Our first result will help us locate extremal curves on the closed cone of curves of a smooth projective surface.
Proposition 5.9 Let X be a smooth projective surface.
a) The class of an irreducible curveC with (C2)≤0 is in∂NE(X).
b) The class of an irreducible curveC with(C2)<0 spans an extremal ray of NE(X).
c) If the class of an irreducible curve C with (C2) = 0 and (KX ·C) < 0 spans an extremal ray of NE(X), the surface X is ruled over a smooth curve,C is a fiber andX has Picard number2.
d) If r spans an extremal ray of NE(X), either r2 ≤ 0 or X has Picard number1.
e) If r spans an extremal ray of NE(X) and r2 < 0, the extremal ray is spanned by the class of an irreducible curve.
IfX has Picard number at least 3, the cone theorem and the lemma imply that anyKX-negative extremal ray is spanned by an exceptional curve.
Proof. Assume (C2) = 0; then [C] has nonnegative intersection with the class of any effective divisor, hence with any element of NE(X). LetH be an ample divisor onX. If [C] is in the interior of NE(X), so is [C] +t[H] for allt small enough; this implies
0≤(C·(C+tH)) =t(C·H) for allt small enough, which is absurd since (C·H)>0.
Assume now (C2)<0 and [C] =z1+z2, wherezi is the limit of a sequence of classes of effectiveQ-divisorsDi,m. Write
Di,m=ai,mC+Di,m0
with ai,m ≥ 0 and Di,m0 effective with (C·D0i,m) ≥ 0. Taking intersections with H, we see that the upper limit of the sequence (ai,m)m is at most 1, so we may assume that it has a limit ai. In that case, ([D0i,m])m also has a limit zi0 = zi −ai[C] in NE(X) which satisfies C·z0i ≥ 0. We have then [C] = (a1+a2)[C]+z10+z02, and by taking intersections withC, we geta1+a2≥1.
But
0 = (a1+a2−1)[C] +z10 +z02
and sinceX is projective, this impliesz10 =z02= 0 and proves b) and a).
Let us prove c). By the adjunction formula (§5.1.1), (KX·C) =−2 andC is smooth rational.
For any divisorD onX such that (D·H)>0 andm > (K(D·H)X·H), the divisor KX−mD has negative intersection withH, hence cannot be equivalent to an effective divisor. It follows thatH0(X, KX−mD) vanishes, hence
H2(X, mD) = 0 (5.5)
by Serre duality. It follows that form sufficiently large, H2(X, mC) vanishes, the Riemann-Roch theorem yields
h0(X, mC)−h1(X, mC) =m+χ(X,OX).
In particular, we may assumeh0(X,(m−1)C)< h0(X, mC). It follows that the linear system|mC|has no base-points: the only possible curve of base-points is C, but it is not, becauseh0(X,(m−1)C)< h0(X, mC), and there are no isolated base-points since (C2) = 0. It defines a morphism fromX to a projective space whose image is a curve. Its Stein factorization yields a morphism fromX onto a smooth curve whose general fiber is numerically equivalent to C hence is rational. All fibers are irreducible sinceR+[C] is extremal. This proves c).
Let us prove d). Let D be a divisor on X with (D2)>0 and (D·H)>0.
Formsufficiently large,H2(X, mD) vanishes by (5.5), and the Riemann-Roch theorem yields
h0(X, mD)≥ 1
2m2(D2) +O(m).
Since (D2) is positive, this proves thatmDis linearly equivalent to an effective divisor formsufficiently large, henceD is in NE(X). Therefore,
{z∈N1(X)R|z2>0 , H·z >0} (5.6) is contained in NE(X); since it is open, it is contained in its interior hence does not contain any extremal ray of NE(X), except ifX has Picard number 1. This proves d).
Let us prove e). Expressr as the limit of a sequence of classes of effective Q-divisorsDm. There exists an integerm0such thatr·[Dm0]<0, hence there exists an irreducible curveCsuch thatr·C <0. Write
Dm=amC+Dm0
witham≥0 andDm0 effective with (C·Dm0 )≥0. Taking intersections with an ample divisor, we see that the upper limit of the sequence (am) is finite, so we may assume that it has a nonnegative limit a. In that case, ([Dm0 ]) also has a limitr0=r−a[C] in NE(X) which satisfies
0≤r0·C=r·C−a(C2)<−a(C2)
It follows that a is positive and (C2) is negative; since R+r is extremal and r=a[C] +r0, the classrmust be a multiple of [C].
Example 5.10 (Abelian surfaces) An abelian surface is a smooth projec-tive surfaceX which is an (abelian) algebraic group (the structure morphisms are regular maps). This implies that any curve on X has nonnegative self-intersection (because (C2) = (C·(g+C))≥0 for anyg∈X). Fixing an ample divisorH onX, we have
NE(X) ={z∈N1(X)R|z2≥0, H·z≥0}
Indeed, one inclusion follows from the fact that any curve on X has nonneg-ative self-intersection, and the other from (5.6). By the Hodge index theorem
(Exercise 5.7.2)), the intersection form on N1(X)R has exactly one positive eigenvalue, so that when this vector space has dimension 3, the closed cone of curves ofX looks like this.
NE(X)
H <0 H >0 z2≥0
0
H = 0
The effective cone of an abelian surfaceX
In particular, it is not finitely generated. Every boundary point generates an extremal ray, hence there are extremal rays whose only rational point is 0:
they cannot be generated by the class of a curve onX.
Example 5.11 (Ruled surfaces) LetXbe aP1-bundle over a smooth curve B of genusg. By Proposition 5.5, NE(X) is a closed convex cone inR2 hence has two extremal rays.
LetF be a fiber; sinceF2= 0, its class lies in the boundary of NE(X) by Lemma 5.9.a) hence spans an extremal ray. Let ξ be the other extremal ray.
Lemma 5.9.d) impliesξ2≤0.
• Ifξ2<0, we may, by Lemma 5.9.d), take forξ the class of an irreducible curveC onX, and NE(X) =R+[C] +R+[F] is closed.
• If ξ2 = 0, decompose ξ in a basis ([F], z) for N1(X)Q as ξ =az+b[F].
Then ξ2 = 0 implies that a/b is rational. However, it may happen that no multiple ofξ can be represented by an effective divisor, in which case NE(X) isnotclosed.
For example, wheng(B)≥2 and the base field isC, there exists a rank-2 lo-cally free sheafE of degree 0 onB, with a nonzero section, all of whose symmet-ric powers arestable.1 For the associated ruled surfaceX =P(E), letEbe a di-visor class representingOX(1). We have (E2) = 0 by (5.3). We first remark that
1For the definition of stability and the construction ofE, see [H2],§I.10.
H0(X,OX(m)(π∗D))vanishes for anym >0and any divisorDonB of degree
≤0. Indeed, this vector space is isomorphic to H0(B,(SymmE)(D)), and, by stability ofE, there are no nonzero morphisms fromOB(−D) to SymmE.
The cone NE(X) is therefore contained in R+[E] +R+∗[F], a cone over which the intersection product is nonnegative. It follows from the discussion above that the extremal ray of NE(X) other thanR+[F] is generated by a class ξwithξ2= 0, which must be proportional toE. Hence we have
NE(X) =R+[E] +R+∗[F]
and this cone is not closed. In particular, the divisorE is not ample, although it has positive intersection with every curve onX.