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Let $X$ be a complex K3 surface and $D$ an effective divisor on $X$.

We shall say: $D$ is connected if its support is connected. $D$ is numerically connected if for any non-trivial effective decomposition $D\sim D_1+D_2$ we have $D_1\cdot D_2\geq1$. Notice that these two notions are not equivalent (if $D$ is not reduced), e.g. $D=2E$ where $E$ is an elliptic curve, is numerically disconnected.

Assume $h^1(D)=0$. The ideal sheaf sequence of $D$ yields $h^0(\mathcal{O}_D)=h^0(\mathcal{O}_X)=1$ in cohomology. Hence $D$ is connected. My question is:

Does $h^1(D)=0$ imply numerical connectedness?

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Interesting question. I think the answer is yes, let me try to prove it.

As you noticed, the ideal sheaf sequence shows that $h^1(D)=0$ is equivalent to the fact that $H^0({\mathcal O}_D)$ is 1-dimensional generated by the constant function $1$.

Considering, for every effective decomposition $D=A+B$, the exact sequence $$ 0 \rightarrow {\mathcal O}_B(-A) \rightarrow {\mathcal O}_D \rightarrow {\mathcal O}_A \rightarrow 0 $$ we deduce then that $H^0({\mathcal O}_D \rightarrow {\mathcal O}_A)$ is injective and therefore, since by Riemann-Roch $B$ has arithmetic genus $1+\frac{B^2}2$, $$0=h^0({\mathcal O}_B(-A))\geq \chi({\mathcal O}_B(-A))=-AB-\frac{B^2}{2}$$ so $B^2 \geq -2AB$ and similarly $A^2\geq -2AB$.

Assume by contradiction $AB\leq 0$: then $A^2B^2 -(AB)^2 \geq 3(AB)^2 \geq 0$. This implies, by the index theorem, that $AB=0$ and $A$ and $B$ are proportional.

Let us then choose a primitive $C \in Pic(X)$ such that $A=aC$, $B=bC$; obviously $C^2=0$ and by Riemann-Roch $\chi(A)=\chi(B)=\chi(C)=\chi(D)=2$. Up to replacing $C$ by $-C$, $C$ is effective, $a$ and $b$ are positive and $h^2(A)=h^2(B)=h^2(C)=h^2(D)=0$.

Then $h^0(C) \geq 2$, so $D=(a+b)C \geq 2C \Rightarrow h^0(D) \geq 3 \Rightarrow h^1(D) \neq 0$, a contradiction.

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  • $\begingroup$ Just a question: I don't really see where do you use the injection $H^0(O_D)\subset H^0(O_A)$. I guess the expression for $\chi(O_B(-A))$ is just Riemann-Roch for singular curves, no? $\endgroup$ – Heitor Mar 3 '14 at 19:11
  • $\begingroup$ You are right, the expression for $\chi({\mathcal O}_B(-A))$ is Riemann-Roch. The injectivity gives me $h^0({\mathcal O}_B(-A))=0$, else $\chi$ could be positive. $\endgroup$ – Roberto Pignatelli Mar 4 '14 at 11:01
  • $\begingroup$ Ah OK, I see. Thank you! I thought that $h^0(\mathcal{O}_B(−A))=0$ would follow from $A$ being effective. Is this not enough? $\endgroup$ – Heitor Mar 4 '14 at 12:08
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    $\begingroup$ No. For instance if $A$ is a a rational curve with $A^2<0$, then $h^0(\mathcal O_A(-A))>0$. $\endgroup$ – rita Mar 4 '14 at 13:18
  • $\begingroup$ Ciao Pigna! welcome to MO! $\endgroup$ – rita Mar 4 '14 at 20:20

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