Stokes' Theorem - Topological Reading; Integration Over Chains

Topological Reading; Integration Over Chains

Let M be a smooth manifold. A smooth singular k-simplex of M is a smooth map from the standard simplex in Rk to M. The free abelian group, S_k, generated by singular k-simplices is said to consist of singular k-chains of M. These groups, together with boundary map, ∂, define a chain complex. The corresponding homology (resp. cohomology) is called the smooth singular homology (resp. cohomology) of M.

On the other hand, the differential forms, with exterior derivative, d, as the connecting map, form a cochain complex, which defines de Rham cohomology.

Differential k-forms can be integrated over a k-simplex in a natural way, by pulling back to Rk. Extending by linearity allows one to integrate over chains. This gives a linear map from the space of k-forms to the k-th group in the singular cochain, S_k*, the linear functionals on S_k. In other words, a k-form defines a functional

on the k-chains. Stokes' theorem says that this is a chain map from de Rham cohomology to singular cohomology; the exterior derivative, d, behaves like the dual of ∂ on forms. This gives a homomorphism from de Rham cohomology to singular cohomology. On the level of forms, this means:

1. closed forms, i.e., have zero integral over boundaries, i.e. over manifolds that can be written as, and
2. exact forms, i.e., have zero integral over cycles, i.e. if the boundaries sum up to the empty set: .

De Rham's theorem shows that this homomorphism is in fact an isomorphism. So the converse to 1 and 2 above hold true. In other words, if {c_i} are cycles generating the k-th homology group, then for any corresponding real numbers, {a_i}, there exist a closed form, such that:

and this form is unique up to exact forms.