About the integral of a positively oriented orientation form for noncompact manifolds.

Question

I'm studying Lee's Intro. to Smooth Manifolds 2 edition, and I have question regarding the situation in the title of this question.

Since he defines integration only for forms with compact support, how should I interpret item c) of proposition 16.6, which asserts that the integral of a positively oriented orientation form $\omega$ is positive. Since a orientation form can't have compact support unless the manifold is compact, should this integral be interpreted in some extended/improper way?

The author does give a brief comment on the possibility of extending the definition of integration to non compact supported forms but doesn't go into any details about how this process could be done.

Answer 1

I can see why you might be confused, but the hypotheses ($\omega$ is compactly supported and an orientation form) imply that $M$ must be compact. So that's the only case to which that statement applies.

Answer 2

Based on how Lee has introduced his definitions, yes, it seems reasonable to assume in addition that in part (c) of that proposition, $M$ is compact. However, the definition of integration on manifolds can be extended even to the case where the forms do not have compact support. Note that the compact support assumption only comes into play to ensure that the integral $\int_M \omega := \sum_i \int_M \psi_i M$ is a finite sum so that everything is well defined (and there are no infinite series to deal with, hence no convergence issues); so to extend the definitions to the non-compact support case, we just have to systematically revisit the definitions and see how to weaken the hypothesis.

Recall that with certain topological restrictions on $M$ (such as second countability and Hausdorffness; which I think are already part of Lee's definition of a manifold), it admits a partition of unity. Suppose we're given a continuous top-degree form $\omega$ on $M$. Now, pick a countable atlas $\mathcal{A}=\{(U_i,\alpha_i)\}_{i=1}^{\infty}$, and a partition of unity $\Psi=\{\psi_i\}_{i=1}^{\infty}$ subordinate to the atlas $\mathcal{A}$ (i.e $\text{supp}(\psi_i)$ is compact and lies in $U_i$). Then, the form $\psi_i\omega$ has compact support contained inside a chart domain $U_i$, so it's integral is well-defined according to Proposition $(16.4)$.

Let us say $\omega$ is integrable with respect to the atlas $\mathcal{A}$ and partition of unity $\Psi$, if the following sum is finite: \begin{align} \sum_{i=1}^{\infty} \int_M |\psi_i \omega| < \infty \tag{*} \end{align} In this case, we define \begin{align} \int_{(M,\mathcal{A}, \Psi)} \omega := \sum_{i=1}^{\infty} \int_M \psi_i \omega \tag{$**$} \end{align}

Note that by the absolute value in $\int_M |\psi_i\omega|$, I just mean we push-forward the form $\psi_i\omega$ using $\alpha_i$, so that we have $(\alpha_i)_*(\psi_i\omega) = f_i\, dx^1 \wedge \dots \wedge dx^n$ for some continuous $f_i:\alpha_i[U_i]\to \Bbb{R}$ with compact support, so we take $\int_M |\psi_i\omega| = \int |f_i| \, dV$; where on the RHS you have a standard integral (either Riemann or Lebesgue; if Riemann then you have to pay more attention to the domain of integration being Jordan-measurable; i.e bounded and boundary having measure zero etc) in $\Bbb{R}^n$ of a non-negative function.

Because this definition of "integrability" of a form requires "absolute convergence" in $(*)$, it follows that the sum in $(**)$ exists. Moreover because of the absolute convergence, we can show that the notion of integrability, and the resulting integral do not depend on the choice of atlas $\mathcal{A}$, nor the partition of unity $\Psi$ (if you have access to it, Spivak's Calculus on Manifolds, Theorem $3.12$ proves something very similar).