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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

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

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    $\begingroup$ I guess you are limited by the Stokes theorem about equality of mixed derivatives, which is basically behind the formula $dd\omega=0$. $\endgroup$ – Loïc Teyssier Apr 30 '14 at 18:42
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    $\begingroup$ Yes. On $\mathbb{R}^n$, a differential form is, more or less, a vector-valued function, and your definition works for vector-valued functions. You just need to show that the definition is preserved under sufficiently smooth changes of variable. The second derivative exists in the weak sense and, using that, $d^2 = 0$. $\endgroup$ – Deane Yang Apr 30 '14 at 18:50
  • $\begingroup$ Are you asking if it makes sense to consider differential forms with distributional coefficients? and if such forms form a complex? $\endgroup$ – Eugene Lerman Apr 30 '14 at 19:25
  • $\begingroup$ @Eugene: no, he's asking if it makes sense to consider a notion of the derivative of a differential form analogous to the above notion of the derivative of a function (in particular, such that the derivative is not necessarily continuous). $\endgroup$ – Qiaochu Yuan Apr 30 '14 at 21:59
  • $\begingroup$ @DeaneYang, I don't understand. It seems to me that the differential of $\mathrm{d}f$ as a vector-valued function is going to be the Hessian matrix of $f$, not zero. $\endgroup$ – Mike Shulman Apr 30 '14 at 23:37
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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 $\displaystyle\sum_{|I| = k} d(f_I) \wedge dx_I$

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  • $\begingroup$ This is not quite the same as the definition of the derivative in the case $k=1$. Conceivably the limit of $\frac{f(p+hv)-f(p)}{h}$ could be zero for all $v$ even if there is no limit of $\frac{|f(p+v)-f(p)|}{|v|}$ as $v\to 0$. $\endgroup$ – Tom Goodwillie May 2 '14 at 1:38
  • $\begingroup$ @TomGoodwillie you are right, I meant to say $k=1, n=1$. Although, I think my proposed definition at the end of the answer does work, and agree with the usual definitions at the level of functions. $\endgroup$ – Steven Gubkin May 2 '14 at 1:46
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    $\begingroup$ I think it's better to forget the scaling factors and say that the integral of $\omega$ over the parallelepiped determined by the $v_i$ is $(d\omega)(v_1,\dots)$ plus an error term that is $\epsilon $ times the volume of the parallelepiped and $\epsilon$ goes to zero as all the $v_i$ go to zero. But I haven't checked that this is invariant under coordinate change. $\endgroup$ – Tom Goodwillie May 2 '14 at 2:07
  • $\begingroup$ hmm...You are probably right. Will think this through. $\endgroup$ – Steven Gubkin May 2 '14 at 3:03
  • $\begingroup$ Tom is right, and I don't think your proposal at the end works in the case $k=1,n>1$ either. But this is the sort of thing I was hoping for. $\endgroup$ – Mike Shulman May 2 '14 at 19:02
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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.

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  • $\begingroup$ Interesting, but this is not any stronger than the existence of partial derivatives (indeed as you say it's even weaker), so I don't think it answers the question. $\endgroup$ – Mike Shulman May 1 '14 at 19:17
  • $\begingroup$ Well, I genuinely thought that I was providing useful information but it's your question so sorry. At the risk of being out of order yet again, I would suggest that if you are interested in subtle notions of differentiability between continous differentiability and the existence of pointwise partial derivatives, you might want to consult the specialists in real analysis (the Bruckners, David Preiss, the czech school---Fabián, Tišer) who have written voluminously on the subtelties of differentiation of functions on finite and even on infinite dimensional spaces. $\endgroup$ – janacek May 1 '14 at 20:16
  • $\begingroup$ It is useful information! I didn't mean to say it was useless, just that it doesn't exactly answer the question as I asked it. Can you give a good reference to read about "currents"? $\endgroup$ – Mike Shulman May 1 '14 at 20:44
  • $\begingroup$ It might be helpful if you could say a little about what you want to do. The distributional derivative is all that's needed in most cases. $\endgroup$ – Deane Yang May 2 '14 at 1:12
  • $\begingroup$ I want to teach my multivariable calculus students about exterior derivatives in a way that looks similar to how I taught them about differentials of functions. (They don't know about distributions yet.) $\endgroup$ – Mike Shulman May 2 '14 at 18:59
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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.

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  • $\begingroup$ I am guessing that, more than looking for "practical application" Mike is interested mostly because he wants a definition which resonates with him as strongly as the definition of the derivative of a function does. Rather than defining exterior derivative by a "formula", he wants a property that that characterizes the exterior derivative, and reduces to the ordinary derivative in the case of functions. $\endgroup$ – Steven Gubkin May 2 '14 at 21:23
  • $\begingroup$ But even the ordinary (pointwise) derivative of a function turns out to be a not very useful definition. $\endgroup$ – Deane Yang May 2 '14 at 21:38
  • $\begingroup$ The derivative of a function is not useful? $\endgroup$ – Steven Gubkin May 2 '14 at 21:40
  • $\begingroup$ The property that a function has a derivative at every point in its domain turns out not to be as useful as Lipschitz or $C^1$. $\endgroup$ – Deane Yang May 2 '14 at 21:45
  • $\begingroup$ Sure, but I am still thankful I know how to define the derivative of a function. $\endgroup$ – Steven Gubkin May 2 '14 at 21:46

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