Timeline for A question on eversion of (odd) spheres
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 8, 2018 at 8:40 | vote | accept | Ali Taghavi | ||
| Oct 6, 2018 at 13:20 | comment | added | Najib Idrissi | Let $f,g : S^n \to \mathbb{R}^{n+1}$. A regular homotopy between $f$ and $g$ induces a homotopy between the two Gauss maps $G_f, G_g : S^n \to S^n$. But the Gauss map of the antipodal is not the antipodal. If $f : S^n \to \mathbb{R}^{n+1}$ is the antipodal, then the degree of the Gauss map is $(-1)^n$. So there is no contradiction... | |
| Oct 6, 2018 at 12:37 | comment | added | Ali Taghavi | I am sorry if my question is elementary. According to the argument in the paper, using Gauss normal map, we conclude that a regular homotopy gives us a self map homotopy. So in the case of $S^2$, existence of an eversion implies that the identity and the amtipodal maps are homotopic self map. What is my mistake? | |
| Oct 4, 2018 at 20:56 | comment | added | Najib Idrissi | @MikeMiller Thanks, I didn't know that actually! | |
| Oct 4, 2018 at 20:56 | history | edited | Najib Idrissi | CC BY-SA 4.0 |
added 170 characters in body
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| Oct 4, 2018 at 20:56 | comment | added | mme | Smale's result is stronger: the only spheres $S^n$ that admit eversions in $\Bbb R^{n+1}$ are the tautological $n = 0$, the famous $n = 2$, and the less-known $n=6$, tied to when spheres are parallelizable. | |
| Oct 4, 2018 at 20:51 | history | answered | Najib Idrissi | CC BY-SA 4.0 |