INTRODUCTION TO ALGEBRAIC GEOMETRY, CLASS 22

RAVI VAKIL

Contents

1. Example: The invertible sheaves O_{ℙ^{n}}(m). 1 1.1. Example: ℙ^{1} 1 1.2. Identification of global sections of O_{ℙ^{n}}(m) with the degree m polynomials

in x_{0}, …, x_{n} (m ≥ 0) 2

2. More background on invertible sheaves 3 2.1. Operations on invertible sheaves 3

Problem Set 8 back at end of lecture. PS9 due today.

PS10 due Thursday December 2. PS11 due Thurs. Dec. 9. PS12 due Monday December 13. If you don’t get a chance to do PS12 by December 13, but still want to think about it, I’ll be around, so you can drop by and we can talk.

1. Example: The invertible sheaves O_{ℙ^{n}}(m).

Caution. I don’t I think I was consistent with transition functions last time. I think everything is correct in the lecture notes. If g_{i} is a section over U_{i} and g_{j} is a section over U_{j}, then g_{i} = f_{ij}g_{j}.

I’ll now describe some invertible sheaves on projective space. I think this description will help you see how to deal with invertible sheaves in general.

Let U_{0}, …, U_{n} be the standard opens of ℙ^{n}. Consider the invertible sheaf (known as O_{ℙ^{n}}(m)) given by the transition functions f_{ij} = (x_{j}/x_{i})^{m}. Recall for this sheaf, the sections corresponding to an open set U are given by the data of sections g_{i} ∈ O(U∩U_{i}) (i.e functions on U∩U_{i}), satisfying the transition information g_{i} = f_{ij}g_{j} on the open set U ∩ U_{i} ∩ U_{j}.

Note that f_{ii} = 1, and f_{ik} = f_{ij}f_{jk}.

We can talk about rational (or meromorphic) sections as well.

Date: Tuesday, November 30, 1999.

1.1. Example: ℙ^{1}. Let’s do an example. To keep with last day’s example, let’s let n = 1, and let’s first take m = −1. Here are the variables we’ll use.

On ℙ^{1}, use projective coordinates [x_{0}; x_{1}].

On open set U_{0}, choose coordinate y_{1} = x_{1}/x_{0}, so U_{0} are the points [1; y_{1}]. (Note that this is equal to [x_{0}; x_{1}]!)

On open set U_{1}, choose coordinate z_{0} = x_{0}/x_{1}, so U_{0} are the points [z_{0}; 1].

Here’s a section over U_{0}: g_{0}(y_{1}) = y_{1} −1. Let me ask you some stupid questions. Where does it have zeroes or poles? Answer: a zero of order 1 at y_{1} = 1, i.e. at [x_{0}; x_{1}] = [1; 1].

Let’s consider this as a rational section over all of ℙ^{1}. What does this section look like over U_{1}? It is given by a function g_{1}(z_{0}) = f_{10}f_{0}(y_{0}) = x_{1}/x_{0}(1/z_{0} − 1) = (z_{0} − 1)/z_{0}^{2}. Zero of order 1 at [1;1], and pole of order 2 at [0;1].

Remark. Note that we calculated that this section had a zero of order 1 at [1; 1] in both open neighbourhoods. But the functions whose order of vanishing we checked were different! Why were we guaranteed to get the same answer? Answer: they differed by a factor that was invertible (i.e. had 0 valuation).

Proposition. O_{ℙ^{1}}(−1) is not isomorphic to the structure sheaf.

Proof 1. The sums of orders of vanishing of this meromorphic section is -1. The sums of orders of vanishing of a meromorphic structure sheaf is 0. (This is the proof I gave last time.

Remark. Notice that In general, the sums of orders of vanishing of a meromorphic section of O_{ℙ^{1}}(m) is m. So they are all non-isomorphic invertible sheaves.

Proof 2. You can check that O_{ℙ^{1}}(−1) has no non-zero global sections. But the structure sheaf has non-zero global sections (the non-zero constants).

1.2. Identification of global sections of O_{ℙ^{n}}(m) with the degree m polynomials in x_{0}, …, x_{n} (m ≥ 0). As another example, I’ll let you calculate the space of global sections of O_{ℙ^{1}}(2).

Using the same variables as before, g_{0}(y_{1}) is a polynomial in y_{1}. g_{1}(z_{0}) is a polynomial in z_{0}, and g_{1}(z_{0}) = z_{0}^{2}g_{0}(1/z_{0}). Hence g_{0} can have degree at most 2. Conversely, any polynomial of degree 2 will work. Hence the space of global sections of O_{ℙ^{1}}(2) has dimension 3. (The same argument will show that the space of global sections of O_{ℙ^{1}}(n) has dimension n + 1 if n ≥ 0.)

You can identify the space of global sections with the vector space of degree 2 polynomials in x_{0} and x_{1}. Here’s the identification. Given a polynomial ax_{0}^{2} + bx_{0}x_{1}+cx_{1}^{2}, to get the corresponding section g_{0}(y_{1}) over U_{0}, just divide by x_{0}^{2}. Hence get a + b(x_{1}/x_{0}) + c(x_{1}/x_{0})^{2} = a + by_{1} + cy_{1}^{2}. To get the corresponding section g_{1}(z_{0}) over U_{1} just divide by x_{1}^{2}, and get a(x_{0}/x_{1})^{2} + b(x_{0}/x_{1}) + c = az_{0}^{2} + bz_{0} + c. Note that g_{0} and g_{1} satisfy the desired transition data: g_{0} = (x_{0}/x_{1})^{2}g_{1}.

In the same way you can prove:

Proposition. Let m ≥ 0. Then the vector space of global sections of O_{ℙ^{n}}(m) can be identified with the vector space of polynomials h(x_{0}, …, x_{n}) of degree m. The identification is this: g_{i} = h/x_{i}^{m}.

Probably an exercise. You’ve actually done this in the case when m = 0, when you proved that the only regular functions on projective space are the constants.

This is really concrete, and to prove to you how concrete it is, let me ask you some questions.

Here’s a section of O_{ℙ^{2}}(3): x^{3} + y^{3} + z^{3}. Where does it have zeroes? How about xy^{2}?

We can extend this to rational sections. Once again, they have to be homogeneous degree m. How about (x^{3} + y^{3} + z^{3})/(x^{2}yz), which is a meromorphic section of O_{ℙ^{2}}(−1). Where are the poles and zeroes?

2. More background on invertible sheaves

2.1. Operations on invertible sheaves. Here are some basic things you can do with invertible sheaves.

i) Pullback. You can pull back invertible sheaves (or line bundles). (Give picture first.) Here’s how. If you have a morphism π : X → Y, and you have an invertible sheaf L on Y defined by open sets U_{i} and transition functions f_{ij} on U_{i} ∩ U_{j}, then define the inverse invertible sheaf π^{∗}L by the open sets π^{−1}(U_{i}), and the functions π^{∗}f_{ij}, which are functions on π^{−1}(U_{i} ∩ U_{j}) = π^{−1}(U_{i}) ∩ π^{−1}(U_{j}).

You can pull back sections as well: if g is a section of L over some open subset U ⊂ Y, then you can define a section π^{−1}g of π^{−1}L over π^{−1}(U) (which is also open). For convenience of notation, I’ll show you how this works for global sections. If g is a global section given by regular functions g_{i} on U_{i} related by g_{i} = f_{ij}g_{j} on U_{i} ∩ U_{j}, when consider the regular functions π^{∗}g_{i} on π^{−1}U_{i}. (Remember that you can pull back regular functions via morhpisms!) Then because g_{i} = f_{ij}g_{j} on U_{i} ∩ U_{j}, we have π^{−1}g_{i} = π^{−1}f_{ij}π^{−1}g_{j} on π^{−1}(U_{i} ∩ U_{j}) = π^{−1}(U_{i}) ∩ π^{−1}(U_{j}). In other words, these regular functions define a section of π^{∗}L.

As an example of how you can use this, consider:

Proposition. O_{ℙ^{n}}(m_{1}) is not isomorphic to O_{ℙ^{n}}(m_{2}) if m_{1} ≠ m_{2}.

Proof. Recall that we’ve proved this for n = 1: any non-zero meromorphic section has “total order of vanishing” m.

Now we’ll deal with the general n. Take the closed immersion ℙ^{1} ↪ ℙ^{m}, sending ℙ^{1} to a line, e.g. (x_{0}; x_{1}) ↦ (x_{0}; x_{1}; 0; …; 0). Check that O_{ℙ^{n}}(m_{i}) pulls back to O_{ℙ^{1}}(m_{i}). But O_{ℙ^{1}}(m_{1}) is not isomorphic to O_{ℙ^{1}}(m_{2}).

Fact. (We probably won’t prove this, but the proof isn’t difficult to follow.) These are all the line bundles on ℙ^{m}. Hence Pic ℙ^{n} ≅ ℤ, with O(1) the generator.

Example: Line bundles on curves. This example is somewhat lame. Consider the curve C y^{2}z = x^{3} + xz^{2}. We can pull back the line bundle O_{ℙ^{2}}(2) to C. We already can see that it has at least 6 linearly-independent sections. Where on the curve does the section x^{2} + y^{2} vanish? Answer: where x^{2} + y^{2} = 0. We can pull back the line bundle O_{ℙ^{2}}(3) to C. We can already see that it has at least 9 linearly independent sections. O_{ℙ^{2}}(3) has 10 sections. But one of them vanishes on C: the section y^{2}z − x^{3} − xz^{2}.

ii) Tensor product of two invertible sheaves. Take 1: Suppose you have two invertible sheaves L and M on X, given by the same open cover U_{i} and (possibly different) transition functions l_{ij} and m_{ij}. Then define the tensor product invertible sheaves L ⊗ M by the same open cover, and the transition function l_{ij}m_{ij}. (You can immediately check that this satisfies the cocyle condition.)

Take 2: Suppose now that you have two invertible sheaves L and M, given with the data of possibly different open covers U_{i} and V_{j} respectively, with transition functions l_{ii′} and m_{jj′} respectively. Then what you want to do is to take a different cover finer than the U_{i} and the V_{j} so you can express L and M using this new cover; then use the construction above. The following does the trick: L ⊗ M is defined using the open sets U_{i} ∩ V_{j}, and the transition functions are given by f_{ij,i′j′} = l_{ii′}m_{jj′}.

Things you might want to check: that this construction is independent of the “representation” of the invertible sheaf. Also, L ⊗ O_{X} is isomorphic to L.

Remark. If a is a global section of L (or more generally, a section over some open set), and b is a section of M, then you can interpret ab as a section of L ⊗ M.

Remark. You can see how this works with O_{ℙ^{1}}(m). Immediately, we have O_{ℙ^{1}}(m + n) ≅ O_{ℙ^{1}}(m) ⊗ O_{ℙ^{1}}(n).

Remark. If a is a section of L (over some open set) and b is a section of M (over the same open set), then ab is naturally a section of L ⊗ M. For example, x^{2} is a global section of O_{ℙ^{1}}(2), and (x + y)^{3} is a global section of O_{ℙ^{1}}(3). What is their product as a global section of O_{ℙ^{1}}(5)? Answer: x^{2}(x + y)^{3}.

iii) Inverse invertible sheaves. Suppose you have an invertible sheaf L defined by the open cover U_{i} and transition functions f_{ij}. Then define the inverse, denoted L^{−1}, by the same open cover, and the transition functions f_{ij}^{−1}. Note that the cocycle condition is satisfied, and also that L ⊗ L^{−1} ≅ O_{X}.

Coming next: The Picard group. Maps to projective space correspond to a vector space of sections of an invertible sheaf. The class group. The canonical invertible sheaf (= the sheaf of differentials). Genus. Riemann-Roch Theorem: statement (no proof) and applications. The Riemann-Hurwitz formula.