# Property of volume form on boundary of manifold

Let $$M$$ be an $$m$$-dimensional oriented manifold with boundary $$\partial M$$, that is connected and oriented. I now have to show that if $$M$$ is compact then if $$\alpha$$ is a volume form on $$\partial M$$, there does not exist a $$\beta \in \Omega^{m-1}(M)$$ such that $$d\beta = 0$$ and $$\iota^*\beta = \alpha$$ where $$\iota$$ is the inclusion map.

I think this proof is correct, but I would like to get some confirmation:

Assume that there is a $$\beta \in \Omega^{m-1}(M)$$ that has these properties. Because $$M$$ is compact, we now that $$\beta$$ has compact support and hence, we may apply Stokes's Theorem, which says that $$\begin{equation*} 0 = \int_M d\beta = \int_{\partial M} \iota^*\beta = \int_{\partial M } \alpha \end{equation*}$$ But $$\alpha$$ is a volume form on so $$\int_{\partial M } \alpha \neq 0$$, which concludes the proof.

I also have to check whether this holds when $$M$$ is not compact but I don't really have an idea on how to start with this statement.

For the noncompact case, what's one of the simplest non-compact manifolds $$M$$ with boundary you can think of? (You can give easy examples with $$\partial M$$ non-compact or compact.)
• Can we take the closed half plane, with $\alpha= \beta = dx$? Dec 28, 2019 at 17:33
• Good. And, analogously, on the cylinder $x^2+y^2=1$, $z\ge 0$? Dec 28, 2019 at 19:07