# Deriving $A^{\mu}_{;\nu}$ from $A_{\mu ; \nu}$

We have a covariant derivative of a covariant tensor: $$A_{\mu ; \nu} = A_{\mu , \nu} - \Gamma^{\alpha}_{\mu \nu} A_{\alpha}$$ The covariant derivative of a contravariant tensor is: $$A^{\mu}_{;\nu} = A^{\mu}_{,\nu} + \Gamma^{\mu}_{\nu \alpha} A^{\alpha}$$

I am trying to use $A^{\mu}_{;\nu} = (g^{\mu \sigma} A_{\sigma})_{;\nu}$ to derive the second equation, given the first.

My attempt is as follows: \begin{align} A^{\mu}_{;\nu} &= (g^{\mu \sigma} A_{\sigma})_{,\nu} - g^{\mu \sigma} \Gamma^{\alpha}_{\sigma \nu}A_{\alpha} \\ &= A^{\mu}_{,\nu} - \Gamma^{\mu \alpha}_{\nu}A_{\alpha} \\ &= A^{\mu}_{,\nu} - (g^{\alpha \epsilon}\Gamma^{\mu}_{\nu \epsilon})A_{\alpha} \\ &= A^{\mu}_{,\nu} - \Gamma^{\mu}_{\nu \epsilon}(g^{\alpha \epsilon}A_{\alpha}) \\ &= A^{\mu}_{,\nu} - \Gamma^{\mu}_{\nu \epsilon}A^{\epsilon} \end{align} Relabelling $\epsilon$ as $\alpha$ we have: $$A^{\mu}_{;\nu}= A^{\mu}_{,\nu} - \Gamma^{\mu}_{\nu \alpha}A^{\alpha}$$

Clearly there is an error in my method here, as I have a minus where I ought to have a plus. Have I missed a step, or am I going about this the completely wrong way?

Thanks, Sean.

• Is the Christoffel symbol symmetric in the upper two indices? – GodotMisogi May 13 '16 at 16:06
• This is the first that I've ever encountered a Christoffel with two up indices. I'm assuming that if it is antisymmetric in the upper indices, then the sign would change between my 2nd and 3rd line of working? – Vielbein May 13 '16 at 16:15
• Covariant derivative of the metric tensor is exactly 0, so you can freely move it in or out the derivative expression. – Alexander May 13 '16 at 16:16

Explicitly working with the components of $g$ and $\Gamma$ can be messy. Here is a nicer way to derive the coordinate expression for the covariant derivative of a vector (or any tensor) without raising or lowering indices.
Consider vector fields $A^\mu, B_\mu$. The covariant derivative of a scalar is the same as the coordinate derivative (in any coordinate system), so $$\nabla_\nu(A^\mu B_\mu)=\partial_\nu(A^\mu B_\mu)= (\partial_\nu A^\mu)B_\mu+A^\mu\partial_\nu B_\mu \ .$$ On the other hand using Leibniz's rule for $\nabla$ and your expression for $\nabla_\nu B_\mu$ $$\nabla_\nu(A^\mu B_\mu)=(\nabla_\nu A^\mu)B_\mu+A^\mu\nabla_\nu B_\mu =(\nabla_\nu A^\mu)B_\mu+A^\mu\partial_\nu B_\mu-A^\mu \Gamma^\alpha_{\mu\nu}B_\alpha \ .$$ Taking the difference of the two equations and relabelling the dumb indices in the Christoffel symbol term we obtain $$B_\mu(\partial_\nu A^\mu+ \Gamma^\mu_{\sigma\nu}A^\sigma-\nabla_\nu A^\mu)=0.$$ Since this must be true for any one-form $B_\mu$ the term in brackets vanishes, and the components of the covariant derivative of a vector in some coordinates are given by
$$\nabla_\nu A^\mu=\partial_\nu A^\mu+ \Gamma^\mu_{\sigma\nu}A^\sigma \ ,$$
$$\nabla_\mu A^\nu = g^{\nu\alpha} \nabla_\mu A_\alpha = g^{\nu\alpha} ( \partial_\mu A_\alpha - \Gamma^\lambda_{\mu\alpha} A_\lambda ) = g^{\nu\alpha} \partial_\mu ( g_{\alpha\beta} A^\beta ) - g^{\nu\alpha} \Gamma^\lambda_{\mu\alpha} g_{\lambda\beta} A^\beta$$ This simplifies to $$\nabla_\mu A^\nu = \partial_\mu A^\nu + ( g^{\nu\alpha} \partial_\mu g_{\alpha\beta} - g^{\nu\alpha} \Gamma^\lambda_{\mu\alpha} g_{\lambda\beta} ) A^\beta$$ Finally, note that $$g^{\nu\alpha} \partial_\mu g_{\alpha\beta} - g^{\nu\alpha} g_{\lambda\beta}\Gamma^\lambda_{\mu\alpha} = g^{\nu\alpha} \partial_\mu g_{\alpha\beta} - \frac{1}{2} g^{\nu\alpha} g_{\lambda\beta} g^{\lambda \rho} ( \partial_\mu g_{\rho\alpha} + \partial_\alpha g_{\rho\mu} - \partial_\rho g_{\mu\alpha} ) = g^{\nu\alpha} \partial_\mu g_{\alpha\beta} - \frac{1}{2} g^{\nu\alpha} ( \partial_\mu g_{\beta\alpha} + \partial_\alpha g_{\beta\mu} - \partial_\beta g_{\mu\alpha} ) = \frac{1}{2} g^{\nu\alpha} ( \partial_\mu g_{\beta\alpha} + \partial_\beta g_{\mu\alpha} - \partial_\alpha g_{\beta\mu} ) = \Gamma^\nu_{\mu\beta}$$ Thus, we find $$\nabla_\mu A^\nu = \partial_\mu A^\nu + \Gamma^\nu_{\mu\beta} A^\beta$$