# Covariant derivative of a contravariant derivative

This is a pretty basic question, but I am having problems understanding the covariant derivative of contravariant vector.

My understanding is that

$$\nabla_{\mu}A^{\nu}=\partial_{\mu}A^{\nu}.\vec{e_{\nu}}+A^{\alpha}\Gamma_{\mu\alpha}^{\nu}.\vec{e_{\nu}}$$

So I decided to write it out longhand for a 3D case. The following is my longhand of the covariant derivative. $$(\frac{\partial A^{x}}{\partial x}+A^{\alpha}\Gamma_{0\alpha}^{0}).\vec{e_{0}}+(\frac{\partial A^{y}}{\partial x}+A^{\alpha}\Gamma_{0\alpha}^{1}).\vec{e_{1}}+(\frac{\partial A^{z}}{\partial x}+A^{\alpha}\Gamma_{0\alpha}^{2}).\vec{e_{2}}\\ (\frac{\partial A^{x}}{\partial y}+A^{\alpha}\Gamma_{1\alpha}^{0}).\vec{e_{0}}+(\frac{\partial A^{y}}{\partial y}+A^{\alpha}\Gamma_{1\alpha}^{1}).\vec{e_{1}}+(\frac{\partial A^{z}}{\partial y}+A^{\alpha}\Gamma_{1\alpha}^{2}).\vec{e_{2}}\\ (\frac{\partial A^{x}}{\partial z}+A^{\alpha}\Gamma_{2\alpha}^{0}).\vec{e_{0}}+(\frac{\partial A^{y}}{\partial z}+A^{\alpha}\Gamma_{2\alpha}^{1}).\vec{e_{1}}+(\frac{\partial A^{z}}{\partial z}+A^{\alpha}\Gamma_{2\alpha}^{2}).\vec{e_{2}}$$

Is this correct?. If not, I would be grateful if someone would give the correct formulae in [longhand please].

On the other hand, If it is correct. Would someone please explain the following. Some of these terms, e.g. $$(\frac{\partial A^{x}}{\partial y}+A^{\alpha}\Gamma_{1\alpha}^{0}).\vec{e_{0}}$$ ,are partial derivatives with respect to one direction, but the corresponding basis is in another direction. I would have expect them to be in the same direction,like a gradient of a scalar. For instance assuming we have a 3D orthogonal system, the second terms disappear and if we have two components Ay and Az go to zero, then maybe you should get a scalar-like result.

The quantities $$\nabla_{\mu}A^{\nu}=\partial_{\mu}A^{\nu}+A^{\alpha}\,\Gamma_{\mu\alpha}^{\nu}$$ are the components of the $$1\choose 1$$-tensor $$\tag{1} (\nabla_{\mu}A^{\nu})(\boldsymbol{\omega}^\mu\otimes \boldsymbol{e}_\nu).$$ Likewise, $$A^\nu$$ are the components of the vector $$\boldsymbol{A}=A^\mu\boldsymbol{e}_\nu\,.$$ If $$\varphi$$ is a scalar then $$\nabla_\mu \varphi=\partial_\mu\phi\,.$$ These are the components of the covector $$\boldsymbol{d}\phi=(\partial_\mu\varphi)\,\boldsymbol{\omega}^\mu\,.$$
• Your long formula is not correct. If we write out (1) we get $$\tag{1'} \left(\partial_\mu A^\nu+ A^0\,\Gamma^\nu_{\mu 0}+ A^1\,\Gamma^\nu_{\mu 1}+ A^2\,\Gamma^\nu_{\mu 2} \right)(\boldsymbol{\omega}^\mu\otimes \boldsymbol{e}_\nu)\,.$$
• Both, $$\nabla_\mu\varphi$$ and $$\nabla_\mu A^\nu$$ contain only the basis covectors $$\boldsymbol{\omega}^\mu$$ (no other directions), respectively $$\boldsymbol{e}^\nu$$ (ditto).
• However, $$\nabla_\mu A^\nu$$ contains all components $$A^0,A^1,A^2$$ unless the corresponding Christoffel symbol is zero.