MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider the manifold $\mathbb{R^2}\setminus \{0\}$, on which the group of rotation acts. The orbits of the group are the circles centered in the origin, and form a foliation of $\mathbb{R^2}\setminus \{0\}$. This foliation will be denoted by $F_1$. The foliation $F_1$ defines univocally another foliation $F_2$, with the following property: the tangent spaces of two leafs of $F_1$ and $F_2$ are orthogonal at the intersection point. In this case $F_2$ is composed by the radial lines from the origin.

My question is the following: to what extent this situation can be generalized, i.e.: assume to have a riemanian manifold $M$, possibly flat, with a foliation $F_1$ defined by the orbits of a group acting on $M$. To what extent does this foliation define univocally an orthogonal foliation $F_2$ with the property that the tangent spaces of any pair of leafs of $F_1$ and $F_2$ are orthogonal at the intersection point(s)?

share|cite|improve this question
You need to learn about the Frobenius integrability criterion, a test for whether a subbundle of the tangent bundle of a manifold does or not correspond to a foliation. – Tom Goodwillie Apr 24 '11 at 20:01
This question is also relevant to your question, though it is more general than group actions:… – Robert Bryant Apr 18 '14 at 20:46
up vote 13 down vote accepted

There's a distinction to be made between two notions: foliations and distributions.

A distribution is the data, at each point m of M, of a subspace of Tm(M). These subspaces are all of the same dimension (say r), and depend smoothly on the point m, which means that they are generated by r smooth vector fields.

A foliation is a partition of the manifold into (not necessarily closed) submanifolds, such that, locally, this partition looks like the standard decomposition of ℝn into translates of ℝd. Ok, there's a caveat in my description since a same leaf could come infinitely many often in the neighborhood of a given point m. Anyways... I'm assuming that you know what a foliation is.

Foliations of M form a subset of distributions on M.

The Frobenius integrability criterion (mentioned by Tom in him remark) states that a distribution D comes from a foliation iff for any vector fields v and w tangent to D, their Lie bracket is again tangent to D.

It turns out that that criterion is always satisfied for one-dimensional distributions, and so one-dimensional distributions are indeed in bijection with one-dimensional foliations. But that's no longer true for r ≥ 2.

The operation of taking orthogonal complement is a very good operation for distributions: it's always well defined, and the orthogonal complement of the orthogonal complement is the distribution you started with.

But the orthogonal complement of a foliation is typically only a distribution. The standard example that illustrates that situation is the vector field sin(z)d/dx + cos(z)d/dy on ℝ3. It defines a perfectly good foliation, but its orthogonal fails to satisfy the Frobenius integrability criterion, and therefore fails to be a foliation (in this particular case, it's a contact structure, another beautiful mathematical notion...).

Ah! You also wanted the foliation to be defined by the orbits of a group acting by isometries... That can be arranged: take the action of S1 on S3 given by the Hopf fibration. The orthogonal distribution is the standard contact structure on S3.

You also said that you wanted you Riemanninan manifold to be flat... In that case, you can take ℝ4=ℂ2 with its S1-action by complex multiplication. That example contains the above S3 example as an invariant submanifold, and therefore reproduces all its features.

share|cite|improve this answer
+1. This is really nicely written--it's an excellent example of how to give a helpful expository (as opposed to problem-solving) answer. – Martin M. W. Apr 25 '11 at 2:15
Thank you, exactly what I looked for. – Bruno Galvan Apr 25 '11 at 16:35

I cannot resist but mention a related concept, which in a sense generalizes the example you quote.

Let a Lie group $G$ act properly and isometrically on the complete Riemannian manifold $M$. The action is called polar if there exists a complete connected submanifold $\Sigma$ that meets all the orbits, and meets them always orthogonally. Such a submanifold $\Sigma$ is called a section. It is easy to see that a section must be totally geodesic. If an action admits a section which is flat in the induced metric, then this action is called hyperpolar. In the case of linear orthogonal actions (or representations), there is no distinction between polar and hyperpolar actions since the complete totally geodesic submanifolds of Euclidean space are its affine subspaces.

One example of polar representation which is very familiar from basic courses in linear algebra is the $SO(n)$-conjugation of $n\times n$ real symmetric matrices. It is well known that every symmetric matrix is orthogonally conjugate to a diagonal matrix, so here the section is given by the subspace of diagonal matrices. More generally,
the standard examples of polar representations are the isotropy representations of symmetric spaces. Conversely, Dadok has shown that these essentially exhaust all the examples.

The orbital foliation of a polar action has many remarkable geometric and topological properties. The story starts with Bott and Samelson in the 1950's and goes on. To mention a recent result, A. Lytchak and G. Thorbergsson have proven that the orbifold points of the orbit space of a proper and isometric action correspond precisely to the points of the manifold where the slice representation is polar (J. Differential Geom.85 (2010), 117-140). Polar actions have also been generalized in many directions, e.g. polar foliations, complex actions, Hilbert space. I leave you with two book references: (link to free book) and J. Berndt, S. Console and C. Olmos, Submanifolds and Holonomy, CRC/Chapman and Hall Research Notes Series in Mathematics 434 (2003), Boca Raton (more recent book).

*Edit:*Ah, I forgot to mention one detail. In your example, you removed the origin to get a second foliation by radial lines. For a polar action, the orbits form a (singular) foliation of course, but the sections form a foliation only if you remove the singular orbits. In fact, a point in the manifold lies in an orbit of lower dimension precisely if it is contained in more than one section.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.