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Let $f:X\to Y$ be a birational morphism, $X, Y$ projective, $X$ smooth (threefold if this helps). Let $Exc(f)\subseteq X$ be the exceptional locus of $f$ and let $E\subseteq Exc(f)$ be an irreducible divisor. Is it true that for any curve $C\subseteq E$ contracted by $f$ one has $C\cdot E<0$? I can see this is true if $C$ is not contained in any other divisor sitting in $Exc(f)$, but what if it is?

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In the case where X is $P^2$ blown up in two points, the two exceptional divisors $E_1$ and $E_2$ satisfy your hypothesis but as they are distinct curves, you have necessairly $(E_1 \cdotp E_2) \geq 0$. Or am I wrong? –  Henri Aug 11 '10 at 16:01
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Yes $E_1\cdot E_2=0$, but the assumptions of the question are not satisfied here. In fact, the question does have an affirmative answer for surfaces using Mumford's about negative definiteness of the intersection matrix. –  Donu Arapura Aug 11 '10 at 16:37
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Mumford's ^result (I wish I could edit comments). –  Donu Arapura Aug 11 '10 at 16:38
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To be more explicit, $E$ is supposed to be irreducible and $C\subseteq E$, which implies $E=C$, when $X$ is a surface. So $C^2 <0$ by Mumford. –  Donu Arapura Aug 11 '10 at 17:23
    
Ok, I didn't notice the irreducibility of the divisor $E$, thanks! –  Henri Aug 11 '10 at 18:10
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up vote 3 down vote accepted

Dear Carlos, the statement is false in general. For example let $Y$ be $\mathbb{C}^3$, let $f_1 : X_1 \rightarrow Y$ be the blowup of a point on $Y$, and $f_2 : X \rightarrow X_1$ the blowup of a point on the exceptional divisor of $f_1$. Let $f : X \rightarrow Y$ be the composition. The exceptional locus of $f$ has two components: a copy $F$ of $\mathbb{P}^2$ (the exceptional divisor of $f_2$) and a copy $E$ of the blowup of $\mathbb{P}^2$ in one point (the strict transform of the exceptional divisor of $f_1$). The intersection $C = E \cap F$ is a copy of $\mathbb{P}^1$, it is a line on $F$ and the $f_2$-exceptional curve on $E$. Now $E \cdot C$ equals $C^2$ computed on $F$ (because $C=E \cap F$), so $E \cdot C = +1$.

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