Suppose you have a proper map between smooth (quasi) projective varieties. Then suppose you have a coherent sheaf on . After viewing that sheaf as an element of the Grothendieck Group of coherent sheaves on , there are two things you could do. The first thing is that you could push forward this element to the Grothendieck Group on , and then take its chern character, to arrive at an element in the Chow Ring of . The second option you have is to first take the chern character of in the Chow Ring of , and then push it forward to the Chow Ring of . Now we have two elements in the Chow Ring of and we could wonder if we have the same element. The Grothendieck Riemann Roch (GRR) Theorem tells us we don’t quite have the same element, but it tells us exactly by how much we are off!
Therem (GRR): Suppose f, X, and Y as above. Let be the Todd classes of the Tangent Bundles of X and Y. (These elements were defined in the last post). Suppose is any element in the Grothendieck Group of coherent sheaves on . Then we have the formula .
Just so we’re clear, on the LHS of the formula, the first product is in the Chow Ring , and then we can push forward to . On the RHS, the pushforward takes place in K-theory, and then the product is in .
We have some easy Corollaries which will also serve as examples for how to work in the K-groups and Chow Rings.
1. Suppose that is a point, is projective of dimension n, and is a coherent sheaf. Then in degree zero and is zero in positive degrees. For a class , the element is simply the degree piece of , referred to as (and is an integer!). The Todd class and is the alternating sum of the proper pushforwards . However, since we are just pushing forward to a point, these are simply the vector spaces . On , the operation of taking the chern character simply reads off the rank of the vector space – then the RHS of the GRR formula reads . So in this case the GRR formula reads , which is commonly called the Hirzebruch Riemann Roch Theorem (HRR).
2. Suppose again that is a point, is a proper curve, and is a line bundle on . We have that and that is the tangent bundle so that . Then the G/HRR formula reads . This is the standard Riemann Roch Theorem for curves. The same argument works if is a rank r vector bundle and we get .
3. Suppose again that is a point, is a proper smooth surfaces, and is a line bundle. Because is the dual of , we again get . We will refer to simply as . Then . We then multiply this by and take degrees to get . When , we get from which we recover the standard Riemann Roch formula for surfaces.
The proof of this theorem will be to prove it for a proper morphism by proving it for a projection and then for an inclusion. We are able to break the proof into these steps because of the following Lemma:
Lemma: Suppose are proper morphisms . Let and set . If GRR is true for and , then it is true for . If GRR is true for and and if is injective, then it is true for .
Proof: The proof must be completely formal because there is nothing much to work with. By GRR for f we have . We apply to both sides and use the fact that to find that . Now applying GRR for the map gives that both of these are equal to , which gives GRR for . The second statement is as formal.
There is one more formal statement to be recorded.
Lemma: Let and be two proper morphisms of smooth varieties. Let and . If GRR is true for and for then it is true for .
Here denotes the image of this element under the natural map induced by the projections.
Proof: Again the proof is completely formal, and quite easy once you realize the following three facts. 1) . To prove this statement, we may assume that x and x’ are classes of coherent sheaves and then use the Kunneth Formula. 2) (this in the Chow Ring). This is true at the level of cycles and then descends to the Chow Ring. 3) . This follows from the multiplicative property of ch. The rest of the proof really is formal.
Given the first Lemma, to prove the GRR formula for , we must prove it for A. the projection (projection onto the first factor) and then for B. a closed immersion .
My plan is to spend one post outlining how each of these cases proceeds and then explain some examples where is not a point so we can see some of the real power of the theorem.