Why is the zeta function of a variety over a finite field not a polynomial? (question about motives) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T04:35:50Z http://mathoverflow.net/feeds/question/58577 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58577/why-is-the-zeta-function-of-a-variety-over-a-finite-field-not-a-polynomial-ques Why is the zeta function of a variety over a finite field not a polynomial? (question about motives) Makhalan Duff 2011-03-15T21:58:13Z 2011-03-16T02:14:34Z <p>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.</p> <p>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?</p> <p>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.</p> <p>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?</p> http://mathoverflow.net/questions/58577/why-is-the-zeta-function-of-a-variety-over-a-finite-field-not-a-polynomial-ques/58579#58579 Answer by Stopple for Why is the zeta function of a variety over a finite field not a polynomial? (question about motives) Stopple 2011-03-15T22:11:44Z 2011-03-15T22:11:44Z <p>Ireland and Rosen <em>A Classical Introduction to Modern Number Theory</em>, give a good explanation of this in Chapter 11. The zeta function is not defined as a rational function, rather it is a generating function of the form $Z(u)=\exp\left(\sum_s N_s u^s/s\right)$ where the $N_s$ is the number of points in the field with $q^s$ elements. It's a miracle that this generating function turns out to be a rational function. </p> http://mathoverflow.net/questions/58577/why-is-the-zeta-function-of-a-variety-over-a-finite-field-not-a-polynomial-ques/58580#58580 Answer by Qiaochu Yuan for Why is the zeta function of a variety over a finite field not a polynomial? (question about motives) Qiaochu Yuan 2011-03-15T22:34:29Z 2011-03-16T01:17:59Z <p>I can't tell what background you have, so here are some things you probably already know.</p> <p>Let's start with a smooth projective curve $C$ over $\mathbb{F}_q$. We want to attach a zeta function to it by analogy with the Dedekind zeta function of a number field. The analogue of an ideal is then an effective rational divisor $D$, and the analogue of the norm is $q^{\deg D}$ (think about the case of a smooth affine curve). So we are led to define </p> <p>$$\zeta_C(s) = \sum_{D \ge 0} \frac{1}{q^{s \deg D}}$$</p> <p>where the sum runs over all effective divisors. <strong>Edit:</strong> I forgot to mention something important! This zeta function has an Euler product. Of course, that's not hard to see because of how divisors are defined, but it's still worth mentioning because it supports the argument that this really is the zeta function we want.</p> <p>A purely combinatorial argument, which is explained in <a href="http://qchu.wordpress.com/2009/08/23/newtons-sums-necklace-congruences-and-zeta-functions/" rel="nofollow">these</a> <a href="http://qchu.wordpress.com/2009/11/04/newtons-sums-necklace-congruences-and-zeta-functions-ii/" rel="nofollow">two</a> blog posts, shows that this is equal to</p> <p>$$\zeta_C(s) = \exp \left( \sum_{k \ge 1} \frac{N_k}{k} q^{ks} \right)$$</p> <p>where $N_k$ is the number of points on $C$ over $\mathbb{F}_{q^k}$. This definition readily extends to an arbitrary variety over $\mathbb{F}_q$ (although I am less sure if there is a direct connection to divisors in this setting), so we can adopt it for an arbitrary variety $X$.</p> <p>This definition has some nice properties that make it rather natural: for example $\zeta_X(s) \zeta_Y(s)$ is just the zeta function of the disjoint union of $X$ and $Y$. </p> <hr> <p>The above definition brings to mind the definition of the <a href="http://en.wikipedia.org/wiki/Lefschetz_zeta_function" rel="nofollow">Lefschetz zeta function</a>. In zeta function language, the Lefschetz fixed point theorem can be stated as follows: if $X$ is a compact polyhedron of dimension $n$ and $f : X \to X$ a continuous function, then</p> <p>$$\zeta_f(t) = \exp \left( \sum_{k \ge 1} \frac{ L(f^k)}{k} t^k \right) = \prod_{i=0}^n \det (1 - tf_{\ast} | H_i(X, \mathbb{Q}))^{(-1)^{i+1}}$$</p> <p>where the RHS is the alternating product of the characteristic polynomials of $f$ acting on homology. Given that $L(f^k)$ is a topologically refined version of "the number of fixed points of $f^k$," Weil was led to the following idea:</p> <ul> <li>The number of points on $X$ over $\mathbb{F}_{q^k}$ is the number of fixed points of the $k^{th}$ power of the Frobenius map acting on the points of $C$ over $\overline{ \mathbb{F}_q }$.</li> <li>So, if there were a good (co)homology theory for varieties over finite fields, one might imagine that a similar product formula could for $\zeta_X(s)$ ($X$ a variety over a finite field) hold based on the induced action of $f$ on (co)homology. </li> </ul> <p>My understanding is that this idea was backed up by the working out of several examples. </p> <hr> <p>As for what the signs are doing in the Lefschetz fixed point theorem, my understanding here is incomplete, but as far as I can tell the basic idea is that if you want to define a notion of trace for a chain map $f : C_{\bullet} \to C_{\bullet}$ (where $C_{\bullet}$ is, say, a chain complex of $\mathbb{Q}$-vector spaces) that is invariant under chain homotopy, I think in some sense the natural definition must be the alternating sum of the traces over the chain groups. This is invariant because it's the same as the alternating sum of the traces over the homology groups.</p> <p>I am positive this is the "correct" definition but I don't really have the background to explain why. </p> http://mathoverflow.net/questions/58577/why-is-the-zeta-function-of-a-variety-over-a-finite-field-not-a-polynomial-ques/58598#58598 Answer by Emerton for Why is the zeta function of a variety over a finite field not a polynomial? (question about motives) Emerton 2011-03-16T02:14:34Z 2011-03-16T02:14:34Z <p>The zeta function of a variety $X$ over a finite field is <em>a priori</em> 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 $\mathbb Z$.)</p> <p>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 $\mathbb F_p$.</p> <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.</p> <p>Of course, one <em>could</em> 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.</p> <p>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.</p> <p>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).</p> <p>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 <em>alternating</em> sum of the dimensions which is additive in exact sequences.</p> <p>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. </p> <p>In conclusion: I think that the "deep reason" that you are looking for is the yoga of Euler characteristics.</p>