Timeline for Why should I prefer bundles to (surjective) submersions?
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 14, 2022 at 12:33 | answer | added | SGS | timeline score: 0 | |
| Jul 26, 2016 at 18:25 | answer | added | Moe Hirsch | timeline score: 6 | |
| S Apr 13, 2016 at 13:22 | history | suggested | D. Corro | CC BY-SA 3.0 |
Edited the differential so that it would read $T_y Y$ instead of $T\_y Y$.
|
| Apr 13, 2016 at 13:11 | review | Suggested edits | |||
| S Apr 13, 2016 at 13:22 | |||||
| Jan 23, 2011 at 19:07 | history | edited | John Klein |
edited tags
|
|
| Jan 23, 2011 at 18:57 | answer | added | John Klein | timeline score: 12 | |
| Aug 3, 2010 at 22:30 | comment | added | Donu Arapura | (This is a late comment, but I hadn't seen this before.) If you are working locally, there's no difference. But globally it's a whole different story. The fibres of a fibre bundle are the same by definition. For a submersion, even the number of connected components of the fibres need not stay constant. And if you start looking at more sophisticated measures of topology, as in the some of the answers, things can only get worse. | |
| Aug 3, 2010 at 20:13 | answer | added | Sean Tilson | timeline score: 4 | |
| Mar 31, 2010 at 2:44 | vote | accept | Theo Johnson-Freyd | ||
| Mar 11, 2010 at 19:23 | answer | added | Sergei Ivanov | timeline score: 15 | |
| Mar 11, 2010 at 18:03 | answer | added | Ben Webster♦ | timeline score: 7 | |
| Mar 11, 2010 at 15:05 | answer | added | Deane Yang | timeline score: 26 | |
| Mar 11, 2010 at 9:46 | comment | added | JBorger | I see where you're coming from about not liking existential quantifiers in certain definitions, but if they're of a local nature (ie there exists a cover such that on each piece blah blah blah), which they are in the case of bundles, then they're really well behaved! This is the whole point of sheaf theory! | |
| Mar 11, 2010 at 6:30 | comment | added | Theo Johnson-Freyd | @Qiaochu: One way to say "smooth manifold" is to talk about maximal atlases, and these are unique. I guess I could use the same device to talk about bundles. So maybe that's not a complaint against them, but it's not a reason to like them any better either. | |
| Mar 11, 2010 at 6:05 | answer | added | some guy on the street | timeline score: 5 | |
| Mar 11, 2010 at 6:02 | comment | added | Tim Perutz | Theo, you answered your own question by saying that you like to work locally. Submersions don't have global structure. Take a smooth fibre bundle and delete any closed subset; it's still a submersion. Now try to say something interesting about its topology. Or integrate a vector field on it. | |
| Mar 11, 2010 at 6:00 | answer | added | Ryan Budney | timeline score: 38 | |
| Mar 11, 2010 at 5:57 | comment | added | Tim Perutz | @Qiaochu. No, you have to specify (say) an equivalence class of atlases to define a smooth manifold. So R with the chart x --> x^3 is a different smooth manifold to R with the obvious chart (though diffeomorphic to it). More interestingly, the action of the homeomorphism group of S^7 on its smooth atlases has 28 orbits. | |
| Mar 11, 2010 at 5:42 | comment | added | Qiaochu Yuan | Doesn't the definition of a smooth manifold demand existence of structure without any uniqueness? (This isn't a rhetorical question - I'm honestly not sure.) | |
| Mar 11, 2010 at 5:32 | history | asked | Theo Johnson-Freyd | CC BY-SA 2.5 |