# Trivialize a cup-product 2-cocycle of $G$ in a larger group $J$

I like to ask a simple question: how to trivialize a cup-product 2-cocycle of $G$ into a 2-coboundary of $J$ in a larger group $J$.

Let us take a nontrivial 2-cocycle $\omega_3^G(g_a, g_b) \in H^2(G,\mathbb{R}/\mathbb{Z})$ in the cohomology group of $G$ with $U(1)=\mathbb{R}/\mathbb{Z}$ coefficient. In otherwords, here the 2-cocycle $\omega_2^G$ is a complex $U(1)=\mathbb{R}/\mathbb{Z}$ function with the norm $|\omega_2^G|=1$ but with a $U(1)$ complex phase satisfying the cocycle condition. Here $g_a, g_b \in G$.

How can we trivialize the 2-cocycle $\omega_2(g_a, g_b)$ of $G$ into 2-coboundary if we lift $G$ into a larger group $J$, and given that we know the group homomorphism $r$:

$$J \overset{r}{\rightarrow} G,$$

so that
$$\omega_2^J(j_a, j_b)=\omega_2^G(r(j_a),r(j_b))=\omega_2^G(g_a, g_b)=\delta v_1(j_a, j_b)=\frac{v_1(j_a) v_1(j_b)}{v_1(j_a j_b)} \text{ is trivial in } H^3(J,\mathbb{R}/\mathbb{Z}).$$

Namely, $\omega_2^J(j_a, j_b) =\delta v_1(j_a, j_b)$ is a 2-coboundary in $J$. I like to take $G=\mathbb{Z}_2 \oplus \mathbb{Z}_2$, and consider the group element $g_a=(g_{a1}, g_{a2}) \in (\mathbb{Z}_2, \mathbb{Z}_2)=\mathbb{Z}_2 \oplus \mathbb{Z}_2 =G$, similarly, $g_b=(g_{b1},g_{b2}) \in G$.

Let me focus on the 2-cocycle

$$\omega_2^G(g_a, g_b)=\exp[i \pi \cdot g_{a1} \cdot g_{b2} ]=(-1)^{g_{a1} \cdot g_{b2} } \in H^2(G,\mathbb{R}/\mathbb{Z}),$$ is in a cup product form of $g_a, g_b$.

Question: What are examples and restrictions of $J$? Whether dihedral group $D_8$ and quaternion $Q_8$ work for $J$? What is the explicit form of 1-cochain $v_1(j)$?

• What more conditions on $J$ would you like so that @EhudMeir 's earlier answer is excluded? – AHusain Dec 2 '16 at 2:49
• I think the dihedral group $D_8$ does not really work, but perhaps the quaternion group works. It is not sue that EhudMeir answer really works out if you tried working through it -- perhaps somewhere there is a loophole. – miss-tery Dec 2 '16 at 2:55
• If you mean that we are free to choose $J$ given $G$ and the cocycle, the question is trivial. If $J$ is given in advance, then I don't understand why your question is in the form "how can we trivialize" when you should ask "is it trivial". – Pierre Dec 3 '16 at 17:58

The given particular 2-cocycle $\omega_2 \in H^2(G,\mathbb{R}/\mathbb{Z})$ of $G=(\mathbb{Z}_2)^2$ is $$\omega_2(g_a , g_b)=(-1)^{[g_{a_1}]_2 [g_{b_2}]_2 }, \;\;\;\;$$ where $g_a=(g_{a_1},g_{a_2}) \in G=(\mathbb{Z}_2)^2$, and similarly for $g_b$. Below I aim to show that both $Q_8 \to (\mathbb{Z}_2)^2$ and $D_8 \to (\mathbb{Z}_2)^2$ provide a 1-cochain trivilize the $\omega_2$ into 2-coboundary in $J$:

Ans 1: True for $D_8 \to (\mathbb{Z}_2)^2$

Write the dihedral group $$D_8=\langle x, R |x^2=R^4=1, x Rx=R^{-1}\rangle$$
We can choose that the $\{1,x,R,xR\}$ in $D_8$ maps to the $(\mathbb{Z}_2)^2$. The $\{R^2 ,xR^2, R^3,x R^3\}$ in $Q_8$ also maps to the $(\mathbb{Z}_2)^2$. This gives us the following group homomorphism: $$D_8 \to (\mathbb{Z}_2)^2.$$

The 1-cochain for $j \in J= D_8$ $$\beta_1(j) =(-1)^{f(j)}$$ satisfies the desired 2-cocycle splitting property. Here we define the function $f$: $$f(1)=f(x)=f(R) =f(xR)=0,$$ $$f(R^2)=f(x \cdot R^2)=f(R \cdot R^2) =f(xR \cdot R^2)=1.$$ I find that $\omega_2= \delta(\beta_1)$.

Ans 2: True for $Q_8 \to (\mathbb{Z}_2)^2$

Write the quaternion $$Q_8=\langle x,y|x^2=y^2, xyx^{-1}=y^{-1},x^4=y^4=1\rangle$$ so that each element in the group we can write uniquely as $x^p y^{q}$ with $p\in\{0,1\}$ and $q \in\{0,1,2,3\}$.

We can choose that the $\{1,x,y,xy\}$ in $Q_8$ maps to the $(\mathbb{Z}_2)^2$. The $\{y^2 ,x\cdot y^2,y\cdot y^2,xy \cdot y^2\}$ in $Q_8$ also maps to the $(\mathbb{Z}_2)^2$. This gives us the following group homomorphism: $$Q_8 \to (\mathbb{Z}_2)^2.$$ The 1-cochain for $j \in J= Q_8$
$$\beta_1(j)=\beta_1(x^{p} y^{q}) = i^{(p+q)}.$$ satisfies the desired 2-cocycle splitting property. I find that $\omega_2= \delta(\beta_1)$.

So to summarize, I give an answer through examples of $J$. I show that dihedral group and quaternion trivialize $G$, with the explicit form of 1-cochain given.

• Thanks, is the 1-cochain for $j \in J= Q_8$, the one you write $\beta_1(j)=\beta_1(x^{p} y^{q}) = i^{(p+q)}$ is the only possible form? How can you determine the 1-cochain? – miss-tery Dec 3 '16 at 21:40
• I think $\beta_1(j)=\beta_1(x^{p} y^{q}) = i^{(p+q)}$, also the other three $i^{(-p+q)}$, $i^{(p+-q)}$, $i^{(-p-q)}$ possible form work as well, due to the $Z_2$-ness nature of the 2-cocycle. You may check. I believe so. – wonderich Dec 3 '16 at 21:42