Sequential products on effect algebras

Vol. 49 (2002)

No. 1

REPORTS ON MATHEMATICAL PHYSICS

SEQUENTIAL

PRODUCTS

ON EFFECT

ALGEBRAS

STAN GUDDER Department of Mathematics, University of Denver Denver, Colorado 80208, USA ([email protected]) and RICHARD GREECHrE Department of Mathematics, Louisiana Tech University Ruston, Louisiana 71272, USA ([email protected]) (Received June 28, 2001 - - Revised September 14, 2001)

A sequential effect algebra (SEA) is an effect algebra on which a sequential product with natural properties is defined. The properties of sequential products on Hilbert space effect algebras are discussed. For a general SEA, relationships between sequential independence, coexistence and compatibility are given. It is shown that the sharp elements of a SEA form an orthomodular poset. The sequential center of a SEA is discussed and a characterization of when the sequential center is isomorphic to a fuzzy set system is presented. It is shown that the existence of a sequential product is a strong restriction that eliminates many effect algebras from being SEA's. For example, there are no finite nonboolean SEA's. A measure of sharpness called the sharpness index is studied. The existence of horizontal sums of SEA's is characterized and examples of horizontal sums and tensor products are presented. Keywords: effect algebras, sequential products, Hilbert space operators, fuzzy sets.

1.

Introduction

Two m e a s u r e m e n t s a a n d b c a n n o t b e p e r f o r m e d s i m u l t a n e o u s l y in g e n e r a l , so they are f r e q u e n t l y e x e c u t e d sequentially. W e d e n o t e b y a o b a s e q u e n t i a l m e a s u r e m e n t in w h i c h a is p e r f o r m e d first a n d b second. W e call a o b the s e q u e n t i a l p r o d u c t o f a a n d b. W e shall restrict o u r a t t e n t i o n to y e s - n o m e a s u r e m e n t s , c a l l e d effects, w h i c h have o n l y two p o s s i b l e results. F o r generality, w e d o n o t a s s u m e that effects are p e r f e c t l y a c c u r a t e m e a s u r e m e n t s . T h a t is, they m a y b e f u z z y o r u n s h a r p . A s w e shall see, the s h a r p effects are t h o s e that satisfy a o a = a . A p a r a d i g m situation is an o p t i c a l b e n c h in w h i c h a b e a m o f p a r t i c l e s p r e p a r e d in a c e r t a i n state is i n j e c t e d at the left a n d then i m p i n g e first u p o n a filter a a n d then u p o n a filter b. P a r t i c l e s that p a s s t h r o u g h b o t h filters e n t e r a d e t e c t i o n d e v i c e at the r i g h t o f b. B e c a u s e o f q u a n t u m interference, the o r d e r o f p l a c e m e n t o f a a n d [87]

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b usually makes a difference and we have aob ~ boa. If it happens that aob = boa we say that a and b are sequentially independent and write a lb. In recent years quantum effects have been studied within a general algebraic framework called an effect algebra. In Section 2 we shall summarize the basic definitions concerning effect algebras and the properties of sequential products on Hilbert space effect algebras. The simplest of these properties are employed as axioms in Section 3 for a sequential effect algebra (SEA). A SEA is an effect algebra on which a sequential product with natural properties is defined. We believe that the axioms for a SEA are physically motivated and can be tested, for example, in the optical bench situation. Various properties of a SEA are proved in Section 3. For instance, relationships between sequential independence, coexistence and compatibility are given. It is also shown that the sharp elements form an orthomodular poset. The sequential center C(E) of a SEA E is the set of elements a C nE such that a Ib for every b • E. In Section 4 it is shown that C(E) coincides with the set of sharp central elements which has previously been studied. Moreover, a characterization is given for when C(E) is isomorphic to a fuzzy set system. Section 5 shows that the existence of a sequential product is a strong restriction that eliminates many effect algebras from being SEA's. It is shown that a Boolean algebra admits a unique sequential product and that certain effect algebras admit a sequential product only if they are Boolean. Moreover, it is proved that if a map preserves the sequential product then it completely preserves the effect algebra structure of the sharp elements. Section 6 defines the sharpness index of an effect. It is demonstrated that if E is isotropically finite, then every unsharp element of E has sharpness index oo. It is shown that a a-SEA is sharply dominating. An example is presented in Section 7 that illustrates various points of the previous sections. Finally, horizontal sums and tensor products of SEA's are considered in Sections 8 and 9. The existence of horizontal sums is characterized and some examples of horizontal sums and tensor products are given.

2. I-lilbert space sequential products This section summarizes the basic definitions concerning effect algebras [1, 6-8, 14, 15] and the properties of sequential products on Hilbert space effect algebras [2, 3, 10, 12, 13]. If ~9 is a partial binary operation, we write a / b if a ~ b is defined. An effect algebra is a system (E, 0, 1, ~ ) where 0, 1 are distinct elements of E and is a partial binary operation on E that satisfies the following conditions. (El) (E2) (E3) (E4)

If a _Lb, then b _La and b ~ a = a @ b . If a .L b and c _L (a ~ b), then b _L c, a _L (b ~ c) and a ~ (b ~ c) = (a ~ b) ~9c. For every a • E there exists a unique a' • E such that a _L a' and a ~ a' = 1. If a _L 1, then a = 0.

In the sequel, whenever we write a ~ b we are implicitly assuming that a _L b. We define a < b if there exists a c • E such that a ~ c = b. If such a c • E

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exists then it is unique and we write c = b e a. It can be shown that (E, <, ') is a partially ordered set with 0 < a < 1 for all a e E, a" = a, and a < b implies b I < a'. Moreover, we have a _J_ b if and only if a < b'. If a _1_ a we call a an isotropic element and when 0 is the only isotropic element of E, then we call E an orthoalgebra. If the n-fold orthosum a • a 0 " . • a is defined in E we denote this element of E by na. If there is a largest n e N such that na is defined, then n is the isotropic index of a and if no such n exists, then n has isotropic index c~. An element a ~ E is sharp if a m a' = 0. Notice that if a ~ 0 is sharp then a has isotropic index 1. We say that E is isotropically finite if every a ~ 0 in E has finite isotropic index. We now give some standard examples of effect algebras. For a Boolean algebra /3, define a _L b if a A b = 0 and in this case a O b = a v b . Then (/3,0, 1, O) is an effect algebra that happens to be an orthoalgebra. In particular, if X is a nonempty set, then (2 x, 0, X, @) is an effect algebra. These effect algebras correspond to classical logic and set theory. For the function space [0, 1] x on the interval [0, 1] c R define the functions f0, fl by fo(x) = 0 , f l ( x ) = 1 for all x ~ X. For f , g _ [0,1] x, we define f _1_ g if f ( x ) + g ( x ) < 1 for all x E X and in this case ( f • g)(x) = f ( x ) + g(x). Then ([0, 1] x, f0, f l , 0 ) is the effect algebra of fuzzy subsets of X. A particularly simple effect algebra is the interval [0, 1] c R. For a, b 6 [ 0 , 1 ] we define a _ L b if a + b < 1 and in this case a O b = a + b . In this section we are mainly concerned with the set E(H) of all self-adjoint operators on a Hilbert space H that satisfy 0 < (Ax, x) < (x, x) for all x 6 H. For A, B 6 C(H) we define A _1_ B if A + B ~ £ ( H ) and in this case A • B = A + B. Then ( g ( H ) , 0, I, 0 ) is an effect algebra that we call a Hilbert space effect algebra. This effect algebra is important in studies of the foundations of quantum physics and quantum measurement theory [2-4, 16, 17]. The quantum effects A ~ C(H) correspond to yes-no measurements that may be unsharp. The set of projection operators 7:'(H) on H form an orthoalgebra that is a sub-effect algebra of E(H). The elements of P ( H ) correspond to sharp quantum effects. If E and F are effect algebras, we say that 4~ • E --+ F is additive if a _1_ b implies 4~(a) _L ~b(b) and ~b(a • b) = q~(a) • q~(b). If 4) " E ~ F is additive and q~(1) = 1, then q~ is a morphism. If ~b • E --+ F is a morphism and ~b(a) _L ~b(b) implies that a _1_ b, then 4~ is a monomorphism. A surjective monomorphism is an isomorphism. It is easy to see that a morphism ~b is an isomorphism if and only if 4~ is bijective and ~b-1 is a morphism. A state on E is a morphism s • E --+ [0, 1]. We interpret s(a) as the probability that the effect a is observed (has answer yes) when the system is in the state s. We denote the set of states on E by f2(E). A set of states S c f2(E) is order determining if s(a) < s(b) for all s 6 S implies that a < b. The sequential product on E(H) is defined by A o B = A1/2BA 1/2 where A 1/2 is the unique positive square root of A [2, 3, 10, 12, 13]. We have that A o B ~ £(H) because

0 < (A1/2BA1/2x, x ) = (BA1/2x, A1/2x) <_(A1/2x, A1/2x)

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= (Ax, x) < Ix, x) for all x 6 H. Notice that B ~ A o B is all B 6 £ ( H ) . We now present some of product on E(H). If A o B = B o A we say and write A l B . The following result is

additive on E(H) and that I o B = B for the important properties of the sequential that A and B are sequentially independent proved in [12, 13].

THEOREM 2.1. (i) For A, B ~ g(H), if A o B ~ P ( H ) then AB = BA. (ii) For A, B ~ E(H), A I B if and only if AB = BA. Applying Theorem 2.1 we obtain the following properties of the sequential product A o B. COROLLARY 2.2. (i) If A o B = O, then B o A = 0. (ii) If A I B, then A I B' and Ao(BoC)=(AoB)oC for all C ~ £ ( H ) . (iii) If C I A and C I B then C I A o B and C I (A ~ B). The next three results are also proved in [13]. Notice that Theorem 2.3 gives the converse of the second part of Corollary 2.2(ii). THEOREM 2.3. For A, B ~ E(H), A o (B o C) = (A o B) o C for every C ~ £(H) if and only if A [ B. THEOREM 2.4. For A, B E E(H) the following statements are equivalent. (ii) B o A = B . (iii) A B = B A = B .

(i) A o B = B .

THEOREM 2.5. For A , B ~ E(H) we have B = A o B ~ A' o B if and only if AIB. We denote the set of positive trace class operators with trace 1 on H by 79(H). The normal states on E(H) have the form P w ( A ) = tr(WA) for some W ~ D(H). We say that A, B ~ £(H) are stochastically independent relative to W ~ 79(H) if Pw(A o B ) = P w ( A ) P w ( B ) . The next result is proved in [13]. THEOREM 2.6. For A, B ~ C(H) the following statements are equivalent. (i) A o ( C o B ) = ( A o C ) o B for every C ~ £ ( H ) . (ii) C o ( A o B ) = ( C o A ) o B for every C ~ E(H). (iii) P w ( A o B ) = P w ( A ) P w ( B ) for every W E 79(H). (iv) A = cI or B = cI for some O < c < 1. We close this section with an application of the interesting work in [18]. This result shows that the sequential product determines the effect algebra structure of E(H) when dim H > 3. THEOREM 2.7. Suppose that dim H > 3. If cp • C(H) --~ £(H) is a bijection satisfying cp(AoB) = cp(A)o(b(B) for all A, B ~ C(H), then cp is an effect algebra isomorphism. Proof: For A 6 E(H) we have tp(A 2) = tp(A o A) = tp(A) o tp(A) = tp(A) 2.

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Hence, for A, B e E ( H ) we have dp(ABA) = q~(A2 o B) = q~(A2) o q~(B) ----~b(A) 2 o ~b(B) = dp(A)dp(B)dp(A).

Applying Theorem 2 [18], ~b has the form ~b(A) = U'AU* where U is either a unitary or an anti-unitary operator on H . It easily follows that ~b is an effect algebra isomorphism. []

3. Sequential effect algebras This section summarizes the basic definitions and results for sequential effect algebras. For a binary operation o, if a o b = b o a we write a I b. A sequential effect algebra (SEA) is a system (E, 0, 1, @, o) where (E, 0, 1, ~ ) is an effect algebra and o • E x E --> E is a binary operation that satisfies the following conditions. (S1) ($2) ($3) ($4) ($5)

b ~ a o b is additive for all a E E. l o a - - - - a for all a • E . If a o b = 0 , then a l b . If a I b, then a I b' and a o (b o c) = (a o b) o c for all c • E. If c I a and c I b, then c I a o b and c I (a @ b).

We call an operation that satisfies (S1)-($5) a sequential product on E. If a I b for all a, b • E we call E a commutative SEA. The effect algebra [0, 1] _ I~ is a SEA with sequential product a o b = ab. Corollary 2.2 shows that C(H) is a SEA under the operation A o B = A U 2 B A U2. More generally, this operation makes the effect algebra of any von Neumann algebra a SEA. A Boolean algebra is a SEA under the operation a o b = a A b. Let X be a nonempty set and let ~ - c [0, 1] x. We call ~- a fuzzy set system on X if f0, f l • ~ - , if f • . T then f l - f • b r, if f , g • ~ " with f + g _ < 1 then f + g • ~ and if f , g • ~- then f g • ~ . Then ~" becomes a SEA when f ~ g = f + g for f + g < 1 and f o g = f g . Except for E ( H ) , all of these examples are commutative. The following lemma summarizes some of the properties of a SEA E. LEMMA 3.1. (i) aoO = Ooa = 0 and a o l = l o a = a f o r all a • E. (ii) a o b <_ a f o r all a, b • E. (iii) I f a <_ b, then c o a <_ c o b f o r all c • E. (iv) I f a <_ b, then co(bOa) =cobOcoa. (v) I r a < b, c l a and c lb, then c l ( b O a ) . Proof: (i) By additivity we have a oO @ O = a o O = a

o (O@O) = a o 0 @ a o 0

and by cancellation we have a o 0 = 0. Applying ($3) gives 0 o a = 0. Since 0 1 a , applying ($2) and ($4) gives a o 1 = 1 o a = a. (ii) Applying (S1) gives a =aoa

=a o(b@b')

=aob~aob'

>aob.

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(iii) I f a < b, there exists a d • E such that a (9 d = b. Hence, cob=coa(gcod

> coa.

(iv) I f a _< b, then by (iii) c o a < c o b. Since a (9 (b @ a ) = b we have that Hence, c o ( b @ a ) = c o b @ c o a . (v) This follows f r o m (S4), ($5) and the identity b @ a = (a (9 b')'. []

coa(gco(b@a)=cob.

We denote the set o f sharp elements o f E by E s . It is clear that 0, 1 • E s and that a' • E s w h e n e v e r a • E s . LEMMA 3.2. The f o l l o w i n g s t a t e m e n t s are equivalent. (i) a • E s . (ii) (iii) a o a = a. P r o o f : To show that (i) implies (ii) suppose that a A a ' =

a o a'=

0.

0. B y L e m m a 3.1(ii)

we have a o a' = a' oa
a'.

0. To show that (ii) implies (iii) suppose that a o a ' =

0. T h e n

a = a oa (ga oa' = a oa.

To show that (iii) implies (i) suppose that a o a = a. T h e n a = a oa (ga oa' = a (ga oa',

so by cancellation a o a ' = 0. I f b _< a, a', then by L e m m a 3.1(iii) we have that a o b < a o a' = 0. Hence, a o b = 0 and similarly a' o b = 0. Hence, b o a = b o a ' = 0

so that b=

boa

(gboa'

=O.

Hence, a m a ' = 0.

[]

LEMMA 3.3. (i) I f a o b = O, then a .L b. (ii) F o r a • E, b • E s , a o b = O i f a n d only i f a 1 b. (iii) F o r a, b • E s , with a I b we have a (9 b • E s . (iv) F o r a , b • E s with a < b we have b @ a • E s . (v) For a , b • E s with a I b we have aob•Es. P r o o f : (i) I f a o b = 0 ,

then b o a = a o b = 0 .

a =a ob(ga

(ii) I f a < b ' then b o a < b o b ' = 0 . (iii) Since a o b = 0 we have

ob'=a

Hence, ob'=

b'oa

< b'.

Hence, a o b = b o a = 0 .

(a (9 b) o (a (9 b) = (a (9 b) o a (9 (a (9 b) o b = a o (a ( g b ) ( g b o (a ( g b ) = a ( g b . (iv) This follows f r o m (iii) and the identity b O a = (a (9 b')'.

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(v) S i n c e a } b w e have (a o b) o (a o b) = a =a

o [b o (a o b)] = a o [b o ( b o a ) ] o [(b o b) o a ] = a o ( b o a )

= a o (a o b ) []

= (a o a ) o b = a o b .

It follows f r o m L e m m a 3.3(iii) that Es is a s u b - e f f e c t a l g e b r a o f E that is an o r t h o a l g e b r a . In g e n e r a l , i f a , b ~ E s then a o b q[ E s so E s is not a s u b - S E A of E. THEOREM 3.4. Let a ~ E and b ~ Es. (i) a < b i f and only i f a o b = b o a = a and b < a i f and only if a o b = b o a = b . (ii) I f a [ b, then a A b = a o b . (iii) I f a _k b, then a @ b = a v b = (a' o b')'.

Proof: (i) I f b o a = a , then a = b o a _< b. Similarly, i f a o b = b then b < a . C o n v e r s e l y , s u p p o s e that a < b. T h e n a _k b', so b y L e m m a 3.3(ii) a o b ' = b'oa = O. H e n c e , a [ b a n d w e have a =aob~aob'=aob. f f b < a then a ' _< b', so f r o m o u r p r e v i o u s work, a ' o b ' = b ' o a ' = a ' . H e n c e , b o a ' = b o (b' o a ' ) = (b o b ' ) o a ' = 0 a n d w e have

b=boa@boa'

=boa

=aob.

(ii) W e have a o b = b o a < a , b. S u p p o s e that c _< a , b . T h e n there exists a d 6 E such that c @ d = a a n d b y (i) we have b o c = c. H e n c e ,

boa =bo(c@d)

=boc~bod

=c~3bod

> c.

W e c o n c l u d e that a A b = a o b. (iii) It is c l e a r that a , b < a @ b. S u p p o s e that a , b < c. T h e n there exists a d ~ E such that a @ d = c a n d b y (i) w e have c o b = b o c = b. B y L e m m a 3.3(ii), aob=boa=O so that

b=boc=bo(a~d)=boa@bod=bod. A p p l y i n g L e m m a 3.1(v) we have b } (c e a ) . H e n c e , b I d a n d b = d o b < d. It follows that a ~ b < a @ d = c. H e n c e , b y (ii) we have that a ~ b = a V b = (a' A b')' = (a' o b')'.

[]

COROLLARY 3.5. Es is a sub-effect algebra o f E that is an orthomodular poset.

Proof: T h i s follows f r o m T h e o r e m 3.4(iii).

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We say that a, b ~ E coexist if there exist c, d, e 6 E such that c ~ d ~ e is defined and a = c ~ d, b = c @ e [16, 17]. THEOREM 3.6. (i) I f a I b then a and b coexist. (ii) For a E E, b E Es, a I b if and only if a and b coexist. Proof: (i) If a I b, then a = a o b ~ a o b' and b =boa~boa'

=aob~a'ob.

Now

1 =a~a'=a~(a'ob~a'ob')=(a~a'ob)~a'ob'. Hence, (a' o b ' ) ' = a ~ a ' o b = a o b ~ a

ob' ~a' ob.

It follows that a and b coexist. (ii) If a and b coexist, then a = c ~ d, b = c ~ e for some c, d, e 6 E where c ~ d @ e is defined. Since c < b we have b lc. Since d 2_ b we have d < b'. Hence, b I d and it follows that b I a. [] It is well known that the converse of Theorem 3.6(i) does not hold [3]. We say that a, b ~ E s are compatible if there exist mutually orthogonal elements c, d, e E s such that a = c v d and b = c v e . COROLLARY 3.7. For a, b ~ Es, a I b if and only if a and b are compatible. Proof: If a I b, then by the proof of Theorem 3.6 and L e m m a 3.3(v), a o b, a o b' and a ' o b are mutually orthogonal elements of Es. By Theorem 3.4(iii) we have a = a ob@a ob' = a ob va ob' and b=aob@a'ob=aobva'ob.

Hence, a and b are compatible. Conversely, suppose that a and b are compatible and a = c v d , b = c v e where c, d, e ~ E s are mutually orthogonal. By Theorem 3.4(iii), a =c~d, b=c~e. Since e 2 _ c , e 2 _ d , we have e 2 _ a . Hence, c ~ d D e is defined. Thus, a and b coexist, so by Theorem 3.6(ii) we have that a lb. [] COROLLARY 3.8. A SEA is a commutative orthoalgebra if and only i f it is a Boolean algebra. In a commutative SEA E, Es is a Boolean algebra. The next result shows that for certain special cases, o and ~ are closely related. COROLLARY 3.9. (i) I f a I b and a 2- b, then a E) b = a o b @ (a' o b')'. (ii) I f a, b ~ E s and a 2- b, then a ~ b = (a' o b')'. (iii) I f a, b ~ E s and a' _l_ b', then a o b = (a' ~ b')'.

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The sequential center of a SEA We define the sequential center of a SEA E as

C(E)={a~E:alb

for all b 6 E } .

The next result follows f r o m our previous work. THEOREM 4.1. (i) C ( E ) is a sub-SEA o f E which is a commutative monoid under o. (ii) C ( E ) f3 Es is a Boolean algebra. An element a E E is principal if b, c < a and b _L c imply that b ~ c < a. It is shown in [9] that principal elements are sharp. We now prove that the converse holds in a SEA. LEMMA 4.2. An element a ~ E is principal if and only if a ~ Es.

Proof: Suppose a is principal and b < a, a'. Then b _1_ a and b, a _< a. Hence, aOb~a=a~O. By cancellation b = 0 so that a A a ' = 0. Conversely, suppose that a ~ Es and b, c < a with b _1_ c. By T h e o r e m 3.4(i) a o b = b and a o c = c. Hence,

a >_ao(b@c)=aobOaoc=b@c.

[]

The next result holds for an arbitrary effect algebra [9]. However, the proof is much simpler for a SEA. COROLLARY 4.3. (i) I f a , b ~ Es and a A b exists, then a A b ~ Es. (ii) I f a , b ~ Es and a v b exists, then a v b E Es.

Proof: (i) By L e m m a 4.2, a and b are principal. Suppose that c, d < a A b and c _ l _ d . Then c < a , b and d < a , b , so that c ( 3 d < a , b . Hence, c @ d < a A b . Thus, a A b is principal, so by L e m m a 4.2, a A b ~ Es. (ii) Since a v b exists, a ' A b ' = (a v b)' exists and is sharp. Hence, a v b ~ Es. [] According to [9], an element a ~ E is central if a, a ' are principal and for every p 6 E there exist q , r ~ E such that q < a, r < a' and p = q (3 r. In [9] the center C ( E ) is defined to be the set of all central elements. THEOREM 4.4. (i) I f a ~ Es, then a I P if and only if there exist q, r ~ E such that q < a, r < a' and p = q ~ r. (ii) C ( E ) = C ( E ) N Es.

Proof: (i) Suppose p = q @ r where q < a, r < a 1. Since a I q and a I r we have that a I P. Conversely, suppose that a I P- Then p = p o a ~ p o a N ' and we have that p o a = a o p < a and p o a t = a ' o p < a'. (ii) I f a ~ C ( E ) , then by L e m m a 4.2, a ~ Es. Moreover, by (i) we have that a E C ( E ) . Hence, C ( E ) c_ C ( E ) n Es. Conversely, if a ~ C ( E ) D Es, then by (i) we have that a E C ( E ) . Hence, C ( E ) (q E s c C ( E ) . []

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s. GUDDER and R. GREECHIE

EXAMPLE. Let .T be the set of differentiable functions f : [0, 1] ~ [0, 1]. Then .T is a fuzzy set system under the previously defined partial operation f @ g = f + g if f + g < 1 and the operation f o g = f g . Hence, ~- is a commutative SEA so that C(~ r) = ~ . However, .T is not lattice ordered because f A g does not exist in ~- in general. This shows that a commutative SEA need not be an MV-algebra. This example showed that C ( E ) need not be lattice ordered. If E and F are SEA's, a map ~b : E --~ F is an isomorphism if ~b is an effect algebra isomorphism satisfying ~b(a o b ) = ~b(a)o ~b(b) for all a, b ~ E. We shall give an example later which shows that C ( E ) need not be isomorphic to a fuzzy set system. Our next result characterizes when C ( E ) is indeed isomorphic to a fuzzy set system. In contrast to C ( E ) , it is interesting to note that C ( E ) is always a Boolean algebra even for an arbitrary effect algebra [9]. A state s on E is called multiplicative if s(a o b) = s ( a ) s ( b ) for all a, b. THEOREM 4.5. The sequential center C ( E ) is isomorphic to a fuzzy set system if and only if C ( E ) admits an order determining set of multiplicative states.

Proof: Suppose .T is a fuzzy set system on f2 and ~b : C ( E ) ~ ~ is an isomorphism. For coe f2, a ~ C ( E ) define co(a) =~b(a)(co). Then co(l) =~b(1)(co) = 1. If a _1_b we have co(a ~ b) = dp(a ~) b)(co) = [~b(a) @ ~b(b)] (co) = qb(a)(co) + dp(b)(co) = co(a) + co(b). Hence, {co:co 6 t } forms a set of states on C ( E ) . These states are multiplicative because

co(a o b) = cp(a o b)(co) = [q~(a)~b(b)] (co) = dp(a)(co)dp(b)(co) = co(a)co(b). To show that this set of states is order determining, suppose that co(a) < co(b) for every co 6 f2. Then cp(a)(co) < dp(b)(co) for every co 6 f2 so that tp(a) < qS(b). Since ~b is an isomorphism, we have that a < b. Conversely, suppose f2 is an order determining set of multiplicative states on C ( E ) . Define q~ • C ( E ) ~ [0, 1] ~ by dp(a)(co) = co(a) and let U c [0, 1] ~ be the range of ~b. Since ~b(1) = f l and q~(0) = f0, f l , f0 ~ jr. For q~(a) c ~" we have [ f l - ~b(a)] (co) = 1 - co(a) = co(d) = dp(a')(co), so that f l - q~(a) 6 U and q~(a') = f l - q~(a). If a _1_b, then q~(a)(co) = co(a) < co(b') = 1 - dp(b)(w). Hence, ~b(a)_1_ q~(b) and we have

dp(a ~ b)(w) = co(a ~) b) = co(a) + co(b) = cp(a)(co) + cp(b)(co)

SEQUENTIAL PRODUCTS ON EFFECT ALGEBRAS

97

for every oo ~ f2. We conclude that q~(a ~ b) = (p(a) ~ (p(b). For (p(a), ~ ( b ) e ~" with dp(a)+dp(b) < 1 we have co(a) < co(b'). Since fa is order determining, we have that a _1_ b. Hence, (p is a m o n o m o r p h i s m and d~(a)+cp(b) ~ ~ . For dp(a),cp(b) ~ we have [(p(a)(p(b)] (o9) = cp(a)(w)cp(b)(w) = co(a)co(b) = co(a o b) = cp(a o b)(w) for every co 6 f2. Hence, cp(a)cp(b)~ Jr and c p ( a o b ) = cp(a)cp(b). We conclude that ~" is a fuzzy set system and (p : C ( E ) ---* jr is an isomorphism. [] COROLLARY 4.6. A SEA E admits an order determining set o f multiplicative states f2 if and only if E is isomorphic to a fuzzy set system f2. Under this isomorphism, Es is a Boolean algebra o f subsets of ~.

Proof: For a ~ Es we have ~b(a) 2 = ~b(a o a) = ~b(a). Hence, ~b(a) is a characteristic function which can be considered to be a subset of f2. The result now follows. []

5.

Existence of sequential products

This section shows that the existence of a sequential product is a strong restriction that eliminates m a n y effect algebras from being SEA's. LEMMA 5.1. For an effect algebra E that is a Boolean algebra there is a unique sequential product a o b = a A b.

Proof: Since E = E s and any two elements of E are compatible, by Corollary 3.7 we have that a I b for every a, b E E and any sequential product o. It follows from T h e o r e m 3.4(ii) that a o b = a A b. [] LEMMA 5.2. Let E be a SEA. (i) I f a E E is an atom then for every b E E either a < b or a <_ b'. (ii) I f a, b ~ E are distinct atoms, then a I b.

Proof: (i) if a o b = 0, we have that a = b which

Since a o b _< a we have a o b = 0 or a o b = a. By L e m m a 3.3(i), then a _< b'. I f a = a o b then a o b' = 0, so again by L e m m a 3.3(i) a _< b. (ii) By (i) we have that a £ b or a _< b. In the second case, is a contradiction. []

THEOREM 5.3. An atomistic orthoalgebra E admits a sequential product if and only if E is Boolean.

Proof: If E is Boolean, we have seen that it admits a sequential product. Conversely, suppose that E admits a sequential product. Since E = Es, by Corollary 3.5, E is an orthomodular poset. Let c, d ~ E. By L e m m a 5.2 we have that c : ( v c i ) v (vei), d = ( v c i ) v (vdi), where ci, di, ei are distinct mutually orthogonal atoms. Since (X/e/)_J_ (X/d/), c and d are compatible. It follows that E is Boolean. [] COROLLARY 5.4. The,re does not exist a sequential product on 7~(H), dim H > 2.

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s. GUDDER and R. GREECHIE

Proof: For dim H > 2, 7~(H) is a nonboolean, atomistic orthoalgebra.

[]

THEOREM 5.5. (i) An orthoalgebra E is a commutative SEA if and only if E is Boolean. (ii) If E is a chain finite SEA, then E is Boolean.

Proof: (i) It is clear that a Boolean algebra is a commutative SEA. Conversely, if an orthoalgebra E is commutative then by Corollary 3.7 any two elements o f E are compatible. It follows that E is Boolean. (ii) I f a e E is an atom, then a = a o a ~ a o a ' implies that a o a - - - 0 or a o a ' = 0. Suppose that a o a -----0. By L e m m a 3.3(i), a ± a so that 2a exists. Now .a o (2a) = a o (a ~ a ) = a o a E)a o a = 0 . Hence, 2a ± a so that 3a exists. Continuing by induction, na exists for all n e 1~. Since a < 2a < 3a < . . - forms an infinite chain, this is a contradiction. Hence, a o a ' = 0, so by L e m m a 3.2 a is sharp. Since the orthosum of sharp elements is sharp, we have E = Es. Hence, E is an atomistic orthoalgebra and the result follows f r o m T h e o r e m 5.3. [] LEMMA 5.6. Let E be a SEA and let dp • E ~ E be an additive function satisfying (i) a = q~(1) • C(E), (ii) if b • E with b < a, then dp(b) = b. Then a • Es and d p ( b ) = a o b for every b • E.

Proof: By (ii) ~ b ( a ) = a. Hence, a = ~b(1) = ~b(a) ~ ~b(a') = a ~ ~b(a'), so by cancellation we have that ~b(a') = 0. I f b < a', then b @ c = a ' for some c • E. Hence, 4~(b) @ 4~(c) = 4,(a') = 0, so that ~b(b) = 0. I f d • E then a ' o d < a', so that 4 ~ ( a ' o d ) = 0. Also, a o d < a so that ~b(a o d ) = a o d. Thus, for every d • E we have that ~b(d) = ~b(d o a ~ d o a ' ) = ~b(a o d) @ ~b(a' o d) = a o d. Since a = ~b(a) = a o a we conclude that a • Es.

[]

L e m m a 5.6 can be used to give another proof of L e m m a 5.1. Indeed, let a o b = a A b be the sequential product on E. For a • E = C ( E ) define t~a : E --+ E by ~ba(b) = a • b where a • b is another sequential product on E. Then t~a satisfies the conditions of L e m m a 5.6 so that a • b = t~a (b) : a o b. Let X be a finite nonempty set and let 0r(X) be the fuzzy set system O r ( X ) = [0, 1] x. A sequential product o on ~ ' ( X ) is homogeneous if ~fl o f = f o ~fl = L f for all f • Or(X), L • [0, 1]. LEMMA 5.7. Or(X) admits a unique homogeneous sequential product f o g

= fg.

Proof: It is clear that f o g = f g is a homogeneous sequential product on Or(X). Let f • g be another homogeneous sequential product on Or(X). Since any two

99

SEQUENTIAL PRODUCTS ON EFFECT ALGEBRAS

elements of ,~(X)s are compatible, it follows f r o m Corollary 3.7 and L e m m a 3.3(v) that .~(X)s is a commutative sub-SEA of (St(X), o). For f • U ( X ) s define Cf : ~ ( X ) s ~ ~'(X)s by C f ( g ) = f • g. Applying L e m m a 5.6, we have that

f •g=¢f(g)=

f og= fg.

Hence, f • g = f g for every f, g • ,~(X)s. Since f • g is homogeneous, we have ( L f ) . ( / z g ) = (lzg)•O~f) for every f, g • ,~(X)s, )~, # • [0, 1]. For any u, v • U ( X ) we have u = Ein=l ~.ifi, "o = Z j % I IdJgj , where £i,/zj • [0, 1], f i , g j • ,~(X)s, i = 1 . . . . . n, j = 1 . . . . . m. It follows that u • g j = g j • u and v . f / = f i . v for every i = 1 . . . . . n, j = 1 . . . . . m. Hence,

u.v=E#ju•gj= = Z

E#jgj.u=

Etzj£igj•

fi

).ildj fi o gj = U O V.

[]

The next result shows that if a map preserves the sequential product then it completely preserves the structure of the sharp elements. THEOREM 5.8. Let E, F be SEA's and let ¢ : E ---> F be a bijection satisfying ¢ ( a o b) = ¢ ( a ) o ¢(b). Then ¢ : Es ---> Fs is an isomorphism. Moreover, if a • Es or b • Es and a < b, then ¢ ( a ) < ¢(b).

Proof: I f a • Es, then ¢ ( a ) = ¢ ( a o a) = ¢ ( a ) o ¢ ( a ) so that ¢ ( a ) • Fs. If b • Fs, then ¢ ( a ) = b for some a • E. But then ¢(a) = b o b = ¢(a) o¢(a) = ¢(a oa). Since ¢ is injective, we have that a = a o a, so a • Es. Hence, ¢ • Es --+ Fs is a bijection. Now ¢ ( a ) = 1 for some a • E. Thus, 4)(1) = ¢ ( 1 ) o 1 = ¢ ( 1 ) o ¢ ( a ) = ¢(1 o a) = ¢ ( a ) = 1. Similarly, ¢ ( b ) =

0 for some b • E and we have ¢ ( 0 ) = ¢ ( 0 o b) = ¢ ( 0 ) o ¢ ( b ) = O.

If a • E s

or b • E s

with a < b ,

then a o b = b o a = a .

Hence,

¢ ( a ) = ¢ ( b o a) = ¢ ( b ) o ¢ ( a ) < ¢ ( b ) . Therefore, ¢ : Es ~

Fs preserves order. I f a • Es, ¢(a) o ¢(a') = ¢ ( a o a') = ¢(0) = O.

Applying L e m m a 3.3(ii) we have that ¢ ( a ' ) < ¢ ( a ) ' . Now, ¢ ( b ) = b • Es. Since ~b(a o b) = ¢ ( a ) o q~(b) = ¢ ( a ) o q~(a)' = 0,

¢ ( a ) ' for some

100

s. GUDDER and R. GREECHIE

we have that a o b = 0. By Lemma 3.3(ii) we have that b < a', so that q~(a)' = ~b(b) < ~b(a'). Hence, q~(a') = ¢(a)'. If a , b ~ E s and a _1_ b then applying Theorem 3.4(iii) gives a @ b = ( a ' o b')'. Since ~b(a)_1_ ¢(b) we have that ~ ( a ~ b)=q~ ( ( a ' o b')') = [q~(a' o b')]' = [q~(a') o ~b(b')]' = [q~(a)' o q~(b)']' = ~b(a) ~ ¢(b). Also, ¢-1 . F s ~

E s preserves order because 4,(a) < ¢(b) implies that

¢(b o a) = ~b(b) o ¢ ( a ) = ~b(a). Hence, a = b o a < b. Thus, ¢ • E s ~ isomorphism.

F s is a monomorphism and therefore is an

[]

The following example show that Theorem 5.8 cannot be strengthened. Let E = [0, 1] x be a fuzzy set system. Define q~: E --, E by q~(f) = f2. Then q~ is a bijection and ~b(f o g) = q~(f) o ~(g). But if f ~ E s then ¢ ( f ' ) = (1 - f ) 2 5~ 1 - f 2 = q~(f),. Moreover, if f , g ¢ E s with f _1_g, then ¢ ( f • g) ~ ¢,(f) • q~(g) in general. 6.

Sharpness index

For n ~ N, a E E we define a n = a o a o • •. o a (n factors). LEMMA 6.1. (i) We h a v e that a n ~ E s a n EEs

if and

only

i f a n+l = a n. (ii) A l s o ,

i f a n d o n l y i f a m = - a n f o r all m > n .

(i) If a n E E s , then a 2n = a n. But Conversely, if a n+l = a n then we have Proof:

a n =a

n+l

=..._~a

a 2n < a n+l < a n,

2n =

SO that a n+l = a n.

(an) 2,

so that a n ~ E s . Moreover, (ii) follows from (i).

[]

If there exists an m such that a m E E s , then the smallest n ~ N such that a n ~ E s is the s h a r p n e s s i n d e x of a. If no such m exists, then the sharpness index of a is c~. We denote the sharpness index of a by s ( a ) . Of course, a ~ E s if and only if s ( a ) = 1. LEMMA 6.2. I f n = s ( a ) < oo, t h e n a n is the largest s h a r p e l e m e n t b e l o w a. P r o o f : Clearly, a n ~ E s and a n < a. Suppose that b < a and b E E s . Then by Theorem 3.4(ii) b o a = a o b. Hence, boa 2 =a2 ob=aob

=b,

and continuing by induction we have that b o a n = an o b = b. Hence, b < a n.

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SEQUENTIAL PRODUCTS ON EFFECT ALGEBRAS

101

A a - e f f e c t algebra is an effect a l g e b r a such that a l > a2 > - - . i m p l i e s that Aai exists. A a - S E A is a S E A that is a a - e f f e c t a l g e b r a E s a t i s f y i n g : (1) I f a l > a2 > . . - , then b o ( A a i ) = A ( b o a i ) for e v e r y b 6 E ; (2) I f a l > a2 > . . . and b [ ai, i = 1, 2 . . . . . then b I Aai. It c a n b e s h o w n that C ( H ) is a a - S E A . THEOREM 6.3. Let E be a a - S E A . I f a ~ E, then there exist b, c E E s such that b is the largest sharp e l e m e n t b e l o w a a n d c is the smallest sharp e l e m e n t above a.

Proof: I f n = s ( a ) < ¢¢, then b y L e m m a 6.2, a n is the largest s h a r p e l e m e n t b e l o w a. S u p p o s e that s ( a ) = oo. S i n c e a > a 2 > . - . , b = A a n exists. N o w b y (1) w e have that a m o b = a m o (A a n) = A a n+m = b. n

S i n c e a m I an for n = 1, 2 . . . . . have that

n

a p p l y i n g (2) g i v e s a m [ b. H e n c e , for all m 6 N w e

b = a mob

= b o a m.

Therefore, b2 =

b o ( A a n) = A ( b o a n) = b,

so that b ~ Es. Clearly, b < a . S u p p o s e that d < a w i t h d ~ Es. T h e n d = d o a = a o d , so that d < a n for all n 6 N. H e n c e , d < A a n = b. W e c o n c l u d e that b is the largest s h a r p e l e m e n t b e l o w a. L e t e b e the largest s h a r p e l e m e n t b e l o w a ' a n d let c = e'. T h e n a < e' = c a n d c ~ Es. I f a < f , w h e r e f ~ E s , then f ' < a ' . H e n c e , f ' < e so that c = e ' < f . H e n c e , c is the s m a l l e s t s h a r p e l e m e n t a b o v e a . [] T h e o r e m 6.3 shows that a a - S E A is s h a r p l y d o m i n a t i n g [6]. I n p a r t i c u l a r , C ( H ) is s h a r p l y d o m i n a t i n g . W e d e n o t e the i s o t r o p i c i n d e x o f an e l e m e n t a b y i(a). THEOREM 6.4. I f 1 < s ( a ) < oo, then there exists an m ~ 1~ such that a ' n o ( a m ) I ~k 0 a n d i [ a m o (am) ' ] = (x).

Proof: S u p p o s e that s ( a ) = 2. S i n c e a ¢ e E s , L e m m a 6.1(ii), a 3 = a 2. T h e n

we have that a o a ' 5~ 0. B y

a 2 = a 3 ~ a 2 o a t = a 2 ~) a 2 o a ' i m p l i e s that a 2 o a ' = 0. Now, a' = (a') 2 G a' o a i m p l i e s that a'oa=aoa'=ao(a')

2~a

z o a' = a o (a') 2 = (a') 2 o a.

Hence, (a') 3 o a = (a') 2 oa = a' oa.

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s. GUDDER and R. GREECHIE

Continuing by induction, we have that (a') n o a = a ' o a for every n ~ N. Now, (a') 2 = (a') 3 @ (a') 2 o a = (a') 3 @ a ' o a. It follows that a'=

[(a') 3 ~ a ' o a ] ~ a ' o a .

Hence, 2 ( a ' o a) is defined and a ' =

(a1)3 f~ 2 ( a l o a). In a similar way

(a') 3 = (a') 4 ~ (a') 3 o a = (a') 4 ~ a ' o a, so that a ' = [(a') 4 ~ a ' o a] • 2 ( a ' o a). Hence, 3 ( a ' o a) is defined and a ' = (a')4 ~ 3 ( a ' o a). Continuing by induction, we conclude that n(a' o a) is defined for all n 6 N. Hence, i(a o a') = ~ . For the general case, assume that s(a) = n, where 1 < n < c~. Suppose that n is even and n = 2m. Then (am) 2 = a 2m E Es and since m < 2m - 1 = n - 1 we have a m q[ ES. H e n c e , s(a m) = 2. It follows f r o m our previous work that am o (am)' ~ 0 and i [ a m o (am) '] = OO. Finally, suppose that n is odd and n = 2m - 1. Then (am) 2 = a 2'n 6 Es. Since m < 2 m - 2 = n - 1, we have a m ¢ Es. Hence, s(a m) = 2 and again a m o ( a m ) ' ~ 0 and i [ a m o (am) '] : oo. [] COROLLARY 6.5. I f E is isotropically finite, then for every a ~ E \ Es we have s(a) = oo. It is easy to see that £ ( H ) is isotropically finite. However, the nonstandard unit interval *[0, 1] with its usual product is a SEA that is not isotropically finite. In fact, every infinitesimal in *[0, 1] has infinite isotropic index. It follows f r o m Corollary 6.5 that every unsharp element of £ ( H ) has infinite sharpness index. The converse of Corollary 6.5 does not hold. For example, in *[0, 1] every unsharp element has infinite sharpness index. In the next section we shall show that there exists a SEA with an element a satisfying 1 < s(a) < c~. COROLLARY 6.6. Let E be an isotropically finite SEA. (i) Every atom o f E is sharp. (ii) I f E is atomic then E is an orthoalgebra.

Proof: (i) I f a is an atom, since a 2 < a we have a 2 = a or a 2 = 0. In the latter case s ( a ) = 2 which contradicts Corollary 6.5. Hence, a is sharp. (ii) Suppose that E is not an orthoalgebra. Then there exists a b ~ E with b A b' ~ 0. Since E is atomic, there exists an atom a with a < b , b ' . But then a < b < a', so that a q~ Es. This contradicts (i). [] THEOREM 6.7. (i) I f a E E s and b E E with b_l_ a, then (a ~ b) n = a ~ b n. (ii) I f E is a a - S E A and a, b ~ E with b _L a, then A(a ~ b ~) : a ~ A b n. n

n

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SEQUENTIAL PRODUCTS ON EFFECT ALGEBRAS

P r o o f : (i) Since b < a' we have b o a ~ = a ' o b = b. Hence, a o b = 0. We now prove the result by induction n. The result certainly holds for n = 1. Assuming the result holds for n we have that

(a ~ b) n+l = (a @ b)" o (a ~3 b) = (a ~ b n) o (a ~ b) = (a ~ b n) o a ~ (a ~ b n) o b = a o (a • b") ~) b o (a ~ b") = a ~ b n+l .

(ii) Since a @ A b n < a ~ b ~, we have that a ~ A b ~ < A ( a ~ b ~ ) . Since a ~ b n > a, we have that A ( a @ b ~) > a. From A ( a ~ b ~) < a ~ b ~ it follows that A ( a G b ~ ) O a < b ~. Hence, m(a @ b ~) O a < Ab ~, so that A(a ~3 b ~) < a @ A b n. [] If E is a o--SEA, we have seen that for any a ~ E there exists a largest sharp element below a. We denote this element by /aJ. COROLLARY 6.8. L e t E be a a - S E A b _L d a n d a ~ c = b ~ d, then

with a, b ~ E s a n d c , d ~ E. (i) I f a _L c,

La @ cJ -- a ~ [cJ = b ~3 LdJ = Lb @ dJ . (ii) I f [cJ = [dJ = 0 , a _ L c, b_L d a n d a O c = b @ d, then a = b (iii) I f b . J _ d , LdJ = 0 a n d a < b @ d , then a < b.

and c=d.

Proof', (i) By Theorem 6.7(i) we have a ~ c ~ = (a @ c)" = (b ~ d ) n : b ~ d".

B y Theorem 6.7(ii) and the proof of Theorem 6.3 we have a~LcJ

= a @ A c n = A ( a ~ c n) = A ( b ~ d ~ ) = b ~ A d

n=b@[dj.

Moreover, [a @ cJ = A(a ~ c) ~ = A(a G c n) = a ~ AC n = a ~ LcJ. (ii) By (i) we have a=a~[cJ

=b@[dJ

=b.

Hence, c = d by cancellation. (iii) Since a < b @ d, there exists a c 6 E such that a ~ c = b @ d. Applying (i) gives a < aO

LcJ =

b~

LdJ =

b.

[]

THEOREM 6.9. I f E is a ~r-SEA then a n y a ~ E has a unique representation a = b @ c, w h e r e b 6 E s a n d LcJ = 0 . P r o o f : It is clear that a = LaJ O ( a O [ a J ) . If b ~ E s and b < a O [aJ, then b < a. Hence, b < LaJ and b < LaJ' so that b = 0. Therefore, t a e [ a J J = 0. This gives the desired representation. To show uniqueness, suppose we have two such representations a = b ~ c = bl @ q , b, bl ~ E s , [cJ = LqJ = 0. Then by Corollary 6.8(ii) we have b = bl and c = q . []

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7.

s. GUDDER and R. GREECHIE

An example

This section presents an example of a SEA that has an element a satisfying s(a) = 2. Let E = w + o9* be the set with elements {0, 1 , a , 2 a . . . . . a', (2a)' . . . . }. By convention, we define 0a = 0. Defining @ on E by (ma) @ (na) = (m + n)a

and when n _< m, (ma)' @ (na) = (na) @ (ma)' = ((m - n)a)',

it is easy to see that (E, 0, 1, @) is an effect algebra [1]. Moreover, j a < (ka)' for every j, k e N because ( j a ) @ ((k + j ) a ) ' = (ka)'. THEOREM 7.1. There is a unique sequential product on the effect algebra E = o9 +03*.

Proof: For x, y 6 E define

xoy=

0

ifx=ma,

y=na,

xAy

if x = m a ,

y=(na)'

((m+n)a)'

if x----(ma)', y = ( n a ) ' .

or x = ( m a ) ' ,

y=na,

It is clear that l o x = x for every x 6 E. Since E is commutative under o, in order to show that o is a sequential product we only need to verify (D) x o ( y @ z ) = ( x o y ) @ ( x o z ) , (A) x o ( y o z ) = ( x o y ) o z .

ylz,

There are eight possibilities for x, y, z: x

y

z

(1)

ra

ma

na

(2)

ra

ma

(na)'

(3)

ra

(ma)'

na

(4)

ra

(ma)'

(na)'

(5)

(ra)'

ma

na

(6)

(ra)'

ma

(na)'

(7)

(ra)'

(ma)'

na

(8)

(ra)'

(ma)'

(na)'

SEQUENTIAL PRODUCTS ON EFFECT ALGEBRAS

105

To verify (D) we only need to consider cases (1), (2), (5) and (6) because (2) and (3) as well as (6) and (7) are symmetric and (4) and (8) never occur because y ,l! z in these cases. In Case (1), both the left and right hand sides are 0. In Case (2), both the left and right hand sides are x. In Case (5), both the left and right hand sides are y ~ z. In Case (6), both the left and right hand sides are ((n - m + r ) a ) ' .

To verify (A), we check all eight cases. (1) r a o ( m a o n a ) = r a o O = O = O o ( n a ) = ( r a o m a ) o n a , (2) r a o ( m a o ( n a ) ' ) = r a o m a = 0 = 0 o ( n a ) ' = ( r a o m a ) o ( n a ) ' , (3) r a o ( ( m a ) ' o n a ) = r a o n a = 0 = ( r a o ( m a ) ' ) o n a , (4) r a o ( ( m a ) ' o ( n a ) ' ) = r a o ( ( m + n ) a ) ' = r a = r a o ( n a ) ' = (ra o (ma)') o (na)',

(5) ( r a ) ' o ( m a o n a ) = ( r a ) ' o 0 = 0 = m a o n a = ( ( r a ) ' o m a ) o n a , (6) ( r a ) ' o ( m a o ( n a ) ' ) = ( r a ) ' o m a = m a = m a o ( n a ) ' = ((ra)' o ma) o (na)',

(7) ( r a ) ' o ( ( m a ) ' o n a ) = ( r a ) ' o n a = n a = ((r + m ) a ) ' o n a = ((ra)' o (ma)') o na,

(8) ( r a ) ' o ( ( m a ) ' o ( n a ) ' ) = ( r a ) ' o ( ( m + n ) a ) ' = ((r + m + n ) a ) ' = ((r + m ) a ) ' o ( n a ) ' = ( ( r a ) ' o ( m a ) ' ) o ( n a ) ' .

It follows that (E, o) is a SEA. For uniqueness, suppose that • • E --+ E is another sequential product on E. Since a is an atom and a • a < a , we have a e a = 0 or a • a = a . But a < a ' so that a ¢ E s . Hence, a e a = 0. Therefore, n a • m a = n m a • a = 0 for every n, m 6 N. Since every x 6 E has the form n a or ( n a ) ' , it is clear that E is commutative under e. For x = m a , y = ( n a ) ' we have that x e. y = m a e ( n a ) ' = ( m a e n a ) @ ( m a e ( n a ) ' ) = m a = x A y .

Since (na)' = (na)' • ma @ (na)' • (ma)' = ma • (na)' • (ma)',

we have that 1 = n a ~ (na)' = (n + m ) a ~3 (na)' • ( m a ) ' .

Hence, ( n a ) ' • ( m a ) ' = ((n + m ) a ) ) ' . x,y~E.

We conclude that x • y = x o y for every []

The elements n a ~ o9 + w*, n ~ O, satisfy s ( n a ) = 2 while the elements ( n a ) ' , n ~ O, satisfy s ( ( n a ) ' ) = co. For this example, if x _L y, then s ( x ~ y ) = m a x ( s ( x ) , s ( y ) ) . O f course, this equation does not hold for an arbitrary SEA. Notice that o9 + o9* is not isotropically finite and is not a ~r-SEA. Since o9 + o9" does not admit an order determining set of states, o9 + o9* is a commutative SEA that is not isomorphic to a fuzzy set system.

106 8.

s. GUDDER and R. GREECHIE

Horizontal sums

Let (Ei, Oi, li, ~ i ) , i ~ I, be a collection of effect algebras with c a r d ( l ) > 1. Their horizontal sum E = H S ( E i , i ~ I) is defined as follows. Identify all the 0i with a single element 0 and all the li with a single element 1. Let E~ = Ei\{Oi, 1i} form the disjoint union 0E~ and let E = {0, 1}UOE;. For a, b ~ Ei for some i a I defined a @ b = a ~]~i b and no other orthosums are define on E. It is then easy to check that (E, 0, 1, @) is an effect algebra. If the Ei, i ~ I, are SEA's we now investigate whether H S ( E i , i ~ I) admits a sequential product and is thus a SEA. For an arbitrary effect algebra E we use the notation E ' = E \ {0, 1} and we denote the trivial effect algebra {0, 1} by 2. THEOREM 8.1. Let El, E2 5~ 2 be effect algebras. (i) I f E1 has an atom, then H S ( E 1 , E2) does not admit a sequential product. (ii) I f E1 is an orthoalgebra, then H S ( E 1 , E2) does not admit a sequential product. Proof: (i) Suppose that H S ( E 1 , E2) admits a sequential product o. Let a ~ E1 be an atom and let b ~ E ~ . Then a o b = 0 or a o b ' = 0 . If a o b = 0 , then a = a o b ' = b ' o a < b', which is a contradiction. Similarly, if a o b ' = O, then a=aob=boa
a = [a o (a o b)]

[a o (a o b) 1] = a o b

[(a o b ) ' o a ] .

Since a = ( a o b ) ~ ( a o b ' ) and aob' ~ O, we have a ~ aob. Hence, 0 < (aob)'oa < 1. It follows that ((a o b ) ' o a ) o b 5~ 0, which is a contradiction. [] Applying Theorem 8.1, if H S ( E 1 , E2) is a SEA then neither E1 nor E2 can be a nontrivial Boolean algebra or oJ + o2*. The next result characterizes horizontal sums of SEA's that admit a sequential product and hence form a SEA. If E, F are effect algebras, an additive map ~b : E ~ F is positive if ~b(a) = 0 implies a = 0. THEOREM 8.2. Let Ei, i E I, be SEA's and let E = H S ( E i , i E I). Then E admits a sequential product if and only if for every a ~ E~ there exists a positive

SEQUENTIAL PRODUCTS ON EFFECT ALGEBRAS

107

additive map eb~i : Ej --+ Ei such that eb]i(1) = a f o r every i, j E I with i ~k j, and if a, b ~ E~, c E Ej with a o b = b o a ~kO, then eb]~b(c) = a oebb(c). Proof: S u p p o s e that E admits a sequential product o. For a 6 E~, define eb~i : Ej --+ Ei, j ~= i, by eb]i(b) = a o b. Notice that a o b ~ Ei b e c a u s e a o b < a. N o w eb)'i is clearly additive and e b y i ( a ) = a. S u p p o s e that a o b = eb]i(b)= 0 and b ¢ 0. T h e n a I b, so that a [ b ' . Hence, a =aob'

=b'oa

~ Ei fqEj.

But then a E {0, 1} which is a contradiction. Hence, eb~i is positive. I f a, b ~ E~, c E Ej with a o b = b o a ¢ O, then

eb;?b (C)

=

(a o b) o c = a o (b o c) = a o cbb (c).

Conversely, suppose eb]i : Ej ---+ El, i, j ~ I, i ~ j , satisfy the given conditions. Define the operation o on E by

aob

if a, b ~ Ei for s o m e i 6 I,

eb)'~(b)

if a 6 E;, b 6 E~, i yk j 6 I.

aob=

We n o w show that o is a sequential product on E. I f a, bl, b2 G_ Ei for s o m e i 6 I with bl -J- b2, then clearly a o (bl ~) b2) = a o bl • a o b2. Otherwise, a 6 E~, bl, b2 E Ej, i ~ j , and we have that

a o (bl ~ b2) = eb]i(bl G b2) = eb~i(bl) ~ ebji(b2) = a o bl @ a o b2. Hence, (S1) holds. that a o b -- 0. I f a, b ~ E j, i ¢ j . T h e n ($4) suppose that a

It is clear that ($2) holds. To show that ($3) holds, suppose b E Ei for s o m e i E I, then b o a = 0, so suppose that a E E~, eb~i(b) = 0, so by positivity, b = 0. Hence, b o a = 0. To verify o b = b o a , where a ~ E~, b ~ E~, i ¢ j . T h e n

a o b = eb}a/(b) = b o a = ebb (a) ~ Ei M Ej. Hence, a o b E {0,1}. I f a o b = 0, then eb]i(b) = 0, so that b = 0 which is a contradiction. If a o b = 1, then eb)~i(b) = 1. Hence, a = eb~i(1) >_ eb~i(b) = 1, so that a = 1 which is again a contradiction. We conclude that a, b ~ Ei for s o m e i ~ I . Hence, a [ b'. Moreover, if a , b ~ E~ and c ~ E j, i ~ j , then a o (b o c) = a o ebbi (C) = eb;?b (C) = (a o b) o c.

108

S. GUDDER and R. GREECHIE

To verify ($5), suppose that c I a and c I b. As before, a, b, c ~ Ei for some i e I and the result follows. [] The next result gives a useful method for constructing a SEA from a horizontal sum of SEA's, E = H S ( E i , i ~ I). Suppose there exist effect algebra morphisms ¢ij : Ei --+ E j, i ~ j ~ I. Define the operation o on E by

a ob = { a ob

if a, b ~ Ei for some i ~ I ,

a o Cji (b) if a 6 E;, b 6 E~, i ¢ j 6 I. COROLLARY 8.3. We have that (E, o) is a SEA if and only if for every a ~ El, b ~ Ej, i ¢ j 6 l, a o b = O implies that a = O or b = O.

Proof: As in the proof of Theorem 8.2, if (E, o) is a SEA and a o b = 0, then a =aob'

=b'oa

~ Ei f)Ej.

Hence, a ~ {0, 1}. If a = 1, then b = 0. Conversely, assume that o satisfies the given condition. For a 6 E~ define the map ¢ja . Ej ~ Ei, i ~ j ~ I, by ¢)a (b) = a o Cji (b). Then ¢]i is additive and ¢]i (1) = a. If ¢~i (b) = 0, then a o b = 0. Since a # 0, by assumption b = 0. Hence, Ca. 31 is positive. Suppose that a, b ~ E~, c ~ Ej with a o b = b o a # 0. Then

¢aob,~.c) ---- (a o b) o ¢ji(c ) = a o (b o ¢ji(c)) = a o ¢b.(c). It follows from the proof of Theorem 8.2 that (E, o) is a SEA.

[]

EXAMPLE. Let E1 = E ( H ) , E2 = [0, 1]. Let W ~ D ( H ) be such that t r ( W A ) = O, A e E ( H ) implies that A = 0. For example, let xi be an orthonormal basis for H and denote the one-dimensional projection onto the span of xi by Pi. Then W = Y~ ~,i Pi, where ~i > 0, ~ ~'i = 1, satisfies this condition. Define ¢12 : E1 ~ E2 by ¢12(A) = t r ( W A ) and ¢21 : E2 --+ E1 by ¢21(3.) = LI. Let E = HS(E1, E2) and define o as above. If A o ~. = 0, then

~A=Ao¢21(X)=O, so that 3 . = 0 or A = 0 . If ~ . o A = 0 , By Corollary 8.3, (E, o) is a SEA.

then ~ . t r ( W A ) = 0

so that 3 . = 0

EXAMPLE. Let E1 = C(H1), E2 = E(H2). Let Wi ~ 79(H/) be such = 0 implies that Ai = 0, i = 1, 2. Define ¢ i j " Ei ~ Ej b y ¢ij(Ai) i , j = 1,2, i ~ j . Let E = HS(E1, E2) and define o as above. If then tr(WeAa)A1 = 0 so that A1 = 0 or Aa = 0 and similarly for A2 Corollary 8.3, (E, o) is a SEA.

or A = 0 .

that tr(WiAi)

----tr(WiAi)I, A~oAa =0, o A1 = 0. By

These examples show that a sequential product need not be unique. Just choose different density operators. In general, when E = HS(Ei, i ~ I) is a SEA then E

SEQUENTIAL PRODUCTS ON EFFECT ALGEBRAS

109

is noncommutative. In fact, if a ~ E~ and b 6 E~, i ~ j , then a o b ~ b o a. Indeed, if a o b = b o a , then a o b < a , b , so that a o b 6 { 0 , 1 } . If a o b = l , then a = 1 which is a contradiction. If a o b = 0, then a I b' and by the previous reasoning a o b ' = 0. But then a = a o b ~ a o b'----0, which is again a contradiction. This also shows that C ( E ) = {0, 1}. EXAMPLE. Contrary to E ( H ) , that a o b = b o a. Take E = i # j 6 {1, 2}, is the natural noncommutative SEA in which

9.

a o (b o c ) =

(a o b ) o c for every c does not imply = [0, 1] and ~ij : Ii ~ Ij, identification. Then by Corollary 8.3, (E, o) is a a o (b o c ) = (a o b ) o c for all a, b, c ~ E.

HS(I1, I2), where Ii

Tensor products

If E and F are SEA's, a SEA-morphism from E to F is a map ~p • E --~ F that is an effect algebra morphism satisfying cp(a o b ) = ~b(a)o ~b(b). If E, F and G are SEA's, a SEA-bimorphism from E × F to G is a map /3 • E × F --~ G that is an effect algebra bimorphism [5, 11] satisfying

/3(a, b) o/3(c, d) =/3(a o c, b o d). The SEA tensor product of two SEA's E, F is a SEA T and a SEA-bimorphism r'ExF--+ T such that (T1) every a ~ T has the form a = q~(al, bl) ~ . . .

~ ok(an, b,,),

(T2) if /3 " E × F --+ G is a SEA-bimorphism, then there exists a SEA-morphism ~b • T --+ G such that /3 = q~ o r. It is easy to see that if the tensor product exists, then it an isomorphism. There are many unanswered questions about example, we do not know whether the SEA tensor product characterization of when the SEA tensor product does exist. shall only give some examples of SEA-bimorphisms and SEA

is unique to within tensor products. For always exists or a In this section we tensor products.

EXAMPLE. Let £ ( H 1 ) ® £(H2) be the standard Hilbert space tensor product and define A o B = A1/ZBA 1/2 as usual. Define 13 • B(Hz) x C(H2) --+ £(H1) ® E(H2) by /3(A1, A2) = A1 ® A2. Then /3 is an effect algebra bimorphism and /3(Al, A2) o/3(Bl, B2) = A1 @ A2 o Bs @ B2

= (A1 ® A2)l/2B1 ® B2(A1 ® A2) 1/2 __ A1/2

,t 1 / 2 o o al/2 A1/2 --'~1 ® za 2 ~1 ® D2za I ® , l 2 a l / 2 D zll/2 a l / 2 D .41/2 za 1 D I , ~ I ® Z a l D2Zal

= fl(A1 o B1, A2 o B2) Hence, fl is a SEA-bimorphism.

110

s. GUDDER and R. GREECHIE

EXAMPLE. For the SEA's, E = 2 x, F = 2 r, G = 2 x × r define the effect algebra bimorphism /~ : E x F ~ G by fl(A, B) = A x B. Then 13 is a SEA-bimorphism because fl(A1, A2) o fl(B1, B2) : (A1 × A2) f) (B1 × B2) ---- (A1 A B1) x (A2 N B2) = fl(A1 o B1, A2 o B2).

THEOREM 9.1. If X ~ 0 is a set and E is a SEA, then the SEA tensor product of 2 x and E exists.

Proof: We call a function f : X --+ E simple if f has a finite number of values and we define T = { f ~ E x " f is simple}. On T define f _1_g if f ( x ) _1_g(x) for all x 6 X, and if f _k g f ( x ) ~ g(x). Defining 0(x) = 0 and l ( x ) = 1 for all x 6 X, that (T, 0, 1, @) is an effect algebra. For f , g E T define f o Again, it is easy to check that (T, 0, 1, @, o) is a SEA. Define

r(A,a)(x)=

l

a 0

define ( f @ g ) ( x ) = it is easy to check

g(x) = f ( x ) o g(x). r : 2 x x E ~ T by

if x ~ A , if x C A .

We use the notation r ( A , a) = Xaa. It is clear that r is an effect algebra bimorphism. Since r ( A , a) o r ( B , b) = (Xaa) o (x~b) = XanBa o b = r ( A o B , a o b) we see that r is a SEA-bimorphism. Moreover, every f ~ T has the unique representation

f =

+

XAiai --~

i=1

r ( A i , ai), i=1

Ai f-] Aj = 0, i ~ j and UAi = X. Let fl : 2 x x E ~ F be a SEA-bimorphism. Define ~b : T ~ F as follows. If f = f~Xaiai is the unique representation of f , then ~b(f) = ~ ( A i , a i ) . Notice that ~ f l ( A i , a i ) is defined because f~fl(Ai, 1) = 1 and fl(Ai, ai) < ~ ( A i , 1). It is straightforward to show that ~b is a SEA-morphism. Moreover, w h e r e ai 5~ a j,

fl(A, a) = dP(xaa ) = dpor(A, a).

[]

A slightly more delicate argument than that used in Theorem 9.1 can be employed to show that the SEA tensor product of a Boolean algebra with an arbitrary SEA always exists. Let X ~ 0 be a set, Q be the rational numbers and define 0r(X) = {u : X ~

Q M [0, 1], u simple}.

Then Or(X) is a fuzzy set system and thus forms a SEA.

SEQUENTIAL PRODUCTS ON EFFECT ALGEBRAS

111

THEOREM 9.2. The SEA tensor product of 3r(X) and £ ( H ) exists.

Proof: Let E = E ( H ) and define the SEA T as in the proof of Theorem 9.1. Define r : ,T(X) x E --~ T by r ( u , a ) ( x ) = u(x)a. Then r is clearly an effect algebra bimorphism. Since r(u, a) o r(v, b) = (ua) o (vb) = uva o b = r ( u o v , a o b), r i s a SEA-bimorphism. As in Theorem 9.1, every f 6 T has the unique representation f = ~ ' r ( X a i , ai). Let fl : $'(X) x E --+ F be a SEA-bimorphism. As in Theorem 9.1, define q~ : T ---, F by ~b(f) = ~fl(Xai, ai). Again, ~b is a SEA-morphism. Moreover, if u 6 $'(X) then u has the unique representation u = Y~.)~iXA~, where ~'i ~ )~j, Ai n A j = 0, i ~ j and U A i = X. It is straightforward to show that fl(ru, a) = fl(u, ra) for every r 6 Q N [0, 1]. We then have that

t~(U, a) : t~( ~))~i XAi , a) = (~)l~(~,i XAi,

a) = ~fl(XAi , Xia)

=(~(~,iXAi,a) =~(~--~,iXAi,a)= q~ o r ( u , a ) .

[]

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