SEQUENTIAL PRODUCTS ON EFFECT ALGEBRAS

Stan Gudder Richard Greechie Department of Department of rvIathematics U niversity of Denver Louisiana Tech University Denver Colorado 80208 Rusto且， Louisiana 71272 [email protected] [email protected]

Abstract A sequential effect algebra (SEA) is an effect algebra on which a sequential product with natural properties is defined. The propertíes of sequential products on Hi1bert 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 ís 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 frorn being SEA's. For example, there are no finite nonboolean SEA's. A measure of sharpness called he sharpness index is studied. The existence of horizontal sums of SEA's is characterized and exarnples of horizontal sums and tensor products are presented.

1 Introd uction

Two measurement5 αand b cannot .be performed 5imultaneously in generaL 50 they are frequently executed 5equentially. 飞斗飞~ denote by α 。 b a sequen­ tial measurement in whichαis performed first and b second. \Ve cal1 α 。 b

l the sequential product of αand b. \Ve shal1 restrict our attention to yes-no measurements. called effects, which have only two possible results. For gen­ erality, we do not 臼sume that effects are perfectly accurate me臼urements. That is. they may be fuzzy or unsharp. :也5 we shall see, the sharp effects are those that satisfy α 。 α=α. A paradigm situation is an optica1 bench in which abeam of partic1es prepared in a certain state is injected at the left and then impinge first upon a filterαand then upon a filter b. Particles that p臼s through both filters enter a detection device at the right of b. Because of quantum interference. the order of placement of αand b usually makes a difference and we have α 。 b#b 。 α. 1f it happens thatα 。 b = boa we say thatαand b are sequentially independent and writeαI b. 1n recent years quantum effects have been studied within a general a1ge­ braic frame 飞;\'ork called an effect algebra. 1n Section 2 we sha11 summarize the basic de且 nitions concerning effect algebras and the properties of sequential products on Hilbert space effect a1gebras. The simplest of these properties are employed as a.xioms in Section 3 for a sequential effect algebra (SE.-\). A SE:\. is an effect algebra on which a sequential product with natural proper­ ties is defined. \Ve be1ieve that the a.xioms for a SEA are physically motivated and can be tested, for example, in the optical bench situation. Various prop­ erties of a SE人 are proved in Section 3. For instance, relationships between sequential independence, coexistence and compatibility are given. 1t is also sho\\'n that the sharp elements form an orthomodular poset. The sequential center C(E) of a SEA E is the set of elements αε E such thatαI b for every b ε E. 1n Section 4 it is shown that C(E) coincides with the set of sharp central elements which has previously been studied. :\loreover. a characterization is given for when C(E) is isomorphic to a fuzzy set system. Section 5 shows that the existence of a sequentia1 product is a strong restriction that eliminates many effect algebras from being SEA 's. It is shown' that a admits a unique sequentia1 product and that certain effect algebras admit a sequential product only if they are Boolean. ~Ioreover. it is prO\'ed that if a map preserves the sequential product then it completely preseryes the effect algebra structure of the sharp elements. Section 6 defines the sharpness index of an effect. 1t is demonstrated that if E is isotropically 且 ni

2 tions 8 and 9. The existence of horizontal sums is characterized and some examples of horizontal sums and tensor products are given.

2 Hilbert Space Sequential Products

This section summarizes the b臼 ic definitions concerning effect algebr臼 [1 ， 6, 7, 8, 14, 15] and the properties of sequential products on Hilbert space effect algebr出 [2. 3. 10, 12, 13]. If EÐ is a partial , we write α -L b ifαEÐ b is defined. .\n effect algebra is a system (E, 0, 1, EÐ) where 0, 1 are distinct elements of E and EÐ is a partial binary operation on E that satisfies the following conditions.

(E1) If a • b, then b 1. αand bEÐ a =αEÐ b. (E2) If α -L b and c -L (αEÐ b) , then b 1. c ， α 1. (b EÐ c) andαEÐ(bEÐc)= (αEÐ b) EÐ c.

(E3) For every αE E there exists a unique a' E E such that α 1. a' and αEÐ a' = 1.

(E4) 1fα -L 1, then α=0.

1n the sequel ，飞，yhenever we writeαEÐ b we are implicit!y assuming that α -L b. 飞飞，~e define α ~ b if there exists a c E E such that αEÐ c = b. If such a c E E exists then it is unique and we write c = b e α. It can be shown that (E ， 三，') is a with 0 ~二 α 三 1 for all αεE.α"α. and α 三 b implies b' ~ a'. Moreover, we have α 1. b if and only ifα 三 b'. If α4α we call αan isotropic element and when 0 is the only isotropic element of E, then we call E an orthoalgebra. If the n-fold orthosum a EÐα@ … @α is defined in E we denote this element of E by nα. If there is a largest ηEN such that. na is defined, then ηis the isotropic index of αand if no such n exists. then ηhas isotropic index ∞. An element αε E is sharp ifα ^a' = O. Xotice that ifα 笋 o is sharp then a has isotropic index 1. \Ve say that E is isotropically finite if everyα 笋 o in E has finite isotropic index. 飞飞'e now give some standard examples of effect algebr臼. For a Boolean al­ gebra 8 , defineα 1. bifα ^b = 0 and in this c臼eαEÐb=αvb. Then (8, 0, 1, EÐ) is an effect algebra that happens to be an orthoalgebra. In particular if X is a nonempty set, then (2X ， 白， X , EÐ) is an effect algebra. These effect algebras correspond to classicallogic and set theory. For the space [0 , 1]''\

3 on the inten.al [0, 1] ç lR define the functions fo , h by fo(x) = O. fI (x) = 1 for all x εX. For f , 9 ç [0, 1]'\, we define f 1. 9 if f(x) + g(x) 三 1 for all Z εX and in this case (J æ g)(x) = f(x) + g(x). Then ([0, l]X,fo ,h ,æ) is the effect algebra of fuzzy subsets of X. A particularly simple effect algebra is the interval [0, 1] ç R. For α ， b ε[0 ， 1] we defineα 1. b ifα 十 b 三 1 and in this case α æb= α +b. In this section we are mainly concerned with the set ê(H) of all self­ adjoint operators on a Hilbert space H that satisfy 0 三 (Ax ， x) 三位 ， x) for all x E H. For .4., B ε ê(H) we define A 1. B if A + B E ê(H) and in this case .4 EÐ B = A + B. Then (ê(H), O,!, æ) 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 , 3. 4 ，而， 17]. The quantum effects A ε ê (H) correspond to yes-no measurements that may be unsharp. The set of projection operators P(H) on H form an orthoalgebra that is a sub-effect algebra of ê(H). The elements of P( H) correspond to sharp quantum effects. If E and F are effect algebras, we say that rþ: E • F is additive ifα 1. b implies rþ( α) 1.功 (b) and rþ( αæ b) = φ(α) EÐ rþ(b). If rþ: E • F is additive and 0(1) = 1, then φis a . If 4>: E • F is a morphism and φ(α) 1. rþ(b) implies that α 1. b, then φis a monomorphism. .\ surjective monomorphism is an isomorphism. It is easy to see that a morphism d> is an isomorphism if and only if rþ is bijective and φ-1 is a morphism. .\ state on E is a morphism s: E → [O~ 1]. \Ve interpret s( α) as the that the effectαis observed (has answer yes) when the system is in the state s. We denote the set of states on E by n(E). A set of states S ç n(E) is order determining if s( α) 三 s(b) for all s ε S implies thatα 三 b. The sequential product on ê(H) is defined by A 0 B = .-1 1/ 2 BA 1/2 1 2 where .-1 / is the unique positive square root of A. [2, 3, 10, 12, 13]. We have that .-1 0 B ε ê (H) because

。三 (.-1 1 / 2 BA 1 / 2 X ， X) = (B 川 2 X ， 川 2X) 三 (A 1/2 ι .4 1 / 2 x) = (Ax, x) 空 (x ， x)

for all x E H. :"J' otice that B 叶 AoB is additive on ê(H) and that IoB = B for all B εε (H). We now present some of the important properties of the sequential product on ê(H). If.-1 0 B = B 0 A we say that .4. and B are sequentially independent and write A I B. The following result is proved in [12. 13]

4 Theorem 2.1. (i) For A. B E E(H) ， 矿.-l o B E P(H) then A. B = B.-l. (ii) For .4, B E E(H), .4 I B 扩 αη d only if AB = BA. Applying Theorem 2.1 we obtain the following properties ofthe sequential product A 0 B.

Corollary 2.2. (i) 扩 A 0 B = 0, then B 0 A = O. (ii) 扩 A I B , then .-l 1 B' α nd .-l o (B 0 C) = (A 0 B) 0 C for α II C E E(H). (iii) 扩 CIA α ndC I B then C I A 0 B α nd C I (A E9 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 .4, B E E(H) , A 0 (B 0 C) (A 0 B) 0 C for every C E E(H) if α nd only if A I B.

Theorem 2.4. For f1 , B ε E (H) the following stαtements are equivalent. (i) A 0 B = B. (ii) B 0 A = B. (iii) AB = BA = B.

Theorem 2.5. For A, B E E(H) ω e have B = A 0 B E9 A' 0 B if and only if

.-1 1 B.

\Ve denote the set of positive trace class operators with trace 1 on H by 1J (H). The normal states on E(H) have the form Pw(.4) = tr(vV.4) for some n'ε 1J (H). \Ve say that A, B E E(H) are stochastically indepen­ dent relative to ~V E 万 (H) if Pw(.4 0 B) = ~以 A)Pw(B). The next result is proved in [13].

Theorem 2.6. For .4, B E E(H) the following statements α陀 equivalent. (i) .-l o ( C 0 B) = (.-l 0 C) 0 B for every C E E ( H). (ii) C 0 (.4 0 B) = (C 0 .4.) 0 B for every C E E(H). (iii) Pw(AoB) = Pw(A)Pw(B) for every lV E 1J (H). (iv) A. = CI or B = CI for some 0 三 C 三1. \Ve close this section 飞，vith 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 之 3. Theorem 2.7. Suppose that dim H ~二 3. 扩 φ: E(H) • E(H) is α bijection sα tisfying cþ( .-l 0 B) = 4>(.4) o 4> (B) for α II A, B E E(H) , then ø is α n effect algebra isomorphism.

3 Proof. For A E ê(H) we have 2 4> (A ) = 4> (A 0 A) = 4> (A) o 4> (A) = 4> (A)2

Hence, for .4, B E ê(H) we have 2 2 4> (ABA) = 4> (A 0 B) = 4> (A ) o 4> (B) = 4>(A)2 0 4> (B) = rþ(A)rþ(B)rþ(A)

Applying Theorem 2 [18] , 4> has the form rþ(A) = U AU. where U is either a unitary or an anti-unitary operator on H. It easily follows that rþ is an effect algebra isomorphism. 口

3 Sequential Effect Algebras

This section summarizes the basic definitions and results for sequential effect algebras. For a binary operation 0 , ifα 。 b=b 。 αwe write a I b. A sequential effect algebra (SEA) is a system (E, 0, L EÐ , 0) where (E, 0, 1, EÐ) is an effect algebra and 0: E x E • E is a binary operation that satisfies the following conditions.

(S1) b 叶 α 。 b is additive for allαε E.

(S2) 1 。 α=αfor all a E E.

(S3) Ifα 。 b = 0, then αI b.

(S4) IfαI b, then αI b' and α 。 (boc) = (α 。 b) 0 c for all c E E.

(S5) If c Iαand c I b, then c Iα 。 b and c I (αEÐ b). 飞毛￥ call an operation that satisfies (S1)-(S5) a sequentia1 product on E. IfαI b for all α . b E E we call E a commutative SEA. The effect algebra (0, 1] ç lR is a SEA with sequential product a 0 b = α b. Corollary 2.2 shows that ê(H) is a SEA under the operation A 0 B = A. 1/2 B A 1/2. NIore generally, this operation makes any von Neumann alge bra a SEA. A Boolean algebra is a SEA under the operation α 。 b= α^ b. Let X be a nonempty set and let :F ç [0, 1]X. We call :F a fuzzy set system on X if 10 ， /1 ε :F， if 1 E :F then !t - f E :F, if f , 9 E :F with 1 + 9 三 l then 1 + 9 E :F and if 1, 9 E :F then f 9 E :F. Then :F becomes a SEA when

6 f $ 9 == f + 9 for f 十 g 三 1 and f 0 9 == f9. 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)α 。 0=0 。 α= 0 and α 。 1 = 1 。 α=α for α II αE E. (ii)α 。 b 三 α for all α ， b E E. (iii) 牙 α 三 b ， then c 。 α 三 co b for αllc ε E. (iv) lf α 三 b ， then co (be α )=cobec 。 α. (v) If α 三 b ， clα and c I b, then c 1 (be α) . Proof. (i) By additivity we have

α 。 0$0=α 。 0=α 。 (0 $ 0) =α 。 0$α 。。 and by cancellation we have α 。 o = O. Applying (S3) gives 0 。 α= O. Since 01α ， applyíng (S2) and (S4) gives α 。 1=1 。 α=α. (ii) Applying (S1) gives

α=α 。 α=α 。 (b $ b') = α 。 b$ α 。 b' 主 α 。 b

(iii) If α 三 b there exists a d E E such that α $ d = b. Hence,

co b = c 。 α $cod~co α

(iv) If α 三 b ， then by (iii) c 。 α 三 c 0 b. Since α $(be α) = b we have that coa$co(be α) == c 0 b. Hence, c 0 (b e α ) =cobeco α. (v) This follows from (S4), (S5) and the identity b e α==(α$ b')'. 口 We denote the set of sharp elements of E by Es. It is clear that 0, 1 E Es and that a' E E s wheneverαε Es. Lemma 3.2. The fol1owi叼 statements a陀 equivalen t. (i)αE Es. (ii)α 。 α， = O. (iii)α 。 α=α. Proof. To show that (i) imp1ies (ii) suppose that α ^a' = O. By Lemma 3.1(ii) we have

α 。 a' =α' 。 α 三 α ， α'

Hence.α 。 a' = O. To show that (ii) implies (iii) suppose thatα 。 α.' = O. Then

α=α 。 α@α 。 α-α 。 α

7 To show that (iii) implies (i) suppose that α 。 α=α. Then

α=α 。 α@α 。 α-α@α 。 d so by cancellation α 。 α ， = O. If b 三 α ， a', then by Lemma 3.1(iii) we have that a 0 b ~二 α 。 a' = O. Hence ， α 。 b = 0 and similarly a' 0 b = O. Hence, b 。 α =b 。 α ， = 0 so that b=b 。 αæb 。 α'=0

Hence.α 〈 α ， = O. 口 Lemma 3.3. (i) lf α 。 b = 0, then α .L b. (ii) For αε E ， b E Es ， α 。 b=O if and only if α .L b. (iii) For α~ b E Es, wíth α .L b we hα ve αæ b E Es. (iv) For a, b E Es wi的 α 三 b we have b e αε Es. (v) For α ， b ε Es wi仇 αI b we have α 。 b E Es.

Proof. (i) If α 。 b = 0, then b 。 α=α 。 b = O. Hence, α=α 。 bæα 。 b' = α 。 b' = b' 。 α 三 b'

(ii) Ifα 三 b' then b 。 α 三 b 0 b' = O. Hencè ， α 。 b=b 。 α=0. (iii) Since α 。 b = 0 we have

(αæ b) 0 (αæ b) = (α æb) 。 αæ(α æb) ob =α 。 (α æb) æbo ( α æb)= αæb

(i\') Thi5 follow5 from iii) and the identity b e α=(αe b')'. (v) Since αIb 飞ve have

(α 。 b) 0 (α 。 b) = α 。 [b 0 (α 。 b)] = α 。 [b 0 (b 。 α)] =α 。 [(bob) 。 α]=α 。 (b 。 α)=α 。 (α 。 b) = (α 。 α) ob = α 。 b 口

It follows from Lemma 3.3(iii) that Es is a sub-~ffect algebra of E that i5 an orthoalgebra. In general, ifα . b ε Es then α 。 b ~ E s 50 E S i5 not a sub-SE人 of E. Theorem 3.4. Let αεE α nd b E Es. (i)α 三 b if and only if α 。 b=b。 α=α αnd b 三 α if α nd only if α 。 b = b 。 α = b. (ii) 1f αI b, then α ^b= α 。 b. ( iii) 扩 α .L b. then α æb= αvb=(α ， 0 b')'.

8 Prool (i) If b 。 α=α. then α = b 。 α 三 b. Similarly: ifα 。 b = b then b 三 α. Conversely, suppose thatα 三 b. Then α 1.. b' so by Lemma 3.3(ii) α 。 b' = b' 0 a = O. Hence ， αI b and we have

α=α 。 bæα 。 b' = α 。 b

If b 三 αthen a' :三 b' so from our previous work ， α ， 0 b' = b' 0 a' = α ， Hence,

b 。 α ， = b 0 (b' 。 α') = (b 0 b') 。 α'=0 and we have

b=b 。 αæb 。 α'=b 。 α=α 。 b

(ii) We have α 。 b=b 。 α 三 α ， b. Suppose that c 三 α ， b. Then there exists a d E E such that c æ d = αand by (i) we have b 0 c = c. Hence,

b 。 α = b 0 (c æ d) = b 0 c æ b 0 d = c æ b 0 d > c

飞~e conclude that α ^b= α 。 b. (iii) It is clear that α ， b 三 αæ b. Suppose that α . b ~二 c. Then there exists a d E E such that αæ d = c and by (i) we have c 0 b = b 0 c = b. By Lemma 3.3(iiLα 。 b=b 。 α= 0 so that

b=boc=bo( α æd) = b 。 α æbod=bod

主 pplying Lemma 3.1(v) we have b I (c e α). Hence, b I d and b = d 0 b 三 d. It follows that a æ b 三 αæ d = c. Hence, by (ii) we have that

α æb= αvb=(α ， ^ b')' = (α ， 0 b')' 口

Corollary 3.5. E A is α sub-effect α 1gebrα， 01 E that is an orthomodular poset.

Proof. This follows from Theorem 3.4(iii). 口

叭;e say that α ， b E E coexist if there exist c, d. e E E such that c æ d æ e is de 自 ned and α =cæd ， b=cffie [16, 17].

Theorem 3.6. (i) 厅 αI b then ααnd b coexist. (ii) For αε E ， b E Es, αIb 矿 and only if α and b coexist.

9 ProoJ. (i) IfαI b, then α=α 。 b E9 α 。 b' and

b=b 。 α E9 b 。 α-α 。 b E9 a' 0 b Now

1=α@α-α E9 (α ， 0 b E9 α ， 0 b') = (α@α ， 0 b) E9 α ， 0 b' Hence.

(α' 。的 -α@α ， 0 b = α 。 b E9 α 。 b' E9 α ， 0 b It follows thatαand b coexist. (ii) If a and b coexist, thenα = cE9 d, b = cEg e for some c, d, e E E where c E9 d E9 e is defìned. Since c 三 b we have b I c. Since d .1 b we have d 三 b'. Hence, b I d and it follows that b Iα. 口 It is well-known that the converse of Theorem 3.6(i) does not hold [3]. We say thatα ， b E Es 缸e compatible if there exist mutually orthogonal elements c, d, e E Es such thatα= c v d and b = c V e. Corollary 3.7. Fora,b E ES1 α Ib 矿 αnd only 矿 α and b α陀 compα tible. ProoJ. If a I b, then by the proof of Theorem 3.6 and Lemma 3.3(v) ， α 。 b ， α 。 b' and a' 0 b are mutually orthogonal elements of Es. By Theorem 3.4(iii) we have α=α 。 b E9 α 。 b' = α 。 bv α 。扩 and

b= α 。 b E9 a' 0 b = α 。 b V a' 0 b Hence, a and b are compatible. Conversely, suppose thatαand b are com­ patible and α= c V d, b = c V e where c, d, e E Es are mutually orthogonal. By Theorem 3 .4 (iii) ， α= c E9 d, b = c E9 e. Since e .1 c, e .1 d, we have e .1 α­ Hence, c E9 d E9 e is defined. Thus ， αand b coexist so by Theorem 3.6(ii) we have thatαI b. 口 Corollary 3.8. A SEA is α commutα tive 0付hoα 1gebra if and only if it is α Boolean algebra. 1n a commu.tative SEA E , Es is α Booleαnα 1gebra.

The next result shows that for certain special cases, 0 and E9 are closely related.

Corollary 3.9. (i) 1f αI b and α .1 b, then a E9 b = α 。 b E9 (α ， 0 b')'. (ii) 1f α . b E Es and α .1 b, then α E9 b=( α， 0 b')'. (iii) 1f α ， b E Es and a' .1 b': then α 。 b = (α， E9 b')'.

10 4 "The Sequential Center of a SEA

马rve define the sequential center of a SEA E as

C(E) = {αε E: αI b for all b E E}

The next result follows from our previous work. Theorem 4.1. (i) C(E) is a sub-SEA of E which is α commutative monoid under o. (ii) C(E) n Es is a Boolean algebra. An element αε E is principal 证i f b, c 三 αand b 1. c imply 也t ha剖tbEÐc~三二 α. It 丽i s 51 conver5e hold5 in a SEA. Lemma 4.2. An element αε E is p时ncipal if and only if a E Es. Proof. Suppose αis principal and b ~二 α ， a'. Then b 1. αand b ， α 三 α. Hence, αEÐ b ~二 α=αEÐO

By cancellation b = 0 so that α^ a' = O. Conversely, suppose that a E Es and b, c ~二 αwith b .L c. By Theorem 3 .4 (i)α 。 b = b and a 0 c = c. Hence,

α 三 α 。 (bEÐc) = α 。 bæ α 。 c=bEÐc 口

The next result holds for an arbitrary effect algebra [9]. However, the proof is much simpler for a SEA. Corollary 4.3. (i) If α ， b E Es and α ^b 臼ists ， then α 八 b E Es. (ii) 厅 α ， bE Es α nd αv b exists, then αvb ε Es. Proof. (i) By Lemma 4.2 ， αand b are principal. Supp05e that c, d 三 α^band c .L d. Then c 三 α ， b and d ~二 α ， b so that c EÐ d ~二 α ， b. Hence, c æ d ~二 α^ b. Thus ， α^ b is principal so by Lemma 4.2 ， α^ b E Es. (ii) Since αv b exist5, a' /\ b':: ( αv b)' exists and is sharp. Hence, a v b E Es. 口 .-\.ccording to [9], an element αE E is central if a, a' are principal and for every p E E tÈere exist q, r E E such that q 三 α r 三 a' and p = q æ r. In [9] the center C(E) is defined to be the set of all central elements. Theorem 4.4. (i) If αε Es ， then αIp 扩 and only 矿 there exist q, r E E such that q 三 α.r 三 dαnd p = q æ r. (ii) C(E) = C(E) n Es.

11 Proof. (i) Suppose p = q 6 r where q 三 α ， r 三 a'. Since αI q and αI r we have that αI p. Conversely, suppose that αI p. Then p = p 。 αæ p 0 a' and we have that p 。 α=α 。 p 三 αand po a' =α ， 0 p 三 a'. (ii) IfαE C(E), then by Lemma 4.2.αε Es. Moreover, by (i) we have thatαε C(E). Hence , C(E) ç C(E) n Es. Conversely, ifαε C(E) n Es, then by (i) we have that αε C(E). Hence, C(E) n Es ç C(E). 口

Example. Let F be the set of differentiable functions f: [0, 1] • [0,1]. Then F is a fuzzy set system under the previously defined partial operation f æ 9 = 1 + 9 if 1 + 9 ~二 1 and the operation 1 0 9 = 1g. Hence, F is a commutative SEA 50 that C(F) = F. However, F is not lattice order because 1 八 9 does not exist in F in general. This shows that a commutative SEA need not be an MV-algebra. 口

This example showed that C(E) need not be lattice ordered. If E and F are SEA、 a map <þ: E • F is an isomorphism if <þ is an effect algebra isomorphism satisfying Ø( α 。 b) = <Þ( α) 0 Ø(b) for all 矶 b E E. \Ve 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 a is called multiplicative if s( α 。 b) = s( α )s(b) for all α ， b. Theorem 4.5. The sequential center C(E) is isomorphic to α fuzzy set sys­ tem if and only 矿 C(E) α dmits an order deterrnining set 01 multiplicative stα tes.

Proof. Suppose F is a fuzzy set system on 0 and <þ: C(E) • F is an isomor­ phism. For ωε0 ， αε C(E) defineω(α) = <Þ( α)(ω). Then 叫 1) = <þ (1)( ω)= 1.If α .1 b we have 马~(α 6b)= φ(α æb)(ω) = [<þ( α) æ <þ (b)] ( ω)=φ(α)(ω) + <þ (b)(ω) =ω(α)+ω (b)

Hence, {山':ωεn} forms a set of states on C(E). These states are multi­ plicative because

ω(α 。 b) = φ(α 。 b)(ω) = [4> (α ) 4> (b)] ( ω) = ø( α)(ω)φ (b)(ω) =ω(α)ω (b)

12 To show that this set of states is order determining, suppose that w( α) 三 ω (b) for every :..J E f2. Then Ø( α)(ω) 三 Ø(b)( ω) for every ωE f2 so that Ø( α) 三伊 (b). Since φis an isomorphism, we have that α 三 b. Conversely, suppose f2 is an order determining set of multiplicative states on C(E). Define ø: C(E) • [0,1]0 by Ø( α)(ω)=ω(α) and let :F ç [O , l]n be the range of ø. Since Ø(1) = h and tþ(O) = 10 , 11 ， /0 ε :F. For Ø( α) E :F we have

[!t - Ø( α)] (ω)=1-ω(α)=ω(α') = Ø( α')(ω) so that !t一 φ(α) E :F and 伊 (α') = 11 - Ø( α). If α J.. b, then

币 (α)(ω)=ω(α) 三 ω (b') = 1 - Ø(b)(ω)

Hence, Ø( α) J.. Ø(b) and we have

功 (α Ei1 b)(ω)=ω(α Ei1 b) = ω(α)+ω (b) = Ø( α)(ω) + Ø(b)(ω)

for every ωε f2.飞，ve conclude that tþ( α Ei1 b) = tþ( α ) EBtþ(b). For 功 (α) ， Ø(b) E :F with φ(α) + ó(b) 三 1 we have ω(α) 三 ω (b'). Since f2 is order determining, we have thatα J.. b. Hence, ø is a monomorphism and Ø( α) + tþ(b) ε :F. For Ø( α) ，仿 (b) E :F we have

[tþ( α) 功 (b)] ( ω) =φ(α)(ω )tþ(b)(ω)=ω(α)ω (b) = ω(α 。 b) = Ø( α 。 b)(ω)

for everωε Sl. Hence, Ø( α )tþ(b) E :F and Ø( α 。 b) = Ø( α )Ø(b). We conclude that :F is a fuzzy set system and ø: C(E) • :F is an isomorphism. 口 Corollary 4.6. A SEA E α dmits αn order determining set 01 multiplicative states n if α nd only if E is 臼 omorphic to a luzzy set system f2. Under this isomorphism. Es is a Booleαn α 1gebra 01 subsets 01 n. Proof. For αE Es we have tþ( α)2 = tþ( α 。 α) = Ø( α). Hence ， φ(α) is a characteristic function which can be considered to be a subset of f2. The result now follows. 口

5 Existence of Sequential Products

This section shows that the existence of a sequential product is a strong restriction that eliminates many effect algebvras from being SE 人 's.

13 Lemma 5.1. For an effect algebra E that is a Boolean algebra there is a unique sequential product α 。 b= α 八 b. Proof. Since E = Es and any two element5 of E are compatible, by Corol­ lary 3. ï we have thatαI b for every α ， b E E and any 5equential product o. It follow5 from Theorem 3.4(ii) thatα 。 b= α 八 b. 口 Lemma 5.2. Let E be"a SEA. (i) 厅 αE E is αn atom then for every b E E either α 三 b orα 三扩. (ii) 厅 α ， b E E are distinct atoms, then α J.. b.

Proof. (i) Since α 。 b 三 αwe have a 0 b = 0 or α 。 b= α. By Lemma 3.3(i), if α 。 b = 0, then α 三扩.If α=α 。 b then α 。 b' = 050 again by Lemma 3.3(i) we have that α 三 b. (ii) By (i) we have that α J.. b orα 三 b. In the second case ， α = b which i5 a contradiction. 口

Theorem 5.3. An atomistic orthoalgebra E admits a sequential product if and only if E is Boolean. Proof. If E i5 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 E. By Lemma 5.2 we have that c = (Vl;) V (Ved, d = (Vl;) V (Vdi) where l;， 吨，向 are distinct mutually orthogonal atoms. Since (Vei) J.. (Vdi), c and d are compatible. It follows that E is Boolean. 口

Corollary 5.4. There does not exist a sequential product on P(H), dim H 2:

Proof. For dim H 2: 2, P(H) is a nonboolean, atomistic orthoalgebra. 口 Theorem 5.5. (i) An orthoalgebra E is a commutative SEA 矿 αnd 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. Con­ versely~ íf an orthoalgebra E is commutative then by Corollary 3. ï any two elements of E are compatible. It follows that E is Boolean. (ii) Ifαε E is an atom, then a =α 。 α@α 。 α ， implies thatα 。 α= 0 or α 。 α ， == o. Suppose that a 。 α= O. By Lemma 3.3(i)α J.. α50 that 2αeXÏsts. Now

α 。 (2α) ==α 。 (α@α) =α 。 α@α 。 α=0

14 Hence. 2α 1. α50 that 3αexi5t5. Continuing by induction ， παexi5t5 for all n E N. Since α< 2α< 3α 〈… form5 an infìnite chain, thi5 i5 a contradiction. Hence ， α 。 α ， = 0 50 by Lemma 3.2 a i5 5harp. Since the orth05um of 5harp element5 i5 5harp, we have E = Es. Hence, E i5 an atomi5tic orthoalgebra and the re5ult follow5 from Theorem 5.3. 口

Lemma 5.6. Let E be α SEA and let

α=

50 by cancellation we have that

功 (b) EÐ <þ (c) =

50 that fþ(b) = O. If d E E then a' 0 d 三 a' 50 that ( α 。 d) = α 。 d. Thu5, for every d E E we have that

Since a =

Lemma 5.6 can be used to give another proof of Lemma 5.1. Indeed, let α 。 b = α^ b be the 5equential product on E. For αE E = C(E) define 4>a: E • E by 仇 (b) = α • b where α • b i5 another 5equential product on E. Then fþa 5ati5fie5 the condition5 of Lemma 5.6 50 thatα .b = 仇 (b) = α 。 b. Let X be a finite nonempty 5et and let .:F(X) be the fuzzy 5et 5y5tem F(X) = [0, l]X. A 5equential product 0 on .:F(X) is homogeneous ifλ fl 。 f=f 。 λ fl = Ãf for all f E F(X) ， λE [0, 1].

Lemma 5.1. .:F (X) α dmits α unique homogeneo 'W3 sequential product f 0 9 = fg.

Proof. It i5 clear that f 0 9 = f 9 is a homogeneou5 sequential product on F(X). Let f. 9 be another homogeneou5 sequential product on .:F(X). Since any two element5 of .:F(X)s are compatible, it follow5 from Corollary 3.7 and Lemma 3.3(v) that :F(X)s is a commutative sub-SEA of (.:F(X), o).

15 For f E F(X)s define 衍: F(X)s • F(X)s by 衍 (g) = f • g. Applying Lemma 5.6, we have that

J.g = 告I (g) = f 0 9 = J 9 Hence, J. 9 = Jg for every J,g E F(X)s. Since J. 9 is homogeneous, we have (λ J) • (μg) = (μg) • (入 J) for every J, 9 E F(X )s， 人 με[0 ， 1]. For any 民 v E F(X) we have U = ~二 λdi ， V = 汇的 gj where λ i ， 的 ε[0 ， 1] , Ji ， 岛 ε F(X )s， i = 1, . .. ， π ， j = 1,... ,m. It follows that U. 岛 = gj.u and v • Ji = Ji • v for every i = 1,. . . ,n, j = 1, . . . , m. Hence,

u.V = 汇 μ川 gj = 汇 μjgj. U = 汇 μjÀigj • Ji

= L >"i J.l. jfi 0 gj = U 0 V

口

The next result shows that if a map preserves the sequential product then it completely preserves the 5tructure of the sharp elements. Theorem 5.8. Let E. F be SEA 's and let cþ: E • F be α bijection sα tisfying ø( α 。 b) = cþ( α) 。 φ (b). Then cþ: Es • Fs is an isomorphism. Moreover， 矿 αε Es or b E Es αnd α 三 b then cþ( α) 三 cþ( b). Proof. If αε Es ， then φ(α) = cþ( α 。 α) = cþ( α) 0 cþ( α) 50 that cþ( α)ε Fs . If b ε Fs then <þ( α) = b for some αE E. But then

cþ( α) = b 0 b = cþ( α) 0 <þ( α) = cþ( α 。 α)

Since φis injective, we have thatα=α 。 αsoαE Es. Hence, cþ: Es • Fs i5 a bijection. Now cþ( α) = 1 for some αE E. Thus,

cþ(l) = <þ (1) 0 1 = cþ (1) 0 cþ( α) = cþ (1 。 α) = cþ( α) = 1

Similarly.φ (b) = 0 for some b E E and we have

φ(0) = cþ(O 0 b) = φ(0) 0 cþ(b) = 0

If αε Es or b E Es withα 三 b then α 。 b=b 。 α=α. Hence.

cþ( α) = cþ(b 。 α) = rþ(b) 0 cþ( α) 三 cþ(b)

16 Therefore, 1>: Es • Fs preserves order. If αE Es,

。 (α) 0 1> (a') = 1> (α 。 α') = 1> (0) = 0

Applying Lemma 3.3(ii) we have that 1> (α') 三1> (α)'. Now 1> (b) = 1> (α)' for sorne b E E s. Since

1> (α 。 b) = 1> (α) 0 1> (b) = 1> (α) 。币 (α)' = 0 we have that α 。 b = O. By Lernrna 3.3(ii) we have that b 三 a' 50 that <Þ( α)' = 1> (b) 三 φ(α'). Hence ，币 (α') = 1> (α)'. Ifα ， b E Es and α 1. b then applying Theorern 3.4 (iii) gives αEÐb=(α ， 0 b')'. Since 1> (α) 1. 1> (b) we have that

1> (αEÐ b) = 1> ((α ， 0 b')') = [1> (α' 。扩)]' = [1> (α') o 1> (b')]'

= [1> (α)' 0 1>( b)']' = 1> (α) EÐ 1>(b)

Also, 1>-1 : Fs • Es preserves order because 1> (α) 三 1> (b) irnplies that 1> (b 。 α) = 1> (b) 0 1> (α) = 1> (α)

Hence:α =b 。 α 三 b. Thus ，币: Es • Fs is a monornorphisrn and therefore is an isornorphisrn. 口

The following exarnple show that Theorem 5.8 cannot be strengthened. Let E = [0 , l]X be a fuzzy set systern. Defìne 1>: E • E by 1>(f) = j2. Then 1> is a bijection and o(f 0 g) = 1>(f) o 1>(g). But if f f/. Es then

。 (f') = (1 - 1) 2 # 1 - ]2 =φ (f)'

:vIoreover, if f , 9 f/. Es with f 1. g, then 1>(f EÐ g) # (f) EÐ 1> (g).

6 Sharpness Index

For ηεN， αε E we defìneα n_ α 。 α 。. ..。 α(ηfactors).

Lemma 6.1. (i) We have that α n E Es if α nd only if an+1 = ♂. (ii) Also, απε Es if αnd only if α m_ α n for all m 之 η.

17 Proof. (i) If α n E Es. then α2n = αn. But α2n <二 αn+l <二 αn so that αn+l = αn Conversely, ifα n+l = αn then we have that

n+l _ _ _2n _ I_n 、 2 α=α= … =α = la'" so thatαn E Es. Moreover, (ii) follows from (i). 口 If there exists an m such thatαm E Es, then the smallestn E N such thatα n E E s is the sharpness index of a. If no such m exists, then the sharpness index of αis ∞. We denote the sharpness index of αby s( α). Of course ， αE Es if 姐d only if s(α) = 1.

Lemma 6.2. 扩 n = s(α) <∞ ， then ♂ is the largest sha叩 element below a.

Proof. Clearly ， αn E Es and ♂三 α. Suppose that b 三 αand b E Es. Then by Theorem 3 叫 ii) b 。 α=α 。 b. Hence,

boa2 = α2 ob = α 。 b=b and continuing by induction we have that b 。 αn = αn 0 b = b. Hence, b< α n 口

Aσ-effect a1gebra is an effect algebra such that 副主 α2> … implies that 八αi exists. A. σ-SEA is a SEA that is a σ-effect algebra E satisfying:

(1) If 副主 α2 主…， then b 0 (八向)=八 (b 。向) for every b E E;

(2) If 副主 α2 主… and b I 句， i = 1, 2, . . ., then b I 八αi. 1t can be shown that &(H) is a σ-SEA.

Theorem 6.3. Let E be aσ -SEA.l1 αε E ， then there exist b, c E Es s1J. ch that b is the largest sharp element below ααnd c is the smallest shα叩 element above a. Proof. If n = s(α) <∞， then by Lemma 6.2 ，♂ is the largest sharp element n belowα. Suppose that s( α) = 00. Since a 2三 α2 2:…， b= 八a exists. No \v by (1) we have that

amob = αm 0 (八α n) = 八αn+m = b n n

18 Since αmlαn for n = 1,2,. . ., applying (2) gives am I b. Hence, for all m E N we have that

b= αmob = b 。 αm

Therefore,

2 b = b 0 (^α") = ^(b 。 α ") = b so that b E Es. Clearly, b 三 a. Suppose that d 三 αwith d E Es. Then d=d 。 α=α 。 d so that d 三 α 11 for all n E N. Hence, d 三〈α 11 = b. 巩'e conclude that b is the largest sharp element belowα. Let e be the largest sharp element below a' and let c =ιThen α 三 e' = c and c E Es. If a ~二 f where f E Es, then f' :三 a' . Hence, f' :三 e so that c = e' 二二 f. Hence, c is the smal1est sharp element above α. 口 Theorem 6.3 shows that a u-SEA is sharply dominating [6]. 1叼 articular ， ε (H) is sharply dominating. We denote the isotropic index of an element a by i(α) . Theorem 6.4. 1f 1 < s( α) <∞ ， then there exists an m E N such that ♂。 (αm)' 笋 o and i [am 0 (αm)']= ∞. Proof. Suppose that s(α) = 2. Since α~ Es we have that α 。 αF 笋 O. By Lemma 6.1(ii) , a3 =α2. Then

2 _ _3 "" _2 _ _2 '" _2 α=α 国 α 。 α=α_, _ 时 α 。 α 2 implies that a 0 a' = O. Now

α'=(α')2 æα' 。 α

imp1ies that α' 。 α=α 。 α-α 。 (α')2 æα2 。 α-α 。 (α')2 = (α')2 。 α

Hence.

(a')3 。 α = (a'? 。 α=α' 。 α

Continuing by induction, we have that (a')n 。 α=α' 。 αfor every n E N. N ow (α')2 = (a')3 EÐ (a')2 。 α = (a')3 ea ' 。 α

19 lt follows that α ， = [(α')3 号 α' 。 α] EÐα' 。 α

Hence. 2(αF 。 α) is defìned and a' = (α')3 EÐ 2(α1 。 α). ln a similar way (a')3 == (a')4 EÐ (α') :i 。 α=(α')4 EÐ a' 。 α

50 that α， == [(a')4 e (['。 α] EÐ 2(α' 。 α) Hence ， 3(α'0α) i5 defìned and a' == (a')4 EÐ 3(αF 。 α). Continuing by induction, we conclude that n( α' 。 α) is defìnecl for all n E N. Hence, i(α 。 α') ==∞. For the general c臼 e ，槌sume that s( α) = n where 1 <η< ∞. Suppose that n is even and n = 2m. Then (αm)2 == a2m E Es and since m 三 2m - 1 == n - 1 we have αm rt Es. Hence, s(αm) == 2. It follow5 frorn our previous work thatαm 0 (αm)' 弄 o and i [am 0 (αm) ，] =∞. Finally, suppose that n i5 odd and n == 2m - 1. Then (α m)2 = α2m E Es. Since m 三 2m - 2 == n - 1, we have atn rt Es. Hence, s(αm) == 2 and again αm 0 (α勺'弄 o and i [am 0 (αm)'] == x. 口 Corollary 6.5. If E is isotropically finite, then for every a E E \ Es we have s( α)== ∞. lt is easy to see that ê(H) is isotropically fìnite. However, the non­ standard unit interval * [0 , 1] with its usual product is a SEA that is not isotropically fìnite. ln fact, every infìllitesimal in *[0, 1] has infìnite isotropic index. It follows frorn Corollary 6.5 that every unsharp element of E(H) hω infìnite sharpness index. The converse of Corollary 6.5 does not hold. For exarnple, in *[0, 1] every unsharp element has infìnite sharpness index. ln the next section we 5hall show that there exists a SEA with an elernent α satisfyi吨 1 < s( α) < 810. Corollary 6.6. Let E be an isotropic1αlly finite SEA. (i) Every atom of E is sh α叩. (ii) 扩 E is αtomic then E is an orthoalgebra. Proof. (i) lfαis an atorn, since a 2 三 α we have a2 =αor α2 == O. ln the latter case s( α) = 2 which contradicts Corollary 6.5. Hence.αis sharp. (ii) Suppose that E is not an orthoalgebra. Then there exists a b E E with b 八 b' 笋 O. Since E is atornic. there exists an atorn αwithα~ b, b'. But then α 三 b 三 α， so that α rt Es. This contradicts (i). 口

20 Theorem 6.7. (i)扩 αE E s and b E E with b 1. α ， then (a æ b)n = α æb n . (ii) If E is ασ -SEA α nd α . b ε E with b 1. α ， then ^(αæ bn) = α æ^bn . n n

l Proof. (i) Since b 三 a we have b 0 a' = a' 0 b = b. Hence.α 。 b = O. We now prove the result by induction n. The result certainly holds for η= 1. Assuming the result holds for ηwe have that n (αæ b) π+1 = (α 67 b)n 0 (α æb) = (αæ b ) 0 (αæ b) n n n =(αæ b ) 。 αæ(α æbn)ob= α 。 (αæ b ) æ b 0 (α æb ) =αæ bn-r 1

n (ii) Since a æ 八俨三 αæ bn, we have that α@ 八 b 三 ^(αæ bn). Since α æbn 主 α ， we have that 八 (α æbn ) ~二 α.From 八 (α EÐbn ) :三 α æbn it follows that 八 (α æbn )8 α 三 bn . Hence ，八 (α æbn )8a 三八 bn so that 八 (α@ 俨)三 α@ 八 bn . 口

If E is a σ-SE.-\. we have seen that for any αε E there exists a largest sharp element belowα. We denote this element by La J .

Corollary 6.8. Let E be a σ -SEA 叨的 α ， b E Es and c. d E E. (i) lf α 1. c. b 1. d and αæ c = b æ d, then

lαæcJ =αæ lcJ = bEÐ ldJ = lbædJ

(ii) 扩 lcJ = LdJ = 0 ， α 1. c, b 1. d αnd αEÐ c = b 67 d. then α = b and c = d. (iii )扩 b 1. d, l dJ = 0 and α 三 b æ d, theηα 三 b.

Proof. (i) By Theorem 6.7(i) we have

αEÐ cn = (a EÐ ct = (b EÐ dt = b æ d"

By Theorem 6.ï(ii) and the proof of Theorem 6.3 we have

αEÐ LcJ =α@ 八 cn = 八 (αEÐ cn) = 八 (bEÐdn ) =bEÐ d" :=bæ ldJ

~Joreover.

n n laEÐcJ= 八 (α 67c) π= 八 (αEÐ c ) = α@ 八 c = αe lcJ

(ii) By (i) we have

α=αEÐlcJ :=bEÐldJ =b

21 Hence, c = d by cancellation. (iii) Since α 三 b EÐ d, there exists a c E E such that αEÐ c = b EÐ d. Applying (i) gives

α 三 αEÐl叫 = b EÐ ldJ = b

口

Theorem 6.9. 扩 E is a σ -SEA then any αE E has a unique 陀P陀sentation a=bEÐc 叫 ere b E Es αnd l 叫= O.

Proof It is clear thatα= laJ EÐ (αe La J ). Ií b E E s and b 三 αe 例， then b 三 α. Hence, b 三 l~J and b 三凶， so that b = O. Therefore, la e laJJ = O. This gives the desired representation. To show uniqueness, suppose we have two such representationsα = b EÐ c = b1 EÐ CI, b, b1 E Es, lcJ = lctJ = o. Then by Corollary 6 圳 i) we have b = b1 and c = Cl' 口

7 An Example

This section presents an example of a SEA that has an element a satisfying s( α) = 2. Let E = ω+ω.. be the set 飞;vi th elements {O ， l ， α ， 2α ， . . . ， α" (2α t... }

By convention, we defìne 0α= O. Defining EÐ on E by

(mα) EÐ (ηα) = (m + n) α

and whenη

(mα)' æ(na) = (na) æ (ma)' = ((m-π)α)'

it is easy to see that (E, 0, 1, EÐ) is an effect algebra [1.]. Moreover, ja 三 (ka)' for every j , k E N because

(j α) EÐ ((k + j)α)' = (kα)'

Theorem 7. 1. There is a unique sequential product on the effect α1gebra E= ω+ω ..

22 ProoJ. For x, y E E define rl--OZHU if x = ma, y = nα ，，、 < z 。 Nud u+ mα ， (η a)' (mα)' ， -- EEl· ifx= y = or x = y = na m n a、‘.• 飞 F if x = (ma)', y = (nα)'

It is clear that 10 x = x for every x E E. Since E is commutative under 0 , in order to show that 0 is a sequential product we only need to verify

(D) xo (yEÐz) = (xoy) EÐ (xoz), y 1. z

(A) X 0 (y 0 z) = (x 0 y) 0 z

There are eight possibilities for x , y, z. x y z (1) ra ma nα (2) ra mα (nα)' (3) ra (mα)' nα (-1) Tα (mα)' (na)' (5) (rα)' mαm (6) (ra)' mα (na)' (7) (ra)' (ma)' na (8) (rα)' (ma)' (n α)' To verify (D) we only need to consider c出es (1) , (2) , (5) and (6) because (2) and (3) a.s well a.s (6) and (7) are symmetric and (4) and (8) never occur because y 1- z in these ca.ses. In Case (1) both the left and right hand sides are O. In Ca.se (2) both the left and right hand sides are x. In Case (5) both the left and right hand sides are y æz. In Case (6) both the left and right hand sides are ((n - m + r)α)' .

To verify (A) , we check all eight cases.

(1) ra 0 (ma 0 na) = ra 0 0 = 0 = 0 0 (nα) = (ra 0 mα) 0 na

(2) ra 0 (mα 。 (nα)') = ra 0 mα= 0 = 00 (n α)' = (rα 。 ma) 0 (n α)'

(3) rα 。 ((mα)' 。 ηα )=raona=O=(rα 。 (mα)') 0 na

23 (4) rα 。 ((mα)' 0 (ηα)') = rα 。 ((m + η)α)' = rα=rα 。 (nα)'

= (rα 。 (mα)') 0 (ηα)'

(5) (rα)' 0 (mα 。 ηα) = (rα)'oO=O=mα 。 nα = ((rα)' omα) 0 na

(6) (rα)' 0 (mα 。 (ηα)') = (rα)' 0 mα=mα= ma 0 (nα)'

= ((r α)' 0 mα) 0 (ηα)'

(7) (rα)' 0 ((ma)' 。 ηα) = (rα)' 。 ηα=ηα = ((r + m)a)' 。 πα

= ((rα)' 0 (mα)') 。 ηα

(8) (rα)'o((mα)' 0 (nα)') = (rα)' 0 ((m + η)α)' = ((r +m +η)α)'

= ((r 十 m)α)' 0 (na)' = ((r α)' 0 (mα)') 0 (ηα)' It follows that (E, 0) is a SEA. For uniqueness, suppose that .: E • E is another sequential product on E. Since αis an atom and α·α 三 α ， we have α·α= 0 or α·α=α. But α 三 a' so thatα~ Es. Hence ， α·α= O. Therefore ， ηα.mα=ηmα·α=0 for every n, m E N. Since every x E E h臼 the form nαor (nα)' ， it ís clear that E is commutative under •. For x = mα ， y = (ηα)' we have that

x.y =mα • (na)' = (mα·πα) EÐ (mα. (ηα)') = ma = x ^ y

Applying Theorem 3.4(iii) we have that

((na)' • (ma)')' = nαEÐmα=(η +m) α

Hence, (nα)'. (mα)' = ((η +m) α))'. \Ve conclude that x.y = xoy for every x,y ε E. 口 The elements nαεω+ω 气 n "# 0, satisfy s(ηα) = 2 while the elements (nα)' ， n "# 0, satisfy s ((nα)') =∞. For this example, ifx 1. y, then s(xEÐy) = max (s(x) , s(y)). Of course, this equation does not hold for an arbitrary SEA. Notice that ω+ω· is not isotropically finite and is not a σ-SEA. Since ω+ω­ does not admit an order determining set of states ， ω+ω· is a commutative SEA that is not isomorphic to a fuzzy set system.

24 8 Horizontal Sums

Let (Ei , Oi , l i, æi), i E 1, be a co11ection of effect algebras 时 th card. (1) > 1. Their horizonta1 sum E = HS ( 丘， i ε1) is defined 臼 fo11ows. Identify all the Oi with a single element 0 and all the 1i with a single element 1. Let EI = Ei \{Oi! 1d, form the disjoint union 山E: and let E = {O, l}uúE:. For α ， b ε Ei for 50me i ε1 defineα æb= αEÐi b and no other orthosums are define on E. It is then easy to check that (E, 0, 1, EÐ) is an effect algebra. If the Ei' i E 1, are SEA's we now investigate whether HS(Ei , i E 1) admits a sequential product and is thus a SEA. For an arbitrary effect algebra E we use the notation E' = E \{O, 1} and we denote the trivial effect algebra {Oll} by 2. Tbeorem 8.1. Let El' E2 笋 2 be effect algebras. (i) 扩 El hωαn atom, then α sequential H S (E1, E2 ) does not admit product. (ii) 11 El is an orthoalgebra, then HS(E1, E2 ) does not admit α sequential product.

Proof. (i) Suppose that H S(El! E2 ) admits a sequential product o. Let a E El be an atom and let b E E~. Then α 。 b = 0 orα 。 b' = o. If α 。 b = 0, then α=α 。 b' = b' 。 α < b' which is a contradiction. Similarly, ifα 。 b' = 0, then

α=α 。 b=b 。 α

25 Applying Theorem 8.1, if HS(E11 E2 ) is a SEA then neither El nor E2 can be a nontrivia1 Boolean a1gebra orω+ω.. The next resul t characterizes horizonta1 sums of SEA's that admit a sequential product and hence form a SEA. If E ， F 缸e effect a1gebras, an additive map 4J: E • F is positive if φ(α) = 0 implies α= o. Theorem 8.2. Let 马， i E 1, be $EA 垣 and let E = HS(且 ， i ε 1). Th en E admits a sequential product 矿 and only if for every αε E; there exists a positive additive map 4J生 :EU → Ei such that 4Jji (1) =α for every i, j E 1 四th i =/; j and if α ， b E E;, c ε 乌四协 α 。 b = bo a =/; 0, then 币;产 (c) = α 。 ejt(c)- Proof Suppose that E admits a sequentia1 product o. For a E E; , define ￠飞 :Ej → EiI j =/; i, by 愕'i(b) = α 。 b. Notice thatα 。 bε Ei becauseα 。 b 三 α­ Now 愕'i is clear1y additive and 愕'i(1) = α. Suppose thatα 。 b= 愕:i(b) = 0 and b =/; o. ThenαI b so thatα111. Hence, α=α 。矿 = b' 。 αE Ei n Ej But then αε{0 ， 1} which is a contradiction. Hence ，句 is positive. If α ， b E E: , c E 乌 with a 0 b = b 。 α=/; 0, then

哝 b(C) = (α 。 b) 0 c=α 。 (b 0 c) =α 。币~i(C)

Conversely, suppose 币2: 马→马， Z,J E 1, i =/; j , satis马r the given conditions. Define the operation 0 on E by Iα 。 b ifα ， b E Ei for some i E 1 α 。 b=< L4J ji(b) ifαE E; , b E Ej , i =/; j E 1

飞鸟飞~ now show that 0 is a sequentia1 product on E. If 矶的， ~ε 马 for some i E 1 with b1 ..L ~ then c1early α 。 (b 1 EÐ~) =α 。 b 1 EÐα 。 b 2 • Otherwise, αε E: ， 仇， ~ε Ej ， i 笋 j ，皿d we have that

α 。 (b 1 EÐ b2 ) = 4J ji(b1 EÐ b2 ) = 币'ji(bd EÐ 币'ji(b2 ) = α 。 b 1 EÐα 。 b2 Hence, (S1) holds. It is clear that (S2) holds. To show that (S3) holds, suppose thatα 。 b = o. Ifα ， b E Ei for some í ε1 ， then b 。 α= 0 so suppose that a E E:, b E 马， i =/; j. Then 币1i(b) = 0 so by positivity, b = o. Hence, b 。 α= o. To veri作 (S4) suppose thatα 。 b = b 。 αwhereαε E: ， b E Ej , i =/; j. Then α 。 b= 得j(b) = b 。 α = 4J~i( α)ε Ein 马

26 Hence.α 。 b ε{0 ， 1}. If a 0 b == 0, then CÞji(b) == 0 50 that b == 0 which is a contradiction. If α 。 b == 1, then cþ主 (b) == 1. Hence,

α= 饨 (1) 主伟 (b) == 1 so thatα= 1 which is again a contradiction. We conclude that α ， b ε 马 for 50me i ε!. Hence ， αlli. Moreover, ifα ， b E E: and c E E斗， i =/: j , then

α 。 (b 0 d) = α 。 CÞ~i(C) = CÞjib(c) = (α 。 b) 0 c

To verify (S5) , 5upp05e th剖 clαand c I b. As before ， α ， b ， c E 马 or some i E ! and the result follow5. 口 The next resu1t gives a useful method for con5tructing a SEA from a horizontal sum of SEA's E == HS( 马 I i E I). Suppose there exist effect algebra 向:j: Ei →马 ， i=/:j ε!. Define the operation 0 on E by

lα 。 b ifα ， b E Ei for some iεI α 。 b==<

lα 。 cþji (b) ifαε E: I b E Ej, i i: j E !

Corollary 8.3. We have that (E,o) is a SEA if αnd only if for eveηαε Ei1 b ε Ej ， i 笋 j E! ， α 。 b = 0 implies that a = 0 or b == O. Proof. .-\s in the proof of Theorem 8.2, if (E, 0) is a SEA and ω b = 0, then

α=α 。 b' == b' 0 a E Ei n Ej

Hence ， αε{0 ， 1}. If α= 1, then b = O. Conversely, assume that 0 satisfies the given condition. For a E E; define the map CÞji: Ej →瓦， t 笋 j E !, by 句j(b) ==α 。 CÞji(b). Then Øji is additi\'e and CÞji(1) =α. If CÞji(b) = 0, then α 。 b = O. Since α 笋 0 ， by assumption b == O. Hence, v is positive. Suppose thatα ， b E E: , c ε Ej with aob = b 。 α 笋 O. Then

。'j:b(C) = (α 。 b) 0 CÞji(C) = α 。 (b 0 cþjj(c)) = α 。司 i( c)

It fo11ows from the proof of Theorem 8.2 that (E, 0) is a SEA. 口

27 Theorem 9.1. 11 X i= ø is a set and E is α SEA ， then the SEA tensor product 012x and E exists. Proof. \Ve call a function 1: X • E simple if 1 h凶 a finite number of values and we define

X T = {I E E : 1 is simple}

On T define 1 1. 9 if I(x) 1. g(x) for a11 x E X and if 1 1. 9 define (f $ g)(x) = I(x) $ g(x). Defining O(x) = 0 and l(x) = 1 for all x E X , it is easy to check that (T, 0, 1, EÐ) is an effect algebra. For 1, 9 E T define 10 g(x) = I(x) 0 g(x). Ag出n ， it is easy to check that (T, 0, 1, $, 0) is a SEA. Define r: 2x x E • T by rtt411· αo ra--A ，，，.飞、、 ifx E A T α Z-aJ = ~ 、 if x A

飞飞马 use the notation r(A ， α) = χAα. It is clear that r is an effect algebra bimorphism. Since

r(A. a) 0 r(B, b) = (χAα) 0 (XBb) = XAnBa 0 b = r(A 0 B ， α 。 b) we see that r is a SEA-bimorphism. Moreover, eve巧r 1 E T has the unique representatlon

1 = EÐ X.-l jaj = EÐ 咐 i ， aj)

内 x where 向#句: .4 i Aj = 0, i 笋 j and uAi = X. Let ß: 2 x E • F be a SEA-bimorphism. Define ø: T • F as follows. If 1 = æXA ‘ αi is the unique representation of 1, then ø(f) = æß(Aù 向 ). Notice that æß(A ， 向) is defined because æß(Ai , 1) = 1 and ß(Aj ， 向)三 ß(A il 1). It is straightforward to show that ø is a SEA-morphism. Moreover,

ß(A ， α) = ø( χAα)=φ 。 r(A ， α) 口

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 arbitrarv SEA alwavs exists.

30 Let X 笋的 be a set, Q be the rational numbers and define :F(X) = {u: X • Qn[O, 1]} Then .r(X) is a fuzzy set system and thus forms a SEA. Theorem 9.2. The SEA tensor product ol .r(X) αnd E(H) exists. Proof. Let E = E (H) and defìne the SEA T 槌 in the proof of Theorem 9. 1. Define r: :F(X) x E • T by r( 吼叫 (x) = u(x) α. Then r is clearly an effect algebra bimorphism. Since

r(u ， α) 0 r(v,b) = (u α) 0 (vb) = ωα 。 b=r(uov ， α 。 b) r is a SEA-bimorphism. As in Theorem 9.1 , every 1 E T h槌 the unique rep­ resentation 1 = EÐr(XAi , ai). Let ß: .r(X) x E • F be a SEA-bimorphism. As in Theorem 9.1 , define r/J: T • F by r/J(f) = EÐß(χ岛，向). Again, cþ is a SE.-\.-morphism. Moreover, if u E :F(X) then u h槌 the unique representa­ tion u = 艺入 iXAi where 沁笋 >"j ， Ai n Aj = ø, i # j and UAi = X. It 臼i S strai地ghtti岛orwar时dtωo s1 then ha毛.ve that ß(u ， α) = ß(EÐ 入 iXAi' α) = EÐß( 入 iXAi' a) = 9ß(XAi' 沁 α) =创 (λ队， α) = r/J (2:: >"iX川) =φ 。 r(u ， a)

口

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