A correspondence between projective representations of $G$ with those of its universal cover

Question

Let $G$ be a connected Lie group and $\mathcal{H}$ be a Hilbert space. Let $U(\mathcal{H})$ denote the the group of all unitary operators on $\mathcal{H}$ with function composition (i.e., $\hat{U}:\mathcal{H}\to \mathcal{H}$ a linear operator satisfying $\langle \hat{U}\Phi |\hat{U}\Psi \rangle =\langle \Phi |\Psi \rangle$) and $U(1)=\lbrace e^{i\theta}:\theta \in \mathbb{R}\rbrace$. One can easily check that $U(1)$ acts on $U(\mathcal{H})$ by $(e^{i\theta},\hat{U})\equiv e^{i\theta}\hat{U}$. Then $U(\mathcal{H})/U(1)$ forms a group with group action $([\hat{U}],[\hat{U}'])\equiv [\hat{U}\circ \hat{U}']$. A projective representation of $G$ is a homoorphism $\mathcal{T}:G\to U(\mathcal{H})/U(1)$.

Assume that $\tilde{G}$ is the universal cover of $G$ with covering map $p:\tilde{G}\to G$. Then $\tilde{G}$ is a central extension of $G$ by $\pi_1 (G)$. If $\mathcal{T}:G\to U(\mathcal{H})/U(1)$ is a projective representation of $G$, then $\mathcal{T}\circ p:\tilde{G}\to U(\mathcal{H})/U(1)$ is a projective representation of $\tilde{G}$.

My question: Does for every projective representation $\mathcal{S}:\tilde{G}\to U(\mathcal{H})/U(1)$ of $\tilde{G}$ there exist a projective representation $\mathcal{T}:G\to U(\mathcal{H})/U(1)$ of $G$ such that $\mathcal{S}=\mathcal{T}\circ p$? In other words, Is there a one-to-one correspondence between projective representations of $G$ with those of $\tilde{G}$?

Answer 1

Let $G=S^1$ which has a universal cover $\mathbb R\to S^1:t\mapsto e^{2\pi it}$.

Now $\mathcal S\colon \mathbb R\to U(\mathbb C^2)/U(1)$ given by $t\mapsto \begin{pmatrix}e^{\pi i t}\\&1\end{pmatrix}$ does not factor through $p$, since if it factored as $\mathcal S=\mathcal T\circ p$ for some $\mathcal T\colon S^1\to U(\mathbb C^2)/U(1)$, then $$\mathcal T\circ p(1)=\mathcal T(e^{2\pi i})=1\ne \mathcal S(1)=\begin{pmatrix}-1\\&1\end{pmatrix}.$$