Integration of $\mathbb{d}\mathbb{d}^{*}+\mathbb{d}^{*}\mathbb{d}$ on manifolds with boundary

53 Views Asked by At

Let $V^p\subset M^n$ be a compact submmanifold with boundary $\partial V$ in a (Riemannian) manifold $M^n$. Let $\boldsymbol{\Delta}=\mathbb{d}\mathbb{d}^{*}+\mathbb{d}^{*}\mathbb{d}$ be the Laplace operator and consider the (continuously differentiable) $p$-form $\omega^p$ on $M^n$.

Is it possible to write the volume integral \begin{align} \int_V \boldsymbol{\Delta}\omega \end{align} as an integral extended over $\partial V$ ? $\int_V\mathbb{d}(\mathbb{d}^{*}\omega)=\int_{\partial V}\mathbb{d}^{*}\omega$, as implied by the Stokes theorem; however, is there a similar result for the second term $\int_V\mathbb{d}^{*}(\mathbb{d}\omega)$ ? More generally, can one say something about

\begin{align} \int_{W^{p-1}} \mathbb{d}^{*}\omega^p \end{align}

Even in the case of a surface and being $\omega$ a function, what would be the correct generalization of Gauss's theorem for $\int_{V^2} \boldsymbol{\Delta}\omega$ ?

1

There are 1 best solutions below

1
On BEST ANSWER

As you know, $d:\Omega^k(V)\to\Omega^{k+1}(V)$. Since $\dim V=p$, we get $\Omega^{p+1}(V)=\{0\}$. Considering that $\iota^*\omega\in\Omega^p(V)$, it follows that $d\iota^*\omega=\iota^*d\omega\in\Omega^{p+1}(V)$ so $\iota^*d^\star d\omega=0$, and it follows that $$\int_Vd^\star d\omega=0\implies \int_V\Delta\omega=\int_{\partial V}d^\star\omega$$