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

Orthocomplemented lattices with a symmetric difference Milan Matouˇ sek 1 ODLs 2 Z 2 -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 △ : X 2 → 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 : (D 1 ) x △ ( y △ z ) = ( x △ y ) △ z , (D 2 ) x △ 1 = x ⊥ , 1 △ x = x ⊥ , (D 3 ) x △ y ≤ x ∨ y . ( X, ∧ , ∨ , ⊥ , 0 , 1) will be denoted by L supp 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 L supp 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 L supp .

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 { L supp ; L is an ODL } , i.e. OML △ = HSP ( { L supp ; 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. (D 2 ): 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 (D 3 ) 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 ⊥ ≤ y ⊥ . But also y ⊥ ≤ y ⊥ . x △ y ⇒ (D 3 ): ( x △ y ⊥ ) △ y ⊥ ≤ y ⊥ . Finally, ( x △ y ⊥ ) △ y ⊥ = x △ ( y ⊥ △ y ⊥ ) = x △ 0 = x .

2 Z 2 -valued states Z 2 ≡ the group { 0 , 1 } with the modulo 2 addition ⊕ Definition 2.1 Let L be an OML. (i) Let s : L → Z 2 be a mapping. Then s is said to be a Z 2 - valued state (abbr., a Z 2 - state ) provided s (1 L ) = 1 and s ( x ∨ y ) = s ( x ) ⊕ s ( y ) whenever x, y ∈ L , x ≤ y ⊥ . (ii) L is called Z 2 - full if ∀ x, y ∈ L, x � = y, x � = 0 L , y � = 1 L ∃ Z 2 -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 0 L ∈ I & ∀ a, b ∈ I : a △ b ∈ I ; I is a proper △ - ideal if I is a △ -ideal and 1 L �∈ 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 I 0 be a proper △ -ideal in L and let b ∈ L , b �∈ I 0 . Then there is a maximal △ -ideal I such that I 0 ⊆ 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 → Z 2 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 Z 2 -state on L supp . Theorem 2.7 Let L be an ODL. Then its support L supp is Z 2 -full.

Proof of Lemma 2.3. The set J is obviously a △ -ideal. Let us see that 1 L �∈ J . Suppose on the contrary that 1 L ∈ J . Then 1 L = a △ x for some element a ∈ I . The But x ⊥ does equality 1 L = a △ x implies that a = x ⊥ . not belong to I which is a contradiction. Thus, 1 L �∈ J . Further, x = 0 L △ x ∈ J . If x ⊥ ∈ J , then 1 L = x △ x ⊥ ∈ J – a contradiction again. ( ⇒ ) Suppose that I is maximal Proof of Prop. 2.4. 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- Then x △ x ⊥ = 1 which means pose that { x, x ⊥ } ⊆ I . 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 1 L = b △ b ⊥ ∈ J . This means that J is not proper. Proof of Prop. 2.5. Write I = { J ⊆ L ; J is a proper △ -ideal, I 0 ⊆ J and b �∈ J } . Then I 0 ∈ I and therefore I � = ∅ . Zorn’s lemma: the set I ordered by inclusion contains a maximal element, I . (1) I is a proper △ -ideal, I 0 ⊆ 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 1 L = 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 Z 2 -state on L supp . 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.

Let L be an ODL. Let x, y ∈ L Proof of Thm. 2.7. with x � = y , x � = 0 and y � = 1. Let us set I 0 = { 0 L , 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 Z 2 -state on L supp .

3 The variety OML △ Let us recall that OML △ = HSP ( { L supp ; L ∈ ODL} ). Theorem 3.1 Let K ∈ OML △ and let x 0 ∈ K , x 0 � = 1 K . Then there is a Z 2 -state, s , on K such that s ( x 0 ) = 0 .

Proposition 3.2 Suppose that L is an OML. Suppose that there are blocks B 1 , B 2 , . . . , B n of L such that the following two conditions are satisfied: (1) each B i , 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 B 1 , B 2 , . . . , B n (i.e., the cardinality of the set { i ; a ∈ B i } is even). Then there is no Z 2 -state on L . Proof. Let s : L → Z 2 be a Z 2 -state. Then ⊕ a ∈ At( B i ) s ( a ) = s (1 L ) = 1 ( i = 1 , . . . , n ). Therefore ⊕ i ∈{ 1 ,...,n } ( ⊕ a ∈ At( B i ) 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).