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?

  • $\begingroup$ What does the author mean by the most general central extension? Do you understand central extensions but don’t understand what “the most general one” means? Or do you not understand what a central extension is? $\endgroup$
    – Ghoster
    Commented Mar 11 at 6:05
  • $\begingroup$ Yes, I know the definiton of central extensions well. But I don't understand what the most general one is. $\endgroup$
    – Mahtab
    Commented Mar 11 at 6:19

1 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 $H_{2n+1}\cong \mathbb{R}^{2n} \oplus \mathbb{R}$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.