Different variations of covariant derivative product rule This is a follow-up question to the accepted answer to this question: Leibniz Rule for Covariant derivatives
The standard Leibniz rule for covariant derivatives is $$\nabla(T\otimes S)=\nabla T\otimes S+T\otimes\nabla S$$
so for $T\otimes\omega\otimes Y$ this would translate to $$\nabla(T\otimes\omega\otimes Y)=(\nabla T)\otimes(\omega\otimes Y)+T\otimes(\nabla\omega\otimes Y)+T\otimes(\omega\otimes\nabla Y).$$
My question is: given a vector field $X$, how do I get from the above that $$\nabla_X(T\otimes\omega\otimes Y)=(\nabla_X T\otimes\omega\otimes Y)+T\otimes\nabla_X\omega\otimes Y+T\otimes\omega\otimes\nabla_XY$$ as written in that answer?
 A: Are you just rearraging the backets? If so remember that
the temsor product is defined to be associative: $ (a\otimes b) \otimes c= a\otimes (b \otimes c)$, so we can write eiher form as simply $a\otimes b \otimes c$.
If you are referring to replacing $\nabla$ by $\nabla_X$ remember that $\nabla$ is always $\nabla_X$ for some $X$.  I.e. $\nabla_\mu\equiv \nabla_{\partial_\mu}$
A: Use that tensor product is associative, so $\nabla(T\otimes \omega \otimes Y)=\nabla[(T \otimes \omega ) \otimes Y]$
Thus you have the Leibniz rule $$\nabla(X\otimes Y)=\nabla(X)\otimes Y+ X\otimes \nabla(Y)$$
that gives you
$$ [\nabla(T\otimes\omega)]\otimes Y + T\otimes \omega \otimes \nabla Y$$
Using again in first term:
$$ \nabla T \otimes \omega \otimes Y+ T\otimes \nabla \omega \otimes Y+T\otimes \omega\otimes \nabla Y$$
Finally just replace $\nabla \rightarrow\nabla_X$.
A: In a chart $U_\alpha : M \rightarrow \mathbb{R}^n$, you have $\nabla_X = X^\mu \nabla_\mu$ so the result follows by patching on overlapping charts.
The last answer in the question cited gives all the details required to be honest.
