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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.

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Please define what $G(k)$ 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 the same definition, 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
As Brian and Jim will know, in books on semi-simple or reductive Lie groups (e.g. Knapp's or Wallach's), or in Harish-Chandra's papers, one finds a definition of what it means for a Lie group to be reductive. Defined thusly, reductive is a priori a property of the Lie group, not the underlying algebraic group (although the result cited in the question shows that it is actually equivalent to the algebraic group being reductive). In any case, it is a property of $G(\mathbb R)$ as a Lie group, not just as an abstract group, so it surely doesn't carry over to $G(k)$ for more general $k$. – Emerton May 19 '10 at 20:04
Right on, Matt. So we need for Ana to clarify the precise meaning of the question over a more general field of char. 0 before one can contemplate giving an answer. – BCnrd May 20 '10 at 0:31
The terminology is still not clear to me. What is the "unipotent radical" of an abstract group? – Jim Humphreys May 20 '10 at 11:16
@Emerton: as far as I can tell, Knapp and Wallach do not give a defintion of reductive as "a property of the Lie group," but as extra structure. – Ben Wieland May 22 '10 at 20:56
up vote 0 down vote accepted

I'm a little hesitant to say anything in the face of all the comments above, but I think Ana is asking for an answer to the following question: If $G$ is a reductive algebraic $k$-group, and we choose:

  1. an algebraic closure $\overline{k}$ of $k$,
  2. a $\overline{k}$-group embedding $G_{\overline{k}} \hookrightarrow GL_n$,
  3. a set-theoretic normal subgroup $U \trianglelefteq G(k)$ such that any element $g \in U$ has unipotent image in $GL_n(\overline{k})$ under the composition $\operatorname{Spec} \overline{k} \to \operatorname{Spec} k \overset{g}{\to} G \to GL_n$,

then is $U$ necessarily trivial?

I think the answer is yes, but I don't know a complete proof. When $k$ is infinite, the Zariski closure of $U$ in $G$ is a normal unipotent subgroup (at least, it sounds plausible). When $k$ is finite, you may have to eliminate small order cases by hand.

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Assume $G$ connected. Question is asking if $G(k)$ can have nontrivial normal subgp $N$ consisting of unipotent elements. The conn'd reductive $G$ is $k$-unirational, so for infinite $k$ the Zar. closure $U$ of $N$ in $G$ is normal. Then char. poly. argument implies $U$ unipotent (maybe disconn'd). Then $U=1$ since $G$ conn'd reductive. So answer to this version is "yes" when $G$ is infinite. For finite $k$ will need to think some more. – BCnrd May 23 '10 at 14:39
OK, say k finite, char. p, max. central k-torus Z in G, and G' the simply conn'd central cover of D(G). Note $Z_ G(k)$ has no nontriv. unip. elts (since $Z_G$ mult. type.). Ignore tiny cases, so G(k) mod center is simple, not p-group. Map $f:Z \times G' \rightarrow G$ is central isogeny, so via preimage on k-pts suffices to show if T is k-torus and G' is s. conn'd ss k-group then no nontrivial normal subgroup N in $(T \times G')(k)$ is p-power order mod center of $(T \times G')(k)$. Toss in T(k) wlog, so wlog T=1. Then clear. For the tiny cases... – BCnrd May 23 '10 at 15:04

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