Timeline for A simple proof that parallelizable oriented closed manifolds are oriented boundaries?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Mar 29, 2018 at 9:59 | comment | added | András Szűcs | Martel. If an n-manifold is stably parallelizable, then it can be immersed in n+1-dimensional Euclidean space. This is a trivial consequence of Hirsch's immersion theorem. | |
| Oct 15, 2014 at 11:41 | vote | accept | Hugo Chapdelaine | ||
| Oct 14, 2014 at 22:59 | history | edited | j.c. | CC BY-SA 3.0 |
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| Oct 14, 2014 at 20:47 | answer | added | András Szűcs | timeline score: 17 | |
| Oct 14, 2014 at 20:25 | answer | added | András Szűcs | timeline score: 35 | |
| Mar 12, 2014 at 17:11 | comment | added | Hugo Chapdelaine | Yes indeed, once one deals with this case, then one may look at a quotient of $G$ by a discrete subgroup $\Gamma\subseteq G$. Then any frame on $G$ may be pushed down to a frame on $G/\Gamma$ | |
| Mar 12, 2014 at 4:52 | comment | added | Richard Montgomery | How about starting off with the apparently simpler problem of M a compact Lie group? Lie groups are parallelizable. Why does every compact Lie group bound? SU(2): check. How about SU(3)? | |
| Mar 8, 2014 at 17:06 | comment | added | JHM | Kirby's lecture notes "The topology of four-manifolds", Ch. VII gives a proof that every orientable $3$-manifold is an orientable boundary. The first observation is that since $M^3$ is parallelizable, it immerses in $R^5$ with a trivial normal bundle. I'm not sure if this is a general consequence of parallelizability. | |
| Mar 7, 2014 at 23:31 | history | asked | Hugo Chapdelaine | CC BY-SA 3.0 |