eement of G has an inverse, and it is cear from the construction that the addition operation satisfies the axioms for a torsionfree abeian group.
10 10 KACH, LANGE, AND SOLOMON Let G i be the subgroup of G consisting of a eement g G with imiting sums of the form q i b i. Since the imiting sums of eements of G are exacty the sums of the form q 0 b q n b n with q n 0 and q i N i H i for i n, it foows that G = i ω G i. Therefore, to show that G = H, it suffices to show that G i = Hi for every i ω. Fix i ω. The group G i is a ran one group which is isomorphic to the subgroup of (Q, + Q ) consisting of the rationas q such that qn i H i. Thus, cacuating the Baer sequence for G i using the rationa 1 Q, we note that for any prime p j, 1/p j G i if and ony if N i /p j H i. Therefore, the entries in the Baer sequences for G i and H i differ ony in the vaues corresponding to the prime divisors of N i and they differ exacty by the powers of these prime divisors. Therefore, by Baer s Theorem, G i = Hi. Part 2. Defining the Computabe Orders on G. We define the computabe ordering of G in stages by specifying a partia binary reation s on G s at each stage s. To mae the ordering reation computabe, we satisfy x s y = ( t s) [x t y] (1) for a x, y G s. Typicay, the reation s wi not describe the ordering between every pair of eements of G s, but it wi have the property that for every pair of eements x, y G s, there is a stage t s at which we decare x t y or y t x, and not both uness x = y. Since we wi be considering severa orderings on G, for an ordering on G, we et (g 1, g 2 ) denote the set {g G g 1 g g 2 }. Moreover, given a 1, a 2 R, we et (a 1, a 2 ) R denote the interva {a R a 1 < R a < R a 2 }. To specify the computabe order on G, we buid a 0 2map from G into R. (Thus our order wi be archimedean.) To describe this order, et {p i } i 1 enumerate the prime numbers in increasing order. We map the basis eement b 0 to r 0 = 1 R. For i 1, we wi assign (in the imit of our construction) a rea number r i to the basis eement b i such that r i is a positive rationa mutipe of p i. We choose the r i in this manner so that they are agebraicay independent over Q. If the eement g G is assigned a imiting sum g = q 0 b q n b n, then our 0 2map into R sends g to the rea q 0 r q n r n. It aso sends 0 G to 0. We need to approximate this 0 2map during the construction. At each stage s, we eep a rea number ri s as an approximation to r i, viewing ri s as our current target for the image of b i. The rea r0 s is aways 1 and the rea ri s is aways a positive rationa mutipe of p i. Exacty which rationa mutipe may change during the course of the diagonaization process. However, if is an even index, then r s wi never change. We coud generate a computabe order on G s by mapping G s into R using a inear extension of the map sending each b s i to ri s. However, this woud restrict our abiity to diagonaize. Therefore, at stage s, we assign each b s i (for i 1) an interva (a s i, âs i ) R where a s i and âs i are positive rationas such that rs i (as i, âs i ) R and â s i as i 1/2s. The image of b s i in R (in the imit) wi be contained in this interva. Because each x G s is assigned a sum describing its reationship to the current approximate basis, we can generate an interva approximating the image of x in R
11 DEGREES OF ORDERS ON TORSIONFREE ABELIAN GROUPS 11 under the 0 2map. That is, suppose x is assigned the sum x = q s 0b s q s nb s n at stage s. The interva constraints on the image of each b s i in R transate into a rationa interva constraint on the image of x in R. The endpoints of this constraint can be cacuated using the coefficients of the sum for x and the rationas a s i and âs i, with the exact form depending on the signs of the coefficients. To define s on G s at stage s, we oo at the interva constraints for each pair of distinct eements x, y G s. If the interva constraint for x is disjoint from the interva constraint for y, then we decare x s y or y s x depending on which inequaity is forced by the constraints. If the interva constraints are not disjoint, then we do not decare any ordering reation between x and y at stage s. Of course, we aso decare x s x for each x G s. To maintain the impication in Equation (1), we wi need to chec that x s y impies x s+1 y. It suffices to ensure that for each x G s, the interva constraint for x at stage s + 1 is contained within the interva constraint for x at stage s. It wi be hepfu for us to now that certain approximate basis eements are mapped to eements of R which are cose to 0 R. Therefore, we wi maintain that 0 a s âs < 1/2 for a stages s and a even indices. (If we wored in a simper context where each H i = Q, or even where each H i Z, we coud sip this step as any archimedean order on such groups H i is dense in R.) We now describe exacty how ri t, at i and ât i are defined at each stage t. Reca that at stage t = 0, the ony eements in G t are 0 G (which is represented by the empty sum and is mapped to 0 R ) and the eement represented by b 0 0 (which is mapped to 1 R ). We set r0 0 := 1 R. At stage t + 1, the definitions of r t+1 i, a t+1 i and â t+1 i for i t depend on whether we add a dependency reation or not. If we do not add a dependency reation, or if i is not an index invoved in an added dependency reation, then we define r t+1 i := ri t (so we maintain our guess at the target rationa mutipe of p i for b i ) and define a t+1 i and â t+1 i so that (a t+1 i, â t+1 i ) R (a t i, â t i) R, r t+1 i (a t+1 i, â t+1 i ) R, and â t+1 i a t+1 i < 1/2 t+1. For the approximate basis eement b t+1 t+1 introduced at this stage, we set rt+1 to be a positive rationa mutipe of p t+1 (requiring r t+1 a t+1 t+1 and ât+1 t+1 t+1 < 1/2t+1 if t+1 is even) and et be positive rationas so that rt+1 (at+1, ât+1 ) R and â t+1 t+1 at+1 < t+1 < 1/2t+1 if t + 1 is even). The diagonaization process may 1/2 t+1 (and aso â t+1 pace some requirements on the rationa mutipe of p t+1 chosen. It remains to hande the indices invoved in a dependency reation of the form b t = qbt+1 or b t = m 1b t+1 j m 2 b t+1. In either case wi be odd and we define r t+1 := p and a t+1, â t+1 Q + such that r t+1 (a t+1, â t+1 ) R and â t+1 a t+1 < 1/2 t+1. For the other indices invoved in an added dependency reation, we spit into cases depending on the type of reation added. (1) If we add a dependency of the form b t = qbt+1, then we set r t+1 := r t. The action of the diagonaization strategy wi ensure that we can choose a t+1, â t+1 Q + such that (a t+1, â t+1 ) R (a t, ât ) R, â t+1 a t+1 < 1/2 t+1 and (qa t+1, qâ t+1 ) R (a t, â t ) R (2)
12 12 KACH, LANGE, AND SOLOMON (2) If we add a dependency of the form b t = m 1b t+1 j m 2 b t+1, then we set r t+1 j := rj t and rt+1 := r t. We wi be in one of two contexts. 2(a). If we are in a context in which (in R) 0 < na t < nâ t < a t < â t < a t j < â t j < (n + 1)a t < (n + 1)â t, (3) then we wi 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, â t ) R. (4) 2(b). If we are in a context in which (in R) 0 < na t < nâ t < a t j < â t j < a t < â t < (n + 1)a t < (n + 1)â t, (5) then we wi 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, â t ) R. (6) By Lemma 3.8, in each of these contexts, there are infinitey many such choices for m 1 and m 2 satisfying the given conditions. Moreover, we can assume that m 1 and m 2 satisfy the divisibiity conditions required by the genera group construction. To expain why appropriate m 1, m 2 N exist for the two contexts above, we rey on the foowing fact about the reas. Lemma 3.7. Let r 1 and r 2 be positive reas that are ineary independent over Q. For any rationa numbers q 1 < q 2 and any integer d 1, there are infinitey many m 1, m 2 N such that m 1 r 1 m 2 r 2 (q 1, q 2 ) R and both m 1 and m 2 are divisibe by d. Lemma 3.8. If we are in the context of (3) (respectivey (5)), then there are infinitey many choices for m 1 and m 2 that are divisibe by any fixed integer d 1 and satisfy (4) (respectivey (6)). Proof. First, suppose we are in the context of (3). We have that b t j and bt are currenty identified with the rationa mutipes rj t and rt of p j and p respectivey, so rj t and rt are ineary independent over Q. Hence, by Lemma 3.7 (requiring m 1 and m 2 to be divisibe by nd where n comes from the context (3) and d comes from the statement of this emma), there are infinitey many choices of m 1, m 2 N such that m 1 rj t m 2r t (at, ât ) R. We et m 2 := m2 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 (4) by shrining the intervas (a t j, ât j ) R and (a t, ât ) R appropriatey. 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 (3). We have reached a contradiction since m 1 rj t m 2nr t (at, ât ) R and rj t (at j, ât j ) R but â t < at j. So, m 1 m2 n = m 2 as desired.
13 DEGREES OF ORDERS ON TORSIONFREE ABELIAN GROUPS 13 Now suppose we are in the context of (5). Since rj t and rt are ineary independent over Q, by Lemma 3.7 (requiring m 1 and m 2 to be divisibe by (n + 1)d) there are infinitey many choices of m 1, m 2 N such that m 1 r t m 2rj t (at, ât ) R. We et m 1 := m1 (n+1). As before, we can choose at+1 j, â t+1 j, a t+1, â t+1 Q satisfying (6). 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 inequaity foows because m 1 1 m 2 and rj t is positive, and the second inequaity foows because rj t < (n + 1)rt by (5). We have reached a contradiction since m 1 r t m 2rj t (at, ât ) R but â t < (n + 1)at. We verify severa properties of this genera ordering construction based on the assumptions that each approximate basis eement b s i eventuay reaches a imit and that we choose our intervas and associated rationas in the manner described above. Lemma 3.9. For every pair of eements x, y G s, if x s y, then x s+1 y. Hence, we have the impication given in (1). Proof. It suffices to show that for each g G s, the interva constraint for g at stage s + 1 is contained in the interva constraint for g at stage s. This fact foows from three observations. Fix g G s. First, if qi sbs i occurs in the sum for g at stage s and the index i is not invoved in an added dependency reation, then q s+1 i = qi s and (a s+1 i, â s+1 i ) R (a s i, âs i ) R. Therefore, the constraint imposed on g by these terms at stage s + 1 is contained in the constraint imposed at stage s. Second, suppose we add a dependency reation of the form b s = qbs+1 and q s 0. From stage s to stage s + 1, the qs bs + qs bs part of the sum for g turns into (q s +qqs )bs+1 +0b s+1 where b s+1 = b s. Since the constraint on rs+1 pays no roe in the constraint on g at stage s+1 and since we have, by (2), that (qa s+1, qâ s+1 ) R (a s, âs ) R, it foows that the constraint imposed by the indices and at stage s + 1 is contained in the constraint imposed at stage s. Third, if we add a dependency reation of the form b s = m 1b s+1 j m 2 b s+1, then a simiar anaysis using (4) and (6) yieds that the constraint imposed by the indices j, and at stage s+1 is contained in the constraint imposed at stage s. Lemma For each i, the imit r i := im s ri s exists and is a rationa mutipe of p i. Furthermore, once ri s reaches its imit, the rationa intervas (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 ony when b s+1 i b s i. Since the atter happens ony finitey often, each ri s reaches a imit. The remainder of the statement is immediate from the construction. Lemma 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 a x G s, we consider distinct eements x, y G s. Let t s be a stage such that x and y have reached their imiting sums and such that
14 14 KACH, LANGE, AND SOLOMON for each b t i occurring in these sums, the rea rt i has reached its imit r i. Because the reas r i are agebraicay independent over Q and the nested approximations (a u i, âu i ) R (for u t) converge to r i, there is a stage at which the interva constraints for x and y are disjoint. At the first such stage, we decare an ordering reation between x and y. We define the order G on G by x G y if and ony if x s y for some s. By Lemma 3.9 and Lemma 3.11, this reation is computabe and every pair of eements is ordered. By construction, the 0 2map from G to R that sends q 0 b 0 + q 1 b 1 + q n b n q 0 + q 1 r q n r n is order preserving. Therefore, our ordered group G is cassicay isomorphic to an ordered subgroup of (R; +, 0 R ) under the standard ordering. Part 3. Buiding C and Diagonaizing. It remains to show how to use this genera construction method to buid the ordered group (G; G ) together with a noncomputabe c.e. set C such that the ony Ccomputabe orders on G are G and G. The requirements S e : Φ e tota = C Φ e to mae C noncomputabe are met in the standard finitary manner. The strategy for S e chooses a arge 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 expain how to meet a singe R e in a finitary manner, eaving it to the reader to assembe the compete finite injury construction in the usua manner. To simpify the notation, we et C e be the binary reation on G computed by Φ C e. We wi assume throughout that C e never directy vioates any of the Π 0 1 conditions in the definition of a group order. For exampe, if we see at some stage s that C e has vioated transitivity, then we can pace a finite restraint on C to preserve these computations and win R e triviay. The strategy to satisfy R e is as foows. For R e, we set the basis restraint K := e. (This restraint is used in the verification that each Ni s reaches a imit.) If G C e and G C e, then there must eventuay be a stage s, an approximate basis eement b s j, a nonnegative integer n, and an even index > K such that: we have decared 0 < s nb s < s b s j < s (n + 1)b s in G s, and the order C e has decared 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. We verify such objects exist in Lemma In the atter case, we wor with the ordering C e, transforming the atter case into the former case. We therefore assume that we are in the former case. Whie waiting for these witnesses, the construction of G proceeds as in the genera description with no dependencies added. 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 0 G, b s j, nbs, and (n + 1)bs.
15 DEGREES OF ORDERS ON TORSIONFREE ABELIAN GROUPS 15 At stage s+1 (without oss of generaity, we assume s+1 is odd), we order the new approximate basis eement b s+1 s+1 depending on whether bs j <C e nb s or bs j >C e (n+1)b s. We say that R e is set up to diagonaize with diagonaization witness b s+1 (D1) If b s j <C e nb s, we order bs+1 so that nbs < s+1 b s+1 s+1 < s+1 b s j, that is, we choose rs+1 s+1 to be a rationa mutipe of p s+1 and rationas a s+1 s+1 and âs+1 so that nâ s < as+1 < rs+1 < âs+1 < as j and âs+1 as+1 < 1/2s+1. (D2) If b s j >C e (n + 1)b s, we order bs+1 so that bs j < s+1 b s+1 s+1 < s+1 (n + 1)b s, that is, we choose r s+1 s+1 to be a rationa mutipe of p s+1 and ra and âs+1 so that âs j < as+1 and s+1 < 1/2s+1. tionas a s+1 s+1 s+1. â s+1 s+1 as+1 We then wait for a stage t + 1 so that C e decares 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. Whie waiting, we assume that no higher priority S i strategy enumerates a number into C and that b u j = bs j, bu = bs, and bu s+1 = b s+1 s+1 at a stages u s + 1 unti R e finds such a stage t + 1 (or for a u s + 1 if R e never sees such a stage). If either of these conditions is vioated, R e is deactivated and returns to ooing for appropriate witnesses to become activated again. If these conditions hod, then we say R e has been activated with potentiay permanent witnesses. We assume that such a stage t + 1 is found, ese R e is triviay satisfied. At stage t + 1, R e acts to diagonaize by restraining C to preserve the computations ordering b t s+1, nb t, and (n + 1)bt under C e and adding a dependency reation as foows. Case 1. If C e decares b t s+1 < C e nb t or bt s+1 > C e (n + 1)b t, then we wi add a reation of the form b t s+1 = qb t+1. Since nb t < t b t s+1 < t (n + 1)b t, we now that nr t < R a t s+1 < R â t s+1 < R (n + 1)r t. There are infinitey many rationas q (n, n + 1) R such that qr t (a s+1, â t s+1) R. For each such q, there are rationas a t+1 and â t+1 such that r t+1 = r t (a t+1, â t+1 ) R (a t, â t ) R, â t+1 a t+1 R 1/2 t+1, and (qa t+1, qâ t+1 ) R (a t s+1, â t s+1) R. Choose q, a t+1, and â t+1 to be the first rationas meeting these conditions such that the assignment of sums to eements of G t remains onetoone. These choices satisfy the necessary requirements for both the group construction and the ordering construction. Furthermore, we have successfuy diagonaized against C e being an ordering of G since any order under which b t+1 = b t is positive must pace b t s+1 between nb t+1 and (n + 1)b t+1. However, 0 G < C e b t+1 and either b t s+1 < C e nb t or bt s+1 > C e (n + 1)b t. Case 2. If C e decares nb t <C e b t s+1 < C e (n + 1)b t, then we now 0 G < C e b t s+1 since 0 G < C e b t. We act depending on whether bt s+1 < t b t j or bt s+1 > t b t j. Case 2(a): If b t s+1 < t b t j, then it is because we acted in (D1) and hence we now that b t+1 j < C e nb t+1 and we are in the context of Equation (3) with = s+1. Let d be the product of a denominators of coefficients qs+1 t for a g G t. We decare b t s+1 = m 1 b t+1 j m 2 b t+1 for positive integers m 1 and m 2 both divisibe by d that satisfy m 1 N m 2 /n and the ordering constraints in
16 16 KACH, LANGE, AND SOLOMON Equation (4) and maintain the onetoone assignment of sums to eements of G t+1. (This choice is possibe by Lemma 3.8.) To see that we have successfuy diagonaized, we show that C e must vioate the order axioms. Since b t s+1 = m 1 b t+1 j m 2 b t+1 and 0 G < C e b t s+1, b t+1, we now 0 G < C e b t+1 j. Because m 1 N m 2 /n and 0 G < C e b t+1 j, we have b t s+1 = m 1 b t+1 j m 2 b t+1 C e (m 2 /n)b t+1 j m 2 b t+1. By our case assumption that b t+1 j < C e nb t+1, we get b t s+1 C e (m 2 /n)b t+1 j m 2 b t+1 < C e (m 2 /n)nb t+1 m 2 b t+1 = 0 G. We have arrived at a contradiction since we have both 0 G < C e b t s+1 (since we are in Case 2) and b t s+1 < C e 0 G by this cacuation. Case 2(b): If b t s+1 > t b t j, then it is because we acted in (D2) and hence we now (n + 1)b t+1 < C e b t+1 j and we are in the context of Equation (5) with = s + 1. Let d be as in Case 2(a) and decare b t s+1 = m 1 b t+1 m 2 b t+1 j for positive integers m 1 and m 2 both divisibe by d that satisfy m 1 N m 2 (n+1) and the ordering constraints in Equation (6) and maintain the onetoone assignment of sums to eements of G t+1 (again by Lemma 3.8.) We show that < C e must vioate the order axioms. Since 0 G < C e b t+1 and m 1 N m 2 (n + 1), we have b t s+1 = m 1 b t+1 m 2 b t+1 j C e m 2 (n + 1)b t+1 m 2 b t+1 j. By our case assumption that (n + 1)b t+1 < C e b t+1 j, we have b t s+1 C e m 2 (n + 1)b t+1 m 2 b t+1 j < C e m 2 b t+1 j m 2 b t+1 j = 0 G. Again, we have arrived at a contradiction since 0 G < C e b t s+1 (since we are in Case 2) and b t s+1 < C e 0 G (by this cacuation). This competes our description of the action of a singe requirement R e. In the fu construction, we set up priorities between S i requirements and R e requirements in the usua way. If i < e, then S i is aowed to enumerate its diagonaizing witness even if it destroys a restraint imposed by R e, but if e i, then S i must pic a new arge witness when R e imposes a restraint. There is aso a potentia confict between different R e requirements. Consider requirements R e and R i invoved in the foowing scenario. Assume that at stage s 0, R i is the highest priority activated requirement with witnesses b s0 j 0, b s0 and n 0. At stage s 0 +1, R i sets up to diagonaize with witness b s0+1 s 0+1 (via either (D1) or (D2)). At stage s 1 > s 0, whie R i is sti waiting to diagonaize, R e is activated with witnesses b s1 j 1, b s1 1, and n 1 such that the index j 1 = s Then R e sets up to diagonaize with b s1+1 s at stage s At stages after s 1 +1, R e is waiting for C e to decare an ordering reation between certain eements (which may never appear) and it needs to maintain b u j 1 = b s1 j 1 (which means b u s 0+1 = b s1 s 0+1 ) to remain in a position to diagonaize. On the other hand, when R i sees C i decare the appropriate order reations, it wants to add a dependency of the form b t s 0+1 = qb t+1 0 or b t s 0+1 = m 1 b t+1 j 0 + m 2 b t+1 0 which woud cause b t+1 s 0+1 (and hence bt+1 j 1 ) to be redefined. 0,
17 DEGREES OF ORDERS ON TORSIONFREE ABELIAN GROUPS 17 In this scenario, if e < i, then when R e sets up to diagonaize at stage s 1 + 1, it cances R i s caim on the diagonaizing witness b s0+1 s 0+1, thus removing the potentia confict. The requirement R i remains activated (since the appropriate C i computations have been preserved) and at the next odd stage s at which R i is the highest priority activated requirement, it wi set up to diagonaize with a new witness b s2+1 s. 2+1 If i < e, then no canceation of setup witnesses taes pace when R e sets up to diagonaize. If R e acts to diagonaize first, there is no confict because R e adds a dependency reation which causes b t+1 s 1+1 to be redefined, but eaves bt+1 j 1 = b t j 1 (and hence b t+1 s = 0+1 bt s 0+1). If R i acts first, then it does cause b t+1 s 0+1 (and hence bt+1 j 1 ) to be redefined, injuring R e. In this case, the witnesses in the activation for R e were not potentiay permanent and R e is deactivated and has to oo for new activating witnesses. Thus, in the fu construction, an R e requirement can be injured by a higher priority S i requirement (which becomes permanenty satisfied) or by a higher priority R i requirement (either because R i diagonaizes and is permanenty satisfied or because R i cances R e s diagonaizing witness and R e can pic a new diagonaizing witness with the same activation witnesses). Thus, the fu construction is finite injury. To verify the construction succeeds, we show that the imits im s b s i and im s Ni s exist and that if C e is an order but is not equa to G or G, then R e is eventuay activated with potentiay permanent witnesses. Lemma The imit b i := im s b s i exists for a i. Proof. The ony approximate basis eements which are redefined are those chosen as diagonaizing witnesses by some R e requirement. Therefore, at stage s + 1, if b s+1 s+1 is not chosen as a diagonaizing witness, then it is never redefined. If b s+1 s+1 is chosen as a diagonaizing witness by R e, then it can be redefined at most once when R e acts to diagonaize. Lemma The imit N i := im s N s i exists for a i. Proof. The ony time N s+1 i Ni s is when we add a dependency reation of the form b s = qbs+1 causing N s+1 = d q dn s. However, in this case, the index is even and a requirement R e can ony add such a dependency if > K = e. Therefore, ony R e with e < can cause N s to changes vaue. Since these requirements ony act finitey often, the vaue of N s changes ony finitey often. Lemma If we fai to find a stage s where R e is activated with potentiay permanent witnesses, 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. Let s be a stage such that a higher priority requirements have finished acting by s. It suffices to show that if we fai to find a stage s s at which R e is activated with some witnesses b s j, n, and, then C e is equa to G or G. First, we caim that if we fai to find a stage s s at which R e is activated, then either 0 G < C e b j for a j or b j < C e 0 G for a j. To prove this caim, suppose that R e is never activated after s and that j 0 and j 1 are indices with b j1 < C e 0 G < C e b j0. Fix a stage s s such that b s j 1 = b j1, b s j 0 = b j0 and b j1 < C e 0 G < C e b j0 is permanenty fixed by stage s. Consider a stage
18 18 KACH, LANGE, AND SOLOMON t s and an even index greater than the basis restraint for R e such that b t = b has reached its imit 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 decares either 0 G < C e b u or bu <C e 0 G permanenty. If 0 G < C e b u, then we must eventuay 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. Aternatey, if b u <C e 0 G, then we must eventuay 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 competes the proof of the caim. To compete the proof of this emma, assume that R e is never activated after s and 0 G < C e b j for a j. We show that C e = G. It foows by a simiar argument that if b j < C e 0 G for a j, then C e = G. By construction, (G; + G, 0 G, G ) can be embedded (as an ordered group) into (R; + R, 0 R, R ) by sending each basis eement 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, 0 G, C e ) into (R; + R, 0 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 equivaent 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 im n 0,r = im (n 0, + 1)r = r 0 = 1 where the imits (and a imits throughout this emma) are taen over even indices. More generay, 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 equivaent to n i, r R r i R (n i, + 1)r and we have Combining these imits, we have and im im n i,r = im (n i, + 1)r = r i. n i, n 0, + 1 = im n i, r (n 0, + 1)r = r i 1 = r i n i, + 1 (n i, + 1)r im = im n 0, n 0, r = r i 1 = r i. We now transate these resuts to (G, C e ). Because R e is never activated after s and 0 G < C e b for a even, the inequaities n i, b C e b i C e (n i, + 1)b hod for a i and a even such that is greater than the basis restraint for R e. In particuar, combining the inequaities 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 inequaity is interpreted as representing the corresponding inequaity after mutipying through by the denominators so a the coefficients are integers.
19 DEGREES OF ORDERS ON TORSIONFREE ABELIAN GROUPS 19 (Aternatey, this inequaity can be viewed in the divisibe cosure of G using the fact that an order on an abeian group has a unique extension to an order on its divisibe cosure.) The imits above show that the map sending b i to r i defines an embedding of (G; + G, 0 G, C e ) into (R; + R, 0 R, R ) as required. 4. Remars and Open Questions Since the construction of the presentation G and the set C is a typica finite injury construction, certain modifications to the constructions are straightforward. Remar 4.1. Rather than buiding G so that there are exacty two computabe orders, it is an easy modification to buid exacty any even number or an infinite number of computabe orders (with no other Ccomputabe orders). For exampe, to buid G with four computabe orders, we doube the number of R e requirements. We buid a computabe order 0 G in which 0 <0 G b 0 < 0 G b 1 and a computabe order 1 G in which 0 <1 G b 1 < 1 G b 0. For each of these orders, we meet a sighty modified requirement for i {0, 1}: R i e: If Φ C e is an ordering of G, then 0 G C e b i C e b 1 i impies C e = i G and b 1 i C e b i C e 0 G impies C e = i G. Note that this requirement suffices because (as shown in Lemma 3.14) if b 0 and b 1 ie on opposite sides of 0 G under C e, then R i e wi be activated and the diagonaization wi guarantee that Φ C e is not an order of G. Since these requirements are sti finitary (both restraint and injury) in nature, these combine easiy to yied the desired resut. The resut in Remar 4.1 contrasts with the cassica situation. As mentioned in Section 1, a countabe torsion free abeian group admits either two or continuum many orders. More generay, it is possibe for a countabe (nonabeian) group to admit either a finite number of orders greater than 2 or countaby many orders. In the finite case, the number of orders must be even and the best nown resuts are that is possibe to have exacty 4n or 2(4n + 3) many orders (see [15] and [18]). It is an open question whether it is possibe to get exacty 2n number of orders for each n. Remar 4.2. We note that the computaby enumerabe set C cannot be compete. The reason is that 0 can compute a basis for any computabe torsionfree abeian group G, and hence G has orders of degree 0. We aso note that, as ong as the construction remains finitary (both restraint and injury), additiona requirements on C can be added. For exampe, owness requirements coud be added, though this woud be counterproductive (the weaer C is computationay, the weaer the resut). Though maing C computationay wea is counterproductive, we as if it is possibe to mae C computationay strong. Question 4.3. Can the set C in Theorem 1.5 have high degree? Question 4.4. Does Theorem 1.5 remain true when G is aowed to be an arbitrary computabe torsionfree abeian group? We end with a resut concerning the genera project of understanding the possibe degree spectra of orders on computabe torsionfree abeian groups.
20 20 KACH, LANGE, AND SOLOMON Proposition 4.5 (With Danie Turetsy). If G is a computabe presentation of a torsionfree abeian group with infinite ran, then deg(x(g)) contains infinitey many ow degrees. Proof. We inductivey show deg(x(g)) must contain at east nmany ow degrees for a n. Fix two ineary independent eements g, h G and et T 0 be a computabe tree such that [T 0 ] (the set of infinite paths through T 0 ) contains exacty 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 cass and hence can be represented in this manner. The Low Basis Theorem appied to T 0 yieds a ow order of some degree d 0. To get a second order of ow degree d 1 d 0, it suffices (as ow over ow is ow) to buid a nonempty d 0 computabe subtree T 1 of T 0 having no d 0 computabe paths. From this, we obtain a ow (ow over d 0 ) order of some degree d 1 not computabe from d 0. The subset T 1 of T 0 is constructed (using an orace of degree d 0 ) by iing paths that agree with the e th (candidate) d 0 computabe order e on the reative ordering of g and h for a sufficienty arge amount of precision. In particuar, to diagonaize against e, we attempt to find positive rationas 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 rationas are found, we i initia 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 and as for every q (1, 4) R and rationa ε > 0, there is an order on G with (q ε)g < h < (q + ε)g. To get a third order of ow degree d 2 {d 0, d 1 }, we repeat this process to construct a (d 0 d 1 )computabe subtree T 2 of T 1 such that T 2 has no d 1 computabe paths. We note that T 2 cannot have any d 0 computabe paths as it is a subtree of T 1. The ony change we need to mae is to require the rationas q 0 and q 1 (being used to diagonaize against the e th (candidate) d 1 computabe 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 yieds the proposition. Note that this proposition aso hods for other casses of degrees which form a basis for Π 0 1 casses and reativize in the appropriate manner. For exampe, deg(x(g)) must contain infinitey many hyperimmunefree degrees. Acnowedgements The first author was partiay supported by a grant from the Pacard Foundation through a PostDoctora Feowship. The second author was partiay supported by NSF DMS and NSF DMS The authors thans Danie Turetsy for aowing them to incude Proposition 4.5. The authors than the anonymous referee for hepfu suggestions. References [1] R. Baer, Abeian groups without eements of finite order, Due Math Journa, vo. 3, 1937, [2] T.C. Craven, The Booean space of orderings of a fied, Transactions of the American Mathematica Society, vo. 209, 1975, [3] M.A. Dabowsa, M.K. Dabowsi, V.S. Harizanov and A.A. Togha, Spaces of orders and their Turing degree spectra, Annas of Pure and Appied Logic, vo. 161, 2010,
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