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I am trying to show that a volume form $\mu$ on a compact manifold $M$ is not exact, i.e. show there is no $\alpha \in \Omega^{n-1}(M)$ such that $d\alpha = \mu$.

My attempt is the following: Suppose, as a contradiction, that $\mu$ is exact. Then, there exist an $(n-1)$-form $\alpha$ such that $d\alpha = \mu$. Then, since compact manifolds are manifolds without boundary, by stokes theorem we know that $\int_M \mu = \int_M d\alpha = \int_{\partial M} \alpha = 0$ since $M$ has no boundary. From here I am not sure how to continue to end my contradiction. I would appreciate any hint or suggestions for the problem. Thanks!

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  • $\begingroup$ You argument is correct. Note that $\int_M \mu$ is positive from the definitions of volume form and integral. $\endgroup$ – Dante Grevino Dec 11 '18 at 16:12
  • $\begingroup$ Oh, ok so the contradiction comes from that fact of positivity. Thanks so much for the clarification! $\endgroup$ – BOlivianoperuano84 Dec 11 '18 at 17:09
  • $\begingroup$ @BOlivianoperuano84 I recommend that you write an answer to your own question math.stackexchange.com/help/self-answer. You should also accept it math.stackexchange.com/help/accepted-answer. The benefit is that it becomes visible at first glance in the question list that is no longer open. $\endgroup$ – Paul Frost Dec 11 '18 at 17:31
  • $\begingroup$ Thanks Paul! I will write the answer! $\endgroup$ – BOlivianoperuano84 Dec 11 '18 at 18:53
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By definition, the integral over a manifold of a top (volume) form must be positive. Thus, since we have that $\int_M \mu = 0$, this is the contradiction we are looking for and so we deduce that a volume form on a compact manifold is not exact.

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