COMPLEX ALGEBRAIC SURFACES CLASS 11

RAVI VAKIL

Contents

1. Geometrically ruled surfaces and projectivizations of rank 2 locally free sheaves 1 2. Geometric facts about geometrically ruled surfaces over C, from geometric facts

about C 4

1. Geometrically ruled surfaces and projectivizations of rank 2 locally free sheaves

Last time, we began analyzing geometrically ruled surfaces by studying projectivizations of rank 2 locally free sheaves.

We proved:

Proposition. Every geometrically ruled surface over C is C-isomorphic to ℙ_{C}(E) for some rank 2 locally free sheaf (vector bundle) over C. The bundles ℙ_{C}(E) and ℙ_{C}(E′) are isomorphic (over C) iff there is an invertible sheaf (line bundle) L on C such that E′ ≅ E ⊗ L.

I then stated the first part of the following lemma.

Lemma: All rank 2 locally free sheaves are filtered nicely by invertible sheaves. Suppose E is a rank 2 locally free sheaf on a curve C.

(i) There exists an exact sequence 0 → L → E → M → 0 with L, M ∈ Pic C. Terminology: E is an extension of M by L.

(ii) If h^{0}(E) ≥ 1, we can take L = O_{C}(D), with D the divisor of zeros of a section of E.

(Hence D is effective, i.e. D ≥ 0.) (iii) If h^{0}(E) ≥ 2 and deg E > 0, we can assume D > 0.

(i) is the most important one.

I mentioned the application of Riemann-Roch for rank 2 vector bundles on a curve:

χ(E) = deg E + 2(1 − g).

Date: Wednesday, November 6.

Second: the filtrations correspond to sections of the projective bundle. Hence all projective bundles over curves have sections.

Proof. (i) We can twist E by some invertible sheaf N so that it has a non-zero section s. Here’s how: take a rational section of E. It has some poles of various orders at various points. Twist by the invertible sheaf allowing these poles.

The section s gives O_{C} → E ⊗ N.

Here’s an argument which doesn’t work. (Find the mistake.) The cokernel is locally free, call it M′, and the kernel of E ⊗ N → M′, which is a subsheaf of O_{C}, is necessarily O_{C}(D) for D the zero-divisor of the section s. Hence we have

0 → O_{C}(D) → E ⊗ N → M′ → 0.

Twist this by N^{∗}, and we get (i).

In fact, the cokernel needn’t be an invertible sheaf! (This is the only flaw.) In general, the cokernel of a morphism of locally free sheaves needn’t be locally free! (This is different behavior than for vector bundles, and is one good reason to keep the two concepts separate in your mind.) You’ve seen an example before: If X is a variety and D a divisor, consider the morphism of invertible sheaves O_{X}(−D) → O_{X}. then we have 0 → O_{X}(−D) → O_{X} → O_{D} → 0. (If you want to see a morphism from a rank 1 locally free to a rank 2 locally free, tweak this to get 0 → O_{X}(−D) → O_{X} ⊕ O_{X} → O_{D} ⊕ O_{X} → 0.) Hence this morphism doesn’t correspond to a morphism of vector bundles. How to patch? We’ll use the useful fact is that any subsheaf of a locally free sheaf on a smooth curve is also locally free. (The proof is omitted. The key is to show that torsion-free sheaves on a smooth curve are locally free, which is an algebraic fact.) Dualize the problematic morphism O_{X} → E ⊗ N to get E^{∗} ⊗ N^{∗} → O_{X}. Take the image of E^{∗} ⊗ N^{∗} in O_{X}. By the useful fact, the image is also an invertible sheaf O_{X}(−D) (check first that the image isn’t 0). The kernel of this morphism is a subsheaf of a locally free sheaf, and is thus also locally free, in fact of rank 1. So we get

0 → M′ → E^{∗} ⊗ N^{∗} → O(D) → 0.

Dualizing gives us 0 → O(D) →^{(∗)} E ⊗ N → (M′)^{∗} → 0.

Useful facts (proofs omitted): (i) the morphism (*) corresponds to a section of E ⊗ N, with poles along D. (ii) to translate to vector bundles, a morphism of vector bundles on a smooth variety corresponds to a morphism of locally free sheaves whose cokernel is locally free (the kernel automatically is). (iii) related fact: sections of the projective bundle correspond to filtrations.

For (ii), note that if h^{0}(E) > 0, then E already has a non-zero section, and we can omit the twist by N. Then that last exact sequence is the one we desire.

For (iii), it suffices to show that there is a section of E that vanishes at some point. Let s and t be two linearly independent sections. As deg E > 0, the section s ∧ t of ∧^{2}E must vanish at some point p ∈ C. That means at that point there are α and β (not both 0) such that αs(p) + βt(p) = 0. So the section αs + βt vanishes at p, proving (iii).

Hence we want to classify extensions of invertible sheaves. Up to twists by invertible sheaves, these correspond to projective bundles with section.

Given an extension 0 → L → E → M → 0, we get a class of H^{1}(C, L ⊗ M^{∗}) as follows. (This has nothing to do with C being a curve, or with L being invertible.) Twist by M^{∗}, take the long exact sequence, and look at the image of 1 ∈ H^{0}(C, O_{C}). Extensions are classified precisely by this cohomology group. In other words, two extensions are isomorphic

0 → L → E → M → 0
|| ↓∼ ||
0 → L → E′ → M → 0

if they induce the same element of H^{1}(C, L ⊗ M^{∗}). One direction is now clear.

Exercise: Check the other direction. (Hint, useful in other circumstances: Given an element of H^{1}(C, L⊗M^{∗}), say in Cech cohomology, recover the extension.) Check that the 0 ∈ H^{1}(C, L ⊗ M^{∗}) corresponds to L ⊕ M. Check that if a = kb where a, b ∈ H^{1}(C, L ⊗ M^{∗}) and k ≠ 0, then E_{a} ≅ E_{b}.

Proposition. Every rank 2 locally free sheaf on ℙ^{1} decomposes into the sum of two invertible sheaves. Hence every geometrically ruled surface over ℙ^{1} is isomorphic to a Hirzebruch surface

𝔽_{n} = ℙ_{ℙ^{1}}(O_{ℙ^{1}} ⊕ O_{ℙ^{1}}(n))

for n ≥ 0.

Hence every geometrically ruled surface is one of the 𝔽_{n}’s described earlier. (We don’t yet know that they are all different yet.)

Proof. We can twist by a line bundle so as to assume that deg E = 0 or 1. By Riemann-Roch, h^{0}(E) ≥ 1, so there is an exact sequence

0 → O_{ℙ^{1}}(k) → E → O_{ℙ^{1}}(d − k) → 0

with k ≥ 0. But these extensions are classified by H^{1}(ℙ^{1}, O_{ℙ^{1}}(2k − d)) ≅ H^{0}(ℙ^{1}, O(−2 − 2k + d))^{∗} = {0}, so this must be a direct sum.

Exercise: Grothendieck’s Theorem. Show that every vector bundle on ℙ^{1} is a direct sum of line bundles.

Similarly, you can prove:

Proposition.

(a) Every rank 2 vector bundle on an elliptic curve is either decomposable, or isomorphic to E ⊗ L, where L ∈ Pic C and E is either (i) the unique non-trivial extension of O_{C} by O_{C}, or (ii) the non-trivial extension of O_{C}(p) by O_{C} for some p. (Exercise.)

(b) For every curve C of genus g, there exist families of rank 2 vector bundles parametrized by some variety S (possibly singular, non-compact) of dimension at least 2g − 3. (See Beauville.)

2. Geometric facts about geometrically ruled surfaces over C, from geometric facts about C

(This should be moved to early in the course notes.) First, we’ll need the definition of O(1) for a projective bundle π : ℙE → B. There is a “tautological” subline bundle of π^{∗}E; this is defined to be O_{P} (−1) (and O_{P} (n) is defined to be the appropriate multiple = tensor power of this).

You can check:

Exercise.

(a) This agrees with the definition of O(1) on a projective space (in the case where E is a point).

(b) In the general case, the restriction of O_{P} (1) to a fiber of π is O(1) on the fiber. (This is basically immediate from (a).)

(c) In the case where dim B = 1 and dim P = 2, O(1) · F = 1 for any fiber F. (This is basically immediate from (b); and it generalizes to projective bundles of arbitrary dimension once we have intersection theory.)

Proposition. Suppose π : S → C is a geometrically ruled surface, corresponding to rank 2 locally free sheaf E. Let h = O_{S}(1) ∈ Pic S or H^{2}(S, ℤ) Then:

(i) Pic S = π^{∗} Pic C ⊕ ℤh, (ii) H^{2}(S, ℤ) = ℤh + ℤf, where f is the class of a fiber, (iii) h^{2} = deg E, (iv) [K] = −2h + (deg E + 2g(C) − 2)f in H^{2}(S, ℤ).

Remarks.

• This tells us about the intersection theory on S.

• (i) and (ii) are true for ℙ^{1}-bundles over an arbitrary base. There are also analogues of (iii) and (iv) over an arbitrary base.

• Note that h · f = 1.

• (ii) follows from (i), as H^{2} is a quotient of Pic.

• Assuming (ii) and (iii), proof of (iv) is an exercise: [K] = ah + bf in H^{2}(S, ℤ). Use the genus formula for a fiber F to get a = −2, and the genus formula for a section to get b.

Proof of (i). We get a map from π^{∗} Pic C⊕ℤh → Pic S. It is injective: suppose (π^{∗}L, nh) ↦ 0; then by restricting to a fiber F, we get n = 0; by restricting to a section, we get L = 0.

Surjectivity: Any element of Pic S is of the form D + mh where D · F = 0. I claim that D = π^{∗}D′ for some D′ on C. Here’s why. Consider D_{n} := D + nF.

I claim that h^{0}(K − D_{n}) = 0 for n ≫ 0. Reason: take a very ample divisor class [H] on S, so [H] · F > 0. Choose n big enough that [H] · (K − D_{n}) < 0. If h^{0}(K − D_{n}) > 0, then there is a non-zero section. There is a section of O(H) whose zero set is an effective curve meeting this zero-set. But H · (K − D_{n}) < 0, contradiction.

Now D_{n}^{2} = D^{2}, and D_{n} · K = D · K − nF · K = D · K − 2n, so by Riemann-Roch:

h^{0}(D_{n}) − h^{1}(D_{n}) + h^{2}(D_{n}) = χ(O_{S}) + \frac{1}{2}(D_{n}^{2} − D_{n} · K)

⇒ h^{0}(D_{n}) ≥ 0 + \frac{1}{2}(D^{2} − D · K + 2n) > 0.

Let E ∈ |D_{n}| be the zero-set of a non-zero section. Then E · F′ = 0 for every fiber F′, so E must be a union of fibers with multiplicity, i.e. E is the pullback of some points on C.

Proof of (iii). ) Define c_{2}(E) := χ(O_{S}) − χ(E) + χ(∧^{2}E). Motivation: If 0 → L → E → M → 0, we want c_{2}(E) = L · M. Well,

L · M = L^{∗} · M^{∗} = χ(O_{S}) − χ(L) − χ(M) + χ(L ⊗ M) = χ(O_{S}) − χ(E) + χ(∧^{2}E) = c_{2}(E).

Apply this now to π^{∗}E on S. There is an exact sequence on C: 0 → L → E → M → 0, so c_{2}(π^{∗}E) = (π^{∗}L · π^{∗}M) = 0. From here on, this argument is “dual” to the one I presented in class, which used a different definition of projective bundle. Also,

0 → O(−1) → π^{∗}E → Q → 0.

Hence h · [Q] = 0. Also, taking the “determinant” of the short exact sequence, we get Q ≅ (π^{∗} ∧^{2} E) ⊗ O_{S}(1), from which [Q] = h + π^{∗}[∧^{2}(E)]. Hence

0 = h · [Q] = h^{2} + h · π^{∗}(∧^{2}E) = h^{2} + deg E,

and we’re done.

Remark. Now that we’ve defined c_{2}, we can describe Noether’s theorem as χ(O_{S}) = \frac{1}{12}(c_{2}(T_{S}) + K_{S}^{2}).