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.



Is there a kind of Poincare duality for Borel equivariant cohomology?
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.


