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Determine the cohomology ring $H^*(\mathbb{R}P^n; \mathbb{Z}).$

I was given a hint to first calculate the cohomology groups $$ H^k(\mathbb{R}P^n;\mathbb{Z}).$$ And that I may use $$ H^*(\mathbb{R}P^n;\mathbb{Z}/2) \cong (\mathbb{Z}/2)[x] /(x ^(n+1)).$$ And that my answer will depend on the parity of $n.$

I also found this link here : integral cohomology ring of real projective space

Still I am unable to fill in all details.

1- How my answer will depend on the parity of $n$?

2- How I And that I may use $$ H^*(\mathbb{R}P^n;\mathbb{Z}/2) \cong (\mathbb{Z}/2)[x] /(x ^(n+1)).$$

Could anyone help me in solving this please?

  • Do you know how to calculate the groups? – Andres Mejia Apr 21 '20 at 03:44
  • you mean cohomology groupsyes I think so @AndresMejia –  Apr 21 '20 at 04:28
  • If you work out the cellular (co)chain complex, you see a dependence on the parity of dimension/degree coming from the fact that the antipodal map $S^k \to S^k$ is orientation preserving iff $k$ is odd. You should find the coboundary operator is $0$ half the time and multiplication by $2$ the other half. In particular the manifold $\mathbb{RP}^n$ is orientable iff $n$ is odd. – William Apr 21 '20 at 13:46
  • In order to use the known ring structure of $H^(\mathbb{RP}^n;\mathbb{Z}/2)$, argue as in Eric Wofsey's answer from the question you linked: use the fact that the coefficient change map $H^(-;\mathbb{Z}) \to H^*(-;\mathbb{Z}/2)$ is a ring homomorphism, and argue that it is injective in positive degrees. – William Apr 21 '20 at 14:05
  • small caveat to my second comment: this coefficient change homomorphism will be injective in positive degrees less than $n$, but will not be injective in degree $n$ when $n$ is odd. – William Apr 21 '20 at 14:19

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