Orthogonal foliations - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T16:12:06Z http://mathoverflow.net/feeds/question/62855 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/62855/orthogonal-foliations Orthogonal foliations Bruno Galvan 2011-04-24T19:42:37Z 2011-06-10T03:20:19Z <p>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.</p> <p>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)?</p> http://mathoverflow.net/questions/62855/orthogonal-foliations/62867#62867 Answer by AndrĂ© Henriques for Orthogonal foliations AndrĂ© Henriques 2011-04-24T22:21:42Z 2011-04-25T00:00:22Z <p>There's a distinction to be made between two notions: <i>foliations</i> and <i>distributions</i>.</p> <p>A <b>distribution</b> is the data, at each point <i>m</i> of <i>M</i>, of a subspace of <i>T<sub>m</sub></i>(<i>M</i>). These subspaces are all of the same dimension (say <i>r</i>), and depend smoothly on the point <i>m</i>, which means that they are generated by <i>r</i> smooth vector fields.</p> <p>A <b>foliation</b> is a partition of the manifold into (not necessarily closed) submanifolds, such that, locally, this partition looks like the standard decomposition of &#8477;<sup><i>n</i></sup> into translates of &#8477;<sup><i>d</i></sup>. Ok, there's a caveat in my description since a same leaf could come infinitely many often in the neighborhood of a given point <i>m</i>. Anyways... I'm assuming that you know what a foliation is.</p> <blockquote> <p>Foliations of <i>M</i> form a subset of distributions on <i>M</i>.</p> <p>The <b>Frobenius integrability criterion</b> (mentioned by Tom in him remark) states that a distribution <i>D</i> comes from a foliation iff for any vector fields <i>v</i> and <i>w</i> tangent to <i>D</i>, their <a href="http://en.wikipedia.org/wiki/Lie_bracket_of_vector_fields" rel="nofollow">Lie bracket</a> is again tangent to <i>D</i>.</p> <p>It turns out that that criterion is <i>always</i> 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 <i>r</i> &ge; 2.</p> </blockquote> <p>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.</p> <p>But the orthogonal complement of a foliation is typically only a distribution. The standard example that illustrates that situation is the vector field sin(<i>z</i>)<i>d/dx</i> + cos(<i>z</i>)<i>d/dy</i> on &#8477;<sup>3</sup>. 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 <a href="http://en.wikipedia.org/wiki/Contact_geometry" rel="nofollow">contact structure</a>, another beautiful mathematical notion...).</p> <hr> <p>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 <i>S</i><sup>1</sup> on <i>S</i><sup>3</sup> given by the Hopf fibration. The orthogonal distribution is the standard contact structure on <i>S</i><sup>3</sup>.</p> <p>You also said that you wanted you Riemanninan manifold to be flat... In that case, you can take &#8477;<sup>4</sup>=&#8450;<sup>2</sup> with its <i>S</i><sup>1</sup>-action by complex multiplication. That example contains the above <i>S</i><sup>3</sup> example as an invariant submanifold, and therefore reproduces all its features.</p> http://mathoverflow.net/questions/62855/orthogonal-foliations/67405#67405 Answer by Claudio Gorodski for Orthogonal foliations Claudio Gorodski 2011-06-10T03:13:58Z 2011-06-10T03:20:19Z <p>I cannot resist but mention a related concept, which in a sense generalizes the example you quote. </p> <p>Let a Lie group $G$ act properly and isometrically on the complete Riemannian manifold $M$. The action is called <strong>polar</strong> 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 <strong>section</strong>. 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 <strong>hyperpolar</strong>. 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. </p> <p>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,<br> the standard examples of polar representations are the isotropy representations of symmetric spaces. Conversely, Dadok has shown that these essentially exhaust all the examples. </p> <p>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: <a href="http://vmm.math.uci.edu/CriticalPointTheory.pdf" rel="nofollow">http://vmm.math.uci.edu/CriticalPointTheory.pdf</a> (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).</p> <p>*<em>Edit:</em>*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. </p>