I would like to know the explicit expression of 2-cocycle from a 2nd cohomology group $H^2[Q_8 \times \mathbb{Z}/2\mathbb{Z}, U(1)]$ with $U(1)\equiv \mathbb{R}/\mathbb{Z}$ coefficient, or namely $H^2[Q_8 \times \mathbb{Z}_2, U(1)]$, here we denote the cyclice group $\mathbb{Z}/2\mathbb{Z}=\mathbb{Z}_2$, and the $Q_8$ is the order 8 quaternion group.

What I had computed (from Kunneth formula, Universal Coefficient Theorem) and what I had known is that: $$H^2[Q_8 \times \mathbb{Z}_2, U(1)]=\mathbb{Z}_2 \times \mathbb{Z}_2.$$ Now I would like to know the two generators of $H^2[Q_8 \times \mathbb{Z}_2, U(1)]=\mathbb{Z}_2 \times \mathbb{Z}_2$ in terms of explicit 2-cocycles. Any Reference and any partial answer are welcome.

Additional remarks and guides:

What I also computed and knew are:$H^1[Q_8,U(1)]=\mathbb{Z}_2 \times \mathbb{Z}_2$, $H^2[Q_8,U(1)]=0$, $H^3[Q_8,U(1)]=\mathbb{Z}_8$. I also know that the explicit 3-cocycle $\omega_3((A,a),(B,b),(C,c)) \in H^3[Q_8,U(1)]=\mathbb{Z}_8$ as: $$\omega_3((A,a),(B,b),(C,c))$$
$$= \exp\left( \frac{2\pi i p}{8} \{ (-)^{B+C}a \left( (-)^C b + c- [(-)^C b + c +2BC] \right) - 2 ABC \} \right)$$ where $p \in \mathbb{Z}_8$. The rectangular brackets means the modulo $4$ in the range $-1,0,1,2$.

Here we denote the elements of $Q_8$ by the 2-tuples $$ (A,a) := X^A R^a \qquad \qquad \mbox{with $A \in 0,1$ and $a \in -1, 0, 1, 2$. } \qquad $$ with $R^{4} = e, \qquad X^2 \; = \; R^2, \qquad XR \; = \; R^{-1} X.$

So for instance the identity is $e=(0,0)$. The multiplication law of $Q_8$ then reads $$ (A,a) \cdot (B,b) = ([A+B], [(-)^B a+b+2AB]), $$

  • 1
    $\begingroup$ I think you can use the ideas in Graham's answer to mathoverflow.net/questions/92684/… to reduce this to a computation that you can do in GAP. From there it should be possible to get an explicit cocycle, though finding a pattern for all $n$ might be a bit difficult. $\endgroup$
    – dvitek
    Sep 6, 2016 at 5:21

1 Answer 1


For any two finite groups $G$ and $H$ the Kuenneth formula will give you a homomorphism $$H^1(G,U(1))\otimes_{\mathbb{Z}} H^1(H,U(1))\to H^2(G\times H,U(1)).$$ We can describe this map explicitly. Let $\phi:G\to U(1)$ and $\psi:H\to U(1)$ be group homomorphisms. For some $n$ and some $n$-th root of unity $\zeta$ we can write $\phi(g) = \zeta^{f(g)}$ and $\psi(h) = \zeta^{r(h)}$ where $f:G\to \mathbb{Z}/n$ and $r:H\to\mathbb{Z}/n$ are group homomorphisms (in the case of $Q_8\times \mathbb{Z}/2$ $\zeta$ will just be $-1$). The cocycle which corresponds to $(\phi,\psi)$ will then be given by $$\alpha((g_1,h_1),(g_2,h_2)) = \zeta^{f(g_2)r(h_1)}.$$ For the example that interests you, you can write explicitly all group homomorphisms from $Q_8$ and from $\mathbb{Z}/2$ to $U(1)$ and get the isomorphism you are interested in.

We have the following ring theoretic interpretation of this: the data of a two cocycle on a group is the same as describing the twisted group algebra $\mathbb{C}^{\alpha} G\times H$. In this group algebra we will have the rules $U_{g_1}U_{g_2} = U_{g_1g_2}$, $U_{h_1}U_{h_2} = U_{h_1h_2}$ and $U_hU_g = \zeta^{f(g)r(h)}U_gU_h$.

In principal, what Kuenneth Formula tells us for the two dimensional case is the following thing: in order to describe a two cocycle on $G\times H$ we need to describe three things:

  1. Its restriction to $G$.
  2. Its restriction to $H$.
  3. The way that ot alters the commutativity between $G$ and $H$.

In this case we know that the second cohomology groups of $G$ and of $H$ are trivial, so we are left only with the third option.

  • $\begingroup$ Thanks Ehud. Say $G=Q_8$, and $H=Z_2$. Do you imply that if we know the explicit 1-cocycle of $H^1[G,U(1)]$ and the explicit 1-cocycle of $H^1[H,U(1)]$, then I am able to determine the explicit 2-cocycle of $H^2[G \times H,U(1)]$? $\endgroup$
    – wonderich
    Sep 7, 2016 at 16:19
  • $\begingroup$ I also know that $H^2[Q_8 \times Z_2, U(1)]=H^1[Q_8, Z_2]=\text{Tor}^Z[Z_2, H^2(Q_8,Z)]=\text{Tor}^Z[Z_2, H^1(Q_8,U(1))]=\text{Tor}^Z[Z_2, Z_2 \times Z_2]=Z_2 \times Z_2= Z_2 \oplus Z_2$ $\endgroup$
    – wonderich
    Sep 7, 2016 at 16:21
  • $\begingroup$ I am puzzled by part of your answer. Suppose we take $G=\mathbb{Z}_M$ and $H=\mathbb{Z}_N$ as an example. Then what will be the group homomorphism be? $f: G=\mathbb{Z}_M \to \mathbb{Z}_n$ and $r: G=\mathbb{Z}_N \to \mathbb{Z}_n$? What is $n$ here gcd$(M,N)$ or lcm$(M,N)$? And should both homomorphisms $f$ and $g$ give the same $n$? $\endgroup$
    – wonderich
    Sep 7, 2016 at 21:20
  • $\begingroup$ you can determine explicitly the two cocycles by the formula written above. In the case of $Q_8$ and $\mathbb{Z}_2$ it is going to be the following: $alpha((g_1,h_1),(g_2,h_2)) = 1$ of $h_1=0$, and $\phi(g_2)$ if $h_1=1$ (we are talking here about residues modulo 2), where $\phi$ is a homomorphism from $Q_8$ to $\mathbb{Z}_2$. Since there are four possible homomorphisms, you will get four possible non-equivalent cocycles. $\endgroup$
    – Ehud Meir
    Sep 8, 2016 at 0:09
  • $\begingroup$ I don't know whether this is consitent with what you said: I know the case that $f: G=Z_M→Z_n$ and $r: H=Z_N→Z_n$, the 1-cocycle for $Z_M$ is $\alpha(g)=\exp[(2\pi i/M) g]$, the 1-cocycle for $Z_N$ is $\alpha(h)=\exp[(2\pi i/N) h]$, the2-cocycle for $Z_M \oplus Z_N$ is $\alpha(g_1,g_2,h_1,h_2)=\exp[(2\pi i/\text{gcd}(M,N)) g_1 h_2]$. $\endgroup$
    – wonderich
    Sep 8, 2016 at 15:04

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