I’m back! So let’s get down to business.
What I really want to talk about is the Riemann Roch Theorem. You may wonder why, since it has already been discussed here , but there have been vast generalizations of this theorem throughout the last century. I want to focus on Grothendieck’s version, but I plan to start slow. The material I’m following will mostly be taken from the classic Serre-Borel paper (recently pointed out to me). Today, we’ll content ourselves with talking about proper maps. I know that Charles already talked about a complete variety here and even proved Chow’s Lemma which is great. It tells us, that all complete varieties are “close” to being projective, so today we’ll focus on the projective ones.
First with the formalities: all varieties will be assumed to be defined over an algebraically closed field. For today, a variety will mean a quasi projective one. Recall that this means it is (isomorphic to one which is) a locally closed subvariety of projective space. A variety is called projective if it (isomorphic to one which is) a closed subvariety of projective space. The letters X and Y will always denote varieties, P and P’ will denote projective spaces. The notation of the ground field will be omitted!
Now, Charles already outlined in his post why the projection takes closed sets to closed sets. Recall this means that projective space is complete. Recall also that given a morphism , we denote to be the graph of f. It’s easy to see that the graph is closed inside the product. Let’s prove two quick facts:
Lemma: Let and be two morphisms, where X and Y are subvarieties of the projective spaces P and P’. Suppose that is closed in and is closed inside . Then is closed inside .
Proof: We have that , and each inclusion is closed in the next. In particular, is closed inside . Note that when we project down to , we get exactly the graph of , which is then closed because projective space is complete.
Lemma: Suppose that is a morphism, and that X can be embedded into two projective spaces P and P’. If is closed, then it is also closed inside .
Proof: Here we apply the previous Lemma to , where the first map is the identity and the second map is f. We must only check that the graph of the identity is closed in , but we’ll leave that as an easy exercise!
This Lemma justifies, Definition: A morphism is called proper if the graph is closed inside for some projective space P containing X.
We immediately have the following equivalence which shows how proper morphisms behave much like complete varieties. (Indeed, it implies that the structure map of a variety is proper if and only if the variety is complete).
Proposition: A morphism is proper if and only if for each variety Z, and each closed subset T of , the image of T in is closed.
Proof: Let P be a projective space containing the embedded X. Suppose f is proper, that is, that is closed. Then also is closed inside and T can be considered closed in this triple product. By the completeness of projective space, the projection of T down to is closed. Conversely, suppose the property is always verified, and now take Z = P and T to be the diagonal of . The image of T inside is none other than , which must be closed, so that f is proper.
We now list some properties of proper morphisms, which are generally quite easy to verify using the definition and/or the propsition:
1. The idenity map is proper.
2. The composition of two proper maps is proper.
3. The product of two proper maps is proper.
4. The image of a closed subvariety under a proper map is proper.
5. An injection is proper if and only if Y is closed in X.
6. Morphisms from a projective variety are proper.
7. A projection is proper if and only if Y is projective.
For example, to verify property 4, suppose that is a proper morphism and T is closed in X. We can apply the proposition where Z is a point so that T is a closed subvariety of . The image, according to the proposition, is closed. The others are as easy. As a corollary, we can prove:
Corollary: A morphism is proper if and only if it factors through where the first map is a closed immersion and the second is the projection map.
Proof: This is necessary by definition (take X embedded in some projective space P) and sufficient by Properties 2,5, and 7.
As a last fact, perhaps you have heard before that proper maps in algebraic geometry are akin to proper maps in topology… that is indeed the case as the following Proposition explains:
Proposition: Suppose that the ground field is the complex numbers. A morphism to be proper (in the algebraic sense) if and only if it is proper (in the topological sense) when X and Y are given their “usual” topologies.
Proof : Suppose that f is proper in the algebraic sense and that K is compact in Y in the topological sense. Let X be embedded in a projective space P, which is well known to be compact. Then we have that , which is again compact, so that f is proper in the topological sense. Suppose conversely that the topological property is satisfied. Suppose T is Zariski closed inside as in the previous proposition (we must show its image is Zariski closed in ). Its image is closed in the usual topology because the map is topologically closed. Then there are some standard comparison facts which allow us to conclude that the image is also Zariski closed. Let W be the Zariski closure of the image. Then there is a Zariski open set U which is Zariski dense in W (this is standard I think?). This open U is also topologically dense in W (this is basically an exercise). This in turn implies that the Zariski and topologicaly closures of the image coincide, but since the image is topologically closed it is also Zariski closed, and so f is (algebraically) proper.
Note that we have only cheated a little bit by using our definition of proper, which depends on having quasi-projective varieties. In general (for example, in Hartshorne’s book), the definition is taken to be the criteria in the first proposition (that is being universally closed, along with separated, and finite type – which are automatic here). All the properties and propositions go through as long as we’re careful about replacing “projective” by “complete” (and applying Chow’s Lemma as explained in the earlier post by Charles).