Past exam of the mathematics course of the University of Cambridge 2005 iii Paper 21 4 c Solution Created 2026-10-03 Updated 2026-10-14
Let with the cup-product pairing of signature . For a K3 surface, the Hodge numbers are . A nonzero holomorphic two-form spans , and its wedge products giveThe remaining filtration is forced by orthogonality:Thus the usual K3 period domain and its quadric compactification areThe quadric is smooth and has complex dimension . Writing , its equations say that are orthogonal positive vectors of equal square. A period therefore determines an oriented positive two-plane in .
There is a polarization qualification: the full cup-product form is not a polarization in the strict sense of part (a). Its real space has signature , so it cannot satisfy the negative-definiteness requirement there. Fixing an ample integral class of square and requiring it to stay of type gives the primitive cohomology latticeIts genuine polarized Griffiths domain isIts compact dual is the corresponding smooth quadric of complex dimension nineteen, and has two components, distinguished by the orientation of the positive two-plane. For ample-polarized geometric K3 moduli one also removes the loci with , , since an ample class cannot be orthogonal to an effective rational curve. Those exclusions concern the ample moduli locus; they are not part of the definition of the polarized Hodge domain.
If one wants a strict polarization on the full of a fixed projective surface, reverse the sign on its rational Kähler line and retain on the orthogonal complement. For example the integral formhas the required positivity and signature . It is a different form from the cup product. The twenty-dimensional full period quadric and the nineteen-dimensional primitive polarized domain should therefore be distinguished. The positivity and signature calculations above explain why these two domains describe different data.