# Stokes theorem and Volume forms

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:

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

If this fails, where does it fail? Are there any assumptions I can add to $$\alpha$$ so that it would work?
• You could have $d\alpha=0$ even though $\alpha$ is everywhere nonzero. – Lord Shark the Unknown Apr 2 at 16:42
• 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 $k$. – achille hui Apr 2 at 16:46
• 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. – Giuseppe Negro Apr 2 at 16:48
• The correct statement is that an exact $n$-form on a compact, oriented $n$-dimensional manifold with no boundary must vanish at some point. – Ted Shifrin Apr 2 at 22:52