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I'm proving a tensor identity and I wonder if it is true that

$$ \nabla_\mu \left( A^\nu B_\nu \right) = \nabla_\mu\left( A^\nu \right) B_\nu + A^\nu \nabla_\mu\left( B_\nu \right) $$

where $A^\nu$ and $B_\nu$ are tensors and $\nabla_\mu$ is the covariant derivative?

It makes sense to me but I would like to confirm since the object here is a scalar due to the contraction and not a more "normal" tensor.

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    $\begingroup$ As far as I know, every kind of derivative in mathematics obeys the product rule. Every derivative is a derivation. $\endgroup$ – G. Smith Mar 3 at 1:51
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The covariant derivative obeys the product rule so yes, that does hold.

Also note that $\nabla_{\mu} (A^{\nu} B_{\nu}) = \partial_{\mu} (A^{\nu} B_{\nu})$, which is the correct covariant derivative for a scalar function - a type $(0,0)$ tensor. This can be seen by simply expanding the covariant derivatives in terms of partial derivatives and the connection, then noting the connection terms cancel: $$ \nabla_{\mu} (A^{\nu} B_{\nu}) = \nabla_{\mu}(A^{\nu}) B_{\nu} + A^{\nu} \nabla_{\mu} B_{\nu} \\ = \partial_{\mu}(A^{\nu})B_{\nu} + \Gamma_{\mu \lambda}^{\nu} A^{\lambda} B_{\nu} + A^{\nu} \partial_{\mu} B_{\nu} - A^{\nu} \Gamma_{\mu \nu}^{\lambda} B_{\lambda}\\ = \partial_{\mu}(A^{\nu}B_{\nu}) \ \ . $$

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