Use the Gysin sequence to show that for a fiber bundle \(S^k \to S^m \xrightarrow{p} S^n\) we must have \(k = n-1\) and \(m = 2n-1\). Then use the Thom isomorphism to show that the Hopf invariant of \(p\) must be \(\pm 1\).
Solution
1. The Gysin sequence for the sphere bundle \(S^k \to S^m \xrightarrow{p} S^n\) is\[\cdots \to H^{i-k-1}(S^n) \to H^i(S^n) \xrightarrow{p^*} H^i(S^m) \to H^{i-k}(S^n) \to \cdots.\]
the Gysin sequence of a sphere bundle.
2. \(H^i(S^m)\) is nonzero only for \(i = 0, m\), and \(H^i(S^n)\) is nonzero only for \(i = 0, n\).
cohomology of spheres.
3. The Gysin sequence forces \(m = n + k + 1\) (so that the total space \(S^m\) has the right dimension).
the Euler class lives in \(H^{k+1}(S^n)\), which is nonzero only if \(k + 1 = n\), i.e. \(k = n - 1\); then \(m = n + k = 2n - 1\).
4. Hence \(k = n - 1\) and \(m = 2n - 1\).
3.
5. The Euler class \(e \in H^n(S^n) = \ZZ\) is a generator (up to sign), since the Gysin sequence is exact and the bundle is nontrivial.
the Euler class of the bundle \(S^{n-1} \to S^{2n-1} \to S^n\) is \(\pm 1\) times the generator of \(H^n(S^n)\).
6. The Hopf invariant of \(p\) equals the Euler class (up to sign), so \(H(p) = \pm 1\).
by the Thom isomorphism, the Hopf invariant of the bundle map \(p\) is the Euler class, which is \(\pm 1\).
7. Q.E.D.
4 and 6.