Rookie questions about k3's

Question

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!

Answer 1

As you said, it's a consequence of Riemann-Roch and the equation $h^1(D)=h^0(D,\mathcal O_D)-1$, which follows from taking cohomology of $$0\to \mathcal O_X(-D)\to \mathcal O_X\to \mathcal O_D\to 0$$ for an effective divisor $D$.

Applying Riemann-Roch to $D=D'+\Delta$, we get $$h^0(D)-h^1(D) = \frac{1}{2}D^2+2 = D'.\Delta-N+2,$$ where $h^2(D)=h^0(-D)=0$ because $D$ is effective. Plugging in the formula for $h^1(D)$, this becomes $$h^0(D)-h^0(D,\mathcal O_D)=D'.\Delta-N+1.$$ Now, $h^0(D)=h^0(D')=h^1(D')+2=h^0(D',\mathcal O_{D'})+1$. Since $D'$ is equivalent to $k$ disjoint irreducible genus 1 curves (the fibers of the linear series $|D'|$), its structure sheaf has $k$ global sections. The equation then becomes $$k-h^0(D,\mathcal O_D)=kE.\Delta-N.$$ The $h^0$ term is the number of connected components of $D$, or equivalently $E\cup \Delta$, where $E$ is a generic fiber of the linear series. Also, $E.\Delta = \sum_i E.\Delta_i$, where each summand is nonnegative because $E$ is movable. To finish, we just do some casework, counting connected components for each incidence possibility.

If $E.\Delta=0$, then $D^2=-2N<0$, which contradicts the initial assumption that $D^2\geq 0$.

If $E.\Delta=1$, then we win.

If $E.\Delta=m>1$, then the only partition of $E.\Delta$ that gives us an integral value of $k$ is $E.\Delta_i=1$ for all $\Delta_i$, and this forces $k=1$ and $N=m$. I couldn't rule out this case a priori, but if you look at Saint-Donat's remark 2.7.4, he assumes $k>2$ before making the claim, so all is well.