Lattice-theoretic decompositions

We can represent a semigroup by a function $f : S^2 \to S$ on ordered pairs. In the case of semilattices, this can naturally be extended to a function on any set: \[ f: \mathcal{P}(S) \to S \] This naturally makes a semilattice operation a mapping between two ordered sets. The most fundamental mappings between ordered sets are ones that are either monotone or antitone. In this case, the two dual concepts of join and meet correspond to monotone and antitone mappings on sets.

• Generalized join: monotone
• Generalized meet: antitone

To see this, consider that joins are upper bounds and so they can only get bigger. Likewise, meets are lower bounds so they can only monotonically decrease. This property holds for any upper-bound producing function or lower-bound producing function regardless of rather or not they are semilattices. Consider that $(\mathbb{N},+)$ produces only upper bounds, and therefore it acts monotonically on sets even though it is not a semilattice. In the case of semilattices, we are interested in inverse images of singletons. \[ f^{-1}(\{x\}) \] The elements of the inverse image are all the semilattice decompositions of the element. These decompositions are naturally partially ordered by inclusion. Indeed, as a set system we will show that they form an upper bounded convex family.

Theorem. semilattice decompositions form an upper bounded convex family of sets

Proof. Suppose the semilattice in question is a join semilattice. Then the principal ideal forms the maximal join representation, as each element $x$ is the join of all of its predecessors. It follows that the inverse image $f^{-1}(\{x\})$ is upper bounded. To see convexity, consider that the join operation is monotone. Therefore, if there is an intermediate decomposition it must have the same join as its predecessor and its parent, because if it did not that would violate monotonicity. Therefore, intermediate sets are in the family of decompositions, which implies that the set system is convex.

Theorem. the decompositions family of an irreducible element forms an interval of sets

Proof. the minimal decomposition of an irreducible element is itself, and that must be contained in every other decomposition because if it is not then the element will not be irreducible. The decompositions are upper bounded convex, so if they are bounded convex then they form an interval.

Corollary. inverse images form a power set partition whose members are upper bounded convex families

Example. consider the weak order [1 3 1].

The following convex upper bounded family consists of all semilattice decompositions of the upper bound element, it is not irreducible so this does not form an interval.

Irreducible decompositions: a special case exists when we seek to decompose semilattice elements into irreducibles. Irreducible decompositions can also be partially ordered, and the maximal decomposition is the one that contains all irreducibles. This can be described by a decomposition mapping from an element to its set of its irreducible components. \[ g: S \to \mathcal{P}(S) \] Decomposition mappings are partial inverses to the set-theoretic generalization of the semilattice operation. In other words, we can go from the decomposition back to the element it was produced from by the semilattice operation.

Proposition. complete irreducible semilattice decomposition form a Moore family

Proof. Suppose that we have that we have a set of irreducibles, then we can always get the complete set of irreducibles by applying the generalized set-theoretic semilattice mapping and then getting the complete set of irreducibles from that. This forms a closure operation, so its lattice of closed sets naturally forms a Moore family.

In general, complete irreducible semilattice decompositions do not form cotopologies. In order for them to do so, the lattice clearly must at least be distributive. The key to demonstrating that complete representations form a cotopology is to prove finite union closure, which is possible for some distributive lattices.

Definition. an element of a semilattice is join-compact if all minimal join decompositions are finite, likewise it is called meet-compact if all minimal meet decompositions are finite.

Compactness can be phrased in terms of partially ordered sets of decompositions of elements. In this sense, compactness simply means that all minimal representations are finite.