# signature of picard lattice

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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## 1 Answer

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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Thanks Sasha, any curve of genus 1 has zero self intersection. Doesnt your answer mean that I will never have a curve of self inter section 0 inside K3 surfaces? i.e., my question is why this curves gives a zero on the diagonal? –  user13559 Sep 16 '11 at 15:14
You will never have such curve AS an element of orthogonal basis. –  Sasha Sep 16 '11 at 17:59
Can you please explain a little bit? I might have an elliptic curve in a K3 surface which has self intersection 0. –  user13559 Sep 16 '11 at 19:49