Timeline for Is the universal covering of an open subset of $\mathbb{R}^n$ diffeomorphic to an open subset of $\mathbb{R}^n$ ?
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 12, 2011 at 1:54 | comment | added | Autumn Kent | Complement of a knot in $S^3$ (and hence $\mathbb{R}^3$) will never work (even if it fails to be hyperbolic), as such a manifold is Haken, and it is a theorem of Waldhausen that the universal cover of a Haken manifold is $\mathbb{R}^3$. So you can avoid Geometrization. | |
| May 12, 2011 at 0:39 | answer | added | Ryan Budney | timeline score: 25 | |
| Mar 15, 2010 at 11:36 | answer | added | Marius Overholt | timeline score: 1 | |
| Mar 15, 2010 at 3:46 | comment | added | Ilya Grigoriev | Complement of a knot won't work: it's usually a hyperbolic manifold, so the universal cover is the hyperbolic 3-space, which is diffeomorphic to $R^3$. | |
| Mar 15, 2010 at 3:37 | comment | added | Ilya Grigoriev | Even crazier: complement of a Hawaiian earring in $R^3$. Cool question, by the way. | |
| Mar 15, 2010 at 3:30 | comment | added | Ilya Grigoriev | I wonder, what happens if you take the complement of a knot in $R^3$? Torus knot in $R^4$? | |
| Mar 15, 2010 at 2:16 | answer | added | Igor Belegradek | timeline score: 3 | |
| Mar 15, 2010 at 0:45 | answer | added | Georges Elencwajg | timeline score: 0 | |
| Mar 14, 2010 at 20:58 | answer | added | Mariano Suárez-Álvarez | timeline score: -1 | |
| Mar 14, 2010 at 19:26 | comment | added | S. Carnahan♦ | If you don't require the differentiable structure on the first copy of $\mathbb{R}^n$ to be standard, then the answer is no, due to the existence of large exotic structures on $\mathbb{R}^4$. | |
| Mar 14, 2010 at 18:38 | history | asked | Fiamma Battaglia - Elisa Prato | CC BY-SA 2.5 |