8
$\begingroup$

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.

Improve this question
$\endgroup$
1
  • $\begingroup$ 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). $\endgroup$
    – Qayum Khan
    Commented Sep 11, 2011 at 0:10

2 Answers 2

Reset to default
6
$\begingroup$

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.

Improve this answer
$\endgroup$
3
  • 2
    $\begingroup$ 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. $\endgroup$
    – Jim Humphreys
    Commented Oct 6, 2014 at 14:12
  • $\begingroup$ @DietrichBude: There seems to be a disconcerting difference in notation between your answer and the question: both question 1 and question 2 are about Lie groups, the only difference being in the representation being either real or complex. There is a third, unnumbered question, about algebraic groups. Also, the Wikipedia page states the Levi-Malcev result only for simply-connected Lie groups, which you don't mention. Is this assumption redundant? $\endgroup$
    – Alex M.
    Commented Mar 5, 2019 at 16:58
  • $\begingroup$ @AlexM. On the contrary, algebraic groups are an important part of the question, see the title of the question. So nothing "disconcerting" here, I think.Secondly, I referred to the Levi decomposition for Lie groups as a "headline". Of course, there are further details, so you are right. $\endgroup$
    – Dietrich Burde
    Commented Mar 5, 2019 at 17:12
6
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • $\begingroup$ 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.) $\endgroup$
    – Jim Humphreys
    Commented Oct 6, 2014 at 17:19
  • $\begingroup$ 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. $\endgroup$
    – Henrik Winther
    Commented Oct 7, 2014 at 8:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .