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(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 $S^{n}$.

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up vote 4 down vote accepted

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.

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I tried to search for it and then tried to read one of the articles, but it is quite involved and confusing to me. So could you please explain to me briefly about this integration along the fibers(see this as a challenge:)) or could you recommend me some references more or less introductory? – Yunfeng Mar 22 '13 at 7:09
A covering map $f\colon X \to Y$ is a local diffeomorphism; a form on $X$ around a point $p$ gives a form on $Y$ around $f(p)$. If you have a form on $X$, for each point $q$ of $Y$ sum over the form you get from each point of $f^{-1}(q)$; the forms you get glue together to a global form on $Y$. – Angelo Mar 22 '13 at 8:18
Ah, I am so stupid! So the result has nothing to do with orientability of $M$, only the finiteness of the covering. I guess there are similar results for the case when the fibre is some compact Lie group with Haar measure for which the summing(integration) also works. I will look it up. Thank you for your patience! – Yunfeng Mar 22 '13 at 8:50
Yunfeng: This construction is also called transfer or shriek map. – Misha Mar 22 '13 at 14:05

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