Timeline for Is every real n-manifold isomorphic to a quotient of $\mathbb{R}^n$?
Current License: CC BY-SA 3.0
19 events
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| May 9, 2014 at 19:52 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
corrected tags; minor editing for correctness and clarity
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| Nov 13, 2010 at 22:00 | vote | accept | Eivind Dahl | ||
| Nov 13, 2010 at 21:59 | comment | added | Ryan Budney | A comment on my comment: one can avoid the use of proper functions by simply writing the manifold as a countable union of compact balls. | |
| Nov 13, 2010 at 21:50 | answer | added | Jeff Strom | timeline score: 33 | |
| Nov 13, 2010 at 21:46 | comment | added | Joel Fine | There is a loosely related open question due to Gromov which asks whether or not every manifold can be realised as the quotient of hyperbolic space by a discrete group of isometries. (He mentions essentially this question on page 12 of a recent paper. See ihes.fr/~gromov/PDF/manifolds-Poincare.pdf ) | |
| Nov 13, 2010 at 21:44 | comment | added | Ryan Budney | Every connected 2nd countable Hausdorff manifold is a quotient of $\mathbb R$. This follows from the fact that every compact connected manifold is a quotient of $[0,1]$ and the fact that non-compact manifolds have proper functions so they're a countable union of compact manifolds. | |
| Nov 13, 2010 at 21:29 | answer | added | David Carchedi | timeline score: 1 | |
| Nov 13, 2010 at 21:23 | answer | added | Bruno Martelli | timeline score: 13 | |
| Nov 13, 2010 at 21:14 | comment | added | David Steinberg | I have an idea for when your manifold M is a CW-complex. If you remove the (n-1)-skeleton of M, then you are left with an open n-ball; this suggests that your manifold M can be obtained as a topological quotient of a closed n-ball, which is certainly the topological quotient of R^n. | |
| Nov 13, 2010 at 21:11 | comment | added | Eivind Dahl | Taking it further: When if ever is it true in the category of differentiable manifolds? I guess I won't go crazy bananas with my first question, heh. | |
| Nov 13, 2010 at 21:03 | answer | added | Neil Strickland | timeline score: 16 | |
| Nov 13, 2010 at 20:59 | history | edited | Eivind Dahl | CC BY-SA 2.5 |
added a linebreak ..
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| Nov 13, 2010 at 20:56 | comment | added | Eivind Dahl | Yes Tilman, that's what I mean! I'm editing the question. | |
| Nov 13, 2010 at 20:37 | comment | added | Tilman | You'll also want to assume that the manifold is connected, otherwise it's obviously false. | |
| Nov 13, 2010 at 20:36 | comment | added | Tilman | You mean: is there a surjective map from $\mathbf{R}^n$ to the manifold such that the manifold has the quotient topology? Then $S^n$ and $\mathbf{R}^n-\{0\}$ are not counterexamples. | |
| Nov 13, 2010 at 20:23 | comment | added | Deane Yang | No. Most manifolds are not. Explicit examples include the $n$-dimensional sphere, where $n \ge 2$ and $R^n$ minus the origin. | |
| Nov 13, 2010 at 20:04 | answer | added | Adam Hughes | timeline score: 1 | |
| Nov 13, 2010 at 19:59 | history | edited | Sean Tilson |
edited tags
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| Nov 13, 2010 at 19:55 | history | asked | Eivind Dahl | CC BY-SA 2.5 |