Divergence theorem for tensor fields on manifolds

Let $M$ be a Riemannian manifold and let $T$ be a $k$-tensor over $M$. I am looking for an analogon of the divergence theorem on manifolds.

I would expect something of the form

$\int_U \operatorname{Div} T = \int_{\partial U} \vec n \lrcorner T$

to be valid for some operator $\operatorname{Div}$ yet to be determined, where $U \subseteq M$ is a smoothly bounded subset of the manifold $M$, where $\vec n \lrcorner T$ is the contraction of $T$ with the outer unit normal along the boundary of $U$.

I am wondering in particular about the nature of the ominous operator $\operatorname{Div}$. A reference to a proof in the literature is sufficient.

• Page 422 of Lee: link.springer.com/book/10.1007%2F978-1-4419-9982-5, page 623 of the 'other' Lee: bookstore.ams.org/gsm-107, and appendix A of Chow & Knopf: bookstore.ams.org/surv-110, should get you started. – g.s Sep 12 '17 at 20:02
• I am aware of the divergence theorem for vector fields on manifolds. How to prove a divergence theorem for tensors is a much more complicated question, though. Do the AMS references contain such a theorem? – shuhalo Sep 12 '17 at 20:53
• So yes starting with Jack Lee's book and then looking at the 2 AMS books, you could deduce what Jack Lee wrote below.. – g.s Sep 12 '17 at 23:27

In order for your equation to make sense, $T$ has to be a tensor field whose divergence is a differential form (alternating tensor) with rank equal to the dimension of the manifold, and the manifold has to be oriented. Moreover, to define the divergence and unit normal, you're going to need a Riemannian metric. So let's assume $(M,g)$ is an oriented Riemannian $n$-manifold, and $T$ is a smooth section of either $\Lambda^n M \otimes TM$ or $\Lambda^n M\otimes T^*M$. Since we have a Riemannian metric, these two bundles are naturally isomorphic via the musical isomorphisms, so let's choose $\Lambda^n M \otimes TM$.
The divergence of such a tensor is defined as $\operatorname{div} T = \operatorname{tr}(\nabla T)$, where the trace is on the last two indices -- the contravariant $TM$ index, and the covariant $T^*M$ index introduced by $\nabla$.
Any section of $\Lambda^n M\otimes TM$ can be written globally as $T = dV_g \otimes X$ for some vector field $X$, and then the fact that $dV_g$ is parallel implies $$\operatorname{div}T = \operatorname{tr} (dV_g \otimes \nabla X) = dV_g \otimes (\operatorname{tr} \nabla X) = (\operatorname{div} X) dV_g.$$ Then the divergence theorem for vector fields on oriented Riemannian manifolds with boundary (see Theorem 16.32 in my Introduction to Smooth Manifolds) implies $$\int_U \operatorname{div}T = \int_U (\operatorname{div} X) dV_g = \int_{\partial U} \langle X, \vec n \rangle\,dV_g = \int_{\partial U} \vec n \lrcorner T.$$ But this is really nothing but the ordinary divergence theorem dressed up in different notation.
• Thank you for the explanation, but I am not d'accord with your initial statement: on $\mathbb R^n$ you can define the divergence of a, say, $k$ tensor and obtain a $(k-1)$ tensor. That's done in tensor analysis all the time. The problem on a manifold is, of course, that you need to compare fibres at different points. For that reason I wonder how structure you need on a manifold to achieve such a generalization - maybe you need the affine structure on $\mathbb R^n$ and there is no hope of a generalization at all? – shuhalo Sep 13 '17 at 16:48
• @shuhalo: The divergence of a $k$-tensor field is perfectly well defined on any Riemannian manifold. But how do you propose to integrate an arbitrary $(k-1)$-tensor field? – Jack Lee Sep 13 '17 at 17:05
• @JackLee: Is the formula $\operatorname{div} T = \operatorname{tr}(\nabla T)$ (with contraction performed on the last two aguments) valid for every kind of tensor? I ask because I am trying to make sense of the formula involving the "sharp" musical isomorphism found on Wikipedia, and something seems to be wrong there. In particular, according to that, the divergence of a $(1,0)$-type tensor field (i.e. a tangent field) should be a... $(1,-1)$ tensor field??? – Alex M. Aug 4 '18 at 19:32
• @AlexM.: It's just a matter of convention. You can define a divergence of any kind of tensor field by taking the trace on the differentiation index along with any other index you like, depending on your purpose. If the original index was covariant, then you need to apply $\sharp$ first to raise it. The formula given in that Wikipedia article doesn't apply to $(1,0)$-tensors (i.e., vector fields), but you can compute the divergence of a vector field just by taking the trace of the total covariant derivative. Since $\nabla T$ has one upper and one lower index, there's no need for $\sharp$. – Jack Lee Aug 4 '18 at 20:25