Theorem 4.13. Let z be a modular element of the geometric lattice L of rank n.

Write χz (t) = χ[ˆ (t). Then

0,z]

®

χL (t) = χz (t) ° µL (y)tn’rk(y)’rk(z) » .

(32)

y : y§z=ˆ

0

Example 4.10. Before proceeding to the proof of Theorem 4.13, let us consider

an example. The illustration below is the a¬ne diagram of a matroid M of rank

3, together with its lattice of ¬‚ats. The two lines (¬‚ats of rank 2) labelled x and y

are modular by Example 4.9(c).

50 R. STANLEY, HYPERPLANE ARRANGEMENTS

y x y

x

Hence by equation (32) χM (t) is divisible by χx (t). Moreover, any atom a of

the interval [ˆ x] is modular, so χx (t) is divisible by χa (t) = t ’ 1. From this it

0,

is immediate (e.g., because the characteristic polynomial χG (t) of any geometric

lattice G of rank n begins xn ’axn’1 +· · · , where a is the number of atoms of G) that

χx (t) = (t ’ 1)(t ’ 5) and χM (t) = (t ’ 1)(t ’ 3)(t ’ 5). On the other hand, since y is

modular, χM (t) is divisible by χy (t), and we get as before χy (t) = (t ’ 1)(t ’ 3) and

χM (t) = (t ’ 1)(t ’ 3)(t ’ 5). Geometric lattices whose characteristic polynomial

factors into linear factors in a similar way due to a maximal chain of modular

elements are discussed further beginning with De¬nition 4.13.

Our proof of Theorem 4.13 will depend on the following lemma of Greene [11].

We give a somewhat simpler proof than Greene.

Lemma 4.5. Let L be a ¬nite lattice with M¨bius function µ, and let z ∈ L. The

o

following identity is valid in the M¨bius algebra A(L) of L:

o

« «

µ(x)x = µ(v)v µ(y)y .

(33) σˆ :=

0

x∈L v¤z y§z=ˆ

0

LECTURE 4. BROKEN CIRCUITS AND MODULAR ELEMENTS 51

Proof. Let σs for s ∈ L be given by (8). The right-hand side of equation (33) is

then given by

µ(v)µ(y)(v ∨ y) = µ(v)µ(y) σs

v¤z v¤z s≥v∨y

y§z=ˆ y§z=ˆ

0 0

= σs µ(v)µ(y)

s v¤s,v¤z

y¤s,y§z=ˆ0

«

«

¬ ·

¬ ·¬ ·

¬ ·

µ(v)· ¬ µ(y)·

= σs ¬

·

¬

s v¤s§z y¤s

y§z=ˆ0

δˆ

0,s§z

«

¬ ·

¬ ·

¬ ·

¬ ·

σs ¬ µ(y)·

= ¬ ·

¬ ·

y¤s

ˆ ¬ ·

s§z=0

y§z=ˆ

0 (redundant)

δˆ

0,s

= σˆ .

0

Proof of Theorem 4.13. We are assuming that z is a modular element of

the geometric lattice L.

Claim 1. Let v ¤ z and y § z = ˆ (so v § y = ˆ Then z § (v ∨ y) = v (as

0 0).

illustrated below).

zv y

z vvy

y

v

0

52 R. STANLEY, HYPERPLANE ARRANGEMENTS

Proof of Claim 1. Clearly z § (v ∨ y) ≥ v, so it su¬ces to show that rk(z § (v ∨

y)) ¤ rk(v). Since z is modular we have

rk(z § (v ∨ y)) = rk(z) + rk(v ∨ y) ’ rk(z ∨ y)

= rk(z) + rk(v ∨ y) ’ (rk(z) + rk(y) ’ rk(z § y))

0

= rk(v ∨ y) ’ rk(y)

¤ (rk(v) + rk(y) ’ rk(v § y)) ’ rk(y) by semimodularity

0

= rk(v),

proving Claim 1.

Claim 2. With v and y as above, we have rk(v ∨ y) = rk(v) + rk(y).

Proof of Claim 2. By the modularity of z we have

rk(z § (v ∨ y)) + rk(z ∨ (v ∨ y)) = rk(z) + rk(v ∨ y).

By Claim 1 we have rk(z § (v ∨ y)) = rk(v). Moreover, again by the modularity of

z we have