Timeline for How can I visualize the nontrivial element of $\pi_4(S^3)$ and $\pi_5(S^3)$ ?
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
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| May 17, 2014 at 22:35 | comment | added | David Roberts♦ | @Chris you can edit it, you know... | |
| May 17, 2014 at 8:09 | comment | added | Chris Schommer-Pries | In the answer it should read "framed k-manifold". For the circle in $S^3$, how many normal framings are there? Well the framing can twist around any number of times, so you get $\mathbb{Z}$ many ($\pi_1 SO(2)$) and this corresponds to $\pi_3 S^2$. However when you suspend then there are only two ways to twists around, the trivial non-twisted way and the other way corresponding to $\pi_1(SO(n))$. | |
| Sep 27, 2011 at 13:26 | vote | accept | Yuji Tachikawa | ||
| Oct 22, 2015 at 5:38 | |||||
| Sep 27, 2011 at 13:26 | comment | added | Yuji Tachikawa | Is it possible to see that this generator when added to it is homotopic to zero? | |
| Sep 27, 2011 at 13:25 | vote | accept | Yuji Tachikawa | ||
| Sep 27, 2011 at 13:25 | |||||
| Sep 22, 2010 at 1:34 | comment | added | j.c. | @Mariano - the $S^1$ arises in the Pontryagin-Thom construction as the inverse image of a regular value of the map from $S^3$ to $S^2$. Homotopy classes of these maps are in correspondence with framed cobordism classes of links. Thus knottedness (or even number of components) of the inverse image is not a homotopy invariant, but "framing number" is. | |
| Sep 20, 2010 at 16:00 | comment | added | Mariano Suárez-Álvarez | The $S^1$ that represents $\eta$ is unknotted, I guess? In that case, what do you get from other knots? | |
| Sep 20, 2010 at 6:09 | history | answered | Dev Sinha | CC BY-SA 2.5 |