Let's assume $X$ is a K3 surface. Why the signature of the picard lattice is $(1,\rho(X)-1)$? We know that (by Hodge index theorem) if we have a curve $D$ in $X$ with self intersection $D^2>0$ then any other curve $D_1$ with $D.D_1=0$ should have $D_1^2\leq 0$. Why $D_1^2=0$ can not happen? If $D_1^2=0$ what happens to the signature of Picard lattice?
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If you diagonalize a quadratic form and a zero on the diagonal appears then the corresponding basis element is contained in the kernel of the form, so the form is degenerate. But for K3 surface Picard coincides with Neron-Severi group which injects into the second cohomology and the intersection form is nondegenerate by Poincare duality. |
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