Simply connected Lie groups homeomorphic to R^n are solvable 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.
 A: 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 finite-dimensional 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. $\Box$
