Totally free arrangements of hyperplanes

A central arrangement $\A$ of hyperplanes in an $\ell$-dimensional vector space $V$ is said to be {\it totally free} if a multiarrangement $(\A, m)$ is free for any multiplicity $ m : \A\to \Z_{>0}$. It has been known that $\A$ is totally free whenever $\ell \le 2$. In this article, we will prove that there does not exist any totally free arrangement other than the obvious ones, that is, a product of one-dimensional arrangements and two-dimensional ones.


Introduction
Let V be an ℓ-dimensional vector space (ℓ ≥ 1) over K with a coordinate system {x 1 , . . . , x ℓ } ⊂ V * . Define S := Sym(V * ) ≃ K[x 1 , . . . , x ℓ ]. Let Der K (S) be the set of all K-linear derivations of S to itself. Then Der K (S) = ⊕ ℓ i=1 S ·∂ x i is a free S-module of rank ℓ. A central arrangement (of hyperplanes) in V is a finite collection of linear hyperplanes in V . In this article we assume that every arrangement is central unless otherwise specified. A multiplicity m is a function m : A → Z >0 and a pair (A, m) is called a multiarrangement. In general, D(A, m) is not necessarily a free S-module. We say that (A, m) is free if D(A, m) is a free S-module. For a fixed arrangement A, a multiplicity m on A is called free if a multiarrangement (A, m) is free. Define The following definition was introduced in [4, Definition 5.4].

Definition 1.1
An arrangement A is called totally free if every multiplicity m : A → Z >0 is a free multiplicity, or equivalently N F M(A) = ∅.
Our main theorem is as follows:

Theorem 1.2 An arrangement A is totally free if and only if it has a decomposition
Ziegler showed in [12,Corollary 7] that (A, m) is a free multiarrangement whenever ℓ ≤ 2. Note that holds true as shown in [3,Lemma 1.4]. Thus A 1 × A 2 × · · · × A s is known to be totally free if each A i is an arrangement in K 1 or K 2 . Theorem 1.2 asserts that the converse is also true. In the next section we will prove Theorem 1.2 in a stronger form: we will show that A is decomposed into one-dimensional arrangements and two-dimensional ones if N F M(A) is a finite set.
Recall that the intersection lattice L(A) is the set {X = H 1 ∩ · · · ∩ H s | H i ∈ A, s ≥ 0} with the reverse inclusion ordering as in [6, Definition 2.1]. Then Theorem 1.2 implies:

Corollary 1.3 Whether an arrangement A is totally free or not depends only on its intersection lattice L(A).
Let A be a nonempty central arrangement and H 0 ∈ A. Define the deletion A ′ and the restriction A ′′ as in [6, Definition 1.14]: Because of the characterization in Theorem 1.2, the total freeness is stable under deletion and restriction: Any subarrangement or restriction of a totally free arrangement is also totally free.
A multiarrangement was introduced and studied by Ziegler in [12]. The third author proved in [9] and [10] that the freeness of a simple arrangement is closely related with the freeness of Ziegler's canonical restriction. Recently the first and second authors and Wakefield developed a general theory of free multiarrangements and introduced the concept of free multiplicity in [3] and [4]. Several papers including [1], [2], [5] and [11] studied the set of free multiplicities for a fixed arrangement A. The main theorem (Theorem 1.2) in this article shows that the set of free multiplicities (or N F M(A)) imposes strong restrictions on the original arrangement A.
Acknowledgements. The first author is supported by the JSPS Research Fellowship for Young Scientists. The second and third authors have been supported in part by Japan Society for the Promotion of Science. The authors thank Professors Sergey Yuzvinsky and Max Wakefield for helpful discussions and comments. We also thank Professor Thomas Zaslavsky for pointing out an error in an earlier version.
If B is a subarrangement of A, then it is easy to see that In fact, Theorem 2.1 is true for any GMP k and LMP k (1 ≤ k ≤ ℓ), see [3,Corollary 4.6]. An arrangement A is said to be reducible if A = A 1 × A 2 for certain arrangements A i in V i (i = 1, 2). We say A is irreducible if it is not reducible.

Lemma 2.2
Let A be an irreducible arrangement in K ℓ with ℓ ≥ 3. Then there exists a subarrangement B with |B| = ℓ + 1 such that: whenever H i , H j , H k are three hyperplanes in B. Moreover the arrangement B is not free.
Proof. Let H 0 ∈ A. Let A ′ and A ′′ be the deletion and the restriction respectively. Then either A ′ or A ′′ is irreducible by Tutte [8] (see also [7,Theorem 4.3.1]). When |A| = ℓ + 1, the arrangement A itself satisfies the condition. We will prove by an induction on |A|. If A ′ is irreducible, then A ′ contains ℓ + 1 hyperplanes satisfying the condition. So we may assume that A ′′ is irreducible. Let A ′ = {H 1 , H 2 , . . . , H n−1 } and H i :  Suppose ℓ ≥ 4. Then, by the induction assumption, there exists a subarrangement {H 1 , H 2 , . . . , H ℓ } of A ′′ satisfying the condition. Then the subarrangement {H 0 , H 1 , . . . , H ℓ } of A satisfies the condition.
Suppose that the arrngement B is free. Then the sum of exponents is equal to ℓ + 1. Thus one has GMP 2 ≤ ℓ 2 ((ℓ + 1)/ℓ) 2 . We also have LMP 2 = ℓ+1 2 . Since Let |A| = n. Then d∈exp(A,m) d = (k − 1)(ℓ + 1) + n and thus with some constants A and B. By Theorem 2.1 we have whenever k is sufficiently large. This is a contradiction because We now prove the following theorem which is stronger than Theorem 1.2.

Theorem 2.4
The following four conditions for a central arrangement A are equivalent: (4) ⇒ (3): Decompose A into irreducible arrangements. Then each of the irreducible arrangements satisfies the assumption (4). Therefore we may assume that A is irreducible from the beginning. Then, by Lemma 2.2, we may conclude ℓ ≤ 2.