Page 92 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 92
Jonathan E. Leech │ Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond
Theorem 3.1.2. Given a quasilattice (Q; ∨, ∧), the map χ: Con(Q) → [D, ∇] × [Δ, D]
defined by χ(θ) = (D∨θ, D∩θ) is an embedding of complete lattices yielding Con(Q) to be a
subdirect product of [D, ∇] and [Δ, D].
Proof. That χ is a homomorphism of complete lattices follows from the above theorem. To see
that χ is 1-1, observe that for any congruence θ on Q,
xθy iff xθx∧y∧x, x∧y∧xθy∧x∧y and y∧x∧yθy.
In general, x∧y∧x θ y∧x∧y iff x∧y∧x θ∩D y∧x∧y as x∧y∧x D y∧x∧y. Also x θ x∧y∧x iff x D∨θ
x∧y∧x, and likewise y∧x∧y θ y iff y∧x∧y D∨θ y. The θ ⇒ θ∨D direction is clear. So suppose
that x D∨θ x∧y∧x. Thus by Theorem 3.1.1(i), u ∈ Q exists such that x D u θ x∧y∧x. But then
x = x∧u∧x θ x∧( x∧y∧x)∧x = x∧y∧x. Similarly y∧x∧y D∨θ y implies y∧x∧y θ y. Consequently:
x θ y if and only if x D∨θ x∧y∧x, x∧y∧x θ∩D y∧x∧y and y∧x∧y D∨θ y.
But this means that θ is determined by D∩θ and D∨θ, so that χ is an embedding. Since [Δ, D]
and [D, ∇] are sublattices of Con(Q) with D∩θ = θ for all θ ∈ [Δ, D] and D∨θ = θ for all
θ ∈ [D, ∇], Con(Q) is embedded as a subdirect product. £
When does Con(Q) factor directly as [D, ∇] × [Δ, D]? To begin, a quasilattice Q splits if
it is isomorphic with the product T × A of a lattice T and an antilattice A. In what follows ρ
denotes the smallest congruence containing ≥.
Theorem 3.1.3. Given a quasilattice Q, the following are equivalent:
i) Q splits.
ii) D ∩ρ = Δ.
iii) A congruence θ exists such that D∩θ = Δ and Doθ = ∇.
iv) For all x, y ∈ Q, x ρ y iff x∨y = y∨x [alternatively, x∧y = y∧x].
v) Con(Q) ≅ [D, ∇] × [Δ,D] under the map θ → (θoD, θ∩D).
Proof. Assume (i). Indeed, let Q = T × A with T and A as above. The D-classes of Q consist of
pairs (t, a) having common T-coordinates and the ρ-classes consist of pairs (t, a) having common
A-coordinates. From this, (ii) follows. Since Doρ = ∇ always holds, (ii) implies (iii). Suppose
that (iii) holds for a congruence θ. Thus each θ-class meets each D-class at a unique element.
Let χ: Q → Q/D × Q/θ be the homomorphism defined by χ(x) = (xD, xθ), where xD and xθ are
the respective congruence classes of x. Since D∩θ = Δ, χ is one-to-one. Since each D -class
meets each θ-class, every possible pair (xD, yθ) is in the image of χ, so that χ is an isomorphism.
Since each θ-class meets each maximal rectangular subalgebra of Q, Q/θ is rectangular. Since
Q/D is always a lattice, we have shown that Q splits. Hence (i) through (iii) are equivalent.
90
Theorem 3.1.2. Given a quasilattice (Q; ∨, ∧), the map χ: Con(Q) → [D, ∇] × [Δ, D]
defined by χ(θ) = (D∨θ, D∩θ) is an embedding of complete lattices yielding Con(Q) to be a
subdirect product of [D, ∇] and [Δ, D].
Proof. That χ is a homomorphism of complete lattices follows from the above theorem. To see
that χ is 1-1, observe that for any congruence θ on Q,
xθy iff xθx∧y∧x, x∧y∧xθy∧x∧y and y∧x∧yθy.
In general, x∧y∧x θ y∧x∧y iff x∧y∧x θ∩D y∧x∧y as x∧y∧x D y∧x∧y. Also x θ x∧y∧x iff x D∨θ
x∧y∧x, and likewise y∧x∧y θ y iff y∧x∧y D∨θ y. The θ ⇒ θ∨D direction is clear. So suppose
that x D∨θ x∧y∧x. Thus by Theorem 3.1.1(i), u ∈ Q exists such that x D u θ x∧y∧x. But then
x = x∧u∧x θ x∧( x∧y∧x)∧x = x∧y∧x. Similarly y∧x∧y D∨θ y implies y∧x∧y θ y. Consequently:
x θ y if and only if x D∨θ x∧y∧x, x∧y∧x θ∩D y∧x∧y and y∧x∧y D∨θ y.
But this means that θ is determined by D∩θ and D∨θ, so that χ is an embedding. Since [Δ, D]
and [D, ∇] are sublattices of Con(Q) with D∩θ = θ for all θ ∈ [Δ, D] and D∨θ = θ for all
θ ∈ [D, ∇], Con(Q) is embedded as a subdirect product. £
When does Con(Q) factor directly as [D, ∇] × [Δ, D]? To begin, a quasilattice Q splits if
it is isomorphic with the product T × A of a lattice T and an antilattice A. In what follows ρ
denotes the smallest congruence containing ≥.
Theorem 3.1.3. Given a quasilattice Q, the following are equivalent:
i) Q splits.
ii) D ∩ρ = Δ.
iii) A congruence θ exists such that D∩θ = Δ and Doθ = ∇.
iv) For all x, y ∈ Q, x ρ y iff x∨y = y∨x [alternatively, x∧y = y∧x].
v) Con(Q) ≅ [D, ∇] × [Δ,D] under the map θ → (θoD, θ∩D).
Proof. Assume (i). Indeed, let Q = T × A with T and A as above. The D-classes of Q consist of
pairs (t, a) having common T-coordinates and the ρ-classes consist of pairs (t, a) having common
A-coordinates. From this, (ii) follows. Since Doρ = ∇ always holds, (ii) implies (iii). Suppose
that (iii) holds for a congruence θ. Thus each θ-class meets each D-class at a unique element.
Let χ: Q → Q/D × Q/θ be the homomorphism defined by χ(x) = (xD, xθ), where xD and xθ are
the respective congruence classes of x. Since D∩θ = Δ, χ is one-to-one. Since each D -class
meets each θ-class, every possible pair (xD, yθ) is in the image of χ, so that χ is an isomorphism.
Since each θ-class meets each maximal rectangular subalgebra of Q, Q/θ is rectangular. Since
Q/D is always a lattice, we have shown that Q splits. Hence (i) through (iii) are equivalent.
90
