Page 80 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 80
Jonathan E. Leech │ Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond
Theorem 2.6.3. For each finite chain T, to within isomorphism every normal skew chain
S with maximal lattice image isomorphic to T is obtained as follows. Take a T-indexed family of
rectangular skew lattices {X(t) | t ∈T}. For each t ∈T, set D(t) = ∏s ≤ t X(s). Then for t1 < t2, the
primitive subalgebra with D-classes D(t2) = ∏ s ≤ t2 X(s) > D(t1) = ∏ s ≤ t1 X(s) is determined
by letting the projection from D(t2) onto D(t1) and the coset bijections from D(t1) into D(t2) be
the coordinate-wise projection and injections between these products. £
Thus given say t1 ≤ t2, (x1, … , xt1) ∧ (y1, … , yt2 ) = (x1 ∧ y1, … , xt1 ∧ yt1) and
(x1, … , xt1) ∨ (y1, … , yt2 ) = (x1 ∨ y1, … , xt1 ∨ yt1, y1+t1,..., yt2 ) .
Alternatively, such a skew chain S is isomorphic to a fibered product of “near constant”
skew chains over a common maximal chain image. In the case of A > B > C above, the skew
chain is isomorphic to the fibered product of the following skew chains over 2 > 1 > 0:
C B′ A′′
≅ ≅
C B′ {0}
≅
C {0} {0}.
We saw in Theorem 2.4.10 that symmetric skew lattices are characterized by the fact that
all skew diamonds have some special properties. For symmetric normal skew lattices we have:
Theorem 2.6.4. A normal skew lattice S is symmetric if and only if for every
incomparable pair of D-classes A and B of S, the skew diamond SA∪B generated from A ∪ B is
isomorphic to a normal skew diamond of the form
X×M×Y
↙↘
A=X×M M×Y=B
↘ ↙
M
where the downward projections and upward coset-bijections are all given in coordinate-wise
fashion. In this case the skew diamond decomposes as a direct product, M × X0 × Y0, where X0
and Y0 are skew chains X > {0} and Y > {0}. In particular, if A∧B = {0}, then SA∪B ≅ A0 × B0.
Proof. By Proposition 2.5.2, the proof reduces to the special case where A∧B = {0}. So what is
the situation in the join class? First it consists of all possible outcomes a∨b. By symmetry, in all
cases a∨b = b∨a. Suppose that a∨b = aʹ∨bʹ. Then since a∨b > both a and aʹ in A, a = aʹ.
Likewise b = bʹ. Thus the a∨b-outcomes are all distinct, with each pair commuting. Thus the
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Theorem 2.6.3. For each finite chain T, to within isomorphism every normal skew chain
S with maximal lattice image isomorphic to T is obtained as follows. Take a T-indexed family of
rectangular skew lattices {X(t) | t ∈T}. For each t ∈T, set D(t) = ∏s ≤ t X(s). Then for t1 < t2, the
primitive subalgebra with D-classes D(t2) = ∏ s ≤ t2 X(s) > D(t1) = ∏ s ≤ t1 X(s) is determined
by letting the projection from D(t2) onto D(t1) and the coset bijections from D(t1) into D(t2) be
the coordinate-wise projection and injections between these products. £
Thus given say t1 ≤ t2, (x1, … , xt1) ∧ (y1, … , yt2 ) = (x1 ∧ y1, … , xt1 ∧ yt1) and
(x1, … , xt1) ∨ (y1, … , yt2 ) = (x1 ∨ y1, … , xt1 ∨ yt1, y1+t1,..., yt2 ) .
Alternatively, such a skew chain S is isomorphic to a fibered product of “near constant”
skew chains over a common maximal chain image. In the case of A > B > C above, the skew
chain is isomorphic to the fibered product of the following skew chains over 2 > 1 > 0:
C B′ A′′
≅ ≅
C B′ {0}
≅
C {0} {0}.
We saw in Theorem 2.4.10 that symmetric skew lattices are characterized by the fact that
all skew diamonds have some special properties. For symmetric normal skew lattices we have:
Theorem 2.6.4. A normal skew lattice S is symmetric if and only if for every
incomparable pair of D-classes A and B of S, the skew diamond SA∪B generated from A ∪ B is
isomorphic to a normal skew diamond of the form
X×M×Y
↙↘
A=X×M M×Y=B
↘ ↙
M
where the downward projections and upward coset-bijections are all given in coordinate-wise
fashion. In this case the skew diamond decomposes as a direct product, M × X0 × Y0, where X0
and Y0 are skew chains X > {0} and Y > {0}. In particular, if A∧B = {0}, then SA∪B ≅ A0 × B0.
Proof. By Proposition 2.5.2, the proof reduces to the special case where A∧B = {0}. So what is
the situation in the join class? First it consists of all possible outcomes a∨b. By symmetry, in all
cases a∨b = b∨a. Suppose that a∨b = aʹ∨bʹ. Then since a∨b > both a and aʹ in A, a = aʹ.
Likewise b = bʹ. Thus the a∨b-outcomes are all distinct, with each pair commuting. Thus the
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