# reductive lie group without $R$-characters compact?

Let $G$ be the real points of an affine algebraic group defined over $R$. If there is no non-trivial characters $G\to R^*$, does it imply $G$ is a compact lie group?

I guess the paper of Borel and Tits on reductive lie groups may have the answer but I do not read Frech.

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$G$ should be the real points of a reductive algebraic group defined over $R$ – Ronggang Nov 14 '10 at 2:00
No. For example SL(2,R) doesn't have any nontrivial characters. – Faisal Nov 14 '10 at 2:10
No: SL_2 is a counterexample – Homology Nov 14 '10 at 2:12
False: for any field $k$ of size $> 3$ (so $(k^{\times})^2 \ne 1$) the $k$-group ${\rm{SL}}_2$ has group of rat'l points ${\rm{SL}}_2(k)$ that is its own derived gp and hence has no nontrivial homomorphism to a commutative gp. Salvage: if $G$ is a conn'd reductive gp over a (possibly archimedean) loc. compact (non-discrete) field $k$ then $G(k)$ is compact if and only if $G$ does not contain ${\rm{GL}}_1$ as a $k$-subgroup (purely algebraic condition!). Generalization: if relax "reductive" to "smooth affine" then the equivalence holds if we also demand no $\mathbf{G}_a$ as a $k$-subgroup. – BCnrd Nov 14 '10 at 2:16
Reference for my proposed salvage: a marvelous proof due to G. Prasad is given in his paper "Elementary proof of a theorem of Bruhat-Tits-Rousseau and of a theorem of Tits" in Bull. Soc. Math. France 110 (1982), pp. 197--202. The proposed generalization lies much deeper. Or rather, if $k$ is perfect (e.g., $\mathbf{R}$) then it's trivially reduced to the reductive case (since geometric unipotent radical descends to $k$ and as such is $k$-split), and probably that's all you might care about. For imperfect $k$ the only proof I know requires everything in the book "Pseudo-reductive groups"... – BCnrd Nov 14 '10 at 2:21