# $G$ cocycle split to a coboundary in $J$, via a group extension

Consider a generic nontrivial $d$-cocycle $\omega_d^G \in H^d(G,U(1))$ in the cohomology group of a group $G$ with $U(1)=\mathbb{R}/\mathbb{Z}$ coefficient. In otherwords, here the $d$-cocycle $\omega_d^G$ is a complex $U(1)=\mathbb{R}/\mathbb{Z}$ function with the norm $|\omega_d^G|=1$ but with a $U(1)$ complex phase satisfying the cocycle condition $\delta\omega_d^G=1$.

question: We like to ask whether there always exists some Abelian group $N$ as a normal subgroup of some bigger group $J$, such that $G$ is the quotient group $$\frac{J}{N}=G$$ and such that we can always trivialize (or split) the $d$-cocycle $\omega_d$ of $G$ into $d$-coboundary if we lift $G$ into a larger group $J$? Given that we know the group homomorphism $r$: $$J \overset{r}{\rightarrow} G.$$ Namely, $$\omega_d^G(g_i,\dots)=\omega_d^G(r(j_i),\dots)= \delta \beta_{d-1}^J(j_i,\dots).$$ with $g \in G, j \in J$.

You are welcome to comment or answer the partial case, for example, when $d=2$, $d=3$ or $d=4$, and when $G$ is a finite group. Partial comments or answers are still welcome!

• In case $d=2$, a two cocycle will give you a short exact sequence of the form $1\to U(1)\to J\to G\to 1$. This $J$ will satisfy the requirement. – Ehud Meir Sep 20 '16 at 17:53
• @Ehud Meir, In this case, can we just consider $J=U(1) \times G$? Please comment on it, thanks. – miss-tery Dec 7 '16 at 3:12

In case d=1, the answer is always negative: 1-cocycles are homomorphisms, 1-coboundaries are always trivial, and inflation is injective. If you do not restrict yourself to the case where $N$ is abelian, the answer is positive: take a short exact sequence $1\to R\to F\to G\to 1$ where $F$ is free. For the abelian case we proceed as follows: you also have the short exact sequence $1\to R_{ab}\to F/[R,R]\to G\to 1$ and you have a morphism of short exact sequences from the first one to the second one. This also gives you a morphism on the corresponding Lyndon-Hochschild-Serre spectral sequences. In particular we have the following commutative diagram: $\require{AMScd}$\begin{CD}H^{d-2}(G,H^1(R_{ab}, U(1)))@> >> H^d(G,U(1))\\ @V V V @V V V \\ H^{d-2}(G,H^1(R,U(1)))@> >> H^d(G,U(1))\end{CD} For $d>2$ the lower horizontal morphism is an isomorphism since the spectral sequence converges to the cohomology of F, which is trivial. The right vertical arrow is the identity, and the left vertical arrow is an isomorphism, since all homomorphisms from $R$ to $U(1)$ factor through $R_{ab}$. But this implies that also the upper horizontal homomorphism is an isomorphism, and that the inflation from $G$ to $F/[R,R]$ is trivial. So this extension trivializes all cocycles at the same time. For $d=2$ we can similarly show that the upper horizontal morphism is surjective, and this is enough for the vanishing of the second cohomology. Another option: a two cocycle will give rise to an extension $$1\to U(1)\to J\to G\to 1.$$ The inflation of the cocycle to $J$ will be zero.
• @Ehud Meir, what is your notation $R_{ab}$? How do you define it? – miss-tery Dec 5 '16 at 19:02
• @miss-tery, $R_{ab}$ is the abelianization of $R$, so $R_{ab}= R/[R,R]$. – Ehud Meir Dec 6 '16 at 8:42
• @Ehud Meir, it looks to me that Lyndon-Hochschild-Serre spectral sequence choose $$H^p(G/N, H^q(N,A)) \to H^{p+q}(G,A),$$ so in your case, shoule one consider $$H^{d-q}(G/N, H^q(N,A)) \to H^{d}(G,A)$$ instead? You have a particular way of choice, I am curious why. – wonderich Dec 7 '16 at 4:10
• The horizontal maps in my diagram are the one arising from the differentials in the $E_2$ page of the spectral sequence. That's why the indices are the way they are. – Ehud Meir Dec 7 '16 at 13:28