The Grothendieck-Riemann-Roch Theorem, Stated

Suppose you have a proper map f:X \rightarrow Y between smooth (quasi) projective varieties.  Then suppose you have a coherent sheaf F on X.  After viewing that sheaf as an element of the Grothendieck Group of coherent sheaves on X, there are two things you could do.  The first thing is that you could push forward this element to the Grothendieck Group on Y, and then take its chern character, to arrive at an element in the Chow Ring of Y.  The second option you have is to first take the chern character of F in the Chow Ring of X, and then push it forward to the Chow Ring of Y.  Now we have two elements in the Chow Ring of Y 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 td(X), td(Y) be the Todd classes of the Tangent Bundles of X and Y.  (These elements were defined in the last post).  Suppose x \in K(X) is any element in the Grothendieck Group of coherent sheaves on X.  Then we have the formula f_*(ch(x) \cdot td(X) ) = ch (f_!(x)) \cdot td(Y) .

Just so we’re clear, on the LHS of the formula, the first product is in the Chow Ring A(X) \otimes \mathbb{Q}, and then we can push forward to A(Y) \otimes \mathbb{Q}.  On the RHS, the pushforward takes place in K-theory, and then the product is in A(Y) \otimes \mathbb{Q}.

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 Y is a point, X is projective of dimension n, and x = F is a coherent sheaf.   Then A(Y) = \mathbb{Z} in degree zero and is zero in positive degrees.  For a class u \in A(X), the element f_*(u) is simply the degree n piece of u, referred to as u_n (and is an integer!).  The Todd class td(Y) = 1 and f_!(x) is the alternating sum of the proper pushforwards R^qf_*(F).  However, since we are just pushing forward to a point, these are simply the vector spaces H^q(X, F).  On Y, the operation of taking the chern character simply reads off the rank of the vector space – then the RHS of the GRR formula reads \Sigma (-1)^q h^q(X, F) = \chi(X, F).  So in this case the GRR formula reads (ch(F) \cdot td(X))_n = \chi(X, F), which is commonly called the Hirzebruch Riemann Roch Theorem (HRR).

2.  Suppose again that Y is a point, X is a proper curve, and F = \mathcal{L}(D) is a line bundle on X.  We have that ch(F) = 1 + D and that T_X = \mathcal{L}(-K) is the tangent bundle so that td(X) = 1 - (K)/2.  Then the G/HRR formula reads \chi(X, F) = deg( (1 + D) \cdot (1 - K/2))_1 = deg(D - K/2) = deg(D) + 1 - g.  This is the standard Riemann Roch Theorem for curves.  The same argument works if F is a rank r vector bundle and we get \chi(X,F) = deg(F) + r(1 - g).

3.  Suppose again that Y is a point, X is a proper smooth surfaces, and F = \mathcal{L}(D) is a line bundle.  Because T_X is the dual of \Omega_X, we again get c_1(T_X) = -K.  We will refer to c_2(T_X) simply as c_2.  Then td(X) = 1 - (1/2)K + (1/12)(K^2 + c_2).  We then multiply this by ch(F) = 1 + D + (1/2)D^2 and take degrees to get \chi(X,F) = (1/2)D\cdot(D - K) + (1/12)(K^2 + c_2).  When D = 0, we get \chi(X, \mathcal{O}_X) = (1/12)(K^2 + c_2) 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 X \rightarrow Y \rightarrow Z are proper morphisms f, g.  Let x \in K(X) and set y = f_!(x).  If GRR is true for (f, x) and (g,y), then it is true for (gf, x).  If GRR is true for (gf, x) and (g,y) and if g_* is injective, then it is true for (f,x).

Proof:  The proof must be completely formal because there is nothing much to work with.  By GRR for f we have f_*(ch(x) \cdot td(X) ) = ch (f_!(x)) \cdot td(Y) .  We apply g_* to both sides and use the fact that (gf)_* = g_*f_* to find that (gf)_*(ch(x) \cdot td(X)) = g_*( ch(y) \cdot td(Y)).  Now applying GRR for the map g gives that both of these are equal to ch((fg)_!(x)) \cdot T(X), which gives GRR for (gf, x).  The second statement is as formal.

There is one more formal statement to be recorded. 

Lemma: Let f: X \rightarrow Y and f': X' \rightarrow Y' be two proper morphisms of smooth varieties.  Let x \in K(X) and x' \in K(X').  If GRR is true for (f,x) and for (f',x') then it is true for (f \times f', x \otimes x').

Here x \otimes x' denotes the image of this element under the natural map K(X) \otimes K(X') \rightarrow K(X \times X') induced by the projections.

Proof:  Again the proof is completely formal, and quite easy once you realize the following three facts.  1) (f \times f')_! (x \otimes x') = f_!(x) \otimes f'_!(x').   To prove this statement, we may assume that x and x’ are classes of coherent sheaves and then use the Kunneth Formula.  2) (f \times f')_*(x \otimes x') = f_*(x) \otimes f'_*(x') (this in the Chow Ring).  This is true at the level of cycles and then descends to the Chow Ring.  3) ch(x \otimes x') = ch(x) \otimes ch(x').    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 f: X \rightarrow Y, we must prove it for A.  the projection Y \times P \rightarrow Y (projection onto the first factor) and then for B. a closed immersion X \rightarrow Y.  

My plan is to spend one post outlining how each of these cases proceeds and then explain some examples where Y is not a point so we can see some of the real power of the theorem.

This entry was posted in Algebraic Geometry, Big Theorems, Cohomology, Intersection Theory. Bookmark the permalink.

6 Responses to The Grothendieck-Riemann-Roch Theorem, Stated

  1. Novice says:

    Could it be argued that this is the most important theorem in Algebraic Geometry?

    Or would the role go to other theorems like the Zariski Main Theorem, Serre Duality Theorem, Abel-Jacobi or the Riemann existence theorems?

  2. Matt DeLand says:

    I think that’s a matter of taste – and of course – depends on what you mean by “important”. It’s certainly an important tool to be able to use – it’s applicable in many contexts.

  3. Novice says:

    The number of application certainly adds to importance; but it is chiefly the role in an intrinsic understanding of the subject that I mean here. Importance as when we say, quadratic reciprocity and other reciprocities are some of the most important theorems in number theory. Or when we say, within mathematics, number theory is important or algebraic geometry is important.

  4. Pingback: The Grothendieck-Riemann-Roch Theorem, a proof-sketch « Rigorous Trivialities

  5. qwerty says:

    Where can an “easy” proof be found (for GRR)?

  6. Pingback: Chern Classes and Generalized Riemann-Roch Theorems | Theories and Theorems

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