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This is not a homework problem and I'm just doing it for fun.

The problem statement is:

Using the cup product structure, show there is no map $\mathbb{R}P^n \to \mathbb{R}P^m$ inducing a nontrivial map $H^1(\mathbb{R}P^m; \mathbb{Z}_2) \to H^1(\mathbb{R}P^n; \mathbb{Z}_2)$ if $n > m$.

Here is my thought: Assume there is such a map $f$. Using naturally of cup product, we have the square that stacks these two lines together commutes: $$ H^1(\mathbb{R}P^m; \mathbb{Z}_2) \times H^1(\mathbb{R}P^m; \mathbb{Z}_2) \to H^2(\mathbb{R}P^m; \mathbb{Z}_2) $$ and $$ H^1(\mathbb{R}P^n; \mathbb{Z}_2) \times H^1(\mathbb{R}P^n; \mathbb{Z}_2) \to H^2(\mathbb{R}P^n; \mathbb{Z}_2) $$ (Forgive me for not knowing how to load tikzcd packages in MSE).

I want to use the fact that $H^{\ast}(\mathbb{R}P^n; \mathbb{Z}_2) \cong \mathbb{Z}_2[\alpha]/(\alpha^{n + 1})$ and similarly for $m$, $H^{\ast}(\mathbb{R}P^m; \mathbb{Z}_2) \cong \mathbb{Z}_2[\beta]/(\beta^{m + 1})$.

This leads me to discover that on each level $k < n + 1$, $H^k(\mathbb{R}P^n; \mathbb{Z}_2) \cong \mathbb{Z}_2(\alpha^k) \cong \mathbb{Z}_2$. I feel like something related to quadratic forms might be related but I'm a beginner so don't know what I'm talking about.

Thanks!!!

nekodesu
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1 Answers1

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let $m<n$. Then since $H^* (\mathbb R P^{k})=\mathbb Z_2[x]/(x^{n+1})$ for any $k$, we can let $\alpha, \beta$ be generators for the cohomology ring of $H^* (\mathbb R P^{m})$ and $H^* (\mathbb R P^{n})$ respectively, then suppose that there is a nontrivial homomorphism $h^*:H^1(\mathbb RP^m) \to H^1(\mathbb RP^n)$. It follows that $h^*(\alpha)=\beta$.

Then $$0={h}^*(\alpha^{m+1})=h^*(\alpha)^{m+1}=\beta^n \neq 0,$$ a contradiction.

Andres Mejia
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  • of course multiplication everywhere is the cup product. – Andres Mejia Apr 17 '18 at 15:52
  • Thanks! I'm confused by one thing you said here: I thought of $H^n(\mathbb{R}P)$ for each $n$ as a $\mathbb{Z}_2$-module generated by $\alpha^n$, so I thought it's isomorphic to $\mathbb{Z}_2$. Why would it have the relation of the whole graded ring? – nekodesu Apr 17 '18 at 17:20
  • see hatcher 3.40 for details, but you are correct. As a graded algebra (with the cup product as multiplication), all you need to see is that $x \smile x^i \neq 0 \in H^{i+1}(\mathbb RP^n)$, so that it is nondegenerate, which allows us to deduce that $x \in H^1$ will generate the ring – Andres Mejia Apr 17 '18 at 17:27
  • was that the confusion? – Andres Mejia Apr 17 '18 at 17:27
  • Hmmm yes that is my confusion but I'm still confused: I understand that $\alpha \in H^1(\mathbb{R}P^n)$ is the generator of the whole ring; in particular, cup product of $\alpha$ with powers of it self would generate generators in all $H^{n}$. My confusion is that: now we are just considering $h^\ast$ from $H^1$ to $H^1$, so does $\alpha^{m+1}$ even live in here (since cup product on just one level is not defined)? – nekodesu Apr 17 '18 at 17:55
  • Aha! But what you said makes perfect sense if $h^{\ast}$ is in fact a ring homomorphism between $H^{\ast}$'s. Am I interpreting this right? – nekodesu Apr 17 '18 at 17:59