Integrating a differential form on a manifold without parametric equations

[saminator910]

I am sure that this can be done, but I haven't been able to figure it out, Is there a way to integrate a differential form on a manifold without using the parametric equations of the manifold? So that you can just use the manifold's charts instead of parametric equations? If you a function mapping a portion of the n-1 plane onto an n manifold M say [itex]c:[-1,1]^{n-1}\rightarrow\ M[/itex] can't you use the pullback to integrate a differential form alpha? [itex]\int_{M}α = \int_{[-1,1]^{n-1}}c^{*}α[/itex]. I read that this could be done, but every time I do the a calculation the integral turns out really weird, maybe an example would help?

 

[jvicens]

Hello there,

Yes, it is possible to integrate a differential form on a manifold without using parametric equations. Instead, we can use the manifold's charts to perform the integration. This method is known as the pullback method.

To explain this method, let's consider a function c:[-1,1]^{n-1}\rightarrow\ M, where M is an n-dimensional manifold. This function maps a portion of the (n-1) plane onto the manifold M. Now, let's say we have a differential form alpha defined on the manifold M. We can use the pullback of c to integrate alpha over M, as shown in the following equation:

\int_{M}α = \int_{[-1,1]^{n-1}}c^{*}α

Here, c^{*}α represents the pullback of alpha using the function c. This means that we are "pulling back" the differential form alpha from the manifold M to the (n-1) plane, where we can perform the integration using the parametric equations of c.

To better understand this concept, let's consider an example. Let's say we have a two-dimensional manifold M, which can be represented by the equation x^2+y^2=1. We can define a function c:[-1,1]\rightarrow\ M as c(t)=(cos(t),sin(t)). This function maps the interval [-1,1] to the unit circle on the manifold M. Now, let's say we have a differential form alpha=xdy-ydx defined on M. Using the pullback method, we can integrate alpha over M as follows:

\int_{M}α = \int_{[-1,1]}c^{*}α = \int_{-1}^{1}c^{*}(xdy-ydx) = \int_{-1}^{1}(cos(t)d(sin(t))-sin(t)d(cos(t)))

= \int_{-1}^{1}(sin(t)cos(t)-cos(t)sin(t))dt = 0

As you can see, the integration using the pullback method is much simpler compared to using the parametric equations directly. I hope this example helps you understand the concept better.

Please let me know if you have any further questions.