Edit: In case that there is no solution for the original question, I modify to enrich the question.

We like to ask a possible specific inflation a $H^3(Q, \mathbb{R} /\mathbb{Z})$ cocycle with a finite group $Q$ into a coboundary in the following two cases in quaternion group or dihedral group:

  1. Inflate the 3-cocycle $\alpha_{1}$ in $Q=Z_2$ via a dihedral group $G=D_8$ of order 8.

  2. Inflate the 3-cocycle $\alpha_{2}$ in $Q=Z_2 \times Z_2$ via a quaternion group $G=H_8$ of order 8.

Consider the cocycle $\alpha_1(g_a,g_b, g_c) \in H^3(Z_2, \mathbb{R} /\mathbb{Z})$ and $\alpha_2((g_{a1},g_{a2}),(g_{b1},g_{b2}),(g_{c1},g_{c2})) \in H^3(Z_2 \times Z_2, \mathbb{R} /\mathbb{Z})$ in the 3rd cohomology group of $Z_2$ and $Z_2 \times Z_2$ respectively, where $g_a,g_b,g_c \in Z_2$ respectively, and $(g_{a1},g_{a2}),(g_{b1},g_{b2}),(g_{c1},g_{c2})\in Z_2 \times Z_2$ respectively.

Let both $\alpha_1$ and $\alpha_2$ to be a cup product form: $$\alpha_1(g_a,g_b, g_c)=(-1)^{g_{a}g_{b}g_{c}}. $$ $$\alpha_2((g_{a1},g_{a2}),(g_{b1},g_{b2}),(g_{c1},g_{c2}))=(-1)^{g_{a1}g_{b2}g_{c2}}. $$

question: How can we trivialize a $H^3(Z_2, \mathbb{R} /\mathbb{Z})$'s cocycle $\alpha_1$ into a coboundary and trivialize a $H^3(Z_2 \times Z_2, \mathbb{R} /\mathbb{Z})$'s cocycle $\alpha_2$ into a coboundary $$\alpha_1= \delta \beta_1$$ $$\alpha_2= \delta \beta_2$$ in a large group dihedral $D_8$ and quaternion $H_8$, respectively, by finding the explicit 2-cochain $\beta_1$ and $\beta_2$?

Where we can regard the group homomorphism $$D_8 \to Z_2 \text{ and } H_8 \to Z_2 \times Z_2. $$

  1. We can choose either the fact that $D_8/Z_4=Z_2$ or $D_8/(Z_2)^2=Z_2$. We can call $D_8=G$, and $Z_4$ or $(Z_2)^2=N$ normal subgroup, and $Z_2=Q$ as the quotient group. We all have $G/N=Q$.

  2. And we can use the fact that $H_8/Z_2=Z_2 \times Z_2$, we can call $H_8=G$, and $Z_2=N$ normal subgroup, and $Z_2 \times Z_2=Q$ as the quotient group. We all have $G/N=Q$.

In particular, not only the explicit 2-cochain, but also I am interested in finding the relations to Lyndon Hochschild Serre spectral sequence and the $d_2$ differential, its homomorphism $d_2:H^1(Q,H^1(N,\mathbb{R} /\mathbb{Z}))\to H^3(Q,\mathbb{R} /\mathbb{Z})$ in the $E_2$ pages. Or whether it requires other $d_n$ differentials to determine the inflation of cocycle.

p.s. The original post in ME received little attention so I decide to try MO.

  • 2
    $\begingroup$ Is it possible that there is no possible solution for this inflation? $\endgroup$ – wonderich Dec 10 '16 at 6:20
  • $\begingroup$ @wonderich There is such an inflation because $H^3(G,U(1))=0$. $\endgroup$ – user43326 Dec 11 '16 at 18:31
  • 2
    $\begingroup$ @user43326, Not sure what is your argument, but $H^3(G=H_{8},R/Z) =Z_8$. $\endgroup$ – wonderich Dec 11 '16 at 19:01
  • $\begingroup$ Sorry, I was thinking of cohomology with the integral coefficients. $\endgroup$ – user43326 Dec 12 '16 at 15:52
  • $\begingroup$ I didn't notice that there was a question on $D_8$. I am not familiar with the computation with the first extension, but if you use the second extension, $H^*(Z/2.Z/2)$ injects to $H^*(D_8.Z/2)$. Since the non-trivial element in $H^3(Z/2,Z/2)$ is actually integral, $H^3(Z/2,U(1))$ probably also injects to $H^3(D_8.U(1))$, which makes such inflation impossible, or am I wrong here? $\endgroup$ – user43326 Dec 12 '16 at 16:09

You can determine the possible inflation of $d$-cocycle in $G$ by lifting to a larger group $G$, from the Lyndon-Hochschild-Serre spectral sequence. For $H/K= G$, with $BG$ path connected and $\pi_1(BG)$ acting trivially on $H^*(K, U(1))$, there is a spectral sequence $\{E^{p,q}_n, d_n\}$ with $E^{p,q}_2 = H^p(G, H^q(K,U(1) ))$. The differential $d_n$: $E^{p,q}_n \to E^{p+n,q-n+1}_n$. And $E^{p,q}_{n+1}=\frac{\text{Ker} d_n}{\text{Im} d_n}$ at $E^{p,q}_{n}$.

We shall focus on the $d_2$ differential of $E_2$ page in LHS spectral sequence $$d_2: E^{p,q}_2 \to E^{p+2,q-1}_2$$ $$\Rightarrow d_2: H^p(Q, H^q(N,U(1) )) \to H^{p+2}(Q, H^{q-1}(N,U(1) )), $$ $$ d_2: H^{d-2}(Q, H^{1}(N,U(1) )) \to H^{d}(Q, H^{0}(N,U(1) ))=H^{d}(Q, U(1)). $$ If we want to trivialize the $d$-cocycle $\omega_d \in H^{d}(Q, U(1))$, we can look for a larger group $G$, where $G/N=Q$ for your $G$. However, all the examples above fail to demonstrate the successful inflation.

So the answer to your questions shall be negative. There may be loopholes, but I could not find any examples either. So this is the best conclusion so far.

  • $\begingroup$ Thanks so much, but I hope that there can be other better answers. Using other ways of trivialization. $\endgroup$ – miss-tery Dec 17 '16 at 22:23
  • $\begingroup$ I accept your answer temporarily for now, but may award to some other people if there are better answers! $\endgroup$ – miss-tery Dec 17 '16 at 22:24

From what I gathered so far (from the comment of user43326 and wonderich), and what I attempted to solve the problem based on direct searching for the trivialization, I found out only negative results --- there may not exist such inflation of $Q$-cocycle to a trivial coboundary in the given $G$.

So my answers to both questions 1 and 2 are unfortunately: No. There is no successful inflation of $Q$-cocycle to a trivial coboundary in the given $G$.

  • $\begingroup$ Please, if there are better answers and explanations afterward, I am happy to accept the later better answer as the correct answer. $\endgroup$ – miss-tery Dec 17 '16 at 20:35

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