In arXiv:9712092 (I discovered this paper from Zurab's answer to this question on MO) the authors find a closed form expression for the metric in terms of the curvature, expressed in normal coordinates -- which can be expanded in powers of the coordinates to reproduce the Riemann Normal Coordinate Expansion.
In this post I want to re-derive this result in a more condensed way (and also fix an issue in their derivation concerning the use of the Riemann tensor in the vielbein basis instead of the coordinate basis). Question: Is the following proof correct?
Proof: Let $x^i$ be normal coordinates around a point $p$. Consider a coordinate point $y^i$ and the geodesic $\gamma^i(t):=t y^i$. Now $J_i(t):=t\partial_i$ is a Jacobi field for each $i$. It satisfies the Jacobi equation $$ \nabla^2_{\dot \gamma}J_i+R(J_i,\dot \gamma)\dot \gamma=0. $$ Consider the vielbeins $e_a$ (and dual vielbeins $e^a$) and expand the Jacobi fields in them: $J_i(t)=te_i^a e_a$.
At $p$, the metric is $g_{ij}=\delta_{ij}$. Therefore we can take the vielbeins at $p$ to be $e_a^i=\delta_a^i$. To obtain the vielbeins at an arbitrary point, $p'$, we simply parallel transport the vielbeins from $p$, along a radial path. Since the metric is preserved by parallel transport, we have $g(e_a,e_b)\rvert_{p'}=g(e_a,e_b)\rvert_p=\delta_{ab}$ as expected. This means we are setting $x^i\nabla_i e_a=0$ (in other words $x^i\omega_{ib}^a=0$ where $\omega$ is the spin connection -- you will find this as the '$SO(d)$ gauge' used in the paper).
This implies that $\nabla_{\dot\gamma}e_a=0$ (since $\dot\gamma$ is radial). Substituting $J_i(t)=te_i^a e_a$ into the Jacobi equation and using $\nabla_{\dot\gamma}e_a=0$ gives $$ \partial_t^2(te^a_i)=ty^ky^jR^l_{kji}e^a_l $$ Then using the fact that $\int_0^1 (1-t)\partial^2_t(tf(t))=f(1)-f(0)$ we have $$ e^a_i(y)=\delta^a_i+\int_0^1 ~dt~(1-t)ty^ky^jR^l_{kji}(ty)e^a_l(ty) $$ Iterating this gives $e^a_i$ as an expansion in $R^i_{jkl}$, which can then be used to construct the metric $g_{ij}=e^a_ie^a_j$.