aa r X i v : . [ m a t h . G R ] M a r LINEAR EXTENSIONS OF PARTIAL ORDERSON ABELIAN GROUPS
TOBIAS SCHLEMMER
Abstract.
Partially ordered groups, also known as po-groups, are groupswith a compatible partial order. Results from
M. I. Zajceva and H.-H. Teh arecombined in order to provide a full characterisation of linear order extensions ofa given order on a group. In contrast to Teh this approach provides a methodto discuss linear orders of different abelian rank in a uniform manner. This willbe achieved by modelling the linear orders using hyperplanes in a real vectorspaces. Among some additional remarks a construction of an archimediandirected order is given for every torsion free abelian group.
Introduction
Linear and lattice ordered groups are well-studied algebraic structures. Thestructure of linearly ordered abelian groups has mainly been discovered by thework of Otto H¨older [12] and Hans Hahn [11]. Later, Friedrich Levi [15] provideda first characterisation of lattice ordered groups and showed that an abelian groupcan be linearly ordered iff it is torsion free. It must be noticed that at the time ofthese early works the theory of modules and vector spaces had not settled down.First works on these topics from the middle of the 19th century have rarely beennoticed in the community. It took nearly 100 years until todays definitions forvector spaces and modules had been fixed.
Анатолий Иванович Мальцев [31] (
A. I. Maľcev ) investigated necessary andsufficient conditions for linear orders on abelian groups. Later,
М. И. Зайцева [28] (
M. I. Zajceva ) published a characterisation of finitely generated archimedianlinearly ordered abelian groups. With a corresponding decomposition into archime-dian subgroups, in that work she discusses also the main properties of a descriptionof linear orders on finitely generated groups. The cardinality of such orders on agiven group has been determined by Shin-Ichi [19] to ℵ . Eventually, Hoon-Heng Teh[25] provided another characterisation of linearly ordered groups based on Hahn’stheorem.The current work provides a characterisation of linear order extensions of abelianpartially ordered groups, which has been developed from scratch, based on theusage and an example from Charles Holland and other authors (cf. e. g., [13]). Levidiscussed the characteristics of lattice ordered groups starting with abelian groupsembedded into one and two dimensional vector spaces and analysed them withrising dimension (bottom-up approach). He provides tools to describe the image ofa positive cone of a lattice ordered group in a vector space by means of rays and Mathematics Subject Classification.
Key words and phrases. po-groups, o-groups, l-groups, linear extension.This work was supported by TU Dresden, Professur f. Angewandte Algebra. edges. We will discuss the same phenomenon by construction of hyperplanes in avector space that are constructed using independent sets of a given group (top-downapproach). Though, this method considers the positive cone in less granularity, itprovides a better bridge to methods from convex geometry than Levi’s approach.
Zajceva ’s equations for archimedian orders on n -generated abelian groups can beinterpreted as certain hyperplanes in n -dimensional real vector spaces. Thus, thiswork can also be considered an extension of her results onto arbitrary abelian groupsand allows to discuss existing orders on them. A further advantage of this viewconsists in the possibility to investigate archimedian and several non-archimedianorders together based on a common mathematical structure. This facet and thediscussion of linear order extensions of given order relations are also an enhancementto Teh’s work, who discusses the archimedian rank of linear orders.The constructive nature of the given article is dedicated to be more accessible fornon-mathematicians and mathematicians with a background different from grouptheory. This is achieved by using injective abelian groups, which need differenttools and provide slightly different insights than the usual upproach using freeabelian ℓ -groups or free vector lattices and the duality theorems provided by W. M.Beynon [2].To achieve this, after a short clearing of notions, mappings between orders ondifferent structures will be investigated. This discussion has the aim to enable usto represent a given order on a torsion free abelian group in a convenient way in thevector space of the direct sum ⊕ E R , where the set of indices E corresponds to amaximal independent set in the given group. There we will use a characterisation oflinearly ordered groups that is a little bit different from those given by Zajceva andTeh (cf. Theorem 24). Finally, with Theorem 28 we will provide a characterisationof linear order extensions on partially ordered abelian groups, with the use of halfspaces and linearly ordered hyperplanes. Thus, we can use methods from convexgeometry to discuss linearly ordered groups. This will be demonstrated in Section7, where some known results have been resembled. This section also shows how toconstruct an archimedian directed order on an arbitrary torsion free abelian group.1.
Preliminaries
In this section we repeat the basics used in this article. The main facts can befound in the usual textbooks about lattice ordered groups (e. g. [1, 3, 5, 32, 10, 14,29, 30]).A group G = ( G, · , − , e ) is called partially ordered iff there exists a partial order ≤ ⊆ G × G such that the following condition holds for all elements x, y, a, b ∈ G : (1) a ≤ b ⇒ xay ≤ xby. It is called linearly ordered iff ≤ is a linear order. A group homomorphism ϕ froma partially ordered group G into another one G ′ is called o-homomorphism iff itis also an order homomorphism. Furthermore it is called an o-isomorphism if it isadditionally an order isomorphism.A very fundamental theorem allows us to characterise each partial order ≤ on agroup G by means of its positive cone G + := { g ∈ G | ≤ g } : The given condition is sufficient to assure that the partial order ≤ is also compatible with theother group operations (inverse element and neutral element). INEAR EXTENSIONES ON ABELIAN PO-GROUPS 3
Theorem 1.
The positive cone G + of a partially ordered group ( G, · , − , e, ≤ ) fulfilsthe following conditions: G + · G + ⊆ G + G + is a semigroup (2) G + ∩ ( G + ) − = { e } G + is a pure subset (3) ∀ g ∈ G : g − G + g ⊆ G + G + is invariant subsemigroup . (4) If G is a directed set, it further fulfils, G = G + · ( G + ) − . (5) and if G is linearly ordered G = G + ∪ ( G + ) − . (6) Conversely, if in G a set P ⊆ G exists, that fulfils (2) to (4) , there exists an orderon G such that P is the set of positive elements of G . If furthermore the Condition (5) is met, the order is directed and, if (6) is true, it is a linear order.Proof. Cf. [14], Theorem 2.1.1 or [30], II.2.1. (cid:3)
An Element a ∈ G + is called infinitesimal with respect to another element b iffthe relation a n ≤ b holds for every integer n ∈ Z . Consequently 0 is infinitesimalwith respect to every other element. If there is no other element infinitesimal withrespect to any other element then the group is called archimedian .It is a well-known fact that each finite group can only be discretely ordered. Thisimplies that only torsion free groups can be linearly ordered. Thus, throughout thispaper we will refer to torsion free groups when ever we talk about groups , unlikeotherwise stated.An independent set E ⊆ G of an abelian group G = ( G, + , − ,
0) is defined as aset of its elements, which fulfils the equation(7) X a ∈ E ′ a ( a ) a = 0 , for each finite subset E ′ ⊆ E and a mapping a ∈ ⊕ E Z iff for all a ∈ E ′ the condition a ( a ) = 0 is met. Zorn’s lemma assures the existence of a maximal independent set.A subset E ⊆ G is independent, iff the subgroup h E i , which is generated by E , isisomorphic to the direct sum of the cyclic groups h a i of all of its elements a ∈ E .A subgroup S ⊆ G is called essential subgroup iff for each subgroup S ′ ⊆ G theintersection S ∩ S ′ = { } is non-trivial. An independent set E is maximal iff itsgenerated subgroup h E i is an essential subgroup of G . For further information werefer to [8] or any other text book on the theory of abelian groups.Given an arbitrary set E we define the direct sum ⊕ E G according to(8) ⊕ E G := n x (cid:12)(cid:12)(cid:12) x : E → G , | supp x | < ∞ o , where supp x := { a ∈ E | x ( a ) = 0 } is the support of the mapping x . The unitvectors will be denoted by(9) e a ( b ) := ( a = b, . T. SCHLEMMER
Figure 1.
Positive cone ina vector space P a a a If the generating set E is not uniquely determined by the context (e.g. when weare using different bases of a vector space) we will write x E ( e ) instead of x ( e ) forany element e ∈ E .A vector space V over an ordered field K = ( K, + , − , , · , − , orderedvector space iff V is an ordered group w.r.t. addition and for all vectors v ∈ V + ,and all positive elements α ∈ K + of the field the condition α v ∈ V + is met.Consequently, the set V + is also a positive cone in the geometrical sense (cf. Fig.1). If A is a set of vectors of a vector space, L A denotes their linear closure.The set of linear combinations of vectors from A with exclusively non-negativecoefficients will be symbolised by L + A . As usual a subset A ⊆ V is called linearsubspace A ≤ V if L A = A is true. A subspace H of a vector space V is called hyperplane if there exists a vector a ∈ V such that the conditions L H = H (cid:12) V and L ( H ∪ { a } ) = V hold. For some base B of the vector space V let S ⊆ B be asubbase. Then for every vector x the sum(10) P B ,S := X s ∈ S x B ( s ) s is called (coordinate) projection of x into L S along B .In Section 4 we need the notion of a neighbourhood. Since vectors in the directsum have finite support, we can use the standard scalar product in the resultingvector space for any ordered field K . To be consistent with the following definitions,we use a generalised version: Let B be a base of the vector space ⊕ E K , x , y ∈ ⊕ E K two vectors, and r : E → K + \ { } a positive mapping. Then(11) h x , y i r, B := X b ∈ B r ( b ) · x B ( b ) y B ( b )denotes a scalar product, as x and y have a finite support, and have values thatdiffer from zero only at finitely many entries. For the same reason given any basis B and any positive integer n ∈ N (12) k x k n,r, B := n s X b ∈ B r ( b ) · | x B ( b ) | n and k x k ∞ ,r, B := max b ∈ B r ( b ) · | x B ( b ) | are norms, and(13) ∆ n,r, B ( x , y ) := k y − x k n,r, B and ∆ ∞ ,r, B ( x , y ) := k y − x k ∞ ,r, B The additive group ( K, + , − ,
0) is an ordered group such that 0 ≤
1, as well as the multiplic-ative group of the positive elements ( K + \ { } , · , − , The complete argumentation be found in the proof of Theorem 8.
INEAR EXTENSIONES ON ABELIAN PO-GROUPS 5 are metrics on ⊕ E K . For the special case n = 2 we have the usual identity p h x , x i r, B = k x k ,r, B .For x ∈ ⊕ E K and r ∈ K we use the following definition for an open ball B r,n, B ( x )(including n = ∞ ):(14) B r,n, B ( x ) := { y ∈ ⊕ E K | ∆ n, r , B ( x , y ) < } . As the generated topologies of these open balls are not necessarily the same, wedefine a finer topology that includes all of them:
Definition 1.
A subset M ⊆ V of a vector space V is called open , iff for each pairof vectors a ∈ M, b ∈ V the set { λ ∈ R | λ a + (1 − λ ) b ∈ M } is open in the set of real numbers R with the standard topology.As all vectors have finite support, this defines a topology on V where all openballs are open sets. The boundary of a set M with respect to this topology isdenoted by ∂ M , and its interior by int M .For two subsets A, B ⊆ G of a group G we use the usual definition of a complexsum(15) A + B := { a + b | a ∈ A, b ∈ B } . If A and B are linear subspaces of a vector space whose intersection A ∩ B = { } is the singleton containing the zero vector , then we denote this by A ⊕ B . Thus,for a hyperplane H ≤ V there exists a vector a such that V = H ⊕ K a .2. Preparation of injective groups
In this section we mainly resemble Walkers theorem [27] for ordered groupsproviding a construction of the embedding into the injective abelian Groups withthe formalism used throughout the current article.We will characterise all linear order extensions of an abelian group G , in thisarticle using the vector space ⊕ E R for a set E , that is constructed depending on G .In order to achieve this we will use properties from the integer module ⊕ E Z andits divisible extension, the rational vector space ⊕ E Q as the latter is the smallestinjective group containing ⊕ E Z . While the first one is isomorphic to a subgroup of G , the latter one is isomorphic to a supergroup of the divisible hull of G . As it isuseful to have a well-defined base in the considered vector spaces, we shortly discussthe necessary steps for the construction of the mentioned algebraic structures onthe basis of the given torsion free abelian group. Furthermore, in this section wediscuss a method to transfer orders between these structures. Let’s start with thefollowing well-known fact:
Lemma 2 (Folklore) . Let G = ( G, · , − , e, ≤ ) be a po-group, G ′ = ( G ′ , · ′ , − ′ , e ′ ) a group and ϕ : G ′ → G an injective homomorphism. Then G ′ forms a po-grouptogether with the relation ⊑ defined by (16) g ⊑ h : ⇔ ϕ ( g ) ≤ ϕ ( h ) . It proved to be useful, to favour monomorphisms over subgroup relations, here.
T. SCHLEMMER ⊕ E Z ⊕ E Q ⊕ E R id idid (a) Monomorphisms ( ⊕ E Z , ≤| Z )( ⊕ E Q , ≤| Q ) ( ⊕ E R , ≤ ) id Q id Z id Z (b) Induced orders Figure 2.
Monomorphisms and induced orders according toLemma 2
Proof.
The mapping ϕ is a group isomorphism from G ′ onto ϕ [ G ′ ] and according toequation (16) it is also an order isomorphism. So it remains to show that the group G ′ is a partially ordered group together with the relation ⊑ . Let a, b, x, y ∈ G ′ elements of the group G ′ . Then the following equivalences hold: a ⊑ b ⇔ ϕ ( a ) ≤ ϕ ( b ) ⇔ ϕ ( x ) · ϕ ( a ) · ϕ ( y ) | {z } = ϕ ( x · ′ a · ′ y ) ≤ ϕ ( x ) · ϕ ( b ) · ϕ ( y ) | {z } = ϕ ( x · ′ b · ′ y ) ⇔ ϕ ( x · ′ a · ′ y ) ≤ ϕ ( x · ′ b · ′ y ) ⇔ x · ′ a · ′ y ⊑ x · ′ b · ′ y. So ϕ is an o-homomorphism from G ′ into G . (cid:3) This lemma allows us to relate possible orders on the three structures ⊕ E Z , ⊕ E Q and ⊕ E R . Figure 2 shows these relationships with respect to the identicalembeddings. In the left hand Figure 2(a) the diagram of the embeddings is shown,while the arrows on the right hand side 2(b) show the possible ways of inductionor transfer of the orders between the structures.In order to add an arbitrary torsion free abelian group into this system of mono-morphisms and induced orders, suppose E is an independent set of the group G .Thus, considering cyclic groups, a positive integer n ∈ N \ { } exists for eachelement g ∈ G such that ng ∈ h E i . Thereby, we found a more or less uniquerepresentation for each Element of G by elements of a maximal independent set E : Lemma 3.
Let G be a torsion free abelian group and E a maximal independent setin G . Then for each element g ∈ G a positive integer q g ∈ N \ { } and a mapping p g : E → Z exist such that the following equation holds: (17) q g g = X a ∈ E p g ( a ) a. The factors q g and p g ( a ) are uniquely defined up to multiplication by a commonrational number.Proof. As h E i is an essential subgroup, for each element g ∈ G the intersection h E i ∩ h g i 6 = ∅ is non-empty. This implies that a positive integer q g exists such that q g g ∈ h E i . This proves the existence of a representation of the form (17). INEAR EXTENSIONES ON ABELIAN PO-GROUPS 7
Let q g g = X a ∈ E p g ( a ) a and q g g = X a ∈ E p ′ g ( a ) a two such representations according to (17). Considering their difference we get0 = ( q g − q g ) g = X a ∈ E ( p g − p ′ g )( a ) a. Since E is independent, we deduce from this equation that for each a ∈ E thecondition p g ( a ) − p ′ g ( a ) = 0 holds. So we proved the identity p g = p ′ g for any fixednumber q g .Let’s assume that we have different representations of the form (17). Since themapping p g is unique for each q g , there must be different numbers on the left handside too. Let q g ∈ N \ { } the smallest integer for which a representation in theform (17) exists. Furthermore, let us assume there exists a number q ′ g ∈ N \ { , } ,which is coprime to q g such that: q ′ g g = X a ∈ E p ′ g ( a ) a. Then the following two equations hold:( q g − q ′ g ) g = X a ∈ E ( p g − p ′ g )( a ) a and( q g + q ′ g ) g = X a ∈ E ( p g + p ′ g )( a ) a. This allows us to use the Euclidian algorithm to find two integers c and d such that cq g + dq ′ g = gcd( q g , q ′ g ) holds. Using these the equationgcd( q g , q ′ g ) g = ( cq g + dq ′ g ) g = X a ∈ E ( c p g + d p ′ g )( a ) a. ( )holds. Since 0 < gcd( q g , q ′ g ) ≤ q g and q g was chosen minimal and q ′ g coprime to q g , there exists a contradiction. So q ′ g is a multiple of q g .Let q ′ g = bq g for some positive integer b ∈ N \ { } . Then we have X a ∈ E b p g ( a ) a = bq g g = q ′ g g = X a ∈ E p ′ g ( a ) a. This proves that both q ′ g and p ′ g are multiples with the same factor of q g and p g ,respectively. (cid:3) Using this Lemma 3 we can find an embedding of G with a maximal independentset E into the direct sum ⊕ E Q (cf. Walker’s theorem [27]): Theorem 4.
Let G = ( G, + , − , be a torsion free abelian group and E ⊆ G amaximal independent set in G . Let for each g ∈ G a mapping p g : E → Z and apositive integer q g ∈ N \ { } defined according to equation (17) . Then the mapping (18) ϕ : G → ⊕ E Q : g p g q g is a monomorphism from G into ⊕ E Q . T. SCHLEMMER
Proof.
Lemma 3 ensures for each g ∈ G the existence of integers q g , p g ( a ) (for all a ∈ E ) that fulfil the following equation: q g g = X a ∈ E p g ( a ) a Let ϕ : G → ⊕ E Q : g p g q g be the mapping as defined in equation (18). Then foreach c ∈ Z the equation ϕ ( g ) = c p g cq g = p g q g holds. From this we infer together withLemma 3 that ϕ is well-defined.Furthermore, having ϕ ( cg ) = c p g q g = cϕ ( g ) we can combine the two representa-tions q g g = X a ∈ E p g ( a ) a, q h h = X a ∈ E p h ( a ) a of elements g, h ∈ G , into one common formula: q g q h ( g + h ) = q h X a ∈ E p g ( a ) a + q g X a ∈ E p h ( a ) a. Thus, the mapping ϕ is a group homomorphism from G into ⊕ E Q with respect tothe addition. Consequently, ϕ ( g + h ) = q h p g q g q h + q g p h q g q h = ϕ ( g ) + ϕ ( h )Let ϕ ( g ) = . Then for each a ∈ E the element p g ( a ) = 0 is zero, because E isan independent set. Thus, for any q g ∈ N \ { } the equality q g g = 0 holds, whichimplies g = 0. So ϕ is injective, which means it is a monomorphism from G into ⊕ E Q . (cid:3) Corollary 5.
The homomorphism ϕ maps each element a ∈ E of the independentset onto a unit vector, i.e., ϕ ( a ) = e a . With Theorem 4 each order in the real (rational) vector space ⊕ E R ( ⊕ E Q )defines an order on the group G . Theorem 6.
Let G be a torsion free abelian group and E a maximal independentset in G . Furthermore, let ϕ : G → ⊕ E Q be defined as in equation (18) . Then themapping (19) ψ : ⊕ E Z → G : x X a ∈ E x ( a ) a is a monomorphism such that for all g ∈ h E i and all x ∈ ⊕ E Z the conditions (20) ψ (cid:0) ϕ ( g ) (cid:1) = g and ϕ (cid:0) ψ ( x ) (cid:1) = x hold.Proof. Firstly we show that ψ is a homomorphism. Since G is commutative, wecan rewrite the sum in the following way: ψ ( x + y ) = X a ∈ E ( x + y )( a ) a = X a ∈ E x ( a ) a + X a ∈ E y ( a ) a = ψ ( x ) + ψ ( y ) . The last identity holds, since G is torsion free. Furthermore, E is an independentset in G and the equation e a ( a ) = 1 holds. Thus, P a ∈ E x ( a ) a = 0 implies x = ,which proves that ψ is an injective homomorphism. INEAR EXTENSIONES ON ABELIAN PO-GROUPS 9 G ⊕ E Z ⊕ E Q ⊕ E R ψϕ id idid (a) Monomorphisms ( G , ⊑ ) ( ⊕ E Z , ≤| Z )( ⊕ E Q , ≤| Q ) ( ⊕ E R , ≤ ) ψϕ id Q id Z id Z (b) Induced orders Figure 3.
Monomorphisms and induced orders according to Lem-mata 2, 4 and 6Since ϕ and ψ are monomorphisms and h E i is a subgroup in G , it suffices toshow the equations (20) for the elements of the generating sets E and { e a | a ∈ E } .For those it is easy to show the following identities: ψ (cid:0) ϕ ( a ) (cid:1) = ψ ( e a ) = a, and ϕ (cid:0) ψ ( e a ) (cid:1) = ϕ ( a ) = e a . Since these identities hold for any element of the corresponding set they are satisfiedfor any element of the corresponding (sub)group, too. (cid:3)
The preceding two theorems prove that the diagram in Figure 3(a) commutes.Thus, we can transfer orders along the arrows in Figure 3(b) between the differentgroups. Together with Theorem 24, this would be sufficient to rework the theoremof
М. И. Зайцева (M. I. Zajceva) [28] in the language of vector spaces. Since wewant to discuss order extensions, we must firstly transfer the order from a givengroup into the corresponding real vector space. The following section will providethis link. 3.
Copying the order into the vector space
Any order in the vector space constructed in the last section implies an order inour group (cf. Figure 2(b)). So far the vector space has no idea about the existingorder in the group G . This section closes this gap. There are some well-knowntheorems such as Hahn’s theorem [11] or the Conrad-Harvey-Holland-Theorem (cf.[9], chap. 4.6) which cover the embedding of ordered abelian groups into real vectorspaces. These theorems provide embeddings for each order, but they do not assurethat it is possible to use the same embedding for all orders.We can transfer the order from the group onto an integer module (see Figure3). However, going this path we loose information about group elements. On theother hand, the integer module ⊕ E Z is a set of size nearly zero in the real vectorspace ⊕ E R , so it is not easy to tell which linear vector space orders correspond toone linear order in the integer module.As the notion convex is already order-theoretically defined in the language ofpartially ordered groups, we choose a different wording, here: Definition 2 (cf. [5], Def. 3.2) . Let G = ( G, + , − , , ≤ ) a partially ordered abeliangroup. The order ≤ is called semiclosed iff for all a ∈ G and n ∈ N \ { } the P a a a ana (a) valid P a a a ana (b) forbidden Figure 4.
Semiclosed order in an ordered groupfollowing implication holds:(21) 0 ≤ na ⇒ ≤ a. We show in this section that semiclosed orders on any abelian torsion free groupcan be easily extended to vector space orders. This will be done by forming theconvex hull. Doing so, we prove that in this particular case the induced order onthe group will be the same as the original one. Fortunately, the class of semiclosedordered torsion free abelian groups is the relevant class of groups to consider, here.Each partially ordered abelian group has a canonical semiclosed order extension,which itself is a suborder of any linear order extension of the group. So the restric-tion to semiclosed groups does not influence the generality of this construction.As shown in Figure 4 the concept of a semiclosed order can be considered asan order on the group, whose positive cone is convex in a geometrical sense (wewill discuss this later). In geometric discussions about convex sets the convex hullplays an important role. As a semiclosed order has a “convex“ positive cone, wecan ask, whether the positive cone of any partial order has a well defined “convexhull“. The following lemma discusses this fact:
Lemma 7 (cf. [5], Cor. 29.10) . Let G = ( G, + , − , , ≤ ) be a partially orderedabelian group and the order relation ≤ ′ defined by (22) 0 ≤ ′ g : ⇔ ∃ m ∈ N \ { } : 0 ≤ mg. Then ≤ ′ is a semiclosed order relation and ( G, + , − , , ≤ ′ ) is a partially orderedgroup. Each linear or lattice order on G is an extension of ≤ ′ .Proof. Firstly, we have to show that P := { g ∈ G | ≤ ′ g } fulfils the conditions(2) to (4). Let g, h ∈ P be two elements of this set. Then there exist two positiveintegers m, n ∈ N \ { } such that 0 ≤ mg and 0 ≤ nh . This implies 0 ≤ mng and0 ≤ mnh . Consequently, we have 0 ≤ mng + mnh = mn ( g + h ) and 0 ≤ ′ g + h .Since G is a commutative group, the set P is a semigroup and invariant. If for anynonzero element g = 0 the inequality 0 ≤ mg holds, there exists no positive number n ∈ N such that ng ≤
0. Otherwise from ng ≤ mng ≤ ≤ mg follows 0 ≤ mng . This would imply g = 0, which has been excluded. Thus, with P ∩ − P = { } the set P is a pure subset of G . This shows that( G, + , − , , ≤ ′ ) is a partially ordered group. Since the condition (22) is always truein lattice ordered groups (cf. [5], Prop. 3.6) and the positive cones of ≤ and ≤ ′ differ by exactly those elements, which contradict this condition, the order ≤ ′ is asuborder of any lattice order extension of ≤ . (cid:3) INEAR EXTENSIONES ON ABELIAN PO-GROUPS 11
Up to now, the transitive closure of the arrows in Figure 3(b) is a partial order.Our aim is to consider the orders on the different groups to be more or less equi-valent. The following theorem extends a given order on a subgroup to a semiclosedorder on the containing vector space. As a result of this theorem, we will be ableto invert the arrows in the lower right triangle of this figure.
Theorem 8.
Let ⊕ E Z an integer module and ⊕ E R a vector space of equal dimen-sion. Then there exists an injective mapping, which maps each semiclosed orderon ⊕ E Z to a vector space order on ⊕ E R . The same is true for the combinations ⊕ E Z / ⊕ E Q and ⊕ E Q / ⊕ E R .Proof. It is sufficient to prove that the positive cone ( ⊕ E Z ) + can be embedded intoa convex invariant subset P of the vector space ⊕ E R that fulfils the Conditions(2) to (4). We will show the injectivity afterwards. As candidate for the set P wechoose the convex hull of ϕ (cid:2) ψ [( ⊕ E Z ) + ] (cid:3) in the real vector space ⊕ E R using themonomorphisms ϕ and ψ defined in (18) and (19). Let P := conv (cid:0) ϕ (cid:2) ψ [( ⊕ E Z ) + ] (cid:3)(cid:1) P ′ := L + (cid:0) ϕ (cid:2) ψ [( ⊕ E Z ) + ] (cid:3)(cid:1) . As P ′ contains any positive linear combinations of its elements it also contains thespecial case of convex combinations. This implies P ⊆ P ′ . Since the zero vector ∈ ⊕ E Z lies in the positive cone and for each positive vector x ∈ ( ⊕ E Z ) + itsmultiple n x ∈ ( ⊕ E Z ) + is positive for every n ∈ N , for each α ∈ R + we also have α x ∈ P . This can be deduced from the equation α n X i =0 β i x i = n X i =0 αβ i x i . Thus, for any two vectors x , y ∈ P and any two positive numbers α, β ∈ R + weget: α x + β y = αα + β (cid:0) ( α + β ) x (cid:1) + (cid:18) − αα + β (cid:19) ( α + β ) y ∈ P, which proves P ′ ⊆ P and together with the preceding results P = P ′ . Furthermore,the set P is a subsemigroup of ⊕ E R . As this vector space is commutative, P isalready an invariant semigroup.The set B := { e a | a ∈ E } is a generating set of the module ⊕ E Z . Thus, B isa basis of ⊕ E R . Since P ⊆ ⊕ E R , the following equation holds: L P = L (cid:16) L + (cid:0) ( ⊕ E Z ) + (cid:1)(cid:17) . Thus, for any vector x ∈ ⊕ E Z ∩ P there exists a linear combination of elementsof a linear independent set B ⊆ ( ⊕ E Z ) + and a positive integer n ∈ N such that x ∈ P ni =1 α i a i where α i ∈ R + and a i ∈ B . Since the range of x is a set of integers,the coefficients α i cannot be irrational, because they are solutions of the system oflinear equations ∀ m ∈ E : n X i =1 α i a i ( m ) = x ( m ) . For α = β = 0 we already proved ∈ Z ⊆ P . G ⊕ E Z ⊕ E Q ⊕ E R ψϕ id idid (a) System of Monomorphisms ( G , ⊑ ) ( ⊕ E Z , ≤| Z )( ⊕ E Q , ≤| Q ) ( ⊕ E R , ≤ ) ψϕ conv R id Q conv R id Z conv Q id Z (b) Implication of induced orders Figure 5.
Transfer of orders on groups with respect to mono-morphisms and convex hulls from an abelian Group G onto a realvector space and backAs a i ( m ) ∈ Z and x ( m ) ∈ Z are integers, the coefficients α i are rational numbers.Let k be the least common denominator of all those numbers α i . Then we canmultiply the system with k and get the modified system of linear equations ∀ m ∈ E : n X i =1 ( kα i ) a i ( m ) = x ′ ( m ) , where x ′ is defined as x ′ := k x . Thus, the vector x ′ is positive in ⊕ E Z and so thevector x is non-negative. Since the order on ⊕ E Z is semiclosed, this also implies x ∈ ( ⊕ E Z ) + . Consequently, we showed that x ∈ P ∩ ⊕ E Z holds iff x ∈ ( ⊕ E Z ) + .So we get P ∩ ⊕ E Z = ( ⊕ E Z ) + and a partial order ≤ ′ ⊆ ⊕ E R × ⊕ E R definedby x ≤ ′ y , iff x − y ∈ P. Finally, we mapped each order ≤ on ⊕ E Z onto an order ≤ ′ on ⊕ E R such that thepositive cone of ≤ ′ reproduces the positive cone of ≤ by the intersection with ⊕ E Z .From that follows that the mapping from the set of partial orders on the group ⊕ E Z into the set of vector space orders on ⊕ E R , is injective.The proof of the other two combinations follows the same path. (cid:3) As already stated, this theorem ensures that the paths in the lower right trianglein Figure 5(b) are equivalent. Due to the additional arrow from ( ⊕ E Z , ≤| Z ) to( ⊕ E Q , ≤| Q ) this is not clear for the upper left triangle in this figure. The followinglemma closes this gap. Lemma 9.
If the group G is semiclosed ordered, E is a maximal independent setin G and the monomorphisms ϕ and ψ are defined according to the Equations (18) and (19) the following equation holds: (23) conv Q (cid:2) ψ − [ G + ∩ h E i ] (cid:3) ∩ ϕ [ G ] = ϕ [ G + ] Proof.
Let x ∈ ϕ [ G + ]. Then there exists an element g ∈ G + such that ϕ ( g ) = x .Since E is a maximal independent set there exists a natural number n ∈ N \ { } such that ng ∈ h E i is an element of the subgroup which is generated by E . Thus ng is contained in the image of the group homomorphism ψ . Considering ⊕ E Q the vector n ψ − ( ng ) is an element of the convex hull conv Q (cid:0) { , ψ − ( ng ) } (cid:1) . Since INEAR EXTENSIONES ON ABELIAN PO-GROUPS 13 the neutral element is always an element of the positive cone, we have shown onedirection of (23): conv Q (cid:2) ψ − [ G + ∩ h E i ] (cid:3) ∩ ϕ [ G ] ⊇ ϕ [ G + ] . Let x ∈ conv Q (cid:2) ψ − [ G + ∩ h E i ] (cid:3) ∩ ϕ [ G ]. Then there exists elements g ∈ G and g , . . . , g n ∈ G + ∩ h E i , positive rational numbers x , . . . , x n ∈ Q + and a positiveinteger n ∈ N \ { } such that the following condition holds: x = ϕ ( g ) = n X i =1 x i ψ − ( g i ) , where n X i =1 x i = 1 . Let q ∈ N be the smallest common denominator of { x , . . . , x n } . Then theproducts { qx , . . . , qx n } ⊆ N are positive integers. From Theorem 6 and Equation(20) we infer that for each i ∈ { , . . . , n } we can use the identity ψ − ( g i ) = ϕ ( g i ).Using these two modifications we get the equation qϕ ( g ) = n X i =1 qx i ϕ ( g i ) . Since ϕ is an injective group homomorphism, we can rewrite this equation as ϕ ( qg ) = ϕ n X i =1 qx i g i ! and get qg = n X i =1 qx i g i . Thus, qg ∈ G + , which implies g ∈ G + , since the order of G is semiclosed. Thisproves the other inclusionconv Q (cid:2) ψ − [ G + ∩ h E i ] (cid:3) ∩ ϕ [ G ] ⊆ ϕ [ G + ]and, thus, the lemma is proved. (cid:3) Linear orders in real vector spaces
Here, we will discuss linear vector space orderings. As already mentioned in thelast section this touches the area of fundamental theorems of the theory of ℓ -groupswith the additional insight how different orders can be represented in the samevector space.We will prove that the positive cone in the generated subspace of a class ofarchimedian equivalent elements of a linearly ordered vector space is defined bylinear half spaces which are bounded by linearly ordered hyperplanes. Using Zorn’slemma, it can be shown that this is equivalent to the description by linearly orderedbases and an extension to Teh’s result [25]. In contrast to that work, we useproperties of normed vector spaces to prove this result for all torsion free abeliangroups.At first we discuss possible linear orders on the vector space ⊕ E R for a givenset E . In each of these orders a hyperplane will contain all infinitesimal elements,and the space will be filled with copies of this hyperplane (c. f., Figure 6). Lemma 10.
Let H be a linearly ordered hyperplane in the vector space ⊕ E R and a ∈ ⊕ E R \ H a vector which is linearly independent from H . Then the set (24) H + ∪ (cid:0) H + ( R + \ { } ) a (cid:1) defines a linear order on the vector space ( ⊕ E R , + , − , ) . Figure 6.
Linear order ina real vectorspace a a a Proof.
Let P := H + ∪ (cid:0) H + ( R + \ { } ) a (cid:1) be the half space, which is defined bythe positive cone of the hyperplane H + and the vector a . Since ⊕ E R is a groupconsidering the addition, We have to prove the conditions (2) to (4) considering P as positive cone of ⊕ E R . This will prove that P is an invariant semigroup in ⊕ E R and for each vector x ∈ ⊕ E R either x ∈ P or − x ∈ P holds.Let x ∈ P be a vector in P . Then either the condition x ∈ H + or x ∈ H + ( R + \{ } ) a holds.(1) The condition x ∈ H + is true iff − x ∈ H − , since H is a vector space itselfand linearly ordered by precondition.(2) If x ∈ H + ( R + \ { } ) a , then a unique vector x ′ ∈ H and a unique positivereal number x ∈ R + \ { } exist such that x = x ′ + x a . This is equivalent to − x = − x ′ + ( − x ) a ∈ H + ( R − \ { } ) a Furthermore, since H is as hyperplane in ⊕ E R and a H , the following holds: ⊕ E R = H ⊕ R a = H + R a = H + ( R + \ { } ) a ˙ ∪ ( H + \ { } ) ˙ ∪ { } ˙ ∪ ( H − \ { } ) ˙ ∪ H + ( R − \ { } ) a = P \ { } ˙ ∪ { } ˙ ∪ − P \ { } In short: ⊕ E R = P ∪ − P , so the Condition (6) is fulfilled. From the two pointsabove follows P ∩ − P = { } , which is the Condition (3).Let’s consider the remaining Conditions (2) and (4). Let x , y ∈ P be two vectorsin P . Then there exist unique decompositions x = x ′ + x a and y = y ′ + y a where x ′ , y ′ ∈ H and x, y ∈ R + . For two non-negative real numbers α, β ∈ R + thisleads to the equation α x + β y = ( α x ′ + β y ′ ) + ( αx + βy ) a . Since the values α, β, x, y ≥ αx + βy ≥ α x ′ + β y ′ ∈ H belongs to the hyperplane. Consequently, for anon-negative sum αx + βy = 0 also the sum of the vectors α x + β y ∈ P is a memberof the set P . On the other hand, the equation αx + βy = 0 induces αx = βy = 0as both values are non-negative. This means that α x ∈ H + , as well as β y ∈ H + .From the semigroup properties of the positive cone of the hyperplane H followsthat also the linear combination α x + β y ∈ H + is an element of this subspace. Sowe have shown for all vectors x , y ∈ P from the half space P and all non-negativereal numbers α, β ∈ R that the linear combination α x + β y ∈ P belongs to P , INEAR EXTENSIONES ON ABELIAN PO-GROUPS 15 too. Thus, P is a subsemigroup of ⊕ E R , so the Condition (2) holds, and – since ⊕ E R is commutative – P is also an invariant subsemigroup of the vector space,which proves condition (4). From Theorem 1 follows that P defines a linear orderon the vector space ⊕ E R considered as a group. Using the unique decomposition x = x ′ + x a for any vector x ∈ P with x ′ ∈ H we infer that α x ′ ∈ H and α x ′ + αx a ∈ P .Thus the vector space ⊕ E R is a linearly ordered vector space. (cid:3) In the opposite direction we can only limit the codimention of the set of infinitesimalelements.
Lemma 11.
Let ⊕ E R = ( ⊕ M R , + , − , , ≤ ) be an ordered vector space. Then theboundary ∂ (cid:0) ( ⊕ E R ) + (cid:1) of the positive cone is a linear subspace of ⊕ E R .Proof. Let x ∈ ∂ (cid:0) ( ⊕ E R ) + (cid:1) a boundary vector of the positive cone. Then for all r ∈ R \ { } the intersections B r ( x ) ∩ ( ⊕ E R ) + and B r ( x ) ∩ ( ⊕ E R ) − are non-empty.Furthermore, for x = either x ∈ ( ⊕ E R ) + or − x ∈ ( ⊕ E R ) + holds.In fact, x ∈ ∂ (cid:0) ( ⊕ E R ) + (cid:1) is satisfied iff − x ∈ ∂ (cid:0) ( ⊕ E R ) + (cid:1) . This follows as x ′ ∈ B r ( x ) ∩ ( ⊕ E R ) + is a member of the subset of the positive elements of a neigh-bourhood of the vector x iff − x ′ ∈ B r ( − x ) \ ( ⊕ E R ) + is located in a correspondingsubset of negative elements. An analogue result is true for ( ⊕ E R ) − . Thus, theneighbourhood B r ( − x ) contains as well positive as well as negative elements.In a next step we show that ∂ (cid:0) ( ⊕ E R ) + (cid:1) is a linear subspace in ⊕ E R . To dothat, we choose two vectors x , y ∈ ∂ (cid:0) ( ⊕ E R ) + (cid:1) . Then for arbitrary positive realnumbers r, α, β ∈ R + \ { } four vectors x ′ , x ′′ , y ′ and y ′′ exist in the following way: x ′ ∈ B r α ( x ) ∩ ( ⊕ E R ) + x ′′ ∈ B r α ( x ) \ ( ⊕ E R ) + y ′ ∈ B r β ( y ) ∩ ( ⊕ E R ) + y ′′ ∈ B r β ( y ) \ ( ⊕ E R ) + Depending on the values of α and β this leads to the following six cases: α > and β > : Here, we get the equations α x ′ + β y ′ = α x + α ( x ′ − x ) | {z } ∈ B r α ( ) ∩ ( ⊕ E R ) + | {z } ∈ B r ( ) ∩ ( ⊕ E R ) + + β y + β ( y ′ − y ) | {z } ∈ B r β ( ) ∩ ( ⊕ E R ) + | {z } ∈ B r ( ) ∩ ( ⊕ E R ) + | {z } ∈ B r ( α x + β y ) ∩ ( ⊕ E R ) + and α x ′′ + β y ′′ = α x + α ( x ′′ − x ) | {z } ∈ B r α ( ) \ ( ⊕ E R ) + | {z } ∈ B r ( ) \ ( ⊕ E R ) + + β y + β ( y ′′ − y ) | {z } ∈ B r β ( ) \ ( ⊕ E R ) + | {z } ∈ B r ( ) \ ( ⊕ E R ) + | {z } ∈ B r ( α x + β y ) \ ( ⊕ E R ) + . Thus, in B r ( α x + β y ) are positive as well as negative elements of the vectorspace ⊕ E R , so α x + β y is also an element of ∂ (cid:0) ( ⊕ E R ) + (cid:1) . α > and β < : This case an be deduced to the last case using the repres-entation α x + β y = α x + ( − β )( − y ) , since y is an element of the boundary iff − y is within the boundary, too. α < and β > : As the addition is commutative this is identical to the pre-deceasing case. α < and β < : With the same idea as in the predeceasing cases, the resultfollows from the first statement. α = 0 and β = 0 : Here, we have β y = . So analogue to the first case follows: α x ′ + β y ′ = α x + α ( x ′ − x ) | {z } ∈ B r α ( ) ∩ ( ⊕ E R ) + ∈ B r ( α x ) ∩ ( ⊕ E R ) + α x ′′ + β y ′′ = α x + α ( x ′′ − x ) | {z } ∈ B r α ( ) \ ( ⊕ E R ) + ∈ B r ( α x ) \ ( ⊕ E R ) + α = 0 : For β = 0 this case is analogue to the predeceasing case. Otherwisewe know α x + β y = ∈ ∂ (cid:0) ( ⊕ E R ) + (cid:1) .So far, we have shown that the relevant operations of the vector space, addition ofvectors, and multiplication with scalars do not exit ∂ (cid:0) ( ⊕ E R ) + (cid:1) . Thus, this set isa linear subspace of ⊕ E R . (cid:3) The next step is to determine the dimension of the boundary
Lemma 12.
Let ⊕ E R = ( ⊕ M R , + , − , , ≤ ) be an ordered vector space. Then forthe codimension of the boundary ∂ (cid:0) ( ⊕ E R ) + (cid:1) the following equation holds: (25) codim ∂ (cid:0) ( ⊕ E R ) + (cid:1) = ( , iff int (cid:0) ( ⊕ E R ) + (cid:1) = ∅ , elseProof. Let us assume that the interior of ( ⊕ E R ) + is non-empty. By Lemma 11 theboundary ∂ (cid:0) ( ⊕ E R ) + (cid:1) is a linear subspace of ⊕ E R . To be a hyperplane, this sethas to be a subspace with codimension 1. To show this, we assume that we haveat least codimension 2 and deduce a contradiction from that assumption. At thebeginning we have to find a vector, which is not in ∂ (cid:0) ( ⊕ E R ) + (cid:1) . Let x , y ∈ ( ⊕ E R ) + be two arbitrary positive elements of ⊕ E R . Then either x ≤ y or y ≤ x follows fromthe linear order. Let w.l.o.g. x ≤ y . This leads to the inequality ≤ y − x and weget for 0 ≤ λ ≤ λ x + (1 − λ ) y = λ x + (1 − λ )( y − x ) + (1 − λ ) x = x + (1 − λ )( y − x ) ≥ x , since 0 ≤ − λ holds, and the positive cone ( ⊕ E R ) + is convex in the sense ofordered groups. Consequently, ( ⊕ E R ) + is convex in the sense of vector space andwith the positive cone, the negative cone ( ⊕ E R ) − = − ( ⊕ E R ) + has this property,too. So we have the identity L (( ⊕ E R ) + ) = L (( ⊕ E R ) − ) = ⊕ E R , since ⊕ E R =( ⊕ E R ) − ∪ ( ⊕ E R ) + . As for each base vector e a with a ∈ E either e a or the vector − e a is positive, and we can replace in the basis each negative basis vector with itsinverse we can assume w.l.o.g. that { e a | a ∈ E } ⊆ ( ⊕ E R ) + . This is also true forits convex hull, which leads to conv { e a | a ∈ E } ∩ int (cid:0) ( ⊕ E R ) + (cid:1) = ∅ .Let a ∈ conv { e a | a ∈ E } ∩ int (cid:0) ( ⊕ E R ) + (cid:1) , and let us assume that the codimen-sion of ∂ (cid:0) ( ⊕ E R ) + (cid:1) is greater than 1. Then besides a another vector x ∈ ( ⊕ E R ) + exists which is linearly independent from { a } ∪ ∂ (cid:0) ( ⊕ E R ) + (cid:1) . Consequently, thereexists an isomorphism ξ , which maps the linear closure of a and x onto R suchthat ξ ( a ) = ( 1 , ξ ( x ) = ( 0 , R is compatible to the order, which is transferred from ⊕ E R to R by ξ . So we canextend the order on R linearly in such a way, that ξ is an o-isomorphism.Using this order extension, we know that ( − , − < ( 0 , < ( 1 , ) holds.Now, we will define a mapping f : R → { − , , } : λ (cid:0) , (cid:1) − λ (cid:0) , (cid:1) > ( 0 , − (cid:0) , (cid:1) − λ (cid:0) , (cid:1) < ( 0 , . Obviously, f (0) = 1 and f (1) = −
1. Thus, f is discontinuous according thenatural topology on R . Let λ be a place of discontinuity of f . Since R is linearlyordered and ( 0 , (cid:18) , (cid:19) − R (cid:18) , (cid:19) , one of the identities f ( λ ) = 1 or f ( λ ) = − ε ∈ R there ex-ists a real number λ ∈ R such that | λ − λ | < ε and f ( λ ) = − f ( λ ). This impliesthat in B ε (cid:16) ϕ − (cid:0) ( 1 , ) − λ ( , ) (cid:1)(cid:17) a positive vector y ′ ∈ ( ⊕ E R ) + and a negativevector y ′′ ( ⊕ E R ) + exist. So ϕ − (cid:0) ( 1 , ) − λ ( , ) (cid:1) ∈ ∂ (cid:0) ( ⊕ E R ) + (cid:1) \ { } , thisis a contradiction to the assumption that this set has no non-zero representativesin the linear closure of x and a . Thus, the boundary ∂ (cid:0) ( ⊕ E R ) + (cid:1) is a hyperplanein ⊕ E R with ∂ (cid:0) ( ⊕ E R ) + (cid:1) ⊕ R a = ⊕ E R .As already shown conv { e a | a ∈ E } can be considered to be a subset of thepositive cone. It also generates the vector space, so the set conv { e a | a ∈ E } ∩ int (cid:0) ( ⊕ E R ) + (cid:1) is empty iff int (cid:0) ( ⊕ E R ) + (cid:1) is empty. In that case the codimension iszero. (cid:3) Consequently the boundary of the positive cone in any ordered vector space iseither a hyperplane or it is the space itself. With this knowledge we can definesome additional notations for a vector a of an ordered vector space: V a := L{ x ∈ ⊕ E R | ≤ x ≤ | a | } and(26) H a := L{ x ∈ ⊕ E R | ≤ x ≪ | a | } (27)In general the subspace V a exists only for such vectors a that have assigneda unique absolute element | a | ≥ . For that it is sufficient that there exists alattice ordered subspace containing a . If the vector space is linearly ordered H a isa hyperplane and Equation (27) can be extended to H a = ∂ (cid:0) ( V a ) + (cid:1) After we have shown that all linearly ordered hyperplanes define linear orders, wecan prove that all linear orders on the vector space ⊕ E R have such a representation. Theorem 13.
Let ⊕ E R = ( ⊕ E R , + , − , , ≤ ) be an ordered vector space. Thepartial order ≤ is linear iff one of the following conditions is true: (1) The positive cone ( ⊕ E R ) + has the form (28) ( ⊕ E R ) + = H + ˙ ∪ (cid:0) H + ( R + \ { } ) a (cid:1) , where H is a linearly ordered hyperplane in ⊕ E R and a ∈ ⊕ E R \ H is apositive vector from the interior of the positive cone. (2) The vector space ⊕ E R is the union of an infinite chain of proper subspacesof the form given in 1. (3) E = ∅ .Proof. As 3 is trivial we have to prove that for every non-trivial vector space either1 or 2 is true. By Lemma 10 the vector space is linearly ordered if the condition 1holds. By Lemma 11 and Lemma 12 the remaining part consists of two questions:(1) Is it sufficient that a vector space is the union of an infinite chain of linearlyordered subspaces in order to get a linear order?(2) Can every linear order with empty interior of the positive cone be describedby such a chain?In order to answer the first question, suppose there are two incomparible elements a , b ∈ ⊕ E R . Then there exist linearly ordered subspaces U a and U b in the chain ofsubspaces that generates the order such that a ∈ U a and b ∈ U b . As both vectorspaces belong to the same chain either U a ≤ U b or U b ≤ U a holds. As a and b are incomparible and the vector spaces are linearly orderded we get the additionalconditons a U b and b U a . Thus both vector spaces do not belong to thesame chain in the subspace lattice, which is a contradiction to the assumption.Consequently, we can answer the first question with “Yes”.To answer the other question let a , b ∈ ⊕ E R be two vectors with a < b . Theneither a ∈ H b or H a ⊕ R a = H b ⊕ R a = H b ⊕ R b holds. Thus, the set W b := { H a ⊕ R a | a ≤ b } is a chain in the subspace lattice of ⊕ E R . Thus, each order ideal in the ordered set( W , ≤ ) with W = [ b ∈⊕ E R W b is a chain. With a similar argumentation we can prove that each order filter in thisordered set is a chain, too. Thus, ( W , ≤ ) is a chain itself. As it describes the linearorder as discussed in the previous question, also this question can be answered with“Yes”. (cid:3) Corollary 14.
Considering the hyperplane H from Lemma 10, Condition 1 and oneof the linear orders, which arise from this hyperplane, in Theorem 13 the followingequation holds: (29) H + = ( ⊕ E R ) + ∩ ∂ (cid:0) ( ⊕ E R ) + (cid:1) . Proof.
The positive cone of the hyperplane is uniquely defined. (cid:3)
The previous lemma shows that linear vector space orders on ⊕ E R can be char-acterised by hyperplanes in sufficiently chosen subspaces. Thus, each linear orderon a given abelian group can be represented by one or an infite chain of linearlyordered hyperplanes in this space. Obviously, for each linearly ordered hyperplane H two linear orders on ⊕ E R exist: A first one, where the open half-space markedby a given vector a ∈ ⊕ E R \ H is positive and a second one where it is negative.Consequently, a given basis of the vector space can be a subset of the positive coneonly in one of these two cases.As the lattice of linear subspaces is inductively ordered and since all subspaces ofa given subspace are subspaces of hyperplanes in the bigger subspace, we can applyZorn’s lemma and receive the result that each linear order on a vector space V isuniquely defined by a given linearly ordered base ( B , ≤ ) according to the followingconstruction: INEAR EXTENSIONES ON ABELIAN PO-GROUPS 19
Let a ∈ B be a base vector. Furthermore, let V a = L{ a ′ | a ′ ≤ a } and H a := L{ a ′ | a ′ < a } a linearly ordered hyperplane. Then the subspace V a can be linearlyordered by Lemma 10 such that a > is positive and V a is itself ordered as describedabove for the vectors a ′′ < a in B .Moreover, the vector spaces V a form a chain in the lattice of subspaces of ⊕ E R that fulfils the following condition: ⊕ E R = [ a ∈ B V a . Thus, ⊕ E R can be linearly ordered, since the order on the hyperplane above couldbe chosen arbitrarily. Obviously, for a given linear order the basis is not uniqueas the subspaces and their order is not influenced by the multiplication of basisvectors with scalars.We will discuss this construction not in further details, as some aspects of thistopic have been used already in different flavours in other literature e.g., the worksof The [25] and Zajceva [28].5. Linear orders on groups
The characterisations of linear orders on groups in The’s [25] and Zajceva’s [28]articles both have drawbacks. The first description needs different vector spaces todescribe all linear orders on a given group, while the inductive nature of the latterrestricts it to mainly countable dimensional vector spaces. Nevertheless, severalauthors have discussed the existence of linear orders on infinitely generated groupsby considering finitely generated subgroups (cf. [7, 18, 22]). So it is convenient toexpect also characterisations of linear orders on groups based on those which havebeen found for finitely generated groups.The approach we use here has been developed independently from the mentionedarticles. With the results of the preceding sections, we are able to describe the linearorders on an abelian group with the help of linearly ordered hyperplanes. Givenan abelian group G = ( G, + , − ,
0) and a maximal independent set E ⊆ G , eachlinearly ordered base in ⊕ E R defines an order on G , and each linear order on thevector space ⊕ E R can be reached that way. Linearly ordered bases define systemsof hyperplanes that are linearly ordered by set inclusion. Each of this systemsdefines a different order on ⊕ E R . This is assured by Lemma 10, Theorem 13, andthe monomorphisms in the Diagram 5(a). Theorem 8 and Lemma 9 assure thateach linear order on the group can be described as linear order on the real vectorspace ⊕ E R .So there is (among others) one question remaining: Which linearly ordered basesdefine the same linear order on the Group G ? We will address this question in thecurrent section.A subspace U of ⊕ E R may be multidimensional, though it contains no rationalvectors ( U ∩⊕ E Q = { } ). On the other hand we know that the linear closure of therational vectors is the complete vector space. Thus, omitting a vector from a basisleads to loss of rational vectors. This means, that every vector in a linearly orderedbase is necessary in order to define a linear order on the rational ”subspace“. Inorder to describe this we will use the following notions. The complete discussion will be made available via [23]
Definition 3.
Let ( B , ≤ ) be a linearly ordered basis of a vector space ⊕ E R . Avector x is called (rationally) active in a subset U ⊆ B iff the linear closures of U and U \ { x } differ in the rational vectors, i.e.(30) L ( U \ { x } ) ∩ ⊕ E Q = L U ∩ ⊕ E Q . Otherwise x is called (rationally) passive .An element a ∈ B is called activator of another element x ∈ B , iff both vectors a and x are active in the principal ideal ↓ ≤ a . If x is its own activator it is called self-activator . The set of activators of an element x is denoted by Act x .Furthermore let V be a subspace of ⊕ E R . Then the set Act V := L ( ⊕ E Q ∩ V )is called the active subspace of V . Corollary 15.
Let ( B , ≤ ) be a linearly ordered basis of ⊕ E R and U ⊆ B a cor-responding subset. Then the active subspace of L U is the linear hull of the activevectors from U in L U . Lemma 16.
Let ( B , ≤ ) the linearly ordered basis of the linearly ordered vectorspace ⊕ E R . Then the set of activators of an element x is an order filter in the setof self-activators of B .Proof. By definition, a vector a is an activator of x iff it is a self-activator and thecondition L ( ↓ ( B , ≤ ) a \ { x } ) ∩ ⊕ E Q = L ( ↓ ( B , ≤ ) a ) ∩ ⊕ E Q . holds. The latter condition is equivalent to the existence of a vector b ∈ V a ∩ ⊕ E Q whose projection along B into L x is non-zero. Such a vector exists, iff there exists abasis vector c such that V c = V b and c is an activator of x . Since in linearly orderedvector spaces b ∈ V a is the same as V b ⊆ V a , we know that c ≤ a . The vector b also proves that this basis vector c is also an activator of x as b H b = H c . On theother hand for each activator c ≤ a of x we can find a vector b whose projection into L x is non-zero. Thus we have proved so far: A self-activator a ∈ B is an activatorof x iff there exists a self-activator c ≤ a that is also an activator of x . This impliesthat for each activator c ∈ Act x of x each self-activator a ≥ c is also an activator of x . (cid:3) Corollary 17.
For every basis vector x and every activator y ∈ Act x the relation Act y ⊆ Act x holds. We consider two linearly ordered bases equivalent iff they generate the same order,i.e. their primary ideals generate the same subspaces ⊕ E Q . Lemma 18.
For every linearly ordered basis B there exists an equivalent basis B ′ such that for each basis vector x ∈ B ′ the condition (31) x ∈ Act ′ x ⇔ x ∈ ⊕ E Q holds.Proof. Let x ∈ B be a self-activator ( x ∈ Act ′ x ). We know by definition that L (cid:0) L ( ↓ ≤ x ) ∩ ⊕ E Q (cid:1) = L ( V x ∩ ⊕ E Q ) and L ( ↓ ≤ x \ x ) ∩ ⊕ E Q = H x ∩ ⊕ E Q hold. Bythe presumption we know ( V x \ H x ) ∩ ⊕ E Q = ∅ . Let us choose a vector f ( x ) ∈ ( V x \ H x ) ∩ ⊕ E Q for every self-activator x ∈ B . In that case we know V f ( x ) = V x .Otherwise choose f ( x ) = x . Then f is a mapping from B into the vector space ⊕ E R . And B ′ := { f ( x ) | x ∈ B } fulfils the condition if f ( x ) ≤ ′ f ( y ) : ⇔ x ≤ y . (cid:3) INEAR EXTENSIONES ON ABELIAN PO-GROUPS 21
Thus, we can develop a standard form for linearly ordered bases.
Definition 4.
A linearly ordered base B is called clarified , iff it fulfils the condition(32) x ∈ Act ′ x ⇔ x ∈ ⊕ E Q . As we can describe every linear order by a clarified base, in the ramaining part ofthis section, we will consider all bases to be clarified if it is not stated otherwise.
Lemma 19.
Let B be a base of ⊕ E R and a , b ∈ B be two different basis vectors.Then the projection of ⊕ E Q along B \ { a , b } into L ( a , b ) is 2-dimensional.Proof. Since both vectors a and b are linear combinations of the standard basis,the projection of the standard base vectors contains a nontrivial vector z . Let P bethe projection along B \ { a , b } . Then from z ∈ P [ ⊕ E Q ] follows the set inclusion L{ z } ⊆ L ( P [ ⊕ E Q ]) ⊆ L{ a , b } .Suppose L{ z } = L ( P [ ⊕ E Q ]). Then we can express every rational vector as alinear combination of z and B \ { a , b } . This leads to the set inclusion ⊕ E Q ⊆ L (cid:0) ( B \ { a , b } ) ∪ { z } (cid:1) ⊆ L B . As the linear hull is a closure operator, we get immediately the set inclusion L ( ⊕ E Q ) ⊆ L (cid:0) ( B \ { a , b } ) ∪ { z } (cid:1) ⊆ L B = L ( ⊕ E Q ) . As z is linear dependend from the vectors a and b , at least one of the equa-tions L{ a , b } = L{ a , z } and L{ a , b } = L{ z , b } holds. W.l.o.g. we can assume L{ a , b } = L{ a , z } . Then we can exchange b in B with z , too. We get L ( ⊕ E Q ) = L ( ⊕ E R ) = L B = L (cid:0) ( B \ { b } ) ∪ { z } (cid:1) As we have seen above, from this follows that a is linear dependent from the set( B \ { a , b } ) ∪ { z } . Thus, we can express a as a linear combination of the vectorsfrom ( B \ { a , b } ) ∪ { z } . As B is a basis this representation has a nonzero z component and can express z as a linear combination of the vectors from B \ { b } .Consequently we get the following equations L (cid:0) ( B \ { a , b } ) ∪ { z } (cid:1) = L ( B \ { b } ) = L ( ⊕ E Q ) = L ( B ) . This shows that either B is not a basis, that a = b , or that the assumption L{ z } = L ( P [ ⊕ E Q ]) is wrong. (cid:3) Using this lemma, we can find a first condition, which explains when two differentorders on ⊕ E R imply the same order on ⊕ E Q . Lemma 20.
Let the vector space ⊕ E R be orderded twice with the linear orderrelations ≤ and ≤ ′ . Furthermore let B a Basis that describes both ≤ and ≤ ′ , and Act respectively
Act ′ the corresponding operators that provide the set of activatorsof ≤ respectively ≤ ′ . Then the induced orders on the rational vector space ⊕ E Q coincide iff for evey element x ∈ B the sets of Activators Act x and Act ′ x are equal.Proof. We have to consider four cases:(1) The two bases are equally ordered. In that case the orders also share thesame sets of activators.(2) There exist two self-activators x and y with x ≤ y and y ≤ ′ x .(3) The sets of self-activators are different and the orders are the same on thoseelements which are self-activators according to both orders. (4) The sets of self-activators are the same and the orders differ only on non-self-activators.Let us consider point 2. Then we can choose two vectors a ∈ ( V x \ H x ) ∩ ⊕ E Q b ∈ ( V ′ y \ H ′ y ) ∩ ⊕ E Q such that the component of a in x direction and the component of b in y directionare non-zero. Then a < b < ′ a holds.In case 3 a vector x ∈ B exists such that x ∈ Act x and x Act ′ x hold. Thenwe choose a rational vector a ∈ ( V x \ H x ) ∩ ⊕ E Q such that the component in x direction is different from zero. Let us choose y ∈ B such that V ′ y = V ′ a . As a hasa non-zero y component, according to B , we have the inequality V y ⊆ V x leadingto Act x ⊆ Act y . As it has a non-zero x component, too, we get Act ′ y ⊆ Act ′ x . ByLemma 19 we can find a vector b ∈ V ′ y ∩ ⊕ E Q whose x component is larger thanthe one of a and whose y component is smaller than the one of a . Then a < b < ′ a proves the assertion.For the remaining part 4 we prove that two inactive basis vectors x , y ∈ B donot influence the result. Suppose the relations x Act ′ x , x Act x , and x < y < ′ x .Suppose there are two vectors a , b ∈ ⊕ E V with a < b . As all vectors have finitesupport with respect to the standard base, their support with respect to B is finite,too. Thus, there exists a largest basis vector z ∈ B such that the z component of a is smaller than the one of b . Then both vectors a , b ∈ V z belong to its vectorspace, while at least one of them is not in the corresponding hyperplane H z leadingto z ∈ Act z . Thus, x = z . Suppose x < z < ′ x and a has a non-zero x component.Then there would be a self-activator z ′ ∈ Act ′ z ′ ∩ Act ′ x such that a has a non-zero z ′ component. As a ∈ V z this implies z ′ < z < ′ z ′ which has been excluded by thepresumption. Consequently, either a has a zero x component or x < z ⇔ x < ′ z holds. The same is true for any other non-self-activator. As the orders on theself-activators coincide between the relations ≤ and ≤ ′ , the z component remainsthe main component to determine the order of a and b leading to a ≤ ′ b . (cid:3) In Theorem 13 we have shown that in ⊕ E R each linearly ordered hyperplane definesa different linear order. On the other hand it is not obvious, which linearly orderedhyperplanes define different linear orders on ⊕ E Q . Lemma 21.
Let ≤ and ≤ ′ be two linear orders on ⊕ E R and a ∈ ⊕ E Q a rationalvector. Furthermore we denote the vector spaces from Equation (26) with V a and V ′ a and the corresponding linearly ordered hyperplanes according to Equation (27)with H a and H ′ a . If the active subspaces Act V a and Act V ′ a are not equal or therestrictions of the hyperplanes H a ∩ Act V a and H ′ a ∩ Act V ′ a to the active subspaces V a and V ′ a differ, then the induced orders w.r.t. Lemma 2 differ on ⊕ E Q .Proof. Suppose at first
Act V a = Act V ′ a . According to the presumption the set U := ( Act V a ) + \ V ′ a + = ( Act V a ) + ∩ V ′ a − \ H ′ a + = ( Act V a ) + ∩ ( V ′ a ) − \ { } is non-empty. Let B the linearly ordered base that describes the order relation ≤ and B ′ the one of ≤ ′ . As the hyperplanes H a and H ′ a differ, there exists a basevector b ∈ B ∩ H a ∩ Act V a \ H ′ a and a base vector b ′ ∈ B ′ ∩ H ′ a ∩ Act V ′ a \ H a . Bothvectors are linear independent. As b ′ H a , for the base vector y ∈ B with V y = V a the set ˆ B = B \ { y } ∪ { b ′ } is another basis that describes ≤ . As a ∈ ⊕ E Q the INEAR EXTENSIONES ON ABELIAN PO-GROUPS 23 vector b ′ is active in V a , by Lemma 19 the projection of ⊕ E Q ∩ V a along ˆ B into L{ b , b ′ } is two-dimensional. Thus it contains for all 4 combinations of positiveor negative (non-zero) scalars with respect to b and b ′ at least one element. Thisimplies that at least one element exists that is positive with respect to both orders,one that is positive with respect to ≤ and negative with respect to ≤ ′ , one that isnegative with respect to both orders and at least one that is negative with respectto ≤ and positive with respect to ≤ ′ . The latter one is sufficient to prove that theorders differ.The remaining part of the proof considers Act V a = Act V ′ a . W.l.o.g. consideringthe definition of Act there exists a vector b ∈ Act V a ∩ ⊕ E Q \ Act V ′ a . Furthermorewe can choose an integer n ∈ Z such that b < a n and as b V ′ a we get the otherinequality a ≪ ′ b . (cid:3) After having shown that different hyperplanes imply different orders, the next ques-tion we have to address, considers the effect of having different orders on ⊕ E R ,which share a common dividing hyperplane. Lemma 22.
Let ≤ and ≤ ′ be two differing linear orders on ⊕ E R , which have thesame hyperplane H a = H ′ a for some rational vector a ∈ ⊕ E Q . If the vector a ispositive w.r.t. each of the orders ( ≤ a and ≤ ′ a ), then the induced orders on V a ∩ ⊕ E Q are identical iff the induced orders on H a ∩ ⊕ E Q and H ′ a ∩ ⊕ E Q areequal.Proof. According to Lemma 2 all elements of ⊕ E Q \ H and all elements of H ∩ ⊕ E Q are either positive with respect to both orders on ⊕ E Q or negative. Thus, the orderon ⊕ E R \ ⊕ E Q does not influence the order on ⊕ E Q . (cid:3) Lemma 23.
Let ≤ and ≤ ′ be two linear orders on ⊕ E R . They are equal on ⊕ E Z iff the coincide on ⊕ E Q .Proof. As ⊕ E Z ⊆ ⊕ E Q the orders are equal on ⊕ E Z if they are equal on ⊕ E Q .For the other direction let a ∈ ⊕ E Q such that a < < ′ a , i.e. it is positive withrespect to one and negative with respect to the other order. Then it has finitelymany rational coordinates with respect to the standard base of ⊕ E Q . Let a ∈ N be the least common denominator of the coordinates of a . Then a a ∈ ⊕ E Z is aninteger vector. Thus we get the inequality a a < < ′ a a . (cid:3) With Corollary 23 and the monomorphisms ϕ and ψ as defined in Theorems 4 and6 we also have a characterisation of linear orders on an arbitrary abelian group: Theorem 24 (Characterisation of linear orders on abelian groups) . Let G = ( G, + , − , be an abelian group, E ⊆ G a maximal independent set in G , and ϕ the canonical embedding of G into ⊕ E R as described in Lemma 4. Letfurther U ⊆ ⊕ E Q be a set that generates ⊕ E R in the following way: (33) ⊕ E R = [ a ∈ U V a . Two linear orders ≤ and ≤ ′ on G are different iff some vector a ∈ U fulfils one ofthe following conditions: (1) H a ∩ Act V a = H ′ a ∩ Act V ′ a (2) Act V a = Act V ′ a (3) ( H a ) + ∩ ⊕ E Q = ( H ′ a ) + ∩ ⊕ E Q . Proof.
As shown in the first part of this article Diagram 5(a) commutes and withCorollary 23 each linear order on ⊕ E Q uniquely defines an order on the group G .Lemma 9 finally assures that there are no more linear orders on G , since each linearorder is semiclosed.Lemma 2 tells us that we can transfer the order from ⊕ E R onto ⊕ E Q and ⊕ E Z .The Theorems 4 and 6 assure that these orders can also be considered as orderson the group G . Let a ∈ U one of the selected vectors. As V a = H a ⊕ R a andsimilar for ≤ ′ , Theorem 13 describes the positive cone of V a using a hyperplane.Lemma 21 shows that Conditions 1 and 2 indicate a difference between the orders ≤ and ≤ ′ . If these conditions do not hold, Lemma 22 proves Condition 3. Thusthe restrictions of the orders ≤ and ≤ ′ to V a ∪ V ′ a are equal iff none of these threeconditions holds.From ⊕ E R = S a ∈ U V a = Act ⊕ E R also the restricted equation ⊕ E Q = S a ∈ U V a ∩⊕ E Q follows as well as the same for the other order relation ≤ . Thus, the orders ≤ and ≤ ′ are equal on ⊕ E Q iff their restrictions to Act V a are equal for each vektor a ∈ U . (cid:3) Corollary 25.
Linear orders on groups can be completely described by linearlyordered hyperplanes and their intersections in certain subspaces of a real vectorspace ⊕ E R . Theorem 24 provides a characterisation of linearly ordered groups independent ofthe ideas described in The’s [25] and Zajceva’s [28] papers. Teh describes linearorders by embedding them into vector spaces of the form ⊕ E R where the cardin-ality | E | is smaller or equal to that of the continuum c . For the cardinality ofthese subsets he defines a prototype of an order and considers all other orderingsto be homomorphic images of the prototypic vector spaces. The dimensions ofthese spaces correspond to the archimedian rank of the generated order in G . Ourcharacterisation does not touch the notion of archimedian orders. The descriptionusing linearly ordered bases provides additional details about in the case of infinitearchimedian rank. The work of Zajceva describes archimedian orders by certainequations. It is not evident that finite linear equations can be used in general forinfinitely generated groups. Nevertheless, these equations can be used in severalcases since they are descriptions of the hyperplanes, which we used in Theorem 24.So far, we didn’t use the properties of ⊕ E R as a scalar product space. This will benecessary for discussing the plane equations.Beforehand we head over to linear order extensions, we can shortly bridge toZajceva’s results. Two elements g, h ∈ G of an abelian linearly ordered group G = ( G, + , − , , ≤ ) with g ≤ h are considered archimedian equivalent iff thereexist two integers m, n ∈ Z such that h m ≤ g ≤ h ≤ g n holds. The group G is archimedian iff all of its elements are archimedian equivalent to each other. Inother words: For each two elements g, h ∈ G with 0 ≤ h − g there exists a number n ∈ Z such that h − ng ≤
0. When we describe the linear order on G by ahyperplane as in Theorem 24, this is equivalent to: for all vectors x , y ∈ ϕ [ G ]with y − x ∈ H + ∪ H + (cid:0) R + \ { } (cid:1) a there exists a number n ∈ Z such that y − n x ∈ H − ∪ H + (cid:0) R − \ { } (cid:1) a . If the intersection H ∩ ϕ [ G ] of the dividinghyperplane of the order and the image of the group is empty, then the vector x is linearly independent from H and, thus, the linear closure L ( H ∪ { x } ) = ⊕ E R is the whole vector space. Thus, there exists an integer n such that y − n x is an INEAR EXTENSIONES ON ABELIAN PO-GROUPS 25 P a a a H + H − (a) valid P a a a (b) forbidden Figure 7.
Positive cone ( P ) and hyperplane H according to thecharacterisation in the Theorems 26 and 28element of the negative half space. On the other hand if x ∈ H is an element ofthe hyperplane and y H is not, then the straight line y + R x is parallel to H anthus y − n x will never be negative. Thus, the group is archimedian iff there is noelement g ∈ G such that ϕ ( g ) ∈ H . According to Corollary 23 this is equivalentto the condition that no vector in H has only rational coordinates with respect tothe standard base defined by the independent set E as described in the precedingsections. This implies that the quotient of any two coordinates of the normal vectorof H must be irrational if it exists. Otherwise we can scale the normal vector bya real number such that it has two rational coordinates. On the other hand if allsuch quotients are irrational, all vectors containing only rational coordinates arenever orthogonal to the normal vector.In [28] Zajceva discusses linear orders on abelian groups with a maximal in-dependent set E by assigning a set of rationally independent numbers E ′ with abijective mapping α : E → E ′ : e α ( e ) to them. So each element g ∈ G with arational representation g = P a ∈ E ξ a ( g ) a ( ξ a : G → Q ) is positive iff the real num-ber P a ∈ E ξ a ( g ) α ( a ) is positive. These equations and inequalities can be consideredas the scalar product of the normal vector to the dividing hyperplane H in the caseof abelian groups with a finite maximal independent set. However if the maximalindependent set E is infinite, the normal vector must have infinite support and thusis not an element of the vector space. In that case using our approach to find sucha representation is less suitable than Zajceva’s method. It is again the hierarchy ofsubspaces generated by the order, that extends this description.6. Linear order extensions
Before we discuss linear order extensions on arbitrary partially ordered torsionfree abelian groups, we start with the special case of ordered vector spaces. So farwe have discussed linearly ordered hyperplanes as representations of linear orderson vector spaces. Now we can relate them to other orders. Figure 7 illustrates ofthe following theorem.
Lemma 26.
Let ( R , ≤ ) be an ordered real vector space, which has the positivecone R + with non-empty interior. A linearly ordered hyperplane H defines a linear order extension ≤ ′ of ≤ iff for its positive cone H + the following condition is true: (34) R + ∩ H ⊆ H + ∩ ∂ R + Proof.
Let ≤ ′ be a linear order on the vector space R defined by a hyperplane H ,which is an order extension of the order ≤ . Moreover, ≤ ′ is also a linear order on H . Then for the positive cone R ′ + := { x ∈ R | ≤ ′ x } the following relation issatisfied:(35) R + ⊆ R ′ + Suppose there exists an element x ∈ int( R + ) ∩ H . Hence, neighbourhood U ( x )exists such that U ( x ) ⊆ R + . On the other hand with H = ∂ R ′ + the inequality U ( x ) \ R ′ + = ∅ holds. So we deduce (cid:0) U ( x ) ∩ R + (cid:1) \ R ′ + = ∅ , which implies R + \ R ′ + = ∅ ( ). This is a contradiction to (35). Consequently, the intersection int R + ∩ H = ∅ is empty. Hence, R + ∩ H ⊆ ∂ R + , which implies R + ∩ H ⊆ ∂ R + ∩ H .Let now x ∈ R + ∩ H − \ { } . In this case the vector x would be positive withrespect to the given order ≤ and with respect to the order ≤ ′ that is generated by thehyperplane H it would be negative. Hence, the orders would not be compatible( ).Since H is linearly ordered, the remaining case is x ∈ H + and thus, R + ∩ H ⊆ H + . So far we have shown that each linear order extension of ≤ fulfils condition (34).If the hyperplane H fulfils the Condition (34), on one side of the hyperplanethere are no elements of R + . In the contrary case two vectors x ∈ int R + and y ∈ H and a positive real number α ∈ R + \ { } exist such that x is on one side of H and y ′ := y − α ( x − y ) is on the other side of H . Then y ′ − x = (1 + α )( y − x ). Let U bean open environment of x . Then the set U ′ := { z ′ ∈ ⊕ E R | z ′ = z + 11 + α ( z − y ′ ) , z ∈ U } is neighbourhood of y and as R + is convex U ′ ⊆ R + is one of its subsets.Thus, we have y ∈ int R + ∩ H ( ) in contradiction to (34). Thus only on oneside of H can be positive elements with respect to the order ≤ .If on both sides of H there are no elements of R + i.e., R + ⊆ H , we can freelychoose the side of the positive cone of ≤ ′ . In this case we choose an arbitrary vector a ∈ R \ H and receive in dependence from a one of the two possible orders.If the set R + \ H is non-empty, we choose a vector a ∈ R + \ H . In this case P := H + ∪ (cid:0) H + ( R + \ { } ) a (cid:1) is the set of positive elements of a linearly orderedgroup. The independence from the chosen representative a has been shown for R + \ H = ∅ in the proof of Theorem 13. (cid:3) As for each vector a ∈ ⊕ E R the subspace V ′ a has a hyperplane that divides itspositive cone according to the order relation ≤ ′ from the negative one, we canimmediately conclude the following corollary: Corollary 27.
Let U ⊆ ⊕ E Q be a set that generates ⊕ E R in the following way: (36) ⊕ E R = [ a ∈ U V a . Then ≤ ′ is a linear extension of ≤ iff for each element a ∈ U the following relationholds: (37) R + ∩ H ′ a ⊆ ( H ′ a ) + ∩ ∂ R +INEAR EXTENSIONES ON ABELIAN PO-GROUPS 27 Figure 8.
Arrangement ofdividinghyperplane andpositive cone P a a a H + H − The main theorem of this article is illustrated in Figure 8. It describes the set oflinear order extensions of a given order on an abelian partially ordered group:
Theorem 28 (Characterisation of linear order extensions on abelian po-groups) . Let G = ( G, + , − , , ≤ ) a po-group, E a maximal independent set in G , ⊕ E R thecorresponding vector space with the embeddings ϕ : G → ⊕ E R and ψ : ⊕ E Z → G as described in (18) and (19) . Then each linear order extension of ≤ can becharacterised by a set U ⊆ ⊕ E Q and for each element a ∈ U a linearly orderedhyperplane H a which fulfils the condition (38) conv( ϕ [ G + ]) ∩ H a ⊆ ( H a ) + ∩ ∂ (cid:16) conv (cid:0) ϕ [ G + ] (cid:1)(cid:17) . For each Element a ∈ U the corresponding hyperplane H a defines a unique linearorder extension in the subspace V a if V a ∩ ϕ [ G + ] \ H a is non-empty, otherwise itcan be used to define exactly 2 linear order extensions in this subspace.Proof. The existence of the monomorphisms has bee proved in the Lemmata 4 and6. With Theorem 8 and Lemma 9 we get the description of the positive cone G + asconv( ϕ [ G + ]) in the vector space ⊕ E R . For non-semiclosed orders this constructionadds only these elements to the positive cone of G , which are positive in every linearorder extension on G , as Lemma 7 shows. Thus this restriction can be loosened.Theorem 24 gives us a description of linear orders on G in the vector space ⊕ E R .Corollary 27 provides the Condition (38) for the characterisations of the linearorders on ⊕ E R as well as the number of orders defined by the hyperplane. Thusall parts of this theorem have been proved elsewhere in this article. (cid:3) With Theorem 28 we have described a geometrical characterisation of linear orderextensions of partially ordered torsion free abelian groups. This allows us, to usemethods from the convex geometry for the construction and investigation of suchlinear order extensions. The following section illustrates this.7.
Further results
As an illustration of the method provided above, in this section some other proofsand additional conclusions are added.So far we have only described existing order extensions. The question has notbeen touched whether such an order extension or hyperplane does exist. The fol-lowing theorem fills this gap. It has been proofed several times (cf. [6, 17, 32, 33]).We add an additional proof:
Theorem 29.
Each partially ordered torsion free group G has a compatible linearorder extension. Proof.
Consider the complete sublattice of the subspace lattice of V that is gen-erated by the set W := { V a | a ∈ V } . This sublattice has a maximal chain K .As each vector a ∈ V is contained in an element of K , the union S K = V is thecomplete vector space V . We define S a := [ { W ∈ K | a W } . As the subspace W a := V a + S a is the join of V a and S a in the subspace lattice, W a is the upper neighbour of S a in K . Then there exists a maximal linear independentset B ⊆ W a that contains a and is liner independent from S a . Then for each vector b ∈ B the vector space V b is the same as V a (otherwise b ∈ S a ).For each vector c ∈ V a ∩ V + there exists a real number c ∈ R such that c c ≤ a which implies ≤ a − c c . Thus there exists a basis B ′ a containing a and a basis of S a such that for all Elements x ∈ V − ∩ V a the distance k a − x k ∞ , B ≥ d ∈ R + \ { } is strictly positive (choose all base vectors to be positive between a and ) and suchthat a is the largest element in the basis B ′ a . Then conv (cid:0) B d , ∞ , B ′ ( a ) ∪ W a (cid:1) hasnon-empty interior and does not hit V − . Then there exists a hyperplane H a suchthat S a ⊆ H a and H a does not hit W a ∩ V + \ S a .Let us chose a base ˆ B a such that it contains a base of S a together with a withthe additional condition that ˆ B a \ ( { a } ∪ S a ) ⊆ H a . Then we define a linear orderrelation ≤ ′ on ˆ B a that has a as largest element and is defined in a way such that forany two elements x ∈ ˆ B a ∩ H \ S a and y ∈ S a the inequality y ≤ ′ x holds. Then ≤ ′ defines a linear order on W a such that the positive cone of W a with respect to theoriginal order ≤ is a subset of the positive cone that is defined by ≤ ′ . Furthermorewe can use ˆ B a \ S a to extend any linearly ordered base of S a to a linearly orderedbase of W a that fulfils the same condition.As the vector a has been chosen arbitrarily, we can find such a subbase ˆ B a forevery subspace W a ∈ K , and the union ˆ B of all these subbases is a basis of theoriginal vector space V . The order on the chain K defines a linear preorder on ˆ B ,which can be turned into a linear order by using the orders of the subbases ˆ B a as developed above. Thus we can define a linear order whose positive cone is asuperset of the one of the original order ≤ , which means that it is a linear orderextension.As every torsion free ordered abelian group G has a closed order extension, wecan use this construction to construct a linear order extension on G . (cid:3) Looking at the proof and the preceding sections it is obvious that every linearorder extension can be found with this construction: Every linear order can beexpressed by a maximal chain in the subspace lattice, and it is an order extensioniff it provides an extension of the positive cone.In general, an order relation on a given set is the intersection of its linear orderextensions (cf. [24]). If the order must be compatible to a given algebraic structure,the set of choices for the linear extensions is much more restricted. For abeliangroups the following theorem has been proved a long time ago (cf. [7], Corollary 6).Also in this case our method provides an alternative proof.
Theorem 30.
Let G = ( G, + , − , , ≤ ) be a partially ordered torsion free abeliangroup. Then the order relation ≤ is the intersection of all of its linear order exten-sions iff it is semiclosed. INEAR EXTENSIONES ON ABELIAN PO-GROUPS 29
Figure 9.
Hyperplane and theimage of a partiallyordered group. Thehyperplane is arrangedin a way that it does notintersect with theimages of the group ele-ments. a a a Proof.
One direction is obvious: Each lattice ordered group is semiclosed, thus alsoeach linearly ordered group. So, the positive cone of an arbitrary order on a givengroup G contains for each n ∈ N \ { } together with na also the element a and,thus, the intersection is also semiclosed (cf. Lemma 7).Considering the other direction, let ≤ be semiclosed. Furthermore, let O the setof all linear order extensions of ≤ which are compatible with G and E a maximalindependent set in G . Then for each monomorphism ϕ : G → ⊕ E R with a e a for all a ∈ E the set P := conv (cid:2) ϕ [ G + ] (cid:3) is the unique representation of the positivecone in ⊕ E R according to Theorem 8 and Lemma 9. Thus it is sufficient to showthat P is the intersection of the corresponding positive half spaces of elements of O . Let P ′ the intersection of all of these half spaces of the linear order extensions.Then P ′ is non-empty because of Theorem 29. Furthermore, P ⊆ P ′ . Now, we willshow the inclusion ⊕ E R \ P ⊆ ⊕ E R \ P ′ .Let x ∈ ⊕ E R \ P . We infer from the convexity of P the equality R x ∩ P = { } .Let y ∈ conv( R + x ∪− P ) ∩ conv( R − x ∪ P ) = L + ( R + x ∪− P ) ∩L + ( R − x ∪ P ). The twosets are equal as both are convex and ∈ R + x ∩ P . Then there exist α, β ∈ R + andvectors p , q ∈ P such that α x − p = y = − β x + q . Consequently, ( α + β ) x = p + q ∈ P ,which implies with the properties of being a cone for α = 0 or β = 0 that also x is in P ( ). Consequently, α = β = 0 and it follows y = p = q = . This impliesconv( R − x ∪ P ) ∩ conv( R + x ∪ − P ) = { } .As proven in Theorem 29, a linear order extension of ≤ exists, for which x isnegative. Consequently, x P ′ . Thus, ⊕ E R \ P ⊆ ⊕ E R \ P ′ follows and so P ′ ⊆ P , from which we get P = P ′ . This means that a semiclosed order ≤ is theintersection of all of its order extensions. (cid:3) H¨older’s theorem has not been used so far, but can be proved easily, now. Themain idea will be formulated as a separate lemma: Lemma 31.
A linearly ordered base of ⊕ E R defines an archimedian order on ⊕ E Q iff it fulfils the following conditions: (1) There exists a largest basis vector b . (2) The largest vector b is the only self-activator in B . In fact, H¨older proved the theorem not for archimedian orders, but for such ordered semigroupsthat provide a Dedekind cut. For a link to the current theorem cf. [3], XIII.12.
Proof.
At first we show that an archimedian order has only one self-activator. Sup-pose there are two self-activators a and b . W.l.o.g. we can assume a , b ∈ ⊕ E Q . Asthe basis is linearly ordered either a < b or b < a is true. Let’s assume the first oneholds. This implies a ∈ H b , which means a is infinitesimal smaller than b ( ). Thusthere is at most one self-activator in B . As ⊕ E Q ⊆ L B holds there is at least oneself-activator in B .The projection along B into the subspace that is generated by two basis vectorsis 2-dimensional as proofed in Lemma 19. This implies that every non-self-activatorhas a self-activator above it. Thus the unique self-activator of an archimedian linearorder must be the maximal element.On the other hand if these two conditions are fulfilled, we can choose the maximalelement b . Then the corresponding hyperplane H b is disjoint from ⊕ E Q . Thus theorder is completely defined by the projection along B into R b . Suppose there thereare two vectors a and a in ⊕ E Q with the same projection into R b . Then thedifference a − a ∈ H b ∩ ⊕ E Q is both infinitesimal smaller than b and rational( ). (cid:3) Geometrically, this implies that every archimedian linear order can be representedby a linearly ordered hyperplane in the complete vector space ⊕ E R that does notintersect with the rational vectors from ⊕ E Q as shown in Figure 9.It has been proved many times that an archimedian o -group is abelian. Thisshall not be repeated, here. Thus, we prove the following theorem only for abeliangroups. Theorem 32 ([3], XIII.12; H¨older, [12]; ) . Every archimedian linearly orderedabelian group is isomorphic to a subgroup of the real numbers.Proof.
Let E be a maximal independent set in the group and B a linearly orderedbase of ⊕ E R . Then, by the preceding lemma the order on the group is defined bythe projection along B into the subspace R b of the maximal basis vector. (cid:3) Corollary 33.
The abelian group ZR with respect to elementwise addition andsubtraction does not permit an archimedian linear order.Proof. For the cardinalities we get the inequality | ZR | ≥ | R | = | P ( R ) | > | R | . Thus, there is no isomorphism from ZR into a subset of the real numbers. (cid:3) The following theorem touches a question that has been published in [4] as Prob-lem 1.7. As it is weaker, it can be considered only as a step into the right direction.
Theorem 34.
Every torsion free abelian group permits an archimedian directedorder.Proof.
Let G be an abelian group. And E a maximal independent set. Then G canbe embedded into ⊕ E Q . Let ϕ : G → ⊕ E Q be the corresponding embedding. Theproduct order ( a ≤ b ⇔ ∀ e ∈ E : a ( e ) ≤ b ( e )) on ⊕ E Q is an archimedian order: Forany two vectors a and b the set supp a ∪ supp b is finite. If both supp a \ supp b = ∅ and supp b \ supp a = ∅ are non-empty either a k b or a k − b holds, thus the twovectors fulfil the archimedian property.For the remaining case we may assume supp b ⊆ supp a . Then there exist anelement e ∈ E and integers a, b ∈ Z such that a a ( e ) = b b ( e ) = 0. If a ( a ) is strictly INEAR EXTENSIONES ON ABELIAN PO-GROUPS 31 positive, then a can be chosen strictly positive, too. This leads to ( a + 1) a ( a ) >b b ( e ) > b ( e ) and ( a + 1) a < b on the one hand, and 2 b b ( e ) = 2 a a ( e ) > a ( e ) and2 b b < a on the other hand, the archimedian property. For a negative coordinate a ( e )and a negative scalar a the inequalities read ( a − a ( e ) > b b ( e ) > b ( e ), ( a − a < b ,2 b b ( e ) = 2 a a ( e ) > a ( e ) and 2 b b < a . As ϕ is an embedding, this proves also thatthe given order is also archimedian on G .Finally the order is a directed order. Let a and b the vectors that correspond totwo arbitrary group elements. Then we define a mapping c : E → Z such that c ( e )is the least integer that fulfils the conditions a ( e ) ≤ c ( e ) and b ( e ) ≤ c ( e ) . With the relation supp c ⊆ supp a ∪ supp b the vector c is a well-defined element of ⊕ E Z and thus it represents an element of G such that a ≤ c and b ≤ c . (cid:3) Conclusion and further topics
In Theorem 24 we have provided a method to describe linearly ordered abeliangroups by means of linearly ordered hyperplanes in a the vector space, arising fromthe set of mappings with finite support from a maximal independent set into thereal numbers. The vector space has been defined as the direct sum ⊕ E R , where E denotes a maximal independent set in the given abelian torsion free group G . Thisdescription enables us to investigate all linear orders on G in one common vectorspace. Thus, we can use it to describe all linear order extensions of a partiallyordered abelian group. We concentrated on torsion free groups, as only those canbe ordered linearly. The description used for linear orders can also be used forsemiclosed partially ordered groups, which have been shown to be compatible withthe attempt to catalogue all compatible linear order extensions of a given partiallyordered group. Finally such a characterisation has been given in Theorem 28.Some additional examples provided an insight, how this method can be applied tomathematical problems arising from the work with partially ordered abelian groups.Though abelian groups have interesting applications, a general description forarbitrary groups would be interesting. The groups considered here are torsionfree abelian groups and have been embedded into some vector space ⊕ E Q whichis always possible as the latter is an injective group. Another idea would be toconsider those groups, which can be embedded into vector spaces over skew fields.Linear order extensions play an important role in the modelling of tone systems(cf. [21]) and ordered generalised interval systems [16] in mathematical music the-ory. In this topic it is necessary to describe factorisations of ℓ -groups by non-convexsubgroups based on linear order extensions. This leads to further interesting ques-tions in the direction of cylindrically and cyclically ordered groups. This theoryhas its application in software development e. g., for understanding just intonationlogics based on the Tonnetz as described by Martin Vogel [26] and provided by Mutabor ([20]). 9.
Acknowledgements
Special thanks to Charles W. Holland for providing his illustration of Zajceva’stheorem and some further references. This has been the initial inspiration for thecurrent work.
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Department of Mathematics, Technische Universit¨at Dresden
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