0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.