Hi everyone,

I am trying to go through parts of Saint-Donat's 1974 paper 'Projective Models of K3-surfaces', and have been stuck on a few claims for a while now - I'd appreciate some help explaining them.

Here is the set-up: $X$ is a (smooth, projective, algebraic) K3 surface. For a divisor $D$ on $X$ the Riemann-Roch theorem reads that

$ h^0(D) + h^0(-D) = 2 + \frac{1}{2}D^2 + h^1(D)$

Since the most troublesome thing here is $h^1(D)$, we should try to say something about it: the 'something' is the following: if $D$ is effective then $h^1(D)=h^0(D, \mathcal{O}_D)-1$.

We furthermore have the following dichotomy for effective divisors $D$ without fixed component: Either:

(1) $D^2 > 0$, $h^1(D)=0$, and the generic member of $D$ is an irreducible curve $1+\frac{1}{2}D^2$ or

(2)$D^2=0$. Then $D$ is linearly equivalent to $kE$ for some irreducible curve of (arithmetic) genus 1, and $k \geq 1$. We furthermore have that $h^1(L)=k-1$ and that every divisor of $|D|$ is equal to a sum of the form $E_1 + ... +E_k$ with each $E_l \in |E|$.

Furthmore, a remark that will probably be relevant is the following: if $\Delta$ is an effective divisor on $X$, then $\dim |\Delta|=0$ if and only if $h^1(\Delta, \mathcal{O}_\Delta)=0$. Furthermore, if $\Delta$ is connected and reduced, then we have $\Delta^2 = -2$.

Consider now an effective divisor $D$ satisfying $D^2 \geq 0$. We write $D \sim D'+\Delta$ where $\Delta$ is the fixed part of $D$. Decompose $\Delta$ into its connected reduced components $\Delta_1, ..., \Delta_N$. Since $D'$ has no fixed part, we are in one case of the dichotomy; assume we are in the second. The claim is that there exists one and only one $\Delta_i$ which satisfies $D' . \Delta_i>0$, and that this $\Delta_i$ in fact satisfies $\Delta_i .E=1$.

Why is the claim true? The author writes that it is an easy consequence of the Riemann-Roch theorem and the equation $h^1(D)=h^0(D, \mathcal{O}_D)-1$ given above.

Thanks a lot!