What is the definition of integration over a manifold and what does it measure?

[JG89]

When we integrate a scalar map over a manifold M, what exactly are we measuring?

If M is the unit circle in R^2, then regular Riemann integration of the function f = 1 over it will yield the volume of a cylinder of height 1. Okay, no problem.

Now, if we integrate f = 1 over the unit circle in R^2 using the definition of the "integral of a scalar map over a manifold" then I get for my answer 2pi.

If f = 2, then I get 2*2pi = 4pi.

What exactly are we measuring here? I'm very confused.EDIT: Ok, I know that if we integrate f = 1 over the manifold M (which in our example is the unit circle), then we are just measuring the 1-dimensional volume of M, which is its length. So no problem there. But what about f = 2? This gives us an answer of 2*2pi = 4pi, so what is that that we are measuring? The circumference of the circle if we go twice around it?

What about the function f(x,y) = x^2 integrated (using the definition of integration over manifolds) over the unit circle? Whatever the answer may be, and I can certainly calculate it, what will that measure?

 

[micromass]

Hi JG89!

There are quite some possible answer for this. Let me explain this with the point of view of "mass".

I'm not going to work with the unit circle because of technicalities, but you'll get the idea.

Basically, let's say that you have a wire. Now, a wire is actually a cylinder with a very small radius, so instead of treating the wire as a cylinder, we might see the wire as a line.

Let's say that our wire is a perfect cylinder of height 1000 and radius 0.1. Then the volume of the wire is

$$V=10 \pi$$

But what if our wire is thinner at some places and thicker at other places? For example, at some places the wire is 0.2 and at some place it is only 0.05. How to calculate the volume then?

Well, we can treat the wire as a perfect line, for example, we can treat the wire as the line [0,1000]. And we can have a radius function

$$r:[0,1000]\rightarrow \mathbb{R}$$

that associates with every point x the instantenious volume. For example, if the radius was uniform, then we would have r as the constant function 0.1 for example.

The instantenious volume at a point x is just $\pi r(x)^2$. To know the total voluma, we need to sum all these instantenious values, this is done by the integral

$\int_0^{1000}{\pi r(x)^2dx}$

Of course, the wire doesn't have to be represented by the line [0,1000]. It can also be a closed wire. Then we would represent it by the circle. A function f on the circle can then represent the radius, and the integral

$\int_{S^1}{\pi f(x)^2dx}$

does then describne the volume of the closed wire.

 

[I like Serena]

When we integrate a scalar map over a manifold M, what exactly are we measuring?

I think you're measuring the properties of your map, which has little to do with the manifold.
Choose a different map, and the integral will yield a different result.

You'll have to define a measure on your manifold (typically through the use of specific maps) before you can make a meaningful integration.

 

[JG89]

Micromass, I understand the example you are given. But when I try to think of the unit circle as a closed wire with my function f(x,y) giving the radius of the closed-wired at the point (x,y) then I run into trouble. Please help me through this.

Let's start with my unit circle and call it S^1. Suppose the radius of the wire is uniform at each point (x,y) of the circle with a radius of 1. Then the volume of the cylinder should be V = pi*r^2*h, where the height is h = 2pi. So V = pi*r^2*2pi = pi*1*2pi = 2pi^2.

Now, the circle can be parametrized by the map g: [0, 2pi) ---> R^2 such that g(x) = (cos x, sin x). Again, we will denote this circle by S^1. So since the radius at any point (x,y) is equal to 1, we can let the function f(x,y) = 1 represent the radius of the wire at any point of the wire. So if we calculate $$\int_{S^1} f$$ we should get 2pi^2 for our answer. But $$\int_{S^1} f = \int_0^{2\pi} f(g(x)) V(Df)$$ where $$V(Df)$$ is the volume of Df (Df is the derivative of f).

Now, f = 1, so f(g(x)) = 1 and $$V(Df) = \sqrt{det[Df^{tr}Df]} = \sqrt{det[sin^2x + cos^2x]} = \sqrt{det[1]} = 1$$. Thus $$\int_{S^1} f = \int_0^{2\pi} 1 = 2 \pi$$.

Where is my mistake?

 

[micromass]

The volume is given by

$$\int_{S^1}{\pi f(x)^2dx}=\pi \int_{S^1}{1dx}=2\pi^2$$

Like expected.

 

[JG89]

I understand why the integral is set up that way to find the volume of the wire, but I am trying to use the definition of integral of scalar map over manifold that is given in my book. Let me give you the definition, but first start it off with another definition:

Definition:

Let k <= n. Let A be open in R^k, and let g: A --> R^n be a map of class C^r. The set Y = g(A), together with the map g, constitute what is called a parametrized-manifold of dimension k. We denote this parametrized-manifold by Y_g; and we define the k-dimensional volume of Y_g by the equation $$v(Y_g) = \int_A V(Dg)$$ where $$V(Dg) = \sqrt{det[Dg^{tr}Dg]}$$.

Note that a k-dimensional parametrized-manifold in R^n is analogous to, say for example, a 2-dimensional parametrized-surface in R^3. The book then goes on to give a plausible argument as to why the volume of the manifold should be defined that way, and I understand all of that.

Ok, now the book defines the integral of a scalar map over a parametrized-manifold:Definition:

Let A be open in R^k; let g: A --> R^n be of class C^r; let Y = g(A). Let f be a real-valued continuous function defined at each point of Y. We define the integral of f over Y_g, with respect to volume , by the equation $$\int_{Y_g} f dV = \int_A (f \circ g)V(Dg)$$

So two questions:

1) What exactly do they mean the integral with respect to volume?

2) if S^1 is my 1-dimensional parametrized-manifold in R^2, and f = 2 is my function, and I integrate f over S^1 using the definition given above, then the answer I get is 2*2pi = 4pi. What exactly is this measurement?

Or if someone could just provide the motivation as to why the integral of a scalar map over a manifold should be defined that way, I would be very happy!