de Rham cohomology is a formal set-up for the analytic problem: If you have a Differential
*k*-Form on a Manifold , is it the Exterior Derivative of another Differential *k*-Form ? Formally, if
then . This is more commonly stated
as , meaning that if is to be the Exterior Derivative of a Differential *k*-Form, a Necessary condition that must satisfy is that its Exterior
Derivative is zero.

de Rham cohomology gives a formalism that aims to answer the question, ``Are all differential -forms on a Manifold with zero Exterior Derivative the Exterior Derivatives of -forms?'' In particular, the th de Rham cohomology vector space is defined to be the space of all -forms with Exterior Derivative 0, modulo the space of all boundaries of -forms. This is the trivial Vector Space Iff the answer to our question is yes.

The fundamental result about de Rham cohomology is that it is a topological invariant of the Manifold, namely: the th de Rham cohomology Vector Space of a Manifold is canonically isomorphic to the Alexander-Spanier Cohomology Vector Space (also called cohomology with compact support). In the case that is compact, Alexander-Spanier Cohomology is exactly singular cohomology.

© 1996-9

1999-05-24