$H_{dR}^k(M/G)\to H_{dR}^k(M)$ is injective

180 Views Asked by At

Let $M$ be a smooth manifold and $G$ a finite group of automorphisms acting properly on it. If $\pi:M\to M/G$ is the projection map, prove that $\pi^*:H_{dR}^k(M/G)\to H_{dR}^k(M)$ is injective.

I know how to prove that $\pi^*:\Omega^k (M/G)\to\Omega^k (M) $ is injective using the fact that $\pi$ is a surjective submersion. But how do I prove injectiveness in cohomology?

1

There are 1 best solutions below

0
On BEST ANSWER

First of all, instead of "action properly", one should say "acting freely", otherwise, the quotient map $q: M\to N=M/G$ cannot be a submersion.

Define the following averaging operator: $$ A: \Omega^*(M)\to \Omega^*(M), A(\lambda)= \frac{1}{|G|}\sum_{g\in G} g^*\lambda. $$ Clearly, $A\circ d= d\circ A$ and $A$ is a retraction from $\Omega^*(M)$ to the subspace of $G$-invariant forms.

Now, suppose that $\omega$ is a closed form on $N$ such that $q^*\omega$ is exact, $q^*\omega=d\eta$. Thus, $A(q^*\omega)=q^*\omega$ and $dA(\eta)= q^*\omega$. Since $A(\eta)$ is $G$-invariant, it projects to a form $\theta$ on $N$ and $d\theta= \eta$. Hence, $q^*$ induces an injective map $H^*_{DR}(N)\to H^*_{DR}(M)$. qed