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By a complex reductive algebraic group I mean the group of complex points of a (possibly disconnected) affine algebraic group defined over $\mathbb{C}$ whose unipotent radical (maximal connected unipotent normal subgroup) is trivial.

I can't seem to find a clear source for the following fact that I believe to be true:

A complex algebraic group is reductive if and only if it is the complexification of a compact Lie group.

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  • $\begingroup$ So when you say the complexification of a Lie group, you mean purely as a group? Since clearly the topologies will not look remotely the same. $\endgroup$
    – Tobias Kildetoft
    Commented Sep 19, 2013 at 7:39
  • $\begingroup$ @Victor: Which direction do you find unclear? $\endgroup$
    – Moishe Kohan
    Commented Sep 19, 2013 at 9:42
  • $\begingroup$ @studiosus I am most concerned with $\implies$. $\endgroup$
    – Maxime
    Commented Sep 19, 2013 at 14:39
  • $\begingroup$ @TobiasKildetoft I believe you can realize a compact Lie group as a real algebraic group in $GL_n\mathbb{R}$ and then complexification can be interpreted as taking the complex zeroes of its defining polynomials in $GL_n\mathbb{C}$. $\endgroup$
    – Maxime
    Commented Sep 19, 2013 at 14:43

2 Answers 2

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This result is true and not easy. You can find this result stated and a proof of "complexification of compact group implies reductive" in chapter 5 of these notes.

I don't know a proof of the converse that doesn't already establish a substantial part of the classification of reductive groups. In the case that $G$ is centerless and simple, you can see a proof as Lemma 2 here.

One sign that it is hard is that you need to use the hypothesis that your complex group is a linear algebraic group. For example, let $E$ be an elliptic curve over $\mathbb{C}$. Then $E$ is a group object in the category of $\mathbb{C}$ varieties which is not the complexification of any compact group. Indeed, if the $j$-invariant of $E$ is not real, then $E$ doesn't even have any anti-holomorphic involutions.

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  • $\begingroup$ (Part 1) Thank you for your answer. One of the lines in the UofT notes is a really good summary of the type of statement found throughout the literature on this question: "I think this is an equivalence of categories". After a lot more searching, I think I may have found what I want in the book "Lie Groups and Algebraic Groups" by Onishchik & Vinberg but there's a problem. In Chapter 5, Section 2, they establish a 1-1 correspondence between compact Lie groups (up to differentiable isomorphism) and reductive complex algebraic groups (up to polynomial isomorphism) given by complexification. $\endgroup$
    – Maxime
    Commented Sep 25, 2013 at 15:19
  • $\begingroup$ (part 2) I would be happy with this, $except$ they seem to define complex reductive groups in terms of their Lie algebra. They define a complex Lie algebra $\mathfrak{g}$ to be reductive if it splits as $\mathfrak{g}=\mathfrak{z}(\mathfrak{g})\oplus[\mathfrak{g},\mathfrak{g}]$ and define a complex algebraic group $G$ to be reductive if its tangent algebra is reductive. But in this case, isn't the additive group $\mathbb{C}$ a counterexample? I feel like I'm missing a basic point here. $\endgroup$
    – Maxime
    Commented Sep 25, 2013 at 15:19
  • $\begingroup$ I agree that the definition of reductive you quote above would imply that $\mathbb{C}$ is reductive; that no algebraic geometer considers $\mathbb{C}$ to be reductive; and that $\mathbb{C}$ is not the complexification of a compact Lie group. I'm not sure what to say beyond that. Perhaps try binary search: Read a statement in the middle of their proof and decide whether you think it is true or false when $G=\mathbb{C}$, then search forward or backward accordingly. $\endgroup$
    – David E Speyer
    Commented Sep 25, 2013 at 16:05
  • $\begingroup$ @DavidSpeyer: should "not hard" in the third paragraph read "hard"? (Not that this is vitally important to your answer.) $\endgroup$
    – user64687
    Commented Sep 25, 2013 at 16:33
  • $\begingroup$ @AsalBeagDubh Fixed, thanks! $\endgroup$
    – David E Speyer
    Commented Sep 25, 2013 at 18:25
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For the $\Rightarrow$ direction you just use Weyl unitary trick in the semisimple part and the fact that $C^*$ is the complexification of $R^*$ for the abelian part.

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    $\begingroup$ I'm not quite sure what you mean. Could you please expand your answer? $\endgroup$
    – Maxime
    Commented Sep 20, 2013 at 4:22

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