Generally when working with differential forms, one assumes that they are continuously differentiable, i.e. $C^r$ for some $1\le r \le \infty$. Under this hypothesis, one can define the exterior derivative in any of the usual ways. And if we regard a continuously differentiable function $f$ as a 0-form, then its exterior derivative $\mathrm{d}f$ is exactly its "total" differential in the usual sense.


However, for a function $f:\mathbb{R}^n\to \mathbb{R}$ we have a notion of *differentiability* which is stronger than the existence of all partial derivatives (or even all directional derivatives), yet weaker than continuous differentiability: $f$ is differentiable at $x$ if there exists a linear map $\mathrm{d}f\_x$ such that


$$ f(x+h) = f(x) + \mathrm{d}f\_x(h) + \epsilon \Vert h\Vert $$


where $\epsilon\to 0$ as $h\to 0$. (Here $x,h\in\mathbb{R}^n$.)


Is there an analogous definition of a not-necessarily-continuous exterior derivative of a differential $p$-form (on $\mathbb{R}^n$, say) for $p>0$?


**Edit:** Apparently the word "analogous" was not clear enough. What I'm hoping for is a definition like this: a $p$-form $\omega$ is *differentiable* if there exists a $(p+1)$-form $\mathrm{d}\omega$ such that


$$ (\text{something involving }\omega) = \mathrm{d}\omega + (\text{something going to } 0) $$



Non-continuous differentiability for differential forms
Let $\omega$ be a differential $k$-form on $\mathbb{R}^n$. Let $v\_1,v\_2,...,v\_{k+1}$ be $k$ vectors in the tangent space to $\mathbb{R}^n$ at the point $p \in \mathbb{R}^n$. 


Given scaling factors $h\_1,h\_2,...,h\_{k+1}$ of the vectors $v\_1,v\_2,...,v\_{k+1}$, we get a parallelepiped $P\_h$ defined by the vectors $h\_1v\_1,h\_2v\_2,...,h\_{k+1}v\_{k+1}$. Let $bP\_h$ be the oriented boundary of this parallelepiped. 


Then the exterior derivative of $\omega$ can be defined by the formula


$$d \omega (p)(v\_1,v\_2,...,v\_{k+1}) = \lim\_{h\_i \to 0} \frac{1}{h\_1h\_2...h\_{k+1}}\int\_{bP\_{h}} \omega $$


Notice that this reduces to the formula for the derivative in the case $k=1$ (In this case integration of a zero form over an oriented collection of points is just signed summation).


So I propose the definition:


Let given $k+1$ vectors $v\_1,v\_2,...,v\_{k+1}$, let $P$ be the oriented parallelepiped spanned by these vectors.


A $k$-form $\omega$ on $\mathbb{R}^n$ is said to be differentiable if there is a $k+1$-form $d\omega$ such that 


$$\int\_{bP} \omega = d\omega(p)(v\_1,v\_2,...,v\_{k+1}) + \epsilon|Vol(P)|$$


where $\epsilon \to 0$ as the $v\_i \to 0$




---


Thinking about this a bit more, what if we try to replace the integration with just evaluation?


Just to see an example play out first, consider a one form $\omega$ on $\mathbb{R}^n$. It seems reasonable to demand the following approximation (to be made precise shortly):


$d\omega(p)(v\_1,v\_2) \approx \omega(p)(v\_1)+\omega(p+v\_1)(v\_2) - \omega(p+v\_2)(v\_1) - \omega(p)(v\_2)$


This is obtained by just making a picture of the parallelogram spanned by $v\_1$ and $v\_2$ and taking signed sum of the boundary.


Example:


If we take a one form $\omega = fdx +gdy$ on $\mathbb{R}^2$ then we should have


$$\begin{align}
d\omega(p)(v\_1,v\_2) &\approx \omega(p)(v\_1)+\omega(p+v\_1)(v\_2) - \omega(p+v\_2)(v\_1) - \omega(p)(v\_2)\\
&=f(p)dx(v\_1)+g(p)dy(v\_1) + f(p+v\_1)dx(v\_2)+g(p+v\_1)dy(v\_2)-f(p+v\_2)dx(v\_1) - g(p+v\_2)dy(v\_1) - f(p)dx(v\_2)-g(p)dy(v\_2)\\
&=[f(p+v\_1)-f(p)]dx(v\_2)-[f(p+v\_2)-f(p)]dx(v\_1)+[g(p+v\_1)-g(p)]dy(v\_2)-[g(p+v\_2)-g(p)]dy(v\_1)\\
&\approx df(p)(v\_1)dx(v\_2)-df(p)(v\_2)dx(v\_1) + dg(p)(v\_1)dy(v\_2)-dg(p)(v\_2)dy(v\_1)\\
&=(df \wedge dx)(v\_1,v\_2)+(dg \wedge dy)(v\_1,v\_2)\\
&=(df \wedge dx+dg \wedge dy)(v\_1,v\_2)
\end{align}$$


This is, of course, the usual formula for the exterior derivative.


So I now propose the following definition:


A $k$-form $\omega$ is said to be differentiable if there is a $k+1$ form $d\omega$ such that 


$$\sum\_{i=1}^{k+1} (-1)^i(\omega(p+v\_i)-\omega(p))(\widehat{v\_i}) = d\omega(p)(v\_1,v\_2,...,v\_{k+1}) + \epsilon \left| \textrm{Vol}(P) \right|$$,


where $\epsilon \to 0$ as $v\_1,v\_2,...,v\_{k+1} \to 0$, $P$ is the parallelepiped based at $p$ and defined by the vectors $v\_1,v\_2,...,v\_{k+1}$. $\widehat{v\_i}$ is the ordered list $v\_1,v\_2,...,v\_{k+1}$ with $v\_i$ left out. 


The expression on the LHS of the expression is "$\omega$ applied to the oriented boundary of $p$".


In the case that $\omega$ is differentiable according to the definition we have given above, we can obtain a formula for the exterior derivative which is similar in spirit to the formula for the derivative. 


Namely, we have 


$$
\begin{align}
d\omega(p)(v\_1,v\_2,...,v\_{k+1}) &= \sum\_{i=1}^{k+1} (-1)^i(\omega(p+v\_i)-\omega(p))(\widehat{v\_i}) - \epsilon \left| \textrm{Vol}(P) \right|
\end{align}
$$


so 


$$
\begin{align}
d\omega(p)(hv\_1,hv\_2,...,hv\_{k+1}) &= \sum\_{i=1}^{k+1} (-1)^i(\omega(p+hv\_i)-\omega(p))(h\widehat{v\_i}) - \epsilon \left| \textrm{Vol}(P\_h) \right|\\
h^{k+1}d\omega &= h^k \sum\_{i=1}^{k+1} (-1)^i(\omega(p+hv\_i)-\omega(p))(\widehat{v\_i})- \epsilon h^{k+1}\left| \textrm{Vol}(P\_1) \right|\\
d\omega &= \sum\_{i=1}^{k+1} (-1)^i\left[\frac{\omega(p+hv\_i)-\omega(p)}{h}\right](\widehat{v\_i})- \epsilon \left| \textrm{Vol}(P\_1) \right|
\end{align}
$$


Taking limits of both sides as $h \to 0$, we have


$$d\omega = \lim\_{h \to 0} \sum\_{i=1}^{k+1} (-1)^i\left[\frac{\omega(p+hv\_i)-\omega(p)}{h}\right](\widehat{v\_i})$$


In the case that $\omega$ is expressed in the form $\displaystyle\sum\_{|I| = k} f\_I dx\_I$ (using multi index notation), you could further reduce this formula to the usual $\displaystyle\sum\_{|I| = k} d(f\_I) \wedge dx\_I$


