INTRODUCTION TO ALGEBRAIC GEOMETRY, CLASS 23

RAVI VAKIL

Contents

1. More background on invertible sheaves 1 1.1. Operations on invertible sheaves 1 1.2. Maps to projective space correspond to a vector space of sections of a

invertible sheaf 3

2. The class group 4

PS10 due today. PS11 out today. PS9 and PS10 back on Tuesday.

PS11 due Thurs. Dec. 9. PS12 due Monday December 13.

Invertible sheaves are manipulated in a really formal way, so it’s hard to see how geometric it is. The earlier major topics (extension of morphisms to projective varieties over nonsingular points of curves; identification of nonsingular projective curves with finitely generated field extensions of transcendence degree 1) involved proving some technically difficult results with relatively few moving parts. Invertible sheaves and line bundles involves much less technical sophistication, but there are many many more ideas involved, so it’s tricky juggling them all.

I really encourage you to play around with invertible sheaves / line bundles in explicit examples. Choose some nice variety, such as ℙ^{1} or ℙ^{2} or ℙ^{2} minus some curve, and choose some nice invertible sheaf like O(3), and work out spaces of global sections.

Remark. An O_{X}-module is an invertible sheaf if there is an open cover U_{1}, …, U_{n} where L_{U_{i}} ≅ O_{U_{i}}. Reason: all we need to find are the transition functions f_{ij}. Let 1_{i} be the section of L over U_{i} corresponding to 1 in O_{U_{i}}. Then f_{ij} = 1_{i}/1_{j} (although I might have this backwards). Perhaps it is best to say this when first introducing invertible sheaves.

You can use a variant of this idea to show that if there is a global section that vanishes nowhere, then L is triival (i.e. ≅ O_{X}). I’ll give the proof later today, but some of you might already see how to show it.

Date: Thursday, December 2, 1999.

1. More background on invertible sheaves

1.1. Operations on invertible sheaves. Last time I described some basic things you can do with invertible sheaves.

i) Pullback. You can pull back invertible sheaves (or line bundles).

Using that, we proved:

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

Fact. Roya mentioned this last time. (We probably won’t prove this for all n, 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.

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: If L and M have possibly different trivializing open covers, you can “refine” both covers to get a common trivializing open cover.

JP asked why this was called tensor product. Answer: this is tensor product in the category of O_{X}-modules.

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. 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}.

Remark. You can even divide by non-zero sections, to get meromorphic sections. The quotient in the above example is x^{2}/(x + y)^{3}, which is a meromorphic section of O_{ℙ^{1}}(−1). Important note: if you have two non-zero rational sections f, g of an invertible sheaf L, then f/g is a non-zero rational section of the trivial sheaf O, i.e. is a rational function. Let me say that again: the ratio of any two (non-zero) rational sections of a line bundle is a rational function.

You can now check that invertible sheaves form an abelian group:

Definition. The (abelian) group of invertible sheaves on X is called the Picard group of X, and is denoted Pic X.

In general, people often write L^{⊗m} for L ⊗ ⋯ ⊗ L (m times, if m > 0), O_{X} (if m = 0), or L ⊗ ⋯ ⊗ L (−m times, if m < 0); you can check that L^{⊗m} ⊗ L^{⊗n} ≅ L^{⊗(m+n)}.

Remark for the arithmetically minded. If you do this construction in the language of schemes, for Spec of the ring of integers in a number field (a number field is a finite extension of ℚ, and the ring of integers are the integral closure of ℤ in the field), then you get the class group of the field. If you’re interested, and know what the class group is, ask me about it.

We’ll see later that Pic ℙ^{1} is ℤ, and I’d mentioned that Pic ℙ^{n} is ℤ as well. Here’s some other behavior: Picard groups can be torsion.

Exercise. Let X be the variety ℙ^{2} minus an irreducible conic. Let O_{X}(m) be the restriction of O_{ℙ^{2}}(m) to X. Show that O_{X}(2) is trivial, but that O_{X}(1) isn’t. Hence O_{X}(1) is a 2-torsion element of Pic X. (Hint: Show that O_{X}(2) has a global section vanishing nowhere, and that every global section of O_{X}(1) vanishes somewhere. Use the following proposition.)

Proposition. Suppose L is an invertible sheaf on a variety X, and that there is a global section s of L vanishing nowhere. Then L is isomorphic to O_{X}, the “trivial sheaf”.

Proof. We give an identification of L(U) with O_{X}(U) (that commutes with restriction maps); this will do the trick. (Do you see why?) To the section a of O_{X}(U) associate the section as of L(U). (More precisely, the section a res_{X,U} s.) Conversely, to the section b of L(U), associate the section bs^{−1} of L(U).

1.2. Maps to projective space correspond to a vector space of sections of a invertible sheaf. Suppose you have a variety X, and an invertible sheaf L, and n + 1 sections s_{0}, …, s_{n} that don’t have a common zero. Then this induces a map to projective space:

(s_{0}, …, s_{n}) : X → ℙ^{n}.

Here’s how.

Let U_{i} be a cover of X trivializing the invertible sheaf, and let f_{ij} be the transition functions of L with respect to this trivialization. I’ll i) define the morphism from U_{i} → ℙ^{n}, and ii) show that they agree (as maps of sets) on the U_{i} ∩ U_{j}.

i) For the morphism from U_{i} → U_{j}, let (g_{0}, …, g_{n}) be functions on U_{i} corresponding to sections (s_{0}, …, s_{n}). Then g_{0}, …, g_{n} don’t have a common zero in U_{i}, and hence define morphism to ℙ^{n}.

ii) Now consider the two morphisms from U_{i} ∩U_{j} to ℙ^{n}, one given by considering it as a subset of U_{i}, and the other by considering it as a subset of U_{j}. As in i), let (g_{0}, …, g_{n}) be the functions on U_{i} that correspond to our sections (s_{0}, …, s_{n}). Similarly, let (h_{0}, …, h_{n}) be the functions on U_{j}. Then under the first morphism, the point p ∈ U_{i} ∩ U_{j} is sent to

(g_{0}(p); …; g_{n}(p)) = (f_{ij}(p)h_{0}(p); …; f_{ij}(p)h_{n}(p)) = (h_{0}(p); …; h_{n}(p))

which is precisely where it is sent to under the second morphism.

Remark. This generalizes the fact stated earlier, that if you have a projective variety X ⊂ ℙ^{l}, and n + 1 polynomials of the same degree m: s_{0}, …, s_{n}, such that they don’t have a common zero on X, then they define a morphism X → ℙ^{n}. Just apply this construction with the line bundle O_{ℙ^{l}}(m), pulled back to X.

Proposition. Under this morphism π, there is a natural isomorphism L ≅ π^{−1}O(1).

This isn’t too hard to prove. It’s harder than an exercise, but certainly something I could explain to you reasonably quickly (using the remark before teh first section of today’s lecture). But we won’t be using it later in the course, so I’ll omit the proof in the interests of time.

2. The class group

Let C be a nonsingular curve. (What I say will work, with relatively simple modification, for any nonsingular variety, so experts might want to make these modifications as I explain things.) Our goal is to descibe the Picard group (the group of line bundles or invertible sheaves) in a different, hopefully more tractable way.

Definition. The Weil divisor group Div(C) is the free group generated by the points of C. Elements of Div(C) are called Weil divisors (or informally, just divisors) of C.

In other words, elements of Div(C) can be written in the form

∑_{p∈C} n_{p}p

where the n_{p} are integers, and only finitely many of them are zero. It’s clear how these objects form a group.

Definition. Elements of Div(C) where all n_{p} are non-negative are called effective (Weil) divisors on C.

Definition. Suppose s is a non-zero rational section of an invertible sheaf L on C, so s ∈ L(C). Let (s) be the “divisors of poles and zeros of s”.

In other words, if (s) = ∑_{p∈C} n_{p}p, then n_{p} is zero if p is neither a zero nor a pole; if p is a zero, then n_{p} is the order of the zero of s at p; if p is a pole, then n_{p} is the negative of the order of the pole.

For example, let C = ℙ^{1}, L be the structure sheaf, and s the rational function x_{0}(x_{0} − x_{1})/x_{1}^{2}. Then (s) = [0; 1] + [1; 1] − 2[1; 0].

Definition/Proposition. The divisors associated to non-zero rational functions (i.e. the case when L is the trivial bundle) is a subgroup of the divisor group, called the subgroup of divisors linearly equivalent to 0, denoted Div_{0}(C).

Proof. We just need to check that this forms a group. If f and g are rational functions, then (f) + (g) = (fg) (explain), and −(f) = (1/f).

Definition. Define the divisor class group of C Cl(C) as the quotient Div(C)/ Lin(C).

Proposition. Lin ℙ^{1} = {∑ n_{p}p where ∑ n_{p} = 0}.

Here’s why. Choose some standard affine 𝔸^{1} with coordinate t, so that no p is at ∞. Then (∏_{p∈𝔸^{1}}(t − p)^{n_{p}}) = ∑ n_{p}p (explain). Thus the right side is contained in the left side. Conversely, we’ve seen that the sums of the orders of vanishing of any rational function on ℙ^{1} is 0, showing that the left side is contained in the right side.

Corollary. Cl(ℙ^{1}) ≅ ℤ.

Theorem. Cl(C) is naturally isomorphic to Pic(C).

In the course of the proof, we’ll see the identification quite clearly.

Corollary. Pic(ℙ^{1}) ≅ ℤ.

The map Pic(C) → Cl(C). First of all, I mentioned earlier that we have a map as sets from data (L, s) (where s is a non-zero rational section of L) to Div(C). Note that those ordered pairs form a group: (L, s)(M, t) = (L ⊗ M, st), (L, s)^{−1} = (L^{∨}, s^{−1}). Note also that this is a group homomorphism; call it ϕ.

Next, if s and t are two rational sections of L. Then (L, s)(L, t)^{−1} = (O_{X}, st^{−1}). Hence ϕ(L, s) ≡ ϕ(L, t) (mod Lin(C)). Hence we get a map Pic(C) → Cl(C).

The map Cl(C) → Pic(C). Similarly, we’ll construct a map Div(C) → Pic(C), and show that Lin(C) goes to the identity in Pic(C), i.e. the invertible sheaf O_{C}. Given a divisor D, we construct an invertible sheaf O_{C}(D).

Define a sheaf as follows. Recall that the sections of the structure sheaf were defined as

O_{C}(U) = ∩_{p∈U}O_{C,p} = {x ∈ k(C)|v_{p}(x) ≥ 0 for all p ∈ U}.

Remember why it was easy to check that this was a sheaf. Define

O_{C}(∑ n_{p}p)(U) = ∩_{p∈U}O_{C,p} = {x ∈ k(C)|v_{p}(x) + n_{p} ≥ 0 for all p ∈ U}.

Example. Suppose C = ℙ^{1}, with coordinates [x_{0}; x_{1}], and D = [1; 0]. Then let’s find the global sections of O_{C}(D). These are rational functions, whose valuations are non-negative everywhere, except possibly -1 at [1; 0]. One such function is 1. Another is x_{0}/x_{1}, which has a simple pole at [1; 0]. Hence O_{C}([1; 0]) has at least 2 global sections.

Exercise. O(p) on ℙ^{1} is O(1). (Hence we’ve found all the global sections; any global section is a linear combination of these 2.) Another consequence: all invertible sheaves on ℙ^{1} are of the form O(m). (For the experts: you can jack of this proof to show the same result for ℙ^{n}. I can give you the details in a minute or two at some other time if you like.) Addition to problem on problem set: the section corresponding to 1 in the previous paragraph actually has a zero (despite what you think) — find it!

Now we’ve gotten this sheaf; I’ll now show that it is invertible, by giving a trivializing cover.

Suppose p_{1}, …, p_{n} are the points appearing in D (i.e. where n_{p} ≠ 0).

Let U_{0} be the open set C \{p_{1}, …, p_{n}}. Then the restriction of the sheaf O_{C}(D) to this open U_{0} is (isomorphic to) the trivial sheaf visibly: For U ⊂ U_{0}, the sections of O_{C}(D)|_{U_{0}} over U are ∩_{p∈U}O_{C,p} = {x ∈ k(C)|v_{p}(x) ≥ 0 for all p ∈ U}, which is precisely O_{C}|_{U_{0}}.

Now define U_{i} (1 ≤ i ≤ n) as follows. Choose any rational function f_{i} ∈ k(C) such that v_{p_{i}}(f_{i}) = n_{i}. Then (f_{i}) = n_{i}p_{i}+ other stuff. Let U_{i} be C minus the other p_{j} minus the other stuff. Hence the “part of the divisor (f_{i}) on U_{i}” is just n_{i}p_{i}. Note that the U_{i}’s and U_{0} cover all of C; so we just need to show that the sheaf O_{C}(D) is trivial on U_{i} (1 ≤ i ≤ n). For U ⊂ U_{i}, the sections of O_{C}(D)|_{U_{i}} over U are ∩_{p∈U}O_{C,p} = {x ∈ k(C)|v_{p}(x/f) ≥ 0 for all p ∈ U}. This is isomorphic to ∩_{p∈U}O_{C,p} = {y ∈ k(C)|v_{p}(y) ≥ 0 for all p ∈ U} (just take y = x/f), which in turn is precisely O_{C}|_{U_{i}}.

So at this point we have a map Div(C) → Pic(C). To construct the desired map Cl(C) → Pic(C), we just need to show that the image of Lin(C) → Pic(C) is 0. (I didn’t do this in class.) Suppose D ∈ Lin(C). Let s be a rational function such that (s) = D. Then

O_{C}(D)(U) = ∩_{p∈U}O_{C,p} = {x ∈ k(C)|v_{p}(x) + n_{p} ≥ 0 for all p ∈ U}
= ∩_{p∈U}O_{C,p} = {x ∈ k(C)|v_{p}(sx) ≥ 0 for all p ∈ U}
= ∩_{p∈U}O_{C,p} = {y ∈ k(C)|v_{p}(y) ≥ 0 for all p ∈ U} = O_{C}(U).

Notice that this identification commutes with restriction maps. Hence O_{C}(D) ≅ O_{C}.

Both of these maps Pic(C) → Cl(C) and Cl(C) → Pic(C) commute.

I didn’t explain this in class. I’ll just sketch both directions. There’s a lot of content packed into a few sentences; ask me about any steps you’re suspicious about.

Start with a divisor D on C. Consider O_{C}(D) as just defined. Choose the section s corresponding to the rational function 1 ∈ k(C). You can check that the divisor of zeros of s is precisely D. Hence the composition Cl(C) → Pic(C) → Cl(C) is the identity.

For the other direction, suppose you have an invertible sheaf L. Let s′ be any rational section, so we get a divisor (s′) = D. We wish to show that L ≅ O_{C}(D). Construct the section s of O_{C}(D) as in the previous paragraph (corresponding to the rational function 1 ∈ k(C)). You can check that (s′) = D as well. Hence s/s′ is a rational section of L ⊗ O_{C}(D)^{∨}; this rational section has associated divisor D − D = 0, hence has no zeros or poles. Hence this is a regular section (as it has no poles) that is invertible (as it has no zeros). But we’ve seen earlier that any invertible sheaf with a nowhere vanishing section is the trivial sheaf, so L ⊗ O_{C}(D)^{∨} ≅ O_{C}, so L ≅ O_{C}(D) as desired.

This completes the proof of the theorem.

Coming up in the last two classes: Degree of a line bundle/invertible sheaf. The sheaf of differentials (= the canonical sheaf) of a nonsingular curve. The celebrated Riemann-Roch theorem, with applications but no proof.