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Let $(M,g)$ be a Riemannian manifold. Assume that $X$ is a non vanishing vector field tangent to $M$.(Or assume that we have a one dimensional foliation of $M$). Under what geometric conditions we are sure that the codimension one distribution on $M$ orthogonal to $X$ (orthogonal to $F$) is integrable? Is there a global geometro-dynamical meaning for such possible conditions?

In particular, what is the answer for the standard metric of $S^{3}$ and its one dimensional foliation by circle arising from Hopf fibration? In this particular case, if this 2 dim. distribution is integrable, to what extent this 2. dim foliation is studied?

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The obstruction against integrability of the orthogonal is called curvature for the (Ehresmann) connection given by orthogonal projection onto the distribution generated by $X$. See section 17 of

  • Peter W. Michor: Topics in Differential Geometry. Graduate Studies in Mathematics, Vol. 93 American Mathematical Society, Providence, 2008. (pdf)

where the situation is slightly more special (the orbit space is a manifold, and the projection is a fiber bundle). Or see the paper for a more general situation (the vertical bundle, here generated by $X$, need not be integrable either, and then you have curvature and cocurvature):

  • Peter W. Michor: Graded derivations of the algebra of differential forms associated with a connection. Differential Geometry, Pe\ niscola, 1988, \eds F.J. Carreras, O. Gil-Medrano, A.M. Naveira Lecture Notes in Math. 1410 (1989), 249--261, Springer-Verlag, Berlin. (pdf)

In the case of $S^3$ and the Hopf fibration, the orbit space is $S^2$, and the connection is a principal $S^1$-connection whose curvature 2-form is a multiple of the volume form on $S^2$, if I remember correctly.

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  • $\begingroup$ Prof. Michor thank you very much for your very interesting answer and your linked references. I 'll try to learn the details of your answer. $\endgroup$ – Ali Taghavi Aug 30 '14 at 22:06
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Actually, it's easier than the general curvature case: Let $X^\flat$ be the $1$-form dual to $X$ via the metric $g$. Then the orthogonal plane field to the integral curves of $X$ is integrable if and only if $$ X^\flat \wedge \mathrm{d} X^\flat = 0. $$

In the case of the $3$-sphere and the vector field $X$ that generates the Hopf fibration, the $3$-form $X^\flat \wedge \mathrm{d} X^\flat$ is nowhere vanishing, so the $2$-plane field is not integrable, so there is no foliation to discuss. In particular, $X^\flat$ defines a contact structure on the $3$-sphere.

Moreover, Haefliger (Comment. Math. Helv. 32 (1958), 249–329) proved that there is no real-analytic foliation of a simply-connected compact $3$-manifold by surfaces. (Reeb famously constructed a smooth foliation of $S^3$, though.)

If you are looking for information about codimension $1$ foliations of manifolds, you should, perhaps, look up references that discuss secondary characteristic classes, such as the Godbillon-Vey class, and their geometric and dynamical meaning. A good place to start would be H. Blaine Lawson's article Foliations (Bulletin of the AMS 80 (1974), 369–418).

Note: In the first version of this answer (written in 2014), I had mistakenly attributed Haefliger's theorem to Arnol'd. My thanks to Dan Asimov for pointing out my error.

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  • $\begingroup$ Prof Bryant Thank you very much for your answer. I 'll try to learn about Godbillon Vey class and its dynamic and geometric meaning. $\endgroup$ – Ali Taghavi Aug 30 '14 at 22:03
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Wasn't it André Haefliger who proved that there exists no real-analytic foliation of the 3-sphere by surfaces?

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  • $\begingroup$ Dear Prof. Asimov. Thank you very much for your answer. I did not heared (or I do not remember) of this theorem .the version I was aware of was the following"Non exsictence of a codimension 1 foliation with non trivial holonomy(in dim 3). Thanks for informing me of this theorem. $\endgroup$ – Ali Taghavi Nov 9 '19 at 14:17
  • $\begingroup$ Dan: Thanks for pointing that out. I wrote my answer off the top of my head and didn't realize that I had mis-remembered the attribution. I'm not sure why I have Arnol'd connected with this result. Maybe it's because I read about it in a book or article by Arnol'd and didn't remember that he attributed it to Haefliger. I'll fix my answer, but put in a note so that your remark won't be orphaned. $\endgroup$ – Robert Bryant Nov 9 '19 at 15:39
  • $\begingroup$ Dear Prof. Asimov thank you again for your answer and your attention to my question. I would like to ask you to give comment on the following question(i am sorry if this question is unrelated to this post but I had intention to ask you via email but I did not find your email address) $\endgroup$ – Ali Taghavi Nov 15 '19 at 15:35
  • $\begingroup$ mathoverflow.net/questions/345685/… $\endgroup$ – Ali Taghavi Nov 15 '19 at 15:35

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