Since there is a fiber bundle \(S^2 \to \mathbb{CP}^5 \to \mathbb{HP}^2\) by Exercise 35 in §4.2, one might ask whether there is an analogous bundle \(S^4 \to \mathbb{HP}^5 \to \mathbb{OP}^2\). Use Steenrod powers for the prime 3 to show that such a bundle cannot exist.
Solution
Assume there were a fiber bundle \[ S^4\longrightarrow\mathbb{HP}^5 \xrightarrow{p}\mathbb{OP}^2. \] Write \[ H^*(\mathbb{HP}^5;\mathbb Z_3)=\mathbb Z_3[x]/(x^6), \qquad |x|=4, \] and \[ H^*(\mathbb{OP}^2;\mathbb Z_3)=\mathbb Z_3[y]/(y^3), \qquad |y|=8. \]
The Gysin sequence of the \(S^4\)-bundle shows that \[ p^*:H^8(\mathbb{OP}^2;\mathbb Z_3) \longrightarrow H^8(\mathbb{HP}^5;\mathbb Z_3) \] is an isomorphism: the adjacent base cohomology groups in degrees \(3\) and \(4\) vanish. Hence, after rescaling \(y\), \[ p^*y=x^2. \tag{1} \]
At the prime \(3\), the reduced power \(P^1\) raises degree by \(4\). Since \[ H^{12}(\mathbb{OP}^2;\mathbb Z_3)=0, \] we have \[ P^1y=0. \tag{2} \] For the generator \(x\in H^4(\mathbb{HP}^\infty;\mathbb Z_3)\), the splitting-principle calculation gives \[ P^1x=x^2. \] Indeed, after restriction to \(\mathbb{CP}^\infty\) one has \(x=-u^2\), and \[ P^1(-u^2)=-2u^4=u^4=x^2\quad\text{in }\mathbb Z_3. \] The Cartan formula therefore yields \[ P^1(x^2)=2x^3\ne0. \tag{3} \]
Naturality applied to (1), together with (2), would give \[ 0=p^*(P^1y)=P^1(p^*y)=P^1(x^2), \] contradicting (3). Hence \[ \boxed{\text{no bundle }S^4\to\mathbb{HP}^5\to\mathbb{OP}^2\text{ exists}.} \]