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The most intuitive example of a gauge symmetry is such where you take a theory that has some global symmetry, and ask what needs to be done for this symmetry to be local. This procedure involves the introduction of new fields. For example, when a global phase symmetry becomes local, you must include the four-potential and the E.M. field (Or EM-like field?).

  1. Is a vector field the only result of this procedure? Can, for example, the metric field and invariance to local change of coordinates, be obtained from a promotion of a global symmetry (global change of coordinates I assume?) to a local one?
  2. What are common examples of gauge symmetries that do not emerge from global symmetries? In particular, examples that do not involve the introduction of new fields? I know that in string theory global symmetries of the world-sheet become local symmetries of space-time.
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    $\begingroup$ 1) GR can be obtained by gauging Lorentz symmetry - the corresponding gauge field is the vielbein. 2) Gauging of discrete symmetries does not lead to introduction of new fields. $\endgroup$ – Prahar Mitra Mar 9 '20 at 22:05
  • $\begingroup$ 1. Can you give a reference for that? You actually get four gauge fields? 2. I think this is not always the case, for example gauge the 1d Ising model. Is there a way do understand when new fields are required? $\endgroup$ – proton Mar 10 '20 at 11:30
  • $\begingroup$ 1) en.wikipedia.org/wiki/Einstein%E2%80%93Cartan_theory 2) Can you explain your comment with 1D Ising model more - I do not know it. My understanding (and I think I am correct) is that gauge fields are needed if you gauge continuous global symmetries. If you gauge discrete symmetries, you might need to introduce additional line operators, but you do not need new local operators. $\endgroup$ – Prahar Mitra Mar 10 '20 at 17:55
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Pure E&M without matter has gauge symmetry ${\bf A}^{\prime}={\bf A}+\mathrm{d}\Lambda$ but is without the global gauge symmetry where $\Lambda={\rm const}.$

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  • $\begingroup$ Not sure I understand. If $\Lambda=const$ then $d\Lambda=0$ so $\boldsymbol{A}$ transforms trivially. $\endgroup$ – proton Mar 10 '20 at 12:03
  • $\begingroup$ $\uparrow$ Right. $\endgroup$ – Qmechanic Mar 10 '20 at 12:06
  • $\begingroup$ So why do you say there is no global symmetry? $\endgroup$ – proton Mar 10 '20 at 12:10
  • $\begingroup$ Because a trivial transformation is by definition not a symmetry. $\endgroup$ – Qmechanic Mar 10 '20 at 12:14
  • $\begingroup$ I see, but it feels as if this is just an extreme case of gauge invariance (where even the connection one-form remains invariant). $\endgroup$ – NDewolf Mar 10 '20 at 13:22
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A theory invariant under some spacetime-dependent transformations is automatically invariant under the constant transformations. So there's always a group of global symmetries that resemble the local ones. Usually these global symmetries are not treated as gauge symmetries. The transformations they induce are physically non-trivial, and their generators act non-trivially on the space of states.

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    $\begingroup$ Local symmetries often preserve boundary conditions, global symmetries do not. This is in fact, the very reason that global symmetries are non-trivial and local symmetries are (often) trivial. Generally speaking, because of such boundary issues, it is not always the case that local symmetries $\implies$ global symmetries. $\endgroup$ – Prahar Mitra Mar 9 '20 at 22:05

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