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!


1 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.