Exercise 63.3.1. Let $X$ be a topological space and $g : X \to X$ a continuous map such that $g^{-1}(U) = U$ for all opens $U$ of $X$. Then $g$ induces the identity on cohomology on $X$ (for any coefficients).

## 63.3 Frobenii

In this section we will prove a “baffling” theorem. A topological analogue of the baffling theorem is the following.

We now turn to the statement for the étale site.

Lemma 63.3.2. Let $X$ be a scheme and $g : X \to X$ a morphism. Assume that for all $\varphi : U \to X$ étale, there is an isomorphism

functorial in $U$. Then $g$ induces the identity on cohomology (for any sheaf).

**Proof.**
The proof is formal and without difficulty.
$\square$

Please see Varieties, Section 33.36 for a discussion of different variants of the Frobenius morphism.

Theorem 63.3.3 (The Baffling Theorem). Let $X$ be a scheme in characteristic $p > 0$. Then the absolute frobenius induces (by pullback) the trivial map on cohomology, i.e., for all integers $j\geq 0$,

is the identity.

This theorem is purely formal. It is a good idea, however, to review how to compute the pullback of a cohomology class. Let us simply say that in the case where cohomology agrees with Čech cohomology, it suffices to pull back (using the fiber products on a site) the Čech cocycles. The general case is quite technical, see Hypercoverings, Theorem 25.10.1. To prove the theorem, we merely verify that the assumption of Lemma 63.3.2 holds for the frobenius.

**Proof of Theorem 63.3.3.**
We need to verify the existence of a functorial isomorphism as above. For an étale morphism $\varphi : U \to X$, consider the diagram

The dotted arrow is an étale morphism and a universal homeomorphism, so it is an isomorphism. See Étale Morphisms, Lemma 41.14.3. $\square$

Definition 63.3.4. Let $k$ be a finite field with $q = p^ f$ elements. Let $X$ be a scheme over $k$. The *geometric frobenius* of $X$ is the morphism $\pi _ X : X \to X$ over $\mathop{\mathrm{Spec}}(k)$ which equals $F_ X^ f$.

Since $\pi _ X$ is a morphism over $k$, we can base change it to any scheme over $k$. In particular we can base change it to the algebraic closure $\bar k$ and get a morphism $\pi _ X : X_{\bar k} \to X_{\bar k}$. Using $\pi _ X$ also for this base change should not be confusing as $X_{\bar k}$ does not have a geometric frobenius of its own.

Lemma 63.3.5. Let $\mathcal{F}$ be a sheaf on $X_{\acute{e}tale}$. Then there are canonical isomorphisms $\pi _ X^{-1} \mathcal{F} \cong \mathcal{F}$ and $\mathcal{F} \cong {\pi _ X}_*\mathcal{F}$.

This is false for the fppf site.

**Proof.**
Let $\varphi : U \to X$ be étale. Recall that ${\pi _ X}_* \mathcal{F} (U) = \mathcal{F} (U \times _{\varphi , X, \pi _ X} X)$. Since $\pi _ X = F_ X^ f$, it follows from the proof of Theorem 63.3.3 that there is a functorial isomorphism

where $\gamma _ U = (\varphi , F_ U^ f)$. Now we define an isomorphism

by taking the restriction map of $\mathcal{F}$ along $\gamma _ U^{-1}$. The other isomorphism is analogous. $\square$

Remark 63.3.6. It may or may not be the case that $F^ f_ U$ equals $\pi _ U$.

We continue discussion cohomology of sheaves on our scheme $X$ over the finite field $k$ with $q = p^ f$ elements. Fix an algebraic closure $\bar k$ of $k$ and write $G_ k = \text{Gal}(\bar k/k)$ for the absolute Galois group of $k$. Let $\mathcal{F}$ be an abelian sheaf on $X_{\acute{e}tale}$. We will define a left $G_ k$-module structure cohomology group $H^ j (X_{\bar k}, \mathcal{F}|_{X_{\bar k}})$ as follows: if $\sigma \in G_ k$, the diagram

commutes. Thus we can set, for $\xi \in H^ j (X_{\bar k}, \mathcal{F}|_{X_{\bar k}})$

where the last equality follows from the commutativity of the previous diagram. This endows the latter group with the structure of a $G_ k$-module.

Lemma 63.3.7. In the situation above denote $\alpha : X \to \mathop{\mathrm{Spec}}(k)$ the structure morphism. Consider the stalk $(R^ j\alpha _*\mathcal{F})_{\mathop{\mathrm{Spec}}(\bar k)}$ endowed with its natural Galois action as in Étale Cohomology, Section 59.56. Then the identification

from Étale Cohomology, Theorem 59.53.1 is an isomorphism of $G_ k$-modules.

A similar result holds comparing $(R^ j\alpha _!\mathcal{F})_{\mathop{\mathrm{Spec}}(\bar k)}$ with $H^ j_ c (X_{\bar k}, \mathcal{F}|_{X_{\bar k}})$.

**Proof.**
Omitted.
$\square$

Definition 63.3.8. The *arithmetic frobenius* is the map $\text{frob}_ k : \bar k \to \bar k$, $x \mapsto x^ q$ of $G_ k$.

Theorem 63.3.9. Let $\mathcal{F}$ be an abelian sheaf on $X_{\acute{e}tale}$. Then for all $j\geq 0$, $\text{frob}_ k$ acts on the cohomology group $H^ j(X_{\bar k}, \mathcal{F}|_{X_{\bar k}})$ as the inverse of the map $\pi _ X^*$.

The map $\pi _ X^*$ is defined by the composition

where the last isomorphism comes from the canonical isomorphism $\pi _ X^{-1} \mathcal{F} \cong \mathcal{F}$ of Lemma 63.3.5.

**Proof.**
The composition $X_{\bar k} \xrightarrow {\mathop{\mathrm{Spec}}(\text{frob}_ k)} X_{\bar k} \xrightarrow {\pi _ X} X_{\bar k}$ is equal to $F_{X_{\bar k}}^ f$, hence the result follows from the baffling theorem suitably generalized to nontrivial coefficients. Note that the previous composition commutes in the sense that $F_{X_{\bar k}}^ f = \pi _ X \circ \mathop{\mathrm{Spec}}(\text{frob}_ k) = \mathop{\mathrm{Spec}}(\text{frob}_ k) \circ \pi _ X$.
$\square$

Definition 63.3.10. If $x \in X(k)$ is a rational point and $\bar x : \mathop{\mathrm{Spec}}(\bar k) \to X$ the geometric point lying over $x$, we let $\pi _ x : \mathcal{F}_{\bar x} \to \mathcal{F}_{\bar x}$ denote the action by $\text{frob}_ k^{-1}$ and call it the *geometric frobenius*^{1}

We can now make a more precise statement (albeit a false one) of the trace formula (63.2.0.1). Let $X$ be a finite type scheme of dimension 1 over a finite field $k$, $\ell $ a prime number and $\mathcal{F}$ a constructible, flat $\mathbf{Z}/\ell ^ n\mathbf{Z}$ sheaf. Then

as elements of $\mathbf{Z}/\ell ^ n\mathbf{Z}$. The reason this equation is wrong is that the trace in the right-hand side does not make sense for the kind of sheaves considered. Before addressing this issue, we try to motivate the appearance of the geometric frobenius (apart from the fact that it is a natural morphism!).

Let us consider the case where $X = \mathbf{P}^1_ k$ and $\mathcal{F} = \underline{\mathbf{Z}/\ell \mathbf{Z}}$. For any point, the Galois module $\mathcal{F}_{\bar x}$ is trivial, hence for any morphism $\varphi $ acting on $\mathcal{F}_{\bar x}$, the left-hand side is

Now $\mathbf{P}^1_ k$ is proper, so compactly supported cohomology equals standard cohomology, and so for a morphism $\pi : \mathbf{P}^1_ k \to \mathbf{P}^1_ k$, the right-hand side equals

The Galois module $H^0 (\mathbf{P}^1_{\bar k}, \underline{\mathbf{Z}/\ell \mathbf{Z}}) = \mathbf{Z}/\ell \mathbf{Z}$ is trivial, since the pullback of the identity is the identity. Hence the first trace is 1, regardless of $\pi $. For the second trace, we need to compute the pullback $\pi ^* : H^2(\mathbf{P}^1_{\bar k}, \underline{\mathbf{Z}/\ell \mathbf{Z}}))$ for a map $\pi : \mathbf{P}^1_{\bar k} \to \mathbf{P}^1_{\bar k}$. This is a good exercise and the answer is multiplication by the degree of $\pi $ (for a proof see Étale Cohomology, Lemma 59.69.2). In other words, this works as in the familiar situation of complex cohomology. In particular, if $\pi $ is the geometric frobenius we get

and if $\pi $ is the arithmetic frobenius then we get

The latter option is clearly wrong.

Remark 63.3.11. The computation of the degrees can be done by lifting (in some obvious sense) to characteristic 0 and considering the situation with complex coefficients. This method almost never works, since lifting is in general impossible for schemes which are not projective space.

The question remains as to why we have to consider compactly supported cohomology. In fact, in view of Poincaré duality, it is not strictly necessary for smooth varieties, but it involves adding in certain powers of $q$. For example, let us consider the case where $X = \mathbf{A}^1_ k$ and $\mathcal{F} = \underline{\mathbf{Z}/\ell \mathbf{Z}}$. The action on stalks is again trivial, so we only need look at the action on cohomology. But then $\pi _ X^*$ acts as the identity on $H^0(\mathbf{A}^1_{\bar k}, \underline{\mathbf{Z}/\ell \mathbf{Z}})$ and as multiplication by $q$ on $H^2_ c(\mathbf{A}^1_{\bar k}, \underline{\mathbf{Z}/\ell \mathbf{Z}})$.

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