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Table of Contents
Integration by parts
I forget how to do integration by parts on a manifold every day of the week. Here's a wiki page to remind me how it should work.
Preliminaries
There's obviously integration by parts on an interval:
$$\int_a^b \phi'(t) \psi(t) dt = [\phi(t)\psi(t)]^b_a - \int_a^b \phi(t) \psi'(t) dt.$$
This is obtained by integrating the product rule $(\phi(t)\psi(t))' = \phi'(t)\psi(t) + \phi(t)\psi'(t)$ and using the fundamental theorem.
More general is Stokes's theorem for an $(n-1)$-form $\omega$ on an $n$-dimenisonal manifold $A$, possibly with boundary.
$$\int_A d \omega = \int_{\partial A} \omega.$$
This recovers the integration by parts above for $A = [a,b]$ and $\omega = \phi(t)\psi(t)$, but is of course more powerful generally.
The divergence theorem
Let $\Omega$ be the volume form on $A$, let $X$ be any vector field on $A$, let $\nu$ be its metric dual and define $\omega := e^*(\nu) \Omega$. By Stokes's theorem, $$ \int_A d (e^*(\nu) \Omega) = \int_{\partial A} \omega.$$ On the other hand, we recognize a Lie derivative in the left-hand side by the Cartan formula and the fact that $\Omega$ is a top form: $d e^*(\nu) \Omega = \{d, e^*(\nu)\} \Omega = L_X \Omega = (\mathrm{div} X) \Omega$.
Thus we have a first form of the divergence theorem: $$ \int_A (\mathrm{div} X) \Omega = \int_{\partial A} e^*(\nu) \Omega.$$ Write in a frame $e_1, \ldots, e_n$ where $e_n$ is normal to $\partial A$ and $\nu = \nu^i e_i$; $e^*(e_n) \Omega$ is the volume form on $\partial A$ and $e^*(e_j) \Omega = 0$ for $j < n$. Thus $$ \int_A (\mathrm{div} X) \Omega = \int_{\partial A} \nu^n \mathrm{vol}_{\partial A}.$$ But $\nu^n = \langle X, e_n\rangle$ so $$ \int_A (\mathrm{div} X) \Omega = \int_{\partial A} \langle X, e_n \rangle \mathrm{vol}_{\partial A}$$ is the familiar form of the divergence theorem.
Integration by parts I
Our goal is to derive some result like the following: for any $\alpha, \beta$ sections of a metric bundle with compatible connection, $$(\nabla_X \alpha, \beta) = (\alpha, -\nabla_X \beta) + \mathrm{(boundary\ terms)}.$$ Round brackets always mean the $L^2(M)$ inner product, while angle brackets mean the fibrewise product.
