Timeline for Can we foliate the punctured space by tori?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Mar 30, 2019 at 14:43 | comment | added | Ali Taghavi | @AllenHatcher I thonk the orientability of the foliation is automatically hold sonce every 2 dim vector bundle over punctured space must be oriantable. | |
| Dec 7, 2013 at 18:16 | vote | accept | Ali Taghavi | ||
| Dec 7, 2013 at 16:21 | comment | added | JHM | It is worthwhile to ensure that the quotient has no boundary, i.e. that every leaf is separating in $R^3-0$. Graciously, Alexander duality (c.f. Bredon's ``Geometry and Topology", GTM 139, Exercise VI.8.1) assures us that "if $M^n$ is compact orientable and $N^{n-1}$ is a smoothly embedded compact orientable submanifold, then $N$ separates $M$". Hence any embedded orientable closed surface separates $S^3$, and hence $S^3-\{p,q\}\simeq R^3-0$. | |
| Dec 7, 2013 at 15:45 | comment | added | Ali Taghavi | "smoothness of the quotient map" is not a consequence of "rank theorem"? | |
| Dec 7, 2013 at 14:00 | comment | added | Ali Taghavi | @J Martel. Thank you very much for your answer. Regarding your last statement I think that every compact codimension one submanifold of R^n is oriantable and separating.If I remember well I saw this theorem some years ago in Hirsch "Differential Topology" | |
| Dec 7, 2013 at 4:11 | comment | added | JHM | Now are there any non-separating embedded tori in $R^3-0 \ldots \ldots$ | |
| Dec 7, 2013 at 3:41 | comment | added | JHM | @AllenHatcher: thank you for the comment. The non-orientability of the total space may be a red herring. The issue with the mobius strip being rather that we have a leaf which is non-separating (namely the centre leaf). My argument above is then only valid so long as every leaf is two-sided, i.e. separating $R^3-0$. | |
| Dec 7, 2013 at 0:43 | comment | added | Allen Hatcher | An orientability hypothesis is needed since an open Möbius band is smoothly foliated by circles. In this case the quotient is a half-open interval, so it's a 1-manifold with boundary. | |
| Dec 6, 2013 at 23:36 | history | edited | JHM | CC BY-SA 3.0 |
clarified some stuff.
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| Dec 6, 2013 at 22:06 | history | answered | JHM | CC BY-SA 3.0 |