3
$\begingroup$

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...?

$\endgroup$
6
+100
$\begingroup$

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: $$ \int_M \operatorname{div}\omega\, \mu = \int_M \operatorname{div}(\omega^\sharp)\, \mu = 0. $$

$\endgroup$
  • $\begingroup$ Does this hold for compact manifolds without boundary? $\endgroup$ – Bellem Jul 27 at 15:49
  • 1
    $\begingroup$ @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. $\endgroup$ – Jack Lee Jul 27 at 16:15
10
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.