# Tensor contraction and covariant derivative

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?

• Note that $\partial_\gamma g_{\tau\mu} A^{\tau\mu} = - \partial_\gamma g^{\tau\mu} A_{\tau\mu}$ Commented Apr 29, 2021 at 19:50

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.