# The integral of a function on manifold and differential form

When we want to integrate a function f over a manifold M, we may meet some problems, for example, the problem showed in the picture below:

Then people used differential form to integrate. But it confused me:does that really solve the problem of integrating the function f on M? How can f be related directly to a k-form $\omega$? Is there a k-form $\omega$ to represent a specific function f on manifold M?

Differential forms are not introduced to answer the question "How do I integrate functions of manifolds?"

Instead, they are introduced to answer a different question, namely "What are the correct objects to integrate on manifolds?"

The answer to that question is: differential forms! More precisely, the correct objects to integrate on an orientable $m$-dimensional manifold are differential $m$-forms.

The reason they are correct is precisely because the integral of a differential form does transform correctly, without its value changing, under a change of coordinates.

Added: You say you still want to know how to integrate a function on a manifold. As I have tried to explain briefly so far, this is a quixotic quest. One should not attempt to integrate functions on manifolds, because it will not be well-defined when you change coordinates.

Here is an example to better illustrate why instead we integrate differential forms, not functions.

In multivariable calculus you learn that total mass equals the integral of density. What are mass and density? Are they functions? If not, what are they?

Let's take a look at a typical mass integral such as $\int_U f \, dx$. The numerical value of this integral represents the total mass. Now, we cannot measure the mass at a point, it does not make mathematical sense. But what we can measure is the infinitesmal ratio of mass per unit volume at a point: that's called the density function, and that's the integrand $f$ in the density integral. Also, the expression $dx$ represents an infintesmal amount of volume. The expression $f \, dx$ then represents the product of the infinitesmal ratio of mass per unit volume, multiplied by infinitesmal volume; so you can think of $f \, dx$ as a measurement of the total mass has been spread around over the whole of the set $U$.

That's one way to think of a differential form such as $f \, dx$ (at least in the special case of a differential $k$-form on a $k$-dimensional manifold). Namely, it is a physical description of a total mass that has been spread over the whole of the manifold.

Note well: the differential form is not the density function $f$; the actual numerical value of density per unit volume depends on the units you choose to measure volume. Instead, the differential form is the density function multiplied by the volume form, namely $f \, dx$. This quantity is well-defined independent of the coordinates you choose.

Let's look at an example to illustrate about what happens under a coordinate change. Start with imagining $U$ as a box $\mathbb{R}^3$ with coordinate functions $x_1,x_2,x_3$, of side length $2$ and volume $8$. Imagine also that we have a total mass $M$ of constant density, the value of that density being $M/8$ at every point. The total mass is $\int_U M/8 \, dx = M/8 \int_U dx = M/8 \times \text{(volume of$U$in$x$-coordinates)} = M/8 \times 8 = M$.

What happens when you change coordinates? Well, suppose you change coordinates by a function of the form $$(y_1,y_2,y_3) = F(x_1,x_2,....,x_k) = (\frac{3}{2} \, x_1,\, \frac{3}{2} \, x_2,\, \frac{3}{2} \, x_3)$$ This corresponds to getting a new ruler, the $y$-ruler, whose unit length is $2/3$ the size of the unit length on the $x$-ruler you started with. Under this coordinate change function $F$, the box with side length $2$, volume $2^k$, and mass $M$ in $x$-coordinates becomes a box with side length $3$ and volume $3^k$ in $y$-coordinates. But the mass $M$ is unchanged.

Mass does not change, just because you are using a different ruler.

In the new $y$-coordinates, the density function is $M/3^3=M/27$ per unit volume. And the new mass integral is $\int_U M/27 \, dy = M/27 \times \text{(volume of$U$in$y$-coordinates)} = M/27 \times 27 = M$.

To summarize this example, the expressions $M/8 \, dx$ and $M/27 \, dy$ represent the same differential form on $U$, expressed in two different coordinate systems. The integral of this differential form over $U$ has a well-defined value, namely $M$, the total mass.

• Thank you, it arises a problem, that is when we integrate a function, it has the geometrical meaning, while I can't see the geometrical meaning of integrating the differential form. Also why do we want to integrate the differential form? What's the motivation of integrating the differential forms? – 6666 Jun 17 '16 at 4:46
• Also I am still curious about how to integrate the function on manifold? – 6666 Jun 17 '16 at 5:17
• Sorry, I still feel confused, in you words, we see the differential form as the total mass has been spread around over the whole of the set U, and the integral of the differential form over U is a the total mass, however, we can do the same thing in calculus, right, then the $dx_1\wedge ...dx_m$ can be replaced by $dx_1d_2...d_m$, and in this sense, they are the same thing? – 6666 Jun 17 '16 at 23:06
• Roughly speaking $dx_1 \wedge dx_2 \wedge \ldots \wedge dx_m$ is the same thing as $dx_1 \, dx_2 \, \ldots \, dx_m$, but you've got to remember, what you learn in a first calculus course needs to be reformulated/abstracted/generalized, before it can be adapted to obtain a good theory of calculus on manifolds. Those $\wedge$ symbols are part of that reformulation/abstraction/generalization. – Lee Mosher Jun 18 '16 at 1:00
• Thank you, it makes sense now. – 6666 Jun 18 '16 at 1:04

In Leibniz notation, even in the beginning, you weren't integrating functions: you were integrating differential forms.

This is true even back in introductory calculus, even though you didn't have words to put to the concept. e.g. you didn't compute $\int f(x)$; you computed $\int f(x) \, \mathrm{d}x$.

And even then is was important not to forget that $\mathrm{d}x$; e.g. if you had some other variable $u$ related to $x$ by $x = g(u)$, then you'd compute $\mathrm{d}x = g'(u) \mathrm{d}u$ and have

$$\int f(x) \, \mathrm{d}x = \int f(g(u)) \, g'(u) \mathrm{d} u$$

Your introductory calculus didn't want to teach differential geometry, so you learned a white lie that you're integrating functions, the point of that diversion is to help break you overcome that lie and realize that differential forms are the right notion.

Generally speaking, when you do talk about integrating functions, it's because you have a fixed volume form in mind, and by "integrate the function" you implicitly mean "integrate the product of that function with the volume form".

Or some equivalent notion is at play; e.g. the hodge star swaps the notion of "function" and "volume form".

That said, you can integrate a function over a zero-dimensional set, but we normally just call that "evaluation". However, it is nice to make this explicit, so that the fundamental theorem of calculus becomes a special case of the generalized Stoke's theorem:

$$\int_0^1 f(x) \, \mathrm{d}x = \int_{\partial[0,1]} f(x) = f(1) - f(0)$$

since the boundary of $[0,1]$ with its usual orientation is $1$ positively oriented and $0$ negatively oriented.

• Thank you, but I still feel confused. We have a geometrical motivation/meaning to lead us to integrate the function, while I can't see the same thing on differential form. Why do we want to integrate a differential form? What does the integral of differential form represent then? Also does that mean we can never integral a function on a manifold? Or maybe it already has a solution to integrate the functions on manifolds? – 6666 Jun 17 '16 at 5:17
• @6666: You're thinking something like "area under a graph"? The function isn't enough -- $f(x)$ only measures height. To get area, you need to measure 'infinitesimal' widths too, e.g. with $\mathrm{d}x$. Or are you thinking about accumulating changes in a function? $\mathrm{d}f(x)$ is already a differential form! You could divide it by $\mathrm{d}x$ if you want to get the derivative with respect to $x$, but you just have to multiply $\mathrm{d}x$ back in to get the 'change' in $f(x)$. – Hurkyl Jun 17 '16 at 5:19
• Sorry, I don't understand, why do we need to measure the width? – 6666 Jun 17 '16 at 6:37
• @6666: Because the area of a rectangle is length times width. – Hurkyl Jun 17 '16 at 13:33