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Define a compact Kähler surface $X$ to be a K3 surface if $X$ is simply connected, $K_X \simeq \mathcal{O}_X$, and $h^{0,1}=0$. If $X$ is projective, then a theorem typically attributed to Bogomolov and Mumford, asserts that $X$ admits a rational curve. It has recently been shown that projective K3 surfaces admit infinitely many rational curves.

  1. Do non-projective K3 surfaces admit rational curves?
  2. Do non-projective K3 surfaces admit infinitely many rational curves?
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    $\begingroup$ A general (non algebraic) K3 surface admits no curve at all, so I suppose you ask for specific examples. The answer to 1. is yes: the Kummer surface associated to a non-algebraic complex torus has 16 (-2)-rational curves. $\endgroup$
    – abx
    Dec 17, 2022 at 8:15
  • $\begingroup$ @abx Thank you! $\endgroup$
    – AmorFati
    Dec 17, 2022 at 21:15

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Some of them do, and some don't.

Indeed, by global Torelli theorem, there is a K3 surface $X$ with $\mathrm{Pic}(X) = 0$. Such $X$ has no curves, in particular no rational curves.

On the other hand, there is a K3 surface $X$ such that $\mathrm{Pic}(X)$ is generated by a single class with square $-2$; such a class (up to sign) is represented by a smooth rational curve.

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  • $\begingroup$ Thank you! This is great! $\endgroup$
    – AmorFati
    Dec 17, 2022 at 21:15

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