For a variety $X/k$, consider the monoid $A$ of classes of ample line bundles in $NS(X)$. What does $A \otimes_\mathbf{Z} \mathbf{R} \subset NS(X)_\mathbf{R}$ look like?
If $X$ is Fano, that is, if $K_X$ is ample, then (the closure of) the ample cone is polyhedral. This follows from Kleiman's criterion saying that it is dual to the cone of effective curves, and from Mori's Cone Theorem saying, roughly, that for any $\epsilon > 0$, the part of the effective cone where $K_X > \epsilon$ is polyhedral. A convenient reference is http://en.wikipedia.org/wiki/Cone_of_curves. The varieties $G/P$ discussed in other answers are Fano. Even in the Fano case, the ample cone still need not be simplicial: toric varieties would furnish convenient counterexamples.
On the other hand, in general the ample cone need not be polyhedral at all. For example, if $C$ is an elliptic curve without complex multiplication, then the NéronSeveri group of $C \times C$ is generated by $C \times p$, $p \times C$, and the diagonal, and in terms of this basis the ample cone is given by $$a+b>0, \qquad b+c > 0, \qquad c+a>0, \quad\mbox{and} \quad ab + bc + ca >0: $$ a circular cone! See, for example, T. Bauer and C. Schulz, Seshadri constants on the selfproduct of an elliptic curve, J. Alg. 320 (2008) 29813005, Section 2.

Tiny quibble: I guess "the part of the effective cone where K_X > epsilon is polyhedral" should be "the part of the effective cone where K_X > epsilon is polyhedral". – user5117 Jun 7 '10 at 11:33


Rob Lazarsfeld pointed out an inaccuracy in my remarks above. Mori's Cone Theorem does not say that the part of the effective cone where K exceeds epsilon is polyhedral. Rather, it says that the effective cone is generated by the part where K is less than epsilon, together with finitely many additional rays. This implies the statement of the previous sentence when the NeronSeveri group has rank ≤ 3, but not otherwise. It also implies that the ample cones of Fano varieties are polyhedral, as stated. – Michael Thaddeus Jul 13 '10 at 15:52

To Michael Thaddeus, Thank you for a nice answer and comment, but I can't imagine that cone with finitely many extremal rays may not be polyhedral. Is there a counterexample or am I missing some obvious picture? – Moon Nov 5 '10 at 1:43

Dear Moon, the point (a little late) is that there are only finitely many extremal rays in any of the halfspaces where K is less than some chosen $\epsilon$, but there may be infinitely (even uncountably) many altogether. To visualise this, imagine taking a transverse "slice" of the cone of curves (as in the picture on p.60 of Lazarsfeld Vol. 1). Suppose that $rho(X)=4$, so this slice lives in 3space. Then we could have the following picture: this slice is the convex set generated by a circle in $K_X^\perp$ together with a single point in $K_X^{<0}$. This doesn't violate the Cone Theorem... – user5117 Dec 2 '10 at 10:49
I guess I should take this opportunity to mention a personal hobbyhorse: the Cone Conjecture of MorrisonKawamata. (See e.g. http://arxiv.org/abs/0901.3361 for precise statements, history, and known cases.)
To give a little background, note that (as mentioned in Michael Thaddeus' answer) for Fano varieties the nef cone (i.e. closure of the ample cone) is polyhedral, and spanned by rational rays. That's the nicest possible answer one can hope for.
Once we leave the Fano realm, however, things get worse. Already for CalabiYau varieties, the cone of curves can fail to be polyhedral, and even be a completely "round" cone. (This happens e.g. for Abelian surfaces of Picard number at least 3.)
That seems pretty bad, but the MorrisonKawamata conjecture predicts that there is still something we can say about the nef cone, in good cases such as CalabiYau varieties. The key observation is that the automorphism group Aut(X) of any variety acts on the nef or ample cone, and one can hope that this group action simplifies the picture somehow.
Morrison's form of the conjecture is roughly the following.
Cone conjecture: Let X be a CalabiYau variety. Then there is a rational polyhedral cone Π inside the nef cone Nef(X) which is a fundamental domain for the action of Aut(X) on Nef(X): in particular, Aut(X).Π = Nef(X).
(To give a more precise statement, one has to consider what happens with nonrational rays on the boundary on the nef cone. Those can never be the image of a rational ray, hence we have to throw out all points of the boundary which are not in the convex hull of the rational rays. There is also an issue of what "fundamental domain" means. But the abbreviated statement above gives the essential flavour, I think.)
So the conjecture predicts that although the nef cone may be very far from being a polyhedral cone, it is "generated" by a polyhedral cone when we take into account the action of automorphims (and, as mentioned, throwing away some boundary points.) If true, that would be a very satisfying description.
The conjecture has been generalised to CalabiYau fibre spaces and then klt CalabiYau pairs. In its most general form, it includes all Fano varieties, all CalabiYau varieties, and many in between.
For K3 surfaces, there is an interesting dichotomy: the cone of curves, and so dually the (closed) cone of ample divisors, is either (locally) polyhedral, or completely circular (has no polyhedral part at all). This was proved in Sándor Kovács's thesis (see "The cone of curves of a K3 surface", Math. Ann. 300). As far as I know, it still remains an amusing open problem whether there exist projective varieties whose cone of ample divisors is partially circular and partially polyhedral, although one expects that there should be plenty of such examples.
The shape of the ample cone can be very different, depending on the variety you are considering. In the preprint "Remarks on the cone of divisors" by Y. Kawamata, for example, you can see an unusual version of the Mori cone theorem that tells us something about the shape of the nef cone. The ample cone, as you very probably know, is the inner of the nef cone, so that they practically look the same.

Thanks for your response. Are there results on examples where the ample cone is $\mathbf{R}_{\geq 0}^\rho$? – TKe Jun 6 '10 at 17:00

Whenever the ample cone is finite rational polyhedral, adding all the primitive integer generators of the cone gives you an ample class which is invariant under automorphisms of your variety. Thus the whole automorphism group embeds in a projective linear group, so e.g. cannot be infinite discrete. – Balazs Jun 7 '10 at 15:27
Remember the ample cone is an open set in $NS(X)\otimes \mathbf{R}$ for the standard euclidean topology. Thus it cannot be $\mathbf{R}_{\geq 0}^\rho $ . But it can be $(\mathbf{R}^+)^{\rho}$. The simplest example is the projective space $\mathbb{P}^n$ over the complex numbers, where $\rho=1$ and the ample cone is simply $\mathbf{R}^+$.

Yes, but are there more examples than, e.g., complete intersections of dimension $\geq 3$? – TKe Jun 6 '10 at 17:17

1For flag manifolds $G/P$, the ample cone is of this form, with $\rho = rank(G')  rank(L')$, for $L$ a Levi subgroup of $P$. – Allen Knutson Jun 6 '10 at 18:04

...and as Damiano mentions in the special case of $({\mathbb P}^1)^\rho$, these are very rarely complete intersections. – Allen Knutson Jun 7 '10 at 2:41
Well, if you have a projective variety you can always find a basis of $NS(X)\otimes \mathbf{R}$ as an $\mathbf{R}$vector space only made by ample $\mathbf{R}$divisors. This implies that every projective variety (say over $\mathbf{C}$),if you suitably choose the basis of the vector space we are speaking about, has ample cone $(\mathbf{R}^+)^{\rho}$. [retracted in comment below]
The reference is in the first chapter of the book "Positivity in algebraic geometry I", by R. Lazarsfeld. I don't have the book now so that I cannot tell you exactly what is the paragraph, but I'm pretty sure it is in the first chapter and it is just an exercise. Does this answer your question?

1Forgive me, my last answer hasn't too much sense. The result on Lazarsfeld's book is true but there is no reason why it should imply the ample cone being $(\mathbf{R}+)^\rho$. It just implyes if the basis it good it contains $(\mathbf{R}+)^\rho$. I've benn too much rash. I don't know an example with the properties you're asking, I'm sorry. – Gianni Bello Jun 6 '10 at 17:55

1An easy example is $(\mathbb{P}^1)^\rho$: it has nef cone with exactly $\rho$ extremal rays, and, if $\rho \geq 3$, it is also not a complete intersection in projective space. Moreover, any projective variety with N\'eronSeveri group of rank two has the required property. – damiano Jun 6 '10 at 18:02