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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?

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

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  • $\begingroup$ 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 $\endgroup$ – Shirish Kulhari Aug 1 at 13:33
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    $\begingroup$ 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 $\nabla_Y\nabla_X$ $\endgroup$ – mike stone Aug 1 at 14:42
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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 $\nabla \rightarrow\nabla_X$.

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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.

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