Timeline for Why exactly do sometimes universal covers, and sometimes central extensions feature in the application of a symmetry group to quantum physics?
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29 events
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| Nov 22 at 9:47 | comment | added | Kai | @Mahtab No there is not a unique one, I believe the bijection is between the set $\{$projective reps of $G\}$ and the set $\{$linear reps of $G_C | C \in H^2(G,U(1))\}$. Each projective rep corresponds uniquely to a linear rep of $G_C$ for some $C$. Different projective reps may correspond to different choices of $C$. | |
| May 13 at 11:56 | comment | added | Mahtab | @ACuriousMind By your explanations, every projective representation $T$ of $G$ is related to a cocycle $C$ and every cocycle $C$ is related to a central extension $G_C$ and finally to a linear representation $\hat{T}$ of $G_C$. Hence as $T$ changes, $G_C$ is changes. My question is that is there a unique central extension $\hat{G}$ of $G$ such that every projective representation of $G$ is in one-to-one correspondence to linear representations of $\hat{G}$? | |
| May 4 at 11:07 | comment | added | ACuriousMind♦ | @Mahtab I'm a bit confused by your comment - that bijection is exactly what this answer attempts to explain, by constructing linear representations from the pairs $(\Sigma, C)$ and vice versa. Which steps are unclear? | |
| May 4 at 10:46 | comment | added | Mahtab | @ACuriousMind Thank you very much for the nice explanation. Could you please explain to me a little bit more this fact: projective representations are in bijection to linear representations of central extensions. This is a big problem for me and I can't prove it. I mean how can I build a linear representation from projective one and converse, and why ther are in a one-to-one corespondence? | |
| Apr 25 at 13:07 | comment | added | SolubleFish | 2. I figured out the link I was looking for : given a finite abelian group and a morphism $\varphi:A\to U(1)$, there is a corresponding map $H^2(G,A) \to H^2(G,U(1))$. The construction of the central extension can be applied with $A$ (constructing some covering spaces). For the $U(1)$ cocycles $C$ which come from $A$, we see that the linear unitary representation restricts from $G_C$ to the covering space associated with the $A$-cocycle. | |
| Apr 25 at 9:50 | comment | added | SolubleFish | 1. I don't think that this is true in general : even if $G$ is simple and compact, there are representations of the trivial extension $G\times U(1) \to U(\mathcal H)$ where the $U(1)$ subgroup is not sent to $U(1)$ via the identity morphism (or where its image is not even central). | |
| Apr 25 at 9:47 | history | edited | SolubleFish | CC BY-SA 4.0 |
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| Apr 24 at 21:59 | comment | added | ACuriousMind♦ | @SolubleFish 1. Yes, that should be the case simply because representations map central subgroups to central subgroups, and the center of $U(n)$ is $U(1)$. 2. The discrete extensions come not from the logic in this answer but simply from the general Lie theory where there's a simply-connected Lie group corresponding to a Lie algebra and any non-simply-connected Lie group is covered by it (and the kernel of the covering map is central); this also gives a projective representation of the covered group and it's a standard Lie theory result that all such finite extensions appear in this way. | |
| Apr 24 at 18:31 | comment | added | SolubleFish | Also, at what point in that derivation do the coverings (i.e. the central extension by discrete/finite abelian groups/subgroups of $U(1)$) appear ? | |
| Apr 24 at 18:29 | comment | added | SolubleFish | When reconstructing the pair $\Sigma,C$ from a unitary representation $\rho$ of a central extension $L$ of $G$ by $U(1)$, we first find the cocycle $C$ from the group extension by identifying $L$ with $G_C$. But then, to find $\Sigma$, we need $\alpha^{-1}\rho(g,\alpha)$ to not depend on $\alpha$, i.e. we need $\rho$ to map the central $U(1)$ in $L\simeq G_C$ isomorphically to the diagonal $U(1)\subset U(\mathcal H)$. Can we prove that this needs to be the case or do we need to add this as an assumption ? | |
| Feb 24 at 14:24 | comment | added | Mahtab | Hello dear @ACuriousMind . Could you please give me a good reference about details of your nice answer? I am really curious about learning projective representations and their relation with central extensions and cohomology in detail. Thanks in advance. | |
| Feb 7 at 11:36 | comment | added | ACuriousMind♦ | @astronautgravity see this answer of mine, in particular the part about the semi-direct product and Maschke's theorem | |
| Feb 7 at 0:26 | comment | added | astronautgravity | This is an excellent answer. Thank you very much! How do things change if the symmetry group is disconnected, e.g. for O(3)? | |
| Jan 17 at 3:23 | comment | added | lucabtz | this Q&A is great, very useful! I think you missed a $c \neq 0$ in the definition of projective Hilbert space though | |
| Jul 28, 2023 at 10:17 | history | edited | ACuriousMind♦ | CC BY-SA 4.0 |
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| Jul 28, 2023 at 10:10 | history | rollback | ACuriousMind♦ |
Rollback to Revision 4
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| S Jul 28, 2023 at 6:39 | history | suggested | mma | CC BY-SA 4.0 |
central extension/semidirect product mismatch has been fixed
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| Jul 28, 2023 at 4:24 | review | Suggested edits | |||
| S Jul 28, 2023 at 6:39 | |||||
| S Jul 27, 2021 at 16:38 | history | suggested | Kvothe | CC BY-SA 4.0 |
I had trouble even recognizing that exact sequence was a technical math term. I think many physicist reading this answer will be unfamiliar with this. Including a link to its definition would help.
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| Jul 27, 2021 at 16:37 | comment | added | ACuriousMind♦ | @Kvothe Yes, look at Schottenloher and Weinberg recommended in the comments to the question | |
| Jul 27, 2021 at 16:33 | comment | added | Kvothe | This is a great answer! Is there perhaps a physics book that treats the central extension of the de Witt algebra to Virasoro in this manner. (I was unfortunate enough to be taught to believe that the central extension was just ad hoc, and was never told about the connection to projective representations.) | |
| Jul 27, 2021 at 16:28 | review | Suggested edits | |||
| S Jul 27, 2021 at 16:38 | |||||
| Jun 11, 2020 at 9:33 | history | edited | CommunityBot |
Commonmark migration
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| Oct 22, 2015 at 17:41 | history | bounty ended | Danu | ||
| Oct 10, 2015 at 19:35 | history | edited | Danu | CC BY-SA 3.0 |
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| Sep 7, 2015 at 23:34 | vote | accept | ACuriousMind♦ | ||
| Sep 5, 2015 at 12:34 | comment | added | ACuriousMind♦ | @WetSavannaAnimalakaRodVance: I say "highly" because the $\Sigma$ together with its $C$ represents a cohomology class (the notation $H^2$ is not an accident, although I didn't explain the connection). I tend to imagine (co)homology classes as "very large" (e.g. in the singular case, the chain groups are indeed absurdly large), but I didn't have a specific cardinality for it in mind. | |
| Sep 5, 2015 at 3:29 | comment | added | Selene Routley | A question on your choice of language: when you say highly nonunique to describe the choice $\sigma\mapsto\Sigma$, are you emphasizing the in general uncountable choice in contrast with the at most countable choice that arises for covers of a finite dimensional Lie group (whose algebra of course doesn't have a central charge)? | |
| Sep 2, 2015 at 21:52 | history | answered | ACuriousMind♦ | CC BY-SA 3.0 |