# Homology modules and symmetry

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.

• I am abit confused. Let's take $B=\mathbb{R}$ with the obvioux $\mathbb{Z}$ action. Then what does your $B'$ look like? – user43326 Oct 27 '19 at 14:29
• 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. – p_k Oct 27 '19 at 14:58
• 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. – HenrikRüping Oct 27 '19 at 18:44

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 - $$H_*(B^{\prime};R[G]) = H_*(B;R)$$.
• 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. – p_k Oct 27 '19 at 17:50