Let $(M, g)$ be a Riemannian manifold endowed with the Riemannian connection, and $p$ be a point on $M$. It is known that $\exp_p:T_pM\to M$ takes a neighborhood of the origin of $T_pM$ diffeomorphically onto a neighborhood of $p$ in $M$. Any such neighborhood of $p$ in $M$ is called a normal neighborhood. Say $U$ be a normal neighborhood of $p$. Choose a linear map $E:\mathbf R^n\to T_pM$ which takes the standard basis of $\mathbf R^n$ to an orthonormal basis of $T_pM$. Then we have a a chart $\varphi=E^{-1}\circ \exp_p^{-1}:U\to \mathbf R^n$. Such a chart is called a normal coordinate chart.
It is also known that if $\gamma_v$ is the geodesic on $M$ starting at $p$ with initial velocity $v\in T_pM$, then $\varphi\circ \gamma_v(t)=(tv^1, \ldots, tv^n)$, where $v^i$ are the coefficients of $v$ in the coordinate $\varphi$.
Using this and using the equation of the geodesic, we can deduce that the Cristoffel symbols $\Gamma^k_{ij}(p)=0$ for all $i, j, k$.
QUESTION. I want to show that the first partial derivatives of the metric coefficients vanish at $p$, that is, $\frac{\partial}{\partial \varphi^k}g_{ij}=0$ at $p$ for all $i, j, k$.
Since
$$\sum_\ell \Gamma^\ell_{ij}g_{\ell k}= \frac{1}{2}\left(\frac{\partial}{\partial \varphi^i}g_{jk}+ \frac{\partial}{\partial \varphi^j}g_{ki}-
\frac{\partial}{\partial \varphi^k}g_{ij}\right)$$
we have the RHS is $0$ at $p$. Putting $j=k$ gives us that $\frac{\partial}{\partial \varphi^i}g_{jj}=0$ at $p$.
But I am not able to show that $\frac{\partial}{\partial \varphi^k}g_{ij}=0$ at $p$.
Best Answer
Using compatibility condition, \begin{equation} \frac{d}{dt} g(V,W) = g(D_tV,W) + g(V,D_tW) \end{equation} where $V,W$ are vector fields along any curve $\gamma$. Take $V=\partial_i, W=\partial_j$ and $\gamma$ is a geodesic through $p$, $\gamma_v (t) = (tv^1,...,tv^n)$ with $v^i = 0 \, (i\neq k)$ and $v^k = 1 $. The formula gives
\begin{equation} \partial_kg_{ij} = \Gamma^m_{ki} g_{mj} + \Gamma^m_{kj} g_{mi} \end{equation} Evaluate at $p$ gives $\partial_kg_{ij}(p)=0$.