Complexification dichotomy for a real simple lie algebra

${\mathfrak{g}}_0$ is semisimple: its radical is an ideal, and the only ideals of the nonabelian simple algebra ${\mathfrak{g}}_0$ are $0$ and ${\mathfrak{g}}_0$, so either the radical is zero and ${\mathfrak{g}}_0$ is semisimple, or the radical is ${\mathfrak{g}}_0$. The latter is impossible: the nonzero derived ideal $[{\mathfrak{g}}_0,{\mathfrak{g}}_0]$ must equal ${\mathfrak{g}}_0$ by simplicity, so the derived series never reaches zero.

$\sigma$ is a conjugate-linear involutive automorphism of $\mathfrak{g}$, hence additive and real-linear with ${\sigma }^2={\mathrm{i}\mathrm{d}}_{\mathfrak{g}}$, its fixed locus is exactly ${\mathfrak{g}}_0$, and $\sigma (iX)=-i\sigma (X)$ for every $X\in \mathfrak{g}$.

$\mathfrak{g}$ is semisimple by [L4] and step 1.1, so it is a nonzero finite direct sum $\mathfrak{g}={\mathfrak{g}}_1\oplus \cdots \oplus {\mathfrak{g}}_n$ of simple ideals. Moreover every ideal $\mathfrak{a}$ of $\mathfrak{g}$ is the sum of the simple ideals it contains: the projection $p_j\text{\hspace{0.17em}⁣}:\mathfrak{g}\to {\mathfrak{g}}_j$ is a Lie algebra homomorphism, so $p_j(\mathfrak{a})$ is an ideal of the simple algebra ${\mathfrak{g}}_j$ and is therefore $0$ or ${\mathfrak{g}}_j$; and $p_j(\mathfrak{a})\mathrm{\ne }0$ forces ${\mathfrak{g}}_j\subseteq \mathfrak{a}$, because for $x\in \mathfrak{a}$ with $x_j:=p_j(x)\mathrm{\ne }0$ one has $[x,{\mathfrak{g}}_j]=[x_j,{\mathfrak{g}}_j]\subseteq \mathfrak{a}\cap {\mathfrak{g}}_j$, which would be zero if $\mathfrak{a}\cap {\mathfrak{g}}_j=0$ and would then put the nonzero $x_j$ in the center $Z({\mathfrak{g}}_j)=0$. Hence the simple ideals are exactly the minimal nonzero ideals, and $[{\mathfrak{g}}_j,{\mathfrak{g}}_j]={\mathfrak{g}}_j\mathrm{\ne }0$ for every $j$.

Let $\pi$ be the permutation of $\{1,\dots ,n\}$ determined by $\sigma ({\mathfrak{g}}_i)={\mathfrak{g}}_{\pi (i)}$, which is well defined because $\sigma$ carries simple ideals to nonzero simple ideals by step 1.2 and step 2.1; then $\pi$ is transitive. Suppose it has at least two orbits, let $O$ be one of them and put $\mathfrak{W}={⨁}_{i\in O}{\mathfrak{g}}_i$ and ${\mathfrak{W}}^{\mathrm{\prime }}={⨁}_{i\mathrm{\notin }O}{\mathfrak{g}}_i$. Both are nonzero ideals of $\mathfrak{g}$, both are $\sigma$-stable because $O$ and its complement are unions of orbits, and $\mathfrak{g}=\mathfrak{W}\oplus {\mathfrak{W}}^{\mathrm{\prime }}$. Every $X\in {\mathfrak{g}}_0$ is $\sigma$-fixed, so by uniqueness of the decomposition its components in $\mathfrak{W}$ and ${\mathfrak{W}}^{\mathrm{\prime }}$ are $\sigma$-fixed as well by step 1.2; hence ${\mathfrak{g}}_0=({\mathfrak{g}}_0\cap \mathfrak{W})\oplus ({\mathfrak{g}}_0\cap {\mathfrak{W}}^{\mathrm{\prime }})$, and each summand is an ideal of ${\mathfrak{g}}_0$ because it is the intersection of the subalgebra ${\mathfrak{g}}_0$ with an ideal of $\mathfrak{g}$. Both summands are nonzero: for any nonzero $x$ in either sigma-stable complex ideal, the two fixed vectors $x+\sigma x$ and $i(x-\sigma x)$ cannot both vanish. Thus ${\mathfrak{g}}_0$ would have two nonzero proper ideals, contradicting simplicity. Hence $\pi$ is transitive.

All orbits of $\pi$ have one or two elements: applying $\sigma$ twice to $\sigma ({\mathfrak{g}}_i)={\mathfrak{g}}_{\pi (i)}$ gives ${\mathfrak{g}}_i={\mathfrak{g}}_{\pi (\pi (i))}$ by step 1.2, so ${\pi }^2=\mathrm{i}\mathrm{d}$ and every orbit of the resulting involution has at most two elements. Transitivity and nonzero $\mathfrak{g}$ therefore give $n=1$ or $n=2$, with the two ideals exchanged in the latter case.

In the case $n=2$ the projection $p\text{\hspace{0.17em}⁣}:\mathfrak{g}\to {\mathfrak{g}}_1$ restricts to an isomorphism of real Lie algebras ${\mathfrak{g}}_0\to ({\mathfrak{g}}_1{)}_{\mathbb{R}}$, where $({\mathfrak{g}}_1{)}_{\mathbb{R}}$ denotes the complex Lie algebra ${\mathfrak{g}}_1$ with scalars restricted to $\mathbb{R}$. Indeed $p$ is a Lie algebra homomorphism; it is injective on ${\mathfrak{g}}_0$ because ${\mathfrak{g}}_0\cap {\mathfrak{g}}_2=0$, an element of that intersection being $\sigma$-fixed and at the same time lying in ${\mathfrak{g}}_1$ by $\sigma ({\mathfrak{g}}_2)={\mathfrak{g}}_1$; and it is surjective because for $x\in {\mathfrak{g}}_1$ the element $x+\sigma x$ lies in ${\mathfrak{g}}_0$ by step 1.2 and is mapped to $x$. Moreover $\sigma$ restricts to a conjugate-linear isomorphism ${\mathfrak{g}}_1\to {\mathfrak{g}}_2$, so ${\mathfrak{g}}_2$ is isomorphic to the complex conjugate Lie algebra $\overline{{\mathfrak{g}}_1}$.

If $n=1$, then $\mathfrak{g}={\mathfrak{g}}_1$ is a complex simple Lie algebra, which is alternative 1 of the Statement. If $n=2$, then $\mathfrak{g}={\mathfrak{g}}_1\oplus {\mathfrak{g}}_2$ with ${\mathfrak{g}}_2=\sigma ({\mathfrak{g}}_1)$ and both ideals simple; by step 4.2 the real Lie algebra ${\mathfrak{g}}_0$ is isomorphic to $({\mathfrak{g}}_1{)}_{\mathbb{R}}$, so ${\mathfrak{g}}_0$ is the complex simple algebra ${\mathfrak{g}}_1$ regarded as a real Lie algebra. The two ideals are isomorphic as complex Lie algebras: by [L6], choose a Chevalley basis $(b_j)$ of ${\mathfrak{g}}_1$ with real structure constants $[b_i,b_j]={\sum }_k{c}_{ij}^k{b}_k$. If $({\bar{b}}_j)$ denotes the corresponding basis of the conjugate algebra $\overline{{\mathfrak{g}}_1}$, then ${\bar{b}}_j\mapsto b_j$ is complex-linear and bracket-preserving because every $c_{ij}^k$ is real. Thus $\overline{{\mathfrak{g}}_1}\cong {\mathfrak{g}}_1$, while the conjugate-linear isomorphism $\sigma :{\mathfrak{g}}_1\to {\mathfrak{g}}_2$ is equivalently a complex-linear isomorphism $\overline{{\mathfrak{g}}_1}\to {\mathfrak{g}}_2$; hence ${\mathfrak{g}}_2\cong {\mathfrak{g}}_1$. This is alternative 2. The alternatives are disjoint because a direct sum of two nonzero ideals is not simple.

Finally, if ${\mathfrak{g}}_0$ is a complex Lie algebra with complex structure $J$, extend $J$ complex-linearly to its complexification. Then $J^2=-1$ and $[JX,Y]=J[X,Y]=[X,JY]$, so the $+i$ and $-i$ eigenspaces are ideals and their sum is the whole complexification. They are interchanged by canonical conjugation since $J$ is defined over $\mathbb{R}$. Both are nonzero: for $0\mathrm{\ne }X\in {\mathfrak{g}}_0$, the vectors $X-iJX$ and $X+iJX$ are nonzero and lie in the respective eigenspaces, by the unique real-imaginary decomposition of [L2]. Hence case 1 cannot occur for a complex algebra regarded as real. Together with step 5.1 this proves the last assertion as well.