Is There an Isomorphism Between G-Invariant Forms and De Rham Cohomology?

[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.

 

[zhentil]

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.