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Cohomology = invariant forms

  1. May 1, 2009 #1
    Prove the following result:

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

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

    I have been suggested the following proof strategy:
    a) if [tex]\omega[/tex] is [tex]G[/tex]-invariant, then d[tex]\omega = 0[/tex]
    b) likewise, d[tex]*\omega = 0[/tex] (Hodge star)
    c) by Hodge theory, [tex]\omega[/tex] 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 [tex]G[/tex]-invariant coordinates (i.e. a local basis of [tex]G[/tex]-invariant vector fields which span the tangent space) - how to prove this? In these coordinates [tex]\omega[/tex] has constant coefficients (why?), so d[tex]\omega = 0[/tex]. How about d[tex]*\omega[/tex]?

    I'd be glad if someone could help with filling in the details.
  2. jcsd
  3. May 1, 2009 #2
    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.
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