[Math] Why is the zeta function of a variety over a finite field not a polynomial? (question about motives)

motivesnt.number-theoryweil-conjectures

I've been doing some light(?) reading on motives and the standard conjectures in an attempt to put various things that I tangentially know in perspective.

The question is this: the Weil conjectures assert that $Z=\frac{P_1(t)…P_{2r-1}(t)}{P_0(t)…P_{2r}(t)}$ where the $P_i$'s are certain polynomials. (the assertion is of course stronger, but the rest of it is besides the point) What is the deep reason that these $P_i$'s alternate between numerator and denominator?

In Milne's notes about motives (http://www.jmilne.org/math/xnotes/MOT.pdf), he asserts the following: let $hX$ be the motive corresponding to $X$, and let $h^0X$,…,$h^{2r}X$ be the conjectured decomposition into pure motives (conjecture C in Milne). He then says: define for a pure motive of weight $k$ the zeta function as the characteristic polynomial of the Frobenius if $k$ is odd, and its inverse if $k$ is even. Then extend the definition to motives by: the zeta function of a direct product of motives goes to the product of the zeta functions of the individual motives. Then indeed:
$Z(X,s)=Z(hX,s)=Z(h^0X,s)…Z(h^{2r}X,s)$, where $Z(h^kX,s)=P_k(t)$ for $k$ odd and $\frac{1}{P_k(t)}$ for $k$ even.

So one can reduce this question to: why are we defining the zeta function of a motive of weight $k$ to be the characteristic polynomial of the Frobenius or its inverse depending on the parity?

Best Answer

The zeta function of a variety $X$ over a finite field is a priori defined to be a point counting function, i.e. it is the following product over the closed points of $X$ (thought of as a scheme): $$\zeta_X(s) = \prod_{x}(1 - | \kappa(x)|^{-s})^{-1},$$ where $\kappa(x)$ is the residue field of $x$ and $|\kappa(x)|$ denotes its order. (This is motivated by analogy with the Riemann zeta function, which is what we get if we apply the same definition with $X$ replaced by Spec $\mathbb Z$.)

Now this will be a Dirichlet series involving only powers of $p^{-s}$ (if $p$ is the char. of the finite field), and so replacing $p^{-s}$ by $T$, we obtain a power series in $T$, whose log can be reinterpreted in the usual way as a generating function counting the number of points of $X$ with values in the various extensions of $\mathbb F_p$.

Now one can count these points by the Lefschetz fixed point formula (applied to the $\ell$-adic cohomology), and this gives the alternating product of char. polys. of Frobenius that you write down in your question.

Of course, one could write down their product, rather than their alternating product, but the resulting power series would not have any particular interpretation; in particular, it wouldn't be related to counting points of $X$ in the same way that the zeta function is.

Milne's definition of the $\zeta$-function directly in terms of $\ell$-adic cohomology is to some extent putting the cart before the horse; as Stopple notes, it is a reasonable definition only because of the back story about counting points and so on.

Nevertheless, if you want to take the definition in terms of cohomology as the basic one, then you can ask yourself: how should you define such a quantity if you want it to behave well under chopping up varieties (which is what motives essentially are --- pieces of varieties cut out by correspondences).

The basic quantity that is defined in terms of cohomology and which is additive with respect to cutting up spaces is the Euler characteristic. And for this additivity to hold, it is crucial that involve an alternating sum, with the sign being dictated by the cohomogical degree. The reason is that the behaviour of cohomology under chopping up and/or gluing is given by the excision and Mayer--Vietoris long exact sequences, and it is the alternating sum of the dimensions which is additive in exact sequences.

Viewed cohomologically, the zeta function is like an enhanced, multiplicative version of the Euler characteristic, and like the Euler characteristic, for it to be multiplicative with respect to cutting up varieties, we must form it via an alternating product.

In conclusion: I think that the "deep reason" that you are looking for is the yoga of Euler characteristics.