# Rational classes of $(-2)$-curves in a minimal surface of general type

Let $$X$$ be a minimal surface of general type over $$\mathbb{C}$$. One can show that if for any set of $$(-2)$$-curves $$C_1,\cdots,C_l$$ on $$X$$, there exists $$k$$, $$1\le k\le l$$ such that $$\sum_{i=1}^k\lambda_iC_i = \sum_{j=k+1}^l\lambda_jC_j,$$ in $$H_2(X,\mathbb{Q})$$ with all $$\lambda_i,\lambda_j\ge 0,$$ then $$\lambda_i=0$$ for $$i=1,\cdots,l$$.

I would like to show that the above result implies there is at most one $$(-2)$$-curve in each rational class. This is claimed in the proof of Proposition VII.2.2.5 of Compact Complex Surfaces (Barth et. al).

My approach was as follows. Suppose there $$C_1$$ and $$C_2$$ are two different $$(-2)$$-curves in the class $$[\Gamma]\in H_2(X,\mathbb{Q})$$. If I could show that $$C_1$$ being homologous to $$C_2$$ implies that there exist $$\lambda_1,\lambda_2>0$$ such that $$\lambda_1C_1=\lambda_2C_2$$, then the above result would give us a contradiction. But I have not been able to do this and I don't know if it's actually true. I know that $$C_1$$ and $$C_2$$ being in the same rational homology class implies $$C_1-C_2$$ is homologous to zero, whence $$C_1-C_2$$ is numerically equivalent to zero.

Questions: In my context,

• Does $$C_1\sim_{\text{hom}} C_2$$ in $$H_2(X,\mathbb{Q})$$ imply $$qC_1 = C_2$$ for some rational $$q>0$$?

• Is there a better way to show that there is at most one $$(-2)$$-curve in each rational class?

• Isn't your first equation stated in the group of rational homology classes, rather than classes up to some finer equivalence? So setting $\lambda_1=\lambda_2=1$, it is true by assumption. Jul 16, 2021 at 19:07
• @WillSawin I have proved the first result and if I'm not wrong, the equation stated for curves and not in group of rational homology classes. Also, to apply it I would first need an equation relating $\lambda_1,\lambda_2$ and the $C_i$s. What can I derive from the fact that $C_1$ and $C_2$ are in the same class? Jul 16, 2021 at 19:43
• So "in $H_2(X,\mathbb Q)$" in the third line of the question should not be there? Jul 16, 2021 at 19:58
• That makes sense; so since the book actually writes "in $H_2(X,\mathbb{Q})$," the first equation must have been stated in the group of rational homology classes, so as you said setting $\lambda_1=\lambda_2=1$ should work. I was wrong about the proof of the first result and overlooked that it's in $H_2(X,\mathbb{Q}).$ Jul 16, 2021 at 20:06
• Welcome to Mathoverflow! Let $C_1$, $C_2$ be distinct $(-2)$-curves on a smooth projective surface; assume that $[C_1] = [C_2]$ as rational homology classes. Then we get a contradiction $-2 = C_1 \cdot C_2 \ge 0$ (the last inequality holds as the curves don't share common components, so intersect effectively). Jul 16, 2021 at 22:35