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I met a question as follows:

Suppose $G$ is a finite group acting freely on a manifold $M$. Show the following: 1) $M/G$ is a manifold. 2) $H^i_{dR}(M/G)=H^i_{dR}(M)^G$. 3) if $N$ is a compact connected $n$- dimensional manifold which is non-orientable, then $H^n_{dR}(n)=0$.

For the first question, I can understand it in the topological manifold case. When it comes to the smooth manifold, I cannot see clearly how the local structure is preserved to be smooth during the quotient. Could someone explain it? Is it necessary to have $G$ to be finite, or is it possible to have infinite $G$ with discrete topology?

The second one should be easier. I would say the relation holds for the algebra of smooth maps which descends to cohomology. The third one is easy using Poincare duality, but could any one provide an solution via part two?

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up vote 8 down vote accepted

For the first question, the assomption is still true if you ask $G$ to be a discret group (or a Lie group) acting smoothly, freely and properly discontinuously on $M$.

This is a technical verification, if you need details this is really well treated in this book : Introduction to Smooth manifolds - J.M. Lee p153

For question 2), there are two points :

  • the covering map $p:M\rightarrow M/G$ induces an injective map (the pull-back) $p^\star:H_{dR}^\bullet(M/G)\rightarrow H_{dR}^\bullet(M)$ :

Since $G$ is finite and $g^\star \circ p^\star=p^\star$ for every $g\in G$, if $p^\star \alpha=d\beta$ then for every $g\in G$, $g^\star \alpha = d\beta$ also. Hence you can write $p^\star \alpha= \dfrac{1}{\# G}\sum_{g\in G}g^\star d\beta=d(\dfrac{1}{\# G}\sum_{g\in G}g^\star \beta)=d\tilde \beta$. But $\tilde \beta$ is invariant under $G$ and $p^\star:\Omega^\bullet(M)\rightarrow \Omega^\bullet (M/G)$ is injective so $\alpha$ is exact.

  • the image of this map is exactly $H_{dR}^\bullet(M)^G$.

For the third question, you can consider the orientable double cover $M^\star\rightarrow M=M^\star/G$ with $G=\mathbb Z/2\mathbb Z$. This is an orientable smooth manifold by construction and $M$ is not orientable iff $M^\star$ is connected.

$\triangleright$ That allows you to use the isomorphism $H_{dR}^n(M^\star)\cong \mathbb R$ given by integration on $M^\star$.

According to point 2), you have to show that $H_{dR}^n(M^\star)^G=\{0\}$ :

Let $\omega\in \Omega^n(M^\star)$ be a closed $n$-form. Note that the only non trivial deck transformation of the orientable double cover is the one that reverses the orientation on $M^\star$, let's denote this map by $f:M^\star\rightarrow M^\star$.

The closed $n$-form $\omega$ is invariant under the action of $G$ iff $f^\star \omega=\omega$ and under this assumption, we get : $$\int_{M^\star}\omega=\int_{M^\star}f^\star\omega = -\int_{M^\star} \omega$$ Hence $\omega$ is an exact n-form and $[\omega]=0\in H_{dR}^n(M^\star)$.

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