Page 27 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 27
I: Preliminaries
Lemma 1.2.12. For all comparable D-classes A ≥ B in a normal band S, if αBA : A → B
is the homomorphism determined for all a ∈ A by a ≥ αBA (a) ∈ B, then:
i) α A = idA for all D-classes A in S, and
A
ii) α B α A = α A for all comparable D-classes A ≥B≥ C in S.
C B C
iii) Given a ∈ A and b ∈ B, ab = αMA (a) α B (b) in the meet-class M of A and B.
M
Proof. (i) is just aaʹa = a for all aʹ ∈ A. (ii) states that (aba)c(aba) = aca for all a ∈ A, b ∈ B
and c ∈ C, a trivial consequence of normality: abacaba = acbcacacbca = accccca = aca.
Finally, (iii) follows from the identity ab = a(ab)ab(ba)b holding in all bands. £
This leads to the following result of Naoki Kimura and Miyuki Yamada.
Theorem 1.2.13. Let T be a meet semilattice poset (T; ≥). Assign a rectangular band Na
to each a ∈ T and a homomorphism Νba : Na → Nb to each pair a ≥ b in T, such that:
i) Given a ≠ b in T, Na and Nb are disjoint.
ii) Νaa is the identity map on Na for all a ∈ T.
iii) Νbc Ν a = Ν ac for all a ≥ b ≥ c in T.
b
Given x ∈ Na and y ∈ Nb, set xy = Νaab (x) Ν b (y) in Nab. Then S =∪a ∈T Na is a normal band
ab
with maximal rectangular sub-bands being the Na, with maximal semilattice image being T and
the canonical homomorphism τ: S → T defined by τ (x) = a if x ∈Na. Conversely, every normal
band arises in this fashion.
Proof. It is easily verified that the above construction produces a band such that x ≥ y for x ∈ Na
a
and y ∈ Nb holds precisely when a ≥ b in T and Ν ab (x) = y. It follows from Lemma 1.2.11 that
S is indeed a normal band. The converse assertion follows from Lemma 1.2.12. £
What is the picture in the case of regular bands? Or is there one? To begin, given
comparable D-classes A ≥ B in a band S and b ∈B, the set AbA = {abaʹ ⎜a, aʹ ∈A} is a coset of
A in B. Similarly the left coset is Ab = {ab ⎜a ∈A} and the right coset is bA = {ba ⎜a ∈ A}.
Proposition 1.2.14. Let S be regular band, with comparable D-classes A > B. Then:
i) For all b, c ∈ B, either AbA = AcA or AbA∩AcA = ∅.
ii) In particular, if AbA = AcA, then abaʹ = acaʹ for all a, aʹ in A.
iii) B is partitioned into a disjoint union of nonempty subsets Bi, each consisting of all
b ∈B inducing the same coset AbA of A in B. In particular, all c ∈ AbA induce
the coset AbA, so that AbA ⊆ Bi ={c ∈ B⎪AcA = AbA}.
iv) For each a ∈ A and each partition cell Bi, a unique bi ∈ Bi exists such that a ≥ bi.
For any b ∈ Bi, aba is this bi and bab = b.
Proof. To begin, suppose AbA ∩ AcA ≠ ∅ for some b, c ∈ C. Thus a1ba2 = a3ca4 for some a1,
a2, a3, a4 ∈A. Corollary 1.2.8 then implies that for all a, aʹ in A,
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Lemma 1.2.12. For all comparable D-classes A ≥ B in a normal band S, if αBA : A → B
is the homomorphism determined for all a ∈ A by a ≥ αBA (a) ∈ B, then:
i) α A = idA for all D-classes A in S, and
A
ii) α B α A = α A for all comparable D-classes A ≥B≥ C in S.
C B C
iii) Given a ∈ A and b ∈ B, ab = αMA (a) α B (b) in the meet-class M of A and B.
M
Proof. (i) is just aaʹa = a for all aʹ ∈ A. (ii) states that (aba)c(aba) = aca for all a ∈ A, b ∈ B
and c ∈ C, a trivial consequence of normality: abacaba = acbcacacbca = accccca = aca.
Finally, (iii) follows from the identity ab = a(ab)ab(ba)b holding in all bands. £
This leads to the following result of Naoki Kimura and Miyuki Yamada.
Theorem 1.2.13. Let T be a meet semilattice poset (T; ≥). Assign a rectangular band Na
to each a ∈ T and a homomorphism Νba : Na → Nb to each pair a ≥ b in T, such that:
i) Given a ≠ b in T, Na and Nb are disjoint.
ii) Νaa is the identity map on Na for all a ∈ T.
iii) Νbc Ν a = Ν ac for all a ≥ b ≥ c in T.
b
Given x ∈ Na and y ∈ Nb, set xy = Νaab (x) Ν b (y) in Nab. Then S =∪a ∈T Na is a normal band
ab
with maximal rectangular sub-bands being the Na, with maximal semilattice image being T and
the canonical homomorphism τ: S → T defined by τ (x) = a if x ∈Na. Conversely, every normal
band arises in this fashion.
Proof. It is easily verified that the above construction produces a band such that x ≥ y for x ∈ Na
a
and y ∈ Nb holds precisely when a ≥ b in T and Ν ab (x) = y. It follows from Lemma 1.2.11 that
S is indeed a normal band. The converse assertion follows from Lemma 1.2.12. £
What is the picture in the case of regular bands? Or is there one? To begin, given
comparable D-classes A ≥ B in a band S and b ∈B, the set AbA = {abaʹ ⎜a, aʹ ∈A} is a coset of
A in B. Similarly the left coset is Ab = {ab ⎜a ∈A} and the right coset is bA = {ba ⎜a ∈ A}.
Proposition 1.2.14. Let S be regular band, with comparable D-classes A > B. Then:
i) For all b, c ∈ B, either AbA = AcA or AbA∩AcA = ∅.
ii) In particular, if AbA = AcA, then abaʹ = acaʹ for all a, aʹ in A.
iii) B is partitioned into a disjoint union of nonempty subsets Bi, each consisting of all
b ∈B inducing the same coset AbA of A in B. In particular, all c ∈ AbA induce
the coset AbA, so that AbA ⊆ Bi ={c ∈ B⎪AcA = AbA}.
iv) For each a ∈ A and each partition cell Bi, a unique bi ∈ Bi exists such that a ≥ bi.
For any b ∈ Bi, aba is this bi and bab = b.
Proof. To begin, suppose AbA ∩ AcA ≠ ∅ for some b, c ∈ C. Thus a1ba2 = a3ca4 for some a1,
a2, a3, a4 ∈A. Corollary 1.2.8 then implies that for all a, aʹ in A,
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