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

$$ \chi(X, L)=\frac{1}{2}(L\cdot L)+2 $$

which can be rewritten as $$ h^0(X, L)+h^0(X, L^\ast)=\frac{1}{2}(L\cdot L)+2+h^1(X, L) $$

(I use Serre's duality to identify $H^2(X, L)$ and $H^0(X, L^\ast)^\ast$)

Assume $(L\cdot L)\geq 2$, then the RHS is bigger or equal than 2, so $H^0(X, L) \neq 0$ or $H^0(X, L^\ast)\neq 0$.

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

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Take $L=2D$ where $D$ is (the line bundle corresponding to) an effective divisor with $D \cdot D = -2$, such as a smooth rational curve. – Noam D. Elkies Mar 26 '13 at 16:08

Since the canonical class of $X$ is trivial, by adjunction one has $$2p_a(L)-2 = L^2,$$ where $p_a$ denotes the arithmetic genus.

Now assume that $L=\mathcal{O}_X(C)$, where $C$ is an effective curve. If $C$ is connected and reduced then $p_a(C) \geq 0$, so $L^2 < -2$ 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 $C^2=-2 n^2$. For instance, one can take as $X$ a quartic surface in $\mathbb{P}^3$ containing a line.

As an example of the second case, one can take a $K3$ surface $X$ wich contains two disjoint smooth rational curves $D_1$ and $D_2$ (a suitable elliptic fibration should do the job) and set $C=D_1+D_2$. Then $C^2=-4$.

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Even simpler than an elliptic fibration: use a "random" smooth quartic surface that contains two skew lines $D_1,D_2$ such as $w=x=0$ and $y=z=0$. – Noam D. Elkies Mar 26 '13 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! – Sándor Kovács Mar 27 '13 at 3:43
@Sandor: don't worry, it can happen sometimes :-) – Francesco Polizzi Mar 27 '13 at 6:30

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 $D\subset X$ 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 $$ \omega_D\simeq \omega_X(D)\otimes \mathscr O_D. $$ If in addition $\omega_X\simeq \mathscr O_X$, then this implies that $$ \deg \omega_D = D^2. $$ If $D^2<0$, then this means that $\chi(\omega_D)=-1$ and by Riemann-Roch $$ 0\leq p_a(D)= \deg\omega_D +1 - \chi(\omega_D)= D^2 +2.$$ In particular, $$D^2\geq -2,$$ and equality holds if and only if $p_A(D)=0$ which holds if and only if $$D\simeq \mathbb P^1.$$

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