Timeline for Is every $S^3$ block bundle over $S^4$ a fiber bundle?
Current License: CC BY-SA 3.0
7 events
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| Aug 10, 2020 at 22:22 | comment | added | Connor Malin | It is true that $p$ is the j homomorphism. $Bp$ classifies taking a vector bundle to its sphere bundle thought of as a block bundle, while $Bj$ classifies taking a vector bundle to its sphere bundle. The equivalence you describe $B\widetilde{TOP}(S^3) \rightarrow BG(S^3)$ classifies taking a spherical block bundle to an equivalent fibration (which is a spherical fibration). So the evident diagram commutes showing p is the J-homomorphism. | |
| Jan 21, 2015 at 22:29 | history | edited | student | CC BY-SA 3.0 |
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| Jan 21, 2015 at 22:11 | comment | added | student | oop!I see your point.$S^n$ is just Topologically rigid,but not smoothly rigid.Now I want to switch from the category from Diff to Top,and the problem still makes sense. | |
| Jan 21, 2015 at 22:06 | comment | added | Oscar Randal-Williams | I don't think this argument can be right: why should $S(M)$ be contractible? (Which is what I suppose you mean by "trivial".) | |
| Jan 21, 2015 at 21:32 | comment | added | student | well,even if this is true,i guess what this argument proved is:For every $S^3$ block bundle over $S^4$,there exists a concordant block bundle which admits fiber bundle structure.This is weaker than the property appeared in the the original problem. | |
| Jan 21, 2015 at 21:13 | history | edited | student | CC BY-SA 3.0 |
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| Jan 21, 2015 at 21:07 | history | answered | student | CC BY-SA 3.0 |