For a vector field $X\in\Gamma(TM)$ on a closed Riemannian manifold $(M,g)$ we have
\begin{align*}
\int_M\text{div}X\;\mu=0,
\end{align*}
where
\begin{align*}
\text{div}X=-g^{ij}g(\nabla_iX,\partial_j).
\end{align*}
Here $\mu$ is the volume form and $\nabla$ is the Levi-Civita connection of the metric $g$.
Question: Is there an analogous statement for differential forms and even for general tensors?
In other words if $\omega\in\Gamma(T^*M)$ is a $1$-form does it hold that
\begin{align*}
\int_M\text{div}\,\omega\;\mu=0\,?
\end{align*}
Here
\begin{align*}
\text{div}\,\omega=-g^{ij}\nabla_i\omega_j.
\end{align*}
We can easily define the divergence for higher-order tensors, so does the 'divergence theorem' also hold here as well?
EDIT: The motivation for this question is as follows. Let $M$ be as above and
\begin{align*}
\dot{R}_{jk}=D\text{Rc}(g)h_{jk}=\frac{1}{2}g^{pq}(\nabla_q\nabla_jh_{pk}+\nabla_q\nabla_kh_{pj}-\nabla_q\nabla_ph_{jk}-\nabla_j\nabla_kh_{pq})
\end{align*}
be the linearisation of the Ricci tensor at $g$ in the direction of the symmetric tensor $h$. Then there is a statement in a paper by Hamilton that
\begin{align*}
\int_M\text{tr}_g\dot R_{jk}\,\mu=\int_Mg^{jk}\dot R_{jk}\,\mu=0.
\end{align*}
I calculate that
\begin{align*}
\text{tr}_g\dot R_{jk}=g^{jk}\dot R_{jk}&=g^{jk}\frac{1}{2}g^{pq}(\nabla_q\nabla_jh_{pk}+\nabla_q\nabla_kh_{pj}-\nabla_q\nabla_ph_{jk}-\nabla_j\nabla_kh_{pq})\\
&=g^{jk}g^{pq}\nabla_q\nabla_jh_{pk}-g^{jk}g^{pq}\nabla_q\nabla_ph_{jk}\\
&=\text{div}(\text{div}\,h)-\Delta\text{tr}_gh.
\end{align*}
The divergence $(\text{div}\,h)_p=g^{jk}\nabla_jh_{pk}$ of the symmetric tensor $h$ is a $1$-form.
QUESTION: The integral of the connection Laplacian is zero but how is the integral of the other term zero? Or am I doing something silly and wrong…?
Best Answer
There is no Stokes's Theorem for general tensors. The general result is that $\displaystyle\int_M d\omega = 0$ for any compact oriented $n$-dimensional manifold $M$ without boundary and $(n-1)$-form $\omega$ on $M$. The divergence theorem you cite is for the case of $\omega = \star\eta$ where $\eta$ is the $1$-form corresponding to $X$ under the isomorphism $T^*M\cong TM$ induced by the Riemannian metric.