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 | |
|---|---|---|---|---|---|
| Mar 9, 2011 at 5:54 | comment | added | Ryan Budney | @Kelly, I think your meta-discussion might be more appropriate for meta.mathoverflow. | |
| Mar 8, 2011 at 18:18 | comment | added | Kelly Davis | @Ryan Maybe I am just missing something, which may be the case, but I don't understand how you're answering the posed question. The question you answer is: "Does there exist an $m < 2n$ and a connected smooth $m$-manifold into which any connected closed smooth $n$-manifold smoothly embeds?". While, as far as I can tell, the posed question is: "Does there exist an $m < 2n$ and a connected closed smooth $m$-manifold into which any connected closed smooth $n$-manifold smoothly embeds?" Notice, another "closed" was introduced into the second question. Are you answering the second question too? | |
| Mar 7, 2011 at 22:36 | comment | added | Ryan Budney | I'm not certain where that result first appears, but the technology is called "smoothing theory". This is a tool that allows you to enumerate all the smooth structures that are compatible with a given PL manifold structure. Hirsch and Mazur "Smoothings of piecewise linear manifolds" and Kirby and Siebenmann's "Foundational essays on topological manifolds, smoothings and triangulations" are standard references. | |
| Mar 7, 2011 at 22:23 | comment | added | Alessandro Sisto | Thanks, I didn't know that result on triangulations. | |
| Mar 7, 2011 at 15:38 | comment | added | Ryan Budney | Hi Alessandro, there's many ways to argue this. One is that all smooth manifolds have triangulations, and there's only countably-many triangulations. Moreover, there's only finitely many smooth structures supported by a given triangulation. | |
| Mar 7, 2011 at 13:08 | comment | added | Alessandro Sisto | The connected sum you describe is second countable only if there are countably many closed n-manifolds up to diffeomorphism. Please forgive me for my ignorance: is this the case? Why? | |
| Mar 6, 2011 at 23:46 | comment | added | Kelly Davis | @Ryan In Edit 2 the OP's requires all manifolds, the "source" manifolds $M_i$ and the "universal" target manifold $U$, are closed, compact without boundary. So, my original objection rules out $(M_1 \times S^1) \# (M_2 \times S^1) \# \cdots$ and as $U$ must be closed $\mathbb R^{n+1} \# (M_1 \times S^1) \# (M_2 \times S^1) \# \cdots$ is ruled out. | |
| Mar 6, 2011 at 23:22 | comment | added | Ben McMillan | That is a very interesting solution, thanks! I wasn't even aware that Whitney embedding could be improved for Euclidean embeddings. | |
| Mar 6, 2011 at 23:19 | vote | accept | Ben McMillan | ||
| Mar 6, 2011 at 23:19 | |||||
| Mar 6, 2011 at 22:05 | comment | added | Kelly Davis | @Ryan I see. So, if we limit ourselves to compact $n$-manifolds $M_i$, then it becomes "interesting" once again as $M_i \times S^1$ is always compact. | |
| Mar 6, 2011 at 22:02 | comment | added | Ryan Budney | @Gil, as Andy mentions this is Massey's Immersion Conjecture. en.wikipedia.org/wiki/Whitney_immersion_theorem The analogous conjecture for embeddings hasn't been formulated let alone proven, as far as I know. | |
| Mar 6, 2011 at 22:01 | comment | added | Ryan Budney | @Kelly: the reason why I'm choosing this non-compact setting is to ensure the infinite connect-sum makes formal sense. Mazur's swindle construction doesn't live entirely in the realm of smooth manifolds. | |
| Mar 6, 2011 at 21:54 | comment | added | Andy Putman | @Gil : That's for immersions, not embeddings. It's the Immersion Conjecture, which was proven by Ralph Cohen. | |
| Mar 6, 2011 at 21:54 | comment | added | Ryan Budney | Kelly, I think there is a mis-communication over what is meant by "infinite connect-sum". Such arguments aren't a problem in the realm I'm working in because the manifolds are non-compact. If you want to make what I said more precise, the idea would be to take $\mathbb R^{n+1}$, and pick a countable discrete collection of disjoint points in $\mathbb R^{n+1}$, and use them as the "attaching points" for the connect sum operation. This doesn't cause any Mazur-swindle problems since the resulting manifold is non-compact. | |
| Mar 6, 2011 at 21:50 | comment | added | Kelly Davis | @Ryan If I understand correctly, your answer is $(M_1 \times S^1) \# (M_2 \times S^1) \# \cdots$. This won't work. An infinite connected sum is not defined. For example, we know there exists an exotic $S^7$, notated as $S_+$ say, and it has an inverse $S_-$ such that $S^7 = S_- \# S_+$. If an infinite connected sum were defined, we would have $S_+ = S_+ \# (S_- \# S_+) \# (S_- \# S_+) \cdots$ $= (S_+ \# S_-) \# (S_+ \# S_-) \# \cdots$ $= S^7$ which is false as $S_+$ is exotic. | |
| Mar 6, 2011 at 21:42 | comment | added | Gil Kalai | Dear Ryan, I vaguely remember a result about embedding in dimension 2n-b(n) where b(n) is the number of 1's in the binary expansion of n. But I dont remember if it was for the same problem. | |
| Mar 6, 2011 at 20:56 | history | answered | Ryan Budney | CC BY-SA 2.5 |