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There have been various other threads about these, but it's been a while, so I'm starting a new one.
It just occurred to me that not only are exterior differential forms and absolute differential forms special cases of coflare differential forms, so are exterior pseudoforms! (Actually, we don't need the full theory of coflare differential forms, but only the alternating ones of up to a given rank $p$: those whose action on a flare depends only on the point and the first $p$ tangent vectors (and none of the higher tangents) and which is zero when any of these tangent vectors are repeated. Note that a form is exterior iff it is both alternating and linear.)
Recall that a top-rank exterior pseudoform is the same as a top-rank absolute form, and we know how to represent absolute forms as coflare forms. In general, you can think of a pseudoquantity (of any sort) as the product of an untwisted quantity and an orientation, so we can represent an orientation as the quotient of a nowhere-zero top-rank exterior form and its absolute value. (Note that a manifold is orientable, meaning that it has an everwhere-defined continuous orientation, iff it has an everywhere-defined continuous nowhere-zero top-rank exterior form, so this makes perfect sense.) Since you can perform any sufficiently-defined real-number operations on coflare forms, this makes an orientation into a coflare form. Finally, any exterior pseudoform (or more generally any coflare pseudoform) of any rank can be represented as a coflare form by multiplying a coflare form by such an orientation.
So for example, if you give $\mathbb{R}^3$ its standard right-handed coordinates, then the standard right-handed orientation is
$\frac{\mathrm{d}x \wedge \mathrm{d}y \wedge \mathrm{d}z}{|\mathrm{d}x \wedge \mathrm{d}y \wedge \mathrm{d}z|} .$If you apply this to a triple $(\mathbf{u}, \mathbf{v}, \mathbf{w})$ of tangent vectors, then the result is $1$ if $(\mathbf{u}, \mathbf{v}, \mathbf{w})$ has a right-handed orientation, $-1$ if they have a left-handed orientation, and undefined ($0/0$) if they are linearly dependent.
Or, if you want to integrate a vector-valued quantity $\mathbf{F}$ on a pseudoriented surface in $\mathbb{R}^3$, then you are really integrating the pseudoform
$\mathbf{F} \cdot đ \mathbf{S} = \mathbf{F} \cdot (\mathrm{d}y \wedge \mathrm{d}z, \mathrm{d}z \wedge \mathrm{d}x, \mathrm{d}x \wedge \mathrm{d}y) \otimes \frac{\mathrm{d}x \wedge \mathrm{d}y \wedge \mathrm{d}z}{|\mathrm{d}x \wedge \mathrm{d}y \wedge \mathrm{d}z|} = \mathbf{F} \cdot (\mathrm{d}z \wedge \mathrm{d}y, \mathrm{d}x \wedge \mathrm{d}z, \mathrm{d}y \wedge \mathrm{d}x) \otimes \frac{\mathrm{d}z \wedge \mathrm{d}y \wedge \mathrm{d}x}{|\mathrm{d}z \wedge \mathrm{d}y \wedge \mathrm{d}x|} .$(I guess that $đ\mathbf{S}$ isn't actually an alternating form; it's a rank-$5$ form that alternates separately in its first $2$ arguments and in its last $3$.)
I guess that you need to use the tensor product in general to represent a pseudoform, since you need $n$ new vectors (where $n$ is the dimension) to apply the orientation to. So even if $\omega$ is a top-rank pseudoexterior form, then $\omega$ is (locally) of the form $\alpha \otimes o$, where $\alpha$ is a top-rank exterior form, rather than $\alpha o$; but since $\alpha$ and $o$ are both rank-$n$, $\alpha \otimes o$ becomes $\alpha o$ under the natural map from $2 n$-forms (those that depend only on the point and the first $2 n$ tangent vectors) to $n$-forms defined by repeating the list of vector arguments.
In general, coflare forms are only integrated on oriented submanifolds/chains. This covers the integration of absolute forms, since either orientation will give the correct result for these. (Technically, you still need to break the domain of integration into orientable parts, I guess.) But other than that, I don't see any way to automatically treat integration of pseudoforms on pseudooriented submanifolds; the ranks doesn't even match the dimension. To integrate the pseudoform $\alpha \otimes o$ on the pseudoriented submanifold $(M, p)$ (where $p$ is a pseudoorientation of $M$), you'll just have to manually apply $o$ to $p$ to produce an orientation $o/p$ of $M$ and then integrate $\alpha$ on $(M, o/p)$. (You also need to break the domain into orientable parts.)
Interesting!
I recently stumbled across this paper, which although they don’t explicitly work with differential forms, do seem to be doing something coflare/cojet-y in order to get a chain rule for higher derivatives that acts like substitution.
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