geometric lattice
A lattice is said to be geometric if it is
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1.
algebraic (http://planetmath.org/AlgebraicLattice),
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2.
semimodular (http://planetmath.org/SemimodularLattice), and
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3.
each compact element is a join of atoms.
By the definition of compactness, the last condition is equivalent to “each compact element is a finite join of atoms”.
Three examples that come to mind are
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•
the power set of a set;
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•
an incidence geometry with the empty set adjoined to form the bottom element; and
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•
a projective geometry (the lattice of subspaces of a vector space).
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
Theorem.
Any lattice interval of a geometric lattice is also geometric.
Proof.
Let be a geometric lattice and a lattice interval of . We first prove that is algebraic, that is, is both complete and that every element is a join of compact elements. Since is complete, both and exist in for any subset . Since for each , and are in fact in . So is a complete lattice.
Now, suppose that . Since is algebraic, is a join of compact elements in : , where each is compact in . Since , the elements are in for each . So . We want to show that each is compact in . Since is compact in , , where are atoms in . Then . Let be a subset of such that . Since and is an atom in and hence compact, there is a finite subset such that . Because , , and so , meaning that is compact in . This shows that , as a finite join of compact elements in , is compact in as well. In turn, this shows that is a join of compact elements in .
Since is both complete and each of its elements is a join of compact elements, is algebraic.
Next, we show that is semimodular. If with ( is covered (http://planetmath.org/CoveringRelation) by ). Since is semimodular, . As is the least upper bound of , , and thus . So is semimodular.
Finally, we show that every compact element of is a finite join of atoms in . Suppose is compact. Then certainly . Consequently, for some finite subset of . But since is atomistic, each element in is a join of atoms in . Take the join of each of the atoms with , we get either or an atom in . Thus, each element in is a join of atoms in and hence is a join of atoms in . ∎
Note that in the above proof, is in fact a finite join of atoms in , for if , then . Otherwise, covers (since is semimodular), which means that is an atom in .
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.
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 |