Show that if \(S^k \to S^m \to S^n\) is a fiber bundle, then \(k = n-1\) and \(m = 2n-1\).
Solution
Theorem.
For a sphere bundle \(S^k\to S^m\to
S^n\), one has \(k=n-1\) and
\(m=2n-1\).
Proof by cohomological spectral sequence.
- In the Serre spectral sequence for \(S^k\to S^m\overset{p}{\to} S^n\), with \(\mathbb Z\)-coefficients, \[ E_2^{p,q}=H^p(S^n;H^q(S^k;\mathbb Z)). \]
- The only nonzero entries are: \[ E_2^{0,0}\cong E_2^{n,0}\cong E_2^{0,k}\cong E_2^{n,k}\cong\mathbb Z. \]
- Since \(S^m\) has cohomology only in degrees \(0,m\), the class at \((n,0)\) must be killed in the spectral sequence.
- The only differential that can hit \((n,0)\) is \[ d_{k+1}:E_{k+1}^{0,k}\to E_{k+1}^{k+1,0}. \] Hence \(k+1=n\).
- Then the class at \((n,0)\) is killed and the class at \((n,k)\) survives to total degree \[ n+k=2n-1, \] giving the top cohomology of \(S^{2n-1}\).
Therefore \(k=n-1\) and \(m=2n-1\). ∎