Cup Product Structure on the Projective Space
Looking at complexes we see that the induced map of cohomology groups is an isomorphism in even degrees and zero in odd degrees (so the notation is slightly misleading: $\alpha$ maps to $0$ and not to $\alpha$). So $\alpha^2=0$ (since $f(\alpha^2)=f(\alpha)^2=0)$ and powers of $\beta$ generate evendimensional cohomology.
The remaining question is why $\alpha\beta^n$ is nonzero. From Gysin exact sequence (see Hatcher 4.D) for the sphere bundle $S^1\to S^\infty\to\mathbb RP^\infty$, $$ \ldots\to H(S^\infty)=0\to H^k(\mathbb RP^\infty)\to H^{k+2}(\mathbb RP^\infty)\to H(S^\infty)=0\to\ldots, $$ we see that multiplication by $\beta$ gives (for $k>0$) an isomorphism $H^k(\mathbb RP^\infty)\to H^{k+2}(\mathbb RP^\infty)$. QED. (But certainly Hatcher had some other argument in mind...)
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adrido
Updated on August 01, 2022Comments

adrido 10 months
I am reading about cup products and am stuck on this exercise in Hatcher (3.2.5). Taking as given that $H^*(\mathbb{R}P^\infty,\mathbb{Z}_2)\simeq\mathbb{Z}_2[\alpha]$, how does one show $H^*(\mathbb{R}P^\infty,\mathbb{Z}_4)\simeq \mathbb{Z}_4[\alpha,\beta]/(2\alpha,2\beta,\alpha^2)$ with $\deg(\alpha)=1,\deg(\beta)=2$?
The cohomology groups are $\mathbb{Z}_4$ in position 0 and $\mathbb{Z}_2$ in every other position. I want to find the cup product structure. To do that, I need to find a cochain map $f:C^\bullet(\mathbb{R}P^\infty,\mathbb{Z}_4)\rightarrow C^\bullet(\mathbb{R}P^\infty,\mathbb{Z}_2)$ induced by the ring map $\mathbb{Z}_4\rightarrow \mathbb{Z}_2$.
The two cochain complexes look like
$ \cdots {\leftarrow}\mathbb{Z}_4\stackrel{0}{\leftarrow} \mathbb{Z}_4\stackrel{2}{\leftarrow} \mathbb{Z}_4\stackrel{0}{\leftarrow} \mathbb{Z}_4\leftarrow 0 $
and
$ \cdots {\leftarrow}\mathbb{Z}_2\stackrel{0}{\leftarrow} \mathbb{Z}_2\stackrel{0}{\leftarrow} \mathbb{Z}_2\stackrel{0}{\leftarrow} \mathbb{Z}_2\leftarrow 0 $
Here are my questions:
$\bullet$ what is the induced cochain map?
$\bullet$ Can someone tell me why $\alpha\cup \alpha=0$?
$\bullet$ Why does $\alpha\cup \beta$ generate in degree 3?
Is there a quick way to answer these geometrically?

user101036 over 8 yearsI think the argument you should use to show that $\alpha \cup \beta$ is in the proof of theorem 3.12, where he shows that for a generator of $H^i(P^n)$ and a generator of $H^{ni}(P^n)$, the cup product is nonzero.

Shubhankar Sahai almost 2 yearsPretty sure that this argument, at least for the $\mathbf{Z}/4$ case will follow from using the coefficient exact sequence $0\to \mathbf{Z}/2\to \mathbf{Z}/4\to \mathbf{Z}/2\to 0$. This will induce an LES on singular cohomology of $\mathbf{RP}^\infty$ and one should be able to deduce the result from this.


adrido over 9 yearsThank you! this makes sense. but I'm very curious about what Hatcher has in mind here. I am thinking if one can get an isomorphism $H^1(\mathbb{R}P^\infty,\mathbb{Z}_4)\times H^2(\mathbb{R}P^\infty,\mathbb{Z}_4)\rightarrow H^3(\mathbb{R}P^\infty,\mathbb{Z}_4)$ out of $\mathbb{Z}_2$ isomorphisms and the module map $\mathbb{Z}_2\stackrel{.2}{\rightarrow}\mathbb{Z}_4$.