Does a chain rule for the covariant derivative exist so that we can evaluate an expression like $$\nabla_c\sqrt{t_{ab}}?$$

where $t_{ab}$ are tensor components?

More generally, how do we take the covariant derivative of a function of tensor components $f(t_{ab})$, i.e. $$\nabla_c f(t_{ab})?$$

  • 1
    $\begingroup$ $\sqrt{t_{ab}}$ are not components of a tensor so your expression is physically meaningless. $\endgroup$
    – G. Smith
    Mar 19, 2021 at 17:57
  • $\begingroup$ @G.Smith why exactly is it not a tensor $\endgroup$
    – Shashaank
    Mar 19, 2021 at 19:10
  • $\begingroup$ @Shashaank It doesn’t satisfy the transformation rule for a tensor with two covariant indices. $\endgroup$
    – G. Smith
    Mar 19, 2021 at 20:34
  • $\begingroup$ @G.Smith I'm thinking of taking the covariant derivative $\nabla_c \sqrt{g}$ where $g$ is the metric determinant. How can I use some sort of chain rule to take care of the square root? $\endgroup$
    – TaeNyFan
    Mar 20, 2021 at 11:15
  • $\begingroup$ Yes. The chain rule works for covariant derivatives. As far as I know, it works for any kind of derivative. $\endgroup$
    – G. Smith
    Mar 20, 2021 at 21:14

1 Answer 1


There is a chain rule if the object is a tensor formed by tensor products and contractions, for instance $$\nabla_c (A^{bde}B_{bdf})=B_{bdf}\nabla_c A^{bde}+A^{bde}\nabla_c B_{bdf}.$$ Something like $f(t_{ab})$ is in general not a tensor so the action of the connection coefficient part of the covariant derivative is undefined on it.


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