Complex algebraic group is reductive $\iff$ it is the complexification of a compact Lie group?

Solution 1:

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.