COMPLEX ALGEBRAIC SURFACES CLASS 17

RAVI VAKIL

Contents

1. Castelnuovo’s Theorem 1 2. Application: The only minimal rational surfaces are ℙ^{2} and 𝔽_{n} (n ≠ 1) 3 3. Albanese variety 4 3.1. The universal property of the Albanese variety 4

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).

Also, I have been repeatedly using a fact that I may not have explicitly mentioned. A singular curve C still has a notion of genus (called the arithmetic genus), g(C) = 1−χ(O_{C}), which is h^{1}(C, O_{C}) if C is connected. (Warning: this is usually denoted p_{a}.) Fact: C irreducible and g(C) = 0 implies that C is smooth and genus 0.

1. Castelnuovo’s Theorem

We spent last time on most of the proof of :

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}).

We first reduced this to:

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.

Date: Wednesday, November 27, 2002. Patch on Fri. Dec. 5.

And then reduced it to:

Castelnuovo”. q = P_{2} = 0 and S minimal implies that there is an effective divisor E on S (i.e. |E| ≠ ∅) such that K · E < 0 and |K + E| = ∅. Keep on board.

The proof was in 3 cases: K^{2} < 0, = 0, and > 0.

We did the cases K^{2} = 0 and K^{2} > 0, so we’re left with:

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

Step 1. Find an effective divisor E on S such that K·E < 0. (2 out of the 3 desired conclusions.) Let H be a hyperplane section of S. If K · H < 0, we can take E = H, and we win. If K · H = 0, then K + nH has sections for n ≫ 0, and we win again: take E ∈ |K + nH|; it is effective, and K · E < 0.

So now assume K · H > 0. Let r_{0} = K · H/(−K^{2}). This is positive. Note that

(H + r_{0}K)^{2} = H^{2} + 2(K · H/(−K^{2}))HK + (K · H)^{2}/(−K^{2})^{2}K^{2} = H^{2} + (K · H)^{2}/(−K^{2}) > 0.

(Don’t be nervous with rational coefficients; we can still formally manipulate them in the same way. We should just be careful about them being actual divisors.) Also (H + r_{0}K) · K = 0. So if p/q (p, q ∈ ℤ^{+}) is a touch greater than r_{0}, we have

(H + \frac{p}{q}K)^{2} > 0, (H + \frac{p}{q}K) · K < 0, (H + \frac{p}{q}K) · H > 0.

Let D_{m} = mq(H + \frac{p}{q}K), so

D_{m}^{2} > 0, D_{M} · K < 0, D_{m} · H > 0.

As m → ∞, h^{0}(K − D_{m}) = 0, as (K − D_{m}) · H < 0. By Riemann-Roch,

h^{0}(D_{m}) ≥ χ(O_{X}) + \frac{1}{2}(D_{m}^{2} − D_{m} · K) → ∞

as m → ∞. Hence h^{0}(D_{m}) > 0 for some large m; take E ∈ |D_{m}|. It is effective, and K · E < 0.

Step 1. Remember our useful remark from a few classes ago: If D is an effective divisor, and C is an irreducible curve such that C^{2} ≥ 0, then D · C ≥ 0.

Suppose we have an effective divisor E on S such that K · E < 0. Then let C be some irreducible component of E such that K · C < 0. By the genus formula,

2g(C) − 2 = K · C + C^{2} ⇒ −2 = K · C + C^{2} − 2g(C) ≤ −1 + C^{2},

so C^{2} ≥ −1, and C^{2} = −1 iff C is an exceptional curve, which is excluded. Hence C^{2} ≥ 0.

Thus (aC + nK) · C < 0 for n ≫ 0. Thus by the useful remark, h^{0}(S, aC + nK) = 0 for n ≫ 0. Clearly h^{0}(S, aC +0K) > 0 (as there is a divisor in the linear system aC, notably aC itself), so there is some n such that |aC + nK| ≠ ∅, |aC + (n + 1)K| = ∅. If D ∈ |aC + nK|, then K · D ≤ −a, and |K + D| = ∅.

2. Application: The only minimal rational surfaces are ℙ^{2} and 𝔽_{n} (n ≠ 1)

Suppose S is a minimal rational surface. Let H be a hyperplane section, i.e. a very ample divisor. Consider the set A of smooth rational curves C with C^{2} ≥ 0. By Castelnuovo’, this set is non-empty.

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. Let m be the smallest possible choice of C^{2} in this family. Among those with this value of C^{2}, choose some C with the smallest possible value of C · H (which is necessarily positive).

Lemma. All curves in |C| are smooth and genus 0.

Proof. Suppose D = ∑ n_{i}C_{i} ∈ |C| isn’t irreducible. K · D < 0, so choose some C_{j} such that K·C_{j} < 0. I’ll show you that C_{j} is a smooth rational curve with C_{j}^{2} ≤ C^{2} and H·C_{j} < H·C, contradicting our choice of C.

Andre pointed out a gap here in the original notes. Here is the patch. By the useful trick, if h^{0}(S, K_{S} + C) > 0, then (K_{S} + C) · C ≥ 0. But by the genus formula, (K_{S} + C) · C = −2.

Recall:

Exercise. (I proved this when showing that Castelnuovo” implies Castelnuovo’, but it’s a good exercise to re-prove it yourself.) If K_{S} ·C < 0 and h^{0}(S, K_{S} +C) = 0 and q = 0, then C is a smooth rational curve with C^{2} ≥ −1. Method: apply Riemann-Roch to K_{S} + C.

Now here q = 0 (as S is rational), and h^{0}(S, O_{S}(K + C_{j})) ≤ h^{0}(S, O_{S}(K + ∑ n_{i}C_{i})) = 0. So C_{j} is a smooth rational curve. Also, C^{2} = (∑ n_{i}C_{i})^{2} ≥ ∑ n_{i}^{2}C_{i}^{2} ≥ C_{j}^{2}. Finally, C · H = (∑ n_{i}C_{i}) · H ≥ C_{j}Ḣ.

Lemma. dim |C| ≤ 2, i.e. h^{0}(S, O(C)) ≤ 3.

Proof. Assume otherwise. Pick some point p ∈ S. Pick an open subset containing p, over which O(C) is trivialized. Then via this trivialization, the h^{0}(S, O(C)) corresponds to some vector space functions, of dimension at least 4. Thus there is some non-zero function that restricts to something vanishing to order 2 at p. In that case, the zero set is singular, contradicting our previous lemma.

Lemma. m = 0 or 1.

Proof. We know m ≥ 0 by construction. Let C_{0} be a fixed curve in that class. Consider the exact sequence 0 → O_{S} → O_{S}(C) → O_{C_{0}}(m) → 0. As q = 0, this is exact on global sections, so 3 ≥ h^{0}(S, O_{S}(C)) = 1 + h^{0}(ℙ^{1}, O(m)) = 2 + m.

Note also that O(C) has no base points. (Explain.)

Theorem. Let S be a minimal rational surface. Then S ≅ ℙ^{2} or some 𝔽_{n} for n ≠ 1.

Proof. If m = 0, then |C| determines a morphism S → ℙ^{1} (as there are no base points), all of whose fibres are smooth rational curves. So S is a rational ruled surface 𝔽_{n}.

If m = 1, then |C| determines a morphism S → ℙ^{2}. The preimages are reduced points (any two sections intersect in m = 1 point, counted with multiplicity), so this is an isomorphism.

3. Albanese variety

I next want to introduce the last important tool you haven’t yet seen in the classification of surfaces, the albanese variety.

As a goal, I will prove (next day):

Theorem. Let S, S′ be two non-ruled minimal surfaces. Then every birational map from S′ to S is an isomorphism. In particular, every non-ruled surface admits a unique minimal model (up to isomorphism); the group of birational maps from a non-ruled minimal surface to itself coincides with the group of automorphisms of the surface.

3.1. The universal property of the Albanese variety. An abelian variety over ℂ is an projective algebraic variety that is a complex torus ℂ^{n}/L (L a lattice of rank 2n); it is a group. In the purely algebraic category, an abelian variety over a field k is a smooth projective group scheme.

Hard theorem. For any (smooth projective) variety X over a field k, there exists an abelian variety A and a morphism X → A with the following universal property: for any abelian variety T and any morphism f : X → T, there exists a unique morphism f̃: A → T such that f̃∘ α = f.

Exercise. A is determined up to isomorphism.

This is called the Albanese variety of X, written Alb(X).

Hard theorem continued. α induces an isomorphism α^{∗} : H^{0}(A, Ω_{A}^{1}) → H^{0}(X, Ω_{X}^{1}).

The proof in general is hard, but over ℂ there is an easier proof using Hodge theory (see Beauville, for example). For simplicity, I’ll work in the complex category.

Some fast consequences.

dim Alb(X) = h^{1}(X, Ω_{X}^{1}). In particular, if X is a surface, dim Alb(X) = q. So as a corollary, if q = 0, every morphism from X to a complex torus is trivial (i.e. every map is reduced to a point). Also, every map from ℙ^{1} to a complex torus is trivial. (Explain why in the complex category.)

The Albanese variety is functorial in nature. If f : X → Y is a morphism of smooth projective varieties, there is a unique morphism F : Alb(X) → Alb(Y) such that the diagram

X ⟶^{f} Y
α_{X} ↓  ↓ α_{Y}
Alb(X)

exists unique F

⟶ Alb(Y)

is commutative.

Moreover, the Abelian variety Alb(X) is generated by α(X) (i.e. there is no abelian subvariety of Alb(X) containing α(X)), as the Abelian subvariety of Alb(X) generated by α(X) satisfies the universal property. In particular, α(X) is not reduced to a point if Alb(X) is not a point. It also follows that if the morphism f is surjective, then so is the morphism F : Alb(X) → Alb(Y).

Fact: If X is a curve, then Alb(X) is the Jacobian Pic^{0}(X) (usually denoted Jac(X)).