# Elliptic K3 surface with a section of infinite order

I apologise for the basic question; I am reading Huybrecht's Lecture Notes on K3 surfaces, and on p.257 it is mentioned an example of K3 surface with infinitely many smooth rational curves. Precisely, suppose $X$ has an elliptic fibration with a section $C$ of infinite order. Then the multiples $nC$ (wrt addition law on the elliptic curve) are also sections, thus smooth rational curves. Now, being sections, they must be disjoint. But this is what seems puzzling to me: I thought that two disjoint smooth rational curves on a K3 surface should give independent classes in the Neron-Severi group, and so how come there can be infinitely many while $\rho(X)$ is bounded? I guess I am missing something.

• To make Sergey's answer even more concrete, try an example such as the Fermat quartic in $\mathbf P^3$. Here an elliptic fibration is given by projecting away from a line on the surface; any other line that is disjoint from the projection centre will give a section. But you can easily find two such lines that are not disjoint from each other. Apr 4, 2017 at 15:53
• @potentiallydense that is quite illuminating, thank you. Please, turn your comment into an answer Apr 4, 2017 at 16:54

To make Sergey's answer even more concrete, try an example such as the Fermat quartic in $\mathbf P^3$. Here an elliptic fibration is given by projecting away from a line on the surface; any other line that is disjoint from the projection centre will give a section. But you can easily find two such lines that are not disjoint from each other.
Here's another way to get a K3 surface with infinitely many smooth rational curves. Let $S$ be the intersection of a $(1,1)$-form and a $(2,2)$-form in $\mathbb P^2\times\mathbb P^2$, so $S$ will be a K3 surface provided it is nonsingular. The two projections $S\to\mathbb P^2$ are degree 2, so give involutions $i_1$ and $i_2$ of $S$. The composition $f:=i_1\circ i_2$ is an automorphism of $S$ of infinite order. Now you just need to choose your forms so that you can find (a reasonably generic) smooth rational curve $C$ on $S$, then the images $f^n(C)$ for $n\in\mathbb Z$ give infinitely many such curves.
As noted by others, these curve will intersect, and it's a nice exercise to compute the intersection indices. The following may help. Let $D_1=H\times\mathbb P^2$ and $D_2=\mathbb P^2\times H$ be in $\text{Pic}(S)$. (Generally, this is the full Picard group.) Let $\alpha=2+\sqrt3$, and let $E^+=\alpha D_1-D_2$ and $E^-=-D_1+\alpha D_2$ be in $\text{Pic}(S)\otimes\mathbb R$. Then $$f^*E^+ \sim \alpha^2 E^+ \quad\text{and}\quad f^*E^-\sim\alpha^{-2}E^-.$$