x∈L(A)

For instance, if A is the arrangement of Figure 3, then

χA (t) = t3 ’ 4t2 + 5t ’ 2 = (t ’ 1)2 (t ’ 2).

Note that we have immediately from the de¬nition of χA (t), where A is in K n ,

that

χA (t) = tn ’ (#A)tn’1 + · · · .

Example 1.4. Consider the coordinate hyperplane arrangement A with de¬ning

polynomial QA (x) = x1 x2 · · · xn . Every subset of the hyperplanes in A has a

di¬erent nonempty intersection, so L(A) is isomorphic to the boolean algebra B n of

all subsets of [n] = {1, 2, . . . , n}, ordered by inclusion.

Proposition 1.2. Let A be given by the above example. Then χA (t) = (t ’ 1)n .

Proof. The computation of the M¨bius function of a boolean algebra is a standard

o

result in enumerative combinatorics with many proofs. We will give here a naive

proof from ¬rst principles. Let y ∈ L(A), r(y) = k. We claim that

µ(y) = (’1)k .

(4)

The assertion is clearly true for rk(y) = 0, when y = ˆ Now let y > ˆ We need to

0. 0.

show that

(’1)rk(x) = 0.

(5)

x¤y

LECTURE 1. BASIC DEFINITIONS 11

The number of x such that x ¤ y and rk(x) = i is k , so (5) is equivalent to the

i

well-known identity (easily proved by substituting q = ’1 in the binomial expansion

k

of (q + 1)k ) i=0 (’1)i k = 0 for k > 0.

i

12 R. STANLEY, HYPERPLANE ARRANGEMENTS

Exercises

We will (subjectively) indicate the di¬culty level of each problem as follows:

[1] easy: most students should be able to solve it

[2] moderately di¬cult: many students should be able to solve it

[3] di¬cult: a few students should be able to solve it

[4] horrendous: no students should be able to solve it (without already knowing how)

[5] unsolved.

Further gradations are indicated by + and ’. Thus a [3“] problem is about the

most di¬cult that makes a reasonable homework exercise, and a [5“] problem is an

unsolved problem that has received little attention and may not be too di¬cult.

Note. Unless explicitly stated otherwise, all graphs, posets, lattices, etc., are

assumed to be ¬nite.

(1) [2] Show that every region R of an arrangement A in Rn is an open convex set.

Deduce that R is homeomorphic to the interior of an n-dimensional ball.

(2) [1+] Let A be an arrangement and ess(A) its essentialization. Show that

tdim(ess(A)) χA (t) = tdim(A) χess(A) (t).

(3) [2+] Let A be the arrangement in Rn with equations

x1 = x2 , x2 = x3 , . . . , xn’1 = xn , xn = x1 .

Compute the characteristic polynomial χA (t), and compute the number r(A) of

regions of A.

(4) [2+] Let A be an arrangement in Rn with m hyperplanes. Find the maximum

possible number f (n, m) of regions of A.

(5) [2] Let A be an arrangement in the n-dimensional vector space V whose normals

span a subspace W , and let B be another arrangement in V whose normals span

a subspace Y . Suppose that W © Y = {0}. Show that

χA∪B (t) = t’n χA (t)χB (t).

(6) [2] Let A be an arrangment in a vector space V . Suppose that χA (t) is divisible

by tk but not tk+1 . Show that rank(A) = n ’ k.

(7) Let A be an essential arrangement in Rn . Let “ be the union of the bounded

faces of A.

(a) [3] Show that “ is contractible.

(b) [2] Show that “ need not be homeomorphic to a closed ball.

(c) [2+] Show that “ need not be starshaped. (A subset S of Rn is starshaped

if there is a point x ∈ S such that for all y ∈ S, the line segment from x to

y lies in S.)

(d) [3] Show that “ is pure, i.e., all maximal faces of “ have the same dimension.

(This was an open problem solved by Xun Dong at the PCMI Summer

Session in Geometric Combinatorics, July 11“31, 2004.)

(e) [5] Suppose that A is in general position. Is “ homeomorphic to an n-

dimensional closed ball?

LECTURE 2

Properties of the intersection poset and graphical

arrangements

2.1. Properties of the intersection poset

Let A be an arrangement in the vector space V . A subarrangement of A is a

subset B ⊆ A. Thus B is also an arrangement in V . If x ∈ L(A), de¬ne the

subarrangement Ax ⊆ A by

Ax = {H ∈ A : x ⊆ H}.

(6)

Also de¬ne an arrangement Ax in the a¬ne subspace x ∈ L(A) by

Ax = {x © H = … : H ∈ A ’ Ax }.

Note that if x ∈ L(A), then

L(Ax ) ∼ Λx := {y ∈ L(A) : y ¤ x}

=

L(Ax ) ∼ Vx := {y ∈ L(A) : y ≥ x}

(7) =

K

x

Ax

K

A

AK

Choose H0 ∈ A. Let A = A ’ {H0 } and A = AH0 . We call (A, A , A ) a

triple of arrangements with distinguished hyperplane H0 .

13

14 R. STANLEY, HYPERPLANE ARRANGEMENTS

H0

A

A™

A"

The main goal of this section is to give a formula in terms of χA (t) for r(A)

and b(A) when K = R (Theorem 2.5). We ¬rst establish recurrences for these two

quantities.

Lemma 2.1. Let (A, A , A ) be a triple of real arrangements with distinguished

hyperplane H0 . Then

r(A) = r(A ) + r(A )

b(A ) + b(A ), if rank(A) = rank(A )

b(A) =

0, if rank(A) = rank(A ) + 1.

Note. If rank(A) = rank(A ), then also rank(A) = 1 + rank(A ). The ¬gure

below illustrates the situation when rank(A) = rank(A ) + 1.

H0

Proof. Note that r(A) equals r(A ) plus the number of regions of A cut into two

regions by H0 . Let R be such a region of A . Then R © H0 ∈ R(A ). Conversely,

if R ∈ R(A ) then points near R on either side of H0 belong to the same region

R ∈ R(A ), since any H ∈ R(A ) separating them would intersect R . Thus R is

cut in two by H0 . We have established a bijection between regions of A cut into

two by H0 and regions of A , establishing the ¬rst recurrence.

The second recurrence is proved analogously; the details are omitted.