Cohomology ring of $\mathbb RP^n$ with integral coefficient.

Question

I know cup product structure on $H^*(\mathbb{R}P^n;\mathbb{Z}_2)= \mathbb{Z}_2[\alpha]/(\alpha^{n+1})$. How to get $H^*(\mathbb{R}P^n;\mathbb{Z})$ from this? I have two cochain complexes for two coefficient rings.

Now my question is what will be the induced map between these two cochain complexes and what will be $H^*(\mathbb{R}P^n;\mathbb{Z})$?

@Thomas, is universal coefficients theorem really telling anything about multiplication in cohomology? — Andrey Ryabichev, Aug 10 '15 at 10:09
This is more a general remark, but unless you're in an exceptional situation, I doubt that you can recover integral cohomology from mod 2 cohomology in any way. You just kill too much information. — Najib Idrissi, Aug 11 '15 at 16:09

Andrey Ryabichev · Answer 1 · 2015-08-14T12:13:41.103

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I think, $H^*(\mathbb RP^{2n+1})=\mathbb Z[\alpha,\beta]/(2\alpha,\alpha^{n+1},\alpha\beta,\beta^2)$, where $\alpha$ has degree $2$ and $\beta$ has degree $2n+1$.

You can see it when you write down spectral sequence for $S^1$-fibration $\mathbb RP^{2n+1}\to\mathbb CP^n$ and look on multiplicative structure (as you doing for calculating ring $H^*(\mathbb CP^{n})$).

And for odd dimension, $H^*(\mathbb RP^{2n})=\mathbb Z[\alpha]/(2\alpha,\alpha^{n+1})$, where $\alpha$ has degree $2$. It makes cleare when you consider inclusion $\mathbb RP^{2n}\to\mathbb RP^{2n+1}$.

EDIT: this spectral sequence just shows that when you project $\mathbb RP^{2n+1}\to\mathbb CP^n$, the pullback maps $\mathbb Z\to\mathbb Z_2$ in cohomology are surjections, and ring structure is induced.

edited Aug 14 '15 at 12:13

answered Aug 10 '15 at 12:18

Andrey Ryabichev

2,682
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Thanks for the answer....I don't think I have understood it clearly....I need some time to think over your solution... – Ripan Saha Aug 10 '15 at 14:20
@Ripan, see the last edit. maybe, my explanation is not pure, but this fact is really easy to understand – Andrey Ryabichev Aug 10 '15 at 15:00
This doesn't give the right answer in the odd case in the top dimension. For example, the degree 1 cohomology of $RP^1$ is $\mathbb{Z}$. (And it doesn't give the right answer in degree 0 regardless.) – John Palmieri Aug 10 '15 at 15:15
@John, i think now i finally fix it – Andrey Ryabichev Aug 10 '15 at 15:33

score 2 · Answer 2 · edited Apr 13 '17 at 12:19

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This is a duplicate of this question, the answer to which is given in a comment. See also this website, which was the first thing I found when I searched for "cohomology of projective space".

edited Apr 13 '17 at 12:19

Community

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answered Aug 10 '15 at 15:17

John Palmieri

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Cohomology ring of $\mathbb RP^n$ with integral coefficient.

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