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Assume that $G$ is a compact Lie group, and $M$ is a smooth $G-$manifold. We consider the equivariant cohomology of $M$, i.e. $H^{*}_{G}(M)$. The famous theorem of Cartan said that if $M$ is compact then the twisted de Rham complex $\Omega^{*}_{G}(M)$ computes the equivariant cohomology of $M$, i.e. $$H^{*}_{G}(M)\simeq H^{*}(\Omega^{*}_{G}(M),d_{eq}).$$ So if $M$ is not compact is the above result aslo true ?

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    $\begingroup$ I think that it is still true. See §2.5 in "Supersymmetry... " by Guillemin-Sternberg. $\endgroup$ – H. Shindoh Jul 19 '13 at 12:24
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Yes, the result is true without the compactness assumption. For a proof that describes explicitly this isomorphism, and does not rely on spectral sequences, see this old note.

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