Page 53 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 53
II: Skew Lattices
Theorem 2.2.8. (Cvetko-Vah) Given a skew lattice the following are equivalent:
1) S has both a left-handed section and a right-handed section.
2) Sub-skew lattices SL and SR exist whose intersection with any D-class is an L-class,
or respectively, an R-class of S.
3) S has a lattice section S0.
When these conditions hold, then:
4) The natural epimorphisms S → S/R and S → S/L induce isomorphisms of SL with
S/R and SR with S/L.
5) Every x ∈ S factors uniquely as a = aʹ∧aʺ with aʹ ∈ SL ∩ Dx, aʺ ∈ SR ∩ Dx.
Under this decomposition
(aʹ∧aʺ)∧(bʹ∧bʺ) = (aʹ∧ bʹ)∧(aʺ∧bʺ) and (aʹ∧aʺ)∨(bʹ∧bʺ) = (aʹ∨ bʹ)∧(aʺ∨bʺ).
6) The functions πL: S → SL and πR: S → SR defined by πL(a) = aʹ and πR(a) = aʺ are
retractions of S upon SL and SR respectively and the commuting composite πLπR is
a retraction of S upon S0; moreover R = ker(πL), L = ker(πR) and D = ker(πLπR).
Proof. (2) clearly implies (1). Given (1), expand Sʹ to the subset SL = ∪{La ⎜a ∈ Sʹ} and expand
Sʺ to the subset SR = ∪{Ra ⎜a ∈ Sʺ}. Since L and R are congruences, SL and SR are as stated in
(2). Given (2) again, since every pair of L and R-classes in a D-class must meet at a single
element in that D-class, (3) follows from (2). Conversely, given (3) one has a minimal case of
(1). Thus (1) – (3) are equivalent. Since SL is an R-class cross-section, the map S → S/R
restricts to an isomorphism of SL upon S/R. Similar remarks apply to SR → S/L and (4) follows.
(5) follows from (4) and the embedding S ≅ S/R ×S/D S/L ⊆ S/R × S/L ≅ SL × SR given by a =
aʹ∧aʺ → (aʹ, aʺ). To see (6), first set aL = a∧a0 and aR = a0∧a. This gives
while a∧b = (a∧a0∧a)∧(b∧b0∧b)
(aL∧bL)∧(aR∧bR) = (a∧a0∧b∧b0)∧(a0∧a∧b0∧b) = (a∧a0∧b)∧(a∧b0∧b).
But both “middles” are equal since
a0∧a∧b∧b0 = a0∧a∧b∧a0∧b0∧a∧b∧b0 = a0∧y∧a0∧b0∧a∧b0 = a0∧b∧a∧b0.
Thus a∧b = (aL∧bL)∧(aR∧bR) follows. Recalling that R(∧) = L(∨) and L(∧) = R(∨), so that
a = aR∨aL and b = bR∨bL gives: a∨b = (aR∨bR) ∨ (aL∨bL) = (aL∨bL) ∧ (aR∨bR). £
In the case where S has a lattice section, Theorem 2.2.8 (4)-(6) describes an internal
(monic) Kimura decomposition of S in contrast to the external (epic) Kimura decomposition of
51
Theorem 2.2.8. (Cvetko-Vah) Given a skew lattice the following are equivalent:
1) S has both a left-handed section and a right-handed section.
2) Sub-skew lattices SL and SR exist whose intersection with any D-class is an L-class,
or respectively, an R-class of S.
3) S has a lattice section S0.
When these conditions hold, then:
4) The natural epimorphisms S → S/R and S → S/L induce isomorphisms of SL with
S/R and SR with S/L.
5) Every x ∈ S factors uniquely as a = aʹ∧aʺ with aʹ ∈ SL ∩ Dx, aʺ ∈ SR ∩ Dx.
Under this decomposition
(aʹ∧aʺ)∧(bʹ∧bʺ) = (aʹ∧ bʹ)∧(aʺ∧bʺ) and (aʹ∧aʺ)∨(bʹ∧bʺ) = (aʹ∨ bʹ)∧(aʺ∨bʺ).
6) The functions πL: S → SL and πR: S → SR defined by πL(a) = aʹ and πR(a) = aʺ are
retractions of S upon SL and SR respectively and the commuting composite πLπR is
a retraction of S upon S0; moreover R = ker(πL), L = ker(πR) and D = ker(πLπR).
Proof. (2) clearly implies (1). Given (1), expand Sʹ to the subset SL = ∪{La ⎜a ∈ Sʹ} and expand
Sʺ to the subset SR = ∪{Ra ⎜a ∈ Sʺ}. Since L and R are congruences, SL and SR are as stated in
(2). Given (2) again, since every pair of L and R-classes in a D-class must meet at a single
element in that D-class, (3) follows from (2). Conversely, given (3) one has a minimal case of
(1). Thus (1) – (3) are equivalent. Since SL is an R-class cross-section, the map S → S/R
restricts to an isomorphism of SL upon S/R. Similar remarks apply to SR → S/L and (4) follows.
(5) follows from (4) and the embedding S ≅ S/R ×S/D S/L ⊆ S/R × S/L ≅ SL × SR given by a =
aʹ∧aʺ → (aʹ, aʺ). To see (6), first set aL = a∧a0 and aR = a0∧a. This gives
while a∧b = (a∧a0∧a)∧(b∧b0∧b)
(aL∧bL)∧(aR∧bR) = (a∧a0∧b∧b0)∧(a0∧a∧b0∧b) = (a∧a0∧b)∧(a∧b0∧b).
But both “middles” are equal since
a0∧a∧b∧b0 = a0∧a∧b∧a0∧b0∧a∧b∧b0 = a0∧y∧a0∧b0∧a∧b0 = a0∧b∧a∧b0.
Thus a∧b = (aL∧bL)∧(aR∧bR) follows. Recalling that R(∧) = L(∨) and L(∧) = R(∨), so that
a = aR∨aL and b = bR∨bL gives: a∨b = (aR∨bR) ∨ (aL∨bL) = (aL∨bL) ∧ (aR∨bR). £
In the case where S has a lattice section, Theorem 2.2.8 (4)-(6) describes an internal
(monic) Kimura decomposition of S in contrast to the external (epic) Kimura decomposition of
51
