Given any graded vector space $V^\bullet$ and any degree 1 linear operator $d\colon V^\bullet\to V^{\bullet+1}$, one gets a complex $(\ker(d^2),d)$. Moreover, if $V^\bullet$ is a graded algebra and $d$ is a graded derivation, then $\ker(d^2)$ is also a subalgebra, and so $(\ker(d^2),d)$ is a dgca.
If we now consider a smooth manifold $M$ with an action of a compact connected Lie group $G$ with Lie algebra $\mathfrak{g}$, we can then consider the graded commutative algebra $$ \Omega^\bullet_{dR}(M)\otimes S^\bullet[\mathfrak{g}^\ast[-2]] $$ and the Cartan equivariant differential $$ d_G= d_{dR}\otimes \mathrm{id}+ i_{\alpha}\otimes u^\alpha $$ where $\{e_\alpha\}$ is a linear basis of $\mathfrak{g}$, the $u^\alpha$ are the dual coordinates placed in degree 2, an $i_\alpha$ is the contraction operator on $\Omega^\bullet_{dR}(M)$ associated with the vector field $X_\alpha$ corresponding to the infinitesimal action of $e_\alpha$. A simple and classical computation shows that the square of $d_G$ is the Euler vector field-type operator $$ (d_G)^2= u^\alpha e_\alpha $$ where the $e_\alpha$-action on $\Omega^\bullet_{dR}(M)\otimes S^\bullet[\mathfrak{g}^\ast[-2]]$ is the $\mathfrak{g}$-module structure induced by the Lie derivative along $X_\alpha$ and by the coadjoint action.
By the above, if I did not make silly mistakes in deriving it, the operator $d_G$ is a degree 1 differential on $\ker((d_G)^2)= \ker(u^\alpha e_\alpha)$. Clearly, $$ \left(\Omega^\bullet_{dR}(M)\otimes S^\bullet[\mathfrak{g}^\ast[-2]]\right)^G=\bigcap_{\alpha} \ker{e_\alpha}\subseteq \ker(u^\alpha e_\alpha), $$ so that the Cartan model for $G$-equivariant cohomology of $M$ is a subcomplex of the complex $\ker(u^\alpha e_\alpha)$.
This, without explicitly mentioning the complex $(\ker((d_G)^2),d_G)$ is the usual argument for showing that $d_G$ squares to zero when restricted to $\left(\Omega^\bullet_{dR}(M)\otimes S^\bullet[\mathfrak{g}^\ast[-2]]\right)^G$. So my question is: what is the precise relation between the Cartan model and $\ker((d_G)^2)$? Are they actually the same complex? is it so at least when $G$ is an $n$-torus (it seems to me the answer is yes in this case, as the coadjoint action is trivial and this simplifies a lot what is going on, but I may be wrong here); if they are generally different, what does the cohomology of $(\ker((d_G)^2),d_G)$ compute?
