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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.

Mikhail Borovoi
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  • 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. – Marc Besson Sep 24 '20 at 16:03
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  • Dear Mikhail, this looks like the closest reference I could possibly expect. Thank you very much. – Marc Besson Sep 24 '20 at 16:16
  • You write: "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." This is not quite correct. – Mikhail Borovoi Sep 24 '20 at 16:21
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    I would say that they are parametrized (in the case $k={\Bbb R}$) by $${\rm ker}[H^1({\Bbb R},N_G(T))\to H^1({\Bbb R},G)].$$ – Mikhail Borovoi Sep 24 '20 at 16:24
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    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. – Mikhail Borovoi Sep 24 '20 at 16:30
  • Do read Section 5 of Chapter I in Serre's book "Galois cohomology". – Mikhail Borovoi Sep 24 '20 at 16:32
  • How does this match with the statement in Kazhdan-Lusztig "Fixed Point Varieties on Affine Flag Manifolds" , where they say that the set of conjugacy classes of maximal tori in $G(\mathcal{K})$ correspond to $H^1(\Gamma, N(\overline{F}))$? My impression is that they were working with the short exact sequence $1 \rightarrow T \rightarrow N_G(T) \rightarrow W \rightarrow 1$; are you using Galois Cohomology for $1 \rightarrow N_G(T) \rightarrow G \rightarrow G/(N_G(T))$, where we have the scheme parametrizing tori on the right? I'm a little confused – Marc Besson Sep 24 '20 at 16:35
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    Read Serre GC I.5, then compute yourself the conjugacy classes in question, and you will see yourself what the correct formula is. – Mikhail Borovoi Sep 24 '20 at 16:40
  • Okay, will do. Thank you very much for your reference and comments. – Marc Besson Sep 24 '20 at 16:41
  • In Serre's book "Galois cohomology", read also Section III.1 and Subsection III.4.5. All this (Sections I.5, III.1, and III.4.5) can be read without reading other parts of Serre's book. You cannot read Kazhdan-Lusztig without reading Serre's book... – Mikhail Borovoi Sep 24 '20 at 18:21
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    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. – LSpice Sep 25 '20 at 13:57
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    (Also, yes to your question; one lets $G$ act transitively on the set of tori, with point stabiliser $N_G(T)$, so, as in general, the parameterisation of rational orbits is $\ker(H^1(\mathbb R, N_G(T)) \to H^1(\mathbb R, G))$. KL are observing that the $N_G(T)$-valued cohomology can be replaced by $W$-valued cohomology, because they are dealing with a field over which connected groups are cohomologically trivial. In fact, I guess that means they don't even need $G$ sc for this purpose ….) – LSpice Sep 25 '20 at 14:02

1 Answers1

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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).

Mikhail Borovoi
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