# Cohomology = invariant forms

• neworder1
In summary, the result being discussed is that for a compact Lie group G with a closed subgroup H, the space of G-invariant differential forms on the quotient space X = G/H is isomorphic to the de Rham cohomology space of X. The suggested proof strategy involves showing that G-invariant forms are closed and coclosed, and using Hodge theory to prove that each cohomology class has a unique harmonic representative. While this may not be an elementary proof, it could possibly be simplified by considering the problem as a fibre bundle and using the Kunneth formula.

#### neworder1

Prove the following result:

let $$G$$ be a compact Lie group, $$H$$ its closed subgroup and $$X = G/H$$. Let $$T(X)$$ denote the space of $$G$$-invariant differential forms on $$X$$ (e.g. $$\omega \in T(X) \Leftrightarrow \forall g \in G g^{*}\omega = \omega$$). Then $$T(X)$$ is isomorphic to $$H^{*}(X)$$, de Rham cohomology space of $$X$$,

Do you know where I can find the proof of this result?

I have been suggested the following proof strategy:
a) if $$\omega$$ is $$G$$-invariant, then d$$\omega = 0$$
b) likewise, d$$*\omega = 0$$ (Hodge star)
c) by Hodge theory, $$\omega$$ is harmonic, and each cohomology class has exactly one harmonic representant

Unfortuately, this is not an elementary proof. But perhaps at least a) and b) can be proved easily? A concept for proving a): locally, we can find $$G$$-invariant coordinates (i.e. a local basis of $$G$$-invariant vector fields which span the tangent space) - how to prove this? In these coordinates $$\omega$$ has constant coefficients (why?), so d$$\omega = 0$$. How about d$$*\omega$$?

I'd be glad if someone could help with filling in the details.

Have you considered looking at it as a fibre bundle and applying the Kunneth formula to it? I must admit that I don't have too much experience with this, but that seems to be the only way to get a quick proof.