TITLE: Divergence of a smooth vector field
QUESTION [2 upvotes]: I was studying about the divergence of a smooth vector field in the book "Calculus of variations and harmonic maps"by Urakawa. For a smooth vector field $X$, the divergence is defined by $div(X)(p) := g(e_i,\nabla_{e_i}X)(p) ; p \in M$ wher $\{e_i\}_{i=1}^n$ is a local orthonormal frame field.
Further, it was written that the definition does not depend on the choice of $\{e_i\}$but i am not able to  prove that  $g(e_i,\nabla_{e_i}X)(p)= g(f_i,\nabla_{f_i}X)(p)$ if $\{e_i\}$ and $\{f_i\}$ are two local orthonormal frame fields.Can someone tell me how to prove this?
Thanks!

REPLY [1 votes]: Let $a_{ij}$ be a matrix such that $f_i = a_{ij} e_j$. Let both $\{e_i\}$ are $\{f_j\}$ are orthonormal basis, we have $\sum_{i=1}^n a_{ij}a_{ik} = \delta_{jk}$ (That is, $A = (a_{ij})\in O(n)$). thus 
$$g(f_i , \nabla_{f_j} X) = a_{ij}a_{ik}\ g(e_j, \nabla_{e_k} X) = \delta_{jk}\ g(e_j, \nabla_{e_k} X) = g(e_j , \nabla_{e_j}X)$$
The answer given by Cook is good (and are just the same), but I think this answer should also be recorded as these arguments (writing out as $a_{ij}$) show up all the time.
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