Page 50 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 50
athan E. Leech │ Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond
Proof. First, if e either ∨-commutes with all x ∈ S or else ∧-commutes with all x ∈ S, then in
particular it does such with all elements in De, which forces De to be trivial. Conversely, if De is
known to be {e}, then by the final assertion of Theorem 2.2.1 e must commute under both
operations with all elements of S. £
While an element that join commutes with all elements in a skew lattice also meet
commutes with all elements (and conversely), in general two elements commuting under one
operation need not commute under the other operation. This is evident in the following example.
Example 2.2.1a. Consider the right-handed skew lattice defined by the Cayley tables
∨ 0 am′ bn′ jp′ ∧ 0 am′ bn′ jp′ j1 − j2
0 0 am′ bn′ jp′ 00 0 0 0
am am am jm jm and am 0 am′ 0 ap′ . a1 − a2 b1 − b2
bn bn jn bn jn bn 0 0 bn′ bp′
jp jp jp jp jp jp 0 am′ bn′ jp′ 0
This skew lattice is jointly determined by being right handed with the displayed Clifford-Mclean
picture and having ≥ given by jn ≥ both an, bn ≥ 0 for n = 1 or 2. While a1 and b2 ∧-commute,
they clearly do not ∨-commute. We denote this example by NSR7 ,0 . (See Section 5.2) £
A skew lattice S is symmetric if for all e, f ∈ S, e∨f = f∨e iff e∧f = f∧e. A symmetric
alternative to the above example is as follows:
Example 2.2.1b. The direct product of the right-handed algebras {a1, a2 } and {b1, b2 } is
{0} {0}
symmetric. Its D-class diagram (with “redundant” 0-coordinates suppressed) is as follows. The
partial ordering between the top class and the intermediate classes given by coordinate projection.
{(a1, b1), (a1, b2 ), (a2, b1), (a2, b2 )}
{a1, a2} {b1, b2} .£
0
Two elements in a skew lattice commute if they commute under both operations. If they
just commute under a single operation, they are said to ∧-commute or ∨-commute, as the case
may be. The following statement is easily verified.
48
Proof. First, if e either ∨-commutes with all x ∈ S or else ∧-commutes with all x ∈ S, then in
particular it does such with all elements in De, which forces De to be trivial. Conversely, if De is
known to be {e}, then by the final assertion of Theorem 2.2.1 e must commute under both
operations with all elements of S. £
While an element that join commutes with all elements in a skew lattice also meet
commutes with all elements (and conversely), in general two elements commuting under one
operation need not commute under the other operation. This is evident in the following example.
Example 2.2.1a. Consider the right-handed skew lattice defined by the Cayley tables
∨ 0 am′ bn′ jp′ ∧ 0 am′ bn′ jp′ j1 − j2
0 0 am′ bn′ jp′ 00 0 0 0
am am am jm jm and am 0 am′ 0 ap′ . a1 − a2 b1 − b2
bn bn jn bn jn bn 0 0 bn′ bp′
jp jp jp jp jp jp 0 am′ bn′ jp′ 0
This skew lattice is jointly determined by being right handed with the displayed Clifford-Mclean
picture and having ≥ given by jn ≥ both an, bn ≥ 0 for n = 1 or 2. While a1 and b2 ∧-commute,
they clearly do not ∨-commute. We denote this example by NSR7 ,0 . (See Section 5.2) £
A skew lattice S is symmetric if for all e, f ∈ S, e∨f = f∨e iff e∧f = f∧e. A symmetric
alternative to the above example is as follows:
Example 2.2.1b. The direct product of the right-handed algebras {a1, a2 } and {b1, b2 } is
{0} {0}
symmetric. Its D-class diagram (with “redundant” 0-coordinates suppressed) is as follows. The
partial ordering between the top class and the intermediate classes given by coordinate projection.
{(a1, b1), (a1, b2 ), (a2, b1), (a2, b2 )}
{a1, a2} {b1, b2} .£
0
Two elements in a skew lattice commute if they commute under both operations. If they
just commute under a single operation, they are said to ∧-commute or ∨-commute, as the case
may be. The following statement is easily verified.
48
