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?

  • $\begingroup$ 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. $\endgroup$
    – Will Sawin
    Jul 16, 2021 at 19:07
  • $\begingroup$ @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? $\endgroup$
    – astana
    Jul 16, 2021 at 19:43
  • $\begingroup$ So "in $H_2(X,\mathbb Q)$" in the third line of the question should not be there? $\endgroup$
    – Will Sawin
    Jul 16, 2021 at 19:58
  • $\begingroup$ 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 $H_2(X,\mathbb{Q}).$ $\endgroup$
    – astana
    Jul 16, 2021 at 20:06
  • 1
    $\begingroup$ 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). $\endgroup$ Jul 16, 2021 at 22:35


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