Saturday, 5 March 2011

at.algebraic topology - De Rham cohomology and antiderivatives

This probably doesn't answer your question, but maybe it is close.



The (Extended) Poincare lemma says that if U is a contractible smooth n-manifold and
omega is a closed differentiable form meaning domega=0 then there is a form phi
with dphi=omega. So, phi is an antiderivative of omega.



More to the point, if omega is a closed form on the smooth manifold M then [omega]inH(M), and omega has an antiderivative if and only if [omega]=0.



More specifically, suppose that M is an oriented smooth n-manifold so that integration
is defined, and omega is a smooth n-form, then omega has an antiderivative if and only if intMomega=0.



If M is in addition compact, and omega is a closed i-form then omega has an antiderivative if and only if for all closed ni forms eta, intMomegawedgeeta=0.



If omega is not closed it can never have an antiderivative.

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