Page 166 of this book says that

any geometry, no matter how curved, is locally flat, at each spatial point we can always construct an infinitesimal patch of a Cartesian coordinate system.

By question is what can be (or how to think about) the local Cartesian coordinates on the surface of an unit sphere? Since it is a Cartesian coordinate the metric tensor must be $\delta_{ij}$. If I use $(x,y,z)$ system, the metric tensor does bot become $\delta_{ij}$ because of the constraint $x^2+y^2+z^2=1$. In spherical polar Coordinates too, the metric is not $\delta_{ij}$.

  • 3
    $\begingroup$ Note the word "infinitesimal". The coordinates will not exactly be "on" the sphere but on the tangent plane to the sphere. You can think of them as generated by a choice of two perpendicular tangent vectors at the point of tangency. "Infinitesimally" tangent plane points will correspond to points of the sphere by exponential map but on any finite patch, no matter how small, sphere's curvature is non-zero, so induced coordinates are not flat. $\endgroup$
    – Conifold
    Commented May 8, 2017 at 19:18
  • $\begingroup$ Minor comment to the post (v1): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$
    – Qmechanic
    Commented Nov 25, 2017 at 15:32

1 Answer 1


Comments to the post (v1):

  1. By the word locally flat Ref. 1 seems to refer the existence of Riemann normal coordinates, i.e. one can arrange at a point that (i) the metric tensor is the Kronecker delta and that (ii) the first partial derivatives of the metric vanish.

  2. This result is not necessarily possible to extend to an open neighborhood because of curvature, cf. e.g. this and this Phys.SE posts.


  1. U. Leonhardt & T. Philbin, Geometry and Light: The Science of Invisibility, 2010; p. 166.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.