# Is there a one-to-one correspondence between vector-valued differential forms and real-valued forms?

One way of thinking about vector $$V$$-differential forms for some real, finite-dimensional vector space $$V$$ is as alternating and $$C^{\infty}(\mathcal{M},\mathbb{R})$$-linear functions of the type

$$\omega:\underbrace{\mathfrak{X}(\mathcal{M})\times \dots\times\mathfrak{X}(\mathcal{M})}_{m -times}\to C^{\infty}(\mathcal{M},V)$$

If I choose a basis $$e_{1},\dots,e_{n}$$ of $$V$$, then I can write

$$\omega(X_{1},\dots,X_{m})(p)=\sum_{i=1}^{n}\omega(X_{1},\dots,X_{m})(p)^{i}e_{i}$$

for all $$X_{1},\dots,X_{m}\in\mathfrak{X}(M)$$ and for all $$p\in\mathcal{M}$$. Are the so-defined component functions

$$\omega^{i}:\underbrace{\mathfrak{X}(\mathcal{M})\times \dots\times\mathfrak{X}(\mathcal{M})}_{m -times}\to C^{\infty}(\mathcal{M},\mathbb{R})$$

then (real-valued)-differential forms on $$\mathcal{M}$$? Is also the other direction true, i.e. choosing $$n$$ real valued differential forms an multiply then point-wise with a basis gives a vector-valued form?

We wish to determine whether specifying a $$V$$-valued differential $$m$$-form $$\omega$$ as you've defined it, i.e. as an alternating $$C^\infty(\mathcal{M})$$-linear map from collections of $$m$$ vector fields into $$C^\infty(\mathcal{M},V)$$, is equivalent, given a basis $$\{e_i\}_{i=1}^n$$ of $$V$$, to specifying $$n$$ real-valued "component forms" $$\omega^i$$ through the relation $$\omega = \sum_i \omega^i e_i$$. Note that the components $$\omega^i$$ to a $$V$$-valued form $$\omega$$ are also alternating by the uniqueness of basis expansions, and so it's clear that starting with either of these and evaluating on a collection of vector fields $$X_1, ..., X_m$$ appropriately yields maps of the other form ($$\mathcal{M} \to \mathbb{R}$$ or $$\mathcal{M} \to V$$), the only question being smoothness.
As this is a question about smoothness, we need to recall the smooth structure with which $$V$$ is typically endowed. If you use the canonical topology on $$V$$ induced by any vector space isomorphism to Euclidean space (the only Hausdorff topology making $$V$$ into a topological vector space), $$V$$ then has a unique smooth structure, also induced by such an isomorphism. The only exception is the particular case that $$V$$ has dimension $$4$$, wherein there are infinitely many additional alternative smooth structures for that same topology, but I'll assume you don't mean to be working with those.
Assuming you use the above standard smooth structure, your result holds, and it's simply the observation (once one passes to a coordinate chart of $$\mathcal{M}$$) that a map $$\mathbb{R}^m \to \mathbb{R}^n$$ is smooth iff its $$n$$ component functions are smooth. This is because the smooth structure is independent of the vector space isomorphism used to induce it (as linear maps in Euclidean space are smooth), and there exists such an isomorphism which maps your chosen basis of $$V$$ to the standard basis of Euclidean space.