vanhees71 said:
The Poincare group allows you to do that, because any unitary ray representation can be equivalently lifted to a unitary representation
Technically, this is not correct. Given a Lie group G and its Lie algebra \mathfrak{g}, then every projective unitary representation \rho : G \to \mbox{U}(\mathbb{P}\mathcal{H}) , lifts to a unique unitary representation U : G \to \mbox{U}(\mathcal{H}) , if the following two conditions hold: 1) G is
simply connected, and 2) the second cohomology group of \mathfrak{g} is
trivial, i.e., \mbox{H}^{2}(\mathfrak{g}, \mathbb{R}) = 0.
The Poincare’ group \mathbb{R}^{(1,3)} \rtimes \mbox{SO}(1, 3) satisfies the second condition (this is why we were able to eliminate the central charges from algebra) but not the first (it is connected but not simply connected). However, the 2 to 1 covering map \varphi : \mathbb{R}^{(1,3)} \rtimes \mbox{SL}(2, \mathbb{C}) \to \mathbb{R}^{(1,3)} \rtimes \mbox{SO}(1, 3) \ \left( \cong \frac{ \mathbb{R}^{(1,3)} \rtimes \mbox{SL}(2, \mathbb{C})}{\{ (0 , \pm I)\}}\right) , is also a homomorphism whose kernel \{ (0 , \pm I)\} coincide with the centre of \mathbb{R}^{(1,3)} \rtimes \mbox{SL}(2 , \mathbb{C}). In other words, we have the following short exact sequence of groups and homomorphisms 1 \rightarrow \{ (0 , \pm I ) \} \overset {i}{\hookrightarrow} \mathbb{R}^{(1,3)} \rtimes \mbox{SL}(2, \mathbb{C}) \overset {\varphi}{\rightarrow} \mathbb{R}^{(1,3)} \rtimes \mbox{SO}(1, 3) \rightarrow 1 . This simply means that the group in the middle (the universal covering group) is the central extension of the Poincare’ group \mathbb{R}^{(1,3)} \rtimes \mbox{SO}(1, 3) by the group \{ (0 , \pm I )\}. Now, any (irreducible) projective unitary representation \pi : \mathbb{R}^{(1,3)} \rtimes \mbox{SO}(1, 3) \to \mbox{U}(\mathbb{P}\mathcal{H}) , induces (an irreducible) projective unitary representation of the universal covering group given by the following composition of homomorphisms \pi \circ \varphi = \hat{\pi} : \mathbb{R}^{(1,3)} \rtimes \mbox{SL}(2, \mathbb{C}) \to \mbox{U}(\mathbb{P}\mathcal{H}) . This, in turn, lifts to (an irreducible) unitary representation U : \mathbb{R}^{(1,3)} \rtimes \mbox{SL}(2, \mathbb{C}) \to \mbox{U}(\mathcal{H}) \ , because \mathbb{R}^{(1,3)} \rtimes \mbox{SL}(2, \mathbb{C}) is simply connected and the second cohomology group of its Lie algebra is trivial. Indeed, there is a
bijective correspondence between the (irreducible) continuous
projective unitary representations of the connected
Poincare’ group \mathbb{R}^{(1,3)} \rtimes \mbox{SO}(1, 3) and the (irreducible) continuous
unitary representation of the simply connected
covering group \mathbb{R}^{(1,3)} \rtimes \mbox{SL}(2, \mathbb{C}).
Finally, recall that the quotient (or canonical projection) map p : \mbox{U}( \mathcal{H}) \to \mbox{U}( \mathbb{P}\mathcal{H}) \ \left( \cong \frac{\mbox{U}( \mathcal{H})}{ \mbox{U}(1)} \right), with the centre \mathcal{Z}\left(\mbox{U}(\mathcal{H})\right) = \big\{ \lambda \ \mbox{id}_{\mathcal{H}}; | \lambda | = 1 \big\} identified with \mbox{U}(1), gives us the following short exact sequence of groups and homomorphisms 1 \rightarrow \mbox{U}(1) \overset{i}{\hookrightarrow} \mbox{U}(\mathcal{H}) \overset{p}{\rightarrow} \mbox{U}(\mathbb{P}\mathcal{H}) \rightarrow 1 . This means that the unitary group of \mathcal{H} \big(i.e., \mbox{U}(\mathcal{H})\big) is the central extension of the projective unitary group \mbox{U}(\mathbb{P}\mathcal{H}) by the group \mbox{U}(1) \big( it is, at the same time, a locally trivial principal \mbox{U}(1)-bundle over \mbox{U}(\mathbb{P}\mathcal{H})\big). Now, if you put the above two exact sequences on top of each other, you obtain a commutative diagram (which I don’t know the correct command for it on here), because one can show that the projective representation \hat{\pi} factors according to \pi \circ \varphi = \hat{\pi} = p \circ U .
