(page 1) Chapter 14

numerical equivalence

The Hodge Index Theorem

In the previous chapter, we identified numerical conditions on a class ξ ∈ Num(X) that would suffice to ensure that every invertible sheaf of type ξ is 0-regular and very ample. In this chapter, we want to show that there always exist classes ξ that satisfy those criteria.

Proposition 14.1. Let c_{2} and ϵ be the constants whose existence is asserted by Cor. 13.12. There exist classes λ ∈ Num(X) satisfying the numerical conditions:

(i) λ.h ≥ c_{2}

(ii) λ.(λ − ω) ≥ (1 − ϵ) [\frac{(λ.h)^{2}}{h.h} ]

Proof: We will, in fact, show that if λ is any numerical class, then the classes λ_{n} = λ + nh satisfy the conditions for n >> 0. We have

λ_{n}.h = (λ + nh).h = nh^{2} + λ.h.

So, by taking n >> 0, we can make this term as large as we wish, thus fulfilling the first criterion. Next, look at

(14.2)  (λ + nh).(λ + nh − ω) − (1 − ϵ) [\frac{((λ + nh).h)^{2}}{h.h} ] .

Expand this expression and write it as a polynomial in n. One finds that the leading term is

(h.h)n^{2} − \frac{(1 − ϵ)}{h.h}(h.h)^{2}n^{2} = ϵ(h.h)n^{2}.

So, (14.2) can also be made positive by taking n >> 0. Therefore, if λ is any numerical equivalence class, then the class λ_{n} satisfies both conditions whenever n >> 0.

Lemma 14.3. Let H be a very ample divisor on X. Let D be a divisor such that D.H > 0 and D^{2} > 0. Then for all n >> 0, the divisor nD is linearly equivalent to an effective divisor.

◯^{c}Coombes 1996  1

(page 2)

2 Chapter 14: The Hodge Index Theorem

Hodge Index Theorem signature index

Proof: When n >> 0, we have nD.H > c_{1}, and so H^{2}(O(nD)) = 0. Thus, the Riemann-Roch Theorem gives

h^{0}(nD) ≥ \frac{1}{2}n^{2}D^{2} − \frac{1}{2}nD.K + χ(O_{X}).

Since D^{2} > 0, the right side becomes positive for large n, and the result follows.

Theorem 14.4. (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.

Proof: We will prove the result by contradiction. We need to handle the cases D^{2} > 0 and D^{2} = 0 separately. So, we begin by assuming that D^{2} > 0. Look at the divisors H_{n} = D + nH. For n >> 0, we know that H_{n} is very ample. Furthermore, D.H_{n} = D^{2} > 0. By the previous lemma, therefore, we see that mD is effective for all m >> 0. But that would force mD.H > 0 and hence D.H > 0, which is a contradiction.

Now suppose that D^{2} = 0. Since D is not numerically trivial, there exists a divisor E with D.E ≠ 0. Replace E by the divisor (H^{2})E − (E.H)H, so we can assume that E.H = 0. Now look at the divisors D_{n} = nD + E. Then D_{n}.H = 0 and D_{n}^{2} > 0 for large n. Thus, we can apply the previous case to finish the proof.

Remark 14.5. By definition, intersection theory on the surface X 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). We’re going to spend a few pages examining nondegenerate bilinear forms with such a signature.

Lemma 14.6. 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 14.7. 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 for all x ∈ S}.

◯^{c}Coombes 1996

Definition 14.8. 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 14.9. 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 14.10. 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 = rb_{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 must both be positive, yielding the result.

◯^{c}Coombes 1996

To proceed, we also need the following standard result. Stein Factorization Theorem

Theorem 14.11. (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.

Proof: The condition is clearly necessary, since if D is ample then some positive multiple mD is very ample. Then m^{2}D^{2} is the degree of the surrface under the corresponding embedding, and mD.C is the degree of C.

Conversely, suppose D^{2} > 0 and D.C > 0 for all irreducible curves C. Fix a very ample divisor H on X, which by Bertini’s Theorem can be represented by an irrreducible curve. So, H.D > 0. By Lem. 14.3, we know that nD is effective for all n >> 0. Replacing D by nD, we can assume that D is effective.

Next, let L = O_{X}(D) be the associated invertible sheaf. We claim that L ⊗ O_{D} is ample on D. To see this, it is enough to show that L ⊗ O_{D_{red}} is ample on the reduced curve D_{red}. In turn, it is enough to show that L ⊗ O_{C} is ample on each irreducible component C of D_{red}. Since the normalization maps f : C̃ → C are finite and surjective, it is even enough to show that f^{∗}(L ⊗ O_{C}) is ample on the nonsingular projective curve C̃. But one has

deg(f^{∗}(L ⊗ O_{C})) = D.C > 0,

since we can represent L as a difference of curves that meet C transversally. Since this degree is positive, the sheaf is ample on the curve C̃. Thus, L ⊗ O_{D} is ample.

Now we will show that L^{n} is generated by global sections for n >> 0. We tensor the standard exact sequence

0 → L^{−1} → O_{X} → O_{D} → 0

with L^{n}. Since L ⊗ O_{D} is ample, we see that H^{1}(L^{n} ⊗ O_{D}) = 0 for n >> 0. Now the long exact sequence of cohomology yields surjections

H^{1}(L^{n−1}) → H^{1}(L^{n}) → 0.

Since the cohomology groups are finite dimensional, their size cannot decrease forever, so it must eventually stabilize. So, for n >> 0, we have surjections

H^{0}(L^{n}) → H^{0}(L^{n} ⊗ O_{D}) → 0.

or n >> 0, the sheaf L^{n} ⊗ O_{D} is generated by global sections. Lifting these sections to L^{n}, they will generate at all points of D. However, L also has a section that vanishes only along D, so it is generated by global section everywhere.

Let ι : X → P^{n} be the morphism defined by an L^{n} that is generated by global sections. If the fibres of ι are not finite, then at least one fibre contains an irreducible curve C that gets collapsed to a point. Pulling back a hyperplane that misses the point gives a divisor E ∼ nD with E.C = 0. But that’s impossible, since E.C = nD.C > 0. By the Stein Factorization Theorem, a projective morphism with finite fibres is actually a finite morphism. Thus L = ι^{∗}(O(1)) is ample on X.

◯^{c}Coombes 1996 Hodge Index Theorem Nakai-Moishezon Criterion

Proposition 14.12. 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 Lem. 14.3, 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 Lem. 14.10 to get

Γ_{+}^{∗} ⊂ Γ^{∗} = Γ.

Translating this statement about cones gives the result.

Finally, we’d like to compare the cones that arise naturally from the consideration of the Hodge Index Theorem and the Nakai-Moishezon Criterion with the numerical conditions that we came up with to ensure that a sheaf is 0-regular and very ample. As we’ve just seen, there are inclusions

ample cone ⊂ positive cone ⊂ effective cone.

We know that if we move far enough out into the ample cone, then we hit classes that have all sorts of good properties, including being generated by global sections and 0-regular. Out first existence result is only slightly stronger than the statement that starting anywhere and moving in the h-direction eventually hits this very ample nappe of the ample cone. To get some more detailed information, we do the following.

Lemma 14.13. Given any divisor D on X, there exists a positive integer n_{D} such that for all curves C on X one has

|D.C| ≤ n_{D} deg(C).

Proof: Let H be a very ample divisor representing O_{X}(1). Choose n sufficiently large that both the divisors D + nH and −D + nH are very ample. Then both the intersection numbers (D + nH).C and (−D + nH).C are positive. So, one has

−n deg(C) ≤ D.C ≤ n deg(C).

Proposition 14.14. We can replace the numericla conditions ensuring that a numerical class λ is 0-regular and very ample with the simpler conditions

(i) deg(λ) ≥ c_{2}

(ii) (λ.λ) ≥ (1 − ϵ)\frac{(deg(λ))^{2}}{h.h}.

Proof: Exercise.

The real point is that these new conditions pick out the frustrum of a cone contained inside the ample cone.

◯^{c}Coombes 1996

(page 6)

◯^{c}Coombes 1996