Page 146 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
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athan E. Leech │ Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond
Theorem 4.4.13. If a normal skew lattice S has finite intersections, then both S/L and
S/R have finite intersections.
Proof. A skew lattice S has finite intersections if and only if each principal ideal S∧x∧S has
finite intersections. For a normal skew lattice S, each ideal S∧x∧S has a lattice section, namely
x∧S∧x, with x∧S = (x∧S∧x)[R] ≅ S/L and S∧x = (x∧S∧x)[L] ≅ S/R. Thus S/L and S/R must
have finite intersections if S does. £
Conversely, one may ask: if both S/L and S/R have finite intersections, must S also? Or
do counterexamples exist? Here is a case of the latter.
Example 4.4.8. Let S be the skew lattice defined on n∈{an, bn} , with the
{an, bn} being a totally ordered family of D-classes. For x, y in a common class {an, bn} ,
x∨y = y∧x = ⎧ x, if n is even.
⎨ y, if n is odd.
⎩
Thus even classes are right-handed and odd classes are left-handed. If say x ∈ {am, bm} but
y ∈ {an, bn} where m < n, then x∨y = y = y∨x and x∧y = x = y∧x. Thus,
both an+1, bn+1 > both an, bn > both an−1, bn−1 for all n ∈ &.
Consequently neither an ∪ bn nor an ∩ bn exist, for all n ∈ &. On the other hand, both S/L and
S/R, have a D-class structure
. . . > { an+2, bn+2 } > { cn+1} > { an, bn } > { cn−1} > { an−2, bn−2 } > …
that is right-handed when n is even and left-handed when n is odd. Both ak ∩ bk = ck−1 and
ak ∪ bk = ck+1 . Cases of pairs of elements from distinct D-classes are trivial, as they involve
pairs x, y where x > y.
Observe that this example is trivially symmetric (since totally quasi-ordered), has many
lattice sections, is distributive and thus is categorical. But for normal skew lattices, and in
particular for skew Boolean algebras, we have the equivalence:
Theorem 4.4.14. A normal skew lattice S has finite intersections if and only if both S/L
and S/R have finite intersections. In particular, a skew Boolean algebra S has finite inter-
sections iff both S/L and S/R do.
Proof. We have already seen (⇒). As for (⇐), without loss in generality we represent S as the
fibred product S/L ×S/D S/R. So let both (xʹ, xʺ) and (yʹ, yʺ) in S be given where xʹ, yʹ ∈S/L and
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Theorem 4.4.13. If a normal skew lattice S has finite intersections, then both S/L and
S/R have finite intersections.
Proof. A skew lattice S has finite intersections if and only if each principal ideal S∧x∧S has
finite intersections. For a normal skew lattice S, each ideal S∧x∧S has a lattice section, namely
x∧S∧x, with x∧S = (x∧S∧x)[R] ≅ S/L and S∧x = (x∧S∧x)[L] ≅ S/R. Thus S/L and S/R must
have finite intersections if S does. £
Conversely, one may ask: if both S/L and S/R have finite intersections, must S also? Or
do counterexamples exist? Here is a case of the latter.
Example 4.4.8. Let S be the skew lattice defined on n∈{an, bn} , with the
{an, bn} being a totally ordered family of D-classes. For x, y in a common class {an, bn} ,
x∨y = y∧x = ⎧ x, if n is even.
⎨ y, if n is odd.
⎩
Thus even classes are right-handed and odd classes are left-handed. If say x ∈ {am, bm} but
y ∈ {an, bn} where m < n, then x∨y = y = y∨x and x∧y = x = y∧x. Thus,
both an+1, bn+1 > both an, bn > both an−1, bn−1 for all n ∈ &.
Consequently neither an ∪ bn nor an ∩ bn exist, for all n ∈ &. On the other hand, both S/L and
S/R, have a D-class structure
. . . > { an+2, bn+2 } > { cn+1} > { an, bn } > { cn−1} > { an−2, bn−2 } > …
that is right-handed when n is even and left-handed when n is odd. Both ak ∩ bk = ck−1 and
ak ∪ bk = ck+1 . Cases of pairs of elements from distinct D-classes are trivial, as they involve
pairs x, y where x > y.
Observe that this example is trivially symmetric (since totally quasi-ordered), has many
lattice sections, is distributive and thus is categorical. But for normal skew lattices, and in
particular for skew Boolean algebras, we have the equivalence:
Theorem 4.4.14. A normal skew lattice S has finite intersections if and only if both S/L
and S/R have finite intersections. In particular, a skew Boolean algebra S has finite inter-
sections iff both S/L and S/R do.
Proof. We have already seen (⇒). As for (⇐), without loss in generality we represent S as the
fibred product S/L ×S/D S/R. So let both (xʹ, xʺ) and (yʹ, yʺ) in S be given where xʹ, yʹ ∈S/L and
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