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In quantum mechanics, unitary projective representations play a crucial role. To be more general, I want to pose the question in sense of projective representations, then everything would follow as a special case for unitary projective representations.

A projective representation is a group homomorphism $\theta:G \to \text{PGL}(V)$. We thus want to classify projective representations, according to ACuriousMind's post. I am trying to see why $H^2(G,\mathbb{C}^{\times})$ classifies projective representations. I know that $H^2(G,\mathbb{C}^{\times})$ classifies central extensions of $G$ by $\mathbb{C}^{\times}$ up to equivalence. Thus, I am inclined to think that a projective representation is in a one-to-one correspondence with a central extension. How could one write this bijection down explicitly?

Moreover, in order to get the classification, we would have to show that equivalence classes of projective representations (for some suitable notion of equivalence- what does equivalence mean here?) are in one-to-one correspondence with equivalence classes of central extensions (here we know what equivalence means). How would one write down this bijection?

Using the above, we can conclude that equivalence classes of projective representations are in one-to-one correspondence with equivalence classes of central extensions of $G$ by $\mathbb{C}^{\times}$, which in turn is in bijective correspondence with $H^2(G,\mathbb{C}^{\times})$, which would conclude the proof.

Note that I have been told that there doesn't exist a bijection between the two following sets:

\begin{equation} \text{ProjRep}:=\{\theta:G \to \text{PGL}(V)| \theta \; \; \text{is a group homomorphism} \} \; \; \text{and} \; \; H^2(G,\mathbb{C}^{\times}). \end{equation} This is why I am looking for a bijection between $H^2(G,\mathbb{C}^{\times})$ and equivalence classes of $\text{ProjRep}$ under a suitable equivalence relation.

ProphetX
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  • I don't understand what you want as an answer to this question. My post shows how to associate a $C\in H^2(G,\mathrm{U}(1))$ with a projective representation, and constructs a linear representation $\sigma_C$ of a central extension $G_C$, so there is an obvious map $C\mapsto (G_C,\sigma_C)$ between projective representations and linear representations of central extensions. What is the question? What do you think is not "explicit" about this? – ACuriousMind Apr 11 '23 at 18:50
  • To every projective representation, we can associate an element in $H^2(G,U(1))$. How to see the converse? Suppose $\alpha \in H^2(G,U(1))$. To which projective representation does $\alpha$ correspond to? Are these associations inverses of each other? – ProphetX Apr 11 '23 at 18:56
  • I am concretely looking for a bijection between (equivalence classes) of projective representations and $H^2(G,\mathbb{C}^{\times})$, as a map, such as:

    $T:\text{ProjRep}/\sim \to H^2(G,\mathbb{C}^{\times})$, $T([\theta])=...$, for some suitable notion of $\sim$.

     – ProphetX Apr 11 '23 at 18:58
  • I also state that in the post: "Conversely, every unitary representation $\rho$ of some $G_C$ gives a pair $\Sigma,C$ by $\Sigma(g) = \alpha^{-1}\rho(g,\alpha)$." (careful, that's the $\alpha$ from my post, not your comment) Obviously, $C$ is still defined by $\Sigma(g)\Sigma(h) = C(g,h)\Sigma(gh)$. It's not hard to show that this is the inverse of the construction $C\mapsto (G_C,\sigma_C)$. – ACuriousMind Apr 11 '23 at 19:00
  • I thought it would be easier to construct a map $Q: \text{ProjRep}/\sim \to \text{CentrExt}/\sim$, where CentrExt denotes the set of all central extensions by $C^{\times}$ and $\sim$ is the notion of equivalence of central extensions. If we had $Q$, we could use the isomorphism between equivalence classes of central extensions and $H^2(G,\mathbb{C}^{\times}$ immediately to obtain the result desired. – ProphetX Apr 11 '23 at 19:01
  • The notion of equivalence is the completely standard notion: Just like for linear representations, two projective representations $\pi_1,\pi_2 : G\to \mathrm{PU}(V)$ are isomorphic when there is an intertwiner $A\in\mathrm{PU}(V)$ such that $A\pi_1(g) = \pi_2(G)A$ for all $g\in G$. – ACuriousMind Apr 11 '23 at 19:02
  • Right. So we have a set: $\text{ProjRep}/\sim$, where $\sim$ is the equivalence relation you just defined. Now we want to look for a bijective map:

    $$\text{ProjRep}/ \sim \to H^2(G,U(1))$$. Could you please write down this map explicitly and leave it as an exercise for me to check this is an isomorphism? I will try to do it.

     – ProphetX Apr 11 '23 at 19:03
  • Again, the map is already in my post, in both directions. For any projective representation, you get the $C\in H^2$ just by choosing any unitary representatives $\Sigma$ and getting $C(g,h)$ as defined in my post. Conversely, for any $C\in H^2$, just construct $(G_C,\sigma_C)$, then define $\Sigma(g) = \sigma_C(g,1)$. $\Sigma : G\to\mathrm{U}(V)$, which yields a projective representation when you concatenate it with the projection $\mathrm{U}(V)\to\mathrm{PU}(V)$. – ACuriousMind Apr 11 '23 at 19:26
  • That's right, after some thought, I understood my confusion. Indeed, everything is in your post. Is there a way to close this post/am I allowed to delete it? – ProphetX Apr 12 '23 at 22:12

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