It is well known how to compute cohomology of a finite cyclic group $C_m=\langle \sigma \rangle$, just using the periodic resolution,


\begin{CD} \cdots @>N>> \mathbb Z C_m @> \sigma -1>> \mathbb Z C_m @>N>> \mathbb Z C_m @> \sigma -1>> \mathbb Z C_m @> >> \mathbb Z \end{CD}

Using this resolution, it easy to see that \begin{align} H^n(C_m; A)= \begin{cases}\{a\in A: Na=0\}/(\sigma-1)A, \qquad &\text{if } n=1, 3, 5, \ldots \\ A^{C_m}/NA, \quad &\text{ if } n = 2, 4, 6, \ldots, \end{cases} \end{align} where $N= 1+ \sigma + \sigma^2 +\cdots +\sigma ^{m-1}$. Now, for some applications of group cohomology is important to work with standard cocycles, that is cocycles respect to the standard (also called Bar) resolutions. A construction of quasi-isomorphism from the periodic resolution to the standard resolution can be done as follows: take a section of $\pi$ in the exact sequence \begin{CD} 0 @>>> \mathbb Z @> m >> \mathbb Z @>\pi>> C_m @>>> 0, \end{CD}so we get a $\gamma\in Z^2(C_m,\mathbb{Z})$. For $Z^1(C_m,A)$ and $Z^2(C_m,A)$ the map can be defined by hand easily. In general we can construct the map $:Z^1(C_m,A)\to Z^{2n+1}(C_m,A)$ just using the cup product $\alpha\mapsto \gamma^{\cup n}\cup \alpha$ and analogously for $:Z^2(C_m,A)\to Z^{2n}(C_m,A)$. Thus, at the end you find the map from the "periodic" cocycles to the standard cocycles.

My question is: How to define in general the quasi-isomorphisms from the standard cocycles to the "periodic" cocycles?


2 Answers 2


Just stumbled across that old question while researching the subject on the internet, ended up doing the computation myself, and thought I would give an answer in case someone finds themself in the same situation.

A possible quasi-isomorphism $u_n: C^n(C_m,A)\to A$ from the standard complex to the periodic one is given by the following formulas: if $n=2r$, and $\tau$ is a generator of $C_m$, $$u_{2r}: f\mapsto (-1)^r\sum_{g_1,\dots,g_r\in C_m}\sum_{\varepsilon_1,\dots,\varepsilon_r\in \{0,1\}} (-1)^{\sum \varepsilon_i} f(g_1,\tau^{\varepsilon_1},\dots,g_r,\tau^{\varepsilon_r}),$$ and if $n=2r+1$, $$u_{2r+1}: f\mapsto (-1)^{r+1}\sum_{g_1,\dots,g_r\in C_m}\sum_{\varepsilon_1,\dots,\varepsilon_{r+1}\in \{0,1\}} (-1)^{\sum \varepsilon_i} f(\tau^{\varepsilon_1},g_1,\tau^{\varepsilon_2},\dots,g_r,\tau^{\varepsilon_{r+1}}).$$

In the example I was interested in, the action of $C_m$ on $A$ was trivial, but I read the computation carefully and I don't think it intervened in this formula.


I am referencing Ken Brown's "Cohomology of Groups" in what follows. Given your two free resolutions, Theorem I.7.5 not only gives the existence of the chain map (that is a homotopy-equivalence), but the proof is constructive because the bar resolution has an explicit basis. For the record: The chain map in the other direction (which you already gave) can also be written down from this Theorem, and is Exercise I.7.1.

When $m=2$ this periodic resolution is precisely the normalized bar resolution (Exercise I.2.2), so your desired quasi-isomorphism should be the quotient map from the bar resolution to the normalized one.

  • 4
    $\begingroup$ Thank. Yes, I know Theorem 1.7.5. But, in general given two resolutions is a not trivial processes to describe explicitly the chain map. I can describe easily for n=1,2, but just for n=3, the formula is not clear for me. My question is if there is a formula for all n's. $\endgroup$ Apr 22, 2016 at 0:02

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