Complex simple lie algebra viewed as a real simple algebra

Given: A finite-dimensional complex simple Lie algebra $\mathfrak{s}$ with multiplication by $i$ written $J$, and the real Lie algebra ${\mathfrak{s}}_{\mathbb{R}}$ obtained by restricting scalars.

${\mathfrak{s}}_{\mathbb{R}}$ is a real Lie algebra whose bracket is the restriction of the bracket of $\mathfrak{s}$; $J$ is $\mathbb{R}$-linear with $J^2=-\mathrm{i}\mathrm{d}$ and $J[X,Y]=[JX,Y]=[X,JY]$, and the complexification $({\mathfrak{s}}_{\mathbb{R}}{)}_{\mathbb{C}}$ carries the bracket extending the one of ${\mathfrak{s}}_{\mathbb{R}}$ (Complexification of a real Lie algebra).

The Killing form of ${\mathfrak{s}}_{\mathbb{R}}$ is $B_{\mathbb{R}}(X,Y)=2\mathrm{Re}{B}_{\mathfrak{s}}(X,Y)$, and a finite-dimensional real Lie algebra is semisimple exactly when its Killing form is nondegenerate (Killing form, Cartan's semisimplicity criterion).

Every ideal of a finite-dimensional semisimple Lie algebra is a direct sum of simple ideals with an ideal complement, hence is itself semisimple; a semisimple Lie algebra equals its own derived algebra (Ideals and quotients of semisimple Lie algebras, Semisimple Lie algebras are centerless and perfect).

The complexification $({\mathfrak{s}}_{\mathbb{R}}{)}_{\mathbb{C}}$ carries the canonical conjugation $\sigma (X+iY)=X-iY$, whose fixed locus is the embedded copy of ${\mathfrak{s}}_{\mathbb{R}}$, and complexification preserves semisimplicity (Complexification has a canonical conjugation with fixed algebra g zero, Complexification of a real Lie algebra, Complexification preserves semisimplicity).

Proof technique: direct computation with ideals and with the explicit isomorphism.

1.1 The real Lie algebra ${\mathfrak{s}}_{\mathbb{R}}$ is semisimple: its Killing form is $B_{\mathbb{R}}(X,Y)=2\mathrm{Re}{B}_{\mathfrak{s}}(X,Y)$ because the adjoint operators of ${\mathfrak{s}}_{\mathbb{R}}$ are the $\mathbb{C}$-linear operators ${\mathrm{ad}}_X$ viewed over $\mathbb{R}$ and the real trace of a complex-linear operator is twice the real part of its complex trace, so if $B_{\mathbb{R}}(X,Y)=0$ for all $Y$, then replacing $Y$ by $JY$ and using $B_{\mathfrak{s}}(X,JY)=i{B}_{\mathfrak{s}}(X,Y)$ gives $\mathrm{Im}{B}_{\mathfrak{s}}(X,Y)=0$ as well, hence $B_{\mathfrak{s}}(X,\cdot )=0$ and $X=0$; thus $B_{\mathbb{R}}$ is nondegenerate and [L2] applies. [given, L1, L2, algebra]

2.1 For every ideal $\mathfrak{a}\subseteq {\mathfrak{s}}_{\mathbb{R}}$ one has $\mathfrak{a}=[\mathfrak{a},{\mathfrak{s}}_{\mathbb{R}}]$: by [L3] applied to the semisimple algebra ${\mathfrak{s}}_{\mathbb{R}}$ of step 1.1, $\mathfrak{a}$ is semisimple and satisfies $\mathfrak{a}=[\mathfrak{a},\mathfrak{a}]$, so $\mathfrak{a}=[\mathfrak{a},\mathfrak{a}]\subseteq [\mathfrak{a},{\mathfrak{s}}_{\mathbb{R}}]\subseteq \mathfrak{a}$. [step 1.1, L3, algebra]

3.1 Every ideal is $J$-stable: if $X\in \mathfrak{a}$ and $X={\sum }_j[X_j,Y_j]$ with $X_j\in \mathfrak{a}$ and $Y_j\in {\mathfrak{s}}_{\mathbb{R}}$ as in step 2.1, then $JX={\sum }_{j}J[X_j,Y_j]={\sum }_j[X_j,J{Y}_j]\in [\mathfrak{a},{\mathfrak{s}}_{\mathbb{R}}]=\mathfrak{a}$ by [L1]. Hence a real ideal of ${\mathfrak{s}}_{\mathbb{R}}$ is a complex subspace and a complex ideal of $\mathfrak{s}$. [step 2.1, L1, algebra]

4.1 Consequently ${\mathfrak{s}}_{\mathbb{R}}$ is simple over $\mathbb{R}$: since $\mathfrak{s}$ is complex simple, a complex ideal is $0$ or $\mathfrak{s}$, so every ideal of ${\mathfrak{s}}_{\mathbb{R}}$ is $0$ or ${\mathfrak{s}}_{\mathbb{R}}$; the algebra is nonabelian because $\mathfrak{s}$ is nonabelian, so it is simple. [step 3.1, algebra]

5.1 Define $\bar{\mathfrak{s}}$ to be the real space $\mathfrak{s}$ with the complex structure $-J$; it is a complex Lie algebra with the same bracket. Then the map $L:({\mathfrak{s}}_{\mathbb{R}}{)}_{\mathbb{C}}\to \mathfrak{s}\oplus \bar{\mathfrak{s}}$, $L(X+iY)=(X+JY,X-JY)$, is a $\mathbb{C}$-linear isomorphism of complex vector spaces: it is additive and $\mathbb{R}$-bilinear in the obvious way, its inverse is $(U,V)\mapsto \frac{1}{2}(U+V)+i\frac{1}{2}{J}^{-1}(U-V)$, and $L(i(X+iY))=L(-Y+iX)=(-Y+JX,-Y-JX)=(J(X+JY),-J(X-JY))$, which is $i$ times $L(X+iY)$ in the complex structure $(J,-J)$ of $\mathfrak{s}\oplus \bar{\mathfrak{s}}$. Dimension counts agree: both sides have complex dimension $2{\dim }_{\mathbb{C}}\mathfrak{s}$. [given, step 4.1, algebra]

6.1 The map $L$ preserves brackets: for $X,Y,X^{\mathrm{\prime }},Y^{\mathrm{\prime }}\in {\mathfrak{s}}_{\mathbb{R}}$ one has $[L(X+iY),L(X^{\mathrm{\prime }}+i{Y}^{\mathrm{\prime }})]=([X+JY,X^{\mathrm{\prime }}+J{Y}^{\mathrm{\prime }}],[X-JY,X^{\mathrm{\prime }}-J{Y}^{\mathrm{\prime }}])$ and $L([X+iY,X^{\mathrm{\prime }}+i{Y}^{\mathrm{\prime }}])=L([X,X^{\mathrm{\prime }}]-[Y,Y^{\mathrm{\prime }}]+i([X,Y^{\mathrm{\prime }}]+[Y,X^{\mathrm{\prime }}]))=([X,X^{\mathrm{\prime }}]-[Y,Y^{\mathrm{\prime }}]+J([X,Y^{\mathrm{\prime }}]+[Y,X^{\mathrm{\prime }}]),[X,X^{\mathrm{\prime }}]-[Y,Y^{\mathrm{\prime }}]-J([X,Y^{\mathrm{\prime }}]+[Y,X^{\mathrm{\prime }}]))$, and the two expressions agree because $[JY,J{Y}^{\mathrm{\prime }}]=J^2[Y,Y^{\mathrm{\prime }}]=-[Y,Y^{\mathrm{\prime }}]$ and $[JY,X^{\mathrm{\prime }}]=J[Y,X^{\mathrm{\prime }}]$ by [L1]. Hence $L$ is an isomorphism of complex Lie algebras. [step 5.1, L1, algebra]

7.1 The canonical conjugation of $({\mathfrak{s}}_{\mathbb{R}}{)}_{\mathbb{C}}$, namely $\sigma (X+iY)=X-iY$, corresponds under $L$ to the swap of the two factors: $L(\sigma (X+iY))=L(X-iY)=(X-JY,X+JY)$, which is the interchange of the entries of $L(X+iY)=(X+JY,X-JY)$; its fixed locus is the image of ${\mathfrak{s}}_{\mathbb{R}}$ under the embedding, in agreement with [L4]. [step 5.1, step 6.1, L4, algebra]

8.1 Combining the steps: ${\mathfrak{s}}_{\mathbb{R}}$ is a simple real Lie algebra by step 4.1, and its complexification is isomorphic to $\mathfrak{s}\oplus \bar{\mathfrak{s}}$ by steps 5.1 and 6.1, with the canonical conjugation of the real form acting as the swap of the two factors by step 7.1. This realizes the complex-as-real alternative of Complexification dichotomy for a real simple lie algebra directly, from the explicit isomorphism $L$ and without using any supplementary clause of that theorem: a complex simple algebra regarded as real has a complexification that is a direct sum of two simple ideals interchanged by conjugation, and the example supplies the isomorphism and the swap. [step 4.1, step 5.1, step 6.1, step 7.1]

9.1 Endpoints and scope: $\mathfrak{s}$ is nonabelian by hypothesis, so ${\mathfrak{s}}_{\mathbb{R}}$ has nonzero bracket and simplicity is not vacuous; for $\mathfrak{s}={\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})$ the real dimension ${\dim }_{\mathbb{R}}{\mathfrak{s}}_{\mathbb{R}}=6$ equals ${\dim }_{\mathbb{C}}(\mathfrak{s}\oplus \bar{\mathfrak{s}})$ computed over $\mathbb{C}$ as $3+3$, in agreement with step 5.1; the zero algebra is excluded because it is not simple, and every step is a finite computation, so the argument uses no choice principle. [given, step 5.1, step 7.1, algebra] ∎