Generalized divergence of tensor in GR Although I've forgotten the proof (and cannot find it in, say, Carroll's book), the following formula holds for the covariant divergence in general relativity:
$$\nabla_{\mu} A^{\mu} = \frac{1}{\sqrt{|g|}} \partial_{\mu} \left( \sqrt{|g|} A^{\mu}\right),$$
where $g = \det(g_{\alpha\beta})$. I was wondering if this formula holds if $A^{\mu}$ is replaced with a general rank $(n,m)$ tensor 
$$T^{\mu \mu_1\mu_2 \cdots \mu_{n-1}}_{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\nu_1\cdots \nu_m}?$$ 
If not, could you point me to any references that have divergence formulas for higher rank tensors? 
 A: No, this does not hold in general for higher-rank tensors.  The general equation for the divergence of a completely contravariant tensor in terms of a coordinate derivative operator $\partial_\mu$ is
$$
\nabla_\mu T^{\mu \nu_1 \dots \nu_n} = \partial_\mu T^{\mu \nu_1 \dots \nu_n} + \Gamma^\mu {}_{\mu \rho} T^{\rho \nu_1 \dots \nu_n} + \sum_{i = 1}^n \Gamma^{\nu_i} {}_{\mu \rho} T^{\mu \nu_1 \dots \rho \dots \nu_n}.
$$
We also have the fact that 
$$
\Gamma^\mu {}_{\mu \rho} = \frac{1}{\sqrt{|g|}} \partial_\mu \sqrt{|g|}.
$$
Thus, 
$$
\nabla_\mu T^{\mu \nu_1 \dots \nu_n} = \frac{1}{\sqrt{|g|}} \partial_\mu \left( \sqrt{|g|} T^{\mu \nu_1 \dots \nu_n} \right) + \sum_{i = 1}^n \Gamma^{\nu_i} {}_{\mu \rho} T^{\mu \nu_1 \dots \rho \dots \nu_n}.
$$
This last sum will not vanish for a general tensor.  However, some or all of the terms may vanish for tensors with a particular symmetry structure.  In particular, if $T^{\mu \nu_1 \dots \nu_n}$ is antisymmetric in all of its indices, then any contraction of two of its indices with the symmetric indices of the Christoffel symbols automatically vanishes;  and thus the entire sum goes away.
For references as to how to take the covariant derivative of a general tensor, see Chapter 5 of Schutz's A First Course in General Relativity for a coordinate-based approach, or Chapter 3 of Wald's General Relativity for a more general approach.
