Timeline for What is the importance of $\pi_{i}G$?
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| Jul 23, 2012 at 15:35 | comment | added | Paul Reynolds | ... if $G$ is a Lie group (since $[S^3,BG] = \pi_2(G) = 0$, a result due to E.Cartan [Car 11]), whereas $S^3$ does admit some non-trivial sphere bundles (see for example [An-Bu-Ka]). For these bundles the total space admits no Riemannian metric such that the projection becomes a Riemannian submersion with totally geodesic fibres." Not totally relevant to the question but an interesting fact none-the-less. | |
| Jul 23, 2012 at 15:30 | comment | added | Paul Reynolds | @Mariano, it was bugging me that I couldn't remember where I read the Serre comment so I typed "noticed by J.P.Serre" into Google and miraculously it came up with exactly one book, a book I am always reading. I had not remembered the situation correctly (it seemed too easy a fact to require a Serre). Remark 9.57 on p249 of Einstein Manifolds by Besse states: "There exist locally trivial fibre bundles whose structural group cannot be reduced to a Lie group, for any family of local trivialisations (this was noticed by J.P.Serre). For example, any principal $G$-bundle over $S^3$ is trivial... | |
| Jul 22, 2012 at 20:11 | comment | added | Paul Reynolds | @Mariano, I can't remember exactly where I read the Serre citation, it was quite a while ago. There are various ways to see it though, the easiest probably as follows. Rank $k$ (smooth) vector bundles over $S^3$ are determined by a clutching map on the equator, i.e. by a homotopy class of maps $S^2 \to O(k)$. | |
| Jul 22, 2012 at 19:24 | comment | added | Mariano Suárez-Álvarez | @Paul, how did he prove it? | |
| Jul 22, 2012 at 11:26 | comment | added | Paul Reynolds | A cool application of the fact $\pi_2G = 0$ is that any principal bundle (and hence any vector bundle) over $S^3$ is trivial, which if I remember correctly was first noticed by Serre. | |
| Jul 22, 2012 at 5:06 | history | answered | Zhaoting Wei | CC BY-SA 3.0 |