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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

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up vote 34 down vote accepted

If you remove a point from $P^2$ you are left with something which looks like a Moebius 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.

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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 '13 at 1:55
The magic is really brilliant! – WishingFish Aug 13 '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 '13 at 4:20
For those that need some help understanding the second proof, you may find this question useful. – Michael Albanese Aug 13 '13 at 8:50

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