Orthogonality in Partial Abelian Monoids and Applications

Pedro Morales and Francisco Garc´ıa Mazar´ıo

Abstract. In this paper we introduce the notion of orthogonality for arbitrary families of elements of a partial abelian monoid, we consider the special case of an eﬀect algebra and we study, as an application, the existence of the support of an uniform semigroup valued measure on an eﬀect algebra.

1. Introduction

The basic algebraic structure considered in our paper is a partial abelian monoid, which is a partial universal algebra endowed with a nullary operation and an associative and commutative binary partial operation. It is worth noticing that these structures are studied from the algebraic standpoint in the papers [10], [24], [39] and [40] and more extensively in the monography [38]. Our interest in a partial abelian monoid L is due to the fact that, for arbitrary families of elements of L, it can be defined an orthogonality structure with nice properties and allowing to say rigorously when L is X-complete for every infinite cardinal number X. We consider, in particular, the orthogonality when L is an effect algebra, we extend easily to L the notion of difference set introduced for orthoalgebras in [14] and [32] and we generalize its main properties. Since L carry also orthogonally additive measures and provide new results in the rapidly devel- oping field of non-commutative measure theory (see [9] and [18]), we consider, as an application, the existence of the support of a positive measure on L with values in an ordered Hausdorff uniform semigroup. The paper is organized as follows. In Section 2 we give an account of the basic theory of partial abelian monoids with several non-trivial examples and a brief presentation of some of its sub- structures: effect algebras, orthoalgebras, orthomodular posets, orthomodular lattices and Boolean algebras. Section 3 studies orthogonal suites in a partial abelian monoid using some ideas of Bourbaki [5] and we deduce several properties, some of them appearing for an special case in [12], and another, like the generalized associative law and the generalized commutative law, appearing in [5] for a commutative semigroup. In Section 4 we present the orthogonality for infinite families of elements of a partial abelian monoid L whose algebraic preordering is antisymmetric, we establish its main properties and we discuss also the X-completeness of L for

2000 Mathematics Subject Classification: 08A55, 06A06, 28B10, 28B15. Key words and phrases: Partial abelian monoid, effect algebra, orthogonal suite, ⊕-join, orthogonal sequence, infinite orthogonal family, X-complete, measure, uniform semigroup, support.

1 2 PEDRO MORALES AND FRANCISCO GARCÍA MAZARÍO any infinite cardinal number X. Section 5 concerns the orthogonality in an effect algebra and, since the compatibility is very difficult to verify in an effect algebra, we use instead an extension of the notion of difference set defined in [14] and [32] for an orthoalgebra, we generalize its main properties and we deduce a relation when the effect algebra is X-complete that allow us to generalize the important Theorem 4.14 of [21]. Finally, in Section 6, after presenting a basic theory of uniform semigroup and some pertinent classes of measures, we define following [33] the notion of support of a positive measure on an effect algebra with values in an ordered Hausdorff uniform semigroup and we establish a twofold generalization of the main results of [29] for an ordered metrizable topological group in the first statement.

2. Preliminaries

For the terminology, notation and details concerning orderings, lattices, universal algebras and abelian groups we refer to [3], [8], [17], [19], [20] and [34]. Let L be a set containing at least one element 0 called zero of L, let ⊥ be a symmetric binary relation on L and let (a, b) → a ⊕ b be a partial binary operation on L of domain ⊥. Then the partial universal algebra hL; ⊕, 0i is called a partial abelian monoid if the following properties hold: (Commutative Law) If a, b ∈ L and a ⊥ b, then a ⊕ b = b ⊕ a. (Associative Law) If a, b, c ∈ L, a ⊥ b and (a⊕b) ⊥ c, then b ⊥ c, a ⊥ (b⊕c) and (a ⊕ b) ⊕ c = a ⊕ (b ⊕ c). (Identity Law) If a ∈ L, then a ⊥ 0 and a ⊕ 0 = a. We note that the identity law implies that ⊥ is not empty. If a, b ∈ L and a ⊥ b, we say that a and b are orthogonal. If the hypotheses of the associative law are veriﬁed, we write a ⊕ b ⊕ c for the element (a ⊕ b) ⊕ c = a ⊕ (b ⊕ c) in L. Clearly, every commutative monoid hM; +, 0i is a partial abelian monoid where ⊥= M ×M and ⊕ = +. For nice examples of commutative monoids see [22, Section 8.4, pp. 259–261]. Henceforth, L = hL; ⊕, 0i denotes a partial abelian monoid. L can be endowed with its algebraic preordering ≤alg deﬁned by the formula

a ≤alg b ⇐⇒ There exists x ∈ L such that a ⊥ x and a ⊕ x = b.

Clearly, 0 is the least element of hL; ≤algi. Lemma 2.1. Let a, b, c, d ∈ L. Then

a) a ≤alg b and c ⊥ b ⇒ c ⊥ a and a ⊕ c ≤alg b ⊕ c. b) a ≤alg b, c ≤alg d and b ⊥ d ⇒ a ⊥ c and a ⊕ c ≤alg b ⊕ d.

Proof. a) Since a ≤alg b, there exists x ∈ L such that a ⊥ x and a ⊕ x = b. Then c ⊥ (a ⊕ x). By the associative law, we have c ⊥ a,(c ⊕ a) ⊥ x and c ⊕ (a ⊕ x) = (c ⊕ a) ⊕ x, and therefore c ⊕ b = (c ⊕ a) ⊕ x. Hence c ⊕ a ≤alg c ⊕ b. b) By a double application of part a). Let u ∈ L. Then we say that

(1) u is a unit in L if a ≤alg u for every a ∈ L. ORTHOGONALITY IN PARTIAL ABELIAN MONOIDS 3

(2) u is cancellable in L if, for all a, b ∈ L such that a ⊥ u, b ⊥ u and a ⊕ u = b ⊕ u, we have a = b. (3) u is directly ﬁnite in L if, for every a ∈ L such that a ⊥ u and a ⊕ u = u, we have a = 0. (This is weaker than to say that u is cancellable in L.) With respect to the partial abelian monoid L we say that

(1) L is unital if L possess a distinguished unit denoted by 1, and then a ≤alg 1 for every a ∈ L. (2) L is cancellative if every element of L is cancellable in L. (3) L is conical if, for all a, b ∈ L such that a ⊥ b and a ⊕ b = 0, we have a = 0 (and therefore b = 0). Lemma 2.2. Assume that L is conical and every element of L is directly ﬁnite. Then

a) hL; ≤algi is a partially ordered set. b) If a, b, c, d ∈ L, a ≤alg b, c ≤alg d, b ⊥ d and a ⊕ c = b ⊕ d, then a = b and c = d. Proof. a) It follows from [24, Lemma 2.3]. b) It follows from Lemma 2.1 b) that a ⊥ c. Since a ≤alg b and c ≤alg d, there exist x, y ∈ L such that a ⊥ x, c ⊥ y, a ⊕ x = b and c ⊕ y = d. Since a ⊕ c = b ⊕ d, we have a ⊕ c = (a ⊕ x) ⊕ (c ⊕ y) and using the commutative and associative laws, we can write a ⊕ c = (x ⊕ y) ⊕ (a ⊕ c). Since every element of L is directly ﬁnite, we get x ⊕ y = 0 and therefore x = y = 0 because L is conical. So a = b and c = d. Remark 2.3. We note that Lemma 2.1 b) and Lemma 2.2 b) contain [21, Lemma 2.1]. Corollary 2.4. If L is conical and every element of L is directly ﬁnite, then L possesses at most one unit.

Assume that L is cancellative. Then if a, b ∈ L and a ≤alg b there exists a unique element c ∈ L such that a ⊥ c and a ⊕ c = b. We write c = b − a and we call c the diﬀerence between b and a. Clearly, a − a = 0 and a − 0 = a for every a ∈ L. Moreover, if a, b ∈ L and a ≤alg b, then a ⊥ (b − a) and a ⊕ (b − a) = b. Lemma 2.5. Assume that L is cancellative and let a, b, c ∈ L. Then

a) a ⊥ b and a ⊕ b ≤alg c ⇒ b ≤alg c and a ≤alg c − b. b) a ≤alg b and c ≤alg b − a ⇒ a ≤alg b − c. c) a ⊥ b ⇒ (a ⊕ b) − a = b.

Proof. a) Since b ≤alg (a ⊕ b), we have b ≤alg c. Moreover, since a ⊕ b ≤alg c, there exists x ∈ L such that x ⊥ (a ⊕ b) and (a ⊕ b) ⊕ x = c. By the commutative law, it follows that (b⊕a)⊕x = c. Then the associative law implies that a ⊥ x, b ⊥ (a⊕x) and b ⊕ (a ⊕ x) = c. Hence a ⊕ x = c − b and therefore a ≤alg c − b. b) Since c ≤alg b − a, there exists y ∈ L such that y ⊥ c and y ⊕ c = b − a. But b = a ⊕ (b − a). So b = a ⊕ (y ⊕ c). By the associative law, we have a ⊥ y, c ⊥ (a ⊕ y) and b = (a ⊕ y) ⊕ c. Then c ≤alg b and b − c = a ⊕ y and this implies that a ≤alg b − c. 4 PEDRO MORALES AND FRANCISCO GARC´IA MAZAR´IO

c) Since a ≤alg (a ⊕ b), we have a ⊕ b = a ⊕ ((a ⊕ b) − a) and by cancellation we get b = (a ⊕ b) − a. Example 2.6. Consider the real field hR; +, ·, 0, 1i endowed with its usual total ordering ≤. Let ⊥ = {(a, b) ∈ R × R : a = 0 or b = 0} ∪ (]0, 1]×]0, 1]) and define a + b a ⊕ b = if a ⊥ b. 1 + a · b Then it is easy to verify that hR; ⊕, 0i is a partial abelian monoid which is conical and every element of R different of 1 is cancellable in hR; ⊕, 0i. Since a, b ∈ R and a ≤alg b ⇐⇒ a = 0 or a = b or (a, b ∈]0, 1] and a < b) it follows that hR; ≤algi is a partially ordered set and hR; ⊕, 0i is not unital. Example 2.7. Consider the abelian group hR; +, 0i endowed with its usual total ordering ≤ and let +∞ be a symbol not belonging to R. Let M = R ∪ {+∞} and we extend the binary operation + on M by putting a + (+∞) = (+∞) + a = +∞ for all a ∈ M. Then M = hM; +, 0i is a commutative monoid which is unital with unit +∞ and every element of R is cancellable in M, but M is not conical. Example 2.8. Let G = hG; +, 0i be a non trivial abelian group and let L(G) be the set of all subgroups of G. Under the relation of inclusion ⊆, L(G) is a partially ordered set. It is actually a lattice if L(G) is endowed with the two binary operations defined by the formulae:

S1 ∧ S2 = S1 ∩ S2

S1 ∨ S2 = {a + b : a ∈ S1 and b ∈ S2}. It is well known that hL(G); ∧, ∨i is a complete modular lattice with least element 0 and greatest element G (see [3, Theorem 11, p. 13] and [17, p. 5]). Define ⊥ = {(S1,S2) ∈ L(G) × L(G): S1 ∧ S2 = 0} and S1 ⊕ S2 = S1 ∨ S2 if S1 ⊥ S2. Then, using [38, Proposition 8.1] it is easy to show that hL(G); ⊕, 0i is a partial abelian monoid which is conical and every element of L(G) is directly finite. Hence Lemma 2.2 implies that hL(G); ≤algi is a partially ordered set (clearly, the relation of inclusion ⊆ is weaker than ≤alg). We consider two important cases: (1) If every finite set of elements of G generates a cyclic subgroup, then it follows from Ore’s Theorem [30, Theorem 4, p. 267] that the partial abelian monoid hL(G); ⊕, 0i is cancellative, but not necessarily unital (this happens, for example, when G is the full rational group). (2) If the order of each element of G is a square-free integer, then it follows from Kertész’s Theorem [27] that the partial abelian monoid hL(G); ⊕, 0i is unital with unit G, but not necessarily cancellative (this happens, for example, when G is the additive group Z(2) × Z(2)). ORTHOGONALITY IN PARTIAL ABELIAN MONOIDS 5

Example 2.9. Let K be any ﬁeld and let V be a vector space over K of dimension greater than or equal to 2. Consider the set L(V ) of all subspaces of V which is partially ordered by the inclusion relation ⊆. Then L(V ) endowed with the two binary operations deﬁned by the formulae:

S1 ∧ S2 = S1 ∩ S2

S1 ∨ S2 = {x + y : x ∈ S1 and y ∈ S2} is a complete modular lattice with least element 0 and greatest element V . If we deﬁne

⊥ = {(S1,S2) ∈ L(V ) × L(V ): S1 ∧ S2 = 0} and

S1 ⊕ S2 = S1 ∨ S2 if S1 ⊥ S2, then it is easy to show that hL(V ); ⊕, 0i is a partial abelian monoid which is conical and every element of L(V ) is directly finite, and therefore hL(V ); ≤algi is a partially ordered set. Also every line in V is not cancellable in L(V ) and by [4, §3, no 3, Proposition 5, p. 37] it follows that hL(V ); ⊕, 0i is unital with unit V . Now suppose that L = hL; ⊕, 0i is a partial abelian monoid. We say that L is an effect algebra if L is conical, cancellative and unital. For example, if hL; ∧, ∨i is a distributive lattice with least element 0 and greatest element 1, then the derived partial abelian monoid hL; ⊕, 0i is an effect algebra. We have the following simple result: Proposition 2.10 ([39, Lemma 1.4]). For every partial abelian monoid L, the following conditions are equivalent: i) L is an effect algebra. ii) There exists an element 1 in L, different of 0, and satisfying the following axioms: (Orthosupplementation Law) For every a ∈ L there exists a unique b ∈ L such that a ⊥ b and a ⊕ b = 1. (Zero-One Law) If a ∈ L and a ⊥ 1, then a = 0. Let L be an effect algebra. If a ∈ L then the unique element b ∈ L such that a ⊥ b and a ⊕ b = 1 is called the orthosupplement of a and denoted by a0. It is clear that the orthosupplement has the following properties: (1) 10 = 0. (2) (a0)0 = a for all a ∈ L. 0 0 (3) If a, b ∈ L and a ≤alg b, then b ≤alg a . 0 (4) Let a, b ∈ L. Then a ⊥ b if and only if a ≤alg b . For a list of nice examples of effect algebras we refer to [13, 4. Examples] and for some of its properties we refer also to [1] and [12]. Another interesting example of an effect algebra is the following: Let I be a non-empty index set and, for every i ∈ I, let Li = hLi; ⊕, 0, 1i be an effect algebra. Form the Cartesian product L = Xi∈I Li and let πi : L → Li be the canonical projection onto the i-th coordinate set. Put 6 PEDRO MORALES AND FRANCISCO GARCÍA MAZARÍO

⊥ = {(a, b) ∈ L × L : πi(a) ⊥ πi(b) for all i ∈ I} and deﬁne ⊕ : ⊥ → L by the formula

πi(a ⊕ b) = πi(a) ⊕ πi(b) for all i ∈ I. If πi(a) = 0 and πi(b) = 1 for all i ∈ I, we put a = 0 and b = 1. It is easy to see that hL; ⊕, 0, 1i is an effect algebra called the Cartesian product of the Q family (Li)i∈I and it is denoted by i∈I Li. An orthoalgebra is an effect algebra L in which the zero-one law is replaced by the stronger (Consistence Law) If a ∈ L and a ⊥ a, then a = 0. For examples and details concerning orthoalgebras we refer to [14], [29] and [32]. An orthomodular poset is an effect algebra L that satisfies the following axiom (Coherence Law) For a, b, c ∈ L, the conditions a ⊥ b, a ⊥ c and b ⊥ c imply (a ⊕ b) ⊥ c. Since the coherence law is stronger than the consistence law, every orthomodular poset is an orthoalgebra. An orthomodular lattice is an orthoalgebra L such that hL; ≤algi is a lattice. It is easy to verify that every orthomodular lattice is an orthomodular poset. A centerium is an orthoalgebra L that satisfies the following axiom

(Law of Compatibility) For all a, b ∈ L there exist a1, a2, c ∈ L such that a1 ⊥ a2, a1 ⊥ c, a2 ⊥ c, a = a1 ⊕ c and b = a2 ⊕ c. A Boolean algebra is an orthomodular poset L that is also a centerium. It follows from [31, Corollary 1.3.14] that L is a Boolean algebra if and only if L is a distributive orthomodular lattice. For terminology and details concerning orthomodular posets, orthomodular lattices and Boolean algebras we refer to [2], [25] and [31].

3. Orthogonal suites

We shall denote by N the set of all natural numbers 0, 1, 2,... and by ≤0 the usual total ordering on N. Moreover, if p, q ∈ N and p ≤0 q we denote by [p, q]0 the order interval {k ∈ N : p ≤0 k ≤0 q} in hN; ≤0i. In this Section L = hL; ⊕, 0i denotes a partial abelian monoid. Let (ai)i∈I be a finite non-empty family of elements of L and let n = card(I). From the n! different total orderings on I we choose one of them denoted by ≤. The finite family (ai)i∈I of elements of L together with the totally ordered set hI; ≤i define an algebraic system called a suite in L of n elements and it is denoted by (ai)i∈hI;≤i. Also we can define i0 = min I and ip = min(I \{i0, i1, . . . , ip−1}) for all p ∈ [1, n − 1]0 if n ≥0 2. A suite in L of n elements (ai)i∈hI;≤i is said to be orthogonal in L either n = 1 or, when n ≥0 2, if

ai0 ⊥ ai1 and aip ⊥ (··· ((ai0 ⊕ ai1 ) ⊕ ai2 ) ··· aip−1 ) ORTHOGONALITY IN PARTIAL ABELIAN MONOIDS 7 for all p ∈ [2, n − 1]0. Further, the element s of L deﬁned respectively by the formulae

s = ai0 and s = (··· ((ai0 ⊕ ai1 ) ⊕ ai2 ) ··· ain−2 ) ⊕ ain−1 L is called the ⊕-join of the orthogonal suite (ai)i∈hI;≤i and it is denoted by i∈hI;≤i ai L or i∈I ai for simplicity. For example, consider the partial abelian monoid L = hL(V ); ⊕, 0i where V is a real infinite-dimensional vector space. Let I = hI; ≤i be a finite totally ordered set with card(I) ≥0 3 and let (Si)i∈I be a family of elements of L. Then the suite (Si)i∈hI;≤i is orthogonal in L if and only if, for every i ∈ I, if S is the subspace of V S L spanned by j∈I\{i} Sj, we have S ∩ Si = 0. In this case, i∈I Si is the subspace S of V spanned by i∈I Si. Since the void subset ∅ of any partially ordered set hJ; ≤i is totally ordered, it follows that (ai)i∈h∅;≤i is a suite in L of 0 elements. Then it is convenient to accept L also that i∈h∅;≤i ai = 0. An important special case of a suite in L is the following (see [12, Definition

4.1] and [38, p. 288]): Let p, q ∈ N be such that p ≤0 q, let I = [p, q]0 and let

(ai)i∈I be a family of elements of L. Since card(I) = q − p + 1, (ai)i∈hI;≤0i is a suite in L of q − p + 1 elements which it will be denoted by (ai)p≤0i≤0q. In the case where (ai)p≤0i≤0q is an orthogonal suite in L, the corresponding ⊕-join is denoted Lq by i=p ai or ap ⊕ ap+1 ⊕ · · · ⊕ aq.

Remark 3.1. Let I = hI; ≤i be a ﬁnite totally ordered set with card(I) = n ≥0 2 and let (ai)i∈I be a family of elements of L. Write i0 = min I, ip = min(I \

{i0, i1, . . . , ip−1}) for all p ∈ [1, n − 1]0 and bk = aik for all k ∈ [0, n − 1]0. It is clear that (ai)i∈hI;≤i is an orthogonal suite in L if and only if (bk)0≤0k≤0n−1 is an L Ln−1 orthogonal suite in L. In either case, we have i∈I ai = k=0 bk. We begin with an obvious lemma which we use sometimes without explicit men- tion:

Lemma 3.2. Let I = hI; ≤i be a ﬁnite totally ordered set with card(I) = n ≥0 2, 0 let in−1 = max I, let I = I \{in−1} and let (ai)i∈I be a family of elements of L. Then the following conditions are equivalent:

i) (ai)i∈hI;≤i is an orthogonal suite in L of n elements. 0 ii) (ai)i∈hI ;≤i is an orthogonal suite in L of n − 1 elements and ain−1 ⊥ L i∈I0 ai . L L Further, i∈I ai = i∈I0 ai ⊕ ain−1 . The following result gives the generalized associative law for the partial binary operation ⊕: Theorem 3.3. Let p ∈ N \{0}, let I = hI; ≤i be a finite non-empty totally ordered set which is the union of non-empty sets Ik (k ∈ [1, p]0) verifying the following condition: i ∈ Ik, j ∈ Il and 1 ≤0 k <0 l ≤0 p imply i < j (1) and let (ai)i∈I be a family of elements of L. If (ai)i∈hI;≤i is an orthogonal suite in L, then 8 PEDRO MORALES AND FRANCISCO GARCÍA MAZARÍO

a) (ai)i∈hIk;≤i is an orthogonal suite in L for all k ∈ [1, p]0. L b) ai is an orthogonal suite in L. i∈Ik 1≤0k≤0p c) L a = Lp L a . i∈I i k=1 i∈Ik i Proof. We proceed by induction on n = card(I). Let n = 1. Since every set Ik is non-empty, it follows that p = 1, and the Theorem is trivial. Let n ≥0 2 and suppose that the Theorem is true for every index set of cardinal n − 1 with a decomposition satisfying (1). We distinguish two cases: I) The set Ip contains a unique element in−1. 0 0 Then in−1 = max I and putting I = I1 ∪ I2 ∪ ... ∪ Ip−1 we have I = I \{in−1}. Hence Lemma 3.2 assures that (ai)i∈hI0;≤i is an orthogonal suite in L of n − 1 L L L elements such that ain−1 ⊥ i∈I0 ai and i∈I ai = i∈I0 ai ⊕ain−1 . Since the decomposition of I0 satisfies the condition (1) the induction hypothesis implies that L (ai)i∈hIk;≤i is an orthogonal suite in L for all k ∈ [1, p − 1] , ai is 0 i∈Ik 1≤0k≤0p−1 L Lp−1 L also an orthogonal suite in L and 0 a = a . Then, concerning i∈I i k=1 i∈Ik i the orthogonal suite (ai)i∈hI;≤i, the properties a) and b) are trivially verified. Since L L a ⊥ 0 a and a = a the property c) follows from Lemma 3.2. in−1 i∈I i i∈Ip i in−1 II) The set Ip contains more than one element. 0 Let in−1 = max I. Then in−1 = max Ip. Let I = {i ∈ I : i < in−1} and 0 0 0 0 let Ip = I ∩ Ip. So card(I ) = n − 1, (ai)i∈hI0;≤i is an orthogonal suite in L, I = 0 I1 ∪...∪Ip−1 ∪Ip and this decomposition satisfies the condition (1). Then the induc- 0 tion hypothesis implies that (ai)i∈hI ;≤i and (ai)i∈hIk;≤i (k ∈ [1, p − 1]0) are orthog- Lp L onal suites in L such that, for bk = ai (k ∈ [1, p − 1] ) and bp = 0 ai, i∈hIk;≤i 0 i∈Ip L Lp we have that (bk)1≤0k≤0p is an orthogonal suite in L and i∈I0 ai = k=1 bk. L Lp−1 L L So 0 ai = ai ⊕ 0 ai . Then with respect to the or- i∈I k=1 i∈Ik i∈Ip Lp−1 L L thogonal suite (ai)i∈hI;≤i in L, since ain−1 ⊥ ai ⊕ 0 ai , k=1 i∈Ik i∈Ip 0 Ip = Ip ∪ {in−1} and in−1 = max Ip, the associative law and Lemma 3.2 imply L that (ai)i∈hIp;≤i is an orthogonal suite in L and therefore ai is i∈Ik 1≤0k≤0p also an orthogonal suite in L with L a = Lp L a , and therefore the i∈I i k=1 i∈Ik i properties a), b) and c) are verified.

Corollary 3.4. Let I = hI; ≤i be a ﬁnite totally ordered set with card(I) = n ≥0 2, 0 let i0 = min I and let I = I \{i0}. If (ai)i∈hI;≤i is an orthogonal suite in L of n elements, then (ai)i∈hI0;≤i is an orthogonal suite in L of n − 1 elements such that L L L ai0 ⊥ i∈I0 ai and i∈I ai = ai0 ⊕ i∈I0 ai .

Theorem 3.5. Let I = hI; ≤i be a ﬁnite totally ordered set with card(I) ≥0 3, let I1 and I2 be two non-empty subsets of I such that I1 ∩ I2 = ∅ and let (ai)i∈I be a family of elements of L. If (ai)i∈hI1;≤i and (ai)i∈hI2;≤i are orthogonal suites in L such that L a ⊥ L a , then (a ) is an orthogonal suite in L and i∈I1 i i∈I2 i i i∈hI1∪I2;≤i L a = L a ⊕ L a . i∈I1∪I2 i i∈I1 i i∈I2 i ORTHOGONALITY IN PARTIAL ABELIAN MONOIDS 9

Proof. Write b = L a , i = min I and i = min(I \{i , i , . . . , i }) for i∈I1 i 0 2 p 2 0 1 p−1 all p ∈ [1, n − 1] where n = card(I ). Then Remark 3.1 implies that L a = 0 2 i∈I2 i ai0 ⊕ ai1 ⊕ ai2 ⊕ · · · ⊕ ain−2 ⊕ ain−1 . Then using the associative law and Corollary 3.4 we can write in a successive process M M ai ⊕ ai = b ⊕ (ai0 ⊕ (ai1 ⊕ · · · ⊕ ain−1 )) i∈I1 i∈I2

= (b ⊕ ai0 ) ⊕ (ai1 ⊕ (ai2 ⊕ · · · ⊕ ain−2 ⊕ ain−1 ))

= ((b ⊕ ai0 ) ⊕ ai1 ) ⊕ (ai2 ⊕ (ai3 ⊕ · · · ⊕ ain−2 ⊕ ain−1 ))

··· = (··· ((b ⊕ ai0 ) ⊕ ai1 ) ⊕ · · · ⊕ ain−2 ) ⊕ ain−1 )) and we get that (a ) is an orthogonal suite in L such that L a = i i∈hI1∪I2;≤i i∈I1∪I2 i L a ⊕ L a . i∈I1 i i∈I2 i

Corollary 3.6. Let I = hI; ≤i be a ﬁnite totally ordered set with card(I) = n ≥0 3, 0 let j ∈ I, let I = I \{j} and let (ai)i∈I be a family of elements of L. If (ai)i∈hI0;≤i L is an orthogonal suite in L of n − 1 elements and aj ⊥ i∈I0 ai, then (ai)i∈hI;≤i is L L an orthogonal suite in L of n elements and i∈I ai = i∈I0 ai ⊕ aj.

Proposition 3.7. Let I = hI; ≤i be a ﬁnite totally ordered set with card(I) = n ≥0 0 3, let j ∈ I, let I = I \{j} and let (ai)i∈I be a family of elements of L. If (ai)i∈hI;≤i is an orthogonal suite in L of n elements, then (ai)i∈hI0;≤i is an orthogonal suite L L in L of n − 1 elements and i∈I ai = i∈I0 ai ⊕ aj.

Proof. Write i0 = min I and ip = min(I \{i0, i1, . . . , ip−1}) for all p ∈ [1, n − 1]0. If j = in−1 (resp. j = i0) then the result follows from Lemma 3.2 (resp. commutative law and Corollary 3.4).

Suppose that i0 < j < in−1. Then there exists h ∈ [1, n − 2]0 such that j = ih. If we put I1 = {i0, i1, . . . , ih−1}, I2 = {ih} and I3 = {ih+1, ih+2, . . . , in−1}, then I1, I2, I3 are non-empty sets, I = I1 ∪ I2 ∪ I3 and this decomposition satisﬁes the condition (1) of Theorem 3.3, and it follows that (ai)i∈hIk;≤i is an orthogonal L suite in L for all k ∈ [1, 3] , ai is also an orthogonal suite in L and 0 i∈Ik 1≤0k≤03 L a = L a ⊕ a ⊕ L a . Then the associative and commutative i∈I i i∈I1 i ih i∈I3 i laws imply that L a ⊥ L a . Since I0 = I ∪ I and I ∩ I = ∅, it i∈I1 i i∈I3 i 1 3 1 3 follows from Theorem 3.5 that (ai)i∈hI0;≤i is an orthogonal suite in L and clearly L L i∈I ai = i∈I0 ai ⊕ aih .

Corollary 3.8. Let I = hI; ≤i be a ﬁnite totally ordered set with card(I) = n ≥0 3. Let J be a non-empty subset of I and let (ai)i∈I be a family of elements of L. If (ai)i∈hI;≤i is an orthogonal suite in L, then (ai)i∈hJ;≤i is also an orthogonal suite L L in L and i∈J ai ≤alg i∈I ai. The following result gives the generalized commutative law for the binary partial operation ⊕:

Theorem 3.9. Let (ai)i∈I be a finite family of elements of L with card(I) = n ≥0 2. Then for any two total orderings on L, the orthogonality of the corresponding suites holds simultaneously and the ⊕-joins have the same value. 10 PEDRO MORALES AND FRANCISCO GARCÍA MAZARÍO

Proof. We proceed by induction on n. Clearly for n = 2 the Theorem is immediate. Let n ∈ N be such that n ≥0 3 and suppose that the Theorem is true for all index set of cardinal n − 1. 0 Let ≤ be a total ordering on I such that (ai)i∈hI;≤0i is an orthogonal suite in L. Let ≤00 be another total ordering on I and suppose that j is the greatest 00 0 element of hI; ≤ i. Put I = I \{j}. Then Proposition 3.7 implies that (ai)i∈hI0;≤0i L is an orthogonal suite in L of n − 1 elements such that aj ⊥ i∈hI0;≤0i ai and L L i∈hI;≤0i ai = i∈hI0;≤0i ai ⊕ aj. Moreover, the induction hypothesis implies L L that (ai)i∈hI0;≤00i is an orthogonal suite in L and i∈hI0;≤00i ai = i∈hI0;≤0i ai. So L aj ⊥ i∈hI0;≤00i ai and Lemma 3.2 implies that (ai)i∈hI;≤00i is an orthogonal suite L L L in L and i∈hI;≤00i ai = i∈hI0;≤00i ai ⊕ aj = i∈hI;≤0i ai.

Corollary 3.10. Let I = hI; ≤i be a finite totally ordered set with card(I) ≥0 2, let σ be a bijection from I onto I, let ≤0 be the total ordering on I defined by the 0 −1 −1 formula: i < j if σ (i) < σ (j) and let (ai)i∈I be a family of elements of L. If (ai)i∈hI;≤i is an orthogonal suite in L, then (ai)i∈hI;≤0i is an orthogonal suite in L L L and i∈hI;≤i ai = i∈hI;≤0i ai. Theorem 3.11. Let p ∈ N\{0}, let I = hI; ≤i be a finite non-empty totally ordered set which is the union of non-empty pairwise disjoint sets Ik (k ∈ [1, p]0) and let (ai)i∈I be a family of elements of L. If (ai)i∈hI;≤i is an orthogonal suite in L, then

a) (ai)i∈hIk;≤i is an orthogonal suite in L for all k ∈ [1, p]0. L b) ai is an orthogonal suite in L. i∈Ik 1≤0k≤0p c) L a = Lp L a . i∈I i k=1 i∈Ik i Proof. It follows from Theorem 3.9 if we choose a total ordering ≤0 on I such that 0 the decomposition of hI; ≤ i satisﬁes the condition (1) of Theorem 3.3.

Theorem 3.12. Let I = hI; ≤i be a ﬁnite totally ordered set with card(I) = n ≥0 2 and let (ai)i∈I and (bi)i∈I be two families of elements of L such that ai ⊥ bi for all i ∈ I. If (ai ⊕ bi)i∈hI;≤i is an orthogonal suite in L, then (ai)i∈hI;≤i and L L L (bi)i∈hI;≤i are orthogonal suites in L such that i∈I ai ⊥ i∈I bi and i∈I (ai ⊕ L L bi) = i∈I ai ⊕ i∈I bi . Proof. We proceed by induction on n. First we prove the Theorem for n = 2. Write I = {i0, i1} where i0 < i1. Since ai1 , bi1 ≤alg (ai1 ⊕ bi1 ) and (ai0 ⊕ bi0 ) ⊥ (ai1 ⊕ bi1 ) it follows from Lemma 2.1 a) that (ai0 ⊕ bi0 ) ⊥ ai1 and (ai0 ⊕ bi0 ) ⊥ bi1 . Then the associative and commutative laws imply that ai0 ⊥ ai1 , bi0 ⊥ bi1 ,(ai0 ⊕ ai1 ) ⊥ bi0 and (ai0 ⊕ bi0 ) ⊕ ai1 =

(ai0 ⊕ ai1 ) ⊕ bi0 . Since (ai0 ⊕ bi0 ) ⊥ (ai1 ⊕ bi1 ) we get (ai0 ⊕ ai1 ) ⊕ bi0 ⊥ bi1 and therefore (ai0 ⊕ai1 ) ⊥ (bi0 ⊕bi1 ) with (ai0 ⊕bi0 )⊕(ai1 ⊕bi1 ) = (ai0 ⊕ai1 )⊕(bi0 ⊕bi1 ). Now let n ∈ N be such that n ≥0 3 and suppose that the Theorem is true for every index set of cardinal n − 1. Assume that card(I) = n, ai ⊥ bi for all i ∈ I and (ai ⊕bi)i∈hI;≤i is an orthogonal suite in L. Choose j ∈ I and let I0 = I \{j}. Then by Proposition 3.7 it follows that ORTHOGONALITY IN PARTIAL ABELIAN MONOIDS 11

(ai ⊕ bi)i∈hI0,≤i is an orthogonal suite in L of n − 1 elements such that (aj ⊕ bj) ⊥ L L L i∈I0 (ai ⊕bi) and i∈I (ai ⊕bi) = i∈I0 (ai ⊕ bi) ⊕(aj ⊕bj). Then the induction hypothesis implies that (ai)i∈hI0;≤i and (bi)i∈hI0;≤i are orthogonal suites in L such L L L L L that i∈I0 ai ⊥ i∈I0 bi and i∈I0 (ai ⊕bi) = i∈I0 ai ⊕ i∈I0 bi . To complete the proof it suﬃces to apply the special case n = 2 and Corollary 3.6.

Corollary 3.13. Let I = hI; ≤i be a ﬁnite totally ordered set with card(I) = n ≥0 2 and let (ai)i∈I and (bi)i∈I be two families of elements of L such that bi ≤alg ai for all i ∈ I. If (ai)i∈hI;≤i is an orthogonal suite in L, then (bi)i∈hI;≤i is an orthogonal L L suite in L and i∈I bi ≤alg i∈I ai.

Proof. Since bi ≤alg ai for all i ∈ I, we get bi ⊥ (ai − bi) and bi ⊕ (ai − bi) = ai for all i ∈ I. Moreover, since (ai)i∈hI;≤i is an orthogonal suite in L, it follows from Theorem 3.12 that (bi)i∈hI;≤i and (ai − bi)i∈hI;≤i are orthogonal suites in L such L L L L L that bi ⊥ (ai − bi) and ai = bi ⊕ (ai − bi) . Whence L i∈I L i∈I i∈I i∈I i∈I i∈I bi ≤alg i∈I ai.

Theorem 3.14. Let n ∈ N \{0} and let (ai)1≤0i≤0n be an orthogonal suite in L. Then there exists a chain b0 ≤alg b1 ≤alg b2 ≤alg · · · ≤alg bn in L such that b0 = 0 Ln and bn = i=1 ai.

Proof. Put b0 = 0 and let p ∈ [1, n]0. By Corollary 3.8 it follows that (ai)1≤0i≤0p Lp is an orthogonal suite in L. Then if we deﬁne bp = i=1 ai we get bp ≤alg bp+1 for all p ∈ [1, n − 1]0, and therefore b0 ≤alg b1 ≤alg b2 ≤alg · · · ≤alg bn.

Theorem 3.15. Assume that L is cancellative. Let n ∈ N \{0}, let b0 ≤alg b1 ≤alg b2 ≤alg · · · ≤alg bn be a chain in L and let ap = bp − bp−1 for all p ∈ [1, n]0. Then Ln (ap)1≤0p≤0n is an orthogonal suite in L such that p=1 ap = bn − b0. Proof. We proceed by induction on n. For n = 1 the Theorem is trivial. Let n ∈ N \{0} and suppose that the Theorem is true for n. Let b0 ≤alg b1 ≤alg b2 ≤alg · · · ≤alg bn ≤alg bn+1 be a chain in L and let ap = bp − bp−1 for all p ∈ [1, n + 1]0. Then the induction hypothesis implies that (ap)1≤0p≤0n Ln is an orthogonal suite in L such that p=1 ap = bn −b0. Since b0 ≤alg bn ≤alg bn+1, we have bn −b0 ⊥ b0, bn = (bn −b0)⊕b0, bn+1 −bn ⊥ bn and bn+1 = (bn+1 −bn)⊕bn. Then bn − b0 ≤alg bn and Lemma 2.1 a) implies that bn+1 − bn ⊥ bn − b0. So Ln an+1 ⊥ p=1 ap and Lemma 3.2 implies that (ap)1≤0p≤0n+1 is an orthogonal suite Ln+1 Ln in L and p=1 ap = p=1 ap ⊕ an+1 = (bn − b0) ⊕ (bn+1 − bn). Since bn+1 = bn ⊕ (bn+1 − bn) = (b0 ⊕ (bn − b0)) ⊕ (bn+1 − bn) = b0 ⊕ ((bn − b0) ⊕ (bn+1 − bn)), we get bn+1 − b0 = (bn − b0) ⊕ (bn+1 − bn) and therefore the Theorem is true for n + 1. Remark 3.16. It is clear that Corollaries 3.10 and 3.13 and Theorems 3.3, 3.5 and 3.11 improve [12, Theorem 4.2]. Also, if E is a commutative monoid, then Theorem 3.3 (resp. 3.9) improves [5, §1, no 3, Théorèm1] (resp. [5, §1, no 5, Théorèm3]). 12 PEDRO MORALES AND FRANCISCO GARCÍA MAZARÍO

4. Inﬁnite orthogonal families

In this Section we assume that L = hL; ⊕, 0i is a partial abelian monoid such that hL; ≤algi is a partially ordered set. This implies, in particular, that L is conical. We extend first the notion of orthogonality to countably infinite families of elements of L. Let M be an infinite subset of N and let (ai)i∈M be a family of elements of L. Then the countably infinite family (ai)i∈M of elements of L together with the totally ordered set hM; ≤0i define an algebraic system called a sequence in L and denoted by (ai)i∈hM;≤0i.

A sequence in L (ai)i∈hM;≤0i is said to be orthogonal in L if, for every n ∈ N, the suite (a ) is orthogonal in L. It is clear that if (a ) is i i∈≤hM∩[0,n]0;≤0i i i∈hM;≤0i an orthogonal sequence in L, then for every inﬁnite subset N of M, the sequence in L (ai)i∈hN;≤0i is also orthogonal in L. WL If (ai)i∈hM;≤0i is an orthogonal sequence in L and there exists i∈M∩[0,n] ai | L 0 n ∈ N in hL; ≤algi, we denote this element by i∈M ai and we call it the ⊕-join of the orthogonal sequence (ai)i∈hM;≤0i. For example, consider the partial abelian monoid L = hR; ⊕, 0i of the Example 2.6, and let a = 1 for all i ∈ . Then it is clear that (a ) is an orthogonal i 2 N i i∈hN;≤0i sequence in L. Consider the family (kj)j∈N of elements of N \{0} such that k0 = 2 and kj+1 = 3kj − 1 for all j ∈ N. Then it is easy to verify by induction on n Ln 1 W 1 that ai = 1 − for all n ∈ . Since (1 − | n ∈ ) = 1 in hL; ≤i i=0 kn N kn N and ≤alg =≤ , we conclude that the ⊕-join of the orthogonal sequence ]0,1] ]0,1] (a ) is equal to 1. i i∈hN;≤0i The following simple result allow us to consider only orthogonal sequences in L \{0}:

Lemma 4.1. Let (ai)i∈hM;≤0i be a sequence in L such that N = {i ∈ M : ai 6= 0} is an inﬁnite subset of M. Then (ai)i∈hM;≤ i is an orthogonal sequence in L if 0 L and only if (ai)i∈hN;≤ i is an orthogonal sequence in L \{0}. Further, ai = L 0 i∈M i∈N ai if one of these ⊕-joins exist in hL; ≤algi.

Proof. Write P = {i ∈ M : ai = 0}. The Lemma is trivial if P = ∅. Suppose P 6= ∅ and let n ∈ N be such that M ∩ [0, n]0 is a non-empty set. Since M ∩ [0, n] = (N ∩ [0, n] ) ∪ (P ∩ [0, n] ), the Lemma follows from Theorems 3.5 0 L 0 L0 and 3.11 noting that ai = ai. i∈M∩[0,n]0 i∈N∩[0,n]0 We say that L is σ-complete if the ⊕-join of every orthogonal sequence in L exists in hL; ≤algi. Corollary 4.2. If the ⊕-join of every orthogonal sequence in L \{0} exists in L \{0}, then L is σ-complete.

For example, consider the partial abelian monoid L = hR; ⊕, 0i of the Example 2.6. We shall show that L is σ-complete. In fact, let (a ) be an orthogonal i i∈hN;≤0i sequence in L. By Corollary 4.2 we may suppose that ai 6= 0 for all i ∈ N. Since ORTHOGONALITY IN PARTIAL ABELIAN MONOIDS 13 x, y ∈ L \{0} and x ⊥ y imply x, y ∈]0, 1] and x ⊕ y ∈]0, 1], it follows that ai ∈]0, 1] Ln for all i ∈ . So ( ai) is an increasing family of elements of h]0, 1]; ≤algi and N i=0 n∈N therefore an increasing family of elements of h]0, 1]; ≤i because ≤alg =≤ . ]0,1] ]0,1] Then the ⊕-join of the orthogonal sequence (a ) exists in hL; ≤ i. i i∈hN;≤0i alg The following result generalizes [24, Theorem 4.9] for m = ℵ0:

Theorem 4.3. Consider the following two statements: W i) There exists the supremum (bi | i ∈ N) for all chain b0 ≤alg b1 ≤alg b2 ≤alg · · · ≤alg bn ≤alg ··· in L. ii) L is σ-complete.

Then i) implies ii). Further, if L is cancellative, then ii) implies i).

Proof. The implication i) ⇒ ii) follows from Theorem 3.14 and the implication ii) ⇒ i) follows from Theorem 3.15.

We extend now the notion of orthogonality to arbitrary infinite families of elements of L. It is well known that to every set X a cardinal number is assigned; it is called the cardinality of X and it is denoted by |X| or card(X) as in the case of finite sets. We denote by ≤c the usual well ordering on every non-empty set of cardinal numbers. If X is a non-empty set, we denote by F(X) the set of all finite subsets of X. Let I be an infinite index set and let (ai)i∈I be a family of elements of L. We say that (ai)i∈I is an orthogonal family in L if, for every F ∈ F(I) \ {∅} and every total ordering ≤ on F , the suite (ai)i∈hF ;≤i is orthogonal in L. In this case, if we L fix the set F , it follows from Theorem 3.9 that the corresponding ⊕-join i∈F ai does not depend of the choice of the total ordering ≤ on F and we can write s(F ) = L ai for all F ∈ F(I) where s(∅) = 0. Moreover, if there exists the supremum W i∈F (s(F ) | F ∈ F(I)) in hL; ≤algi, we call it the ⊕-join of the orthogonal family L L W L (ai)i∈I in L and we denote it by i∈I ai. So i∈I ai = i∈F ai | F ∈ F(I) . Clearly, every infinite subfamily of an orthogonal family in L is also orthogonal in L. Let X be an infinite cardinal number. We say that L is X-complete if the ⊕-join of every orthogonal family in L (ai)i∈I with card(I) ≤c X exists in hL; ≤algi. We say that L is complete if L is X-complete for every infinite cardinal X. Clearly, L is σ-complete if and only if L is ℵ0-complete. Hence the partial abelian monoid of the Example 2.6 is ℵ0-complete. Consider also the partial abelian monoid L = hL(V ); ⊕, 0i of the Example 2.9 where V is an infinite-dimensional real vector space. We shall show first that the partial orderings ≤alg and ⊆ on L(V ) coincide. Clearly, S1,S2 ∈ L(V ) and S1 ≤alg S2 imply S1 ⊆ S2. To establish the converse, let us suppose that S1,S2 ∈ L(V ) and S1 ⊆ S2. Since L is unital there exists S ∈ L(V ) such that S ∧ S1 = 0 and S ⊕ S1 = V . Choose S3 = S ∧ S2. Then S3 ∈ L(V ) and 14 PEDRO MORALES AND FRANCISCO GARCÍA MAZARÍO clearly S1 ∧ S3 = 0. Moreover, since L(V ) is a modular lattice, we get

S1 ⊕ S3 = S1 ∨ (S ∧ S2)

= (S1 ∨ S) ∧ S2

= V ∧ S2

= S2 and therefore S1 ≤alg S2. Now we shall show that L is complete. Let X be an inﬁnite cardinal number and let (Si)i∈I be an orthogonal family in L with card(I) ≤c X. Then, for every F ∈ F(I)\{∅} and every total ordering ≤ on F , the suite (Si)i∈hF ;≤i S is orthogonal in L whose ⊕-join is the subspace S(F ) of V generated by i∈F Si. Since L(V ) is a complete lattice, the supremum W(S(F ) | F ∈ F(I)) exists in hL(V ); ⊆i and therefore in hL(V ); ≤algi. Theorem 4.4. Assume that L is cancellative. Let I be an inﬁnite index set and let (ai)i∈I and (bi)i∈I be two orthogonal families in L such that ai ≤alg bi for all L L i ∈ I. If there exist i∈I ai and i∈I bi and they are equal, then ai = bi for all i ∈ I.

Proof. For every i ∈ I there exists ci ∈ L such that ai ⊥ ci and ai ⊕ ci = bi. Fix an index j in I. Let F ∈ F(I), let F0 = F ∪ {j} and let ≤ be a total ordering in F0. Then the suites (ai)i∈hF0;≤i and (bi)i∈hF0;≤i are orthogonal in L.

Therefore, from Theorem 3.12, it follows that (ci)i∈hF0;≤i is also an orthogonal suite L L L L L in L with i∈F ai ⊥ i∈F ci and i∈F ai ⊕ i∈F ci = i∈F bi. Since L 0 L 0 L 0 0 0 cj ≤alg i∈F ci and i∈F ai ≤alg i∈F ai, it follows from Lemma 2.1 b) that L 0 L L0 L cj ⊥ i∈F ai and cj ⊕ i∈F ai ≤alg i∈F bi ≤alg i∈I bi. Hence Lemma 2.5 L L 0 L a) implies that i∈F ai ≤alg i∈I bi − cj for all F ∈ F(I). So i∈I ai ≤alg L L L i∈I bi − cj, and from Lemma 2.5 b) we get that cj ≤alg i∈I bi − i∈I ai and therefore cj = 0 for all j ∈ I. So aj = bj for all j ∈ I. Theorem 4.5. Assume that L is cancellative. Let I be an inﬁnite index set which is the union of disjoint inﬁnite sets I1 and I2 and let (ai)i∈I be a family of elements of L verifying the conditions:

i) The families (ai)i∈I1 and (ai)i∈I2 are orthogonal in L. ii) There exist L a and L a in hL; ≤ i. i∈I1 i i∈I2 i alg iii) L a ⊥ L a . i∈I1 i i∈I2 i Then the family (a ) is orthogonal in L having L a ⊕L a as ⊕-join. i i∈I i∈I1 i i∈I2 i Proof. Write a = L a and b = L a . i∈I1 i i∈I2 i Let F ∈ F(I) \ {∅}. We may suppose that F1 = F ∩ I1 and F2 = F ∩ I2 are non-empty. Let ≤ be a total ordering on F . Then (ai)i∈hF1;≤i and (ai)i∈hF2;≤i are orthogonal suites in L and, using the hypothesis iii), Lemma 2.1 b) implies that L L i∈F ai ⊥ i∈F ai. Hence it follows from Theorem 3.5 that the suite (ai)i∈hF ;≤i 1 2 L L L is orthogonal in L and i∈F ai = i∈F ai ⊕ i∈F ai . So (ai)i∈I is an L 1 2 orthogonal family in L such that ai ≤alg (a ⊕ b) by Lemma 2.1 b) for all F ∈ i∈F L F(I). Let c ∈ L be an upper bound of the set { i∈F ai : F ∈ F(I)} in hL; ≤algi. ORTHOGONALITY IN PARTIAL ABELIAN MONOIDS 15

Let F1 ∈ F(I1) \ {∅}, F2 ∈ F(I2) \ {∅} and let ≤ be a total ordering on F1 ∪ F2. So L a ≤ c and by Theorem 3.5 we have L a ⊕ L a ≤ c. i∈F1∪F2 i alg i∈F1 i i∈F2 i alg Then, if we fix F , Lemma 2.5 a) implies that L a ≤ c − L a for 2 i∈F1 i alg i∈F2 i all F ∈ F(I ) and therefore a ≤ c − L a . So L a ≤ c − a for all 1 1 alg i∈F2 i i∈F2 i alg F2 ∈ F(I2) and consequently b ≤alg c − a. Since a ⊥ b and a ⊕ (c − a) = c, Lemma 2.1 a) implies that a ⊕ b ≤alg c. Hence the ⊕-join of the orthogonal family (ai)i∈I exists in hL; ≤algi and it is equal to a ⊕ b. Corollary 4.6. Assume that L is cancellative and complete. Let I be an infinite index set, let (ai)i∈I be an orthogonal family in L and let J be an infinite subset L L of I such that I \ J is also an infinite set. If a = j∈J aj, b = k∈I\J ak and L c = i∈I ai, then a ⊥ b and c = a ⊕ b. Proof. It suffices by Theorem 4.5 to show that a ⊥ b. Let F1 ∈ F(J) \ {∅} and let F2 ∈ F(I \ J) \ {∅}. Take F = F1 ∪ F2 and let ≤ be a total ordering on F . By Theorem 3.11 we get L a ⊥ L a i∈F1 i i∈F2 i and L a ⊕ L a = L a ≤ c. Then Lemma 2.5 a) implies that i∈F1 i i∈F2 i i∈F i alg L a ≤ c − L a for all F ∈ F(J) and therefore a ≤ c − L a . i∈F1 i alg i∈F2 i 1 alg i∈F2 i So L a ≤ c − a for all F ∈ F(I \ J), and consequently b ≤ c − a. Since i∈F2 i alg 2 alg a ⊥ (c − a), Lemma 2.1 a) implies that a ⊥ b.

Theorem 4.7. Assume that L is σ-complete. Let I be a set of cardinal ℵ0 and let (ai)i∈I be an orthogonal family in L. If ϕ is a bijection from N onto I and bj = aϕ(j) for all j ∈ N, then the sequence (bj)j∈h ;≤0i is orthogonal in L and L L N bj = ai. j∈N i∈I Proof. We shall show first that the sequence (b ) is orthogonal in L. Let j j∈hN;≤0i 0 0 n ∈ N and let F = {ϕ(j): j ∈ [0, n]0}. Then F is a finite non-empty subset 0 0 0 of I. Define a total ordering ≤ on F putting ϕ(j) < ϕ(k) if j, k ∈ [0, n]0 and 0 j <0 k. Then (ai)i∈hF 0;≤0i is an orthogonal suite in L. Since ϕ(0) = min F and 0 ϕ(k) = min(F \{ϕ(0), ϕ(1), . . . , ϕ(k − 1)}) for all k ∈ [1, n]0 and bj = aϕ(j) for all j ∈ [0, n]0, it follows from Remark 3.1 that (bj)0≤0j≤0n is an orthogonal suite in L, and therefore the sequence (b ) is orthogonal in L. j j∈hN;≤0i Moreover Mn M M bj = ai ≤alg ai j=0 i∈F 0 i∈I L L for all n ∈ . Whence bj ≤alg ai. N j∈N i∈I To show the other inequality, let F ∈ F(I) \ {∅} and let ≤ be a total ordering on F . Then ϕ−1(F ) is a finite non-empty subset of N. Let n = max ϕ−1(F ) in 0 0 0 hN; ≤0i. Put F = {ϕ(j): j ∈ [0, n]0}. Then F ∈ F(I) and F ⊇ F . Define as above the total ordering ≤0 on F 0. Then, using Theorem 3.9, Corollary 3.8 and Remark 3.1, we get M M M Mn M a = a ≤ a = b ≤ b i 0 i alg 0 0 i j alg j i∈hF ;≤i i∈hF ;≤ i i∈hF ;≤ i j=0 j∈N L L for all F ∈ F(I), and therefore ai ≤alg bj. i∈I j∈N 16 PEDRO MORALES AND FRANCISCO GARCÍA MAZARÍO

5. Orthogonality in eﬀect algebras

In this Section, L = hL; ⊕, 0, 1i denotes an eﬀect algebra. The following result will be useful for our presentation: Lemma 5.1. Let a, b, c ∈ L. Then

a) c ≤alg a

Proof. a) Since a = c⊕(a−c) and b = c⊕(b−c), we get c⊕(a−c)

Let a ∈ L \{0}. We define 0a = 0 and 1a = a. If n ∈ N \{0, 1} and (ai)1≤0i≤0n is an orthogonal suite in L of n elements such that ai = a for all i ∈ [1, n]0 we write Mn na = ai i=1 and we say that na is defined. If n ∈ N\{0, 1}, na is defined and na is not orthogonal to a, then n is called the isotropic index of a. If na is defined for all n ∈ N \{0, 1}, we say that a has an infinite isotropic index, and, in this case if ai = a for all i ∈ N, then the sequence (a ) is orthogonal in L. i i∈hN;≤0i For example, in L = R+[0, 1] every isotropic element has a finite isotropic index ≥0 2. For an example of an effect algebra with elements of infinite isotropic index, see [13, Example 4.6]. A non-empty subset L1 of L is said to be a sub-effect algebra of L if the following axioms hold:

(1) 0 and 1 belong to L1. 0 (2) If a ∈ L1, then a ∈ L1. (3) If a, b ∈ L1 and a ⊥ b, then a ⊕ b ∈ L1.

Clearly, in the induced structure, L1 is an eﬀect algebra in its own right. As such, if L1 is a Boolean algebra (resp. orthomodular poset), we call it a Boolean subalbegra (resp. sub-orthomodular poset) of L. + For example, if L = R [0, 1], then L1 = Q ∩ [0, 1] is a sub-eﬀect algebra of L and L2 = {0, 1} is a Boolean subalgebra of L. A non-empty subset C of L is say to be a pairwise orthogonal set in L if a, b ∈ C and a 6= b imply a ⊥ b. √ For example, if L = +[0, 1] and C = {0, √1 , √2−1 }, then C is a pairwise R 2 2 orthogonal set in L. Assume that L is an orthoalgebra and let C be a non-empty subset of L. Recall that ORTHOGONALITY IN PARTIAL ABELIAN MONOIDS 17

(1) C is a compatible set in L if there exists a Boolean subalgebra L1 of L such that C ⊆ L1. (2) C is a jointly orthogonal set in L if C is a pairwise orthogonal and compatible set in L.

For example, if L is an orthoalgebra, n ∈ N \{0, 1} and (ai)i≤0i≤0n is an orthogonal suite in L \{0}, then C = {ai : i ∈ [1, n]0} is a jointly orthogonal set in L by [29, Theorem 2.21]. Theorem 5.2. Assume that L is an orthoalgebra and let C be a pairwise orthogonal set in L. Then C is a jointly orthogonal set in L if and only if C is contained in a sub-orthomodular poset of L. Proof. The necessity is trivial. To prove the sufficiency, suppose that there exists a sub-orthomodular poset L1 of L such that C ⊆ L1. Let F ∈ F(C) \ {∅}. Then F is a finite pairwise orthogonal set in L1 and [21, Corollary 2.11] implies that the sub-orthoalgebra L2 of L1 generated by F is Boolean. So L2 is a Boolean subalgebra of L such that F ⊆ L2, and therefore F is a jointly orthogonal set in L. By [21, Theorem 3.4] it follows that C is a jointly orthogonal set in L. Corollary 5.3. Assume that L is an orthomodular poset. Then a non-empty subset C of L is a jointly orthogonal set in L if and only if C is a pairwise orthogonal set in L. Example 5.4. Consider the set L = {0, 1}N with zero 0 = (0, 0, 0,..., 0,...) and unit 1 = (1, 1, 1,..., 1,...). For two elements e = (ei)i∈N and f = (fi)i∈N of L we define e ⊥ f if eifi = 0 for all i ∈ N and, in this case, we put e ⊕ f = (ei + fi)i∈N. Then it is easy to show that L = hL; ⊕, 0, 1i is an orthomodular poset. For every n ∈ N consider the element an of L defined by the formula an = (δni)i∈N where δni = 0 if i 6= n and δnn = 1. It is clear that an 6= am if and only if n 6= m. We shall show that C = {an : n ∈ N} is a jointly orthogonal set in L \{0}. In fact, let n, m ∈ N be such that n 6= m. If i = n, then i 6= m and therefore δmi = 0; if i = m, then i 6= n and therefore δni = 0; if i 6= n and i 6= m, then δni = 0 and δmi = 0. Since this implies that δniδmi = 0 for all i ∈ N, it follows that an ⊥ am. Then Corollary 5.3 implies that C is a jointly orthogonal set in L and it is trivial that an 6= 0 for all n ∈ N. Now we extend the notion of difference set considered in [14] and [32] for orthoalgebras and we establish its main properties. Let n ∈ N \{0} and let D be a finite subset of L \{0} such that card(D) = n. We say that D is a difference set in L of cardinality n if there exists a finite strictly increasing family (b ) in hL; ≤ i such that D = {b − b : i ∈ [1, n] } and, i i∈[0,n]0 alg i i−1 0 in this case, we say that the family (b ) yields the difference set D. i i∈[0,n]0 + 1 For example, if L = R [0, 1] and D = {a, 2a, 3a, 4a} where a = 10 , then D is a difference set in L of cardinality four, because

Lemma 5.5. Every diﬀerence set in L is a pairwise orthogonal set in L \{0}.

Proof. Let n ∈ N \{0} and let D be a difference set in L of cardinality n yielded by the finite strictly increasing family (b ) in hL; ≤ i. Then D = {b − b : i i∈[0,n]0 alg i i−1 i ∈ [1, n]0}. By Theorem 3.15 it follows that (bi − bi−1)1≤0i≤0n is an orthogonal suite in L \{0} and since card(D) = n we have bi − bi−1 6= bj − bj−1 if i 6= j. Then if J = {i, j}, Corollary 3.8 implies that the suite (bk − bk−1)k∈hJ;≤0i is orthogonal in L, and therefore bi − bi−1 ⊥ bj − bj−1. Lemma 5.6. Let n ∈ N \{0, 1}, let D be a difference set of cardinality n and let d ∈ D. Then D \{d} is a difference set in L of cardinality n − 1. Proof. Suppose that D is yielded by the finite strictly increasing family (b ) i i∈[0,n]0 in hL; ≤algi. Then there exists k ∈ [1, n]0 such that d = bk − bk−1. Define

cj = bj if j ∈ [0, k − 1]0 and cj = bj+1 − d if j ∈ [k, n − 1]0 with [k, n − 1] = ∅ if k = n. By Lemma 5.1, (c ) is a finite strictly 0 j j∈[0,n−1]0 increasing family in hL; ≤algi and D \{d} = {cj − cj−1 : j ∈ [1, n − 1]0}. Hence D \{d} is a difference set in L of cardinality n − 1. Corollary 5.7. Every non-empty subset of a difference set in L is a difference set in L.

Lemma 5.8. Let n ∈ N \{0} and let D be a difference set of cardinality n. If (b ) and (c ) are finite strictly increasing families in hL; ≤ i both of i i∈[0,n]0 j j∈[0,n]0 alg which yield the difference set D, then bn − bo = cn − c0.

Proof. We may suppose that n ≥0 2. Write di = bi − bi−1 and ei = ci − ci−1 for all i ∈ [1, n]0. By Theorem 3.15 Ln (di)1≤0i≤0n and (ei)1≤0i≤0n are orthogonal suites in L \{0} such that i=1 di = Ln bn − b0 and i=1 ei = cn − c0. Since

D = {di : i ∈ [1, n]0} = {ei : i ∈ [1, n]0} there exists a bijection σ from [1, n]0 onto [1, n]0 such that ei = dσ(i) for all i ∈ −1 −1 [1, n]0. Define a total ordering ≤ on [1, n]0 putting i < j if σ (i) <0 σ (j). Then by Corollary 3.10 it follows that (di)i∈h[1,n] ;≤i is an orthogonal suite in L Ln L 0 and i=1 di = i∈h[1,n] ;≤i di. Since [1, n]0 = {σ(i): i ∈ [1, n]0} and σ(1) < 0 L Ln Ln σ(2) < ··· < σ(n), we have di = dσ(i) = ei = cn − c0 and i∈h[1,n]0;≤i i=1 i=1 therefore bn − b0 = cn − c0. Let n ∈ N \{0} and let D be a difference set in L of cardinality n yielded by the finite strictly increasing family (b ) in hL; ≤ i. We define i i∈[0,n]0 alg M D = bn − b0 noting that L D is well-defined by Lemma 5.8. If we consider the void subset ∅ of L as a difference set in L of cardinality zero, we can define M ∅ = 0. ORTHOGONALITY IN PARTIAL ABELIAN MONOIDS 19

+ 1 1 5 For example, if L = R [0, 1] and D = { 4 , 3 , 12 }, then it is easy to see that D is a diﬀerence set in L of cardinality three with L D = 1.

Lemma 5.9. Let n ∈ N \{0, 1} and let D be a difference set in L of cardinality n. If d ∈ D, then (D \{d}) ⊥ d and L D = L(D \{d}) ⊕ d. Proof. Let (b ) be a finite strictly increasing family in hL; ≤ i which yields i i∈[0,n]0 alg the difference set D in L and let k ∈ [1, n]0 be such that d = bk − bk−1. Suppose that k = n. Then D \{d} = {bi − bi−1 : i ∈ [1, n − 1] } and therefore L 0 (D \{d}) = bn−1 − b0. Since b0

D \{d} = {cj − cj−1 : j ∈ [1, n − 1]0} where c0 = b0, c1 = b1,..., ck−1 = bk−1, ck = bk+1 − d,..., cn−1 = bn − d. So L (D \{d}) = (bn − d) − b0. Since d ⊥ bn − d and (bn − d) − b0 ≤alg bn − d, we get L by Lemma 2.1 a) that d ⊥ (bn − d) − b0 and therefore d ⊥ (D \{d}). Moreover, by [12, Lemma 4.8 (iv)] it follows that (bn − d) − b0 = (bn − b0) − d and therefore L L (D \{d}) ⊕ d = D. Lemma 5.10. Let n ∈ N \{0}, let D be a difference set in L of cardinality n and let d be an element of L \{0} such that d 6∈ D and d ⊥ L D. Then D ∪ {d} is a difference set in L of cardinality n + 1 and L(D ∪ {d}) = (L D) ⊕ d. Proof. Let (b ) be a finite strictly increasing family in hL; ≤ i which yields i i∈[0,n]0 alg the difference set D. Then

D = {bi − bi−1 : i ∈ [1, n]0} L and D = bn − b0. Deﬁne c0 = 0, c1 = b1 − b0,..., cn = bn − b0 and cn+1 = L (bn − b0) ⊕ d because d ⊥ D. Then ci−1

Lemma 5.11. Let I = hI; ≤i be a finite total ordered set with card(I) = n ≥0 2 and let (ai)i∈I be a family of pairwise different elements of L \{0}. Then D = {ai : i ∈ I} is a difference set in L of cardinality n if and only if (ai)i∈hI;≤i is an L L orthogonal suite in L. Furthermore, D = i∈F ai.

Proof. It remains to prove the necessity. Write i1 = min I and ip = min(I \

{i1, i2, . . . , ip−1}) for all p ∈ [2, n]0 and ck = aik for all k ∈ [1, n]0. By Remark 3.1, Ln L (ck)1≤0k≤0n is an orthogonal suite in L\{0} such that k=1 ck = i∈I ai. Clearly,

D = {ck : k ∈ [1, n]0}. Put Mk b0 = 0 and bk = ci for all k ∈ [1, n] . i=1 0 20 PEDRO MORALES AND FRANCISCO GARC´IA MAZAR´IO

Then (b ) is a finite strictly increasing family in hL; ≤ i such that c = i i∈[0,n]0 alg k bk − bk−1 for k ∈ [1, n] . So D = {bk − bk−1 : k ∈ [1, n] } is a difference set in L of 0 L L 0 cardinality n such that D = i∈I ai. Lemma 5.12. If D and E are two difference sets in L such that D ∩ E = ∅ and L D ⊥ L E, then D ∪ E is also a difference set in L such that L(D ∪ E) = (L D) ⊕ (L E).

Proof. Let n, m ∈ N \{0} and suppose that card(D) = n and card(E) = m. Then we can write D = {d1, d2, . . . , dn} and E = {e1, e2, . . . , em} and since D ∩ E = ∅, the n + m elements of L \{0} d1, d2, . . . , dn, e1, e2, . . . , em are pairwise diﬀerent and its union gives the set D ∪ E. Moreover, the suites (di)1≤0i≤0n and (ej)1≤0j≤0m are orthogonal in L as was indicated in the proof of Lemma 5.5. Deﬁne

c1 = d1, . . . , cn = dn, cn+1 = e1, . . . , cn+m = em and put I1 = [1, n]0 and I2 = [n + 1, n + m]0. Then [1, n + m]0 = I1∪I2, I1∩I2 = ∅, (c ) and (c ) are orthogonal suites in L \{0} such that L c ⊥ i i∈hI1;≤0i i i∈hI2;≤0i i∈I1 i L c . By Theorem 3.5 it follows that (c ) is an orthogonal suite i∈I2 i i i∈h[1,n+m]0;≤0i in L such that Mn+m M M ci = ci ⊕ ci . i=1 i∈I1 i∈I2 Then Lemma 5.11 implies that D ∪ E is a difference set in L of cardinality n + m L L L such that (D ∪ E) = ( D) ⊕ ( E). Theorem 5.13. Let X be an infinite cardinal number. Consider the following statements: i) L is X-complete. ii) For every pairwise orthogonal set C in L\{0} with card(C) ≤c X and whose finite subsets are difference sets in L there exists the supremum W(L D | D ∈ F(C)) in hL; ≤algi. Then i) implies ii). Further, if L is an orthoalgebra, then ii) implies i).

Proof. i) ⇒ ii). Let C be a pairwise orthogonal set in L \{0} with card(C) ≤c X and whose finite subsets are difference sets in L. For every i ∈ C put ai = i. Then (ai)i∈C is a family of elements of L \{0} such that C = {ai : i ∈ C}. Let D ∈ F(C) \ {∅} and let ≤ be a total ordering on D. Then D = {ai : i ∈ D} is a difference set in L, and therefore the suite (ai)i∈hD;≤i is orthogonal in L. So (ai)i∈C is an orthogonal family in L and since L is X-complete, we deduce that the W L supremum ( D | D ∈ F(C)) exists in hL; ≤algi. ii) ⇒ i). Let (ai)i∈I be an orthogonal family in L \{0} such that card(I) ≤c X. Then the set C = {ai : i ∈ I} is a pairwise orthogonal set in L \{0} and since L is an orthoalgebra it follows that (ai)i∈I is a family of pairwise different elements of L \{0}. Then it follows from Lemma 5.11 that every finite subset of C is a difference set in L and therefore, by the hypothesis, there exists the supremum W L ( D | D ∈ F(C)) in hL; ≤algi. W L It remains to show that the ⊕-join of the orthogonal family (ai)i∈I is ( D | W L L D ∈ F(C)). Clearly ( D | D ∈ F(C)) is an upper bound of the set { i∈F ai : ORTHOGONALITY IN PARTIAL ABELIAN MONOIDS 21

L F ∈ F(I)}. Suppose u ∈ L \{0} and i∈F ai ≤alg u if F ∈ F(I) and ≤ is a total ordering of F . Then D = {ai : i ∈ F } is a difference set in L such that L L W L D = i∈F ai, and therefore ( D | D ∈ F(C)) ≤alg u. Note that if L is an orthoalgebra and C is a finite non-empty jointly orthogonal set in L, then D = C \{0} is a difference set in L, and we can define M M C = D. Let X be an infinite cardinal number. Following [21] an orthoalgebra L is said to be X-orthosummable if, for every jointly orthogonal set C in L with card(C) ≤c W L X, there exists the supremum ( F | F ∈ F(C)) in hL; ≤algi. If L is ℵ0- orthosummable we call it a σ-orthoalgebra. Corollary 5.14. Assume that L is an orthoalgebra and let X be an infinite cardinal number. Then the following statements are equivalent: i) L is X-complete. ii) L is X-orthosummable

Proof. It follows from Theorems 5.2 and 5.13 and [32, Theorem 2.1]. Theorem 5.15. Assume that L is an orthoalgebra. Then the following statements are equivalent: i) L is a σ-orthoalgebra. ii) L is σ-complete. iii) Every countable chain in L has a supremum in hL; ≤algi.

Proof. It follows from Corollary 5.14 and Theorem 4.3. Remark 5.16. Theorem 5.15 contains [21, Theorem 4.14].

As an illustration we prove that the effect algebra L = R+[0, 1] is complete. Let X be an infinite cardinal number and let (ai)i∈I be an orthogonal family in L \{0} with card(I) ≤c X. Let F ∈ F(I) \ {∅} and let ≤ be a total ordering on F . Then L P (ai)i∈hF ;≤i is an orthogonal suite in L, and therefore i∈F ai = i∈F ai ≤alg 1. P o So ai ≤ 1 for all F ∈ F(I). By [7, Chapitre IV, §7, n 1, Théorème 1, p. i∈F L P 171], the family (ai)i∈I is summable in R and i∈I ai = i∈I ai ≤alg 1. So the ⊕-join of the orthogonal family (ai)i∈I exists in hL; ≤algi. Then L is X-complete.

6. The support of a measure

Let L = hL; ⊕, 0i be a partial abelian monoid and let M = hM; +, 0i be a commutative monoid. An element µ of M L is said to be a measure on L with values in M if µ(0) = 0 and µ(a ⊕ b) = µ(a) + µ(b) whenever a, b ∈ L and a ⊥ b. We denote by a(L, M) the set of all measures on L with values in M. If µ ∈ a(L, M), we note the following additional properties of µ: (1) µ is finitely additive, that is, if hI; ≤i is a finite non-empty totally ordered set L P and (ai)i∈hI;≤i is an orthogonal suite in L, then µ i∈I ai = i∈I µ(ai). 22 PEDRO MORALES AND FRANCISCO GARCÍA MAZARÍO

(2) Suppose that L is cancellative. If a, b ∈ L and a ≤alg b, then µ(b) = µ(a) + µ(b − a). We consider the following interesting examples:

Example 6.1. Consider the partial abelian monoid L = hR; ⊕, 0i of Example 2.6 and the commutative monoid M = hM; +, 0i of Example 2.7. Deﬁne µ : L → M by the formulae: µ(a) = th−1 a if a ∈ [0, 1[ µ(1) = +∞ µ(a) = 0 if a ∈ R \ [0, 1]. Then the additive formula for hyperbolic tangent implies that µ is a measure on L with values in M. Example 6.2. Recall that a reﬁnement monoid is a commutative monoid M = hM; +, 0i satisfying the following axiom

(Refinement Law) For all a0, a1, b0, b1 ∈ M such that a0 +a1 = b0 +b1 there exist elements cij ∈ M (i, j = 0, 1) with ai = ci0 + ci1 and bi = c0i + c1i for all i = 0, 1. For example, the usual commutative monoid hN ∪ {+∞}; +, 0i is a refinement monoid. Following [10] an element µ ∈ a(L, M) is a V -measure if M is a conical refinement monoid and the following axioms hold: (1) If a ∈ L and µ(a) = 0, then a = 0. (2) If a ∈ L, x1, x2 ∈ M and µ(a) = x1 + x2, then there exist a1, a2 ∈ L such that a1 ⊥ a2, a = a1 ⊕ a2 and µ(ai) = xi for i = 1, 2. For example, if X is any non-empty set, then the counting measure over X is a V -measure (see [11, p. 54]). Following [3] we call po-monoid a commutative monoid M = hM; +, 0i endowed with a partial ordering ≤ on M such that a, b, c ∈ M and a ≤ b imply a + c ≤ b + c, and we write M = hM; +, 0, ≤i. For example, every ordered abelian group is a po- monoid. Moreover, consider the submonoid hR+ ∪ {+∞}; +, 0i of the commutative monoid given in the Example 2.7. Then hR+ ∪ {+∞}; +, 0, ≤algi is a po-monoid by Lemma 2.1. Let M = hM; +, 0, ≤i be a po-monoid and let µ ∈ a(L, M). Then we say that µ is a positive measure if 0 ≤ µ(a) for all a ∈ L. Clearly, if µ is a positive measure on L, then µ is an increasing function from hL; ≤algi into hM; ≤i. Unexplained topological terminology is that of [26]. Following [35] a uniform semigroup is a commutative monoid S = hS; +, 0i endowed with a uniformity U for S such that the function (a, b) → a+b is uniformly continuous from S × S into S, and it is denoted by hS; +, 0, Ui. Examples 6.3. 1) Every abelian topological group is a uniform semigroup by [7, Chapitre III, §3, no 3, Théorème 2]. ORTHOGONALITY IN PARTIAL ABELIAN MONOIDS 23

2) Consider the commutative monoid S∞ = hR+ ∪{+∞}; +, 0i and let U∞ be the uniformity for S∞ generated by the metric p∞ deﬁned by the formula p∞(x, y) = x y +∞ | 1+x − 1+y | with the convention 1+(+∞) = 1. Noting that p∞ satisﬁes the following 0 0 0 0 0 0 semi-invariant property: p∞(x+y, x +y ) ≤ p∞(x, x )+p∞(y, y ) for all x, x , y, y ∈ S∞, we can show that hS∞; +, 0, U∞i is a uniform semigroup. In fact, let U ∈ U∞. We may suppose that U = U = {(x, y): p∞(x, y) < } for some > 0. Since the elements of a base for the product uniformity for S∞ × S∞ are of the form W (U, V ) = {(x, y), (x0, y0)) : (x, x0) ∈ U and (y, y0) ∈ V }

0 0 it suffices to choose W = W (U 1 ,U 1 ) to have f(W ) ⊆ U where f((x, y), (x , y )) = 2 2 (x + y, x0 + y0). 3) Let G be an abelian topological group and let N (0) be the neighbourhood system of the identity 0. On the set S = 2G \ {∅} consider the binary operation S1 + S2 = {x + y : x ∈ S1 and y ∈ S2}. Then S = hS; +, {0}i is a commutative monoid. For every V ∈ N (0) put U(V ) = {(S1,S2) ∈ S × S : Si ⊆ S3−i for i = 1, 2}. Then it is easy to show that the set {U(V ): V ∈ N (0)} is a base for a uniformity U for S such that hS; +, {0}, Ui is a uniform semigroup. Let S = hS; +, 0, Ui be a uniform semigroup. It is well known that there exists a family P of pseudo-metrics p on S which are semi-invariant and if U(p, ) = {(x, y) ∈ S × S : p(x, y) < }, then the set {U(p, ): p ∈ P, > 0} is a subbase for the uniformity U (see [15], [36] and [37]). Clearly if τ denotes the topology generated by U, then the topological space (S, τ) is Hausdorff if and only if the condition p(x, y) = 0 for all p ∈ P and x, y ∈ S imply x = y. In this case we say that the uniform semigroup S is Hausdorff. Let S = hS; +, 0, Ui be a Hausdorff uniform semigroup and let τ be the topology generated by U. Following [16], a family (ai)i∈I of elements of L is summable in P P S if the net ( ai)F ∈hF(I);⊇i is τ-convergent and we write τ-limF ai = P i∈F i∈F i∈I ai. It is easy to prove that a family (ai)i∈I in S is summable to an element s ∈ S if and only if, for every p ∈ P and every > 0, there exists F0 ∈ F(I) P such that F ∈ F(I) and F ⊇ F0 imply p( i∈F ai, s) < . Moreover, if (ai)i∈I and (bi)i∈I are summable families in S, then the family (ai + bi)i∈I is summable in S P P P and i∈I (ai + bi) = i∈I ai + i∈I bi.

Lemma 6.4. Let (ai)i∈I be a family of elements of S and let I1 and I2 be non- empty and disjoint sets whose union is I. If the subfamilies (ai)i∈I1 and (ai)i∈I2 are summable in S, then (a ) is also summable in S and P a = P a + i i∈I i∈I i i∈I1 i P a . i∈I2 i Proof. Let p ∈ P and let > 0. Put X sk = ai for k = 1, 2. i∈Ik 0 0 Then there exists Fk ∈ F(Ik) such that Fk ∈ F(Ik) and Fk ⊇ Fk imply X 1 p ai, sk < for k = 1, 2. i∈Fk 2 24 PEDRO MORALES AND FRANCISCO GARC´IA MAZAR´IO

0 0 Let F0 = F1 ∪ F2 and let F ∈ F(I) be such that F ⊇ F0. Then F ∩ Ik ∈ F(Ik), 0 F ∩ Ik ⊇ Fk (k = 1, 2), F = (F ∩ I1) ∪ (F ∩ I2) and (F ∩ I1) ∩ (F ∩ I2) = ∅. So X X X p ai, s1 + s2 = p ai + ai, s1 + s2 i∈F i∈F ∩I1 i∈F ∩I2 X X ≤ p ai, s1 + p ai, s2 i∈F ∩I1 i∈F ∩I2 1 1 < + 2 2 = . P Hence the family (ai)i∈I is summable in S and i∈I ai = s1 + s2.

Lemma 6.5. Let (xi)i∈I be a summable family in S, let σ be a bijection from an index set K onto I and let yk = xσ(k) for all k ∈ K. Then (yk)k∈K is a summable P P family in S and k∈K yk = i∈I xi. Proof. Let p ∈ P and let > 0. Put X s = xi. i∈I P Then there exists F0 ∈ F(I) such that F ∈ F(I) and F ⊇ F0 imply p i∈F xi, s < −1 . Let H0 = σ (F0). Then H0 ∈ F(K). Let H ∈ F(K) be such that H ⊇ H0. Then σ(H) ∈ F(I) and σ(H) ⊇ σ(H0) = F0, and therefore X p xi, s < . i∈σ(H)

Write H = {k0, k1, . . . , kn} where n ∈ N. Then σ(H) = {σ(k0), σ(k1), . . . , σ(kn)} and the generalized commutative law for hS; ⊕, 0i imply that X Xn Xn X yj = yk = xσ(k ) = xi j∈H i=0 i i=0 i i∈σ(H) and therefore X p yj, s < . j∈H P Consequently, the family (yk)k∈K is summable in S and k∈K yk = s. Let L = hL; ⊕, 0i be a partial abelian monoid, let S = hS; +, 0, Ui be a Hausdorff uniform semigroup and let µ ∈ a(L, S). We say that (1) µ is s-bounded if τ-lim µ(a ) = 0 for every orthogonal sequence (a ) n n n n∈hN;≤0i in L. (2) µ is countably additive if, for every orthogonal sequence (ai)i∈h ;≤ i such L N 0 that the ⊕-join i∈ ai exists in hL; ≤algi, the family (µ(ai))i∈N is sum- LN P mable in S and µ ai = µ(ai). i∈N i∈N (3) µ is completely additive if, for every orthogonal family (ai)i∈I in L such that L the ⊕-join i∈I ai exists in hL; ≤algi, the family (µ(ai))i∈I is summable in L P S and µ i∈I ai = i∈I µ(ai). For a nice example of an element of a(L, S), where L is an effect algebra and S is a Hausdorff abelian topological group, see [9, Example 4.4 (2)]. ORTHOGONALITY IN PARTIAL ABELIAN MONOIDS 25

Remarks 6.6. (1) It is clear that every element of a(L, S) which is completely additive is also countably additive. (2) If S is a Hausdorff abelian topological group and L is a σ-complete effect algebra, then every element of a(L, S) which is countably additive is also s-bounded (see [18, Lemma 3.1]). But it is not true if S is a uniform semigroup. For example, consider the commutative monoid hN ∪ {+∞}; +, 0i and let U be the uniformity for N∪{+∞} determined by the semi-invariant x y +∞ metric p(x, y) = | 1+x − 1+y | with the convention 1+(+∞) = 1. Then hN ∪ {+∞}; +, 0, Ui is a Hausdorff uniform semigroup. Let the usual complete Boolean algebra L = h2N; ∩, ∪i and consider the counting measure µ over N. Then µ is countably additive, but not s-bounded because µ({n}) = 1 for any n ∈ and ({n}) is an orthogonal sequence in N n∈hN;≤0i L. (3) There are also elements of a(L, S) which are s-bounded but not countably additive, even if L is a complete effect algebra and S is a Hausdorff abelian topological group (see [18, Example 3.2]). A uniform semigroup S = hS; +, 0, Ui is said to be ordered if there exists a partial ordering ≤ on S such that hS; +, 0, ≤i is a po-monoid and we write S = hS; +, 0, U, ≤i. For example, hS∞; +, 0, U∞, ≤algi is an ordered uniform semigroup. Recall that, following [23], an ordered topological group is an ordered abelian group which is also a topological group. For nice examples of positive measures on an orthoalgebra with values in an ordered Hausdorff topological group, see [29, Examples 5.1]. In the remainder of this Section, L = hL; ⊕, 0, 1i denotes an effect algebra and S = hS; +, 0, U, ≤i denotes an ordered Hausdorff uniform semigroup. If µ ∈ a(L, S), then the subset of L ker µ = {a ∈ L : µ(a) = 0} is called the kernel of µ and clearly 0 ∈ ker µ. Let µ be a positive element of a(L, S). Following [33], we say that (1) µ is a Jauch-Piron measure if every finite subset M of ker µ has an upper bound in ker µ. (2) µ has a support if there exists an element s ∈ L such that ker µ = [0, s0]. The support of µ, when it exists, is unique, and will be denoted by supp(µ).

+ Example 6.7. For every i ∈ N consider the effect algebra Li = R [0, 1] and let Q L = Li with zero 0 = (0, 0, 0,...) and unit 1 = (1, 1, 1,...). Choose the usual i∈N ordered Hausdorff uniform semigroup S = S∞. For every element a = (ai)i∈N of L put ∞ X µ(a) = ai. i=0 Clearly, µ is a positive measure of a(L, S). Moreover, since ker µ = {0}, it follows that µ is a Jauch-Piron measure with support 1. Example 6.8. Let H be a complex Hilbert space of dimension not less than 3 with inner product (· | ·), let I be the identity operator on H, let Bsa(H) be the 26 PEDRO MORALES AND FRANCISCO GARCÍA MAZARÍO additive abelian group of all self-adjoint operators on H ordered by the positive cone + + Bsa(H) = {A ∈ Bsa(H):(Aϕ | ϕ) ≥ 0 for all ϕ ∈ H}. Clearly, 0,I ∈ Bsa(H) and 0 < I. Then we can define the effect algebra + E(H) = Bsa(H) [0,I] called the standard effect algebra on H. Let L = G(H) be the sub-effect algebra of E(H) consisting of all idempotent elements in E(H). Let ϕ ∈ H be such that kϕk = 1 and let S be the usual ordered additive topological group of real numbers. For every operator P in L put µ(P ) = (P ϕ | ϕ). Then it is easy to see that µ is a positive measure on L with values in S. Moreover, ker µ = {P ∈ L :(P ϕ | ϕ) = 0} and since (P ϕ | ϕ) = (P 2ϕ | ϕ) = (P ϕ | P ∗ϕ) = (P ϕ | P ϕ) we get ker µ = {P ∈ L : P ϕ = 0}. Let Q be the element of L defined by the formula Qψ = (ψ | ϕ)ϕ for all ψ ∈ H. Let P ∈ ker µ. Since (QP )ψ = Q(P ψ) = (P ψ | ϕ)ϕ = (ψ | P ϕ)ϕ = 0 we get that (I − Q)P = P . Then [2, Theorem 4.5, p. 112] implies that P ≤ I − Q, and therefore ker µ = [0,I − Q]. So the support of µ exists and it is equal to Q, which is a one-dimensional projector acting in H. The main results of this Section are the following: Theorem 6.9. Assume that L is complete and the uniform space (S, U) is metrizable. If µ is a positive s-bounded countably additive element of a(L, S) with support, then µ is a completely additive Jauch-Piron measure. Proof. Write s = supp(µ). Let M be a finite subset of ker µ. So M ⊆ ker µ = [0, s0], 0 0 and therefore µ(s ) = 0 and m ∈ M implies m ≤alg s . Hence µ is a Jauch-Piron measure. ORTHOGONALITY IN PARTIAL ABELIAN MONOIDS 27

Let (ai)i∈I be an orthogonal family in L \{0}. Since L is a complete effect algebra, there exists the ⊕-join of the orthogonal family (ai)i∈I in hL; ≤algi. Write L a = i∈I ai and J = {i ∈ I : µ(ai) 6= 0}. We may suppose that J is infinite. Since the uniform space (S, U) is metrizable, it follows from [26, 13 Metrization Theorem, p. 186] that the uniformity U has a countable base {Un : n ∈ N}. Then {Un[0] : n ∈ N} is a countable base for the neighbourhood system N (0) of the identity 0. For every n ∈ N, let Jn = {i ∈ J : µ(ai) ∈/ Un[0]}. We shall show that every Jn is finite. Suppose that there exists m ∈ N such that Jm is infinite. By [6, o §6, n 1, Lemma 1, p. 67] there exists a countably infinite subset K of Jm. Let σ be a bijection from N onto K. Write bk = aσ(k) for all k ∈ N. Then (bk)k∈N is a subfamily of (a ) , and therefore (b ) is an orthogonal sequence in L\{0}. i i∈I k k∈hN;≤0i Since µ is s-bounded, it follows that τ-limk µ(bk) = 0, a contradiction with the fact that µ(bk) ∈/ Um[0] for all k ∈ N. T∞ Since S is a Hausdorff uniform semigroup, it follows that n=0 Un[0] = {0}, and S∞ L therefore J ⊆ Jn. Hence J is a countably infinite set. Write b = ai and L n=0 i∈J c = i∈I\J ai. By Corollary 4.6 we have b ⊥ c and a = b⊕c. So µ(a) = µ(b)+µ(c). Let F ∈ F(I \ J). By the finite additivity of µ it follows that M X µ ai = µ(ai) = 0. i∈F i∈F L 0 W L Hence i∈F ai ≤alg s for all F ∈ F(I\J). So c = i∈F ai | F ∈ F(I \ J) ≤alg 0 0 s . Since µ is positive and µ(s ) = 0, we have µ(c) = 0. Moreover, since µ(ai) = 0 for all i ∈ I \ J, it follows that the family (µ(ai))i∈I\J is trivially summable in S P and i∈I\J µ(ai) = 0 = µ(c). On the other hand, let ϕ be a bijection from N onto J and put bk = aϕ(k) for all k ∈ N. Then by Theorem 4.7 it follows that (bk)k∈h ;≤0i is an orthogonal sequence L L N in L \{0} such that k∈ bk = j∈J aj = b. Since µ is countably additive, we N P get that the family (µ(bk))k∈ is summable in S and µ(bk) = µ(b). Put N k∈N −1 xk = µ(bk) for all k ∈ N and σ = ϕ . Then σ is a bijection from J onto N. Let yj = xσ(j) for all j ∈ J. So yj = µ(bσ(j)) = µ(aj) for all j ∈ J. By Lemma 6.5 it P follows that the family (µ(aj))j∈J is summable in S and j∈J µ(aj) = µ(b). Then P Lemma 6.4 implies that the family (µ(ai))i∈I is summable in S and i∈I µ(ai) = µ(b) + µ(c) = µ(a) proving that µ is completely additive. Theorem 6.10. Assume that L is complete and let µ be an element of a(L, S). If µ is a positive completely additive Jauch-Piron measure, then µ has a support. Proof. We may suppose that there exists a ∈ L \{0} such that µ(a) = 0. Consider the partially ordered set (C, ⊆), where C = {C ∈ 2L\{∅} : C is a pairwise orthogonal set in L \{0} whose finite subsets are difference sets in L such that µ(b) = 0 for all b ∈ C}. Since C = {a} ∈ C, the set C is non-empty. Let C0 be a totally S ordered subset of C and let C0 = {C : C ∈ C0}. Clearly C0 is a non-empty subset of L \{0} such that µ(b) = 0 for all b ∈ C0. Let D be a finite subset of C0. Since hC0; ⊆i is a totally ordered set, there exists C ∈ C0 such that D ⊆ C, and therefore D is a difference set in L. To show that C0 is a pairwise orthogonal set in L \{0}, let a1, a2 ∈ C0 be such that a1 6= a2. Then there exist C1,C2 ∈ C0 28 PEDRO MORALES AND FRANCISCO GARCÍA MAZARÍO such that a1 ∈ C1, and a2 ∈ C2. We may suppose that C1 ⊆ C2. Then a1, a2 ∈ C2 and therefore a1 ⊥ a2. Consequently, C0 is an upper bound of the set C0 and by Zorn Lemma, the partially ordered set hC; ⊆i contains a maximal element E0.

For every i ∈ E0 put ai = i. Then (ai)i∈E0 is a family of elements of L \{0} such that E0 = {ai : i ∈ E0}. Since L is complete, the proof of Theorem 5.13 implies that the family (a ) is orthogonal in L and there exists s = L a i i∈E0 i∈E0 i in hL; ≤algi. Since µ is completely additive, the family (µ(ai))i∈E0 is summable in S and P µ(a ) = µ(s). Since µ(a ) = 0 for all i ∈ E we get µ(s) = 0, and i∈E0 i i 0 therefore s ∈ ker µ. We shall show that s0 = supp(µ). For this we must prove that ker µ = [0, s]. Let c ∈ ker µ. Choose M = {s, c}. Since µ is a Jauch-Piron measure, there exists d ∈ ker µ such that s ≤alg d and c ≤alg d. Write e = d − s. Then e ⊥ s. Since ai ≤alg s for all i ∈ E0, it follows from Lemma 2.1 a) that e ⊥ ai for all i ∈ E0. Moreover, since µ is positive, µ(d) = 0 and e ≤alg d, it follows that µ(e) = 0. Since c ≤alg d, to prove that c ≤alg s, it suffices to establish that e = 0. Suppose that e 6= 0 and let E = E0 ∪ {e}. Then E is a pairwise orthogonal set in L \{0} such that µ(b) = 0 for all b ∈ E. Moreover, let D ∈ F(E). Then L D \{e} ∈ F(E0) is a difference set in L such that (D \{e}) ≤alg s and therefore L(D \{e}) ⊥ e. By Lemma 5.10 it follows that D is a difference set in L. Then the set E belongs to C and strictly contains E0. This contradicts the maximality of E0. So e = 0 and the proof is complete. Remark 6.11. It is clear that Theorem 6.10 is a twofold generalization of [29, The- orem 5.7] and Theorem 6.9 is a twofold generalization of [29, Theorem 5.6] for the case of an ordered metrizable topological group. Consequently, they are improve- ments of an interesting result of Maeda [28, Proposition 1.11, p. 239].

References [1] M. K. Bennet and D. J. Foulis, Phi-symmetric effect algebras, Found. Phys. 25 (1995), 1699–1722. [2] L. Beran, Orthomodular Lattices-Algebraic Approach, Academia, Praha, 1984. [3] G. Birkhoff, Lattice Theory, Third (New) Edition, Amer. Math. Soc. Coll. Publ. 25, Providence, R.I., 1967. [4] N. Bourbaki, Eléments´ de Mathématique, Algèbre, Chapitre II , Hermann, Paris, 1955. [5] N. Bourbaki, Eléments´ de Mathématique, Algèbre, Chapitre 1 , Hermann, Paris, 1958. [6] N. Bourbaki, Eléments´ de Mathématique, Thèorie des Ensemples, Chapitre 3 , Hermann, Paris, 1963. [7] N. Bourbaki, Eléments´ de Mathématique, Topologie Générale, Chapitres 1-4 , Hermann, Paris, 1971. [8] P. Crawley and R. P. Dilworth, Algebraic Theory of Lattices, Prentice Hall, Englewood Cliffs, N.J., 1973. [9] P. de Lucia and P. Morales, Measures on effect algebras, Atti Sem. Mat. Fis. Univ. Modena 51 (2003), 273–293. [10] H. Dobbertin, Refinement monoids, Vaught monoids, and Boolean algebras, Math. Ann. 265 (1983), 473–487. [11] H. Federer, Geometric Measure Theory, Springer Verlag, New York, 1969. [12] D. J. Foulis and M. K. Bennet, Effect algebras and unsharp quantum logics, Found. Phys. 24 (1994), 1331–1352. ORTHOGONALITY IN PARTIAL ABELIAN MONOIDS 29

[13] D. J. Foulis, R. J. Greechie and M. K. Bennet, Sums and products of interval algebras, Intern. J. Theoret. Phys., 33 (1994), 2114–2136. [14] D. J. Foulis, R. J. Greechie and G. T. Rüttimann, Filters and supports in orthoalgebras, Intern. J. Theoret. Phys., 31 (1992), 789–807. [15] G. Fox and P. Morales, Uniform semigroup-valued measures I , Rapport de recherche 80-17, Universitéde Montréal(1980), pp. 1–20. [16] G. Fox and P. Morales, On uniform semigroup-valued additive set fuctions, Canad. Math. Bull. 26 (1983), 26–34. [17] L. Fuchs, Infinite Abelian Groups, Vol. I , Academic Press, London, 1970. [18] F. Garc´ıaMazar´ıo, Convergence theorems for topological group valued measures in effect algebras, Bull. Austral. Math. Soc. 64 (2001), 213–231. [19] G. Grätzer, General Lattice Theory, Second Edition, Bikhäuser, Basel, 1998. [20] G. Grätzer, Universal Algebra, Van Nostrand, Princeton, N.J., 1968. [21] E. D. Habil, Orthosummable orthoalgebras, Internat. J. Theoret. Phys. 33 (1994), 1957–1984. [22] E. Hille and R. S. Phillips, Fuctional Analysis and Semi-groups, American Mathematical Society, Providence, R.I., 1957. [23] G. Jameson, Ordered Linear Spaces, Lect. Notes Math. 141, Springer Verlag, New York, 1970. [24] G. Jenˇcaand S. Pulmannová, Quotients of partial abelian monoids and the Riesz decomposition property, Algebra Universalis 47 (2002), 443–477. [25] G. Kalmbach, Orthomodular Lattices, Academic Press, London, 1983. [26] J. L. Kelley, General Topology, Springer Verlag, New York, 1955. [27] A. Kertész, On groups every subgroup of which is a direct summand, Publ. Math. Debrecen 2 (1951), 74–75. [28] S. Maeda, Probability measures on projections in von Neumann algebras, Reviews Math. Phys. 1 (1990), 235–290. [29] P. Morales and F. Garc´ıaMazar´ıo, The support of a measure in ordered topological groups, Atti Sem. Mat. Fis. Univ. Modena 45 (1997), 179–221. [30] O. Ore, Structures and group theory II , Duke Math. J. 4 (1938), 247–269. [31] G. T. Rüttimann, Non-commutative Measure Theory, Habilitationsschrift, University of Berne, Switzerland, 1979. [32] G. T. Rüttimann, The approximate Jordan-Hahn decomposition, Canad. J. Math. XLI (1989), 1124–1146. [33] G. T. Rüttimann, On σ-finite elements in orthomodular posets, Atti Sem. Mat. Fis. Univ. Modena 42 (1994), 285–300. [34] H. H. Schaefer, Banach Lattices and Positive Operators, Springer-Verlag, New York, 1974. [35] M. Sion, A Theory of Semigroup Valued Measures, Lect. Notes Math. 355, Springer-Verlag, Berlin, 1973. [36] H. Weber, Fortsetzung von Massen mit Werten in uniformen Halbgruppen, Arch. Math. 27 (1976), 412–423. [37] H. Weber, Compacteness in spaces of group-valued contents, the Vitali-Hahn-Saks theorem and Nikodým’sboundedness theorem, Rocky Mountain J. Math. 16 (1986), 253–275. [38] F. Wehrung, The dimension monoid of a lattice, Algebra Universalis 40 (1998), 247–411. [39] A. Wilce, Perspectivity and congruence in partial abelian semigroups, Math. Slovaca 48 (1998), 117–135. [40] A. Wilce, Partial abelian semigroups, Internat. J. Theoret. Phys. 34 (1995), 1807–1812.

Departement´ de Mathematiques´ et d’Informatique, Universite´ de Sherbrooke, Sher- brooke, Quebec,´ J1K 2R1, Canada

Departamento de Matematica´ Aplicada (E. U. Informatica),´ Universidad Politecnica´ de Madrid, 28031-Madrid, Spain E-mail address: [email protected]