Surfaces

Report on Meeting 10

Example 1. Let E be an elliptic curve, so that its étale cohomology is

dim_{Q_{l}} H^{i}(E, Q_{l}) = \{ \begin{matrix}1 & if i = 0, 2, \\ 2 & if i = 1. \end{matrix}

If F is another elliptic curve, then we can compute the étale cohomology of the product using the Künneth Formula

H^{i}(E × F, Q_{l}) = ⨁_{p+q=i} H^{p}(E, Q_{l}) ⊗ H^{q}(F, Q_{l}).

It follows that the Betti numbers of the product are b_{1} = 4, b_{2} = 6. Moreover, the canonical class on the product is obtained by tensoring the pullbacks of the canonical classes of the factors, so K ∼ 0. It follows that the surface E × F corresponds to the fifth line of the table.

Proposition 2. Let X be a surface with κ = 0 and b_{2} = 6. Then X is an abelian surface.

Proof: From the table last time, we see that Δ = 0 and s = q = 2. So the Picard scheme is reduced and two dimensional, and its dual Alb(X) is also two dimensional. Consider the Albanese mapping α : X → Alb(X). First, we claim that α is surjective. If not, then its image is a curve C of genus g ≥ 2 on Alb(X). Pullback the endomorphism of the Albanese defined by multiplication by n to get a commutative diagram

Y ⟶^{β} Alb(X)
↓^{f}  ↓^{n}
X ⟶^{α} Alb(X)

Now the image D = β(Y) is an unramified cover of C of degree n^{2q}. Moreover, D must be connected, being the inverse image of an ample curve in the Albanese. So, the natural map Alb(Y) → Alb(D) must be a surjection to an abelian variety of dimension g > 2. However, Y is an étale cover of X, so K_{Y} ∼ 0 and Y must appear in the table. But there are no such surfaces in the table, and we have verified the first claim.

Now things get tricky. If α is a separable map (in characteristic 0, say), we can proceed as follows: Let ω be a translation invariant differential form on Alb(X). Then ω has no zeros or poles, and its pullback (because of separability!) is a nonzero regular differential form on X. The zeroes of α^{∗}(ω) lie along the ramification of the map. Since (α^{∗}(ω)) = K_{X} = 0, the map is unramified, and so α is a finite étale cover. But every finite étale cover of an abelian variety is itself an abelian variety.

The trick in characteristic p > 0 is to reduce to the case of a finite field, and use an idea of Tate. The first point is that X lies in a family of smooth projective surfaces, all with the same invariants (namely, K ∼ 0 and q = 0), and this family can be defined over a finite field. It folows from Thm. 6.14 of Mumford’s Geometric Invariant Theory that we only need to show that the members of this family over closed points of the base are abelian varieties. Of course, the closed points are finite fields.

So, assume that the ground field F is a finite field. Consider the pullback diagrams

X_{l} ⟶^{α_{l}} Alb(X)
↓^{f_{l}}  ↓^{[l]}
X ⟶^{α} Alb(X)

Since f_{l} is an étale morphism, each X_{l} is a surface of the same type as X. That is, K ∼ 0 and q ≥ 2. Since X_{l} is a surface in the table, we must have q = 2. We now claim that α : X → Alb(X) (and hence each α_{l}) is a finite morphism. We already know it is a projective morphism. So, by Hartshorne [Ex. III.11.2], we only need to check that it is has finite fibres. Since it is a surjection between varieties of the same dimension, the only possible problem arises from a curve E on X whose image is a point e = α(E) ∈ Alb(X). Now write the Stein factorization of α as X → Y → Alb(X), where the first map is projective with connected fibres and the second is finite. The first map is, therefore, a birational isomorphism that collapses the curve E. However, for each l, the inverse image of e in the Albanese consists of l^{4} points, each of which is the contraction of a curve in X_{l}. So, X_{l} has at least l^{4} disjoint curves with negative self-intersection. It follows that the Picard group has rank at least l^{4}, and thus b_{2}(X_{l}) ≥ l^{4}. Looking at the table, we see that b_{2}(X_{l}) = b_{2}(X) = 6. This is a contradiction, so there can be no curves collapsed by α. Thus, as claimed, α is a finite morphism.

Let d = deg(α) = deg(α_{l}). Fix an ample divisor D on Alb(X). Pulling back gives ample divisors D_{l} = α_{l}^{∗}(D) on X_{l}. The Hilbert polynomial is given by

χ(O_{X_{l}}(nD_{l})) = dχ(O_{X}(D)),

which is independent of l. Thus, there exists an integer N (also independent of l) such that all of the divisors ND_{l} are very ample. So, the infinite set of varieties X_{l} can all be embedded into a single projective space P^{M} with a fixed degree. However, since the ground field is finite, there are only finitely many subvarieties of P^{M} of fixed dimension and degree.

By construction, the group A[l] of l-torsion points in the Albanese variety acts on the surface X_{l} by translation. Moreover, each g ∈ A[l] takes the sheaf O_{X_{l}}(D_{l}) into a sheaf that is algebraically equivalent to it. Let G_{l} denote the group of automorphisms of X_{l} that take D_{l} to a numerically equivalent divisor. Not only is A[l] ⊂ G_{l}, but A[m] ⊂ G_{l} for infinitely many m (by the repvious paragraph). So, G_{l} is infinite. Since G_{l} is an algebraic group, it has only finitely many connected components. Thus, the connected component of the identity, G_{l}^{0}, must be infinite. If G_{l}^{0} contains a nontrivial linear subgroup, then it follow that X_{l} is ruled. This is impossible since K ∼ 0. So, G_{l}^{0} is an abelian variety. Since A[m] = (Z/m)^{4}, we see that the dimension of this abelain variety is at least 2. Thus X_{l} is a coset space G_{l}^{0}/H, and it must also be an abelain variety. Now we have X caught in the middle between two abelian surfaces:

X_{l}⟶^{etale}X⟶Alb(X).

It follows that X itself is an abelian surface.

Remark 3. It’s time again to recap where we stand. We first considered minimal surfaces X on which we could find an effective curve C with K.C < 0. We showed that if K^{2} > 0, then X is rational, and if K^{2} ≤ 0, then X is ruled. We next considered the situation when K.C ≥ 0 for all effective C, and showed that this forces K^{2} > 0. We then added the assumption that K.H = 0 for some ample divisor H, and used this to show that K is a torsion class in the Picard group (and hence K is numerically equivalent to 0 and κ = 0). Next, we used the standard formula relating surface invariants to produce a table for all possible invariants when K is torsion. There were five theoretically possible values for the second Betti number: 22, 14, 10, 6, or 2. We have shown:

(i) There are no surfaces with K torsion and b_{2} = 14. (ii) If K is torsion and b_{2} = 22, then X is a K3 surface. (iii) If K is torsion and b_{2} = 6, then X is an abelain surface.

The remaining two cases are more delicate, since they can potnetially subdivide into distinct cases in positive cahracteristic. Our next goal is to produce a proof of the following theorem:

Theorem 4. Let X be a minimal surface with K^{2} = 0 and with K.C ≥ 0 for all effective divisors C. Then either 2K ∼ 0 or there exists a pencil of curves on X with arithmetic genus p_{a} = 1.

Remark 5. Notice that this theorem applies more widely than just described: it includes surfaces on which K^{2} = 0 but K.H > 0 for every ample divisor H. Such surfaces will eventually be seen to have Kodaira dimension κ = 1.

Definition 6. An effective divisor C = ∑ n_{i}E_{i} on X is said to be of canonical type if K.E_{i} = C.E_{i} = 0 for all i. If C is also connected and if gcd(n_{i}) = 1, then we say that C is indecomposable of canonical type.

Proposition 7. Assume K^{2} = 0 and K.C ≥ 0 for all effective C. Then either 2K ∼ 0 or X contains at least one indecomposable curve ofg canonical type.

Proof: Suppose first that |2K| ≠ ∅. Let C ∈ |2K|. We will show that if C is nonzero, then it is of canonical type. So, write C = ∑ n_{i}E_{i}. We have

0 = 2K^{2} = K.C = ∑ n_{i}K.E_{i}.

Since all K.E_{i} ≥ 0, they must actually equal 0. Hence, C.E_{i} = 2K.E_{i} = 0 as well. Decomposing C, we can find an indecomposable curve of canonical type.

We still need to see what happens when |2K| = ∅. In this case, p_{g} = P_{2} = 0. So Δ = 0 and q = s. If q = 0, then the surface would be rational by Castelnuovo’s Criterion. But that is impossible since K^{2} = 0. So, q > 0. We have

χ = 1 − q + p_{g} = 1 − q

and

12χ = K^{2} + c_{2} = c_{2} = 2 − 4q + b_{2} ≥ 3 − 4q.

Thus, 12(1 − q) ≥ 3 − 4q or 9 ≥ 8q. So, we must have q = 1.

Now consider the Albanese morphism α : X → E, where E is an elliptic curve. let F_{0} be an irreducible fibre of α. If p_{a}(F_{0}) = 1, we are done. Otherwise, p_{a}(F_{0}) ≥ 2, so K.F_{0} = 2p_{a}(F_{0}) − 2 ≥ 2. Let F be any other fibre of α. We have an exact sequence

0 → O_{X}(2K + F − F_{0}) → O_{X}(2K + F) → O_{F_{0}}(2K.F_{0}) → 0,

where the last term uses the fact that F.F_{0} = 0 in the Chow ring. Note that

H^{2}(O_{X}(2K + F − F_{0}) = H^{0}(O_{X}(F_{0} − F − K) = 0

because otherwise we would have K.F_{0} < 0. By Riemann-Roch, we have

χ(2K + F − F_{0}) = \frac{1}{2}(2K + F − F_{0})(K + F − F_{0}) + χ(O_{X})

= \frac{1}{2}(2K^{2} + 3K.(F − F_{0}) + (F − F_{0})^{2}

= 0,

since F − F_{0} is numerically equivalent to 0. We have two possibilities:

(i) |2K + F − F_{0}| ≠ ∅|, or (ii) H^{i}(O_{X}(2K + f − F_{0})) = 0 for all i. In the first case, any effective divisor in that inear system must be of canonical type, by the same argument as before. We now show how to reduce the second case to the first. Suppose every fibre F falls into the second case. Fix a nonzero section s ∈ H^{0}(O_{F_{0}}(2K.F_{0})) with associated effective divisor A on E = Alb(X). For each P ∈ E, we get a unique lift to a section s_{P} ∈ H^{0}(O_{X}(2K + α^{∗})(P) − F_{0}), with a corresponding effective divisor D_{P}. This yields a one-dimensional family of divisors such that D_{P} .F_{0} = 0 for all P ≠ P_{0}. Moreover, all the D_{P} are distinct (since they are not even linearly equivalent!). Therefore

X = closure of ∪_{P ≠P_{0}} D_{P} .

In particular, as P → P_{0}, the curves D_{P} must specialize to a divisor D_{0} that conatins the entire fibre F_{0}. But then

D_{0} = F_{0} + C

and

D_{0} ∈ |2K + F_{0}|.

Thus |2K| ≠ ∅. Since that is a contradiction, we can always find ourselves in the first case.