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.

  • $\begingroup$ 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 Sep 19 '13 at 7:39
  • $\begingroup$ @Victor: Which direction do you find unclear? $\endgroup$ – Moishe Kohan Sep 19 '13 at 9:42
  • $\begingroup$ @studiosus I am most concerned with $\implies$. $\endgroup$ – Maxime Sep 19 '13 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 Sep 19 '13 at 14:43

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.

  • $\begingroup$ (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 Sep 25 '13 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 Sep 25 '13 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 Sep 25 '13 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 Sep 25 '13 at 16:33
  • $\begingroup$ @AsalBeagDubh Fixed, thanks! $\endgroup$ – David E Speyer Sep 25 '13 at 18:25

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.

  • 2
    $\begingroup$ I'm not quite sure what you mean. Could you please expand your answer? $\endgroup$ – Maxime Sep 20 '13 at 4:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.