Manifold with no Finsler structure? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T18:07:26Z http://mathoverflow.net/feeds/question/90424 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90424/manifold-with-no-finsler-structure Manifold with no Finsler structure? ssquidd 2012-03-07T04:18:10Z 2012-03-07T08:42:04Z <p>Is there a good example for a smooth manifold to which one cannot give a Finsler structure in any meaningful way? Ideal the example should be of low dimension and not too bizarre.</p> http://mathoverflow.net/questions/90424/manifold-with-no-finsler-structure/90434#90434 Answer by Ben McKay for Manifold with no Finsler structure? Ben McKay 2012-03-07T08:42:04Z 2012-03-07T08:42:04Z <p>As Lee Mosher pointed out, every manifold admits a Riemannian metric, by a partition of unity argument. One can easily rescale any Riemannian metric by a positive function to give a complete Riemannian metric. Every Riemannian metric is a Finsler metric. Every Riemannian metric on any compact manifold can be rescaled to have constant scalar curvature (the Yamabe problem, proved by Aubin, Schoen and Yau). But the metric might not be very meaningful. There doesn't appear to be any natural choice of metric on an abstract manifold, say up to isometry, except in low dimension. It might be interesting to study Finsler metrics with some type of constant curvature, or some other natural class of Finsler metrics.</p>