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I was trying to understand this interesting question by example.

Let me follow their previous discussion and ask: Let a generic nontrivial 2-cocycle $\omega_2^G(g_1,g_2) \in H^2(G,\mathbb{R}/\mathbb{Z})$ 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.

I like to ask how can we trivialize the 2-cocycle $\omega_2(g_1,g_2)$ of $G$ into 2-coboundary if we lift $G$ into a larger group $J$, and given that we know the group homomorphism $r$: $$J \overset{r}{\rightarrow} G.$$ In particular I like to focus on:

$$SU(2) \overset{r}{\rightarrow} SO(3).$$

So that
$$\omega_2^J(j_1,j_2)=\omega_2^G(r(j_1),r(j_2))=\omega_2^G(g_1,g_2) \text{ is trivial in } H^2(J,U(1)).$$ Namely $\omega_2^G(r(j_1),r(j_2))$ becomes a 2-coboundary in $H^2(J,\mathbb{R}/\mathbb{Z})$ for the cohomology group of $J$, but $\omega_2^G(g_1,g_2)$ originally was not a 2-coboundary but was a 2-cocycle for the cohomology group of $G$. We can explicitly write $$ \omega_2^G(g_1,g_2)=\omega_2^G(r(j_1),r(j_2))= \frac{\beta_1^J(j_2)\beta_1^J(j_1)}{\beta_1^J(j_1 j_2)}. $$ Here $\beta_1^J(j_1)$ is a 1-cochain for $j_1, j_2 \in J$, and that $g_1=r(j_1)$, $g_2=r(j_2)\in G$.

Is that true that the 2-cocycle in $$H^2[SO(3),\mathbb{R}/\mathbb{Z}]=\mathbb{Z}/2\mathbb{Z}$$ for $G=SO(3)$ can be trivialize in $J=SU(2)$? How can it be shown? Here we consider the cohomology group of the Lie group $SO(3)$ and $SU(2)$.

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    $\begingroup$ What sort of cohomology are you taking? Do you mean continuous (or smooth) cohomology, or the group cohomology of the underlying group? $\endgroup$ – David Roberts Nov 4 '16 at 12:52
  • $\begingroup$ The question you link to seems to consider arbitrary discrete groups, whereas you have Lie groups. This is kinda different! $\endgroup$ – David Roberts Nov 4 '16 at 12:53
  • $\begingroup$ thanks, I mean the "group cohomology of the underlying group." $\endgroup$ – miss-tery Nov 4 '16 at 14:02
  • $\begingroup$ Isn't it the case that the extension $$1\to \mathbb{Z}/2\to SU(2)\to SO(3)\to 1$$ corresponds to the nontrivial element in $H^2(SO(3),\mathbb{R}/\mathbb{Z})$? If so, it is more or less tautological that the cocycle trivializes, and it is also quite clear how to write the cochain $\beta$. $\endgroup$ – Ehud Meir Nov 12 '16 at 12:34
  • $\begingroup$ Does your example only work in $H^2$ or does it work for $H^n$ for other $n$? If it is clear, in either cases, you can write it as an answer even if it is trivial to you, it is non trivial to me still! Thanks. $\endgroup$ – miss-tery Nov 12 '16 at 20:53
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Let $G$ be a group, and let $$1\to A\to J\to G\to 1$$ be an extension of groups with an abelian kernel. Choose a set-theoretical lifting $s:G\to J$ of the quotient map $p:J\to G$. Now define a function $\beta:G^2\to A$ by the formula $$\alpha(g,h) = s(g)s(h)s(gh)^{-1}.$$ This formula defines a two cocycle, and this is the two cocycle which corresponds to the above extension. Let us now show that the inflation of $\alpha$ to $J$ is trivial: define $\beta(j) = j\cdot((sp)(j))^{-1}\in A$. Then a direct calculation shows that $\delta\beta=inf(\alpha)$.

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  • $\begingroup$ how would you express the 2-cocycle for $H^2[SO(3),R/Z]$? $\endgroup$ – miss-tery Nov 14 '16 at 0:54
  • $\begingroup$ What is $sp$ in your expression? $\endgroup$ – miss-tery Dec 2 '16 at 3:00
  • $\begingroup$ I am not sure whether you can explain this $$\beta(j) = j\cdot((sp)(j))^{-1}\in A$$, your notation confuses me -- but I will accept it as an answer -- I hope you can fill in sometime. $\endgroup$ – miss-tery Dec 2 '16 at 3:01
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    $\begingroup$ what I mean was the following: $sp$ was just the concatenation of the functions s and p. The inverse is the inverse in the group $J$. $\endgroup$ – Ehud Meir Dec 2 '16 at 10:48
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    $\begingroup$ A related but harder challenging question: mathoverflow.net/questions/265953 $\endgroup$ – wonderich Mar 30 '17 at 21:06

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