Definition of the associated bundle Let $P(M,G)$ be a principal $G$-bundle. The action of $G$ on a manifold $F$ is defined from left, the action of $G$ on $P\times F$ is defined as:
$$G:\quad(u,f) \to (ug,g^{-1}f)$$
The associated fiber bundle $E=P\times_\rho F$ is defined then as a quotient of $P\times F$ by the relation $\sim$:
$$E = P\times F / \sim$$
where
$$\quad(u,f) \sim (ug,g^{-1}f)$$
Could you please comment on why one cannot simply define $(u,f) \sim (ug,f)$? My natural guess is that this would necessarily lead to the trivial bundle $M\times F$. But probably there are more serious arguments.
 A: One possible way to see "why" we define the associated bundle with $g^{-1}$ acting on $F$ is that it offers a way to remain invariance.
A: The important point in the construction is, that it equips the associated bundle with the transition functions of the principal bundle. To see that recall the following facts:

*

*Since G acts freely and transitively on the Fibers $P_p$, every point in $E=P\times F \Big/\tilde{}$ can be represented as $[s_\alpha(p), x]$ where $s_\alpha \in \Gamma(P\restriction_{U_\alpha})$ is a local section and $x\in F$ is uniquely determined.


*A local trivialization $\Phi_\alpha$ on a principal bundle is equivalent to a local section $s_\alpha \in \Gamma(P\restriction_{U_\alpha})$ by:
$$ s_\alpha(p)g = \Phi^{-1}(p,g) $$
(Given the trivialization you obtain the section by setting g = e the identity element, and given the section you get the trivialization - This fact is btw very useful for example in seeing that TM isn't trivial by the Hairy Ball theorem).


*Loval sections of P are related by the transition functions:
$$ s_\alpha = s_\beta g_{\beta \alpha} $$


*The local trivializations of $\pi^\rho: E \rightarrow M$ are defined by local sections (<-> local triv of the original principal bundle):
$$ (\Phi^\rho_\alpha)^{-1}(p,x) = [s_\alpha(p),x] $$
We thus have:
$$ (\Phi_\alpha^\rho)^{-1} (p,x) = [s_\alpha(p),x] = [s_\beta(p)g_{\beta \alpha}, x] = [s_\beta(p), g_{\alpha \beta} x] = (\Phi_\beta^\rho)^{-1}(p, g_{\beta \alpha}x) $$
$$ \Rightarrow \Phi^\rho_\beta \circ (\Phi^\rho_\alpha)^{-1}(p,x) = (p, g_{\beta \alpha} x) $$
So the bundle has the same transition functions as the principal bundle.
It is also instructive to study this construction on the Frame bundle $P= F^GE \rightarrow M$ of which gives you the original vector bundle $E\rightarrow M$
