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The exterior derivative was introduced to me with the following axioms:

  • $df$  is the differential of  f  for smooth functions  f
  • $d(df) = 0$ for any smooth function  f
  • $d(α ∧ β) = dα ∧ β + (−1)^p (α ∧ dβ)$ where $α$ is a p-form

I wondered if it was possible to generalize the concept to arbitrary functionals. The problem is obvious: In the above formulation, we only deal with $p$-forms where $p$ has to be an integer. When dealing with arbitrary functionals, this can not be enough.

To clarify, the above definition of the ext. derivative is only suitable for functionals involving vector spaces with countable dimension. I am searching for a generalization where the vector space that is the domain of the functional might be uncountably dimensional. An example of such a functional is the $L^2$ norm:

$f \mapsto I(f) = \int_\Omega |f(x)|^2 dx$

I'm trying to define a meaningful exterior derivative $d I$.

Is it possible to define an outher derivate much in the same sense of above, that maintains most (all?) of the properties of the standard exterior derivative?

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    I honestly do not understand what you're talking about here. Can you be specific about what you mean by "functionals"? – 2017-02-06
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    @TedShifrin I hope I cleared it up a bit – 2017-02-06
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    You can certainly do differential forms on Banach spaces. You just can't do proofs by choosing local coordinates. Note that the $1$-form $df$ is determined by knowing directional derivatives $df_p(v)$, which you can certainly compute in your setting. After all, that's what calculus of variations is doing from the outset. – 2017-02-06
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    It's not clear to me what problem you have with the exterior algebra being graded by integers. – 2017-02-06
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    In fact, my initial impression is, as long as you can define the spaces of $0$ and $1$ forms and have a derivation $d$ from $0$-forms to $1$-forms, everything else follows naturally. – 2017-02-06

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