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Suppose I have a group ${ G }$ acting freely on the sphere ${ S^n }$ for ${ n \geq 2 }$. Does this somehow get me a free resolution of ${ \mathbb{Z} }$ as a ${ \mathbb{Z}[G] }$-module?

Forgive me if the question is obvious. I was reading Brown's book on Cohomology of Groups, and I saw there that the previous statement holds true if we replace ${ S^n }$ with a contractible CW space ${ X }$ (Or anything which is acyclic).

I saw someone mention this statement as a comment on this MSE question Proving that there doesn't exist a free $\mathbb{Z}_2\times \mathbb{Z}_2$ action on $S^n$. which makes me think that a proof of it would be trivial, but I just can't figure it out!

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Suppose you have a $G$-CW complex structure on $S^n$ for your free action. Then cellular chains are a complex of free $G$-modules. This chain complex fails to be a resolution of $H_0(S^n)=\mathbb{Z}$ because $H_n(S^n)=\mathbb{Z}$. But we can just glue together another copy of the chains to kill the kernel $C_n(S^n)\rightarrow C_{n-1}(S^n)$, ad infinitum. Hence, we get a periodic resolution of $\mathbb{Z}$ as a $G$-module.

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  • $\begingroup$ Please may you add a few more details?, how does procedure use the free $G$-action? $\endgroup$
    – Nick L
    Oct 16 at 8:49
  • $\begingroup$ The free G action is what makes the chain complex consist of free G modules. You pick a representative of each orbit to form a basis. $\endgroup$ Oct 16 at 14:47
  • $\begingroup$ @ConnorMalin What exactly do you mean by "glue together" another copy of ther chains? $\endgroup$ Oct 18 at 16:14

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