Page 154 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
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Jonathan E. Leech │ Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond
4.5 Omega algebras and skew Boolean covers
Given a lattice L = (L; ∨, ∧), upon setting ω(L) = {(a, b) ∈ L × L⎪a ≥ b} we obtain a
sublattice of L × L that clearly is distributive when L is. But what if L is a (possibly generalized)
Boolean lattice with zero 0, so that for each a ∈ L, the principal poset ideal ⎡a⎤ = {b ∈ L⎮a ≥ b}
is a Boolean lattice?
Proposition 4.5.1. Given a lattice L with minimal element 0, ω(L) forms a generalized
Boolean lattice if and only if L, and hence ω(L), is trivial.
Proof. If a > 0 for some a ∈L, then (a, 0) has no complement in ⎡(a, a)⎤. £
Given a generalized Boolean algebra B, alternative twisted meet and join operations can
be defined on the underlying set of ω(B) to give a Boolean skew lattice as follows:
(a, aʹ) ∧ (b, bʹ) = (a∧b, aʹ∧b) and (a, aʹ) ∨ (b, bʹ) = (a∨b, (aʹ \ b)∨bʹ).
Here all component operations are in B with a \ b being the relative complement of a∧b in ⎡a⎤.
Lemma 4.5.2. Under the operations above, ω(B) is left-handed skew lattice with a zero.
Proof. That ∧ is idempotent and associative is easily checked, as is the fact that ∨ is idempotent.
Associativity of ∨ reduces to comparing the second coordinates of [(a, aʹ) ∨ (b, bʹ)] ∨ (c, cʹ) and
(a, a′) ∨ [(b, bʹ) ∨ (c, cʹ)] to see if {[(aʹ \ b) ∨ bʹ] \ c} ∨ cʹ = [aʹ \ (b ∨ c)] ∨ (bʹ \ c) ∨ cʹ. But
[(aʹ \ b) ∨ bʹ] \ c = [a ʹ \ (b∨c)] ∨ (bʹ \ c)
holds for generalized Boolean algebras. Next, expanding gives:
(a, aʹ) ∧ [(a, aʹ)∨(b, bʹ)] = (a, aʹ) ∧ (a∨b, (aʹ \ b) ∨ bʹ) = (a∧(a ∨ b), a ʹ∧ (a∨b)) = (a, aʹ)
and
(a, aʹ) ∨ [(a, aʹ)∧(b, bʹ)] = (a, aʹ) ∨ (a∧b, aʹ∧b) = (a ∨ (a∧b), [aʹ \ (a∧b)] ∨ (aʹ∧b)) = (a, aʹ).
In similar fashion, the remaining pair of absorption identities are verified. Clearly (0, 0) is the
zero element. Finally, (a, aʹ) ∧ (b, bʹ) ∧ (a, aʹ) = (a, aʹ) ∧ (b, bʹ) since both reduce to (a∧b, aʹ∧b).
Thus the identity x∧y∧x = x∧y holds, as well as its equivalent dual, x∨y∨x = y∨x. o
Lemma 4.5.3. (ω(B); ∨, ∧) is a symmetric skew lattice. If (a, aʹ) and (b, bʹ) commute,
then (a, aʹ) ∧ (b, bʹ) = (a∧b, aʹ∧bʹ) and (a, aʹ) ∨ (b, bʹ) = (a∨b, aʹ∨bʹ).
Proof. Given (aʹ, a), (bʹ, b) ∈ ω(B), (a, aʹ) ∧ (b, bʹ) = (b, bʹ) ∧ (a, aʹ) if and only if aʹ∧b = a∧bʹ
in the second coordinate of the outcomes, with both equaling aʹ∧bʹ. Likewise, (a, aʹ) ∨ (b, bʹ) =
(b, bʹ) ∨ (a, aʹ) if and only if (aʹ \ b) ∨ bʹ = (bʹ \ a) ∨ aʹ in the second coordinate of the outcomes
with both equaling aʹ∨bʹ. Assuming aʹ∧b = a∧bʹ we get
152
4.5 Omega algebras and skew Boolean covers
Given a lattice L = (L; ∨, ∧), upon setting ω(L) = {(a, b) ∈ L × L⎪a ≥ b} we obtain a
sublattice of L × L that clearly is distributive when L is. But what if L is a (possibly generalized)
Boolean lattice with zero 0, so that for each a ∈ L, the principal poset ideal ⎡a⎤ = {b ∈ L⎮a ≥ b}
is a Boolean lattice?
Proposition 4.5.1. Given a lattice L with minimal element 0, ω(L) forms a generalized
Boolean lattice if and only if L, and hence ω(L), is trivial.
Proof. If a > 0 for some a ∈L, then (a, 0) has no complement in ⎡(a, a)⎤. £
Given a generalized Boolean algebra B, alternative twisted meet and join operations can
be defined on the underlying set of ω(B) to give a Boolean skew lattice as follows:
(a, aʹ) ∧ (b, bʹ) = (a∧b, aʹ∧b) and (a, aʹ) ∨ (b, bʹ) = (a∨b, (aʹ \ b)∨bʹ).
Here all component operations are in B with a \ b being the relative complement of a∧b in ⎡a⎤.
Lemma 4.5.2. Under the operations above, ω(B) is left-handed skew lattice with a zero.
Proof. That ∧ is idempotent and associative is easily checked, as is the fact that ∨ is idempotent.
Associativity of ∨ reduces to comparing the second coordinates of [(a, aʹ) ∨ (b, bʹ)] ∨ (c, cʹ) and
(a, a′) ∨ [(b, bʹ) ∨ (c, cʹ)] to see if {[(aʹ \ b) ∨ bʹ] \ c} ∨ cʹ = [aʹ \ (b ∨ c)] ∨ (bʹ \ c) ∨ cʹ. But
[(aʹ \ b) ∨ bʹ] \ c = [a ʹ \ (b∨c)] ∨ (bʹ \ c)
holds for generalized Boolean algebras. Next, expanding gives:
(a, aʹ) ∧ [(a, aʹ)∨(b, bʹ)] = (a, aʹ) ∧ (a∨b, (aʹ \ b) ∨ bʹ) = (a∧(a ∨ b), a ʹ∧ (a∨b)) = (a, aʹ)
and
(a, aʹ) ∨ [(a, aʹ)∧(b, bʹ)] = (a, aʹ) ∨ (a∧b, aʹ∧b) = (a ∨ (a∧b), [aʹ \ (a∧b)] ∨ (aʹ∧b)) = (a, aʹ).
In similar fashion, the remaining pair of absorption identities are verified. Clearly (0, 0) is the
zero element. Finally, (a, aʹ) ∧ (b, bʹ) ∧ (a, aʹ) = (a, aʹ) ∧ (b, bʹ) since both reduce to (a∧b, aʹ∧b).
Thus the identity x∧y∧x = x∧y holds, as well as its equivalent dual, x∨y∨x = y∨x. o
Lemma 4.5.3. (ω(B); ∨, ∧) is a symmetric skew lattice. If (a, aʹ) and (b, bʹ) commute,
then (a, aʹ) ∧ (b, bʹ) = (a∧b, aʹ∧bʹ) and (a, aʹ) ∨ (b, bʹ) = (a∨b, aʹ∨bʹ).
Proof. Given (aʹ, a), (bʹ, b) ∈ ω(B), (a, aʹ) ∧ (b, bʹ) = (b, bʹ) ∧ (a, aʹ) if and only if aʹ∧b = a∧bʹ
in the second coordinate of the outcomes, with both equaling aʹ∧bʹ. Likewise, (a, aʹ) ∨ (b, bʹ) =
(b, bʹ) ∨ (a, aʹ) if and only if (aʹ \ b) ∨ bʹ = (bʹ \ a) ∨ aʹ in the second coordinate of the outcomes
with both equaling aʹ∨bʹ. Assuming aʹ∧b = a∧bʹ we get
152
