Divergence computation using a geodesic frame

Question

I have no idea to start this problem, could anyone help me? thankyou very much guys!

I don´t know how to write in LaTex, so here is a image of the problem

https://i.stack.imgur.com/3vj7c.jpg

Thankyou very much!

Answer 1

Item (b) is an immediate corollary of (a) by taking $E_i = \partial/\partial x^i$ in $\Bbb R^n$. That said, we use orthonormal expansion: $${\rm grad}(f) = \sum_{i=1}^n \langle {\rm grad}(f),E_i\rangle E_i = \sum_{i=1}^n {\rm d}f(E_i)E_i = \sum_{i=1}^n E_i(f)E_i.$$For the divergence, we compute the trace of $\nabla X$ using the definition and orthonormal expansion again, as follows: $$\nabla X(E_j) = \nabla_{E_j}X = \sum_{i=1}^n \langle \nabla_{E_i}X,E_j\rangle E_j. $$But if $X = \sum_{k=1}^n f_kE_k$ we have that $$\nabla_{E_i}X = \nabla_{E_i}\left(\sum_{k=1}^n f_kE_k\right) = \sum_{k=1}^n E_i(f_k)E_k + \sum_{k=1}^n f_k \nabla_{E_i}E_k.$$Since we assume that $(E_i)$ is a geodesic frame, $(\nabla_{E_i}E_k)|_p = 0$ for all choices of $i$ and $k$, and so we have $$(\nabla X)_p(E_j|_p) = \sum_{i=1}^n \left\langle \sum_{k=1}^n (E_i)_p(f_k)E_k|_p, E_j|_p\right\rangle E_j|_p = \sum_{i=1}^n \sum_{k=1}^n (E_i)_p(f_k) \delta_{kj}E_j|_p = \sum_{i=1}^n (E_i)_p(f_j)E_j|_p.$$This means that $${\rm div}(X)(p) = \sum_{i=1}^n (E_i)_p(f_i),$$as wanted.