PlanetMath: geometric lattice

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

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 $\bigvee S$ and $\bigwedge S$ exist in

for any subset $S\subseteq L$ . Since $x\le s\le y$ for each $s\in S$ , $\bigvee S$ and $\bigwedge S$ are in fact in

. So

is a complete lattice.

Now, suppose that $a\in I$ . Since is algebraic, is a join of compact elements in : $a=\bigvee_i a_i$ , where each is compact in . Since $a_i\le y$ , the elements $b_i:=a_i\vee x$ are in for each . So $a=a\vee x=(\bigvee_i a_i)\vee x=\bigvee_i (a_i\vee x)=\bigvee_i b_i$ . We want to show that each is compact in . Since is compact in , $a_i=\bigvee_{k=1}^{m}\alpha_k$ , where $\alpha_k$ are atoms in . Then $b_i=(\bigvee_{k=1}^{m}\alpha_k)\vee x=\bigvee_{k=1}^{m}(\alpha_k\vee x)$ . Let be a subset of such that $\alpha_k\vee x\le \bigvee S$ . Since $\alpha_k\le \bigvee S$ and $\alpha_k$ is an atom in and hence compact, there is a finite subset $F\subseteq S$ such that $\alpha_k\le \bigvee F$ . Because $F\subseteq I$ , $x\le \bigvee F$ , and so $\alpha_k\vee x\le \bigvee F$ , meaning that $\alpha_k\vee x$ 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 $c,d\in I$ with $c\wedge d\prec c$ ( $c\wedge d$ is covered by ). Since is semimodular, $d\prec c\vee d$ . As $c\vee d$ is the least upper bound of $\lbrace c,d\rbrace$ , $c\vee d\le y$ , and thus $c\vee d\in I$ . So is semimodular.

Finally, we show that every compact element of is a finite join of atoms in . Suppose $a\in I$ is compact. Then certainly $a\le \bigvee I$ . Consequently, $a\le \bigvee J$ 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 . $\qedsymbol$

Note that in the above proof,

is in fact a finite join of atoms in

, for if $\alpha_k\le x$ , then $\alpha_k\vee x=x$ . Otherwise, $\alpha_k\vee x$ covers

(since

is semimodular), which means that $\alpha_k\vee x$ 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.