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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!

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You want to add that $M$ is oriented (otherwise the integral doesn't make sense) and connected (otherwise the statement isn't true.) –  David Speyer Feb 19 '12 at 17:26
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1 Answer 1

(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.

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M is orientable. Can we use another way to solve this problem without using De Rham theory? –  henry Feb 20 '12 at 10:12
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