geometric lattice

A latticeMathworldPlanetmathPlanetmath is said to be geometric if it is

  1. 1.

    algebraic (,

  2. 2.

    semimodular (, and

  3. 3.

    each compact element is a join of atoms.

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

Three examples that come to mind are

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

To generate geometric lattices from existing ones, one has the following


Any lattice interval of a geometric lattice is also geometric.


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 completePlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath and that every element is a join of compact elements. Since L is complete, both S and S exist in L for any subset SL. Since xsy for each sS, S and S are in fact in I. So I is a complete latticeMathworldPlanetmath.

Now, suppose that aI. Since L is algebraic, a is a join of compact elements in L: a=iai, where each ai is compactPlanetmathPlanetmath in L. Since aiy, the elements bi:=aix are in I for each i. So a=ax=(iai)x=i(aix)=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(αkx). Let S be a subset of I such that αkxS. Since αkS and αk is an atom in L and hence compact, there is a finite subset FS such that αkF. Because FI, xF, and so αkxF, meaning that αkx 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,dI with cdc (cd is covered ( by c). Since L is semimodular, dcd. As cd is the least upper bound of {c,d}, cdy, and thus cdI. So I is semimodular.

Finally, we show that every compact element of I is a finite join of atoms in I. Suppose aI is compact. Then certainly aI. Consequently, aJ 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 αkx, then αkx=x. Otherwise, αkx covers x (since L is semimodular), which means that αkx is an atom in I.

Remark. In matroidMathworldPlanetmath 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.

Title geometric lattice
Canonical name GeometricLattice
Date of creation 2013-03-22 15:57:32
Last modified on 2013-03-22 15:57:32
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 12
Author CWoo (3771)
Entry type Definition
Classification msc 05B35
Classification msc 06C10
Classification msc 51D25