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I asked this in MSE, but without success, so I hope, it will be suitable here.

E.B.Vinberg and A.L.Onishchik in their book give the following two definitions.

  1. For a complex Lie group $G$ its real Lie subgroup $H$ is called a real form of $G$, if

    a) the Lie algebra $L(H)$ of $H$ is a real form of the Lie algebra $L(G)$ of $G$ (this means that $L(G)$ is a complexification of $L(H)$), and

    b) $H$ has a non-empty intersection with each connected component of $G$

  2. A (continuous) homomorphism $S:G\to G$ (of a complex Lie group $G$) is called a real structure on $G$, if

    a) $S(S(g))=g$, for $g\in G$, and

    b) the corresponding mapping of the Lie algebra $d S:L(G)\to L(G)$ is a real structure (another term: an involution) on $L(G)$.

It is known that on a connected complex Lie group (or on an irreducible algebraic group) $G$ for each real structure $S:G\to G$ the subgroup of stable elements $$ H=\{g\in G:\ Sg=g\} $$ is a real form on $G$, and the Lie algebra $L(H)$ coincides with the set of stable elements for $dS$: $$ L(H)=\{x\in L(G):\ dS(x)=x\}. $$

My question:

Is this a one-to-one correspondence between real forms and real structures (for a reasonable class of groups)?

In particular,

if $H$ is a real form of a complex a Lie group $G$, does there always exist a real structure $S:G\to G$ such that $H$ is the subgroup of stable elements for $S$ in $G$, and $L(H)$ is the subspace of stable elements for $dS$ in $L(G)$?

This is strange, I can't find a reference, even for the case, when

$H$ is a compact real Lie group, and $G$ is its complexification.

(Actually, that would be enough for me.)

I would be grateful for anybody's help.

Improve this question
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    $\begingroup$ On the Lie algebra level, yes. On the Lie group level, if $G$ is simply connected. See Ch III, Section 6 (p 178 ff) of Helgason: Differential Geometry, Lie groups, and symmetric spaces. $\endgroup$
    – Peter Michor
    Commented May 22, 2015 at 6:58
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    $\begingroup$ 1. If $G$ is a complex Lie group, and $S\colon G\to G$ is an anti-holomorphic involutive (i.e., $S^2=1$) automorphism, then the subgroup of fixed points $G^S$ of $S$ in $G$ is the desired real Lie group (the real form corresponding to the real structure $S$). $\endgroup$
    – Mikhail Borovoi
    Commented May 22, 2015 at 23:50
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    $\begingroup$ 2. Conversely, if $G_0$ is a compact Lie group, then it is a real algebraic group, i.e., it is the set of real zeros in ${\rm GL}(n,\mathbb{R})$ of a finite set polynomials with real coefficients in the $n^2$ matrix elements $g_{i,j}$. Let $G$ denote the complex Lie group of complex zeros of these polynomials in ${\rm GL}(n,\mathbb{C})$, it has a canonical real structure $S\colon g\mapsto \bar g$ (the complex conjugation on the matix elements). This is the desired complexification of a real algebraic group $G_0$. $\endgroup$
    – Mikhail Borovoi
    Commented May 23, 2015 at 0:03
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    $\begingroup$ This is a bijection between the isomorphism classes of: (A) affine real algebraic groups $G_0$ (resp. affine real varieties) and (B) affine complex algebraic groups $G$ (resp. varieties) endowed with a anti-holomorphic involutive automorphism $S$. The map from (A) to (B): $G_0\mapsto G_0\times_R C$. The map from (B) to (A): you take the ring ($\mathbb{C}$-algebra) of regular functions $R=\mathbb{C}[G]$ on $G$, consider the ring ($\mathbb{R}$-algebra) of $S$-invariants $R^S$, and set $G_0={\rm Spec}\,R^S$. $\endgroup$
    – Mikhail Borovoi
    Commented May 24, 2015 at 1:45
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    $\begingroup$ This is called "Galois descent", a reference is Serre's book "Algebraic Groups and Class Fields", Ch. V-20, Prop. 12 (page 142 of Russian edition). This is a reference for any Galois extension $k_1/k$. You should take $k=\mathbb{R}$, $k_1=\mathbb{C}$. $\endgroup$
    – Mikhail Borovoi
    Commented May 24, 2015 at 2:05

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