Timeline for Is it possible to improve the Whitney embedding theorem?
Current License: CC BY-SA 2.5
17 events
| when toggle format | what | by | license | comment | |
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| Mar 6, 2011 at 23:19 | vote | accept | Ben McMillan | ||
| Mar 7, 2011 at 8:53 | |||||
| S Mar 6, 2011 at 23:19 | vote | accept | Ben McMillan | ||
| Mar 6, 2011 at 23:19 | |||||
| Mar 6, 2011 at 23:19 | vote | accept | Ben McMillan | ||
| S Mar 6, 2011 at 23:19 | |||||
| Mar 6, 2011 at 23:12 | answer | added | John Klein | timeline score: 10 | |
| Mar 6, 2011 at 21:16 | answer | added | Bruno Martelli | timeline score: 14 | |
| Mar 6, 2011 at 20:56 | answer | added | Ryan Budney | timeline score: 24 | |
| Mar 6, 2011 at 20:15 | history | edited | Ben McMillan | CC BY-SA 2.5 |
added 179 characters in body; added 4 characters in body; added 92 characters in body
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| Mar 6, 2011 at 19:35 | comment | added | John Klein | I think that, in addition to assuming the domain manifolds to be connected we should also assume "closed," that is, compact without boundary. | |
| Mar 6, 2011 at 17:39 | comment | added | Aaron Mazel-Gee | 4. From the way Stiefel-Whitney classes control embedding results about $\mathbb{RP}^n\rightarrow \mathbb{R}^N$ I'd imagine that the answer might be bound by characteristic classes, although because of that example I think we'd find it impossible to get any complete results. Does anyone know if the sharp answers there were obtained through general methods, or were they ad hoc? | |
| Mar 6, 2011 at 17:35 | comment | added | Aaron Mazel-Gee | 1. You should probably just sprinkle the word "connected" all over this question to keep it interesting. 2. We already have the "upper bound" of $\mathbb{R}^{2n}$ for $n$-manifolds, so I think the answer to whether we can do this sort of has to be yes. 3. Certainly we can compare $n$-universal manifolds if their dimensions are different, but a priori there might be two $n$-universal $N$-manifolds and neither covers the other. Or perhaps it's provable that one must cover the other; that'd be a neat result. | |
| Mar 6, 2011 at 16:52 | comment | added | Martin Brandenburg | @Ben: Perhaps a silly question, but in which manifold does every possibly non-connected smooth surface embed? | |
| Mar 6, 2011 at 12:35 | comment | added | Martin Brandenburg | Another common finiteness condition is paracompactness. This is equivalent to the 2nd countability of each component. Perhaps Ben should add "2nc countable" to the question. | |
| Mar 6, 2011 at 10:40 | comment | added | Kelly Davis | @Johannes $U$ is not a $n$-manifold as it's not second countable. Definitions of "manifold" that I've seen all require second countability. | |
| Mar 6, 2011 at 10:31 | history | edited | Ben McMillan | CC BY-SA 2.5 |
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| Mar 6, 2011 at 10:25 | comment | added | Johannes Ebert | This is an extremely silly comment; sorry about that. Let $S$ be a set of $n$-manifolds containing one manifold out of each diffeomorphism class. Then $U=\coprod_{M \in S} M$ is an $n$-manifold and each $n$-manifold embeds into $U$. Of course, $U$ is not second countable. | |
| Mar 6, 2011 at 9:42 | comment | added | Zack | A nitpick: the second map in the example isn't well defined. $\mathbb{RP}^3$ doesn't contain two $\mathbb{RP}^2$s (the complement of one of them is $\mathbb{R}^3$). $\mathbb{RP}^3\sharp\mathbb{RP}^3$ should work, though. | |
| Mar 6, 2011 at 8:38 | history | asked | Ben McMillan | CC BY-SA 2.5 |