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Modular Elements
Let L be a lattice. Recall that L is modular if it is ranked, and its rank function r satisfies
(1) r(x) + r(y) = r(x ∨ y) + r(x ∧ y)
for every x, y ∈ L. (This is not how we first defined modular lattices, but we proved that it is an equivalent condition; see notes from 1/30 and 2/1.)
Definition 1. An element x ∈ L is a modular element if (1) holds for every y ∈ L. Thus L is modular if and only if every element of L is modular.
- The elements ˆ0 and ˆ1 are clearly modular in any lattice.
- If L is geometric, then every atom x is modular. Indeed, for y ∈ L, if y ≥ x, then y = x ∨ y and x = x ∧ y, while if y 6 ≥ x then y ∧ x = ˆ0 and y ∨ x m y.
- The coatoms of a geometric lattice, however, need not be modular. Let L = Πn; recall that Πn has rank function r(π) = n − |π|. Let x = 12|34, y = 13| 24 ∈ Π 4. Then r(x) = r(y) = 2, but r(x ∨ y) = r(ˆ1) = 3 and r(x ∧ y) = r(ˆ0) = 0. So x is not a modular element.
Proposition 1. The modular elements of Πn are exactly the partitions with at most one nonsingleton block.
Proof. Suppose that π ∈ Πn has one nonsingleton block B. For σ ∈ Πn, let
X = {C ∈ σ | C ∩ B 6 = ∅}, Y = {C ∈ σ | C ∩ B = ∅}.
Then
π ∧ σ =
C ∩ B | C ∈ X
{i} | i 6 ∈ B
π ∨ σ =
C∈X
C
∪ Y
so
|π ∧ σ| + |π ∨ σ| = (|X| + n − |B|) + (1 + |Y |) = (n − |B| + 1) + (|X| + |Y |) = |π| + |σ|,
proving that π is a modular element.
For the converse, let B, C be nonsingleton blocks of π, then let σ have the two nonsingleton blocks {i, k}, {j, }, where i, j ∈ B and k, ∈ C. Then r(σ) = 2 and r(π ∧ σ) = r(ˆ0) = 0, but
r(π ∨ σ) = r(π) + 1 < r(π) + r(σ) − r(π ∧ σ)
so π is not a modular element.
The usefulness of a modular element is that if one exists, we can factor the characteristic polynomial of L.
Theorem 2. Let L be a geometric lattice of rank n, and let z ∈ L be a modular element. Then
(2) χL(k) = χˆ 0 ,z ·
y: y∧z=ˆ 0
μL(ˆ 0 , y)kn−r(z)−r(y)
I’ll skip the proof, which uses calculation in the M¨obius algebra; see Stanley, HA, pp. 50–52.
Corollary 3. Let L be a geometric lattice, and let a ∈ L be an atom. Then
χL(k) = (k − 1)
x: x 6 ≥a
μL(ˆ 0 , x)kr(L)−^1 −r(x).
(We already knew that k − 1 had to be a factor of χL(k), because χL(1) =
x∈L μL(ˆ^0 , x) = 0. Still, it’s nice to see it another way.)
Corollary 4. Let L be a geometric lattice, and let z ∈ L be a coatom that is a modular element. Then
χL(k) = (k − e)χˆ 0 ,z,
where e is the number of atoms a ∈ L such that a 6 ≤ z.
Example 1. Corollary 4 provides another way of calculating the characteristic polynomial of Πn. Let z be the coatom with blocks [n − 1] and {n}, which is a modular element by Proposition 1. There are n − 1 atoms a 6 ≤ z, namely the partitions whose nonsingleton block is {i, n} for some i ∈ [n − 1], so we obtain
χΠn (k) = (k − n + 1)χΠn− 1 (k)
and by induction χΠn (k) = (k − 1)(k − 2) · · · (k − n + 1).
Supersolvable Lattices
Let L be a geometric lattice with atoms A. Recall from (2) that if z is a modular element of L, then the characteristic polynomial of L factors:
χL(k) = χˆ 0 ,z ·
y: y∧z=ˆ 0
μL(ˆ 0 , y)kn−r(z)−r(y)
Of course, we can always apply this for an atom z (Corollary 3). But, as we’ve seen with Πn, something even better happens if z is a coatom: we can express χL(k) as the product of a linear form (the bracketed sum) with the characteristic polynomial of a smaller geometric lattice, namely [ˆ 0 , z].
If we are extremely lucky, L will have a maximal chain of modular elements
ˆ0 = x 0 l x 1 l · · · l xn− 1 l xn = ˆ 1.
In this case, we can apply Corollary 4 successively with z = xn− 1 , z = xn− 2 ,... , z = x 1 to split the characteristic polynomial completely into linear factors:
χL(k) = (k − en− 1 )χˆ 0 ,xn− 1 = (k − en− 1 )(k − en− 2 )χˆ 0 ,xn− 2 =... = (k − en− 1 )(k − en− 2 ) · · · (k − e 0 ),
where
ei = #{atoms a of [ˆ 0 , xi+1] | a 6 ≤ xi} = #{a ∈ A | a ≤ xi+1, a 6 ≤ xi}.
Definition 2. A geometric lattice L is supersolvable if it has a modular maximal chain, that is, a maximal chain ˆ0 = x 0 l x 1 l · · · l xn = ˆ1 such that every xi is a modular element. A central hyperplane arrangement A is called supersolvable if L(A) is supersolvable.
- Any modular lattice is supersolvable, because every maximal chain is modular.
- Πn is supersolvable. because we can take xi to be the partition whose unique nonsingleton block is [i + 1]. Thus the braid arrangement Brn is supersolvable.
