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Not all real Lie groups have a complexification, but the universal complexification always exists and is unique.

My question is, when is a complexification also the universal complexification?

Edit: Let me make my question more precise: Let $G$ be a Lie group and $\hat{G}$ a complex Lie group such that $G\subset \hat{G}$ as an embedded Lie subgroup, and that the Lie algebra of $\hat{G}$ is isomorphic to the complexification of the Lie algebra of $G$.

I want to prove the following: If $G_{\mathbb{C}}$ is a complex Lie group such that there is a real analytic homomorphism $\varphi: G\to G_{\mathbb{C}}$, such that for any real analytic homomorphism $\phi: G\to H$ from $G$ to a complex Lie group $H$, there exists a unique complex analytic homomorphism $\hat{\phi}:G_\mathbb{C} \to H$ and $\hat{\phi} \circ \varphi = \phi$, then $\hat{G}\cong G_\mathbb{C}$.

By the universal property, all such $G_\mathbb{C}$ are isomorphic as complex Lie groups. Also I think the converse holds if $\varphi$ is an embedding. If the statement above does not follow, what additional requirement do we need on $\hat{G}$? Could someone please give me a hint as of how to prove it?

Qidi
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    What does "the complexification" mean if it means something different from "the universal complexification"? – Qiaochu Yuan May 25 '15 at 22:45
  • @QiaochuYuan By complexification I mean the complex Lie group $G$ is a complexification of $G_0$ if $G$ contains $G_0$ as a Lie subgroup and its Lie algebra is the complexification of the Lie algebra of $G_0$ – Qidi May 25 '15 at 22:48
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    Yes, that's what "a complexification" means. What does "the complexification" mean? – Qiaochu Yuan May 25 '15 at 22:50
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    @QiaochuYuan Oh I meant when does "a" complexification coincide with "the" universal complexification, was that what you were asking me to clarify? – Qidi May 25 '15 at 22:55

1 Answers1

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Since the question is old, it possible that you've found the answer by now, but the answer is no. This is essentially because complexifications are not unique. Consider the multiplicative group $R^*_0$ of positive real numbers. Then, the universal complexification is the logarithm map into the complex numbers $\log: \mathbb{R}^*_0\to \mathbb{R} \subset \mathbb{C}$. You can see this because the logarithm map is a diffeomorphism between $\mathbb{R}^*_0$ and $\mathbb{R}$, and the inclusion $\mathbb{R}\subset \mathbb{C}$ is the complexification of the latter. On the other hand, the inclusion $\mathbb{R}^*_0\to \mathbb{C}^*$ is a complexification, but clearly not isomorphic to the universal complexification.

A more interesting question, perhaps what you meant is the following: suppose $G$ admits a complexification. Then, is the universal complexification also a complexification? The answer is no. It is true that if we let $g^+:G\to G^+$ be the universal complexification, and $g^\mathbb{C}:G\to G^\mathbb{C}$ a complexification, then since $g^\mathbb{C}$ is an embedding and factors through $g^+$, it follows that $g^+$ is an embedding.

EDIT: I have actually sat down, and concluded that in fact the answer is yes, if there is a complexification, then the universal complexification is a complexification. We already know that the complexification map is necessarily an embedding. If you look at the construction of the universal complexification, you can see that in fact the Lie algebra of the complexification has to be isomorphic to the Lie algebra of a complexification, and so also the complexification of Lie algebra of the original group.

Pax
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Artur Araujo
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  • Thanks for your answer! One question: other than universality, is there any other criteria one can check to verify if a complexification is universal? Analogous to universal covers being simply connected – Qidi Apr 30 '16 at 18:33
  • I am afraid not. There is, however, an explicit construction of the universal complexification using Tannakian duality, which could theoretically be useful. – Artur Araujo May 01 '16 at 21:32