COMPLEX ALGEBRAIC SURFACES CLASS 3

RAVI VAKIL

Contents

1. Another extended example: ℙ^{n} 1 1.1. Useful interpretation of rational sections of O_{ℙ^{n}}(d) 2 1.2. The canonical sheaf of ℙ^{n} 3 1.3. Sections of O_{ℙ^{n}}(d) and maps to projective space; more generally, invertible

sheaves and maps to projective space 4

Recap of last time.

The times have changed to Wednesdays and Fridays 2:10–3:25. We will likely meet on a couple of Mondays as well for 50 minutes to catch up if need be; if that happens, I’ll warn you in advance.

We’ve been discussing line bundles, a.k.a. invertible sheaves; they form a group called Pic(X), a.k.a. H^{1}(X, O_{X}^{∗}).

We related invertible sheaves and divisors. In particular, for a smooth variety, we showed that Pic(X) ≅ Div(X) modulo linear equivalence. The divisors linearly equivalent to 0 were the divisors of rational functions. This involved the construction of the invertible sheaf O(D), where D is a divisor.

If X is a (proper, nonsingular) curve, then there is a degree map Pic(X) → ℤ.

As examples, we showed that Pic(𝔸^{1}) ≅ {1}, by showing directly that any point was rationally equivalent to 0. Secondly, we showed that Pic(ℙ^{1}) ≅ ℤ by showing that any two points were rationally equivalent, so Pic(ℙ^{1}) was generated by O(pt). Using the degree map, for example, we showed that O(n(pt)) ≠ O.

1. Another extended example: ℙ^{n}

This actually generalizes.

Date: Wednesday, October 9.

First, Pic(𝔸^{n}) = {1}. Reason: Let 𝔸^{n} = Spec[x_{1}, …, x_{n}]. Then the irreducible divisors correspond to irreducible polynomials (up to scalars). This requires the fact that all codimension one loci are the zero sets of polynomials in the n variables. (This isn’t hard, but involves the algebraic theory of dimension, so we won’t go into it here.) Thus any divisor is the divisor of a rational function, e.g. (y − x^{2}) − 3x − 2y = div((y − x^{2})/(x^{3}y^{2})).

Next: on to ℙ^{n} = Proj k[x_{0}, …, x_{n}]. (Should I say something about Proj notation? The x_{i} are projective coordinates.) The irreducible divisors correspond to irreducible homogeneous polynomials in the x_{i} of positive degree. Reason: For future use, let U_{0}, …, U_{n} be the basic affine open sets; U_{i} is where x_{i} ≠ 0. Note that ℙ^{n} − U_{0} = (x_{0} = 0) = (x_{0}), so the only irreducible divisor missing U_{0} is the hyperplane x_{0} = 0. Given an irreducible divisor meeting U_{0}, it is irreducible in U_{0}, and hence corresponds to an irreducible polynomial p(u_{1}, …, u_{n}) in the coordinates of U_{0} (where u_{i} = x_{i}/x_{0}), the full divisor is the closure of its restriction to U_{0}. In confusing math-ese: D = \overline{D|_{U_{0}}}. Then check that D is the vanishing set of the polynomial x_{0}^{deg p}p(u_{1}, …, u_{n}), and that this is an irreducible homogeneous in the x_{i}.

Finally, Theorem. Pic(ℙ^{n}) ≅ ℤ.

Remark. The generator is the class of the hyperplane H. As with ℙ^{1}, we use the notation O_{ℙ^{n}}(d) for O_{ℙ^{n}}(dH). (Hence we have a degree map here too.)

Proof. We need to show (i) that any divisor is linearly equivalent to a multiple of H, and (ii) that O(dH) ≠ 0.

(i) Given an irreducible divisor p(x⃗) = 0, note that p/x_{0}^{d:=deg p} is a rational function, and its divisor is (p) − d(x_{0}), so (p) = dH in the Picard group.

(ii) Interpret H as O_{ℙ^{n}}((x_{0})). Consider the immersion of a line in projective space ℙ^{1} ↪ ℙ^{n}, [x; y] ↦ [x; y; 0; ...; 0]. Pullback gives a map Pic(ℙ^{n}) → Pic(ℙ^{1}). Take a hyperplane meeting the line at a point, e.g. (x_{0}). Check that O_{ℙ^{n}}(H) pulls back to O_{ℙ^{1}}(pt), which has infinite order.

I next want to give you a way of interpreting sections of O_{ℙ^{n}} over any open set.

1.1. Useful interpretation of rational sections of O_{ℙ^{n}}(d). Rational sections of O_{ℙ^{n}}(d) correspond to rational homogeneous functions P (x_{0}, …, x_{n})/Q(x_{0}, …, x_{n}) with degree d. This behaves well with respect to multiplication, i.e. if you have a rational sections s resp. t of O_{ℙ^{n}}(d) resp. O_{ℙ^{n}}(e), then st is a rational section of O_{ℙ^{n}}(d + e), and if t ≠ 0 then s/t is a rational section of O_{ℙ^{n}}(d − e), and this interpretation respects multiplication and division.

If you want to see actual sections over an open set U, you allow poles away from U. For example, global sections correspond to the vector space of degree d in the n + 1 projective co-ordinates x_{0}, …, x_{n}. (As a corollary, h^{0}(ℙ^{n}, O_{ℙ^{n}}(d)) = \binom{n+d}{d}.) For example, O_{ℙ^{2}}(2) is a 6-dimensional vector space, with generators x_{0}^{2}, …, x_{1}x_{2}. What’s the divisor corresponding to global section x_{0}x_{1} + x_{0}x_{2}? Ans: (x_{0}) + (x_{1} + x_{2}).

Proof: Exercise.

I’ll do it for global sections of ℙ^{2}, so you can see the bijection.

Let me pick coordinates on the three open sets:

• U_{0} = {x_{0} ≠ 0}. Coordinates (u_{1}, u_{2}) where (x_{0}; x_{1}; x_{2}) = (1; u_{1}; u_{2}), so u_{1} = x_{1}/x_{0}.

• U_{1} = {x_{1} ≠ 0}. Coordinates (v_{0}, v_{2}) where (x_{0}; x_{1}; x_{2}) = (v_{0}; 1; v_{2}).

• U_{2} = {x_{2} ≠ 0}. Coordinates (w_{0}, w_{1}) where (x_{0}; x_{1}; x_{2}) = (w_{0}; w_{1}; 1).

Then on the overlap, you can check things like w_{0} = v_{0}w_{1} etc.

Let’s consider O_{ℙ^{2}}(2) in its guise of O_{ℙ^{2}}((x_{0})^{2}).

• Restrict first to U_{0}. We’re allowed to have polynomials in u_{1} and u_{2}.

• Restrict to U_{1}. We’re allowed to have polynomials in v_{0} and v_{2}, and poles of order up to 2 in v_{0}.

• Restrict to U_{2}. We’re allowed to have polynomials in w_{0} and w_{1}, and poles of order up to 2 in w_{0}.

Then we have gluing data. Now you do the algebra, and when the dust settles, what this corresponds to are polynomials in u_{1} and u_{2} of degree up to 2:

?+?u_{1}+?u_{2}+?u_{1}^{2}+?u_{1}u_{2}+?u_{2}^{2} = \frac{1}{x_{0}^{2}} (x_{0}^{2}+?x_{0}x_{1}+?x_{0}x_{2}+?x_{1}^{2}+?x_{1}x_{2}+?x_{2}^{2}).

So we made a choice of manifestation of O_{ℙ^{2}}(2) by picking the divisor x_{0}^{2}, and we got homogeneous polynomials with denominator x_{0}^{2}. Now you check that if you picked a different manifestation O_{ℙ^{2}}(2) ≅ O_{ℙ^{2}}((p)) where p is degree 2, then the last line would have been

\frac{1}{p}(x_{0}^{2}+?x_{0}x_{1}+?x_{0}x_{2}+?x_{1}^{2}+?x_{1}x_{2}+?x_{2}^{2}),

and in fact the isomorphism between the two would preserve the degree 2 polynomial. Hence this correspondence between the vector space of global sections and the vector space of degree 2 polynomials is well-defined.

This argument extends to (i) rational sections and (ii) ℙ^{n} without change.

Let me make this very explicit. Suppose I have a section of O_{ℙ^{2}}(2) that I’m calling x_{0}^{2} − x_{1}x_{2}. You would like to see it as an element of O_{ℙ^{2}}(D), where D has degree 2, for example O_{ℙ^{2}}(x_{0}x_{2}). Then the corresponding element of O_{ℙ^{2}}(D) is (x_{0}^{2} − x_{1}x_{2})/(x_{0}x_{2}). The content of this “theorem” is that this is a well-defined bijection, independent of your choice of D.

1.2. The canonical sheaf of ℙ^{n}. Now let’s find the canonical sheaf of ℙ^{n}. My goal here is (i) to show you that the canonical sheaf isn’t scary, and (ii) to actually get the number.

Theorem. K_{ℙ^{n}} ≅ O_{ℙ^{n}}(−n − 1). Remember this!

Proof. I’ll just do the case n = 2 for notational convenience. The sections of K over U_{0} are p(u_{1}, u_{2})du_{1} ∧ du_{2}. Let’s take the section du_{1}du_{2} over this open set, and see what its poles and zeroes are over all of ℙ^{2}. There’s only one divisor away from U_{0}, x_{0} = 0, so we need only check this divisor.

Let’s look over U_{1}, with coordinates v_{0} and v_{2}. Remember that (1; u_{1}; u_{2}) = (v_{0}; 1; v_{2}), so u_{1} = 1/v_{0} and u_{2} = v_{2}/v_{0}.

Thus in terms of the coordinates of U_{1} the section transforms into:

du_{1} ∧ du_{2} = (−\frac{1}{v_{0}^{2}} dv_{0}) ∧ (\frac{v_{0}dv_{2} − v_{2}dv_{0}}{v_{0}^{2}} ) = −\frac{1}{v_{0}^{3}} dv_{0} ∧ dv_{2}.

So we see a pole along v_{0} = 0 of order 3, as desired.

1.3. Sections of O_{ℙ^{n}}(d) and maps to projective space; more generally, invertible sheaves and maps to projective space. One of the central facts about invertible sheaves on proper schemes X is loosely that global sections give maps to projective space. I want to show this to you first in the case of O_{ℙ^{2}}(d) to show you that this isn’t hard, and then I’ll return to a general discussion of varieties.

First, let me take 4 sections of O_{ℙ^{2}}(2). For convenience, let me take the projective coordinates to be x, y, z, rather than x_{0}, x_{1}, x_{2}. I’ll choose the 4 sections: x^{2}, y^{2}, z^{2}, xy. Then for any point of ℙ^{2}, the point [x^{2}, y^{2}, z^{2}, xy] is a well-defined point of ℙ^{3}. (Explain: (i) even though x etc. isn’t well-defined, this is a well-defined point of projective space, and (ii) these are never all zero.)

More generally: Definition. For an arbitrary scheme X with invertible sheaf L, a vector space of global sections with basis s_{0}, …, s_{n} is said to be base-point free if they have no common zeros on X. Then a basepoint free vector space V of n + 1 global sections gives a map to ℙ^{n}. If you unwind the definition carefully, you’ll see that this gives X → ℙV^{∗}.

I should also then define base points: Definition. given a vector space of global sections, their locus of common zeros is called the base locus, or base points. (Normally you take the scheme-theoretic intersection.)

Important fact: there is a converse to this construction. If X is proper (not necessarily nonsingular): there is a bijection between π : X → ℙ^{n} and (X, L, (s_{0}, …, s_{n})/k^{∗}) where s_{i} ∈ Γ(X, L). The bijection from right to left was described before. In the other direction: L = π^{∗}O(1), and s_{i} = π^{∗}x_{i}.

To see if you understand this fact, here’s an immediate consequence: Exercise. The only morphisms from ℙ^{n} to ℙ^{m} if m < n are the constant maps.

Back to the example of ℙ^{2} and O_{ℙ^{2}}(2): The sheaf O_{ℙ^{2}}(2). Six sections.

[x; y; z] → [x^{2}; y^{2}; z^{2}; xy; yz; zx].

(Draw a picture.) Hyperplane sections correspond to conics. The degree of a subvariety X of ℙ^{n} can be defined as the number of intersection points of X with a general linear space of complementary dimension. Hence the degree of this embedded ℙ^{2} is 4. (Common terminology that won’t come up in this course: This is an example of the famous Veronese embedding of ℙ^{2}. In general, use ℙ^{n} and its space of global sections O_{ℙ^{n}}(d) to get a map to a big projective space ℙ^{N}, where N = \binom{d+n+1}{d} − 1 (I think!). Exercise. The degree of the Veronese-embedded space is d^{n} by a similar argument.

Definition. An invertible sheaf L is very ample if the global sections of L gives a closed immersion into projective space.

Fact. equivalent to: “separates points and tangent vectors”. (Explain why, loosely. Separating points means there is a hyperplane passing through one point but not the other; that means that they don’t map to the same point in projective space. Separating tangent vectors loosely means that by the implicit function theorem, you have a local isomorphism. Complex geometers might buy this.)

Again, hyperplane sections correspond to H^{0}(X, L).

Definition. The corresponding map to projective space is called a linear system. (I’m not sure if I’ll use this terminology, but I want to play it safe.)

|L| : X → ℙ^{n} = ℙH^{0}(X, L)^{∗}.

Definition. An invertible sheaf is ample if some power of it is very ample.

Note: A very ample sheaf on a curve has positive degree. Hence an ample sheaf on a curve has positive degree. We’ll soon see that this is an “if and only if”.

Fact (Serre vanishing). Suppose M is any coherent sheaf e.g. an invertible sheaf, or more generally a locally free sheaf (essentially, a vector bundle), and L is ample. Then for n >> 0, H^{i}(X, M ⊗ L^{n}) = 0 for i > 0.

Next day: Serre duality; Riemann-Roch; lots of applications to curves. Then on to surfaces!