# Reference: forms invariant under Lie group action give the de Rham cohomology?

I'm looking for a reference for a proof of the following fact:

Let $G$ be a compact, connected Lie group acting on a smooth manifold $M$. Then inclusion of the differential forms invariant under the action of $G$ give an isomorphism with the de Rham cohomology group.

I could not find this in Lee or Spivak, which are my go-to references.

• Invariant cohomology is actually discussed in Volume 5 of Spivak, among other places. – Ted Shifrin Dec 15 '13 at 23:07
• Ah, thank you! (I only looked in Volume 1) – user908123 Dec 15 '13 at 23:19
• Basic idea, not sure where it is written down: Forms are cohomologous to its shifts under the group action. Averaging all of these shifts gives an invariant form cohomologous to the original. – Steven Gubkin Dec 15 '13 at 23:34
• To be precise, in Spivak vol. 5, Look at chapter 13, section 16 pg 308,309 – Warlock of Firetop Mountain Dec 20 '18 at 15:18