# What is the generalization of Gauss's Theorem to a manifold?

In a (pseudo-)Riemannian manifold with constant basis vectors, one certainly has that the integral of the divergence of a tensor field $T$ over a submanifold $\Omega$ is equal to the integral over the boundary $\partial\Omega$ of the inner product of $T$ with the outward-directed unit normal vector. To be more explicit, in component form in Minkowski space, $$\int\limits_\Omega \partial_\mu {{T^\alpha}_\beta}^\mu \ \mathrm{d}x^0 \wedge \mathrm{d}x^1 \wedge \mathrm{d}x^2 \wedge \mathrm{d}x^3 = \int\limits_{\partial\Omega} {{T^\alpha}_\beta}^\mu \varepsilon_{\mu|\alpha\beta\gamma|} \ \mathrm{d}x^\alpha \wedge \mathrm{d}x^\beta \wedge \mathrm{d}x^\gamma,$$ where I'm using the convention in Misner, Thorne, & Wheeler that $|\alpha\beta\gamma|$ runs over all indices such that $0 \leq \alpha < \beta < \gamma \leq 3$, and $\epsilon$ is the Levi-Civita symbol/tensor (there being no difference in the lowered-index form in Minkowski space).

The problem is, the proof given in that book only makes sense in Cartesian coordinates. What if I have some arbitrary coordinates in a curved manifold? Is there an analogous equation relating $$\int\limits_\Omega \nabla_\mu {{T^\alpha}_\beta}^\mu \sqrt{|g|} \ \mathrm{d}^4x \leftrightarrow \int\limits_{\partial\Omega} {{T^\alpha}_\beta}^\mu n_\mu \sqrt{|\gamma|} \ \mathrm{d}^3x?$$ (Here $\nabla$ is the covariant derivative compatible with the metric, $g$ is the determinant of the metric, and $\gamma$ is the determinant of the induced metric on the surface.)

Wald1 shows in an appendix that $$\int\limits_\Omega \nabla_\mu v^\mu = \int\limits_{\partial\Omega} v^\mu n_\mu,$$ where I've followed his lead and suppressed the volume elements. This would seem to indicate that I can follow a naive "partial to covariant" prescription to make my first equation turn into my second. But does this work for arbitrary tensors? I worry because rank $(1,0)$ tensors can have fortuitous cancellations, such as how $\nabla_\mu v^\mu = \partial_\mu (\sqrt{|g|} v^\mu) / \sqrt{|g|}$ holds without extra connection terms.

1 Yes, another physics book. It occurs to me that I don't own any pure math differential geometry books.

• I friend of mine (whom I passed your question) tells me that the answer to your question is on p. 147 & 148 of “Geometrical methods of mathematical physics” by Bernard Schutz. Regards – chris May 26 '14 at 19:54

Your statement for general tensors is not true, I think. Assume that $T^{\alpha \mu}_{\beta} = S^{\alpha}_{\beta} v^{\mu}$, where $\nabla S$ is not identically zero. Then, dropping indices, $\nabla T = (\nabla S)v+ S\nabla v$. The second term of the right-hand side, $S\nabla v$, after being integrated, gives the right-hand side of your hypothesized formula, by Gauss's theorem. But then you also need to integrate the first term, $(\nabla S)v$, and there is no reason to expect this to always be 0.