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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 $\omega=d\omega'$ then $d\omega=0$. This is more commonly stated as $d \circ d = 0$, 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 $k$th 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 $k$th de Rham cohomology Vector Space of a Manifold $M$ is canonically isomorphic to the Alexander-Spanier Cohomology Vector Space $H^k(M;\Bbb{R})$ (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, Differential k-Form, Exterior Derivative, Vector Space

© 1996-9 Eric W. Weisstein