# curl$(\mathbf{F} \times \mathbf{G})$ with Einstein Summation Notation

I'm aware of another post on this vector identity at Proving vector calculus identity $\nabla \times (\mathbf a\times \mathbf b) =\cdots$ using Levi-Civita symbol, but I have a question on the derivation based on P47 of Source: http://www.unl.edu.ar/ceneha/uploads/Cartesian_tensors_Index_notation_&_summation_convention.pdf.

$$[\color{green}{\nabla} \color{brown}{\times} (\mathbf{F} \times \mathbf{G})]_i = \color{brown}{\epsilon_{ijk}}\color{green}{\partial_j}(\epsilon_{lmk}F_l G_m)$$ $$= \underbrace{\color{brown}{\epsilon_{ijk}}\epsilon_{klm}}_{\Large= { \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}}}\color{green}{\partial_j}F_l G_m$$ Next, since we're working with the $i$th component, change $l \rightarrow i \text{ & } j \rightarrow m$ in the first term and change $m \rightarrow i \text{ & } j \rightarrow l$ in the second term. $$= \color{red}{(\partial_mF_i)G_m} + F_i(\partial_mG_m) - (\partial_lF_l)G_i - F_l(\partial_lG_i)$$ $$= \color{red}{?} + F_i(\nabla \centerdot \mathbf{G}) - (\nabla \cdot \mathbf{F})G_i - (\mathbf{F} \cdot \nabla)G_i$$

$\Large{\text{Question:}}$ How do I simplify $\color{red}{(\partial_mF_i)G_m}$? How do I decide between
$1.(\partial_mF_i)G_m$
$2.(\partial_mG_m)F_i$
$3.(G_m\partial_m)F_i$ & other multiplicative rearrangements of these three objects...?

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Of course, (1) is no change. (2) would mean differentiating $G_m$ instead of $F_i$, so it's not equivalent. (3) is equivalent, as $F_i$ is being differentiated still. Remember, all these components are just functions, not vectors or anything. This is one of the "benefits" of index notation: everything commutes, more or less, with the caveat that partial derivatives still have to act on something, and usually by convention we take that they act on whatever is to their right.
You already saw that $F_\ell(\partial_\ell G_i) = (F \cdot \nabla )G_i$. What you have with $(\partial_m F_i) G_m$ is exactly a counterpart to this term, just with $F$ and $G$'s roles reversed. You can rearrange to get $G_m \partial_m F_i$, and no parentheses are necessary--again, these are functions, not vectors. If need be, write out the sums explicitly to verify this is legal.
Thank you for your response. Actually, since you referred to it, I don't understand why $$F_l(\partial_l G_i) = (\mathbf{F} \cdot \nabla)G_i.$$ On the LHS, $\partial_l$ acts on $G_i$ so $G_i$ is being differentiated. Based on your answer, wouldn't then $$\color{magenta}{F_l(\partial_l G_i) = F_l(\nabla \cdot \mathbf{G})}?$$ –  LePressentiment May 6 '13 at 20:18
The repeated index $\ell$ tells you that $F$ must somehow be dotted with $\nabla$, yet clearly $F$ is not being differentiated. This is satisfied by $(F \cdot \nabla) G_i$. Here, $F$ is dotted with $\nabla$, but $\nabla$ still acts on $G_i$. I think you're getting confused by the parentheses into thinking $\nabla$ is acting on $F$ instead. It's not. In some places, I've seen the use of overdots or other marks to denote what $\nabla$ acts upon, regardless of what is nearby it. You could write this then as $(F \cdot \dot\nabla ) \dot G_i$ to be unambiguous. –  Muphrid May 6 '13 at 21:29
Thank you very much for your second response. I know that a repeated subscript (ie $l$ here) implies summation of this subscript. Are you saying that if a function and a derivative share the same subscript, then they must be dotted together? Where does this rule come from? Unfortunately, I haven't seen it in my notes. –  LePressentiment May 7 '13 at 21:28
@LaPrevoyance The derivative is no different from any other vector in terms of its properties with respect to summation. $A_i B_i = A\cdot B$ for two vectors $A, B$. That's the general way of converting between dot products and index notation. You can put $F, \nabla$ in there instead and the rule is similarly valid. –  Muphrid May 7 '13 at 21:42