Show there is no fiber bundle \(S^7 \to S^{23} \to \mathbb{OP}^2\).
Solution
Suppose there were a fiber bundle \[ S^7\longrightarrow S^{23}\xrightarrow{p}\mathbb{OP}^2. \] Let \[ C_p=\mathbb{OP}^2\cup_p CS^{23} \] be the mapping cone of the projection.
Coning each fiber \(S^7\) produces a disk \(D^8\). Thus \(C_p\) is the Thom space of the associated cone/disk bundle over \(\mathbb{OP}^2\). Since its boundary is the sphere \(S^{23}\), collapsing this boundary to the cone point gives a closed \(24\)-manifold. The Thom isomorphism (equivalently Poincaré duality in \(C_p\)) gives \[ H^*(C_p;\mathbb Z) \cong \mathbb Z\{1,x,x^2,x^3\}, \qquad |x|=8, \] with \[ x^3\ne0\in H^{24}(C_p;\mathbb Z). \tag{1} \] More explicitly, \(x\) restricts to a generator of \(H^8(\mathbb{OP}^2)\), \(x^2\) generates degree \(16\), and Poincaré duality forces the pairing of degrees \(8\) and \(16\) to be unimodular, so \(x^3\) is a generator up to sign.
Reduce mod \(3\). The instability axiom for reduced powers says that for a degree-eight class, \[ P^4x=x^3. \tag{2} \] But the mod-3 Adem relation with \(a=1,b=3\) is \[ P^1P^3=P^4. \tag{3} \] Since \[ H^{20}(C_p;\mathbb Z_3)=0, \] we have \[ P^3x=0. \] Equations (2) and (3) then give \[ x^3=P^4x=P^1P^3x=0, \] contradicting (1). Therefore \[ \boxed{\text{there is no fiber bundle }S^7\to S^{23}\to\mathbb{OP}^2.} \]