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Suppose that $f : Y \to X$ is a birational morphism, and $E$ is some effective exceptional divisor on $Y$. The negativity lemma says that $E \cdot C \leq 0$ for any curve $C$ contracted by $f$. I'm wondering about a sort of converse: suppose $C$ is a curve such that $E \cdot C < 0$. Must $C$ be contracted by $f$? Would any additional conditions guarantee that it is?

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2 Answers 2

up vote 6 down vote accepted

No. $E\cdot C<0$ implies that $C\subseteq E$, but it may not be contracted. In fact, if $E$ is not contracted to a point, then many curves in $E$ will be like that.

Assume that $f$ is the blow up of a smooth subvariety of a smooth variety $\Sigma\subset X$ and $E=f^{-1}(\Sigma)$. Then $f|_E:E\to\Sigma$ is a $\mathbb P^n$-bundle and $\mathscr O_Y(E)|_E\simeq \mathscr O_{E/\Sigma}(-1)$. In other words, $-E|_E\sim H$, where $H$ is an $f$-ample effective divisor. Now if $C\subset E$ is such that $f|_C$ is non-constant but $C\cap H\neq\emptyset$, then $E\cdot C<0$ but $C$ is not contracted.

To see that this can actually happen, just take $X$ to be a smooth threefold, $\Sigma\subset X$ a smooth proper curve. Then $E\to\Sigma$ is a ruled surface and $H\subset E$ is a section. If $C$ is another (multi-)section that intersects $H$, then these satisfy the above criterion.

By the way... The negativity lemma does not imply what you are saying. If there are more than one exceptional divisors, say $E_1$ and $E_2$, then for any contracted curve chosen so that $C_1\subseteq E_1$ and $C_1\not\subseteq E_2$, it follows easily that $E_2\cdot C_1\geq 0$.

In particular, if say $X$ is a surface with an $A_2$ singularity, then the exceptional set consists of two curves $E_1$ and $E_2$, they both have self-intersection $-2$, but $E_1\cdot E_2=1$.

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Artie's answer is pretty much the same as mine, and we must have posted them virtually at the same time, so he deserves your up-vote, too!! –  Sándor Kovács Jul 6 '11 at 5:26

Dear Anonymous, in general the answer to you first question is no. The condition E.C<0 of course guarantees that C is contained in E, but there's no reason that it should be contracted (unless f contracts E to a point, which I guess gives an answer to the second question.)

For an example, let X be a Calabi-Yau variety of dimension 3, C a smooth rational curve in X with normal bundle O(-1)^2, and Y the blowup of X along C. Then the exceptional divisor E is isomorphic to P^1 x P^1, and the restriction of f to E is just projection. If now $\Gamma$ is a section of the projection $f_{|E}$, by adjunction one calculates that $E.\Gamma = K_E.\Gamma = -2$, but f maps $\Gamma$ isomorphically onto C --- in particular, does not contract it.

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Artie's answer is pretty much the same as mine, and we must have posted them virtually at the same time, so he deserves your up-vote, too!! –  Sándor Kovács Jul 6 '11 at 5:26
    
:) Thanks for the plug, Sándor. Indeed, when I wrote my answer your (superior) one had not yet appeared. –  Artie Prendergast-Smith Jul 6 '11 at 6:32

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