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?


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.