Fibre Bundles Part 1

Problem 5.15 Let $\pi: P \rightarrow M$ be a surjective submersion and let $P \times G \rightarrow P$ be an action of $G$ on $P$ satisfying the following two properties: (i) $G$ acts freely, i.e. $u \cdot g=u, u \in P, g \in G$, implies $g=e$. (ii) The fibres of $\pi$ coincide with the orbits of $G$, i.e. $$ \pi^{-1}(x)=u \cdot G, \quad u \in \pi^{-1}(x), x \in M $$ Prove: 1. If $\pi$ admits a global section (i.e. there exists a smooth map $s: M \rightarrow P$ such that $\pi \circ s=\operatorname{id}_{M}$ ), then $\pi$ is trivialisable, i.e. there exists a $G$-equivariant diffeomor$\operatorname{phism} \Phi: M \times G \rightarrow P$. 2. Every $x \in M$ admits an open neighbourhood $U$ such that $\pi^{-1}(U)$ is $G$ equivariantly diffeomorphic to $U \times G$.

Problem 5.16 A morphism of principal bundles $\pi: P \rightarrow M, \pi^{\prime}: P^{\prime} \rightarrow M^{\prime}$, with structure groups $G, G^{\prime}$, respectively, is a smooth map $\Phi: P \rightarrow P^{\prime}$ and a Lie group homomorphism $\gamma: G \rightarrow G^{\prime}$ such that $$ \Phi(u \cdot g)=\Phi(u) \cdot \gamma(g), \quad u \in P, g \in G $$ (i) If $\Phi: P \rightarrow P^{\prime}$ is a morphism, then prove that there exists a unique smooth map $\phi: M \rightarrow M^{\prime}$ making the following diagram commutative: (ii) Suppose now that $M^{\prime}=M, G^{\prime}=G, \pi=\pi^{\prime}$, and $\gamma=\mathrm{id}_{G}$. Then a principalbundle morphism $\Phi: P \rightarrow P$, is said to be an automorphism if $\Phi$ is a diffeomorphism. Prove that all the automorphisms constitute a group with respect to composition. This group is denoted by Aut $P$. Prove that the map Aut $P \rightarrow$ Diff $M, \Phi \mapsto \phi$, with $\phi$ given as in (i) above, is a group homomorphism. Remark The kernel of the last homomorphism is called the gauge group of $P$ and it is denoted by $\operatorname{Gau} P$.

Problem 5.17 (Hopf Bundles) Set $$ \begin{aligned} & S^{1}=\{x \in \mathbb{C}:|x|=1\}, \\ & S^{2}=\left\{(x, t) \in \mathbb{C} \times \mathbb{R}:|x|^{2}+t^{2}=1\right\}, \\ & S^{3}=\left\{(x, y) \in \mathbb{C}^{2}:|x|^{2}+|y|^{2}=1\right\}, \\ & S^{7}=\left\{(x, y) \in \mathbb{H}^{2}:|x|^{2}+|y|^{2}=1\right\} . \end{aligned} $$ The spheres $S^{1}$ and $S^{3}$ are Lie groups with respect to the multiplication induced from $\mathbb{C}$ and $\mathbb{H}$, respectively (see Problem 4.103). Let $S^{1}$ act on $S^{3}$ (resp., $S^{3}$ on $S^{7}$ ) by the formula $$ (x, y) \cdot z=(x z, y z), \quad(x, y) \in S^{3}, z \in S^{1}\left(\operatorname{resp} .(x, y) \in S^{7}, z \in S^{3}\right) $$ Let $$ \pi_{\mathbb{C}}: S^{3} \rightarrow \mathbb{C} \times \mathbb{R}, \quad \pi_{\mathbb{H}}: S^{7} \rightarrow \mathbb{H} \times \mathbb{R} $$ be the maps given by $$ \begin{array}{ll} \pi_{\mathbb{C}}(x, y)=\left(2 y \bar{x},|x|^{2}-|y|^{2}\right), & (x, y) \in S^{3}, \\ \pi_{\mathbb{H}}(x, y)=\left(2 y \bar{x},|x|^{2}-|y|^{2}\right), & (x, y) \in S^{7} \end{array} $$ Prove: (i) $\pi_{\mathbb{C}}\left(S^{3}\right)=S^{2}$. (ii) $\pi_{\mathbb{H}}\left(S^{7}\right)=S^{4}$ (iii) The induced map $\pi_{\mathbb{C}}: S^{3} \rightarrow S^{2}$ is a principal $S^{1}$-bundle with respect to the action of $S^{1}$ on $S^{3}$ defined above. (iv) The induced map $\pi_{\mathbb{H}}: S^{7} \rightarrow S^{4}$ is a principal $S^{3}$-bundle with respect to the action of $S^{3}$ on $S^{7}$ defined above. (v) $\mathbb{C P}^{1} \cong S^{2}$. (vi) $\mathbb{H} \mathrm{P}^{1} \cong S^{4}$.