Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be a smooth projective variety over an algebraically closed field $k$. Let $NS(X)$ denote the group $[Pic(X)/Pic^a(X)]\otimes_{\mathbb{Z}}\mathbb{Q}$, where $Pic^a(X)$ denotes the divisors which are algebraically equivalent to 0. If $X$ is a surface, then the pairing on $NS(X)$ given by $(D,D')\mapsto D\cdot D'$ is non-degenerate.

For higher dimensions, let $\Theta$ denote an ample divisor on $X$. Let $d$ be the dimension of $X$. Is it true that the pairing on $NS(X)$ given by $(D,D')\mapsto D\cdot D'\cdot \Theta^{d-2}$ is non-degenerate.

Is there a good reference for the above result.

share|improve this question

1 Answer 1

up vote 6 down vote accepted

I don't have know a reference off hand. So let me just give you a proof when $char\\, k=0$. I expect it's true in general.

We can assume without loss of generality that $k=\mathbb{C}$. Then via the exponential sequence and the Lefschetz $(1,1)$ theorem, $NS(X)_\mathbb{Q}$ (your $NS(X)$) can be identified the space of $(1,1)$ classes in $H^2(X,\mathbb{Q})$. Given a nonzero $D\in NS(X)_\mathbb{Q}$, there exists $D''\in H^{d-1,d-1}(X)\cap H^{2d-2}(X,\mathbb{Q})$ such that $D\cup D''\not=0$ by Poincare duality. Hard Lefschetz tells us that $D'' = \Theta^{d-2}\cdot D'$ for some $D'\in NS(X)_\mathbb{Q}$.

Here's a different argument which works over any algebraically closed field. Let $Y\subset X$ be surface given as intersection of $d-2$ general hyperplanes with respect to the embedding given by $N\Theta$, $N\gg 0$.

Claim: The restriction $NS(X)_\mathbb{Q}\to NS(Y)_\mathbb{Q}$ is injective.

Proof: We can see, from the Kummer sequence, that $NS(X)_\mathbb{Q}\subset NS(X)_\mathbb{Q_\ell}$ embeds into $\ell$-adic cohomology $H^2(X_{et},\mathbb{Q}_\ell)$. Therefore the claim follows from weak Lefschetz [Milne, Etale cohomology]

The pairing on $NS(X)_\mathbb{Q}$ is nonzero multiple of the restriction of the pairing on $Y$. The result now follows from the Hodge index theorem for surfaces [Hartshorne, Alg. Geom.]

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.