Timeline for Can we foliate the punctured space by tori?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 9, 2013 at 10:53 | comment | added | Ali Taghavi | @Bin Yu, thank you for your comments | |
| Dec 9, 2013 at 4:40 | comment | added | Bin Yu | trivial. But the fact that a torus embedded into $S^3$ has to be separable forbidden it is $Z_2$. (3) A suggestion: what I can explain is as above, more you can found in "Cesar Camacho, Alcides Lins Neto, "Geometry Theory of Foliations". | |
| Dec 9, 2013 at 4:34 | comment | added | Bin Yu | @Taghavi, (1) if a torus $T$ in $S^3$ is nonseprating, for a given point $P\in T$, you can find an arc $l$ in $W=S^3 -T$ approching to $P$ on both sides of $T$, then $c= l\cup \{P\}$ is a closed curve intersecting with $T$ exactly one point ($P$). Therefore, in cohomology, the cup product of $[c]\in H^1 (S^3)$ an $[T]\in H^2 (S^3)$ has to be nonzero. This conflicts the fact that $H^1 (S^3) =0$. (2) I think that I have explain detaily why the holonomy is trivial in "... we claim that $P_0 =P_1$, otherwise there are two possibilites: ". Roughly, the finite subgroup of $SO(1)$ is either $Z_2$ or | |
| Dec 7, 2013 at 14:16 | comment | added | Ali Taghavi | @Bin Yu, thanks a lot for your answer I learn a lot from yor explaination. But I do not underestand some thing for example "there is a closed curve in S^3 which intersect torus in one point, this is a contradiction since H^1(S^3) is trivial".I have another question about your answer:with a compactness argument you said that the holonomy orbit of each point on transversal section, is a finite set, but it does not implies that the holonomy is trivial. Am I missing something? | |
| Dec 7, 2013 at 9:08 | comment | added | Bin Yu | (3)if a torus bundle over a 1-manifold (denoted by $M$) is homeomorphic to $R^3−\{0\}$, then they 1-manifold has to be $R$. But $R$ is contractable, so $M$ has to be trivial bundle (this is a fundamental fact of fiber bundle). (4)Sorry, before, I think such things are well-known. Actually "my proof" is finished before the line. (if I add a reference, for instance "Cesar Camacho, Alcides Lins Neto, "Geometry Theory of Foliations") . But I feel it is not complicated, so I try to explain all... it seems not successful:) | |
| Dec 7, 2013 at 9:05 | comment | added | Bin Yu | @Taghavi, (1) if the fundamental group of a leaf is trivial, then the foliation restricted to a small neighborhood of the leaf is trvial.(it can be followed by the def of foliation). (2) every torus embedded in $R^3−\{0\}$ is separating (there are several ways to explain this, for instance, using homology: if there exists a nonseparating torus in $R^3−\{0\}$, then automatically there exists a nonseparating torus in $S^3$(two points compactness to $R^3−\{0\}$), then there is a closed curve in $S^3$ intersect with the torus exactly one point, it is impossible since $H_1 (S^3)=0$. | |
| Dec 6, 2013 at 16:21 | comment | added | Ali Taghavi | @J.Martel as a combination of your idea and Bin's proof it seems that we can conclude " A codimension one foliation of an open manifold with compact leaves, gives a trivial bundle" Am i write? is it a known theorem in foliation theory? However some part of Bin's proof is not clear for me: How is the open set U(T) constructed? what was the role of contractibility of T-l\cup m? | |
| Dec 6, 2013 at 15:06 | comment | added | JHM | I find this argument to be unclear. As i see it, there should be a uniform argument applicable to tori and closed higher genus surfaces. If indeed any such foliation is actually a locally trivial fibration, then within the smooth category it will be a trivial bundle over (the only noncompact smooth 1-manifold) $R$. | |
| Dec 6, 2013 at 12:39 | comment | added | Ali Taghavi | As a consequence of leary hirsch theorem? | |
| Dec 6, 2013 at 9:22 | comment | added | Bin Yu | @Taghavi, I don't prove the fiber bundle is trivial. I just say "it is a torus bundle over a 1-manifold". But every torus bundle over a 1-manifold can't be homeomorphic to $R^3 -\{0\}$. | |
| Dec 6, 2013 at 8:33 | comment | added | Ali Taghavi | Thank you for the answer.you proved that the foliation gives us a fibre bundle, since the holonomy is trivial. Why this fibre bundle is globally trivia? | |
| Dec 6, 2013 at 5:34 | history | answered | Bin Yu | CC BY-SA 3.0 |