Given a principal $G$ bundle $P(M,G)$ and a manifold $F$ with an action of $G$ on it from left, we construct a fibre bundle over $M$ with fiber $F$ and call this the associated fiber bundle for $P(M,G)$.
I do not get the motivation behind the construction given in Kobayashi and Nomizu which I will write down below.
Idea is to construct a fibre bundle with fibre $F$ i.e., we need to construct a smooth manifold $E$ and a smooth map $\pi_E:E\rightarrow M$ that gives a fiber bundle with fiber $F$.
Kobayashi's proof goes as follows :
They consider the product manifold $P\times F$ with an action of $G$ as $g.(u,\xi)=(ug,g^{-1}\xi)$. Then they consider the quotient space $(P\times F)/G$ and call this $E$.
Consider the map projection map $P\times F\rightarrow M$ defined as $(p,\xi)\mapsto \pi(p)$.
This induces a map $\pi_E:E=(P\times F)/G\rightarrow M$. As $P\rightarrow M$ is a principal $G$ bundle given $x\in M$ there exists an open set $U$ containing $x$ and a local trivialization $\pi^{-1}(U)\rightarrow U\times G$. They then give a bijection $\pi_E^{-1}(U)\rightarrow U\times F$ and give a smooth structure on $E$ so that these bijections are difeomorphisms. Then, they cal $(E,\pi_E,M,P,F)$ the fiber bundle associated to principal $G$ bundle.
I am trying to understand the motivation for the above construction.
Suppose $F=H$, a Lie group and the action of $G$ on $H$ is given by a morphism of Lie grroup $\phi:G\rightarrow H$ with $G\times H\rightarrow H$ given by $(g,h)\mapsto \phi(g)^{-1}h$ do get a principal $H$ bundle in above construction?
Edit : I thinnk above content looks like it is asking why do we need the construction of associated fiber bundles. No, what I am asking is, suppose I have a Principal $G$ bundle $P(M,G)$ with an action of $G$ on a manifold $F$ and I want to associate some fiber bundle on $M$ with fiber $F$. Then, what suggests you to think of above construction. How does it occur naturally? Are there any other properties of Fiber bundle I should have in mind which suggest this way of construction.
Best Answer
This construction reverses the construction of the frame bundle from a vector bundle (e.g., the tangent bundle). The idea is that each point $f \in F_p$ in the frame bundle of a vector bundle $E$ is, by definition a basis of $E_p$. This therefore defines a natural map of $F \times \mathbb{R}^k \rightarrow E$, where $k$ is the rank of $E$. The inverse image of any $e_p \in E$ is naturally isomorphic to $G$, so the quotient $(F\times \mathbb{R}^k)/G$ is a vector bundle isomorphic to $E$.