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In our course of differential geometry we defined the integral $\int_{U} \omega$ of a differential form $\omega=f dx_1\wedge \ldots \wedge dx_n: T^nU \rightarrow \mathbb R$ with $U\subseteq \mathbb R^n$, $U$ open and compact support $\operatorname{supp}(\omega)$ via $$\int_U \omega := \underbrace{\int_U f(x_1,\ldots,x_n) dx_1 \ldots dx_n}_\text{Riemann integral}$$

This is the same definition Manfredo do Carmo used in his book "Differential Forms and Applications" [chapter 4.1, page 55].

Unfortunately there was no motivation given for this definition in our course. For me as a student it seemed that our professor set $\int_U f dx_1\wedge \ldots \wedge dx_n$ to be $\int_U f(x_1,\ldots,x_n) dx_1 \ldots dx_n$ because both notions are similar. But the reasoning must be vice versa: Because both concepts of integration lead to the same result the notations are similar.

So why does the above definition make sense? Is there some kind of intuitive idea behind the integral of $\omega$ from which I can see that it is the usual Riemann integral? What is the concept/idea behind integrating differential forms? (For example I can think of the Riemann integral as the limit of the area under step functions)

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2 Answers 2

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Once I heard differential forms being introduced in a classroom as "the stuff we want to integrate". Take a look at the change of variables formula for integration in $\mathbb{R}^n$ (the Jacobian formula) and you will see it behaves the way differential forms do. The impression one gets is that differential forms were created to simplify integration. I think the motivation is clear if you look at the properties that define a differential form as being tailor made to be used in integration. I agree the motivation is opaque if you think of forms by themselves and then ask why their integral is defined in that way. Also of course, once a mathematical object is defined, mathematicians will explore its properties independent of the reason that leads to its creation. Complex numbers were created to express solutions of polynomial equations but obviously their use has gone beyond that, with forms is the same story.

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There are a couple confusing things going on here.

First, basis covectors are often denoted something like $\mathrm dx^i$ which is visually similar to $dx^i$ (I'm writing them like so, differently, to emphasize these are different notions; it wouldn't surprise me to see $dx^i$ as a basis covector either though), so the relationship between a basis covector and a differential appears clear, even obvious.

And yet these things are not really even alike.

When you integrate a $k$-covector, what are you really doing? You're integrating on some (usually) $k$-dimensional manifold. If $e_1, e_2, \ldots$ are basis vectors of this manifold*, then an infinitesimal patch of this manifold is described using a $k$-vector $dV = (e_1 \wedge e_2 \wedge \ldots \wedge e_k) dx^1 dx^2 \ldots dx^k$.

What do $k$-covectors do? Well, usually we're told they eat $k$ vectors in all to give a scalar (or a scalar field, if in fact we have a $k$-covector field). Alternatively, they can be seen to eat a single $k$-vector.

So what you're really doing when you integrate a $k$-covector is this:

$$\int_U \omega \equiv \int_U \omega(e_1 \wedge e_2 \wedge \ldots \wedge e_k) \, dx^1 dx^2 \ldots dx^k$$

For some reason, people seldom even talk about the existence of $k$-vectors. Knowing they exist at all is really important. It turns the Riemann tensor into a map from $2$-vectors to $2$-vectors, for instance. Anyway, the object $\omega(e_1 \wedge \ldots \wedge e_k)$ is a scalar function and as such you clearly a classic Riemannian integral now.

*I don't denote that these basis vectors depend on the point of the manifold you take them at, but they do have this dependence.

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  • $\begingroup$ By $dx^i$ you mean the exterior derivative of the function $x^i:U\rightarrow \mathbb R:p\mapsto \phi_i(m)$ whereby $(U,\phi)$ is a chart of the manifold and $\phi_i(m)$ is the $i$th component of vector $\phi(m)$? $\endgroup$
    – Stephan Kulla
    Jan 30, 2014 at 10:33
  • $\begingroup$ when you say "$e_1,e_2,\ldots$ are basis vectors of this manifold", then you mean the section defined on each trivalization $\psi:TU\rightarrow U \times \mathbb R^k$ ($U$ open set of manifold) by $e_i : U \rightarrow TU: m \mapsto \psi^{-1}(m,\tilde e_i)$ whereby $\tilde e_i$ is the ith standard basic vector of $\mathbb R^k$? $\endgroup$
    – Stephan Kulla
    Jan 30, 2014 at 10:42
  • $\begingroup$ I use $\mathrm d$ for the exterior derivative here, not $d$. Otherwise, yes. As far as the basis vectors, I personally prefer to think of them merely as elements of the tangent space, but I think your map definition is correct, sure. $\endgroup$
    – Muphrid
    Jan 30, 2014 at 16:19

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