Page 106 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 106
Jonathan E. Leech │ Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond
Multiply both sides of (1) on the right by bcb to get babcbabcb = bacbabcb. Simplifying on the
left gives
(3) babcb = bacbabcb.
Using (2), the right side of (3) simplifies as follows: bacbabcb =(2) bacbacb = bacb. Thus (3)
becomes babcb = bacb, which is regularity.
So assume that (S, •) is regular. Then a•Lb•La = (aba)•La = a(aba)a = aba = a•Lb,
making (S, •L) left-regular. If e ≤ f in (S, •), then efe = f = fef, that is e•Lf = f = f•Le, so that e ≤ f
in (S, •L). Conversely e•Lf = f = f•Le reduces to efe = f = fef so that ef = efefef = f = fefefe = fe
in (S, •). In similar fashion, e ≺ f in (S, •) iff e ≺ f in (S, •L), since f•Le•Lf = fefef which
immediately reduces to fef. Thus fef = f iff f•Le•Lf = f. £
Thus, if (S, •) is regular, then it is isotopic as a band to both (S, •L) and (S, •R). Our next
theorem rests on the lemma and a special case of Corollary 3.4.4. We give an alternative proof.
Lemma 3.4.13. If (S, ∧, ∨) is a refined quasilattice that is left-handed in that
x∧y∧x = x∧y and x∨y∨x = y∨x, then it is a skew lattice.
Proof. Being a quasilattice, x∧(x∨y) = x∧(x∨y)∧x = x. Being a paralattice,
(y∨x)∧x = (x∨y∨x)∧x = x.
The dual identities are similarly seen. £
This leads to Kinyon’s observation (via email) about refined quasi-lattices.
Theorem 3.4.14. Given a quasilattice (S, ∨, ∧), it is a refined quasilattice if and only if
(S, ∨R, ∧L) is a left-handed skew lattice, in which case (S, ∨, ∧) and (S, ∨R, ∨L) are isotopic under
the identity map on S. In addition, they share all instances of commutation in which case the
outcomes agree for the corresponding pairs of operations.
Proof. As seen in the proof above, a∧Lb∧La expanded reduces to a∧b∧a. Likewise, b∨Ra∨Rb
reduces to b∨a∨b. Thus whenever (S, ∨R, ∧L) is a quasilattice, both algebras share a common
natural quasi-order. Also in general, a∧b = b∧a iff a∧Lb = b∧La with both outcomes being equal,
and a∨b = b∨a iff a∨Rb = b∨Ra, with both outcomes being equal. Thus (S, ∨R, ∧L) has a
coherent natural partial order iff (S, ∨, ∧) does, in which case both algebras share the same natural
partial order. It follows that if (S, ∨, ∧) is a refined quasilattice if and only if (S, ∨R, ∧L) is a left-
handed skew lattice, thanks to the previous lemmas. £
Every refined quasilattice is thus just a “scrambled skew lattice”. Indeed given a left-
handed skew lattice (S, ∨, ∧) with S/D denoted by T, various refined quasi-lattices (S, ∨*, ∧*)
may be recovered from it by (1) doing a fibered product factorization (S1, ∨1) ×T (S2, ∨2) of the
right regular band (S, ∨), replacing ∨2 by its left-handed dual operation and finally shifting the
resulting operation on S1 ×T S2 back to S to get ∨*; and (2) likewise factoring the left-regular band
(S, ∧) as say (S3, ∧3) ×(T, ∧) (S4, ∧4) and replacing ∧4 by its right-handed dual and then shifting the
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Multiply both sides of (1) on the right by bcb to get babcbabcb = bacbabcb. Simplifying on the
left gives
(3) babcb = bacbabcb.
Using (2), the right side of (3) simplifies as follows: bacbabcb =(2) bacbacb = bacb. Thus (3)
becomes babcb = bacb, which is regularity.
So assume that (S, •) is regular. Then a•Lb•La = (aba)•La = a(aba)a = aba = a•Lb,
making (S, •L) left-regular. If e ≤ f in (S, •), then efe = f = fef, that is e•Lf = f = f•Le, so that e ≤ f
in (S, •L). Conversely e•Lf = f = f•Le reduces to efe = f = fef so that ef = efefef = f = fefefe = fe
in (S, •). In similar fashion, e ≺ f in (S, •) iff e ≺ f in (S, •L), since f•Le•Lf = fefef which
immediately reduces to fef. Thus fef = f iff f•Le•Lf = f. £
Thus, if (S, •) is regular, then it is isotopic as a band to both (S, •L) and (S, •R). Our next
theorem rests on the lemma and a special case of Corollary 3.4.4. We give an alternative proof.
Lemma 3.4.13. If (S, ∧, ∨) is a refined quasilattice that is left-handed in that
x∧y∧x = x∧y and x∨y∨x = y∨x, then it is a skew lattice.
Proof. Being a quasilattice, x∧(x∨y) = x∧(x∨y)∧x = x. Being a paralattice,
(y∨x)∧x = (x∨y∨x)∧x = x.
The dual identities are similarly seen. £
This leads to Kinyon’s observation (via email) about refined quasi-lattices.
Theorem 3.4.14. Given a quasilattice (S, ∨, ∧), it is a refined quasilattice if and only if
(S, ∨R, ∧L) is a left-handed skew lattice, in which case (S, ∨, ∧) and (S, ∨R, ∨L) are isotopic under
the identity map on S. In addition, they share all instances of commutation in which case the
outcomes agree for the corresponding pairs of operations.
Proof. As seen in the proof above, a∧Lb∧La expanded reduces to a∧b∧a. Likewise, b∨Ra∨Rb
reduces to b∨a∨b. Thus whenever (S, ∨R, ∧L) is a quasilattice, both algebras share a common
natural quasi-order. Also in general, a∧b = b∧a iff a∧Lb = b∧La with both outcomes being equal,
and a∨b = b∨a iff a∨Rb = b∨Ra, with both outcomes being equal. Thus (S, ∨R, ∧L) has a
coherent natural partial order iff (S, ∨, ∧) does, in which case both algebras share the same natural
partial order. It follows that if (S, ∨, ∧) is a refined quasilattice if and only if (S, ∨R, ∧L) is a left-
handed skew lattice, thanks to the previous lemmas. £
Every refined quasilattice is thus just a “scrambled skew lattice”. Indeed given a left-
handed skew lattice (S, ∨, ∧) with S/D denoted by T, various refined quasi-lattices (S, ∨*, ∧*)
may be recovered from it by (1) doing a fibered product factorization (S1, ∨1) ×T (S2, ∨2) of the
right regular band (S, ∨), replacing ∨2 by its left-handed dual operation and finally shifting the
resulting operation on S1 ×T S2 back to S to get ∨*; and (2) likewise factoring the left-regular band
(S, ∧) as say (S3, ∧3) ×(T, ∧) (S4, ∧4) and replacing ∧4 by its right-handed dual and then shifting the
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