Let $X$ be a complex surface (i.e., complex dimension 2) and $C$ a $(-1)$-curve in $X$, i.e., a reduced, compact, connected curve $C$ with self-intersection $-1$. I'm trying to understand the proof of the equivalent characterization of $(-1)$-curves, namely, $$C^2 < 0 \ \ \text{and} \ \ (C, K_X) < 0.$$ Indeed, if $C$ is a $(-1)$-curve, then $C^2 =-1$ by definition and by adjunction we have $$K_C = K_X \otimes \mathcal{O}_X(C) \vert_C.$$ Then $$(C, K_C) = (C, K_X) + (C, \mathcal{O}_X(C) \vert_C),$$ since the intersection product is bilinear with respect to the tensor product of line bundles. How do we deduce from this that $$(C,K_X) <0?$$


The adjunction formula allows us to write $$\text{deg}(K_C) = (K_X + C)C = (K_X \cdot C) + C^2.$$ Now we use that $C$ is a $(-1)$-curve:

  • It is isomorphic to $\mathbb{P}^1$, so that $\text{deg}(K_C) = -2$

  • We have $C^2 = -1$

Hence we get $$-2 = (K_X \cdot C) -1$$ by the formula above, i.e. $(K_X \cdot C) = -1$.

  • $\begingroup$ Thank you very much, this is exactly what I was looking for. $\endgroup$
    – AmorFati
    Apr 21 '20 at 23:42
  • $\begingroup$ Glad that I could help. $\endgroup$
    – Con
    Apr 21 '20 at 23:43

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