Timeline for $2$ dimensional foliations of space whose leaves contain the trajectories of a given vector field
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 12, 2019 at 15:31 | comment | added | Tom Goodwillie | Yes, that's a simpler way to explain it. | |
| Aug 11, 2019 at 13:25 | history | bounty ended | Ali Taghavi | ||
| Aug 11, 2019 at 13:25 | vote | accept | Ali Taghavi | ||
| Aug 6, 2019 at 12:00 | comment | added | Ali Taghavi | Dear Prof. Goodwillie, according to your valuable comment I just realized that my revious comment was not correct. In fact every such a foliation must be invariant under $S^1$ action since each leaf is a union of Hopf fibers. So we easiy get a foliation of $S^2$, a contradiction. Thanks again for your very interesting answer! | |
| Aug 5, 2019 at 23:02 | comment | added | Tom Goodwillie | For each point in $S^2$ its preimage in $\mathbb R^3$ would be contained in one of the leaves. Therefore each leaf would be the preimage of some subset of $S^2$. In order for the preimages of these subsets to be the leaves of a foliation, the sets themselves would have to be the leaves of a foliation. | |
| Aug 4, 2019 at 21:55 | comment | added | Ali Taghavi | My apology if my question is elementary. How the foliation on total space, which is not necessarily $S^1$ invariant, give us a foliation of the bases space? | |
| Oct 1, 2017 at 18:28 | comment | added | Tom Goodwillie | Yes, that was my reasoning. No, the Reeb foliation has a torus leaf, but no torus in $S^3$ can be transverse to the Hopf fibration because that would make it a covering space of $S^2$. In fact, no $2$-dimensional foliation of $S^3$ that has a compact leaf can be transverse to the Hopf fibration. | |
| Oct 1, 2017 at 11:20 | comment | added | Ali Taghavi | So a natural question is that "Is the Reeb foliation transverse to the hopf fibration, at all points?* | |
| Oct 1, 2017 at 11:16 | comment | added | Ali Taghavi | Thank you very much for this interesting answer. I understand the tangent vector field to the Hopf foliation is not contained in any 2 dimensional smooth distribution $D$ otherwise since any such distribution would be map to a $1$ dimensional foliation of $S^2$.Am I correct? | |
| Oct 1, 2017 at 11:11 | vote | accept | Ali Taghavi | ||
| Aug 11, 2019 at 13:25 | |||||
| Oct 1, 2017 at 1:38 | history | answered | Tom Goodwillie | CC BY-SA 3.0 |