Finite-dimensional vector-valued forms and their exterior derivative

Let $M$ be a smooth manifold, let $V$ be a finite-dimensional real vector space, and let $k\ge 0$. A smooth $V$-valued differential $k$-form on $M$ is a family of alternating $k$-linear maps

${\alpha }_p:(T_{p}M{)}^k⟶V$

such that $\lambda \circ \alpha$ is a scalar smooth $k$-form in the sense of A smooth differential $k$-form for every $\lambda \in V^{\ast }$. Equivalently, for one (and hence every) basis $(v_1,\dots ,v_r)$ of $V$, the unique components in $\alpha ={\sum }_i{v}_i{\alpha }^i$ all belong to ${\mathrm{\Omega }}^k(M)$.

For such an $\alpha$, its componentwise exterior derivative is the unique $V$-valued $(k+1)$-form $d_V\alpha$ characterized by

$\lambda \circ d_V\alpha =d(\lambda \circ \alpha )$

for every $\lambda \in V^{\ast }$. Indeed, in a basis with dual basis $({\lambda }^1,\dots ,{\lambda }^r)$, set

$d_V\alpha ={\sum }_i{v}_{i}d{\alpha }^i,{\alpha }^i={\lambda }^i\circ \alpha \mathrm{.}$

The scalar formula defining $d$ is real-linear term by term The exterior derivative by the invariant vector-field formula, and its output is a smooth form The invariant exterior-derivative formula is $C^{\mathrm{\infty }}$-multilinear. Thus the displayed construction has the stated characterization. It is independent of the chosen basis because linear functionals separate points of $V$, and the characterization also proves uniqueness.

The two smoothness descriptions are equivalent in both directions: testing all $\lambda$ includes the dual basis components, while every $\lambda \circ \alpha$ is a fixed real linear combination of the components in one basis. If $M$ is empty, these assignments and identities are vacuous. If $V=0$, there is only the zero-valued form and its derivative is zero; if $\dim V=1$, the definition is exactly the scalar definition after choosing one nonzero basis vector. No metric, nondegeneracy, manifold boundary, or endpoint is involved. A single finite basis exists by finite-dimensionality; the definition is basis-independent and chooses no basis or family, so no choice axiom is used.