Reductive Lie Groups and Complexification 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!
 A: 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.) 
