General Conditions for Stokes' Theorem

[Mandelbroth]

What is the least restrictive set of conditions needed to utilize the formula $\int\limits_{\Omega}\mathrm{d}\alpha=\int\limits_{\partial\Omega} \alpha$?

 

[Ben Niehoff]

I think the only conditions are those needed to define the integrals (i.e. the same kinds of conditions used to define 1-dimensional integrals). I don't think there are any extra geometrical conditions, provided you're on a differentiable manifold.

You can even relax smoothness of the manifold (and the forms!) if you are careful about using Dirac delta functions. Generally, since a form is something you integrate, they should be thought of as distributions.

The boundary operator can be used in a distributional sense as well. For example, the boundary of a sphere is zero. But it might be helpful to think of a sphere with a point removed, whose boundary is therefore a point; then you can use Stokes' theorem to integrate forms over the sphere.

 

[Mandelbroth]

You can even relax smoothness of the manifold (and the forms!) if you are careful about using Dirac delta functions. Generally, since a form is something you integrate, they should be thought of as distributions.

Could you please expand on this? Thinking of forms as distributions feels foreign, and I don't see where that line of thought would go.

 

[Ben Niehoff]

Expand on it how? Surely you can figure out how to integrate something like

\delta(x,y,z) \, dx \wedge dy \wedge dz

Do you have a specific question?

 

[Ben Niehoff]

This might be a better example of what I'm talking about. Say we want to find the area of a sphere. The form we want to integrate is

\omega = \sin \theta \, d \theta \wedge d \phi
Now, the sphere $\Omega$ is a closed surface, so $\partial \Omega = 0$. However, the coordinate patch $\tilde \Omega$ covered by the coordinates $\phi \in (0, 2\pi), \; \theta \in (0, \pi)$ is not a closed surface, and is in fact contractible. We have that $\partial \tilde \Omega$ is the union of the north and south poles of the sphere, and a segment of a great circle that runs between them.

Now, it so happens that

\omega = d \big( - \cos \theta \, d \phi \big) = d \alpha
so we can use Stokes' theorem. So

\int_{\tilde \Omega} \omega = \int_{\partial \tilde \Omega} \alpha = - \int_{\partial \tilde \Omega} \cos \theta \, d \phi
To integrate around the "cut" between the north and south poles, we draw a loop around it. On either side there is a vertical part where $d \phi = 0$, and so these parts do not contribute. Then around the north and south poles, there are tiny circles, at which $\cos \theta = \pm 1$ and $\phi$ runs from 0 to $2 \pi$. The tiny circles go opposite directions, so each part contributes positively:

- \int_{\partial \tilde \Omega} \cos \theta \, d \phi = 2 \pi + 2 \pi = 4 \pi
So you see, if you are careful about how you cut up a manifold, you can apply Stokes' theorem in all sorts of situations.

In this case, we took a closed surface and removed a set of measure zero to turn it into a surface with boundary. The reason this worked is because the form $\omega$ is smooth on the set of measure zero that we removed. If that were not the case (say $\omega$ had a delta-function-like contribution on the "cut"), then you would have to include an extra piece to account for that.