Non-recursive freeness and non-rigidity

Abstract

In the category of free arrangements, inductively and recursively free arrangements are important. In particular, in the former, Terao's open problem asking whether freeness depends only on combinatorics is true. A long standing problem whether all free arrangements are recursively free or not was settled by the second author and Hoge very recently, by giving a free but non-recursively free plane arrangement consisting of 27 planes. In this paper, we construct free but non-recursively free plane arrangements consisting of 13 and 15 planes, and show that the example with 13 planes is the smallest in the sense of the cardinality of planes. In other words, all free plane arrangements consisting of at most 12 planes are recursively free. To show this, we completely classify all free plane arrangements in terms of inductive freeness and three exceptions when the number of planes is at most 12. Several properties of the 15 plane arrangement are proved by computer programs. Also, these two examples solve negatively a problem posed by Yoshinaga on the moduli spaces, (inductive) freeness and, rigidity of free arrangements. 1 Corollary 1.3 ). Freeness of arrangements depends only on their combinatorics for central arrangements A in C 3 with |A| ≤ 12.By investigating the structure of the classification in Theorem 1.1, we can say that almost all the free arrangements with small exponents are inductively free.then A is either inductively free or a dual Hesse arrangement appearing in Theorem 1.1.Another formulation of Problem 1.2 is using the moduli space of all arrangements whose intersection lattice is a given lattice, see Definition 2.16 for details:Open problem 1.5. Does a lattice L exist such that the moduli space V(L) contains a free and a non-free arrangement?Not much is known about this moduli space regarding freeness; Yuzvinsky [16] proved that the free arrangements form an open subset in V(L) (see also [15, Theorem 1.50] for the case of dimension three). The first intuition one gets when working with these notions is, that either the moduli space is very big and the free arrangements are inductively free, or the moduli space is zero dimensional and all its elements are Galois conjugates (let us call such an arrangement rigid). Yoshinaga proposed the following problem which is stronger than Terao's open problem since the property of being free is invariant under Galois automorphisms: NON-RECURSIVELY FREENESS AND NON-RIGIDITY 3 Problem 1.6 (Yoshinaga,[15, p. 20,(11)]). Is a free arrangement either inductively free, or rigid? In other words, does the following inclusion hold?{Free arrangements}⊂{Inductively free} ∪ {Rigid}.Beyond the fact that we find new small arrangements which are free but not recursively free, our examples with 13 and 15 planes provide a negative answer to Yoshinaga's problem 1.6, i.e., the following holds.Theorem 1.7. Let L be the intersection lattice of our 13 or 15 plane arrangements. Then the moduli spaces V(L) of them are one-dimensional (hence not rigid), and not inductively free, but free. I...

Divisionally free arrangements of hyperplanes

“…There are only three arrangements known which are free but non-recursively free. The first one is found by Cuntz and Hoge in [11], and the other two are in [4]. Though some free arrangements (like A(G 31 ) in the previous section) are not known whether it is recursively free or not (see [10], Corollary 4.3), by these known results on RF , it is worth considering Conjecture 4.1 for recursively free arrangements.…”

“…For details on divisionally free arrangements, see Theorem 4. 4. The most famous and important class of free arrangements with the same properties is the class of inductively free arrangements introduced by Terao in [19] (see Definition 4.2).…”

Line and rational curve arrangements, and Walther’s inequality

“…If m(C) ≤ 3, the claim follows Theorem 3.2. If m(C) = 4, then in both cases A and A ′ we get mdr(f ) ≥ (d − 2)/2 = 3.5, and the claim follows from Theorem 1.1 (2) and Theorem 3.2 (2).…”

“…Written down explicitly, this means that for d = 2d ′ even, one has ν(C) = 3(d ′ ) 2 − 3d ′ + 1 − τ (C), while for d = 2d ′ + 1 odd, one has ν(C) = 3(d ′ ) 2 − τ (C). Now we address the claim (2). Let C : f = 0 be a line arrangement satisfying mdr(f ) ≥ (d − 2)/2.…”

Recursively free reflection arrangements

“…One of them is the reflection arrangement of the exceptional complex reflection group G 27 . Very recently, Abe, Cuntz, Kawanoue, and Nozawa [ACKN14] found smaller examples (with 13 hyperplanes, being the smallest possible, and with 15 hyperplanes) for free but not recursively free arrangements in characteristic 0.…”