y≥x

M¨bius inversion yields

o

(’1)dim(x) r(Ax ) = (’1)dim(y) µ(x, y).

y∈L(A)

y≥x

Putting x = Rn gives

(’1)n r(A) = (’1)dim(y) µ(y) = χA (’1),

y∈L(A)

thereby proving (11).

The relatively bounded case (equation (12)) is similar, but with one technical

complication. We may assume that A is essential, since b(A) = b(ess(A)) and

χA (t) = tdim(A)’dim(ess(A)) χess(A) (t).

In this case, the relatively bounded regions are actually bounded. Let

Fb (A) = {F ∈ F(A) : F is relatively bounded}

“= F.

F ∈Fb (A)

The di¬culty lies in computing ψ(“). Zaslavsky conjectured in 1975 that “ is

star-shaped, i.e., there exists x ∈ “ such that for every y ∈ “, the line segment

LECTURE 2. PROPERTIES OF THE INTERSECTION POSET 21

b

a a

b

c d c d

Figure 2. Two arrangements with the same intersection poset

joining x and y lies in “. This would imply that “ is contractible, and hence (since

“ is compact when A is essential) ψ(“) = 1. A counterexample to Zaslavsky™s

conjecture appears as an exercise in [5, Exer. 4.29], but nevertheless Bj¨rner and

o

Ziegler showed that “ is indeed contractible. (See [5, Thm. 4.5.7(b)] and Lecture 1,

Exercise 7.) The argument just given for r(A) now carries over mutatis mutandis

to b(A). There is also a direct argument that ψ(“) = 1, circumventing the need to

show that “ is contractible. We will omit proving here that ψ(“) = 1.

Corollary 2.1. Let A be a real arrangement. Then r(A) and b(A) depend only on

L(A).

Figure 2 shows two arrangements in R2 with di¬erent “face structure” but

the same L(A). The ¬rst arrangement has for instance one triangular and one

quadrilateral face, while the second has two triangular faces. Both arrangements,

however, have ten regions and two bounded regions.

We now give two basic examples of arrangements and the computation of their

characteristic polynomials.

Proposition 2.4. (general position) Let A be an n-dimensional arrangement of m

hyperplanes in general position. Then

m n’2 m

χA (t) = tn ’ mtn’1 + ’ · · · + (’1)n

t .

2 n

In particular, if A is a real arrangement, then

m m

+···+

r(A) = 1+m+

2 n

m m

b(A) = (’1)n 1 ’ m + ’ · · · + (’1)n

2 n

m’1

= .

n

Proof. Every B ⊆ A with #B ¤ n de¬nes an element xB = H of L(A).

H∈B

Hence L(A) is a truncated boolean algebra:

L(A) ∼ {S ⊆ [m] : #S ¤ n},

=

22 R. STANLEY, HYPERPLANE ARRANGEMENTS

Figure 3. The truncated boolean algebra of rank 2 with four atoms

ordered by inclusion. Figure 3 shows the case n = 2 and m = 4, i.e., four lines in

general position in R2 . If x ∈ L(A) and rk(x) = k, then [ˆ x] ∼ Bk , a boolean

0, =

k

algebra of rank k. By equation (4) there follows µ(x) = (1) . Hence

(’1)#S tn’#S

χA (t) =

S⊆[m]

#S¤n

m

= tn ’ mtn’1 + · · · + (’1)n .2

n

Note. Arrangements whose hyperplanes are in general position were formerly

called free arrangements. Now, however, free arrangements have another meaning

discussed in the note following Example 4.11.

Our second example concerns generic translations of the hyperplanes of a lin-

ear arrangement. Let L1 , . . . , Lm be linear forms, not necessarily distinct, in the

variables v = (v1 , . . . , vn ) over the ¬eld K. Let A be de¬ned by

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

where a1 , . . . , am are generic elements of K. This means if Hi = ker(Li (v) ’ ai ),

then

Hi1 © · · · © Hik = … ” Li1 , . . . , Lik are linearly independent.

For instance, if K = R and L1 , . . . , Lm are de¬ned over Q, then a1 , . . . , am are

generic whenever they are linearly independent over Q.

nongeneric generic

It follows that if x = Hi1 © · · · © Hik ∈ L(A), then [ˆ x] ∼ Bk . Hence

0, =

(’1)#B tn’#B ,

χA (t) =

B

LECTURE 2. PROPERTIES OF THE INTERSECTION POSET 23

1 6 3 12 4 12

Figure 4. The forests on four vertices

where B ranges over all linearly independent subsets of A. (We say that a set of hy-

perplanes are linearly independent if their normals are linearly independent.) Thus

χA (t), or more precisely (’t)n χA (’1/t), is the generating function for linearly

independent subsets of L1 , . . . , Lm according to their number of elements. For in-

stance, if A is given by Figure 2 (either arrangement) then the linearly independent

subsets of hyperplanes are …, a, b, c, d, ac, ad, bc, bd, cd, so χA (t) = t2 ’ 4t + 5.

Consider the more interesting example xi ’ xj = aij , 1 ¤ i < j ¤ n, where the

aij are generic. We could call this arrangement the generic braid arrangement G n .

Identify the hyperplane xi ’ xj = aij with the edge ij on the vertex set [n]. Thus

a subset B ⊆ Gn corresponds to a simple graph GB on [n]. (“Simple” means that

there is at most one edge between any two vertices, and no edge from a vertex to

itself.) It is easy to see that B is linearly independent if and only if the graph GB

has no cycles, i.e., is a forest. Hence we obtain the interesting formula

(’1)e(F ) tn’e(F ) ,

(14) χGn (t) =

F

where F ranges over all forests on [n] and e(F ) denotes the number of edges of

F . For instance, the isomorphism types of forests (with the number of distinct

labelings written below the forest) on four vertices are given by Figure 4. Hence