Nowhere vanishing 1-form on a compact manifold

Question

I need to prove that a nowhere vanishing 1-form on a compact manifold can not be exact.

I know it should be an application of Stoke's Theorem, but I'm not sure how to reach a contradiction. I think if we let the 1-form be $\omega$ and integrate it over the manifold M, then we will use the exactness to get the integral to be 0. But I believe this is only true if the manifold has empty boundary. Also I'm not sure why the integral being zero would be a contradiction.

Any help is understanding this problem would be appreciated

Answer 1

Let $\omega$ be a $1$-form on a compact manifold $M.$ We have that $\omega$ is exact if and only if there exists a smooth function $f:M\to \mathbb{R}$ such that $\omega =df.$

Now, since $M$ is compact and $f$ is continuous there exist $p,q\in M$ such that $f(p)\le f(x)\le f(q), \forall x\in M.$ But it is $(df)(p)=(df)(q)=0.$ That is, $\omega(p)=\omega(q)=0.$