Let $X$ be a K3 surface and $D$ an effective divisor such that $h^0(D)\geq2$ and $h^1(D)=0$.
Is this enough to show that $D$ is connected?
Any reference would also be appreciated (I looked in Saint-Donat' thesis but did not get the answer)
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3$\begingroup$ (Sorry; I originally misread the question, and left an incorrect comment.) Serre duality shows $h^1(D)=h^1(-D)$, and then the ideal sheaf sequence for $D$ shows that $h^0(O_D)=1$. $\endgroup$– user5117Commented Feb 14, 2014 at 11:17
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$\begingroup$ Pardon me @ArtiePrendergast-Smith but I am a bit 'slow'. So does $h^0(O_D)=1$ imply that $D$ is connected? $\endgroup$– HeitorCommented Feb 14, 2014 at 11:48
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$\begingroup$ Yes, if there were $n$ connected components then $h^0(O_D) = n$ (each component gets its own vector space of constant global sections). $\endgroup$– Karl SchwedeCommented Feb 14, 2014 at 13:18
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1$\begingroup$ Artie, why don't you write your comment as a proper answer (instead suggesting to relocate this question to M.SE), so that everyone can understand that your first comment is actually a complete answer :) ? $\endgroup$– aglearnerCommented Feb 14, 2014 at 14:18
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1$\begingroup$ @aglearner: because I think the question should be on M.SE, and I am a stubborn old mule. $\endgroup$– user5117Commented Feb 14, 2014 at 14:33
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1 Answer
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I think the answer is yes. Here is a sketch proof.
Keep in mind that by Serre duality that $h^2(D)=0$.
Suppose that the divisor is not connected. Then there are two possibilities:
1) There is an isolated exceptional subdivsor in $D$ (by this I mean a divisor $E$ that is a connected collection of $-2$ curves with $E^2<0$). It is clear that if you remove this divisor from $D$ then $h^0(D)$ will not change. At the same time $D^2$ will increase. From Riemann Roch it would follow that $\chi (D)$ increase as well. But since $h^0$ and $h^2$ don't change, this means that $h^1$ decreased. Hence it could not be zero in the beginning.
2) $D$ is a union of $\ge 2$ fibres of an elliptic fibration. In this case $\chi(D)=2$, at the same time if $D$ is not connected then $h^0(D)\ge 3$. This gives by Riemann-Roch that $h^1(D)\ge 1$.
answered Feb 14, 2014 at 12:26
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$\begingroup$ In fact after I wrote this answer I realised that the first comment of Artie solves the problem. $\endgroup$ Commented Feb 14, 2014 at 20:36