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I think I have seen more than one reference in which the cone of numerically effective curves can be 'not polyhedral', i.e. with an infinite number of extremal rays

I cannot remember where I read that the usual example is the blow-up of the plane in the $9$ points of intersection of $2$ general cubics. This should give an infinite number of $-1$-curves, but I don't manage to see why it should be infinite! As far as I see the strict transform $C$ of any curve in the pencil would give me $C^2=\pi*(C)+ \sum E_i^2=9-9=0$ where $E_i$ are the exceptional divisors.

What am I missing?

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    $\begingroup$ Have you consulted Lazarsfeld's book (Positivity in Algebraic Geometry, Volume I)? This example is almost certainly in there. $\endgroup$
    – Michael Joyce
    Commented Jan 19, 2012 at 18:31
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    $\begingroup$ It is also possible to exhibit infinitely many exceptional curves on $X$ by looking at the system of Diophantine equations $D^2=-1$ $-K.D=1$. One possibility is $$ 3k(k+1)\pi^*L-k(k+2)E_1-k^2E_2-k(k+1) (E_3+\ldots+E_9) \qquad k=0,1,2,\ldots $$ $\endgroup$
    – J.C. Ottem
    Commented Jan 20, 2012 at 0:13
  • $\begingroup$ Unfortunately my library has a lock on that book. I should just 'recall' it. Thanks for letting me know :) $\endgroup$
    – Jesus Martinez Garcia
    Commented Jan 23, 2012 at 11:10

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Yes, this is the standard example of a variety whose cone of curves has infinitely many extremal rays. (A reference is Koll\'ar--Mori, p.22).

To see why there are infinitely many (-1)-curves: each of the 9 points you blow up gives a (-1)-curve E_i, as you know. But the E_i are also sections of the elliptic fibration determined by the pencil. So one can use the group structure of the generic fibre (which is an elliptic curve over the function field k(P^1)) to translate any of the E_i to any other section of the fibration; since the action is by automorphisms of the surface, the other sections must be (-1)-curves also. As long as there are infinitely many sections (which is guaranteed by the assumption that the cubics are general) one gets infinitely many (-1)-curves this way.

Edit: It's worth mentioning that there are other examples of surfaces with non-finitely generated cone of curves, which differ from the example above in an interesting way. For example, suppose X is a K3 surface which has Picard rank at least 3 and no (-2) curves. Then the (closed) cone of curves of X consists of all classes in N^1(X) satisfying x^2 ≥ 0 and x.H ≥ 0 for any fixed ample class H. This is the standard "round cone" in N^1, with uncountably many extremal rays. (A similar example is obtained by taking X to be an abelian surface with Picard rank at least 3.)

As mentioned in the comments, Volume 1 of Lazarsfeld's great book Positivity in Algebraic Geometry is the best reference for these kinds of questions.

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    $\begingroup$ Maybe I should remark: in particular, the (-1)-curves are not coming from curves in the pencil! Indeed, one can see that without doing any calculation at all, since a (-1)-curve is always smooth rational, while a curve in the pencil is a cubic in P^2. $\endgroup$
    – user5117
    Commented Jan 19, 2012 at 18:44
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    $\begingroup$ You can also construct the (-1)-curves by looking at the (pre)image by quadratic Cremona transforms (based at subsets of the nine points) of the E_i. You get (a line through 2 of the points), (a conic through five of the points), (a quartic with three double points going through five more), etc. All of these are (-1)-curves on the blowup, and it is not difficult to see that they have arbitrarily large degree. $\endgroup$
    – quim
    Commented Jan 20, 2012 at 21:50
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    $\begingroup$ quim: Yes, and that point of view has the advantage that it doesn't require the points to be the base locus of a pencil of cubics. This shows that the blowup of P^2 in any sufficiently general set of 9 points has infinitely generated cone of curves. $\endgroup$
    – user5117
    Commented Jan 20, 2012 at 22:01
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    $\begingroup$ Artie, since you mentioned K3's one should perhaps point to the more interesting case when $X$ has Picard rank at least 3 and does contain (-2)-curves and has an infinite automorphism group, then the cone is infinite polyhedral as in the case of $\mathbb P^2$ blown-up at 9 points, but the extremal rays are (the) (-2) curves. For instance the cone of curves of a K3 surface with Picard number 20 is always like that. For more see springerlink.com/content/r654368027w422r2 $\endgroup$
    – Sándor Kovács
    Commented Jan 21, 2012 at 0:57
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    $\begingroup$ Oh, wait, I see, they are the ones that quim mentions. In the case of a quartic passing through 5 marked points, 4 of them with multiplicity 2 I get infinite number of such curves and the proper transforms are -1 curves. Thanks for all the answers and sorry it took me so long to reply! $\endgroup$
    – Jesus Martinez Garcia
    Commented Apr 2, 2012 at 9:10

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