On wikipedia I came across the formula $\text{div}(X)\text{vol} = \mathcal{L}_X(\text{vol})$ where $X$ is a vectorfield on a Riemannian manifold with volume form $\text{vol}$.
I am wondering how this formula might be proven. I have tried to find a proof in the literature but with no success. I have tried to do it myself: I started by applying Cartan's magic formula
$$
\mathcal{L}_X(\text{vol}) = \iota_Xd\text{vol} + d(\iota_X\text{vol}).
$$
Since $\text{vol}$ is a form of top dimension its exterior derivative is zero. Hence we have
$$
\mathcal{L}_X(\text{vol}) = d(\iota_X\text{vol}).
$$
I am aware of two definitions of the divergence, namely (in local coordinates)
$$
\text{div}(X) = dx^k(\nabla_{\partial_k}X)
$$
or
$$
\text{div}(X) = \frac{1}{\sqrt{g}}\partial_k\left(\sqrt{g}X^k\right).
$$
But I don't know how to go from there, since I don't really know what to do with the term $\iota_X\text{vol} = \text{vol}(X,\cdot,…,\cdot)$. Any help would be highly appreciated.
Best Answer
Thanks to the comments of peek-a-boo I was able to come up with a proof. I will write it out, for the sake of completeness.
Proposition The divergence of a vectorfield $X$ on a Riemannian manifold is $$\tag{1} \text{div}(X) = \frac{1}{\sqrt{g}}\partial_i\left(\sqrt{g} X^i\right). $$
Proposition $\text{div}(X)\text{vol} = \mathcal{L}_X(\text{vol})$.
Proof. As explained in the question, we have by Cartan magic that $\mathcal{L}_X(\text{vol}) = d(\iota_X\text{vol})$. Hence we want to show that $d(\iota_X\text{vol})$ equals the right hand side in $(1)$.
In a local chart, let $X = X^i \partial_i$ and $\text{vol} = \sqrt{g}dx^1 \wedge...\wedge dx^n$. By the linearity of $\iota$ in its first argument we have $$ \iota_X\text{vol} = \iota_{X^i\partial_i}\text{vol} = X^i\iota_{\partial_i}\text{vol} $$ and so $$ \iota_X\text{vol} = X^i\iota_{\partial_i}(\sqrt{g}dx^1\wedge...\wedge dx^n). $$ By the product rule for $\iota$ (see this post by peek-a-boo) this is $$ X^i\iota_{\partial_i}(\sqrt{g}) dx^1\wedge...\wedge dx^n + X^i\sqrt{g} \iota_{\partial_i}(dx^1\wedge...\wedge dx^n). $$ Since $\sqrt{g}$ is a 0-form its interior product vanishes identically by definition. We will thus examine the term $\iota_{\partial_i}(dx^1\wedge ... \wedge dx^n)$. In the the same post peek-a-boo uses the product rule to expand this term, arriving at (in our case) $$ \iota_{\partial_i}(dx^1\wedge ... \wedge dx^n) = \sum_{k=1}^{n}(-1)^{k-1}\delta^k_i dx^1\wedge...\wedge \widehat{dx^i}\wedge...\wedge dx^n. $$ Hence $$ \iota_X\text{vol} = \sum_{i=1}^{n}(-1)^{i-1}X^i\sqrt{g}dx^1\wedge...\widehat{dx^i}\wedge...\wedge dx^n. $$ Now we can evaluate the exterior derivative of this. Again, we will used the rule established by peek-a-boo in the same post, which says that for $\omega = \omega_I \alpha^I$ where the $\alpha$ are basis forms and $I$ is a multi-index we have $$ d\omega = \frac{\partial \omega_I}{\partial x^i}dx^i\wedge \alpha^I. $$ Applying this and using the linearity of $d$ yields $$ d(\iota_X\text{vol})=\sum_{i=1}^{n}(-1)^{i-1}\frac{\partial(\sqrt{g}X^i)}{\partial x^k}dx^k\wedge dx^1\wedge...\widehat{dx^i}\wedge...\wedge dx^n. $$ Since the wedge products in this sum are of top dimension they are only non-zero when $k=i$. Therefore $$ d(\iota_X\text{vol})=\sum_{i=1}^{n}(-1)^{i-1}\frac{\partial(\sqrt{g}X^i)}{\partial x^i}dx^i\wedge dx^1\wedge...\widehat{dx^i}\wedge...\wedge dx^n. $$ Using that $\wedge$ is alternating we can move the $dx^i$ from the beginning to the place where it is left out and get $$ d(\iota_X\text{vol})=\sum_{i=1}^{n}\frac{\partial(\sqrt{g}X^i)}{\partial x^i}dx^1\wedge...\wedge dx^n. $$ Finally, using $\text{vol} = \sqrt{g}dx^1 \wedge...\wedge dx^n$, we arrive at $$ d(\iota_X\text{vol})=\sum_{i=1}^{n}\frac{1}{\sqrt{g}}\frac{\partial(\sqrt{g}X^i)}{\partial x^i}\text{vol} $$ which is the desired result.$\Box$