[Math] De Rham cohomology groups of projective real space

Question

I would like to calculate the de De Rham cohomology groups of projective real space $\mathbb{RP}^{n}$. Well, i know all groups of De Rham cohomology os $n$-sphere $\mathbb{S}^{n}$ and that the map $\pi:\mathbb{S}^{n}\to \mathbb{RP}^{n}$ (the restriction to the projection map on $\mathbb{S}^{n}$) is a smooth submersion. With these tools, can I compute the cohomology groups of real projective space? How can I do this?

Answer 1

The following approach has the benefit that it is elementary in the sense that it only uses basic (homological and linear) algebra and the tools you allude to. More on this later*. I will just expand on the relevant parts of what is in this link.

You can use the antipode $A:S^n \to S^n$ to decompose $\Omega^k(S^n)$ in a direct sum of two subspaces $\Omega^k(S^n)_+ \oplus \Omega^k(S^n)_-$, since the induced map $A^*$ is such that $A^*{}^2=\mathrm{Id}$ (the subspaces are the eigenspaces of $1$ and $-1$, respectively). By naturality of $A^*$, $d$ respects that decomposition. It follows that $H^k(S^n) \simeq H^k(\Omega(S^n)_+)\oplus H^k(\Omega(S^n)_-)$ (the right side is "algebraic" cohomology originated from the complex $\Omega(S^n)_+$, resp. $\Omega(S^n)_-$), and it is clear that each one is the eigenspace of $A^*$ (on the cohomology level) of eigenvalue $+1$ and $-1$, respectively.

Since $A$ has degree $(-1)^{n+1}$ (it is a diffeomorphism that reverses/preserves orientation accordingly with the parity of $n$) and $H^n(S^n;\mathbb{R})=\mathbb{R}$, we have that $H^n(S^n)=H^n(\Omega(S^n))_+$ if $n$ is odd, and $H^n(S^n)=H^n(\Omega(S^n))_-$ if $n$ is even.

If we show that $\pi^*: \Omega^k(\mathbb{R}P^n) \to \Omega^k(S^n)_+$ is a isomorphism (the fact that the image indeed lies on $\Omega^k(S^n)_+$ follows from the fact that $\pi A=\pi$, because then $A^*\pi^*=\pi^*$), since it is natural it will follow that $H^k(\mathbb{R}P^n) \simeq H^k(\Omega(S^n))_+$, and hence we will have calculated the de Rham cohomology of $\mathbb{R}P^n$, since we know who are the $H^k(\Omega(S^n))_+$ (this is a good moment to see if you are following up: what is $H^k(\Omega(S^n))_+$, in terms of $n,k$? The only one to think about really is when $n=k$, the rest must be trivial - except $H^0$, of course).

To show that it is an isomorphism, you can check in the pdf I linked or use the answers by Mariano and user432847 here, which deal with a more general statement. Maybe seeing both would be best, since the general outlook that Mariano gives makes the fact that $\pi^*$ is an isomorphism with $\Omega^k(S^n)_+$ more intuitive.

*There are other methods to compute the cohomology of the projective space. Cellular homology + universal coefficients + deRham theorem is one example which you may want to check out.