{H1 , . . . , Hp } ⊆ A, p ¤ n ’ dim(H1 © · · · © Hp ) = n ’ p

{H1 , . . . , Hp } ⊆ A, p > n ’ H1 © · · · © Hp = ….

For instance, if n = 2 then a set of lines is in general position if no two are parallel

and no three meet at a point.

Let us consider some interesting examples of arrangements that will anticipate

some later material.

Example 1.2. Let Am consist of m lines in general position in R2 . We can compute

r(Am ) using the sweep hyperplane method. Add a L line to Ak (with AK ∪ {L} in

general position). When we travel along L from one end (at in¬nity) to the other,

every time we intersect a line in Ak we create a new region, and we create one new

region at the end. Before we add any lines we have one region (all of R2 ). Hence

r(Am ) = #intersections + #lines + 1

m

= + m + 1.

2

Example 1.3. The braid arrangement Bn in K n consists of the hyperplanes

xi ’ xj = 0, 1 ¤ i < j ¤ n.

Bn :

Thus Bn has n hyperplanes. To count the number of regions when K = R, note

2

that specifying which side of the hyperplane xi ’ xj = 0 a point (a1 , . . . , an ) lies

on is equivalent to specifying whether ai < aj or ai > aj . Hence the number of

regions is the number of ways that we can specify whether ai < aj or ai > aj for

1 ¤ i < j ¤ n. Such a speci¬cation is given by imposing a linear order on the

ai ™s. In other words, for each permutation w ∈ Sn (the symmetric group of all

permutations of 1, 2, . . . , n), there corresponds a region Rw of Bn given by

Rw = {(a1 , . . . , an ) ∈ Rn : aw(1) > aw(2) > · · · > aw(n) }.

Hence r(Bn ) = n!. Rarely is it so easy to compute the number of regions!

Note that the braid arrangement Bn is not essential; indeed, rank(Bn ) = n ’ 1.

When char(K) = 2 the space W ⊆ K n of equation (1) can be taken to be

W = {(a1 , . . . , an ) ∈ K n : a1 + · · · + an = 0}.

The braid arrangement has a number of “deformations” of considerable interest.

We will just de¬ne some of them now and discuss them further later. All these

arrangements lie in K n , and in all of them we take 1 ¤ i < j ¤ n. The reader who

LECTURE 1. BASIC DEFINITIONS 5

likes a challenge can try to compute their number of regions when K = R. (Some

are much easier than others.)

• generic braid arrangement: xi ’ xj = aij , where the aij ™s are “generic”

(e.g., linearly independent over the prime ¬eld, so K has to be “su¬ciently

large”). The precise de¬nition of “generic” will be given later. (The prime

¬eld of K is its smallest sub¬eld, isomorphic to either Q or Z/pZ for some

prime p.)

• semigeneric braid arrangement: xi ’xj = ai , where the ai ™s are “generic.”

• Shi arrangement: xi ’ xj = 0, 1 (so n(n ’ 1) hyperplanes in all).

• Linial arrangement: xi ’ xj = 1.

• Catalan arrangement: xi ’ xj = ’1, 0, 1.

• semiorder arrangement: xi ’ xj = ’1, 1.

• threshold arrangement: xi + xj = 0 (not really a deformation of the braid

arrangement, but closely related).

An arrangement A is central if H∈A H = …. Equivalently, A is a translate

of a linear arrangement (an arrangement of linear hyperplanes, i.e., hyperplanes

passing through the origin). Many other writers call an arrangement central, rather

than linear, if 0 ∈ H∈A H. If A is central with X = H∈A H, then rank(A) =

codim(X). If A is central, then note also that b(A) = 0 [why?].

There are two useful arrangements closely related to a given arrangement A.

If A is a linear arrangement in K n , then projectivize A by choosing some H ∈ A

n’1

to be the hyperplane at in¬nity in projective space PK . Thus if we regard

n’1

PK = {(x1 , . . . , xn ) : xi ∈ K, not all xi = 0}/ ∼,

where u ∼ v if u = ±v for some 0 = ± ∈ K, then

H = ({(x1 , . . . , xn’1 , 0) : xi ∈ K, not all xi = 0}/ ∼) ∼ PK .

= n’2

The remaining hyperplanes in A then correspond to “¬nite” (i.e., not at in¬nity)

n’1

projective hyperplanes in PK . This gives an arrangement proj(A) of hyperplanes

n’1

in PK . When K = R, the two regions R and ’R of A become identi¬ed in

1 2

proj(A). Hence r(proj(A)) = 2 r(A). When n = 3, we can draw PR as a disk with

antipodal boundary points identi¬ed. The circumference of the disk represents the

hyperplane at in¬nity. This provides a good way to visualize three-dimensional real

linear arrangements. For instance, if A consists of the three coordinate hyperplanes

x1 = 0, x2 = 0, and x3 = 0, then a projective drawing is given by

1

2

3

The line labelled i is the projectivization of the hyperplane xi = 0. The hyperplane

at in¬nity is x3 = 0. There are four regions, so r(A) = 8. To draw the incidences

among all eight regions of A, simply “re¬‚ect” the interior of the disk to the exterior:

6 R. STANLEY, HYPERPLANE ARRANGEMENTS

12 24

14

34

23

13

Figure 1. A projectivization of the braid arrangement B4

2

1

2

3

1

Regarding this diagram as a planar graph, the dual graph is the 3-cube (i.e., the

vertices and edges of a three-dimensional cube) [why?].

For a more complicated example of projectivization, Figure 1 shows proj(B 4 )

(where we regard B4 as a three-dimensional arrangement contained in the hyper-

plane x1 + x2 + x3 + x4 = 0 of R4 ), with the hyperplane xi = xj labelled ij, and

with x1 = x4 as the hyperplane at in¬nity.

LECTURE 1. BASIC DEFINITIONS 7

We now de¬ne an operation which is “inverse” to projectivization. Let A be

an (a¬ne) arrangement in K n , given by the equations

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

Introduce a new coordinate y, and de¬ne a central arrangement cA (the cone over

A) in K n — K = K n+1 by the equations

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

For instance, let A be the arrangement in R1 given by x = ’1, x = 2, and x = 3.

The following ¬gure should explain why cA is called a cone.

’1 2

3

It is easy to see that when K = R, we have r(cA) = 2r(A). In general, cA has

the “same combinatorics as A, times 2.” See Exercise 1.