Given a Riemannian manifold with a Riemannian connection, let $e_i$ be a moving frame, $\omega^i$ be its dual frame, $\omega_i^j$ be its connection 1-form. Here is a way to define 1-st and 2-nd partial derivative: $$ \begin{cases} f_i\omega^i:=\mathrm{d}f\\ f_{ij}\omega^j:=\mathrm{d}f_i-f_j\omega_i^j\end{cases}$$
My question is:
- Why do we define the 2-nd derivative in this way? Are there any intuition inside? (I think probably one intuition is that we can differentiate the both side, apply $\mathrm{\omega_i}=\omega_j^i\wedge \omega_j$ to get: $$ (\mathrm{d}f_i-f_j\omega_j^i)\wedge \omega^i=0$$ But I cannot see any further.)
Edit: From here we deduce that the definition in this way can ensure $f_{ij}=f_{ji}$. But I wonder if there is a more natural reason to define in this way.
- For example, if we just define $f_{ij}\omega^j:=\mathrm{d}f_i$, what is the problem? Does this definition leads to any contradiction?
Any hints or comments are welcomed. Thank you very much!
For the second question. If we apply $e_i$ to the first equation we obtain, for any smooth function $\varphi$ $$ \varphi \circ f_i= e_i (\varphi \circ f) $$ If now $f_{ij} \omega^j= df_i$ we have $$ \varphi \circ f_{ij}= e_j e_i(\varphi \circ f) $$ Notice that $f_{ij}=f_{ji}$ implies that $[e_i,e_j]=0$. In general moving frames doesn't need to be formed by commuting vector fields, so this is false.
I think the whole reason to define partial derivatives in that way is to assure that there is symmetry between the second partial derivatives. It looks pretty "natural" as a condition to impose to me, but I'm only guessing.