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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 to me because the derivatives transform exactly like a tensor 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.

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You made your mistake in the first line. Between the second and the third equal signs you raised the index through a partial derivative, but $$ g^{ip} A_{p|k} \neq A^{i}_{|k} .$$ (It would be true for the covariant derivative.)

I think you'll find this is different by exactly the term that you're trying to cancel at the end, although you may need to simply further to see it explicitly.

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Thanks to @Brick I could finish my calculation using:

\begin{equation} (g^{ip} A_p)_{\mid k}=A^i_{\mid k} \end{equation} \begin{equation} g^{ip}_{\mid k} g_{pd} A^d + g^{ip} A_{p \mid k} \end{equation} Since $(g^{ip} g_{pd})_{\mid k} = 0$ we can conclude: \begin{equation} A^i_{\mid k} = -g_{pd\mid k} g^{ip} A^d + g^{ip} A_{p \mid k} \end{equation} Which solves my problem.

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