Skip to main content
MathOverflow

Return to Question

added 17 characters in body
Source Link
Misha Verbitsky
Misha Verbitsky
  • 9.2k
  • 1
  • 28
  • 48

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 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.

-2 -> $-2$, and consistent hyphenation
Source Link
LSpice
LSpice
  • 12.9k
  • 4
  • 45
  • 69

K3 surfaces with no -2−2 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$-2$)-vector if its square is -2$-2$. 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$-2$)-vectors? I think I can prove that there exists an indefinite sublattice of rank 11$11$ which has no -2($-2$)-vectors. Is it true that an indefinite sublattice of rank $>11$ always has ($-2$)-2 vectorsvectors? Is there any sharp bound on rankranks of sublattices not admitting ($-2$)-2 vectorsvectors? I would be very grateful for any answers or references to papers where something similar was considered.

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

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.

Source Link
Misha Verbitsky
Misha Verbitsky
  • 9.2k
  • 1
  • 28
  • 48

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