Let $G$ be a finite (or discrete) group, $M$ a $d$-dimensional manifold with smooth $G$-action (I am interested in the case where the action is not free, so $M/G$ is not a manifold). For an Abelian group $A$, let $\mathcal{C}^n(M,A)$ be the group of $n$-cochains on $X$ with $A$ coefficients. We can treat this as a $G$-module (Abelian group with compatible $G$-action) with the $G$-action inherited from the $G$-action on $M$.

Now for any $G$-module $B$, we can introduce the Abelian group of "group cochains" $\mathcal{C}^m(G, B)$ that are used to define group cohomology. For example, we can define $\mathcal{C}^m(G, B) = \mathrm{Hom}_G(F_n, B)$, where $$\cdots F_n \to F_{n-1} \to \cdots \to F_0 \to \mathbb{Z} \to 0$$ is a projective $\mathbb{Z}[G]$-resolution of the integers.

In particular, $\mathcal{Q}^{m,n} = \mathcal{C}^m(G, \mathcal{C}^n(X,A))$ defines a double complex. The total cohomology of this double complex computes the equivariant cohomology $H^{\bullet}( (M \times EG)/G, A)$.

Now suppose I instead define $\widetilde{\mathcal{Q}}^{m,n} = \mathcal{C}^m(G, \mathcal{C}_{d-n}(M,A))$, where $\mathcal{C}_{\bullet}(M,A)$ denotes the group of $(d-n)$-chains on $X$ with $A$ coefficients. What is the interpretation of the total cohomology of this complex? It's not equivariant homology, because we are still doing "cohomology in G". Under what circumstances is there a Poincare duality relating this "equivariant (?)-ology" to the equivariant cohomology? At the level of cellular chains/cochains of $M$ it seems to hold. But there are subtleties about what kind of cellulations should be allowed, given the $G$ action.