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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.

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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.

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  • $\begingroup$ 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? $\endgroup$
    – leob
    Commented Feb 8, 2022 at 8:48
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    $\begingroup$ @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. $\endgroup$
    – DanielC
    Commented Feb 8, 2022 at 10:33
  • $\begingroup$ 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? $\endgroup$
    – leob
    Commented Feb 8, 2022 at 11:25
  • $\begingroup$ @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. $\endgroup$
    – J. Murray
    Commented Feb 8, 2022 at 15:35
  • $\begingroup$ @J. Murray Great, thank you! $\endgroup$
    – leob
    Commented Feb 9, 2022 at 2:24

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