Hypersurfaces orthogonal to a cone - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T16:46:35Z http://mathoverflow.net/feeds/question/38064 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/38064/hypersurfaces-orthogonal-to-a-cone Hypersurfaces orthogonal to a cone Willie Wong 2010-09-08T14:45:49Z 2010-09-08T17:59:33Z <p>This question is somewhat related to <a href="http://mathoverflow.net/questions/37524/differential-inclusions-for-distributions" rel="nofollow">http://mathoverflow.net/questions/37524/differential-inclusions-for-distributions</a> but I am asking for something rather more specific, so I hope it is alright to leave this as a separate, new question. </p> <p>Let $M$ be a smooth manifold, then given a one-form $\omega$ on $M$, Frobenius' theorem gives a simple way to test whether the distribution defined by the kernel of $\omega$ in $TM$ is integrable. </p> <p>Now let $F\subset T^*M$ be a subset of the cotangent bundle such that the restriction of the canonical projection $\pi$ is onto, and such that $\pi^{-1}(p)$ is an open convex cone for every $p$ (I don't know whether this condition will affect the answer; I'm just including the information on what I know). I am interested in knowing conditions which will guarantee that there exists (locally) an integrable distribution in $F$. </p> <p>Trvially some restrictions must apply. One may imagine that $F$ is in some sense not continuous, such that for any $\omega_p\in F_p$, there does <em>not</em> exist any smooth extension $\omega$ to any neighborhood of $p$. An example would be taking $M$ to be $\mathbb{R}^2$, and $F_p$ to be the first quadrant for all $p\neq 0$, and the second quadrant for $p = 0$. </p> <p>So one specific question is: is this lack of freedom the only difficulty? Is the following statement true?</p> <blockquote> <p>Suppose $F$ has the property that, for any $p$, and any $\omega_p \in F_p$, there exists some smooth one-form $\omega$ that is a section of $F$, such that $\omega |_p = \omega_p$, and $d\omega |_p = 0$, then for any $q\in M$, we can find an open neighborhood $U$ of $q$ and some one-form $\eta$ over $U$ such that $\eta$ is a section of $F|_U$ and $d\eta = 0$.</p> </blockquote> <p>Feel free to ask for clarifications.</p> http://mathoverflow.net/questions/38064/hypersurfaces-orthogonal-to-a-cone/38070#38070 Answer by Michael for Hypersurfaces orthogonal to a cone Michael 2010-09-08T16:11:22Z 2010-09-08T17:59:33Z <p>In the case where <code>$F \subset T^*(M)$</code> is open the argument is roughly as follows: locally closed 1-forms on $M$ are in one to one correspondence with Lagrangian submanifolds of <code>$T^*(M)$</code> (with its canonical symplectic structure) which are transversal to the projection to $M$. Given a point $p\in F$ you can then choose a Lagrangian plane <code>$\Pi \subset T_p(T^*M)$</code> transversal to $\pi$ and a extend it to a Lagrangian submanifold through $p$. This will be still transversal and contained in $F$ after restricting to a sufficiently small portion.</p> <p>I'm a bit in a hurry at the moment, but you can ask me for more details or probably find them in any text on symplectic geometry.</p> http://mathoverflow.net/questions/38064/hypersurfaces-orthogonal-to-a-cone/38078#38078 Answer by Bill Thurston for Hypersurfaces orthogonal to a cone Bill Thurston 2010-09-08T17:10:50Z 2010-09-08T17:10:50Z <p>There are a global obstrutions in some cases when local solutions exist, even in the case that the cone is an open subset of the cotangent bundle. For example: let $M^3$ be the tangent line bundle to any Riemannian surface $N^2$. The Levi-Civita connection defines a 2-plane field, where curves tangent to these planes represent transportation of a tangent line that is parallel under the connection. This is not integrable except where the Gaussian curvature is 0. Let's relax the condition, allowing the open cone of motions of lines where they drift from parallelism by a rate say &lt; .1 degree per unit of arc length. By the h-principle, there are local solutions --- actually, you can get local solutions by parallel transport of lines along geodesics from a central point, up to a small radius.</p> <p>However: note that for any global solution, the leaves of the foliation are covering spaces of $N^2$. If the surface is a sphere, this would mean they project homeomorphically, so each leaf would give a global line field --- but we all know you can't comb a billiard ball, so this is impossible. It's also known that if the surface has negative Euler characteristic, it can't be done, by a theory developed by John Milnor and John Wood. A circle bundle over a surface other than $S^2$ admits a foliation transverse to the fibers if and only if the Euler class of the bundle has absolute value that does not exceed -Euler characteristic of base. For the tangent sphere bundle, this just barely works: some significant examples are the Anosov foliations of the geodesic flow for any metric of negative curvature. For the tangent line bundle, it doesn't work, since its Euler class is doubled from the tangent sphere bundle.</p> <p>Quite a bit more is known about existence and non-existence of foliations, but I think this answers your specific question.</p>