Does the divergence theorem require the covariant derivative to be metric compatible?

Question

I know this is more of a mathematical question, but it arises in the context of general relativity and uses its language so I thought it would be best to ask it here. I understand that the divergence theorem can be expressed as

\begin{equation} \int_M d^n x \sqrt{\left| g \right|} \nabla_a X^a = \int_{\partial M} d^{n-1} x \sqrt{\left| h \right|} n_a X^a \end{equation}

where $X^a$ is a vector field on $M$ and $h$ denotes the determinant of the metric on $\partial M$ induced by pulling back the metric on $M$. As I understand it, the covariant derivative must be metric compatible, in the sense that it must use the Levi-Civita connection calculated from the metric tensor $g$ on $M$. The question is: Does this hold if the covariant derivative is calculated using other Christoffel symbols? Maybe the Levi-Civita connection of another metric $\hat{g}$ on the same manifold $M$?

Answer 1

Suppose $M$ to be orientable and oriented and let $\mu\in\Omega^m(M)$ be a volume form ($m=\dim M$). Let $\nabla$ be a linear connection on $\tau:TM\rightarrow M$.

Given a vector field $X\in\mathcal D(M)$, the divergence of $X$ with respect to the volume form $\mu$ is $$ \mathrm{Div}_\mu(X):=(\mathscr L_X\mu)/\mu, $$ where $\mathscr L_X$ is the Lie derivative w.r.t. $X$. The divergence of $X$ with respect to the connection $\nabla$ is $$ \mathrm{Div}_\nabla(X):=\mathrm{Tr}(\nabla X). $$

Gauss' theorem always holds for divergence with respect to a volume form. Specifically, if $\Omega\subseteq M$ is a compact $m$ dimensional submanifold with boundary, then $$ \int_\Omega\mathrm{Div}_\mu(X)\mu=\int_\Omega\mathscr L_X\mu=\int_\Omega\mathrm{d}(X\rfloor\mu)=\int_{\partial\Omega}X\rfloor\mu. $$

Now if $(U,x)$ is a local chart for $M$, and if $\mu=\rho\, \mathrm dx^1\wedge\dots\wedge\mathrm dx^m$, then $$ \mathrm{Div}_\mu(X)=\frac{1}{\rho}\frac{\partial}{\partial x^i}\left(\rho X^i\right),\quad X=X^i\frac{\partial}{\partial x^i}. $$

Meanwhile, let us use the convention for covariant differentiation that the index corresponding to the direction of differentiation is the last one, $$ \nabla_k X^i=\frac{\partial X^i}{\partial x^k}+\Gamma^i_{\ jk}X^j. $$ Then $$ \mathrm{Div}_\nabla(X)=\frac{\partial X^i}{\partial x^i}+\Gamma^i_{\ ji}X^j.$$ The covariant derivative of the volume form is $$ \nabla_i\rho=\frac{\partial\rho}{\partial x^i}-\Gamma^j_{\ ji}\rho. $$

We now want to look for a condition where the two notions of divergence coincide. Expanding the divergence with respect to volume form, we get $$ \mathrm{Div}_\mu(X)=\frac{\partial X^i}{\partial x^i}+X^i\frac{\partial}{\partial x^i}\ln\rho=\frac{\partial X^i}{\partial x^i}+\Gamma^j_{\ ji}X^i+\rho^{-1}X^i\nabla_i\rho. $$

Now, the torsion of the connection is $$ T^i_{\ jk}=\Gamma^i_{\ kj}-\Gamma^i_{\ jk}, $$ and taking a trace gives $$ T_j=\Gamma^k_{\ kj}-\Gamma^k_{\ jk},\quad T_j=T^k_{\ jk}, $$ with which we have $$ \Gamma^j_{\ ji}X^i=T_i X^i+\Gamma^j_{\ ij}X^i, $$ which then gives $$ \mathrm{Div}_\mu(X)=\mathrm{Div}_\nabla(X)+\left(T_i+\rho^{-1}\nabla_i\rho\right)X^i. $$

We then obtain that the condition that the two notions of divergence coincide for all vector fields is that the connection and volume form should satisfy the equation $$ T_i=-\rho^{-1}\nabla_i\rho. $$ In this case, a Gauss theorem applies to the divergence $\mathrm{Div}_\nabla$.

When the vector part ($T_i$) of the torsion vanishes (which of course is the case when the torsion altogether vanishes), then this condition reduces to $\nabla_i\rho=0$, or equivalently $\nabla\mu=0$, i.e. the connection must be volume-preserving.