(’1)#Btn’rank(B) .

(10) χA (t) =

B⊆A

B central

Example 2.6. Let A be the arrangement in R2 shown below.

c d

a

b

18 R. STANLEY, HYPERPLANE ARRANGEMENTS

The following table shows all central subsets B of A and the values of #B and

rank(B).

B #B rank(B)

… 0 0

a 1 1

b 1 1

c 1 1

d 1 1

ac 2 2

ad 2 2

bc 2 2

bd 2 2

cd 2 2

acd 3 2

It follows that χA (t) = t2 ’ 4t + (5 ’ 1) = t2 ’ 4t + 4.

Proof of Theorem 2.4. Let z ∈ L(A). Let

Λz = {x ∈ L(A) : x ¤ z},

the principal order ideal generated by z. Recall the de¬nition

Az = {H ∈ A : H ¤ z (i.e., z ⊆ H)}.

By the Crosscut Theorem (Theorem 2.2), we have

(’1)k Nk (z),

µ(z) =

k

where Nk (z) is the number of k-subsets of Az with join z. In other words,

(’1)#B .

µ(z) =

B⊆Az

T

z= H∈B H

Note that z = H∈B H implies that rank(B) = n ’ dim z. Now multiply both sides

by tdim(z) and sum over z to obtain equation (10).

We have now assembled all the machinery necessary to prove the Deletion-

Restriction Lemma (Lemma 2.2) for χA (t).

Proof of Lemma 2.2. Let H0 ∈ A be the hyperplane de¬ning the triple

(A, A , A ). Split the sum on the right-hand side of (10) into two sums, depending

on whether H0 ∈ B or H0 ∈ B. In the former case we get

(’1)#B tn’rank(B) = χA (t).

H0 ∈B⊆A

B central

In the latter case, set B1 = (B’{H0 })H0 , a central arrangement in H0 ∼ K n’1 and

=

H0

a subarrangement of A = A . Since #B1 = #B’1 and rank(B1 ) = rank(B)’1,

we get

(’1)#B tn’rank(B) (’1)#B1 +1 t(n’1)’rank(B1 )

=

H0 ∈B⊆A B1 ∈A

B central

= ’χA (t),

and the proof follows.

LECTURE 2. PROPERTIES OF THE INTERSECTION POSET 19

2.2. The number of regions

The next result is perhaps the ¬rst major theorem in the subject of hyperplane

arrangements, due to Thomas Zaslavsky in 1975.

Theorem 2.5. Let A be an arrangement in an n-dimensional real vector space.

Then

= (’1)n χA (’1)

(11) r(A)

b(A) = (’1)rank(A) χA (1).

(12)

First proof. Equation (11) holds for A = …, since r(…) = 1 and χ… (t) = tn .

By Lemmas 2.1 and 2.2, both r(A) and (’1)n χA (’1) satisfy the same recurrence,

so the proof follows.

Now consider equation (12). Again it holds for A = … since b(…) = 1. (Recall

that b(A) is the number of relatively bounded regions. When A = …, the entire

ambient space Rn is relatively bounded.) Now

χA (1) = χA (1) ’ χA (1).

Let d(A) = (’1)rank(A) χA (1). If rank(A) = rank(A ) = rank(A ) + 1, then d(A) =

d(A ) + d(A ). If rank(A) = rank(A ) + 1 then b(A) = 0 [why?] and L(A ) ∼ L(A )

=

[why?]. Thus from Lemma 2.2 we have d(A) = 0. Hence in all cases b(A) and d(A)

satisfy the same recurrence, so b(A) = d(A).

Second proof. Our second proof of Theorem 2.5 is based on M¨bius inversion

o

and some instructive topological considerations. For this proof we assume basic

knowledge of the Euler characteristic ψ(∆) of a topological space ∆. (Standard

notation is χ(∆), but this would cause too much confusion with the character-

istic polynomial.) In particular, if ∆ is suitably decomposed into cells with fi

i-dimensional cells, then

ψ(∆) = f0 ’ f1 + f2 ’ · · · .

(13)

We take (13) as the de¬nition of ψ(∆). For “nice” spaces and decompositions, it is

¯

independent of the decomposition. In particular, ψ(Rn ) = (’1)n . Write R for the

closure of a region R ∈ R(A).

¯

De¬nition 2.4. A (closed) face of a real arrangement A is a set … = F = R © x,

where R ∈ R(A) and x ∈ L(A).

¯

If we regard R as a convex polyhedron (possibly unbounded), then a face of

¯

A is just a face of some R in the usual sense of the face of a polyhedron, i.e., the

¯ ¯

intersection of R with a supporting hyperplane. In particular, each R is a face of

A. The dimension of a face F is de¬ned by

dim(F ) = dim(a¬(F )),

where a¬(F ) denotes the a¬ne span of F . A k-face is a k-dimensional face of A.

For instance, the arrangement below has three 0-faces (vertices), nine 1-faces, and

seven 2-faces (equivalently, seven regions). Hence ψ(R2 ) = 3 ’ 9 + 7 = 1.

20 R. STANLEY, HYPERPLANE ARRANGEMENTS

Write F(A) for the set of faces of A, and let relint denote relative interior. Then

Rn = relint(F ),

F ∈F(A)

where denotes disjoint union. If fk (A) denotes the number of k-faces of A, it

follows that

(’1)n = ψ(Rn ) = f0 (A) ’ f1 (A) + f2 (A) ’ · · · .

Every k-face is a region of exactly one Ay for y ∈ L(A). Hence

r(Ay ).

fk (A) =

y∈L(A)

dim(y)=k

Multiply by (’1)k and sum over k to get

(’1)n = ψ(Rn ) = (’1)dim(y) r(Ay ).

y∈L(A)

Replacing Rn by x ∈ L(A) gives

(’1)dim(x) = ψ(x) = (’1)dim(y) r(Ay ).