geometric lattice (original) (raw)

By the definition of compactness, the last condition is equivalent to “each compact element is a finite join of atoms”.

From the last two examples, one sees how the name “geometric” lattice is derived.

Proof.

Let L be a geometric lattice and I=[x,y] a lattice interval of L. We first prove that I is algebraic, that is, I is both complete and that every element is a join of compact elements. Since L is complete, both ⋁S and ⋀S exist in L for any subset S⊆L. Since x≤s≤y for each s∈S, ⋁S and ⋀S are in fact in I. So I is a complete lattice.

Now, suppose that a∈I. Since L is algebraic, a is a join of compact elements in L: a=⋁iai, where each ai is compact in L. Since ai≤y, the elements bi:=ai∨x are in I for each i. So a=a∨x=(⋁iai)∨x=⋁i(ai∨x)=⋁ibi. We want to show that each bi is compact in I. Since ai is compact in L, ai=⋁k=1mαk, where αk are atoms in L. Then bi=(⋁k=1mαk)∨x=⋁k=1m(αk∨x). Let S be a subset of I such that αk∨x≤⋁S. Since αk≤⋁S and αk is an atom in L and hence compact, there is a finite subset F⊆S such that αk≤⋁F. Because F⊆I, x≤⋁F, and so αk∨x≤⋁F, meaning that αk∨x is compact in I. This shows that bi, as a finite join of compact elements in I, is compact in I as well. In turn, this shows that a is a join of compact elements in I.

Since I is both complete and each of its elements is a join of compact elements, I is algebraic.

Next, we show that I is semimodular. If c,d∈I with c∧d≺c (c∧d is covered (http://planetmath.org/CoveringRelation) by c). Since L is semimodular, d≺c∨d. As c∨d is the least upper bound of {c,d}, c∨d≤y, and thus c∨d∈I. So I is semimodular.

Finally, we show that every compact element of I is a finite join of atoms in I. Suppose a∈I is compact. Then certainly a≤⋁I. Consequently, a≤⋁J for some finite subset J of I. But since L is atomistic, each element in J is a join of atoms in L. Take the join of each of the atoms with x, we get either x or an atom in I. Thus, each element in J is a join of atoms in I and hence a is a join of atoms in I. ∎

Note that in the above proof, bi is in fact a finite join of atoms in I, for if αk≤x, then αk∨x=x. Otherwise, αk∨x covers x (since L is semimodular), which means that αk∨x is an atom in I.

Remark. In matroid theory, where geometric lattices play an important role, lattices considered are generally assumed to be finite. Therefore, any lattice in this context is automatically complete and every element is compact. As a result, any finite lattice is geometric if it is semimodular and atomistic.