# 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?

• It does not. It holds when $A^\mu$ is replaced with a general $p$-form (which is a totally antisymmetric $(p,0)$ tensor. Oct 1, 2018 at 14:40

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.
• I believe that for general Christoffel symbols and non-zero $T$, total antisymmetry is the only case where the extra sum vanishes.