What is a simple way of understanding associated bundles?

Question

I am currently struggling to understand the definition of an associated bundle. The one I have been given is

Let $(P,M,\pi)$ be a $G$-principal bundle (with right G action $R$) over some manifold $M$. Given another manifold $F$ equipped with a left $G$ action, $R$. We define the equivalence relation over $P\times F $ as $(p,f)\sim (R(g,p),L(g^{-1},f))$ (where of course we mean by this that there is a $g\in G$... etc ) and then the set $P_F=P\times F /\sim$.
Then, $(P_F,M,\pi_F)$, with $\pi_F$ sending $[p,f] \mapsto \pi(p)$, is called an associated bundle.

I would like someone to help me navigate this definition. I have understood conceptually principal bundles as bundles which can have groups as fibers in a non trivial way, so to speak . I can see how the definition of a principal bundle formalizes this idea.
But what is the idea behind an associated bundle? What are we trying to do? I kind of get why this bundle is associated with the principal bundle; after all, we constructed it with the total space, the projection map and the group action of the principal bundle. However, it's this construction that is not clear. I can see why one would consider $P \times F$; but why would one mod this set out by that equivalence relation? Why that one specifically? Why are the actions kind of """inverses"""??

I have seen the example of the GL- $\mathbb{R}^n$ associated bundle to the frame bundle, which is isomorphic as a bundle to the tangent bundle; in this example, I can see how somehow these "actions working in reverse" and taking the quotient by that relation conspire to give something isomorphic to the tangent bundle. But beyond that I can't see what this construction was designed to achieve.

Answer 1

The definition you are working with is the "elegant" one. One can also construct associated bundles in another way as follows, which for the first encounter could be a little bit more illuminating:

Since $P$ is locally trivial, there exists an open cover $(U_i)$ of $M$ and principal bundle isomorphisms

$$ \varphi_i:P_{|U_i}\to U_i\times G $$

Then, since the right $G$-equivariant functions on $G$ are given by the left multiplications, the transition functions are of the form

$$\varphi_j\circ\varphi_i^{-1}:U_i\cap U_j\times G\to U_i\cap U_j\times G,\;\;(x,h)\to(x,g_{ji}(x)h)$$ for unique maps $g_{ij}:U_i\cap U_j\to G$. So you can view $P$ as build up from the parts $U_i\times G$ glued by the maps $g_{ji}$ (see Fiber bundle construction theorem for more details). Now to construct the associated bundle, simply replace $G$ with $F$ in each of the parts $U_i\times G$. Hence the associated bundle is build up from the parts $U_i\times F$ glued by the same maps $g_{ji}$. The so constructed bundle is then isomorphic to the bundle from your definition, which can be shown by constructing local trivializations $\psi_i:{P_F}_{|U_i}\to U_i\times F$ whose transition functions are also given by the maps $g_{ji}$.

Answer 2

I think that the right way to address this is to look at the example of the frame bundle of a vector bundle $E\to M$ in a nice description. If $V$ is the standard fiber of $E$, then the frame bundle $PE\to M$ is a principal $GL(V)$-bundle, whose fiber $P_xE$ over $x\in M$ can be identified with the set of all linear isomorphism $u:V\to E_x$, where $E_x$ is the fiber of $E$ over $x$. Then the principal right action of $GL(V)$ is simply given by composition from the right, i.e. $R(u,A)=u\circ A$ for $A\in GL(V)$.

Now in this example it is easy to see how to recover $E$ from $PE$: You simply consider the map $PE\times V\to E$ which sends $u\in P_xE$ (i.e. $u:V\to E_x$) and $v\in V$ to $u(v)\in E_x$. Now it just remains to understand when two pairs $(u_1,v_1)$ and $(u_2,v_2)$ lead to the same element of $E$. Of course, this is only possible if $u_1$ and $u_2$ sit over the same point $x\in M$, in which case there is a unique $A\in GL(V)$ such that $u_2=R(u_1,A)=u_1\circ A$. But then $u_2(v_2)=u_1(A(v_2))$ which equals $u_1(v_1)$ if and only if $v_1=Av_2$ or equivalently iff $v_2=A^{-1}v_2$. So you exactly recover the relation that $(u_2,v_2)=(R(u_1,A),L(A^{-1},v_1))$ for some $A\in GL(V)$. The inversion in one of the actions is unavoidable since you deal with a left action and a right action at the same time and this is needed to obtain a well defined right action on $PE\times V$.