3

Let L be a line bundle on an (algebraic) K3 surface over a field k. The Riemann-Roch theorem specializes to

χ(X,L)=12(LL)+2

which can be rewritten as h0(X,L)+h0(X,L)=12(LL)+2+h1(X,L)

(I use Serre's duality to identify H2(X,L) and H0(X,L))

Assume (LL)2, then the RHS is bigger or equal than 2, so H0(X,L)0 or H0(X,L)0.

I guess the converse statement is not true, that is: there exists a line bundle L such that H0(X,L)0 but (LL)<2. But I don't manage to give an example. Could anybody help me?

1
  • 4
    Take L=2D where D is (the line bundle corresponding to) an effective divisor with DD=2, such as a smooth rational curve. Mar 26, 2013 at 16:08

2 Answers 2

7

Since the canonical class of X is trivial, by adjunction one has 2pa(L)2=L2, where pa denotes the arithmetic genus.

Now assume that L=OX(C), where C is an effective curve. If C is connected and reduced then pa(C)0, so L2<2 implies that C is either non-reduced or non-connected.

As an example of the first case, one can take C=nD, where D is a smooth rational curve (as suggested by Noam Elkies in his comment), so that C2=2n2. For instance, one can take as X a quartic surface in P3 containing a line.

As an example of the second case, one can take a K3 surface X wich contains two disjoint smooth rational curves D1 and D2 (a suitable elliptic fibration should do the job) and set C=D1+D2. Then C2=4.

3
  • 2
    Even simpler than an elliptic fibration: use a "random" smooth quartic surface that contains two skew lines D1,D2 such as w=x=0 and y=z=0. Mar 26, 2013 at 17:55
  • @Francesco: I've just realized that you have the same statement here that I made in my answer. I only scanned your answer for the first time and mainly just read the last two paragraphs. My apologies! Mar 27, 2013 at 3:43
  • @Sandor: don't worry, it can happen sometimes :-) Mar 27, 2013 at 6:30
6

Noam and Francesco have already pointed out that in order to get such a line bundle you can always take either a multiple of a (2)-curve or the union of disjoint (2)-curves, both of which is possible on many K3's.

On the other hand, if you were looking for an L that has a non-zero section whose zero locus is irreducible, then the self-intersection has to be at least 2. This follows easily from the adjunction formula and Riemann-Roch:

If DX is an irreducible curve on a smooth surface X, then D is a Cartier divisor and hence Gorenstein, so it has a dualizing sheaf which is a line bundle and this dualizing sheaf satisfies the formula ωDωX(D)OD. If in addition ωXOX, then this implies that degωD=D2. If D2<0, then this means that χ(ωD)=1 and by Riemann-Roch 0pa(D)=degωD+1χ(ωD)=D2+2. In particular, D22, and equality holds if and only if pA(D)=0 which holds if and only if DP1.

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.