Integration of closed forms

1. Aug 10, 2011

arestes

Hi guys!
I am reading a paper which uses closed forms $$\omega$$ on a p-dimensional closed submanifold $$\Sigma$$ of a larger manifold $$M$$. When we integrate $$\omega$$ we get a number
$$Q(\Sigma) =\int _{\Sigma}\omega$$ which, in principle, depends on the choice of $$\Sigma$$ but because $$\omega$$is closed, $$Q(\Sigma)$$ is said to be unchanged by continuous deformations of $$\Sigma$$. The converse is supposed to be true as well.

Now, I understand this as a generalization of Gauss' law in electrostatics (three dimensions) but I only remember that for this particular case, the demonstration I saw was purely geometrical. I know that this has to use the fact that $$\Sigma$$ does not have a boundary (closed submanifold). How exactly is this proved?

Last edited: Aug 10, 2011
2. Aug 10, 2011

Ben Niehoff

If $\Sigma$ is continuously deformed to $\Sigma'$, then we have that

$$\Sigma - \Sigma' = \partial \mathcal{R}$$

where $\mathcal{R}$ is the (p+1)-dimensional region enclosed by $\Sigma$ and $\Sigma'$, and $\partial$ is the boundary operator. Applying the boundary operator again and using $\partial^2 = 0$, we get

$$\partial \Sigma - \partial \Sigma' = 0$$

which means that the continuous deformations of $\Sigma$ must leave its boundary unchanged (if it has a boundary). This is analogous to Ampere's law: the B flux through a loop is proportional to the current enclosed, and the B flux can be evaluated on any surface having the loop as its boundary.

Then we can apply the generalized Stokes' theorem:

$$\int_{\Sigma - \Sigma'} \omega = \int_{\partial \mathcal{R}} \omega = \int_{\mathcal{R}} d\omega = 0$$

Finally, we have, by linearity,

$$\int_{\Sigma} \omega = \int_{\Sigma'} \omega$$

3. Aug 12, 2011

arestes

hi thanks! that is partly what I wanted. However, this argument still doesn't use the fact that $$\Sigma$$ is a closed submanifold, that is, its boundary is the empty set. I think it should be important.

4. Aug 13, 2011

Ben Niehoff

It's not important, as you can see in my argument above. It is simpler to assume that $\Sigma$ has no boundary, but it is not necessary, as long as $\Sigma'$ has the same boundary. That is, any continuous deformation which leaves the boundary unchanged is allowed.

5. Aug 13, 2011

lavinia

The general theorem is Stoke's theorem if the deformation is smooth. The two bounding manifolds of the deformation are homologous so a cocycle will have the same value on each. The closed form is a cocycle in the de Rham complex of the larger manifold.