Complexification preserves semisimplicity

Given: A finite-dimensional real Lie algebra ${\mathfrak{g}}_0$ with complexification ${\mathfrak{g}}_{\mathbb{C}}={\mathfrak{g}}_0{\otimes }_{\mathbb{R}}\mathbb{C}$ and canonical real embedding $\epsilon (X)=X\otimes 1$.

For a finite-dimensional Lie algebra $\mathfrak{l}$ over a field of characteristic zero (in particular over $\mathbb{R}$ or $\mathbb{C}$), the algebra is semisimple if and only if $B_{\mathfrak{l}}$ is nondegenerate; the Killing form is the trace form of the adjoint representation, $B_{\mathfrak{l}}(X,Y)=\mathrm{tr}({\mathrm{ad}}_X\circ {\mathrm{ad}}_Y)$ (Cartan's semisimplicity criterion, Killing form).

The complexification carries the bracket $[X\otimes z,Y\otimes w]=[X,Y]\otimes zw$ and every element has a unique expression $X\otimes 1+iY\otimes 1$ with $X,Y\in {\mathfrak{g}}_0$; the embedded copy of ${\mathfrak{g}}_0$ is a real form, in particular a real Lie subalgebra (Complexification of a real Lie algebra).

Proof technique: direct.

1.1 We compute the Killing form of ${\mathfrak{g}}_{\mathbb{C}}$ on embedded elements. Let $X,Y\in {\mathfrak{g}}_0$ and choose a real basis $u_1,\dots ,u_n$ of ${\mathfrak{g}}_0$. This is also a complex basis of ${\mathfrak{g}}_{\mathbb{C}}$: it spans over $\mathbb{C}$ because $X\otimes z=z\epsilon (X)$ and it is independent over $\mathbb{C}$ by [L2]; consequently $\epsilon$ is injective, since $\epsilon X=0$ for $X={\sum }_k{t}_k{u}_k$ means ${\sum }_k{t}_k(u_k\otimes 1)=0$, which forces $t_k=0$ for every $k$ by that independence. In this basis the matrix of ${\mathrm{ad}}_{\epsilon (X)}$ has entries determined by $[\epsilon X,\epsilon u_k]=\epsilon [X,u_k]$, the $\mathbb{R}$-linear extension of ${\mathrm{ad}}_X$. Hence $B_{{\mathfrak{g}}_{\mathbb{C}}}(\epsilon X,\epsilon Y)=\mathrm{tr}({\mathrm{ad}}_{\epsilon X}{\mathrm{ad}}_{\epsilon Y})={\mathrm{tr}}_{\mathbb{R}}({\mathrm{ad}}_X{\mathrm{ad}}_Y)=B_{{\mathfrak{g}}_0}(X,Y),$ using that the trace of the complexification of a real endomorphism equals its real trace computed in the real basis $\{u_k\}$. [L1, L2, algebra]

2.1 The restriction map $Z\mapsto iZ$ is $\mathbb{R}$-linear and injective on ${\mathfrak{g}}_{\mathbb{C}}$; by [L2] the $\mathbb{R}$-span of $\epsilon ({\mathfrak{g}}_0)$ and $i\epsilon ({\mathfrak{g}}_0)$ is all of ${\mathfrak{g}}_{\mathbb{C}}$, so every element is $X\otimes 1+iY\otimes 1$ and the assignment $X+iY\mapsto (X,Y)$ is a real vector-space isomorphism ${\mathfrak{g}}_{\mathbb{C}}\to {\mathfrak{g}}_0\times {\mathfrak{g}}_0$. In particular, if a real form of the complex bilinear form $B_{{\mathfrak{g}}_{\mathbb{C}}}$ has no radical outside $0$ then it is nondegenerate as a real form, and conversely. [L1, L2, step 1.1, algebra]

2.2 Suppose first that ${\mathfrak{g}}_{\mathbb{C}}$ is semisimple, so that $B_{{\mathfrak{g}}_{\mathbb{C}}}$ is nondegenerate by [L1]. If $X\in {\mathfrak{g}}_0$ satisfies $B_{{\mathfrak{g}}_0}(X,Y)=0$ for all $Y\in {\mathfrak{g}}_0$, then by step 1.1, $B_{{\mathfrak{g}}_{\mathbb{C}}}(\epsilon X,\epsilon Y)=0$ for all $Y\in {\mathfrak{g}}_0$. Complex bilinearity gives $B_{{\mathfrak{g}}_{\mathbb{C}}}(\epsilon X,Z)=0$ for every $Z\in {\mathfrak{g}}_{\mathbb{C}}$: writing $Z=\epsilon Y+i\epsilon Y^{\mathrm{\prime }}$ with $Y,Y^{\mathrm{\prime }}\in {\mathfrak{g}}_0$, the pairing is $B_{{\mathfrak{g}}_{\mathbb{C}}}(\epsilon X,\epsilon Y)+i{B}_{{\mathfrak{g}}_{\mathbb{C}}}(\epsilon X,\epsilon Y^{\mathrm{\prime }})=0$. Nondegeneracy forces $\epsilon X=0$, hence $X=0$, so $B_{{\mathfrak{g}}_0}$ is nondegenerate and ${\mathfrak{g}}_0$ is semisimple by [L1]. [L1, L2, step 1.1, algebra]

3.1 Conversely suppose ${\mathfrak{g}}_0$ is semisimple, so $B_{{\mathfrak{g}}_0}$ is nondegenerate. Let $Z\in {\mathfrak{g}}_{\mathbb{C}}$ satisfy $B_{{\mathfrak{g}}_{\mathbb{C}}}(Z,W)=0$ for all $W$. Write $Z=\epsilon X+i\epsilon Y$ with $X,Y\in {\mathfrak{g}}_0$. Taking first $W=\epsilon U$ for $U\in {\mathfrak{g}}_0$ and using complex linearity in the second argument together with step 1.1 gives $0=B_{{\mathfrak{g}}_{\mathbb{C}}}(\epsilon X,\epsilon U)+i{B}_{{\mathfrak{g}}_{\mathbb{C}}}(\epsilon Y,\epsilon U)=B_{{\mathfrak{g}}_0}(X,U)+i{B}_{{\mathfrak{g}}_0}(Y,U),$ so both $B_{{\mathfrak{g}}_0}(X,U)=0$ and $B_{{\mathfrak{g}}_0}(Y,U)=0$ for all $U\in {\mathfrak{g}}_0$. Nondegeneracy gives $X=Y=0$ and hence $Z=0$, so $B_{{\mathfrak{g}}_{\mathbb{C}}}$ is nondegenerate and ${\mathfrak{g}}_{\mathbb{C}}$ is semisimple by [L1]. The two implications together prove the equivalence. [L1, L2, step 1.1, step 2.2, algebra] ∎