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.)

reductive Lie groupin 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 – Jim Humphreys Mar 20 '13 at 20:31commutativecompact 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... – user28172 Mar 20 '13 at 23:46