If $G$ is a connected Lie group with Lie algebra $\mathcal{G}$, then de Rham cohomology of left invariant différential forms $H_L^*(G)$ is isomorphic to the Chevalley–Eilenberg cohomology $H^*(\mathcal{G})$ for the trivial action of $\mathcal{G}$ on $\mathbb{R}$.

This is obvious, if we associate to each left invariant differential form on $G$ its value at the neutral element $e$ of $G$, after the identification of $\mathcal{G}^*$ with $T_e^*G$.

A less obvious, is the fact that for a compact Lie group $G$ its de Rham cohomology $H^*(G)$ is naturally isomorphic to $H_L^*(G)$ (the subcomplex of left invariant differential forms).

May you point me to a nice reference to this subjet: for the de Rham cohomology of a compact Lie group?


1 Answer 1


The intuition is that since any closed form $\omega$ is cohomologous to $L_g^* \omega$ (where $L_g$ is left-translation, and the claim is because there is path between $g$ and the identity that "homotopes" $L_g^* \omega$ to $\omega$), so $\omega$ is cohomologous to the "average" of all $L_g^* \omega$ (i.e. the Haar integral $\int_G L_g^* \omega$). This average is clearly left-invariant.

This is not a rigorous argument, though. You can find more details in these notes, including a proof of the additional fact that the $i$th cohomology can be computed as simply the space of bi-invariant $i$-forms.

  • 5
    $\begingroup$ In fact, everyone should read the original paper of Chevalley and Eilenberg, as it is a masterpiece of readablity! $\endgroup$ Sep 2, 2011 at 14:53
  • $\begingroup$ Many thanks Akhil, for your answer and for the nice notes! $\endgroup$
    – amine
    Sep 2, 2011 at 20:59
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    $\begingroup$ The link seems no more work. $\endgroup$
    – user148212
    Aug 6, 2018 at 2:54
  • 1
    $\begingroup$ @user148212 use time machine web.archive.org/web/20160804023527/http://… $\endgroup$ Oct 3, 2021 at 18:14

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