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




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.
