"A proof" that Riemann tensor is zero Riemann tensor is zero in flat space, and well, it is tensor. Thus we have the tensor equation R=0 which means that Riemann tensor is zero in all the coordiantes systems, which is completely a lie. Where is my mistake here?
 A: It's zero in all coordinate systems in that flat space, which is completely true. But if you're in a curved space, there is no coordinate system where $R=0$ to begin with.
A: Being a physicist I think I know where the questioner is coming from. If we can evaluate a tensor in a local inertial frame and write it in tensor form then we know  it in any reference frame. The questioner is saying: in a local inertial frame R=0, but this is a tensor equation therefore R is always zero. The problem with this argument is that the Riemann tensor, although a "local" property of space-time, is not "sufficiently local" to be evaluated by making the approximation of flat space-time (i.e. an inertial frame) around a point. Specifically, the Riemann tensor involves second-order derivatives of the metric tensor, while the local inertial frame is only a first-order order approximation. That is, it contains only information about the first-order derivatives of the metric tensor. So a local inertial frame can be used to evaluate tensors that depend only on first order derivatives of the metric tensor, but cannot be used to evaluate the Riemann tensor.
A: The Riemann tensor transforms under a coordinate transformation as $R'^{\sigma}_{\rho \mu \nu}= (S^{−1})^{\sigma}_{\alpha} S^{\beta}_{\rho}S^{\gamma}_{\mu}S^{\delta}_{\nu}R^{\alpha}_{\beta \gamma \delta}$ which means that the components in the new coordinate system are linear combinations of its components in the old coordinate system. 
So if the components are all zero in the old coordinate system, they'll also be zero in the new coordinate system. That's why you only need to prove they're zero in only one coordinate system to prove that the space is flat.
