Vanish christoffel symbols implies $g_{ij,k}=0$

Let $(M,g)$ be a pseudo-riemann manifold and $(U,\psi=(x^1,\ldots,x^n))$ a local chart around some point $p$ in $M$. It is easy to show that if $\partial g_{ij}/\partial x^k=0$ in $p$ for all $i,j,k$ then $\Gamma_{ij}^k(p)=0$ for all $i,j,k$ because: $$\Gamma_{ij}^k =\frac{1}{2}\sum_{h=1}^n \Big(\frac{\partial g_{ih}}{\partial x^j}+\frac{\partial g_{jh}}{\partial x^i}-\frac{\partial g_{ij}}{\partial x^h}\Big) g^{kh}$$ is the converse also true?

• are you asking whether $g_{ij,k}=0$? Jun 19, 2016 at 19:07
• My question is: if $\Gamma_{ij}^k(p)=0$ then $g_{ij,k}(p)=0$? Jun 19, 2016 at 19:09
• This is not what you've written in the headline Jun 19, 2016 at 19:09
• Is M specifically a 4-manifold, or should the first line say $\psi = (x^1, \dots, x^n)$? Jun 19, 2016 at 20:54

The defining equation of the metric connection is $\nabla g = 0$, or in coordinates
$$\nabla_i g_{jk} = \partial_i g_{jk} - \Gamma_{ij}^l g_{lk} - \Gamma_{ik}^l g_{jl} = 0.$$
Evaluating this at $p$ where $\Gamma = 0$ gives $\partial g=0$ at $p$.
Alternatively, if you want to work with nothing but the coordinate expression for $\Gamma$ in terms of $\partial g$, it's easier to work with $$\gamma_{ijk} = g_{kl} \Gamma^l_{ij}=\frac12\left(-\partial_k g_{ij} + \partial_i g_{jk} + \partial_jg_{ki}\right).$$ Note that $$\gamma_{ijk} + \gamma_{kji} = \partial_jg_{ik}.$$
Since $\Gamma = 0$ at $p$ we must have $\gamma = 0$ there also; so we get $\partial_j g_{ik} = 0$.