General relativity algebraic manipulation help I'm having difficulty understanding a lot of the fundamentals behind the algebra of general relativity. I have a specific question I'm trying to understand but any pointers about how any of it works with index contraction and covariant derivatives and stuff would be amazing.
The question is as follows:
Consider the expression:
$$T^{\mu\nu} = F^{\mu\rho} F_\rho^\nu-\frac{1}{4}g^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}$$
Where $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$
satisfies
$$\partial^\mu F_{\mu\nu} = \eta^{\mu\rho}\partial_\rho F_{\mu\nu}$$
Show $T^{\mu\nu}$ is a energy-momentum tensor using the identity:
$$\nabla_\mu F_{\nu\rho}+\nabla_\nu F_{\rho\mu}+\nabla_\rho F_{\mu\nu} = 0$$
I gather that this is equivalent to asking for a proof of the identities:
$$ 1) T^{\mu\nu} = T^{\nu\mu} $$
$$ 2) \nabla_\mu T^{\mu\nu} = 0 $$
The answer writes off the first as obvious (which I don't see) and gives the following proof of the second:
$$ \nabla_\mu T^{\mu\nu} = \nabla_\mu (F^{\mu\rho} F_\rho^\nu-\frac{1}{4}g^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta})$$
$$ = (\nabla_\mu F^{\mu\rho}) F_\rho^\nu + F^{\mu\rho}(\nabla_\mu F_\rho^\nu)-\frac{1}{2}F_{\alpha\beta}(\nabla^\nu F^{\alpha\beta})$$
$$ = F_{\mu\rho}(\nabla^\mu F^{\nu\rho} - \frac{1}{2}\nabla^\nu F^{\mu\rho})$$
$$ = \frac{F_{\mu\rho}}{2}(\nabla^\mu F^{\nu\rho} - \nabla^\rho F^{\nu\mu} - \nabla^\nu F^{\mu\rho})$$
$$ = \frac{F_{\mu\rho}}{2}(\nabla^\mu F^{\nu\rho} + \nabla^\rho F^{\mu\nu} + \nabla^\nu F^{\rho\mu}) = 0$$
Any light anyone can shed onto any of the methods used to obtain the results here would be amazing. I understand to a certain level but I just really need some guidance on how to generally manipulate expressions such as this one.
 A: First of all, $F_\rho^\nu$ should be $F^\nu$$_\rho$. Since $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ is antisymmetric, you need to know which index has been lowered. $\partial^\mu F_{\mu\nu}=\eta^{\mu\rho}\partial_\rho F_{\mu\nu}$ should also be $\partial^\mu F_{\mu\nu}=g^{\mu\rho}\partial_\rho F_{\mu\nu}$ since you are considering curved spacetime.

*

*Since $g^{\mu\nu}=g^{\nu\mu}$, the second term of the energy-stress tensor is symmetric as well. For the first term, we simply have:
\begin{equation}
F^{\mu\rho}F_{\;\;\rho}^\nu
=g_{\rho\sigma}F^{\mu\rho}F^{\nu\sigma}
=F^{\nu\sigma}F_{\;\;\sigma}^\mu.
\end{equation}

*What basically happens here, is that the covariant derivation $\nabla$ satisfies the Leibniz rule and applying it to the covariant metric tensor results in zero:
\begin{equation}
\nabla_\rho g_{\mu\nu}
=\partial_\rho g_{\mu\nu}
-\Gamma_{\rho\mu}^\sigma g_{\sigma\nu}
-\Gamma_{\rho\nu}^\sigma g_{\mu\sigma}
=\partial_\rho g_{\mu\nu}
-\Gamma_{\nu\rho\mu}
-\Gamma_{\mu\rho\nu}=0.
\end{equation}
Using this result, applying it to the contravariant metric tensor also results in zero:
\begin{equation}
\nabla_\rho g^{\mu\nu}
=\delta_\lambda^\mu\nabla_\rho g^{\lambda\nu}
=g^{\kappa\mu}g_{\kappa\lambda}\nabla_\rho g^{\lambda\nu}
=g^{\kappa\mu}\nabla_\rho\left(g_{\kappa\lambda}g^{\lambda\nu}\right)
=g^{\kappa\mu}\nabla_\rho\delta_\kappa^\nu=0.
\end{equation}
Therefore you can just take metric tensors in and out of the covariant differentiation, which happens in the second term for example. The two terms of the Leibniz rule can be shown to be identical using this, so the $-\frac{1}{4}$ turns into a $-\frac{1}{2}$. The rest is just simple index manipulation using $A^\sigma B_\sigma=A_\sigma B^\sigma$.

