Let $\mathfrak{L}$ be a real lie algebra for which I know the multiplication table. And let $\mathfrak{L}'$ be another real lie algebra (of the same dimension).
In my way of showing that $\mathfrak{L}'\cong \mathfrak{L}$, I ended by constructing a complex basis of $\mathfrak{L}'$ such that in this basis $\mathfrak{L}'$ has the same multiplication table as $\mathfrak{L}$.
Question: Can we say that $\mathfrak{L}'$ is isomorphic to $\mathfrak{L}$ ?
Thank you !
No, we cannot say that $L$ and $L'$ are isomorphic as real Lie algebras when we only know that their complexifications are isomorphic. For example, let $L=\mathfrak{so}_3(\Bbb R)$ and $L'=\mathfrak{sl}_2(\Bbb R)$. Then, over $\Bbb C$ we have $$ L\cong L'\cong \mathfrak{sl}_2(\Bbb C), $$ but $L$ and $L'$ are not isomorphic as real Lie algebras, since one algebra has a $2$-dimensional real subalgebra, the other one hasn't.
References:
Lie algebra isomorphism between ${\rm sl}(2,{\bf C})$ and ${\bf so}(3,\Bbb C)$
Lie algebras ${\rm sl}(2,{\bf R})$ and $({\bf R}^3,\wedge)$ are not isomorphic
