# Equivariant Levi subalgebras.

Suppose $\mathfrak g$ is a finite dimensional Lie algebra over a field on characteristic zero and $G$ is a finite group of automorphisms of $\mathfrak g$.

Does there necessarily exist a Levi subalgebra of $\mathfrak g$ which is $G$-invariant?

By Levi subalgebra I mean a semisimple complement of the solvable radical, as in the Levi-Malcev theorem. My field is $\mathbb Q$... but if needed I could probably deal with extensions of scalars.

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"Levi subalgebra" has more than one meaning nowadays, so it's important to include a precise definition. (For instance, "Levi subalgebra" sometimes means a complement to the nilradical of an arbitrary parabolic subalgebra.) Also, does it matter whether the field is assumed to be algebraically closed? – Jim Humphreys Mar 2 '12 at 13:56

## 1 Answer

Yes. This results and its variations are contained in papers by E.J. Taft:

• Invariant Wedderburn factors, Illinois J. Math. 1 (1957), N4, 565-573 http://projecteuclid.org/euclid.ijm/1255380679 .
• Invariant Levi factors, Michigan Math. J. 9 (1962), N1, 65-68 DOI:10.1307/mmj/1028998623
• Orthogonal conjugacies in associative and Lie algebras, Trans. AMS 113 (1964), No.1, 18-29 DOI:10.2307/1994088
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Taft's results from 1957+ still seem to be optimal, though I'd forgotten about them. Note that the first paper is most important for the question here, though done in wider generality than Lie algebras. (Also, the other two papers are freely available online, via Project Euclid projecteuclid.org/… or via AMS e-math page.) – Jim Humphreys Apr 4 '12 at 17:32
Ah! Great! (Thank you for going through my list of questions, by the way :=) ) – Mariano Suárez-Alvarez Apr 4 '12 at 18:08