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Taking 3-D Minkowski spacetime line element in General Relativity:

$$ds^2=-c^2dt^2+dx^2+dy^2+dz^2, $$

when considering a change into spherical coordinates leads to:

$$ds^2=-c^2dt^2+dr^2+r^2\left(d\theta^2+\sin^2\theta\,d\phi^2\right).$$

In several books, it is said that this is still Euclidean flat-space time, for it is only a change of coordinates speaking about the same as in Euclidean plane... but my big inquiry is under what point of view is this still flat, since the Levi-Civita connection $\Gamma^{\alpha}_{\,\,\beta\lambda}$ for this new space-time is not zero for some components. Are these symbols equal to zero a necessary condition for giving flat space-time?

I have not computed the components of the Riemann tensor for the polar coordinates spacetime, yet. But it is easy to see that for Cartesian coordinates they are equal to zero. If they were nonzero, does this assume that the deviation of geodesics equal to zero is still obeyed? From since I can remember, if the components of the Riemann tensor $R^{\alpha}_{\,\,\beta\mu\nu}$ are all zero, you get deviation zero and you can talk about Euclidean, flat space-time. Also, I can remember that if the Ricci scalar $R=0$ if and only if flat space-time is given. Am I correct?

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    $\begingroup$ Curvature is given by the Riemann tensor, so you should try to calculate it. Christoffel's symbols are not 0 in polar coordinates on an Euclidean space either. You need a connection in usual polar coordinates. If a space is flat in some coordinates, it must be flat, but that doesn't mean your coordinates don't need a connection. $\endgroup$
    – FGSUZ
    Commented Jun 26, 2018 at 20:46
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    $\begingroup$ The curvature is derived from the metric. If you haven't changed the metric, you haven't changed the curvature. $\endgroup$
    – WillO
    Commented Jun 26, 2018 at 20:47
  • $\begingroup$ There are terms such as, Riemann flat, Ricci flat, and scalar flat, each meaning slightly different things (somewhat obvious by name). Just because Christoffel is non zero doesn't mean that the manifold is flat. $\endgroup$
    – user196418
    Commented Jun 26, 2018 at 21:00

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Under a coordinate change, the metric may change form, but it is fundamentally the same manifold you are dealing with, and curvature scalars are diffeomorphism invariants.

While $\Gamma^a_{bc} \neq 0$, Minkowski space in any set of coordinates has $R^a_{bcd} = 0$. To convince yourself without calculating, see a coordinate change as a relabelling of positions. Rather than a grid, you might use a spherical coordinate system, but the points you are labelling on the surface are not being moved. The distance between any two is still the same.

The notion of curvature has to be independent of any coordinate system, since that is something we impose on the manifold and is not an intrinsic property.

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