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geometric lattice
A lattice is said to be geometric if it is
1. 2. semimodular, and
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

an incidence geometry with the empty set adjoined to form the bottom element; and

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.
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 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.
Mathematics Subject Classification
05B35 no label found06C10 no label found51D25 no label found Forums
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