A linear system of a curve on a K3 surface.

Question

Let $S$ be a K3 surface and $C \subset S$ be a smooth curve of genus $g$ (assume it represents a primitive homology class). Is it possible to compute the dimension of the linear system $|C|$ only from these data?

Answer 1

Yes.

First, the adjunction formula tells you that $$K_C = (K_S + O_S(C))_{|C} = O_S(C)_{|C}$$

so $C^2=\text{deg }K_C=2g-2$.

On the other hand, Riemann–Roch says that $$\chi(O_S(C))=\chi(O_S)+\frac12 C \cdot \left(K_S +C \right)\\=2+\frac12 C^2.$$

Now the left-hand side is $$h^0(O_S(C))-h^1(O_S(C))+h^2(O_S(C))$$ so to get what we want we need to be able to say something about the $h^1$ and $h^2$ terms. In general this might be hard, but the fact that we're on a $K3$ surfaces saves the day:

• first, Serre duality says $h^2(O_S(C))=h^0(K_S-O_S(C))=h^0(O_S(-C))=0$ since $-C$ is not effective.
• second, Serre duality also says $h^1(O_S(C))=h^1(O_S(-C))$. How does that help? Well, we have the ideal sheaf sequence for $C$:

$$ 0 \rightarrow O_S(-C) \rightarrow O_S \rightarrow O_C \rightarrow 0;$$ taking the associated long exact sequence of cohomology, and using $h^1(O_S)=0$ (because $S$ is $K_3$) we get $h^1(O_S(-C))=0$ as we want.

So in the end $h^0(O_S(C))=\chi(O_S(C))=2+\frac12 C^2 = 2+ (g-1)=g+1$. Or, if you prefer to phrase it in terms of linear systems, $\text{dim }|C|=g$.