I would like to ask if there is a way to know how to find out if a space is flat or curved given a metric that could describe a flat space in curvilinear coordinates or just curved space.

For example, given the metric how can I differentiate between a flat 3D space with the metric given in spherical coordinates and the curved surface of a sphere. Wouldn't these two cases give the same non-vanishing Riemann curvature tensor components?

P.S. I am only just a beginner on general relativity.

  • $\begingroup$ It might be instructive (if tedious) to actually calculate the Riemann tensor in the two cases you mention. You will find that, even in spherical coordinates, every component of the Riemann tensor for flat 3D space will vanish. This is not so for the sphere. This shouldn't be surprising --- it's generally possible to 'slice up' a flat space into a stack of curved (hyper)surfaces (in this case spheres of radius $r$). Just because these surfaces are themselves curved, doesn't mean the whole space is! $\endgroup$
    – gj255
    Jan 2, 2017 at 21:10
  • $\begingroup$ @gj255 This is a pretty good insight, thanks! $\endgroup$
    – user99651
    Jan 2, 2017 at 21:15

1 Answer 1


These two metrics are not the same. The first metric is $$ds^2 = dr^2 + r^2 d\Omega^2$$ while the second metric is $$ds^2 = r_0^2 d\Omega^2$$ where $r_0$ is a constant, the radius of the sphere. These are different quantities; the spaces involved don't even have the same number of dimensions. The Riemann curvature of the former metric is zero, while the curvature of the latter is not.

You're correct to conclude that the first metric should have zero curvature, since it's related to Cartesian coordinates by a coordinate change and the curvature is a tensor. But this doesn't say anything about the second metric.

You might think the second case should be the same, because the sphere can be embedded into flat space. The geometrical difference is that vectors on the sphere must lie tangent to the sphere. That means that parallel transport on the sphere is not the same as parallel transport in the embedding space, since in the former case, we have to project the vector back down to the sphere after every step.


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