Page 230 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 230
athan E. Leech │ Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond
Theorem 6.2.16 If s is associative on all primitive subalgebras of a s-band S, then s
is associative on S, and conversely.
Proof. Granted the assumption, we need only show that s is associative when S is generated
from two incomparable D-classes, say A and B. Let their meet class be M and join class be J.
Let a lattice section of S be given by T = {a0, b0, m0, j0}. Take a ∈ A, b ∈ B. We show that asb
= a ∨T b. To begin
a ∨T b = (aa0 + bb0 − aa0bb0)(a0a + b0b − a0ab0b)
= a + am0b − ab0b + bm0a + b − bm0ab − abm0a − aa0b + ab
= a + bm0a + b − bm0ab − abm0a
since aa0b + ab0b − am0b = aj0b = ab by regularity. Of course, asb = a + b + ba − aba − bab.
Denoting the difference (a ∨T b) − (asb) by Δ(a, b), we have
Δ(a, b) = bm0a − bm0ab − abm0a − ba + aba + bab.
The associativity of s on the primitive s-band M∪B implies that
Δ(abm0a, b) = bm0a − bm0ab − abm0a − bam0a + abm0a + bab
= bm0a − bm0ab − bam0a + bab = 0.
Subtracting from Δ(a, b) yields the refinement Δ(a, b) = bam0a + aba − abm0a − ba. Using the
latter to calculate Δ(a, bam0b) in the associative context of M∪A gives:
0 = Δ(a, bam0b) = bam0a + aba − abm0a − ba = Δ(a, b) .
The converse is trivial. £
Since s is associative on a s-band only if it is thus on all primitive subalgebras, two of
our earlier associativity criteria can be fine-tuned for the primitive case as follows.
Corollary 6.2.17. Given a primitive s-band P with D-classes A > B, the following are
equivalent:
i) s is associative on S.
ii) For all a ∈ A and b, c ∈ B, a[b, c]2 = [b, c]2a.
iii) For all a ∈ A and b, c ∈ B, both
a(bc – cbc)a = a(bc – cbc) and a(bc – bcb)a = (bc – bcb)a.
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Theorem 6.2.16 If s is associative on all primitive subalgebras of a s-band S, then s
is associative on S, and conversely.
Proof. Granted the assumption, we need only show that s is associative when S is generated
from two incomparable D-classes, say A and B. Let their meet class be M and join class be J.
Let a lattice section of S be given by T = {a0, b0, m0, j0}. Take a ∈ A, b ∈ B. We show that asb
= a ∨T b. To begin
a ∨T b = (aa0 + bb0 − aa0bb0)(a0a + b0b − a0ab0b)
= a + am0b − ab0b + bm0a + b − bm0ab − abm0a − aa0b + ab
= a + bm0a + b − bm0ab − abm0a
since aa0b + ab0b − am0b = aj0b = ab by regularity. Of course, asb = a + b + ba − aba − bab.
Denoting the difference (a ∨T b) − (asb) by Δ(a, b), we have
Δ(a, b) = bm0a − bm0ab − abm0a − ba + aba + bab.
The associativity of s on the primitive s-band M∪B implies that
Δ(abm0a, b) = bm0a − bm0ab − abm0a − bam0a + abm0a + bab
= bm0a − bm0ab − bam0a + bab = 0.
Subtracting from Δ(a, b) yields the refinement Δ(a, b) = bam0a + aba − abm0a − ba. Using the
latter to calculate Δ(a, bam0b) in the associative context of M∪A gives:
0 = Δ(a, bam0b) = bam0a + aba − abm0a − ba = Δ(a, b) .
The converse is trivial. £
Since s is associative on a s-band only if it is thus on all primitive subalgebras, two of
our earlier associativity criteria can be fine-tuned for the primitive case as follows.
Corollary 6.2.17. Given a primitive s-band P with D-classes A > B, the following are
equivalent:
i) s is associative on S.
ii) For all a ∈ A and b, c ∈ B, a[b, c]2 = [b, c]2a.
iii) For all a ∈ A and b, c ∈ B, both
a(bc – cbc)a = a(bc – cbc) and a(bc – bcb)a = (bc – bcb)a.
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