Isomorphism between the cohomology of projective spaces and spheres

Question

An invariant form on $S^{n}$ is a form $\omega$ such that $i^{*} \omega=\omega$,where $i$ is antipodal map. The vector space of invariant forms on $S^{n}$, denoted $\Omega^{*}(S^{n})^{I}$, is a differential complex and so the invariant cohomology $H^{*}(S^{n})^{I}$ of $S^{n}$ is defined. Show that $H^{*}(RP^{n})$ isomorphic to $H^{*}(S^{n})^{I}$?

Answer 1

One can show that there is an isomorphism of co-chain complexes $\Omega^*(\mathbb{RP}^n) \to \Omega^*(S^n)^I$, whence the isomorphism in cohomology follows. Let $\pi : S^n \to \mathbb{RP}^n$ denote the canonical projection. Given $k \in \{0,\ldots,n\}$, define the linear map $$f_k : \Omega^k(\mathbb{RP}^n) \to \Omega^k(S^n)^I$$ via the rule $$f_k(\theta) = \pi^*(\theta)$$ for $\theta \in \Omega^k(\mathbb{RP}^n)$. Since $\pi\circ i = \pi$, we have $i^*(\pi^*\theta) = (\pi\circ i)^*\theta = \pi^*(\theta)$, so indeed $f_k(\theta) \in \Omega^k(S^n)^I$. Since $\pi^*(d\theta) = d(\pi^*\theta)$, the collection of the $f_k$'s forms a homomorphism of co-chain complexes $\Omega^*(\mathbb{RP}^n) \to \Omega^*(S^n)^I$.

Since $\pi : S^n \to \mathbb{RP}^n$ is a covering map, the induced map $T_x\pi : T_xS^n \to T_{\pi(x)}\mathbb{RP}^n$ is an isomorphism for all $x \in S^n$. This allows us to define the linear map $$g_k : \Omega^k(S^n)^I \to \Omega^k(\mathbb{RP}^n) $$ given by $$(g_k(\omega))_y = \omega_x \circ \wedge^k((T_x\pi)^{-1})$$ for $\omega \in \Omega^k(S^n)^I$, $y \in \mathbb{RP}^n$ and $x \in S^n$ any of the two points in $\pi^{-1}(y)$. Here, $\wedge^k((T_x\pi)^{-1}) : \wedge^k(T_y\mathbb{RP}^n) \to \wedge^k(T_xS^n)$ denotes the $k$th exterior power of the map $(T_x\pi)^{-1} : T_y\mathbb{RP}^n \to T_xS^n$. Using the invariance of $\omega$ under $i$, i.e. the fact that $\omega_x = \omega_{i(x)} \circ \wedge^k(T_x i)$ for $x \in S^n$, one can show that the right hand side of the defining equation for $g_k(\omega)$ depends only on $y \in \mathbb{RP}^n$, but not on the choice of the point $x \in \pi^{-1}(y) \subset S^n$. Then, using a calculation in local charts, one can show that $g_k(\omega)$ is a smooth differential $k$-form.

By construction, $f_k$ and $g_k$ are mutually inverse linear maps. Therefore, the co-chain complexes $\Omega^*(\mathbb{RP}^n)$ and $\Omega^*(S^n)^I$ are isomorphic. Moreover, since $\pi^*(\theta \wedge \varphi) = \pi^*(\theta) \wedge \pi^*(\varphi)$ for any $\theta \in \Omega^k(\mathbb{RP}^n)$, $\varphi \in \Omega^\ell(\mathbb{RP}^n)$, we have also established an isomorphism of $\Omega^*(\mathbb{RP}^n)$ and $\Omega^*(S^n)^I$ as algebras. The induced isomorphism $H^*(\mathbb{RP}^n) \to H^*(S^n)^I$ is an isomorphism of algebras (under the cup product) as well.