de Rham Cohomology

de Rham cohomology is a formal set-up for the analytic problem: If you have a differential k-form omega on a manifold M, is it the exterior derivative of another differential k-form omega^'? Formally, if omega=domega^' then domega=0. This is more commonly stated as d degreesd=0, meaning that if omega is to be the exterior derivative of a differential k-form, a necessary condition that omega must satisfy is that its exterior derivative is zero.

de Rham cohomology gives a formalism that aims to answer the question, "Are all differential k-forms on a manifold with zero exterior derivative the exterior derivatives of (k-1)-forms?" In particular, the kth de Rham cohomology vector space is defined to be the space of all k-forms with exterior derivative 0, modulo the space of all boundaries of (k-1)-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 kth de Rham cohomology vector space of a manifold M is canonically isomorphic to the Alexander-Spanier cohomology vector space H^k(M;R) (also called cohomology with compact support). In the case that M is compact, Alexander-Spanier cohomology is exactly singular cohomology.

See also

Alexander-Spanier Cohomology, Change of Variables Theorem, Cohomology, Differential k-Form, Exterior Derivative, Vector Space

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Cite this as:

Weisstein, Eric W. "de Rham Cohomology." From MathWorld--A Wolfram Web Resource.

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