COMPLEX ALGEBRAIC SURFACES CLASS 16

RAVI VAKIL

Contents

1. Castelnuovo’s Theorem 1 1.1. Motivation: Minimal rational surfaces 1 1.2. Motivation Luroth’s theorem (in characteristic 0) 2 2. Proof of Castelnuovo’s criterion (part 1) 2

On board beforehand:

• Useful trick. |D| ≠ ∅ (i.e. h^{0}(D) > 0), C irreducible, C^{2} ≥ 0 implies DC ≥ 0.

• Genus formula. 2g(C) − 2 = C · (K_{S} + C).

• Riemann-Roch: χ(D) = χ(O) + \frac{1}{2}D · (D − K).

• Riemann-Roch: In the case when h^{1}(O) = 0 and h^{0}(K − D) = 0, we have h^{0}(D) ≥ 1 + \frac{1}{2}D · (D − K) (with equality iff h^{1}(D) = h^{2}(O) = 0).

1. Castelnuovo’s Theorem

We saw how tricky it was to show that a surface is rational.

Theorem: Castelnuovo’s Rationality Criterion. Let S be a surface with q = P_{2} = 0. Then S is rational.

Reminder. q = h^{1}(S, O_{S}) = h^{0}(S, Ω_{S}) = h^{2}(S, Ω_{S}) = h^{1}(S, K_{S}) (draw Hodge diamond). This is called the irregularity of a surface.

P_{2} = h^{0}(S, K_{S}^{⊗2}).

It was once believed that this could be weakened to q = P_{1} = 0, which is somehow more attractive (as P_{1} is an entry in the Hodge diamond), but this false, and we may see examples before the end of the course (Enriques surfaces, Godeaux surfaces).

1.1. Motivation: Minimal rational surfaces. We know lots of rational surfaces now: ℙ^{2}, 𝔽_{n}, and blow-ups of these. At this point, we may suspect that we’ve found them all. How can we show this? We’ll use Castelnuovo’s criterion.

Date: Friday, November 22.

1.2. Motivation Luroth’s theorem (in characteristic 0). A variety V of dimension n is unirational if there is a dominant map (i.e. one with dense image) ℙ^{n} −−→ V.

Lüroth’s Theorem. Every unirational curve is rational.

Proof. This is true in arbitrary characteristic, but here’s a proof that works only in characteristic 0. Suppose ℙ^{1} −−→ C, where C is a curve, possibly singular and not proper. Then we also get a rational map ℙ^{1} −−→ C′, where C′ is a smooth compactification of a smoothing of C. By our lemma from long ago, any rational map from a smooth curve to anything projective extends to a morphism, so we have ℙ^{1} → C′. Dominant implies surjective. So we can apply the Riemann-Hurwitz formula, to see that

2 − 2g(ℙ^{1}) = d(2 − 2g(C′)) − ramification contribution.

The left side is 2, but if g(C′) > 0 the right side can’t be positive.

Theorem. In characteristic 0, every unirational surface is rational.

In positive characteristic, the theorem is false! Ask Ted Hwa for an example.

Question: where does the following argument break down in positive characteristic?

Proof. Suppose S is a unirational surface. If there was any doubt, let’s say that it is smooth and compact. (Otherwise, there is a way of producing a smooth and compact birational model.) So we have ℙ^{2} −−→ S. By the elimination of indeterminacy, we can blow up ℙ^{2} and get a morphism Bl ℙ^{2} → S. This morphism is dominant and hence surjective. Interpret q(S) as H^{0}(S, Ω_{S}), and recall P_{2}(S) = H^{0}(S, K_{S}^{⊗2}). If q > 0 or P_{2} > 0, then pullback the nonzero form (i.e. section of either Ω_{S} or K_{S}^{⊗2}) to get a non-zero section of the corresponding bundle on Bl(ℙ^{2}). This would give q(Bl(ℙ^{2})) > 0 or P_{2}(Bl(ℙ^{2})) > 0.

Hence q(S) = P_{2}(S) = 0. Then by Castelnuovo, S is rational.

Remark. Even in characteristic 0, there are 3-folds that are unirational but not rational, and they are not even that exotic! It is not hard to show that smooth cubic threefolds in ℙ^{4} are all unirational; Clemens and Griffiths showed that none of them are rational! Iskovskih and Manin did the same for quartic threefolds as well.

2. Proof of Castelnuovo’s criterion (part 1)

We’ll make a couple of reduction steps.

Castelnuovo’. Let S be a minimal surface with q = P_{2} = 0. Then there exists a smooth rational curve C on S such that C^{2} ≥ 0. Keep on board.

Proof that Castelnuovo’ implies Castelnuovo’s criterion.

O_{S}(C) clearly has a section, one whose zero set is C. We’ll see that in fact h^{0}(S, O_{S}(C)) ≥ 2, so “the curve moves”. Consider 0 → O_{S} → O_{S}(C) → O_{C}(C) → 0. Now q = h^{1}(S, O_{S}) = 0, so when we take global sections, the sequence remains exact, so

h^{0}(S, O_{S}(C)) = h^{0}(S, O_{S}) + h^{0}(C, O_{C}(C)))
= 1 + C^{2} − g(C) + 1 + h^{1}(C, O_{C}(C))
= 2 + C^{2}  (as C ≅ ℙ^{1}, and O_{C}(C) has positive degree)
≥ 2

So taking 2 sections, C and one other, we get a rational map S −−→ ℙ^{1}. After blowing up, this becomes a morphism S̃ −−→ ℙ^{1}. One of its fibers is isomorphic to C. By the Noether-Enriques theorem, it follows that S is rational.

So now we want to prove Castelnuovo’. Instead we’ll prove

Castelnuovo”. q = P_{2} = 0 implies that there is an effective divisor E on S such that K · E < 0 and |K + E| = ∅. Keep on board: We seek |E| ≠, |E + K| = ∅, K · E < 0.

Castelnuovo” implies Castelnuovo’. For then some component C of E satisfies K·C < 0, and any component satisfies h^{0}(S, K + C) = 0. Applying Riemann-Roch to K + C we get

0 = h^{0}(K + C)
≥ h^{0}(K + C) − h^{1}(K + C) + h^{0}(−C)
= χ(K + C)
= χ(O_{X}) + \frac{1}{2}((K + C) − K) · (K + C)
> h^{0}(O_{X}) − h^{1}(O_{X}) + h^{2}(O_{X}) + \frac{1}{2}(C + K) · C
≥ 1 + \frac{1}{2}(C + K) · C
= g(C).

Hence g(C) = 0. (C + K) · C = −2, hence C^{2} ≥ −1. If C^{2} = −1, then C is an exceptional curve, and we hypothesized that there weren’t any. So Castelnuovo’ follows.

Proof of Castelnuovo” in the case K^{2} = 0.

How can we possibly use P_{2} = 0? Only one reasonable way: Our hypothesis P_{2} = 0 gives h^{2}(−K) = 0 (Serre duality). Hence by Riemann-Roch (and q = 0):

h^{0}(−K) ≥ h^{0}(−K) − h^{1}(−K) + h^{2}(−K) = h^{0}(O) − h^{1}(O) + h^{2}(O) + K^{2} ≥ 1 + K^{2}.

(We’ll use this in the K^{2} > 0 case too.)

So | − K| ≠ ∅. Let H be a hyperplane section of S. Then H · K < 0. Note:

• If n = 0, then |H + nK| ≠ ∅.

• If n ≫ 0 then |H + nK| = ∅ (as (H + nK) · H < 0)

Thus there is an n ≥ 0 such that |H + nK| ≠ ∅, but |H + (n + 1)K| = ∅ as |H| ≠ ∅, and (H + nK) · H < 0 for n ≫ 0). Let D be an element. |K + D| = ∅, and K · D = −(−K) · H < 0.

Proof of Castelnuovo” in the case K^{2} > 0.

Recall h^{0}(−K) = 1 + K^{2}, so h^{0}(−K) ≥ −2. Suppose D ∈ | − K|.

Three cases:

(1) There is a reducible choice of D, i.e. A, B effective with A + B ∈ | − K|.

(2) Pic(C) = ℤK. (This implies that there is no reducible choice of D (why?), but we don’t care.)

(3) All divisors in | − K| irreducible, and Pic(C) ≠ ℤK.

Case 1: There is a reducible choice of D, i.e. A, B effective with A + B ∈ | − K|. Then A · K or B · K < 0, say the former. Then A is an effective divisor on S such that A · K < 0, and |A + K| = | − B| = ∅.

Case 2: Pic(C) = ℤK. This is the only case where characteristic 0 comes up! From the exact sequence

H^{1}(S, O_{S}) → Pic S → H^{2}(S, ℤ) → H^{2}(S, O_{S})

we have H^{2}(S, ℤ) ≅ Pic S = ℤK. Thus b_{2} = 1. By Poincare duality, the intersection form on H^{2}(S, ℤ) is unimodular, so K^{2} = 1. By Noether’s formula,

1 = χ(O_{S}) = \frac{1}{12}(K^{2} + 2 − 2b_{1} + b_{2})

from which b_{1} = −4, contradiction.

Case 3: All divisors D in | − K| irreducible and Pic(C) ≠ ℤK. Suppose H were an effective divisor. As | − K| ≠ ∅, there exists n > 0 such that |H + nK| ≠ ∅ and |H + (n + 1)K| = ∅. If (H + nK) · K < 0, we’d be done.

Take an H such that H + nK ≠ 0. Let E ∈ |H + nK|, E = ∑ n_{i}C_{i}. Then K · E = −D · E, and by the useful remark D · E ≥ 0 since D is irreducible. We are painfully close to being done: we have K · E ≤ 0, and we want K · E < 0!

Thus K · C_{i} ≤ 0 for some C = C_{i}. Hence |K + C| = ∅, from which 0 = h^{0}(K + C) ≥ 1 + \frac{1}{2}(C^{2} + CK) = g(C). g(C) = 0, and C^{2} = −2 − K · C (genus formula). We have gained exactly one thing in this paragraph: our divisor C is irreducible, whereas our divisor E was not necessarily. We know that |C| ≠ ∅, |K + C| = ∅, and K · C ≤ 0, and we want to show that K · C < 0.

So we’ll assume K · C = 0, and find a contradiction. From the genus formula, C^{2} = −2. We’ll calculate h^{0}(−K − C). Note that h^{0}(2K + C) = h^{0}(2K + (−D)) ≤ h^{0}(K + C) = 0.

Thus

h^{0}(−K − C) ≥ χ(−K − C) = χ(O_{X}) + \frac{1}{2}((K + C)^{2} + K(K + C))
= 1 + \frac{1}{2}(C^{2} + 3KC + 2K^{2})
≥ K^{2}
≥ 1

Since C^{2} = −2, we have C ≠ −K, so there exists a nonzero effective divisor A such that A + C ∈ | − K|. This contradicts our hypothesis that | − K| has no reducible divisors.

All that’s left is: Proof of Castelnuovo” in the case K^{2} < 0.