Section 10.4 Examples and Computations

Example 10.15. Complex Hopf bundles.

Let \(S^1 \subset \mathbb{C}\) act on \(S^{2n+1} \subset \mathbb{C}^{n+1}\) by scalar multiplication. The quotient map

\begin{equation*} p \colon S^{2n+1} \to S^{2n+1}/S^1 \cong \mathbb{C}P^n, \qquad z \mapsto [z_0 : \cdots : z_n], \end{equation*}

is a fibre bundle with fibre \(S^1\text{.}\) The construction also applies to \(S^\infty = \bigcup_n S^{2n+1}\text{,}\) giving a fibre bundle

\begin{equation*} S^1 \to S^\infty \to \mathbb{C}P^\infty. \end{equation*}

Since \(S^\infty\) is contractible and \(S^1\) is connected, the long exact sequence gives, for \(i\geq1\text{,}\)

\begin{equation*} \pi_i(\mathbb{C}P^\infty) \cong \pi_{i-1}(S^1) = \begin{cases} \mathbb{Z} & \text{if } i = 2, \\ 0 & \text{otherwise.} \end{cases} \end{equation*}

Example 10.16. The Hopf fibration.

Specialising to \(n = 1\) and identifying \(\mathbb{C}P^1 \cong \mathbb{C} \cup \{\infty\} \cong S^2\) via \([z_0 : z_1] \mapsto z_0/z_1\) yields the Hopf fibration

\begin{equation*} S^1 \to S^3 \xrightarrow{\;\eta\;} S^2. \end{equation*}

In its long exact sequence

\begin{equation*} \cdots \to \pi_n(S^1) \to \pi_n(S^3) \xrightarrow{\eta_*} \pi_n(S^2) \xrightarrow{\partial} \pi_{n-1}(S^1) \to \pi_{n-1}(S^3) \to \cdots \end{equation*}

we know \(\pi_1(S^1) \cong \mathbb{Z}\) (Theorem 2.16), \(\pi_k(S^1) = 0\) for \(k \geq 2\) (Proposition 5.14, using \(\mathbb{R} \to S^1\)), and \(\pi_1(S^3) = \pi_2(S^3) = 0\) (Corollary 9.6). Hence

\begin{equation*} \pi_2(S^2) \cong \pi_1(S^1) \cong \mathbb{Z}, \qquad \eta_* \colon \pi_n(S^3) \xrightarrow{\;\cong\;} \pi_n(S^2) \text{ for } n \geq 3. \end{equation*}

Example 10.17. Quaternionic Hopf bundles.

Identifying \(S^{4n+3}\) with the unit sphere in \(\mathbb{H}^{n+1}\text{,}\) right multiplication by unit quaternions \(S^3 \subset \mathbb{H}\) gives, just as in Example 10.15, a fibre bundle

\begin{equation*} S^3 \to S^{4n+3} \to \mathbb{H}P^n. \end{equation*}

For \(n = 1\) this is a second Hopf fibration \(S^3 \to S^7 \to \mathbb{H}P^1 \cong S^4\text{.}\) The relevant part of the long exact sequence is

\begin{align*} \pi_6(S^7)=0 &\longrightarrow \pi_6(S^4) \longrightarrow \pi_5(S^3) \longrightarrow \pi_5(S^7)=0\\ &\longrightarrow \pi_5(S^4) \longrightarrow \pi_4(S^3) \longrightarrow \pi_4(S^7)=0\\ &\longrightarrow \pi_4(S^4) \longrightarrow \pi_3(S^3) \longrightarrow \pi_3(S^7)=0 . \end{align*}

Thus

\begin{equation*} \pi_n(S^3)\cong\pi_{n+1}(S^4)\qquad\text{for }3\leq n\leq5. \end{equation*}

Example 10.18. Octonionic Hopf bundle.

Using the octonions, one can construct another Hopf bundle

\begin{equation*} S^7\longrightarrow S^{15}\longrightarrow S^8. \end{equation*}

Its long exact sequence gives \(\pi_n(S^7)\cong\pi_{n+1}(S^8)\) for \(7\leq n\leq13\text{.}\)

Example 10.19. Orthogonal and unitary groups.

The map \(A \mapsto A e_n\) sends \(O(n)\) onto \(S^{n-1}\text{,}\) and the stabiliser of \(e_n\) is \(O(n-1)\text{.}\) One can show (using Gram–Schmidt to build local trivialisations) that this, and its unitary analogue, are fibre bundles

\begin{equation*} O(n-1) \hookrightarrow O(n) \to S^{n-1}, \qquad U(n-1) \hookrightarrow U(n) \to S^{2n-1}. \end{equation*}

Since \(\pi_k(S^{n-1}) = 0\) for \(k \leq n-2\text{,}\) the long exact sequence shows that \(\pi_i(O(n-1)) \to \pi_i(O(n))\) is an isomorphism for \(i \leq n-3\text{,}\) and similarly \(\pi_i(U(n-1)) \to \pi_i(U(n))\) is an isomorphism for \(i \leq 2n-3\text{.}\) So \(\pi_i(O(n))\) and \(\pi_i(U(n))\) are independent of \(n\) for \(n\) large.

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