COMPLEX ALGEBRAIC SURFACES CLASS 12

RAVI VAKIL

Contents

1. Geometric facts about a geometrically ruled surface π : S = ℙ_{C}E → C from geometric facts about C 2

2. The Hodge diamond of a ruled surface 3 3. The surfaces 𝔽_{n} 3 3.1. Getting from one 𝔽_{n} to another by elementary transformations 4 4. Fun with rational surfaces (beginning) 4

Last day:

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.

We showed that extensions 0 → L → E → M → 0 are classified by H^{1}(C, L ⊗ M^{∗}). The element 0 corresponded to a splitting. If one element is a non-zero multiple of the other, they correspond to the same E, although different extensions.

As an application, we proved: Proposition. Every rank 2 locally free sheaf on ℙ^{1} is a direct sum of invertible sheaves.

I can’t remember if I stated the implication:

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.

Date: Friday, November 8.

(We don’t yet know that they are all different yet, but we will soon.)

We then began to study :

1. Geometric facts about a geometrically ruled surface π : S = ℙ_{C}E → C from geometric facts about C

In particular, we found the intersection theory of S in terms of Pic C. One of the players was the class O(1) of the projective bundle. Let me repeat the definition of this.

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.

(c) In the case where dim C = 1 and dim P = 2, O(1) · F = 1 for any fiber F. (This generalizes to projective bundles of arbitrary dimension once we have intersection theory.)

(d) Hence if L is any invertible sheaf on C, then π^{∗}L · O(1) = deg L.

Once you check (a), the rest follow quickly in order.

With this definition, we have:

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).

I gave a proof of this, but my proof of (iv) needs to be edited. The corrected version is in the Class 11 notes. (Remark: The reason for the discrepancy is because there are two possible definitions of the projectivization of a vector space. The traditional one is that of one-dimensional subspaces of a vector space. That’s the one that most of the world uses. An alternate one is that of one-dimensional quotients of a vector space. That’s the one that Grothendieck used, because it makes certain statements cleaner, and as a result much but not of the algebraic geometry community uses this definition. So be warned.)

2. The Hodge diamond of a ruled surface

Recall that the outer entries of a Hodge diamond are birational invariants. I should have proved this earlier, and will now. (Move to earlier notes at some point.)

Theorem. h^{0}(S, Ω_{S}) = h^{1,0} is a birational invariant. (Similarly for h^{2,0}, and also so for pluricanonical forms P_{n} = H^{0}(S, K_{S}^{⊗n}).) (This works for smooth projective varieties of any dimension.)

Proof. Suppose ϕ : S′ −−→ S is a birational map, which is a morphism from S′ − F → S, where F is a finite set. Thus we have a map H^{0}(S, Ω_{S}) → H^{0}(S′ − F, Ω_{S′−F}). In fact, this extends: the poles of a differential form are pure codimension 1. Thus we have H^{0}(S, Ω_{S}) → H^{0}(S′, Ω_{S′}) which takes a preserves the restriction of the form to their common open set. The same argument works for the rational map S → S′, so we get an isomorphism.

Hence the numerical invariants of a ruled surface π : S → C are as follows. The outer entries of the Hodge diamond are the same as for C × ℙ^{1}, which can be checked directly to be

h^{0,0} = 1
h^{1,0} = g  h^{0,1} = g
h^{2,0} = 0 h^{1,1} h^{0,2} = 0
h^{2,1} = g  h^{1,2} = g
h^{2,2} = 1

The central number is the rank of the Picard group (by the Lefschetz (1,1)-theorem), or over ℂ can be found using the Euler characteristic.

Also, P_{n} = H^{0}(S, K_{S}^{⊗n}) = 0. Important fact: this characterizes ruled surfaces. (I doubt we’ll get to this fact in this course, but we might.)

Second important fact: if q = P_{2} = 0, then S is rational. (Castelnuovo’s Rationality Criterion, to be discussed in a few classes.)

3. The surfaces 𝔽_{n}

We already know that Pic 𝔽_{n} = ℤh ⊕ ℤf (n ≥ 0) with f^{2} = 0, f · h = 1, and h^{2} = n. All geometrically ruled surfaces over ℙ^{1} are of this sort. But are they all different? Yes!:

Proposition. If n > 0, there is a unique irreducible curve E on 𝔽_{n} with negative selfintersection. If e is its class in Pic 𝔽_{n}, then e = h − nf.

(This curve is often called E.)

Corollary. e^{2} = −n. Note that 𝔽_{0} = ℙ^{1} × ℙ^{1} has no curves of negative self-intersection, as they all “move” (under the automorphisms of both ℙ^{1}’s). Hence 𝔽_{n} and 𝔽_{m} are not isomorphic unless n = m. Also, all 𝔽_{n} are minimal except for 𝔽_{1}, which has a (−1)-curve. In fact, 𝔽_{1} is the blow-up of ℙ^{2} at a point. Reason: the blow-up of ℙ^{2} at a point is a rational geometrically ruled surface with a (-1)-curve, so it must be 𝔽_{1}!

Proof of the proposition. First, I’ll produce the curve E of negative self intersection, which is a section. Consider the section E of π : 𝔽_{n} → ℙ^{1} corresponding to the subline bundle O_{ℙ^{1}}. Let e be its class of this curve E in Pic 𝔽_{n}. e = h + rf for some r. Since O_{𝔽_{n}}(1)|_{E} = O_{ℙ^{1}}, we know that h · e = deg O_{𝔽_{n}}(1)|_{E} = 0, from which r = −n, and then e^{2} = (h − nf)^{2} = −n.

Next, I’ll show that this is the only curve of negative self-intersection. If C is irreducible on 𝔽_{n}. [C] = ah + bf. C · f = 0 implies a ≥ 0. C · E ≥ 0 implies (bf · (h − nf)) = b ≥ 0. Finally, C · C = (ah + bf) = a^{2}n + 2ab ≥ 0.

3.1. Getting from one 𝔽_{n} to another by elementary transformations. Consider 𝔽_{n}. Blow up a point on E, and blow down the proper transform of the fiber. We again have a rational ruled surface. Show that you have 𝔽_{n+1}.

Instead, blow up a point not on E, and blow down the proper transform of the fiber. Show that you have 𝔽_{n−1}. What goes wrong when n = 0?

4. Fun with rational surfaces (beginning)

The surface F_{0}. This is ℙ^{1} × ℙ^{1}. By the above determination of the intersection theory of 𝔽_{n}, the intersection theory of 𝔽_{0} is ℤ[h, f]/h^{2} = f^{2} = 0, hf = 1. h and f are the classes of fibers of the two projections to ℙ^{1}. Let ([x; y], [u; v]) be co-ordinates to ℙ^{1} × ℙ^{1}. Then xu, xv, yu, yv are sections of O(h + f), that separate points and tangent vectors, and hence give a closed immersion into ℙ^{3}. The degree of the immersed surface is (h + f)^{2} = h^{2} + 2hf + f^{2} = 0 + 2 + 0 = 2, so we have a smooth quadric in ℙ^{3}. Conversely, any two smooth quadrics over an algebraically closed field are isomorphic. The reason is linear algebra: quadratic forms in 4 variables are classified, and are (up to change of basis):

0, x_{0}^{2}, x_{0}^{2} + x_{1}^{2}, x_{0}^{2} + x_{1}^{2} + x_{2}^{2}, x_{0}^{2} + x_{1}^{2} + x_{2}^{2} + x_{3}^{2}.

Only the last one is smooth. (The rest are: all of projective space, a double plane, the union of two planes, and a quadric cone.)

Thus all smooth quadrics in ℙ^{3} (over an algebraically closed field) are isomorphic to ℙ^{1} × ℙ^{1}.

Many of the other surfaces corresponds to blow-ups of ℙ^{2} at a certain number of points. Before discussing them, here is a

Useful proposition. Consider the blow-up of ℙ^{2} at n general points, giving exceptional divisors E_{1}, …, E_{n}. Then the intersection ring on ℙ^{2} is given by

ℤ[H, E_{1}, …, E_{n}]/H^{2} = 1, HE_{i} = 0, E_{i} · E_{j} = 0, E_{i}^{2} = −1.

We can understand divisors and sections of divisors in terms of divisors on ℙ^{2} with certain multiplicities at the E_{j}. More precisely: the vector space of sections of aH − b_{1}E_{1} − ⋯ − b_{n}E_{n} is naturally isomorphic to the vector space of degree a polynomials in ℙ^{2} vanishing with multiplicity at least b_{i} on E_{i}.

I’ll prove this next class.

The surface 𝔽_{1}. As observed before, 𝔽_{1} is (isomorphic to) the blow-up of ℙ^{2} at a point.

Consider the divisor class 2H − E_{1}. This corresponds to conics in ℙ^{2} through p_{1}, which gives a five-dimensional vector space. It separates points and tangent vectors (somewhat obviously away from E_{1}, and not so obviously along E_{1}). Thus we get an immersion of 𝔽_{1} into ℙ^{4}. Its degree is (2H − E) · (2H − E) = 3. We get a cubic surface in ℙ^{4}.

Blow up ℙ^{2} at two points. Now blow up ℙ^{2} at two points. Where are the (−1)-curves? There are obviously two: our two exceptional divisors. But there is one more: the proper transform of the line joining the two points. It has self-intersection number (H − E_{1} − E_{2})^{2} = −1.

When you blow down that “bonus” rational curve, what do you get? In fact: ℙ^{1} × ℙ^{1}. Eric saw this by interpreting what we just did as an elementary transformation of 𝔽_{1}.

Here’s another way of seeing it. Conic through 2 points (2H − E_{1} − E_{2}) gives a map to ℙ^{3}, and it is degree 2. Get smooth quadric surface.

What are the maps to the two ℙ^{1}’s? Answer: projection from each of two points. You can see why the proper transform of the line gets sent to a point under these two projections.