Is every complex Lie algebra a complexification?

I'm wondering if every finite-dimensional complex Lie algebra can be written as a complexification of a real Lie algebra. At the vector space level, clearly every $\mathbb{C}^n$ is a complexification of $\mathbb{R}^n$, but at the level of bracket structure, it's not clear to me if we can consistently "decomplexify" the Lie algebra. Putting it another way, for any given $n$-dimensional complex Lie algebra, can we find a basis $\{e_1,\ldots,e_n\}$ such that the structure constants are all real?

Solution 1:

The answer is no.

A "minimal" example of a complex Lie algebra without a real form is given in example 1.36 of these lecture notes of Karl-Hermann Neeb's. To spell it out, it is the three-dimensional Lie algebra over $\mathbb C$ with basis $x,y,z$ and commutators

$$[x,y]=0, [z,x]=2x, [z,y] = i y.$$

It is minimal in the sense that every $0$- , $1$-, or $2$-dimensional (https://math.stackexchange.com/a/976271/96384, Classsifying 1- and 2- dimensional Algebras, up to Isomorphism) complex Lie algebra does have a real form.

Note that the above Lie algebra is solvable. On the other end of the spectrum, it is a well-known (although maybe not suffciently celebrated) result of the structure theory that all semisimple complex Lie algebras do have real forms -- actually, they can be defined over $\mathbb Q$ or even $\mathbb Z$, a result usually attributed to Chevalley (see "Chevalley basis").

Finally, note that the above example does have a form over the number field $\mathbb Q(i)$. The answers to Are there finite-dimensional Lie algebras which are not defined over the integers? (especially user YCor's comment to Qiaochu Yuan's answer) exhibit examples of complex Lie algebras which do not have forms over any number field, nor $\mathbb R$.