Finite universal covering induces injective maps on cohomology

Question

I am trying to prove the following:

Suppose $M$ is a smooth, connected manifold with finite fundamental group and $f : \widetilde{M} \rightarrow M$ is its (smooth) universal cover. Show that $f^* : H_{dR}^k(M) \rightarrow H_{dR}^k(\widetilde{M})$ is injective for every $k$.

For $k = 0$ this is clear by the connectedness of $M$, and for $k = 1$ this is clear because $\lvert \pi_1(M)\rvert < \infty$ implies that $H_{dR}^1(M) = 0$. Beyond that, I can determine that $f$ is a finite cover since $[\pi_1(M) : f_*(\pi_1(\widetilde{M}))] = \lvert \pi_1(M)\rvert < \infty$ is the number of covering sheets. This tells me that $f$ is a proper local diffeomorphism, and indeed that $M \cong \widetilde{M}/\pi_1(M)$. I have not been able to gain any steam from this though. My thought is that I should be able to find an inductive argument, but unfortunately I am not able to progress from $k = 1$ to $k = 2$. Can anyone give me some guidance on this one please?

Answer 1

In general, if a finite group $G$ acts on a manifold $M$ without fixed points and we let $N=M/G$, which is also a manifold, and $p:M\to N$ is the canonical projection map, then $p$ induces a map $p^*:H^*(N)\to H^*(M)$ which is injective. In fact, the action of $G$ on $M$ induces one on $H^*(M)$, and the image of $p^*$ is precisely $H^*(M)^G$, the set of $G$-invariant elements of $H^*(M)$.

To check the injectivity of $p^*$, which is what you want, you need to check that if $\omega$ is closed form on $N$ such that the pullback $p^*\omega$ is exact, then $\omega$ itself is exact. Now, that $p^*\omega$ be exact means that there is a form $\eta$, one degree lower, such that $p^*\omega=d\eta$.

We want, at this point, to show that there is a form $\xi$ on $N$ such that $\omega=d\xi$, and we have $\eta$ to work with, so one might imagine that we can construct $\xi$ from $\eta$ somehow. The key point is that we need some criterion to solve the following problem:

given a form on $M$, when is it the pullback of a form on $N$?

For example, one can easily check that a necessary condition is that the form on $M$ you start with be invariant under the action of $G$.

If you come up with a criterion which can decide this and it applies to $\eta$, then you can "descend" $\eta$ to $N$ and so on. I suggest you try to do this.

Answer 2

Based on Mariano's suggestion (and terminology), I believe I have a solution for this problem:

First, we can see that $p^*\omega$ must be $G$-invariant because for each $g \in G$ we have $p\circ g = p$, which means at the level of cohomology we have $g^*\circ p^* = p^*$. Hence $g^*(p^*\omega) = p^*\omega$ and so $p^*\omega$ is $G$-invariant. In fact we can then see that for any form $\theta \in \Omega^*(M)$ which is $G$-invariant, the form $\overline{\theta}$ defined by $\overline{\theta}_y(X) := \theta_x(\widetilde{X}_x)$ (for $x \in p^{-1}(y)$ and $dp_x(\widetilde{X}_x) = X$ where $\widetilde{X}_x$ exists because $dp_x$ is an isomorphism) is well-defined because $x_1,x_2 \in p^{-1}(y)$ implies there is a $g \in G$ with $gx_1 = x_2$ and $p\circ g = p$ implies $dp_{x_2}(dg_{x_1}(\widetilde{X}_{x_1})) = dp_{x_2}(\widetilde{X}_{x_2})$. So $dg_{x_1}(\widetilde{X}_{x_1}) = \widetilde{X}_{x_2}$ since $dp_{x_2}$ is an isomorphism and hence $\theta_{x_1}(\widetilde{X}_{x_1}) = \theta_{x_2}(\widetilde{X}_{x_2})$ by $G$-invariance.

Now suppose $\omega \in H^k(N)$ is such that when we pull it back to $M$ we get $p^*\omega = d\eta$ for some $\eta \in H^{k-1}(M)$. From the discussion above, we see that $d\eta$ is $G$-invariant. i.e. $g^*(d\eta) = d\eta$. This implies $d(g^*\eta - \eta) = 0$ so $g^*\eta - \eta$ is only closed (not $0$ as we would like). Getting around this can be done by making the following definition: let $\varphi := \frac{1}{\lvert G \rvert}\sum_{g \in G}g^*\eta$. With this definition, $\varphi$ is $G$-invariant and $d\varphi = \frac{1}{\lvert G \rvert}\sum_{g \in G}d(g^*\eta) = \frac{1}{\lvert G \rvert}\sum_{g\in G}g^*(d\eta) = d\eta$. So $\varphi$ descends to the form $\overline{\varphi}$ on $N$.

We can see that $d\overline{\varphi} = \omega$ by the following: \begin{align*} d\overline{\varphi}_y(X) & = d\varphi_x(\widetilde{X}_x) \hspace{10pt}\mbox{(for $p(x) = y$ and $\widetilde{X}_x$ as defined above)}\\ & = d\eta_x(\widetilde{X}_x)\\ & = (p^*\omega)_x(\widetilde{X}_x)\\ & = \omega_{p(x)}(dp_x(\widetilde{X}_x))\\ & = \omega_y(X) \end{align*}

So $p^*\omega$ exact implies $\omega$ is exact. Thus $p^* : H^k(N) \rightarrow H^k(M)$ is injective for each $k$. This implies the result for my original question in the case that $G$ is the fundamental group of the codomain.