construct nonvanishing one form on $\mathbb{RP}^3$ I met such a problem in my homework: construct an everywhere nonvanishing 1-form on $\mathbb{RP}^3$; can your construction be generalized to $\mathbb{RP}^n$?
I know nothing about the background of the exercise and have no idea how to deal with such a problem (I'm a beginner on manifolds and know very little about algebraic topology). I guess the concrete construction may push out some 1-form on $\mathbb{R}^3$. Can anyone give me some suggestions? Thanks!
 A: As observed in the comments, we have the following proposition:
Proposition:  Suppose $\pi:\overline{X}\rightarrow X$ is a normal covering map of smooth manifolds ("normal" in the sense that $\pi_\ast(\pi_1(\overline{X}))$ is a normal subgroup of $\pi_1(X)$).  Then a $k$-form $\overline{\omega}$ on $\overline{X}$ is of the form $\pi^\ast \omega$ for a $k$-form $\omega $ on $X$ if and only if $\overline{\omega}$ is invariant under the group of deck transformations.
Proof.  First, assume $\overline{\omega} = \pi^\ast \omega$ for some $k$-form $\omega$ on $X$.  Let $f$ be any deck group transformation, meaning that $\pi = \pi\circ f$.  Then \begin{align*} f^\ast\overline{\omega} &= f^\ast (\pi^\ast\omega)\\ &= (\pi \circ f)^\ast \omega\\&= \pi^\ast\omega\\&=\overline{\omega}.\end{align*}  Thus, $\overline{\omega}$ is invariant under the action of the deck group.  (The proof we just completed doesn't require a normal covering - any covering is ok.)
On the other hand, now assume that $\overline{\omega}$ is invariant under the deck group.  We define $\omega$ on $X$ as follows.  Pick $x\in X$ and let $U\subseteq X$ denote any evenly covered neighborhood of $x$, meaning that $\pi^{-1}(U)$ is a disjoint union of $U_\alpha$ where $\pi$ maps each $U_\alpha$ to $U$ diffeomorphically.  I will let $\pi_{\alpha}$ denote the restriction of $\pi$ to $U_\alpha$.  Then each $\pi_{\alpha}$ is a diffeomorphism from $U_\alpha$ to $U$.
Picking any single one of the $U_\alpha$ (which I'll call $U_0$), we define $\omega$ on $U$ by $\omega = (\pi_0^{-1})^\ast \overline{\omega}$.
What happens if we pick a different $U_\alpha$, say $U_1$?  Well, because $\pi$ is a normal covering, there is a deck group element $f$ which maps $U_0$ to $U_1$.  Then, $ \pi_0 = \pi_1\circ f$.  Thus, because $\overline{\omega}$ is invariant under the deck group action, we see that \begin{align*} \omega &= (\pi_0^{-1})^\ast \overline{\omega}\\ &= (f^{-1}\circ \pi_1^{-1})^\ast \overline{\omega} \\ &= (\pi_1^{-1})^\ast(f^{-1})^\ast \overline{\omega}\\ &= (\pi_1^{-1})^\ast \overline{\omega}.   \end{align*}   So, the definition of $\omega$ is independent of which $U_\alpha$ we pick.  In a similar fashion (with ommitted calculation), $\omega$ is also independent of the choice of trivializing neighborhood.
To conclude the proof, we need only demonstrate that $\pi^\ast \omega = \overline{\omega}$.  To see this, note that any $\overline{x}\in \overline{X}$ lies in some open set $U_0$ which is mapped diffeomorphically to an open set $U$ in $X$.  Then $\pi = \pi_0$ on $U_0$, so $\pi^\ast(\omega(\pi(\overline{x}))) = \pi_0^\ast( \pi_0^{-1})^\ast \overline{\omega} = \overline{\omega}$.  $\square$
Armed with this, to find a non-vanishing form on $\mathbb{R}P^3$, you can instead find a form on $S^3$ which is preserved by the antipodal map. As you wrote, the form $ydx - xdy + zdt - t dz$ does the trick (interpreting $(x,y,z,t)\in \mathbb{R}^4$).  And, as you noted, this generalizes to any odd dimensional $\mathbb{R}P^n$.
As mentioned in the comments, the even dimensional $\mathbb{R}P^n$s do not have a non-vanishing $1$-form on them, as a consequence of the Hairy Ball theorem for $S^{n}$.
