The generic splitting type of the logarithmic bundle of an arrangement of at most sixteen lines is combinatorial

Let $\mathcal{A}$ be an arrangement of $d$ lines in ${\mathbb{P}}_{\mathbb{C}}^2$ and let $\mathcal{T}⟨\mathcal{A}⟩$ be its bundle of logarithmic vector fields. On a general line, $\mathcal{T}⟨\mathcal{A}⟩$ splits as $\mathcal{O}(-k)\oplus \mathcal{O}(k+1-d)$ with $k=\min \{\mathrm{m}\mathrm{d}\mathrm{r}(\mathcal{A}),\lfloor \frac{d-1}{2}\rfloor \}$, where $\mathrm{m}\mathrm{d}\mathrm{r}$ is the minimal degree of a Jacobian syzygy. We prove that $k$ is determined by the intersection lattice whenever $d\le 16$, answering in this range a question of Cook, Harbourne, Migliore and Nagel. Since freeness can be read off from $k$ and $c_2(\mathcal{T}⟨\mathcal{A}⟩)$, this strengthens Terao’s freeness conjecture in this range; in particular we obtain a proof of Terao’s conjecture for $14$ lines that uses it only for $13$ lines. For any number $d$ of lines, it follows that $\mathrm{m}\mathrm{d}\mathrm{r}$ is determined by the lattice whenever $\mathrm{m}\mathrm{d}\mathrm{r}\le 6$ and $2\mathrm{m}\mathrm{d}\mathrm{r}\le d-3$, and that Terao’s conjecture holds for every free arrangement whose smaller exponent is at most $7$. The proof reduces the question to critical arrangements: $2s+3$ lines, each containing at most $s+1$ intersection points, with $\mathrm{m}\mathrm{d}\mathrm{r}=s$. For these, the inflection polynomial of the minimal derivation factors as $Q_{\mathcal{A}}R$ with $R\in H^0({\mathcal{I}}_Z^2(s-3))$, where $Z$ is the zero scheme of the derivation, of length $\nu =c_2(\mathcal{T}⟨\mathcal{A}⟩)-s(s+2)$. For $s\le 5$ this forces $\nu \le 1$, and Chern class arguments then reduce everything to Terao’s conjecture for $13$ lines. For $s=6$ we show $\nu \le 3$, using a Riccati equation along multiple components of $R$, Ziegler’s multirestriction, and local computations verified by computer algebra; an exhaustive enumeration of the 26,265,318 candidate lattices with $\nu \in \{2,3\}$ that satisfy Bojanowski’s inequality completes the proof.