How to construct the forms of $\mathbb{G}_m$? This is a sort of sequel to a question I asked before: Forms of the multiplicative group.
My question is how to construct the forms of $\mathbb{G}_m$ using Galois cohomology.
If $k$ is a field (assume it's of characteristic 0), and $X$ is a $k$-scheme, a form of $X$ is defined to be a $k$-scheme $X'$ such that $X'_{\overline{k}}$ is isomorphic to $X_{\overline{k}}$. Let $\Gamma=Gal(\overline{k}/k)$. Then there is a natural bijection between isomorphism classes of forms of $X$ with $H^1(\Gamma,Aut_{\overline{k}}(X))$. 
I want to understand both directions of the correspondence and how it works for $X=\mathbb{G}_m$, and how to construct explicitly the forms of $\mathbb{G}_m$. I understand how the map works in one direction. Given a form $X'$ with an isomorphism $\phi:X_{\overline{k}} \rightarrow X'_{\overline{k}}$, we want to define a cocycle from $\Gamma$ to $Aut_{\overline{k}}(X)$. Let $\sigma \in \Gamma$ be given. Then $\sigma$ induces another isomorphism $\sigma \phi:X_{\overline{k}} \rightarrow X'_{\overline{k}}$. To get a cocycle, send $\sigma$ to $\phi^{-1} \circ (\sigma \phi)$. (Here I'm thinking of $\sigma \phi$ as "upstairs" and I'm starting from the lower left and going around clockwise.) I don't know how the reverse direction goes, though: given a cocycle, how do you get a form? How does it work for $\mathbb{G}_m$? Looking back at the question I asked before (the link is above), I still don't see how one gets $T_a$. Basically, it seems that there should be algorithm to get the forms but I don't know how it works for this example, which should be the simplest case (non-split rank one tori)
 A: This is the general theory of descent (here Galois descent). It is a bit long and I won't reproduce the proofs here. There are plenty of places where this is done, and it will be explained more clearly than I ever could.
But let us see what is going on for $\mathbb{G}_m$. I hope it will give you an idea of the general facts.
Let $(a_\sigma)$ be a cocycle with value in $\operatorname{Aut}(\mathbb{G}_{m,\overline{k}})$. (So really an homomorphism $G_k\rightarrow\{\pm 1\}$ but for the sake of generality I will only use the cocycle condition).
Let us construct an action of $G_k$ on $\overline{k}[X,X^{-1}]$ by putting $\sigma.P=(a_\sigma^{-1})^\#\sigma(P)$ where $(a_\sigma^{-1})^\#$ is the ring homomorphism induced by $a_\sigma\in\operatorname{Aut}(\mathbb{G}_m)$ and $\sigma(P)$ is the action of Galois on the coefficient of the Laurent series $P$.
This is indeed an action :
$$(\sigma\tau).P=(a_{\sigma\tau}^{-1})^\#\sigma\tau(P)=((a_\sigma\sigma(a_\tau))^{-1})^\#\sigma\tau(P)=(a_\sigma^{-1})^\#\sigma(a_\tau^{-1})^\#\sigma\tau(P)$$
$$ \sigma.(\tau.P)=(a_\sigma^{-1})^\#\sigma((a_\tau^{-1})^\#\tau(P))$$
so these two Laurent series are equal (recall the action of $G_k$ on automorphisms).
Note that this action acts by semi-linear automorphism : $\sigma.(\lambda P)=\sigma(\lambda)\sigma.P$
Then the construction will be $T_a\simeq \operatorname{Spec}\overline k[T,T^{-1}]^G$, the algebra of invariant under the action.
Concretely, we have seen that a cocycle $(a_\sigma)$ is just an homomorphism $f:G_k\rightarrow\operatorname{Aut}(\mathbb{G}_m)=\{\pm 1\}$ such that $f(\sigma)=-1$ iff $\sigma(\sqrt{a})=-\sqrt{a}$. Hence the morphism $a_\sigma^\#$ maps $T$ to $T$ $\sigma(\sqrt{a})=\sqrt{a}$ and $T$ to $T^{-1}$ if $\sigma(\sqrt{a})=-\sqrt{a}$. Or directly $a_\sigma^\#(T)=T^{f(\sigma)}$.
From this, it follows that the action of $G_k$ is $\sigma.P(T)=\sigma(P(T^{f(\sigma)}))$. It follows that the fixed algebra consist of polynomial of the form $\sum \lambda_i T^i$ where $\lambda_{2i}\in k$, $\lambda_{2i+1}-\lambda_{-2i-1}\in k$ and $\lambda_{2i+1}+\lambda_{-2i-1}\in k.\sqrt{a}$. With $U=\frac{1}{2}(T+T^{-1})$ and $V=\frac{\sqrt{a}}{2a}(T-T^{-1})$, this algebra can be given the following description $k[U,V]/(U^2-aV^2-1)$. This is exactly the algebra of the twisted multiplicative group $\mathbb{G}_{m,a}$.
