I recently came across a video by prof Fredrick Schuller on general relativity where he defines the leibniz rule to be,

$\nabla_X (T(\omega,Y))=\nabla_XT(\omega,Y)+T(\nabla_{X} \omega,Y)+T(\omega,\nabla_{X} Y)$

Where $X$ and $Y$ are vector fields, $\omega$ is a covector field and $T$ is a $(1,1)$ tensor. The rule can be generalised for $(p,q)$ tensors similarly.

I cannot find a way to show that it is equivalent to the leibniz rule expressed as $\nabla_X$ acting on tensor product of two tensor fields.

i.e. $\nabla_{X}(T \otimes S) = \nabla_XT\otimes S + T\otimes \nabla_XS$

How do I proceed to show the equivalence between the two?

  • 3
    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Sep 30, 2019 at 16:07
  • $\begingroup$ One ingredient is likely compatibility with contraction. (The first equation's lhs seems to be the covariant derivative wrt $X$ of the contraction of $T$ and $\omega$ and $Y$?) $\endgroup$
    – Emil
    Sep 30, 2019 at 16:51
  • $\begingroup$ It seems my wording was wrong commute with contraction seems to be more common. $\endgroup$
    – Emil
    Sep 30, 2019 at 17:00
  • $\begingroup$ In the lecture he doesn't talk about contraction, neither the commutativity of contraction with derivative. Perhaps this article could clarifying: en.wikipedia.org/wiki/Tensor_contraction?wprov=sfla1 $\endgroup$
    – nadapez
    Nov 19, 2020 at 18:13

2 Answers 2


The covariant derivative is defined to obey the Leibnitz rule. If the ${\bf e}_i$ are a vielbein basis then We define the action of $\nabla_X$ on any function $f(x)$ by
$$ \nabla_Xf= Xf = X^\mu \partial_\mu f, $$ and on the elements ${\bf e}_i$ of a vielbein basis by
$$ \nabla_X {\bf e}_i = {\bf e}_j {\omega^j}_{i\mu}X^\mu. $$ We extend to any other object by demanding that both linearity and Liebnitz rule hold. So, on a vector field $Y= Y^i {\bf e}_i$, we have $$ \nabla_X Y= (\nabla_X Y^i){\bf e}_i + Y^i (\nabla_X {\bf e}_i)\\ = (X^\mu\partial_\mu Y^i) {\bf e}_i+ Y^i ({\bf e}_j{\omega^j}_{i\mu}X^\mu)\\ = X^\mu (\partial_\mu Y^i + Y^j {\omega^i}_{j\mu}){\bf e}_i. $$ Note that the position-dependent numerical components $Y^i(x)$ of a vector are still just functions. We do the same for a tensor $$ \nabla_X (T^{ij}{\bf e}_i\otimes {\bf e}_j)= (\nabla_X T^{ij}){\bf e}_i\otimes {\bf e}_j+ T^{ij}(\nabla_X{\bf e}_i)\otimes {\bf e}_j+ T^{ij}{\bf e}_i\otimes (\nabla_X {\bf e}_j)\\ =X^\mu (\partial_\mu T^{ij}+ {\omega^i}_{k\mu} T^{kj} + {\omega^j}_{k\mu} T^{ik}){\bf e}_{i}\otimes {\bf e}_j $$ It should now be clear that for Liebnitz holds for any tensor product or contraction.


I would think of it like this, using the fact that covariant derivative commutes with contractions (use C for contraction) and Liebniz rule

$\nabla_X (T(\omega,Y)) =\nabla_X (C C ( T\otimes \omega \otimes Y)) = CC \nabla_X(T\otimes \omega \otimes Y) = CC((\nabla_X T)\otimes \omega \otimes Y+ T\otimes (\nabla_X\omega) \otimes Y+ T\otimes \omega \otimes( \nabla_X Y)) = \nabla_XT(\omega,Y)+T(\nabla_{X} \omega,Y)+T(\omega,\nabla_{X} Y)$

NOTE: one might have to define what slots C contracts first but I think this just works (maybe some care is needed if you use the clifford product or have antisymmetric tensors?)

  • $\begingroup$ Yes, but in the lecture he doesn't talk about that commutativity property. I wonder how that would be prooved $\endgroup$
    – nadapez
    Nov 19, 2020 at 18:17
  • 1
    $\begingroup$ This question says it commutes and quotes a source, maybe you could find something there: math.stackexchange.com/q/3665934 (I hope my reasoning wasn't wrong in this answer :S). Maybe this one too math.stackexchange.com/q/482645. $\endgroup$
    – Emil
    Nov 19, 2020 at 19:43

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