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

Solution 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.