Timeline for Finding a 1-form adapted to a smooth flow
Current License: CC BY-SA 3.0
23 events
| when toggle format | what | by | license | comment | |
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| Jul 30, 2017 at 10:57 | comment | added | Ali Taghavi | @LeeMosher Dear Prof. Mosher, thank you very much for this very helpful picture in wikipedia. | |
| Jul 30, 2017 at 7:32 | comment | added | Ali Taghavi | @RobertBryant Thank you very much for this example and consideration of my request. So I learn that closed orbits are not the only obstruction for geodesibility, provided we enlarge the domension to n=3. | |
| Jul 30, 2017 at 6:50 | history | bounty ended | Ali Taghavi | ||
| Jul 29, 2017 at 20:42 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Simplified the argumen for the second example.
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| Jul 29, 2017 at 19:18 | comment | added | Robert Bryant | @AliTaghavi: I have added a paragraph to this answer that constructs an example without closed orbits that does not admit an adapted $1$-form. | |
| Jul 29, 2017 at 15:05 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Added the requested example without closed orbits
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| Jul 19, 2017 at 12:52 | comment | added | Lee Mosher | This flow, thought of as a foliation, is a very popular one in low dimensional topology. I entered "Reeb foliation" into wikipedia and found an appropriate picture of the flow restricted to the annulus defined by $0 \le x \le \pi$: en.wikipedia.org/wiki/Reeb_foliation#/media/… | |
| Jul 8, 2017 at 5:22 | comment | added | Ali Taghavi | Your example and this very interesting question of @TerryTao is a motivation to some other questions which I am asking in some separate questions. | |
| Jul 8, 2017 at 5:20 | comment | added | Ali Taghavi | @RobertBryant Thank you very much for this very interesting example. Now My misunderstanding is removed. Actually your example provide a counterexample to a question which I was interested in it, since many years ago, I had intention to look at limit cycles as closed geodesic but your very interesting example shows that it is not always possible. | |
| Jul 7, 2017 at 3:49 | comment | added | Ali Taghavi | @VítTuček Thank you very much for the picture. | |
| Jul 5, 2017 at 16:33 | comment | added | Vít Tuček | Picture's worth a thousand words. goo.gl/chZMZS (It was surprisingly difficult to use WolframAlpha for this.) | |
| Jul 5, 2017 at 10:48 | comment | added | Robert Bryant | @AliTaghavi You should just draw a picture of the flow lines of X in the plane; then it should be clear. I don't have any better advice to help you see what is happening. | |
| Jul 5, 2017 at 10:42 | comment | added | Ali Taghavi | @RobertBryant thank you. My apology for this extra question: but what is my mistake to think that two (consecutive) limit cycles must have the same orientation.One can compare the situation with the punctured plane(an annular region). i mean that WLOG we may assume two limit cycles are not far from each other. | |
| Jul 5, 2017 at 10:31 | comment | added | Robert Bryant | @AliTaghavi Yes, it is well-defined on the torus. Both the coefficients and the coordinate vector fields are invariant under translation by $(2\pi,0)$ and $(0,2\pi)$, so, of course, the vector field is invariant and hence drops to the torus. | |
| Jul 5, 2017 at 7:08 | comment | added | Ali Taghavi | @RobertBryant Is the vector field in your example, really well defined on the torus?I think its flow is not necessarily invariant under the $\mathbb{Z}^2$ action on the plane (is not this invariance necessary to have a well define flow on torus? ). On the other hand it is very difficult to imagine that two closed orbit on the torus has opposite direction. Lets compare the situation with a non vanishing vector field on the punctured plane, obvuoysly every two closed orbit have the same orientation. May be I am missing some thing? | |
| Jul 5, 2017 at 6:42 | comment | added | Robert Bryant | @AliTaghavi Yes, that is correct. | |
| Jul 5, 2017 at 6:10 | comment | added | Robert Bryant | @TerryTao Yes, similarly, one can construct a vector field $X$ on $S^3$ that has two closed orbits, $C_0$ and $C_1$, while all the other orbits $\alpha$-limit to $C_0$ and $\omega$-limit to $C_1$ and intersect an separating torus transversely. Again, any function on $S^3$ that is constant on the $X$-orbits must be constant, and, again, one can show that $\theta$ must be closed (though the argument is a little more subtle), which, since $S^3$ is simply connected, implies that $\theta(X)$ cannot be positive everywhere. | |
| Jul 5, 2017 at 5:55 | comment | added | Ali Taghavi | @RobertBryant Am I correct to conclude that there is no a Riemannian metric on the torus such that trajectories of the vector field in your example are (unparametrized) geodesics? | |
| Jul 5, 2017 at 5:11 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Simplified some statements and cleaned up the grammar.
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| Jul 5, 2017 at 2:13 | comment | added | Terry Tao | The key here seems to be that there are no non-constant $X$-invariant functions. My intuition previously was that this basically only happened when the flow was ergodic with respect to some volume form, but now I see that there are many other vector fields with this property also. | |
| Jul 5, 2017 at 1:58 | vote | accept | Terry Tao | ||
| Jul 5, 2017 at 1:58 | comment | added | Terry Tao | Very nice! I had suspected that the counterexample had to have a somewhat chaotic flow; this example isn't quite as wild as I had thought necessary, but it looks like it is still enough. | |
| Jul 4, 2017 at 21:34 | history | answered | Robert Bryant | CC BY-SA 3.0 |
