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Proof of a conjecture of W. Veys. (English) Zbl 0857.14019
In this article the authors prove a conjecture of W. Veys about the complement of curves in the complex projective plane $$\mathbb{P}^2 (\mathbb{C})$$: Let $$C_i$$, $$i=1, \dots,n$$, be distinct irreducible curves in $$\mathbb{P}^2 (\mathbb{C})$$; if the topological Euler characteristic $$e(\mathbb{P}^2 (\mathbb{C}) \backslash \bigcup^n_{i=1} C_i)\geq 0$$, then each $$C_i$$ is a rational curve. In the first part of the article the authors study a one-parameter family of plane curves. Let $$f:X\to S$$ be a proper flat holomorphic mapping from a reduced complex surface $$X$$ to the unit disc $$S$$, such that the restriction $$X\backslash X_0 \to S\backslash \{0\}$$ is a topological fibre bundle. By using perversity of the sheaf complex of nearby cycles $$\psi_f (\mathbb{C}_X)[1]$$ and of vanishing cycles $$\varphi_f(\mathbb{C}_X)[1]$$ of $$\mathbb{C}_X[2]$$ for $$f$$, the authors can prove the following inequality between Euler characteristics, which is the main argument to obtain the conjecture.
Proposition: Let $$(X_t)_{t\in L}$$ be a pencil of plane curves. Suppose that the general member $$F$$ of this pencil is irreducible. Then $$e(X_t)\geq e(F)$$ for all $$t\in T$$.
If the conjecture was not verified, we could suppose there exist a minimal counterexample with the following properties:
(1) $$C_1$$ is not a rational curve.
(2) $$n\geq 3$$, and for $$2\leq i\leq n$$ we have $$e(C^\circ_i)=1$$, where $$C^\circ_i=C_i \backslash \bigcup_{j\neq i}C_i$$.
We can define the pencil of curves of degree $$d$$ spanned by $$n_1C_1$$ and $$n_2C_2$$: $f:X= \{(x, (\lambda_1: \lambda_2)) \in\mathbb{P}^2 (\mathbb{C}) \times \mathbb{P}^1(\mathbb{C})|\;\lambda_1 F_1(x)^{n_1} + \lambda_2F_2(x)^{n_2}=0\} \to\mathbb{P}^1 (\mathbb{C})$ where $$F_i$$ is the polynomial of degree $$d_i$$ defining $$C_i$$, and lcm$$(d_1,d_2) = d=n_1d_1=n_2d_2$$. The general fiber of the pencil $$f:X \to\mathbb{P}^1(\mathbb{C})$$ is irreducible. Let $$g:Y \to\mathbb{P}^1 (\mathbb{C})$$ be a smooth minimal model for $$f$$, as $$C_1$$ is not rational, its normalization $$\overline C_1$$ occurs as an irreducible component of $$g^{-1} (0)$$ and $$H^1(Y_0, \mathbb{Q})\neq 0$$. By the proposition and the invariant cycle theorem we deduce $$H^1(Y,\mathbb{Q}) \neq 0$$, and we get a contradiction because $$Y$$ is birational to $$X$$, so is rational.
Reviewer: M.Vaquie (Paris)

##### MSC:
 14H45 Special algebraic curves and curves of low genus 14M20 Rational and unirational varieties 14H10 Families, moduli of curves (algebraic) 14F45 Topological properties in algebraic geometry
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##### References:
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