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Some (probably very easy) questions on intersection theory on surfaces...

Say $S$ is a smooth projective surface over $\mathbb{C}$ with canonical divisor $K_S$.

  • If $S$ is not ruled and $H$ is a hyperplane section (for an arbitrary embedding), do we always have that $K_S \cdot H > 0$? I know that $K_S \cdot H \geq 0$, but why would $K_S \cdot H = 0$ be impossible? EDIT: OK, this can happen if $K_S = 0$, of course (see QiL's answer). But if I moreover assume that $K_S^2 > 0$, can we still have $K_S \cdot H = 0$?

  • If $D \cdot H < 0$ for some divisor $D$ and some hyperplane section $H$, does it follow that $H^0(D,\mathcal{O}_S(D)) = 0$? It seems reasonable but also too simple, so I'm not so sure.

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up vote 3 down vote accepted
  1. It can happen that $K_S=0$ (i.e. $S$ is a K3 surface), then $K_S\cdot H=0$.

  2. If $H^0(D, O_S(D))\ne 0$, then up to linear equivalence, $D\ge 0$, so $D\cdot H\ge 0$ because $H$ is ample.

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Thank you, QiL. Concerning (1), I added an extra hypothesis - does the result hold with this hypothesis? Concerning (2), what is the formal argument showing that $D \cdot H \geq 0$ if $D$ is effective and $H$ is ample? Intuitively I feel that this has to be right, but I don't see a formal proof... – Evariste May 30 '12 at 22:07
$K3$-surfaces aren't ruled so it can still happen that $K_S=0$ with your extra hypothesis. Moreover, even if $K_S = 0$ you can have that $K_S\cdot H =0$, I think. – Harry May 31 '12 at 7:54
Yes, but I added the hypothesis $K_S^2 > 0$... – Evariste May 31 '12 at 16:14
@Evariste, then $K_S\cdot H>0$. See Hodge index theorem (Hartshorne, V.1.9). For $D\cdot H>0$: there is a hyplane $H'$ which doesn't contain $D$. So $D\cdot H=D\cdot H'>0$. See also Hartshorne V.1.10 for the inverse implication. – user18119 May 31 '12 at 22:16
Thank you very much! – Evariste Jun 1 '12 at 22:00

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