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.

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.

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 rank of sublattices not admitting -2 vectors? I would be very grateful for any answers or references to papers where something similar was considered.

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.