When is an $n$-form on $\mathbb S^n$ exact

Question

In John Lee's Introduction to Smooth Manifolds, he leaves the following as exercise 17.22: Show that $\eta \in \Omega^n(\mathbb S^n)$ is exact if and only if $$\int_{\mathbb S^n}\eta = 0.$$ The "if" part is easy: if $\eta$ is exact, then by Stokes' theorem (and because $\partial \mathbb S^n = \emptyset$), then $\int_{\mathbb S^n}\eta = 0$. My struggle begins when trying to prove the converse. Some of my ideas include using $\mathbb S^n$ as the boundary of the closed unit ball or trying to use the fact that $\mathbb S^n$ is connected and that any $C^{\infty}(M)$ on a connected smooth manifold $M$ has the same sign, but I am at the moment stuck. Any ideas?

Answer 1

Just want to provide details on the hint, and any corrections welcome.

Let $\omega$ be a smooth orientation form on $S^n$. Theorem 17.21 tells us that $[\omega]$ spans $H_{dR}^n \cong \Bbb{R}$. Assume $[\eta] \neq 0$. Then $[\eta] = c[\omega]$ for some scalar $c$. Thus, $[\eta] = [c\omega]$, which is to say $\eta = c\omega + \alpha$ for $\alpha$ exact. Then, since we've already shown exactness implies an integral of zero, we get $$\int_{S^n}\eta = c\int_{S^n} \omega + \int_{S^n} \alpha = c\int_{S^n}\omega \neq 0.$$ Hence, $\int_{S^n}\eta = 0 \Rightarrow [\eta] = 0$, which means $\eta$ is exact.