A question about Riemann curvature tensor and metric tensor

Question

The Riemann curvature tensor can be expressed as:

$$R^\rho{}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho{}_{\nu\sigma} - \partial_\nu\Gamma^\rho_{\mu\sigma} + \Gamma^\rho{}_{\mu\lambda}\Gamma^\lambda{}_{\nu\sigma} - \Gamma^\rho{}_{\nu\lambda}\Gamma^\lambda{}_{\mu\sigma}$$

where the $$\Gamma^{k}{}_{ij}$$ are the Christoffel symbols. $$\begin{align} \Gamma^m{}_{ij}&= g^{mk}\Gamma_{kij}\\[0.2em] & =\frac12\, g^{mk} \left( \frac{\partial}{\partial x^j} g_{ki} +\frac{\partial}{\partial x^i} g_{kj} -\frac{\partial}{\partial x^k} g_{ij} \right)\\ & \equiv\frac12\, g^{mk} \left( g_{ki,j} + g_{kj,i} - g_{ij,k} \right) \,. \end{align}$$ with $g_{ij}$ metric tensor of the manifold.

My question is:

given a manifold with metric tensor $g_{\mu\nu}$ we can calculate the Riemann tensor. But, given a $R^\rho{}_{\sigma\mu\nu}$, does exist only a $g_{\mu\nu}$ having that Riemann curvature tensor or there ar many metric tensors with the given $R^\rho{}_{\sigma\mu\nu}$? Thanks in advance.

Answer 1

A simple counterexample is a flat Euclidean space of any dimension. The Riemann curvature is uniformly zero. A diffeomorphism of the space back to itself stretching/shrinking/shearing the space can be thought of as a change of metric on the original space, but the curvature remains zero. This argument should generalize to any manifold with constant fixed curvature (n-spheres & hyperbolic geometries), but when the curvature is not constant an automorphism may not preserve the curvature at some points.