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-2 -> $-2$, and consistent hyphenation
LSpice
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K3 surfaces with no −2 curves

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 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.

Misha Verbitsky
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