# Complementary integrable vector fields

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?

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.

• 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. – Ali Taghavi Aug 30 '14 at 22:06

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.

• Prof Bryant Thank you very much for your answer. I 'll try to learn about Godbillon Vey class and its dynamic and geometric meaning. – Ali Taghavi Aug 30 '14 at 22:03

Wasn't it André Haefliger who proved that there exists no real-analytic foliation of the 3-sphere by surfaces?

• 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. – Ali Taghavi Nov 9 '19 at 14:17
• 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. – Robert Bryant Nov 9 '19 at 15:39