# 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?

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. $$\nabla_\mu\equiv \nabla_{\partial_\mu}$$

• Thanks man! But.. by googling I found out that $\nabla_XT$ is actually the object $\nabla T$ somehow acting on $X$, i.e., $\nabla_XT=(\nabla T)(X)$. This is what's bugging me. Basically I have to "act" the LHS and RHS in the second equation on $X$. I'm not sure how $X$ makes it way through the brackets and tensor products to the derivative term. Specifically how does $(\nabla T\otimes\omega\otimes Y)(X)$ become $(\nabla_XT\otimes\omega\otimes Y)$? This might sound like a ridiculously naive question but I don't want to take anything for granted – Shirish Kulhari Aug 1 '20 at 13:33
• The fact that some texts write the $(X)$ after the $\nabla$ is just notational "$f(x)$" habit. What is meant is better written as $\nabla_X=X^\mu\nabla_\mu$. The $X$ is not being differentiated. It does get differetiated in a second derivative $\nabla_Y\nabla_X$ – mike stone Aug 1 '20 at 14:42

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 $$\nabla \rightarrow\nabla_X$$.

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.