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.
