Let $g_{ij}$ be a Riemannian metric in normal coordinates at $p$. Why is $\partial_k g_{ij}(p)=0$? Let $(M^n,g)$ be a Riemannian manifold, fix a point $p\in M$, let $e_1,\ldots,e_n$ be an orthonormal basis for $\mathrm{T}_p M$, let $$\exp_p:\mathrm{B}_{\mathrm{T}_p M}(0,\varepsilon)\to\mathrm{B}_M(p,\varepsilon)$$ be a diffeomorphism defining a normal neighborhood, and use the coordinates $e_1,\ldots,e_n$ on $\mathrm{B}_{\mathrm{T}_p M}(0,\varepsilon)$ to induce coordinates $x^1,\ldots,x^n$ on $M$. Then we have the following properties:


*

*$p = (0,\ldots,0)$ in coordinates

*Geodesics starting at $(0,\ldots,0)$ are of the form $\gamma(t) = t\,v$ for some $v\in\mathrm{T}_p M$

*$g_{ij}(0,\ldots,0) = \delta_{ij}$


I know how to prove these three properties, but I don't see how to easily compute $g_{ij}$ in all of $\mathrm{B}_{\mathrm{T}_p M}(0,\varepsilon)$. How would I show that $\partial_k g_{ij}=0$?

If $\gamma (t)=\exp_p\ te_i$ and $E_k$ is normal coordinate vector
field, then $ E_k(t)=(d\exp_p)_{te_i}\ e_k$, and $ (\gamma
'(t),E_k(t))=0$ by Gauss lemma
Hence $$0=\gamma'(\gamma',E_k)=(E_i,\nabla_{E_i} E_k) $$
Hence $$ 0=(E_j+E_k,\nabla_{E_j+E_k} E_m)
=(E_j,\nabla_{E_k}E_m)+(E_k,\nabla_{E_j}E_m) = E_m (E_k,E_j)$$
