Real forms correspond to conjugate-linear involutions

Given: A finite-dimensional complex Lie algebra $\mathfrak{g}$; a real form ${\mathfrak{g}}_0$ of $\mathfrak{g}$ with $\mathfrak{g}={\mathfrak{g}}_0\oplus i{\mathfrak{g}}_0$; a conjugate-linear involution $\sigma$ of $\mathfrak{g}$; and a complex automorphism $u$ of $\mathfrak{g}$.

The real form condition means that the complex-linear extension $e\text{\hspace{0.17em}⁣}:{\mathfrak{g}}_0{\otimes }_{\mathbb{R}}\mathbb{C}\to \mathfrak{g}$, $e(X\otimes z)=zX$, of the inclusion is an isomorphism, equivalently ${\mathfrak{g}}_0\cap i{\mathfrak{g}}_0=0$ and $\mathfrak{g}={\mathfrak{g}}_0+i{\mathfrak{g}}_0$, with ${\mathfrak{g}}_0$ closed under the bracket; the conjugate-linear involution condition is stated in the same definition (Real form of a complex Lie algebra).

For a real Lie algebra ${\mathfrak{h}}_0$ with complexification ${\mathfrak{h}}_0{\otimes }_{\mathbb{R}}\mathbb{C}$, the map ${\sigma }_0(X\otimes z)=X\otimes \bar{z}$ is a well-defined conjugate-linear bracket-preserving involution with fixed locus ${\mathfrak{h}}_0\otimes 1$, and the canonical embedding is an injective real Lie-algebra homomorphism (Complexification has a canonical conjugation with fixed algebra g zero).

Proof technique: direct.

1.1 Every $Z\in \mathfrak{g}$ has a unique expression $Z=X+iY$ with $X,Y\in {\mathfrak{g}}_0$ by [L1], so ${\sigma }_{{\mathfrak{g}}_0}$ is well-defined and additive; it is conjugate-linear because ${\sigma }_{{\mathfrak{g}}_0}((a+ib)(X+iY))=(aX-bY)-i(bX+aY)=\overline{(a+ib)}(X-iY)$ for real $a,b$, and it is involutive with fixed locus exactly ${\mathfrak{g}}_0=\{X+i0\}$, since $X-iY=X+iY$ forces $2Y=0$. If ${\sigma }_{{\mathfrak{g}}_0}={\sigma }_{{\mathfrak{g}}_0^{\mathrm{\prime }}}$, then their fixed loci ${\mathfrak{g}}_0,{\mathfrak{g}}_0^{\mathrm{\prime }}$ coincide, so the assignment is injective. [L1, algebra]

1.2 For a conjugate-linear involution $\sigma$, the fixed locus ${\mathfrak{g}}^{\sigma }$ is a real subspace: it is closed under addition and under real scalars, and if $Z\in {\mathfrak{g}}^{\sigma }$ with $Z\mathrm{\ne }0$ then $iZ\mathrm{\notin }{\mathfrak{g}}^{\sigma }$ because $\sigma (iZ)=-i\sigma Z=-iZ\mathrm{\ne }iZ$; hence ${\mathfrak{g}}^{\sigma }\cap i{\mathfrak{g}}^{\sigma }=0$. Every $Z\in \mathfrak{g}$ decomposes as $Z=\frac{1}{2}(Z+\sigma Z)+\frac{1}{2}(Z-\sigma Z)$ with $\frac{1}{2}(Z+\sigma Z)$ fixed and $\frac{1}{2}(Z-\sigma Z)$ anti-fixed, so $\mathfrak{g}={\mathfrak{g}}^{\sigma }\oplus i{\mathfrak{g}}^{\sigma }$. The fixed locus is a real Lie subalgebra because $\sigma [X,Y]=[\sigma X,\sigma Y]=[X,Y]$ for $X,Y\in {\mathfrak{g}}^{\sigma }$. [L1, algebra]

2.1 ${\sigma }_{{\mathfrak{g}}_0}$ preserves brackets: writing $Z=X+iY$, $W=U+iV$ with $X,Y,U,V\in {\mathfrak{g}}_0$ and using $\mathbb{C}$-bilinearity, $[Z,W]=([X,U]-[Y,V])+i([X,V]+[Y,U])$ with both components in the real subalgebra ${\mathfrak{g}}_0$ by [L1], so ${\sigma }_{{\mathfrak{g}}_0}[Z,W]=([X,U]-[Y,V])-i([X,V]+[Y,U])=[X-iY,U-iV]=[{\sigma }_{{\mathfrak{g}}_0}Z,{\sigma }_{{\mathfrak{g}}_0}W]$. [L1, step 1.1, algebra]

2.2 The involution associated with ${\mathfrak{g}}^{\sigma }$ is $\sigma$: since $\sigma$ is conjugate-linear and fixes $X,Y\in {\mathfrak{g}}^{\sigma }$, we have $\sigma (X+iY)=\sigma X+\bar{i}\sigma Y=X-iY={\sigma }_{{\mathfrak{g}}^{\sigma }}(X+iY)$ for all $X,Y\in {\mathfrak{g}}^{\sigma }$, and by step 1.2 every element of $\mathfrak{g}$ is of this form. The composite assignments are therefore inverse on the objects displayed: $({\mathfrak{g}}_0{)}^{{\sigma }_{{\mathfrak{g}}_0}}={\mathfrak{g}}_0$ by step 1.1 and ${\sigma }_{{\mathfrak{g}}^{\sigma }}=\sigma$ by the displayed computation. [step 1.1, step 1.2, algebra]

3.1 Let $u$ be a complex automorphism. For $Z\in \mathfrak{g}$, one has $Z\in u({\mathfrak{g}}_0)$ exactly when $u^{-1}Z\in {\mathfrak{g}}_0$, equivalently when ${\sigma }_{{\mathfrak{g}}_0}{u}^{-1}Z=u^{-1}Z$, and this is equivalent to $u{\sigma }_{{\mathfrak{g}}_0}{u}^{-1}Z=Z$. Hence $u({\mathfrak{g}}_0)={\mathfrak{g}}^{u{\sigma }_{{\mathfrak{g}}_0}{u}^{-1}}$. Applying step 2.2 to this involution gives ${\sigma }_{u({\mathfrak{g}}_0)}=u\circ {\sigma }_{{\mathfrak{g}}_0}\circ u^{-1}$. Conversely, if ${\sigma }_{{\mathfrak{g}}_0^{\mathrm{\prime }}}=u{\sigma }_{{\mathfrak{g}}_0}{u}^{-1}$, then their fixed loci give ${\mathfrak{g}}_0^{\mathrm{\prime }}=u({\mathfrak{g}}_0)$. [L1, step 1.1, step 1.2, step 2.2, algebra]

4.1 The two constructions are bijective at the level of isomorphism classes. Steps 1.1 and 2.2 first show that they are mutually inverse on objects. If $f:{\mathfrak{g}}_0\to {\mathfrak{g}}_0^{\mathrm{\prime }}$ is an isomorphism of real Lie algebras, let $e_0:{\mathfrak{g}}_0{\otimes }_{\mathbb{R}}\mathbb{C}\to \mathfrak{g}$ and $e_0^{\mathrm{\prime }}:{\mathfrak{g}}_0^{\mathrm{\prime }}{\otimes }_{\mathbb{R}}\mathbb{C}\to \mathfrak{g}$ be the complex-linear isomorphisms furnished by [L1]. Then $u=e_0^{\mathrm{\prime }}\circ (f\otimes {\mathrm{id}}_{\mathbb{C}})\circ e_0^{-1}$ is a complex Lie-algebra automorphism of $\mathfrak{g}$ and carries ${\mathfrak{g}}_0$ onto ${\mathfrak{g}}_0^{\mathrm{\prime }}$. Conversely, a complex automorphism carrying ${\mathfrak{g}}_0$ onto ${\mathfrak{g}}_0^{\mathrm{\prime }}$ restricts to a real Lie-algebra isomorphism of those fixed real subalgebras. Thus ordinary real-isomorphism classes of real forms are exactly the ${\mathrm{Aut}}_{\mathbb{C}}(\mathfrak{g})$-orbits. Step 3.1 identifies these orbits with conjugacy classes of involutions, proving the asserted class bijection. [L1, step 1.1, step 2.2, step 3.1, algebra] ∎