Timeline for How does all of the bundles over a certain manifold characterize the homotopy class of the base manifold?
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| Apr 20, 2011 at 12:25 | comment | added | Tom Goodwillie | If lethe would prefer an example $M\to N$ where $M$ and $N$ are closed manifolds of the same dimension (rather than $S^3$ and a point) then there is also $S^3\times S^3\to S\wedge S^3=S^6$. | |
| Apr 20, 2011 at 12:07 | comment | added | Igor Belegradek | By the way, there do exist nontrivial smooth sphere bundles over $S^3$. Their structure group of course is not reducible to a Lie group. This boils down to finding a sphere whose diffeomorphism group has nontrivial $\pi_2$. I think, examples can be found be in [Antonelli- Burghelea-Kahn, "The non-finite homotopy type of some diffeomorphism groups", Topology 11 (1972), 1–49]. | |
| Apr 20, 2011 at 11:20 | comment | added | algori | lethe -- first things first: the $\pi_2$ of any Lie group (that is, a finite-dimensional one) is 0: see e.g., mathoverflow.net/questions/8957/homotopy-groups-of-lie-groups for a sketch of a proof. One could rephrase the question like so: is there a fibre $F$ and group $G$ of transformations of $F$ such that for all bases the homotopy classes of maps to the $BG$ form a complete invariant of the homotopy type of the base? I can't answer that off hand, but somehow I doubt this has a positive answer. | |
| Apr 20, 2011 at 11:00 | comment | added | Honglu | Why does these homotopy group calculation conclude all the bundles are trivial? Sorry I didn't even finish learning Hatcher's Algebraic Topology. And why $\pi_2(G_0)=0$? At the beginning I meant vector bundles of finite rank with structure group $GL(n,\mathbb R)$ because I only know the homotopy invariance in this case. But when I realized I didn't specify what kind of bundles, I decided not to change it since such restriction seems unnatural. Thus, is it possible to extend the statement to fiber bundles? I mean, does homotopy invariance still hold in fiber bundles? | |
| Apr 20, 2011 at 6:21 | comment | added | algori | David -- those disclaimers never help, do they;)? | |
| Apr 20, 2011 at 6:14 | comment | added | David Roberts♦ | This true if $G$ is a finite dimensional Lie group. Loop groups give simple counterexamples otherwise. But then most vector bundles are of finite rank ;-) | |
| Apr 20, 2011 at 6:07 | comment | added | algori | Mark -- thanks. Smooth would do just as fine. I decided to put some kind of a disclaimer since I wasn't completely sure which bundles lethe had in mind. | |
| Apr 20, 2011 at 6:01 | comment | added | Mark Grant | Great answer! Why do you specify topological, and not smooth, vector bundles? | |
| Apr 20, 2011 at 5:48 | history | answered | algori | CC BY-SA 3.0 |