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Let $X$ be a complex projective surface of general type, $K_X$ be the very ample canonical divisor (which is automatically minimal). Is there an example for such a $X$ that there exist a sequence of negative curves $\{C_i\}_{i=1}^{\infty}$ with $C_i^2$ being fixed, but the geometric genus $g(C_i)$ going to infinity?

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Yes. Start with a rational surface $S$ containing infinitely many $(-1)$-curves $E_n$ (for instance $\mathbb{P}^2$ blown up along 9 general points). Choose a very ample divisor $H$ on $S$, a smooth curve $B$ in the linear system $|2H|$, and consider the double covering $\pi : X\rightarrow S$ branched along $B$. Then $K_X\cdot \pi ^*E_n=2(K_S+H)\cdot E_n=-2+2H\cdot E_n\ $ goes to infinity with $n$, and so does $g(\pi ^*E_n)$. Note that if $\pi ^*E_n$ is not irreducible, it is the union of two smooth rational curves, and $X$ contains only finitely many such curves (Lu-Miyaoka); so $\pi ^*E_n$ is irreducible for all but finitely many $n$.

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    $\begingroup$ It is not clear indeed, but I think it follows from the fact that $X$ contains only finitely many smooth rational curves (Lu-Miyaoka). If $\pi ^*E_n$ is not irreducible, it is the union of two smooth rational curves. I will edit my answer. $\endgroup$
    – abx
    Commented Dec 24, 2017 at 6:11
  • $\begingroup$ Sorry let me add my comments again: Thank you for your example Prof. Beauville. For the negative curve I always assume it is integral. I am not worrying about the geometric genus going to infinity, since I can prove that for surface of general type with ample canonical line bundle, there are only finitely many negative integral curves with geometric genus less than any given integer. I was worrying about that $C_i^2$ can be a fixed number. Are almost all the pullback curve $\pi^*E_n$ irreducibe in $X$? Thank you. $\endgroup$
    – Feng Hao
    Commented Dec 24, 2017 at 6:13
  • $\begingroup$ Sorry I have one more issue. Why have the two components to be smooth? , since it is branched covering. $\endgroup$
    – Feng Hao
    Commented Dec 24, 2017 at 6:29
  • $\begingroup$ $\pi$ induces on each component a one-to-one map onto the (smooth) curve downstairs. That map is necessarily an isomorphism (think of the normalization). $\endgroup$
    – abx
    Commented Dec 24, 2017 at 6:36

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