A question about contractibility of a Lie group

Question

In Lawson's book Spin Geometry, chapter II, before Proposition 1.11, it mentions a fact that if the 1st, 2nd and 3rd homotopy groups of a Lie group $G$ vanish (although $\pi_2(G)=0$ automatically), then $G$ is contractible. I wonder why?

Answer 1

Let $G$ be a 3-connected finite-dimensional real lie group.

First let's reduce to the compact case. By the Cartan-Iwasawa-Malcev theorem, $G$ has a maximal compact subgroup $H$ and is homeomorphic to a topological product $H\times\mathbb{R}^n$. Since products are preserved by homotopy groups, $H$ is also a 3-connected finite-dimensional real lie group and if we show that $H$ is contractible, $G$ obviously is contractible.

One can even show that there are no 3-connected compact finite dimensional lie groups as follows:

If $H$ is 3-connected and abelian, it's lie algebra is abelian and hence $H$ is isomorphic to $\mathbb{R}^n$, what is a contradiction as $H$ is compact. So the abelian case was easy. Suppose $H$ is non-abelian. Since it's 3-connected, the third cohomology group with real coefficients vanishes. That implies that the third deRham-cohomology of $H$ vanishes too. Now have a look at Jason DeVito's great answer of this question, where he explains with the help of the Maurer-Cartan form, that every compact connected non-abelian lie group has nontrivial third deRahm cohomology. This means there are no 3-connected compact finite-dimensional lie groups and $G$ is diffeomorphic to $\mathbb{R}^n$, hence contractible.

The reason I mentioned finite dimensionality is that this is not true in the world of infinite dimensional lie groups. In the case of $O(n)$ the 0-connected cover is $SO(n)$, the 1-connected (and also two-connected) cover is $Spin(n)$ and the three connected cover is $String(n)$, which cannot be a finite dimensional lie group by the previous reasoning.