15
$\begingroup$

I am looking for a reference to the following simple statement; it must be classical. (It is easy to proof, but I want to have a reference.)

A metric space $X$ that corresponds to a Riemannian manifold $(M,g)$ completely determines the underlying smooth manifold $M$ and the metric tensor $g$.

Improve this question
$\endgroup$
0

2 Answers 2

Reset to default
21
$\begingroup$

Isn't this the Myers-Steenrod theorem? "If $(M,g)$ and $(N,h)$ are connected Riemannian manifolds and $f:(M,d_g)\to(N,d_h)$ is an isometry, then $f:(M,g)\to(N,h)$ is a smooth isometry"

Improve this answer
$\endgroup$
14
$\begingroup$

It was proven by Dick Palais.

MR0088000 (19,451a) Reviewed Palais, Richard S. On the differentiability of isometries. Proc. Amer. Math. Soc. 8 (1957), 805–807. 53.2X

MathSciNet

@article {MR88000, AUTHOR = {Palais, Richard S.}, TITLE = {On the differentiability of isometries}, JOURNAL = {Proc. Amer. Math. Soc.}, FJOURNAL = {Proceedings of the American Mathematical Society}, VOLUME = {8}, YEAR = {1957}, PAGES = {805--807}, ISSN = {0002-9939}, MRCLASS = {53.2X}, MRNUMBER = {88000}, MRREVIEWER = {K. Krickeberg}, DOI = {10.2307/2033302}, URL = {https://doi-org.ucc.idm.oclc.org/10.2307/2033302}, }

According to Palais, if I read his paper correctly, Myers and Steenrod proved the differentiability of isometries but Palais obtained an explicit description of smooth functions on the manifold from the metric geometry.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ A small precision: what Palais shows is that one can reconstruct canonically the riemannian metric from the distance. The theorem about isometries as stated by @Quarto Bendir is indeed due to Myers-Steenrod. $\endgroup$
    – abx
    Commented Apr 19, 2020 at 5:07
  • $\begingroup$ The theorems seem to be equivalent. $\endgroup$
    – Anton Petrunin
    Commented Apr 19, 2020 at 16:56

You must log in to answer this question.

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