Questions on smoothness of Riemann metrics I've heard assertions of the sort:


*

*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.

*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.


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?
 A: 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.
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 conformally flat metrics such that the conformal coefficient depends on one variable only and the line is where this variable is a constant.
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.
A: 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:
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 $C^\infty$.
A: *

*NO. Given a Riemannian manifold, it might be possible to improve smoothness by changing atlas. The atlas with harmonic functions as coordinates is the best [proved by Samuil Shefel (1979) and rediscovered by Dennis DeTurck and Jerry Kazdan (1981)]. But, the obtained metric might be worse than $C^\infty$.


*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.
