Timeline for Quotient of trivial bundles
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| 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 13, 2013 at 15:41 | answer | added | Tom Goodwillie | timeline score: 6 | |
| Mar 12, 2013 at 20:08 | comment | added | Ricardo Andrade | @Alex_K: You are welcome. @Chris Gerig: I cannot understand whether you are implying that every manifold is stably parallelizable. Nevertheless, that is certainly not true: any non-orientable manifold is not stably parallelizable. | |
| Mar 12, 2013 at 19:38 | comment | added | Chris Gerig | Taking that one step further, you can always stabilize by adding trivial summands: $TM\oplus\epsilon^m= \epsilon^N$ and then do your division. | |
| Mar 12, 2013 at 12:33 | answer | added | Ben McKay | timeline score: 6 | |
| Mar 12, 2013 at 12:08 | comment | added | Alex_K | Oh yes...thank you for pointing out my mistake...every stably trivial bundle is a counter example to what I said... | |
| Mar 12, 2013 at 9:15 | comment | added | Ricardo Andrade | @Alex_K: The normal bundle of $S^2$ in $\mathbb{R}^3$ is trivial. So the tangent bundle of $S^2$ is the quotient of a trivial bundle (the tangent bundle of $\mathbb{R}^3$ restricted to $S^2$) by a trivial 1-dimensional sub-bundle (the normal bundle of $S^2$). Yet $S^2$ is not parallelizable. | |
| Mar 12, 2013 at 8:58 | comment | added | Alex_K | If you equip the bundle with the constant metric, then the quotient can be identified with the orthogonal complement and thus is trivial... | |
| Mar 12, 2013 at 8:45 | history | edited | kalafat | CC BY-SA 3.0 |
typo
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| Mar 12, 2013 at 8:13 | comment | added | IMeasy | At least in the algebraic setting, I feel like the quotient is always a trivial bundle because a map between two trivial bundles is just a vector of scalars... | |
| Mar 12, 2013 at 7:59 | history | asked | kalafat | CC BY-SA 3.0 |