Let $W$ be a subspace of the vector space $\mathbb{K}^n$, where $\mathbb{K}$ is a field of characteristic $0$. The support of a vector $v = (v_1,\ldots, v_n) \in \mathbb{K}^n$ is given by $\text{supp}(v) = \left\{i : v_i\neq 0\right\}$. Let $L$ denote the set of supports of all vectors in $W$, ordered by reverse inclusion.

I want to show that $L$ is a semimodular lattice. By Proposition 3.3.2 in Stanley's Enumerative Combinatorics, vol. 1, we can either show that $L$ is graded (every maximal chain has the same length) and that its rank functions $\rho:L\rightarrow \left\{0,1,\ldots, n \right\}$ satisfies $$ \rho(s)+\rho(t) \geq \rho(s\wedge t)+\rho(s\vee t)$$

for any $s,t \in L$ or we can show that if $s$ and $t$ both cover $s\wedge t$, then $s\vee t$ covers both $s$ and $t$.

Here's the problem...I know that the meet of any $s,t \in L$ is $s\cup t$, but I don't know what the join is. Ultimately what I'm trying to show is the $L$ is a geometric lattice, which requires showing that $L$ is semimodular. Perhaps there is a better way of proving this.

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    $\begingroup$ The lattice $L$ is (isomorphic to) the intersection lattice of a hyperplane arrangement (or maybe its dual). Namely, let $\left(e_1,e_2,\ldots,e_k\right)$ be a basis of the vector space $W$. For each $i \in \left\{1,2,\ldots,n\right\}$, let $F_i$ be the set of all $\left(x_1,x_2,\ldots,x_k\right) \in \mathbb{K}^k$ such that the $i$-th coordinate of $x_1 e_1 + x_2 e_2 + \cdots + x_k e_k$ is $0$. This $F_i$ is either the whole $\mathbb{K}^k$ or a hyperplane. The intersections of these hyperplanes correspond precisely to the supports of vectors in $W$. $\endgroup$ – darij grinberg Apr 10 '15 at 2:45
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    $\begingroup$ The hyperplane arrangement formed by the $F_i$ that aren't the whole $\mathbb{K}^k$ is, of course, central. The fact that intersection lattices of central hyperplane arrangements are geometric lattices is, I think, well-known. $\endgroup$ – darij grinberg Apr 10 '15 at 2:47
  • $\begingroup$ @darijgrinberg Why is the hyperplane arrangement formed by the $F_i$ that aren't all of $\mathbb{K}^k$ central? Aren't the $F_i$s pairwise disjoint to begin with? $\endgroup$ – T. Wilkins Apr 10 '15 at 18:02
  • $\begingroup$ All $F_i$ contain $0$. Maybe you misunderstood my definition of them? $\endgroup$ – darij grinberg Apr 10 '15 at 18:04
  • $\begingroup$ @darijgrinberg My linear algebra stinks, sorry. I understood that $F_i$ is the set of all points in $\mathbb{K}^k$ such that in the sum $x_1e_1+\ldots +x_ke_k$ we have $x_i=0$. Is that incorrect? $\endgroup$ – T. Wilkins Apr 10 '15 at 18:09

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