# Riemann curvature tensor as second derivative of the metric

1. Aug 23, 2008

### jdstokes

It is a standard fact that at any point $p$ in a Riemannian space one can find coordinates such that $\left.g_{\mu\nu}\right|_p = \eta_{\mu\nu}$ and $\left.\partial_\lambda g_{\mu\nu}\right|_p$.

Consider the Taylor expansion of $g_{\mu\nu}$ about p in these coordinates:

$g_{\mu\nu} = \eta_{\mu\nu} + \frac{1}{2!} (\partial_\lambda\partial_\sigma g_{\mu\nu})(x^\lambda - x_p^\lambda)(x^\sigma-x^\sigma_p) + \cdots$.

The claim is that in fact $R_{\mu\lambda \nu\sigma} = \partial_\lambda\partial_\sigma g_{\mu\nu}$. The problem is that I'm not sure the Riemann curvature tensor has these symmetries.

2. Aug 23, 2008

### jdstokes

The expression I gave before is not quite complete, I should have written

$(\partial_\lambda\partial_\sigma g_{\mu\nu})_p = \frac{2}{3} R_{\mu\lambda \nu\sigma}|_p$

3. Aug 23, 2008

### jdstokes

After thinking about this a little more I realize that the correct expression is in fact

$(\partial_\lambda\partial_\sigma g_{\mu\nu})_p = \frac{1}{3} (R_{\mu\lambda \nu\sigma}+ R_{\mu\sigma\nu\lambda})|_p$

which implies

$g_{\mu\nu} = \eta_{\mu\nu} + \frac{1}{3} (R_{\mu\lambda\nu\sigma})_p(x^\lambda - x_p^\lambda)(x^\sigma-x^\sigma_p) + \cdots$