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Computing the cohomology ring of $\mathbb RP^2 \times \mathbb S^3$ with $\mathbb Z_2$ coefficients.

Here are my thoughts:

To use kunneth theorem and then knowing that the cohomology ring of $\mathbb RP^2$ is $\mathbb Z_2[\alpha]/ (\alpha^3)$ where $|\alpha=1|$ but what is the cohomology ring of $\mathbb S^3$ with $\mathbb Z_2$ coefficients?

Any clarification will be greatly appreciated!

Brain
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  • Either use the Universal Coefficient Theorem or compute the homology (by your favorite method) of $S^3$ with $\Bbb Z_2$ chains. Also, please stop using the group-cohomology tag; that has a completely different meaning. – Ted Shifrin Dec 20 '23 at 05:41
  • what do you mean? Is the cohomology ring of the sphere is just its homology? @TedShifrin I do not understand your point, could you clarify please? – Brain Dec 20 '23 at 05:44
  • @Brain Clarification in these subjects is not possible in my opinion. You should google it and agree many times. – Bob Dobbs Dec 20 '23 at 07:08

1 Answers1

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I couldn't find a reference in the internet. For $n$-dimensional sphere, I think, $$H^*(S^n,\Bbb Z_2)=\Bbb Z_2[x]/(x^2)$$ where $x\in H^n(S^n,\Bbb Z_2)$ is a generator.

Bob Dobbs
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    should not it be $n + 1$ in the denominator? Is not the $RP^2$ defined as $S^n$ with antipodal points identified? Does this help us in anything? – Brain Dec 20 '23 at 12:02
  • @Brain $x^2\in H^{2n}(S^n,\Bbb Z_2)=0.$ For second question: If $n=2$ – Bob Dobbs Dec 20 '23 at 12:13
  • so you are saying that this is the cohomology ring for any n sphere right? But how do you know this? – Brain Dec 20 '23 at 12:25
  • @Brain This is a long story to read. See Hatcher's book for example, Algebraic Topology. It is a popular book. There are better books. I am not so good. Abstract things. – Bob Dobbs Dec 20 '23 at 13:02
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    You can know that this is true for the following reason: On the group level, for any abelian group $A$, we have $H^j(S^n;A) = A$ if $j=0,n$ and $0$ otherwise (use your favorite method, e.g. excision + long exact sequence and induct on dimension). Now if $A$ is a ring ($A = \mathbb{Z}/2\mathbb{Z}$ in this question), we have $H^n(S^n;A) = A$ generated by some class $x$. The only possible multiplication that requires determining in $H^*(S^n;A)$ is $x^2$. But $x^2$ lies in $H^{2n}(S^n;A)$ which is zero. The ring that expresses these two rules is $A[x]/(x^2)$. Now set $A = \mathbb{Z}/2\mathbb{Z}$. – kamills Dec 20 '23 at 14:25
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    @Brain Will you notify us when you computed $H^*(\Bbb RP^3\times \Bbb S^3; \Bbb Z_2)$ in terms of generators and relations? – Bob Dobbs Dec 20 '23 at 16:49
  • hahaha :) sure why not? – Brain Dec 20 '23 at 17:31