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I have the following short argument which seems to say that there are no non-vanishing, (n-1)-forms on a closed manifold of dimension n but I am not very confident in my understanding of a "volume form" because I am used to only working with $\textit{the}$ volume form (induced by a metric) so I don't know if this applies:

Suppose that $M$ is an oriented, n-dimensional manifold which is also closed (i.e. $\partial M$ is empty). Suppose also that $\alpha$ is a non-vanishing (n-1)-form. Then $d\alpha$ is an n-form.

$\textbf{Question 1:}$ Does that mean that $d\alpha$ is a "volume form"? Can I conclude that $\int_{M}d\alpha$ is nonzero? Let's suppose the answer is yes. Let $d\alpha = Vol_{\alpha}$.

$\textbf{Question 2:}$ In this case, it seems like Stokes theorem and the fact that $M$ is closed says:

$$0 \neq \int_{M}Vol_{\alpha} = \int_{M}d \alpha = \int_{\partial M} \alpha = 0$$

Which is a contradiction.

If this fails, where does it fail? Are there any assumptions I can add to $\alpha$ so that it would work?

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  • $\begingroup$ You could have $d\alpha=0$ even though $\alpha$ is everywhere nonzero. $\endgroup$ – Lord Shark the Unknown Apr 2 at 16:42
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    $\begingroup$ Even when $d\alpha \ne 0$, there is no need for it to be a constant multiple of the Volume form. What you have done is ruled out the possiblitity that $d\alpha = k Vol$ for non-zero constant $k$. $\endgroup$ – achille hui Apr 2 at 16:46
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    $\begingroup$ Here's a counterexample. Consider the torus $\mathbb T^2=\mathbb R^2 / \mathbb Z^2$. This is a closed and compact $2$-manifold. You have the global coordinate system $(x_1, x_2)$, and $dx_1$ is a $1$-form that does not vanish at any point. $\endgroup$ – Giuseppe Negro Apr 2 at 16:48
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    $\begingroup$ (The name is Stokes, not Stoke.) $\endgroup$ – Hans Lundmark Apr 2 at 18:36
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    $\begingroup$ The correct statement is that an exact $n$-form on a compact, oriented $n$-dimensional manifold with no boundary must vanish at some point. $\endgroup$ – Ted Shifrin Apr 2 at 22:52

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