# 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##?

Gold Member
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.

Gold Member
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?

Gold Member
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.