Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
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 –  Sean Tilson Jun 3 '12 at 3:35
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! –  Tom Church Jun 3 '12 at 5:02
If you consider only the mappings $\Pi \rightarrow \mathbb{Z}$ with finite support, then the two definitions of group ring ought to coincide. –  James Miller Jun 3 '12 at 5:44
add comment

1 Answer 1

up vote 2 down vote accepted

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

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.