1. Trivial example. First, suppose $X$ is finite. Then we have a finite set $S := X(\overline{\mathbb{F}}\_q)$ with an action of $\text{Fr}\_q$. How can one explain why the rationality of the zeta function should be true in this case (it is still nonobvious from the definition of $Z(X, t)$)? Let $\mathbb{Q}[S]$ be the vector space spanned by $S$; then we get an induced linear action of $\text{Fr}\_q$ on $\mathbb{Q}[S]$. We have$$Z(X, t) = \det(1 - t \cdot \text{Fr}\_q; \mathbb{Q}[S])^{-1}.\tag\*{$(1)$}$$This is seen by showing that the logarithmic derivatives of the two sides are the same. Let $\{\alpha\_i\}$ denote the collection of the eigenvalues of $\text{Fr}\_q$ acting on $\mathbb{Q}[S]$, counted with their algebraic multiplicities. Then$$\det(1 - t \cdot \text{Fr}\_q) = \prod\_i (1 - \alpha\_i t).$$Applying $t \cdot {{\text{d}\log}\over{\text{d}t}}$ to both sides of $(1)$, we obtain$$\sum\_{n = 1}^\infty |X(\mathbb{F}\_{q^n})|t^n = \sum\_i \sum\_{n = 1}^\infty (\alpha\_i t)^n = \sum\_{n = 1}^\infty \text{Tr}(\text{Fr}\_q^n) \cdot t^n.$$However, it is very easy to see that$$\text{Tr}(\text{Fr}\_q^n; \mathbb{Q}[S]) = \left|S^{\text{Fr}\_q^n}\right|,$$and so we get $(1)$ as desired.
2. A very different, but morally analogous, story. Let$$X = \text{compact }C^\infty\text{ manifold},$$$$\text{Fr} = \text{a self-diffeomorphism of }X,\text{ such that }\forall n \in \mathbb{N}, \text{ }\left|X^{\text{Fr}^n}\right| < \infty.$$Assume that $X$ is orientable, and that for any fixed point $x \in X$ of $\text{Fr}^n$ (for some $n \in \mathbb{N})$, the condition $\det(1 - \text{Fr}^n; T\_xX) > 0$ holds. Then $\deg(1 - \text{Fr}^n; T\_xX) \neq 0$ for all $x \in X^{\text{Fr}^n}$ is equivalent to all fixed points of $\text{Fr}^n$ being nondegenerate, i.e. the corresponding intersection points of $\Delta\_X$ and $\Gamma(\text{Fr}^n)$ are transverse (here, $\Delta$ means diagonal, $\Gamma$ means graph). Now we can define the zeta function $Z((X, \text{Fr}), t)$ in the same way as above:$$Z((X, \text{Fr}), t) = \prod\_{i = 0}^{\dim X} \det(1 - t \cdot \text{Fr}^\*; H^i(X, \mathbb{Q}))^{(-1)^{i + 1}}.\tag\*{$(2)$}$$If $X$ is finite, then $H^0(X, \mathbb{Q}) \cong \mathbb{Q}[X]^\*$, and we are essentially reduced to the previous example.
3. Anyways, applying $t \cdot {{\text{d}\log}\over{\text{d}t}}$ to both sides of $(2)$, we can reduce it to the following identity, the well-known Grothendieck-Lefschetz fixed point formula: for all $n \in \mathbb{N}$,$$\left|X^{\text{Fr}^n}\right| = \sum\_{i = 0}^{\dim X} (-1)^i \text{Tr}\left(\text{Fr}^n; H^i(X, \mathbb{Q})\right).\tag\*{(3)}$$


**What I want.** I want to see how one can guess the Lefschetz fixed point formula, which entails an informal argument which can be turned into a rigorous proof. The idea to imitate the argument given above in the case where $X$ is finite. I want to use the fact that cohomology can be computed as the de Rham cohomology.


**Question.** Pretend that all the spaces of differential forms $\Omega^i(X)$ are finite dimensional, replace $H^u(X)$ by $\Omega^i(X)$. How do I see $(2)$ by using the same method that I used in the case where $X$ is finite? What are the extra nuances I have to take into consideration/take care of?



Rationality of zeta function and Grothendieck-Lefschetz fixed point formula, cohomology can be computed as the de Rham cohomology
The following is sections 3-4 of van der Put's [The cohomology of Monsky and Washnitzer](https://eudml.org/doc/94862) with all $p$-adic issues and analytic subtleties (necessary to make the argument correct, of course!) removed. 


Let $X$ be a smooth manifold of dimension $n$ and let
$F: X \to X$ be a degree $q^n$ finite map. (You can just view $q$ as a real number, the $n$-th root of the degree; of course, I am calling it that because it is the order of the ground field in the Frobenius setting.) We want to compute
$$\sum (-1)^j \mathrm{Tr}\ F^{\ast} H^j\_c(X) \to H^j\_c(X)$$
where $H^j\_c$ is compactly supported cohomology.


Using Poincare duality, we may instead compute
$$q^n \sum (-1)^j \mathrm{Tr}\ (F^{\ast})^{-1} H^j(X) \to H^j(X).$$
In van der Put, this is given as the starting goal without motivation (equation 1.2).


(This part is section 3.2.) Let $\psi : \Omega^j(X) \to \Omega^j(X)$ be push down along $F$: The value of $\psi(\omega)$ near a point $x \in X$ is the sum of $\omega$ over the $q$ preimages of $x$ (computed with multiplicity, if there is ramification. We observe: $\psi \circ d = d \circ \psi$ (because $d$ is linear) so $\psi$ passes to cohomology. We have $\psi(F^{\ast} \omega) = q^n \omega$ (pulling back and pushing forward adds up the same thing $\deg F = q^n$ times). So $q^{-n} \psi$ is a left inverse of $F^{\ast}$ on $H^{\ast}(X)$. As $\dim H^{\ast}(X)$ is finite, this just means that $\psi = q^n (F^{\ast})^{-1}$. (At the time van der Put was writing, Monsky-Washnitzer cohomology was not known to be finite dimensional, so he has a trick to get around this -- Theorem 3.2.(iv). Finiteness was proved by [Berthelot](http://link.springer.com/article/10.1007%2Fs002220050143) and [Mebkhout](http://muse.jhu.edu/article/762/pdf).) So we can rewrite the desired sum as
$$ \sum (-1)^j \mathrm{Tr}\ \psi: H^j(X) \to H^j(X).$$


For future use, we note that $\psi(F^{\ast}(a) \omega)=a \psi(\omega)$ for $a \in \mathcal{O}$ and $\omega \in \Omega^j$. This is also clear: If $a(x)=\alpha$, then multiplying $\omega$ by $F^{\ast}(\omega)$ multiplies by $\alpha$ at each point in $F^{-1}(x)$, so the sum defining $\psi(F^{\ast} \omega)$ is multiplied by $\alpha$ at $x$.




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I pause for an example:


Let $F : \mathbb{C}^{\ast} \to \mathbb{C}^{\ast}$ be the $q$-th power map, so the fixed points are the $q-1$ many nonzero roots of $z^q=z$. Let $\zeta$ be a primitive $q-1$ root of unity. Then $\psi(g)(z) = \sum\_{k=1}^{q-1} g(\zeta^k z)$ for any function $g$ on $\mathbb{C}^{\ast}$. In particular
$$\psi(z^a) = \begin{cases} q z^{a/q} & a \equiv 0 \bmod q \\ 0 & a \not \equiv 0 \bmod q \end{cases}.$$
Working a bit harder,
$$\psi(z^a dz/z) = \begin{cases} z^{a/q} dz/z & a \equiv 0 \bmod q \\ 0 & a \not \equiv 0 \bmod q \end{cases}.$$
No, I am not missing a factor of $q$. The push forward of $dz/z$ along the $q$-th power map is $dz/z$. Let's write $w$ for the coordinate on the source space and $z$ on the target, so $F^{\ast} z=w^q$ and $F^{\ast} (dz/z)=q dw/w$. Then $$\psi(dw/w) = \psi((1/q) dz/z) = \sum (1/q) d(\zeta^k z)/(\zeta^k z) = q\cdot(1/q) dz/z=dz/z.$$


Morally, the trace of $\psi$ on $\Omega^0$ wants to be $q$, and on $\Omega^1$ wants to be $1$. When we pass to $H^0$ and $H^1$, this is true with no analytic subtlties. (So $q-1$ is the number of fixed points, as desired.) To make the argument on $\Omega^j$, one uses the word "nuclear" a lot. I don't understand this part, but I believe that one of the reasons to use $\psi$ instead of $F^{\ast}$ is that $\psi$ is nuclear.




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So, we continue with the general argument. We now reduce to the case where there are no fixed points. Let the fixed points be $z\_1$, ..., $z\_N$ and let $U = X \setminus \{ z\_1, \ldots, z\_N \}$. In the case of Frobenius, $F$ maps $U$ to $U$. There is a Gysin sequence relating $H^{\ast}(X)$, $H^{\ast}(U)$ and a simple contribution for each fixed point, and we can see that the alternating sums of traces of $F$ on $H^{\ast}(U)$ and $H^{\ast}(X)$ differ by $N$. 


I think (this part isn't in van der Put) that one can still make this argument even if $F(U)$ isn't $U$. Let $U' = F^{-1}(U)$. We can map $\Omega^j(U) \to \Omega^j(U') \overset{\psi}{\longrightarrow} \Omega^j(U)$ where the first map is restriction and the second makes sense because $U' \to U$ is finite. I think we can make the same argument with this map. This seems to be another advantage of the pushforward $\psi$ over the pull back $F^{\ast}$. 


I was also glossing over the difference between the characteristic $p$ variety and its $p$-adic lift two paragraphs back. (Because I decided to shift to the case of a manifold, where we don't have this issue.) In the $p$-adic lift, we are removing $F$-invariant balls around each fixed point, not just points. 


(This is the proof of 4.10) Assume that $F$ has no fixed points. So, as $a$ ranges over $\mathcal{O}(X)$, there is no place where all the functions $F^{\ast}(a) - a$ vanish. So the $F^{\ast}(a)-a$ generate the unit ideal; say $\sum (F^{\ast} a\_i -a\_i) b\_i =1$. 


So, for any $\omega \in \Omega^j$, we have
$$\sum \psi((F^{\ast} a\_i) b\_i \omega - b\_i a\_i \omega) = \psi(\omega).$$
(We have used that multiplication is commutative.) 
Since $\psi(F^{\ast}(a) \eta) = a \psi(\eta)$, we deduce 
$$\sum a\_i \psi(b\_i \omega) - \psi(b\_i a\_i \omega) = \psi(\omega).$$
In other words, the summand is the commutator of the operations $\omega \mapsto a\_i \omega$ and $\omega \to \psi(b\_i \omega)$. Since commutators always have trace zero, we "deduce" that $\psi: \Omega^j \to \Omega^j$ has trace zero, and hence the alternating sum of traces is zero.


