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4

No, the gauge group of electromagnetism is still $\mathrm{U }(1)$. This is because there is a difference between the gauge group $G$ of a physical theory on a manifold $M$ and the group of gauge transformations $\mathcal{G}$ consisting of local functions $M\to G$ and the algebra of infinitesimal gauge transformations consisting of local functions ...


1

A massless vector is the gauge field $A_\mu$ and the massless symmetric traceless tensor is the metric perturbation $h_{\mu\nu}$. They can couple only to conserved currents, namely the $U(1)$ vector current $j_\mu$ and the stress tensor $T_{\mu\nu}$ respectively. The requirement for this arises from gauge invariance. Coupling a gauge field to a current ...


4

Yes, you missed the whole point of gauge invariance which clearly has to be a symmetry – the Hamiltonian has to be and is invariant under these transformations. In the case of a global symmetry, a Hamiltonian may afford "not to be symmetric" under it. But once a group of transformations is said to be a gauge symmetry, the failure of $H$ to be invariant would ...



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