Link between covariant divergence and Lie derivative

Question

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.

Answer 1

Following the comments of Didier and studying his answer I was able to prove it coordinate free with the trace definition. Because I like doing Riemannian geometry coordinate free and find the trace definition of the divergence much more natural I will also give this proof. We start by restating the definition.

Definition Let $X$ be a vectorfield on $M$ and $A_1,...,A_n$ a basis of $TM$ with duals $A^1,...,A^n$ such that $A^i(A_j) = \delta^i_j$. Then the covariant divergence of $X$ is $$ \text{div}(X) = A^i(\nabla_{A_i}X). $$

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Next we will prove an auxiliary claim that is used in the computation of the proof.

Claim If $E_1,...,E_n$ is an orthonormal basis (where the duals $E^1,...,E^n$ are $E^i = \langle E_i, \cdot\rangle)$ then $$ \mathcal{L}_X(E^i) = \langle \nabla_XE_i,\cdot\rangle + E^i(\nabla_\cdot X). $$ Proof. Let $Y$ be any vectorfield. By the Leibnitz rule we have that \begin{align} \mathcal{L}_X(E^i)(Y) &= \mathcal{L}_X(E^i(Y)) - E^i(\mathcal{L}_XY)\\ &=X(E^i(Y)) - E^i([X,Y])\\ &=X(E^i(Y)) - E^i(\nabla_XY-\nabla_YX)\\ &=X\langle E_i,Y \rangle - \langle E_i,\nabla_XY\rangle + \langle E_i,\nabla_YX\rangle\\ &=\langle \nabla_XE_i,Y\rangle + \langle E_i,\nabla_XY\rangle -\langle E_i,\nabla_XY\rangle + \langle E_i,\nabla_YX\rangle\\ &=\langle \nabla_XE_i,Y\rangle +\langle E_i,\nabla_YX\rangle. \end{align} Thus $$ \mathcal{L}_X(E^i) = \langle \nabla_XE_i,\cdot\rangle +\langle E_i,\nabla_\cdot X\rangle $$ which was to show.$\Box$

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Proposition Let $\text{vol}$ be the volume form on $M$. Then $$ \mathcal{L}_X(\text{vol})=\text{div}(X)\text{vol} $$ Proof. We choose an orthonormal basis $E_1,...,E_n$. The duals $E^1,...,E^n$ are $E^i = \langle E_i, \cdot\rangle$ and the volume form is $\text{vol} = E^1 \wedge ... \wedge E^n$. Now we compute \begin{align} \mathcal{L}_X(\text{vol}) &= \mathcal{L}_X(E^1 \wedge ... \wedge E^n)\\ &= \sum_{i=1}^{n}E^1 \wedge ...\wedge \mathcal{L}_X(E^i)\wedge... \wedge E^n\\ &= \sum_{i=1}^{n}E^1 \wedge ...\wedge (\langle\nabla_XE_i,\cdot\rangle + E^i(\nabla_\cdot X))\wedge... \wedge E^n\\ \end{align} where we used the claim. Note that applying $X$ to $1 = \langle E_i,E_i\rangle$ yields $$ 0 = X\langle E_i,E_i\rangle = 2\langle\nabla_XE_i,E_i\rangle $$ and therefore $\langle\nabla_XE_i,E_i\rangle = 0$. This means that when we write this 1-form in the basis $\langle\nabla_XE_i,\cdot\rangle = \alpha_j E^j$ the i-th coefficient is zero and thus in the above wedge product we have a term $E^k \wedge E^k$ for each non-vanishing coefficient $\alpha_k \neq 0$ and therefore this 1-form does not contribute to the wedge product and so $$ \mathcal{L}_X(\text{vol}) = \sum_{i=1}^{n}E^1 \wedge ...\wedge E^i(\nabla_\cdot X)\wedge... \wedge E^n. $$ The form at hand is of top dimension. Hence there exists a coefficient $\alpha \in C^\infty(M)$ such that $$ \sum_{i=1}^{n}E^1 \wedge ...\wedge E^i(\nabla_\cdot X)\wedge... \wedge E^n = \alpha E^1 \wedge ... \wedge E^n. $$ We will determine this coefficient by evaluating the right and left hand sides on $E_1,...,E_n$. For the RHS we get $$ \alpha E^1 \wedge ... \wedge E^n(E_1,...,E_n) = \alpha. $$ The LHS evaluates to \begin{align} \text{LHS} & = \sum_{i=1}^n\det\left(\begin{array}{ccc}E^1(E_1) & ... & E^n(E_1)\\ ...& E^i(\nabla_{E_i}X)& ...\\ E^1(E_n) & ... & E^{n}(E_n) \\\end{array}\right)\\ &=\sum_{i=1}^{n}E^i(\nabla_{E_i}X)\\ &=\text{div}(X). \end{align} Hence $\alpha = \text{div}(X)$ and so we have shown that \begin{align} \mathcal{L}_X(\text{vol}) &= \sum_{i=1}^{n}E^1 \wedge ...\wedge E^i(\nabla_\cdot X)\wedge... \wedge E^n\\ & = \text{div}(X)E^1\wedge ... \wedge E^n\\ & = \text{div}(X)\text{vol} \end{align} which is the result.$\Box$

Answer 2

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.$\Box$