On the de Morgan Property of the Standard BrouwerZadeh Poset 1

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1 Foundations of Physics, Vol. 30, No. 10, 2000 On the de Morgan Property of the Standard BrouwerZadeh Poset 1 G. Cattaneo, 2 J. Hamhalter, 3 and P. Pta k 3 Received March 10, 1999 The standard BrouwerZadeh poset 7(H) is the poset of all effect operators on a Hilbert space H, naturally equipped with two types of orthocomplementation. In developing the theory, the question occured if (when) 7(H) fulfils the de Morgan property with respect to both orthocomplementation operations. In Ref. 3 the authors proved that it is the case provided dim H<, and they conjectured that if dim H=, then the answer is in the negative. In this note, we first give a somewhat simpler proof of the known result for dim H<, and then we give a proof to the conjecture: We show that if dim H=, then the de Morgan property is not valid. RESULTS In this paper we shall be exclusively interested in the standard Brouwer Zadeh poset 7(H) on a Hilbert space H. This poset endowed with two natural types of orthocomplementation operation constitutes a prominent model in the so called unsharp approach to the foundation of quantum mechanics (see Refs. 1 and 7, and for more general investigation and the explanation of the link with quantum axiomatic and fuzzy set theory; see, e.g., Refs. 3, 6, and 2). 1 The second author acknowledges the support of the Alexander von Humboldt Foundation, Bonn, the Grant Agency of the Czech Republic, Grant , and the Grant J of the Czech Technical University. The third author acknowledges the support of the grant VS of the Czech Ministry of Education. 2 Dipartimento di Informatica, Sistemistica e Comunicazione, Universita di Milano-Bicocca, Via Bicocca degli Arcimboldi 8, I Milano, Italy; cattangdisco.unimib.it. 3 Department of Mathematics, Faculty of Electrical Engineering, Czech Technical University, Zikova 4, Praha 6, Czech Republic; hamhaltemath.feld.cvut.cz and ptak math.feld.cvut.cz Plenum Publishing Corporation

2 1802 Cattaneo, Hamhalter, and Pta k Definition 1. The standard BrouwerZadeh poset is a quadruple (7(H),, t, $), where (i) H is a complex Hilbert space and 7(H) is the set of all linear operators P on H such that 0(. P(.))&.& 2 for each. # H (ii) is the partial ordering on 7(H) defined by putting PQ (. P(.)) (. P(.)) for all. # H (iii) (iv) t : 7(H) 7(H) is the (Brouwer) orthocomplementation defined by P t :=Proj Ker(P), where Proj Ker(P) is the projection of H onto the null subspace Ker(P) ofp, $: 7(H) 7(H) is the (Zadeh) orthocomplementation defined by P$=1&P, where 1 is the unit operator on H. We shall simply denote the above quadruple by 7(H). Obviously, the zero operator 0 and the unit operator 1 belong to 7(H) and, moreover, 7(H) is exactly the set of all operators P which satisfy 0P1. Observe that 7(H) is almost never a lattice with respect to. In analyzing the de Morgan property of 7(H), one easily finds that the operation $ always does satisfy this property. Indeed, the validity of de Morgan law means then the validity of the following two implications: (de Mo$ 1): (de Mo$ 2): If P 6Q exists in 7(H), then (1&P) 7 (1&Q) exists in 7(H), and (1&P) 7 (1&Q)=(P6Q)$. If P 7Q exists in 7(H), their (1&P) 6 (1&Q) exists in 7(H), and (1&P) 6 (1&Q)=(P7Q)$. The above two implications can be verified easily (see Ref. 3). If we consider t and ask about the validity of (de Mo t 1) and (de Mo t 2), the case of dim H<+ allows for a positive answer, too. This has been shown in Ref. 3 and will be also given a proof in the sequel. If dim H=, it will be shown that the de Morgan property is never satisfied. (It should be noted that the paper (4) treats this problem in the von Neumann algebra setting, classifying the ``de Morgan'' von Neumann algebras.) Theorem 1. Let H be a Hilbert space. Then 7(H) satisfies the de Morgan law if and only if dim H<.

3 On the de Morgan Property of the Standard BrouwerZadeh Poset 1803 Proof. We are interested in the de Morgan property with respect to the orthocomplementation t. Thus, we ask if (when) the following properties are satisfied: (de Mo t 1) If P 7 Q exists in 7(H), then Proj Ker(P) 6Proj Ker(Q) exists in 7(H) and Proj Ker(P 7 Q) =Proj Ker(P) 6 Proj Ker(Q) (de Mo t 2) If P 6 Q exists in 7(H), then Proj Ker(P) 7Proj Ker(Q) exists in 7(H), and Proj Ker(P 6 Q) =Proj Ker(P) 7 Proj Ker(Q) As one easily sees, the property (de Mo t 2) is always satisfied (see also Ref. 3). Thus, what remains to be investigated is the property (de Mo t 1). We shall show that if dim H<, then (de Mo t 1) is satisfied, and if dim H=, then (de Mo t 1) is not satisfied. Let dim H<. Since for each operator P of 7(H) we obviously have Ker(P)=R(P) =, where R(P) = is the orthogonal complement in H of the range, R(P), of P, we see that (de Mo t 1) for P, Q is equivalent to verifying the following property: R(P 7 Q)=R(P) 7 R(Q). Let us prove the latter equality. If P=0 or Q=0, then it is trivial. Suppose that both P and Q are nontrivial. Let (P i ), resp. (Q i ), be the orthogonal spectral projection operators of P and Q, respectively. So n P= : m * i P i and Q= : + j Q j (0<* i 1, 0<+ j 1 for each in, jm). Put k=min i, j (* i, + j ). Then k>0 and we obtain n Pk } : m Qk } : j=1 P i =k } R(P) Thus P, Qk }(R(P) 7 R(Q)) and so Q j =k } R(Q) R(P 7 Q)=R(k(R(P) 7 R(Q))=R(P) 7 R(Q) Since the inequality R(P) 7R(Q)R(P 7 Q) is trivial, the proof of the case of dim H< is complete.

4 1804 Cattaneo, Hamhalter, and Pta k Assume that dim H= and show that (de Mo t 1) is never satisfied. Suppose first that H is separable. Let [e n ]/H be an orthonormal basis of H. Take the vector x= n=1 (1n) e n and consider the projection operators P en, P x on the respective one-dimensional subspaces of H spanned by x and e n, correspondingly. Write P=P x and Q= : n=1 1 n 3 P e n Then P, Q # 7(H). We shall prove that (P 7 Q) t {P t 6Q t. Let us first show that Q t =0. In other words, let us prove that Ker Q=0. Fix v # H. Then v= j=1 (e j v) e j, and therefore Q(v)= : n=1 1 n 3 (e n v) e n We see that Q(v)=0 if and only if v=0. Let us go on, showing that P 7 Q=0. Suppose that T # 7(H) is a lower bound of P and Q. Take w # Ker(P). Then 0(w T(w)) (w P(w)) =0, and therefore &T 12 (w)&=0. This implies that T(w)= T 12 (T 12 w)=0. As a result, Ker P/Ker T. This means that (Ker(T )) = / (Ker(P)) = and this further implies that Proj (Ker(T )) =Proj (Ker(P)) == Proj R(P). Since Proj R(P) =P and since P is a one-dimensional projector, we see that T # [*P * # (0, 1)]. Suppose that T=*P for *>0. We know that 0*P=TQ. Hence, for all n # N, we infer that (e n *Pe n )=*(e n Proj x e n ) =*(e n (x e n )x) =* (x e n ) 2 =* 1 n 2 1 n 3 It follows that for any n # N we have 0*1n. This means that *=0 and this is absurd. Hence P 7 Q=0. We therefore have (P 7 Q) t =1. But P t 6Q t =P t =1&P{1, and we conclude that (P 7 Q) t {P t 6Q t. Let us finally consider a nonseparable H. If we mimic the proof taking for [e n ], (n # N ) an orthonormal system, we still obtain P 7Q=0 (or, equivalently, (P 7 Q) t =1). We may not have Q t =0. Instead, if we take closed linear span E of [e n ], we obtain Q t =Proj E =. Since x # E, we see that P=Proj x Proj E. Thus, Q t P$=P t. We see that P t 6 Q t either does not exist or, in the case it exists, P t 6 Q t P$ 6P$=P${1. The proof is completed.

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