# Why are these maps of $G$-modules?

A $G$-complex is a CW-complex $X$ together with an action of $G$ on $X$ which permutes the cells (I think the action should be continuous). This action induces an action on the cellular chain complex $C_*(X)$. My question is: why the differential is a map of G-modules?

i.e.: why $\partial(ge^\alpha_n)=g\partial(e^\alpha_n)$?

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Notice that, with your definition, the map $m_g: X \rightarrow X$ is a cellular map, and so induces a map of chain complexes $(m_g)_*: C_*(X) \rightarrow C_*(X)$. By definition, maps of chain complexes commute with the boundary map. So this is really just a consequence of the more general fact that cellular maps induce chain maps on the cellular chain complex. –  Dylan Wilson Sep 3 '11 at 22:24
Are we using continuity somewhere? –  Berry Sep 3 '11 at 22:51
cellular ---> continuous by the definition of the topology of the complex –  Dylan Wilson Sep 3 '11 at 23:13
I meant the continuity of the action –  Berry Sep 3 '11 at 23:43
At least in Hatcher and every time I have seen this the continuity of cellular maps is usually implied (the book/chapter says all maps are continuous in the beginning or something like it somewhere) but not explicitly said. –  Sven Sep 6 '11 at 4:57

## 1 Answer

If you want to have a group acting on a structure, then the action must preserve whatever that structure is. For a group acting on a topological space, each $g\in G$ must act by homeomorphisms. If you want to have a notion of an action on a CW-complex, it is necessary to have the action preserve the CW structure. The question is, what should this entail?

First, and foremost, since every CW complex is a topological space, we must have that the action is a topological action. Second, we would like for the group action to respect our decomposition into cells. Therefore, for each open $n$-cell $N_{\alpha}\subset X$, we should like that the restriction of the action of $g$ to $N_{\alpha}$ induces a homeomorphism $g_{\alpha}:N_{\alpha}\stackrel{\sim}{\longrightarrow}N_{g\alpha}$ (implicit in this is that we do have a permutation of the cells, as you indicated, but this is a much stronger condition). Moreover, we need these maps to be compatible with the attaching maps. The question is, what does "compatibility" mean?

Let us assume that we have defined the action on the $(n-1)$ skeleton $X^{(n-1)}$, and that we wish to extend the action to $X^{(n)}$. We have that the disjoint union of all the open $n$-cells has an action of $G$, which by continuity gives an action on the disjoint union on the closed $n$-cells, which yields an action on the disjoint union of the boundaries of the closed $n$-cells. The attaching maps thus give a map between $G$-spaces, and this map yields an action on $X^{(n)}$ if and only if the map is $G$-equivariant.

Thus, we have defined what it means to be a CW complex with a $G$-action. We wish to see why the action passes to the level of the cellular chain complex. Let $A$ be the boundary of an $n$-cell, and let $B$ be an $(n-1)$-cell in $X^{(n-1)}/X^{(n-2)}$, which is a wedge of spheres. If $f$ is the (quotient of the) attaching map from $A$ to $X^{(n-1)}$, let $f_B$ be the quotient of $f$ when we collapse all the spheres except for $B$. From equivariance (our compatibility between attaching maps), we have that $\deg f_B=\deg g (f_B)$, where $$g(f_B):g(\partial A)\to X^{(n-1)} \to X^{(n-1)}/X^{(n-2)} \to gB.$$

Looking at the definition of the differential in the cellular chain complex, this is exactly the condition that the differential is a map of $G$-modules.

N.B. If you have an action on $X$ (viewed as just a topological space) that permutes the cells and restricts to homeomorphisms, then you must satisfy automatically satisfy the compatibility condition for the attaching maps (which I leave as an easy exercise). However, I think that if you want to view CW complexes as combinatorial-type objects built up from constituent pieces, it is better to build the extra condition into the definition of $G$-action.

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