# Cellular homology of the universal cover

Let $$X$$ be a connected pointed CW complex. Let $$\tilde{X}$$ be its universal covering space and $$G=\pi_{1}(X)$$.

Lets denote $$(C^{Cell}_{\ast}(\tilde{X}),d)$$ the cellular chain complex associated to $$\tilde{X}$$. By construction each $$C^{Cell}_{n}(\tilde{X})$$ is a free (left) $$G$$-module and the differential operators $$d:C^{Cell}_{n+1}(\tilde{X})\rightarrow C^{Cell}_{n}(\tilde{X})$$ are (left) $$G$$-equivariant.

Question: Since each $$C^{Cell}_{n}(\tilde{X})$$ is a free (left) $$G$$-module, there is also a "natural" action of $$G$$ on $$C^{Cell}_{n}(\tilde{X})$$ on the right by multiplication. I was wondering if $$d:C^{Cell}_{n+1}(\tilde{X})\rightarrow C^{Cell}_{n}(\tilde{X})$$ is also right $$G$$-equivariant ?

• I guess you mean the right action defined by $x.g:={g^{-1}x}$. From $d(g.x)=g.dx$ for all $g$ you get in particular $d(g^{-1}.x)=g^{-1}.dx$ for all g. So the answer is yes. Mar 1, 2021 at 12:06
• @ThiKu no, by right action a do mean the following: let A be a ring, then any direct sum of $A$ is an A-bimodule in obvious way. Mar 1, 2021 at 12:28
• @cellular The action you describe depends on the choice of basis, i.e. in this case by the choice of a lift for every cell of $X$. Mar 1, 2021 at 12:39
• @cellular I don't know, but the fact that the action is not, in fact, "natural" should go in the question, I think Mar 1, 2021 at 12:59
• In the case of the universal covering space of a wedge of 2-circles, if you use the usual basis to define the right action it will not be equivariant. Mar 1, 2021 at 13:13

As Benjamin Steinberg says, it does not work in general, and you can see the problem already for 1-dimensional complexes. Suppose you take a Cayley graph for $$G$$. This is a graph (or 1-dimensional CW-complex) with vertex set $$G$$ and edge set $$G\times S$$, where $$S\subseteq G$$ is a set of elements that generates $$G$$. The edge $$(g,s)$$ joins the two vertices $$g$$ and $$gs$$. I have set this up so that $$G$$ acts on the left: for $$h\in G$$, $$h(g,s)=(hg,s)$$ defines the action. As we expected, the edge $$h(g,s)$$ joins the two vertices $$hg$$ and $$hgs$$. Since the edge set and the vertex set are both free as $$G$$-sets, it is possible to make them into $$G-G$$-bisets. But there isn't usually a way to do this that preserves the incidence relation between vertices and edges. The first thing you might try is to define the right action on edges by $$(g,s)k)=(gk,s)$$. But then the edge $$(g,s)k$$ joins the vertices $$gk$$ and $$gks$$, whereas the images under $$k$$ of the ends of the edge $$(g,s)$$ are $$gk$$ and $$gsk$$. If the subset $$S$$ is closed under conjugation, then you could define $$(g,s)k=(gk,k^{-1}sk)$$, but if $$S$$ is not closed under conjugation (which it won't usually be) this won't work. In the case of the universal cover of the wedge of two circles, $$S$$ is a set of two elements that freely generate the fundamental group.
In terms of $$G$$-sets, you can think about the vertices and directed edges as free $$G$$-sets, and the map taking a directed edge to its initial vertex as a $$G$$-map. The left $$G$$-maps from $$G$$ to $$G$$ are exactly the right multiplications by elements of $$G$$, and unless $$G$$ is abelian, many of these will fail to commute with the right $$G$$-action. So although you can make any free $$G$$-set into a $$G-G$$-biset, most left $$G$$-maps will not be $$G-G$$-bimaps.