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.

  1. 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.

  1. 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.

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    $\begingroup$ Have a look at this preprint. $\endgroup$ – Mikhail Borovoi Sep 24 '20 at 16:07
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    $\begingroup$ I would say that they are parametrized (in the case $k={\Bbb R}$) by $${\rm ker}[H^1({\Bbb R},N_G(T))\to H^1({\Bbb R},G)].$$ $\endgroup$ – Mikhail Borovoi Sep 24 '20 at 16:24
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    $\begingroup$ How to use it? You compute both sets, and you compute the map. Thus you get the kernel. In the case $G={\rm SU}_2$ the kernel is trivial, while in the case ${\rm SL}(2,{\Bbb R})$ it is nontrivial. $\endgroup$ – Mikhail Borovoi Sep 24 '20 at 16:30
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    $\begingroup$ Read Serre GC I.5, then compute yourself the conjugacy classes in question, and you will see yourself what the correct formula is. $\endgroup$ – Mikhail Borovoi Sep 24 '20 at 16:40
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    $\begingroup$ The matching with the Kazhdan–Lusztig statement (Fixed-point varieties in affine flag manifolds) is that their group $G$ is simply connected, so the $G$-valued cohomology is trivial (I think … at least it's true $p$-adically). @MikhailBorovoi's reference: Borovoi and Timashev - Galois cohomology of real semisimple groups via Kac labelings. $\endgroup$ – LSpice Sep 25 '20 at 13:57

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).


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