Real forms of complex reductive groups I have a collection of related (to me) questions, which stem from the fact that I feel like I have a bunch of pieces, but not a full clear picture. I'm curious about forms of reductive groups in general, so I'm only asking about $\mathbb{C}/\mathbb{R}$ for simplicity's sake and for explicit examples.
As a first fact, I know that $k$ forms of algebraic varieties $X_{k'}$ are classified by $H^1(\operatorname{Gal}(k'/k), \operatorname{Aut}_{k'}(X))$. There's an abstract (to me) way of producing the desired forms by twisting by cocycles.
However, the explicit ways I have of constructing different forms feel different to me.

*

*Tori. Here I immediately reach for $\operatorname{Res}_{k'/k}(T)$, or perhaps a norm torus $\operatorname{Res}_{k'/k}^{(1)}(T)$. For instance, two real forms of $\mathbb{G}_{m}(\mathbb{C})$ are precisely $\mathbb{R}^*$ and $\operatorname{Res}_{\mathbb{C}/\mathbb{R}}^{(1)}(\mathbb{G}_m(\mathbb{C})) = \mathbb{R}[x,y]/(x^2+y^2-1)$.

I'm not clearly aware of how to view this second construction of a non-split (actually anisotropic?) torus as coming from twisting with a cocycle.


*Semisimple groups. Here the natural example is $\operatorname{SL}_2(\mathbb{C})$. The split real form is $SL_2(\mathbb{R})$, so I search for a way to construct $\operatorname{SU}_2(\mathbb{R})$. In my head, here I'm doing something much more cocycle-y, when I take the fixed points of $(x, (\overline{x}^{-1})^t)$ where $S_2$ is acting by exchanging coordinates: here I'm aware that I'm taking an automorphism of $\operatorname{SL}_2(\mathbb{C})$ given by inverse transpose, and composing it with the Galois action of complex conjugation, and taking fixed points. It should be clear that my understanding of this is pretty ad-hoc, but at least I'm aware that something of this sort is related to descent.

So my questions are as follows:
A) How does restriction of scalars (and maybe taking norms) fit in with the more general cohomological machinery of constructing forms via twisting?
B) Let's say that I constructed the two real forms $\operatorname{SL}_2(\mathbb{R})$ and $\operatorname{SU}_2(\mathbb{R})$. Is there any way to predict or understand which forms of tori will appear? In $\operatorname{SL}_2(\mathbb{R})$ we get both forms, $\mathbb{R}^*$ embedded diagonally and $S^1$ embedded via $$\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix}.$$
In $\operatorname{SU}_2$, however, we only get the latter. Is there some more abstract way to parametrize which forms of tori will appear in a given form of a reductive group? I know that conjugacy classes of tori should be parametrized by $H^1(\operatorname{Gal}(k'/k), N_G(T))$ (at least I think this) but I'm not sure how to use this.
Sorry for the convoluted question, I just feel as though I have the pieces of the puzzle in hand...
I would also be delighted if anyone felt like there was a good reference (even if it only deals with $\mathbb{C}/\mathbb{R}$) for this material.
 A: I answer Question 1. It is just a calculation.
Instead of a real torus, say ${\bf T}$, I consider a pair $(T,\sigma)$,
where $T$ is a complex torus and $\sigma\colon T\to T$ is an anti-holomorphic involution.
See this question and YCor's answer.
For a complex torus $T$, consider the cocharacter group
$${\sf X}_*(T)={\rm Hom}(T, {\Bbb G}_{m,{\Bbb C}}).$$
To a real torus ${\bf T}=(T,\sigma)$ we associate a pair ${\sf X}_*({\bf T}):=({\sf X}_*(T),\sigma_*)$,
where $\sigma_*\in {\rm Aut\,}\,{\sf X}_*(T)$ is the induced automorphism.
It satisfies  $\sigma_*^2=1$.
We denote $\Gamma={\rm Gal}({\Bbb C}/{\Bbb R})=\{1,\gamma\}$, where $\gamma$ is the complex conjugation.
We obtain an action of $\Gamma$ on ${\sf X}_*(T)$ (namely, $\gamma$ acts via $\sigma_*$).
In this way we obtain an equivalence between the  category of ${\Bbb R}$-tori
and the category of $\Gamma$-lattices (finitely generated ${\Bbb Z}$-free $\Gamma$-modules):
$$ {\bf T}\rightsquigarrow {\sf X}_*({\bf T}). $$
Moreover, this is an exact functor:  a short exact sequence of real tori
$$ 1\to{\bf T}'\to{\bf T}\to{\bf T}''\to 1$$
induces a short exact sequence of $\Gamma$-lattices
$$ 0\to {\sf X}_*({\bf T}') \to {\sf X}_*({\bf T}) \to {\sf X}_*({\bf T}'')\to 0.$$
Now consider the torus ${\Bbb G}_{m,{\Bbb R}}=({\Bbb C}^\times,\,z\mapsto\bar z)$ and the corresponding $\Gamma$-lattice  $({\Bbb Z},1)$.
Moreover, consider the torus
$$R_{{\Bbb C}/{\Bbb R}}{\Bbb G}_{m,{\Bbb C}}=(\,{\Bbb C}^{\times\,2},\, (z_1,z_2)\mapsto (\bar z_2,\bar z_1)\,)$$
and the corresponding $\Gamma$-lattice $({\Bbb Z}^2,J)$,
where
$$ J=\begin{pmatrix}0&1\\1&0\end{pmatrix}. $$
Consider the norm homomorphism
$$N\colon R_{{\Bbb C}/{\Bbb R}}{\Bbb G}_{m,{\Bbb C}}\to {\Bbb G}_{m,{\Bbb R}},\quad (z_1,z_2)\mapsto z_1z_2$$
and the corresponding morphism of $\Gamma$-lattices
$$N_*\colon ({\Bbb Z}^2,J)\to ({\Bbb Z},1),\quad (x_1,x_2)\mapsto x_1+x_2.$$
By definition,
$$R_{{\Bbb C}/{\Bbb R}}^{(1)}{\Bbb G}_{m,{\Bbb C}}=\ker N,$$
and so its cocharacter group is $\ker N_*=\{(x, -x)\mid x\in{\Bbb Z}\}.$
The complex conjugation $\gamma$ acts on $\ker N_*$ by $J$, that is,
$$(x,-x)\mapsto (-x, x).$$
We see that $\ker N_*\simeq ({\Bbb Z},-1)$, and hence
$$R_{{\Bbb C}/{\Bbb R}}^{(1)}{\Bbb G}_{m,{\Bbb C}}\simeq ({\Bbb C}^\times, z\mapsto \bar z^{\,{-1}}).$$
Since
$$ (z\mapsto \bar z^{\,{-1}})\,=\,(z\mapsto z^{-1})\,\circ\,(z\mapsto \bar z),$$
we see that $R_{{\Bbb C}/{\Bbb R}}^{(1)}{\Bbb G}_{m,{\Bbb C}}$ can be obtained from ${\Bbb G}_{m,{\Bbb R}}=({\Bbb C}^\times,\,z\mapsto\bar z)$
by twisting by the cocycle $\gamma\mapsto (z\mapsto z^{-1})$, as required.
Note that these three $\Gamma$-lattices
$({\Bbb Z},1),\ ({\Bbb Z}^2,J),$, and $({\Bbb Z},-1)$  are the only indecomposable $\Gamma$-lattices (up to isomorphism);
see this answer.
It follows that these three real tori ${\Bbb G}_{m,{\Bbb R}}$,
$R_{{\Bbb C}/{\Bbb R}}{\Bbb G}_{m,{\Bbb C}}$, and
$R_{{\Bbb C}/{\Bbb R}}^{(1)}{\Bbb G}_{m,{\Bbb C}}$
are the only indecomposable real tori (again, up to isomorphism).
