Timeline for Fundamental group of the complement of a codimension two submanifold
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
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| S 7 hours ago | history | bounty ended | ThorbenK | ||
| S 7 hours ago | history | notice removed | ThorbenK | ||
| 17 hours ago | comment | added | HJRW | @ThorbenK: done! Thanks for the suggestion. | |
| 17 hours ago | answer | added | HJRW | timeline score: 3 | |
| yesterday | comment | added | ThorbenK | @HJRW do you want to post your comment as an a sweetheart? I would prefer to grab you the bounty than Ryan. | |
| Dec 12 at 16:10 | history | edited | ThorbenK | CC BY-SA 4.0 |
added 12 characters in body
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| Dec 12 at 16:10 | comment | added | ThorbenK | @MoisheKohan Oh, I see. I never specified the manifolds to be smooth since I didn't see it as necessary. So feel free to assume that they are smooth or have a handlebody decomposition! | |
| Dec 12 at 16:06 | comment | added | Moishe Kohan | Are you assuming existence of a handle decomposition (which is true in dimensions $\ne 4$)? | |
| Dec 12 at 8:19 | history | edited | ThorbenK | CC BY-SA 4.0 |
added 13 characters in body
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| S Dec 11 at 11:57 | history | bounty started | ThorbenK | ||
| S Dec 11 at 11:57 | history | notice added | ThorbenK | Draw attention | |
| S Dec 10 at 10:42 | history | suggested | Julian Seipel | CC BY-SA 4.0 |
Identified typos and corrected those
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| Dec 10 at 9:49 | review | Suggested edits | |||
| S Dec 10 at 10:42 | |||||
| Dec 10 at 7:30 | history | edited | ThorbenK | CC BY-SA 4.0 |
Made it less ambigous
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| Dec 9 at 22:19 | comment | added | Ryan Budney | Now that I understand your question I see it's very close to the question of asking which manifolds have handle decompositions such that the 2-skeleton is homotopy-equivalent to a wedge of 2-spheres. The main difference is you are concerned with fundamental groups rather than homotopy-type. | |
| Dec 9 at 20:48 | history | became hot network question | |||
| Dec 9 at 20:44 | answer | added | Ryan Budney | timeline score: 3 | |
| Dec 9 at 12:49 | comment | added | ThorbenK | Yes I was expecting a similar argument, since 3-manifolds with free fundamental group are so restricted, so that one has to end up with something that can be build from $S^1 \times S^2$. | |
| Dec 9 at 12:45 | comment | added | HJRW | As you suspect, in dimension 3 the answer to your first question should be "no". Indeed, if one deletes any collection of circles from $M$ then the complement is a 3-manifold with toroidal boundary. Since the fundamental group is free, no boundary component can $\pi_1$-embed. But then, by Dehn's lemma, every boundary component is compressible. There's something to check here, but I think it follows that the complement is a connect sum of solid tori and $S^1\times S^2$'s, whence $M$ has to be a connect sum of lens spaces and $S^1\times S^2$'s. | |
| Dec 9 at 10:46 | history | asked | ThorbenK | CC BY-SA 4.0 |