Assume the fact that \(H^*(\mathbb{C}P^n;\mathbb{Z}) = \mathbb{Z}[\alpha]/(\alpha^{n+1})\), where \(\alpha\) has degree \(2\), and \(H^*(\mathbb{R}P^n;\mathbb{Z}_2) = \mathbb{Z}_2[\beta]/(\beta^{n+1})\), where \(\beta\) has degree \(1\). (These are “truncated polynomial rings”.) Show that if \(k < n\) there is no retraction from \(\mathbb{C}P^n\) to its standard submanifold \(\mathbb{C}P^k\), and similarly for \(\mathbb{R}P^n\).
Solution
1. Let\[ i:\mathbb CP^k\hookrightarrow\mathbb CP^n \]be the standard inclusion and suppose a retraction \(r:\mathbb CP^n\to\mathbb CP^k\) existed. Then\[ i^*\circ r^*=\operatorname{id}_{H^*(\mathbb CP^k;\mathbb Z)}. \]
A retraction satisfies \(r\circ i=\operatorname{id}_{\mathbb CP^k}\). Cohomology is contravariant, so \((r\circ i)^*=i^*r^*\).
2. If \(\alpha_k\in H^2(\mathbb CP^k)\) and \(\alpha_n\in H^2(\mathbb CP^n)\) are the standard generators, then\[ r^*(\alpha_k)=\alpha_n. \]
The standard inclusion satisfies \(i^*(\alpha_n)=\alpha_k\). Since \(H^2(\mathbb CP^n)\cong\mathbb Z\), write \(r^*(\alpha_k)=m\alpha_n\). Applying \(i^*\) and using 1 gives \(m\alpha_k=\alpha_k\), hence \(m=1\).
3. This is impossible when \(k<n\).
In \(H^*(\mathbb CP^k)\) one has \(\alpha_k^{k+1}=0\). Therefore \[ 0=r^*(\alpha_k^{k+1})=r^*(\alpha_k)^{k+1}=\alpha_n^{k+1}. \] But \(k+1\le n\), so \(\alpha_n^{k+1}\ne0\) in \[ H^*(\mathbb CP^n;\mathbb Z)=\mathbb Z[\alpha_n]/(\alpha_n^{n+1}), \] a contradiction.
4. The identical argument over \(\mathbb F_2\) rules out a retraction\[ \mathbb RP^n\to\mathbb RP^k \]for \(k<n\).
Let \(\beta_k,\beta_n\) be the degree-one generators. A retraction would force \(r^*(\beta_k)=\beta_n\), but then \[ 0=r^*(\beta_k^{k+1})=\beta_n^{k+1}\ne0 \] because \(k+1\le n\). Thus no such retraction exists.