# De Rham Differential for Vector Valued Forms?

Let $M$ be a smooth manifold of dimension $n$ and let $V$ be a $\mathbb R$-vector space of finite dimension $\ell$. A $k$-form on $M$ with values on $V$ is a map $\omega$ on $M$ such that: $$\omega_p:=\omega(p):\underbrace{T_pM\times \ldots\times T_pM}_{k}\longrightarrow V,$$ is $k$-multilinear map over $\mathbb R$ and $\omega_p$ is alternating.

Given a basis $\{x_1, \ldots, x_\ell\}$ for $V$ and $v_1, \ldots, v_k\in T_pM$ we might write $$v=\sum_{j=1}^\ell \omega_j(p, v_1, \ldots, v_k) x_j,$$ for unique scalars $\omega_j(p, v_1, \ldots, v_k)\in\mathbb R$. This way we might define $k$-forms $\omega_j$ on $M$ (with values on $\mathbb R$) setting $$(\omega_j)_p:T_pM\times \ldots \times T_pM\longrightarrow \mathbb R, (v_1, \ldots, v_k)\longmapsto \omega_j(p, v_1, \ldots, v_k).$$ This way we might write informally $$\omega=\sum_{j=1}^\ell \omega_j x_j,$$ and we define the de Rham differential of $\omega$ as $$d\omega=\sum_{j=1}^\ell d\omega_j x_j,$$ where $d$ in $d\omega_j$ is the usual de Rham differential.

How can I justify $d\omega$ does not depend on the choice of basis for $V$?

If $y_1, \cdots y_l$ is another basis for $V$, then $x_i = \sum_{j} A_{ij} y_j$ and
$$\omega=\sum_{j=1}^\ell \omega_j x_j = \sum_{j,k} \omega_j A_{jk} y_k.$$
As $dA_{jk} = 0$,
$$\sum_{k} d\left(\sum_j\omega_j A_{jk}\right) y_k = \sum_{j,k} d \omega_j A_{jk} y_k = \sum_j d\omega_j x_j.$$
Thus $d$ is independent of the basis.
The same argument shows that one can define $d$ on a vector bundle $V$ where the transition matrices are constant.