# Can we state the Riemann Hypothesis part of the Weil conjectures directly in terms of the count of points?

For algebraic curves we can state the Riemann hypothesis part of the Weil conjectures directly as a formula for the number of points on the curves, sidestepping the zeta function. Namely, given a smooth algebraic curve $$X$$ of genus $$g$$ defined over $$\mathbb{F}_q$$ elements, let $$N_n(X)$$ be its number of points over $$\mathbb{F}_{q^n}$$. Then

$$N_n(X) = q^n - \alpha_1^n - \cdots - \alpha_{2g}^n + 1$$

where all the numbers $$\alpha_i \in \mathbb{C}$$ have $$|\alpha_i| = \sqrt{q}$$.

Can we give a roughly similar formula for $$N_n(X)$$ when $$X$$ is a smooth higher-dimensional variety, say of dimension $$d$$? I know we can write

$$N_n(X) = \sum_{k = 0}^{2d} (-1)^k \mathrm{tr}\left(\mathrm{Frob}^n \colon H^k(X) \to H^k(X) \right)$$

where $$H^k$$ is etale cohomology, but I want to go a bit further, use the Riemann Hypothesis, and see numbers of magnitude $$q^{k/2}$$ showing up, presumably the eigenvalues of the Frobenius.

In fact what I'm really hoping for is a sum over pure motives: I hear pure motives defined using numerical equivalence give a semisimple category, so we can decompose $$X$$ uniquely as a sum of simple pure motives, and I'm hoping each one of these contributes a term to $$N_n(X)$$. If so, does each motive of dimension $$k$$ give a term whose absolute value grows like $$q^{nk/2}$$?

(By the way, I know we can state the Riemann hypothesis part of the Weil conjectures as a bound on the number of points, which is great, but I'm not interested in that here.)

(1) We have $$N_n(X) = \sum_{k = 0}^{2d} (-1)^k \mathrm{tr}\left(\mathrm{Frob}^n \colon H^k(X) \to H^k(X) \right) = \sum_{k = 0}^{2d} (-1)^k \sum_{i=1}^{ h^k(X)} \lambda_{k,i}^n$$

where $$\lambda_{k,1},\dots,\lambda_{k,h^k(X)}$$ are the eigenvalues of $$\mathrm{Frob}$$ on $$H^k(X)$$, counted with multiplicity.

Furthermore, we have $$|\lambda_{k,i}| =q^{k/2}$$ so $$\left| \lambda_{k,i}^n\right| = q^{nk/2}$$ as long as $$X$$ is a smooth projective variety (or even smooth proper). (If $$X$$ is not proper, we need to use compactly-supported cohomology to obtain the formula for the number of points, and then we only obtain an upper bound for the size of the eigenvalue.)

Technically speaking, the correct statement (see Lemma 1.7 of La Conjecture de Weil I by Pierre Deligne) is that the eigenvalues are algebraic numbers all whose Galois conjugates in $$\mathbb C$$ have size $$q^{k/2}$$. This deals with the issue that the eigenvalues are naturally $$\ell$$-adic integers and thus aren't naturally complex numbers, by considering the complex numbers that are Galois conjugates in the sense of satisfying the same minimal polynomial.

This notational difficulty is removed if we avoid étale cohomology and merely state the Riemann hypothesis in the form that

$$N_n(X) = \sum_{k = 0}^{2d} (-1)^k \sum_{i=1}^{ h^k(X)} \alpha_{k,i}^n$$ where $$\alpha_{k,i}$$ are complex numbers of absolute value $$q^{k/2}$$ and $$h^i(X)$$ is the $$i$$th Betti number of the de Rham cohomology of a characteristic $$0$$ lift.

This form is actually totally equivalent to the original formulation of Weil. We recover Weil's statement, in terms of the zeta function, by multiplying both sides by $$U^n$$, dividing by $$n$$, summing over $$n$$, and exponentiating. (In fact, this is exactly how Weil proves his conjecture for Fermat hypersurfaces in the paper where he originally formulates it, Number of Solutions fo Equations in Finite Fields.) We can recover this statement from Weil's by taking the logarithm of both sides, extracting the coefficient of $$U^n$$, and multiplying by $$n$$.

Of course, Weil could not give the statement in terms of étale cohomology because it hadn't been invented yet!

(2) This works in at least one notion of motives.

I think this is easiest to do if we use homological equivalence ($$\ell$$-adic, crystalline, whichever) as our equivalence relation for motives.

The idea here is that motives are supposed to have, for each cohomology theory, a realization functor that sends the motive to its cohomology groups, which on varieties $$X$$ recovers the cohomology groups of $$X$$, and is compatible with direct sums.

So if we express $$X$$ as a sum of (indecomposable, to avoid worrying about semisimplicity) motives, we can write the (signed) trace of $$\mathrm{Frob}^n$$ on the cohomology of $$X$$ as the sum of the (signed) trace of $$\mathrm{Frob}^n$$ on the cohomology of the motives.

To make the contribution of each motive of size $$q^{kn/2}$$, we just need each indecomposable motive to have cohomology only in a particular degree.

If we define motives using correspondences up to homological equivalence, say for $$\ell$$-adic homology, then the $$\ell$$-adic realization functor exists by construction.

Furthermore, because the eigenvalues of Frobenius on each cohomology group are distinct, we can define by linear algebra a polynomial in Frobenius, with rational coefficients, which acts by $$1$$ on the $$k$$th cohomology group and $$0$$ on each other cohomology group (the Künneth type standard conjecture). If we work with cycles up to homological equivalence, this cycle is idempotent, so defines a splitting of the motive of $$X$$.

Using these splittings, we can split $$X$$ into motives each of which has cohomology only in a single degree, as desired.

For Chow motives with respect to which other equivalence relations this can be made to work, as well as for which other notions of motives, I will leave to experts.

• Thanks! Just as a sanity check, do those eigenvalues $\lambda_1, \dots, \lambda_{h^k(X)}$ all have absolute value $q^{k/2}$? May 13 at 21:18
• @JohnBaez That is correct. May 13 at 21:21
• @JohnBaez I have added more detail, including a bit of history, to my answer. May 13 at 22:14
• Thanks, that helps! I figured something like this should be true, but somehow I didn''t see it spelled out in any introductions: it seems everyone speaks in terms of the zeta function. It seems to me that speaking in terms of $N_n(X)$ makes the statement more easily understood by beginners (like me). May 14 at 4:53