Counterexample showing that G-invariant de Rham cohomology different from cohomology of G-invariant sub-complex? – mathoverflow.net

Counterexample showing that G-invariant de Rham cohomology different from cohomology of G-invariant sub-complex? – mathoverflow.net

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If $G$ is a discrete or a Lie Group acting smoothly on a manifold $M$, we can define the algebra of $G$-invariant de Rham classes, $H(M)^G$, and we can also consider the cohomology of the sub-complex ...

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