# Integration over parametrizations in the border of a manifold with corners (John M. Lee Introduction to Smooth Manifolds exercise 16.22)

In exercise $$16.22$$ of Lee's Introduction to Smooth Manifolds, the author wants us to translate the result of proposition $$16.8$$, the integration over parametrizations, to the border of a manifold with corners. I think the precise statement is the following one:

Let $$\partial M$$ be the border a compact, oriented smooth $$(n+1)-$$manifold with corners $$M$$, and let $$\omega$$ be a compactly supported $$n-$$form on $$\partial M$$. Suppose $$D_1,\dots,D_k$$ are open domains of integration in $$\mathbb{R}^n$$ (this means that each $$D_i$$ is bounded and its (topological) boundary $$\partial D_i$$ has measure zero as a subset of $$\mathbb{R}^n$$), and for $$i=1,\dots,k$$ we are given smooth maps $$F_i:\bar D_i\rightarrow\partial M$$ satisfying:

$$(i)\;F_i$$ restricts to an orientation-preserving diffeomorphism from $$D_i$$ onto an open subset $$W_i\subseteq \partial M$$;

$$(ii)\;W_i\cap W_j=\varnothing$$ when $$i\not=j$$;

$$(iii)\;\text{supp }\omega\subseteq\bar W_1\cup\dots\cup\bar W_k$$.

Then

$$\int_{\partial M}\omega=\sum_{i=1}^k\int_{D_i} F_i^*\omega$$

However, I don't understand the following points:

• First $$-$$ why is $$M$$ required to be compact, while the statement of proposition $$16.8$$ doesn't require the manifolds in question to be compact?
• Second $$-$$ is my restatement of $$16.8$$ complete? I feel like something is missing, but I am not sure. After seeing an application of this result in the demonstration of Stokes' Theorem for chains, I can't help thinking that what I have written is good enough
• Third $$-$$ is there any point in the proof of $$16.8$$ where any step has to be completely rewritten? After reading the proof several times, I feel like there is nothing wrong with it when we change the sets and maps to adapt this demonstration to the case we are studying. However, there must be some technical detail I am not seeing at all.

PS1: First point is somewhat obvious, since one can construct spaces where no finite decomposition $$D_1,\dots,D_k$$ and maps $$F_i:\bar D_i\rightarrow \partial M$$ exists (for instance, think of $$M$$ as a "snake" made of solid parallelepipeds that extends indefinitely across $$\mathbb{R}^3$$)

PS2: Needless to say, the sets $$W_i$$ cover the whole space $$\partial M$$ with the exception of some subset of measure zero in $$\partial M$$

• Since I am stuck on the same exercise at the moment, one thing I was wondering is whether it really makes sense to talk about smooth $F_i: \bar{D_i} \to \partial M$. Since $\partial M$ is not necessarily a smooth manifold (nor a smooth manifold with boundary/corners), what does smoothness of $F_i$ mean in that case? Commented Mar 11, 2023 at 15:56

It doesn't really need to be compact. I stated it that way because it was the only case I intended to use, but your question made me realize that there's no simplification in assuming $$M$$ is compact. (Even if $$\partial M$$ doesn't have a finite decomposition into nicely parametrized subsets, we can always find finitely many such parametrizations whose images cover $$\operatorname{supp}\omega$$. And in any case, the existence of such parametrizations is part of the hypothesis of the theorem.) I've added a correction to my online list.