I'm trying to cook up a non-vanishing vector field on $\mathbb{R}P^{2n+1}$. I know that $S^{2n+1}$ admits one, namely $(x_1,\dots,x_{2n+2})\mapsto (-x_2,x_1,\dots,-x_{2n+2},x_{2n+1})$. Moreover, I know that $S^{2n+1}$ is a smooth double cover of $\mathbb{R}P^{2n+1}$ via the map $x\mapsto \{x,-x\}$. Since this vector field is odd, $X(p)=-X(-p)$, I was hoping there might be a way to cook up a vector field on $\mathbb{R}P^{2n+1}$. So, this motivates the two following questions:
Specifically, how may one explicitly construct a non-vanishing vector field on $\mathbb{R}P^{2n+1}$ (using the route above or not).
Say $\tilde M$ and $M$ are smooth manifolds, and $p:\tilde{M}\to M$ is a smooth covering map. If $X(p)$ is a smooth vector field on $\tilde{M}$, under what conditions is there a natural way to cook up a vector field on $M$? (I don't mean natural in the rigorous sense).
Thanks!
More generally. Suppose that a group $G$ acts properly discontinuously on a manifold $M$ and that you have a vector field $X$ on $M$ which is invariant under $G$, so that for all $g\in G$ and all $p\in M$ we have $$d_pg(X_p)=X_{gp}.$$ Then the quotient $M/G$ is a manifold, the canonical projection $\pi:M\to M/G$ is smooth and locally a diffeo, and there is a vector field $Y$ on $M/G$ such that $d_p\pi(X_p)=Y_{\pi(p)}$ for all $p\in M$.
In particular, if the field $X$ happens to be everywhere non-zero, the field $Y$ has the same property.