# Cohomology and motives

This summer I am trying to write a little bit about Weil cohomology theories for the Stacks project. My motivation is that I want to add the theorem that over a field of characteristic zero de Rham cohomology is one. And in turn, I became motivated to do this, because of the discussions I had with our graduate students in my course Ask me anything. Because after all: what is a “good cohomology theory” for algebraic varieties?

While working on this, I found that I had never properly understood the reason for all the conditions imposed by Kleiman in his paper. I still haven’t understood, I think. In this post I will discuss a basic question I had as I worked through this material and the answers I came up with so far.

Let k be an algebraically closed field; this will be the base field for our smooth projective varieties X, Y, Z. Let F be a field of characteristic 0; this will be the coefficient field for our cohomologies. Feel free to assume F is algebraically closed too. A Weil cohomology theory H^* assigns a graded commutative F-algebra H^*(X) to X in a contravariant manner. It comes with additional structure, namely trace maps and cycle classes, and it has to satisfy a bunch of axioms:

(A) Poincare duality: if dim(X) = d the trace map H^{2d}(X)(d) —> F is an isomorphism, H^i nonzero only for 0 ≤ i ≤ 2d, H^i is finite dimensional and the pairing H^i(X) x H^{2d – i}(X)(d) —> F defined using cup product and the trace map is nondegenerate,

(B) Kunneth formula: H^*(X x Y) = ⨁ H^i(X) ⊗ H^j(Y) and this is compatible with trace maps.

(C) Cycle classes are compatible with pullbacks, pushforwards, intersection products, and trace of the cycle class of a point on Spec(k) is 1.

Remark 1: I’ve written the axioms slightly differently from Kleiman in his paper “Algebraic cycles and the Weil conjecture”. In particular I am keeping track of Tate twists and I am adding the axiom that trace is compatible with Kunneth (which you can deduce from the other axioms — but will become important later).

Remark 2: In some references compatibility of cycle classes and pushforwards is omitted. But without this compatibility I immediately get stuck, for example I don’t know how to compute the class of the diagonal without this axiom. If you know a proof of this compatibility from the other axioms, please drop me an email.

Let M_k be the symmetric monoidal category of Chow motives over k (using rational equivalence). Let h(X) be the motive of X and recall that h(-) is contravariant. Let us write 1(1) for the Tate motive, i.e., the inverse of the invertible object h^2(P^1) of M_k.

A key consequence of the axioms (A, (B), (C) is that H^*(X) = G(h(X)) for some symmetric monoidal functor G from M_k to the symmetric monoidal category of graded F-vector spaces such that G(1(1)) is nonzero only in degree -2.

In fact, let’s start with a G as above and consider the contravariant functor H^*(X) = G(h(X)) from smooth projective varieties to graded F-vector spaces. What can we say about H^*? It turns out that you get cup products, trace maps, cycles classes and all the axioms (A), (B), (C) as formulated above except for possibly “the trace map … isomorphism” and “H^i nonzero only for 0 ≤ i ≤ 2d”.

Question: Can we make an example of a G where the corresponding H^*(X) does not have the “correct” betti numbers?

For example, we can ask whether there could be a G such that H^*(X) is nonzero in negative degrees for some X? Or we could ask whether we can find a G such that H^0(X) has dimension > 1. I think the answer is “yes” when k is the complex numbers and conjecturally “no” when the ground field k is the algebraic closure of a finite field.

Remark 3: When k is the algebraic closure of a finite field, a paper of Katz and Messing proves the betti numbers are the same for any Weil cohomology theory satisfying more axioms (namely some Lefschetz type thing). As far as I can tell, this result doesn’t apply to answer my question above, since I’m even weakening the assumptions of a Weil cohomology theory.

Construction of a weird G when both k and F are the field of complex numbers. Namely, let (H’)^* be the functor which sends X to (H’)^*(X) = ⨁ H^{p, q}(X) viewed as a bigraded vector space over F. This is a Weil cohomology theory: by Hodge theory it is just the same as sending X to its usual cohomology with complex coefficients. By the discussion above we get G’ such that for every motive M we have a bigrading G'(M) = ⨁ (G’)^{p, q}(M). (One could say that the target category is the category of graded complex Hodge structures, suitably defined as a symmetric monoidal category.) Of course as a graded F-vector space we have that the degree of the summand (G’)^{p, q}(X) is p + q, in other words, we have

(G’)n(M) = ⨁ n = p + q (G’)p, q(M)

Now I am going to define a weird G by setting G(M) equal to the graded F-vector space with degree n part equal to

Gn(M) = ⨁ n = 3p – q (G’)p, q(M)

Observe that G(1(1)) is still in degree -2! (A technical point is that because I chose 3 and -1 odd the functor G is still compatible with the commutativity constraints which you have to check in order to see that G is a symmetric monoidal functor.) Now, if E is an elliptic curve then H^*(E) = G(h(E)) has betti numbers dim H^{-1}(E) = 1, dim H^0(E) = 1, dim H^1(E) = 0, dim H^2(E) = 1, dim H^3(E) = 1 and everything else zero. Cool!

On the other hand, let p be a prime number and suppose that k is the algebraic closure of the field with p elements. Then I think conjecturally all objects in M_k ⊗ F are semi-simple and all simple objects are “1 dimensional”. Moreover, I think that those simple guys are classified by Weil p-numbers α in F up to multiplication by roots of unity (these α are algebraic numbers, but aren’t necessarily algebraic integers, and have a weight which is an integer which can be negative). Say M(α) is the motive corresponding to α. Now the point is that given an α of weight zero, its nth root for any n > 1 is also a Weil p-number of weight zero. Thus the 1-dimensional F-vector space G(M(α)) had better be in degree 0. And this then fixes the degrees for all motives because the motive corresponding to α = p^{-1} is 1(1) which we are requiring to be in degree -2.

Enjoy!