Timeline for Is the unit tangent bundle of $S^{n}$ parallelizable?
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Nov 5, 2014 at 20:12 | vote | accept | Ali Taghavi | ||
| Nov 5, 2014 at 18:10 | answer | added | Ryan Budney | timeline score: 12 | |
| Nov 5, 2014 at 17:39 | comment | added | Ryan Budney | The unit tangent bundle of a sphere is usually just called a Stiefel manifold (of 2-frames). | |
| Nov 5, 2014 at 17:22 | comment | added | Ryan Budney | When referring to a bundle, yes that's what it means. In reference to a manifold, sure it is talking about the tangent bundle of that manifold. | |
| Nov 5, 2014 at 17:20 | comment | added | Steven Landsburg | @RyanBudney (and Peter Crooks): I don't think you are using language in a standard way. The statement that "$M$ is parallelizable" always (as far as I'm aware) means that the tangent bundle to $M$ is trivial. I've never (outside of your comments) seen "parallelizable" used to mean "trivial a as a vector bundle". | |
| Nov 5, 2014 at 17:17 | comment | added | Ryan Budney | I would like to second Peter Crooks's suggestion. You are literally asking if a bundle is parallelizable, which usually is interpreted as the question of if the bundle is trivial. But as we can see, you are talking about the total space of the bundle, if that is parallelizable as a manifold. You should re-word your question, and the title. | |
| Nov 5, 2014 at 17:07 | comment | added | Marco Golla | It's also parallelisable for $n=1,3,7$, since in this case your manifold is a product of spheres, one of which is odd-dimensional (see mathoverflow.net/questions/52871/… ) | |
| Nov 5, 2014 at 16:12 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
added 24 characters in body
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| Nov 5, 2014 at 16:11 | comment | added | ThiKu | The unit tangent bundle of the 2-sphere is parallelisable. In fact, every orientable 3-manifold is parallelisable. The latter can be proven by Computing $w_2=0$. | |
| Nov 5, 2014 at 16:11 | comment | added | Ali Taghavi | @PeterCrooks according to your first comment: It is true that $TS^{2}$ is not a trivial bundle. But it is not true that $TS^{2}$ is not parallelizable (As a manifold) | |
| Nov 5, 2014 at 16:07 | comment | added | Ali Taghavi | @PeterCrooks I think my question is not unclear. what is the unclear point? | |
| Nov 5, 2014 at 16:04 | comment | added | Ali Taghavi | @PeterCrooks I consider $M=$ the unit tangent bundle of $S^{n}$. Is $M$, as a manifold, parallelizable? Note that there is no a primary obstruction since the euler charactristic iz $0=0\times 2$ | |
| Nov 5, 2014 at 16:02 | comment | added | Peter Crooks | @AliTaghavi, it would probably help to mention the ambiguity that naturally arises here. You might include your qualification in the statement of the question. | |
| Nov 5, 2014 at 16:01 | comment | added | Ali Taghavi | @LiviuNicolaescu I am not talking about sphere. I talk about the unit tangent bundle of sphere | |
| Nov 5, 2014 at 16:00 | comment | added | Ali Taghavi | @TommasoCenteleghe However $S^{n}$ is not parallelizable for almost all $n$ but its tangent bundle is always parallizable. see here | |
| Nov 5, 2014 at 15:59 | comment | added | Liviu Nicolaescu | A simple google search with the question which spheres are parallelizable yielded this paper of Bott. ams.org/journals/bull/1958-64-03/S0002-9904-1958-10166-4/… | |
| Nov 5, 2014 at 15:55 | comment | added | Tommaso Centeleghe | I thought the tangent bundle $TS^n$ is parallelizable if and only if $n\in\{1;3;7\}$. For $n$ even you won't even find a nowhere vanishing vector field on $S^n$! | |
| Nov 5, 2014 at 15:48 | history | asked | Ali Taghavi | CC BY-SA 3.0 |