Degrees of orders on torsion-free Abelian groups

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1 Weesey Coege Weesey Coege Digita Schoarship and Archive Facuty Research and Schoarship Degrees of orders on torsion-free Abeian groups Asher M. Kach Karen Lange Reed Soomon Foow this and additiona wors at: Recommended Citation Degrees of orders on torsion-free abeian groups, A. Kach, K. Lange, and D. Soomon, Annas of Pure and Appied Logic, 164(7-8) (2013), This Artice is brought to you for free and open access by Weesey Coege Digita Schoarship and Archive. It has been accepted for incusion in Facuty Research and Schoarship by an authorized administrator of Weesey Coege Digita Schoarship and Archive. For more information, pease contact

2 DEGREES OF ORDERS ON TORSION-FREE ABELIAN GROUPS ASHER M. KACH, KAREN LANGE, AND REED SOLOMON Abstract. We show that if H is an effectivey competey decomposabe computabe torsion-free 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 torsion-free 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 torsion-free 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: August 4, Mathematics Subject Cassification. Primary: 03D45; Secondary: 06F20. Key words and phrases. ordered abeian group, degree spectra of orders. 1

3 2 KACH, LANGE, AND SOLOMON In this paper, we present a resut concerning computabiity-theoretic 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 non-negative 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 one-to-one correspondence between positive cones and orderings. Furthermore, an arbitrary subset X G is the positive cone of an ordering on G if and 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 torsion-free, 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 torsion-free 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 torsion-free 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 torsion-free 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 torsion-free 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

4 DEGREES OF ORDERS ON TORSION-FREE ABELIAN GROUPS 3 torsion-free 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 torsion-free 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 torsion-free abeian group is orderabe. Because torsion-free abeian groups are Z-modues, notions such as inear independence pay a arge roe in studying these groups. Definition 1.3. Let G be a torsion-free 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 torsion-free 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 torsion-free abeian group has a computabe copy with a computabe basis. Therefore, every computabe torsion-free 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 torsion-free 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 torsion-free abeian groups. Definition 1.4 (Khisamiev and Krypaeva [14]). A computabe infinite ran torsion-free 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.

5 4 KACH, LANGE, AND SOLOMON Theorem 1.5. Let H be an effectivey competey decomposabe infinite ran torsion-free 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 C-computabe 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 torsion-free abeian group fais. Proposition 1.6. Let H be an infinite ran torsion-free 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 torsion-free 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 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 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

6 DEGREES OF ORDERS ON TORSION-FREE 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 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 p-height 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 nonzero eements, then B H,h B H, ĥ. The Baer sequence B H of the group H is (any representative of) this equivaent cass. Levi 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 C-computabe 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

7 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 zero 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 Q-inear sum over the stage s approximate basis of the form q s 0b s q s nb s n where n s, qi s Q for i n, and qs n 0. (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 one-to-one, and the zero 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. 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. For each i ω, we fix a nonzero 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. 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 Q-inear 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 Levi s Theorem) that in the imit, the i-th 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 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

8 DEGREES OF ORDERS ON TORSION-FREE ABELIAN GROUPS 7 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 one-to-one. 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 one-to-one assignment of sums to eements of G s+1. 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.)

9 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 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. 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. 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 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, 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 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 Z. The anaysis for the index is identica. Let g G. Suppose there is a stage t such that g is assigned a sum q0b t t 0+ +qnb t t n that is not ater changed in the sense that, for a stages u t, the eement g is 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. Proof. Proof of (1a). 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

10 DEGREES OF ORDERS ON TORSION-FREE ABELIAN GROUPS 9 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 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. 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 torsion-free 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

11 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 2-map 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 2-map 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 2-map 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 under the 0 2-map. That is, suppose x is assigned the sum x = q s 0b s q s nb s n at stage s. The 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.

12 DEGREES OF ORDERS ON TORSION-FREE ABELIAN GROUPS 11 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) (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 3.5 (given beow), in each of these contexts, there are infinitey many such choices for m 1 and m 2 satisfying the given conditions. Moreover,

13 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)). 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 3.4 (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. Now suppose we are in the context of (5). Since rj t and rt are ineary independent over Q, by Lemma 3.4 (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.

14 DEGREES OF ORDERS ON TORSION-FREE ABELIAN GROUPS 13 We define G on G by x G y if and ony if x s y for some s. We verify that G is a computabe order under 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 3.6. 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 r s i exists and is a rationa mutipe of p i. Furthermore, once r s i 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 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 (as+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. Proof of (2). We have r s+1 i ri s ony when bs+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. Proof of (3). 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 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. Proof of Lemma. By Statements (1) and (3), G is computabe and every pair of eements is ordered. By construction, the 0 2-map 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. 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

15 14 KACH, LANGE, AND SOLOMON noncomputabe c.e. set C such that the ony C-computabe 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. After expaining one requirement in isoation, we examine the interaction between R e strategies in detai to carify the finitey nature of the construction. 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 3.9. 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. 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 â s+1 s+1 as+1 s+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

16 DEGREES OF ORDERS ON TORSION-FREE ABELIAN GROUPS 15 strategy enumerates a number into C beow the restraint and that b u j = bs j, bu = bs, 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 for a u s + 1 if R e never sees such a stage. (We discuss how to hande R e if either of these conditions is vioated beow when we examine the interaction between strategies.) 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 one-to-one. 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 Equation (4) and maintain the one-to-one assignment of sums to eements of G t+1. (This choice is possibe by Lemma 3.5.) 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, 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

17 16 KACH, LANGE, AND SOLOMON 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 one-to-one assignment of sums to eements of G t+1 (again by Lemma 3.5.) 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 with b s1+1 s at stage s , and n 1 with j 1 = s Then R e sets up to diagonaize 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. 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 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 0, to be redefined, but eaves bt+1 j 1 = b t j 1 (and (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.