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Everything on $\Bbb{C}$.

Let $S$ be ruled surface over a curve $C$ and $K$ a canonical divisor on $S$. What is $K^2$ ?

In particular I would like to understand if $K^2=0$ when $C$ is either rational or elliptic.

Thanks.

Heitor Fontana
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    If $S$ is geometrically ruled, then $K^2=8(1-g(C))$. (from Beauville's "Complex Algebraic Surfaces", III.21) – Yuchen Liu May 17 '13 at 14:37
  • You need to be careful, since a ruled surface is anything birational to some $\mathbb P^1\times C.$ In particular, if we take $\mathbb P^1\times C$ and blow up at some finite number of points to get $S,$ then $K_S^2$ will decrease by the number of blowups. – Andrew May 17 '13 at 14:47
  • @Andrew Yes, that's why I use geometrically ruled rather than ruled, which implicate that every fiber is isomorphic to $\mathbb{P}^1$. – Yuchen Liu May 18 '13 at 02:38
  • Dear @jerrysciencemath, I agree, when I said "you need" I really only meant "one needs," i.e. it was a general comment, not necessarily directed at your valid remark! Cheers – Andrew May 19 '13 at 16:18
  • @Andrew Thanks, your answer is very clear and helpful! – Yuchen Liu May 20 '13 at 02:46

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Consider the case $C$ rational. For example, we could let $S$ be the blowup of $\mathbb P^2$ at a point, which equals the Hirzebruch surface $\mathbb F_1.$ Then we know that $K_S^2=K_{\mathbb P^2}^2-1=9-1=8$ so in this case $K^2\neq0.$ (Actually, I believe we can show $K_{\mathbb F_n}^2=8$ for every Hirzebruch surface, so in this case we will have $K^2=0$ if and only if $S$ is the blowup of a minimal surface at $8$ (for $\mathbb F_n$) or $9$ (for $\mathbb P^2$) points.)

If $C$ is elliptic and $S=\mathbb P^1\times C$ or if $S$ is geometrically ruled over $C,$ then $K_S^2=8(1-g)=0.$ But again, if we take any such surface and blow up some points, we get a surface with $K_S^2<0.$ So to satisfy the desired condition, we must know that $S$ is minimal.

Andrew
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