Let $G$ be a compact Lie group and let $H$ be a closed subgroup of $G$, with Lie algebras $\mathfrak{g}$ and $\mathfrak{h}$.


We denote $G\times\_H \mathfrak{g} / \mathfrak{h}$: the set of orbits $(G \times \mathfrak{g} / \mathfrak{h})/H $ of the right action of $H$ on $ G \times \mathfrak{g} / \mathfrak{h}$ ($H$ acts on $G$ by multiplication and acts on $\mathfrak{g} / \mathfrak{h}$ by the adjoint action).


We have this identification: $T(G/H) \simeq G\times\_H \mathfrak{g} / \mathfrak{h}$.


My question is why the space of differential forms of $G/H$, $A^\bullet(G/H)$, satisfies
$$
A^\bullet(G/H) \simeq \Gamma (G/H, G\times\_H {\bigwedge}^\bullet{(\mathfrak{g} / \mathfrak{h})}^\*) \simeq {(C^\infty (G) \otimes {\bigwedge}^\bullet {(\mathfrak{g} / \mathfrak{h})}^\*)}^H.
$$



Description of $A^\bullet(G/H)$
By definition, $k$-forms are sections of the bundle $\bigwedge{}^kT^\*(G/H)$, which is the associated bundle $G\times\_H\bigwedge^{k}(\mathfrak{g}/\mathfrak{h})^\*$. You then apply the general formula that the sections of the associated bundle for any $H$-representation $V$ is $(C^{\infty}(G)\otimes V)^H$: the tensor product $C^{\infty}(G)\otimes V$ is the sections of the trivial bundle $G\times V \to G$, and such a section is pulled back from a section of $G\times\_H V\to G/H$ if and only if it is invariant.


