Page 49 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 49
II: Skew Lattices
Expanding (a ∨ b ∨ a) ∧ (a ∨ c ∨ a) and then reducing, we get:
(a ∨ b ∨ a) ∧ (a ∨ c ∨ a) = a + bc – bca – abc – bac +baca + abac.
Equating both outcomes and then cancelling corresponding pairs of identical terms we are left
with abca = – bac +baca + abac. Rearranging gives abca + bac = baca + abac, which
is true, since it is a case of Theorem 2.1.7(3) with x = baca and y = abac, so that
xy = bac, yx = abca, xyx = baca again and yxy = abac again. £
An alternative proof that D4 holds follows from D3 holding and the fact that skew
lattices in rings are symmetric (Theorem 2.2.6). This forces D4 to hold also. (See Section 5.2.)
2.2 Instances of commutative behavior
Skew lattices, like many noncommutative structures, can possess abundant instances of
commutativity. Indeed, selective instances of commutativity play an important role in their basic
theory. We begin a pair of results to this effect.
Theorem 2.2.1. Let S be a skew lattice and let A and B be D-classes in S with join class
J = A∨B and meet class M = A∧B. Then given v ∈ J with v ≥ a ∈ A and v ≥ b ∈ B,
a∨b = v = b∨a. Similarly, given m ∈ M such that a ≥ m and b ≥ m for a ∈ A and b ∈ B,
a∧b = m = b∧a. Thus
J = {a∨b⎮a ∈ A, b ∈ B & a∨b = b∨a} and M = {a∧b⎮a ∈ A, b ∈ B & a∧b = b∧a}.
Moreover, for every a ∈ A there exist b, bʹ ∈ B such that a∨b = b∨a in J and a∧bʹ = bʹ∧a in M.
Proof. Given v ∈ J, let a ∈ A and b ∈ B be such that v ≥ a, b. (Both a and b exist since for any x
in A and y in B, v ≥ v∧x∧v in A and v ≥ v∧y∧v in B.) For this a and b we have a∨b ∈ J so that
a∨b = a∨b∨v∨a∨b = a∨v∨b = v.
Similarly, b∨a equals v also and the assertion about J is seen. The case for M is similar. For the
final assertion, pick a in A and let v ∈ J and m ∈ M be such that v ≥ a ≥ m. Now apply remarks
from the first part of the proof. £
This theorem has the important corollary:
Theorem 2.2.2. Given an element e in a skew lattice S, the following are equivalent:
1) De = {e}.
2) For all x ∈ S, e∨x = x∨e.
3) For all x ∈ S, e∧x = x∧e.
The subset of such elements forms a sublattice Z(S) of S (called the center of S).
47
Expanding (a ∨ b ∨ a) ∧ (a ∨ c ∨ a) and then reducing, we get:
(a ∨ b ∨ a) ∧ (a ∨ c ∨ a) = a + bc – bca – abc – bac +baca + abac.
Equating both outcomes and then cancelling corresponding pairs of identical terms we are left
with abca = – bac +baca + abac. Rearranging gives abca + bac = baca + abac, which
is true, since it is a case of Theorem 2.1.7(3) with x = baca and y = abac, so that
xy = bac, yx = abca, xyx = baca again and yxy = abac again. £
An alternative proof that D4 holds follows from D3 holding and the fact that skew
lattices in rings are symmetric (Theorem 2.2.6). This forces D4 to hold also. (See Section 5.2.)
2.2 Instances of commutative behavior
Skew lattices, like many noncommutative structures, can possess abundant instances of
commutativity. Indeed, selective instances of commutativity play an important role in their basic
theory. We begin a pair of results to this effect.
Theorem 2.2.1. Let S be a skew lattice and let A and B be D-classes in S with join class
J = A∨B and meet class M = A∧B. Then given v ∈ J with v ≥ a ∈ A and v ≥ b ∈ B,
a∨b = v = b∨a. Similarly, given m ∈ M such that a ≥ m and b ≥ m for a ∈ A and b ∈ B,
a∧b = m = b∧a. Thus
J = {a∨b⎮a ∈ A, b ∈ B & a∨b = b∨a} and M = {a∧b⎮a ∈ A, b ∈ B & a∧b = b∧a}.
Moreover, for every a ∈ A there exist b, bʹ ∈ B such that a∨b = b∨a in J and a∧bʹ = bʹ∧a in M.
Proof. Given v ∈ J, let a ∈ A and b ∈ B be such that v ≥ a, b. (Both a and b exist since for any x
in A and y in B, v ≥ v∧x∧v in A and v ≥ v∧y∧v in B.) For this a and b we have a∨b ∈ J so that
a∨b = a∨b∨v∨a∨b = a∨v∨b = v.
Similarly, b∨a equals v also and the assertion about J is seen. The case for M is similar. For the
final assertion, pick a in A and let v ∈ J and m ∈ M be such that v ≥ a ≥ m. Now apply remarks
from the first part of the proof. £
This theorem has the important corollary:
Theorem 2.2.2. Given an element e in a skew lattice S, the following are equivalent:
1) De = {e}.
2) For all x ∈ S, e∨x = x∨e.
3) For all x ∈ S, e∧x = x∧e.
The subset of such elements forms a sublattice Z(S) of S (called the center of S).
47
