From convex geometry to contact topology - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T08:34:43Z http://mathoverflow.net/feeds/question/116356 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116356/from-convex-geometry-to-contact-topology From convex geometry to contact topology alvarezpaiva 2012-12-14T09:54:44Z 2012-12-14T09:54:44Z <p>Here is a problem in contact topology that was suggested by Petya's answer to <a href="http://mathoverflow.net/questions/116200/a-problem-on-convex-geometry" rel="nofollow">this mathoverflow question</a> of mine.</p> <p>Let $S^* \mathbb{R}^n$ be the space of cooriented contact elements of $\mathbb{R}^n$. I will think of contact elements as pairs of point-cooriented hyperplane, where the point lies on the hyperplane. <strong>It will then make sense to say that two elements of $S^* \mathbb{R}^n$ are parallel.</strong></p> <p><strong>Question.</strong> Let $i: S^{n-1} \rightarrow S^* \mathbb{R}^n$ be a Legendrian embedding that is Legendrian isotopic to the manifold of all cooriented hyperplanes passing through a point. Does there necessarily exist a point $x \in S^{n-1}$ such that the contact elements $i(x)$ and $i(-x)$ are parallel?</p> <p><em>Remark.</em> Because Petya's proof is a simple application of critical point theory for the support function and the support function is just a simple instance of a generating function for a Legendrian submanifold, I'm guessing off the top of my head that his proof extends. The condition of being Legendrian isotopic to the manifold of all cooriented hyperplanes passing through a point garantees the existence of a generating function quadratic at infinity.</p>