Timeline for Any 3-manifold can be realized as the boundary of a 4-manifold
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 28, 2020 at 14:19 | comment | added | Connor Malin | 1 follows for orientable manifolds from the fact that every 3-manifold is smoothable and every orientable smooth 3-manifold is parallelilizable since by Thom's theorem about characteristic numbers it must be null bordant. | |
| Jul 21, 2018 at 17:02 | answer | added | Igor Rivin | timeline score: 3 | |
| Jul 21, 2018 at 4:02 | answer | added | Ian Agol | timeline score: 15 | |
| Jul 21, 2018 at 3:35 | comment | added | wonderich | Thanks Andy for the nice comment - I asked a more basic question at MS a week ago but only a few comments (helpful though) math.stackexchange.com/q/2850317/79069 any-closed-3-manifold-is-a-boundary-of-some-compact-4-manifold but there are no answers. | |
| Jul 21, 2018 at 3:16 | review | Close votes | |||
| Jul 27, 2018 at 3:05 | |||||
| Jul 21, 2018 at 2:57 | comment | added | Andy Putman | For your first question, all the proofs I know that 3-manifolds are the boundaries of 4-manifolds show that in fact they are the boundaries of smooth 4-manifolds (well, except for the ones that just spit out a triangulated 4-manifold immediately!), and in dimension 4 smooth and PL are the same. The second question is trivial since all 3-manifolds can be triangulated. This question is more appropriate for math.se, and I have voted to close. | |
| Jul 21, 2018 at 2:48 | history | asked | wonderich | CC BY-SA 4.0 |