10
$\begingroup$

I've heard assertions of the sort:

  1. 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.
  2. 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?

Improve this question
$\endgroup$
6
  • 2
    $\begingroup$ For #1, see: DeTurck, Dennis M.; Kazdan, Jerry L. Some regularity theorems in Riemannian geometry. Ann. Sci. École Norm. Sup. (4) 14 (1981), no. 3, 249–260. You need assumptions on the curvature tensor (and its covariant derivatives) if you want higher regularity. I don't know what you mean by #2. Could you explain why the three-dimensional sphere has global co-ordinates? $\endgroup$
    – Deane Yang
    Commented Apr 20, 2011 at 14:07
  • 2
    $\begingroup$ In fact, for #1, in the DeTurck-Kazdan paper you find a counterexample in the first paragraph. Note that in this case "changing coordinates" actually corresponds to changing atlas (as noted by Anton below). I wonder if for #1 you intend them to be Einstein manifolds? In which case the result is true using elliptic regularity. $\endgroup$
    – Willie Wong
    Commented Apr 20, 2011 at 16:21
  • $\begingroup$ On #2: if you believe that the best coordinates one can use is the harmonic ones, then in fact on any compact, closed manifold, you will not be able to extend the coordinates globally... $\endgroup$
    – Willie Wong
    Commented Apr 20, 2011 at 16:46
  • 1
    $\begingroup$ @Igor: you posted a link to the Georgia Tech proxy, which most of us cannot go through :-p. The link Igor meant to post is MR: ams.org/mathscinet-getitem?mr=MR2204038 article: dx.doi.org/10.1090/S0002-9947-06-04090-6 $\endgroup$
    – Willie Wong
    Commented Apr 20, 2011 at 17:34
  • 1
    $\begingroup$ A most recent discussion of these issues is the following paper [Taylor, Trans. Amer. Math. Soc. 358 (2006), 2415-2423] avaialble at ams.org/journals/tran/2006-358-06/S0002-9947-06-04090-6/… $\endgroup$
    – Igor Belegradek
    Commented Apr 20, 2011 at 18:21

3 Answers 3

Reset to default
14
$\begingroup$

  1. 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 $C^\infty$.

  2. 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.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Anton, what's the reference for Shefel? $\endgroup$
    – Deane Yang
    Commented Apr 20, 2011 at 15:21
  • $\begingroup$ @Deane: Shefel, S. Z. --- 1979 and 1982. both in Russian the second one is translated. $\endgroup$
    – Anton Petrunin
    Commented Apr 20, 2011 at 15:40
  • $\begingroup$ Anton, thanks. I'm still not sure which papers you're citing. Is it: "Smoothness of a conformal mapping of Riemannian spaces. (Russian) Sibirsk. Mat. Zh. 23 (1982), no. 1, 153–159, 222."? And how do his results compare to DeTurck and Kazdan? Did he prove the same results either earlier or independently? Or does he prove more? $\endgroup$
    – Deane Yang
    Commented Apr 20, 2011 at 15:54
  • 1
    $\begingroup$ @Deane: looking at the MR, the results seems to be roughly comparable. The main lemma mentioned in the MR is equivalent to theorem 2.1 of DeTurck-Kazdan. And it looks like from the review of '79 you get an a priori estimate from the connection coefficients back up to the metric: Theorem 2 controls the regularity of the conformal map by the regularity of the conformal factor, compare that to Theorem 3.4 of DeTurck-Kazdan. So I would guess "roughly the same result, slightly earlier, not communicated well to the 'west' for the obvious reasons." $\endgroup$
    – Willie Wong
    Commented Apr 20, 2011 at 16:41
  • 1
    $\begingroup$ I will make sure to cite Shefel from now on. $\endgroup$
    – Deane Yang
    Commented Apr 21, 2011 at 2:02
8
$\begingroup$

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.

Improve this answer
$\endgroup$
7
$\begingroup$

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 $C^\infty$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .