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?
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 nonconstant 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 selfintersection $2$, but $E_1\cdot E_2=1$. 


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 CalabiYau 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. 

