Positivity question on K3 surfaces

Question

Let $X$ be a smooth projective complex K3 surface and $L, D$ two effective divisors, $L^2\geq0$ and $D^2\geq0$.

(Q1). do we have $L\cdot D\geq0$ ?

If either one has positive self-intersection, the answer is yes by index theorem. Assume $L^2=D^2=0$. If either one is basepoint free then it is a multiple of an elliptic curve and so the answer is yes.

It remains to consider the case when $L$ and $D$ are both not basepoint free.

I did not manage to show this, so I tried to construct a simple counterexample where $L$ and $D$ both share the same rational curve as fixed part, but I could not find any. Is there any?

Answer 1

Let me prove that $(L,D)\geq 0$.

Consider the set of all (1,1)-classes which satisfy $\eta^2\geq 0$. It is a union of two components $P_+$ and $P_-$, intersecting in 0. If $\eta, \eta'$ are in the same component ($P_+$ or $P_-$), they intersect non-negatively, by Hodge index theorem. Let $P_+$ be a component containing a Kahler class. It remains to prove that any effective class that satisfies $\eta^2=0$ lies in $P_+$.

However, for each $a\in P_-$ and $b\in P_+$, one has $(a,b) \leq 0$, because vectors in $P_+$ intersect positively by Hodge index theorem. Therefore, an intersection of any class in $P_-$ and the Kaehler class is non-positive. Therefore, $P_-$ does not contain effective classes.