I have found the following claim in many proofs "Simply connected Lie groups homeomorphic to $\mathbb{R}^n$ are solvable". But the universal covering of $SL(2,\mathbb{R})$ satisfies the hypothesis of this claim and it is far from being solvable. Could someone give me the right topological hypothesis on a Lie group which lead to its solvability and where I can find a proof of this fact.

2$\begingroup$ A topological group is homeomorphic to a Euclidean space if and only if it is a semidirect product $S\ltimes R$ with $S\simeq\tilde{SL}_2(\mathbf{R})^k$ for some $k$ and $R$ a simply connected solvable Lie group. $\endgroup$ – YCor Mar 27 '15 at 10:31

1$\begingroup$ @YCor : Nice! Do you have a reference for that? $\endgroup$ – abx Mar 27 '15 at 10:58

$\begingroup$ @YCor Thank you for your answer, I will be pleased if you give me a reference for this result. On other hand, in the paper 'HOMOGENEOUS SPACETIMES OF ZERO CURVATURE' in PROCEEDINGS OF THE AMERICAN MATHEMATICALSOCIETY Volume 107, Number 3, November 1989, to prove the corollary 3.6 the authors affirm that a simply connected Lie group homeomorphic to $R^n$ is solvable. Does that mean that their proof is false? $\endgroup$ – user61471 Mar 27 '15 at 11:45

2$\begingroup$ @user61471: It would help to have more precise references for the "many proofs" you mention, for example in the short paper you mention in your comment: ams.org/mathscinetgetitem?mr=975639 $\endgroup$ – Jim Humphreys Mar 27 '15 at 12:59

$\begingroup$ See the book by Onishnik and Vinberg: Lie Groups and Lie Algebras III: Structure of Lie Groups and Lie Algebras. Encyclopaedia of Math. Sciences, Springer. p52 (theorem 3.2) books.google.fr/books?id=l8nJCNiIQAAC&pg=PA51&lpg=PA51 $\endgroup$ – YCor Mar 27 '15 at 20:16
I think you just need the extra hypothesis that the Lie group is a matrix group. That is,
Proposition: Let $G$ be a Lie group with a faithful finitedimensional complex representation $V$. If $G$ is homeomorphic to $\mathbb{R}^k$ (which implies that it's simply connected), then it is solvable.
In fact we can use the (apparently?) weaker hypothesis that $G$ has trivial homology.
Proof. It suffices to show that the semisimple part $\mathfrak{g}_{ss}$ of the Levi decomposition of $\mathfrak{g}$ vanishes. First, recall that every connected Lie group $G$ deformation retracts onto its maximal compact subgroup, which, being in particular a compact oriented manifold, has nontrivial top homology. Hence if $G$ has trivial homology then its maximal compact vanishes, and so it can have no compact subgroups.
Now consider the Cartan decomposition $\mathfrak{g}_{ss} = \mathfrak{k} \oplus \mathfrak{p}$. By hypothesis, $\mathfrak{g}_{ss} \otimes \mathbb{C}$ acts on $V$, and hence so does its compact real form $\mathfrak{g}_c = \mathfrak{k} \oplus i \mathfrak{p}$, which integrates to a compact Lie group $G_c$ also acting on $V$. In particular, $\mathfrak{k}$ integrates to a compact subgroup of $G_c$ and hence of $G$. But $G$ has no nontrivial compact subgroups, and so $\mathfrak{k}$ vanishes. It follows that the Killing form of $\mathfrak{g}_{ss}$ is positive definite, and so $\mathfrak{g}_{ss}$ must also vanish. $\Box$

$\begingroup$ The reason you need $V$ in this argument is to guarantee that $\mathfrak{k}$ integrates to something compact; the argument fails for $G = \widetilde{SL}_2(\mathbb{R})$ precisely because $\mathfrak{k}$ does not integrate to a compact subgroup of $G$ in that case. You'd like to argue starting from the fact that the Killing form on $\mathfrak{g}_{ss}$ is negative definite on $\mathfrak{k}$, but it doesn't follow that the Killing form on $\mathfrak{k}$ itself is negative definite. $\endgroup$ – Qiaochu Yuan Mar 29 '15 at 3:57