Let $M$ be a compact manifold, de Rham theorem asserts that this is an isomorphism between de Rham cohomology and singular cohomology. Suppose that $M$ admits a smooth group $G$, where $G$ is a compcat Lie group. We can define the complex of $G$-invaraint forms with differential whose cohomology is denoted by $H^{*,G}_{dR}(M)$; and complex of $G$-invariant cochain with real coefficient, whose cohomology is denoted by $H^{*,G}(M)$.

I guess that $H^{*,G}_{dR}(M)\cong H^{*,G}(M)$.

**Q** 1. Did someone already show this(any reference), or its proof is exactly the same as the de Rham Theorem? (I can not image any obstruction)

or

- It is wrong, with a counter-example?

path connected. The action of a finite group on itself is a counter example otherwise. $\endgroup$ – S. carmeli Dec 18 '19 at 20:37