How is the theorem of Stoke's proved for closed submanifolds without boundaries?

[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 \omegais 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?

 

[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

 

[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.

 

[Ben Niehoff]

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.

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.

 

[lavinia]

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 \omegais 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?

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.