De Rham Cohomology of $\mathbb{P}_\mathbb{R}^2$

Question

I'm trying to show that every closed 1-form on $\mathbb{P}_\mathbb{R}^2$ is exact. Towards this end, let $\alpha:S^2\longrightarrow S^2$ be the antipodal map, and $\pi:S^2\longrightarrow \mathbb{P}_\mathbb{R}^2$ the canonical quotient map. If $\omega\in\Omega^1(\mathbb{P}_\mathbb{R}^2)$, the naturality of $d$ shows that $\pi^*\omega$ is closed. Because $H^1(S^2)=0$, we also have $\pi^*\omega=df$ for some $f\in C^\infty(S^2)$. Then, using the fact that $\pi\circ \alpha=\pi$, I want to "average" $f$ to produce a well-defined map $g:\mathbb{P}_\mathbb{R}^2\longrightarrow \mathbb{R}$. Fix $p\in S^2$, and define $$g(p)=\frac{1}{2}\left(f(p)+f(-p)\right)=\frac{1}{2}\left(f(p)+(\alpha^*f)(p)\right).$$ Since $g(p)=g(-p)$, this map is also well-defined on $\mathbb{P}_\mathbb{R}^2$ (and clearly smooth). Hence $g$ descends to a smooth map $g:\mathbb{P}_\mathbb{R}^2\longrightarrow \mathbb{R}$. Now, \begin{align*} dg_p(v)&=\frac{1}{2}\left(df_p(v)+\alpha^*d_pf(v)\right)\tag{naturality of $d$}\\ &=\frac{1}{2}\left(\pi^*\omega_p(v)+\alpha^*\pi^*\omega_p(v)\right)\\ &=\frac{1}{2}\left(\pi^*\omega_p(v)+(\pi\circ\alpha)^*\omega_p(v)\right)\\ &=\frac{1}{2}\left(\pi^*\omega_p(v)+\pi^*\omega_p(v)\right)\tag{$\pi\circ \alpha=\pi$}\\ &=\pi^*\omega_p(v)\\ &=\omega_{\pi(p)}(d\pi_p(v)). \end{align*} I would like to conclude that $dg=\omega$, but it seems like my definitions are set up so that $v\in T_pS^2$, because we are plugging in $d\pi_p(v)$ into $\omega\in\Omega^1(\mathbb{P}_\mathbb{R}^2)$. Would anyone please let me know where I have gone wrong, or what I can do to properly modify my argument?

Answer 1

Here's your mistake: when writing $g(p) = (f(p)+f(-p))/2$, this lives on the sphere, not on the projective plane. For $g\colon S^2 \to \Bbb R$ so defined, let $\widetilde{g} \colon \mathbb{P}^2_{\mathbb{R}}\to \Bbb R$ be the induced function, so that $\widetilde{g} \circ \pi = g$. Then ${\rm d}g = \pi^*\omega$ reads $\pi^*({\rm d}\widetilde{g}) = \pi^*\omega$. As $\pi$ is a surjective submersion, $\pi^*$ is injective and therefore ${\rm d}\widetilde{g} = \omega$, as required.

By the way, your idea of averaging $f$ actually shows a more general result: if $M$ is any smooth manifold with $H^1(M)=0$ and $\Gamma$ is a finite group acting freely and properly discontinuously on $M$ with a smooth quotient $M/\Gamma$, then $H^1(M/\Gamma)=0$. (Maybe even replacing $\Gamma$ with a compact Lie group works, replacing the average with integration against the Haar measure.)