Let $C \subset S$ be a smooth curve in a K3 surface $S$. Why is the dimension of the linear system $|C|$ the genus of $C$?
Here is what I tried: $\dim |C| = \dim H^0(\mathcal{O}(C)) – 1$, and this appears in the Euler characteristic $\chi(\mathcal{O}(C))$. Riemann-Roch on K3 surfaces is $$\chi(\mathcal{O}(C)) = \frac{1}{2} C^2 + 2,$$ and we can connect $C^2$ to the genus $g = g(C)$ using the adjunction formula, which on a K3 surface is $$2g – 2 = C^2.$$
Combining Riemann-Roch and the adjunction formula then leads to
$$\chi(\mathcal{O}(C)) = g + 1,$$
so it remains to show $H^0(\mathcal{O}(C)) = \chi(\mathcal{O}(C))$, i.e. $H^1(\mathcal{O}(C)) = 0 = H^2(\mathcal{O}(C))$. By Serre duality $H^2(\mathcal{O}(C) = H^0(\mathcal{O}(-C))^*$, and $\mathcal{O}(-C)$ is the ideal sheaf of $C$, which does not have any global sections. Hence $H^2(\mathcal{O}(C)) = 0$.
But why does the superabundance $\dim H^1(\mathcal{O}(C))$ vanish? Or is that not true in general? In the application I'm interested in, $C$ also generates the Picard groups of $S$.
Best Answer
I found the missing piece in Huybrecht's Lectures on K3 surfaces: The above argument shows $$\dim H^0(\mathcal{O}(C)) \geq g + 1.$$ Taking global sections of the sequence $$ 0 \to \mathcal{O}(C) \otimes \mathcal{O}(-C) \to \mathcal{O}(C) \to \mathcal{O}(C)|_C \to 0$$ yields an exact sequence $$0 \to H^0(\mathcal{O}_S) \to H^0(\mathcal{O}(C)) \to H^0(\mathcal{O}(C)|_C) \to H^1(\mathcal{O}_S) = 0$$ Hence $\dim H^0(\mathcal{O}(C)) = H^0(\mathcal{O}(C)|_C) + 1$. Also, $\omega_C = \omega_S|_C \otimes \mathcal{O}(C)|_C = \mathcal{O}(C)|_C$, so $\dim H^0(\mathcal{O}(C)) = \dim H^0(\omega_C) + 1 = g + 1$.