Algebraic Topology Hatcher Chapter 3.2 Problem 3

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!

• Do you know how to compute cup products on $\mathbb{R}P^n$? – Jason DeVito Aug 27 '20 at 20:24
• Correct me if I'm wrong but in this section it only gives a very general formula for this product. I'm not sure how to apply it to $\mathbb{R}P^n$. – Math_Day Aug 27 '20 at 20:38
• Use Hatcher's Theorem 3.12. – John Palmieri Aug 27 '20 at 20:41

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.

• How do you know we cannot have a homomorphism between $\mathbb{Z}(2)[x]/(x^{m+1}) \rightarrow \mathbb{Z}(2)[x]/(x^{n+1})$? If $m+1|n+1$ then we could map it like $x \rightarrow x^\frac{n+1}{m+1}$. – Math_Day Aug 29 '20 at 0:19
• It is possible to get such a homomorphism, but $f^*$ can never be this morphism since it is forced to be graded. In this case this means that $x\mapsto \lambda_1 x$, $x^2\mapsto \lambda_2 x^2$, $x^3 \mapsto \lambda_3 x^3,\ldots$, where $\lambda_i\in \mathbb{Z}/(2)$. – Frederik Aug 29 '20 at 16:47
• Also recall that $x$ in this case is the generator of the cohomology group $H^1(\mathbb{R}P^m;\mathbb{Z}/(2))$ in the domain and the generator of $H^1(\mathbb{R}P^n;\mathbb{Z}/(2))$ in the codomain. That $f$ induces isomorphism on $H^1$ precisely means that $f^*:x\mapsto x$. – Frederik Aug 29 '20 at 16:49