Let $(P,\pi,M,G)$ be a principal bundle, where $P$, $M$, $G$ are smooth manifolds (total space, base space and fiber=structure group respectively, so $G$ is both fiber and structure group)
- the projection map is $\pi:u\in P\rightarrow p\in M$ $\Rightarrow$ $\pi^{-1}(U_i)\equiv \{ u\in P: \pi(u)\in U_i \}$, where $U_i$ is an open set in $M$
- the local trivialization diffeo is $\phi_i:(p,g_i)\in U_i\times G\rightarrow \phi_i(p,g_i)\equiv u\in \pi^{-1}(U_i)$
- $G$ acts on the right on $P$ globally, $(u,a)\in P\times G\rightarrow ua =\phi_i(p,g_ia)=\phi_j(p,g_ja)\in P$ $\Rightarrow$ $\pi(u)=\pi(ua)=p \ \ \forall a \in G$
Then we can construct a fiber bundle $(E=P\times F /_G,\pi_E,M,F,G)$ associated to this principal bundle, which has the same base space $M$ and the same structure group $G$, a new fiber $F$ and
- elements of $E$ are the equivalence classes $[(u,f)]\equiv \{(u',f'): u'=ug, \ f'=g^{-1}f\}$, where we have used the global right action of $G$ on $P$ and we have defined a left action of $G$ on the new fiber $F$
- the projection map is $\pi_E:[(u,f)] \in E\rightarrow \pi_E([(u,f)])=\pi(u)=p\in M$
My problem is about the local trivialization of the associated bundle. I know that this diffeo is a map
$$
\psi_i: (p,f_i)\in U_i\times F\rightarrow \psi_i(p,f_i)\in\pi_E^{-1}(U_i)
$$
but I don't know how to proceed (I'm studying on Nakahara, but this is not very bright on the book). I try to explain myself better: my guess is that $\psi_i(p,f_i)$ is the element (=the equivalence class) $[(u,f_i)]$, so if we consider another open set $U_j$ such that $U_i \cap U_j \neq \emptyset$ and a point $p\in U_i\cap U_j$, then we can construct the transition function $h_{ij}:p\in U_i\cap U_j \rightarrow h_{ij}(p)\equiv g \in G$
$$
\psi_i: (p,f_i)\in U_i\cap U_j\times F\rightarrow \psi_i(p,f_i)=[(u,f_i)]\in\pi_E^{-1}(U_i\cap U_j)\\
\psi_j: (p,f_j)\in U_i\cap U_j\times F\rightarrow \psi_j(p,f_j)=[(u,f_j)]\in\pi_E^{-1}(U_i\cap U_j)\\
\Rightarrow h_{ij}(p)\equiv \psi^{-1}_i(p,-)\circ\psi_j(p,-):f_j\in F \rightarrow f_i=h_{ij}(p)f_j=gf_j\in F
$$
but from this last relation, it follows that
$$
\psi_i(p,gf_j)=\psi_i(p,f_i)=\psi_j(p,f_j)
\Rightarrow [(u,gf_j)]=[(u,f_i)]=[(u,f_j)]
$$
Is this conclusion correct? Because I don't think it is. So I think that probably I'm interpreting the local trivialization diffeo of an associated bundle incorrectly. Please be indulgent, I'm not well versed in this topic. Thank you.
Best Answer
I think I found an answer to my question: the correct definition of the local trivialization diffeo should be
$$ \psi_i : (p,f_i)\in U_i\times F\rightarrow \psi_i(p,f_i)=[(\phi_i(p,e),f_i)]\in\pi_E^{-1}(U_i) $$
where $e$ is the identity of the structure group $G$ (which is also the fiber of $P$). Here we are using the canonical local trivialization $\phi_i$ of the principal bundle, so $\phi_i(p,e)=s_i(p)$ is the section.
This definition allows us to construct the following transition function
$$ h_{ij}(p)\equiv \psi^{-1}_i(p,-)\circ\psi_j(p,-):f_j\in F \rightarrow f_i=h_{ij}(p)f_j=gf_j\in F $$
therefore
$$ \psi_i(p,gf_j)=\psi_i(p,f_i)=\psi_j(p,f_j) \Rightarrow [(\phi_i(p,e),gf_j)]=[(\phi_i(p,e),f_i)]=[(\phi_j(p,e),f_j)] $$
but we know that $t_{ij}(p)=\phi_{i,p}^{-1}\circ \psi_{j,p}$ and $t_{ij}(p)\equiv a \in G$
$$ [(\phi_{j,p}(e),f_j)]=[(\phi_{i,p}\circ \phi_{i,p}^{-1}\circ \phi_{j,p}(e),f_j)]=[(\phi_{i,p}\circ t_{ij}(p)e,f_j)]=[(\phi_{i,p}(e)a,f_j)]=[\phi_{i}(p,e),af_j]\\ [(\phi_{j,p}(e),f_j)]=[(\phi_j(p,e),f_j)]=[(\phi_i(p,e),f_i)]=[(\phi_i(p,e),gf_j)]\\ \Rightarrow [\phi_{i}(p,e),af_j]=[(\phi_i(p,e),gf_j)] $$
so we obtain that the transition functions of $P$ and $E$ are the same $a=g$, correctly.