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I've recently asked a question on Physics SE about how one should interpret a particular object in the local expression for the spin covariant derivative. Namely, the object was $\mathrm{d}\psi(X)$, where $\psi \colon U \to \Delta$ and $U$ is a contractible open set on a manifold (typically the domain of some chart) and $\Delta$ is the vector space for the Dirac representation. $X$ is some vector field on the manifold.

In an answer to my question, it was said that $\psi$ should be interpreted as a $0$-form (with values on the vector space $\Delta$), and hence one can take the exterior derivative of $\psi$ as one usually does for forms. Nevertheless, a new question was raised: I know that the wedge product is not defined in general for forms with values on vector spaces, but is the exterior derivative always well-defined?

In short, is the exterior derivative well-defined for forms with values on vector spaces?

Níckolas Alves
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    Yes. Write it in terms of components with respect to a basis, and then check the result is independent of the choice of basis. For wedge product, you need a multiplicative structure on the vector space (e.g., a Lie bracket). – Ted Shifrin Apr 23 '24 at 20:52
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    ^ or if you want a basis-free explanation, then it boils down to the fact that all of the “basic” differential calculus one learns can be generalized to maps from an open subset of a Banach space into another Banach space (or simply normed vector spaces). If the domain is a manifold, then this theory can be “transported”verbatim as long as the target space is a fixed vector space. Another way to say it is you simply take the tangent map, then project onto the vectorial factor (using triviality of the tangent bundle of a vector space). – peek-a-boo Apr 23 '24 at 22:14
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    If your target space is a vector bundle then of course complications arise and you’ll need a connection to covariantly exterior differentiate (i.e the covariant bit is referring to the target, and the exterior bit is referring to the domain being a manifold). See sections 4 and 5 of my answer here (though you may want to read the whole thing because I talk about wedge products in general and what’s needed for them). – peek-a-boo Apr 23 '24 at 22:18
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    And just as a fun “application” of some of these ideas in a situation more familiar in physics, see this PhySE answer of mine on describing rotational kinematics in the language of (vector-valued) forms. – peek-a-boo Apr 23 '24 at 22:25
  • Indeed you can see all this on the wiki page for vector-valued differential forms. In fact, the wedge product is always defined as well, at the cost of it taking values in the tensor product of the vector space instead. – Callum Apr 24 '24 at 13:13

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