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Let $G$ be a compact Lie group and let $H$ be a closed subgroup of $G$, with Lie algebras $\mathfrak{g}$ and $\mathfrak{h}$.

We denote $G\times_H \mathfrak{g} / \mathfrak{h}$: the set of orbits $(G \times \mathfrak{g} / \mathfrak{h})/H $ of the right action of $H$ on $ G \times \mathfrak{g} / \mathfrak{h}$ ($H$ acts on $G$ by multiplication and acts on $\mathfrak{g} / \mathfrak{h}$ by the adjoint action).

We have this identification: $T(G/H) \simeq G\times_H \mathfrak{g} / \mathfrak{h}$.

My question is why the space of differential forms of $G/H$, $A^\bullet(G/H)$, satisfies $$ A^\bullet(G/H) \simeq \Gamma (G/H, G\times_H {\bigwedge}^\bullet{(\mathfrak{g} / \mathfrak{h})}^*) \simeq {(C^\infty (G) \otimes {\bigwedge}^\bullet {(\mathfrak{g} / \mathfrak{h})}^*)}^H. $$

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    $\begingroup$ For a principal $H$-bundle $P\to M$, the differential forms on $M$ can be identified with $H$-invariant forms on $P$ which are basic, i.e. vanish on the vector fields generating the $H$-action. In your example $P = G$, and forms on $G$ can be identified with functions to $\Lambda^*\mathfrak g^*$ using the trivialization of $TG$. The basic forms are then given by exterior powers of the annihilator of $\mathfrak h$, i.e. $(\mathfrak g/\mathfrak h)^*$. $\endgroup$
    – Bertram Arnold
    Mar 7, 2021 at 15:27
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    $\begingroup$ This is a good exercise which you should work hard to complete yourself. $\endgroup$
    – mme
    Mar 7, 2021 at 16:06
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    $\begingroup$ Your other recent question suggests you are working through a reference on differential forms. In addition to the excellent suggestion to work these exercises yourself, if you do ask about them, then you should mention the referece you are using. \\ TeX note: please use TeX, like $G\times H$ G \times H, rather than Unicode, like $G × H$ G × H. I have edited accordingly. $\endgroup$
    – LSpice
    Mar 9, 2021 at 19:55

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By definition, $k$-forms are sections of the bundle $\bigwedge{}^kT^*(G/H)$, which is the associated bundle $G\times_H\bigwedge^{k}(\mathfrak{g}/\mathfrak{h})^*$. You then apply the general formula that the sections of the associated bundle for any $H$-representation $V$ is $(C^{\infty}(G)\otimes V)^H$: the tensor product $C^{\infty}(G)\otimes V$ is the sections of the trivial bundle $G\times V \to G$, and such a section is pulled back from a section of $G\times_H V\to G/H$ if and only if it is invariant.

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  • $\begingroup$ thanks a lot for your answer. $\endgroup$
    – Samia
    Mar 9, 2021 at 21:02

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