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Let $G$ be a Lie group acting on a manifold $M$.

Consider the transformation groupoid $\mathcal{G}=(G\times M\rightrightarrows M)$. We have the notion of de Rham cohomology of a Lie groupoid by considering de Rham cohomology of simplicial manifold associated to it, denoted by $\mathcal{G}_\bullet$.

Is there any relation between de Rham cohomology of Lie groupoid $\mathcal{G}$ (eg page 11 of Laurent-Gengoux–Tu–Xu, Chern-Weil map for principal bundles over groupoids, Math. Z. 255 (2007) pp451–491, arXiv:math/0401420) and the de Rham (??) cohomology of the Lie group $G$ and the de Rham cohomology of the manifold $M$?

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  • $\begingroup$ What is DeRham Cohomology of a Lie groupoid ? Do you mean groupoid cohomology defined the same way as group cohomology but restricting to composable chains ? $\endgroup$ Commented Jul 11, 2019 at 22:11
  • $\begingroup$ Please see edit @InfiniteLooper $\endgroup$ Commented Jul 11, 2019 at 22:17
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    $\begingroup$ In this special case your de Rham cohomology associated to the Lie groupoid is classically known as equivariant (Borel) cohomology. There is a Serre spectral sequence that has the flavour of what you want and is often used to compute equivariant cohomology; see (1.2.1) of math.ias.edu/~goresky/pdf/equivariant.jour.pdf. Note that first page of the Serre spectral sequence involves cohomology over $BG$, in general with twisted coefficient group. The relation to $H^*(G)$ is more subtle (this is the "Koszul duality" in the title of the linked paper). $\endgroup$ Commented Jul 12, 2019 at 0:00
  • $\begingroup$ @DanielPomerleano Thanks for the link. Can you give a reference for “Lie groupoid cohomology same thing as Equivariant cohomology”. Thank you :) $\endgroup$ Commented Jul 12, 2019 at 8:07

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Proposition $13$ and Remark $16$ in page $10$ of Cohomology or Stacks says that, there is a natural isomorphism

$$H^i_G(X)\rightarrow H^i(X\times G\rightrightarrows X)$$ where, $H^i_G(X)$ is the $i^{\text{th}}$ equivariant cohomology of $X$ with respect to action of $G$ and $H^i(X\times G\rightrightarrows X)$ is the $i^{\text{th}}$ deRham cohomology of the transformation groupoid $(X\times G\rightrightarrows X)$.

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    $\begingroup$ The existence of such an isomorphism does not depend on $G$ compact, as stated in Remark 16 of the paper you link. $\endgroup$
    – mme
    Commented Jul 12, 2019 at 11:58
  • $\begingroup$ @MikeMiller edited. Thanks $\endgroup$ Commented Jul 12, 2019 at 12:38

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