# Proof in Hamilton: Divergence theorem for differential forms?

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...?

On a Riemannian manifold, the divergence theorem applies to $1$-forms as well as to vector fields. The simplest way to see this is by using the "musical isomorphisms" between $1$-forms and vector fields. This are the inverse isomorphisms $\flat\colon TM\to T^*M$ and $\sharp\colon T^*M\to TM$ defined by raising and lowering indices. If $X$ is a vector field and $\omega$ is a $1$-form, $$(X^\flat)_i = g_{ij}X^j, \qquad (\omega^\sharp)^j = g^{jk}\omega_i.$$ Then the divergence of $\omega$ is simply defined to be $\operatorname{div}\omega := \operatorname{div}(\omega^\sharp)$. In components, $\operatorname{div}\omega = g^{jk}\nabla_j\omega_k$.
The divergence theorem for $1$-forms then follows directly from the one for vector fields: $$\int_M \operatorname{div}\omega\, \mu = \int_M \operatorname{div}(\omega^\sharp)\, \mu = 0.$$
• @Bellem: Yes, that was the OP's stipulation, so I wrote my answer under that assumption. If $M$ has nonempty boundary, then there's a boundary integral on the right-hand side. If $\omega$ is a $1$-form, it's $\int_{\partial M} \omega(N) \hat\mu$, and if $X$ is a vector field, it's $\int_{\partial M} \langle X,N\rangle \hat \mu$. Here $\hat \mu$ is the induced volume form on the boundary, and $N$ is the outward unit normal. – Jack Lee Jul 27 at 16:15
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.
Addition to Jack Lee's answer: Since $\nabla_i\omega^\sharp=(\nabla_i\omega)^\sharp$ (the covariant derivative respects the musical isomorphisms) and we also calculate that \begin{align*} \text{div}\,\omega&:=\text{div}\,\omega^\sharp\\ &=-g^{ij}g(\nabla_i\omega^\sharp,\partial_j)\\ &=-g^{ij}((\nabla_i\omega)^\sharp)^\flat_j\\ &=-g^{ij}\nabla_i\omega_j. \end{align*}