A $2$-form on $S^2$ is exact if it integrates to zero.

Question

I'm trying to show that a $2$-form on $S^2$ is exact if and only if it integrates to zero, without appealing to de Rham's theorem (basically only using the Poincaré lemma [that every closed form on a contractible manifold is exact]).

One direction is easy, since $\partial S^2=0$, Stokes's theorem shows that if $\omega=d\psi$ for some $(n-1)$-form $\psi$ on $S^2$, then $\int_{S^2}\omega=\int_{S^2}d\psi=\int_{\emptyset}\psi=0$.

I know the usual way to go, to decompose $S^2$ into it's northern and southern hemispheres, each of which is contractible. So if $\int_{S^2}\omega=0$, this gives two $(n-1)$-forms $\psi^+$ and $\psi^-$ on the northern and southern hemispheres, respectively with $d\psi^{\pm}=\omega$ on their domains. Moreover, Stokes's theorem shows again that

$0=\int_{S^2}\omega=\int_{\{x_3\ge 0\}}d\psi^++\int_{\{x_3\le 0\}}d\psi^-=\int_{\{x_3=0\}}\iota_{\{x_3=0\}}^*(\psi^+)-\iota_{\{x_3=0\}}^*(\psi^-)$

But now I have no idea how to proceed. Thanks in advance for your help!

Also, I realize that this question has been answered before, but keep in mind I'm looking for a solution that does not use de Rham's theorem.

Answer 1

I believe I almost have a direct solution starting where your original post left off. I'll write $S^1$ for the equator and leave off all the ugly $\iota^*$s. Since $$\int_{S^1}( \psi^+ - \psi^- )=0,$$ we know by the 1-dimensional analog of the theorem we are trying to prove (thankfully that case is easy) that $\psi^+-\psi^-$ is exact, so denote it by $df$. Smoothly extend the function $f$ to the full top hemisphere while preserving $d\tilde f|_{S^1} = df|_{S^1}$.

Now define a 1-form $\tilde\psi^+$ on the top hemisphere by $$\tilde\psi^+ = \psi^+ - d\tilde f$$ so that $\tilde\psi^+ = \psi^-$ on the equator and $d\tilde\psi^+ = \omega$ on the top hemisphere.

The part I'm not sure about is stitching together $\tilde\psi^+$ and $\psi^-$ along the equator - while their values and their exterior derivatives coincide, as far as I can tell it's possible that the other components of the differentials (i.e. the "divergence" part) would have a discontinuity. If anyone knows how to rule this out please leave a comment!

Answer 2

Here is one other potential solution.

If $\omega$ is uniformly $0$, then we are obviousy finished. Otherwise, we have that $\int_{S^2} \omega = 0$, which implies that at some $p$ and $q$ on $S^2$, $\omega$ is positive and negative respectively. By the intermediate value theorem, there exists some point $r \in S^2$ at which $\omega = 0$. It is therefore sufficient to integrate over $S^2 - \{r\}$. Since $\omega$ is a top form, $d\omega = 0$. The punctured sphere $S^2 - \{r\}$ is contractible (since it's a disk), implying by Poincare's Lemma that $\omega$ is exact.