# Must the adjoint representation of a semisimple Lie group be algebraic?

Suppose G is a real semisimple Lie group without compact factor, then Aut(g) is obviously an algebraic subgroup of SL(n,R) where g is the Lie algebra and n is the dimension. Regard the adjoint representation of G as a subgroup of Aut(g), does it always have finite-index, or at least is it always Zariski closed?

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A couple of comments: 1) In the 1950s Chevalley (and some others like Hochschild) tried to develop a notion of "linear algebraic group" over an arbitrary field of characteristic 0. But this leads to technical problems and doesn't generalize to prime characteristic. So your notion of "algebraic subgroup" should start with the special linear group over the reals being viewed as the group of rational points of an algebraic group over the complex field. 2) Is there really a need to avoid compact factors here? –  Jim Humphreys Mar 12 '12 at 22:12
Isn´t it is already clear that $Aut(\mathfrak g)/Inn(\mathfrak g)$ is finite for compact semisimple $G$? –  Claudio Gorodski Mar 13 '12 at 2:27