# 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.