# Trivialize a cup-product 3-cocycle of $G$ in a larger group $J$

Inspired by this question, let us take a nontrivial 3-cocycle $\omega_3^G(g_a, g_b, g_c) \in H^3(G,\mathbb{R}/\mathbb{Z})$ in the cohomology group of $G$ with $U(1)=\mathbb{R}/\mathbb{Z}$ coefficient. In otherwords, here the 3-cocycle $\omega_3^G$ is a complex $U(1)=\mathbb{R}/\mathbb{Z}$ function with the norm $|\omega_3^G|=1$ but with a $U(1)$ complex phase satisfying the cocycle condition. Here $g_a, g_b, g_c \in G$.

We like to ask how can we trivialize the 3-cocycle $\omega_3(g_a, g_b, g_c)$ of $G$ into 3-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
$$\omega_3^J(j_a, j_b, j_c)=\omega_3^G(r(j_a),r(j_b),r(j_c))=\omega_3^G(g_a, g_b, g_c) \text{ is trivial in } H^3(J,\mathbb{R}/\mathbb{Z}).$$

I like to take $G=\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$, and consider the group element $g_a=(g_{a1}, g_{a2}, g_{a3}) \in (\mathbb{Z}_2, \mathbb{Z}_2, \mathbb{Z}_2)=\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2 =G$, similarly, $g_b=(g_{b1},g_{b2},g_{b3}) \in G$ and $g_{c}=(g_{c1},g_{c2},g_{c3}) \in G$. Let me focus on the 3-cocycle

$$\omega_3^G(g_a, g_b, g_c)=\exp[i \pi \cdot g_{a1} \cdot g_{b2} \cdot g_{c3}]=(-1)^{g_{a1} \cdot g_{b2} \cdot g_{c3}} \in H^3(G,\mathbb{R}/\mathbb{Z}),$$ is in a cup product form of $g_a, g_b, g_c$.

Here $\cdot$ is a usual product of multiplication. And $g_{a1} \cdot g_{b2} \cdot g_{c3} \in \{0,1\}= \mathbb{Z}_2$.

We wish to find that $\omega_3^G(r(j_a),r(j_b),r(j_c))$ becomes a 3-coboundary in $H^3(J,\mathbb{R}/\mathbb{Z})$ for the cohomology group of $J$, but $\omega_3^G(g_a,g_b,g_c)$ originally was not a 3-coboundary but was a 3-cocycle for the cohomology group of $G$. We hope to explicitly write $$\omega_3^G(g_a,g_b,g_c)=\omega_3^G(r(j_a),r(j_b),r(j_c))= \frac{\beta_2^J(j_b,\; j_c)\beta_2^J(j_a,\; j_b j_c)}{\beta_2^J(j_a j_b,\; j_c) \beta_2^J(j_a,\; j_b)}.$$ Here $\beta_2^J(j_a,\; j_b)$ is a 2-cochain for $j_a, j_b, j_c \in J$, and that $g_a=r(j_a)$, $g_b=r(j_b)$, $g_c=r(j_c) \in G$.

Question: For example, can we find explicitly what is the minimal $J$, so we can trivialize a cup-product 3-cocycle $(-1)^{g_{a1} \cdot g_{b2} \cdot g_{c3}}$ of $G=\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$ in a larger group $J$? Here $G$ is order 8 group. Perhaps $J$ is a group of order 16 at least?

## 1 Answer

The group $J$ has to be of order divisible by 16, due to the fact that if the order of the group $J$ is $8n$ where $n$ is an odd number, then the inflation of the cocycle will not be trivial (for example due to the fact that the quotient map $r:J\to G$ will split).

There is no one minimal $J$. The reason for this is the following: You can consider your cocycle in the following way: on $\mathbb{Z}_2$ take the nontrivial element $c$ in $H^1(\mathbb{Z}_2,\mathbb{Z}_2)$, lift to to $G$ three times (each time use a different copy of $\mathbb{Z}_2$), and denote the results by $c_1, c_2$ and $c_3$. Take the cup product $c_1c_2c_3\in H^3(G,\mathbb{Z}_2)$ and apply the natural embedding of $\mathbb{Z}_2$ in $\mathbb{R}/\mathbb{Z}$. In this way you get your cocycle $\omega$. Now, if you consider the cocycle $c_1c_2\in H^2(\mathbb{Z}_2\times \mathbb{Z}_2,\mathbb{Z}_2)$ This corresponds to an extension of the form $$1\to \mathbb{Z}_2\to D_8\to \mathbb{Z}_2\times\mathbb{Z}_2\to 1,$$ and the inflation of $c_1c_2$ to $D_8$ is then trivial. For that reason, the map $D_8\times \mathbb{Z}_2\to G$ trivializes the cocycle. Of course that you could have chosen $\mathbb{Z}_2\times D_8$, with the cocycle $c_2c_3$ instead. This also gives us a concrete way to construct $\beta$: After following the trivialization to $c_1c_2$ arising from the above extension and by writing elements in $D_8$ as triples $(x,y,z)$ with $x,y,z\in\mathbb{Z}_2$ (so that the multiplication is given by $$(x_1,y_1,z_1)(x_2,y_2,z_2) = (x_1+x_2+y_1z_2,y_1+y_2,z_1+z_2),$$ we end up with the formula $$\beta(((x_1,y_1,z_1),w_1)),((x_2,y_2,z_2),w_2)) = (-1)^{x_1w_2}$$

• The fact about trivialization of 2 cocycles in the relevant extension can be found in most textbooks on cohomology of groups or even of homological algebra. The suggested solution will be of the form $$1\to Z_2\to D_8\times Z_2\to Z_2^3\to 1.$$ – Ehud Meir Nov 11 '16 at 15:57
• What book/papers do you recommend me the most for this context? Thank you. – miss-tery Nov 11 '16 at 18:24
• you are welcome :) I think that in Cohomology of groups of Brown, or in Homological algebra by Rotman or Homological algebra of Weibel you will some good answers – Ehud Meir Nov 11 '16 at 22:20
• Thanks, What about the continuous group, :), mathoverflow.net/questions/253887, any idea? (p.s. I will accept your answer in a few days if no one gives other better reply.) – miss-tery Nov 12 '16 at 2:23
• @Ehud Meir, do you see that your example $D_8 \to (\mathbb{Z}_2)^2$also works for split all the 3-cocycles in $$\omega_3\in H^3(\mathbb{Z}_2 \times \mathbb{Z}_2, \mathbb{R}/\mathbb{Z})?$$ For example, the cup-product of $a1 \cup a2 \cup a2$, and also $a1 \cup a1 \cup a1$? – wonderich Dec 4 '16 at 19:52