Degrees of orders on torsion-free abelian groups

21 Pages·2012·0.35 MB·English

DEGREES OF ORDERS ON TORSION-FREE ABELIAN GROUPS
ASHERM.KACH,KARENLANGE,ANDREEDSOLOMON
Abstract. WeshowthatifHisaneffectivelycompletelydecomposablecom-
putabletorsion-freeabeliangroup,thenthereisacomputablecopyGofHsuch
thatG hascomputableordersbutnotordersofevery(Turing)degree.
1. Introduction
A recurring theme in computable algebra is the study of the complexity of rela-
tionsoncomputablestructures. Forexample,fixanaturalmathematicalrelationR
on some class of computable algebraic structures such as the successor relation in
the class oflinear orders orthe atom relation inthe class of Boolean algebras. One
canconsiderwhethereachcomputablestructureintheclasshasacomputablecopy
inwhichtherelationisparticularlysimple(saycomputableorloworincomplete)or
whether there are structures for which the relation is as complicated as possible in
every computable presentation. For the successor relation, Downey and Moses [9]
show there is a computable linear order L such that the successor relation in every
computablecopyofLisascomplicatedaspossible,namelycomplete. Ontheother
hand, Downey [5] shows every computable Boolean algebra has a computable copy
inwhichthesetofatomsisincomplete. Alternately,onecanexploretheconnection
between definability and the computational properties of the relation R.
More abstractly, one can start with a set S of Turing (or other) degrees and
ask whether there is a relation R on a computable structure A such that the set of
degreesoftheimagesofRinthecomputablecopiesofAisexactlyS. Forexample,
Hirschfeldt [13] proved that this is possible if S is the set of degrees of a uniformly
c.e. collection of sets.
One can also consider relations such as “being a k-coloring” for a computable
graph or “being a basis” for a torsion-free abelian group. In these examples, for
eachfixedcomputablestructure,therearemanysubsetsofthedomain(orfunctions
on the domain) satisfying the property. It is natural to ask whether there are
computable structures for which all of these instantiations are complicated and
whether this complexity depends on the computable presentation. In the case
of k-colors of a planar graph, Remmel [22] proves that one can code arbitrary Π0
1
classes(uptopermutingthecolors)bythecollectionofk-colorings. Fortorsion-free
abelian groups, there is a computable group G such that every basis computes 0(cid:48).
However,foranycomputableH,onecanfindacomputablecopyofthegivengroup
inwhichthereisacomputablebasis(seeDobritsa[4]). Therefore,whileeverybasis
can be complicated in one computable presentation, there is always a computable
presentation having a computable basis.
Date:July27,2012.
2010 Mathematics Subject Classification. Primary: 03D45;Secondary: 06F20.
Key words and phrases. orderedabeliangroup,degreespectraoforders.
1
2 KACH,LANGE,ANDSOLOMON
In this paper, we present a result concerning computability-theoretic properties
of the spaces of orderings on abelian groups. To motivate these properties, we
compare the known results on computational properties of orderings on abelian
groupswiththoseforfields. Wereferthereaderto[11]and[14]foramorecomplete
introductiontoorderedabeliangroupsandto[16]forbackgroundonorderedfields.
Definition1.1. AnorderedabeliangroupconsistsofanabeliangroupG =(G;+,0)
and a linear order ≤ on G such that a ≤ b implies a+c ≤ b+c for all c ∈ G.
G G G
An abelian group G that admits such an order is orderable.
Definition 1.2. The positive cone P(G;≤ ) of an ordered abelian group (G;≤ )
G G
is the set of non-negative elements
P(G;≤ ):={g ∈G|0 ≤ g}.
G G G
Because a ≤ b if and only if b−a ∈ P(G;≤ ), there is an effective one-to-one
G G
correspondence between positive cones and orderings. Furthermore, an arbitrary
subsetX ⊆GisthepositiveconeofanorderingonG ifandonlyifX isasemigroup
such that X ∪X−1 = G and X ∩X−1 = {0 }, where X−1 := {−g | g ∈ X}. We
G
let X(G) denote the space of all positive cones on G. Notice that the conditions for
being a positive cone are Π0.
1
The definitions for ordered fields are much the same, and we let X(F) denote
the space of all positive cones on the field F. We suppress the definitions here as
the results for fields are only used as motivation. As in the case of abelian groups,
the conditions for a subset of F to be a positive cone are Π0.
1
Classically, a field F is orderable if and only if it is formally real, i.e., if −1
F
is not a sum of squares in F; and an abelian group G is orderable if and only if
it is torsion-free, i.e., if g ∈ G and g (cid:54)= 0 implies ng (cid:54)= 0 for all n ∈ N with
G G
n>0. In both cases, the effective version of the classical result is false: Rabin [21]
constructed a computable formally real field that does not admit a computable
order, and Downey and Kurtz [6] constructed a computable torsion-free abelian
group (in fact, isomorphic to Zω) that does not admit a computable order.
Despitethefailureoftheseclassificationsintheeffectivecontext,wehaveagood
measure of control over the orders on formally real fields and torsion-free abelian
groups. Because the conditions specifying the positive cones in both contexts are
Π0,thesetsX(F)andX(G)areclosedsubsetsof2F and2G respectively,andhence
1
under the subspace topology they form Boolean topological spaces. If F and G
are computable, thenthe respectivespaces of ordersform Π0 classes, and therefore
1
computable formally real fields and computable torsion-free abelian groups admit
orders of low Turing degree.
Forfields,onecansayconsiderablymore. Craven[2]provedthatforanyBoolean
topologicalspaceT,thereisaformallyrealfieldF suchthatX(F)ishomeomorphic
to T. Translating this result into the effective setting, Metakides and Nerode [20]
provedthatforanynonemptyΠ0classC,thereisacomputableformallyrealfieldF
1
such that X(F) is homeomorphic to C via a Turing degree preserving map. Fried-
man, Simpson, and Smith [10] proved the corresponding result in reverse mathe-
matics that WKL is equivalent to the statement that every formally real field is
0
orderable.
Most of the corresponding results for abelian groups fail. For example, a count-
able torsion-free abelian group G satisfies either |X(G)| = 2 (if the group has
rank one) or |X(G)| = 2ℵ0 and X(G) is homeomorphic to 2ω. For a computable
DEGREES OF ORDERS ON TORSION-FREE ABELIAN GROUPS 3
torsion-free abelian group G, even if one only considers infinite Π0 classes of sepa-
1
rating sets (which are classically homeomorphic to 2ω) and only requires that the
map from X(G) into the Π0 class be degree preserving, one cannot represent all
1
such classes by spaces of orders on computable torsion-free abelian groups. (See
Solomon[25]foraprecisestatementandproofofthisresult.) However,theconnec-
tion to Π0 classes is preserved in the context of reverse mathematics as Hatzikiri-
1
akouandSimpson[12]provedthatWKL isequivalenttothestatementthatevery
0
torsion-free abelian group is orderable.
Becausetorsion-freeabeliangroupsareageneralizationofvectorspaces,notions
such as linear independence play a large role in studying these groups.
Definition 1.3. Let G be a torsion-free abelian group. Elements g ,...,g are
0 n
linearly independent (or just independent) if for all c ,...,c ∈Z,
0 n
c g +c g +···+c g =0
0 0 1 1 n n G
impliesc =0for0≤i≤n. Aninfinitesetofelementsisindependent ifeveryfinite
i
subset is independent. A maximal independent set is a basis and the cardinality of
any basis is the rank of G.
Solomon [25] and Dabkowska, Dabkowski, Harizanov, and Tonga [3] established
that if G is a computable torsion-free abelian group of rank at least two and B is a
basis for G, then G has orders of every Turing degree greater than or equal to the
degree of B. Therefore, the set
deg(X(G)):={d|d=deg(P) for some P ∈X(G)}
contains all the Turing degrees when the rank of G is finite (but not one) and con-
tains cones of degrees when the rank is infinite. As mentioned earlier, Dobritsa [4]
proved that every computable torsion-free abelian group has a computable copy
with a computable basis. Therefore, every computable torsion-free abelian group
has a computable copy that has orders of every Turing degree, and hence has a
copy in which deg(X(G)) is closed upwards.
Our broad goal, which we address one aspect of in this paper, is to better un-
derstand which Π0 classes can be realized as X(G) for a computable torsion-free
1
abelian group G and how the properties of the space of orders changes as the com-
putable presentation of G varies. Specifically, is deg(X(G)) always upwards closed?
If not, does every group H have a computable copy in which it fails to be upwards
closed? We show that if H is effectively completely decomposable, then there is a
computable G ∼=H such that deg(X(G)) contains 0 but is not closed upwards. We
conjecture that this statement is true for all computable infinite rank torsion-free
abelian groups.
Definition 1.4. A computable infinite rank torsion-free abelian group H is ef-
fectively completely decomposable if there is a uniformly computable sequence of
rank one groups H , for i∈ω, such that H is equal to ⊕ H (with the standard
i i∈ω i
computable presentation).
There are a number of recent results concerning computability theoretic proper-
tiesofclassicallycompletelydecomposablegroupsin,forexample,[7],[8],and[19].
Our main result is the following theorem.
4 KACH,LANGE,ANDSOLOMON
Theorem 1.5. Let H be an effectively completely decomposable computable infinite
rank torsion-free abelian group. There is a computable presentation G of H and a
noncomputable, computably enumerable set C such that:
• The group G has exactly two computable orders.
• Every C-computable order on G is computable.
Thus, the set of degrees of orders on G is not closed upwards.
If H is effectively completely decomposable, then deg(X(H)) contains all Tur-
ing degrees because H has a computable basis formed by choosing a nonidentity
element h from each H . Therefore, although the group G in Theorem 1.5 is
i i
completely decomposable in the classical sense, it cannot be effectively completely
decomposable.
In general, one does not expect the collection of degrees realizing a relation on a
fixedcomputablecopyofanalgebraicstructuretobeupwardsclosedandhencethis
result is not surprising from that perspective. However, the corresponding result
for the basis of a computable torsion-free abelian group fails.
Proposition 1.6. Let H be an infinite rank torsion-free abelian group with a
computable basis B. For every set D, there is a basis B of H such that
D
deg(B )=deg(D).
D
Proof. Let B ={b0,b1,...} be effectively listed such that bi <N bi+1. Fix a set D.
Let BD = {n0b0,n1b1,...} where the ni ∈ N are chosen so that nibi <N ni+1bi+1
and n is even if and only if i ∈ D. It is clear that B is a basis for H and that
i D
B ≤ D. To compute D from B , let B = {c ,c ,...} be listed in increasing
D T D D 0 1
order. Foreachi,wecanfindc effectivelyinB ,andthenwecaneffectively(with
i D
no oracle) find b and n such that c =n b . By testing whether n is even or odd,
i i i i i i
we can determine whether i∈D. (cid:3)
In Section 2, we present background algebraic information. In Section 3, we
give the proof of Theorem 1.5. In Section 4, we state some generalizations of our
results, present some related open questions, and finish with remarks concerning
the following general question.
Question1.7. DescribethepossibledegreespectraofordersX(G)onacomputable
presentation G of a computable torsion-free abelian group.
Our notation is mostly standard. In particular we use the following convention
from the study of linear orders: If ≤ is a linear order on G, then ≤∗ denotes the
G G
linear order defined by x ≤∗ y if and only if y ≤ x. Note that if (G;≤ ) is an
G G G
ordered abelian group, then (G;≤∗) is also an ordered group.
G
2. Algebraic background
In our proof of Theorem 1.5, we will need two facts from abelian group theory.
Thefirstfactisthateverycomputablerankonegroupcanbeeffectivelyembedded
into the rationals. To define this embedding for a rank one H, fix any nonidentity
elementh∈H. Everynonidentityelementg ∈H satisfiesauniqueequationofthe
form nh = mg where n ∈ N, m ∈ Z, n,m (cid:54)= 0, and gcd(n,m) = 1. Map H into Q
by sending 0H to 0Q, sending h to 1Q, and sending g satisfying nh = mg (with
constraints as above) to the rational n. Because this map is effective, the image
m
of H in Q is computably enumerable and hence we can view H as a computably
DEGREES OF ORDERS ON TORSION-FREE ABELIAN GROUPS 5
enumerable subgroup of Q. Although the image need not be computable, it does
contain Z and, more generally, is closed under multiplication by any integer.
IfH=⊕ H iseffectivelycompletelydecomposable,wecaneffectivelymapH
i∈ω i
intoQω =⊕ Q(withitsstandardcomputablepresentation)byfixinganoniden-
i∈ω
tity element h ∈ H for each i and mapping H into Q as above. Therefore, we
i i i
will often treat H as a computably enumerable subgroup of Qω, and, in particular,
treat elements in each H subgroup as rationals.
i
The second fact we need is Baer’s Theorem (see [1]) giving classical algebraic
invariantsforrankonegroups. TheBaersequenceofarankonegroupisafunction
of the form f : ω → ω∪{∞} modulo the equivalence relation ∼ defined on such
functions by f ∼g if and only if f(n)(cid:54)=g(n) for at most finitely many n and only
when neither f(n) nor g(n) is equal to ∞.
To define the Baer sequence of a rank one group H, fix a nonidentity element
h ∈ H and let {p } denote the prime numbers in increasing order (later, for
i i∈ω
notational convenience, we alter the indexing to start with one). For a prime p, we
say pk divides h (in H) if pkg = h for some g ∈ H. We define the p-height of an
element h by
(cid:40)
k if k is greatest such that pk divides h,
ht (h):=
p ∞ otherwise, i.e., if pk divides h for all k.
The Baer sequence of h is the function B (n) = ht (h). If h,hˆ ∈ H are non-
H,h pn
identity elements, then B ∼ B . The Baer sequence B of the group H is
H,h H,hˆ H
(any representative of) this equivalent class. Baer’s Theorem states that for rank
one groups, H ∼=H if and only if B ∼B .
0 1 H0 H1
3. Proof of Theorem 1.5
Fix an effectively completely decomposable group H = ⊕ H as in the state-
i∈ω i
ment of Theorem 1.5. We divide the proof into three steps. First, we describe
our general method of building the computable copy G =(G;+ ,0 ) which is ∆0-
G G 2
isomorphic to H. Second, we describe how the computable ordering ≤ on G is
G
constructed. (The second computable order on G is ≤∗.) Third, we give the con-
G
struction of C and the diagonalization process to ensure the only C-computable
orders on G are ≤ and ≤∗.
G G
Part 1. General Construction of G.
ThegroupG isconstructedinstages, withG denotingthefinitesetofelements
s
(cid:83)
inGattheendofstages. WemaintainG ⊆G andletG:= G . Wedefine
s s+1 s s
a partial binary function + on G giving the addition facts declared by the end of
s s
stage s. To make G a computable group, we do not change any addition fact once
it is declared, so we maintain
x+ y =z =⇒ (∀t≥s)[x+ y =z]
s t
for all x,y,z ∈G . Furthermore, for any pair of elements x,y ∈G , we ensure the
s s
existence of a stage t and an element z ∈G such that we declare x+ y =z.
t t
To define the addition function, we use an approximation {bs,bs,...,bs} ⊆ G
0 1 s s
to an initial segment of our eventual basis for G. During the construction, each
approximate basis element bs will be redefined at most finitely often, so each will
i
eventually reach a limit. We let b := lim bs denote this limit. If k is an even
i s i
6 KACH,LANGE,ANDSOLOMON
index then the approximate basis element bs will never be redefined, so although
k
weoftenusethenotationbs (foruniformity),wehaveb =bs foralls. AlthoughG
k k k
will not be effectively decomposable, the group G will decompose classically into a
countable direct sum using the basis B ={b ,b ,b ,...}.
0 1 2
At stage 0, we begin with G :={0,1}. We let 0 denote the identity element 0
0 G
and we assign 1 the label b0. We declare 0 + 0 = 0 , 0 + b0 = b0, and
0 G 0 G G G 0 0 0
b0+ 0 =b0.
0 0 G 0
More generally, at stage s, each element g ∈G is assigned a Q-linear sum over
s
the stage s approximate basis of the form
qsbs+···+qsbs
0 0 n n
where n≤s, qs ∈Q for i≤n, and qs (cid:54)=0. (Later there will be further restrictions
i n
onthevaluesofqs toensurethatG isisomorphictoH.) Thisassignmentisrequired
i
to be one-to-one, and the identity element 0 is always assigned the empty sum.
G
It will often be convenient to extend such a sum by adding more approximate
basis elements on the end of the sum with coefficients of zero. For example, the
element 0 can be represented by any sum in which all the coefficients are zero.
G
We trust that using such extensions will not cause confusion.
Wedefinethepartialfunction+ onG bylettingx+ y =z (forx,y,z ∈G )if
s s s s
the assigned sums for x and y add together to form the assigned sum for z. Thus,
the function + is commutative and associative (on the elements for which it is
s
defined) and satisfies x+ 0 =x for all x∈G .
s G s
For each i ∈ ω, we fix a nonidentity element h ∈ H and embed H into Q by
i i i
sendinghi to1Q asdescribedinSection2. WeequateHi withitsimageinQinthe
sense of treating elements of H as rationals. For example, if a∈H and q ∈Q, we
i i
letqa∈Qdenotetheproductofqwiththeimageofaunderthis(fixed)embedding
of H . (Recall that while we cannot determine effectively whether the rational qa
i
is in H , if qa is in H , then we will eventually see this fact.) In particular, since h
i i i
is mapped to 1Q, if a∈Hi and a=qhi, we view a as being the rational q.
At each stage s, we maintain positive integers Ns for i ≤ s. These integers
i
restrain the (nonzero) coefficients qs of bs allowed in the Q-linear sum for each
i i
element g ∈ G by requiring that qsNs ∈ H and that we have seen this fact by
s i i i
stage s. Using the fact that N :=lim Ns exists and is finite for all i, we will show
i s i
(using Baer’s Theorem) that in the limit, the i-th component of G is isomorphic
to H , and hence that G is a computable copy of H. (Later we will introduce a
i
basis restraint K ∈ω that will prevent us from changing Ns too often.)
i
During stage s+1, we do one of two things – either we leave our approximate
basisunchangedorweaddadependencyrelationforasinglebs forsomeoddindex
(cid:96)
(cid:96)≤s. The diagonalization process dictates which happens.
Case 1. If we leave the basis unchanged, then we define bs+1 := bs for all i ≤ s.
i i
For each g ∈G (viewed as an element of G ), we define qs+1 :=qs and assign g
s s+1 i i
thesamesumwithbs+1 andqs+1 inplaceofbs andqs, respectively. Itfollowsthat
i i i i
x+ y =z (for x,y,z ∈G ) if x+ y =z. We set Ns+1 :=Ns for all i≤s and
s+1 s s i i
Ns+1 :=1.
s+1
We add two new elements to G , labeling the first by bs+1 and labeling the
s+1 s+1
second by qs+1bs+1 +···+qs+1bs+1, where (cid:104)qs+1,...,qs+1(cid:105) is the first tuple of
0 0 n n 0 n
rationals (under some fixed computable enumeration of all tuples of rationals) we
find such that n ≤ s, qs+1 (cid:54)= 0, qs+1Ns+1 ∈ H at stage s for all i ≤ n, and this
n i i i
DEGREES OF ORDERS ON TORSION-FREE ABELIAN GROUPS 7
sum is not already assigned to any element of G . (We can effectively search for
s+1
such a tuple.) This completes the description of G in this case.
s+1
Case 2. If we redefine the approximate basis element bs (for the sake of diagonal-
(cid:96)
izing) by adding a new dependency relation, then we proceed as follows. We define
bs+1 := bs for all i ≤ s with i (cid:54)= (cid:96). The diagonalization process will tell us either
i i
to set bs = qbs+1 for some rational q, or to set bs = m bs+1 +m bs+1 for some
(cid:96) k (cid:96) 1 j 2 k
integers m and m . (We will specify properties of these integers below.) In either
1 2
case, the index k will be even and greater than the basis restraint K and j,k <(cid:96).
We assign g ∈ G the same sum except we replace each bs by bs+1 (for i ≤ s and
s i i
i (cid:54)= (cid:96)) and we replace bs by either qbs+1 or m bs+1 +m bs+1 (as dictated by the
(cid:96) k 1 j 2 k
diagonalization process).
For example, if the diagonalization process tells us to make
bs =m bs+1+m bs+1, then the sum for g ∈G changes from
(cid:96) 1 j 2 k s
qsbs+···qsbs+···+qsbs +···+qsbs+···+qsbs
0 0 j j k k (cid:96) (cid:96) s s
at stage s (where we have added zero coefficients if necessary) to
qsbs+1+···+qsbs+1+···+qsbs+1+···+qs(m bs+1+m bs+1)+···+qsbs+1
0 0 j j k k (cid:96) 1 j 2 k s s
=qsbs+1+···+(qs+qsm )bs+1+···+(qs+qsm )bs+1+···+qsbs+1
0 0 j (cid:96) 1 j k (cid:96) 2 k s s
atstages+1. Therefore,wesetqs+1 :=qs+qsm ,qs+1 :=qs+qsm ,andqs+1 :=0,
j j (cid:96) 1 k k (cid:96) 2 (cid:96)
whileleavingqs+1 :=qs foralli(cid:54)∈{j,k,(cid:96)}. Similarly,ifthediagonalizationprocess
i i
tells us to make bs = qbs+1, then we set qs+1 := qs +qqs and qs+1 := 0 while
(cid:96) k k k (cid:96) (cid:96)
leaving qs+1 =qs for all i(cid:54)∈{k,(cid:96)}.
i i
WedefineNs+1,fori≤s,asfollows. Ifbs =m bs+1+m bs+1,thenNs+1 :=Ns
i (cid:96) 1 j 2 k i i
for all i ≤ s. If bs = qbs+1, then Ns+1 := Ns for all i ≤ s with i (cid:54)= k and
(cid:96) k i i
Ns+1 := d dNs where d is the denominator of q (when written in lowest terms)
k q k q
and d is the product of all the (finitely many) denominators of coefficients qs for
(cid:96)
g ∈G . In either case, set Ns+1 :=1.
s s+1
WeaddthreenewelementstoG ,labelingthefirstbybs+1,labelingthesecond
s+1 (cid:96)
by bs+1, and labeling the third by qs+1bs+1+···+qs+1bs+1 where (cid:104)qs+1,...,qs+1(cid:105)
s+1 0 0 n n 0 n
is the first tuple of rationals we find such that n≤s, qs+1 (cid:54)=0, qs+1Ns+1 ∈H at
n i i i
stage s for all i≤n, and this sum is not already assigned to any element of G .
s+1
This completes the description of G in this case.
s+1
WenoteseveraltrivialpropertiesofthetransformationsofsumsinCase2. First,
theapproximatebasiselementbs+1 doesnotappearinthenewsumforanyelement
(cid:96)
of G viewed as an element of G . Second, for any element g ∈ G , if qs = 0,
s s+1 s (cid:96)
then the coefficients qs+1 and qs+1 satisfy qs+1 =qs and qs+1 =qs. Third, by the
j k j j k k
linearity of the substitutions, if x+ y =z, then x+ y =z.
s s+1
We also require two additional properties which place some restrictions on the
rational q or the integers m and m . The first property is that the assignment
1 2
of sums to elements of G (viewed as elements of G ) remains one-to-one. The
s s+1
diagonalization process will place some restrictions on the value of either q or m
1
and m , but as long as there are infinitely many possible choices for these values
2
(whichwewillverifywhenwedescribethediagonalizationprocess),wecanassume
theyarechosentomaintaintheone-to-oneassignmentofsumstoelementsofG .
s+1
8 KACH,LANGE,ANDSOLOMON
The second property is that for each g ∈G , we need each coefficient qs+1 to
s+1 i
satisfy qs+1Ns+1 ∈ H . We will verify this property below under the assumption
i i i
thatwhenwesetbs =m bs+1+m bs+1,theintegersm andm arechosensothat
(cid:96) 1 j 2 k 1 2
theyaredivisibleby thedenominatorofeach qs coefficientofeachg ∈G . (Again,
(cid:96) s
we will verify this property of m and m in the description of the diagonalization
1 2
process.)
We now check various properties of this construction under these assumptions
and the assumption that the limits b := lim bs and N := lim Ns exist for all i
i s i i s i
(which will be verified in the diagonalization description).
Lemma 3.1. For g ∈ G , the coefficients in the assigned sum qsbs +···+qsbs
s 0 0 n n
satisfy qsNs ∈H .
i i i
Proof. The proof proceeds by induction on s. If g is added at stage s, then the
result for g follows trivially. Therefore, fix g ∈G and assume the condition holds
s
at stage s. We show that the condition continues to hold at stage s+1. Note that
if we do not add a dependency relation (i.e., we are in Case 1), then the condition
at stage s+1 follows immediately since qs+1 = qs and Ns+1 = Ns. Assume we
i i i i
add a new dependency relation; we split into cases depending on the form of this
dependency.
If bs = qbs+1, then for all i (cid:54)∈ {k,(cid:96)}, the condition holds since qs+1 = qs and
(cid:96) k i i
Ns+1 = Ns. For the index (cid:96), we have qs+1 = 0 and hence the condition holds
i i (cid:96)
trivially. For the index k, we have qs+1 =qs+qqs and Ns+1 =d dNs. Therefore,
k k (cid:96) k q k
qs+1Ns+1 =(qs+qqs)d dNs =qsd dNs+qqsd dNs.
k k k (cid:96) q k k q i (cid:96) q k
Since qsNs ∈H and d d∈Z, we have qsd dNs ∈H . By definition, qd ∈Z and
k k k q k q k k q
qsd ∈ Z, and hence qqsd dNs ∈ Z ⊆ H . Therefore, we have the desired property
(cid:96) (cid:96) q k k
when bs =qbs+1.
(cid:96) k
If bs = m bs+1 +m bs+1, then for all i (cid:54)∈ {j,k} the condition holds as above.
(cid:96) 1 j 2 k
For the index j, we have qs+1 = qs+qsm and Ns+1 = Ns. By assumption, the
j j (cid:96) 1 j j
integer m is divisible by the denominator of qs and hence qsm ∈Z. Therefore,
1 (cid:96) (cid:96) 1
qs+1Ns+1 =(qs+qsm )Ns =qsNs+qsm Ns ∈H
j j j (cid:96) 1 j j j (cid:96) 1 j j
since qsNs ∈ H by the induction hypothesis and qsm Ns ∈ Z. The analysis for
j j j (cid:96) 1 j
the index k is identical. (cid:3)
Lemma 3.2. For each g ∈ G, there is a stage t such that g is assigned a sum
qtbt +···+qtbt that is not later changed in the sense that, for all stages u ≥ t,
0 0 n n
the element g is assigned the sum qubu+···+qubu with bu = bt and qu = qt for
0 0 n n i i i i
all i≤n.
Proof. When g enters G, it is assigned a sum. The coefficients in this sum only
change when a diagonalization occurs. In this case, some approximate basis ele-
ment bs with nonzero coefficient in the sum for g is made dependent via a relation
(cid:96)
oftheformbs =qbs+1 orbs =m bs+1+m bs+1 withj,k <(cid:96). Therefore,eachtime
(cid:96) k (cid:96) 1 j 2 k
the sum for g changes, some approximate basis element with nonzero coefficient
is replaced by rational multiples of approximate basis elements with lower indices.
This process can only occur finitely often before terminating. (cid:3)
DEGREES OF ORDERS ON TORSION-FREE ABELIAN GROUPS 9
We refer to the sum in Lemma 3.2 as the limiting sum for g and denote it by
q b +···+q b . ItfollowsfromLemma3.1andLemma3.2thateachcoefficientq
0 0 n n i
in a limiting sum satisfies q N ∈H .
i i i
Lemma 3.3. For each rational tuple (cid:104)q ,...,q (cid:105) such that q (cid:54)= 0 and q N ∈ H
0 n n i i i
for all i ≤ n, there is an element g ∈ G such that the limiting sum for g is
q b +···+q b .
0 0 n n
Proof. For a contradiction, suppose there is a rational tuple violating this lemma.
Fixtheleastsuchtuple(cid:104)q ,...,q (cid:105)inourfixedcomputableenumerationofrational
0 n
tuples. Lets≥nbeastagesuchthatbs,...,bs andNs,...,Ns havereachedtheir
0 n 0 n
limits,eachtuplebefore(cid:104)q ,...,q (cid:105)whichsatisfiestheconditionsinthelemmahas
0 n
appearedasthelimitingsumofanelementinG ,andwehaveseenbystagesthat
s
q N ∈ H for each i ≤ n. By our construction, at stage s+1, either there is an
i i i
elementthatisassignedthesumq bs+1+···+q bs+1 orelseweaddanewelement
0 0 n n
to G and assign it this sum. In either case, this element has the appropriate
s+1
limitingtuplesincebs+1,...,bs+1 havereachedtheirlimits(andthusweobtainour
0 n
contradiction). (cid:3)
By Lemma 3.3 and the remarks following Lemma 3.2, the limiting sums of ele-
ments of G are exactly the sums q b +···+q b with q (cid:54)= 0 and q N ∈ H for
0 0 n n n i i i
all i≤n.
Lemma3.4. Ifx+ y =z,thenx+ y =z forallt≥s. Inparticular,ifx+ y =z,
s t s
then the limiting sums for x and y add to form the limiting sum for z.
Proof. In both cases of stage s+1 of our construction, we checked that x+ y =z
s
implies x+ y =z. Thus, the result follows by induction. (cid:3)
s+1
Lemma 3.5. For each pair x,y ∈G , there is a stage t≥s and an element z ∈G
s t
such that x+ y =z. For each x∈G , there is a stage t≥s and an element z ∈G
t s t
such that x+ z =0 .
t G
Proof. For the first statement, fixing x,y ∈ G , let u ≥ s be a stage at which x
s
and y have been assigned their limiting sums
x=qubu+···+qubu and y =qˆubu+···+qˆubu,
0 0 n n 0 0 n n
adding zero coefficients if necessary to make the lengths equal. By Lemma 3.1,
for all t ≥ u and i ≤ n, we have that qtNt ∈ H and qˆtNt ∈ H . Therefore,
i i i i i i
(qt+qˆt)Nt ∈ H . As in the proof of Lemma 3.3, there must be a stage t ≥ u and
i i i i
an element z ∈G assigned to the sum
t
z =(qt +qˆt)bt +···+(qt +qˆt)bt.
0 0 0 n n n
Then x+ y =z.
t
The proof of the second statement is similar. (cid:3)
Using Lemma 3.4 and Lemma 3.5, we define the addition function + on G by
G
putting x+y =z if and only if there is a stage s such that x+ y =z.
s
Lemma 3.6. The group G is a computable copy of H.
Proof. The domain and addition function on G are computable. By Lemma 3.5,
every element of G has an inverse, and it is clear from the construction that the
addition operation satisfies the axioms for a torsion-free abelian group.
10 KACH,LANGE,ANDSOLOMON
Let G be the subgroup of G consisting of all element g ∈ G with limiting sums
i
of the form q b . Since the limiting sums of elements of G are exactly the sums of
i i
the form q b +···+q b with q (cid:54)= 0 and q N ∈ H for i ≤ n, it follows that
0 0 n n n i i i
G ∼= ⊕ G . Therefore, to show that G ∼= H, it suffices to show that G ∼= H for
i∈ω i i i
every i∈ω.
Fixi∈ω. ThegroupG isarankonegroupwhichisisomorphictothesubgroup
i
of (Q,+Q) consisting of the rationals q such that qNi ∈ Hi. Thus, calculating
the Baer sequence for Gi using the rational 1Q, we note that for any prime pj,
1/pk ∈ G if and only if N /pk ∈ H . Therefore, the entries in the Baer sequences
j i i j i
for G and H differ only in the values corresponding to the prime divisors of N
i i i
and they differ exactly by the powers of these prime divisors. Therefore, by Baer’s
Theorem, G ∼=H . (cid:3)
i i
Part 2. Defining the Computable Orders on G. We define the computable
ordering of G in stages by specifying a partial binary relation ≤ on G at each
s s
stage s. To make the ordering relation computable, we satisfy
x≤ y =⇒ (∀t≥s)[x≤ y] (1)
s t
for all x,y ∈G . Typically, the relation ≤ will not describe the ordering between
s s
every pair of elements of G , but it will have the property that for every pair of
s
elements x,y ∈G , there is a stage t≥s at which we declare x≤ y or y ≤ x, and
s t t
not both unless x = y. Since we will be considering several orderings on G, for an
ordering (cid:52) on G, we let (g1,g2)(cid:52) denote the set {g ∈G|g1 ≺g ≺g2}. Moreover,
given a1,a2 ∈R, we let (a1,a2)≤R denote the interval {a∈R|a1 <R a<R a2}.
TospecifythecomputableorderonG,webuilda∆0-mapfromGintoR. (Thus
2
our order will be archimedean.) To describe this order, let {p } enumerate the
i i≥1
prime numbers in increasing order. We map the basis element b0 to r0 = 1R. For
i≥1, we will assign (in the limit of our construction) a real number r to the basis
√ i
element b such that r is a positive rational multiple of p . We choose the r
i i i i
in this manner so that they are algebraically independent over Q. If the element
g ∈G is assigned a limiting sum
g =q b +···+q b ,
0 0 n n
then our ∆0-map into R sends g to the real q r +···+q r . It also sends 0 to 0.
2 0 0 n n G
We need to approximate this ∆0-map during the construction. At each stage s,
2
we keep a real number rs as an approximation to r , viewing rs as our current
i i i
target for the image of b . The real rs is always 1 and the real rs is always a
i √ 0 i
positive rational multiple of p . Exactly which rational multiple may change
i
during the course of the diagonalization process. However, if k is an even index,
then rs will never change.
k
We could generate a computable order on G by mapping G into R using a
s s
linear extension of the map sending each bs to rs. However, this would restrict
i i
our ability to diagonalize. Therefore, at stage s, we assign each bs (for i ≥ 1) an
i
interval (as,as) where as and as are positive rationals such that rs ∈(as,as)
i (cid:98)i ≤R i (cid:98)i i i (cid:98)i ≤R
and as−as ≤ 1/2s. The image of bs in R (in the limit) will be contained in this
(cid:98)i i i
interval.
Because each x∈G is assigned a sum describing its relationship to the current
s
approximate basis, we can generate an interval approximating the image of x in R