# Why should a projective representation of a classical symmetry lift to a non-projective representation?

Background

Take a classical system with symmetry $$G$$. Suppose we can quantize this to a quantum system with Hilbert space $$\mathcal{H}$$. The state space of the quantum system is given by the projective Hilbert space $$\mathbb{P} \mathcal{H}$$. A quantization of the symmetry $$G$$ (if it exists) is a projective representation $$\rho: G \rightarrow U \left( \mathbb{P} \mathcal{H} \right)$$ into the unitary operators on $$\mathbb{P} \mathcal{H}$$.

Question

From what I have read, it seems like we want (or physically expect?) this $$\rho: G \rightarrow U \left( \mathbb{P} \mathcal{H} \right)$$ to lift to a representation $$\hat{\rho}: G \rightarrow U \left( \mathcal{H} \right)$$. Why is this?

Motivation

I am trying to understand why, in certain circumstances, we quantize the central extension of $$G$$, rather than $$G$$ itself. For instance in CFT, the symmetry group $$G = \text{Conf} \left( \mathbb{R}^{1, 1} \right)$$ quantizes to a group whose Lie algebra is the Witt algebra. However in this situation we quantize the central extension of $$G$$ to arrive at the Virasoro algebra instead.

A quantization of the symmetry $$G$$ (if it exists) is a projective representation $$\rho:G\rightarrow U(\mathbb P\mathcal H)$$ into the unitary operators on $$\mathbb P\mathcal H$$.

To clarify - $$\mathbb P\mathcal H$$ is not a vector space, and so I'm not so sure what a unitary operator on $$\mathbb P\mathcal H$$ would be. My assumption is that you are referring to the ray product $$(\Phi,\Psi)$$ between elements of $$\mathbb P\mathcal H$$, defined via $$(\Phi,\Psi):= \frac{|\langle\phi,\psi\rangle|}{\Vert\phi\Vert \ \Vert\psi\Vert}$$ where $$\phi,\psi\in \mathcal H$$ are any two representatives of the equivalence classes $$\Phi$$ and $$\Psi$$. A symmetry transformation is then a map $$T:\mathbb P\mathcal H \rightarrow \mathbb P\mathcal H$$ such that $$(T\Phi,T\Psi)=(\Phi,\Psi)$$. This looks rather like the definition of a unitary operator on a Hilbert space, but there is no notion of linearity present here.

Wigner's theorem tells us that any such $$T$$ can be induced by some unitary or antiunitary operator on $$\mathcal H$$, which descends to a transformation on $$\mathbb P\mathcal H$$ in the obvious way. By extension, if we want to represent the action of a symmetry group $$G$$ rather than just a single transformation, we might seek linear representations $$\rho:G\rightarrow U(\mathcal H)$$. However, requiring $$\rho$$ to be a true group homomorphism is generally too strong; if it is a homomorphism up to a phase - i.e. $$\rho(g)\rho(h)=\rho(gh)C(g,h)$$ for some phase factor $$C(g,h)$$ - then it also induces a well-defined group action on $$\mathbb P\mathcal H$$ because the extra phase factor is lost during the projectivization from $$\mathcal H$$ to $$\mathbb P\mathcal H$$.

The takeaway is that the action of a symmetry group $$G$$ on $$\mathbb P\mathcal H$$ can be understood through a projective unitary representation of $$G$$ on $$\mathcal H$$, so we need to understand the latter if we want to understand how to implement symmetry groups in our theories.

From what I have read, it seems like we want (or physically expect?) this $$\rho$$ to lift to a representation $$\hat{\rho}: G \rightarrow U \left( \mathcal{H} \right)$$

If $$\rho:G\rightarrow \mathrm{Aut}(\mathbb P\mathcal H)$$ is a group action whose elements are symmetry transformations as defined above, then Wigner's theorem tells us that that it lifts to a projective unitary representation $$\hat \rho:G\rightarrow \mathbb PU(\mathcal H)$$. It may lift to a genuine representation - i.e. one might be able to arrange that the phase factors $$C(g,h)$$ referenced above are all equal to $$1$$ - but in general this cannot be done.

I am trying to understand why, in certain circumstances, we quantize the central extension of $$G$$, rather than $$G$$ itself.

As it turns out, projective representations of $$G$$ can be put into one-to-one correspondence with (genuine) linear representations of its central extensions (the same is true for the Lie algebra $$\mathfrak g$$). Since linear representations are far nicer to work with, this is what we tend to study.

ACuriousMind has written a wonderful answer on this subject which you can find here. I doubt any summary I could provide would do it justice, so I suggest reading it directly if you haven't already done so.

• Thanks @J. Murray. This explains why I couldn't find any explanation of why $\rho$ should lift in general. I know that if every projective rep of $G$ doesn't necessarily lift to a genuine rep, we can instead quantize the central extension $\tilde{G}$ of $G$ to resolve this annoyance. So is moving from a theory with symmetry $G$ to $\tilde{G}$ simply a case of convenience? Are there no other motivating reasons?
– leob
Feb 8, 2022 at 8:48
• @leob Quantum theories reside in the superposition principle which we cannot attain on projective space. We are forced to seek representations on linear (topological) spaces, thus we need Bargmann's-Mackey theory of lifting. Feb 8, 2022 at 10:33
• Ok, so let me see if I understand. The projective space $\mathbb{P} \mathcal{H}$ only gives us pure states of the quantum system. A projective unitary representation $\hat{\rho}: G \rightarrow U \left( \mathbb{P} \mathcal{H} \right)$ describes a symmetry of these pure states. However because we can have arbitrary superpositions of pure states, this symmetry must ALWAYS lift to a unitary representation $\rho: G \rightarrow U \left( \mathcal{H} \right)$. Is this correct?
– leob
Feb 8, 2022 at 11:25
• @leob I think I misunderstood some aspects of your question upon first reading it, so I've reworked my answer. Let me know if the edit does not address your question. Feb 8, 2022 at 15:35
• @J. Murray Great, thank you!
– leob
Feb 9, 2022 at 2:24