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Motive-ating the Weil Conjecture Proof June 10, 2010

Posted by David Speyer in Algebraic Geometry, Category Theory.



The Ring of Correspondences

Let X be a smooth, proper algebraic variety. In previous posts, we have seen the power of studying X by introducing the ring R whose elements are classes of subvarieties in H^*(X \times X) and where multiplication is given by \alpha \cdot \beta = (\pi_{13})_* \left( \pi_{12}^* \alpha \cup \pi_{23}^* \beta \right), with \pi_{ij} the three projections from X \times X \times X to X \times X. We are now going to create a category where R will be the endomorphism ring of X.

The first thing we have to think about is what we mean by the cohomology of X \times X. Since motives are supposed to be a universal cohomology theory, we don’t want to bias definitions by building in a particular cohomology theory at the beginning.

The most naive solution would be just to take the vector space of all cycles on X \times X, and replace cup product with actual intersection. The reason we can’t do this is that actual intersection will give the wrong answers. For example, consider a line in the projective plane. We want its self intersection to be a point, not
itself. So we need to pass to some sort of theory where intersection products work correctly, which means something like cohomology.

Fortunately, Fulton, MacPherson and others did a magnificent job creating a purely algebraic theory of intersection of algebraic cycles, as summarized in Fulton’s book Intersection Theory. Following the ideas of that book, for a smooth proper algebraic variety Z, let C(Z) be the \mathbb{Q}-vector space spanned by algebraic cycles in Z. Let A(Z) be a quotient of C(Z) which is small enough that the operations of cup-product, pull-back and push-forward are well defined. There are four standard choices of A(Z), presented in order of increasingly coarse equivalence relations:

  • Rational Equivalence: two cycles are equivalent if they can be linked by a family parametrized by \mathbb{P}^1.
  • Algebraic Equivalence: two cycles are equivalent if they can be linked by a family parameterized by some connected algebraic variety (equivalently, by some connected algebraic curve).
  • Cohomological Equivalence: For your favorite cohomology theory H^*, two cycles are the same if they have the same image in H^*(Z).
  • Numerical Equivalence: Two cycles C_1 and C_2 are equivalent if, for all algebraic cycles D of complementary dimension, \int C_1 D = \int C_2 D.

The theory of motives can be built using any of these, and they all have strengths and weaknesses. If rational or algebraic equivalence is used, then the endomorphism rings of our objects will contain negligible morphisms. (See Noah and my posts on this in a different context.) Cohomological equivalence builds into the theory a particular choice of cohomology theory. Numerical equivalence is extremely hard to test, as it is hard to deal with the quantification over all cycles. Grothendieck’s Conjecture D states that numerical and cohomological equivalence are, in fact, the same. The others are all known to be different.

A(Z) is graded by dimension of cycle. We’ll write A^{i}(Z) for the subspace spanned by algebraic cycles of codimension i. (Here and throughout the post, dimensions are algebraic. So, if Z is a variety over \mathbb{C}, we are talking about cycles whose real codimension is 2i.)

Now, let X and Y be smooth proper varieties, of dimensions d and e. The category of motives will have objects [X] and [Y] (there will be other objects, which we will introduce later). We define \mathrm{Hom}([X],[Y]) to be A^{d}(X \times Y). We define the composition map \mathrm{Hom}([X],[Y]) \times \mathrm{Hom}([Y], [Z]) \to \mathrm{Hom}([X], [Z]) by \alpha \circ \beta = (\pi_{13})_* \left( \pi_{12}^* \alpha \cup \pi_{23}^* \beta \right), where \pi_{ij} is the projection from X \times Y \times Z onto the i-th and j-th factors.

The category of motives is defined to be, roughly, the category whose objects are smooth, proper algebraic varieties and whose morphisms are given by \mathrm{Hom} as above. We’ll fill in the details below. For any cohomology theory H^* and any index i, the map X \mapsto H^i(X) is a functor from motives to abelian groups. There is a contravariant functor from smooth proper varieties to motives; a map f: Y \to X is sent to its graph, considered as a cycle in X \times Y.

One thing that I wondered about when I saw this definition: Why not take \mathrm{Hom}(X,Y) to be all of A(X \times Y), rather than just the codimension d part? If we did this, then X \mapsto H^*(X) would be a functor, but the individual components X \mapsto H^i(X) would not. Since (in this series of posts) our motivating goal is to understand the action of Frobenius on various cohomology groups, it would be unfortunate if we could not study one cohomology group at a time.

Idempotent Completion

The first technical point is that we want the category of motives to idempotently complete. That means that, if X is a variety, and \pi is an idempotent in the ring \mathrm{Hom}([X],[X]), then we adjoin a formal image of the map \pi. I wrote more about this process here.

Top and bottom dimensional cohomology, tensor product structure

Let X be smooth and projective of dimension d. There are two important classes in \mathrm{Hom}([X],[X]): we define e_0 to be \{ \mathrm{pt} \} \times X and e_d to be \{ X \} \times \{ \mathrm{pt} \}. The maps e_0 and e_d are easily seen to be idempotent. We temporarily define M^0(X) and M^d(X) to be their images.

For any X and Y, the cycle \{ \mathrm{pt} \} \times Y gives a map from M^0(X) to M^0(Y). Moreover, this map is an isomorphism with inverse \{ \mathrm{pt} \} \times X \subset Y \times X. In short, M^0(X) and M^0(Y) are canonically isomorphic for all X and Y.
For this reason, we will engage in the standard abuse of notation and talk about THE object \boldsymbol{1}, which is M^0(X) for every X.

If X and Y have the same dimension, then X \times \{ \mathrm{pt} \} gives a similar canonical isomorphism between M^d(X) and M^d(Y). Again, we will abuse notation and talk about THE object L^d, which is M^d(X) for every d-dimensional X.

We can describe these structures more concisely by putting a tensor structure on the category of motives. We define [X_1] \otimes [X_2] to be the object [X_1\times X_2]. Given f_i \in \mathrm{Hom}(X_i, Y_i), the map f_1 \otimes f_2 is the obvious product cycle. (One must also extend this definition to the case where X and Y are defined as the images of some idempotents; we leave this to the reader.) Then one can check that \boldsymbol{1} is the tensor identity, and L^d is L^{\otimes d}, where we abbreviate L^1 to L.

If we are working over \mathbb{F}_p, then Frobenius acts on L^{d} by p^d. Many of the classical spaces of algebraic geometry decompose, in the category of motives, as a direct sum of copies of L^{\otimes k} for various k‘s; this is why the number of points on them is given by polynomials in p. I love spaces like this, and have spent a lot of time thinking about them but, for number theorists, the interesting parts of the theory only come when we get beyond these examples.


We have put a tensor structure on the category of motives, which means we know how to formally take the tensor product of two motives. We would like to also take the dual of a motive, so that we can formulate statements like Poincare duality in the theory of motives.

Guided by Poincare duality, we can write down the intersection product. Let \Delta be the class of the diagonal in (X \times X) \times X; this gives a map [X] \otimes [X] \to [X] in the category of motives. In particular, [X] acquires a ring structure, as a cohomology theory should have, but this is not what I want to focus on right now. Rather, I want to look at \Delta \circ e_d, where d=\dim X. This is a map [X] \otimes [X] \to L^{d}. In classical algebraic topology, one would choose an isomorphism between the top cohomology of X and the bottom (an orientation) and this map would be the Poincare duality pairing.

We’d rather not choose an isomorphism between L^{\otimes d} and \boldsymbol{1}. (Among reasons, Frobenius acts on them differently.) Instead, what we do is to formally invert L (with respect to tensor product). More specifically, we define \mathrm{Hom}(X \otimes L^{a}, Y \otimes L^b) to be whichever of \mathrm{Hom}(X, Y \otimes L^{b-a}) and \mathrm{Hom}(X \otimes L^{a-b}, Y) involves a positive exponent. One can show that this is also A^{\dim X + b-a}(X \times Y). So this provides another answer to the question “where did the other graded pieces of A^*(X \times Y) go?”

We then define Poincare duality to be true, declaring that the dual of [X] \otimes L^{a} is [X] \otimes L^{-d-a}. Extending to classes which are defined as images of idempotents is, again, left to the reader.


An object of the category of motives is a triple (X, \pi, k), where X is a smooth proper variety, \pi an idempotent in \mathrm{Hom}([X],[X]) and k an integer. One should think of this as “the image of \pi:[X] \to [X], tensored with L^k.” See Milne’s article for precise definitions of morphisms and tensor structure.

A major issue: The missing grading

We have talked about the idempotents e_0 and e_d, which project onto top and bottom cohomology. One of the great missing results of the category of motives is that no one can show that projectors onto the other cohomology groups exist. Specifically, this is Grothendieck’s Conjecture C. Until then, we have the possibility that the universal cohomology theory is not graded, which would seem peculiar. I think this is one of the things that made people hesitate to embrace motives.

I think (please correct me if I am wrong) that such projectors are known to exist for varieties over \mathbb{F}_p, using homological equivalence motives, and using the Weil Conjectures. The point is that the characteristic polynomials of Frobenius on the different H^i(X) are relatively prime, since their eigenvalues have different norms. So one can find some polynomial e_i(F) which acts by 1 on H^i(X) and by 0 on the other H^j(X). This polynomial, interpreted as an endomorphism of the motive [X], is the desired projector. Of course, this is not helpful to those who want to find a new proof of the Weil conjectures.

Review Question

If you want to test your understanding of this series of posts, you should attempt to rephrase all of the Standard Conjectures as statements about the category of motives.

Thanks for reading to the end!

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