Timeline for flexibility of almost contact ``Reeb'' vector fields
Current License: CC BY-SA 3.0
12 events
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| Mar 28, 2012 at 12:35 | history | edited | Sam Lisi | CC BY-SA 3.0 |
clarified points raised by alvarezpaiva and BS
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| Mar 28, 2012 at 12:33 | comment | added | Sam Lisi | @BS, alvarezpaiva, sorry, I did mean minimal and the manifold is closed. I am editing the question to clarify this. | |
| Mar 28, 2012 at 10:50 | comment | added | BS. | By the way, you should precise your question (as alvarezpaiva suggets). Does "invariant sets" mean "minimal (closed invariant) sets"? Are your manifolds closed ? | |
| Mar 28, 2012 at 10:45 | comment | added | BS. | Just a remark : in 3 dimensions, your vector field can be any non-vanishing vector field on any orientable 3-manifold, since you can set the 2-form to be the interior product of $R$ and a volume form. In higher dimensions, you have to add that the normal hyperplane field has an almost complex structure, a condition on the homotopy class of the vector field, thus invariant by deformation. | |
| Mar 28, 2012 at 8:21 | comment | added | alvarezpaiva | Am I right in thinking that a very weak version of your question is: Given a nowhere zero vector field $X$ on a compact manifold, can it be deformed through nowhere zero vector fields to a vector field $Y$ for which the only invariant sets of its flow are non-degenerate periodic orbits? I still feel the problem is the meaning of "the only invariant sets are ..." Aren't there stable and unstable manifolds ? Maybe I'm just slow this morning. | |
| Mar 27, 2012 at 20:07 | history | edited | Sam Lisi | CC BY-SA 3.0 |
title better matches the question
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| Mar 22, 2012 at 17:33 | comment | added | Sam Lisi | @alvarezpaiva: thanks for this observation. I think it means that I am trying to approach this lemma the wrong way. | |
| Mar 22, 2012 at 17:08 | comment | added | alvarezpaiva | What you're asking sounds hard: if you take your hypersurface to be the unit co-sphere bundle of some Riemannian or Finsler metric, then just perturbing it to a metric where all closed geodesics are non-degenerate is the bumpy metric theorem. Of course, the theorem says that such metrics are residual, which is much stronger that what you need. But in exchange, you would need the bumpy metric to have ergodic geodesic flow in a very strong way. | |
| Mar 22, 2012 at 15:05 | comment | added | Sam Lisi | @Dmitri: I want $\Sigma$ to be a co-oriented hypersurface in $M$, so I have chosen a vector field $\nu$ in a neighbourhood of $\Sigma$ that is transverse to $\Sigma$. I use it to define $R$ by the relationship that $\omega(\nu, R) = 1$ and $R$ is in the kernel of $\omega|_{\Sigma}$. I'm sorry I was a bit terse in my question. | |
| Mar 22, 2012 at 13:55 | comment | added | Dmitri Panov | Could you please explain how you use $\nu$ to define $\Sigma$? Is $\nu$ the kernel of the restriction of $\omega$ to $M$? | |
| Mar 22, 2012 at 12:42 | comment | added | alvarezpaiva | Could you please give a reference to Giroux's paper? | |
| Mar 22, 2012 at 12:21 | history | asked | Sam Lisi | CC BY-SA 3.0 |