# Commutativity of pullbacks and the exterior derivative as an unbounded operator on $L^2$

Let $d_c, \delta_c$ be operators with domains $D(d_c) = D(\delta_c) = C_{c}^\infty(\wedge T^\ast M)$. We let $d_c$ be the usual exterior derivative on compactly supported smooth forms, ie., $d_c\omega = dx^k \wedge \nabla_k \omega$ and $\delta_c = dx^k \llcorner \nabla_k \omega$.

We define $d:D(d) \subset L^2 \to L^2$ as the adjoint of $\delta_c$. Ie, the operator with largest domain $D(d)$ satisfying $(d\omega,\eta) = (\omega,\delta_c \eta)$ for all $\omega \in D(d)$ and $\eta \in C_{c}^\infty$.

How do I show that for some smooth coordinate chart $\phi:U \subset M \to \phi(U) \subset \mathbb{R}^n$ that $d(\phi^{-1})^\ast\omega = (\phi^{-1})^\ast d\omega$ for $\omega \in D(d)?$

Note that I don't know that $\bar{d_c}= d$, so I can't simply approximate by smooth forms. In fact, this arises in trying to prove exactly that statement.

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Fix $u,v \in C_c^\infty$. Then, let $\psi = \phi^{-1}$. Then, let $(\psi^\ast)^\ast$ be the adjoint of $\psi^\ast$. So, $$\langle u, (\psi^\ast)^\ast\delta_c v \rangle = \langle \psi^\ast u, \delta_c v \rangle = \langle d\psi^\ast u, v \rangle = \langle \psi^\ast d u, v \rangle = \langle d \psi^\ast u , v \rangle = \langle u, \delta_c (\psi^\ast)^\ast v \rangle.$$
By density of $C_c^\infty$ in $L^2$, we have that $(\psi^\ast)^\ast \delta_c v = \delta_c (\psi^\ast)^\ast v$ for all $v \in C_c^\infty$ in $L^2$.
Applying this to the adjoint $d$ of $\delta_c$ gives the desired result.
Should the expression between the fourth and fifth 'equal to' symbols be $\langle du, (\psi^*)^*v\rangle$? –  Michael Albanese May 18 '12 at 16:23