Tensor contraction and covariant derivative

Question

I have naive question about GR and Covariant derivative. You know \begin{align} \nabla_{\gamma} g_{\alpha \beta}=\nabla_{\gamma} g^{\alpha \beta}=0 \end{align}

And I would like to compute covariant derivative of $g_{\mu \nu}A^{\mu \nu}$, where $A^{\mu \nu}$ is a suitable tensor. \begin{align} \nabla_{\gamma} (g_{\mu \nu}A^{\mu \nu}) \end{align} Now, $g_{\mu \nu}A^{\mu \nu}$ is just scalar, it's possible to expect this answer $\partial_{\gamma} (g_{\mu \nu}A^{\mu \nu})$. This can be checked explicitly. \begin{align} \nabla_{\gamma} (g_{\mu \nu}A^{\mu \nu})&=g_{\mu \nu}\nabla_{\gamma} A^{\mu \nu}\\ &=g_{\mu \nu}(\partial_{\gamma} A^{\mu \nu}+\Gamma^{\mu}_{\ \lambda\gamma}A^{\lambda\nu}+\Gamma^{\nu}_{\ \lambda\gamma}A^{\mu\lambda})\\ &=g_{\mu \nu}\partial_{\gamma} A^{\mu \nu}+\Gamma_{\nu\lambda\gamma}A^{\lambda\nu}+\Gamma_{\mu \lambda\gamma}A^{\mu\lambda}\\ &=g_{\mu \nu}\partial_{\gamma} A^{\mu \nu}+2\Gamma_{\nu\lambda\gamma}A^{\lambda\nu}\\ &=g_{\mu \nu}\partial_{\gamma} A^{\mu \nu}+(\partial_{\gamma}g_{\nu\lambda}+\partial _{\lambda}g_{\nu \gamma}-\partial_{\nu}g_{\lambda\gamma })A^{\lambda\nu}\\ &=g_{\mu \nu}\partial_{\gamma} A^{\mu \nu}+\partial_{\gamma}g_{\nu\lambda}A^{\lambda\nu} \end{align} where I used the fact that $A^{\mu \nu}$ is a symmetric tensor because it is contracted with $g^{\mu \nu}$. However, when I tried to calculate it in another way, I could not. \begin{align} \nabla_{\gamma} (g^{\mu \nu}A_{\mu \nu})&=g^{\mu \nu}\nabla_{\gamma} A_{\mu \nu}\\ &=g^{\mu \nu}(\partial_{\gamma} A_{\mu \nu}-\Gamma^{\lambda}_{\ \mu\gamma}A_{\lambda\nu}-\Gamma^{\lambda}_{\ \nu\gamma}A_{\mu\lambda})\\ &=g^{\mu \nu}(\partial_{\gamma} A_{\mu \nu}-2\Gamma^{\lambda}_{\ \mu\gamma}A_{\lambda\nu})\\ &=g^{\mu \nu}(\partial_{\gamma} A_{\mu \nu}-g^{\lambda \tau}(\partial_{\gamma}g_{\tau \mu}+\partial_{\mu} g_{\tau \gamma}-\partial_{\tau} g_{\mu \gamma})A_{\lambda \nu})\\ &=g^{\mu \nu}\partial_{\gamma} A_{\mu \nu} -(\partial_{\gamma}g_{\tau \mu}+\partial_{\mu} g_{\tau \gamma}-\partial_{\tau} g_{\mu \gamma})A^{\tau \mu}\\ &=g^{\mu \nu}\partial_{\gamma} A_{\mu \nu}-\partial_{\gamma}g_{\tau \mu}A^{\tau \mu}\\ \end{align} However this is not $\partial_{\gamma} (g^{\mu \nu}A_{\mu \nu})$ in general. What can I do from here? Or is there a rule that the metric must have lower indexes in this case?

Answer 1

As you said, the contraction is a scalar so that $$ \begin{align} \nabla_{\gamma} (g^{\mu \nu} A_{\mu \nu}) &= \partial_{\gamma}(g^{\mu \nu}A_{\mu \nu}) \\ &= A_{\mu \nu} \, \partial_{\gamma}g^{\mu \nu} + g^{\mu \nu} \partial_{\gamma} A_{\mu \nu} \ \end{align} $$ and the same thing for $\nabla_{\gamma} (g_{\mu \nu} A^{\mu \nu})$. Your equations are correct, but the second term in your second calculation (with the minus sign) has $A^{\mu \nu} \partial_{\lambda} g_{\mu \nu}$ rather than $A_{\mu \nu} \partial_{\gamma}g^{\mu \nu}$, which is probably what you're looking for. The solution is that swapping them over gives a minus sign, i.e: $$ A_{\mu \nu} \partial_{\gamma}g^{\mu \nu} = - A^{\mu \nu} \partial_{\lambda} g_{\mu \nu} \ . $$ You can expand this out if you want to check this.