Correspondence between complex and real subalgebras

Question

Let $\mathfrak g$ be a real Lie algebra and let $\mathfrak g^\mathbb C$ be its complexification. Is every complex subalgebra of $\mathfrak g^\mathbb C$ a complexification of some subalgebra of $\mathfrak g$? If not, what is a counterexample?

Answer 1

Counterexample: $$\mathfrak{g} = \mathfrak{su}_2 =\lbrace \pmatrix{ai & b+ci\\ -b+ci & -ai} : a,b,c \in \Bbb R \rbrace$$ $$\mathfrak{g}_{\Bbb C} \simeq \mathfrak{sl}_2(\Bbb C) = \lbrace \pmatrix{x & y\\ z & -x} : x,y,z \in \Bbb C \rbrace$$ Then the subalgebra $\mathfrak{b} :=\lbrace \pmatrix{x & y\\ 0 & -x} : x,y,z \in \Bbb C \rbrace \subsetneq \mathfrak{g}_{\Bbb C}$ does not come from a subalgebra of $\mathfrak{g}$. (In general, many real forms of semisimple Lie algebras do not contain subalgebras which become Borel subalgebras after complexification; only the so-called quasi-split ones do. There's certainly tons of other examples for other classes of Lie algebras.)