DEGREES OF ORDERS ON TORSIONFREE ABELIAN GROUPS


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1 1 DEGREES OF ORDERS ON TORSIONFREE ABELIAN GROUPS 2 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 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 [25] 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: December 13, 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 [16] for a more compete introduction to ordered abeian groups and to [18] 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 [24] 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 [23] 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 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 [28] 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 Zmodues, 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 [28] 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 (Khisamiev and Krypaeva [14]). A computabe infinite ran torsionfree abeian group H is effectivey competey decomposabe if there is a uniformy computabe sequence of ran one subgroups H i of H, 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], [15], and [22]. Our main resut is the foowing theorem.
4 4 KACH, LANGE, AND SOLOMON Theorem 1.5. Let H be an effectivey competey decomposabe 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 nonzero 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 153 from the study of inear orders: If G is a inear order on G, then G denotes the 154 inear order defined by x G y if and ony if y G x. Note that if (G; G ) is an 155 ordered abeian group, then (G; G ) is aso an ordered group 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 nonzero eement h H. Every nonzero 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 163 n constraints as above) to the rationa m. Because this map is effective, the image 164 of H in Q is computaby enumerabe and hence we can view H as a computaby
5 DEGREES OF ORDERS ON TORSIONFREE ABELIAN GROUPS 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 nonzero 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 Levi s Theorem (see [19] and [1]) giving cassica agebraic invariants for ran one groups caed Baer sequences. 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 nonzero 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. 182 The Baer sequence of h is the function B H,h (n) = ht pn (h). If h, ĥ H are nonzero 183 eements, then B H,h B H,ĥ. The Baer sequence B H of the group H is (any 184 representative of) this equivaent cass. Levi s Theorem states that for ran one 185 groups, H 0 = H1 if and ony if B H0 B H Proof of Theorem Fix an effectivey competey decomposabe group H = i ω H i as in the state 188 ment of Theorem 1.5. We divide the proof into three steps. First, we describe 189 our genera method of buiding the computabe copy G = (G; + G, 0 G ) which is isomorphic to H. Second, we describe how the computabe ordering G on G is 191 constructed. (The second computabe order on G is G.) Third, we give the con 192 struction of C and the diagonaization process to ensure the ony Ccomputabe 193 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] 200 for a x, y, z G s. Furthermore, for any pair of eements x, y G s, we ensure the 201 existence of a stage t and an eement z G t such that we decare x + t y = z. 202 To define the addition function, we use an approximation {b s 0, b s 1,..., b s s} G s 203 to an initia segment of our eventua basis for G. During the construction, each 204 approximate basis eement b s i wi be redefined at most finitey often, so each wi 205 eventuay reach a imit. We et b i := im s b s denote this imit. If is an even i
6 6 KACH, LANGE, AND SOLOMON 206 index then the approximate basis eement b s wi never be redefined, so athough 207 we often use the notation b s (for uniformity), we have b = b s for a s. Athough G 208 wi not be effectivey decomposabe, the group G wi decompose cassicay into a 209 countabe direct sum using the basis B = {b 0, b 1, b 2,...}. 210 At stage 0, we begin with G 0 := {0, 1}. We et 0 denote the zero eement 0 G and 211 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 More generay, at stage s, each eement g G s is assigned a Qinear sum over 213 the stage s approximate basis of the form q s 0b s q s nb s n 214 where n s, qi s Q for i n, and qs n 0. (Later there wi be further restrictions 215 on the vaues of qi s to ensure that G is isomorphic to H.) This assignment is required 216 to be onetoone, and the zero eement 0 G is aways assigned the empty sum. It 217 wi often be convenient to extend such a sum by adding more approximate basis 218 eements on the end of the sum with coefficients of zero. We define the partia 219 function + s on G s by etting x + s y = z (for x, y, z G s ) if the assigned sums for x 220 and y add together to form the assigned sum for z. 221 For each i ω, we fix a nonzero eement h i H i and embed H i into Q by 222 sending h i to 1 Q as described in Section 2. We equate H i with its image in Q in 223 the sense of treating eements of H i as rationas. In particuar, since h i is mapped 224 to 1 Q, if a H i and a = qh i, we view a as being the rationa q. 225 At each stage s, we maintain positive integers Ni s for i s. These integers 226 restrain the (nonzero) coefficients qi s of b s i aowed in the Qinear sum for each 227 eement g G s by requiring that qi sn i s H i and that we have seen this fact by 228 stage s. Using the fact that N i := im s Ni s exists and is finite for a i, we wi show 229 (using Levi s Theorem) that in the imit, the ith component of G is isomorphic 230 to H i, and hence that G is a computabe copy of H. (Later we wi introduce a 231 basis restraint K ω that wi prevent us from changing Ni s too often.) 232 During stage s + 1, we do one of two things either we eave our approximate 233 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. 236 For each g G s (viewed as an eement of G s+1), we define q s+1 i := qi s and assign g 237 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 238 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 239 Ns+1 s+1 := We add two new eements to G s+1, abeing the first by b s+1 s+1 and abeing the 241 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 242 rationas (under some fixed computabe enumeration of a tupes of rationas) we 243 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 244 sum is not aready assigned to any eement of G s+1. (We can effectivey search for 245 such a tupe.) This competes the description of G s+1 in this case. 246 Case 2. If we redefine the approximate basis eement b s (for the sae of diagona 247 izing) by adding a new dependency reation, then we proceed as foows. We define 248 b s+1 i := b s i for a i s with i. The diagonaization process wi te us either 249 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 250
7 DEGREES OF ORDERS ON TORSIONFREE ABELIAN GROUPS 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 257 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 := qi s for a i {j,, }. Simiary, if the diagonaization process 258 i 259 tes us to mae b s 260 eaving q s+1 i = qi s We define N s = qbs+1, then we set q s+1 := q s + qqs for a i {, }. and qs+1 := 0 whie 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) 264 and d is the product of a the (finitey many) denominators of coefficients q s for 265 g G s. In either case, set Ns+1 s+1 := We add three new eements to G s+1, abeing the first by b s+1, abeing the second 267 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 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 269 stage s for a i n, and this sum is not aready assigned to any eement of G s This competes the description of G s+1 in this case. 271 We note severa trivia properties of the transformations of sums in Case 2. First, 272 the approximate basis eement b s+1 does not appear in the new sum for any eement 273 of G s viewed as an eement of G s+1. Second, for any eement g G s, if q s = 0, 274 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 275 inearity of the substitutions, if x + s y = z, then x + s+1 y = z. 276 We aso require two additiona properties which pace some restrictions on the 277 rationa q or the integers m 1 and m 2. The first property is that the assignment 278 of sums to eements of G s (viewed as eements of G s+1 ) remains onetoone. The 279 diagonaization process wi pace some restrictions on the vaue of either q or m and m 2, but as ong as there are infinitey many possibe choices for these vaues 281 (which we wi verify when we describe the diagonaization process), we can assume 282 they are chosen to maintain the onetoone assignment of sums to eements of G s The second property is that for each g G s+1, we need each coefficient q s+1 i to 284 satisfy q s+1 i N s+1 i H i. We wi verify this property beow under the assumption 285 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 286 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.)
8 8 KACH, LANGE, AND SOLOMON 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 q0b s s qnb s s n satisfy qi sn i s H i Proof. The proof proceeds by induction on s. If g is added at stage s, then the 295 resut for g foows triviay. Therefore, fix g G s and assume the condition hods at 296 stage s. Note that if we do not add a dependency reation (i.e., we are in Case 1), 297 then the condition at stage s + 1 foows immediatey. Assume we add a new 298 dependency reation; we spit into cases depending on the form of this dependency. 299 If b s = qbs+1, then for a i {, }, the condition hods since q s+1 i = qi s and 300 N s+1 i = Ni s. For the index, we have qs+1 = 0 and hence the condition hods triviay. For the index, we have q s+1 = q s + qqs s+1 and N = d q dn s. Therefore, 302 q s+1 N s+1 = (q s + qq s )d q dn s = qd s q dni s + 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 303 q sd Z, and hence qqs d qdn s Z H. Therefore, we have the desired property 304 when b s = qbs 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 307 integer m 1 is divisibe by the denominator of q s 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 308 since qj sn j s H j by the induction hypothesis and q sm 1Nj s Z. The anaysis for 309 the index is identica. 310 Let g G. Suppose there is a stage t such that g is assigned a sum q0b t t 0+ +qnb t t n 311 that is not ater changed in the sense that, for a stages u t, the eement g is 312 assigned the sum q0 u b u qnb u u n with b u i = bt i and qu i = qi t for a i n. In this case, we refer to this sum as the imiting sum for g and denote it by q 0 b 0 + +q n b n Lemma 3.2 (Basic properties of the construction). (1) (a) Each g G has a imiting sum with coefficients q i satisfying q i N i H i. (b) 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. (2) (a) 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. (b) 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. (c) For each x G s, there is a stage t s and an eement z G t such that x + t z = 0 G. 325 Proof. Proof of (1a). When g enters G, it is assigned a sum. The coefficients in 326 this sum ony change when a diagonaization occurs. In this case, some approximate 327 basis eement b s with nonzero coefficient in the sum for g is made dependent via a 328 reation of the form b s = qbs+1 or b s = m 1b s+1 j + m 2 b s+1 with j, <. Therefore, 329 each time the sum for g changes, some approximate basis eement with nonzero 330 coefficient is repaced by rationa mutipes of approximate basis eements with
9 DEGREES OF ORDERS ON TORSIONFREE ABELIAN GROUPS ower indices. This process can ony occur finitey often before terminating. The ast property of the imiting sum foows from Lemma 3.1. Proof of (1b). 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). Proof of (2). Property (2a) foows by induction and the fact that x + s y = z impies x + s+1 y = z at each stage s of the construction. For Property (2b), 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 qi tn i t H i and ˆq i tn i t H i. Therefore, (qi t + ˆqt i )N i t H i. By (1b), there is 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 Property (2c) is simiar. By Properties (1b) and (1a) in 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. Using Properties (2a) and (2b) in Lemma 3.2, 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 3.3. The set G is a computabe copy of H. Proof. The domain and addition function on G are computabe. By Property (2c) in Lemma 3.2, 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. 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 Levi 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
10 10 KACH, LANGE, AND SOLOMON 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, 392 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, 390 then r s wi never change. 391 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 393 our abiity to diagonaize. Therefore, at stage s, we assign each b s i (for i 1) an 394 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 395 and â s i as i 1/2s. The image of b s i in R (in the imit) wi be contained in this 396 interva. 397 Because each x G s is assigned a sum describing its reationship to the current 398 approximate basis, we can generate an interva approximating the image of x in R 399 under the 0 2map. That is, suppose x is assigned the sum x = q0b s s qnb s s n 400 at stage s. The interva constraints on the image of each b s i in R transate into a 401 rationa interva constraint on the image of x in R. The endpoints of this constraint 402 can be cacuated using the coefficients of the sum for x and the rationas a s i and âs i, 403 with the exact form depending on the signs of the coefficients. 404 To define s on G s at stage s, we oo at the interva constraints for each pair 405 of distinct eements x, y G s. If the interva constraint for x is disjoint from the 406 interva constraint for y, then we decare x s y or y s x depending on which 407 inequaity is forced by the constraints. If the interva constraints are not disjoint, 408 then we do not decare any ordering reation between x and y at stage s. Of course, 409 we aso decare x s x for each x G s. 410 To maintain the impication in Equation (1), we wi need to chec that x s y 411 impies x s+1 y. It suffices to ensure that for each x G s, the interva constraint 412 for x at stage s + 1 is contained within the interva constraint for x at stage s.
11 DEGREES OF ORDERS ON TORSIONFREE ABELIAN GROUPS It wi be hepfu for us to now that certain approximate basis eements are 414 mapped to eements of R which are cose to 0 R. Therefore, we wi maintain that a s âs < 1/2 for a stages s and a even indices. (If we wored in a 416 simper context where each H i = Q, or even where each H i Z, we coud sip this 417 step as any archimedean order on such groups H i is dense in R.) 418 We now describe exacty how ri t, at i and ât i are defined at each stage t. Reca that 419 at stage t = 0, the ony eements in G t are 0 G (which is represented by the empty 420 sum and is mapped to 0 R ) and the eement represented by b 0 0 (which is mapped 421 to 1 R ). We set r0 0 := 1 R. 422 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 423 we add a dependency reation or not. If we do not add a dependency reation, 424 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 425 p i for b i ) 426 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 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 428 p t+1 (requiring rt+1 t+1 < 1/2t+1 if t+1 is even) and et 429 a t+1 t+1 and ât+1 be positive rationas so that rt+1 (at+1, ât+1 ) R and â t+1 t+1 at+1 < 430 1/2 t+1 (and aso â t+1 t+1 < 1/2t+1 if t + 1 is even). The diagonaization process may pace some requirements on the rationa mutipe of 431 p t+1 chosen. It remains to 432 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 := 433 p and 434 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 For the other indices invoved in an added dependency reation, we spit into 436 cases depending on the type of reation added. 437 (1) If we add a dependency of the form b t = qbt+1, then we set r t+1 := r t. 438 The action of the diagonaization strategy wi ensure that we can choose 439 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 and (qa t+1, qâ t+1 ) R (a t, â t ) R (2) 441 (2) If we add a dependency of the form b t = m 1b t+1 j m 2 b t+1, then we set 442 r t+1 j := rj t and rt+1 := r t. We wi be in one of two contexts (a). If we are in a context in which (in R) < 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 3.5 (given beow), in each of these contexts, there are infinitey many such choices for m 1 and m 2 satisfying the given conditions. Moreover,
12 12 KACH, LANGE, AND SOLOMON 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 3.4. 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 3.5. 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)). 460 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 461 p re 462 spectivey, so rj t and rt are ineary independent over Q. Hence, by Lemma (requiring m 1 and m 2 to be divisibe by nd where n comes from the context (3) 464 and d comes from the statement of this emma), there are infinitey many choices of 465 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 466 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 467 (4) by shrining the intervas (a t j, ât j ) R and (a t, ât ) R appropriatey. 468 It remains to see why we must have m 1 m2 n = m 2. Suppose m 1 > m2 n = m 2, 469 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 470 because rj t nrt > 0 by (3). We have reached a contradiction since 471 m 1 rj t m 2nr t (at, ât ) R and rj t (at j, ât j ) R 472 desired. but â t < at j. So, m 1 m2 n = m 2 as 473 Now suppose we are in the context of (5). Since rj t and rt are ineary independent 474 over Q, by Lemma 3.4 (requiring m 1 and m 2 to be divisibe by (n + 1)d) there are 475 infinitey many choices of m 1, m 2 N such that m 1 r t m 2rj t (at, ât ) R. We et 476 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). 477 It remains to see why m 1 = m 1 (n + 1) m 2 (n + 1). Suppose 478 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. 479 The first inequaity foows because m 1 1 m 2 and rj t is positive, and the second 480 inequaity foows because rj t < (n + 1)rt by (5). We have reached a contradiction 481 since m 1 r t m 2rj t (at, ât ) R but â t < (n + 1)at.
13 DEGREES OF ORDERS ON TORSIONFREE ABELIAN GROUPS We define G on G by x G y if and ony if x s y for some s. We verify 483 that G is a computabe order under the assumptions that each approximate basis 484 eement b s i eventuay reaches a imit and that we choose our intervas and associated 485 rationas in the manner described above Lemma 3.6. The reation G is a computabe order on G. Furthermore, G is cassicay isomorphic to an ordered subgroup of (R; +, 0 R ) under the standard ordering. Proof. We begin by verifying the foowing properties of the construction. (1) For every pair of eements x, y G s, if x s y, then x s+1 y. (2) 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. (3) For each pair x, y G s, there is a stage t s for which either x t y or y t x Proof of (1). It suffices to show that for each g G s, the interva constraint for g 496 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 498 at stage s and the index i is not invoved in an added dependency reation, then 499 q s+1 i = qi s and (as+1 i, â s+1 i ) R (a s i, âs i ) R. Therefore, the constraint imposed on g 500 by these terms at stage s + 1 is contained in the constraint imposed at stage s. 501 Second, suppose we add a dependency reation of the form b s = qbs+1 and 502 q s 0. From stage s to stage s + 1, the qs bs + qs bs part of the sum for g turns into 503 (q s )bs+1 +0b s+1 where b s+1 = b s. Since the constraint on rs+1 pays no roe in 504 the constraint on g at stage s+1 and since we have, by (2), that (qa s+1, qâ s+1 ) R 505 (a s, âs ) R, it foows that the constraint imposed by the indices and at stage 506 s + 1 is contained in the constraint imposed at stage s. 507 Third, if we add a dependency reation of the form b s = m 1b s+1 j m 2 b s+1, 508 then a simiar anaysis using (4) and (6) yieds that the constraint imposed by the 509 indices j, and at stage s + 1 is contained in the constraint imposed at stage s. 510 Proof of (2). We have r s+1 i ri s ony when bs+1 i b s i. Since the atter happens 511 ony finitey often, each ri s reaches a imit. The remainder of the statement is 512 immediate from the construction. 513 Proof of (3). Since x s x for a x G s, we consider distinct eements x, y G s. 514 Let t s be a stage such that x and y have reached their imiting sums and such 515 that for each b t i occurring in these sums, the rea rt i has reached its imit r i. Because 516 the reas r i are agebraicay independent over Q and the nested approximations 517 (a u i, âu i ) R (for u t) converge to r i, there is a stage at which the interva constraints 518 for x and y are disjoint. At the first such stage, we decare an ordering reation 519 between x and y. 520 Proof of Lemma. By Statements (1) and (3), G is computabe and every pair of 521 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 522 is order preserving 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
14 14 KACH, LANGE, AND SOLOMON 525 noncomputabe c.e. set C such that the ony Ccomputabe orders on G are G 526 and G. 527 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 534 to assembe the compete finite injury construction in the usua manner. After 535 expaining one requirement in isoation, we examine the interaction between R e 536 strategies in detai to carify the finitey nature of the construction. 537 To simpify the notation, we et C e be the binary reation on G computed by Φ C e. 538 We wi assume throughout that C e never directy vioates any of the Π 0 1 conditions 539 in the definition of a group order. For exampe, if we see at some stage s that C e 540 has vioated transitivity, then we can pace a finite restraint on C to preserve these 541 computations and win R e triviay. 542 The strategy to satisfy R e is as foows. For R e, we set the basis restraint 543 K := e. (This restraint is used in the verification that each Ni s reaches a imit.) 544 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: 546 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 548 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. 549 We verify such objects exist in Lemma 3.9. In the atter case, we wor with the 550 ordering C e, transforming the atter case into the former case. We therefore 551 assume that we are in the former case. 552 Whie waiting for these witnesses, the construction of G proceeds as in the genera 553 description with no dependencies added. When such s, b s j, n, and are found, we 554 say R e is activated, and we restrain C to preserve the computations ordering 0 G, 555 b s j, nbs, and (n + 1)bs. 556 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 (D1) If b s 559 j <C e nb s choose r s , we order bs+1 s+1 so that nbs < s+1 b s+1 s+1 < s+1 b s j s+1. s+1 to be a rationa mutipe of p s+1 and rationas a s+1 s+1 and âs+1 as+1 < 1/2s+1. so that nâ s < as+1 < rs+1 < âs+1 < as j (n + 1)b s, we order bs+1, that is, we and âs+1 s (D2) If b s j >C e s+1 so that bs j < s+1 b s+1 s+1 < s+1 (n + 1)b s, that is, we choose rs+1 s p s+1 and rationas a s+1 s+1 and âs+1 so that âs j < as+1 and 565 â s+1 s+1 as 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 567 (n+1)b s or bt s+1 > C e (n+1)b s. Whie waiting, we assume that no higher priority S i
15 DEGREES OF ORDERS ON TORSIONFREE ABELIAN GROUPS strategy enumerates a number into C beow the restraint and that b u j = bs j, bu = bs, 569 and b u s+1 = b s+1 s+1 at a stages u s + 1 unti R e finds such a stage t + 1 or 570 for a u s + 1 if R e never sees such a stage. (We discuss how to hande R e 571 if either of these conditions is vioated beow when we examine the interaction 572 between strategies.) If these conditions hod, then we say R e has been activated 573 with potentiay permanent witnesses. 574 We assume that such a stage t + 1 is found, ese R e is triviay satisfied. At 575 stage t + 1, R e acts to diagonaize by restraining C to preserve the computations 576 ordering b t s+1, nb t, and (n + 1)bt under C e and adding a dependency reation as 577 foows. 578 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 579 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, 582 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, 583 and â t+1 to be the first rationas meeting these conditions such that the assignment 584 of sums to eements of G t remains onetoone. 585 These choices satisfy the necessary requirements for both the group construction 586 and the ordering construction. Furthermore, we have successfuy diagonaized 587 against C e being an ordering of G since any order under which b t+1 = b t is positive 588 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 589 b t s+1 < C e nb t or bt s+1 > C e (n + 1)b t. Case 2. If C 590 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 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. 592 Case 2(a): If b t s+1 < t b t j, then it is because we acted in (D1) and hence we 593 now that b t+1 j < C e nb t+1 and we are in the context of Equation (3) with 594 = s+1. Let d be the product of a denominators of coefficients qs+1 t for a 595 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 both divisibe by d that satisfy m 1 N m 2 /n and the ordering constraints in 597 Equation (4) and maintain the onetoone assignment of sums to eements 598 of G t+1. (This choice is possibe by Lemma 3.5.) 599 To see that we have successfuy diagonaized, we show that C e must vio 600 ate 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, 601 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, 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. j
16 16 KACH, LANGE, AND SOLOMON 605 Case 2(b): If b t s+1 > t b t j, then it is because we acted in (D2) and hence we 606 now (n + 1)b t+1 < C e b t+1 j and we are in the context of Equation (5) with 607 = 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 608 positive integers m 1 and m 2 both divisibe by d that satisfy m 1 N m 2 (n+1) 609 and the ordering constraints in Equation (6) and maintain the onetoone 610 assignment of sums to eements of G t+1 (again by Lemma 3.5.) 611 We show that < C e must vioate the order axioms. Since 0 G < C e b t+1 and 612 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 618 requirements in the usua way. If i < e, then S i is aowed to enumerate its diago 619 naizing witness even if it destroys a restraint imposed by R e, but if e i, then S i 620 must pic a new arge witness when R e imposes a restraint. 621 There is aso a potentia confict between different R e requirements. Con 622 sider requirements R e and R i invoved in the foowing scenario. Assume that 623 at stage s 0, R i is the highest priority activated requirement with witnesses b s0 j 0, b s0 0, 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) 625 or (D2)). At stage s 1 > s 0, whie R i is sti waiting to diagonaize, R e is activated 626 with witnesses b s1 j 1, b s1 1, and n 1 with j 1 = s Then R e sets up to diagonaize 627 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 629 certain eements (which may never appear) and it needs to maintain b u j 1 = b s1 j (which means b u s 0+1 = b s1 other 631 hand, when R i sees Cs i decare the appropriate order reations, it wants to add a 632 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 633 cause b t+1 s 0+1 (and hence bt+1 j 1 ) to be redefined. 634 In this scenario, if e < i, then when R e sets up to diagonaize at stage s 1 + 1, it 635 cances R i s caim on the diagonaizing witness b s0+1 s 0+1, thus removing the potentia 636 confict. The requirement R i remains activated (since the appropriate C i com 637 putations have been preserved) and at the next odd stage s at which R i is 638 the highest priority activated requirement, it wi set up to diagonaize with a new witness b s2+1 s If i < e, then no canceation of setup witnesses taes pace when R e sets up to 641 diagonaize. If R e acts to diagonaize first, there is no confict because R e adds a 642 dependency reation which causes b t+1 s 1+1 to be redefined, but eaves bt+1 j 1 = b t j 1 (and 643 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 644 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.
17 DEGREES OF ORDERS ON TORSIONFREE ABELIAN GROUPS 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. 656 Lemma 3.7. The imit b i := im s b s i exists for a i. 657 Proof. The ony approximate basis eements which are redefined are those chosen as 658 diagonaizing witnesses by some R e requirement. Therefore, at stage s + 1, if b s+1 s is not chosen as a diagonaizing witness, then it is never redefined. If b s+1 s+1 is chosen 660 as a diagonaizing witness by R e, then it can be redefined at most once when R e 661 acts to diagonaize. 662 Lemma 3.8. The imit N i := im s N s i exists for a i. 663 Proof. The ony time N s+1 i Ni s is when we add a dependency reation of the 664 form b s = qbs+1 causing N s+1 = d q dn s. However, in this case, the index is even 665 and a requirement R e can ony add such a dependency if > K = e. Therefore, 666 ony R e with e < can cause N s to changes vaue. Since these requirements ony 667 act finitey often, the vaue of N s changes ony finitey often. 668 Lemma 3.9. If we fai to find a stage s where R e is activated with potentiay 669 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 671 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, 673 then C e is equa to G or G. 674 First, we caim that if we fai to find a stage s s at which R e is activated, 675 then either 0 G < C e b j for a j or b j < C e 0 G for a j. 676 To prove this caim, suppose that R e is never activated after s and that j 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, 678 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 679 t s and an even index greater than the basis restraint for R e such that b t = b 680 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. 681 Since C e is an order, there must be a stage u t at which it decares either G < C e b u or bu <C e 0 G permanenty. 683 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. 684 Therefore, R e is activated at stage v (with j = j 1, =, and n = n 1 ) for the desired 685 contradiction. Aternatey, if b u <C e 0 G, then we must eventuay see n 0 b v <C e G < C e b v j 0 for some v u. Again, R e is activated at stage v (with j = j 0, =, 687 and n = n 0 ) for the desired contradiction. This competes the proof of the caim. 688 To compete the proof of this emma, assume that R e is never activated after s 689 and 0 G < C e b j for a j. We show that C e = G. It foows by a simiar argument 690 that if b j < C e 0 G for a j, then C e = G.
18 18 KACH, LANGE, AND SOLOMON 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. (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
19 DEGREES OF ORDERS ON TORSIONFREE ABELIAN GROUPS 19 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 3.9) 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 [17] and [21]). 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. 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
20 20 KACH, LANGE, AND SOLOMON 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, [4] V. Dobritsa, Some constructivizations of abeian groups, Siberian Journa of Mathematics, vo. 24, 1983, [5] R. Downey, Every recursive booean agebra is isomorphic to one with incompete atoms, Annas of Pure and Appied Logic, vo. 60, 1993, [6] R. Downey and S.A. Kurtz, Recursion theory and ordered groups, Annas of Pure and Appied Logic, vo. 32, 1986, [7] R. Downey, A.M. Kach, S.S. Goncharov, J.F. Knight, O.V. Kudinov, A.G. Meiniov and D. Turetsy, Decidabiity and computabiity of certain torsionfree abeian groups, Notre Dame Journa of Forma Logic, vo. 51 (1), 2010, [8] R. Downey and A.G. Meniov, Effectivey categorica abeian groups, submitted. [9] R.G. Downey and M.F. Moses, Recursive inear orders with incompete successivities, Transactions of the American Mathematica Society, vo. 326 (2), 1991, [10] H.M. Friedman, S.G. Simpson and R.L. Smith, Countabe agebra and set existence axioms, Annas of Pure and Appied Logic, vo. 25, 1983, [11] L. Fuchs, Partiay ordered agebraic systems, Pergamon Press, Oxford, 1963.
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