Considering a complex algebraic group G defined over the reals, one knows from an article of Borel and Harish-Chandra (Arithmetic subgroups of algebraic groups, Annals of Mathematics 75 (1962)) that G is reductive (as a complex group) if and only if G(R), the subgroup of its real points, is reductive (as a real group). One natural question is whether the same is true replacing the reals by an arbitrary field (say of characteristic 0) and the complexes by its algebraic closure?

Clarifying: If $G$ is an affine reductive algebraic group defined over $k$ we know that $G$ can be seen as a subgroup of $Gl(n,\bar{k})$, with $\bar{k}$ the algebraic closure of k. Let $G(k)=G\cap Gl(n,k)$. My question is if $G(k)$ is also reductive? Meaning reductive when the unipotent radical of the group is trivial.

`$G(k)$ is reductive" means (with $G(k)$ an abstract group, throwing away information of $k$-variety $G$ which gives substance to theory of algebraic groups). You seem to also have in mind a definition of`

$G(\overline{k})$ is reductive''; do you really mean ``$G_ {\overline{k}}$ is reductive''? Please clarify if the conditions you have in mind over $k$ and over $\overline{k}$ are instances of thesamedefinition, or if the one over $\overline{k}$ is meant to involve the algebro-geometric structure of $G_ {\overline{k}}$ in a way that your condition over $k$ does not. – BCnrd May 19 '10 at 12:43