Timeline for Is it true that all sphere bundles are some double of disk bundle?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 22, 2016 at 2:27 | vote | accept | Shinichiro Nakamura | ||
| May 20, 2016 at 16:38 | comment | added | John Klein | Here's a spherical fibration version of the statement you're after: If a spherical fibration is an unreduced fiberwise suspension of another one, then that fibration comes equipped with two sections which are vertically homotopic. Conversely, if $E \to B$ is a spherical fibration equipped with two sections, represented by a fiberwise map $s:B\times S^0 \to E$, then in a certain metastable range the fibration fiberwise desuspends relative to $s$. This is explained in the paper: Klein, John R.; Williams, E. Bruce Homotopical intersection theory. I. Geom. Topol. 11 (2007), 939–977. | |
| May 20, 2016 at 12:36 | comment | added | Neil Strickland | @SebastianGoette I guess I was just thinking about the linear case, I agree that there might be more to say in general. | |
| May 20, 2016 at 11:39 | comment | added | Sebastian Goette | @Neil Is it clear that you have a strict double, that is, are both halves automatically isomorphic? Then I guess, another equivalent formulation is that the sphere bundle is a fibrewise suspension of another sphere bundle. And somehow this seems to imply that all sphere bundles with a section are linear, but that seems to be too strong a conclusion. | |
| May 20, 2016 at 10:44 | comment | added | Neil Strickland | More precisely, it is not hard to see that a sphere bundle is the double of a disc bundle if and only if it has a section. | |
| May 20, 2016 at 8:40 | history | answered | Sebastian Goette | CC BY-SA 3.0 |