Let $G=\mathbb{Z^2}$ be a group acting on a set $Y=\mathbb{Z}$ defined as

$n\cdot(a,b)=n+a+b, ~\textrm{for all}~ n \in Y \textrm{ and } (a,b)\in G .$

Consider chain complexes $(C_*(Y), \partial)$ and $(C_*(Y)_G : = C_*(Y) \otimes_{\mathbb{Z[G]}} \mathbb{Z}, \partial \otimes id_{\mathbb{Z}})$, where $C_n(Y)$ is the free $\mathbb{Z}$-module generated by $(n+1)$-tuples $(y_0, y_1, \ldots,y_n)$ of elements of $Y$, $\partial = \sum_i (-1)^i \partial_i$ and $\partial_i(y_0, y_1, \ldots,y_n)=(y_0, \ldots,y_{i-1}, y_{i+1}, \ldots, y_n)$. I have to prove that $H_i(C_*(Y)_G)=0$ for all $i \geq 2$ and $H_1(C_*(Y)_G)=\mathbb{Z}$. I don't know how to proceed. Can someone suggest direction to solve this?

  • $\begingroup$ In the tensor product defining $C_*(Y)_G$, does $\mathbb{Z}$ have the trivial $G$-module structure? $\endgroup$ Commented Feb 1, 2020 at 16:31
  • $\begingroup$ @Lama:: Yes, there $\mathbb{Z}$ has trivial $G$-module structure. And $C_{*}(Y)$ has natural $G$-action induced from the action of $G$ on set $Y$. $\endgroup$
    – eyp
    Commented Feb 2, 2020 at 3:17


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