Associated bundles: isomorphism between spaces of differential forms. I think this will be an easy question for numerous people.
Let $\pi:P\rightarrow M$ be a principal bundle and $\rho:G\rightarrow GL(V)$ a representation.
The space of $k$ forms on $M$ with values in $P\times_G V$ (denote as $\Omega^k(M;P\times_G V)$  can be identified with with the space of horizontal, right invariant $k$-forms on $M$ (denote as $\Omega^k_G(P;V))$. 
Ie, there is an isomorphism: 
$\Omega^k_G(P;V)\cong \Omega^k(M;P\times_G V)$.
I am reading through some lecture notes which say
Let $\overline{\zeta}\in \Omega^k_G(P;V)$. Define $\zeta_{\alpha}=s_{\alpha}^*\overline{\zeta}\in \Omega^k(U_{\alpha};V)$. ($s_{\alpha}$ is the local section $s_{\alpha}:U_{\alpha}\rightarrow P$).
It then asks to show that $\{\zeta_{\alpha}\}$ define a form in $\Omega^k(M;P\times_G V)$ by showing that the 'gluing' equation is satisfied
$\zeta_{\alpha}=\rho(g_{\alpha\beta})\circ \zeta_{\beta}$.
Here $g_{\alpha\beta}$ is the transition functions related to the local trivialisations which satisfy $s_{\beta}(m)=s_{\alpha}(m)g_{\alpha\beta}(m)$.
I have managed to show that the required equation holds. My question is - why do the constructed $\zeta_{\alpha}\in \Omega^k(U_{\alpha};V)$ define forms in $\Omega^k(M;P\times_G V)$? I understand that $P\times_G V$ has the structure of a fibre bundle with typical fibre $V$, but I am not sure why the gluing equation is important. I am guessing it has something to do with the fact that because the equation holds, one is able to extend the local definition go a global one. I'm not sure. If someone can help that would be great.
 A: This is a situation where generalizing makes things clearer. Suppose that $X$ is a manifold and $G$ is a lie group acting freely and properly discontinuously on $X$. It follows that the quotient $X/G$ exists in the category of manifolds and we have a smooth map $ \rho : X \to X/G$. Assume that $ \pi : E \to X $ is a $G$-equivariant vector bundle. That is, $E$ is a vector bundle with a $G$ action such that $ \pi $ is $G$-equivariant and $ g : E_x \to E_{gx}$ is a linear isomorphism. Note that if $s$ is a section of $E$, then $g s g^{-1}$ is a section of $E$. Therefore $G$ acts on the sections of $E$.
The quotient $E/G$ is a vector bundle on $X/G$. In the quotient $E/G$, $E_x$ and $E_{gx}$ get identified via multiplication by $g$. Moreover, if $ U \subseteq X/G$ is open, then
$$ H^0(E/G,U) = H^0(E,\pi^{-1}(U))^G $$
If this is not clear, try and draw some cartoons in the case when $G$ is a finite group. This is a special case of the following more general fact: suppose that $E$ and $F$ are $G$-equivariant vector bundles on $X$. Then 
$$ {\rm Hom} \, (E/G,F/G) = {\rm Hom}(E,F)^G $$
Now consider the following exact sequence of $G$-equivariant vector bundles on $X$:
$$ 0 \to V \to TX \xrightarrow{\rho'} \rho^* T(X/g) \to 0 $$
We call $V$ the vertical subbundle: It consists of vectors which are tangent to the orbits. The isomorphism $ \rho' : TX / V \to \rho^* T(X/G)$ induces an isomorphism $(TX/V)/G \cong T(X/G)$. If $E$ is a $G$-equivariant vector bundle on $X$ then
$$ {\rm Hom} \, (T(X/G),E/G) = {\rm Hom} \, ((TX/V)/G,E/G) = {\rm Hom} \, (TX/V,E)^G $$
This says that $E/G$-valued 1-forms on $X/G$ are the same as $E$-valued 1-forms $\eta$ on $X$ such that 


*

*$\eta(v) = 0 $ for tangent vectors $v \in V$

*$L_g^* \eta = g \cdot \eta $


extending this to $p$-forms is easy: the first condition becomes $\eta(v_1,\dots,v_p) = 0 $ whenever one of the $v_i$ is tangent to a orbit and the second does not change. 
Now let us apply this framework to your problem. Assume that $M$ is a manifold, $G$ is a Lie group, $ \pi : P \to G$ is a principal $G$-bundle and $V$ is a $G$-representation. Then $ P / G = M $. The vector bundle $P \times V$ is $G$-equivariant with $G$-action given by
$$ (p,v) \cdot g = (pg,g^{-1}v) $$
The quotient bundle $P \times V / G = P \times_G V $ is a vector bundle on $M$. A $P \times_G V$-valued $p$-form on $M$ is the same as a $V$-valued $p$-form $\eta$ on $P$ such that 


*

*$\eta(v_1,\dots,v_p) = 0 $ when one of the $v_i$ is tangent to the fiber

*$R_g^* \eta = g^{-1} \cdot \eta $


The formulas look a little different because $G$ acts on $P$ from the right instead of the left. This is exactly this isomorphism $\Omega^p_G(P,V) \cong \Omega^p( M, P\times_G V) $ which you were asking about.
A: Same thing as your other question, here is a note I took:
https://www.evernote.com/shard/s318/sh/e4354637-e8a0-4a89-9043-4507303f7006/af3513ed77440494665094c5f4270cc3
