Does the covariant derivative of the derivative of the metric tensor exist? if so, how do you evaluate it? $$\nabla_a (\partial_b g^{\mu v})=?$$ It would seem natural to assume that this cannot exist, stemming from the fact that the covariant derivative of the Christoffel symbol does not exist, and it is made up of the derivatives of the metric.

  • $\begingroup$ In non-flat space, the bracketed expression is not a tensor, hence the covariant derivative is not defined. $\endgroup$
    – DanielC
    Commented Jul 9, 2022 at 14:10

1 Answer 1


No. Covariant derivatives act on tensor fields of arbitrary rank, but the object $\partial g$ whose components in an arbitrary coordinate chart are given by $$\big(\partial g\big)_b^{\ \ \mu \nu}:= \partial_b g^{\mu\nu}$$ is not a tensor field (nor do the Christoffel symbols $\Gamma^i_{jk}$ constitute the components of a tensor, as you say).

  • $\begingroup$ What if you were to multiply the derivative of the metric by some contravariant vector with the same index b (the index of the derivative), and by that summing over b, would the resulting term satisfy a tensor field? if so, is taking the covariant derivative of that term valid? $\endgroup$
    – Tachyon
    Commented Jul 9, 2022 at 14:35
  • 2
    $\begingroup$ @Tachyon Have you tried working out the transformation properties of that object to see? That would be a useful exercise, and you’ll find the answer to be no, that is still not a tensor field so you cannot take its covariant derivative. $\endgroup$
    – J. Murray
    Commented Jul 9, 2022 at 15:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.