# geometric lattice

1. 1.

algebraic (http://planetmath.org/AlgebraicLattice),

2. 2.

semimodular (http://planetmath.org/SemimodularLattice), and

3. 3.

each compact element is a join of atoms.

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 $L$ 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 $\bigvee S$ and $\bigwedge S$ exist in $L$ for any subset $S\subseteq L$. Since $x\leq s\leq y$ for each $s\in S$, $\bigvee S$ and $\bigwedge S$ are in fact in $I$. So $I$ is a complete lattice  .

Now, suppose that $a\in I$. Since $L$ is algebraic, $a$ is a join of compact elements in $L$: $a=\bigvee_{i}a_{i}$, where each $a_{i}$ is compact  in $L$. Since $a_{i}\leq y$, the elements $b_{i}:=a_{i}\vee x$ are in $I$ for each $i$. 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 $b_{i}$ is compact in $I$. Since $a_{i}$ is compact in $L$, $a_{i}=\bigvee_{k=1}^{m}\alpha_{k}$, where $\alpha_{k}$ are atoms in $L$. Then $b_{i}=(\bigvee_{k=1}^{m}\alpha_{k})\vee x=\bigvee_{k=1}^{m}(\alpha_{k}\vee x)$. Let $S$ be a subset of $I$ such that $\alpha_{k}\vee x\leq\bigvee S$. Since $\alpha_{k}\leq\bigvee S$ and $\alpha_{k}$ is an atom in $L$ and hence compact, there is a finite subset $F\subseteq S$ such that $\alpha_{k}\leq\bigvee F$. Because $F\subseteq I$, $x\leq\bigvee F$, and so $\alpha_{k}\vee x\leq\bigvee F$, meaning that $\alpha_{k}\vee x$ is compact in $I$. This shows that $b_{i}$, 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 $c,d\in I$ with $c\wedge d\prec c$ ($c\wedge d$ is covered (http://planetmath.org/CoveringRelation) by $c$). Since $L$ is semimodular, $d\prec c\vee d$. As $c\vee d$ is the least upper bound of $\{c,d\}$, $c\vee d\leq y$, and thus $c\vee d\in I$. So $I$ is semimodular.

Finally, we show that every compact element of $I$ is a finite join of atoms in $I$. Suppose $a\in I$ is compact. Then certainly $a\leq\bigvee I$. Consequently, $a\leq\bigvee J$ 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, $b_{i}$ is in fact a finite join of atoms in $I$, for if $\alpha_{k}\leq x$, then $\alpha_{k}\vee x=x$. Otherwise, $\alpha_{k}\vee x$ covers $x$ (since $L$ is semimodular), which means that $\alpha_{k}\vee x$ 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.

Title geometric lattice GeometricLattice 2013-03-22 15:57:32 2013-03-22 15:57:32 CWoo (3771) CWoo (3771) 12 CWoo (3771) Definition msc 05B35 msc 06C10 msc 51D25