# The Weil conjectures and the problem of coefficients

I have been discussing the Weil conjectures. In this post, I want to discuss one of the main difficulties in their proof. You should be able to follow this even if you have not been reading the earlier posts.

The main claim of the Weil conjectures can be summarized as follows:

Theorem Let $X$ be a smooth projective variety of dimension $d$ over $\mathbb{F}_q$. Then there are complex numbers $\alpha_{i,j}$, for $0 \leq i \leq 2d$ and $1 \leq j \leq b_i(X)$, and obeying $|\alpha_{i,j}| = q^{i/2}$, such that $\displaystyle{ \# X(\mathbb{F}_{q^k}) = \sum_{i=1}^{2d} (-1)^i \sum_{j=1}^{b_i} \alpha_{i,j}^k.}$

The analogous claim in characteristic zero, which we proved last time, is

Theorem Let $X$ be a smooth projective variety over $\mathbb{C}$, with $\omega \in H^2(X)$ the hyperplane class. Let $F$ be an endomorphism of $\omega$ such that $F^* \omega = q \omega$. Then there are complex numbers $\alpha_{i,j}$ as above, such that $\displaystyle{ \# (\mbox{fixed points of}\ F^k) = \sum_{i=1}^{2d} (-1)^i \sum_{j=1}^{b_i} \alpha_{i,j}^k.}$

The way we proved this was to essentially establish:

Theorem Let $X$, $F$ and $\omega$ be as above. Then there is a positive definite Hermitian inner product on $H^i(X, \mathbb{C})$ such that $q^{-i/2} F^* : H^i \to H^i$ is a unitary operator.

The required theorem then follows from the Lefschetz trace formula, and the fact that the eigenvalues of a unitary matrix have norm $1$.

If we wanted to prove the characteristic $p$ theorem in the same way, we might hope to follow the same process: define $H^i(X, \mathbb{C})$, place a Hermitian structure on it, and prove a Lefschetz trace formula.

There are cohomology theories with Lefschetz trace formulas. But they do not have coefficients in $\mathbb{C}$. Their coefficient ring is $\mathbb{Q}_{\ell}$ (etale cohomology) or the ring $W(\mathbb{F}_q)$ of Witt vectors (various $p$-adic theories).

In particular, while we can talk about the eigenvalues of $F^*: H^i(X) \to H^i(X)$, there will be no natural way to embed these eigenvalues in $\mathbb{C}$. There is an unnatural way: If we could prove that the characteristic polynomial of $F^*$ has coefficients in $\mathbb{Q}$, then we could consider the roots of the same polynomial in $\mathbb{C}$. It would be enough to prove that these complex roots have norm $q^{i/2}$. But it is unclear how to show that the polynomial has coefficients in $\mathbb{Q}$, let alone how to bound the roots. In particular, it is unclear what could be the analogue of a Hermitian structure for a $\mathbb{Q}_{\ell}$ vector space.

In future posts, we will tackle these problems. For now, below the fold, I reproduce an argument of Serre showing that there is no way to define a reasonable cohomology theory for finite characteristic varieties, with coefficients in $\mathbb{R}$. (And therefore, not in any subring of $\mathbb{R}$ either.)

# An example of Serre

Let $q=p^2$, for $p$ a prime which is $3 \mod 4$. Let $i$ denote one of the square roots of $-1$ in $\mathbb{F}_q$. Let $E$ be the elliptic curve $y^2 = x^3-x$ over $\mathbb{F}_q$ (including the point at infinity.)

Let $F$ be the automorphism $(x,y) \mapsto (x^p, y^p)$ of $E$, and let $J$ be $(x,y) \mapsto (-x, iy)$. Then $F^2=-p$, $J^2=-1$ and $FJ=-JF$. Let $Q$ be the non-commutative ring $\mathbb{Z} \langle F, J \rangle / (F^2+p, \ J^2+1, \ FJ+JF)$. So $H^1(E, \mathbb{R})$, should such a thing exist, should be an $Q \otimes \mathbb{R}$-module, and a $2$ dimensional $\mathbb{R}$-vector space.

But $Q \otimes \mathbb{R}$ is isomorphic to the quaternions. (Take $F \to \sqrt{p} \ i$ and $J \to j$.) And any module for the quaternions has dimension divisible by $4$, a contradiction.

The intriguing thing is that there is $2$ dimensional complex representation of $Q$. The action of $F$ and $J$ are given by $F=\left( \begin{matrix} 0 & \sqrt{p} \\ -\sqrt{p} & 0 \end{matrix} \right)$ and $J=\left( \begin{matrix} i & 0 \\ 0 & -i \end{matrix} \right)$.

Every element of $Q$ has integral trace and determinant in this representation. For any $a \in Q$, we have $\displaystyle{ \# (\mbox{fixed points of}\ a^k) = 1 - \mathrm{Tr}(a^k) + \mathrm{Det}(a^k) }$

and the eigenvalues of $a$ have norm $\sqrt{\mathrm{Det}(a)}$. It is just as if $Q$ had a representation in $\mathrm{Mat}_2(\mathbb{Z})$ which rescales a positive definite Hermitian form!

Those of you who like tensor categories may like to start thinking about how to axiomatize this example; it will come back when we get to motives.

## 4 thoughts on “The Weil conjectures and the problem of coefficients”

1. Jared Weinstein says:

Good post! This argument also shows that you can’t have coefficients in Q_p either, because Q_p doesn’t split Q. You really need to take your coefficient field to be the ring of Witt vectors of F_q in order to get an action of J on p-adic cohomology. This is a very different scenario from etale cohomology with l-adic coefficients, where you have an action of J on H^1(E, Q_l) regardless of whether Q_l contains a square root of -1. I hadn’t thought about this subtlety before reading your post.