# Does any complex representation of $\mathfrak{g}_\mathbb{C}$ define a real representation of $\mathfrak{g}$?

Proposition 3.39 of Hall's Lie Groups, Lie Algebras And Representations:

"Let $\mathfrak{g}$ be a real Lie algebra, $\mathfrak{g}_\mathbb{C}$ its complexification, and $\mathfrak{h}$ an arbitrary complex Lie algebra. Then every real Lie algebra homomorphism of $\mathfrak{g}$ into $\mathfrak{h}$ extends uniquely to a complex Lie algebra homomorphism of $\mathfrak{g}_\mathbb{C}$ into $\mathfrak{h}$."

In particular this means that any real representation of $\mathfrak{g}$ defines a complex representation of $\mathfrak{g}_\mathbb{C}$.

Question: Does the converse hold? Does any complex representation of $\mathfrak{g}_\mathbb{C}$ define a real representation of $\mathfrak{g}$? Are there any conditions for when this may or may not hold?

Of course it does. If the Lie algebra $$\mathfrak{g}_{\mathbb C}$$ acts on a complex vector space $$V$$, simply consider the restriction of this action to $$\mathfrak g$$. To be more precise, if $$X\in\mathfrak g$$ and if $$v\in V$$, then define $$X.v$$ as $$(X\otimes 1).v$$. I am assuming here that Hall defined $$\mathfrak{g}_{\mathbb C}$$ as $$\mathfrak{g}\bigotimes_{\mathbb{R}}\mathbb{C}$$. If he used another definition, please say which one did he use.

• So then there is a bijection between the real representations of the algebra and complex representations of the complexification? Or can,say, two complex representations restrict to the same real representation? May 22, 2017 at 18:02
• I would not use the word bijection in this context. But, yes, every representaion of $\mathfrak g$ on a complex vector space $V$ induces a representation of $\mathfrak{g}_{\mathbb C}$ on $V$ and every representation of $\mathfrak{g}_{\mathbb C}$ on $V$ is induced by some representation of $\mathfrak g$ on $V$. May 22, 2017 at 18:32
• Nonisomorphic real representations of $\mathfrak{g}$ can, using the process Hall describes, give isomorphic complex representations of $\mathfrak{g}_{\mathbb{C}}$. So, this functor from real representations of $\mathfrak{g}$ to complex representations of $\mathfrak{g}_{\mathbb{C}}$ is not an equivalence of categories. ("Equivalence" is the right generalization of "bijection" in this context.) Feb 28, 2021 at 3:32
• @JohnBaez: What's an example for this, and am I right in my hope that it involes non-semisimple Lie algebras, i.e. for semisimple LAs there is an equivalence? This question also quotes a proposition that we at least have a one-to-one correpsondence on irreducible representations: math.stackexchange.com/q/1408894/96384. Nov 18, 2021 at 17:01
• The proposition in that question asserts an equivalence between complex irreducible representations of a real Lie algebra $\mathfrak{g}$ and complex irreducible representations of $\mathfrak{g}_{\mathbb{C}}$. I was claiming nonisomorphic real representations of $\mathfrak{g}$ can, upon complexification, give isomorphic complex representations of $\mathfrak{g}_{\mathbb{C}}$. So, what are you actually interested in? Nov 18, 2021 at 22:10

The comment section to the other answer brought up an issue which maybe deserves another answer.

What is true:

For a real Lie algebra $$\mathfrak g$$, there is a one-to-one correspondence between

1. representations of $$\mathfrak g$$ on complex vector spaces; that is, $$\mathbb R$$-linear Lie algebra homomorphisms $$\mathfrak g \rightarrow \mathrm{End}_{\mathbb C}(V)$$ for a complex vector space $$V$$, and
2. representations of the complexification $$\mathfrak g_\mathbb C := \mathbb C \otimes_{\mathbb R} \mathfrak g$$ on complex vector spaces; that is, $$\mathbb C$$-linear Lie algebra homomorphisms $$\mathfrak g_\mathbb C \rightarrow \mathrm{End}_{\mathbb C}(V)$$ for a complex vector space $$V$$,

Namely, the quote from Hall's book in the OP explicitly states that $$\mathfrak h$$ is a complex Lie algebra, and Hall and José Carlos Santos' answer apply this to the case $$\mathfrak h = \mathrm{End}_{\mathbb C}(V)$$ for a complex vector space $$V$$.

Where one has to be careful:

If in the title and the question body the phrase "real representation of $$\mathfrak g$$" is supposed to mean a representation of $$\mathfrak g$$ on a real vector space $$W$$, i.e. a Lie algebra homomorphism $$\mathfrak g \rightarrow \mathrm{End}_{\mathbb R}(W)$$, then

1. such a representation also does define a complex representation of $$\mathfrak g_\mathbb C$$, but that needs an extra step absent from the construction in the quote from Hall;
2. it is not true that this process is "reversible" in general; indeed, not only can it happen that non-isomorphic real representations of $$\mathfrak g$$ give out isomorphic complex representations of $$\mathfrak g_\mathbb C$$ (as John Baez pointed out), but more strikingly, not every complex representation of $$\mathfrak g_{\mathbb C}$$ comes from a real representation of $$\mathfrak g$$ via such procedure.

Ad 1: The extra step is that we first have to complexify the real vector space $$W$$ to $$W_\mathbb C := W \otimes_\mathbb R \mathbb C$$, and extend our real representation $$\mathfrak g \rightarrow \mathrm{End}_{\mathbb R}(W)$$ via the natural $$\mathbb R$$-linear inclusion $$\mathrm{End}_{\mathbb R}(W) \hookrightarrow \mathrm{End}_{\mathbb C}(W_\mathbb C)$$. Only then do we have a map $$\mathfrak g \rightarrow \mathrm{End}_{\mathbb C}(V)$$ (namely, $$V := W_\mathbb C$$) as in what was true above, and can proceed like there.

Ad 2: But it is precisely this extra step which is not reversible. Note also that in José Carlos Santos' answer, what comes out is a complex representation of $$\mathfrak g$$, i.e. a map $$\mathfrak g \rightarrow \mathrm{End}_{\mathbb C}(V)$$ with a complex vector space $$V$$. There is no reason, and in general it is not true, that that complex vector space $$V$$ has a real subspace $$W$$ which is stable under the Lie algebra action and whose complex span is all of $$W$$.

As a standard example, consider the obvious complex representation of the real Lie algebra $$\mathfrak{su}_2$$ on $$V:=\mathbb C^2$$ via writing $$\mathfrak{su}_2$$ as the matrices $$\pmatrix{ai & b+ci\\-b+ci &-ai}$$ with $$a,b,c \in \mathbb R$$. If there were a two-(real-)dimensional subspace $$W$$ of $$V$$ stable under the $$\mathfrak{su}_2$$-action, we would have a nonzero map $$\mathfrak{su}_2 \rightarrow \mathfrak{gl}_\mathbb R(W) \simeq \mathfrak{gl}_2(\mathbb R)$$. It's easily seen that such a map would induce an isomorphism $$\mathfrak{su}_2 \simeq \mathfrak{sl}_2(\mathbb R)$$, which is absurd for a plethora and more of reasons.

It is, rather, an interesting question to decide for a given real Lie algebra $$\mathfrak g$$, which of its complex representations "come" from real representations in such way, which ones are "truly complex", and as it turns out there is a third case, namely they can come from "quaternionic" representations. Cf. part A of https://math.stackexchange.com/a/4026224/96384 as well as What property of the root system means a Lie algebra has complex structure?, or the discussion in https://math.stackexchange.com/a/3712110/96384.