Timeline for Poincaré dodecahedron space
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Dec 30, 2015 at 10:43 | history | suggested | John B | CC BY-SA 3.0 |
minor adjustments
|
| Dec 30, 2015 at 9:58 | review | Suggested edits | |||
| S Dec 30, 2015 at 10:43 | |||||
| Nov 7, 2013 at 9:34 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
replaced deprecated tags 'geometry' and 'topology'
|
| Nov 7, 2013 at 8:45 | answer | added | R. Quehenberger | timeline score: 0 | |
| Oct 4, 2012 at 16:28 | vote | accept | Caramba | ||
| Apr 1, 2012 at 3:53 | answer | added | user22556 | timeline score: 3 | |
| Mar 24, 2012 at 0:37 | history | edited | Misha |
edited tags
|
|
| Mar 24, 2012 at 0:36 | answer | added | Misha | timeline score: 5 | |
| Mar 21, 2012 at 5:17 | answer | added | Alex Suciu | timeline score: 6 | |
| Mar 21, 2012 at 3:28 | comment | added | Tom Goodwillie | I believe that there is the following approach: In $SO(3)$ there is a group $G$ of order $60$, symmetries of the dodecahedron. Thinking of $SO(3)$ as a Riemannian manifold, the twelve elements of $G$ closest to the identity are the $36$ degree rotations in $G$. The elements of $SO(3)$ that are closer to the identity than to any other element of $G$ form a domain shaped like a dodecahedron, and the manifold $SO(3)$ can be assembled by gluing together the sixty translates of this by the elements of $G$. The $3$-sphere, double cover of $SO(3)$, is made of $120$ of these. | |
| Mar 20, 2012 at 23:51 | comment | added | Mark Grant | Have you looked at en.wikipedia.org/wiki/Binary_icosahedral_group and tried to map $a,\ldots,f$ to some unit quaternions? | |
| Mar 20, 2012 at 22:53 | comment | added | Paul | Fernando: I think the question is not "how does one compute a presentation of $\pi_1(X)$", but why the presentation is a presentation of the binary icosahedral group. In particular, why it finite (and of order 120). I don't know any simple way to see this. | |
| Mar 20, 2012 at 22:52 | comment | added | Fernando Muro | Well, I don't remember now whether the inherited cell structure has exactly one vertex, or maybe more, but this is not really important for the computation. | |
| Mar 20, 2012 at 22:33 | comment | added | Fernando Muro | This is an easy exercise I used to give to undergraduate students of algebraic topology. The quotient space inherits a cell structure with only one vertex from the dodecahedron. The fundamental group can therefore be easily computed from the cell decomposition in the usual way. | |
| Mar 20, 2012 at 22:07 | history | asked | Caramba | CC BY-SA 3.0 |