3
$\begingroup$

If $\omega$ is an $n$-form on a compact $n$-dimensional manifold $M$ without boundary, then $\omega $ is exact if and only if $\int\limits_{M}{\omega }=0$.

Maybe there are two ways - use de Rham theory, and another way is to prove this directly.I don't know both. Help!

  • 6
    You want to add that $M$ is oriented (otherwise the integral doesn't make sense) and connected (otherwise the statement isn't true.)2012-02-19

1 Answers 1

9

(Is $M$ orientable?)

If $\omega$ ist exact, then $ \int\limits_{M}{\omega }= \int\limits_{M}{d\theta }=\int\limits_{\partial M}{\theta } = \int\limits_{\emptyset}\theta=0 $ If $\int\limits_{M}{\omega }=0$ (or for the above direction as well), you can apply the fact that $\int\limits_{M}:H^n\rightarrow \mathbb R$ is an isomorphism.

  • 0
    M$ $is orientable. Can we use another way to solve this problem without using De Rham theory?2012-02-20