Galois action on the character group of a torus I am trying to understand how the Galois action on the $\bar{F}$-points of a connected, reductive group is related to the Galois action on the group of characters $X^*(T)$ of a maximal torus. A closely related question has already been asked at this post: Galois action on character group
The answer given there almost makes sense to me, but the Galois action on $X^*(T)$ is defined in terms of a Galois action on $\mathbb{G}_m$; is this the usual Galois action on $F$-points, or does the action on $\mathbb{G}_m$ arise from the (established) Galois action on $T(\bar{F})$? A non-trivial example (say of a unitary group) would also be greatly appreciated.
 A: $\newcommand{\Hom}{\mathrm{Hom}}$$\newcommand{\Z}{\mathbb{Z}}$$\newcommand{\Spec}{\mathrm{Spec}}$$\newcommand{\ov}[1]{\overline{#1}}$$\newcommand{\bb}[1]{\mathbb{#1}}$
When I write $\overline{F}$ below I mean the separable closure of $F$.
So, by definition $X^\ast(T):=\mathrm{Hom}_{\ov{F}-\mathrm{grps}}(T_{\ov{F}},\mathbb{G}_{m,\ov{F}})$. The action of $\sigma\in\Gamma_F$ (I'll use $\Gamma_F$ for the absolute Galois group) is defined as follows:
$$\sigma\cdot f:= \sigma_{\bb{G}_{m,\ov{F}}}\circ f\circ \sigma_{T_{\ov{F}}}^{-1}$$
where for any $F$-scheme $X$ we define $\sigma_{X_{\ov{F}}}$ to be the natural map $X_{\ov{F}}\to X_{\ov{F}}$ which, when one writes $X_{\ov{F}}:= X\times_{\Spec(F)}\Spec(\ov{F})$, is trivial on the $X$-component and is $\Spec(\sigma^{-1})$. 
So, let's do an example very explicitly. Let's set $F$ to be any field and $E/F$ to be a degree $2$ Galois extension. Then, we can form the group $U(1)_{E/F}$, or just $U(1)$, defined as the kernel of the natural map $\mathrm{Res}_{E/F}\bb{G}_{m,E}\to \bb{G}_{m,F}$ given by the norm. This is a one-dimensional torus. If we write $E=F(\sqrt{d})$ where $d\in F$, then as an affine scheme we can write $U(1)\cong \Spec(F[a,b]/(a^2-db^2-1))$ where Hopf algebra structure is the map 
$$F[a,b]/(a^2-db^2-1)\to F[a_1,b_1,a_2,b_2]/(a_1^2-db_1^2-1,a_2^2-db_2^2-1)$$
(where we have made the identification of $F[a,b]/(a^2-db^2-1)\otimes_F F[a,b]/(a^2-db^2-1)$ with $F[a_1,b_1,a_2,b_2]/(a_1^2-db_1^2-1,a_2^2-db_2^2-1)$) given by 
$$a\mapsto a_1a_2+db_1b_2,\quad b\mapsto ab_2+ba_2$$
Let us calculate its character lattice. Well, we have a natural isomorphism $\varphi:U(1)_{\ov{F}}\to \bb{G}_{m,\ov{F}}$ given on coordinate rings (using $t$ for the paramter of $\bb{G}_{m,F}$) by 
$$\ov{F}[t]\to \ov{F}[a,b]/(a^2-db^2-1):t\mapsto a+b\sqrt{d}$$
with inverse 
$$\Spec(\varphi^{-1}):\ov{F}[a,b]/(a^2-db^2-1)\to \ov{F}[t]$$
given by 
$$a\mapsto \frac{1}{2}(t+t^{-1}),\quad b\mapsto \frac{1}{2\sqrt{d}}(t-t^{-1})$$
It's then not hard to see that $\Hom_{\ov{F}-\mathrm{grps}}(U(1)_{\ov{F}},\bb{G}_{m,\ov{F}})=\Z\cdot\varphi\cong \Z$. Let's now see how $\Gamma_F$ acts on $\Z\cdot \varphi$. Of course, it suffices to specify how it acts on $\varphi$. Now, by definition we have that $\sigma\cdot \varphi$ is the map $ \sigma_{\bb{G}_{m,\ov{F}}}\circ \varphi\circ \sigma_{T_{\ov{F}}}^{-1}$. Let's see how this acts on coordinate rings. Namely, this map corresponds to a map
$$\ov{F}[t]\to \ov{F}[a,b]/(a^2-db^2-1)$$
which can be thought of as factorized as
\begin{equation}
$$\ov{F}[t]\xrightarrow{\sigma^{-1}}\ov{F}[t]\xrightarrow{\Spec(\varphi)}\ov{F}[a,b]/(a^2-db^2-1)\xrightarrow{\sigma}\ov{F}[a,b]/(a^2-db^2-1)$$
and we'll call these arrows 1, 2, and 3 (labeled left to right). Well, let's see where $t$ goes. In the first map it maps to $t$ again. Under the second map it maps to $a+b\sqrt{d}$ and under the last map it maps to $a+b\sigma(\sqrt{d})$. 
In particular, note that $\Gamma_F$ acts through the quotient $\mathrm{Gal}(E/F)$ and if we denote by $\sigma$ the non-trivial element of $\mathrm{Gal}(E/F)$ this satisfies that $\sigma\cdot \varphi$ sends $t$ to $a-b\sqrt{d}$. Note though that this is just $(a+b\sqrt{d})^{-1}$ and so it's not hard to see that $\sigma\cdot \varphi=-\varphi$ (our groups are multiplicatively written so it might be more natural to write $\varphi^{-1}$ but this could be confused with the inverse of $\varphi$ so I wrote it additively).
In other words, $X^\ast(U(1))$ is the $\Gamma_F$-module $\mathbb{Z}$ where the $\Gamma_F$ structure is the composition
$$\Gamma_F\twoheadrightarrow \mathrm{Gal}(E/F)\to \mathrm{Aut}(\mathbb{Z})\cong \mathbb{Z}/2\mathbb{Z}$$
where the arrow $\mathrm{Gal}(E/F)\to\mathbb{Z}/2\mathbb{Z}$ is the obvious isomorphism.

To see the relationship between $\sigma_{T_{\ov{F}}}$ and $T(\ov{F})$ for a torus $T$ over $F$ note that $T(\ov{F})$ is nothing other than $\mathrm{Hom}_F(\mathrm{Spec}(\ov{F}),T)$. But, this is the same as $\Hom_{\ov{F}}(\Spec(\ov{F}),T_{\ov{F}})$. By the above we then know how $\Gamma_F$ should act on $f\in\Hom_{\ov{F}}(\Spec(\ov{F}),T_{\ov{F}})$: 
$$\sigma\cdot f:= \sigma_{T_{\ov{F}}}\circ f\circ \sigma_{\Spec(F)_{\ov{F}}}^{-1}$$
Let's do two examples to see that this is what you should get:


*

*If $T$ is $\mathbb{G}_{m,F}$ then $T(\ov{F})=\ov{F}^\times$ and for $\alpha\in T(\ov{F})$ you expect that $\sigma_{T_{\ov{F}}}(\alpha)=\sigma(\alpha)$. Well, note that in scheme world this $\alpha\in\ov{F}$ corresponds to the map $\Spec(\ov{F})\to T_{\ov{F}}$ that corresponds to $\ov{F}$-algebra map $\ov{F}[t,t^{-1}]\to \ov{F}$ that takes $t$ to $\alpha$. Note then that $\Spec(\sigma\cdot f)=\Spec(\sigma_{\Spec(F)_{\ov{F}}}^{-1})\circ \Spec(f)\circ\Spec(\sigma_{T_{\ov{F}}})$ takes (using the sequence of maps on the right hand side) $t$ iteratively to $t$, then $\alpha$, then finally $\sigma(\alpha)$. So you get what you expect.

*If $T=U(1)_{E/F}$ then $T(\ov{F})=\{(\alpha,\beta)\in\ov{F}^2:\alpha^2-d\beta^2=1\}$. You expect that $\sigma$ acts on $(a,b)$ as $(\sigma(a),\sigma(b))$. To see this, note that $(\alpha,\beta)$ corresponds to the $\ov{F}$-algebra map $\ov{F}[a,b]/(a^2-db^2-1)\to \ov{F}$ taking $a$ to $\alpha$ and $b$ to $\beta$. Note then that if you consider $\Spec(\sigma\cdot f)=\Spec(\sigma_{\Spec(F)_{\ov{F}}}^{-1})\circ \Spec(f)\circ\Spec(\sigma_{T_{\ov{F}}})$ this takes $a$ to $\sigma(\alpha)$ and $b$ to $\sigma(\beta)$, just as you'd expect.

