Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G$ be a Lie group and $R$ be the largest connected solvable normal subgroup of $G$.

Question 1

Is there a Lie subgroup $S$ such that: (1) $G=SR$; (2) every real representation of $S$ is semisimple?

Question 2

Is there a Lie subgroup $S$ such that: (1) $G=SR$; (2) every complex representation of $S$ is semisimple?

Let $G$ be an algebraic group and $R$ be the largest connected solvable normal subgroup of $G$. Is there a algebraic subgroup $S$ such that: (1) $G=SR$; (2) every representation of $S$ is semisimple?

I want to know the formal statement and references.

share|improve this question
I believe the answer to (1) is yes; see the Springer Encyclopedia of Mathematics, eom.springer.de/L/l058590.htm, Lie groups, section on their global structure. There it states that $S \cap R$ is trivial if $G$ is simply connected. Also, maybe you should edit the third question and make a Part (3). –  Qayum Khan Sep 11 '11 at 0:10
add comment

1 Answer

Question $1$: The theorem of Mostow says that every connected algebraic group $G$ over a field $K$ of characteristic zero has a Levi decomposition. This means, $G$ has a reductive algebraic subgroup $S$ such that $G$ is the semidirect prodcut of $S$ and $R_u(G)$, the unipotent radical of $G$. Moreover, any reductive algebraic subgroup $S'$ of $G$ is conjugate to $S$. The result is not true in general in prime characteristic. There is a nice article by Jim Humphreys, EXISTENCE OF LEVI FACTORS IN CERTAIN ALGEBRAIC GROUPS, in the pacific journal of mathematics $1967$.

Question $2$: The Levi decomposition for Lie groups was shown by Levi and Malcev.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.