I was wondering if Stokes' theorem could be formulated in a setting which could be easily applied in situations where the traditional form cannot, such as on manifolds with corners like a rectangle or on a cone. I was thinking of something like:

If $M$ is a n-dimensional oriented Lipschitz-manifold with boundary and $\omega$ a compactly supported locally Lipschitz $n-1$-form on $M$, then $$\int_{M}d\omega=\int_{\partial M}\omega.$$

The notion of the (exterior) derivative of a form would of course have to include some notion of almost everywhereness on $M$, like applying Rademacher's theorem to the functions $\omega\circ\phi$ for a countable cover with charts $\phi$. I wonder if this has been done or can be done at all.

  • $\begingroup$ I don't know about the more general case, but can't Stokes' theorem be applied to manifolds with corners using the usual notions anyway? I think so. $\endgroup$
    – Sam
    Commented Apr 24, 2011 at 1:02
  • 5
    $\begingroup$ Yes, as Sam mentions Stokes' theorem applies without any trouble to manifolds with cubical (or even worse) corners. For Lipschitz manifold it holds as well. See for example: math.neu.edu/sites/default/files/salvi.h/Stokes%20theorem.pdf a Google search will give you many more references. $\endgroup$ Commented Apr 24, 2011 at 1:20
  • $\begingroup$ related mathoverflow.net/questions/414309/…. @RyanBudney, unfortunately the link you posted is not working anymore. $\endgroup$
    – No-one
    Commented Jan 22, 2022 at 13:57
  • $\begingroup$ @Titti: There are textbook proofs of Stokes' theorem that are essentially in the cubical corner category. Usually they are not written up in generality, but the proof in Hubbard's textbook is one approach. I think he got the idea from Arnold. $\endgroup$ Commented Jan 22, 2022 at 19:50
  • $\begingroup$ @RyanBudney I have just checked the proof in Hubbard's book and I don't think it generalizes to Lipschitz manifolds. In fact, I don't think that it is possible to prove the Stokes theorem in the Lipschitz case without referring to currents, since in this context it is not even clear what is a differential form, not to mention the exterior differential. $\endgroup$
    – No-one
    Commented Jan 23, 2022 at 0:14

2 Answers 2


There appear to be plenty of references on google (as Ryan pointed out). Here is one on Stokes theorem for Lipschitz forms on a smooth manifold:


Does that help?


Yes, the Stokes theorem can be formulated and holds for compactly-supported Lipschitz forms on oriented Lipschitz manifolds with boundary. The proof I know, however, is not obvious and uses techniques of the theory of currents (https://en.wikipedia.org/wiki/Current_(mathematics)). The paper linked by user230715 is good, but see also the more complete https://link.springer.com/article/10.1007%2FBF00971687. Notice that, once everything has been made rigorous, it is enough to prove the result for Lipschitz forms on smooth manifolds because we can take pullbacks and work in coordinates (as we do in the proof of the smooth case).

I don't think that elementary (i.e. avoiding currents) proofs of the Stokes theorem exist in the general Lipschitz context, but they do in some special cases. You can find, for example, a relatively easy one that holds for Lipschitz domains in $\mathbb{R}^n$ and $C^1$ forms in the appendix of the book Measure Theory and Integration by Michael Taylor.

See also my question on mathoverflow for a far-reaching generalization of the Stokes theorem to countably rectifiable sets (of which Lipschitz manifolds are an instance) https://mathoverflow.net/questions/414309/measure-theoretic-boundary-and-federers-theorem-in-arbitrary-codimension.


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