Transformations induced by geodesics of boundary - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T08:23:19Z http://mathoverflow.net/feeds/question/6165 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/6165/transformations-induced-by-geodesics-of-boundary Transformations induced by geodesics of boundary lemega 2009-11-19T20:25:20Z 2009-11-19T21:36:40Z <p>I have a general question in Riemannian geometry: Let M be a compact manifold and $\partial M \neq \emptyset$. Then shoot a geodesic from any boundary point perpendicularly into the interior of M. How can one prove it will end at boundary? If so, it induces a transformation of $\partial M$, does anyone know any result about this transformation? For example, one can ask rigidity property, i.e. if the transformation is an isometry, can one say anything about M?</p> http://mathoverflow.net/questions/6165/transformations-induced-by-geodesics-of-boundary/6167#6167 Answer by Joel Fine for Transformations induced by geodesics of boundary Joel Fine 2009-11-19T20:47:44Z 2009-11-19T20:47:44Z <p>It doesn't seem true to me that the geodesic will always return to the boundary. It may end up accumulating around a closed geodesic in the interior. </p> <p>For example, consider the hyperboloid of one sheet $$x^2+y^2-z^2=1$$ in $\mathbb{R}^3$. This is a surface of revolution and so one can talk of the angular momentum of a path with respect to the axis. For a geodesic this must be constant (Clairault's theorem). If a geodesic has momentum exactly one (in some well-chosen units!) it will accumulate on the "waist" of the surface, where $z=0$. </p> <p>Of course this is a non-compact example, but if a geodesic accumulates on the waist, it stays inside some compact region. So it is easy to imagine a compact surface with boundary which contains the relevant compact region and for which this accumulating geodesic meets the boundary at a right angle. This gives an example where the geodesic doesn't return to the boundary. (I think it's even possible to find an example of a metric on the disc of this sort.)</p> <p>(In case you're curious, a geodesic on the hyperboloid starting with $z>0$, pointing "down" and with momentum less than 1 will pass through the waist and end up asymptotic to a meridian at z = -infinity. If the momentum is bigger than 1 it will not reach the waist and swing back upwards, ending up asymptotic to a meridian at z = +infinity.)</p> http://mathoverflow.net/questions/6165/transformations-induced-by-geodesics-of-boundary/6169#6169 Answer by Dmitri for Transformations induced by geodesics of boundary Dmitri 2009-11-19T20:54:20Z 2009-11-19T21:13:43Z <p>In fact it is not true that the geodesic will shoot again the boundary. You can already construct such a contexample on a disc. First thing that you should do is to understand that if we have a hyperbolic annulus i.e. H^2/Z it has one closed geodesics and there are other (inifinite) geodesics that spriral and finally tend to the closed one. Such a geodesic is contained in a half of the cylinder. Now you can make a cut on the cylinder that will intersect the geondesic by the angle pi/ end through the rest. Finally replace the other infinite half cylinder by a disk.</p> <p>There are some cases when this map is a piescewise isomtetry -- there is a whole science about this topic, called interval exchange transformations. This happen for surfaces of any genus >1. </p> <p>It is not clear how much you can say about M if the map is well defined. For example you can take MxI -- a direct product. But in the case when the boundary of M is connetced I don't see for the moment any other example than just a standard disk D^n (say with metric of constant curvature)</p> http://mathoverflow.net/questions/6165/transformations-induced-by-geodesics-of-boundary/6176#6176 Answer by Richard Kent for Transformations induced by geodesics of boundary Richard Kent 2009-11-19T21:36:40Z 2009-11-19T21:36:40Z <p>You can also see this indirectly in some examples:</p> <p>Take a hyperbolic surface $S$ with a single geodesic boundary component $\gamma$. Choosing inward pointing normals to $\gamma$, you can lift $\gamma$ to the unit tangent bundle. If all paths perpendicular to $\partial S$ ended in $\partial S$, the geodesic flow (suitably interpreted) would produce a nontrivial homotopy from $\partial S$ into $\partial S$. This would produce a nontrivial conjugacy $t\gamma t^{-1} = \gamma^k$ in the fundamental group, which can't happen. </p>