# Cohomology of projective plane

How I can compute cohomology de Rham of the projective plane $P^{2}(\mathbb{R})$ using Mayer vietoris or any other methods?

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I retagged your question because I don't (as of now) see how it relates to algebraic geometry. –  user641 Mar 28 '11 at 1:12
If you remove a point from $P^2$ you are left with something which looks like a Moebious band. You can use this to compute $H^\bullet(P^2)$.
Let $p\in P^2$, let $U$ be a small open neighborhood of $p$ in $P^2$ diffeomorphic to an open disc centered at $p$, and let $V=P^2\setminus\{p\}$. Now use Mayer-Vietoris.
The cohomology of $U$ you know. The open set $V$ is diffeomorphic to an open moebious band, so that tells you the cohomology; alternatively, you can check that it deformation-retracts to the $P^1\subseteq P^2$ consiting of all lines orthogonal to the line corresponding to $p$ (with respect to any inner product in the vector space $\mathbb R^3$ you used to construct $P^2$), and the intersection $U\cap V$ has also the homotopy type of a circle. The maps in the M-V long exact sequence are not hard to make explicit; it does help to keep in mind the geometric interpretation of $U$ and $V. Later: alternatively, one can do a bit of magic. Since there is a covering$S^2\to P^2$with$2$sheets, we know that the Euler characteristics of$S^2$and$P^2$are related by$\chi(S^2)=2\chi(P^2)$. Since$\chi(S^2)=2$, we conclude that$\chi(P^2)=1$. Since$P^2$is of dimension$2$, we have$\dim H^p(P^2)=0$if$p>2$; since$P^2$is non-orientable,$H^2(P^2)=0$; finally, since$P^2$is connected,$H^0(P^2)\cong\mathbb R$. It follows that$1=\chi(P^2)=\dim H^0(P^2)-\dim H^1(P^2)=1-\dim H^1(P^2)$, so that$H^1(P^2)=0$. Even later: if one is willing to use magic, there is lot of fun one can have. For example: if a finite group$G$acts properly discontinuously on a manifold$M$, then the cohomology of the quotient$M/G$is the subset$H^\bullet(M)^G$of the cohomology$H^\bullet(M)$fixed by the natural action of$G$. In this case, if we set$M=S^2$,$G=\mathbb Z_2$acting on$M$so that the non-identity element is the antipodal map, so that$M/G=P^2$: we get that$H^\bullet(P^2)=H^\bullet(S^2)^G$. We have to compute the fixed spaces: •$H^0(S^2)$is one dimensional, spanned by the constant function$1$, which is obviously fixed by$G$, so$H^0(P^2)\cong H^0(S^2)^G=H^0(S^2)=\mathbb R$. • On the other hand,$H^2(S^2)\cong\mathbb R$, spanned by any volume form on the sphere; since the action of the non-trivial element of$G$reverses the orientation, we see that it acts as multiplication by$-1$on$H^2(S^2)$and therefore$H^2(P^2)\cong H^2(S^2)^G=0$. • Finally, if$p\not\in\{0,2\}$, then$H^p(S^2)=0$, so that obviously$H^p(P^2)\cong H^p(S^2)^G=0$. Luckily, this agrees with the previous two computations. - That's some clever magic, I like it. – Aaron Mazel-Gee Mar 28 '11 at 3:57 Ah, great, wonderful wonderful Mariano! – WishingFish Aug 13 at 1:55 The magic is really brilliant! – WishingFish Aug 13 at 3:54 Can I ask that why$\chi(P^2)=\dim H^0(P^2)-\dim H^1(P^2)\$? Thank you. –  WishingFish Aug 13 at 4:20