- #1
jdstokes
- 523
- 1
It is a standard fact that at any point [itex]p[/itex] in a Riemannian space one can find coordinates such that [itex]\left.g_{\mu\nu}\right|_p = \eta_{\mu\nu}[/itex] and [itex]\left.\partial_\lambda g_{\mu\nu}\right|_p[/itex].
Consider the Taylor expansion of [itex]g_{\mu\nu}[/itex] about p in these coordinates:
[itex]g_{\mu\nu} = \eta_{\mu\nu} + \frac{1}{2!} (\partial_\lambda\partial_\sigma g_{\mu\nu})(x^\lambda - x_p^\lambda)(x^\sigma-x^\sigma_p) + \cdots[/itex].
The claim is that in fact [itex]R_{\mu\lambda \nu\sigma} = \partial_\lambda\partial_\sigma g_{\mu\nu}[/itex]. The problem is that I'm not sure the Riemann curvature tensor has these symmetries.
Consider the Taylor expansion of [itex]g_{\mu\nu}[/itex] about p in these coordinates:
[itex]g_{\mu\nu} = \eta_{\mu\nu} + \frac{1}{2!} (\partial_\lambda\partial_\sigma g_{\mu\nu})(x^\lambda - x_p^\lambda)(x^\sigma-x^\sigma_p) + \cdots[/itex].
The claim is that in fact [itex]R_{\mu\lambda \nu\sigma} = \partial_\lambda\partial_\sigma g_{\mu\nu}[/itex]. The problem is that I'm not sure the Riemann curvature tensor has these symmetries.