Why can we only integrate top forms?

Question

I understand some of the mechanics behind integration over manifolds, but I am lacking on some of the intuition.

In Tu's An Introduction to Manifolds, when introducing integration of differential forms (Section 23.4), he says:

Our approach to integration over a general manifold has several distinguishing features:

1. The manifold must be oriented
2. On a manifold of dimension $n$, one can integrate only $n$-forms, not functions
3. The $n$-forms must have compact support.

My current understanding is that (1) is necessary to give a sign to the integral, just as we have in $\mathbb{R}^n$.

(3) is necessary to eliminate cases where the domain is infinite (for example, to rule out $\int_{-\infty}^\infty dx$).

However, I am less certain about why (2) is necessary. In Lee's An Introduction to Smooth Manifolds, he gives some intuition about how top forms provide a way of measuring "signed volume" at each point of a manifold. Thus I see why they're useful in integration, but I don't see why this rules out integrating differential forms that are not top forms (including $0$ forms, i.e. functions). What is the intuition behind this?

Answer 1

The central idea motivating integration is to provide a way of "summing up" all the values of some function over a given region. This must be done through some limiting process, e.g. a Riemann sum, and in order for this limit to make any sense, we must have a concept of how much to weight an infinitesimally small "cube" in the region. This is done by defining some notion of volume. For an $n$-dimensional manifold, $n$-forms are a natural way to define volume. Therefore a reasonable notion of integration on a manifold is inherently tied to specifying some $n$-form.

Maybe a more lowbrow reason: think about intuition from multivariable calculus. To integrate over an $n$-dimensional cube, you iteratively take $n$ single variable integrals corresponding to the $n$ coordinate directions, and for this to work you really do need to be integrating an $n$-form. I can't imagine a reasonable definition where you could integrate with fewer or more forms, since that would no longer capture the idea of "integrating over an $n$-dimensional cube."

Answer 2

tl;dr If you could integrate 0-forms on a smooth manifold, you'd automatically get a natural definition of volume of all (say, open) subsets $U\subset M$ by $\textrm{vol}(U):=\int_U 1$. But as you know, angles, lengths, volumes require an additional structure - a metric, so this should be impossible. The differentiable (smooth) structure does not distinguish between the single-chart manifolds "$\mathbb R$ with a chart $\varphi(x)=x$" and "$\mathbb R$ with a chart $\varphi(x)=3x$" - they're smoothly isomorphic. On the other hand, the set $[0,1]$ has a length 1 in the former and a length 3 in the latter.

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I hope a hands-on example can build some intuition that scalar functions (0-forms) are in a sense insensitive to changes of coordinates. Lee actually mentions in paragraph The Geometry of Volume Measurement that their integral would not be invariant under coordinate transformations; let me elaborate with an explicit example.

Consider an open real interval $M\subset\mathbb R$ as a manifold and the constant function $f(x)=1$ on it. I do not specify the interval on purpose, because it doesn't matter here (neither its length nor its endpoints).

1. Within a chart $((0,1),\varphi)$ (so $\varphi:M\to(0,1)$) our $f$ has exactly the same values: $f(t)=1$ for all $t\in(0,1)$. Define here (in this chart) the constant 1-form $\omega=dt$.

2. Now look at $f$ and $\omega$ in a 3 times wider coordinate system $\psi:M\to(0,3)$ (with a change of variables $(\psi\circ\varphi^{-1})(t)=3t$).
   Here $f(t)=1$ (again!) for all $t\in(0,3)$, but (!) the coordinate tangent vectors $\partial t$ are 3 times shorter (a $\partial t$ under $\varphi$ is $3\partial t$ under $\psi$), so (!) $\omega=\frac 1 3 dt$ (and not $\omega=dt$ like above).

That is, $\omega$ "felt" the change but $f$ did not!

What about the integral? Under $\varphi$, \begin{align} \int_M f&=\int_{(0,1)} 1dx=1\\ \int_M \omega&=\int_{(0,1)} dt=1, \end{align} while under $\psi$, \begin{align} \int_M f&=\int_{(0,3)} 1dx=3\\ \int_M \omega&=\int_{(0,3)} \frac 1 3dt=3\times \frac 1 3 = 1. \end{align}

That is, $\omega$ correctly adjusted under the change and the integral remained the same, while $f$ blindly remained a constant 1 in both coordinate systems and its integral was dependent on the euclidean size of the arbitrarily chosen chart image ($\varphi(M)$ or $\psi(M)$ above).

For this reason $\int_M f$ is meaningless, while $\int_M \omega$ is well-defined.