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A smooth, complex, projective surface, such that the canonical bundle is trivial and the irregularity is equal to zero is called a K3 surface. Recently I received feedback regarding work I had done. In my work I had assumed that if $ M $ was a variety which was not uniruled, then I assumed that it was possible to find an open neighborhood $ U \subseteq M $ such that $ U $ did not intersect any rational curve of $ M $ non-trivially.

I received an email from someone who had read this work in which they said that this was not always possible for a K3 surface.

I later found a paper by Fedor Bogomolov, Brendan Hasset and Yuri Tschinkel in which the authors of this paper proved that for a very general K3 surface with Picard group generated by a class of degree two, there are an infinite number of rational curves. So it seemed like the input I was given that it might not be possible to find an open sub-variety of a K3 surface which did not intersect any rational curves non-trivially could be true.

Let $ W $ be the set of points of a K3 surface $ S $ which are not contained in a rational curve and let $ \overline{W} $ be its closure. If $ W $ is not dense in $ S $ there should be a dominant, rational map $ \phi: \mathbb{P}^{1}_{\mathbb{C}} \times \operatorname{Mor}(\mathbb{P}^{1}_{\mathbb{C}},S) \dashrightarrow S $ since any point of $ S \setminus \overline{W} $ is contained in a rational curve. So there should be a curve $ C $ contained in $ \operatorname{Mor}(\mathbb{P}^{1}_{\mathbb{C}},S) $ such that $ \phi: \mathbb{P}^{1}_{\mathbb{C}} \times C \dashrightarrow S $ is generically finite and dominant. This would mean that $ S $ is uniruled. However, the genus of a uniruled variety over a field of characteristic zero vanishes, while the genus of a K3 surface is one. Can anyone tell me what the error is in this p"r"oof that for any K3 surface there is an open sub-variety which does not intersect a rational curve non-trivially?

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    $\begingroup$ Maybe someone told you that K3s can have a dense set of rational curves, but didn't say, or didn't mean to say, that the set was an open subvariety? $\endgroup$
    – Will Sawin
    Commented Aug 29 at 0:02
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    $\begingroup$ Anyway, it is better to say where you learned that from. $\endgroup$
    – Sasha
    Commented Aug 29 at 5:09
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    $\begingroup$ That is definitely not true over $\mathbb{C}$. Maybe they meant "entire curves" rather than "rational curves". It is true over fields of positive characteristic that there are unirational K3 surfaces. $\endgroup$
    – Jason Starr
    Commented Aug 29 at 9:57
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    $\begingroup$ Your argument in the last paragraph seems to assume that any dense subset of $S$ contains a nonempty open set. That is not true. $\endgroup$
    – Lazzaro Campeotti
    Commented Aug 29 at 13:00
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    $\begingroup$ Yes $W$ is not constructible. Over the complex numbers, the dense set of rational curves is a countable union of rational curves. Countable unions of closed subsets are not constructible in general so the complement $W$ is also not constructible. In particular, every Zariski open set meets infinitely many of these rational curves. This is a reflection of the fact that Zariski opens are very large (analogous say to the fact that every Zariski open set of $\mathbb{P}^1$ contains countably many points with integer coordinates so this set is dense even though it is a countable union of closeds). $\endgroup$
    – Dori Bejleri
    Commented Aug 29 at 14:17

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The OP requested that I add my comment as an answer:

Your argument in the last paragraph seems to assume that any dense subset of $S$ contains a nonempty open set. That is not true.

Specifically in this case, if $S$ is a $K3$ surface with infinitely many rational curves, and if $W \subset S$ is the complement of the union of all rational curves, then $W$ is dense but it does not contain any nonempty open set. The last claim is easy to see: If $U \subset W$ were a nonempty open set, then the complement $S \setminus U$ would be a proper closed subset of $S$, therefore of dimension at most 1, and since a closed set has finitely many components, $S \setminus U$ could contain at most finitely many curves.

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Rational curves are dense on K3 for all K3 outside of a Baire set (countable union of closed, nowhere dense). Here is the reference: https://arxiv.org/abs/1004.5167, Density of Rational Curves on K3 Surfaces, Xi Chen, James D. Lewis

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