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 ?

  • $\begingroup$ 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. $\endgroup$
    – ThiKu
    Mar 1, 2021 at 12:06
  • $\begingroup$ @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. $\endgroup$
    – cellular
    Mar 1, 2021 at 12:28
  • $\begingroup$ @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 $X$. $\endgroup$ Mar 1, 2021 at 12:39
  • 1
    $\begingroup$ @cellular I don't know, but the fact that the action is not, in fact, "natural" should go in the question, I think $\endgroup$ Mar 1, 2021 at 12:59
  • 3
    $\begingroup$ 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. $\endgroup$ Mar 1, 2021 at 13:13

1 Answer 1


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.


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