Classifying projective representations

Question

Blagoje Oblak in their thesis "BMS particles in three dimensions” says that

"Given a group $G,$ suppose we wish to find all its projective unitary representations. The above considerations [in the thesis] provide an algorithm that allows us, in principle, to solve the problem:

1. First find the universal cover $\tilde{G}$ of $G$ to take care of topological central extensions.

2. Then find the most general central extension $\hat{\tilde{G}}$ of $\tilde{G}$ in order to take care of differentiable central extensions.

3. Finally, consider an exact unitary representation of $\hat{\tilde{G}}$; any projective unitary representation of $G$ may be seen as a representation of that type.

What I know:

(a) If $T:G\to U(PH)$ is a unitary projective representation (i.e. a map $T:G\to U(H)$ with $T(f)T(g)=e^{iC(f,g)}T(fg)$ for all $f,g\in G$), then $C:G\times G\to \mathbb{R}$ is a real 2-cocycle on $G$, i.e. $[C]\in H^2 (G;\mathbb{R})$.

(b) $H^2 (G;\mathbb{R})$ classifies central extensions of $G$, i.e. for every $[C]\in H^2 (G;\mathbb{R})$, $\hat{G}_C \equiv G\times \mathbb{R}$ with group action $(f,\lambda)\cdot (g,\mu ):=(fg,\lambda +\mu +C(f,g))$ is a central extension of $G$ by $\mathbb{R}$, if $\hat{G}$ is a central extension of $G$ by $\mathbb{R}$, then there exists $[C]\in H^2 (G;\mathbb{R})$ such that $\hat{G}=\hat{G}_C$, and $[C]=[C']$ iff $\hat{G}_C$ and $\hat{G}_{C'}$ are equivalent as central extensions. The same holds for $H^2 (\mathfrak{g};\mathbb{R})$ for a Lie algebra $\mathfrak{g}$.

(c) If $T:G\to U(PH)$ is a unitary projective representation of $G$ (with 2-cocycle $C$ as a phase), then there exists unitary representation $\hat{T}:\hat{G}_C \to U(H)$ such that $T\pi =\gamma \hat{T}$ in which $\pi :\hat{G}_C \to G$ and $\gamma :U(H)\to U(PH)$.

(d) By Bargmann’s theorem, if $H^2 (\tilde{\mathfrak{g}};\mathbb{R})=0$ (where $\tilde{\mathfrak{g}}$ is the Lie algebra of $\tilde{G}$), then there is one-to-one correspondence between unitary representations of $\tilde{G}$ and unitary projective representations of $G$.

My questions:

(i) What does the author mean by the most general central extension $\hat{\tilde{G}}$ of $\tilde{G}$? And how we can do steps (2) and (3)?

(ii) If $C$ is a real 2-cocycle on $G$, can we build a unitary representation $T:G\to U(PH)$? (I mean does the converse of (a) hold?)

(iii) If $H^2 (\tilde{\mathfrak{g}};\mathbb{R})=\neq 0$ in (d), what can we do?

Answer 1

(i): $\hat{\tilde{G}}$ here means universal central extensions of $\tilde{G}$. That is, if $\hat{\tilde{G'}}$ if another central extension of $\tilde{G}$ (by the same group $\mathbb{R}$) then there exists a unique isomorphism $\hat{\tilde{G}} \xrightarrow{\sim} \hat{\tilde{G'}}$

(ii): If $[C] \sim [0] \in H^2 (G;\mathbb{R})$ then we can always take $C(g,h)=\alpha(gh) − \alpha( g ) − \alpha(h)$. for $\alpha: G \to \mathbb{R}$. Then $e^{i\alpha}T$ will give a (non-projective) unitary representation of $G$

(iii): Let's answer why vanishing 2nd cohomology is important for the proof of Bargmann’s theorem first: Given $G$ a connected, simply connected, finite-dimensional Lie group (or one could also pass to the universal cover). Define the $1$-dimensional central extension $\hat{G}$ of $G$: $$ \hat{G}=\{(g,A)\mid\pi(A)=T(g)\} $$ where $\pi:GL(H) \to PGL(H)$ the canonical surjection and $T:G\to U(PH)$ an unitary projective rep of $G$. Now the map $\hat{T}: \hat{G} \to GL(H)$ with $\hat{T}(g,A)=A$ is a non-projective representation of $\hat{G}$.

At the level of Lie Algebra, $\hat{\mathfrak{g}}$ is also an $1$-dimensional central extension of $\mathfrak{g}$. Now since $H^2(\mathfrak{g},\mathbb{R})$ classifies central extensions of $\mathfrak{g}$ by $\mathbb{R}$, we use the fact $H^2(\mathfrak{g},\mathbb{R})=0$ to show that the central extension by must be $\hat{\mathfrak{g}}=\mathfrak{g} \oplus \mathbb{R}$. By Lie group–Lie algebra correspondence, then $\hat{G}=G \times \mathbb{R}$ and the short exact sequence splits which now we are given the lift $\sigma: G \to \hat{G}$. We see now then $\hat{T} \circ \sigma$ is a non-projective representation of $G$.

If $H^2(\mathfrak{g},\mathbb{R}) \neq 0$ then we the central extension above will not be trivial and we can not find the lift that we wanted. For an example in physics, take $G=\mathbb{R}^2$ then also $\mathfrak{g}=\mathbb{R}^2$ with $H^2(\mathfrak{g},\mathbb{R})= \mathbb{R} \neq 0$. Consider the unitary projective representation which is translation in position and momentum in $1$-D for quantum particles: $$T:\mathbb{R}^2 \to PGL(L^2(\mathbb{R}^2))$$ such that $T(p,q) \psi(x)=e^{ipx} \psi(x-q)$ for $(p,q) \in \mathbb{R}^2$. This $T$ is projective with $C \big( (p,q),(p',q') \big) =e^{-ip'q}$. Now this can't be lifted to a non-projective representation of $\mathbb{R}^2$ because otherwise $T(p,q)$ and $T(p',q')$ must commute since $\mathbb{R}^2$ is commutative, but we know that translation in position and momentum can't commute or simply by checking the expression.

That's why in QM we just take the Heisenberg Lie algebra $H_3 \cong\mathbb{R}^2 \oplus \mathbb{R}$ which is a trivial extension of $\mathbb{R}^2$ such that it's possible to find a lift to non-projective representation. The same argument can be generalized to $H_{2n+1}\cong \mathbb{R}^{2n} \oplus \mathbb{R}$