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Let $B$ be a cellular (simplicial, semi-simplicial etc) complex having $\mathbb{Z}^n$-symmetry (that is, there is a free action of $\mathbb{Z}^n$ on $B$, commuting with the boundary operators) and let $R$ be a commutative ring. In this case it is possible to establish an isomorphism (as $R$-modules) of the homology modules $H_*(B, R)$ and $H_*(B', R[\mathbb{Z}^n])$, where the "quotient" complex $B'$ is constructed in the following way:

1) In each dimension $d$, choose a representative cell in each orbit of $\mathbb{Z}^n$-action on $B$. Define the chain modules $C_d(B')$ as the free $R[\mathbb{Z}^n]$-modules on the basis of these representatives.

2) Let $e_l$ be a cell of dimension $d$ in $B$ and let $e$ be the corresponding representative in its $\mathbb{Z}^n$-orbit, with $l \in \mathbb{Z}^n$ bringing $e$ to $e_l$. Let also $f$ be the respective basis element in $C_d(B')$. Then the mapping of $e_l$ to $1l \cdot f$ extends to an $R$-module isomorphism $\alpha: C_d(B) \to C_d(B')$ (here $1l \in R[\mathbb{Z}^n]$ stands for the element of the group ring corresponding to $l \in \mathbb{Z}^n$).

3) Since the boundary operators $\partial$ in $B$ commute with the action of $\mathbb{Z}^n$, the operators $\alpha \circ \partial \circ \alpha^{-1}$ are $R[\mathbb{Z}^n]$-linear, and make $B'$ into a differential complex of $R[\mathbb{Z}^n]$-modules. Its homology modules are isomorphic (as $R$-modules) to those of $B$.

My question is: is there a more elegant and natural way (without choosing representatives) to construct this "quotient" complex $B'$ and the corresponding isomorphism of homology modules? Also, I have a feeling that this is a particular case of some more general theory, but do not see which one.

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  • $\begingroup$ I am abit confused. Let's take $B=\mathbb{R}$ with the obvioux $\mathbb{Z}$ action. Then what does your $B'$ look like? $\endgroup$ – user43326 Oct 27 '19 at 14:29
  • $\begingroup$ Let us consider the case $R=\mathbb{Z}$. In this case $B'$ consists of 2 free modules of rank 1 (in dimensions 1 and 0) over $\mathbb{Z}[\mathbb{Z}]$. If we represent $\mathbb{Z}[\mathbb{Z}]$ as a ring of Laurent polynomials over a variable $X$, then the differential operator is (in the obvious basis) just the multiplication by $X-1$. Since this operation is injective, we have $H_1=0$, and since its image contains only Laurent polynomials with zero sum of coefficients, $H_0=\mathbb{Z}$, as it should be. $\endgroup$ – p_k Oct 27 '19 at 14:58
  • $\begingroup$ Yes $C_*(B)$ is naturally a free $R[G]$- module, but to explicitly choose a basis one has to pick one cell in each orbit. However it comes equipped with a natural equivalence class of bases given by the construction above. The base change matrices are always permutation matrices with where entries are allowed to be elements from $G\subset R[G]$. Note that such an equivalence class is really an additional structure; not all bases lie in that class. If you want to look further into those ideas, the term Whitehead-group might be helpful. $\endgroup$ – HenrikRüping Oct 27 '19 at 18:44
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Suppose a group $G$ acts `properly discontinuously' on a friendly topological space $B$, so that $G$ is acting freely on $B$, and, if we let $B^{\prime} = B/G$, then $B \rightarrow B^{\prime}$ is a covering map. It is a nice exercise in the standard lifting theorem of covering space theory to show that cellular structures on $B^{\prime}$ correspond to $G$--equivariant cellular structures on $B$. (Similarly for simplicial structures, etc.)

If $M$ is a $R[G]$--module, $H_*(B^{\prime};M)$, the homology of $B^{\prime}$ with twisted coefficients $M$, is defined to be the homology of the complex $C_*(B) \otimes_{R[G]} M$. There are two extreme cases. Let $M = R$. As $C_*(B) \otimes_{R[G]} R = C_*(B^{\prime})$, we have $H_*(B^{\prime};R) = H_*(B^{\prime};R)$. (phew!). At the other extreme, let $M = R[G]$. As $C_*(B) \otimes_{R[G]} R[G] = C_*(B)$, we have - yes - $H_*(B^{\prime};R[G]) = H_*(B;R)$.

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    $\begingroup$ Oh, I just learned two things: the term "homology with twisted coefficients" and the fact that $C_*(B)$ is naturally a free $R[G]$-module. My problem was that I tried to reason in terms of a basis of this module, but this is an unnecessary compication. $\endgroup$ – p_k Oct 27 '19 at 17:50

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