Orthocomplemented Lattices with a Symmetric Difference

Orthocomplemented lattices with a symmetric diﬀerence

Milan Matouˇsek

1 ODLs

2 Z2-valued states

3 The variety OML4 1 ODLs

Deﬁnition 1.1 Let L = (X, ∧, ∨,⊥ , 0, 1, 4), where (X, ∧, ∨,⊥ , 0, 1) is an orthocomplemented lattice and 4 : X2 → X is a binary operation. Then L is said to be an orthocomplemented diﬀerence lattice (abbr., an ODL) if the following formulas hold in L: (D1) x 4 (y 4 z) = (x 4 y) 4 z, ⊥ ⊥ (D2) x 4 1 = x , 1 4 x = x , (D3) x 4 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, 4) be an ODL. Then the following statements hold true: (1) x 4 0 = x, 0 4 x = x, (2) x 4 x = 0, (3) x 4 y = y 4 x, (4) x 4 y⊥ = x⊥ 4 y = (x 4 y)⊥ , (5) x⊥ 4 y⊥ = x 4 y , (6) x 4 y = 0 ⇔ x = y , (7) (x ∧ y⊥) ∨ (y ∧ x⊥) ≤ x 4 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 4 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 4 b = (a ∧ b⊥) ∨ (b ∧ a⊥) = (a ∨ b) ∧ (a ∧ b)⊥; (A corollary: For each block B of L, the operation 4 on L acts on B as the standard symmetric diﬀerence.) (ii) a ⊥ b (i.e. a ≤ b⊥) ⇔ a 4 b = a ∨ b. A NATURAL QUESTION: Which OMLs are embed- dable into ODLs ?

A MORE GENERAL QUESTION: Let us consider the variety OML4 of OMLs generated by the class {Lsupp; L is an ODL}, i.e. OML4 = HSP ({Lsupp; L is an ODL}). Which OMLs belong to the variety OML4 ? The main aim of this note is to show that the class OML4 does not contain all OMLs. ⊥ Proof of Prop. 1.2. (D2): 1 4 1 = 1 = 0. (1) x40 = x4(141) = (x41)41 = x⊥41 = (x⊥)⊥ = x; the identity 0 4 x = x has an analogous proof. (2) Let us ﬁrst show that x⊥ 4 x⊥ = x 4 x. We con- secutively obtain x⊥ 4 x⊥ = (x 4 1) 4 (1 4 x) = (x 4 (1 4 1)) 4 x = (x 4 0) 4 x = x 4 x. Moreover, we have x 4 x ≤ x as well as x 4 x = x⊥ 4 x⊥ ≤ x⊥. This implies that x 4 x ≤ x ∧ x⊥ = 0. (3) (x 4 y) 4 (x 4 y) = 0 ⇒ (x 4 y 4 x 4 y) 4 y = 0 4 y ⇒ x 4 y 4 x = y ⇒ (x 4 y 4 x) 4 x = y 4 x ⇒ x 4 y = y 4 x. (4) x 4 y⊥ = x 4 (y 4 1) = (x 4 y) 4 1 = (x 4 y)⊥. (5) Applying (4), we obtain x⊥ 4 y⊥ = (x⊥ 4 y)⊥ = (x 4 y)⊥⊥ = x 4 y. (6) If x = y, then x 4 y = 0 by the condition (2). Con- versely, suppose that x 4 y = 0. Then x = x 4 0 = x 4 (y 4 y) = (x 4 y) 4 y = 0 4 y = y. (7) The property (D3) together with the properties (4), (5) imply that x 4 y ≤ x ∨ y, x 4 y ≤ x⊥ ∨ y⊥ = (x ∧ y)⊥, x ∧ y⊥ ≤ x 4 y, x⊥ ∧ y ≤ x 4 y.

Proof of Prop. 1.4. (i) It follows from Prop. 1.2. (ii) (⇒) We use Prop. 1.2. (⇐) x 4 y = x ∨ y ⇒ y ≤ ⊥ ⊥ ⊥ ⊥ x 4 y ⇒ x 4 y ≤ y . But also y ≤ y . (D3): (x4y⊥)4y⊥ ≤ y⊥. Finally, (x4y⊥)4y⊥ = x4(y⊥4y⊥) = x 4 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 6= y, x 6= 0L, y 6= 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 4-ideal if 0L ∈ I & ∀a, b ∈ I : a 4 b ∈ I;

I is a proper 4-ideal if I is a 4-ideal and 1L 6∈ I;

I is a maximal 4-ideal if I is proper 4-ideal and for any proper 4-ideal J with I ⊆ J we have I = J. Lemma 2.3 Suppose that L is an ODL and I is a proper 4-ideal in L. Suppose that x ∈ L and neither x nor x⊥ belongs to I. Let us write J = I ∪ {a 4 x; a ∈ I}. Then J is also a proper 4-ideal in L and, moreover, x ∈ J and x⊥ 6∈ J.

Proposition 2.4 Let L be an ODL and let I be a proper 4-ideal in L. Then I is a maximal 4-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 4-ideal in L and let b ∈ L, b 6∈ I0. Then there is a maximal 4-ideal I such that I0 ⊆ I and b 6∈ I. Proposition 2.6 Let L be an ODL and I be a maximal 4-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 6∈ I). Then s(x 4 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 4-ideal. Let us see that 1L 6∈ J. Suppose on the contrary that 1L ∈ J. Then 1L = a 4 x for some element a ∈ I. The ⊥ ⊥ equality 1L = a 4 x implies that a = x . But x does not belong to I which is a contradiction. Thus, 1L 6∈ J. ⊥ ⊥ Further, x = 0L 4 x ∈ J. If x ∈ J, then 1L = x 4 x ∈ J – a contradiction again.

Proof of Prop. 2.4. (⇒) Suppose that I is maximal and x ∈ L. Suppose further that x 6∈ I and, also, x⊥ 6∈ I. Then (Lemma 2.3) there is a 4-ideal J such that I ⊆ J and I 6= J. As a result, at least one element of the set {x, x⊥} belongs to I. Looking for a contradiction, suppose that {x, x⊥} ⊆ I. Then x 4 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 4-ideal J. J is strictly larger than I ⇒ ∃b ∈ L : b ∈ J, b 6∈ I. card({b, b⊥} ∩ I) = 1 and b 6∈ I ⇒ b⊥ ∈ I. Now, both the elements b and b⊥ belong to J and there- ⊥ fore 1L = b 4 b ∈ J. This means that J is not proper. Proof of Prop. 2.5. Write I = {J ⊆ L; J is a proper 4-ideal, I0 ⊆ J and b 6∈ J}. Then I0 ∈ I and therefore I 6= ∅. Zorn’s lemma: the set I ordered by inclusion contains a maximal element, I. (1) I is a proper 4-ideal, I0 ⊆ I and b 6∈ I. (2) b⊥ ∈ I (otherwise the 4-ideal I0 = I ∪ {c 4 b⊥; c ∈ I} extends I, Lemma 2.3, and I0 belongs to the system I). (3) I is a maximal 4-ideal (Suppose therefore that I ⊂ J for a proper 4-ideal J. Thus, J is strictly larger than I and therefore J 6∈ I. Therefore b ∈ J and since b⊥ ∈ I ⊆ ⊥ J, we see that 1L = b 4 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(x4y) = s(x)⊕s(y). We will argue by cases. For example, let us suppose that x ∈ I and y 6∈ I. Then x 4 y 6∈ I (indeed, should x 4 y be an element of I, then y = x 4 (x 4 y) ∈ I which is a contradiction). Hence s(x 4 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 4 y = x ∨ y. Then s(x ∨ y) = s(x 4 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 6= y, x 6= 0 and y 6= 1. Let us set I0 = {0L, y}. According to Prop. 2.4 there is a maximal 4-ideal I in L such that y ∈ I and x 6∈ I. Let us set s(a) = 0 for a ∈ I and s(a) = 1 for a ∈ L, a 6∈ I. Then, according to Prop. 2.6, the mapping s is a Z2-state on Lsupp. 3 The variety OML4 Let us recall that OML4 = HSP ({Lsupp; L ∈ ODL}).

Theorem 3.1 Let K ∈ OML4 and let x0 ∈ K, x0 6= 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 ⊕ s(a) = s(1 ) = 1 (i = 1, . . . , n). a∈At(Bi) L Therefore ⊕ (⊕ s(a)) = 1 ⊕ ... ⊕ 1. i∈{1,...,n} a∈At(Bi) 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 4 of Prop. 3.2 and therefore L(R ) 6∈ OML4. Proof. Notations: instead of −1 we will write 1,¯ instead of (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 1111¯ 111¯1¯ 1111¯ 1111¯ 0100 0010 0010 111¯1¯ 111¯ 1¯ 1111¯ 111¯ 1¯ 1111¯ 1111¯ 0011 0101 1001 1100¯ 1010¯ 1100¯ 1001 1010¯ 1001¯ 0011¯ 0101¯ 1001¯ 0011¯ 0101¯ 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. = 2: OML ( 2) is isomorphic with MO and it is n L R 2ℵ0 a support of an ODL (see [4]). 3 n = 3: There is at least one Z2-state on OML L(R ) (if we set s(a) = 1 for each atom a). The question whether 3 exists a Z2-state, s, on L(R ) 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.

@ ss s @ @ @ s L s @ s @ @ L s @ L s L @ s @ L s @sL s @L @ @ s @ s @ s @ s @ s s s s Proof of Thm 3.1. If Li (i ∈ I) are ODLs, then Y Y (Li)supp = ( Li)supp i∈I i∈I

⇒ OML4 = HSP ({Lsupp; L ∈ ODL}) = HS({Lsupp; L ∈ ODL}).

K ∈ OML4 ⇒ ∃ ODL L ∃ OML L1: L1 is a sub-OML of 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

−1 I0 = {x ∈ L; x ≤ a ∨ c for some a ∈ f (0K)}. Then I0 is a proper 4-ideal in L (indeed, (i) 0L ∈ I0; (ii) if x, y ∈ I0, then x ≤ a ∨ c and y ≤ b ∨ c for some −1 a, b ∈ f (0K) and therefore x 4 y ≤ x ∨ y ≤ (a ∨ b) ∨ c – −1 but the element a ∨ b belongs to f (0K); (iii) if 1L ∈ I0 −1 then 1L = a ∨ c for some a ∈ f (0K) and therefore 1K = f(a∨c) = f(a)∨f(c) = d, which is a contradiction). Stone’s lemma: there is a maximal 4-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. ⊥ −1 It means that (x1 ∨ x2) ∧ (x1 ∧ x2) ∈ f (0K). Since ⊥ x1 4 x2 ≤ (x1 ∨ x2) ∧ (x1 ∧ x2) (see Prop. 1.4), we have x1 4 x2 ∈ I. But then 0 = σ(x1 4 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 ≤ y1 . Then f(x1 ∨ y1) = f(x1) ∨ f(y1) = x ∨ y. Because x1 ⊥ y1 it holds x1 ∨ y1 = x1 4 y1 (see Prop. 1.4). Now, s(x ∨ y) = σ(x1 ∨ y1) = σ(x1 4 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 n L(R ) which satisﬁes s(1) = 1. A trivial Z2-measure on n L(R ) 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 OML4 ? (All OMLs MOκ belong to this variety.)

(2) Is the variety OML4 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., DvureˇcenskijA., L¨angerH.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- oretical Physics 44 (2005), 885-896.

3. Havl´ıkF., Ortokomplement´arn´ıdiferenˇcn´ısvazy (Czech, Mgr Thesis, Department of Logic, Faculty of Arts, Charles University in Prague, 2007).

4. MatouˇsekM., Orthocomplemented lattices with a symmetric difference, Algebra Universalis 60 (2009), 185-215. 5. MatouˇsekM., Pták Orthocomplemented posets with a symmetric difference, Order (2009), 26:1-21.

6. MatouˇsekM., Pt´akP., On identities in orthocomplemented diﬀerence lattices, Mathematica Slovaca, to appear.

7. MatouˇsekM., Pt´akP., Symmetric diﬀerence on orthomodular lattices and Z2-valued states, CMUC, to appear.

8. MatouˇsekM., PtákP., Orthocomplemented difference lattices with few generators, to appear. 9. Navara M., PtákP., For n ≥ 5 there is no nontrivial n Z2-measure on L(R ), International Journal of The- oretical Physics 43 (2004), 1595-1598.

10. Park E., Kim M.M., Chung J.Y., A note on symmetric diﬀerences of orthomodular lattices, Commun. Korean Math. Soc. 18 (2003), No 2, 207-214.

11. Pt´akP., Pulmannov´aS., 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 Matouˇsek

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