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How do we compute the Cohomology algebra of the real projective space: $ H^{\bullet} (\mathbb{P}^n (\mathbb{R}), \mathbb{Q}) $, by induction, and by using Mayer Vietoris sequence ?

Thanks in advance for your help.

YoYo12
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  • @Steve D : I'm trying to write $ \mathbb{P}^n (\mathbb {R}) $ in the form: $ \mathbb{P}^n (\mathbb{R}) = U \cup V $ with $ U $ and $ V $ are two contractible subsets, and $ U \cap V = \mathbb{P}^{n-1} (\mathbb{R}) $, but I do not know how to find $ U $ and $ V $. – YoYo12 Sep 12 '18 at 14:25
  • Have you considered the example of $\mathbb P^1(\mathbb R)$? Here, you can even draw a picture and see what open subsets to choose. If you don't see it immediately, have a look at "stereographic projection". –  Sep 12 '18 at 15:04
  • Also there are similar questions on MSE (https://math.stackexchange.com/questions/413802/mayer-vietoris-for-mathbb-c-mathbb-pn?rq=1 ) and searching google with "de rham cohomology mayer vietoris real projective space" yields a lot of results. –  Sep 12 '18 at 15:07
  • @James : For : $ \mathbb{P}^1 ( \mathbb{R} ) $, we have : $ \mathbb{P}^1 ( \mathbb{R} ) = U_0 \cup U_1 $ with : $ U_0 = { \ [x] \in \mathbb{P}^1 ( \mathbb{R} ) \ | \ x_0 \neq 0 \ } $ and $ U_1 = { \ [x] \in \mathbb{P}^1 ( \mathbb{R} ) \ | \ x_1 \neq 0 \ } $, right ? but, is $ U_0 \cap U_1 = \mathbb{P}^0 ( \mathbb{R} ) $ ? – YoYo12 Sep 12 '18 at 15:27
  • If you remove a point from a disk interior,the rest retracts to its boundary. What does that look like in $\mathbb{P}^n(\mathbb{R})$, considered as a disk with antipodal boundary points identified? That should give you one open set. The other can be taken as a neighborhood of that point you removed. What does that look like in $\mathbb{P}^n(\mathbb{R})$? – Steve D Sep 12 '18 at 15:51
  • Note that not both $U$, $V$ will be contractible if $n\geq 2$. – Tyrone Sep 12 '18 at 15:57
  • @Steve D : If we remove a point from a disk interior , the rest retracts to its boundary. So, it looks like $ \displaystyle \bigvee_{ i =1,2 } S_{i}^{n} $ in $ \mathbb{P}^n ( \mathbb{R} ) $, right ? – YoYo12 Sep 12 '18 at 16:01
  • @BoulizOmar The intersection will be $\mathbb R$ with a point removed for $\mathbb P^1(\mathbb R)$. More general you can cover your $\mathbb P^n(\mathbb R)$ by the affine coordinate patches $U_i={x_i\neq 0}\cong \mathbb R^n$. The intersection of two such coordinate patches is $\mathbb R^n$ minus a point. Working our the cohomology for this and for the affine pieces is then simple. Let me know if you need help with that. –  Sep 13 '18 at 08:52
  • Ok. Thank you very much @James. I understand now. Thank you. :-) – YoYo12 Sep 13 '18 at 12:00
  • I have found the solution here : https://math.stackexchange.com/questions/2914629/homotopy-type-of-a-disk-and-homotopy-type-of-a-sphere – YoYo12 Sep 13 '18 at 12:05

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