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Riemann curvature tensor as second derivative of the metric

  1. Aug 23, 2008 #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.
  2. jcsd
  3. Aug 23, 2008 #2
    The expression I gave before is not quite complete, I should have written

    [itex](\partial_\lambda\partial_\sigma g_{\mu\nu})_p = \frac{2}{3} R_{\mu\lambda \nu\sigma}|_p[/itex]
  4. Aug 23, 2008 #3
    After thinking about this a little more I realize that the correct expression is in fact

    [itex](\partial_\lambda\partial_\sigma g_{\mu\nu})_p = \frac{1}{3} (R_{\mu\lambda \nu\sigma}+ R_{\mu\sigma\nu\lambda})|_p[/itex]

    which implies

    [itex]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[/itex]
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