Why $\nabla \omega = e^i \otimes \nabla _{e_i}\omega $?

Question

For understanding the formal adjoint of div, I read the book, but I don't know how to get $$ \nabla \omega = e^i \otimes \nabla _{e_i}\omega \tag{1} $$ In my view, there is $$ \nabla \omega(e_i, e_j,...)=(\nabla_{e_i}\omega)(e_j,...) = e_i(\omega(e_j,...))-\omega(\nabla_{e_i}e_j,...) -... \tag{2} $$ I can't get (1) from (2). I feel that I am on a wrong way, seemly, there are some things I didn't learn.

Answer 1

Note that in the equation (1) we have $$\nabla \omega = \color{red}{e^i} \otimes \nabla _{\color{blue}{e_i}}\omega\tag{$\star$}\label{a}$$ that is $e^i$s are covectors dual of ONB vectors $e_i$s. And if we want to calculate $\nabla \omega$ at $e_2$ for example we have: $$\nabla \omega(e_2,- )=(\nabla_{e_2}\omega)(-)$$

and from \eqref{a} we have (note that $e^2(e_2)=1$ and $e^2(e_3)=0.$) $$(\nabla \omega)(e_2,- ) = (\color{red}{e^i} \otimes \nabla _{\color{blue}{e_i}}\omega)(e_2,- )=1\otimes (\nabla _{\color{blue}{e_2}}\omega)(-)$$

In short the eq \eqref{a} just says that if you want to compute $\nabla\omega$ at $e_k$, roughly just look at the $k-$th place of first factor of $e^i \otimes \nabla _{e_i}\omega$.

If you evaluate both of equation that you want to prove they are equal then you will see that they are really equal.