Timeline for Quotient of trivial bundles
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 17, 2015 at 11:03 | comment | added | Andrea | Could you please clarify the relation between the quotient bundle and the homotopy class of the map $X\to S^{n-1}$ representing the subbundle? | |
| Mar 19, 2013 at 3:12 | vote | accept | kalafat | ||
| Mar 19, 2013 at 3:11 | vote | accept | kalafat | ||
| Mar 19, 2013 at 3:12 | |||||
| Mar 15, 2013 at 4:06 | comment | added | kalafat | Orientability assumption makes it into $S^{n−1}$ rather than $P^{n−1}$ I guess. | |
| Mar 13, 2013 at 22:28 | comment | added | Tom Goodwillie | I meant of course no cohomology in degrees $n-k$ or higher. | |
| Mar 13, 2013 at 22:27 | comment | added | Tom Goodwillie | Yes. In fact, if $X$ has the homotopy type of a complex of dimension less than $n-k$, or if it has no cohomology in degrees less than that (even with twisted coefficients), that's enough. | |
| Mar 13, 2013 at 21:42 | comment | added | Ricardo Andrade | A small comment since the question is about manifolds. When $X$ is a smooth manifold of dimension $d$ whose interior has no compact component, you can get away with only $d\leq n-k$. This follows from Tom's answer because $X$ is then homotopy equivalent to a CW-complex of dimension $d-1$: $X$ admits a handlebody decomposition with no handles of dimension $d$. | |
| Mar 13, 2013 at 17:04 | comment | added | Mariano Suárez-Álvarez | Ah, right! ${}{}$ | |
| Mar 13, 2013 at 16:27 | comment | added | Tom Goodwillie | No. We are given a trivial rank one subbundle. | |
| Mar 13, 2013 at 15:49 | comment | added | Mariano Suárez-Álvarez | You want $P^{n-1}$, no? | |
| Mar 13, 2013 at 15:48 | history | edited | Tom Goodwillie | CC BY-SA 3.0 |
added 39 characters in body
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| Mar 13, 2013 at 15:41 | history | answered | Tom Goodwillie | CC BY-SA 3.0 |