From Chern: Lectures on Differential Geometry, page 147
Chern claimed that a torsion-free connection is completely determined locally by the curvature tensor. To show that he considered a geodesic normal neighborhood at a fixed point $O$. He choose a natural frame at $O$, then parallel transport it along the radial lines to form a frame field $\{ e_{i}\}$ in the neighborhood. If we denote the dual frame field by $\{\theta^{i}\}$ such that $\theta^{i}(e_{j})=\delta_{ij}$, and let the direction vector of the radial line to be $\alpha^{i}_{0}$ at the origin, then Chern claims that $$ \theta^{i}\equiv a^{i}dt \pmod{d\alpha^{k}}, \theta^{j}_{i}\equiv 0 \pmod{d\alpha^{k}}, a^{k}=t\alpha^{k}_{0} $$ because the frame field is parallel along the geodesic curve $a^{i}t$. Here $\theta^{j}_{i}$ is the connection matrix. However, it is not clear to me how to derive it. It is clear that we can write the frame field by $$ X=x^{i}\frac{\partial}{\partial u_{i}}, x_{i}=ta^{i}_{0} $$ Now substitute this into the equation $DX=0$ we get $$ \frac{dx^{i}}{dt}+x^{j}\Gamma^{i}_{jk}\frac{du^{k}}{dt}=0\rightarrow a^{i}_{0}+ta^{j}_{0}\Gamma^{i}_{jk}a^{k}_{0}=0\rightarrow a^{i}_{0}=0? $$ But I do not see how this help me to get $\theta^{i}\equiv a^{i}dt \pmod{d\alpha^{k}}$. I assume it should be something either self-evident or trivial. One way I thought about deriving it is to regard $\{\theta^{i}\}$ to be a covector field, thus we can use the equation $D\theta^{i}=-\sum^{q}_{\alpha=1}\theta^{i}_{j}\theta^{j}=0$, but it does not give me the result I wanted.