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Let $G$ be a complex Lie group (not necessarily connected) with reductive Lie algebra $\frak{g}$. (We may assume that $G$ has finitely many connected components and is linear-algebraic.) Of course, $G$ need not be the complexification of a compact Lie group (ex. $G=\mathbb{C}$). To what extent, however, is $G$ "close" to being the complexification of a compact Lie group? Does $G$ belong to some kind of extension involving the complexification of a compact Lie group? Is $G$ some reasonably nice quotient of the complexification of a compact Lie group? I would appreciate any answers to questions of this nature. Also, I would appreciate any and all references.

Thanks!

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  • $\begingroup$ The functor $G \rightarrow G(\mathbf{C})$ is an equivalence from linear algebraic $\mathbf{C}$-groups $G$ with reductive $G^0$ to complex Lie groups $H$ with reductive Lie algebra and finite $\pi_0(H)$ such that $Z_{H^0}$ is a power of $\mathbf{C}^{\times}$. For any such $H$ and maximal compact subgroup $K$, denote by $K'$ the unique linear algebraic $\mathbf{R}$-group with $K'(\mathbf{R})=K$ meeting every component of $K'$. Then $K'(\mathbf{C})=H$ and $H$ is the complexification of $K'$ as defined in Bourbaki. See D.3.2 and D.3.3 in the Luminy notes on reductive group schemes (use Google). $\endgroup$
    – user28172
    Commented Mar 20, 2013 at 15:08
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    $\begingroup$ @PDC: The expression reductive Lie group in the header already raises questions about how you would define this notion. For linear algebraic groups the concept depends on the Jordan decomposition rather than the Lie algebra. Your 1-dimensional example shows the complication here, while your earlier question is relevant: mathoverflow.net/questions/124418 $\endgroup$
    – Jim Humphreys
    Commented Mar 20, 2013 at 20:31
  • $\begingroup$ Correction to my previous comment: I should have written $Z^0_{H^0}$ rather than $Z_{H^0}$. $\endgroup$
    – user28172
    Commented Mar 20, 2013 at 20:42
  • $\begingroup$ These are good points. For me, the relevant notion of reductivity for complex Lie groups is that of linear reductivity (or equivalently, being the complexification of a compact Lie group). Is the issue with example $G=\mathbb{C}$ in some sense the only way that $G$ can fail to be the complexification of a compact Lie group? Are there any structure theorems for complex Lie groups $G$ with reductive Lie algebras that relate such groups $G$ to linearly reductive groups? $\endgroup$
    – Peter Crooks
    Commented Mar 20, 2013 at 20:53
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    $\begingroup$ @PDC: There are commutative compact complex Lie groups which have nothing to do with linear algebraic groups, namely the "complex tori" in the sense of $V/L$ for a finite-dimensional complex vector space $V$ and full rank lattice $L$ in $V$. So the structure of the center needs to be brought out in the analytic theory to "rule out" problematic cases. The description I gave with the Lie algebra and the center (and a bit for the component group) seems as "good" as one can hope to say over $\mathbf{C}$ (life is harder over $\mathbf{R}$), or maybe someone else has a better idea... $\endgroup$
    – user28172
    Commented Mar 20, 2013 at 23:46

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Suppose that $ G $ is a connected affine algebraic group over $ \mathbb C $. Then there is a short exact sequence $ 1 \rightarrow U \rightarrow G \rightarrow L \rightarrow 1 $, where $ U $ is the unipotent radical of $ G $. $ L $ is a reductive group, i.e. the complexification of a connected compact Lie group. $ U $ is a unipotent group: a successive extension of copies of $\mathbb C $ (the additive group).

On the Lie algebra level, we have a similar extension $ 0 \rightarrow \mathfrak u \rightarrow \mathfrak g \rightarrow \mathfrak l \rightarrow 0 $.

Now, the question assumes that $ \mathfrak g $ is reductive, which means that $ \mathfrak u $ is an abelian Lie algebra and thus $ U = \mathbb C^n $. Also the fact that $ \mathfrak g $ is reductive means that the extension splits and we have $ \mathfrak g = \mathfrak u \oplus \mathfrak l $. I think that this implies that $ G = L \times \mathbb C^n $.

So if my reasoning is correct, then any connected $ G $ with reductive Lie algebra is just the product of a reductive group with $ \mathbb C^n $.

(If $ G $ is disconnected, the situation is more complicated. For example, suppose we have a finite group $ L $ which acts linearly on $ \mathbb C^n $. Then we can form the semidirect product $ G = L \ltimes \mathbb C^n $. It's Lie algebra will be just abelian and thus reductive, but $ G $ is not a product.)

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  • $\begingroup$ If the Lie algebra is reductive, it must be the direct sum of a semisimple Lie algebra and its center = radical (one of Bourbaki's equivalent conditions). This is stronger than just having an abelian unipotent radical in the group $G$. $\endgroup$
    – Jim Humphreys
    Commented Dec 16, 2014 at 23:40

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