The covariant derivative of a contravector field is given by: \begin{equation} D_{k} A^{i} \equiv A^{i}_{\parallel k} = A^{i}_{\mid k} + \Gamma^{i}_{kp} A^{p} \end{equation} With $A^{i}_{\mid k} = \partial_{k} A^{i}$. While the covariant derivative of a covector field is: \begin{equation} A_{i\parallel k} = A_{i\mid k} - \Gamma^{p}_{ik} A_{p} \end{equation} This of course makes perfect sense because the derivatives transform exactly like tensors of rank (1,1) and (0,2). However, since they are tensors, they should transform to each other, once the metric tensor $g$ is contracted with them. \begin{equation} A_{i \parallel k} = g_{ip} A^{p}_{\parallel k} \qquad A^{i}_{\parallel k} = g^{ip} A_{p\parallel k} \end{equation} This is where I struggle. I have tried it from bouth directions and failed but here is what I came up with. \begin{equation} A^{i}_{\parallel k} = g^{ip} A_{p\parallel k} = g^{ip} A_{p \mid k} - g^{ip} \Gamma^{b}_{pk} A_{b}= A^{i}_{\mid k} - g^{ip} \frac{g^{bd}}{2} \left(\frac{\partial g_{pd}}{\partial x^k} + \frac{\partial g_{kd}}{\partial x^p} - \frac{\partial g_{pk}}{\partial x^d} \right) A_b \end{equation} \begin{equation} =A^{i}_{\mid k} + \frac{g^{ip}}{2} \left(\frac{\partial g_{pk}}{\partial x^d} + \frac{\partial g_{pd}}{\partial x^k} - \frac{\partial g_{kd}}{\partial x^p} \right) A^d - g^{ip} \frac{\partial g_{pd}}{\partial x^k} A^d = A^i_{\mid k} + \Gamma^i_{kd} A^d - g^{ip} \frac{\partial g_{pd}}{\partial x^k} A^d \end{equation} So it is the last term that is troubling me. I would expect it to be zero but I just can not see why this should be the case.
Any help whould be appreciated.
The mistake is when you asserted
$$ g^{ip}A_{p|k} = A^{i}{}_{|k}$$
You should have
$$ A^i_{|k} = (g^{ip} A_p)_{|k} = g^{ip} A_{p|k} + (\partial_k g^{ip}) A_p $$
which would cancel out the final term that you are worrying about.