1.2. The intersection poset

Recall that a poset (short for partially ordered set) is a set P and a relation ¤

satisfying the following axioms (for all x, y, z ∈ P ):

(P1) (re¬‚exivity) x ¤ x

(P2) (antisymmetry) If x ¤ y and y ¤ x, then x = y.

(P3) (transitivity) If x ¤ y and y ¤ z, then x ¤ z.

Obvious notation such as x < y for x ¤ y and x = y, and y ≥ x for x ¤ y will be

used throughout. If x ¤ y in P , then the (closed) interval [x, y] is de¬ned by

[x, y] = {z ∈ P : x ¤ z ¤ y}.

Note that the empty set … is not a closed interval. For basic information on posets

not covered here, see [18].

De¬nition 1.1. Let A be an arrangement in V , and let L(A) be the set of all

nonempty intersections of hyperplanes in A, including V itself as the intersection

over the empty set. De¬ne x ¤ y in L(A) if x ⊇ y (as subsets of V ). In other words,

L(A) is partially ordered by reverse inclusion. We call L(A) the intersection poset

of A.

Note. The primary reason for ordering intersections by reverse inclusion rather

than ordinary inclusion is Proposition 3.8. We don™t want to alter the well-established

de¬nition of a geometric lattice or to refer constantly to “dual geometric lattices.”

The element V ∈ L(A) satis¬es x ≥ V for all x ∈ L(A). In general, if P is a

poset then we denote by ˆ an element (necessarily unique) such that x ≥ ˆ for all

0 0

8 R. STANLEY, HYPERPLANE ARRANGEMENTS

Figure 2. Examples of intersection posets

x ∈ P . We say that y covers x in a poset P , denoted x y, if x < y and no z ∈ P

satis¬es x < z < y. Every ¬nite poset is determined by its cover relations. The

(Hasse) diagram of a ¬nite poset is obtained by drawing the elements of P as dots,

with x drawn lower than y if x < y, and with an edge between x and y if x y.

Figure 2 illustrates four arrangements A in R2 , with (the diagram of) L(A) drawn

below A.

A chain of length k in a poset P is a set x0 < x1 < · · · < xk of elements of

P . The chain is saturated if x0 x1 · · · xk . We say that P is graded of rank

n if every maximal chain of P has length n. In this case P has a rank function

rk : P ’ N de¬ned by:

• rk(x) = 0 if x is a minimal element of P .

• rk(y) = rk(x) + 1 if x y in P .

If x < y in a graded poset P then we write rk(x, y) = rk(y) ’ rk(x), the length

of the interval [x, y]. Note that we use the notation rank(A) for the rank of an

arrangement A but rk for the rank function of a graded poset.

Proposition 1.1. Let A be an arrangement in a vector space V ∼ K n . Then the

=

intersection poset L(A) is graded of rank equal to rank(A). The rank function of

L(A) is given by

rk(x) = codim(x) = n ’ dim(x),

where dim(x) is the dimension of x as an a¬ne subspace of V .

ˆ

Proof. Since L(A) has a unique minimal element 0 = V , it su¬ces to show that

(a) if x y in L(A) then dim(x)’dim(y) = 1, and (b) all maximal elements of L(A)

have dimension n ’ rank(A). By linear algebra, if H is a hyperplane and x an a¬ne

subspace, then H © x = x or dim(x) ’ dim(H © x) = 1, so (a) follows. Now suppose

that x has the largest codimension of any element of L(A), say codim(x) = d. Thus

x is an intersection of d linearly independent hyperplanes (i.e., their normals are

linearly independent) H1 , . . . , Hd in A. Let y ∈ L(A) with e = codim(y) < d. Thus

y is an intersection of e hyperplanes, so some Hi (1 ¤ i ¤ d) is linearly independent

from them. Then y © Hi = … and codim(y © Hi ) > codim(y). Hence y is not a

maximal element of L(A), proving (b).

LECTURE 1. BASIC DEFINITIONS 9

’2

1

2 1 1

’1 ’1

’1 ’1

1

Figure 3. An intersection poset and M¨bius function values

o

1.3. The characteristic polynomial

A poset P is locally ¬nite if every interval [x, y] is ¬nite. Let Int(P ) denote the

set of all closed intervals of P . For a function f : Int(P ) ’ Z, write f (x, y) for

f ([x, y]). We now come to a fundamental invariant of locally ¬nite posets.

De¬nition 1.2. Let P be a locally ¬nite poset. De¬ne a function µ = µP :

Int(P ) ’ Z, called the M¨bius function of P , by the conditions:

o

= 1, for all x ∈ P

µ(x, x)

=’

(2) µ(x, y) µ(x, z), for all x < y in P.

x¤z<y

This second condition can also be written

µ(x, z) = 0, for all x < y in P.

x¤z¤y

If P has a ˆ then we write µ(x) = µ(ˆ x). Figure 3 shows the intersection poset

0, 0,

3

L of the arrangement A in K (for any ¬eld K) de¬ned by QA (x) = xyz(x + y),

together with the value µ(x) for all x ∈ L.

A important application of the M¨bius function is the M¨bius inversion for-

o o

mula. The best way to understand this result (though it does have a simple direct

proof) requires the machinery of incidence algebras. Let I(P ) = I(P, K) denote

the vector space of all functions f : Int(P ) ’ K. Write f (x, y) for f ([x, y]). For

f, g ∈ I(P ), de¬ne the product f g ∈ I(P ) by

f g(x, y) = f (x, z)g(z, y).

x¤z¤y

It is easy to see that this product makes I(P ) an associative Q-algebra, with mul-

tiplicative identity δ given by

1, x = y

δ(x, y) =

0, x < y.

De¬ne the zeta function ζ ∈ I(P ) of P by ζ(x, y) = 1 for all x ¤ y in P . Note that

the M¨bius function µ is an element of I(P ). The de¬nition of µ (De¬nition 1.2) is

o

10 R. STANLEY, HYPERPLANE ARRANGEMENTS

equivalent to the relation µζ = δ in I(P ). In any ¬nite-dimensional algebra over a

¬eld, one-sided inverses are two-sided inverses, so µ = ζ ’1 in I(P ).

Theorem 1.1. Let P be a ¬nite poset with M¨bius function µ, and let f, g : P ’ K.

o

Then the following two conditions are equivalent:

g(y), for all x ∈ P

f (x) =

y≥x

µ(x, y)f (y), for all x ∈ P.

g(x) =

y≥x

Proof. The set K P of all functions P ’ K forms a vector space on which I(P )

acts (on the left) as an algebra of linear transformations by

(ξf )(x) = ξ(x, y)f (y),

y≥x

where f ∈ K P and ξ ∈ I(P ). The M¨bius inversion formula is then nothing but

o

the statement

ζf = g ” f = µg.

We now come to the main concept of this section.

De¬nition 1.3. The characteristic polynomial χA (t) of the arrangement A is de-

¬ned by

µ(x)tdim(x) .