Questions on smoothness of Riemann metrics - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T11:08:59Z http://mathoverflow.net/feeds/question/62393 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/62393/questions-on-smoothness-of-riemann-metrics Questions on smoothness of Riemann metrics Andrew 2011-04-20T11:01:32Z 2011-04-20T16:53:16Z <p>I've heard assertions of the sort:</p> <ol> <li>Let there be a Riemann metric (not very smooth, say of class $C^1$ or $C^2$ or maybe $C$?) in a neighbourhood of a point on a manifold. Then it is possible to choose coordinates so that the metric is $C^\infty$ or even analytic in them.</li> <li>In case of 3-dimensional manifolds it is possible to choose such coordinates globally, so the manifold becomes a smooth one. In the case of higher dimensions $n\ge4$ it is not true.</li> </ol> <p>Are those assertions true? I've heard them some time ago and not sure I remember all the details. Is it a well-known thing? Are there some detailed references?</p> http://mathoverflow.net/questions/62393/questions-on-smoothness-of-riemann-metrics/62421#62421 Answer by Anton Petrunin for Questions on smoothness of Riemann metrics Anton Petrunin 2011-04-20T15:16:02Z 2011-04-20T15:46:00Z <ol> <li><p>NO. Given a Riemannian manifold, it might be possible to improve smoothness by changing atlas. It is proved by Shefel, that the atlas with harmonic functions as coordinates is the best. But, the obtained metric might be worse than $C^\infty$.</p></li> <li><p>There is no local-global issue here, harmonic atlas is defined locally and it is the best one globally. So you get problems starting with dimension 2.</p></li> </ol> http://mathoverflow.net/questions/62393/questions-on-smoothness-of-riemann-metrics/62431#62431 Answer by Vladimir S Matveev for Questions on smoothness of Riemann metrics Vladimir S Matveev 2011-04-20T16:18:34Z 2011-04-20T16:18:34Z <p>I confirm the Anton's answer (No, and the phenomenon is essentially local), but I suggest another explanation which works for C^1 2-dimensional metrics. </p> <p>We will look for a counterexample in the class of metrics such that they are C^2 everywhere except for some line, where they are C^1. Then, it is possible and relatively easy to cook an example such that the curvature of the metric is discontinuous at this special line; you can do it in the class of confomally flat metrics such that the conformal coefficient depends on one variable only and the line is where this variable is a constant. </p> <p>Since in order to determine the curvature of a metric you only need the distance function corresponding to this metric, and distance function does not depend on how smooth is your atlas, you can not make this metric smooth by the change of the atlas. </p> http://mathoverflow.net/questions/62393/questions-on-smoothness-of-riemann-metrics/62432#62432 Answer by Deane Yang for Questions on smoothness of Riemann metrics Deane Yang 2011-04-20T16:20:15Z 2011-04-20T16:20:15Z <p>If you combine the work of Jost-Karcher on almost linear co-ordinates with the work of DeTurck-Kazdan and Shefel on harmonic co-ordinates (I recommend a paper of Stefan Peters on a proof of the Gromov convergence theorem), you get the following:</p> <p>If there exist local co-ordinates in which a Riemannian metric $g$ is $C^1$ and has bounded sectional curvature, then there exist local (harmonic) co-ordinates in which the metric is $C^{1,\alpha}$ for every $\alpha > 0$. If, in addition to this, the covariant derivatives of the Ricci tensor up to order $k$ are locally bounded, then there exist local harmonic co-ordinates in which the metric is $C^{k+1,\alpha}$ for any $\alpha > 0$. If, in particular, the covariant derivatives of Ricci of all orders are bounded, then there exist local harmonic co-ordinates in which the metric is $C^\infty$.</p>