Freeness and multirestriction of hyperplane arrangements (Q2894204)

A central hyperplane arrangement \({\mathcal A}\) in a vector space \(V\) of dimension \(\ell\) is said to be free if the module \(\Omega^1({\mathcal A})\) of logarithmic differential \(1\)-forms associated with is free. Terao's conjecture is that freeness is a combinatorial property (of simple arrangements). \(\mathcal A\) is called weakly tame if \(\Omega^1({\mathcal A})\) has projective dimension at most~\(1\). The result here is that, if \({\mathcal A}\) is weakly tame (or weakly dually tame, which is the same condition for vector fields instead of differentials), or if \(\ell\leq 4\), then \({\mathcal A}\) is free if its restriction to a hyperplane \(H\) is free and the characteristic polynomial restricts in the expected way.NEWLINENEWLINEThe converse to this is known (for all \(\ell\geq 3\)) and the result here was proved for \(\ell=3\) by \textit{M. Yoshinaga} [Bull. Lond. Math. Soc. 37, No. 1, 126--134 (2005; Zbl 1071.52019)]. The further progress comes from applying other results of Yoshinaga and from the crucial observation that the logarithmic differentials with poles along \(H\) split off from \(\Omega^1({\mathcal A})\): that is, if \(H=\{\alpha_H=0\}\) and \(S=K[V]\) then NEWLINE\[NEWLINE \Omega^1({\mathcal A})\cong S{{d\alpha_H}\over{\alpha_H}}\oplus {{d\alpha_H}\over{\alpha_H}}\wedge\Omega^1({\mathcal A}). NEWLINE\]NEWLINE This is described by the author as ``trivial'' but it is what gives the condition of tameness its good inductive properties.NEWLINENEWLINEThe condition on the characteristic polynomial can be interpreted more geometrically in terms of characteristic polynomials of non-central arrangements (the intersections of \({\mathcal A}\) with \(\{\alpha_H=\epsilon\}\)) and central multiarrangements (i.e.\ with multiplicities) such as the restriction of \({\mathcal A}\) to \(H\). However, there does not seem to be a satisfactory way to treat characteristic polynomials for both these cases together. Moreover, the multiplicities in multiarrangements are in general not combinatorial, so (again according to the author) the relation with Terao's conjecture is at least not straightforward.