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I recently got interested in representation theory in quantum mechanics and I read the following theorem:

Let $G$ be a simply-connected Lie group with $H^2(\mathfrak{g},\mathbb{R})=0$ and let $\mathcal{H}$ be a complex Hilbert space. Then every projective representation $\rho:G\to \text{Aut}(\mathbb{P}(\mathcal{H}))$ lifts to a unitary representation $\pi:G\to U(\mathcal{H})$.

I am looking for a proof of the theorem above, does anyone have a reference where it is proven?

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    $\begingroup$ A projective representation gives an extension of G by U(1), and so a an extension of the Lie algebra of G by R. I think the vanishing of the cohomology as shown means the Lie algebra extension is trivial, hence the group extension splits (probably using simple connectedness here), hence the projective representation lifts to an honest representation. But this needs a proper proof/reference. $\endgroup$
    – David Roberts ♦
    Commented Jun 8, 2019 at 10:27
  • $\begingroup$ The simplest example of failure without simple connectedness would be to consider the 2-dimensional obvious faithful representation of $SU(2)$ as a 2-dimensional projective representation of $G=SO(3)=SU(2)/\pm$. $\endgroup$
    – YCor
    Commented Jun 8, 2019 at 10:37
  • $\begingroup$ @DavidRoberts I think you're right about the trivial Lie algebra extension, but I am uncertain about your last point. Do you have proof of that? $\endgroup$
    – Lucas Smits
    Commented Jun 8, 2019 at 11:05
  • $\begingroup$ @LucasSmits I think it's an exercise to fill in the details. $\endgroup$
    – YCor
    Commented Jun 8, 2019 at 11:10
  • $\begingroup$ @YCor the fibre Z/2 of SU(2)->SO(3) isn't connected, and the classifying space of Z/2 is not simply connected, so the fundamental group plays a role here. But for U(1)-extensions the story is a bit different. Even if the Lie algebra intension is split one could have a flat U(1)-bundle. $\endgroup$
    – David Roberts ♦
    Commented Jun 10, 2019 at 5:54

2 Answers 2

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The original reference is Thms 3.2 and 7.1 in

Bargmann, V., On unitary ray representations of continuous groups, Ann. Math. (2) 59, 1-46 (1954). ZBL0055.10304.

Added: For a nice shortened proof, I just noticed this very commendable paper (see linked review):

Simms, D. J., A short proof of Bargmann’s criterion for the lifting of projective representations of Lie groups, Rep. Math. Phys. 2, 283-287 (1971). ZBL0232.22021.

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Since $H^2(\mathfrak{g},\mathbb{R})=0$ and since $G$ is simply connected, it follows from the van Est theorems that $H^2(G,S^1)=0,$ which means that all $S^1$- extensions of $G$ are trivial. But a lifting of a projective representation $\rho$ of $G$ is equivalent to the data of a trivialization of the $S^1$ - extension of $G$ given by the pullback (by $\rho$) of the extension $1\to S^1\to U(\mathcal{H})\to \mathbb{P}(U(\mathcal{H}))\to 1,$ which from the above van Est argument must exist.

For the relevant van Est theorems, see https://mathscinet.ams.org/mathscinet-getitem?mr=59285 or Theorem 2.3 in https://arxiv.org/abs/1909.12100

Edit: In the finite dimensional case the Lie algebra extension associated to the extension $1\to S^1\to U(\mathcal{H})\to \mathbb{P}(U(\mathcal{H}))\to 1$ splits, and the proof I gave can be tweaked to show that projective unitary representations of simply connected Lie groups always lift to unitary representations. It probably doesn't split in the case that the hilbert space is infinite dimensional and this is why the assumption that $H^2(\mathfrak{g},\mathbb{R})=0$ is needed. In this case there is a slight gap in my proof since I didn't show that the pullback of the extension is actually a smooth $S^1$ - extension of $G$, but maybe this isn't hard to show.

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