While reading the appendix of Friedrich's "Dirac Operators in Riemannian Geometry" I met the following definition.
Let $P$ be a $G$-principal bundle over $M$, and $\rho:G\to GL(V)$ a representation.
A $q$ form $w \in \Lambda^q(P,V)$ with values in $V$, is called tensorial of type $\rho$ if
1) $R_g^*w = \rho(g^{-1})w$ (pseudo-tensorialty, $R_g:P\to P$ is the right action )
2) $w_p(X_1,\dots, X_q) = 0$ if at least one of the vectors $X_i$ is vertical i.e. $d\pi(X_i) = 0$ ($\pi:P\to M$ is the projection).
Then the author states the following very interesting proposition without proof
The vector space of tensorial $q$-forms of type $\rho$ on $P$ with values in $V$ is isomorphic to the vector space of $\Lambda^q(M;P\times_\rho V)$ of $q$-forms on $M$ with values in the associated vector bundle $P\times_\rho V$.
The question is how is this isomorphism defined?
It doesn't seems straightforward to me. Given $w\in\Lambda^q(M;P\times_\rho V)$, we have something that takes tangent vectors to $M$, but we want something that takes tangent vectors to $P$, So I thought of simply pulling back $w$ using $\pi$. But the form $\pi^*w $ satisfy 2) but not 1) since $R_g^*\pi^* w = (\pi\circ R_g)^* w$ and $\pi\circ R_g = \pi$ because the right action preserve the fiber. So we would have that $R_g^*\pi^* w = \pi^* w $.