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 $M$?

  • $\begingroup$ 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 ? $\endgroup$ Jul 11, 2019 at 22:11
  • $\begingroup$ Please see edit @InfiniteLooper $\endgroup$ Jul 11, 2019 at 22:17
  • 5
    $\begingroup$ 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). $\endgroup$ 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 :) $\endgroup$ Jul 12, 2019 at 8:07

1 Answer 1


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 $(X\times G\rightrightarrows X)$.

  • 1
    $\begingroup$ The existence of such an isomorphism does not depend on $G$ compact, as stated in Remark 16 of the paper you link. $\endgroup$
    – mme
    Jul 12, 2019 at 11:58
  • $\begingroup$ @MikeMiller edited. Thanks $\endgroup$ Jul 12, 2019 at 12:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.