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?
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I can't indicate a full proof, but it's not so easy: at the least one needs to know no spheres are Lie groups beyond three dimensions, which requires the classification of real division algebras. – Kevin Carlson Oct 06 '14 at 07:01
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@KevinCarlson: Thanks! And could you please recommend any references which contains the proof? – user181227 Oct 06 '14 at 07:21
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Well, if you can convince yourself that you could extend a Lie group structure on $S^k$ to a division algebra structure on $\mathbf{R}^{k+1},$ all you need is the theorem of Frobenius you'll find on the Wiki page about real division algebras. – Kevin Carlson Oct 06 '14 at 20:30
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There are ways which do not need division algebras to show, which spheres are lie groups, see e.g. http://math.stackexchange.com/questions/12453/is-there-an-easy-way-to-show-which-spheres-can-be-lie-groups – archipelago Oct 09 '14 at 00:14
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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.
archipelago
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Yes, I have seen the result of $G$ having a nontrivial $H^3_{dR}(G,\mathbb{R})$ in the third volume of Novikov's Modern Geometry, but isn't this fact only holds for semi-simple Lie groups? – user181227 Oct 09 '14 at 18:29
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@user181227: since the group is nonabelian, use the fact that it's quotient by maximal abelian subgroup is nontrivial semisimple. – Moishe Kohan Oct 09 '14 at 23:26