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

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

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

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 $$M$$?

• Why don't you go through all the steps carrying the general Stokes thorem in the language of de Rham complex into this variant you posted? Which are the assumptions in the general theorem vs which assumptions are needed in the derivation. Commented Apr 8 at 16:29

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.