Page 57 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 57
II: Skew Lattices
Similarly (b ∨ a) ∧ b ≠ b implies that (5) fails in general. Hence symmetry must also follow from
(4) and (5) combined.
Conversely, suppose that S is distributive, symmetric and normal. Assume in addition
that S is right-handed. Thus both distributive laws
a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c) and (b ∨ c) ∧ d = (b ∧ d) ∨ (c ∧ d)
hold by the right-handed version of the standard distributive laws. We consider the status of
a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c).
First, b ∧-commutes with both a∧(b∨c) and (a∧b) ∨ (a∧c) with a∧b being their common meet.
b∧a∧(b∨c) = b∧a∧b∧(b∨c) = b∧a∧b = a∧b. a∧(b∨c)∧b = a∧b∧(b∨c)∧b = a∧b.
b∧[(a∧b) ∨ (a∧c)] =R b∧[a∧(b∨c)]∧[(a∧b) ∨ (a∧c)] = b∧(b∨c)∧a∧[(a∧b) ∨ (a∧c)]
= b∧a∧[(a∧b) ∨ (a∧c)] = b∧(a∧b)∧[(a∧b) ∨ (a∧c)] = b∧(a∧b) = a∧b.
[(a∧b) ∨ (a∧c)] ∧ b = (a∧b) ∨ (a∧c∧b) = (a∧b) ∨ (a∧b∧c∧b) = (a∧b)
Next, the necessarily commuting join of b with either a∧(b∨c) or (a∧b)∨(a∧c) is (b∨a)∧(b∨c).
Indeed
b ∨ [a∧(b∨c)] = (b∨a) ∧ (b∨b∨c) = (b∨a) ∧ (b∨c)
and
b ∨ [(a∧b)∨(a∧c)] = b ∨ (b∧a∧b) ∨ (a∧c) = b ∨ (a∧c) = (b∨a)∧(b∨c).
Hence a∧(b∨c) and (a∧b) ∨ (a∧c) are D-equivalent and both are ≤ (b∨a) ∧ (b∨c). By normality
they are equal. We have seen that (4) and (5) above hold when S is distributive, symmetric,
normal and right-handed. This must also true in the left-handed case. It follows that they must
hold when S is distributive, symmetric and normal. £
Corollary 2.3.5. A normal skew lattice (S; ○, •) in a ring is strongly distributive. £
Proof. Indeed, any skew lattice (S; ○, •) is already distributive and symmetric.
The simplest class of normal bands are the rectangular bands that satisfy abc = ac.
⎧ ⎡ 0 A AB ⎤ ⎫
⎪ ⎢ I j× j ⎪ nn
Example 2.3.1. The set of matrices ⎨ ⎢ 0 B ⎥ A ∈ F i× j, B ∈ F j×k ⎬ in F ×
⎩⎪ ⎣⎢ 0 0 0 ⎥
⎦⎥ ⎭⎪
with i + j + k = n and i + j < n, for i, j, k fixed is a rectangular band. For such bands
55
Similarly (b ∨ a) ∧ b ≠ b implies that (5) fails in general. Hence symmetry must also follow from
(4) and (5) combined.
Conversely, suppose that S is distributive, symmetric and normal. Assume in addition
that S is right-handed. Thus both distributive laws
a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c) and (b ∨ c) ∧ d = (b ∧ d) ∨ (c ∧ d)
hold by the right-handed version of the standard distributive laws. We consider the status of
a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c).
First, b ∧-commutes with both a∧(b∨c) and (a∧b) ∨ (a∧c) with a∧b being their common meet.
b∧a∧(b∨c) = b∧a∧b∧(b∨c) = b∧a∧b = a∧b. a∧(b∨c)∧b = a∧b∧(b∨c)∧b = a∧b.
b∧[(a∧b) ∨ (a∧c)] =R b∧[a∧(b∨c)]∧[(a∧b) ∨ (a∧c)] = b∧(b∨c)∧a∧[(a∧b) ∨ (a∧c)]
= b∧a∧[(a∧b) ∨ (a∧c)] = b∧(a∧b)∧[(a∧b) ∨ (a∧c)] = b∧(a∧b) = a∧b.
[(a∧b) ∨ (a∧c)] ∧ b = (a∧b) ∨ (a∧c∧b) = (a∧b) ∨ (a∧b∧c∧b) = (a∧b)
Next, the necessarily commuting join of b with either a∧(b∨c) or (a∧b)∨(a∧c) is (b∨a)∧(b∨c).
Indeed
b ∨ [a∧(b∨c)] = (b∨a) ∧ (b∨b∨c) = (b∨a) ∧ (b∨c)
and
b ∨ [(a∧b)∨(a∧c)] = b ∨ (b∧a∧b) ∨ (a∧c) = b ∨ (a∧c) = (b∨a)∧(b∨c).
Hence a∧(b∨c) and (a∧b) ∨ (a∧c) are D-equivalent and both are ≤ (b∨a) ∧ (b∨c). By normality
they are equal. We have seen that (4) and (5) above hold when S is distributive, symmetric,
normal and right-handed. This must also true in the left-handed case. It follows that they must
hold when S is distributive, symmetric and normal. £
Corollary 2.3.5. A normal skew lattice (S; ○, •) in a ring is strongly distributive. £
Proof. Indeed, any skew lattice (S; ○, •) is already distributive and symmetric.
The simplest class of normal bands are the rectangular bands that satisfy abc = ac.
⎧ ⎡ 0 A AB ⎤ ⎫
⎪ ⎢ I j× j ⎪ nn
Example 2.3.1. The set of matrices ⎨ ⎢ 0 B ⎥ A ∈ F i× j, B ∈ F j×k ⎬ in F ×
⎩⎪ ⎣⎢ 0 0 0 ⎥
⎦⎥ ⎭⎪
with i + j + k = n and i + j < n, for i, j, k fixed is a rectangular band. For such bands
55
