Page 156 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 156
Jonathan E. Leech │ Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond

Algebras of the form ω(B) with operations as defined above are called omega algebras.
These algebras lie in the varieties of left-handed skew Boolean algebras and left-handed skew
Boolean ∩-algebras, with the particular emphasis indicated by the context.

Characterizing omega algebras

Given a skew lattice S, recall that a lattice section of S is any sublattice S0 that intersects
every D-class of S. Any such lattice must be isomorphic to the maximal lattice image S/D of S.

Lemma 4.5.6. Given an omega algebra ω(B), S0 = {(a, a)⎪a ∈B} is a lattice section. £

Lemma 4.5.7. Given (a, aʹ) ∈ ω(B), the map (a, bʹ) → (a, aʹ) ∩ (a, bʹ) is a bijection of
the D-class of (a, aʹ) with the principal ideal ⎡(a, aʹ)⎤ of (ω(B), ≥).

Proof. In ω(B) the ideal ⎡(a, aʹ)⎤ is just the set {(b, aʹ∧b)⎪a ≥ b}. Given aʹ ≤ a, then for all x ≤ a,
(a, aʹ) ∩ (a, x) = ([aʹ∧x] + [a \ (aʹ∨x)], aʹ∧x). Given any b ≤ a, we seek some x ≤ a such that
([aʹ∧x] + [a \ (aʹ∨x)], aʹ∧x) = (b, aʹ∧b) in ⎡(a, aʹ)⎤. Setting x = aʹ∧b ∨ [a \ (aʹ∨b)] we get,
aʹ∧x = (aʹ∧ b) ∨ 0 = aʹ∧ b and thus

[aʹ∧x] + [a \ (aʹ∨x)] = aʹ∧b + [a \ (aʹ ∨ [a \ (aʹ∨b)])]

= aʹ∧b + [a \ (aʹ ∨ [(a \ aʹ) ∧ (a \ b)])]

= aʹ∧b + [(a \ aʹ ) ∧ (aʹ ∨ b)] = b.

Is x ≤ a unique? Suppose that y ≤ a also satisfies the desired conditions. Then aʹ∧x = aʹ∧b = aʹ∧y
and [aʹ∧x] + [a \ (aʹ∨x)] = b = [aʹ∧y] + [a \ (aʹ∨y)], from which also follows first a \ (aʹ∨x) = a \
(aʹ∨y) and then aʹ∨x = aʹ∨y. Cancelling aʹ∧x = aʹ∧y and aʹ∨x = aʹ∨y in B gives x = y. £

We thus have the following characterization of omega algebras:

Theorem 4.5.8. A left-handed skew Boolean ∩-algebra S is isomorphic to ω(B) for
some generalized Boolean algebra B if and only if S has a lattice section S0 and for all e ∈S the
map a → e ∩ a is a bijection of the D-class De with the principal poset ideal ⎡e⎤. Under these
conditions S is isomorphic to both ω(S0) and ω(S/D).

Proof. The conditions are clearly necessary. Conversely, given these conditions, for each e ∈ S0
let βe: De → ω(S0) be given by βe(f) = (e, e∩f). Next we define β: S → ω(S0) to be the union
∪eβe. β is at least a bijection from S to ω(S0). It also preserves ∧. Given e1 and e2 in S0, then for
all f1 D e1 and f2 D e2 in S we have

β[f1∧f2] = (e1∧e2, e1∧e2 ∩ f1∧f2) = (e1∧e2, e1∧e2 ∩ f1∧f2∧e2) = (e1∧e2, e1∧e2 ∩ f1∧e2)
and

β[f1]∧β[f2] = (e1, e1 ∩ f1) ∧ (e2, e2 ∩ f2) = (e1∧e2, [e1 ∩ f1] ∧ e2) = (e1∧e2, e1 ∩ f1 ∩ e2).

154
   151   152   153   154   155   156   157   158   159   160   161