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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?

Jia Yiyang
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    I think this is equivalent to having a real form. This claims not every complex Lie algebra has a real form http://www.encyclopediaofmath.org/index.php/Complexification_of_a_Lie_algebra – Tim kinsella Jan 16 '15 at 22:47
  • @Timkinsella, I did come across that, but the definition of a real form seems more stringent, that every element of the Lie algebra must be uniquely representable in terms of the real form. So if one can prove every complexfication has a real form, then I guess the answer to my question is no. – Jia Yiyang Jan 17 '15 at 02:29
  • @Timkinsella, you are right, after a closer look it is equivalent to having a real form, I misunderstood the definition of a real form. So indeed the answer is no, however I'd be happy to see a concrete example. – Jia Yiyang Jan 17 '15 at 12:37
  • Try a Google search for "complex real algebra with no real form" – Tim kinsella Jan 17 '15 at 13:10
  • @Timkinsella, alright thanks. – Jia Yiyang Jan 17 '15 at 17:07

1 Answers1

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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 $\mathbb R$.

Torsten Schoeneberg
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  • "It is a well-known result of the structure theory that all semisimple complex Lie algebras do have real forms". Could you expand a bit more on that? Is this just a consequence of the classification theorem for semisimple Lie algebra? – toyr99 Jun 10 '23 at 08:52
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    @toyr99: As said, it is a consequence of the existence of a Chevalley basis or more precisely, that all numbers / coefficients that occur in such a Chevalley basis for a simple complex Lie algbera are actually $\in \mathbb Q$. Because now, for any field $\mathbb Q \subset k \subset \mathbb C$, you can just look at the vector space spanned by that basis over that field, and by the relations it is a Lie algebra which, when yo complexify it, gives out the simple complex one you started with. Cf. also https://math.stackexchange.com/a/3618376/96384 – Torsten Schoeneberg Jun 11 '23 at 00:27