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Given a theory with gauge symmetry. After gauge fixing where does the symmetry go?

Does the gauge symmetry turn into a global symmetry?

For example there is a way to quantize fields theory with BRST fields. These fields cancel the additional degrees of freedom. But does the new action (original action+BRST fields) have some new internal symmetry?

As an example, local supersymmetry tells us a graviton must have a gravitino super-partner. After gauge fixing this gravitino doesn't just disappear (I presume), so some residual symmetry must remain.

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  • $\begingroup$ The part about BRST doesn't really seem to connected with the rest of this question, but in any case the relation between gauge symmetry and BRST symmetry is already discussed at physics.stackexchange.com/q/255135/50583 $\endgroup$ – ACuriousMind Oct 15 at 18:01
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After you've fixed a complete gauge (i.e one that really picks precisely one representant of every gauge orbit), the gauge symmetry is gone. If your gauge condition is incomplete, then some residual gauge symmetry might remain (e.g. the Lorenz gauge condition leaves a residual gauge symmetry for harmonic functions).

Gauge symmetries cannot "turn into global symmetries". The former signify unphysical redundancies in our choice of description of the system under consideration, the latter are properties of the system itself with real physical consequences, such as the existence of conserved quantities due to Noether's theorem.

But it may well be the case that the "global" version of a (local) gauge symmetry is not gauge to begin with, such as in the case of the $\mathrm{U}(1)$ symmetry of electromagnetism with charged matter. The local version of this symmetry is gauge, but the global $\mathrm{U}(1)$ symmetry acts only on the charged fields and not on the gauge field, hence does not vanish through fixing points in the gauge orbits since it does not act on the orbits to begin with. It is physical and enforces conservation of charge - not a gauge symmetry at all.

Likewise, for the graviton/gravitino, it is global supersymmetry that connected particles and their super-partners, and your gauge-fixing condition likely only eliminates the local part.

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  1. Gauge-fixing of gaugesymmetry is necessary to avoid singular volume factors in the path integral, cf. e.g. my Phys.SE answer here. The gauge-fixing condition must as a minimum accomplish this task.

  2. The global part of gaugesymmetry (and other global symmetries) should not be gauge-fixed as they don't render the path integral ill-defined.

  3. It seems relevant to OP's question to mention the background field method, which is a way to allow the background field to maintain gaugesymmetry even after gauge-fixing the corresponding quantum field. (This doesn't spoil the path integral since we don't integrate over the background field.)

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