Let $T_{ij}$ be a tensor on a Riemannian manifold. Is it true that $$ \nabla_l g^{ij}T_{ij} = g^{ij} \nabla_l T_{ij} $$ ?
I guess that this is true if I show that $$ \nabla_l g^{ij} = 0. $$ I started computing and I have: $$ \nabla_l g^{ij} = \partial_l (g^{ij} ) + \Gamma^{i}_{lk} g^{kj} + \Gamma^j_{lk} g^{ki} $$ but I don't know how to carry on the computation... Any hint or suggestion will be very appreciated!
It is a requirement and is equivalent to the analog of the leibniz rule.
For instance, consider \begin{equation} d g(v,u) = \nabla g(u, v)+ g(\nabla u, v)+ g(u, \nabla v) \hspace{1cm} \end{equation} $g$ is a $(0,2)$-tensor and $u,v$ are two vector fields. In index notation this is gien by
\begin{equation} {(g_{ij} v^i u^j)}_ {,\ k}= g_{ij, k} v^i u^j + g_{ij} v^i_{\ ,k} u^j + g_{ij,k} v^i u^j_{\ ,k} \end{equation}
The left hand side here is saying first contract and then differentiate, the right hand side is you first differentiate using the Leibniz rule for the product and then contract.