Algebraic Topology Hatcher Chapter 3.2 Problem 3

Question

The Problem: Using the cup product structure, show there is no map $\mathbb{R}P^n \rightarrow \mathbb{R}P^m$ inducing a nontrivial map $H^1(\mathbb{R}P^m; \mathbb{Z_2}) \rightarrow H^1(\mathbb{R}P^n; \mathbb{Z_2})$ if $n > m$. What is the corresponding result for maps $\mathbb{C}P^n \rightarrow \mathbb{C}P^n$?

Notes: For cochains $\phi \in C^k(X; R)$ and $\psi \in C^l(X; R)$ we can define the cup product $\phi \smile \psi \in C^{k+l}(X;R)$ is the cochain defined on the singular simplex $\sigma: \Delta^{k+l} \rightarrow X$ is given by the formula: $$\phi \smile \psi(\sigma) = \phi(\sigma|[v_0,...,v_k])\psi(\sigma|[v_k,...,v_{k+l}])$$ There are some properties of these cochairs for instance we have the induced cup product $H^k(X; R) \times H^l(X; R) \xrightarrow{\smile} H^{k+l}(X; R)$. I'm not sure if the induced map in this problem is the same as this one. I suspect this problem has something to do with Theorem 3.12 of this section because that theorem is about the projective space but it only contains $H^*$ which is defined to be the direct sum of all homology groups. I'm having trouble finding in this section information that relates the first cohomology group of $\mathbb{R}P^m$ and $\mathbb{R}P^n$.

Thank you!

Answer 1

The only nontrivial map $H^1(\mathbb{R}P^m;\mathbb{Z}/(2))\to H^1(\mathbb{R}P^n;\mathbb{Z}/(2))$ is the identity since both domain and codomain is isomorphic to $\mathbb{Z}/(2)$. Assume for contradiction that $f:\mathbb{R}P^n\to \mathbb{R}P^m$ induces this nontrivial map on $H^1$.

Recall that $H^*(\mathbb{R}P^n;\mathbb{Z}/(2))\cong \mathbb{Z}/(2)[x]/(x^{n+1})$ with $|x|=1$, and that the induced map $f^*:H^*(\mathbb{R}P^m;\mathbb{Z}/(2))\to H^*(\mathbb{R}P^n;\mathbb{Z}/(2))$ is a ring homomorphism. This gives us that we have ring homomorphism $f^*:\mathbb{Z}/(2)[x]/(x^{m+1})\to \mathbb{Z}/(2)[x]/(x^{n+1})$ which maps $x\to x$. However, this is not a ring homomorphism since $n>m$, hence we have arrived at a contradiction. We conclude that no such $f$ exists.

The similar result for $\mathbb{C}P^n$ is that there exists no map $\mathbb{C}P^n\to \mathbb{C}P^m$ inducing a nontrivial map $H^2(\mathbb{C}P^m;\mathbb{Z})\to H^2(\mathbb{C}P^n;\mathbb{Z})$ when $n>m$. The argument for this is the same mutatis mutandis. Note in this case the nontrivial map on $H^2$ does not need to be the identity since we are working with coefficients over $\mathbb{Z}$ now. However, it is still multiplication by a non-zero integer which still leads to the contradiction.