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Let $G$ be a finite group and let $D(g)$ be a projective representation of $G$ i.e. \begin{equation} D(g) D(h) = e^{i \omega(g,h)} D(gh) \end{equation} These can be classified by the equivalence relation $\omega(g,h) \sim \omega(g,h)+\theta(g)+ \theta(h) - \theta(gh)$ subject to the condition $\omega(g,h)+\omega(gh,l)-\omega(h,l)-\omega(g,hl) = 0$. The distinct equivalent classes of projective representations are labeled by elements of $H^2(G,U(1))$. I know that for every such finite group, there is atleast one finite covering group C with the property that every projective representation of G can be lifted to an ordinary representation of C [1].

My question is about the relationship between the irreducible representations (irreps) of C and the group $H^2(G,U(1))$. Specifically, is the following statement true?:

Every irrep $\Gamma_i$ of C can be associated an element of $\nu \in H^2(G,U(1))$ like $\Gamma^\nu_i$. The group property of $H^2(G,U(1))$ is reflected in the Clebsch-Gordan decomposition of tensor product of irreps of C: \begin{equation} \Gamma^\nu_i \otimes \Gamma^\mu_j \cong \bigoplus_k \Gamma^{\nu+\mu}_k \end{equation}

I have noticed that this is true for all cases I have seen when $ H^2(G,U(1)) \cong \mathbb{Z}_2$ like $G = \mathbb{Z}_2 \times \mathbb{Z}_2$, $C = D_8$ but I am unsure if this is true in general.

[1] https://en.wikipedia.org/wiki/Schur_multiplier#Relation_to_projective_representations

PS: I am a physicist and my interest in the above question comes from its use in condensed matter physics.

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    $\begingroup$ I do not understand your question at all. Could you explain it in the case $H^2(G,U(1))=0$? There are a number of finite groups with that property. $\endgroup$
    – abx
    Jul 22 '18 at 18:50
  • $\begingroup$ Consider $G= \mathbb{Z}_2$ which has $H^2(G,U(1)) =0$. In this case, the covering group is $\mathbb{Z}_2$ itself. All irreducible representations of $\mathbb{Z}_2$ are labeled by the trivial element of $H^2(G,U(1))$. $\endgroup$
    – sawd
    Jul 22 '18 at 19:10
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The answer to your question is Yes. Consider your covering group $C$ as a central extension: $$1 \to N \to C \to G \to 1$$ and suppose it is given by a 2-cocycle $\alpha \in H^2(G, N)$. Then for any homomorphism $f: N \to \mathbb{C}^\times$, we get a 2-cocycle $f \circ \alpha \in H^2(G,\mathbb{C}^\times)$ by composition of maps, and this yields a homomorphism $$\operatorname{Hom}(N,\mathbb{C}^\times) \to H^2(G,\mathbb{C}^\times).$$ You are assuming that $C$ is a "representation group", which amounts to the property that this homomorphism is an isomorphism. Thus, any projective representation of $G$ lands in a commutative diagram $$\require{AMScd} \begin{CD} 1 @>>> N @>>> C @>>> G @>>> 1 \\ @. @VVV @VVV @VVV \\ 1 @>>> \mathbb{C}^\times @>>> GL(V) @>>> PGL(V) @>>> 1 \end{CD}$$ The 2-cocycle for your projective representation corresponds to the 1-dimensional representation of $N$ given by the leftmost vertical arrow. Any tensor product of representations of $C$ induces a tensor product of 1-dimensional representations of $N$. Tensor product of 1-dimensional representations of $N$ yields the group law on $\operatorname{Hom}(N,\mathbb{C}^\times)$, and by our assumption, this is the group law on $H^2(G, \mathbb{C}^\times)$.

I found all the statements I needed in this exposition by Mendonca.

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