General Relativity professor told us that if the divergence of a tensor is zero, thence the tensor is symmetric.

How can one prove that?

I know that, for example,

$$ \nabla^\mu F_{\mu\nu} = g_{\nu\rho} \nabla_\mu F^{\mu\rho} = \frac{1}{\sqrt{g} } \partial_\mu \left( \sqrt{g} g^{\mu\alpha} g^{\rho\beta} F_{\alpha\beta} \right) $$

For a $(0,2)$ rank tensor. How to proceed? I'm still studying tensor calculus hence I am not a pro, and I often miswrite.


Either your professor is wrong, or you heard him/her wrong.

Divergence of a tensor places no constraint on the algebraic symmetries of a tensor.

Consider the simplest case where you work on Minkowski space. The tensor field $$ X = \left( \frac{\partial}{\partial x^1} \right) \otimes \left( \frac{\partial}{\partial t} \right) $$ is divergence free (in fact, it is parallel and its covariant derivative $\nabla X \equiv 0$). But $X$ is not symmetric.

  • $\begingroup$ My professor told that (I also written on the notebook). Probably he made an error, or he got confused. Anyway thank you!! $\endgroup$ – Les Adieux Apr 19 '18 at 13:56

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