When Big cone= Ample cone ?

Let $X$ be a complex projective variety (you may assume it is smooth, or has mild singularities, is simply connected or any other assumption which makes you happy)

Let $Big(X)$ and $Amp(X)$ be the cone of big and ample divisors respectively. Both are open cones in $N^1(X)$.

What can we say if we know they are equal? Note that in general $Amp(X) \subseteq Big(X)$ and the former is smaller when there are effective divisors which are not nef.

• Dear Mohammad, I don't have a general answer to your question: I just wanted to mention that one situation in which Amp=Big is the case of abelian varieties (of any dimension). In that case, every effective divisor is nef because of translations, so the nef cone and (pseudo-)effective cone have equal interiors: that is to say, Amp=Big. – user5117 Jun 20 '11 at 19:08
• That is exactly what I feel. I expect that for high Picard numbers and under some additional condition, all example should be abelian varieties or abelian variety fibration or may be a quotient of them. – Mohammad Farajzadeh-Tehrani Jun 21 '11 at 13:12
• Thinking a little more, what I said works just as well for any homogeneous variety. Do those fit into your classification? – user5117 Jun 21 '11 at 13:51
• I don't have any classification in mind, I am looking for one. – Mohammad Farajzadeh-Tehrani Jun 21 '11 at 15:23
• Yes, I was just referring to your statement in the previous comment: "I expect that for high Picard numbers and under some additional condition, all example should be abelian varieties or abelian variety fibration or may be a quotient of them." – user5117 Jun 21 '11 at 15:30

If you consider the case of smooth surfaces then $Big(X)=Amp(X)$ if and only if $X$ has no curves with negative self-intersection. The point is that for surfaces the closers of these two cones are dual.
• Yes, for the surfaces the answer is easier to understand and somehow (may be not precisely) such surfaces are minimal surfaces. Things get more difficult and more interesting in dimension three. The things I like to know is whether this condition means $X$ is minimal? How big the Picard number can be in this case? What if $X$ is also smooth? – Mohammad Farajzadeh-Tehrani Jun 20 '11 at 17:11
• For example, if $X$ is Calabi-Yau 3-fold, this would imply that the boundary of ample cone is a subset of cubic cone (similar to dimension 2 for which boundary of ample cone would lie on conic cone) and as far as I now there is no known example of such C.Y of Picard number bigger than two. – Mohammad Farajzadeh-Tehrani Jun 20 '11 at 17:21
• As soon as $X$ admits a proper morphism $f:X\to Y$ which is generically finite but not finite, then $Big(X)\neq Amp(X)$. Indeed, for any ample line bundle $L$ on $Y$, $f^*L$ is big, but not ample. – ACL Jun 20 '11 at 21:29
• @Mohammad The Picard number can be arbitrarily large (without the CY) condition: to get Picard number $n$ it suffices to consider a generic complete intersection of large degree (and dimension at least $2$) in $(\mathbb{P}^1)^n$. – naf Jun 21 '11 at 10:28