# $G$-invariant de Rham theorem

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

1. It is wrong, with a counter-example?
• The complex of invariant $H^{*,G}_{dR}(M)$ is isomorphic to $H^*_{dR}(M)$, at least if $G$ is compact: You prove this by averaging a form over the group action. – Thomas Rot Dec 18 '19 at 14:32
• If $G$ is not compact, clearly the result is not true, as we can see from the real number line. – Ben McKay Dec 18 '19 at 15:21
• @Thomas Rot for this you need the group to be compact and path connected. The action of a finite group on itself is a counter example otherwise. – S. carmeli Dec 18 '19 at 20:37
• Thanks Ben McKay and S. Carmeli. – Thomas Rot Dec 18 '19 at 21:19
• @ThomasRot So, do you think it is true/ trivial? – DLIN Dec 19 '19 at 3:05