# Minimal number of nodes in a complex line arrangement.

Let $\mathcal{A}$ be a collection of $n$ lines. Assume that $\mathcal{A}$ is not a pencil. It is known (see http://www.springerlink.com/content/320p742475v6q746/) that if all lines are in $\mathbb{RP}^2$, then there are at least $6n/13$ nodes.

What is the minimal number of nodes, if all lines are in $\mathbb{CP}^2$?

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## 1 Answer

In a paper by Hirzebruch "Singularities of algebraic surfaces and characteristic numbers", Cont. Math. vol. 58, part I, 1986, there is an inequality for the number of double and triple points of an arrangement of $n$ complex lines. Denote by $t_k$ the number of points lying on precisely $k$ lines; if $t_n=t_{n-1}=t_{n-2}=0$, then $$t_2+3t_3/4\ge n+\sum_{r\ge 5}(2r-9)t_r.$$ This inequality is obtained from the Bogomolov-Miyaoka-Yau inequality for surfaces of general type: taking a suitable cover of the plane branched on an arrangement not satisfying the inequality above and desingularizing one would obtain a surface of general type with $c_1^2>3c_2$.

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Thanks a lot for the reference. – Fei YE Aug 15 '11 at 8:13
You're welcome. You should also have a look at Giancarlo Urzua's Ph.D.'s thesis, he has a section about line arrangements. – rita Aug 15 '11 at 13:26