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.
Best Answer
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.