Divergence is coordinate independent I am working on a project about Spectral Geometry. One of the main goals of the project is to be able to define the Laplacian on a Riemannian Manifold.
As such, Let $(M,g)$ be a Riemannian Manifold, then the Laplacian is given by $\triangle_g= -\operatorname{div}_g\circ \operatorname{grad}_g$. Given a vector field X on M, we define the divergence as being
$$d(\iota_X \omega)=(\operatorname{div}_g X) \omega,$$
where $\omega$ is the volume form and $\iota_X \omega(Y_1, ..., Y_{n-1}) = \omega (X, Y_1, ..., Y_{n-1})$. I have already shown that given coordinates $x_1,...,x_n$ then we can write the divergence of $X=\sum_{i=1}^n X_i \frac{d}{dx_i}$ as:
$$\frac{1}{\sqrt{\det g}}\sum_{i=1}^n\frac{\partial}{\partial x_i}(X_i\sqrt{\det g })$$
I can clearly see that this definition $d(\iota_X \omega)=(\operatorname{div}_g X) \omega$ is independent of the choice of coordinates system. However, if instead of been given that first definition I was only given the last one, how can I prove the independence?
I have tried this:
Let $x_i$ and $y_i$ be two charts of $M$. Let  $g$ be the metric of $M$ with respect to the coordinates $x_i$ and $\widetilde{g}$ the one with respect to $y_i$. Then thinking in terms of matrices we have that $\widetilde{g}=J^t g J$, where $J$ is the jacobian associated with the change of variables from $x_i$ to $y_i$ so we get that:
$$\sqrt{\det(\widetilde{g})}=\sqrt{\det (g)} \times \det (J)$$
Let $X=\sum_{i=1}^n X_i \frac{d}{dx_i}$ be a vector field in $x_i$ coordinates then it can be written as $\sum_{i,a=1}^n X_i \frac{dy_a}{dx_i}\frac{d}{dy_a}$ so let $ X_i \frac{dy_a}{dx_i}= Y_i$.Then ,
\begin{align*}
    \frac{1}{\sqrt{\det \widetilde{g}}}\sum_{i=1}^n \frac{\partial}{\partial y_i}(Y_i\sqrt{\det \widetilde{g}})&=\frac{1}{\sqrt{\det g} \det J}\sum_{i=1}^n \frac{\partial}{\partial y_i}(Y_i\sqrt{\det g} \det(J))\\
    &=\frac{1}{\sqrt{\det g} \det J}\sum_{i,a=1}^n \frac{\partial x_a}{\partial y_i}\frac{\partial}{\partial x_a}(X_i\frac{\partial y_a}{\partial x_i}\sqrt{\det g}\det(J))\\
&=\frac{1}{\sqrt{\det g} \det J}\sum_{i,a=1}^n \det(J)\frac{\partial y_a}{\partial x_i} \frac{\partial x_a}{\partial y_i}\frac{\partial}{\partial x_a}(X_i\sqrt{\det g})+ \frac{1}{\sqrt{\det g} \det J}\sum_{i,a=1}^n X_i\sqrt{\det(g)} \frac{\partial x_a}{\partial y_i}\frac{\partial}{\partial x_a}(\frac{\partial y_a}{\partial x_i}\det J)\\
\end{align*}
Now I am stuck and not sure how to proceed. On the left side, I have what I want but on the right side, I don't know how to proceed. Any ideas?
 A: Before we start, let's use $a, b, \cdots$ for indices of $x's$ and $i, j, \cdots$ for indices of $y$'s. Also we use the Einstein summation convention, that repeated indices are summed over from $ 1$ to $n$.
\begin{align*}
    \frac{1}{\sqrt{\det \widetilde{g}}}& \frac{\partial}{\partial y^i}(Y^i\sqrt{\det \widetilde{g}})\\
& =\frac{1}{\sqrt{\det {g}} \det J} \frac{\partial x^b}{\partial y^i}\frac{\partial}{\partial x^b}(X^a \frac{\partial y^i}{\partial x^a} \sqrt{\det {g}} \det J) \\
&= \frac{1}{\sqrt{\det {g}} \det J} \frac{\partial x^b}{\partial y^i}\frac{\partial y^i}{\partial x^a} \frac{\partial}{\partial x^b}(X^a  \sqrt{\det {g}} \det J) \\ 
&\ \ \ \ + \frac{1}{\sqrt{\det {g}} \det J} \frac{\partial x^b}{\partial y^i}\frac{\partial^2 y^i}{\partial x^a \partial x^b} (X^a  \sqrt{\det {g}} \det J) \\
&= \frac{1}{\sqrt{\det {g}} \det J} \frac{\partial}{\partial x^a}(X^a  \sqrt{\det {g}} \det J) + \frac{\partial x^b}{\partial y^i}\frac{\partial^2 y^i}{\partial x^a \partial x^b} X^a  \\
&= \frac{1}{\sqrt{\det {g}}} \frac{\partial}{\partial x^a}(X^a  \sqrt{\det {g}}) \\
&\ \ \ + \frac{1}{ \det J} X^a \frac{\partial}{\partial x^a}( \det J) + \frac{\partial x^b}{\partial y^i}\frac{\partial^2 y^i}{\partial x^a \partial x^b} X^a \\
&= \frac{1}{\sqrt{\det {g}}} \frac{\partial}{\partial x^a}(X^a  \sqrt{\det {g}}) \\
&\ \ \ + X^a \left( \frac{\partial}{\partial x^a}(\log \det J) + \frac{\partial x^b}{\partial y^i}\frac{\partial^2 y^i}{\partial x^a \partial x^b} \right) \\
\end{align*}
Thus it suffices to check
$$ \frac{\partial}{\partial x^a}(\log \det J) + \frac{\partial x^b}{\partial y^i}\frac{\partial^2 y^i}{\partial x^a \partial x^b}  = 0.$$
We rewrite it as
$$ \frac{\partial}{\partial x^a}(\log \det J)= \frac{\partial}{\partial x^a}\left(\frac{\partial x^b}{\partial y^i}\right)\frac{\partial y^i}{ \partial x^b} $$
using
$$\frac{\partial x^b}{\partial y^i}\frac{\partial^2 y^i}{\partial x^a \partial x^b}  = - \frac{\partial}{\partial x^a}\left(\frac{\partial x^b}{\partial y^i}\right)\frac{\partial y^i}{ \partial x^b}$$
but this is just the famous Jacobi formula, since $J = (\frac{\partial x}{\partial y})$.
