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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?
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    $\begingroup$ 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. $\endgroup$ – Thomas Rot Dec 18 '19 at 14:32
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    $\begingroup$ If $G$ is not compact, clearly the result is not true, as we can see from the real number line. $\endgroup$ – Ben McKay Dec 18 '19 at 15:21
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    $\begingroup$ @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. $\endgroup$ – S. carmeli Dec 18 '19 at 20:37
  • $\begingroup$ Thanks Ben McKay and S. Carmeli. $\endgroup$ – Thomas Rot Dec 18 '19 at 21:19
  • $\begingroup$ @ThomasRot So, do you think it is true/ trivial? $\endgroup$ – DLIN Dec 19 '19 at 3:05
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A reference for the proof (from the n-lab entry): Eckhard Meinrenken, Equivariant cohomology and the Cartan model, in: Encyclopedia of Mathematical Physics, Pages 242-250 Academic Press 2006 (pdf, doi:10.1016/B0-12-512666-2/00344-8) http://www.math.toronto.edu/mein/research/enc.pdf

The proof is as indicated by Thomas Rot, and requires the group to be compact and connected.

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  • $\begingroup$ I’m not sure this has the (dR-side) result and proof alluded to by @ThomasRot. On the other hand, I think Chevalley-Eilenberg (1948) have it as Theorem 2.3, attributing the method to, as you said, Cartan (1930). $\endgroup$ – Francois Ziegler Dec 19 '19 at 20:02

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