Timeline for Jordan Curve Theorem for Manifolds
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 11, 2011 at 18:26 | comment | added | Ryan Budney | @James: the target space being Euclidean isn't terribly important for Jordan-Brouwer, it's an expository simplification -- the proof is pretty much always more general than the statement of the theorem. | |
| Jun 11, 2011 at 17:58 | answer | added | Theo Johnson-Freyd | timeline score: 0 | |
| Jun 11, 2011 at 17:51 | vote | accept | Dilitante | ||
| Jun 11, 2011 at 17:33 | comment | added | Dilitante | Should be simply connected, yes. Ryan, This differs from the Jordan-Brouwer theorem in that now we are talking about mappings into any simply connected Manifold, not just R^n | |
| Jun 11, 2011 at 15:44 | comment | added | Sergey Melikhov | Ryan: I don't think transversality and Guillemin-Pollack suffice. The question is about topological embeddings. | |
| Jun 11, 2011 at 15:35 | comment | added | Ryan Budney | James, IMO this question is a great math.stackexchange.com question. | |
| Jun 11, 2011 at 15:16 | comment | added | Ryan Budney | The general theorem of the form of (1) is called the Jordan-Brouwer Separation theorem. See en.wikipedia.org/wiki/Jordan_curve_theorem also the Differential Topology text of Guillemin and Pollack. General theorems of the type (2) follow directly from elementary transversality theorems, see also Guillemin and Pollack. | |
| Jun 11, 2011 at 13:51 | answer | added | Angelo | timeline score: 8 | |
| Jun 11, 2011 at 13:40 | comment | added | George Lowther | For (1) I think you need $f$ to act trivially on the nth (co)homology group. Think about embedding a circle in a torus. I need not break it into two components. | |
| Jun 11, 2011 at 13:37 | comment | added | Francesco Polizzi | If you take as $C$ a meridian of a $2$-torus $M\subset \mathbb{R}^3$, it seems to me that $M−C$ is connected | |
| Jun 11, 2011 at 13:03 | history | asked | Dilitante | CC BY-SA 3.0 |