This seems to be a trivial question, but I am genuinely confused about it.


The notion of weights as in [Deligne's Weil II](http://www.numdam.org/item/PMIHES_1980__52__137_0/) are defined in terms of eigenvalues of automorphisms that Frobenius morphisms induce on stalks. The following is a definition that is found in many literatures:



> 
> Let $k$ be a finite field of $q = p^d$ elements, and $X\_0$ be a scheme over $k$.
>  For an étale morphism $\phi:U\_0 \to X\_0$, it is a theorem that its pullback under the absolute Frobenius $Fr\_{X\_0}:X\_0 \to X\_0$ is isomorphic to $\phi$ itself. Via this isomorphism, for any sheaf $\mathcal{G}\_0$ on (the étale site of) $X\_0$ one can define a morphism $Fr\_{X\_0}^{\ast} \mathcal{G}\_0 \to \mathcal{G}\_0$, denoted by $Fr\_{\mathcal{G}\_0}$.
> 
> 
> Let $X$ be a pullback of $X\_0$ via $k \to \bar{k}$, and $\mathcal{G}$ be a pullback of $\mathcal{G}\_0$ via $X\to X\_0$. Then pulling back $Fr\_{\mathcal{G}\_0}$ defines a morphism $F\_X^{\ast}\mathcal{G} \to \mathcal{G}$, denoted by $F\_{\mathcal{G}}$. 
> 
> 
> For any geometric point $\bar{x}$ of $X$, whose underlying point $x$ has residue field of elements $q^d$, the $d$-th iteration of $F\_X$ fixes $\bar{x}$, so pulling back the $d$-th iteration of $F\_{\mathcal{G}}$ via $\bar{x}$ induces a morphism $\mathcal{G}\_{\bar{x}}\to \mathcal{G}\_{\bar{x}}$, which is called the geometric Frobenius. 
> 
> 
> The weights of $\mathcal{G}$ at $\bar{x}$ are eigenvalues of this morphism.
> 
> 
> 


Now in most of literatures, it is stated without proof that the weight of $\mathbb{Q}\_l(1)$, the inverse limit of $\mu\_{l^n}$s is $-2$. This looks intuitively clear: the arithmetic Frobenius, which can be proven to be inverse to the geometric Frobenius, **seems to be** acting on $\mu\_{l^n}$ via $\xi \to \xi^{q^d}$, so its inverse has the weight $-2$.


However, I can't deduce this formally from the definition of the Frobenius map; it looks as if every maps in the sections of the Frobenius are canonical isomorphisms, so maps among stalks are the canonical isomorphisms as well. Can somebody formally deduce the computation of the weight of $\mathbb{Q}\_l(1)$ from definitions?



Computing weights of $\bar{\mathbb{Q}}_l(1)$ from the definition
First, there is no point in including $X$ in the definition. We're interested in the stalk at $x$. We can also view $x$ as a geometric point of $X\_0$. Pulling back from $X\_0$ to $X$ and then taking the stalk is the same as taking the stalk at $x$, almost by definition.


By definition, the stalk of $\mathcal G\_0$ at $x$ is the limit over all etale neighborhoods $U$ of $x$ of $\mathcal G\_0(U)$. Recall that these are pairs of maps maps $a: \operatorname{Spec} \kappa \to U, b: U \to X$ such that $b \circ a=i$ for a fixed map $i: \operatorname{Spec} \kappa \to X$.


On the other hand, the stalk of $Fr^\* \mathcal G\_0$ at $x$ is the limit over all etale neighborhoods $U'$ of $Fr(x)$ of $\mathcal G\_0(U')$. In other word this is a limit over pairs of maps $a': \operatorname{Spec} \kappa \to U', b':U' \to X$ such that $b'\circ a'= Fr\_X \circ i$.


The natural isomorphism is noting that given data $U,b,a$ we have data $U', b',a'$ by taking $U'=U, b'=b$, $a' = Fr\_U \circ a = a \circ Fr\_{\kappa}$ because $b \circ Fr\_U = Fr\_X \circ b$. And we can reverse this by noting that $Fr\_X$ has an inverse.


Now if $x$ happens to be stable under $Fr\_X$, we get another isomorphism between these two where we simply take $(U',b',a')=(U,b,a)$. 


Now if $\mathcal G\_0$ happens to be $\mu\_{\ell^n}$, the stalk of $\mathcal G\_0$ at $x$ is naturally isomorphic to $\mu\_{\ell^n}(\kappa)$. The isomorphism sends $s \in \mu\_{\ell^n}(U)$ to $a^\* s \in \mu\_{\ell^n}(\kappa)$. This isomorphism is compatible with the second isomorphism $Fr^\* \mathcal G\_{0,x} \cong \mathcal G\_{0,x}$, because that isomorphism sends $a^\*s $ to $a^\*s$. 


But using this natural isomorphism to calculate the first isomorphism, it sends $a^\* s \to Fr\_{\kappa}^\* a^\* s$. In other words, Frobenius acts on the stalk $\mu\_{\ell^n}(\kappa)$ by sending $\zeta \in \mu\_{\ell^n}(\kappa)$ to $Fr^\* \zeta = \zeta^q$.


The same thing works for $q^d$.




---


Why is the natural isomorphism what I said?


It's helpful to remember that $Fr^\* \mathcal G ( U)$ is a limit over commutative squares


$\require{AMScd}$
\begin{CD}
 X @>Fr\_X>> X\\
 @A{b}AA @AA{b'}A\\
 U @>>j> V
\end{CD}


where we then sheafify, and in this case the limit collapses to simply $V =U, b'=b, j = Fr\_U$, and the sheafification is unnecessary. This shows that $Fr^\* \mathcal G(U) =\mathcal G(U)$, and is the unique isomorphism in this level of generality.


If we then take a stalk at $x$, we'll take the limit of $\mathcal G(U)$ over triples $U, a,b$ as mentioned earlier. When we take the stalk of the pullback $(Fr\_X^\*\mathcal G)\_x$, we get a limit over diagrams like this:


\begin{CD}
 X @>Fr\_X>> X\\
 @A{b}AA @AA{b'}A\\
 U @>>j> V\\
@A{a}AA\\
\operatorname{Spec} \kappa
\end{CD}
and the natural isomorphism with $(\mathcal G)\_{Fr(x)}$ is by viewing it as a limit over diagrams like 


\begin{CD}
 @. X\\
 @. @AA{b'}A\\
 U @>>j> V\\
@A{a}AA\\
\operatorname{Spec} \kappa
\end{CD}


where $V$ plays the role $U$ did before and $j \circ a$ plays the role $a$ did before.


Our canonical isomorphism between these two things involves the identification $V= U, b'=b, j = Fr\_U,$ and thus $j \circ a = Fr\_U \circ a$. So that justifies what I said earlier.


