to be inserted.

Note (for readers with some knowledge of topology). (a) Let M be a matroid

on the linearly ordered set u1 < u2 < · · · < um . Note that F ∈ BC(M ) if and only

if F ∪ {um } ∈ BC(M ). De¬ne the reduced broken circuit complex BCr (M ) by

BCr (M ) = {F ∈ BC(M ) : um ∈ F }.

Thus

BC(M ) = BCr (M ) — um ,

the join of BCr (M ) and the vertex um . Equivalently, BC(M ) is a cone over BCr (M )

with apex um . As a consequence, BC(M ) is contractible and therefore has the ho-

motopy type of a point. A more interesting problem is to determine the topological

nature of BCr (M ). It can be shown that BCr (M ) has the homotopy type of a wedge

LECTURE 4. BROKEN CIRCUITS AND MODULAR ELEMENTS 47

of β(M ) spheres of dimension rank(M ) ’ 2, where (’1)rank(M )’1 β(M ) = χM (1)

(the derivative of χM (t) at t = 1). See Exercise 21 for more information on β(M ).

(b) [to be inserted]

As an example of the applicability of our results on matroids and geometric

lattices to arrangements, we have the following purely combinatorial description of

the number of regions of a real central arrangement.

Corollary 4.7. Let A be a central arrangement in Rn , and let M be the matroid

de¬ned by the normals to H ∈ A, i.e., the independent sets of M are the linearly

independent normals. Then with respect to any linear ordering of the points of M ,

r(A) is the total number of subsets of M that don™t contain a broken circuit.

Proof. Immediate from Theorems 2.5 and 4.12.

4.2. Modular elements

We next discuss a situation in which the characteristic polynomial χM (t) factors in

a nice way.

De¬nition 4.12. An element x of a geometric lattice L is modular if for all y ∈ L

we have

rk(x) + rk(y) = rk(x § y) + rk(x ∨ y).

(31)

Example 4.9. Let L be a geometric lattice.

(a) ˆ and ˆ are clearly modular (in any ¬nite lattice).

0 1

(b) We claim that atoms a are modular.

Proof. Suppose that a ¤ y. Then a § y = a and a ∨ y = y, so equation

(31) holds. (We don™t need that a is an atom for this case.) Now suppose

a ¤ y. By semimodularity, rk(a ∨ y) = 1 + rk(y), while rk(a) = 1 and

rk(a § y) = rk(ˆ = 0, so again (31) holds.

0)

(c) Suppose that rk(L) = 3. All elements of rank 0, 1, or 3 are modular by

(a) and (b). Suppose that rk(x) = 2. Then x is modular if and only if for

all elements y = x and rk(y) = 2, we have that rk(x § y) = 1.

(d) Let L = Bn . If x ∈ Bn then rk(x) = #x. Moreover, for any x, y ∈ Bn we

have x § y = x © y and x ∨ y = x ∪ y. Since for any ¬nite sets x and y we

have

#x + #y = #(x © y) + #(x ∪ y),

it follows that every element of Bn is modular. In other words, Bn is a

modular lattice.

(e) Let q be a prime power and Fq the ¬nite ¬eld with q elements. De¬ne

Bn (q) to be the lattice of subspaces, ordered by inclusion, of the vector

space Fn . Note that Bn (q) is also isomorphic to the intersection lattice

q

of the arrangement of all linear hyperplanes in the vector space Fn (q).

Figure 4 shows the Hasse diagrams of B2 (3) and B3 (2).

Note that for x, y ∈ Bn (q) we have x § y = x © y and x ∨ y = x + y

(subspace sum). Clearly Bn (q) is atomic: every vector space is the join

(sum) of its one-dimensional subspaces. Moreover, Bn (q) is graded of rank

n, with rank function given by rk(x) = dim(x). Since for any subspaces

x and y we have

dim(x) + dim(y) = dim(x © y) + dim(x + y),

48 R. STANLEY, HYPERPLANE ARRANGEMENTS

010 011

101

110 001

100 111

B2(3)

B3(2)

Figure 4. The lattices B2 (3) and B3 (2)

it follows that L is a modular geometric lattice. Thus every x ∈ L is

modular.

Note. A projective plane R consists of a set (also denoted R) of

points, and a collection of subsets of R, called lines, such that: (a) every

two points lie on a unique line, (b) every two lines intersect in exactly one

point, and (c) (non-degeneracy) there exist four points, no three of which

are on a line. The incidence lattice L(R) of R is the set of all points

and lines of R, ordered by p < L if p ∈ L, with ˆ and ˆ adjoined. It

0 1

is an immediate consequence of the axioms that when R is ¬nite, L(R)

is a modular geometric lattice of rank 3. It is an open (and probably

intractable) problem to classify all ¬nite projective planes. Now let P and

Q be posets and de¬ne their direct product (or cartesian product) to be

the set

P — Q = {(x, y) : x ∈ P, y ∈ Q},

ordered componentwise, i.e., (x, y) ¤ (x , y ) if x ¤ x and y ¤ y . It is easy

to see that if P and Q are geometric (respectively, atomic, semimodular,

modular) lattices, then so is P — Q (Exercise 7). It is a consequence of the

“fundamental theorem of projective geometry” that every ¬nite modular

geometric lattice is a direct product of boolean algebras Bn , subspace

lattices Bn (q) for n ≥ 3, lattices of rank 2 with at least ¬ve elements

(which may be regarded as B2 (q) for any q ≥ 2) and incidence lattices of

¬nite projective planes.

(f) The following result characterizes the modular elements of Πn , which is

the lattice of partitions of [n] or the intersection lattice of the braid ar-

rangement Bn .

Proposition 4.9. A partition π ∈ Πn is a modular element of Πn if

and only if π has at most one nonsingleton block. Hence the number of

modular elements of Πn is 2n ’ n.

Proof. If all blocks of π are singletons, then π = ˆ which is modular by

0,

(a). Assume that π has the block A with r > 1 elements, and all other

blocks are singletons. Hence the number |π| of blocks of π is given by

LECTURE 4. BROKEN CIRCUITS AND MODULAR ELEMENTS 49

n ’ r + 1. For any σ ∈ Πn , we have rk(σ) = n ’ |σ|. Let k = |σ| and

j = #{B ∈ σ : A © B = …}.

Then |π § σ| = j + (n ’ r) and |π ∨ σ| = k ’ j + 1. Hence rk(π) = r ’ 1,

rk(σ) = n ’ k, rk(π § σ) = r ’ j, and rk(π ∨ σ) = n ’ k + j ’ 1, so π is

modular.

Conversely, let π = {B1 , B2 , . . . , Bk } with #B1 > 1 and #B2 > 1.

Let a ∈ B1 and b ∈ B2 , and set

σ = {(B1 ∪ b) ’ a, (B2 ∪ a) ’ b, B3 , . . . , Bk }.

Then

|π| = |σ| = k

π§σ = {a, b, B1 ’ a, B2 ’ b, . . . , B3 , . . . , Bk } ’ |π § σ| = k + 2

π∨σ = {B1 ∪ B2 , B3 , . . . , Bl } ’ |π ∨ σ| = k ’ 1.

Hence rk(π) + rk(σ) = rk(π § σ) + rk(π ∨ σ), so π is not modular.

In a ¬nite lattice L, a complement of x ∈ L is an element y ∈ L such that

x § y = ˆ and x ∨ y = ˆ For instance, in the boolean algebra Bn every element has

0 1.

a unique complement. (See Exercise 3 for the converse.) The following proposition

collects some useful properties of modular elements. The proof is left as an exercise

(Exercises 4“5).

Proposition 4.10. Let L be a geometric lattice of rank n.

(a) Let x ∈ L. The following four conditions are equivalent.

(i) x is a modular element of L.

(ii) If x § y = ˆ then rk(x) + rk(y) = rk(x ∨ y).

0,

(iii) If x and y are complements, then rk(x) + rk(y) = n.

(iv) All complements of x are incomparable.

(b) (transitivity of modularity) If x is a modular element of L and y is modular

in the interval [ˆ x], then y is a modular element of L.

0,

(c) If x and y are modular elements of L, then x § y is also modular.

The next result, known as the modular element factorization theorem [16], is

our primary reason for de¬ning modular elements ” such an element induces a