It is well known (see e.g. K. Brown, "Cohomology of groups") that a degree-3 cohomology class of a group G with coefficients in a module A can be thought of as an equivalence class of crossed modules, in the sense that every crossed module gives a double extension of G by A. Now, it is also well known that degree-3 cohomology of Z/n with coefficients in R/Z (with a trivial action of Z/n) is isomorphic to Z/n. My question is, how does one actually construct a crossed module corresponding to an element of Z/n?


1 Answer 1


Here is a general procedure to construct a crossed module $\mathcal{G} = (G_1 \stackrel{d}{\to} G_0)$ from a normalized 3-cocycle $\alpha \in Z^3(G; \mathbb{R}/\mathbb{Z})$ which is additive in the third variable.

First, we get a cocycle $\omega \in Z^2(G; G^{\ast})$, where $G^{\ast}$ is the Pontryagin dual of $G$ equipped with the trivial $G$-action, by $\omega(a,b)(c) = \alpha(a,b,c)$. We use this cocycle to get an extension $$0 \to G^{\ast} \to G_0 \to G \to 0.$$

We will construct the $G_0$-module $G_1$ as an extension of $G_0$-modules $$0 \to \mathbb{R}/\mathbb{Z} \to G_1 \to G^{\ast} \to 0.$$ Since $\mathbb{R} / \mathbb{Z}$ is divisible, this sequence splits as an exact sequence of abelian groups. Thus, $G_1 \cong \mathbb{R} / \mathbb{Z} \times G^{\ast}$ as abelian groups. The action of $G_0$ factors through an action of $G$: $$a: G \times (\mathbb{R} / \mathbb{Z} \times G^{\ast}) \to \mathbb{R} / \mathbb{Z} \times G^{\ast}$$ $$a(g, (\theta, \gamma)) = (\theta + \gamma(g), \gamma)$$

The map $G_1 \stackrel{d}{\to} G_0$ is just the composite $G_1 \to G^{\ast} \to G_0$.

OK, let's specialize now to the case $G = \mathbb{Z} / n\mathbb{Z}$. We can realize the class in $H^3(\mathbb{Z} / n\mathbb{Z}; \mathbb{R}/\mathbb{Z})$ corresponding to $m \in \mathbb{Z} / n\mathbb{Z}$ by the normalized cocycle $$\alpha(a, b, c) = \begin{cases} 0 & \text{if $a + b < n$,}\\ cm & \text{otherwise.} \end{cases}$$ (Here we are taking the image of the cocycle as living in $\mathbb{Z} / n\mathbb{Z}$ as a subgroup of $\mathbb{R}/\mathbb{Z}$, and we will likewise identify $\mathbb{Z} / n\mathbb{Z}$ with its Pontryagin dual by sending $a \in \mathbb{Z} / n\mathbb{Z}$ to $\frac{a \cdot -}{n} \in (\mathbb{Z} / n\mathbb{Z})^\ast$.) So explicitly, the group $G_0$ is the set $\mathbb{Z} / n\mathbb{Z} \times \mathbb{Z} / n\mathbb{Z}$ equipped with the operation $$(a, b) \oplus (c, d) = \begin{cases} (a + c, b + d) & \text{if $b + d < n$,}\\ (a + c + m, b + d) & \text{otherwise.} \end{cases}$$ The group $G_1$ is just $\mathbb{R} / \mathbb{Z} \times \mathbb{Z} / n\mathbb{Z}$. The action of $G_0$ on $G_1$ is given by $$(a, b) \rhd (\theta, c) = (\theta + \frac{bc}{n}, c).$$ The map $G_1 \stackrel{d}{\to} G_0$ is the composite of projection onto the second factor followed by inclusion into the first.


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