Some preliminary definitions:

Let $\Pi = \langle \alpha | \alpha^2 = 1\rangle$ be the cyclic group of order $2$ and let $\mathbb{Z}\Pi$ denote the group ring of $\Pi$ over $\mathbb{Z}$. Embed $\Pi$ in $\mathbb{Z}\Pi$ by letting $1:\Pi \rightarrow \mathbb{Z}$ be the map defined by $1(1) = 1$ and $1(\alpha) = 0$ and letting $\alpha:\Pi \rightarrow \mathbb{Z}$ be the map defined by $\alpha(1) = 0$ and $\alpha(\alpha) = 1$.

Now let $(W,\partial)$ be the chain complex defined as follows. Set $W_n$ as the free $\mathbb{Z}\Pi$-module on the single generator $e_n$ if $n \geq 0$ and set $W_n = 0$ otherwise. Let $\partial_n$ be the zero homomorphism for all $n \leq 0$ and let $\partial_n$ be the homomorphism defined by $\partial_n(r) = (\alpha+(-1)^n)(r)$ for all $n > 0$. Lastly, let $\varepsilon:W_0 \rightarrow \mathbb{Z}$ be the augmentation map defined by setting $\varepsilon(1) = \varepsilon(\alpha) = 1$.

The problem:

I have managed to show that the augmented complex $(W,\partial,\varepsilon)$ is acyclic and am currently trying to find a chain map $\nabla:W \rightarrow W \otimes_{\mathbb{Z}} W$ which lifts the identity $\mathbb{Z} \rightarrow \mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}$. I have been told that it is possible to take $\nabla(e_0) = e_0 \otimes e_0$ and $\nabla(e_1) = e_1 \otimes \alpha e_0 + e_0 \otimes e_1$ and so on, but I can't seem to figure out how this works. Alexander-Whitney maps and Eilenberg-Zilber maps seem like they might be relevant, but I am uncertain how to use them here. Would someone help orient me in the right direction here?

  • $\begingroup$ so you need to define a chain map by specifying its values on each $e_i$. Use the fact that is a chain map so you know... this should help. Also, this question is more appropriate for math.stackexchange.com $\endgroup$ Jun 3, 2012 at 3:35
  • $\begingroup$ Incidentally, it seems you are confusing the group ring $\mathbb{Z}\Pi$ (the group of formal linear combinations of elements of $\Pi$) with its dual (the group of functions from $\Pi$ to $\mathbb{Z}$). For a finite group like $\Pi=\mathbb{Z}/2\mathbb{Z}$ there is not a huge difference, but for infinite groups the distinction is very important. For example, if $\Pi$ is countably infinite, the group ring $\mathbb{Z}\Pi$ will be countably infinite, while its dual is uncountable! $\endgroup$
    – Tom Church
    Jun 3, 2012 at 5:02
  • $\begingroup$ If you consider only the mappings $\Pi \rightarrow \mathbb{Z}$ with finite support, then the two definitions of group ring ought to coincide. $\endgroup$ Jun 3, 2012 at 5:44

1 Answer 1


Chapter XI "Products" of Cartan and Eilenberg's 1956 classic book "Homological Algebra" includes an explicit formula on page 219.


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