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 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 at 16:51
  • $\begingroup$ It seems my wording was wrong commute with contraction seems to be more common. $\endgroup$ – Emil Sep 30 at 17:00

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?)


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.


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