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?


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.

| cite | improve this answer | |
  • 4
    $\begingroup$ In fact, everyone should read the original paper of Chevalley and Eilenberg, as it is a masterpiece of readablity! $\endgroup$ – Mariano Suárez-Álvarez Sep 2 '11 at 14:53
  • $\begingroup$ Many thanks Akhil, for your answer and for the nice notes! $\endgroup$ – amine Sep 2 '11 at 20:59
  • 1
    $\begingroup$ The link seems no more work. $\endgroup$ – user148212 Aug 6 '18 at 2:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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