Does $L$ having a real form imply that $L_{\mathbb C}\simeq L\oplus L$? Assume we are given a complex Lie algebra $L$ having a real form. That is, there is a real Lie algebra $g$ which has a complexification $g_\mathbb{C}\cong L$.
I have seen several examples where the following holds, but is $L_\mathbb{C}\cong L\oplus L$ always true in this situation?
 A: For clarity, I denote by $Res_{\mathbb C\vert \mathbb R}A$ the scalar restriction of a complex Lie algebra $A$ to $\mathbb R$, i.e. $A$ viewed as real Lie algebra.
If $L \simeq g \otimes_{\mathbb R} \mathbb C$, then there are isomorphisms
$$ (Res_{\mathbb C\vert \mathbb R}L) \otimes_{\mathbb R} \mathbb C \simeq Res_{\mathbb C\vert \mathbb R} (g \otimes_{\mathbb R} \mathbb C) \otimes_{\mathbb R} \mathbb C \simeq g \otimes_{\mathbb R} (\mathbb C \otimes_{\mathbb R} \mathbb C) \stackrel{!}\simeq g \otimes_{\mathbb R} (\mathbb C \oplus\mathbb C) \simeq (g \otimes_{\mathbb R} \mathbb C) \oplus (g \otimes_{\mathbb R} \mathbb C) \simeq L \oplus L$$
of $\mathbb C$-Lie algebras (complex scalars always acting on the right here). The crucial step uses 
$$\mathbb C \otimes_{\mathbb R} \mathbb C \stackrel{!}\simeq \mathbb C \oplus\mathbb C$$
as $(\mathbb R, \mathbb C)$-modules.

Note that in general for a field extension $E \vert K$ of degree $n$ and an $E$-Lie algebra $L$, one has an obvious map
$$ (Res_{E\vert K}L) \otimes_{K} E \simeq (Res_{E\vert K} L) \otimes_{K} K^n  \simeq \underbrace{(Res_{E\vert K}L) \oplus \dots \oplus (Res_{E\vert K}L)}_{n}$$
but only as $K$-Lie algebras, as such maps (just given by choosing a $K$-basis of $E$) are not $E$-linear unless in trivial cases, and are different from the map above.
As for a concrete counterexample in case $L$ has no real form, I do not know, but I suspect already the easiest cases of such Lie algebras (which would be solvable three-dimensional) would give counterexamples.
