A Symmetric-Difference-Closed Orthomodular Lattice That Is Stateless

This paper carries on the investigation of the orthomodular lattices that are endowed with a symmetric difference. Let us call them ODLs. Note that the ODLs may have a certain bearing on “quantum logics” - the ODLs are close to Boolean algebras though they capture the phenomenon of non-compatibility. The initial question in studying the state space of the ODLs is whether the state space can be poor. This question is of a purely combinatorial nature. In this note, we exhibit a finite ODL whose state space is empty (respectively, whose state space is a singleton).


Definition 1
Let L = (X, ≤, , , 0, 1), where (X, ≤, , 0, 1) is an orthomodular lattice and : X 2 → X is a binary operation called a symmetric difference. Then L is said to be an orthomodular difference lattice (an ODL) if the following conditions are satisfied (x, y, z ∈ X): i. x (y z) = (x y) z, ii.
The class of ODLs forms a variety of algebras and we note that not every orthomodular lattice can be endowed with a symmetric difference. A basic example of ODLs is a Cartesian product of a Boolean algebra and the "Chinese lantern" MO 3 (see [1] for a thorough algebraic analysis of ODLs). In Proposition 1 that follows, we encounter several other examples. We would need the following construction (a generalization of Construction 8.5 of [1]). Let us first recall the standard construction of the horizontal sum of orthomodular lattices (see e.g., [12], p.59). If L α , α ∈ I are orthomodular lattices, then the horizontal sum Hor(L α , α ∈ I ) is defined by identifying the 0s and 1s in the disjoint union α∈I L α .
The Hor(L α , α ∈ I ) is obviously an orthomodular lattice. Let us formulate the following definition.
Definition 2 Let B be a Boolean algebra. Let B α , α ∈ I be a collection of Boolean algebras such that each B α , α ∈ I , is a subset of B (not necessarily a Boolean subalgebra of B) and, moreover, their minimal and maximal elements in each B α are identical to 0 and 1 of B. Let us further require that the symmetric difference α of each B α equals the symmetric difference of B (i.e., if a, b ∈ B α , α ∈ I then a α b = a b). Let us say that the collection B α , α ∈ I covers B if the following conditions are fulfilled (α, β ∈ I ):

Results
Proposition 1 Let B be a Boolean algebra with symmetric difference and let {B α , α ∈ I } cover B. Then the horizontal sum Hor(B α , α ∈ I ) endowed with becomes an ODL.
Proof Since Hor(B α , α ∈ I ) is an OML it is sufficient to verify that the required properties of hold true. The first two properties hold because they hold in B. The third one (x y ≤ x ∨ y) holds when x, y are in the same Boolean algebra. Otherwise, it holds trivially since x ∨ y = 1.
Let us call the above construction the ODL horizontal sum. It should be noted that the essential property of a collection B α , α ∈ I is the property 1. of Definition 2. Indeed, if B α , α ∈ I fulfils the property 1. of Definition 2 and if there is a b ∈ B such that b ∈ B \∪ α∈I B α , we can always add the Boolean algebra B b = {0, b, b , 1} to the collection B α , α ∈ I as far as we reach the collection that covers B. The following lemma will be also used in the desired construction of a stateless ODL.

Lemma 1 Let B be a Boolean algebra and let be the (unique) symmetric difference of
Assume that 0 < c i < 1 for all i ∈ {1, 2, 3} and assume further that c 1 c 2 c 3 = 1. Consider the set D = {0, c 1 , c 2 , c 3 , c 1 , c 2 , c 3 , 1} and endow D with the structure of a Boolean algebra by using the Boolean isomorphism onto the Boolean algebra exp{1, 2, 3} (thus, the elements c 1 , c 2 and c 3 will become the atoms of D). Then the Proof Observe first that all elements of D are distinct. Indeed, suppose for instance that c 1 = c 2 . But c 2 c 2 = 1 (see [1]) and therefore c 1 c 2 c 2 = c 2 c 2 c 3 = 1 c 3 = c 3 = 1. This is absurd. In showing that D agrees with the original ∈ B, let us argue by cases. Then and we are done.
The definition of a state on an ODL reads as follows. The state space S(L) of an ODL L is a convex compact set, and hence S(L) is big if L has an abundance of pure states. This does not have to be the case (observe that L does not have to be set-representable; see [1]). In fact, it does not have to have any state at all, as we are going to show. The following construction would serve the purpose. Proof The required properties can be checked by a slightly tedious but relatively straightforward verification (see also [3]).

Theorem 3 Let L be the ODL horizontal sum of the collection of Boolean subalgebras of the Construction 1. Then L does not have any state.
Proof For the intuition, let us use the graphical representation of L in Fig. 1 (a variant of the Greechie paste job, see [10]). The horizontal lines represents maximal Boolean subalgebras of L of the cardinality higher than 4. The dotted (vertical) lines indicate that the symmetric difference of connected vertices is 1; e.g., a 1 b 1 c 1 = 1, and similarly a 3 b 3 d 1 = 1, etc. Hence the construction of L as the ODL horizontal sum gives us the following set of equations: Looking for a contradiction, suppose that s is a state on L. By the property of s we see that This implies that Analogously, we obtain Summing up the inequalities, we infer that 6 i=1 (s(a i ) + s(b i )) ≥ 3. But it also holds that 7 i=1 s(a i ) = 7 i=1 s(b i ) = 1. Together with the non-negativity of s, we have derived a contradiction. This completes the proof.
In concluding the paper, let us observe that Theorem 3 allows one to spell out another slightly bizarre result.

Theorem 4 There is a (non-Boolean) ODL with exactly one state.
Proof Let L be an ODL with no state. Let K = L × {0, 1}, where {0, 1} is meant as a Boolean algebra. Then K has exactly one state. Indeed, let us set s(e, 0) = 0 and s(e, 1) = 1 for any e ∈ L. Then s is a two-valued state on K. It is easily seen that this s is the only state on K