Covariant derivative of a contravariant vector

[bolbteppa]

An easier way to remember these calculations is to include the basis vectors and to treat this as a basic calculus problem, then for a contravariant vector field you have

\begin{align}
\vec{A}_{;j} &= \partial_j (A^i \vec{e}_i) = A^i_{,j}\vec{e}_i + A^i \vec{e}_{i,j} = A^i_{,j} \vec{e}_i + A^i \vec{e}_{i,j} \vec{e}^k \cdot \vec{e}_k \\

&= A^i_{,j} \vec{e}_i + A^i (\vec{e}_{i,j} \cdot \vec{e}^k) \vec{e}_k = A^i_{,j} \vec{e}_i + A^k (\vec{e}_{k,j} \cdot \vec{e}^i) \vec{e}_i = A^i_{,j} \vec{e}_i + A^k \Gamma^i_{kj} \vec{e}_i \\

& = (A^i_{,j} + \Gamma^i_{kj} A^k) \vec{e}_i = A^i_{;j} \vec{e}_i
\end{align}

while for a covariant vector field we have

\begin{align}
\vec{A}_{;j} &= \partial_j (A_i \vec{e}^i) = A_{i,j}\vec{e}^i + A_i \vec{e}^i_{,j} = A_{i,j} \vec{e}^i + A_i \vec{e}^i_{,j} \vec{e}_k \cdot \vec{e}^k \\

&= A_{i,j} \vec{e}^i + A_i (\vec{e}^i_{,j} \cdot \vec{e}_k) \vec{e}^k = A_{i,j} \vec{e}^i - A_i (\vec{e}^i \cdot \vec{e}_{k,j}) \vec{e}^k = A_{i,j} \vec{e}^i - A_k (\vec{e}^k \cdot \vec{e}_{i,j}) \vec{e}^i \\

&= A_{i,j} \vec{e}^i - A_k \Gamma^k_{ij} \vec{e}^i = (A_{i,j} - \Gamma^k_{ij} A_k) \vec{e}_i = A_{i;j} \vec{e}_i

\end{align}

Where I used $$(\vec{e}_i \cdot \vec{e}^k)_{,j} = (\delta_i^k)_{,j} = 0 \rightarrow \vec{e}_{i,j} \cdot \vec{e}^k = - \vec{e}_i \cdot \vec{e}^k_{,j}$$

Note you can do both of these without randomly adding $$1 = \vec{e}_k \cdot \vec{e}^k$$ in the calculation, that just helps me get the indices right, you can ignore them and switch the indices yourself to get the answer, as is done here

http://www.physicspages.com/2013/02/16/covariant-derivative-and-connections/