Galois covering induces an isomorphism on the level of (co)homology

Question

The setting is the following : we have a smooth Galois cover of manifolds $p : Y \to X$, with (Galois) automorphism group $G$. Denote by $\Omega^*(X)$ and $\Omega^*(Y)$ the spaces of differential forms on $X$ and $Y$ respectively. Somehow we want to express the cohomology of $X$ and $Y$ in terms of each other (motivating example : expressing the cohomology of projective space in terms of that of the sphere). This question is strongly related, but here we are concerned with the technical details of the claim. For now we are only interested in homology.

Firstly we want a nice action of $G$ on $\Omega^*(Y)$. I think we can define one as follows : if $g$ is a covering space automorphism of $Y$, we can associate an automorphism $g^* : \Omega^*(Y) \to \Omega^*(Y)$. The action induced by $G$ is an action by such automorphisms.

Now denote by $\Omega^*(Y)^G$ the subalgebra of differential forms on $Y$ that are fixed by all automorphisms $g^*$.

Claim : The induced map $p^*:\Omega^*(X) \to \Omega^*(Y)^G$ is an isomorphism.

This is where I encounter some problems. Firstly I can show that $p^*(\Omega^*(X)) \subset \Omega^*(Y)^G$ (the covering map $p$ and some automorphism $g$ of covering spaces form a commutative diagram. This gives us a "reverse" commutative diagram in homology with $p^*$ and $g^*$ from which we can deduce the inclusion). However the reverse inclusion seems difficult. We may have to use at some point the fact that $p$ is Galois ($Y=X/G$) but it gets more confusing as I go forward.

Moreover we should be able to find an explicit inverse for $p^*$ but I am unable to do so. Is my group action even reasonable ? Any help in either direction would be greatly appreciated.

Answer 1

Here is a very quick sketch of the -hopefully correct- proof of the isomorphism (criticism is welcome and highly encouraged !). To avoid confusion "the action of $G$" will denote the standard action by automorphisms on Y, and "the action of $G^*$" will denote the action by the induced automorphisms $g^*$ on $\Omega^*(Y)^G$. We will try to build the inverse explicitly (locally, then gluing everything together).

• Take $U$ in $X$ a trivializing open set for $p$. Denote by $V$ one of the sheets in $Y|_U$. Take a form $\omega \in \Omega^*(V)^G$.
• Recall that forms in $\Omega^*(Y)^G$ are fixed by the action of $G^*$. Moreover, since $p$ is Galois, the action of $G$ on $Y$ is transitive. In particular, this gives us that the choice of $V$ doesn't matter, because $\omega$ will be the same on all the sheets.
• Now $p$ is a homeomorphism $V \to U$, so $p^{-1}|_U : U \to V$ exists and so does $(p^{-1}|_U)^*:\Omega^*(V)^G \to \Omega^*(U)$. By the previous point, this yields an isomorphism $\Omega^*(Y|_U)^G \to \Omega^*(U)$.
• We can repeat the construction for another trivializing open $U'$ in $X$. Forms upstairs in $Y$ will glue because they are defined globally, and thus they will glue downstairs in $X$. Thus we obtain an isomorphism $\Omega^*(Y)^G \to \Omega^*(X)$.

This result is interesting because it allows us to prove "half" of the following, more difficult result in cohomology.

Theorem : The inclusion $i :\Omega^*(Y)^G \hookrightarrow \Omega^*(Y)$ induces a map $i^* : H^*(\Omega^*(Y)^G) \to H^*(Y)^G$ in cohomology. Moreover, if the group $G$ is finite, we have an isomorphism $$ H^*(X) \cong H^*(\Omega^*(Y)^G) \cong H^*(Y)^G $$

We can prove this with some technical work (considering "weighted" forms $\frac{1}{|G|}\sum_{g\in G}g.\omega$). But this is for another time. As a direct application of this result, we can calculate $H^*(\mathbb{P}^n(\mathbb{R}))$ thanks to that of $S^n$, as explained in the linked question.