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A nice property of $\mathbb P^n$ is:

Property 1: Two subvarieties $U,V$ such that $\operatorname{dim} U +\operatorname{dim} V \geq n$ always intersect.

(for example, any 2 curves in $\mathbb P^2$ intersect)

There are other smooth varieties $X$ when Properties 1 holds. For example, a sufficient condition is that the ranks of $\text{CH}^i_{num}(X)$ are $1$ for $i\leq n/2$. Here $n = \operatorname{dim} X$ and $\text{CH}^i_{num}(X)$ is the Chow group of codimension $i$ modulo numerical equivalences.

My question is whether some converse is true:

Question: Let $X$ be a smooth projective variety satisfying Property 1. Does that impose some upper bounds on the ranks of $\text{CH}^i_{num}(X)$ for $i\leq n/2$?

Let's assume we are over $\mathbb C$, but I am also interested in results over any ground fields. One can ask the same questions for the ranks of $\text{CH}^i_{hom}(X)$ (I think they are conjectured to be the same). The baby case is $i=1$, where the question asks if Property 1 tells us something about the rank of the Neron-Severi group of $X$.

I am aware that the question is a little vague (upper bound as function of what?), but that was because of my ignorance, so comments to improve the question are welcome.

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This is a very interesting question and I guess that a general answer is unknown, already in the case $i=1$. Let me just make the following

Remark. There exists no upper bound on $\textrm{rank } NS(X)$ which is independent on the dimension.

In fact, let us consider a complex Abelian variety $X$ of dimension $g$ such that $End_{\mathbb{Q}}(X)$ is a totally real number field of degree $g$ over $\mathbb{Q}$. These varieties do exist and the general one is simple, see [Birkenhake - Lange, Chapters 5 and 9].

Therefore it is known that

$\rho(X) =\textrm{rank } \textrm{NS}(X) = \textrm{rank } End^s_{\mathbb{Q}}(X)=g$,

where $End^s_{\mathbb{Q}}(X)$ denotes the subgroups of elements in $End_{\mathbb{Q}}(X)$ which are symmetric with respect to the Rosati involution.

On the other hand, in a simple Abelian variety any effective divisor is ample, so two effective divisors always intersect and $X$ satisfies Property 1.

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  • $\begingroup$ Dear Francesco, thanks, this is very interesting. So I gather from it that there is no known bounds even when X is a surface? $\endgroup$
    – Hailong Dao
    Commented Oct 10, 2010 at 14:44
  • $\begingroup$ And +1. Also, can you explain your last sentence a bit, why two effective divisors always intersect implies Prop 1? $\endgroup$
    – Hailong Dao
    Commented Oct 10, 2010 at 14:49
  • $\begingroup$ Dear Hailong, I do not know any bounds for surfaces, but maybe other people could give more precise answers. About Prop 1, since Abelian varieties have the group of translations, if you take two cycles $B$ and $B'$ of complementary dimension, you can always move $B$ in its rational equivalence class in such a way that it intersects $B'$ properly. The only exception is when $B$ is a translate of $B'$, which does not intersect $B'$... $\endgroup$
    – Francesco Polizzi
    Commented Oct 10, 2010 at 16:14
  • $\begingroup$ ...But then $B$ and $B'$ are contained in fibres of a morphism $f \colon X \to Y$. By adjunction formula, the fibres of $f$ must be Abelian subvarieties of $X$, so this is not possible when $X$ is simple. $\endgroup$
    – Francesco Polizzi
    Commented Oct 10, 2010 at 16:14

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