One way to define a gauge theory is that whenever the Lagrangian is invariant under some local transformations, we say these local transformations are local gauge transformations and the theory is a gauge theory.

However, I think a better way to understand gauge theory is not about the local symmetry of the Lagrangian, but about the structure of the Hilbert space: a gauge theory is a theory where all states in the Hilbert space are invariant under the local transformations. This constraint on Hilbert space already encodes that the Hamiltonian or Lagrangian should be gauge invariant, otherwise under time evolution a gauge invariant state may become a gauge non-invariant state.

Also, I think in principle one can have a theory which does have a local symmetry but it is not a gauge theory. For example, we can take the electromagnetism Lagrangian without imposing gauge equivalence, namely, $A_\mu$ and $A_\mu+\partial_\mu\chi$ label different states. This theory has a Lagrangian that is invariant under a local transformation, but it is not a gauge theory. Of course, this theory does not describe relativistic massless spin-1 particle since it has more degrees of freedom than required.

  1. Now here is a question: is the global gauge transformation a symmetry transformation (a transformation acting on states and mapping one state to another)? Suppose the global gauge transformation is a special case of local gauge transformation, as it seems to be, it should not be a symmetry transformation.

  2. On the other hand, according to Noether's theorem, a symmetry of the Lagrangian, which may or may not be a symmetry of the theory, leads to local conserved current. However, Weinberg-Witten theorem states that there is no gauge invariant conserved current in non-Abelian gauge theory, although the total charge is conserved. Is this related to the observation that a global gauge transfomation is also not a symmetry transformation?

  3. And why does this theorem only apply to non-Abelian gauge theories?

  • $\begingroup$ Related: physics.stackexchange.com/q/13870/2451 and links therein. $\endgroup$ – Qmechanic Nov 26 '14 at 18:57
  • $\begingroup$ How can you take the EM Lagrangian without imposing gauge equivalence? The EM Lagrangian has gauge equivalence. There is no physical way to distinguish states $A_\mu$ and $A_\mu + \partial_\mu\chi$. $\endgroup$ – David Vercauteren Dec 17 '14 at 7:38

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