Free arrangements with low exponents

Abstract

In this article we show that any free hyperplane arrangement with exponents 1's and 2's is a supersolvable arrangement. We conjecture that any free arrangement with exponents 1's, 2's and exactly one 3, is also supersolvable, and we show this conjecture for hyperplane arrangements of ranks 4 and 5, and for inductively free arrangements of any rank.

In sections 4.2.1 and 4.2.2 calculations were done by hand and there is no guarantee that we covered all possible cases. Proof of Prop 4.8, case (a) the conclusion that g=1 or h=1 is not justified: take the hyperplane arrangement with defining polynomial xy(x+y)(x+2y)(x+z)zw(z+w), with \ell=x+z