# A question about Riemann curvature tensor and metric tensor

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 symobols. \begin{align} \Gamma^m{}_{ij}= g^{mk}\Gamma_{kij}\\ =\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) =\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.

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1) See M. Berger's wonderful "A panoramic view of Riemannian geometry" for a detailed discussion of your question. The short answer is that many metrics can share a curvature tensor. 2) Please never use the Christoffel symbols, but strive to express all calculations in coordinate invariant notation (for your own sanity's sake). There's a lovely discussion of the evolution of notations and concepts in Riemannian geometry in Vol. 2 of Spivak's opus, which is put to great use in Gallot, Hulin & Lafontaine's "Riemannian geometry", which is very good for a beginner. – Gunnar Þór Magnússon Apr 24 '12 at 15:34
This isn't how the align environment works; you need to use & to mark the spots where the alignment should take place. – joriki Apr 24 '12 at 17:10