# Stokes theorem for manifolds with boundary as disjoint union of submanifolds

Looking at the generalizations of Stokes theorem, I did find a version for manifold with corners, but I was surprised this generalization doesn't contain a simple example such as the cone. So my question is the following :

Is the following version of Stokes theorem correct?

Let $M$ be a smooth submanifold without boundary of $\mathbb{R}^N$ of dimension $n\leq N$, and suppose that the topological boundary $\partial M$ of $M$ is the finite disjoint union of (smooth) submanifolds of $\mathbb{R}^N$, each of them of dimension $\leq n-1$. Then for all $n-1$ form on $\mathbb{R}^N$ with compact support, we have $\int_M d\omega = \sum_i \int_{B_i} \omega$, where $(B_i)$ are all the submanifolds of dimension $n-1$ in the decomposition of $\partial M$.

I am aware of a generalization of Stokes theorem that contains the cone, that is in the book of Partial Differential Equations 1 of Sauvigny, where the notion of "singular set with capacity zero" is described, but I admit I don't understand what it means. I suppose (but I am not sure) that the previous statement is true if one can show that submanifolds of dimension $\leq n-2$ in the topological boundary of $M$ are with capacity zero with respect to $M$ in the sense of Sauvigny. Is that correct?

TL;DR Yes, i think this is correct and should follow directly from a general version of Stokes's theorem in Geometric Measure Theory.

Recall that a rectifiable set of dimension $m$ in ${\bf R}^n$ is a set of finite $m$-dimensional Hausdorff measure almost contained (wrt this measure) in the union of images of countably many Lipschitz functions from ${\bf R}^m$ to ${\bf R}^n$. Manifolds with boundaries, cones, etc are of course rectifiable.

The Stokes's theorem is valid for any rectifiable set with compact support. Actually this is a tautology because Stokes' formula is used to define the boundary of such a set as a rectifiable current (i.e as a linear form on the sets of differential forms, or equivalently as a differential form with distribution coefficients).

So the question is, when does the boundary of a rectifiable set is associated to a rectifiable set itself, such as to give a geometric meaning to the boundary? The Closure Theorem from Geometric Measure Theory implies that the boundary is a rectifiable set if and only if it has finite mass.

A general reference is the book "Geometric Measure Theory" by H. Federer. If you are not interested in the lengthy and technical proofs, you can instead read the shorter book "Geometric Measure Theory, a beginner's guide" by Frank Morgan (chapter 4) which surveys the whole theory.

The first step in the theory is to define a notion of a tangent cone at a given point of a measurable set and to show that if the set is m-rectifiable, then this cone is actually an m-dimensional plane almost everywhere.

• Thank you, that helps a lot. Just so that I understand a little bit more the details, I have a few questions. 1) What do you mean by rectifiable set with compact support ? 2) What does it mean for a rectifiable set to have finite mass? 3) How to show that a finite union of submanifolds of dimension $\leq n-1$ has finite mass? 4) Does this imply that we must rectrict to boundaries with finite hausdorff measure ? – Jon-S Oct 31 '16 at 22:59
• For a set, compat support just means that the closure is compact. The mass of a rectifiable set is the Haussdorff measure of the set counting multiplicities. So for a finite union of manifolds, you just have to show that the n-1 Haussdorff measure of each pieces is finite. – coudy Nov 1 '16 at 11:05