# Wedge product of closed $2$-forms not exact

Problem: Suppose $w$ is a closed $2$-form on a compact manifold $M$ without boundary, with $\dim(M) = 2n$ for some $n$. Denoting $w^k = w \wedge \dots \wedge w$ to be the wedge product of $w$ with itself for $k$ times, we also suppose $w^n$ is a volume form on $M$. Prove that $w^k$ is not exact, for $k = 1, 2, \dots, n$.

Attempt: I'm not sure how to approach the problem, but I suspect it has something to do with Stoke's theorem. For instance, if $w^n$ is exact, write $w^n = d \alpha,$ then $0 \neq \int_M w^n = \int_M d \alpha = 0$, but what about $w^k$ for $k \lt n$? Any help is appreciated!

• Without the condition that $M$ has no boundary, your assertion is false, even for $n=1$. Take $\omega = dx\wedge dy (=d(xdy))$ and $M=B$ is an open ball. – user99914 Apr 19 '18 at 17:11
• hmm thanks, I was suspecting the question is lacking conditions. I will add this condition in (this is a problem in a practice problem set so there could be typos...). – Longti Apr 19 '18 at 17:13
• Your new question is basically anwered here – user99914 Apr 19 '18 at 17:20

Suppose that $\omega^k=du$, $\omega^n=\omega^k\wedge\omega^{n-k}=du\wedge \omega^{n-k}$.
$d(u\wedge\omega^{n-k})=du\wedge\omega^{n-k}+(-1)^{2k-1}u\wedge d\omega^{n-k}=du\wedge\omega^{n-k}$, since $d\omega=0$ recursively, $d\omega^p=0$. We deduce that $\int_M\omega^n=\int_Md(u\wedge\omega^{n-k})=0$ contradiction.