Timeline for A simple proof that parallelizable oriented closed manifolds are oriented boundaries?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Oct 15, 2014 at 11:41 | vote | accept | Hugo Chapdelaine | ||
| S Oct 15, 2014 at 7:29 | history | suggested | Marco Golla | CC BY-SA 3.0 |
edited formatting (\mathbb's, mostly).
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| Oct 15, 2014 at 7:13 | review | Suggested edits | |||
| S Oct 15, 2014 at 7:29 | |||||
| Oct 14, 2014 at 23:38 | comment | added | Neil Strickland | @AndréHenriques: the boundary of $N$ is morally $S(TM)$; that is only the same as $M\times S^{m-1}$ because $M$ is parallelizable. | |
| Oct 14, 2014 at 22:49 | comment | added | André Henriques | Where did you use that $M$ is parallelizable? | |
| Oct 14, 2014 at 22:47 | history | edited | André Henriques | CC BY-SA 3.0 |
added 77 characters in body
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| Oct 14, 2014 at 22:01 | comment | added | András Szűcs | If m is odd, then this gives even oriented null-cobordism (because if we choose q to be odd, then both RP^q and RP^{q-m+1} are orientable). | |
| Oct 14, 2014 at 21:35 | history | edited | András Szűcs | CC BY-SA 3.0 |
added 1 character in body
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| Oct 14, 2014 at 20:53 | history | edited | André Henriques | CC BY-SA 3.0 |
fixed Latex
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| Oct 14, 2014 at 20:47 | history | answered | András Szűcs | CC BY-SA 3.0 |