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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.

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    $\begingroup$ Perhaps try to deform $\psi ^+$ by a closed form so that it matches up smoothly with $\psi ^-$? $\endgroup$ – Anthony Carapetis Aug 5 '13 at 3:58
  • $\begingroup$ Can't you use the fact that $H^2(S^2)$=$\mathbb Z$ ,which has a single generator, and, I think, $H^2$ is generated by the volume form, which does not integrate to $0$ , nor does any non-zero multiple of it; i.e., the only non-trivial 2-forms on $S^2$ are volume forms, which do not integrate to 0. $\endgroup$ – DBFdalwayse Aug 5 '13 at 5:19
  • $\begingroup$ @DBF: The OP is looking for a proof without using de Rham's theorem; i.e. without knowing that $H^2_{dR}(S^2) = H^2(S^2;\mathbb{R}).$ One would have to prove that $H^2_{dR}(S^2) = \mathbb{R}$ without appealing to de Rham's... $\endgroup$ – Anthony Carapetis Aug 5 '13 at 6:30
  • $\begingroup$ @Anthony; my bad. How about something of this sort: using orientability of $S^2$ so that there is a top nowhere-zero form, which cannot be exact, or it would integrate to zero. And we know the space of top forms is 1-dimensional. Then we must show that it is isomorphic to $\mathbb R$ $\endgroup$ – DBFdalwayse Aug 5 '13 at 6:37
  • $\begingroup$ I think we can show the orientability of $S^2$ by using, say a normal vector field, theorem, and we can show orientability is equivalent to having a nowhere-zero top form, without using DeRham's theorem for neither. In the worse case, I think we may be able to show $H^2_{dR}(S^2)=\mathbb R$ using Mayer-Vietoris. $\endgroup$ – DBFdalwayse Aug 5 '13 at 7:02
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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!

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    $\begingroup$ Consider open neighborhoods $U_+$, $U_-$ of the closed upper and lower hemisphere, then $\psi^+ -\psi^-$ is a $1$-form defined on $U_+\cap U_-$, satisfying $\int_{S^1}(\psi^+ -\psi^-)=0$. By the Poincare lemma and the $1$-dimensional analog, $\psi^+ -\psi^-=df$ for some smooth $f$ defined on $U_+\cap U_-$. Let $\{\rho_+,-\rho_-\}$ be the partition of unity associated to $\{U_+, U_-\}$, then $f=\rho_+ f-\rho_- f$. Hence $\psi^+ + d(\rho_- f) = \psi^- +d(\rho_+ f)$ on $U_+\cap U_-$, so we glue them together to get $\psi$ s.t. $\omega=d\psi$. QED $\endgroup$ – Yuchen Liu Aug 5 '13 at 14:18
  • $\begingroup$ @YuchenLiu: looks good! Just to confirm I understand: you are taking the same $\psi^\pm$ as before and extending them smoothly to neighbourhoods? And you get $\psi^+ - \psi^-$ exact on the annulus $U_+ \cap U_-$ since the homotopy invariance of $H_{dR}^1$ implies it is generated by the "rotation" form $d\phi$ ($\phi$ the azimuthal angle coordinate) which has non-zero integral on $S^1$, yes? $\endgroup$ – Anthony Carapetis Aug 6 '13 at 4:32
  • $\begingroup$ I am not extending $\phi^\pm$ to $U_\pm$. Instead, I choose $U_\pm$ to be contractible, then choose $\phi^\pm$ directly on $U_\pm$ satisfying $d\phi^\pm=\omega$. $\endgroup$ – Yuchen Liu Aug 9 '13 at 2:49
  • $\begingroup$ If $\omega$ is a $1$-form on $S^1$ then $\int_{S^1} \omega =0$ implies that $\omega$ is exact. If $\omega$ is a $1$-form on a manifold $V$ that is homotopy equivalent to $S^1$ (like $U_+\cap U_i$), and $\int_V \omega=0$, then how do we draw the same conclusion that there exists a function $f$ on $V$ (not on $S^1$) such that $\omega=df$? $\endgroup$ – Dimitris Aug 24 '14 at 0:42

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