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In Schottentloher's book, a theorem is stated:

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Later on, a remark is made:

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My confusion comes: Is E always of the form made in the remark, namely are the following to sets equal(up to set theoretic iso)?: \begin{equation} \{(U,g) \in U(\mathbb{H}) \times G \,:\;\; \hat{\gamma} (U)=Tg\}=U(1) \times_{\omega} G \end{equation} Or does this equation hold only for $\mathbb{H}=C$?

EDIT: Put more precisely, according to ACouriousMind's reply:

Why is the subgroup of $U(\mathbb{H}) \times G$, which obeys that $\hat{\gamma}(U)=Tg$ isomorphic to the semidirect product $U(1) \times_{\omega} G$?

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    $\begingroup$ I don't understand the question and I think you have misunderstood what the text is saying: It's defining $E$ as a subgroup of $U(H)\times G$ - the subgroup that obeys $\hat{\gamma}(U) = Tg$, and this subgroup is then always isomorphic to the semi-direct product in the remark. I don't know where the idea that $E$ is "of the form $U(H)\times G$" comes from. $\endgroup$
    – ACuriousMind ♦
    Mar 31 at 15:06
  • $\begingroup$ Precisely. This is my question. You claim that the subgroup that obeys the condition you wrote is always isomorphic to the semi-direct product in the remark. Could you please explicitly write this isomorphism out? $\endgroup$
    – ProphetX
    Mar 31 at 15:09
  • $\begingroup$ Are they only set theoretically isomorphic,or also group theoretically isomorphic? $\endgroup$
    – ProphetX
    Mar 31 at 15:17
  • $\begingroup$ Did you try just spelling out the definition? The pre-image of any single element under $\hat{\gamma}$ is a $\mathrm{U}(1)$ coset just because it's the kernel of $\hat{\gamma}$. The definition of $E$ just says "all $(U,g)$ such that $U$ is in the pre-image of $Tg$". So the set-theoretic structure of $E$ as $\mathrm{U}(1)\times G$ follows immediately. That the group product inherited as a subgroup is the same as the semi-direct product from the remark is similarly just unpacking definitions (note that the "Wigner's theorem" in the remark is also just taking a pre-image). $\endgroup$
    – ACuriousMind ♦
    Mar 31 at 15:47

1 Answer 1

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As ACuriousMind said, the preimage of $Tg$ under $\hat{\gamma}$ must be a $U(1)$-coset for each $g$. So $$E =\cup_{g\in G} \{(U(1)U_g, g)\}$$ where $U_g$ is some choice of representatives for each coset. You can then simply identify $(e^{i\phi}, g)\mapsto (e^{i\phi}U_g, g)$

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