# Intersection positivity for curves and surfaces

Let $X$ be a smooth complete variety over an algebraically closed field of dimension $\geq3$. Given a divisor $D_1$ on $X$ with $D_1 \cdot C>0$ for every curve $C \subset X$, and a divisor $D_2$ on $X$ satisfying $D_2^2 \cdot S>0$ for every surface $S \subset X$, does there exist a divisor $D$ on $X$ satisfying $D \cdot C>0$ and $D^2 \cdot S>0$ for every curve and surface respectively? I am willing to make any assumption on $X$, except that $X$ be projective.

As I understand it, since $D_1$ is nef, we have that $D_1^2 \cdot S\geq 0$, so even if for $m>>0$ we manage to have $(mD_1+D_2) \cdot C>0$ for every curve, the other requirement becomes $(mD_1+D_2)^2 \cdot S = m^2(D_1^2 \cdot S) + 2m(D_1\cdot D_2 \cdot S) + D_2^2 \cdot S >0$. The first term is non-negative, the last term is positive, but what needs to happen to ensure the middle term is non-negative also?

## 1 Answer

I am not sure whether your original question is true, but it seems to me that your proposed solution does not work. Here is why:

The main problem is that you know very little about $D_2$. Given what we know, it does not even have to be effective!

Example Let $X$ be a smooth complete variety with divisors $A,B$ such

• $A\cdot B=0$

• $(A^2+B^2)\cdot S>0$ for any surface $S$, and

• $B$ and $B^2$ are effective classes.

Then $D_2=A-B$ has the property that $D_2^2\cdot S>0$ for any $S$, but $D_1\cdot D_2\cdot B\cdot H <0$ for any $D_1$ divisor satisfying the criterion given in the question and an appropriate $H$ such that $B\cdot S$ is an effective surface and $B^2\cdot H$ is an effective curve.

To make this complete I should give an example that such an $X$ with $A,B$ exists. I think I can construct an example (a product of two surfaces blown up along a surface) satisfying these properties, but I don't know a simple one. In any case I think it is reasonable to expect that such an example exists which shows that $D_1\cdot D_2\cdot S$ is not necessarily non-negative even on a projective variety.

• Yes, you're right Sandor. It seems one cannot get around this without making some critical assumptions on $D_1 \cdot D_2$. Thank you. – Parsa Oct 28 '11 at 11:27