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Is the following obvious? I think it is, but wanted to make sure before an exam tomorrow!

"There is a bijection between the complex representations of a real Lie algebra and the complex representations of its complexification."

I think this is true (and very useful) because the complex representation of the complexification has the extra requirement of being complex linear, which fixes any extra freedom you'd get from complexifying.

Morally though, this seems weird - particularly as a complex Lie algebra may have non-isomorphic real forms. Does anyone have any comments on this which might make it seem more intuitive?

Many thanks in advance.

glS
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Edward Hughes
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  • Well, I have an obvious comment that since you are only taking complex representations they don't see the difference in the real forms (which after all comes precisely from not being able to use complex numbers to change the signature of the Killing form). Not sure if you want something more than this. – Marek Sep 25 '13 at 16:51
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    This is true (technically it's not a bijection for set-theoretical issues, but an "equivalence of categories"), but often people are too sloppy about this. For example one should watch out that this equivalence is not well-behaved w.r.t. conjugation on the real side (that's where different real forms behave differently). Compare in particular the "Final footnote" in my answer https://math.stackexchange.com/a/3258221/96384. – Torsten Schoeneberg Jul 18 '19 at 22:09

1 Answers1

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It's 7 years too late for your exam but hopefully anyone else who is also wondering will appreciate an answer: yes, this is true (as long as you are careful to either say "bijection between isomorphism classes" or to talk about an equivalence of categories as Torsten says in the comments), and it follows directly from the universal property of complexification, namely that if $\mathfrak{g}$ is a real Lie algebra and $\mathfrak{h}$ is a complex Lie algebra then

$$\text{Hom}_{\mathbb{R}}(\mathfrak{g}, \mathfrak{h}) \cong \text{Hom}_{\mathbb{C}}(\mathfrak{g}_{\mathbb{C}}, \mathfrak{h}).$$

Then take $\mathfrak{h} = \mathfrak{gl}_n(\mathbb{C})$.

Morally though, this seems weird - particularly as a complex Lie algebra may have non-isomorphic real forms. Does anyone have any comments on this which might make it seem more intuitive?

Yes, and all of those real forms have the same complex representation theory, because they all have the same complex representation theory as their complexification. For example $\mathfrak{su}(2)$ and $\mathfrak{sl}_2(\mathbb{R})$ are non-isomorphic real forms of $\mathfrak{sl}_2(\mathbb{C})$ and they all have the same complex representation theory. Everything that the complex representation theory can detect about a Lie algebra factors through its complexification, again by the universal property of complexification.

Qiaochu Yuan
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