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In Weinberg QFT Vol.1, it says one can enlarge the symmetry group $H$ to the universal covering group $C$ such that one obtains a trivial cocycle or $C$ is simply connected whereas $H$ is not. I get the mathematical meaning of the statement. I wonder whether this also requires $C$ being an extended symmetry of the system somehow though I thought $H$ is the largest symmetry group of the system. I am kind of confused here how can one further extend the system invariant under the action of the universal covering group unless one define the action of elements beyond $H$ being trivial.

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marked as duplicate by ACuriousMind, Martin, Qmechanic Oct 29 '15 at 16:45

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    $\begingroup$ It's a general principle that we must consider the projective representations of $H$, which are often precisely the linear representations of $C$, see this Q&A of mine $\endgroup$ – ACuriousMind Oct 28 '15 at 3:26
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    $\begingroup$ The post of @ACuriousMind is more complete than this one of mine (physics.stackexchange.com/q/96045), but it's nonetheless closely related and might be useful to you even though it doesn't directly answer your questions. $\endgroup$ – joshphysics Oct 28 '15 at 3:29