Surfaces

Report on Meeting 6

Remark 1. We will use the following notational conventions:

• F is a field (for today, algebraically closed).

• X is a smooth projective minimal surface over F.

• K is the canonical divisor on X.

• D is any divisor on X.

• C is an effective divisor (or curve) on X.

• H is an ample divisor (or hyperplane section) on X.

Theorem 2. (Nakai-Moishezon Criterion) A divisor D on X is ample if and only if D^{2} > 0 and D.C > 0 for all irreducible curves C.

Proof: See [Hartshorne, V.1.10].

Lemma 3. If there exists a curve C with C.K < 0, then there exists an ample H with H.K < 0.

Proof: Write C = ∑ m_{i}C_{i} where the C_{i} are irreducible curves and the m_{i} > 0. For at least one of the components, we must have C_{i}.K < 0. So, we may assume that C is irreducible. If C^{2} < 0, then we have

C.(K + C) = C.K + C^{2} = 2g − 2 ≥ −2.

Since both summands are negative, we conclude that C^{2} = C.K = −1, and so the curve is exceptional of the first kind. This is impossible (by minimality), and therefore we must have C^{2} ≥ 0.

Now let H_{0} be an arbitrary ample divisor on X. We will apply the Nakai-Moishezon Criterion:

(nC + H_{0})^{2} = n^{2}C^{2} + 2n(C.H_{0}) + H_{0}^{2} > 0

and

(nC + H_{0}).C′ = n(C.C′) + H_{0}.C′ > 0

for all irreducible C′. So, we see that nC + H_{0} must be ample for all n > 0. However, (nC + H_{0}).K < 0 for n >> 0.

Corollary 4. If there exists a curve C with C.K < 0, then P_{n} = 0 for all n ≥ 1.

Proof: Recall that P_{n} = dim_{F} H^{0}(X, O_{X}(nK)), and that nonzero sections in this cohomology group give rise to effective divisors linearly equivalent to nK. By the lemma, there is an ample divisor H such that (nK).H = n(K.H) < 0 for all n ≥ 1. Since ample divisors intersect all effective divisors positively, this means that nK can never be linearly equivalent to an effective divisor. Thus, there are no global section of O_{X}(nK), and the plurigenera must all be 0.

Remark 5. Under these hypotheses, the Kodaira dimension is κ = −1, since the Kodaira ring (being the sum of the groups of global sections discussed in the previous proof) is zero.

Remark 6. In particular, we have that p_{g} = P_{1} = 0. From meeting 4, we recall that this forces Δ = 0 and s = q. So, the magic formula

10 + 12p_{g} = K^{2} + b_{2} + 8s + 2Δ

simplifies to just

10 = K^{2} + b_{2} + 8q.

Proposition 7. If there exists a curve C such that C.K < 0, and if K^{2} > 0, then q = 0.

Proof: Using the previous remark, we see that q must be either 0 or 1. If dim(Alb_{X}) = q = 1, then the Albanese mapping is a morphism from X onto an elliptic curve. Fibers and (multi-)sections of such a morphism are always linearly independent in the Néron-Severi group, and hence the existence of such a morphism forces b_{2} ≥ 2. But then the magic formula cannot be satisfied with K^{2} > 0.

Remark 8. We have shown so far that the hypotheses:

• There exists a curve C with C.K < 0, and

• The canonical divisor has positive self-intersection (K^{2} > 0),

force P_{2} = q = 0. In addition, we note that p_{g} = 0 and that 10 = K^{2} + b_{2}, so 1 ≤ K^{2} ≤ 9. It is a classical result of Castelnuovo that X is rational whenever P_{2} = q = 0. For the remainder of this section, we assume that P_{2} = q = 0 and K^{2} > 0.

Lemma 9. Let C be an effective divisor on a surface with p_{g} = q = 0. Then

dim |K + C| ≥ p_{a}(C) − 1.

Proof: On any surface, we can apply the Riemann-Roch Theorem to the divisor K + C. One gets

l(K + C) − s(K + C) + l(−C) = \frac{1}{2}(K + C)(K + C − K) + 1 − s + p_{g}.

Now using the genus formula on the effective curve C, one gets

l(K + C) − s(K + C) + l(−C) = p_{a}(C) − s + p_{g}.

Since C is effective, l(−C) = 0. The hypotheses now yield

l(K + C) ≥ p_{a}(C),

which is equivalent to the statement of the result.

Remark 10. Classically, the linear system |K + C| is called the adjoint of the linear system |C|.

Lemma 11. The anticanonical class −K has effective representatives.

Proof: By the Riemann-Roch Theorem for Surfaces, we have

l(−K) − s(−K) + l(2K) = \frac{1}{2}(−K).(−K − K) + 1 − q + p_{q} = K^{2} + 1.

Now use the fact that P_{2} = l(2K) = 0 to get the result.

Lemma 12. p_{a}(−K) = 1.

Proof: We take an effective divisor representing −K, and apply the general genus formula to get

2p_{a}(−K) − 2 = (−K).(−K + K) = 0.

Lemma 13. If C is an irreducible curve with |K +C| = ∅, then C is a nonsingular rational curve.

Proof: It is enough to show that the arithmetic genus is zero. But this follows immediately from Lemma 9.

Lemma 14. Let −K be an effective divisor. If it is not irreducible, then each component has arithmetic genus zero.

Proof: Let C be an irreducible component, and write −K = C + D, where D is effective and nonzero. Then |C + K| = | − K − D + K| = | − D| must be empty. Apply the previous lemma.

Proposition 15. If | − K| contains a reducible linear subsystem δ, then X is a rational surface.

Proof: Let B be the fixed component of δ, and write δ = B + δ_{1}, where δ_{1} is a linear system without fixed part. If δ_{1} is irreducible, take A to be a generic (and hence irreducible) member of this linear system. If δ_{1} itself is reducible, then it is either composite with an irreducible pencil or all cycles in δ_{1} are of the form p^{e}δ_{2} for some irreducible linear system δ_{2}. In the first case, take A to be a general member of the pencil, and in the second case, take it to be a general member of δ_{2}. In all cases, A is an irreducible curve, and it is a component of an effective reducible anticanonical cycle. By the previous lemma, A has arithmetic genus zero. We either have an irreducible linear system of positive dimension consisting of rational curves (in which case the surface is rational), or a pencil of rational curves, in which case the surface is ruled. In the latter case, since q = 0, it must in fact be a rational ruled surface.