# 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.

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.

• Have a look at this preprint. Sep 24, 2020 at 16:07
• I would say that they are parametrized (in the case $k={\Bbb R}$) by $${\rm ker}[H^1({\Bbb R},N_G(T))\to H^1({\Bbb R},G)].$$ Sep 24, 2020 at 16:24
• 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. Sep 24, 2020 at 16:30
• Read Serre GC I.5, then compute yourself the conjugacy classes in question, and you will see yourself what the correct formula is. Sep 24, 2020 at 16:40
• 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. Sep 25, 2020 at 13:57

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).