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.



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


