Orthocomplemented lattices with a symmetric difference Milan Matou - - PowerPoint PPT Presentation

Orthocomplemented lattices with a symmetric difference

Milan Matouˇ sek 1 ODLs 2 Z2-valued states 3 The variety OML△

1 ODLs

Definition 1.1 Let L = (X, ∧, ∨,⊥ , 0, 1, △), where (X, ∧, ∨,⊥ , 0, 1) is an orthocomplemented lattice and △ : X2 → X is a binary operation. Then L is said to be an

• rthocomplemented difference lattice (abbr., an ODL) if

the following formulas hold in L: (D1) x △ (y △ z) = (x △ y) △ z, (D2) x △ 1 = x⊥, 1 △ x = x⊥, (D3) x △ y ≤ x ∨ y. (X, ∧, ∨, ⊥, 0, 1) will be denoted by Lsupp and called the support of L.

Proposition 1.2 Let L = (X, ∧, ∨,⊥ , 0, 1, △) be an ODL. Then the following statements hold true: (1) x △ 0 = x, 0 △ x = x, (2) x △ x = 0, (3) x △ y = y △ x, (4) x △ y⊥ = x⊥ △ y = (x △ y)⊥ , (5) x⊥ △ y⊥ = x △ y , (6) x △ y = 0 ⇔ x = y , (7) (x ∧ y⊥) ∨ (y ∧ x⊥) ≤ x △ y ≤ (x ∨ y) ∧ (x ∧ y)⊥.

Theorem 1.3 Let L be an ODL. Then its support Lsupp is an OML.

• Proof. We will prove

x, y ∈ L, x ≤ y, y ∧ x⊥ = 0 ⇒ x = y . x ≤ y ⇒ (x∧y⊥)∨(y∧x⊥) = y∧x⊥ = 0, (x∨y)∧(x∧y)⊥ = y ∧ x⊥ = 0.

• Prop. 1.2: x △ y = 0 (by (7)) and therefore x = y (by

(6)). Let L be an ODL. Then any OML notion can be referred to L as well by applying this notion to the corresponding OML Lsupp.

Proposition 1.4 Let L be an ODL, a, b ∈ L. (i) a C b ⇒ a △ b = (a ∧ b⊥) ∨ (b ∧ a⊥) = (a ∨ b) ∧ (a ∧ b)⊥; (A corollary: For each block B of L, the operation △ on L acts on B as the standard symmetric difference.) (ii) a ⊥ b (i.e. a ≤ b⊥) ⇔ a △ b = a ∨ b.

A NATURAL QUESTION: Which OMLs are embed- dable into ODLs ? A MORE GENERAL QUESTION: Let us consider the variety OML△ of OMLs generated by the class {Lsupp; L is an ODL}, i.e. OML△ = HSP({Lsupp; L is an ODL}). Which OMLs belong to the variety OML△ ? The main aim of this note is to show that the class OML△ does not contain all OMLs.

Proof of Prop. 1.2. (D2): 1 △ 1 = 1⊥ = 0. (1) x△0 = x△(1△1) = (x△1)△1 = x⊥△1 = (x⊥)⊥ = x; the identity 0 △ x = x has an analogous proof. (2) Let us first show that x⊥ △ x⊥ = x △ x. We con- secutively obtain x⊥ △ x⊥ = (x △ 1) △ (1 △ x) = (x △ (1 △ 1)) △ x = (x △ 0) △ x = x △ x. Moreover, we have x △ x ≤ x as well as x △ x = x⊥ △ x⊥ ≤ x⊥. This implies that x △ x ≤ x ∧ x⊥ = 0. (3) (x △ y) △ (x △ y) = 0 ⇒ (x △ y △ x △ y) △ y = 0 △ y ⇒ x △ y △ x = y ⇒ (x △ y △ x) △ x = y △ x ⇒ x △ y = y △ x. (4) x △ y⊥ = x △ (y △ 1) = (x △ y) △ 1 = (x △ y)⊥. (5) Applying (4), we obtain x⊥ △ y⊥ = (x⊥ △ y)⊥ = (x △ y)⊥⊥ = x △ y. (6) If x = y, then x △ y = 0 by the condition (2). Con- versely, suppose that x △ y = 0. Then x = x △ 0 =

x △ (y △ y) = (x △ y) △ y = 0 △ y = y. (7) The property (D3) together with the properties (4), (5) imply that x △ y ≤ x ∨ y, x △ y ≤ x⊥ ∨ y⊥ = (x ∧ y)⊥, x ∧ y⊥ ≤ x △ y, x⊥ ∧ y ≤ x △ y. Proof of Prop. 1.4. (i) It follows from Prop. 1.2. (ii) (⇒) We use Prop. 1.2. (⇐) x △ y = x ∨ y ⇒ y ≤ x △ y ⇒ x △ y⊥ ≤ y⊥. But also y⊥ ≤ y⊥. (D3): (x△y⊥)△y⊥ ≤ y⊥. Finally, (x△y⊥)△y⊥ = x△(y⊥△y⊥) = x △ 0 = x.

2 Z2-valued states

Z2 ≡ the group {0, 1} with the modulo 2 addition ⊕ Definition 2.1 Let L be an OML. (i) Let s : L → Z2 be a mapping. Then s is said to be a Z2-valued state (abbr., a Z2-state) provided s(1L) = 1 and s(x ∨ y) = s(x) ⊕ s(y) whenever x, y ∈ L, x ≤ y⊥. (ii) L is called Z2-full if ∀x, y ∈ L, x = y, x = 0L, y = 1L ∃ Z2-state s on L: s(x) = 1 & s(y) = 0.

Definition 2.2 Let L be an ODL and let I ⊆ L. I is a △-ideal if 0L ∈ I & ∀a, b ∈ I : a △ b ∈ I; I is a proper △-ideal if I is a △-ideal and 1L ∈ I; I is a maximal △-ideal if I is proper △-ideal and for any proper △-ideal J with I ⊆ J we have I = J.

Lemma 2.3 Suppose that L is an ODL and I is a proper △-ideal in L. Suppose that x ∈ L and neither x nor x⊥ belongs to I. Let us write J = I ∪ {a △ x; a ∈ I}. Then J is also a proper △-ideal in L and, moreover, x ∈ J and x⊥ ∈ J. Proposition 2.4 Let L be an ODL and let I be a proper △-ideal in L. Then I is a maximal △-ideal in L iff card({x, x⊥} ∩ I) = 1 for any x ∈ L. Proposition 2.5 (Stone’s lemma) Let L be an ODL, let I0 be a proper △-ideal in L and let b ∈ L, b ∈ I0. Then there is a maximal △-ideal I such that I0 ⊆ I and b ∈ I.

Proposition 2.6 Let L be an ODL and I be a maximal △-ideal in L. Let us define a mapping s : L → Z2 as follows: s(a) = 0 (resp., s(a) = 1) if a ∈ I (resp., a ∈ I). Then s(x △ y) = s(x) ⊕ s(y) for any x, y ∈ L. A consequence: The mapping s is a Z2-state on Lsupp. Theorem 2.7 Let L be an ODL. Then its support Lsupp is Z2-full.

Proof of Lemma 2.3. The set J is obviously a △-ideal. Let us see that 1L ∈ J. Suppose on the contrary that 1L ∈ J. Then 1L = a △ x for some element a ∈ I. The equality 1L = a △ x implies that a = x⊥. But x⊥ does not belong to I which is a contradiction. Thus, 1L ∈ J. Further, x = 0L △ x ∈ J. If x⊥ ∈ J, then 1L = x △ x⊥ ∈ J – a contradiction again. Proof of Prop. 2.4. (⇒) Suppose that I is maximal and x ∈ L. Suppose further that x ∈ I and, also, x⊥ ∈ I. Then (Lemma 2.3) there is a △-ideal J such that I ⊆ J and I = J. As a result, at least one element of the set {x, x⊥} belongs to I. Looking for a contradiction, sup- pose that {x, x⊥} ⊆ I. Then x △ x⊥ = 1 which means that 1 ∈ I – a contradiction (I is supposed proper). (⇐) Let us suppose that card({x, x⊥} ∩ I) = 1 for any

x ∈ L. Suppose further that I ⊂ J for a proper △-ideal J. J is strictly larger than I ⇒ ∃b ∈ L : b ∈ J, b ∈ I. card({b, b⊥} ∩ I) = 1 and b ∈ I ⇒ b⊥ ∈ I. Now, both the elements b and b⊥ belong to J and there- fore 1L = b △ b⊥ ∈ J. This means that J is not proper. Proof of Prop. 2.5. Write I = {J ⊆ L; J is a proper △-ideal, I0 ⊆ J and b ∈ J}. Then I0 ∈ I and therefore I = ∅. Zorn’s lemma: the set I ordered by inclusion contains a maximal element, I. (1) I is a proper △-ideal, I0 ⊆ I and b ∈ I. (2) b⊥ ∈ I (otherwise the △-ideal I′ = I ∪ {c △ b⊥; c ∈ I} extends I, Lemma 2.3, and I′ belongs to the system I). (3) I is a maximal △-ideal (Suppose therefore that I ⊂ J

for a proper △-ideal J. Thus, J is strictly larger than I and therefore J ∈ I. Therefore b ∈ J and since b⊥ ∈ I ⊆ J, we see that 1L = b △ b⊥ ∈ J. This means that J is not proper.)

Proof of Prop. 2.6. Let us consider x, y ∈ L. We are to prove the equality s(x△y) = s(x)⊕s(y). We will argue by

• cases. For example, let us suppose that x ∈ I and y ∈ I.

Then x △ y ∈ I (indeed, should x △ y be an element of I, then y = x△(x△y) ∈ I which is a contradiction). Hence s(x △ y) = 1 = 0 ⊕ 1 = s(x) ⊕ s(y). It remains to show that s is a Z2-state on Lsupp. Of course, s(1) = 1. Let us take x, y ∈ L with x ⊥ y.

• Prop. 1.4: x △ y = x ∨ y.

Then s(x ∨ y) = s(x △ y) = s(x) ⊕ s(y) by the analysis

• above. The proof of Prop. 2.6 is complete.

Proof of Thm. 2.7. Let L be an ODL. Let x, y ∈ L with x = y, x = 0 and y = 1. Let us set I0 = {0L, y}. According to Prop. 2.4 there is a maximal △-ideal I in L such that y ∈ I and x ∈ I. Let us set s(a) = 0 for a ∈ I and s(a) = 1 for a ∈ L, a ∈ I. Then, according to

• Prop. 2.6, the mapping s is a Z2-state on Lsupp.

3 The variety OML△

Let us recall that OML△ = HSP({Lsupp; L ∈ ODL}). Theorem 3.1 Let K ∈ OML△ and let x0 ∈ K, x0 = 1K. Then there is a Z2-state, s, on K such that s(x0) = 0.

Proposition 3.2 Suppose that L is an OML. Suppose that there are blocks B1, B2, . . . , Bn of L such that the following two conditions are satisfied: (1) each Bi, 1 ≤ i ≤ n is finite and n is an odd number, (2) if a ∈ L is an atom in L, then a lies in an even number of blocks B1, B2, . . . , Bn (i.e., the cardinality of the set {i; a ∈ Bi} is even). Then there is no Z2-state on L.

• Proof. Let s : L → Z2 be a Z2-state.

Then ⊕a∈At(Bi)s(a) = s(1L) = 1 (i = 1, . . . , n). Therefore ⊕i∈{1,...,n}(⊕a∈At(Bi)s(a)) = 1 ⊕ . . . ⊕ 1. The right-hand side = 1 (contains the element 1 exactly n-many times and n is odd). The left-hand side = 0 (contains any expression s(a) (where a ∈ At(A)) even number of times).

Proposition 3.3 OML L(R4) satisfies the assumptions

• f Prop. 3.2 and therefore L(R4) ∈ OML△.
• Proof. Notations: instead of −1 we will write ¯

1, instead

• f (a, b, c, d) (where a, b, c, d ∈ {0, 1, ¯

1}) we will only write abcd. Let us consider the following blocks

B1 B2 B3 B4 B5 B6 B7 B8 B9 1000 1000 0100 1111 1111 111¯ 1 11¯ 1¯ 1 111¯ 1 11¯ 11 0100 0010 0010 11¯ 1¯ 1 1¯ 11¯ 1 11¯ 11 1¯ 11¯ 1 1¯ 111 1¯ 111 0011 0101 1001 1¯ 100 10¯ 10 1¯ 100 1001 10¯ 10 100¯ 1 001¯ 1 010¯ 1 100¯ 1 001¯ 1 010¯ 1 0011 0110 0101 0110

The above table is taken from [9].

Proposition 3.4 If n ≥ 4 is even, then there is no Z2- state on L(Rn). If n ≥ 5 is odd, then there is exactly one Z2-state on L(Rn). n = 1: OML L(R1) is two element Boolean algebra. n = 2: OML L(R2) is isomorphic with MO2ℵ0 and it is a support of an ODL (see [4]). n = 3: There is at least one Z2-state on OML L(R3) (if we set s(a) = 1 for each atom a). The question whether exists a Z2-state, s, on L(R3) such that s(a) = 0 for some atom a is open. We only know that L(R3) is not a support of any ODL (see [5]).

Proposition 3.5 The OML portrayed below by its Gree- chie diagram satisfies the assumptions of Prop. 3.2.

• ❅

❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ s s s

• s

s s s ❅ ❅ ❅ ❅ s s s s

• s

s s s ❅ ❅ ❅ ❅ s s ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ s s s s

Proof of Thm 3.1. If Li (i ∈ I) are ODLs, then

• i∈I

(Li)supp = (

• i∈I

Li)supp ⇒ OML△ = HSP({Lsupp; L ∈ ODL}) = HS({Lsupp; L ∈ ODL}). K ∈ OML△ ⇒ ∃ ODL L ∃ OML L1: L1 is a sub-OML

• f Lsupp and K is a homomorphic image of L1.

Let f be a surjective morphism L1 → K. Let us choose an element c ∈ L1 such that f(c) = x0 and let us set I0 = {x ∈ L; x ≤ a ∨ c for some a ∈ f−1(0K)}. Then I0 is a proper △-ideal in L (indeed, (i) 0L ∈ I0; (ii) if x, y ∈ I0, then x ≤ a ∨ c and y ≤ b ∨ c for some a, b ∈ f−1(0K) and therefore x △ y ≤ x ∨ y ≤ (a ∨ b) ∨ c –

but the element a ∨ b belongs to f−1(0K); (iii) if 1L ∈ I0 then 1L = a ∨ c for some a ∈ f−1(0K) and therefore 1K = f(a∨c) = f(a)∨f(c) = d, which is a contradiction). Stone’s lemma: there is a maximal △-ideal I in L with I0 ⊆ I. Let σ be the Z2-state on L determined by I (i.e., σ(x) = 0 for x ∈ I, σ(x) = 1 otherwise). For y ∈ K let us set s(y) = σ(x), where x ∈ L1 is such an element that f(x) = y. The definition of s is correct: To this end, suppose that y = f(x1) = f(x2) for x1, x2 ∈ L1. Then f((x1∨x2)∧(x1∧ x2)⊥) = (f(x1)∨f(x2))∧(f(x1)∧f(x2))⊥ = y ∧y⊥ = 0K. It means that (x1 ∨ x2) ∧ (x1 ∧ x2)⊥ ∈ f−1(0K). Since x1 △ x2 ≤ (x1 ∨ x2) ∧ (x1 ∧ x2)⊥ (see Prop. 1.4), we have x1 △ x2 ∈ I. But then 0 = σ(x1 △ x2) = σ(x1) ⊕ σ(x2)

and therefore σ(x1) = σ(x2). s is a Z2-state: (i) s(1K) = σ(1L) = 1; (ii) Suppose that x, y ∈ K, x ≤ y⊥. Then there are x1, y1 ∈ L1 such that f(x1) = x, f(y1) = y and x1 ≤ y⊥

1 .

Then f(x1 ∨ y1) = f(x1) ∨ f(y1) = x ∨ y. Because x1 ⊥ y1 it holds x1 ∨ y1 = x1 △ y1 (see Prop. 1.4). Now, s(x ∨ y) = σ(x1 ∨ y1) = σ(x1 △ y1) = σ(x1) ⊕ σ(y1) = s(x) ⊕ s(y). s(x0) = 0: s(d) = σ(c) = 0 since c ∈ I.

Sketch of proof of Prop. 3.4. In [9] it is shown that if n ≥ 4, then there is no nontrivial Z2-measure, s, on L(Rn) which satisfies s(1) = 1. A trivial Z2-measure on L(Rn) gives us Z2-state only for n odd (in this case we set s(a) = 1 for each atom a).

OPEN PROBLEMS (1) Does any (modular) set-representable OML belong to the variety OML△ ? (All OMLs MOκ belong to this variety.) (2) Is the variety OML△ finitely based, i.e. is there a finite set of identities which determines this variety ? (3) Is the class Z2-full OMLs closed on homomorphic images ?

• 1. Dorfer G., Dvureˇ

censkij A., L¨ anger H.M., Symmetric difference in orthomodular lattices, Math. Slovaca 46 (1996), 435-444.

• 2. Dorfer G., Non-commutative symmetric differences

in orthomodular lattices, International Journal of The-

• retical Physics 44 (2005), 885-896.
• 3. Havl

´ ık F., Ortokomplement´ arn ´ ı diferenˇ cn ´ ı svazy (Czech, Mgr Thesis, Department of Logic, Faculty of Arts, Charles University in Prague, 2007).

• 4. Matouˇ

sek M., Orthocomplemented lattices with a symmetric difference, Algebra Universalis 60 (2009), 185-215.

• 5. Matouˇ

sek M., Pt´ ak Orthocomplemented posets with a symmetric difference, Order (2009), 26:1-21.

• 6. Matouˇ

sek M., Pt´ ak P., On identities in orthocomple- mented difference lattices, Mathematica Slovaca, to appear.

• 7. Matouˇ

sek M., Pt´ ak P., Symmetric difference on or- thomodular lattices and Z2-valued states, CMUC, to appear.

• 8. Matouˇ

sek M., Pt´ ak P., Orthocomplemented differ- ence lattices with few generators, to appear.

• 9. Navara M., Pt´

ak P., For n ≥ 5 there is no nontrivial Z2-measure on L(Rn), International Journal of The-

• retical Physics 43 (2004), 1595-1598.
• 10. Park E., Kim M.M., Chung J.Y., A note on symmet-

ric differences of orthomodular lattices, Commun. Korean Math. Soc. 18 (2003), No 2, 207-214.

• 11. Pt´

ak P., Pulmannov´ a S., Orthomodular Structures as Quantum Logics, Kluwer Academic Publishers, Dordrecht/Boston/London, 1991.

• 12. Svozil K., Tkadlec J., Greechie diagrams, noexis-

tence of measures in quantum logics, and Kochen– Specker-type constructions, J. Math. Phys. 37 (1996), 5380-5401.

Milan Matouˇ sek Rusk´ a 22 101 00 Prague 10 Czech Republic e-mail: matmilan@email.cz