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According to the answer to this post, the Christoffel symbols in Riemann normal coordinates are approximated by

$$\Gamma^{k}_{ij}(x)~\sim~\frac{1}{2} R^k{}_{ilj}(x_0) \xi^l \tag{5.10}$$

which came from eq. (5.10) in D. Friedan and P. Windey, Supersymmetric Derivation of the Atiyah-Singer Index and the Chiral Anomaly, Nucl.Phys. B235 (1984) 395.

But using the standard formula

$$g_{ij}= \delta_{ij}-\frac{1}{3}R_{ijkl}\xi^k\xi^l $$

we obtain that

$$\Gamma^{k}_{ij} =-\frac{1}{3}(R^k{}_{ijl}+R^k{}_{jil})\xi^l $$

Please, let me know what is the correct expression.

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In Riemann Normal Coordinates; you can take advantage of the special symmetry \begin{equation} g_{ab, cd} = g_{cd, ab} \end{equation} which allows you to express the curvature (very simply) as \begin{equation} R_{\mu \nu \alpha \beta} = g_{\mu \beta, \alpha \nu} - g_{\mu \alpha, \nu \beta} \end{equation} If you transpose the first two indices of $R_{mnab}$ you reverse the sign of the quantity, and likewise for the last two indices. Also, if we swap the first and last pairs of indices we leave it unchanged. OfThis rationale leads to skew symmetry on three indices as; \begin{equation} R_{abcd} + R_{acdb} = -R_{abdc} = R_{cdab} \end{equation} Leading to \begin{equation} R_{abcd}+R_{acdb} +R_{abdc} =0. \end{equation} I suspect these relations will help you in your quest.

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  • $\begingroup$ Many thanks for your comment but unfortunately I do not see how to conciliate the two expressions for the Christoffel symbols in Riemann normal coordinates. $\endgroup$ – Juan Ospina Nov 10 '14 at 16:06
  • $\begingroup$ OK, no problem; using the metric tensor $g_{ab}$ you can raise an index on the curvature relations at the bottom of my hint and once you have re-written the summation of the skew-symmetric version you can plug them into your last finding; simplify and you should see some interesting results. $\endgroup$ – Autolatry Nov 10 '14 at 16:17
  • $\begingroup$ Dear Autolatry, please excuse me but I do not understand that you are trying to explain me. Please could you perform explicitly your computations? Many thanks. $\endgroup$ – Juan Ospina Nov 10 '14 at 17:13
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The formula you have for the metric is not quite right. The curvature tensor is skew in the first pair of indices and the last pair but your metric has it being symmetric.

Noneltheless, I agree with you calculation of the Christoffel symbol. I note that our formula for ${\Gamma^\lambda}_{\mu\nu}$ is symmetric in the last two indices as befits a torsion-free connection, while the "$\Gamma$" with the -1/2 is not obviously symmetric in i j, but would be skew symmetric if the k index were lowered. I suspect therefore that it is the "spin connection" rather than the Cristoffel symbol. I naively thought that these quntities would coincide in normal co-ordinates, but perhaps not. I will check!

Yes I have checked. The vielbeins get quadratic corrections, and this makes the spin connection urn out to be $$ {\omega^b}_{a\mu}= -\frac 12 R_{ba\mu\tau}x^\tau + O(|x|^2). $$ Thus the claimed Christoffel symbol is in fact the vierlbein spin connection.

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