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?
Take the 2minute tour
×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $E \cdot C = +1$. 

