has only double intersection points.

Call $m=(m_1, \dots, m_r)$ the [*combinatorial type*]{} of ${{\mathcal A}}$. Obviously the type $m$ completely describes the combinatorics of the arrangement. A concrete type $m$ split solvable arrangement is constructed in [@CDP] in the following way. Consider the arrangement $\hat{{{\mathcal A}}}(m)$ in ${\mathbb{C}}{\mathbb{P}}^r$ of equation $$\label{longeq}
 z_0(z_1-z_0)(z_1-2z_0) \dots (z_1-m_1z_0)(z_2-z_0)\dots (z_2-m_2z_0) \dots (z_r-m_rz_0)=0.$$ Then the arrangement ${{\mathcal A}}(m)$ obtained as the intersection of $\hat{{{\mathcal A}}}(m)$ with a generic $2$-plane $U \cong {\mathbb{C}}{\mathbb{P}}^2$ in ${\mathbb{C}}{\mathbb{P}}^r$ is a type $m$ split solvable arrangement. The lines of ${{\mathcal A}}(m)$ are the traces on $U$ of the hyperplanes of $\hat{{{\mathcal A}}}(m)$.

\[rem:hat\] It follows from [@Ha] that an arrangement ${{\mathcal A}}$ is generic if and only if it is the intersection of $\hat{{{\mathcal A}}}(m)$ with a generic linear subspace of ${\mathbb{C}}^{r+1}$ of dimension at least $3$, for $m=(1,\dots, 1)$.

Split solvable arrangements are remarkable for illustrating a perfect equivalence between combinatorics and topology.

\[cor17\] For an arrangement ${{\mathcal A}}\subset {\mathbb{P}}^2$ the following are equivalent:

1.  \[comb\] ${{\mathcal A}}$ has the same combinatorics as a split solvable arrangement of type $m$.

2.  \[toptype\] The complement $M_{{{\mathcal A}}}$ is homeomorphic to the complement of the type $m$ arrangement ${{\mathcal A}}(m)$.

This is 