# Volume form on a compact manifold is not exact

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!

• You argument is correct. Note that $\int_M \mu$ is positive from the definitions of volume form and integral. – Dante Grevino Dec 11 '18 at 16:12
• Oh, ok so the contradiction comes from that fact of positivity. Thanks so much for the clarification! – BOlivianoperuano84 Dec 11 '18 at 17:09
• @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. – Paul Frost Dec 11 '18 at 17:31
• Thanks Paul! I will write the answer! – BOlivianoperuano84 Dec 11 '18 at 18:53

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.