It follows that rk(v) + rk(y) = rk(v ∨ y), as claimed.

Now substitute µ(v)v ’ µ(v)trk(z)’rk(v) and µ(y)y ’ µ(y)tn’rk(y)’rk(z) in the

right-hand side of equation (33). Then by Claim 2 we have

vy ’ tn’rk(v)’rk(y) = tn’rk(v∨y) .

Now v ∨ y is just vy in the M¨bius algebra A(L). Hence if we further substi-

o

n’rk(x)

tute µ(x)x ’ µ(x)t in the left-hand side of (33), then the product will be

preserved. We thus obtain

«

·«

¬

¬ ·

¬ ·

µ(v)trk(z)’rk(v) · µ(y)tn’rk(y)’rk(z) ,

µ(x)tn’rk(x) = ¬

¬ ·

v¤z y§z=ˆ

x∈L 0

χL (t) χz (t)

as desired.

Corollary 4.8. Let L be a geometric lattice of rank n and a an atom of L. Then

µ(y)tn’1’rk(y) .

χL (t) = (t ’ 1)

y§a=ˆ

0

Proof. The atom a is modular (Example 4.9(b)), and χa (t) = t ’ 1.

Corollary 4.8 provides a nice context for understanding the operation of coning

de¬ned in Chapter 1, in particular, Exercise 2.1. Recall that if A is an a¬ne

arrangement in K n given by the equations

L1 (x) = a1 , . . . , Lm (x) = am ,

then the cone xA is the arrangement in K n —K (where y denotes the last coordinate)

with equations

L1 (x) = a1 y, . . . , Lm (x) = am y, y = 0.

LECTURE 4. BROKEN CIRCUITS AND MODULAR ELEMENTS 53

Let H0 denote the hyperplane y = 0. It is easy to see by elementary linear algebra

that

L(A) ∼ L(cA) ’ {x ∈ L(A) : x ≥ H0 } = L(A) ’ L(AH0 ).

=

Now H0 is a modular element of L(A) (since it™s an atom), so Corollary 4.8 yields

µ(y)t(n+1)’1’rk(y)

χcA (t) = (t ’ 1)

y≥H0

= (t ’ 1)χA (t).

There is a left inverse to the operation of coning. Let A be a nonempty linear

arrangement in K n+1 . Let H0 ∈ A. Choose coordinates (x0 , x1 , . . . , xn ) in K n+1

so that H0 = ker(x0 ). Let A be de¬ned by the equations

x0 = 0, L1 (x0 , . . . , xn ) = 0, . . . , Lm (x0 , . . . , xn ) = 0.

De¬ne the deconing c’1 A (with respect to H0 ) in K n by the equations

L1 (1, x1 , . . . , xn ) = 0, . . . Lm (1, x1 , . . . , xn ) = 0.

Clearly c(c’1 A) = A and L(c’1 A) ∼ L(A) ’ {x ∈ L(A) : x ≥ H0 }.

=

4.3. Supersolvable lattices

For some geometric lattices L, there are “enough” modular elements to give a

factorization of χL (t) into linear factors.

De¬nition 4.13. A geometric lattice L is supersolvable if there exists a modular

maximal chain, i.e., a maximal chain ˆ = x0 x1 · · · xn = ˆ such that each xi

0 1

is modular. A central arrangement A is supersolvable if its intersection lattice L A

is supersolvable.

Note. Let ˆ = x0 x1 · · · xn = ˆ be a modular maximal chain of the

0 1

geometric lattice L. Clearly then each xi’1 is a modular element of the interval

[ˆ xi ]. The converse follows from Proposition 4.10(b): if ˆ = x0 x1 · · · xn = ˆ

0, 0 1

is a maximal chain for which each xi’1 is modular in [ˆ xi ], then each xi is modular

0,

in L.

Note. The term “supersolvable” comes from group theory. A ¬nite group “

is supersolvable if and only if its subgroup lattice contains a maximal chain all of

whose elements are normal subgroups of “. Normal subgroups are “nice” analogues

of modular elements; see [17, Example 2.5] for further details.

Corollary 4.9. Let L be a supersolvable geometric lattice of rank n, with modular

maximal chain ˆ = x0 x1 · · · xn = ˆ Let T denote the set of atoms of L, and

0 1.

set

ei = #{a ∈ T : a ¤ xi , a ¤ xi’1 }.

(34)

Then χL (t) = (t ’ e1 )(t ’ e2 ) · · · (t ’ en ).

Proof. Since xn’1 is modular, we have

y § xn’1 = ˆ ” y ∈ T and y ¤ xn’1 , or y = ˆ

0 0.

54 R. STANLEY, HYPERPLANE ARRANGEMENTS

By Theorem 4.13 we therefore have

®

ˆ

χL (t) = χxn’1 (t) µ(a)tn’rk(a)’rk(xn’1 ) + µ(ˆ n’rk(0)’rk(xn’1 ) .

0)t

° »

a∈T

a¤xn’1

Since µ(a) = ’1, µ(ˆ = 1, rk(a) = 1, rk(ˆ = 0, and rk(xn’1 ) = n ’ 1, the

0) 0)

expression in brackets is just t ’ en . Now continue this with L replaced by [ˆ xn’1 ]

0,

(or use induction on n).

Note. The positive integers e1 , . . . , en of Corollary 4.9 are called the exponents

of L.

(a) Let L = Bn , the boolean algebra of rank n. By Exam-

Example 4.11.

ple 4.9(d) every element of Bn is modular. Hence Bn is supersolvable.

Clearly each ei = 1, so χBn (t) = (t ’ 1)n .

(b) Let L = Bn (q), the lattice of subspaces of Fq . By Example 4.9(e) every

n

element of Bn (q) is modular, so Bn (q) is supersolvable. If k denotes the

j

number of j-dimensional subspaces of a k-dimensional vector space over

Fq , then

= [1] ’ [i’1]

i

ei 1

q i ’ 1 q i’1 ’ 1

’

=

q’1 q’1

= q i’1 .

Hence

χBn (q) (t) = (t ’ 1)(t ’ q)(t ’ q 2 ) · · · (t ’ q n’1 ).

In particular, setting t = 0 gives

n

µBn (q) (ˆ = (’1)n q ( 2 ) .

1)

Note. The expression k is called a q-binomial coe¬cient. It is a

j

polynomial in q with many interesting properties. For the most basic

properties, see e.g. [18, pp. 27“30].

(c) Let L = Πn , the lattice of partitions of the set [n] (a geometric lattice of

rank n ’ 1). By Proposition 4.9, a maximal chain of Πn is modular if and