Stokes theorem with corners

Question

I've found the following version of Stokes' theorem in G. Stolzenberg's lecture notes 19:

Notation:

for $1 \le n \le m$

$\Lambda(m, n) = \{ \lambda: \{1,...,n\} \rightarrow \{1,...,m\} \ | \ \lambda(1) < ... \lambda(n) \}$

$p_{\lambda}: \mathbb{R}^m \ni (x_1, ..., x_m) \rightarrow (x_{\lambda(1)}, ..., x_{\lambda(n)}) \in \mathbb{R}^n$

$\mathcal{L}^n$ is $n$-dimensional Lebesgue measure and $\mathcal{H}^n$ is Hausdorff measure

Now, the theorem itself:

Assume that

$(1)$ $N$ is an $n$-dimensional, oriented class $C^1$ submanifold of $\mathbb{R}^m$, $2 \le n \le m$

$(2) \ M \subset N$ is open, $int _N (\overline{M}^N) = M$ and $\partial _N M$ is either empty or is an $n-1$-dimensional class $C^1$ submanifold in $\mathbb{R}^{m}$

$(3) \ \overline{M}$ is compact, $\mathcal{H}^n(M) < \infty$, $\mathcal{H}^{n-1}(\partial_NM) < \infty$

$(4)$ Let's define $\delta: = \overline{M} \setminus N$ and assume that $(*) \ \ \forall \lambda \in \Lambda(m, n-1) : \ \mathcal{L}^{n-1}(p_{\lambda}(\delta)) = 0$

$(5) \ \omega \in \Omega_{n-1}^{(1)} (\mathbb{R}^m)$

$(6)$ We induce the orientation on $M$ and on the boundary: $\partial _NM$ from $N$

Then we have that:

$d \omega $ absolutely integrable on $M$, $\omega$ on $\partial_NM$ if it is not empty

and $\int_M d \omega = \int_{\partial_NM} \omega$

While proving this theorem, we assume, wlog, that $p_{\lambda} : \mathbb{R}^m \ni (x_1, ..., x_m) \rightarrow (x_1, ..., x_{n-1}) \in \mathbb{R}^{n-1}$

My question is:

Apparently, we can replace $(*)$ with $\mathcal{H}^{n-1}(\delta)=0$ which implies $(*)$.

I was wondering whether you could explain to me why this is true.

Also, could you recommend a book or a paper in which I can find something more about Stokes' theorem with corners (singular points)? I've already read

• Stokes theorem for manifolds with corners? and consequently:

• Brian Conrad's notes on differential geometry: math.stanford.edu/~conrad/diffgeomPage/handouts.html (but the problem is that I need a source which has been published)

• a chapter dedicated to Stokes' theorem in Sauvigny's "Partial
  Differential Equations" (here, we simlarly consider the set of
  singular boundary points which has capacity zero, although at the
  moment I'm not able to decide which assumption is more general )

• I've also read this article but it's not connected to my main
  theorem (presented above)

Could you recommend some books, papers in which I can find something about Stokes theorem which will "agree" with the theorem I wrote down here?

I would be extremely grateful for all your insight.

Answer 1

• MR0760450 de Rham, Georges Differentiable manifolds. Forms, currents, harmonic forms. Translated from the French by F. R. Smith. With an introduction by S. S. Chern. Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], 266. Springer-Verlag, Berlin, 1984. x+167 pp. ISBN: 3-540-13463-8 58A14 (55N99 58-01). See page 23.

A direct proof is not difficult: Remove the corners $C$ (codim $\ge 2$ in $M$) from the boundary. Then Stokes theorem holds for differential forms with compact support in $M\setminus C$. But you can approximate differential forms on $M$ with compact support by differential forms on $M\setminus C$ with compact support and the integrals over the manifold and over the boundary converge.

EDIT:

Namely, step on a differential form on $M$ with a smooth bump function which vanishes on an $\epsilon$-neighborhood of $C$ and which is 1 off the $2\epsilon$-neighborhood of $C$. Since $C$ is a union of codim $\ge 2$ strata, the measure of these neighborhoods goes to 0 with $\epsilon\to0$

Answer 2

This book http://www.math.wustl.edu/~sk/books/root.pdf (Geometric Integration Theory, by S.G. Krantz and H.R. Parks) is a self-contained introduction to geometric measure theory. See also Hassler Whitney's classic "Geometric Integration Theory".

http://www.ams.org/journals/bull/1993-29-02/S0273-0979-1993-00429-4/ (Stokes' theorem for nonsmooth chains, by J. Harrison) provides a generalization of the Stokes' theorem.