(A qual problem) Let $\pi:S^{n}\rightarrow M$ be a covering map, $M$ being an orientable manifold. Show that $H\_{deR}^{k}(M)=0$ for $1\leq k < n $. 


We can show $H\_{deR}^{1}(M)=0$ by the following argument. For a closed 1-form $\omega$ on $M$, $\pi^{\*}\omega $ is closed on $S^{n}$ thus exact, so $\pi^{\*}\omega$ is equal to some $df$, $f$ being a smooth function on $S^{n}$. For any deck transformation $g$ on $S^{n}$, $g^{\*}\pi^{\*}\omega=\pi^{\*}\omega$, thus we deduce that $d(g^{\*}f-f)=0$, so $g^{\*}f-f=C$ for some constant $C$. Then we apply $g^{\*}$ on $g^{\*}f-f=C$, we get` $(g^{2})^{\*}f-g^{\*}f=C$. Continue applying $g^{\*}$, and noting that $g$ is of finite order, say $m$, we finally get $f-(g^{m-1})^{\*}f=C$. Adding these $m$ equalities together, we get $C=0$. Thus $f$ is deck transformation invariant, then $h=f\circ \pi^{-1}$ is well defined on $M$ and $\omega=dh$. Thus $\omega$ is exact.


This argument only uses the fact that $S^{n}$ is of $H\_{deR}^{1}$ zero, and compact(to guarantee that $g$ is of finite order). I try to use induction for higher order deRham cohomology, but I seems to meet an essential obstacle. What special structures of $S^{n}$ should we use to carry on the argument, or is $M$ being orientable playing a role here? 


I am also interested in examples(other than $\mathbb{R}\mathbb{P}^{n}$) or even a classification of manifolds with covering space $S^{n}$.



deRham cohomology of a manifold with covering space $S^{n}$
A finite covering map induces an injection on de Rham cohomology. Try searching for "integration along the fibers"; yours is an easy case, as you integrate on a finite set of points, which means you just sum.


