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Is there some relation between De Rham cohomology group of Lie group as a manifold and group cohomology of Lie group?

At first glance, they are two different things. De Rham Cohomology group is defined by differential form on manifold. While group cohomology is used to classify the group extension.

My question:

1.For group cohomology $H^n_\sigma(G,A)$, we need group $G$, abelian group $A$ and $\sigma : G\rightarrow \mathrm{Aut}(A)$. If I fixed $G$ is some Lie group, $A=\mathbb{R}$ and $\sigma$ as trivial homomorphism. Is there some relation between group cohomology $H^n_0(G,\mathbb{R})$ and De Rham cohomology $H^{n}_{\mathrm{dR}}(G)$?

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    $\begingroup$ There is a spectral sequence involving both, in view of the fibration $G\to EG\to BG$ with contractible total space, and since group cohomology of $G$ is "almost the same" as cohomology of $BG$. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Nov 30, 2017 at 3:27
  • $\begingroup$ @მამუკაჯიბლაძე Thanks. Could you give me some reference? $\endgroup$
    – maplemaple
    Commented Nov 30, 2017 at 4:28
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    $\begingroup$ This is fairly classical material. A most general and most recent treatment that I could find is in A Cocycle Model for Topological and Lie Group Cohomology by Wagemann and Wockel; the treatment I personally like most is in the book COHOMOLOGY OF INFINITE-DIMENSIONAL LIE ALGEBRAS by Fuks (from page 289 on) $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Nov 30, 2017 at 5:00
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    $\begingroup$ Crosspost: math.stackexchange.com/questions/2543535/… $\endgroup$
    – Qiaochu Yuan
    Commented Nov 30, 2017 at 6:38
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    $\begingroup$ Actually, the group cohomology of $G$ isn't almost the same as the cohomology of $BG\,.$ Their relationship is similar to that of the cohomology of a manifold with coefficients in $\mathcal{O}$ and de Rham cohomology...that is, not much. $\endgroup$
    – Josh Lackman
    Commented May 13, 2022 at 19:35

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The answer to the question is: not really.

Consider $G=\mathbb{R}^n\,.$ This has nontrivial group cohomology in all degrees $\le n$ (by the van Est isomorphism theorem, for example), while the classifying space is the point $*\,,$ which has vanishing de Rham cohomology in positive degrees. Therefore, there is certainly no surjection from the de Rham cohomology to the group cohomology.

On the other extreme, consider $G=T^n\,.$ This has vanishing group cohomology in positive degrees (since it's compact), while it has nontrivial de Rham cohomology in all degrees $\le n\,,$ therefore there is no such injection either.

The only relation I know of is that if your group $G$ is compact and has vanishing homotopy groups up to and including degree $n\,,$ then both cohomologies vanish up to degree $n$ (again, by the van Est isomorphism theorem). However, if $n\ge 3$ then this implies that $G$ is a point, by the result in this question: https://math.stackexchange.com/questions/960372/a-question-about-contractibility-of-a-lie-group

There isn't even a strong relationship between group cohomology of $G$ and the cohomology of $BG\,.$ There is one, in a sense, but it's more complicated. If $G$ is discrete however, the group cohomology of $G$ is isomorphic to the cohomology of $BG$ with coefficients in the discrete group $\mathbb{R}\,.$ For nondiscrete Lie groups $G$, there is an analogue to this isomorphism, but it's more complex and involves $EG\,.$

Note: when I say group cohomology I am strictly speaking of the cohomology of group cocycles valued in the Lie group $\mathbb{R}\,,$ however you can consider group cohomology valued in any abelian group.

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  • $\begingroup$ I'm confused...I thought $H^k(G, \mathbb{Z}) \cong H^k(BG, \mathbb{Z})$ (LHS=group cohomology, RHS=simplicial cohomology) for all groups. I know this question is about real coefficients since that's where the de Rham isomorphism holds, but doesn't the universal coefficient theorem give a pretty close relation between (the free part of) $H_k(BG, \mathbb{Z})$ and $H^k(BG, \mathbb{R})$? $\endgroup$
    – Joshua Grochow
    Commented Sep 27 at 15:43
  • $\begingroup$ The isomorphism you wrote is correct. It's also true when you're computing the group cohomology in the discrete group $\mathbb{R},$ but I took the question to mean the Lie group cohomology with cochains being continuous or smooth functions. I think this is more common. $\endgroup$
    – Josh Lackman
    Commented Oct 3 at 19:36

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