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I seem to remember that a K3 surface with big Picard rank always has smooth rational curves.

This question is equivalent to the following question about integral quadratic lattices. Let us call a vector in a lattice ($-2$)-vector if its square is $-2$. Consider a primitive, non-degenerate, indefinite lattice in $H^2(K3, {\Bbb Z})$, which has rank $\geq d$. Does it follow that it always has integer ($-2$)-vectors? I think I can prove that there exists an indefinite sublattice of rank $11$ which has no ($-2$)-vectors. Is it true that an indefinite sublattice of rank $>11$ always has ($-2$)-vectors? Is there any sharp bound on ranks of sublattices not admitting ($-2$)-vectors? I would be very grateful for any answers or references to papers where something similar was considered.

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    $\begingroup$ the question is, for which $d$ we can be sure there are no -2 vectors? What about $d=12$? What about $d=20$? $\endgroup$
    – Misha Verbitsky
    Commented Jan 31, 2023 at 22:56
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    $\begingroup$ Did you look at McMullen’s nonprojective K3 surfaces associated to Salem numbers? I will look now to see their ranks . . . $\endgroup$
    – Jason Starr
    Commented Feb 1, 2023 at 1:01
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    $\begingroup$ . . . The rank is zero. $\endgroup$
    – Jason Starr
    Commented Feb 1, 2023 at 1:05
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    $\begingroup$ For projective K3s, there are -2 curves whenever the Picard rank is at least 12. See, for example, Huybrechts's Lectures on K3 surfaces, Chapter 14, Corollary 3.8. $\endgroup$
    – Samir Canning
    Commented Feb 1, 2023 at 7:09
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    $\begingroup$ Jason: for McMullen's nonprojective K3 surfaces, the Picard lattice is either degenerate or negative definite, and such lattices are easy to construct in any rank. $\endgroup$
    – Misha Verbitsky
    Commented Feb 1, 2023 at 9:04

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