Can differential forms be generalized to (separable) Banach spaces?

This thought occurred to me earlier and I'm surprised I hadn't considered it previously. I get the feeling that no meaningful generalization can occur in a non-separable Banach space but on the surface it seems like a meaningful generalization can be had if the Banach space is separable. Even if it's possible, convergence and other issues likely severely complicate the matter.

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Just thinking off the top of my head, it seems to me there are at least two sources of headache here: 1) Unless you're working with Hilbert spaces, you need to pick some flavour of Banach space tensor product. 2) Once you've chosen your favourite Banach space tensor product, you need to decide whether $k$-forms live in $\wedge^k B^\ast$ or, more generally, in $(\wedge^k B)^\ast$. –  Branimir Ćaćić Mar 30 at 23:36
Yeah that's a good point. I guess Hilbert spaces are the most natural setting anyway. –  Cameron Williams Mar 30 at 23:44

Yes, it is possible. For example, Serge Lang exhibits the basics of a theory of differential forms on Banach manifolds in Chapter V of his Differential and Riemannian Manifolds. (Non-)Separability is not an issue.

According to Lang, a $p$-form on a Banach space $E$ is simply a continuous alternating $p$-linear map on $E$. This yields a notion of $p$-forms on any Banach manifold and Lang shows how to define exterior and Lie derivative and derives their fundamental algebraic properties in this context.

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Neat! Thanks for the reference. I'll check it out. –  Cameron Williams Jul 1 at 14:39