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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.

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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

2 Answers 2

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.

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1  
Note that George McNinch has developed the ideas further for algebraic groups: front.math.ucdavis.edu/1007.2777 (and my earlier note is online here: projecteuclid.org/…). Concerning linear reductivity of representations for reductive groups in characteristic 0, there are numerous references. –  Jim Humphreys Oct 6 at 14:12

I was unsure if I should post this because the question has not been active for a year. I decided to go ahead anyway, because people might have the same question in the future.

The following paragraph is taken from Structure of Lie Groups and Lie Algebras by V. V. Gorbatsevich, A. L. Onishchik and E. B. Vinberg.

Let $A$ be a simply-connected covering group for $SL_2(\mathbb{R})$. As is known, $Z(A) \simeq \mathbb{Z}$. Let $z_0$ be a generator of the group $Z(A)$, and let $t_0 \in \mathbb{T}=SO_2$ be a rotation through an angle incommensurable with $\pi$. Then $\Gamma = \langle(t_0, z_0)\rangle$ is a discrete normal subgroupof $\mathbb{T} \times A$. Let $G = (\mathbb{T} \times A)/\Gamma$. Then the image $L$ of the subgroup A under the natural homomorphism $\mathbb{T} \times A \rightarrow G$ is a Levi subgroup of G. The fact that the subgroup $\langle t_0 \rangle$ is dense in $\mathbb{T}$ implies that $L$ is dense in $G$, i.e. it is not a Lie subgroup.

This shows that question (1), which is true for algebraic groups in characteristic 0 as explained in the other answer, is not generally true for Lie groups, as the Levi factor can fail to be closed.

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Actually, there need not exist a Levi decomposition for an algebraic group in prime characteristic. But in characteristic 0 Chevalley's old framework is applicable. (In either case, the Jordan decomposition for algebraic groups makes it somewhat more natural to deal with the unipotent radical and a possible reductive Levi factor.) –  Jim Humphreys Oct 6 at 17:19
    
I edited in "in characteristic 0". My thought was that people interested in algebraic groups would read the answer by Dietrich, where this is explained, but I see now that it was confusing. –  Henrik Winther Oct 7 at 8:46

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