Computing sections of $\mathcal{O}(mE)$, $E$ elliptic curve, on a K3 surface

Question

Let $X$ be a K3 surface and $E$ a smooth genus 1 curve on $X$. Then $E^2=0$ and $h^0(\mathcal{O}(E))=2$ by Riemann-Roch. Let $$L:=\mathcal{O}(mE)$$ for some integer $m\geq2$.

Question: how does one compute $h^0(L)=m+1$ in the simplest way?

By Riemann-Roch we get $h^0(L) = 2 + h^1(L)$, so the question is equivalent to the fact that $h^1(L)=m-1$. Since $L$ is base-point free it gives a morphism $X\to \Bbb{P}^N$, $N=h^0(L)-1$, having a curve $\Delta$ as image (since it must contract all the curve in the pencil $|E|$). I would like to say something like that the general hyperplane section of $\Delta$ corresponds to $E_1+\cdots+E_{m}$ with each $E_i$ linearly equivalent to $E$, but I am not sure how to proceed. Any hint?

Answer 1

One way is to use elliptic pencil map $f:X \to \mathbb{P}^1$ and to note that $f_*O_X \cong O_{\mathbb{P}^1}$ and $R^1f_*O_X \cong \mathbb{P}^1(-2)$. Then $$ H^\bullet(X,O(mE)) = H^\bullet(X,f^*O_{\mathbb{P}^1}(m)) = H^\bullet(\mathbb{P}^1,Rf_*f^*O_{\mathbb{P}^1}(m)). $$ By projection formula $f_*f^*O_{\mathbb{P}^1}(m) \cong O_{\mathbb{P}^1}(m) \otimes f_*O_X \cong O_{\mathbb{P}^1}(m)$ and $R^1f_*f^*O_{\mathbb{P}^1}(m) \cong O_{\mathbb{P}^1}(m) \otimes R^1f_*O_X \cong O_{\mathbb{P}^1}(m-2)$. This gives $$ H^0(X,O(mE)) = H^0(\mathbb{P}^1,O_{\mathbb{P}^1}(m)), \qquad H^1(X,O(mE)) = H^0(\mathbb{P}^1,O_{\mathbb{P}^1}(m-2)), $$ so their dimensions are $m+1$ and $m-1$ respectively.