nLab skew lattice

Contents

Idea

The notion of skew lattices is that of lattices (as an algebraic structure (L,∧,∨)(L, \wedge, \vee)), except that commutativity axioms are dropped.

Definition

A skew lattice is a set LL equipped with two associative operations ∨,∧\vee, \wedge that satisfy the following versions of absorption laws:

x∧(x∨y)=x=(y∨x)∧x,x∨(x∧y)=x=(y∧x)∨x. x \wedge (x \vee y) = x = (y \vee x) \wedge x, \qquad x \vee (x \wedge y) = x = (y \wedge x) \vee x \,.

Observe that the notion of skew lattice is self-dual: a statement in the language of skew lattices holds iff the dual statement, obtained by swapping all instances of ∧\wedge with ∨\vee and vice-versa, also holds.

Properties

Proposition

In a skew-lattice, every element xx satisfies x∧x=xx \wedge x = x. Dually, x∨x=xx \vee x = x.

Proof

We have

x=x∧(x∨(x∧x))=x∧xx = x \wedge (x \vee (x \wedge x)) = x \wedge x

where the first equation is from the first absorption law and the second equation is from the second absorption law.

It follows that a skew lattice is a band under either operation ∧\wedge, ∨\vee. Accordingly (see there), there is a natural preorder structure definable from either operation; call them ⪯ ∧\preceq_\wedge and ⪯ ∨\preceq_\vee. These two preorder structures are opposite to one another, by the following results (as adapted from a relevant MathStackExchange discussion):

Lemma

In a skew lattice, x=x∧yx = x \wedge y iff y=x∨yy = x \vee y. Similarly, y=x∧yy = x \wedge y iff x=x∨yx = x \vee y.

Proof

Immediate from the absorption laws.

Proposition

In a skew lattice, we have x⪯ ∧yx \preceq_\wedge y iff y⪯ ∨xy \preceq_\vee x.

Proof

The two relations mean, by definition, x=x∧y∧xx = x \wedge y \wedge x and y=y∨x∨yy = y \vee x \vee y. We are to show that these are equivalent. By duality, only the forward implication needs to be proven.

By hypothesis x=x∧(y∧x)x = x \wedge (y \wedge x) and the lemma, we have y∧x=x∨(y∧x)y \wedge x = x \vee (y \wedge x). Denote the last equation by EE.

Then

y∨x = (y∨x)∧((y∨x)∨(y∧x)) (absorption) = (y∨x)∧(y∨(x∨(y∧x)) (associativityof∨) = (y∨x)∧(y∨(y∧x)) (byE) = (y∨x)∧y (absorption)\array{ y \vee x & = & (y \vee x) \wedge ((y \vee x) \vee (y \wedge x)) & (absorption) \\ & = & (y \vee x) \wedge (y \vee (x \vee (y \wedge x)) & (associativity\; of\; \vee) \\ & = & (y \vee x) \wedge (y \vee (y \wedge x)) & (by\; E) \\ & = & (y \vee x) \wedge y & (absorption) }

Then, from y∨x=(y∨x)∧yy \vee x = (y \vee x) \wedge y and the lemma, we have (y∨x)∨y=y(y \vee x) \vee y = y, as was to be shown.

References

Last revised on June 12, 2025 at 20:55:59. See the history of this page for a list of all contributions to it.