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 is 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 $$ (A,a) \cdot (B,b) = ([A+B], [(-)^B a+b+2AB]), $$

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 $$ (A,a) \cdot (B,b) = ([A+B], [(-)^B a+b+2AB]), $$

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 is 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 $R^{2N} = e, \qquad X^2 \; = \; R^N, \qquad XR \; = \; R^{-1} X.$

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

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 is 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 $$ (A,a) \cdot (B,b) = ([A+B], [(-)^B a+b+2AB]), $$

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 is 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 $p \in \mathbb{Z}_8$.

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 $R^{2N} = e, \qquad X^2 \; = \; R^N, \qquad XR \; = \; R^{-1} X.$

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

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 is 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 $R^{2N} = e, \qquad X^2 \; = \; R^N, \qquad XR \; = \; R^{-1} X.$

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