Show that if \(S^k \to S^m \xrightarrow{p} S^n\) is a fiber bundle, then \(m = 2n-1\), \(k = n-1\), and, when \(n > 1\), \(H(p) = \pm 1\).
Solution
Suppose \[ S^k\longrightarrow S^m\xrightarrow{p}S^n \] is a fiber bundle.
First compare dimensions. Since a fiber bundle is locally a product, the total space is locally modeled on \[ \mathbb R^n\times\mathbb R^k, \] so \[ m=n+k. \tag{1} \]
Assume first that \(n>1\). Then the base is simply connected. The fiber cannot be \(S^0\), since a bundle with discrete two-point fiber over a simply connected base is the trivial double cover, whose total space is disconnected. Hence \(k>0\), and by (1), \(m>n\) and \(m>k\).
The long exact homotopy sequence of the bundle contains \[ \pi_n(S^m)\longrightarrow\pi_n(S^n) \longrightarrow\pi_{n-1}(S^k). \] Since \(m>n\), the first group is zero, while \[ \pi_n(S^n)\cong\mathbb Z. \] Thus \(\pi_{n-1}(S^k)\ne0\), forcing \[ k\le n-1. \tag{2} \] On the other hand, if \(k<n-1\), then the segment \[ \pi_{k+1}(S^n)\longrightarrow\pi_k(S^k) \longrightarrow\pi_k(S^m) \] would be \[ 0\longrightarrow\mathbb Z\longrightarrow0, \] a contradiction. Hence \[ k\ge n-1. \tag{3} \] From (2) and (3), \[ \boxed{k=n-1}, \] and then (1) gives \[ \boxed{m=2n-1}. \] For \(n=1\), the same dimension relation and the homotopy sequence force \(k=0\), hence \(m=1\), which is the exceptional double-cover case.
Now suppose \(n>1\), so the bundle has the form \[ S^{n-1}\longrightarrow S^{2n-1}\xrightarrow{p}S^n. \] Consider the mapping cone \[ C_p=S^n\cup_p CS^{2n-1}. \] This is a closed \(2n\)-manifold. Indeed, away from the copy of \(S^n\) and the cone point this is clear. Near a point of the copy of \(S^n\), choose a bundle trivialization over a small ball \(U\subset S^n\). Coning each fiber \(S^{n-1}\) produces a disk \(D^n\), so a neighborhood is \[ U\times D^n\cong\mathbb R^{2n}. \] A neighborhood of the cone point is the cone on \(S^{2n-1}\), hence a \(2n\)-disk.
The cellular cohomology of the mapping cone has \[ H^n(C_p;\mathbb Z)\cong\mathbb Z, \qquad H^{2n}(C_p;\mathbb Z)\cong\mathbb Z. \] Let \(\alpha\) and \(\beta\) be generators in these degrees. By definition of the Hopf invariant, \[ \alpha^2=H(p)\,\beta. \] Since \(C_p\) is a closed orientable \(2n\)-manifold, Poincaré duality says that the middle-dimensional intersection pairing \[ H^n(C_p;\mathbb Z)\times H^n(C_p;\mathbb Z) \longrightarrow H^{2n}(C_p;\mathbb Z)\cong\mathbb Z \] is unimodular. As the middle cohomology has rank one, its matrix is the \(1\times1\) matrix \((H(p))\). Unimodularity therefore forces \[ \boxed{H(p)=\pm1}. \]
Thus every such sphere bundle satisfies \[ \boxed{m=2n-1,\qquad k=n-1,} \] and for \(n>1\) its projection has Hopf invariant \(\pm1\).