# Divergence of a tensor

On pg.70 of Dalarsson's "Tensors, Relativity and Cosmology"

For a mixed tensor of contravariant order 2 and covariant order 1 $$(T^{mn}_{p,m})$$, the divergence with respect to m is defined as:$$T^{mn}_{p,m}=\frac{1}{\sqrt{g}}\frac{\partial}{\partial x^m}(\sqrt{g} T^{mn}_{p})$$(1)

which I thought is equivalent to $$T^{mn}_{p,m}=\frac{\partial T^{mn}_p}{\partial x^m}+\Gamma^m_{rm}T^{rn}_p$$(2)

But since $$T^{mn}_{p,m}=\frac{\partial T^{mn}_p}{\partial x^m}+\Gamma^m_{rm}T^{rn}_p+\Gamma^n_{rm}T^{mr}_p-\Gamma^r_{pm}T^{mn}_r$$ (3)

Doesn't (2) imply that the last two terms on the RHS of (3) vanish? I tried to express the last two Christoffel symbols on the RHS in terms of the metric tensors but they do not seem to cancel?

• Yup, that equation is just straight-up wrong. – Michael Seifert Jun 11 '19 at 13:07
• Does that book really use a comma to indicate a covariant derivative? – G. Smith Jun 11 '19 at 15:51
• @G. Smith No, the author prefers to use $D_m$ but I find it more efficient to use a comma instead. – Chern-Simons Jun 11 '19 at 17:09
• In the books and papers I’ve seen, a comma always indicates an ordinary partial derivative and a semicolon indicates a covariant derivative. – G. Smith Jun 11 '19 at 17:11

The formula $$\nabla_\mu T^\mu = \frac{1}{\sqrt g} \partial_\mu \sqrt g T^\mu$$ only holds for contravariant vectors. It is not true for higher order tensors, so your equations (1) and (2) are not valid.
• Note that if you contract the $n$ and $p$ indices in the OP's eq. (3), you do get the desired result. This is because the contracted tensor $T^{mn}_n$ is a contravariant vector, and contraction commutes with the covariant derivative. – Michael Seifert Jun 11 '19 at 13:01