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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.

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This kind of thing shows up quite naturally in parameterised stable homotopy theory. Let me translate an idea I know from there into the language in this question.

Cap product gives a map $$C^{p}(M ; A) \otimes C_q(M, \mathbb{Z}) \to C_{q-p}(M;A),$$ which is $G$-equivariant and is a chain map when we sum over $(p,q)$ and give the two sides their natural differential. With the cup product in group cohomology this yields maps $$C^n(G ; C^p(M ; A)) \otimes C^m(G ;C_q(M, \mathbb{Z})) \to C^{n+m}(G ; C^p(M ; A) \otimes C_q(M, \mathbb{Z})) \to C^{n+m}(G;C_{q-p}(M;A))$$ and again summing up over all indices, and flattening (n,m) and (p,q), it gives a map of double complexes.

Thus if I give you a cycle $ \xi \in C^0(G ;C_d(M, \mathbb{Z}))$ then there is an induced map of double complexes $$\xi_\# : \mathcal{Q}^{n,p} \to \widetilde{\mathcal{Q}}^{n,p}.$$ Furthermore, if $\xi$ is a fundamental cycle for $M$ under the evaluation map $C^0(G ;C_d(M, \mathbb{Z})) \to C_d(M, \mathbb{Z})$, then the map induced by $\xi_\#$ on spectral sequences for these double complexes is, on $E_2$-pages, $$H^n(G ; H^p(M;A)) \to H^n(G ; H_{d-p}(M;A))$$ induced by the map on coeffciients which caps with the fundamental class: this is an isomorphism by ordinary Poincare duality for $M$, and so $\xi_\#$ is a quasiisomorphism.

It remains for me to give you such a cycle $\xi$. The spectral sequence for the double complex $C^*(G ;C_\bullet(M, \mathbb{Z}))$ has $$E_2^{p,q} = H^p(G ; H_q(M;\mathbb{Z}))$$ so as long as $M$ is oriented and the $G$-action preserves this orientation then we can form the class $\xi' :=[M] \in H^0(G ; H_d(M;\mathbb{Z}))$. Some contemplation shows that the differentials in this spectral sequence go $$d_r : E_r^{p,q} \to E_r^{p+r, q+r-1}$$ from which it follows that this $\xi'$ is a permanent cycle. If $\xi$ is any cycle in the double complex representing $\xi'$, then it has the properties I used in the third paragraph.

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