The Stacks project

63.3 Frobenii

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

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).

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

\[ \xymatrix{ U \ar[rd]_\varphi \ar[rr]^-\sim & & {U \times _{\varphi , X, g} X} \ar[ld]^{\text{pr}_2} \\ & X } \]

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 $j\geq 0$,

\[ F_ X^* : H^ j (X, \underline{\mathbf{Z}/n\mathbf{Z}}) \longrightarrow H^ j (X, \underline{\mathbf{Z}/n\mathbf{Z}}) \]

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

\[ \xymatrix{ U \ar@{-->}[rd] \ar@/^1pc/[rrd]^{F_ U} \ar@/_1pc/[rdd]_\varphi \\ & {U \times _{\varphi , X, F_ X} X} \ar[r]_-{\text{pr}_1} \ar[d]^{\text{pr}_2} & U \ar[d]^\varphi \\ & X \ar[r]^{F_ X} & X. } \]

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

\[ \xymatrix{ U \ar[rd]_{\varphi } \ar[rr]_-{\gamma _ U} & & U \times _{\varphi , X, \pi _ X} X \ar[ld]^{\text{pr}_2} \\ & X } \]

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

\[ \mathcal{F} (U) \longrightarrow {\pi _ X}_* \mathcal{F} (U) = \mathcal{F} (U \times _{\varphi , X, \pi _ X} X) \]

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

\[ \xymatrix{ X_{\bar k} \ar[rd] \ar[rr]^{\mathop{\mathrm{Spec}}(\sigma ) \times \text{id}_ X} & & X_{\bar k} \ar[ld] \\ & X } \]

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

\[ \sigma \cdot \xi := (\mathop{\mathrm{Spec}}(\sigma ) \times \text{id}_ X)^*\xi \in H^ j(X_{\bar k}, (\mathop{\mathrm{Spec}}(\sigma ) \times \text{id}_ X)^{-1} \mathcal{F}|_{X_{\bar k}}) = H^ j (X_{\bar k}, \mathcal{F}|_{X_{\bar k}}), \]

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

\[ (R^ j\alpha _*\mathcal{F})_{\mathop{\mathrm{Spec}}(\bar k)} \cong H^ j (X_{\bar k}, \mathcal{F}|_{X_{\bar k}}) \]

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

\[ H^ j(X_{\bar k}, \mathcal{F}|_{X_{\bar k}}) \xrightarrow {{\pi _ X}_{\bar k}^*} H^ j(X_{\bar k}, (\pi _ X^{-1} \mathcal{F})|_{X_{\bar k}}) \cong H^ j(X_{\bar k}, \mathcal{F}|_{X_{\bar k}}). \]

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 frobenius1

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

63.3.10.1
\begin{equation} \label{trace-equation-trace-formula-second} \sum \nolimits _{x \in X(k)} \text{Tr}(\pi _ x | \mathcal{F}_{\bar x}) = \sum \nolimits _{i = 0}^2 (-1)^ i \text{Tr}(\pi _ X^* | H^ i_ c(X_{\bar k}, \mathcal{F})) \end{equation}

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

\[ \sum \nolimits _{x \in X(k)} \text{Tr}(\varphi | \mathcal{F}_{\bar x}) = \# \mathbf{P}^1_ k(k) = q+1. \]

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

\[ \text{Tr}(\pi ^* | H^0 (\mathbf{P}^1_{\bar k}, \underline{\mathbf{Z}/\ell \mathbf{Z}})) + \text{Tr}(\pi ^* | H^2 (\mathbf{P}^1_{\bar k}, \underline{\mathbf{Z}/\ell \mathbf{Z}})). \]

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

\[ \text{Tr}(\pi _ X^* | H^2 (\mathbf{P}^1_{\bar k}, \underline{\mathbf{Z}/\ell \mathbf{Z}})) = q \]

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

\[ \text{Tr}(\text{frob}_ k^* | H^2 (\mathbf{P}^1_{\bar k}, \underline{\mathbf{Z}/\ell \mathbf{Z}})) = q^{-1}. \]

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 $H^2_ c(\mathbf{A}^1_{\bar k}, \underline{\mathbf{Z}/\ell \mathbf{Z}})$.

[1] This notation is not standard. This operator is denoted $F_ x$ in [SGA4.5]. We will likely change this notation in the future.

Comments (12)

Comment #421 by Rex on

In the proof of Lemma 44.69.6, there is a missing subscript in one of the fiber products: \mathcal{F} (\mathcal{U} \times{\varphi, X, \pi_X} X)

Also, is there any reason to use f as the exponent in p^f = q? This made me a bit confused about the notation F_X^f later on. Wouldn't m be a more natural choice for an exponent?

Comment #423 by on

Thanks for pointing out the type. Fixed here. I think it is customary to write for the number of elements of a finite field. Yes, there are lots of f's floating around...

Comment #1069 by David Zureick-Brown on

Typo after Remark 44.80.7

"Consider the We will define a left"

Comment #1864 by Michael Harris on

There's a typo on the second line of the proof of Theorem 50.79.4: S should be X.

Comment #2603 by on

references on the Arithmetic Frobenius are: Y. Flicker, "Drinfeld modui schemes and automorphiic forms: The theory of elliptic modules with applications" (2013). R.Kiehl and R. Weissauer, "Weil Conjectures, Perverse sheaves and l'adic Fourier Transforms" (2001). H.Esnault.

Comment #2628 by on

Dear Dennis Lieberman, I am going to leave this as it is for now. Maybe the book by Kiehl and Weissauer should be mentioned in the introduction to this chaper...

Comment #2989 by Wen-Wei Li on

Probably a typo in the formula before Lemma 53.86.7: the should be .

Comment #3044 by on

Re: formula 53.85.0.1 appears in wikipedia under Grothendieck trace formula (Formal statement for L-functions) and in the Encyclopedia of mathematics under the Lefschetz trace formula. Both sites agree that the formula applies to a constant sheaf of rationals; consequently, the left-hand side gives the number of points of the "field extension of the scheme." (wiki) Similarly, in the Encylopedia of Mathematics ,the left hand side gives the number of "points of the scheme X with values in k."

Comment #5113 by pippo on

Typo: in (03SX) the first should be .


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