Connecting two pictures of the Zeta function

Question

Lets consider two views of zeta functions of curves.

For the following, let $\mathbb{F}_p$ be the field with $p$ elements where $p$ is prime, and let $\overline{\mathbb{F}_p}$ be the algebraic closure of $\mathbb{F}_p$. Let $X$ be a smooth curve over $\text{Spec}(\mathbb{F}_p)$.

View 1: Using a Weil cohomology theory, obtain a chain complex $C^*$. Applying the cohomology construction to the frobenius map $F : X \rightarrow X$, we get a map $f : C^* \rightarrow C^*$. Chain complexes do not have determinant in the ordinary sense, but there is a unique generalization, which is $f_* \mapsto \text{Det}(f_*) := \prod_{i = 0}^n \text{det}(f_n)^{(-1)^{i+1}}$ (this only works for certain complexes). We consider the generalized characteristic polynomial. $$ \text{Det}(1 - t f_*) = \prod_{i = 1}^n \text{det}(1 - ft)^{-1^{i+1}}$$ A Poincare duality for the étale cohomology leads quickly to a functional equation. Viewing the zeta function this way seems to accord well with the proofs of the weil conjectures, which are higher analogues of a simple theorem on finite dimensional vector spaces: the lefshetz fixed point theorem.

View 2: The global zeta function $Z(f, \psi)$ is $$ \int_{\mathbb{I}} f(x) \psi(x) d \mu^\times (x)$$ This view is accompanied by several variants, each expressing the L-function as the Fourier transform of some character.

In this view (the view of class field theory) there is a correspondence between characters on the absolute Galois group $\pi_1(\mathbb{Q})$ and characters on $\mathbb{A}^\times / \mathbb{Q}^\times$. This view is explained here and here.

Both of these views have toy versions in finite dimensional vector spaces using sets instead of schemes. But it is not clear from the second view that zeta functions should arise as characteristic polynomials.

Can anyone give an explanation of why we should expect the (higher analogue of the) characteristic polynomial to be a character? Is there a toy-model where the two views can be unified here? I am searching for some relatively simple observations about the relationships between characteristic polynomials and characters to the effect that these views would seem like the higher analogue of a natural observation about char polys on finite dimensional vector spaces.

Answer 1

The quickest answer I can give is that the Lefschetz formula gives the identity $$\prod_{i }\det (1 - ft, H^i(X))^{ (-1)^{i+1}} = \prod_{v \in |X| } \frac{1}{1 - t^{\deg v}} $$ and then the product on the right side can be expanded out into a sum over the divisors $D$ on $X$ of the function $t^{\deg D}$ (i.e. a character), which can be further represented as an integral over the ideles by replacing each divisor with a corresponding set of ideles.

The first step, the product identity, can be proved as follows: Using the identity $(1- u)^{-1} = e^{ \sum_{m=1}^{\infty} u^m /m } $ and the closely related $ \det (1-ft)^{-1} = e^{ \sum_{n=1}^{\infty} \operatorname{tr}(f^n) t^n /n }$, it suffices to prove

$$\sum_i (-1)^i \sum_{n=1}^{\infty} \operatorname{tr}(f^n, H^i(X)) t^n/n = \sum_{v \in |X|} \sum_{m=1}^{\infty} t^{ m \deg v} /m$$

which, extracting the coefficients of $t^n$, follows from the Lefschetz formula by

$$ \sum_i (-1)^i \operatorname{tr}(f^n, H^i(X)) = \# X^{ f^n} = \#X( \mathbb F_{q^n}) = \sum_{ v \in |X| , \deg v \mid n} \deg v$$ since each closed point of degree $\deg v$ corresponds to $\deg v$ points over $\mathbb F_{q^n}$ whenever $n$ is a multiple of $v$.

The second step, $$\prod_{v \in |X|} \frac{1}{ 1- t^{\deg v}} = \sum_{D \geq 0} t^{ \deg D}$$ follows from the definition of effective divisors as formal sums of closed points.

The third step,

$$ \sum_{D \geq 0} t^{ \deg D} = \int_{ a \in \mathbb I_F} 1_{a \textrm{ integral}} t^{ \deg a} d\mu $$

follows since each idele has a divisor, the divisor is effective if and only if the idele is integral (locally at each place), the degree of that divisor equals the degree of the idele, and the measure of the adeles with a given divisor is independent of the divisor (since the measure is multiplicatively invariant.

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But I'm pretty sure for the question that you don't just want this, the standard proof of the formula, but something more intuitive. Why should we expect something like this is the case?

I think that the right answer is not to seek some totally separate argument but to concentrate on each step of this and try to figure out why it is intuitive and natural from the right perspective. For example, in the case of finite schemes, the Lefschetz formula, involving only $H^0$, as trivial. We can thus think of this formula as the natural generalization of that simple identity to the higher-dimensional case.

But if you only want to know why the higher analogue of the characteristic polynomial is the sum of a fixed function against a varying character, there is another approach. I kind of think the best way to present it is to answer your question with a question, i.e.

Why is the sum over all tilings of $1 \times n$ rectangles with $1 \times 1$ and $1 \times 2$ blocks of the character $t^n$ equal to $\frac{1}{ \det (1- Mt)}$, where $M$ is the $2 \times 2$ matrix $ \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$?

to which one answer could be

Grouping terms, this sum is the generating function $\sum_n F_n t^n$ of the Fibonacci series $F_n$. The Fibonacci series satisfy the recurrence relation $F_{n+2} = F_{n+1}+ F_n$, which can be expressed with the matrix $M$ as $\begin{pmatrix} F_{n+1} \\ F_{n+2} \end{pmatrix} = M \begin{pmatrix} F_n \\ F_{n+1} \end{pmatrix}$, and the identity $\sum_n F_n t^n = \frac{1}{1-t-t^2} = \frac{1}{ \det (1- Mt)}$ can be easily proven by multiplying both sides by $\det (1- Mt)$ and applying the recurrence relation.

but to which another answer could be

Generating functions tend to be related to characteristic polynomials of matrices (and other nice mathematical functions) by a number of identities. The generating function $\sum_n F(n) t^n$, where $F(n)$ counts objects of a particular type, can be expressed as a sum over objects of that type of $t^n$, and when objects of that type have a natural notion of combination, composition, or multiplication, the function $t^n$ is frequently a character.