This is a follow-up question to the one I made here. On the wiki page, the divergence of a vector field $X$, denoted $\nabla\cdot X$, is defined as the function satisfying $\left(\nabla\cdot X\right)\text{vol}_n=L_X\text{vol}_n$. The page gives $\nabla\cdot X=\frac{1}{\sqrt{\vert g\vert}}\partial_i\left(\sqrt{\vert g\vert}X^i\right)$, where $X^i$ is the $i^{\text{th}}$ component of the vector field $X$.
If I evaluate $\text{vol}_n$ at $\left(\partial_1,\dots,\partial_n\right)$ then $\text{vol}_n\left(\partial_1,\dots,\partial_n\right)=\sqrt{\vert g\vert}$. This leaves $\left(\nabla\cdot X\right)\text{vol}_n=\partial_i\left(\sqrt{\vert g\vert}X^i\right)$.
On the other hand, if I use the fact that the $L_X\left(f\right)=X\left(f\right)$ for smooth functions, then $L_X\left(\text{vol}_n\left(\partial_1,\dots,\partial_n\right)\right)=L_X\left(\sqrt{\vert g\vert}\right)=X\left(\sqrt{\vert g\vert}\right)=X^i\partial_i\left(\sqrt{\vert g\vert}\right)$.
Clearly something has gone wrong. Can anyone help?
Best Answer
Almost all of your statements are correct. Especially is the LHS of $(\nabla \cdot X) \mathrm{vol}_n = L_X \mathrm{vol}_n$ evaluated at $(\partial_1, \dots, \partial_n)$ equal to $\partial_i(\sqrt{|g|} X_i)$. (But take care, although this is probably what you meant, you wrote another (wrong) equality.)
The RHS of $(\nabla \cdot X) \mathrm{vol}_n = L_X \mathrm{vol}_n$ evaluated at $(\partial_1, \dots, \partial_n)$ equals $$(L_X \mathrm{vol}_n) (\partial_1, \dots, \partial_n)) $$$$= X(\mathrm{vol}_n (\partial_1, \dots, \partial_n)) - \mathrm{vol}_n(L_X \partial_1, \partial_2, \dots, \partial_n) - \dots - \mathrm{vol}_n(\partial_1, \dots, L_X \partial_n)$$
Substituting your second statement and $L_X \partial_k = [X^i \partial_i, \partial_k] = X^i[\partial_i, \partial_k] - \partial_k(X^i) \partial_i = - \partial_k(X^i) \partial_i $ in gives: $$(L_X \mathrm{vol}_n) (\partial_1, \dots, \partial_n)) = X^i \partial_i (\sqrt{|g|}) + \partial_i (X^i) (\sqrt{|g|}) = \partial_i (X^i \sqrt{|g|}) $$
where the last equality comes from the product rule.