In the context of vector fields in $\mathbb{R}^3$ with the usual metric, it is the well known operator giving a function from a vector field $X$ by the expression
$$ \operatorname{div} X=\sum_{i=1}^3\frac{\partial X^{i}}{\partial x^{i}}, $$with its usual meaning. In physical terms, the divergence of a vector field is the extent to which the vector field flux behaves like a source at a given point. It is a local measure of its "outgoingness" – the extent to which there are more of the field vectors exiting an infinitesimal region of space than entering it. A point at which the flux is outgoing has positive divergence, and is often called a "source" of the field. A point at which the flux is directed inward has negative divergence, and is often called a "sink" of the field.
It can be generalized to a vector field $X$ in any dimension and with any Riemannian metric (indeed we only need to highlight a volume form in this space) in the following way. If we perform the Lie derivative of the Riemannian volume form we obtain another volume form. The coefficient of proportionality is the divergence. That is, if the distinguished volume form is $\Omega$ then
$$ \mathcal{L}_X \Omega=\mbox{div}_{\Omega}(X) \Omega $$Equivalently, $d(X\,\lrcorner\,\Omega)=\mbox{div}(X)\Omega$ (using Cartan formula and that $\Omega$ is closed). Moreover, also for Cartan formula we have that for any smooth function $f$
$$ \mathcal{L}_X (f\Omega)=d(X\,\lrcorner\,f\Omega)=d(fX\,\lrcorner\,\Omega)= $$ $$ =\mathcal{L}_{fX}(\Omega) $$And therefore, by one of the formulas for Lie derivative, exterior derivatives, bracket, interior product (the number 2)
$$ \mbox{div}(fX)=f\mbox{div}(X)+X(f) $$since $\mathcal{L}_{fX}(\Omega)=f\mathcal{L}_X \Omega+df\wedge X\,\lrcorner\,\Omega=f\mbox{div}(X) \Omega+X(f)\Omega= \mbox{div}(fX)\Omega$
(I don't know yet how to fill the gap $df\wedge X\,\lrcorner\,\Omega=X(f)\Omega$...).
A function $f$ such that $\mbox{div}(fX)=0$ is called a Jacobi last multiplier for $X$.
Suppose we have in $M$ two volume forms, $\Omega$ and $g\Omega$, with $g$ any smooth function. By definition $\mbox{div}_{g\Omega}(X)$ is the function which satisfies
$$ \mathcal{L}_X(g\Omega)=\mbox{div}_{g\Omega}(X) g\Omega $$But since $\mathcal{L}_{gX}(\Omega)=\mbox{div}_{\Omega}(gX) \Omega$ and $\mathcal{L}_{gX}(\Omega)=\mathcal{L}_{X}(g\Omega)$ we conclude
$$ \mbox{div}_{g\Omega}(X)=\frac{\mbox{div}_{\Omega}(gX)}{g} \tag{1} $$(Named as Jacobi's lemma in Muriel_2014)
Lemma. Suppose the coordinate change in $\mathbb{R}^n$:
$$ \varphi: (x_1,\ldots,x_n) \mapsto (y_1,\ldots,y_n) $$Suppose, also, that we have a vector field $X=\sum X_i \partial_{x_i}$ that gets transformed to $Y=\sum Y_i \partial_{y_i}=\varphi_*(X)$. Then
$$ \sum \frac{\partial X_i}{x_i}=H\sum \frac{\partial}{y_i}\left(\frac{Y_i}{H}\right) $$where $H$ is the Jacobian of the coordinate change. $\blacksquare$
Proof
Denote $\Omega=dx_1\wedge \cdots \wedge dx_n$ and $\tilde{\Omega}=dy_1\wedge \cdots \wedge dy_n$. We have that $(\varphi^{-1})^*(\Omega)=\frac{1}{H}\tilde{\Omega}$.
Now, observe that
$$ \mbox{div}_{\Omega} (X)=\varphi^*(\mbox{div}_{(\varphi^{-1})^*(\Omega)}(Y)) $$(we are simply computing the same thing in other coordinates).
So therefore
$$ \mbox{div}_{\Omega} (X)=\varphi^*(\mbox{div}_{\frac{1}{H}\tilde{\Omega}}(Y)) \tag{1} $$and according to $(1)$:
$$ \mbox{div}_{\Omega} (X)=\varphi^*\left(H\mbox{div}_{\tilde{\Omega}}\left(\frac{Y}{H}\right)\right) $$which is the desired formula $\blacksquare$
In particular, if $\mathcal{L}_X \Omega=0$ then $\mathcal{L}_Y \frac{1}{H}\bar{\Omega}=0$ (which is concluded from (1)).
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Author of the notes: Antonio J. Pan-Collantes
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