Do elongated convex objects all have long simple geodesics? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T09:42:43Z http://mathoverflow.net/feeds/question/51759 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/51759/do-elongated-convex-objects-all-have-long-simple-geodesics Do elongated convex objects all have long simple geodesics? Joseph O'Rourke 2011-01-11T13:40:57Z 2011-12-26T13:30:31Z <p>Let $S$ be a closed convex surface, the boundary of a compact convex body in $\mathbb{R}^3$. I am interested in whether there are conditions on its shape that ensure that it supports a long, simple (non-self-crossing) geodesic. The <em>length</em> of a geodesic for my purposes is the longest distance you can travel along the geodesic before returning to your starting point. Some condition is necessary for the type of result I seek, for all the geodesics on a sphere have the same length.</p> <p>Define the <em>elongation</em> $L$ of $S$ as the largest height to diameter ratio, $h/d$, of a cylinder of height $h$ and diameter $d$ in which $S$ is tightly inscribed. By <em>tightly inscribed</em> I mean that $S$ touches the top, bottom, and sides of the cylinder in such a manner that neither the height nor diameter can be reduced.</p> <p>I could use a theorem of this type:</p> <blockquote> <p>If $S$ has elongation $L \ge k$, then there is a simple geodesic on $S$ of length $\ge f(k)$, where $f(k)$ is some increasing function of $k$, e.g., $c k$ for a constant $c > 0$.</p> </blockquote> <p>Perhaps such a theorem cannot exist. Or maybe a theorem of this ilk exists, but only with certain smoothness assumptions? There are always at least three simple closed geodesics on $S$, by a theorem of Lyusternik and Schnirelmann, but perhaps they might all be short?</p> <p>For an ellipsoid, the three simple closed geodesics follow the major and minor axes, and the longest of those satisfies the type of relationship I seek. (Elongation could as well be defined in terms of an enclosing ellipsoid rather than cylinder.) And a cylindrical $S$ supports a long spiral geodesic: <br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<img src="http://people.csail.mit.edu/~orourke/MathOverflow/BarberPole.jpg" alt="Barber Pole"><br /> Such spirals are exactly the type of geodesic I seek. Thanks for any ideas or pointers!</p> <p><b>Edit</b>. This may not add much, but here is how I view a long geodesic on a cylinder: starting at $a$, crossing the bottom in a segment $x x'$, crossing the top in $y y'$, and stopping at $b$ just before it is about to cross itself. <br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<img src="http://people.csail.mit.edu/~orourke/MathOverflow/CylinderGeodesic.jpg" alt="Cylinder"></p> http://mathoverflow.net/questions/51759/do-elongated-convex-objects-all-have-long-simple-geodesics/51760#51760 Answer by Anton Petrunin for Do elongated convex objects all have long simple geodesics? Anton Petrunin 2011-01-11T14:09:06Z 2011-01-11T16:36:50Z <p>Your estimates are not scale invariant, so I am trying to guess what you want from the picture. </p> <p>A closed geodesic cuts your surface into two discs. Both have geodesic as a boundary, positive curvature and area $\le$ than area of your original surface. If geodesic is long, then (with the intrinsic metric) these discs look almost like segments. It has to have curvature near $\pi$ in concentrated form near the ends.</p> <p>Thus if long geodesic exist then almost all curvature can be covered by 4 fingers on your surface...</p> <p>For example, </p> <ul> <li>you can not have it if Gauss curvature $\ge 1$. (In this case you can still have long shapes: say a doubling of a slice of unit shpere between meridians can be embedded into $\mathbb R^3$ as a convex surface, one can smooth singularities on the poles.)</li> <li>you can not have it on a polyhedral surface with more than 4 vertexes. If you have an arbitrary long simple geodesics on the surface of tetrahedral, the sum of angles around each vertex has to be $=\pi$.</li> </ul> http://mathoverflow.net/questions/51759/do-elongated-convex-objects-all-have-long-simple-geodesics/84313#84313 Answer by Igor Rivin for Do elongated convex objects all have long simple geodesics? Igor Rivin 2011-12-26T13:30:31Z 2011-12-26T13:30:31Z <p>There is a <a href="http://dl.dropbox.com/u/5188175/calabicao.pdf" rel="nofollow">highly relevant paper of Gene Calabi and J. Cao</a>, where they show that there always exists a geodesic of length at least twice the diameter (in the metric space sense) of your surface. I think this answers your question.</p>