1 forms on the sphere

Question

Let $\pi:\mathbb{S}^2\to\mathbb{R}P^2$ be the canonical projection. Show that an $1$-form $\beta$ in $\mathbb{S}^2$ satisifies $\beta = \pi^*\rho$, for some $\rho$ $1$-form in $\mathbb{R}P^2$ if and only if $a^*\beta= \beta$, where $a:\mathbb{S}^2\to \mathbb{S}^2$ is the antipodal map $a(p) = -p$.
I'm at a loss for what to do in both directions. I've tried writing the definitions, but I'm not getting anywhere. For example, for the forward direction, I have that if $X\in T_p\mathbb{S}^2$ then $$\beta(X) = \beta_p(X) = (\pi^*\rho)_p(X) = \rho_{\pi(p)}((d\pi)_p(X)) = \rho_{\pi( a(p))}((d\pi)_p(X))$$ But I don't know there to go from there. Any tips?
Edit: Ok, so I figured out the forward implication, here's what I've come up for the other one:
So we must define $\rho$ a 1 form in $\mathbb{R}P^2$. Let $X\in T_{\pi(p)}\mathbb{R}P^2$. Since $(d\pi)_p$ is surjective, $\exists Y\in T_p\mathbb{S}^2$ such that $X = (d\pi)_p(Y)$, so define $$\rho_{\pi(p) }(X) = \beta_p(Y)$$ Then $\rho$ clearly satisfies $\beta = \pi^*\rho$. I'm just having a hard time trying to prove $\rho$ is well defined. Clearly if $\pi(q) = \pi(p)$ then $q=p$ or $q=-p=a(p)$, so that $\beta_q=\beta_{a(p)}$, but I'm not sure how to deal with the $Y$.

Answer 1

It's overkill time: let $G$ be a Lie group acting freely and properly discontinuously on a smooth manifold $M$, so that the quotient $M/G$ is a smooth manifold for which the quotient projection $\pi\colon M\to M/G$ is a surjective submersion.

Claim: $\Omega^k(M/G) \cong \{ \alpha \in \Omega^k(M) \mid \alpha\mbox{ is $G$-invariant and $G$-horizontal} \}$, where $G$-invariant means that $g^*\alpha = \alpha$ for all $g\in G$, where we abuse notation and write $g\colon M\to M$ for the action map given by $g(x) = g\cdot x$, and $G$-horizontal means that $\alpha_x(v_1,\ldots, v_k) = 0$ whenever there is $1\leq i \leq k$ such that $v_i\in \ker {\rm d}\pi_x$.

Before we move on to justify this, note that $T_{G\cdot x}(M/G) \cong T_xM/\ker{\rm d}\pi_x$ for every $x\in M$ (this is the first isomorphism theorem from linear algebra), so $G$-horizontality is automatically satisfied whenever $G$ is discrete (so $\ker {\rm d}\pi_x$ is trivial, as $\pi$ becomes a smooth covering and hence a local diffeomorphism). The example you have there is just when $$M = \Bbb S^2,\quad G = \Bbb Z_2 = \{{\rm Id}_{\Bbb S^2}, a\},\quad M/G = \Bbb R{\rm P}^2,\quad\mbox{and}\quad k=1.$$ Understand the general mechanism and you'll never have to do any ad hoc analysis whenever a quotient manifold appears in front of you again.

Proof of claim: assume that $\alpha\in \Omega^k(M)$ is $G$-invariant and $G$-horizontal, and define $\widetilde{\alpha} \in \Omega^k(M/G)$ by $$\widetilde{\alpha}_{G\cdot x}(\hat{v}_1,\ldots, \hat{v}_k) = \alpha_x(v_1,\ldots, v_k),$$where $\hat{v}_1,\ldots,\hat{v}_k \in T_{G\cdot x}(M/G)$ and $v_1,\ldots, v_k\in T_xM$ are such that ${\rm d}\pi_x(v_i) = \hat{v}_i$ for all $1\leq i \leq k$. The issue here is showing that $\widetilde{\alpha}$ is well-defined, as once this is in place, $\widetilde{\alpha}$ is the unique $k$-form with $\pi^*\widetilde{\alpha}=\alpha$. So, replace $x$ with $g\cdot x$ and $v_1,\ldots, v_k$ with vectors $v_1',\ldots, v'_k \in T_{g\cdot x}(M)$ such that ${\rm d}\pi_{g\cdot x}(v_i') = \hat{v}_i$ for all $1\leq i \leq k$. Then we have that $\pi\circ g = \pi$ and the chain rule together imply that the difference $v_i' - {\rm d}g_x(v_i)$ is in $\ker {\rm d}\pi_{g\cdot x}$, for all $1\leq i \leq k$. Now, telescope: $$\begin{align} \alpha_{gx}(v_1',\ldots,v_k') - \alpha_x(v_1,\ldots,v_k) &= \alpha_{gx}(v_1',\ldots,v_k') - \alpha_{gx}({\rm d}g_x(v_1),\ldots,{\rm d}g_x(v_k)) \\ &= \sum_{i=1}^{k} \alpha_{gx}({\rm d}g_x(v_1),\ldots,{\rm d}g_{x}(v_{i-1}), v_i'-{\rm d}g_x(v_i),v_{i+1}',\ldots, v_k') \\ &=0, \end{align}$$where in the first equality $G$-invariance enters, and in the last $G$-horizontality enters. You may want to expand this sum for $k=3$ to get a feeling for what's happening.

Then, $\alpha \mapsto \widetilde{\alpha}$ is a well-defined map from the set of $G$-invariant and $G$-horizontal $k$-forms on $M$, and it is a bijection as its inverse is the pull-back map $\pi^*$; it's easy to see that if we start with $\widetilde{\alpha} \in \Omega^k(M/G)$, then $\pi^*\widetilde{\alpha}$ is $G$-invariant and $G$-horizontal. This concludes the argument.

Observe that $\alpha$ being alternating was irrelevant: this tells you precisely when covariant tensor fields on $M$ survive in $M/G$, in general.