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We have to prove that $H^{*}(\mathbb{R}P^n, \mathbb{Z}_2) \simeq \mathbb{Z}_{2}[x]/(x)^{n+1}$ as ring. So we have to find an isomorphism $$ \phi: \mathbb{Z}_{2}[x]/(x)^{n+1} \rightarrow H^{*}(\mathbb{R}P^n, \mathbb{Z}_2) $$ We can send $x$ in a sub-manifold of $\mathbb{R}P^n$ of codimention $1$ in $H_{*}(\mathbb{R}P^n, \mathbb{Z}_2)$ (? $\mathbb{R}P^{n-1}$, why isn't it a $0$-class in homology?). How can we build this isomorphism? And why can we use duality among $H^{*}$ and $H_{*}$ (if $n$ is odd $\mathbb{R}P^n$ isn't oriented)?

Olivier Bégassat
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ArthurStuart
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  • There is a generalized notion of orientation that depends on the coefficient group. For $G = \mathbb{Z}_2$, there is only one orientation class. In other words, every manifold is orientable. – Sammy Black May 20 '13 at 19:17
  • Also, with coefficients $G = \mathbb{Z}$ or $G = \mathbb{R}$ (or some other group with units $G^\times = {-1, +1})$, $\mathbb{R}P^n$ is orientable iff $n$ is odd. – Sammy Black May 20 '13 at 19:21
  • @SammyBlack But how can I find the isomorphism? – ArthurStuart May 20 '13 at 19:26
  • In order to calculate $x \smile x \in H^{n-2}(\mathbb{R}P^n, \mathbb{Z}_2)$, you have to consider the transverse intersection of the submanifold $\mathbb{R}P^{n-1}$ with itself. This requires that you perturb one copy of the manifold so that the intersection is of minimal dimension $d = (n-1) + (n-1) - n = n - 2$. – Sammy Black May 20 '13 at 19:40
  • In terms of an actual computation, you need a combinatorial cell structure on $\mathbb{R}P^n$ that is compatible with the decomposition $\mathbb{R}P^n = e^n \cup \mathbb{R}P^{n-1} = \cdots = e^n \cup e^{n-1} \cup \cdots \cup e^0$. – Sammy Black May 20 '13 at 19:43
  • @SammyBlack But why $H^{n-2}$? and why $\mathbb{R}P^{n-1}$ isn't the $0-$class in homology? – ArthurStuart May 20 '13 at 19:46
  • The Gysin sequence + induction is a nice way to do this and see the multiplicative structure – Dylan Wilson May 21 '13 at 00:23

1 Answers1

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Here is a quick way to get what you want:

Note that $\mathbb{R}P^{n-1}\hookrightarrow\mathbb{R}P^n$ induces isomorphisms on $H_i(-;\mathbb{Z}_2)$ for $i\le n-1$.
Now $x$ generates $H_1(\mathbb{R}P^n;\mathbb{Z}_2)$ and we want to show that $H^*(\mathbb{R}P^n;\mathbb{Z}_2)\cong\mathbb{Z}_2[x]/(x^{n+1})$, so by induction we only need to show that $x\cup x^{n-1}\ne 0$.

Now just plug-n-chug through Poincare-duality with cup/cap products: $[\mathbb{R}P^n]\cap(x\cup x^{n-1})=([\mathbb{R}P^n]\cap x)\cap x^{n-1}=1$, where this last equality is because $[\mathbb{R}P^n]\cap x$ is a generator of $H_{n-1}(\mathbb{R}P^n;\mathbb{Z}_2)$! Thus $x^n\ne 0$.

Chris Gerig
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    Can you please explain me how you mean to use induction there? I cannot understand it –  May 22 '18 at 21:26