MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 0 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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