Let $(M,g)$ be a smooth Riemannian manifold. There exist a connection $∇ $ that is compatible with the Riemannian structure, and this connection is called the Levi-Civita connection of the Riemmannian metric. "Compatible" means that $∇g=0$. And as usual $g^{ij}$ are the coefficients of the dual metric, i.e. the entries of the inverse of the matrix $(g_{ij})$.
For the Christoffel symbols, I have$$\Gamma_{ij}^{k}=\frac{1}{2} g^{kl}\bigg(\frac { \partial g_{il} }{ \partial x^j }+\frac { \partial g_{lj} }{ \partial x^i }-\frac { \partial g_{ij} }{ \partial x^l }\bigg).$$ $∇g$ is sometimes written as $$∇g=g_{ij,k}dx^i \otimes dx^j \otimes dx^k,$$ where $g_{ij,k}=\frac { \partial g_{ij} }{ \partial x^k }-g_{lj}\Gamma^l_{ik}-g_{il}\Gamma^l_{jk}$.
I want to calculate $g^{ij}_{\,\,\,,k}$，Is it zero? Will someone be kind enough to give me some hints on this problem?Thank you very much!
$g^{il}g_{lj}=\delta^i_j$ is constant, so using Leibniz, $$ 0 = g^{il}{}_{,k} g_{lj} + g^{il}g_{lj,k} $$ Contracting with $g^{mj}$, $$ g^{im}{}_{,k} = - g^{mj}g^{il} g_{lj,k}, $$ and then you can insert your previous expression for $g_{lj,k}$ (and in particular, it's not zero). The covariant derivative $g^{im}{}_{;k}$ is zero, by essentially the same argument; the point is that both derivatives are set up to have the Leibniz property over contraction.