Non-continuous differentiability for differential forms

Question

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

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

Answer 1

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 $\displaystyle\sum_{|I| = k} d(f_I) \wedge dx_I$

Answer 2

Every differentiable (even continuous) function has a distributional derivative, even of all orders. Also the theorem on equality of mixed partial derivatives holds when the derivatives are taken in distributional sense. So the formula $dd\omega=0$ is still valid as is a suitable version of Stokes' theorem. There are several advantages in this approach since distributional derivatives can detect physically meaningful jumps in the derivatives of functions. This theory can be derived at a fairly elementary level but, not surprisingly, has been developed at a high level of sophistication decades ago--- in the theory of currents ("courants") by Georges de Rham.

Disclaimer. This aspect was mentioned in comments above but, since it seems to have found no resonance, I decided to write it up as an answer with a little more detail.

Answer 3

It suffices to assume that the coefficients of a differential form, written with respect to a local co-ordinate system, are Lipschitz. This property is invariant under $C^1$ changes of co-ordinates. Moreover, the weak partial derivatives of the coefficients are $L^\infty$ and also remain so under $C^1$ changes of co-ordinates. The pointwise derivative is defined equal to the weak derivative almost everywhere.

You can therefore define the exterior derivative of a Lipschitz differential form using the standard formula, and it is an $L^\infty$ form. You can also define the exterior derivative of an $L^\infty$ form with respect to co-ordinates using distributional partial derivatives, but if you want to make this independent of co-ordinates, you need to use currents. In any case, $d^2 = 0$.

You appear to be asking for something stronger than this, because you want the pointwise derivative to exist everywhere instead of almost everywhere. As for functions, this turns out to be a difficult property to characterize and use. On the other hand, for most applications assuming either the stronger assumption of $C^1$ or the weaker assumption of Lipschitz suffices.