In describing part of the geometry of the cone of curves for an algebraic surface $S$, we need to find $(-1)$ curves within $S$. Once we've done that, then we can say that the "negative" part of the cone of curves has as many extremal rays as $(-1)$ curves. Here I am using the cone theorem: $$\overline{\mathrm{NE}}(S)=\overline{\mathrm{NE}}(S)_{K_S\geq 0}+\sum_i \mathbb{R}^+[C_i]$$ where $C$ are such negative self-intersection rational curves.

However, is there an intuitive argument showing that if we have a curve of negative self-intersection, then such a curve is going to generate a extremal ray in the cone of curves?

What is known about this "positive" part $\overline{\mathrm{NE}}(S)_{K_S\geq 0}$ of the cone describe in the theorem?

  • $\begingroup$ That looks like an interesting theorem! Could you provide a link for those new to it, please? $\endgroup$ Mar 2, 2010 at 9:52
  • $\begingroup$ Sure, Wikipedia has two excellent references en.wikipedia.org/wiki/Cone_of_curves $\endgroup$ Mar 2, 2010 at 15:42
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    $\begingroup$ Your second sentence is not right as stated. Every rational curve of negative self-intersection yields an extremal ray, not only (-1)-curves. $\endgroup$ Apr 14, 2010 at 13:50
  • $\begingroup$ That's right...got it. Thks! $\endgroup$ Apr 14, 2010 at 14:29

3 Answers 3


EDIT: We may assume that the Picard number is at least two, as otherwise the cone is simply a ray generated by any effective curve. In particular, every effective curve is extremal. I will also assume that "curve" means "effective curve". (This edit was prompted by Damiano's comment that is now (sadly) deleted. It was a useful contribution.)

A curve on a surface is simultaneously a curve and a divisor and assuming the surface is smooth or at least $\mathbb Q$-factorial, then the curve, as a divisor, induces a linear functional on $1$-cycles. This works better if the surface is proper, so let's assume that.

So, if $C$ is such a curve, then the corresponding linear function on the space where $NE(S)$ lives is best represented by the hyperplane on which it vanishes and remembering which side is positive and which one is negative.

If $C$ is reducible, then it may have negative self-intersection, but it is not extremal. For an example, blow up two separate points on a smooth surface and take the sum of the exceptional divisors. My guess is that you meant irreducible, so let's assume that.

Now we have $3$ cases:

  1. $C^2>0$. In this case $C$ is in the interior of the cone and it cannot be extremal, can't even be on the boundary (Use Riemann-Roch to prove this).

  2. $C^2=0$. Since $C$ is irreducible, it follows that it is nef and hence a limit of ample classes, so it is effective, but as Damiano pointed out I have already assumed that. (It is left to the reader to rephrase this if $C$ is assumed to be nef instead of effective). In this case the hyperplane corresponding to $C$ as a linear functional is a supporting hyperplane of the cone, intersecting it at least in the ray generated by $C$. So $C$ is definitely on the boundary, but it may or may not be extremal depending on the surface. For example any curve of self-intersection $0$ on an abelian surface is extremal, but for instance a member of a fibration that also has reducible fibers is not extremal despite being irreducible. For the latter think of a K3 surface with an elliptic fibration that has some $(-2)$-curves contained in some fibers.

  3. $C^2<0$. If $C$ is effective, then $C\cdot D>0$ for any irreducible curve $D\neq C$. This means that $C$ and all other irreducible curves lie on different sides of the hyperplane corresponding to $C$ as a linear functional, so the convex cone they generate must have $C$ generating an extremal ray.

Observe that we did not use the Cone Theorem. In fact one gets a different "cone theorem" this way:

Theorem Let $S$ be a smooth projective surface $H$ an arbitrary ample divisor on $S$ and let $$ Q^+=\{\sigma\in N_1(S) \vert \sigma^2 >0, H\cdot\sigma \geq 0 \} $$ be the "positive component" of the interior of the quadric cone defined by the intersection pairing. Then $$ \overline{NE}(S) = \overline{Q^+} + \sum_{C^2<0} \mathbb R_+[C] $$

There is also one for $K3$'S, using the above notation:

Theorem Let $S$ be a smooth algebraic K3 surface and assume that its Picard number is at least $3$. (If the Picard number is at most $2$, then there are not too many choices for a cone). Then one of the following holds:

(i) $$ \overline{NE}(S) = \overline{Q^+}, or $$

(ii) $$ \overline{NE}(S) = \overline{\sum_{C\simeq \mathbb P^1, C^2<0} \mathbb R_+[C]}. $$ The two cases are distinguished by the fact whether there exists a curve in $S$ with negative self-intersection. If the Picard number is at least $12$, then only (ii) is possible.

For proofs and more details, see The cone of curves of a K3 surface. (There is also a newer version which is less detailed, but works in arbitrary characteristic. See here or here.)

  • $\begingroup$ Hi damiano, yes, you are right, but... 1) The question is not interesting otherwise. If the Picard number is one, the cone is just a ray and every effective curve is extremal (belonging to the same extremal ray). But I should have said so. 2) I suppose you are referring to the comment that effective is not needed. When I wrote that I was going to write something different and did not go back to fix it. I will edit it now. $\endgroup$ Oct 19, 2010 at 9:28
  • $\begingroup$ I agree that I was being very picky, and I had given you my vote! I have also erased my previous comment, as no longer relevant. d $\endgroup$
    – damiano
    Oct 19, 2010 at 9:47
  • $\begingroup$ could you update the link to the paper. it does no longer works $\endgroup$
    – user267839
    Dec 27, 2020 at 18:33
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    $\begingroup$ @katalaveino: Apparently, Springer changed their URLs and didn't think about backward compatibility. I updated the link. I also added the title of the paper, so if this happens again, then one will still be able to find it. $\endgroup$ Dec 29, 2020 at 0:59

One further comment (which I can't add as a comment due to lack of reputation): in the case where S is the blowup of the projective plane in $r \geq 10$ very general points (for smaller r, the K-nonnegative part of the cone of curves is either empty or a single ray), Nagata's conjecture on curves predicts the following upper bound: if there exists a curve C whose projection to the plane has degree d and passes through the blown-up points p_1,...,p_r with multiplicities m_1,...,m_r, then we have

$$ d > \frac{1}{\sqrt r} \ \sum_{i=1}^r m_i$$

To take a very simple case, this says that there cannot exist a cubic curve passing through 10 very general points in the plane (which of course we already know), or equivalently that the cone of curves of the blowup doesn't contain the K-positive vector 3L-L_1-...-L_10. (Here L is the class in N^1(S) of the line in the plane, and L_i the class of the exceptional curve of the blowup of p_i).


A curve (irreducible reduced divisor) C with negative selfintersection is always an extremal ray. To see it, first observe that C is the only effective divisor in its complete linear system |C| (if D~C, then C·D=C²<0 so C is a component of D, and therefore D=C), and for the same reason nC is the only effective divisor in |nC|. Now, if C were not extremal it would be possible to express it as the sum of two things in the Mori cone, which means you could write nC~D+E for some n, with D and E nontrivial effective divisors. This is a contradiction.

On the positive part, I believe the most difficult case is rational surfaces, see a recent preprint by Tommaso de Fernex arXiv:1001.5243. I also think in Lazarsfeld's book (Positivity in Algebraic Geometry) there are a few examples, including ones in which the positive part is "round" (part of its border is defined by the quadratic equation $C²\ge 0$).


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