Surfaces

Report on Meeting 9

We have completely analyzed the situation when there exists an effective divisor C with C.K < 0. We showed for a minimal surface that if K^{2} > 0, then X is rational, and if K^{2} ≤ 0, then X is ruled. From now on, we will assume that no such curve exists. More precisely, we have: If C is any effective divisor on X, then C.K ≥ 0. We will need the following consequence of the Riemann-Roch Theorem for Surfaces.

Theorem 1. (Hodge Index) Let H be an ample divisor on the surface X, and suppose that D is a divisor that is not numerically equivalent to 0 for which D.H = 0. Then D^{2} < 0.

Remark 2. Let P ic^{n}(X) denote the group of divisor classes that are numerically equivalent to 0 and let Num(X) = P ic(X)/P ic^{n}(X) = NS(X)/torsion. Intersection theory defines a nondegenerate bilinear form on Num(X). The Hodge Index Theorem is equivalent to the assertion that the induced form on Num(X) ⊗ R has its signature or index equal to (1, m).

Lemma 3. Let V be a real vector space with a nondegenerate bilinear form of signature (1, m). Let x ∈ V be such that x.x > 0. If x.y = 0, then y.y <= 0.

Proof: We can find a basis for V of the form {e_{0} = x, e_{1}, …, e_{m}}, where e_{i}.e_{j} = 0 for i ≠ j and e_{i}.e_{i} < 0 for i > 0. Writing y = ∑_{i=0}^{m} r_{i}e_{i} in terms of this basis, we see that the hypothesis x.y = 0 is equivalent to r_{0} = 0. Then

y.y = ∑_{i=1}^{m} r_{i}^{2}e_{i}^{2} <= 0.

Definition 4. Let V be a finite dimensional real vector space with a nondegenerate bilinear form. The dual of a subset S ⊂ V is defined to be

S^{∗} = {y ∈ V |x.y ≥ 0∀x ∈ S}.

Definition 5. Let V be a real vector space with a nondegenerate bilinear form of signature (1, m). Fix h ∈ V such that h.h > 0. Let Γ ⊂ V denote the “positive cone” with respect to h. In other words,

Γ = {x ∈ V |x.x ≥ 0 and x.h ≥ 0}.

Remark 6. A subset of V is a cone if it is closed under addition and under scalar multiplication by positive constants. That Γ is closed under positive scalar multiplication is immediate from the definition. That it is closed under addition follows from the triangle inequality (applied to the orthogonal complement of the one dimensional subspace spanned by h).

Lemma 7. Let V be a real vector space with a nondegenerate bilinear form of signature (1, m). Fix h ∈ V such that h.h > 0, and let Γ denote the positive cone with respect to h. Then Γ is self-dual.

Proof: Choose a basis {e_{0}, e_{1}, …, e_{m}} where (i) e_{0} = (1/r)h is a positive real multiple of h such that e_{0}^{2} = 1, (ii) e_{i}^{2} = −1 for i ≥ 1, and (iii) e_{i}.e_{j} = 0 for i ≠ j. Now write x = ∑_{i=0}^{m} a_{i}e_{i}. We find that x ∈ Γ if and only if a_{0} ≥ 0 and ∑_{i=1}^{m} a_{i}^{2} < a_{0}^{2}. Now let y ∈ Γ^{∗} and write y = ∑_{i=0}^{m} b_{i}e_{i}. We have

y.h = r.b_{0} ≥ 0

since h ∈ Γ. Hence b_{0} ≥ 0. Next, let

c = \sqrt{∑_{i=1}^{m} b_{i}^{2}}.

By construction, z = ce_{0} + ∑_{i=1}^{m} b_{i}e_{i} lies in Γ. So,

0 ≤ y.z = cb_{0} − ∑_{i=1}^{m} b_{i}^{2} = cb_{0} − c^{2} = c(b_{0} − c).

Thus b_{0} ≥ c, and it follows that y ∈ Γ. So, we have shown that Γ ⊂ Γ^{∗}.

For the converse, we need to check that the inner product of two elements x, y ∈ Γ is nonnegative. Write x and y as above. By the Cauchy-Schwartz Inequality, we have

(∑_{i=1}^{m} a_{i}b_{i})^{2} ≤ (∑_{i=1}^{m} a_{i}^{2}) (∑_{i=1}^{m} b_{i}^{2}) ≤ a_{0}^{2}b_{0}^{2}.

Consequently,

(a_{0}b_{0} + ∑_{i=1}^{m} a_{i}b_{i}) (a_{0}b_{0} − ∑_{i=1}^{m} a_{i}b_{i}) ≥ 0.

Since at least one of the factors on the left is positive, they m,ust both be positive, yielding the result.

To procedd, we also need the following standard result.

Theorem 8. (Nakai-Moishezon Criterion) A divisor D on a surface X is ample if and only if D^{2} > 0 and D.C > 0 for all irreducible curves C on X.

Proposition 9. Let D be a divisor on a surface X. Assume that D.C ≥ 0 for all effective divisors C on X. Then D^{2} ≥ 0.

Proof: Fix an ample divisor H on X. Let Γ ⊂ NS(X) ⊗_{Z} R denote the “positive cone” with respect to H. Let Γ_{+} denote the cone spanned by all effective divisors. By the Nakai-Moishezon Criterion, the interior of Γ is contained inside Γ_{+}. Since they are both cones, we have Γ ⊂ Γ_{+}. Since dualizing is inclusion reversing, we can apply the previous lemma to get

Γ_{+}^{∗} ⊂ Γ^{∗} = Γ.

Translating this statement about cones gives the result.

Corollary 10. If K.C ≥ 0 for all effective divisors C, then K^{2} ≥ 0.

Remark 11. We have reached another decision point. We need to decide if every ample divisor intersects the canonical divisor positively. As we did with our first major decision, we will first determine the consequences of assuming the existence of a divisor with some special property, and oly later look at the general case. In particular, we will assume: There exists an ample divisor H satisfies K.H = 0.

Proposition 12. Suppose K^{2} ≥ 0 and there exists an ample divisor H such that K.H = 0. Then for every effective divisor C, we have K.C = 0.

Proof: Since K.H = 0, we conclude from the Hodge Index Theorem that K^{2} ≤ 0. Since we have assumed that K^{2} ≥ 0, it follows that K^{2} = 0. So, K lies on the boundary of the positive cone Γ. Since Γ is contained in Γ_{+}, therer exists a rational number r such that rK is an effective divisor. But then rK.H = 0 forces rK ∼ 0. Therefore, K is a torsion element of the Néron-Severi group, and it must be numerically equivalent to 0.

Corollary 13. Suppose K^{2} ≥ 0 and there exists an ample divisor H such that K.H = 0. Then the Kodaira dimension is κ = 0.

Proof: Let m be the smallest positive integer such that K is m-torsion in the Néron-Severi group. Then

P_{n} = dim(Γ(X, O_{X}(nK))) = \{ \begin{matrix}0 & if gcd(m, n) = 1, \\ 1 & if m|n. \end{matrix}

It follows that the Kodaira ring has transcendence degree 1 (with a nontrivial element of degree m providing a transcendence basis), so the Kodaira dimension is 0.

Remark 14. Let us recall our famous magic formula

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

that provides so much information about the structure of surfaces. In our case, we have K^{2} = 0 and p_{g} = P_{1} ≤ 1. Combining these facts with the standard constraints (Δ is even with 0 ≤ Δ ≤ 2p_{g}, b_{2} is positive, and s is nonegative), we can enumerate the possible numerical invariants of a numerical surface that satisfies our hypotheses. Here are the possibilities:

\begin{matrix}κ & p_{g} & s & Δ & b_{2} & K^{2} & q & b_{1} & χ & p_{a} & c_{2} \\ 0 & 1 & 0 & 0 & 22 & 0 & 0 & 0 & 2 & 1 & 24 \\ 0 & 1 & 1 & 0 & 14 & 0 & 1 & 2 & 1 & 0 & 12 \\ 0 & 0 & 0 & 0 & 10 & 0 & 0 & 0 & 1 & 0 & 12 \\ 0 & 1 & 1 & 2 & 10 & 0 & 0 & 0 & 1 & 0 & 12 \\ 0 & 1 & 2 & 0 & 6 & 0 & 2 & 4 & 0 & −1 & 0 \\ 0 & 0 & 1 & 0 & 2 & 0 & 1 & 2 & 0 & −1 & 0 \\ 0 & 1 & 2 & 2 & 2 & 0 & 1 & 2 & 0 & −1 & 0 \end{matrix}

Example 15. Let X be a smooth projective surface of degree 4 in P = P^{3}. The general formula for the canonical sheaf on a smooth hypersurface of degree d in P^{n} is ω_{X} = O_{X}(d − n − 1). In our case, we have ω_{X} = O_{X}, so K ∼ 0. Thus K^{2} = 0 and p_{g} = 1. From the exact sequence

0 → O_{P} (−4) → O_{P} → O_{x} → 0,

and the standard computation of cohomology on projective space, one can also compute directly that s = 0. By looking at our table, we see thatif it is minimal, then this surface belongs to the class in the first row, with b_{2} = 22. However, since the canonical class is linearly equivalent to 0, it must be a minimal surface.

Definition 16. A minimal surface with κ = 0 and b_{2} = 22 is called a K3 surface.

Proposition 17. If X is K3 surface, then

(a) X has no nontrivial étale coverings. In other words, π_{1}(X) = 0.

(b) If D is a numerically trivial divisor on X, then D is linearly equivalent to 0. In other words, P ic_{X}^{0} = 0.

Proof: (a) Suppose f : Y → X is a connected étale covering of degree d. Since étale is smooth of relative dimension 0, we would then have K_{Y} ∼ f^{∗}(K_{X}) ∼ 0. So, Y must also be a surface in the table. However, deg(c_{2}(Y)) = deg(f^{∗}(c_{2}(X))) = d ∗ deg(c_{2}(X)) = 24d. Since 24 is the largest value of c_{2} in the table, it follows that d = 1 and f is an isomorphism.

(b) Since H^{1}(X, O_{X}) = 0 is the tangent space to the Picard scheme P ic^{0}(X), it follows that P ic_{X}^{0} is a finite discrete group. Take a line bundle L = O_{X}(D) that represents a point in P ic_{X}^{0}. Then D^{2} = D.K = 0. By Riemann-Roch, we have χ(L) = χ(O_{X}) = 2. Thus, we either have H^{0}(X, L) ≠ 0 or H^{0}(X, L^{−1}) = H^{2}(X, L) ≠ 0. So either L or its inverse is represented by an effective divisor C. Since both sheaves live in P ic_{X}^{0}, we must have C = 0, and the result follows.

Proposition 18. There are no surfaces with κ = 0 and b_{2} = 14.

Proof: Since p_{g} = 1 and K is numerically trivial, we in fact have K ∼ 0. Since b_{1} = 2 > 0, we know that X has a positive dimensional Picard scheme. In particular, there exists a divisor D on X that is numerically equivalent to 0 but not linearly equivalent to 0. So, χ(O_{X}(D)) = χ(O_{X}) = 1. As before, Serre duality shows that either D or −D is equivalent to an effective divisor. But the only effective divisor that is numerically trivial is 0, which contradicts our assumption.