It is well known that in geodesic coordinates we have $$ g_{ij}=\delta_{ij}-\frac{1}{3}\sum_{k,l}R_{ijkl}x^{k}x^{l}+O(|x|^{3}) $$ I have been trying to find a rigorous proof of it, but I cannot find a readable proof online (see this one, for example) or derive it myself. From the compatibility with metric we have $$ \partial_{i}g_{jk}=\sum_{\alpha}(\Gamma^{\alpha}_{ij}g_{ak}+\Gamma^{\alpha}_{ik}g_{aj}) $$ Since we know that $\nabla_{X}X=0$ for any vector from the origin, we have $\nabla_{i}\partial_{j}=0,\forall i,j$. This then implies all $\Gamma^{k}_{ij}=0$. So the first order derivative vanishes. However, it is not clear to me how to compute the second derivative. We have: $$ \partial_{l}\partial_{i}g_{jk}=\sum_{\alpha}(\partial_{l}\Gamma^{\alpha}_{ij}g_{\alpha k}+\partial_{l}\Gamma^{\alpha}_{ik}g_{\alpha j}) $$ This can be simplified further by noticing that in the origin we have $g_{ij}=\delta_{ij}$. Therefore the above sum is in fact $$ \partial_{l}\partial_{i}g_{jk}=\partial_{l}\Gamma^{k}_{ij}+\partial_{l}\Gamma^{j}_{ik} $$ And we would have proved the statement if we can show that $$ \frac{1}{2}(\partial_{l}\Gamma^{k}_{ij}+\partial_{l}\Gamma^{j}_{ik})=-\frac{1}{3}(R_{ijkl}+R_{ljki}) $$ By definition we have $$ R_{abcd}=g_{ae}R^{e}_{bcd}=\sum_{e}g_{ae}R^{e}_{bcd}=R^{a}_{bcd}=\partial_{c}\Gamma^{a}_{db}-\partial_{d}\Gamma^{a}_{bc} $$ Therefore we have $$ R_{ijkl}+R_{ljki}=\partial_{k}\Gamma^{i}_{jl}-\partial_{l}\Gamma^{i}_{jk}+\partial_{k}\Gamma^{l}_{ij}-\partial_{i}\Gamma^{l}_{jk} $$ But I do not know why we would have $$ \frac{1}{2}(\partial_{l}\Gamma^{k}_{ij}+\partial_{l}\Gamma^{j}_{ik})=-\frac{1}{3}(\partial_{k}\Gamma^{i}_{jl}-\partial_{l}\Gamma^{i}_{jk}+\partial_{k}\Gamma^{l}_{ij}-\partial_{i}\Gamma^{l}_{jk}) $$ I do not know if I missed something obvious like using the Bianchi identity.


I asked this many years ago as a beginning graduate student. One general method of solving it is using Jacobi fields, and it can be found for example in Schoen-Yau's book. If I recall correctly, it is listed as an execrise problem in Morgan-Tian's book. The first one proving it is apparently Riemann himself.

I do not know if my original method has any hope (it has been four years, so maybe not). Hopefully this helps other people.

  • $\begingroup$ I think you have to use the geodesic equation in your original method. $\endgroup$
    – ibnAbu
    Jul 4 '19 at 8:16
  • $\begingroup$ For anyone interested, I think do Carmo says (chapter 4 of "Riemannian Geometry") that the only formula Riemann proved in relation to the metric is the one appearing on page 9 of (this translation by Clifford) of his original paper. emis.de/classics/Riemann/WKCGeom.pdf In Riemann's formula the sectional curvature is constant so it's a lot less general than the one here. Do Carmo has a guided exercise to prove it in chapter 8. Also, I read that Riemann never worked with the the Riemann curvature tensor. That was introduced by Christoffel. $\endgroup$ Jun 23 '21 at 19:43

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