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Let $M$ be a Riemannian manifold. The Hodge decomposition tells that $$ \Omega^*(M) = \mathrm{im} \ d \oplus \mathrm{im} \ d^* \oplus \mathscr H^*(M) $$ where $d^*$ is the adjoint operator of the exterior derivative $d$. Namely, a differential form $\omega$ has a unique decomposition as a sum in the form $\omega=d\alpha_1+d^*\alpha_2+\gamma$ where $\gamma$ is a harmonic form in the sense that $\Delta \gamma=0$ for $\Delta=dd^*+d^*d$.

We further assume that there is an action of a compact Lie group $G$ with Lie algebra $\mathfrak g$, and we can consider the space $ \Omega^*_G(M)= (\mathbb C[\mathfrak g]\otimes \Omega^*(M))^G $ of $G$-equivariant differential forms (i.e. $G$-invariant polynomial maps $\mathfrak g\to \Omega^*(M)$). Also, we can define the equivariant exterior differential $d_{\mathfrak g}$. Then, the equivariant cohomology $H^*_G(M)$ is the cohomology of the complex $(\Omega^*_G(M), d_{\mathfrak g})$.

By analogy, we could probably define $\Delta_{\mathfrak g}=d_{\mathfrak g}d^*_{\mathfrak g}+d^*_{\mathfrak g}d_{\mathfrak g}$ and the space of $G$-equivariant harmonic forms $\mathscr H^*_{\mathfrak g}(M)$; then, perhaps we could also show $\mathscr H_{\mathfrak g}^*(M)\cong H_G^*(M)$ and a similar Hodge decomposition $$ \Omega^*_G(M)=\mathrm{im} \ d_{\mathfrak g}\oplus \mathrm{im} \ d_{\mathfrak g}^* \oplus \mathscr H^*_{\mathfrak g}(M)$$

This looks natural and standard. If what I guess was true, does anyone know a good reference? Thanks.

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    $\begingroup$ I don't have a good reference, but I know a good explanation of why this should work: the classifying space of any compact group can be written as a direct limit of Riemannian (actually, Kahler) manifolds: for example, $S^1$ gives $\mathbb{CP}^{\infty}$. So you can think of equivariant cohomology as an inverse limit of the cohomology of smooth Riemannian manifolds whose union is the Borel space. $\endgroup$
    – Ben Webster
    Commented Feb 4, 2021 at 3:05

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