Let $S$ be a smooth projective algebraic surface over $\mathbb C$ and $C$ be a smooth curve on $S$. Is it always true that $dim_{\mathbb C} H^0(S,O_{S}(-C))=0$ ? In particular, is it zero when $S$ is a K3-surface ?
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6$\begingroup$ Is S connected? In the connected case, can you remind me what the space H^0(S,O_S) is? $\endgroup$– Kevin BuzzardCommented Feb 17, 2010 at 12:28
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1$\begingroup$ $H^0(S,\mathcal{O}_S)=\mathbb{C}$. $\endgroup$– Fei YECommented Feb 17, 2010 at 16:32
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4$\begingroup$ Right. And now what about the subspace of those functions that vanish on C?? $\endgroup$– Kevin BuzzardCommented Feb 17, 2010 at 20:56
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I'm not sure about the general case, but this is at least true if the divisor associated to the curve C is ample:
In this case the line bundle L associated to the divisor is ample, so the Kodaira-Nakano vanishing theorem applies. By Serre duality we get
$$ 0 = H^{2,1}(S,L) = H^1(S, \Omega^2 \otimes L) = H^1(S,L*)^* = H^1(S,O(-C))^* $$
as the dual of $L$ is the line bundle associated to the divisor $-C$. Remember that $\mathcal I_C = O(-C)$, then the long exact sequence associated to
$$ 0 \to O_S(-C) \to O_S \to O_C \to O $$
gives the exact sequence
$$ 0 \to H^0(S,O_S(-C)) \to \mathbb C \to \mathbb C \to 0 $$
as both $O_S$ and $O_C$ only have constant global sections. The vanishing of $H^0$ follows.
The general case would hold in the same way if $H^1(S,O(-C)) = H^1(S,\mathcal I_C) = 0$ for any smooth curve $C$ in $S$.
answered Feb 17, 2010 at 16:43
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4$\begingroup$ As Kevin hinted in his comment above, one doesn't need to go to so much trouble. Simply note that $H^0(S, O_S(-C))$ is the subspace of $H^0(S,O_S)$ consisting of sections which vanish along $C$. If $S$ is connected, then $H^0(S,O_S)$ is just the constants, $\mathbb C$. If $C$ is non-empty, then since a constant which vanishes at a point in $C$ must vanish everywhere (because it is constant!), we see that $H^0(S, O_S(−C))=0$. (There is no need for any hypotheses other than that $S$ is connected and $C$ is a non-empty curve.) $\endgroup$– EmertonCommented Feb 17, 2010 at 17:28
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$\begingroup$ Thanks Emerton. I realized on the way home that I was being thick about things. I hope this'll teach me to never break out the cohomology until I've thought about what I'm doing. $\endgroup$ Commented Feb 17, 2010 at 18:19